3rd edition neamen solutions

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Chapter 1 Exercise Problems EX1.1 ⎛ − Eg ⎞ ni = BT 3 / 2 exp ⎜ ⎟ ⎝ 2kT ⎠ GaAs: ni = ( 2.1× 1014 ) ( 300 ) Ge: ni = (1.66 × 1013 ) ( 300 ) 3/ 2 3/ 2 ⎛ ⎞ −1.4 ⎟ or ni = 1.8 × 106 cm −3 exp ⎜ ⎜ 2 ( 86 × 10−6 ) ( 300 ) ⎟ ⎝ ⎠ ⎛ ⎞ −0.66 ⎟ or ni = 2.40 × 1013 cm −3 exp ⎜ ⎜ 2 ( 86 × 10−6 ) ( 300 ) ⎟ ⎝ ⎠ EX1.2 (a) majority carrier: holes, po = 1017 cm −3 minority carrier: electrons, n 2 (1.5 × 10 no = i = 1017 po ) 10 2 = 2.25 × 103 cm −3 (b) majority carrier: electrons, no = 5 × 1015 cm −3 minority carrier: holes, n 2 (1.5 × 10 ) = 4.5 × 104 cm −3 po = i = 5 × 1015 no 10 2 EX1.3 For n-type, drift current density J ≅ eμn nE or 200 = (1.6 × 10−19 ) ( 7000 ) (1016 ) E which yields E = 17.9 V / cm EX1.4 Diffusion current density due to holes: dp J p = −eD p dx ⎛ −1 ⎞ ⎛ −x ⎞ = −eD p (1016 ) ⎜ ⎟ exp ⎜ ⎟ ⎜L ⎟ ⎜L ⎟ ⎝ p⎠ ⎝ p⎠ (a) At x = 0 (1.6 ×10 ) (10 ) (10 ) = 16 A / cm = −19 Jp 16 2 10−3 −3 (b) At x = 10 cm ⎛ −10−3 ⎞ J p = 16 exp ⎜ −3 ⎟ = 5.89 A / cm 2 ⎝ 10 ⎠ EX1.5 ⎡N N Vbi = VT ln ⎢ a 2 d ⎣ ni ⎡ (1016 )(1017 ) ⎤ ⎤ ⎥ or Vbi = 1.23 V = ( 0.026 ) ln ⎢ ⎥ ⎢ (1.8 × 106 )2 ⎥ ⎦ ⎣ ⎦ EX1.6 ⎛ V ⎞ C j = C jo ⎜1 + R ⎟ ⎝ Vbi ⎠ and −1/ 2 3. ⎡N N ⎤ Vbi = VT ln ⎢ a 2 d ⎥ ⎣ ni ⎦ ⎡ (1017 )(1016 ) ⎤ ⎥ = 0.757 V = ( 0.026 ) ln ⎢ ⎢ (1.5 × 1010 )2 ⎥ ⎣ ⎦ 5 ⎞ ⎛ Then 0.8 = C jo ⎜ 1 + ⎟ ⎝ 0.757 ⎠ or C jo = 2.21 pF −1/ 2 = C jo ( 7.61) −1/ 2 EX1.7 ⎡ ⎛v ⎞ ⎤ iD = I S ⎢exp ⎜ D ⎟ − 1⎥ ⎢ ⎝ VT ⎠ ⎥ ⎣ ⎦ ⎡ ⎛ v ⎞ ⎤ so 10−3 = (10−13 ) ⎢ exp ⎜ D ⎟ − 1⎥ ⎝ 0.026 ⎠ ⎦ ⎣ ⎡ 10−3 ⎤ Solving for the diode voltage, we find vD = ( 0.026 ) ln ⎢ −13 + 1⎥ ⎣10 ⎦ or vD ≅ ( 0.026 ) ln (1010 ) which yields vD = 0.599 V EX1.8 ⎛V ⎞ VPS = I D R + VD and I D ≅ I S exp ⎜ D ⎟ ⎝ VT ⎠ ( 4 − VD ) so 4 = I D ( 4 ×103 ) + VD ⇒ I D = 4 ×103 and ⎛ V ⎞ I D = (10 −12 ) exp ⎜ D ⎟ ⎝ 0.026 ⎠ By trial and error, we find I D ≅ 0.864 mA and VD ≅ 0.535 V EX1.9 (a) ID = (b) ID = Then R = (c) VPS − Vγ R VPS − Vγ R 5 − 0.7 ⇒ I D = 1.08 mA 4 VPS − Vγ ⇒R= ID = 8 − 0.7 = 6.79 kΩ 1.075 4. ID(mA) Diode curve 1.25 1.08 Load lines (b) (a) 0 0.7 2 4 VD(v) 6 8 EX1.10 PSpice analysis EX1.11 Quiescent diode current I DQ = VPS − Vγ = 10 − 0.7 = 0.465 mA 20 R Time-varying diode current: V 0.026 We find that rd = T = = 0.0559 kΩ I DQ 0.465 Then id = vI 0.2sin ω t (V ) = ⋅ or id = 9.97sin ω t ( μ A) rd + R 0.0559 + 20 ( kΩ ) EX1.12 ⎛I ⎞ ⎛ 1.2 × 10−3 ⎞ or VD = 0.6871 V For the pn junction diode, VD ≅ VT ln ⎜ D ⎟ = ( 0.026 ) ln ⎜ −15 ⎟ ⎝ 4 × 10 ⎠ ⎝ IS ⎠ The Schottky diode voltage will be smaller, so VD = 0.6871 − 0.265 = 0.4221 V ⎛V ⎞ Now I D ≅ I S exp ⎜ D ⎟ ⎝ VT ⎠ or 1.2 × 10−3 IS = ⇒ I S = 1.07 × 10−10 A 0.4221 ⎞ ⎛ exp ⎜ ⎟ ⎝ 0.026 ⎠ EX1.13 P = I ⋅ VZ ⇒ 10 = I ( 5.6 ) ⇒ I = 1.79 mA Also I = 10 − 5.6 = 1.79 ⇒ R = 2.46 kΩ R Test Your Understanding Exercises TYU1.1 (a) T = 400K ⎛ − Eg ⎞ Si: ni = BT 3 / 2 exp ⎜ ⎟ ⎝ 2kT ⎠ ni = ( 5.23 × 1015 ) ( 400 ) or ni = 4.76 × 1012 cm −3 3/ 2 ⎡ ⎤ −1.1 ⎥ exp ⎢ −6 ⎢ 2 ( 86 × 10 ) ( 400 ) ⎥ ⎣ ⎦ 5. Ge: ni = (1.66 × 1015 ) ( 400 ) 3/ 2 ⎡ ⎤ −0.66 ⎥ exp ⎢ −6 ⎢ 2 ( 86 × 10 ) ( 400 ) ⎥ ⎣ ⎦ or ni = 9.06 × 1014 cm −3 GaAs: ni = ( 2.1× 1014 ) ( 400 ) 3/ 2 ⎡ ⎤ −1.4 ⎥ exp ⎢ −6 ⎢ 2 ( 86 × 10 ) ( 400 ) ⎥ ⎣ ⎦ or ni = 2.44 × 109 cm −3 (b) T = 250 K Si: ni = ( 5.23 × 1015 ) ( 250 ) 3/ 2 ⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢ 2 ( 86 × 10−6 ) ( 250 ) ⎥ ⎣ ⎦ or ni = 1.61× 108 cm −3 Ge: ni = (1.66 × 1015 ) ( 250 ) 3/ 2 ⎡ ⎤ −0.66 ⎥ exp ⎢ −6 ⎢ 2 ( 86 × 10 ) ( 250 ) ⎥ ⎣ ⎦ or ni = 1.42 × 1012 cm −3 GaAs: ni = ( 2.10 × 1014 ) ( 250 ) 3/ 2 ⎡ ⎤ −1.4 ⎥ exp ⎢ ⎢ 2 ( 86 × 10−6 ) ( 250 ) ⎥ ⎣ ⎦ or ni = 6.02 × 103 cm −3 TYU1.2 (a) n = 5 × 1016 cm −3 , p 0 Vγ = 0 Voltage across RL + R1 = vi ⎛ RL ⎞ 1 Voltage Divider ⇒ v0 = ⎜ ⎟ vi = vi 2 ⎝ RL + R1 ⎠ 38. ␯0 20 2.13 For vi > 0, (Vγ = 0 ) ␯i R1 ϩ Ϫ ␯0 R2 RL a. ⎛ R2 || RL ⎞ v0 = ⎜ ⎟ vi ⎝ R2 || RL + R1 ⎠ R2 || RL = 2.2 || 6.8 = 1.66 kΩ ␯0 ⎛ 1.66 ⎞ v0 = ⎜ ⎟ vi = 0.43 vi ⎝ 1.66 + 2.2 ⎠ 4.3 v0 ( rms ) = b. v0 ( max ) 2 ⇒ v0 ( rms ) = 3.04 V 2.14 3.9 ⇒ I 2 = 0.975 mA 4 20 − 3.9 IR = = 1.3417 mA 12 I Z = 1.3417 − 0.975 ⇒ I Z = 0.367 mA IL = PT = I Z ⋅ VZ = ( 0.367 )( 3.9 ) ⇒ PT = 1.43 mW 2.15 (a) 40 − 12 = 0.233 A 120 P = ( 0.233)(12 ) = 2.8 W IZ = (b) IR = 0.233 A, IL = (0.9)(0.233) = 0.21 A 12 So 0.21 = ⇒ RL = 57.1Ω RL P = ( 0.1)( 0.233)(12 ) ⇒ P = 0.28 W (c) 2.16 Ri VI II IZ ϩ VZ Ϫ V0 RL IL VI = 6.3 V, Ri = 12Ω, VZ = 4.8 39. a. 6.3 − 4.8 ⇒ 125 mA 12 I L = I I − I Z = 125 − I Z II = 25 ≤ I L ≤ 120 mA ⇒ 40 ≤ RL ≤ 192Ω b. PZ = I Z VZ = (100 )( 4.8 ) ⇒ PZ = 480 mW PL = I LV0 = (120 )( 4.8 ) ⇒ PL = 576 mW 2.17 a. 20 − 10 ⇒ I I = 45.0 mA 222 10 IL = ⇒ I L = 26.3 mA 380 I Z = I I − I L ⇒ I Z = 18.7 mA II = b. PZ ( max ) = 400 mW ⇒ I Z ( max ) = ⇒ I L ( min ) = I I − I Z ( max ) = 45 − 40 ⇒ I L ( min ) = 5 mA = 400 = 40 mA 10 10 RL ⇒ RL = 2 kΩ (c) For Ri = 175Ω I I = 57.1 mA I L = 26.3 mA I Z = 30.8 mA I Z ( max ) = 40 mA ⇒ I L ( min ) = 57.1 − 40 = 17.1 mA RL = 10 ⇒ RL = 585Ω 17.1 2.18 a. From Eq. (2-31) 500 [ 20 − 10] − 50 [15 − 10] I Z ( max ) = 15 − ( 0.9 )(10 ) − ( 0.1)( 20 ) 5000 − 250 4 I Z ( max ) = 1.1875 A I Z ( min ) = 0.11875 A = From Eq. (2-29(b)) Ri = 20 − 10 ⇒ Ri = 8.08Ω 1187.5 + 50 b. PZ = (1.1875 )(10 ) ⇒ PZ = 11.9 W PL = I L ( max ) V0 = ( 0.5 )(10 ) ⇒ PL = 5 W 2.19 (a) As approximation, assume I Z ( max ) and I Z ( min ) are the same as in problem 2.18. V0 ( max ) = V0 ( nom ) + I Z ( max ) rZ = 10 + (1.1875)(2) = 12.375 V V0 ( min ) = V0 ( nom ) + I Z ( min ) rZ = 10 + (0.11875)(2) = 10.2375 V 40. % Reg = b. 2.20 % Reg = = 12.375 − 10.2375 × 100% ⇒ % Reg = 21.4% 10 VL ( max ) − VL ( min ) VL ( nom ) × 100% VL ( nom ) + I Z ( max ) rz − (VL ( nom ) + I Z ( min ) rz ) VL ( nom ) ⎡ I Z ( max ) − I Z ( min ) ⎤ ( 3) ⎦ =⎣ = 0.05 6 So I Z ( max ) − I Z ( min ) = 0.1 A 6 6 = 0.012 A, I L ( min ) = = 0.006 A 500 1000 VPS ( min ) − VZ Now I L ( max ) = Now Ri = or 280 = I Z ( min ) + I L ( max ) 15 − 6 ⇒ I Z ( min ) = 0.020 A I Z ( min ) + 0.012 Then I Z ( max ) = 0.1 + 0.02 = 0.12 A and Ri = or 280 = VPS ( max ) − 6 0.12 + 0.006 VPS ( max ) − VZ I Z ( max ) + I L ( min ) ⇒ VPS ( max ) = 41.3 V 2.21 Using Figure 2.21 a. VPS = 20 ± 25% ⇒ 15 ≤ VPS ≤ 25 V For VPS ( min ) : I I = I Z ( min ) + I L ( max ) = 5 + 20 = 25 mA Ri = b. VPS ( min ) − VZ II = 15 − 10 ⇒ Ri = 200Ω 25 For VPS ( max ) ⇒ I I ( max ) = 25 − 10 ⇒ I I ( max ) = 75 mA Ri For I L ( min ) = 0 ⇒ I Z ( max ) = 75 mA VZ 0 = VZ − I Z rZ = 10 − ( 0.025 )( 5 ) = 9.875 V V0 ( max ) = 9.875 + ( 0.075 )( 5 ) = 10.25 V0 ( min ) = 9.875 + ( 0.005 )( 5 ) = 9.90 ΔV0 = 0.35 V c. % Reg = ΔV0 × 100% ⇒ % Reg = 3.5% V0 ( nom ) 2.22 From Equation (2.29(a)) VPS ( min ) − VZ 24 − 16 or Ri = 18.2Ω Ri = = I Z ( min ) + I L ( max ) 40 + 400 Also Vr = VM VM ⇒C = 2 fRC 2 fRVr R ≅ Ri + rz = 18.2 + 2 = 20.2Ω Then 41. C= 24 ⇒ C = 9901 μ F 2 ( 60 )(1)( 20.2 ) 2.23 VZ = VZ 0 + I Z rZ VZ ( nom ) = 8 V 8 = VZ 0 + ( 0.1)( 0.5 ) ⇒ VZ 0 = 7.95 V Ii = VS ( max ) − VZ ( nom ) Ri = 12 − 8 = 1.333 A 3 For I L = 0.2 A ⇒ I Z = 1.133 A For I L = 1 A ⇒ I Z = 0.333 A VL ( max ) = VZ 0 + I Z ( max ) rZ = 7.95 + (1.133)( 0.5 ) = 8.5165 VL ( min ) = VZ 0 + I Z ( min ) rZ = 7.95 + ( 0.333)( 0.5 ) = 8.1165 ΔVL = 0.4 V ΔVL 0.4 = ⇒ % Reg = 5.0% % Reg = V0 ( nom ) 8 Vr = VM VM ⇒C = 2 fRC 2 fRVr R = Ri + rz = 3 + 0.5 = 3.5Ω Then C = 12 ⇒ C = 0.0357 F 2 ( 60 )( 3.5 )( 0.8 ) 2.24 (a) For −10 ≤ vI ≤ 0, both diodes are conducting ⇒ vO = 0 For 0 ≤ vI ≤ 3, Zener not in breakdown, so i1 = 0, vO = 0 vI − 3 mA 20 1 ⎛ v −3⎞ vo = ⎜ I ⎟ (10 ) = vI − 1.5 20 ⎠ 2 ⎝ At vI = 10 V, vo = 3.5 V, i1 = 0.35 mA For vI > 3 i1 = ␯O(V) 4 3.5 Ϫ10 (b) 3.0 For vI < 0, both diodes forward biased 0 − vI . At vI = −10 V , i1 = −1 mA 10 v −3 For vI > 3, i1 = I . At vI = 10 V , i1 = 0.35 mA 20 −i1 = 10 ␯I(V) 42. i1(mA) 0.35 Ϫ10 10 ␯I(V) 3 Ϫ1 2.25 (a) 1K ␯I ␯0 1K 2K ϩ15 V1 1 V1 = × 15 = 5 V ⇒ for vI ≤ 5.7, v0 = vI 3 For vI > 5.7 V vI − (V1 + 0.7 ) 15 − V1 V1 + = , v0 = V1 + 0.7 1 2 1 15 − ( v0 − 0.7 ) v0 − 0.7 vI − v0 + = 1 2 1 vI 15.7 0.7 ⎛ 1 1 1⎞ + + = v0 ⎜ + + ⎟ = v0 ( 2.5 ) 1 2 1 ⎝ 1 2 1⎠ 1 vI + 8.55 = v0 ( 2.5 ) ⇒ v0 = vI + 3.42 2.5 vI = 5.7 ⇒ v0 = 5.7 vI = 15 ⇒ v0 = 9.42 ␯0(V) 9.42 5.7 0 5.7 15 ␯I (V) (b) iD = 0 for 0 ≤ vI ≤ 5.7 Then for vI > 5.7 V ⎛ v ⎞ vI − ⎜ I + 3.42 ⎟ vI − vO ⎝ 2.5 ⎠ or i = 0.6vI − 3.42 For v = 15, i = 5.58 mA = iD = I D D 1 1 1 43. iD(mA) 5.58 5.7 15 ␯I (V) 2.26 ⎛ 20 ⎞ For D off, vo = ⎜ ⎟ (20) − 10 = 3.33 V ⎝ 30 ⎠ Then for vI ≤ 3.33 + 0.7 = 4.03 V ⇒ vo = 3.33 V For vI > 4.03, vo = vl − 0.7; For vI = 10, vo = 9.3 (a) ␯O(V) 9.3 3.33 0 (b) 4.03 10 ␯I (V) For vI ≤ 4.03 V , iD = 0 10 − vo vo − ( −10 ) = 10 20 3 Which yields iD = vI − 0.605 20 For vI = 10, iD = 0.895 mA For vI > 4.03, iD + iD(mA) 0.895 0 2.27 Ϫ30 4.03 ␯O 12.5 10.7 10.7 30 ␯I Ϫ30 30 − 10.7 = 0.175 A 100 + 10 v0 = i(10) + 10.7 = 12.5 V For vI = 30 V, i = 10 ␯I(V) 44. b. ␯O 12.5 10.7 0 Ϫ30 2.28 ϩ 5Ϫ ␯I ␯O R ϭ 6.8 K Vγ = 0.6 V vI = 15sin ω t ␯O Ϫ4.4 Ϫ19.4 2.29 a. Vγ = 0 0 Ϫ3 V Vγ = 0.6 0 Ϫ2.4 b. Vγ = 0 20 5 Vγ = 0.6 19.4 5 2.30 ␯O 45. 10 6.7 0 Ϫ4.7 Ϫ10 2.31 One possible example is shown. Ri Ii ϩ Ϫ DZ Vign L D ϩ VZ ϭ 14 V Ϫ ϩ Ϫ RADIO VRADIO L will tend to block the transient signals Dz will limit the voltage to +14 V and −0.7 V. Power ratings depends on number of pulses per second and duration of pulse. 2.32 ␯O(V) 40 (a) 0 ␯O(V) 35 (b) 0 Ϫ5 2.33 C ␯I ␯O ϩ Vx Ϫ a. For Vγ = 0 ⇒ Vx = 2.7 V b. For Vγ = 0.7 V ⇒ Vx = 2.0 V 2.34 C ϩ ␯I Ϫ 2.35 ϩ Ϫ 10 V ϩ ␯O Ϫ 46. 20 ␯O 10 VB ϭ 0 0 ␯I Ϫ10 20 ␯O 13 10 3 0 VB ϭ ϩ3 V ␯I ϩ VB Ϫ7 20 ␯O 10 7 VB ϭ Ϫ3 V 0 Ϫ3 ␯ I ϩ VB Ϫ13 2.36 For Figure P2.32(a) 10 ␯I 0 Ϫ10 ␯O Ϫ20 2.37 a. 10 − 0.6 ⇒ I D1 = 0.94 mA 9.5 + 0.5 V0 = I D1 ( 9.5 ) ⇒ V0 = 8.93 V I D1 = ID2 = 0 b. 5 − 0.6 ⇒ I D1 = 0.44 mA 9.5 + 0.5 V0 = I D1 ( 9.5 ) ⇒ V0 = 4.18 V I D1 = c. d. 10 = ID2 = 0 Same as (a) (I ) 2 ( 0.5 ) + 0.6 + I ( 9.5 ) ⇒ I = 0.964 mA V0 = I ( 9.5 ) ⇒ V0 = 9.16 V I D1 = I D 2 = 2.38 a. b. I ⇒ I D1 = I D 2 = 0.482 mA 2 I = I D1 = I D 2 = 0 V0 = 10 47. 10 = I ( 9.5 ) + 0.6 + I ( 0.5 ) ⇒ I = I D 2 = 0.94 mA I D1 = 0 V0 = 10 − I ( 9.5 ) ⇒ V0 = 1.07 V c. 10 = I ( 9.5 ) + 0.6 + I ( 0.5 ) + 5 ⇒ I = I D 2 = 0.44 mA I D1 = 0 V0 = 10 − I ( 9.5 ) ⇒ V0 = 5.82 V d. 10 = I ( 9.5 ) + 0.6 + I ( 0.5) ⇒ I = 0.964 mA 2 I ⇒ I D1 = I D 2 = 0.482 mA 2 V0 = 10 − I ( 9.5 ) ⇒ V0 = 0.842 V I D1 = I D 2 = 2.39 a. V1 = V2 = 0 ⇒ D1 , D2 , D3 , on V0 = 4.4 V 10 − 4.4 ⇒ I = 0.589 mA 9.5 4.4 − 0.6 = ID2 = ⇒ I D1 = I D 2 = 7.6 mA 0.5 = I D1 + I D 2 − I = 2 ( 7.6 ) − 0.589 ⇒ I D 3 = 14.6 mA I= I D1 I D3 b. V1 = V2 = 5 V D1 and D2 on, D3 off I 10 = I ( 9.5 ) + 0.6 + ( 0.5 ) + 5 ⇒ I = 0.451 mA 2 I I D1 = I D 2 = ⇒ I D1 = I D 2 = 0.226 mA 2 I D3 = 0 V0 = 10 − I ( 9.5 ) = 10 − ( 0.451)( 9.5 ) ⇒ V0 = 5.72 V V1 = 5 V, V2 = 0 D1 off, D2, D3 on V0 = 4.4 V c. 10 − 4.4 ⇒ 9.5 4.4 − 0.6 = ⇒ 0.5 I= I D2 I = 0.589 mA I D 2 = 7.6 mA I D1 = 0 I D 3 = I D 2 − I = 7.6 − 0.589 ⇒ I D 3 = 7.01 mA V1 = 5 V, V2 = 2 V D1 off, D2, D3 on V0 = 4.4 V d. 10 − 4.4 ⇒ 9.5 4.4 − 0.6 − 2 = ⇒ 0.5 I= I D2 I = 0.589 mA I D 2 = 3.6 mA I D1 = 0 I D 3 = I D 2 − I = 3.6 − 0.589 ⇒ I D 3 = 3.01 mA 2.40 (a) D1 on, D2 off, D3 on So I D 2 = 0 Now V2 = −0.6V , I D1 = 10 − 0.6 − ( −0.6 ) R1 + R2 = 10 ⇒ I D1 = 1.25 mA 2+6 48. V1 = 10 − 0.6 − (1.25 )( 2 ) ⇒ V1 = 6.9 V I R3 = I D3 −0.6 − ( −5 ) = 2.2 mA 2 = I R 3 − I D1 = 2.2 − 1.25 ⇒ I D 3 = 0.95 mA (b) D1 on, D2 on, D3 off So I D 3 = 0 V1 = 4.4 V , I D1 = 10 − 0.6 − 4.4 5 = 6 R1 or I D1 = 0.833 mA I R2 = 4.4 − ( −5 ) R2 + R3 = 9.4 = 0.94 mA 10 I D 2 = I R 2 − I D1 = 0.94 − 0.833 ⇒ I D 2 = 0.107 mA V2 = I R 2 R3 − 5 = ( 0.94 )( 5 ) − 5 ⇒ V2 = −0.3 V All diodes are on V1 = 4.4V , V2 = −0.6 V (c) I D1 = 0.5 mA = 10 − 0.6 − 4.4 ⇒ R1 = 10 k Ω R1 I R 2 = 0.5 + 0.5 = 1 mA = I R 3 = 1.5 mA = 4.4 − ( −0.6 ) −0.6 − ( −5 ) R3 R2 ⇒ R2 = 5 k Ω ⇒ R3 = 2.93 k Ω 2.41 ⎛ 0.5 ⎞ For vI small, both diodes off vO = ⎜ ⎟ vI = 0.0909vI ⎝ 0.5 + 5 ⎠ When vI − vO = 0.6, D1 turns on. So we have vI − 0.0909vI = 0.6 ⇒ vI = 0.66, vO = 0.06 vI − 0.6 − vO vI − vO vO 2v − 0.6 + = which yields vO = I 5 5 0.5 12 2vI − 0.6 When vO = 0.6, D2 turns on. Then 0.6 = ⇒ vI = 3.9 V 12 v − 0.6 − vO vI − vO vO vO − 0.6 + = + Now for vI > 3.9 I 5 5 0.5 0.5 2vI + 5.4 ; For vI = 10 ⇒ vO = 1.15 V Which yields vO = 22 For D1 on 2.42 ϩ10 V 10 K D2 D1 ␯I ␯0 D3 D4 10 K Ϫ10 V 10 K 49. For vI > 0. when D1 and D4 turn off 10 − 0.7 = 0.465 mA 20 v0 = I (10 kΩ ) = 4.65 V I= ␯0 4.65 Ϫ10 Ϫ4.65 4.65 10 ␯I Ϫ4.65 v0 = vI for − 4.65 ≤ vI ≤ 4.65 2.43 a. R1 D2 ϩ10 V V0 D1 ID1 R1 = 5 kΩ, R2 = 10 kΩ D1 and D2 on ⇒ V0 = 0 R2 Ϫ10 V 10 − 0.7 0 − ( −10 ) − = 1.86 − 1.0 5 10 = 0.86 mA I D1 = I D1 b. R1 = 10 kΩ, R2 = 5 kΩ, D1 off, D2 on I D1 = 0 I= 10 − 0.7 − ( −10 ) = 1.287 15 V0 = IR2 − 10 ⇒ V0 = −3.57 V 2.44 If both diodes on (a) VA = −0.7 V, VO = −1.4 V I R1 = IR2 I R1 + I D1 10 − ( −0.7 ) = 1.07 mA 10 −1.4 − ( −15 ) = = 2.72 mA 5 = I R 2 ⇒ I D1 = 2.72 − 1.07 I D1 = 1.65 mA D1 off, D2 on 10 − 0.7 − ( −15 ) I R1 = I R 2 = = 1.62 mA 5 + 10 VO = I R 2 R2 − 15 = (1.62 )(10 ) − 15 ⇒ VO = 1.2 V (b) VA = 1.2 + 0.7 = 1.9 V ⇒ D1 off , I D1 = 0 2.45 50. (a) D1 on, D2 off 10 − 0.7 I D1 = = 0.93 mA 10 VO = −15 V (b) D1 on, D2 off 10 − 0.7 I D1 = = 1.86 mA 5 VO = −15 V 2.46 15 − (V0 + 0.7 ) V0 + 0.7 V0 + 10 20 20 15 0.7 0.7 1 1 1 ⎞ ⎛ ⎛ 4.0 ⎞ − − = V0 ⎜ + + ⎟ = V0 ⎜ ⎟ 10 10 20 ⎝ 10 20 20 ⎠ ⎝ 20 ⎠ V0 = 6.975 V ID = = V0 ⇒ I D = 0.349 mA 20 2.47 10 K V1 Va Ϫ VD ϩ 10 K Vb V2 ID 10 K 10 K a. V1 = 15 V, V2 = 10 V Diode off Va = 7.5 V, Vb = 5 V ⇒ VD = −2.5 V ID = 0 b. V1 = 10 V, V2 = 15 V Diode on V2 − Vb Vb Va Va − V1 = + + ⇒ Va = Vb − 0.6 10 10 10 10 15 10 ⎛1 1⎞ ⎛1 1⎞ ⎛ 1 1⎞ + = Vb ⎜ + ⎟ + Vb ⎜ + ⎟ − 0.6 ⎜ + ⎟ 10 10 10 10 ⎠ 10 10 ⎠ ⎝ ⎝ ⎝ 10 10 ⎠ ⎛ 4⎞ 2.62 = Vb ⎜ ⎟ ⇒ Vb = 6.55 V ⎝ 10 ⎠ 15 − 6.55 6.55 ID = − ⇒ I D = 0.19 mA 10 10 VD = 0.6 V 2.48 vI = 0, D1 off, D2 on 10 − 2.5 = 0.5 mA 15 vo = 10 − ( 0.5 )( 5 ) ⇒ vo = 7.5 V for 0 ≤ vI ≤ 7.5 V I= For vI > 7.5 V , Both D1 and D2 on vI − vo vo − 2.5 vo − 10 = + or vI = vo ( 5.5 ) − 33.75 15 10 5 When vo = 10 V, D2 turns off vI = (10 )( 5.5 ) − 33.75 = 21.25 V For vI > 21.25 V, vo = 10 V 51. 2.49 a. V01 = V02 = 0 b. V01 = 4.4 V, V02 = 3.8 V c. V01 = 4.4 V, V02 = 3.8 V Logic “1” level degrades as it goes through additional logic gates. 2.50 a. V01 = V02 = 5 V b. V01 = 0.6 V, V02 = 1.2 V c. V01 = 0.6 V, V02 = 1.2 V Logic “0” signal degrades as it goes through additional logic gates. 2.51 (V1 AND V2 ) OR (V3 AND V4 ) 2.52 10 − 1.5 − 0.2 I= = 12 mA = 0.012 R + 10 8.3 R + 10 = = 691.7Ω 0.012 R = 681.7Ω 2.53 10 − 1.7 − VI =8 0.75 VI = 10 − 1.7 − 8 ( 0.75 ) ⇒ VI = 2.3 V I= 2.54 ϩ VR Ϫ ϩ VPS Ϫ Rϭ 2 K VR = 1 V, I = 0.8 mA VPS = 1 + ( 0.8 )( 2 ) VPS = 2.6 V 2.55 I Ph = η eΦA 0.6 × 10−3 = (1) (1.6 × 10−19 )(1017 ) A A = 3.75 × 10−2 cm 2 52. Chapter 3 Exercise Solutions EX3.1 VTN = 1 V , VGS = 3 V , VDS = 4.5 V VDS = 4.5 > VDS ( sat ) = VGS − VTN = 3 − 1 = 2 V Transistor biased in the saturation region I D = K n (VGS − VTN ) ⇒ 0.8 = K n ( 3 − 1) ⇒ K n = 0.2 mA / V 2 2 2 (a) VGS = 2 V, VDS = 4.5 V Saturation region: I D = ( 0.2 )( 2 − 1) ⇒ I D = 0.2 mA 2 (b) VGS = 3 V, VDS = 1 V Nonsaturation region: 2 I D = ( 0.2 ) ⎡ 2 ( 3 − 1)(1) − (1) ⎤ ⇒ I D = 0.6 mA ⎣ ⎦ EX3.2 VTP = −2 V , VSG = 3 V VSD ( sat ) = VSG + VTP = 3 − 2 = 1 V (a) (b) (c) VSD = 0.5 V ⇒ Nonsaturation VSD = 2 V ⇒ Saturation VSD = 5 V ⇒ Saturation EX3.3 ⎛ R2 ⎞ ⎛ 160 ⎞ VG = ⎜ ⎟ (VDD ) = ⎜ ⎟ (10 ) = 3.636 V = VGS ⎝ 160 + 280 ⎠ ⎝ R1 + R2 ⎠ I D = 0.25 ( 3.636 − 2 ) = 0.669 mA 2 VDS = 10 − ( 0.669 )(10 ) = 3.31 V P = I DVDS = ( 0.669 )( 3.31) = 2.21 mW EX3.4 I DQ = K P (VSG + VTP ) 2 1.2 = 0.4 (VSG − 1.2 ) ⇒ VSG = 2.932 V 2 ⎛ R1 ⎞ 1 VSG = ⎜ ⋅ VTTN − VDD ⎟ VDD = R2 ⎝ R1 + R2 ⎠ Note K = kΩ 1 2.932 = ( 200 )(10 ) ⇒ R2 = 682 K R2 682 R1 = 200 ⇒ R1 = 283 K 682 + R1 RD = 10 − 4 =5K 1.2 EX3.5 ⎛ R2 ⎞ ⎛ 40 ⎞ VG = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 = −1 V R1 + R2 ⎠ ⎝ 40 + 60 ⎠ ⎝ V − ( −5 ) 2 ID = S = K n (VGS − VTN ) RS (a) VS = VG − VGS 53. 2 ( 5 − 1) − VGS = ( 0.5 )(1) (VGS − 2VGS + 1) 2 2 0.5VGS − 3.5 = 0 VGS = 7 VGS = 2.646 V I D = ( 0.5 )( 2.646 − 1) ⇒ I D = 1.354 mA VDS = 10 − (1.354 )( 3) = 5.937 V 2 4 − VGS = K n (1)(VGS − VTN ) (b) 2 (1) K n = (1.05 )( 0.5 ) = 0.525 (2) K n = ( 0.95)( 0.5 ) = 0.475 (3) VTN = (1.05 )(1) = 1.05 V (4) VTN = ( 0.95 )(1) = 0.95 V (1)-(3) 2 4 − VGS = 0.525 (VGS − 2.1VGS + 1.1025 ) 2 0.525VGS − 0.1025VGS − 3.421 = 0 VGS = 0.1025 ± 0.010506 + 7.1841 = 2.652 V 2 ( 0.525 ) I D = 0.525 ( 2.652 − 1.05 ) = 1.348 mA VDS = 10 − (1.348 )( 3) = 5.957 V (2)-(4) 2 4 − VGS = 0.475 (VGS − 1.9VGS + 0.9025 ) 2 2 0.475VGS + 0.0975VGS − 3.5713 = 0 VGS = −0.0975 ± 0.00950625 + 6.78547 2 ( 0.475 ) VGS = 2.641 V I D = 0.475 ( 2.641 − 0.95 ) = 1.359 mA VDS = 10 − (1.359 )( 3) = 5.924 V (1)-(4) 2 4 −VGS = ( 0.525) (VGS −1.9VGS + 0.9025) 2 2 0.525 VGS + 0.0025VGS − 3.5262 = 0 VGS = −0.0025 ± 0.00000625 + 7.40502 2 ( 0.525) = 2.5893 V I D = ( 0.525)( 2.5893 − 0.95) = 1.411 2 VDS = 10 − I D ( 3) = 5.7678 V (2)-(3) 2 4 − VGS = 0.475 (VGS − 2.1VGS +1.1025 ) 2 0.475VGS + 0.0025VGS − 3.4763 = 0 −0.0025 ± 0.00000625 + 6.60499 2(0.475) = 2.7027 VGS = VGS I D = (0.475)(2.7027 − 1.05) 2 = 1.2973 mA VDS = 10 − I D (3) = 6.108 V 1.297 ≤ I DQ ≤ 1.411 mA 5.768 ≤ VDS ≤ 6.108 V EX3.6 54. ⎛ R2 ⎞ VG = ⎜ ⎟ (10 ) − 5 ⎝ R1 + R2 ⎠ ⎛ 200 ⎞ =⎜ ⎟ (10 ) − 5 = 0.714 V ⎝ 350 ⎠ VS = 5 − I D RS = 5 − (1.2 ) I D So VSG = VS − VG = 5 − (1.2 ) I D − 0.714 = 4.286 − (1.2 ) I D ID = 4.286 − VSG 1.2 I D = K p (VSG + VTP ) 2 ( 2 4.286 − VSG = (1.2 )( 0.25 ) × VSG − 2VSG ( −1) + ( −1) 2 4.286 − VSG = ( 0.3) V − 0.6VSG + 0.3 ) 2 SG 2 0.3VSG + 0.4VSG − 3.986 = 0 VSG = ( 0.4 ) −0.4 ± + 4 ( 0.3)( 3.986 ) 2 2 ( 0.3) Must use + sign ⇒ VSG = 3.04 V I D = ( 0.25 )( 3.04 − 1) ⇒ I D = 1.04 mA 2 VSD = 10 − I D ( RS + RD ) = 10 − (1.04 )(1.2 + 4 ) ⇒ VSD = 4.59 V VSD > VSD ( sat ) , Yes EX3.7 VSD = 10 − I DQ ( RS + RP ) VSD = 10 − K P (VSG + VTP ) ( RS + RP ) 2 Set VSD = VSG + VTP VSG + VTP = 10 − ( 0.25 )(VSG + VTP ) ( 5.2 ) 2 1.3 (VSG + VTP ) + (VSG + VTP ) − 10 = 0 2 (VSG + VTP ) = −1 ± 1 + 4 (1.3)(10 ) 2 (1.3) = 2.415 V ( 3.42 V ) VSD = 2.415 V ( 2.42 V ) 2 I D = ( 0.25 )( 2.415 ) = 1.46 mA VSG = 3.415 V EX3.8 ⎛ R2 ⎞ ⎛ 240 ⎞ VG = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 R1 + R2 ⎠ ⎝ 240 + 270 ⎠ ⎝ VG = −0.294 V ID = VS − ( −5 ) RS = VG − VGS + 5 2 = K n (VGS − VTN ) RS ⎛ 0.08 ⎞ 2 4.706 − VGS = ⎜ ⎟ ( 4 )( 3.9 ) (VGS − 2.4VGS + 1.44 ) ⎝ 2 ⎠ 2 0.624VGS − 0.4976VGS − 3.80744 = 0 55. VGS = 0.4976 ± 0.2476 + 9.50337 2 ( 0.624 ) VGS = 2.90 V 2 ⎛ 0.08 ⎞ ID = ⎜ ⎟ ( 4 )( 2.90 − 1.2 ) ⇒ I D = 0.463 mA 2 ⎠ ⎝ VDS = 10 − I D ( 3.9 + 10 ) ⇒ VDS = 3.57 V EX3.9 10 − VSG 2 ID = and I D = K p (VSG + VTP ) RS 0.12 = ( 0.050 )(VSG − 0.8 ) VSG = 2.35 V 2 10 − 2.35 ⇒ RS = 63.75 kΩ 0.12 VSD = 8 = 20 − I D ( RS + RD ) 8 = 20 − ( 0.12 )( 63.75 ) − ( 0.12 ) RD RS = RD = (1) (2) (3) (4) 20 − ( 0.12 )( 63.75 ) − 8 KP KP VTP VTP ID ⇒ RD = 36.25 kΩ 0.12 = ( 0.05 )(1.05 ) = 0.0525 = ( 0.05 )( 0.95 ) = 0.0475 = −0.8 (1.05 ) = −0.84 V = −0.8 ( 0.95 ) = −0.76 V 10 − VSG 2 = = K P (VSG + VTP ) RS (1)-(3) 2 10 − VSG = ( 63.75 )( 0.0525 ) ⎡VSG − 1.68VSG + 0.7056 ⎤ ⎣ ⎦ 2 3.347VSG − 4.623VSG − 7.6384 = 0 VSG = 4.623 ± 21.372 + 102.263 2 ( 3.347 ) VSG = 2.352 V ⇒ I D ≅ 0.120 mA VSD ≈ 8.0 V (2)-(4) 2 10 − VSG = ( 63.75 )( 0.0475 ) ⎡VSG − 1.52VSG + 0.5776 ⎤ ⎣ ⎦ 2 3.028VSG − 3.603VSG − 8.251 = 0 VSG = 3.603 ± 12.9816 + 99.936 2 ( 3.028 ) VSG = 2.35 V I D ≈ 0.120 VSD ≈ 8.0 (1)-(4) 2 10 − VSG = ( 63.75 )( 0.0525 ) ⎡VSG − 1.52VSG + 0.5776 ⎤ ⎣ ⎦ 2 3.347VSG − 4.087VSG − 8.06685 = 0 56. VSG = 4.087 ± 16.7036 + 107.999 2 ( 3.347 ) VSG = 2.279 V I D = 0.121 mA VSD = 7.89 V (2)-(3) 2 10 − VSG = ( 63.75 )( 0.0475 ) ⎡VSG − 1.68VSG + 0.7056 ⎤ ⎣ ⎦ 2 3.028VSG − 4.0873VSG − 7.8634 = 0 VSG = 4.0873 ± 16.706 + 95.242 2 ( 3.028 ) VSG = 2.422 V I D = 0.119 mA VSD = 8.11 V Summary 0.119 ≤ I D ≤ 0.121 mA 7.89 ≤ VSD ≤ 8.11 V EX3.10 V − VGS , I D = DD RS I D = K n (VGS − VTN ) 2 2 2 10 − VGS = (10 )( 0.2 ) (VGS − 2VGSVTN + VTN ) 2 10 − VGS = 2VGS − 8VGS + 8 2 2VGS − 7VGS − 2 = 0 VGS = 7± (7) + 4 ( 2) 2 2 ( 2) 2 Use + sign: VGS = VDS = 3.77 V 10 − 3.77 ⇒ I D = 0.623 mA 10 Power = I DVDS = ( 0.623)( 3.77 ) ⇒ Power = 2.35 mW ID = EX3.11 (a) VI = 4 V, Driver in Non ⋅ Sat. K nD ⎡ 2 (VI − VTND ) VO − VO2 ⎤ = K nL [VDD − VO − VTNL ] ⎣ ⎦ 2 2 5 ⎡ 2 ( 4 − 1) VD − VD ⎤ = ( 5 − VD − 1) = ( 4 − VO ) = 16 − 8VO + VO2 ⎣ ⎦ 2 2 2 6VD − 38VO + 16 = 0 VD = 38 ± 1444 − 384 2 ( 6) VD = 0.454 V (b) VI = 2 V Driver: Sat K nD [VI − VTND ] = K nL [VDD − VO − VTNL ] 2 2 5 [ 2 − 1] = [5 − VO − 1] 2 2 5 = 4 − VO ⇒ VO = 1.76 V EX3.12 If the transistor is biased in the saturation region 57. I D = K n (VGS − VTN ) = K n ( −VTN ) 2 2 I D = ( 0.25 )( 2.5 ) ⇒ I D = 1.56 mA 2 VDS = VDD − I D RS = 10 − (1.56 )( 4 ) ⇒ VDS = 3.75 VDS > VGS − VTN = −VTN 3.75 > − ( −2.5 ) Yes — biased in the saturation region Power = I DVDS = (1.56 )( 3.75 ) ⇒ Power = 5.85 mW EX3.13 (a) For VI = 5 V, Load in saturation and driver in nonsaturation. I DD = I DL K nD ⎡ 2 (VI − VTND ) VO − VO2 ⎤ = K nL ( −VTNL ) ⎣ ⎦ 2 K nD ⎡ K 2 2 5 − 1)( 0.25 ) − ( 0.25 ) ⎤ = 4 ⇒ nD = 2.06 ⎣ ( ⎦ K nL K nL (b) I DL = K nL ( −VTNL ) ⇒ 0.2 = K nL ⎡ − ( −2 ) ⎤ ⎣ ⎦ 2 2 K nL = 50 μ A / V 2 and K nD = 103 μ A / V 2 EX3.14 For M N I DN = I DP K n (VGSN − VTN ) = K p (Vscop + VTP ) 2 2 VGSN = 1 + ( 5 − 3.25 − 1) = 1.75 V = VI Vo = VDSN ( sat ) = 1.75 − 1 ⇒ Vo = 0.75 V For M P : VI = 1.75 V VDD − VO = VSD ( sat ) = VSGP + VTP = ( 5 − 3.25 ) − 1 = 0.75 V So Vot = 5 − 0.75 ⇒ Vot = 4.25 V EX3.15 For RD = 10 k Ω, VDD = 5 V, and Vo = 1 V 5 −1 = 0.4 mA 10 2 I D = K n ⎡ 2 (VGS − VTN ) VDS − VDS ⎤ ⎣ ⎦ ID = 2 I D = 0.4 = K n ⎡ 2 ( 5 − 1)(1) − (1) ⎤ ⇒ K n = 0.057 mA / V 2 ⎣ ⎦ P = I D ⋅ VDS = ( 0.4 )(1) ⇒ P = 0.4 mW EX3.16 a. V1 = 5 V, V2 = 0, M 2 cutoff ⇒ I D 2 = 0 I D = K n ⎡ 2 (VI − VTN ) VO − VO2 ⎤ = ⎣ ⎦ 5 − VO RD ( 0.05 )( 30 ) ⎡ 2 ( 5 − 1)V0 − V02 ⎤ = 5 − V0 ⎣ ⎦ 58. 1.5V02 − 13V0 + 5 = 0 V0 = 13 ± (13) − 4 (1.5 )( 5) ⇒ V0 = 0.40 V 2 (1.5 ) 2 5 − 0.40 ⇒ I R = I D1 = 0.153 mA 30 V1 = V2 = 5 V I R = I D1 = b. 5 − VO = 2 K n ⎡ 2 (VI − VTN ) VO − VO2 ⎤ ⎣ ⎦ RD { } 5 − V0 = 2 ( 0.05 )( 30 ) ⎡ 2 ( 5 − 1)V0 − V02 ⎤ ⎣ ⎦ 3V02 − 25V0 + 5 = 0 V0 = 25 ± ( 25) − 4 ( 3)( 5 ) ⇒ V0 = 0.205 V 2 ( 3) 2 5 − 0.205 ⇒ I R = 0.160 mA 30 = I D 2 = 0.080 mA IR = I D1 EX3.17 M 2 & M 3 watched ⇒ I Q1 = I REF 1 = 0.4 mA 0.4 = 0.3 (VGS 3 − 1) ⇒ VGS 3 = VGS 2 = 2.15 V 2 0.4 = 0.6 (VGS 1 − 1) ⇒ VGS1 = 1.82 V 2 EX3.18 2 ⎛ 0.04 ⎞ 0.1 = ⎜ ⎟ (15 )(VSGC − 0.6 ) ⎝ 2 ⎠ VSGC = 1.177 V = VSGB 2 ⎛ 0.04 ⎞ ⎛ W ⎞ 0.2 = ⎜ ⎟ ⎜ ⎟ (1.177 − 0.6 ) ⎝ 2 ⎠ ⎝ L ⎠B ⎛W ⎞ ⎜ ⎟ = 30 ⎝ L ⎠B 2 ⎛ 0.04 ⎞ 0.2 = ⎜ ⎟ ( 25 )(VSGA − 0.6 ) ⎝ 2 ⎠ VSGA = 1.23 V EX3.19 (a) I REF = K n 3 (VGS 3 − VTN ) = K n 4 (VGS 4 − VTN ) VGS 3 = 2 V ⇒ VGS 4 = 3 V K K 1 2 2 ( 2 − 1) = n 4 ( 3 − 1) ⇒ n 4 = K n3 K n3 4 (b) I Q = K n 2 (VGS 2 − VTN ) But VGS 2 = VGS 3 = 2 V 2 0.1 = K n 2 ( 2 − 1) ⇒ 2 (c) 2 K n 2 = 0.1 mA / V 2 0.2 = K n 3 ( 2 − 1) ⇒ K n 3 = 0.2 mA / V 2 2 0.2 = K n 4 ( 3 − 1) ⇒ K n 4 = 0.05 mA / V 2 2 EX3.20 2 59. VS 2 = 5 − 5 = 0 RS 2 = I D 2 = K n 2 (VGS 2 − VTN 2 ) 5 = 16.7 K 0.3 2 0.3 = 0.2 (VGS 2 − 1.2 ) ⇒ VGS 2 = 2.425 V ⇒ VG 2 = VGS 2 + VS = 2.425 V 2 5 − 2.425 = 25.8 K 0.1 VS 1 = VG 2 − VDSQ1 = 2.425 − 5 = −2.575 V RD1 = RS 1 = −2.575 − ( −5 ) 0.1 ⇒ RS 1 = 24.3 K I D1 = K n1 (VGS 1 − VTN 1 ) 2 0.1 = 0.5 (VGS1 − 1.2 ) ⇒ VGS 1 = 1.647 V VG1 = VGS 1 + VS 1 = 1.647 + ( −2.575 ) ⇒ VG1 = −0.928 V 2 ⎛ R2 ⎞ 1 VG1 = ⎜ ⎟ (10 ) − 5 = ⋅ RTN ⋅ (10 ) − 5 R1 ⎝ R1 + R2 ⎠ 1 −0.928 = ( 200 )(10 ) − 5 ⇒ R1 = 491 K R1 491 R2 = 200 ⇒ R2 = 337 K 491 + R2 EX3.21 VS1 = I D RS − 5 = (0.25)(16) − 5 = −1 V I DQ = K n (VGS1 − VTN ) 2 ⇒ 0.25 = 0.5(VGS 1 − 0.8) 2 ⇒ VGS 1 = 1.507 V VG1 = VGS 1 + VS 1 = 1.507 − 1 = 0.507 V ⎛ ⎞ R3 R3 (5) ⇒ R3 = 50.7 K VG1 = ⎜ ⎟ (5) ⇒ 0.507 = 500 ⎝ R1 + R2 + R3 ⎠ VS 2 = VS 1 + VDS1 = −1 + 2.5 = 1.5 V VG 2 = VS 2 + VGS = 1.5 + 1.507 = 3.007 V ⎛ R2 + R3 ⎞ ⎛ R2 + R3 ⎞ VG 2 = ⎜ ⎟ (5) ⇒ 3.007 = ⎜ ⎟ (5) R1 + R2 + R3 ⎠ ⎝ 500 ⎠ ⎝ R2 + R3 = 300.7 R2 = 300.7 − 50.7 ⇒ R2 = 250 K R1 = 500 − 250 − 50.7 ⇒ R1 = 199.3 K VD 2 = VS 2 + VDS 2 = 1.5 + 2.5 = 4 V 5−4 ⇒ RD = 4 K RD = 0.25 EX3.22 VDS ( sat ) = VGS − VP = −1.2 − ( −4.5 ) ⇒ VDS ( sat ) = 3.3 V ⎛ ( −1.2 ) ⎞ ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ = 12 ⎜ 1 − ⇒ I D = 6.45 mA ⎜ ( −4.5 ) ⎟ ⎟ VP ⎠ ⎝ ⎝ ⎠ 2 2 EX3.23 Assume the transistor is biased in the saturation region. 60. ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ 2 2 ⎛ ⎞ V 8 = 18 ⎜ 1 − GS ⎟ ⇒ VGS = −1.17 V ⇒ VS = −VGS = 1.17 ⎜ ( −3.5 ) ⎟ ⎝ ⎠ VD = 15 − ( 8 )( 0.8 ) = 8.6 VDS = 8.6 − (1.17 ) = 7.43 V VDS = 7.43 > VGS − VP = −1.17 − ( −3.5 ) = 2.33 Yes, the transistor is biased in the saturation region. EX3.24 I D = 2.5 mA ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ 2 2 ⎛ V ⎞ 2.5 = 6 ⎜1 − GS ⎟ ⇒ VGS = −1.42 V ⎜ ( −4 ) ⎟ ⎝ ⎠ VS = I D RS − 5 = ( 2.5 )( 0.25 ) − 5 VS = −4.375 VDS = 6 ⇒ VD = 6 − 4.375 = 1.625 5 − 1625 RD = ⇒ RD = 1.35 kΩ 2.5 ( 20 ) 2 R1 + R2 = 2 ⇒ R1 + R2 = 200 kΩ VG = VGS + VS = −1.42 − 4.375 = −5.795 ⎛ R2 ⎞ VG = ⎜ ⎟ ( 20 ) − 10 ⎝ R1 + R2 ⎠ ⎛ R ⎞ −5.795 = ⎜ 2 ⎟ ( 20 ) − 10 ⇒ R2 = 42.05 kΩ → 42 kΩ ⎝ 200 ⎠ R1 = 157.95 kΩ → 158 kΩ EX3.25 VS = −VGS . I D = 0 − VS VGS = RS RS ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ 2 ⎛ V VGS V2 ⎞ ⎛ V ⎞ = 6 ⎜ 1 − GS ⎟ = 6 ⎜ 1 − GS + GS ⎟ 1 4 ⎠ 2 16 ⎠ ⎝ ⎝ 2 0.375VGS − 4VGS + 6 = 0 2 VGS = 4 ± 16 − 4 ( 0.375 )( 6 ) 2 ( 0.375 ) VGS = 8.86 or VGS = 1.806 V impossible ID = VGS = 1.806 mA RS 61. VD = I D RD − 5 = (1.81)( 0.4 ) − 5 = −4.278 VSD = VS − V0 = −1.81 − ( −4.276 ) ⇒ VSD = 2.47 V VSD ( sat ) = VP − VGS = 4 − 1.81 = 2.19 So VSD > VSD ( sat ) EX3.26 Rin = R1 R2 = R1 R2 = 100 kΩ R1 + R2 I DQ = 5 mA, VS = − I DQ RS = − ( 5 )(1.2 ) = −6 V VSDQ = 12 V ⇒ VD = VS − VSDQ = −6 − 12 = −18 V RD = −18 − ( −20 ) 5 ⇒ RD = 0.4 kΩ 2 ⎛ V ⎞ ⎛ V ⎞ I DQ = I DSS ⎜ 1 − GS ⎟ ⇒ 5 = 8 ⎜ 1 − GS ⎟ VP ⎠ 4 ⎠ ⎝ ⎝ VGS = 0.838 V 2 VG = VGS + VS = 0.838 − 6 = −5.162 ⎛ R2 ⎞ VG = ⎜ ⎟ ( −20 ) ⎝ R1 + R2 ⎠ 1 −5.162 = (100 )( −20 ) ⇒ R1 = 387 kΩ R1 R1 R2 = 100 ⇒ ( 387 ) R2 = 100 ( 387 ) + 100 R2 R1 + R2 ( 387 − 100 ) R2 = (100 )( 387 ) ⇒ R2 = 135 kΩ TYU3.1 VTN = 1.2 V , VGS = 2 V (a) V DS ( sat ) = VGS − VTN = 2 − 1.2 = 0.8 V (i) VDS = 0.4 ⇒ Nonsaturation (ii) VDS = 1 ⇒ Saturation (iii) VDS = 5 ⇒ Saturation (b) VTN = −1.2 V , VGS = 2 V V DS ( sat ) = VGS − VTN = 2 − ( −1.2 ) = 3.2 V (i) VDS = 0.4 ⇒ Nonsaturation (ii) VDS = 1 ⇒ Nonsaturation (iii) VDS = 5 ⇒ Saturation TYU3.2 (a) W μ n Cox 2L −14 ∈ox ( 3.9 ) ( 8.85 × 10 ) Cox = = = 7.67 × 10−8 F / cm −8 tox 450 × 10 Kn = Kn = (100 )( 500 ) ( 7.67 ×10−8 ) ⇒ K n = 0.274 mA / V 2 2 (7) (b) VTN = 1.2 V, VGS = 2 V 62. (i) VDS = 0.4 V ⇒ Nonsaturation (ii) 2 I D = ( 0.274 ) ⎡ 2 ( 2 − 1.2 )( 0.4 ) − ( 0.4 ) ⎤ ⇒ I D = 0.132 mA ⎣ ⎦ VDS = 1 V ⇒ Saturation I D = ( 0.274 )( 2 − 1.2 ) ⇒ I D = 0.175 mA 2 (iii) VDS = 5 V ⇒ Saturation I D = ( 0.274 )( 2 − 1.2 ) ⇒ I D = 0.175 mA 2 VTN = −1.2 V , VGS = 2 V (i) VDS = 0.4 V ⇒ Nonsaturation (ii) 2 I D = ( 0.274 ) ⎡ 2 ( 2 + 1.2 )( 0.4 ) − ( 0.4 ) ⎤ ⇒ I D = 0.658 mA ⎣ ⎦ VDS = 1 V ⇒ Nonsaturation 2 I D = ( 0.274 ) ⎡ 2 ( 2 + 1.2 )(1) − (1) ⎤ ⇒ I D = 1.48 mA ⎣ ⎦ (iii) VDS = 5 V ⇒ Saturation I D = ( 0.274 )( 2 + 1.2 ) ⇒ I D = 2.81 mA 2 TYU3.3 (a) VSD (sat) = VSG + VTP = 2 − 1.2 = 0.8 V (i) Non Sat (ii) Sat (iii) Sat (b) VSD (sat) = 2 + 1.2 = 3.2 V (i) Non Sat (ii) Non Sat (iii) Sat TYU3.4 (a) (3.9)(8.85 × 10−14 ) ⎛ W ⎞ ⎛ μ p Cox ⎞ KP = ⎜ ⎟ ⎜ ⎟ Cox = 350 × 10−8 ⎝ L ⎠⎝ Z ⎠ = 9.861× 10−8 KP = (40) ⎛ ( 300 ) ( 9.861× 10 ⎜ (2) ⎜ 2 ⎝ K P = 0.296 mA / V (b) (i) −8 )⎞ ⎟ ⎟ ⎠ 2 I D = (0.296) ⎡ 2(2 − 1.2)(0.4) − (0.4) 2 ⎤ ⎣ ⎦ = 0.142 mA (ii) I D = (0.296) [ 2 − 1.2] ⇒ I D = 0.189 mA 2 (iii) ID = 0.189 mA 2 (i) I D = (0.296) ⎡ 2 ( 2 + 1.2 )( 0.4 ) − ( 0.4 ) ⎤ ⎣ ⎦ = 0.710 mA (ii) 2 I D = (0.296) ⎡ 2 ( 2 + 1.2 )(1) − (1) ⎤ ⎣ ⎦ =1.60 mA (iii) I D = ( 0.296 )( 2 + 1.2 ) = 3.03 mA TYU3.5 2 63. (a) λ = 0, VDS ( sat ) = 2.5 − 0.8 = 1.7 V For VDS = 2 V , VDS = 10 V ⇒ Saturation Region I D = ( 0.1)( 2.5 − 0.8 ) ⇒ I D = 0.289 mA 2 (b) λ = 0.02 V −1 I D = K n (VGS − VTN ) (1 + λVDS ) For VDS = 2 V 2 I D = ( 0.1)( 2.5 − 0.8 ) ⎡1 + ( 0.02 )( 2 ) ⎤ ⇒ I D = 0.300 mA ⎣ ⎦ VDS = 10 V 2 (c) 2 I D = ( 0.1) ⎡( 2.5 − 0.8 ) (1 + ( 0.02 )(10 ) ) ⎤ ⇒ I D = 0.347 mA ⎣ ⎦ For part (a), λ = 0 ⇒ ro = ∞ For part (b), λ = 0.02 V −1 , −1 2 2 ro = ⎡λ K n (VGS − VTN ) ⎤ = ⎡( 0.02 )( 0.1)( 2.5 − 0.8 ) ⎤ ⎣ ⎦ ⎣ ⎦ −1 or ro = 173 k Ω TYU3.6 VTN = VTNO + γ ⎡ 2φ f + VSB − 2φ f ⎤ ⎣ ⎦ 2φ f = 0.70 V , VTNO = 1 V (a) VSB = 0 ⇒, VTN = 1 V (b) ⎡ ⎤ VSB = 1 V , VTN = 1 + ( 0.35 ) ⎣ 0.7 + 1 − 0.7 ⎦ ⇒ VTN = 1.16 V VSB = 4 V , VTN = 1 + ( 0.35 ) ⎡ 0.7 + 4 − 0.7 ⎤ ⇒ VTN = 1.47 V ⎣ ⎦ (c) TYU3.7 I D = K n (VGS − VTN ) 2 0.4 = 0.25 (VGS − 0.8 ) ⇒ VGS = 2.06 V 2 ⎛ R2 ⎞ VGS = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠ ⎛ R ⎞ 2.06 = ⎜ 2 ⎟ ( 7.5 ) ⇒ R2 = 68.8 kΩ ⎝ 250 ⎠ R1 = 181.2 kΩ VDS = 4 = VDD − I D RD 7.5 − 4 RD = ⇒ RD = 8.75 kΩ 0.4 VDS > VDS ( sat ) , Yes TYU3.8 64. VS − ( −5 ) ID = and VS = −VGS RS So RS = 5 − VGS 0.1 I D = K n (VGS − VTN ) 2 0.1 = ( 0.080 )(VGS − 1.2 ) ⇒ VGS = 2.32 V 5 − 2.32 So RS = ⇒ RS = 26.8 kΩ 0.1 VDS = VD − VS ⇒ VD = VDS + VS = 4.5 − 2.32 2 VD = 2.18 5 − VD 5 − 2.18 RD = = ⇒ RD = 28.2 kΩ ID 0.1 VDS > VDS ( sat ) , Yes TYU3.9 For VDS = 2.2 V 5 − 2.2 ⇒ I D = 0.56 mA ID = 5 I D = K n (VGS − VTN ) 0.56 = K n ( 2.2 − 1) 2 2 K n = 0.389 mA / V = W μ n Cox ⋅ 2 L W ( 389 )( 2 ) W = ⇒ = 19.4 L L ( 40 ) TYU3.10 (a) The transition point is ( VDD − VTNL + VTND 1 + K nD / K nL VIt = = ) 1 + K nD /K nL ( 5 − 1 + 1 1 + 0.05/ 0.01 ) 1 + 0.05/ 0.01 7.236 = ⇒ VIt = 2.236 V 3.236 VOt = VIt − VTND = 2.24 − 1 ⇒ VOt = 1.24 V (b) We may write I D = K n D (VGSD − VTND ) = ( 0.05 )( 2.236 − 1) ⇒ I D = 76.4 μ A 2 TYU3.11 2 ( VDD − VTNL + VTND 1 + K nD /K nL VIt = ) 1 + K nD /K nL 2.5 = ( 5 − 1 + 1 1 + K nD /K nL ) 1 + K nD /K nL 2.5 + 2.5 K nD /K nL = 5 + K nD /K nL ⇒ K nD /K nL = b. For VI = 5, driver in nonsaturated region. 5 − 2.5 = 1.67 ⇒ K nD /K nL = 2.78 1.5 65. I DD = I DL K nD ⎡ 2 (VI − VTND ) VO − VO2 ⎤ = K nL (VGSL − VTNL ) ⎣ ⎦ 2 K nD 2 ⎡ 2 (VI − VTND ) VO − VO2 ⎤ = [VDD − VO − VTNL ] ⎦ K nL ⎣ 2.78 ⎡ 2 ( 5 − 1) V0 − V02 ⎤ = [5 − V0 − 1] ⎣ ⎦ 2 22.24V0 − 2.78V02 = ( 4 − V0 ) 2 = 16 − 8V0 + V02 3.78V02 − 30.24V0 + 16 = 0 V0 = 30.24 ± ( 30.24 ) − 4 ( 3.78 )(16 ) ⇒ V0 = 0.57 V 2 ( 3.78 ) 2 TYU3.12 We have VDS = 1.2 V < VGS − VTN = −VTN = 1.8 V Transistor is biased in the nonsaturation region. V − VDS 5 − 1.2 2 ⎡ ⎤ = ⇒ I D = 0.475 mA I D = K n ⎣ 2 (VGS − VTN ) VDS − VDS ⎦ and I D = DD 8 RS 0.475 = K n ⎡ 2 ( 0 − ( −1.8 ) ) (1.2 ) − (1.2 ) ⎤ ⎣ ⎦ 0.475 = K n ( 2.88 ) ⇒ K n = 0.165 mA/V 2 2 W μ n Cox ⋅ 2 L 165 )( 2 ) W ( W = ⇒ = 9.43 35 L L Kn = TYU3.13 (a) Transition point for the load transistor – Driver is in the saturation region. I DD = I DL K nD (VGSD − VTND ) = K nL (VGSL − VTNL ) 2 2 VDSL ( sat ) = VGSL − VTNL = −VTNL ⇒ VDSL = VDD − VOt = 2 V Then VOt = 5 − 2 = 3 V , VOt = 3 V K nD (VIt − 1) = ( −VTNL ) K nL 0.08 (VIt − 1) = 2 ⇒ VIt = 1.89 V 0.01 (b) For the driver: VOt = VIt − VTND VIt = 1.89 V , VOt = 0.89 V TYU3.14 2 I D = K n ⎡ 2 (VGS − VTN ) VDS − VDS ⎤ ⎣ ⎦ 2 = ( 0.050 ) ⎡ 2 (10 − 0.7 )( 0.35 ) − ( 0.35 ) ⎤ ⎣ ⎦ I D = 0.319 mA RD = VDD − Vo 10 − 0.35 = ⇒ RD = 30.3 kΩ ID 0.319 TYU3.15 (a) Transistor biased in the nonsaturation region 66. 5 − 1.5 − VDS = 12 R 2 I D = K n ⎡ 2 (VGS − VTN ) VDS − VDS ⎤ ⎣ ⎦ ID = 2 12 = 4 ⎡ 2 ( 5 − 0.8 ) VDS − VDS ⎤ ⎣ ⎦ 2 4VDS − 33.6VDS + 12 = 0 ⇒ VDS = 0.374 V Then R = 5 − 1.5 − 0.374 ⇒ R = 261 Ω 12 TYU3.16 a. ID = 5 − VO = K n ⎡ 2 (V2 − VTN ) VO − VO2 ⎤ ⎣ ⎦ RD 5 − ( 0.10 ) b. 2 = K n ⎡ 2 ( 5 − 1)( 0.10 ) − ( 0.10 ) ⎤ ⇒ K n = 0.248 mA / V 2 ⎣ ⎦ 25 5 − V0 = 2 ( 0.248 ) ⎡ 2 ( 5 − 1) V0 − V02 ⎤ ⎣ ⎦ 25 2 5 − V0 = 12.4 ⎡8V0 − V0 ⎤ ⎣ ⎦ 12.4V02 − 100.2V0 + 5 = 0 V0 = 100.2 ± (100.2 ) − 4 (12.4 )( 5 ) ⇒ V0 = 0.0502 V 2 (12.4 ) 2 TYU3.17 2 2 I DQ = K (VGS − VTN ) ⇒ 5 = 50 (VGS − 0.15 ) ⇒ VGS = 0.466 V VS = ( 0.005 )(10 ) = 0.050 V ⇒ VGG = VGS + VS = 0.466 + 0.050 ⇒ VGG = 0.516 V VD = 5 − ( 0.005 )(100 ) ⇒ VD = 4.5 V VDS = VD − VS = 4.5 − 0.050 ⇒ VDS = 4.45 V TYU3.18 2 I D = K ⎡ 2 (VGS − VTN ) VDS − VDS ⎤ ⎣ ⎦ 2 = 100 ⎡ 2 ( 0.7 − 0.2 )( 0.1) − ( 0.1) ⎤ ⎣ ⎦ ID = 9 μA RD = 2.5 − 0.1 ⇒ RD = 267 kΩ 0.009 67. Chapter 3 Problem Solutions 3.1 ⎛ W ⎞ ⎛ k ′ ⎞ ⎛ 10 ⎞ ⎛ 0.08 ⎞ 2 Kn = ⎜ ⎟ ⎜ n ⎟ = ⎜ ⎟⎜ ⎟ = 0.333 mA/V L ⎠ ⎝ 2 ⎠ ⎝ 1.2 ⎠ ⎝ 2 ⎠ ⎝ For VDS = 0.1 V ⇒ Non Sat Bias Region (a) VGS = 0 ⇒ I D = 0 (b) 2 VGS = 1 V I D = 0.333 ⎡ 2 (1 − 0.8 )( 0.1) − ( 0.1) ⎤ = 0.01 mA ⎣ ⎦ (c) VGS = 2 V 2 I D = 0.333 ⎡ 2 ( 2 − 0.8 )( 0.1) − ( 0.1) ⎤ = 0.0767 mA ⎣ ⎦ (d) VGS = 3 V 2 I D = 0.333 ⎡ 2 ( 3 − 0.8 )( 0.1) − ( 0.1) ⎤ = 0.143 mA ⎣ ⎦ 3.2 All in Sat region ⎛ 10 ⎞⎛ 0.08 ⎞ 2 Kn = ⎜ ⎟⎜ ⎟ = 0.333 mA/V ⎝ 1.2 ⎠⎝ 2 ⎠ (a) ID = 0 (b) I D = 0.333[1 − 0.8] = 0.0133 mA (c) I D = 0.333[ 2 − 0.8] = 0.480 mA (d) I D = 0.333[3 − 0.8] = 1.61 mA 2 2 2 3.3 (a) Enhancement-mode (b) From Graph VT = 1.5 V Now 2 0.03 = K n ( 2 − 1.5 ) = 0.25 K n ⇒ K n = 0.12 0.15 = K n ( 3 − 1.5 ) = 2.25 K n K n = 0.0666 0.39 = K n ( 4 − 1.5 ) = 6.25 K n K n = 0.0624 0.77 = K n ( 5 − 1.5 ) = 12.25 K n K n = 0.0629 2 2 2 From last three, K n (Avg) = 0.0640 mA/V 2 (c) 3.4 a. iD (sat) = 0.0640(3.5 − 1.5) 2 ⇒ iD (sat) = 0.256 mA for VGS = 3.5 V iD (sat) = 0.0640(4.5 − 1.5) 2 ⇒ iD (sat) = 0.576 mA for VGS = 4.5 V VGS = 0 VDS ( sat ) = VGS − VTN = 0 − ( −2.5 ) = 2.5 V i. VDS = 0.5 V ⇒ Biased in nonsaturation ii. 2 I D = (1.1) ⎡ 2 ( 0 − (−2.5) )( 0.5 ) − ( 0.5 ) ⎤ ⇒ I D = 2.48 mA ⎣ ⎦ VDS = 2.5 V ⇒ Biased in saturation I D = (1.1) ( 0 − ( −2.5 ) ) ⇒ I D = 6.88 mA 2 iii. VDS = 5 V Same as (ii) ⇒ I D = 6.88 mA b. VGS = 2 V VDS ( sat ) = 2 − ( −2.5 ) = 4.5 V i. VDS = 0.5 V ⇒ Nonsaturation I D = (1.1) ⎡ 2(2 − (−2.5))(0.5) − (0.5) 2 ⎤ ⇒ I D = 4.68 mA ⎣ ⎦ 68. VDS = 2.5 V ⇒ Nonsaturation ii. I D = (1.1) ⎡ 2(2 − (−2.5))(2.5) − (2.5) 2 ⎤ ⇒ I D = 17.9 mA ⎣ ⎦ VDS = 5 V ⇒ Saturation iii. I D = (1.1) ( 2 − ( −2.5 ) ) ⇒ I D = 22.3 mA 2 3.5 VDS > VGS − VTN = 0 − ( −2 ) = 2 V Biased in the saturation region k′ W 2 I D = n ⋅ (VGS − VTN ) 2 L 2 W ⎛ 0.080 ⎞ ⎛ W ⎞ = 9.375 1.5 = ⎜ ⎟ ⎜ ⎟ ⎡ 0 − ( −2 ) ⎤ ⇒ ⎣ ⎦ L ⎝ 2 ⎠⎝ L ⎠ 3.6 ′ kn = μ n Cox = μ n ∈ox tox = ( 600 )( 3.9 ) (8.85 ×10−14 ) tox (a) 500 A 250 ′ kn = 82.8 μ A/V 2 (c) 100 ′ kn = 207 μ A/V 2 (d) 50 ′ kn = 414 μ A/V 2 (e) 25 2.071× 10−10 tox ′ kn = 41.4 μ A/V 2 (b) = ′ kn = 828 μ A/V 2 3.7 a. Cox = Kn = ∈ox ( 3.9 ) ( 8.85 × 10 = t0 x 450 × 10−8 −14 )⇒∈ ox t0 x = 7.67 ×10−8 F/cm 2 μ n Cox W 2 ⋅ L 1 64 ( 650 ) ( 7.67 ×10−8 ) ⎛ ⎞ ⎜ ⎟ 2 ⎝ 4 ⎠ K n = 0.399 mA / V 2 = b. VGS = VDS = 3 V ⇒ Saturation I D = K n (VGS − VTN ) = ( 0.399 )( 3 − 0.8 ) ⇒ I D = 1.93 mA 2 3.8 2 ⎛ ω ⎞⎛ k′ ⎞ I D = ⎜ ⎟ ⎜ n ⎟ (VGS − VTN ) 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎛ ω ⎞ ⎛ 0.08 ⎞ 1.25 ⎜ ⎟⎜ ⎟ ( 2.5 − 1.2 ) ⇒ ω = 23.1 μ m ⎝ 1.25 ⎠ ⎝ 2 ⎠ 3.9 ∈ Cox = ox t0 x = ( 3.9 ) (8.85 ×10−14 ) 400 × 10−8 = 8.63 × 10−8 F/cm 2 2 69. Kn = μ n Cox W ⋅ L 1 ⎛W ⎞ = ( 600 ) ( 8.63 × 10−8 ) ⎜ ⎟ 2 ⎝ 2.5 ⎠ 2 K n = (1.036 × 10−5 ) W I D = K n (VGS − VTN ) 2 1.2 × 10 −3 = (1.036 × 10 −5 ) W ( 5 − 1) ⇒ W = 7.24 μ m 2 3.10 Biased in the saturation region in both cases. ′ kp W 2 I D = ⋅ (VSG + VTP ) 2 L 2 ⎛ 0.040 ⎞⎛ W ⎞ (1) 0.225 = ⎜ ⎟⎜ ⎟ ( 3 + VTP ) 2 ⎠⎝ L ⎠ ⎝ 2 ⎛ 0.040 ⎞ ⎛ W ⎞ 1.40 = ⎜ (2) ⎟ ⎜ ⎟ ( 4 + VTP ) ⎝ 2 ⎠⎝ L ⎠ Take ratio of (2) to (1): (4 + VTP ) 2 1.40 = 6.222 = 0.225 (3 + VTP ) 2 6.222 = 2.49 = 4 + VTP ⇒ VTP = −2.33 V 3 + VTP W 2 ⎛ 0.040 ⎞ ⎛ W ⎞ Then 0.225 = ⎜ = 25.1 ⎟ ⎜ ⎟ ( 3 − 2.33) ⇒ L ⎝ 2 ⎠⎝ L ⎠ 3.11 VS = 5 V, VG = 0 ⇒ VSG = 5 V VTP = −0.5 V ⇒ VSD ( sat ) = VSG + VTP = 5 − 0.5 = 4.5 V a. VD = 0 ⇒ VSD = 5 V ⇒ Biased in saturation I D = 2 ( 5 − 0.5 ) ⇒ I D = 40.5 mA 2 b. VD = 2 V ⇒ VSD = 3 V ⇒ Nonsaturation c. 2 I D = 2 ⎡ 2 ( 5 − 0.5 )( 3) − ( 3) ⎤ ⇒ I D = 36 mA ⎣ ⎦ VD = 4 V ⇒ VSD = 1 V ⇒ Nonsaturation d. 2 I D = 2 ⎡ 2 ( 5 − 0.5 )(1) − (1) ⎤ ⇒ I D = 16 mA ⎣ ⎦ VD = 5 V ⇒ VSD = 0 ⇒ I D = 0 3.12 (a) (b) Enhancement-mode From Graph VTP = + 0.5 V 0.45 = k p ( 2 − 0.5 ) = 2.25 K p ⇒ K p = 0.20 1.25 = k p ( 3 − 0.5 ) = 6.25 K p 0.20 2.45 = k p ( 4 − 0.5 ) = 12.25 K p 0.20 2 2 2 4.10 = k p ( 5 − 0.5 ) = 20.25 K p 2 (c) 0.202 Avg K p = 0.20 mA/V 2 iD (sat) = 0.20 (3.5 − 0.5) 2 = 1.8 mA iD (sat) = 0.20 (4.5 − 0.5) 2 = 3.2 mA 70. 3.13 VSD ( sat ) = VSG + VTP (a) VSD ( sat ) = −1 + 2 ⇒ VSD ( sat ) = 1 V (b) VSD ( sat ) = 0 + 2 ⇒ VSD ( sat ) = 2 V (c) VSD ( sat ) = 1 + 2 ⇒ VSD ( sat ) = 3 V ID = (a) (b) (c) ′ kp W k′ W 2 2 p ⋅ (VSG + VTP ) = ⋅ ⋅ ⎡VSD ( sat ) ⎤ ⎣ ⎦ 2 L 2 L 2 ⎛ 0.040 ⎞ ID = ⎜ ⎟ ( 6 )(1) ⇒ I D = 0.12 mA ⎝ 2 ⎠ 2 ⎛ 0.040 ⎞ ID = ⎜ ⎟ ( 6 )( 2 ) ⇒ I D = 0.48 mA ⎝ 2 ⎠ 2 ⎛ 0.040 ⎞ ID = ⎜ ⎟ ( 6 )( 3) ⇒ I D = 1.08 mA ⎝ 2 ⎠ 3.14 VSD (sat) = VSG + VTP = 3 − 0.8 = 2.2 V ⎛ 15 ⎞⎛ 0.04 ⎞ 2 KP = ⎜ ⎟⎜ ⎟ = 0.25 mA/V ⎝ 1.2 ⎠⎝ 2 ⎠ a) b) c) d) e) 2 VSD = 0.2 Non Sat I D = 0.25 ⎡ 2 ( 3 − 0.8 )( 0.2 ) − ( 0.2 ) ⎤ = 0.21 mA ⎣ ⎦ 2 VSD = 1.2 V Non Sat I D = 0.25 ⎡ 2 ( 3 − 0.8 )(1.2 ) − (1.2 ) ⎤ = 0.96 mA ⎣ ⎦ 2 VSD = 2.2 V Sat I D = 0.25(3 − 0.8) = 1.21 mA VSD = 3.2 V Sat ID = 1.21 mA VSD = 4.2 V Sat ID = 1.21 mA 3.15 ′ k p = μ p Cox = μ p ∈ox t0 x = ( 250 )( 3.9 ) (8.85 ×10−14 ) t0 x (a) tox = 500Å ⇒ k ′ = 17.3 μ A/V 2 p (b) 250Å ⇒ k ′ = 34.5 μ A/V 2 p (c) 100Å ⇒ k ′ = 86.3 μ A/V 2 p (d) ′ 50Å ⇒ k p = 173 μ A/V 2 (e) = 25Å ⇒ k ′ = 345 μ A/V 2 p 3.16 −14 ∈ox ( 3.9 ) ( 8.85 × 10 ) = = 6.90 × 10−8 F/cm 2 Cox = −8 t0 x 500 × 10 ′ kn = ( μ n Cox ) = ( 675 ) ( 6.90 × 10−8 ) ⇒ 46.6 μ A/V 2 k ′ = ( μ p Cox ) = ( 375 ) ( 6.90 × 10−8 ) ⇒ 25.9 μ A/V 2 p PMOS: 8.629 × 10−11 t0 x 71. ID = k′ ⎛ W ⎞ 2 p ⎜ ⎟ (VSG + VTP ) 2 ⎝ L ⎠p 2 ⎛ 0.0259 ⎞⎛ W ⎞ ⎛W ⎞ 0.8 = ⎜ ⎟⎜ ⎟ ( 5 − 0.6 ) ⇒ ⎜ ⎟ = 3.19 ⎝ 2 ⎠⎝ L ⎠ p ⎝ L ⎠p L = 4 μ m ⇒ W p = 12.8 μ m ⎛ 0.0259 ⎞ 2 Kp = ⎜ ⎟ ( 3.19 ) ⇒ K p = 41.3 μ A/V = K n ⎝ 2 ⎠ Want Kn = Kp k′ ⎛ W ⎞ ′ kn ⎛ W ⎞ p ⎜ ⎟ = ⎜ ⎟ = 41.3 2 ⎝ L ⎠N 2 ⎝ L ⎠p ⎛ 46.6 ⎞ ⎛ W ⎞ ⎛W ⎞ ⎜ ⎟ ⎜ ⎟ = 41.3 ⇒ ⎜ ⎟ = 1.77 2 ⎠ ⎝ L ⎠N ⎝ ⎝ L ⎠N L = 4 μ m ⇒ WN = 7.09 μ m 3.17 VGS = 2 V, I D = ( 0.2 )( 2 − 1.2 ) = 0.128 mA 1 1 r0 = = ⇒ r0 = 781 kΩ λ I D ( 0.01)( 0.128 ) 2 VGS = 4 V, I D = ( 0.2 )( 4 − 1.2 ) = 1.57 mA 1 r0 = ⇒ r = 63.7 kΩ ( 0.01)(1.57 ) 0 2 VA = 1 λ = 1 ⇒ VA = 100 V ( 0.01) 3.18 2 2 ⎛ 0.080 ⎞ ID = ⎜ ⎟ ( 4 )( 3 − 0.8 ) = ( 0.16 )( 3 − 0.8 ) ⇒ I D = 0.774 mA 2 ⎠ ⎝ 1 1 1 ⇒λ = = ⇒ λ (max) = 0.00646 V −1 r0 = λ ID r0 I D ( 200 )( 0.774 ) VA ( min ) = 1 λ ( max ) = 1 ⇒ VA ( min ) = 155 V 0.00646 3.19 VTN = VTNO + γ ⎡ 2φ f + VSB − 2φ f ⎤ ⎣ ⎦ ΔVTN = 2 = ( 0.8 ) ⎡ 2φ f + VSB − 2 ( 0.35 ) ⎤ ⎣ ⎦ 2.5 + 0.837 = 2 ( 0.35 ) + VSB ⇒ VSB = 10.4 V 3.20 VTN = VTNo + r ⎡ 2φ f + VSB − 2φ f ⎤ ⎣ ⎦ ⎡ 2 ( 0.37 ) + 3 − 2 ( 0.37 ) ⎤ = 0.75 + 0.6 ⎣ ⎦ = 0.75 + 0.6 [1.934 − 0.860] VTN = 1.39 V VDS (sat) = 2.5 − 1.39 = 1.11 V 72. 2 ⎛ 0.08 ⎞ Sat Region I D = (15 ) ⎜ ⎟ ( 2.5 − 1.39 ) ⎝ 2 ⎠ I D = 0.739 mA (a) 2 ⎛ 0.08 ⎞ ⎡ Non-Sat I D = (15 ) ⎜ ⎟ 2 ( 2.5 − 1.39 )( 0.25 ) − ( 0.25 ) ⎤ ⎦ 2 ⎠⎣ ⎝ I D = 0.296 mA (b) 3.21 a. VG = %ox t0 x = ( 6 × 106 )( 275 × 10−8 ) VG = 16.5 V VG = b. 16.5 ⇒ VG = 5.5 V 3 3.22 Want VG = ( 3)( 24 ) = %ox t0 x = ( 6 × 106 ) t0 x t0 x = 1.2 ×10−5 cm = 1200 Angstroms 3.23 ⎛ R2 ⎞ ⎛ 18 ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ (10 ) = 3.6 V ⎝ 18 + 32 ⎠ ⎝ R1 + R2 ⎠ Assume transistor biased in saturation region V V − VGS 2 = K n (VGS − VTN ) ID = S = G RS RS 3.6 − VGS = ( 0.5 )( 2 )(VGS − 0.8 ) 2 = VGS − 1.6VGS + 0.64 2 2 VGS − 0.6VGS − 2.96 = 0 VGS = ID = 0.6 ± VG − VGS RS ( 0.6 ) 2 + 4 ( 2.96 ) ⇒ VGS = 2.046 V 2 3.6 − 2.046 = ⇒ I D = 0.777 mA 2 VDS = VDD − I D ( RD + RS ) = 10 − ( 0.777 )( 4 + 2 ) ⇒ VDS = 5.34 V VDS > VDS ( sat ) 3.24 73. ID(mA) 4 (a) Q-pt Q-pt 1.67 (b) 4 5 V (V) DS VGS = 4 V VDS (sat) = 4 − 0.8 = 3.2 V (a) If Sat I D = 0.25 ( 4 − 0.8 ) = 2.56 2 VDS = 1.44 × Non-Sat 2 4 = I D RD + VDS = K n RD ⎡ 2 (VGS − VT ) VDS − VDS ⎤ + VDS ⎣ ⎦ 2 4 = ( 0.25 )(1) ⎡ 2 ( 4 − 0.8 ) VDS − VDS ⎤ + VDS ⎣ ⎦ 2 4 = 2.6VDS − 0.25VDS 2 0.25VDS − 2.6VDS + 4 = 0 2.6 ± 6.76 − 4 = 1.88 V 2 ( 0.25 ) VDS = 4 − 1.88 = 2.12 mA 1 (b) Non-Sat region 2 5 = I D RD + VDS = K n RD ⎡ 2 (VGS − VT )VDS − VDS ⎤ + VDS ⎣ ⎦ ID = 2 5 = ( 0.25 )( 3) ⎡ 2 ( 5 − 0.8 ) VDS − VDS ⎤ + VDS ⎣ ⎦ 2 5 = 7.3VDS − 0.75VDS 2 0.75 VDS − 7.3VDS + 5 = 0 VDS = 7.3 ± 53.29 − 15 2 ( 0.75 ) VDS = 0.741 V 5 − 0.741 ID = = 1.42 mA 3 3.25 ID(mA) 2.92 (a) Q-pt 1.25 (b) 3.5 5 V (V) SD 74. VSG = VDD = 3.5 (a) VSD ( sat ) = 3.5 − 0.8 = 2.7 V If biased in Sat region, I D = ( 0.2 )( 3.5 − 0.8 ) = 1.46 mA VSD = 3.5 − (1.46 )(1.2 ) = 1.75 V 2 × Biased in Non-Sat Region. 2 3.5 = VSD + I D RD = VSD + K p RD ⎡ 2 (VSG + VTP ) VSD − VSD ⎤ ⎣ ⎦ 2 3.5 = VSD + ( 0.2 )(1.2 ) ⎡ 2 ( 3.5 − 0.8 ) VSD − VSD ⎤ ⎣ ⎦ 2 3.5 = VSD + 1.296 VSD − 0.24 VSD 2 0.24 VSD − 2.296 VSD + 3.5 = 0 VSD = +2.296 ± 5.272 − 3.36 use − sign VSD = 1.90 V 2 ( 0.24 ) 2 I D = ( 0.2 ) ⎡ 2 ( 3.5 − 0.8 )(1.9 ) − (1.9 ) ⎤ = 0.2 [10.26 − 3.61] ⎣ ⎦ 3.5 − 1.90 ID = = 1.33 mA 1.2 I D = 1.33 mA VSG = VDD = 5 V VSD ( sat ) = 5 − 0.8 = 4.2 V (b) If Sat Region I D = ( 0.2 )( 5 − 0.8 ) = 3.53 mA, VSD < 0 2 Non-Sat Region. 2 5 = VSD + K p RD ⎡ 2 (VSG + VTP ) VSD − VSD ⎤ ⎣ ⎦ 2 5 = VSD + ( 0.2 )( 4 ) ⎡ 2 ( 5 − 0.8 ) VSD − VSD ⎤ ⎣ ⎦ 2 5 = VSD + 6.72 VSD − 0.8 VSD 2 0.8 VSD − 7.72 VSD + 5 = 0 VSD = ID = 7.72 ± 59.598 − 16 use − sign VSD = 0.698 V 2 ( 0.8 ) 5 − 0.698 ⇒ I D = 1.08 mA 4 3.26 10 − VS 2 = K p (VSG + VTP ) RS Assume transistor biased in saturation region ⎛ R2 ⎞ VG = ⎜ ⎟ ( 20 ) − 10 ⎝ R1 + R2 ⎠ ID = ⎛ 22 ⎞ =⎜ ⎟ ( 20 ) − 10 ⇒ VG = 4.67 V ⎝ 8 + 22 ⎠ VS = VG + VSG 10 − ( 4.67 + VSG ) = (1)( 0.5 )(VSG − 2 ) 2 2 5.33 − VSG = 0.5 (VSG − 4VSG + 4 ) 2 0.5VSG − VSG − 3.33 = 0 VSG = 1± (1) 2 + 4 ( 0.5 )( 3.33) 2 ( 0.5 ) ⇒ VSG = 3.77 V 75. VSD 10 − ( 4.67 + 3.77 ) ⇒ I D = 3.12 mA 0.5 = 20 − I D ( RS + RD ) = 20 − ( 3.12 )( 0.5 + 2 ) ⇒ VSD = 12.2 V ID = VSD > VSD ( sat ) 3.27 VG = 0, VSG = VS Assume saturation region 2 I D = 0.4 = K p (VSG + VTP ) 0.4 = ( 0.2 )(VS − 0.8 ) 2 0.4 + 0.8 ⇒ VS = 2.21 V 0.2 VD = I D RD − 5 = ( 0.4 )( 5 ) − 5 = −3 V VSD = VS − VD = 2.21 − ( −3) ⇒ VSD = 5.21 V VS = VSD > VSD ( sat ) 3.28 VDD = I DQ RD + VDSQ + I DQ RS 2 ⎛ k ′ ⎞⎛ W ⎞ (1) 10 = I DQ ( 5 ) + 5 + VGS and I DQ = ⎜ n ⎟ ⎜ ⎟ (VGS − VTN ) ⎝ 2 ⎠⎝ L ⎠ 2 ⎛ 0.060 ⎞ ⎛ W ⎞ or (2) I DQ = ⎜ ⎟ ⎜ ⎟ (VGS − 1.2 ) ⎝ 2 ⎠⎝ L ⎠ Let VGS = 2.5 V Then from (1), 10 = I DQ ( 5 ) + 5 + 2.5 ⇒ I D = 0.5 mA W 2 ⎛ 0.060 ⎞⎛ W ⎞ Then from (2), 0.5 = ⎜ = 9.86 ⎟⎜ ⎟ ( 2.5 − 1.2 ) ⇒ 2 ⎠⎝ L ⎠ L ⎝ V 2.5 I DQ RS = VGS ⇒ RS = GS = ⇒ RS = 5 k Ω I DQ 0.5 IR = 10 = ( 0.5 )( 0.05 ) = 0.025 mA R1 + R2 Then R1 + R2 = 10 = 400 k Ω 0.025 ⎛ R2 ⎞ ⎛ R2 ⎞ ⎜ ⎟ (VDD ) = 2VGS ⇒ ⎜ ⎟ (10 ) = 2 ( 2.5 ) ⇒ R1 = R2 = 200 k Ω R1 + R2 ⎠ ⎝ 400 ⎠ ⎝ 3.29 ⎛ 75 ⎞ K n = ( 25 ) ⎜ ⎟ ⇒ 0.9375 mA/V 2 ⎝ 2⎠ ⎛ 6 ⎞ VG = ⎜ ⎟ (10 ) − 5 = −2 V ⎝ 6 + 14 ⎠ (VG − VGS ) − ( −5) 2 = I D = K n (VGS − VTN ) RS −2 − VGS + 5 = ( 0.9375 )( 0.5 )(VGS − 1) 2 3 − VGS = 0.469 (VGS − 2VGS + 1) 2 76. 2 0.469 VGS + 0.0625 VGS − 2.53 = 0 VGS = −0.0625 ± 0.003906 + 4.746 ⇒ VGS = 2.26 V 2 ( 0.469 ) I D = 0.9375 ( 2.26 − 1) ⇒ I D = 1.49 mA 2 VDS = 10 − (1.49 )(1.7 ) ⇒ VDS = 7.47 V 3.30 20 = I DQ RS + VSDQ + I DQ RD (1) 20 = VSG + 10 + I DQ RD ⎛ k′ ⎞⎛ W ⎞ 2 p I DQ = ⎜ ⎟ ⎜ ⎟ (VSG + VTP ) 2 ⎠⎝ L ⎠ ⎝ 2 ⎛ 0.040 ⎞ ⎛ W ⎞ (2) I DQ = ⎜ ⎟ ⎜ ⎟ (VSG − 2 ) ⎝ 2 ⎠⎝ L ⎠ For example, let I DQ = 0.8 mA and VSG = 4 V ⎛ 0.040 ⎞ ⎛ W Then 0.8 = ⎜ ⎟⎜ ⎝ 2 ⎠⎝ L I DQ RS = VSG ⇒ ( 0.8 ) RS W 2 ⎞ = 10 ⎟ ( 4 − 2) ⇒ L ⎠ = 4 ⇒ RS = 5 k Ω From (1) 20 = 4 + 10 + ( 0.8 ) RD ⇒ RD = 7.5 k Ω IR = 20 = ( 0.8 )( 0.1) ⇒ R1 + R2 = 250 k Ω R1 + R2 ⎛ R1 ⎞ ⎜ ⎟ ( 20 ) = 2VSG = ( 2 )( 4 ) ⎝ R1 + R2 ⎠ R1 ( 20 ) = 8 ⇒ R1 = 100 k Ω, R2 = 150 k Ω 250 3.31 (a) (i) I Q = 50 = 500 (VGS − 1.2 ) ⇒ VGS = 1.516 V 2 VDS = 5 − ( −1.516 ) =⇒ VDS = 6.516 V (ii) I Q = 1 = ( 0.5 )(VGS − 1.2 ) ⇒ VGS = 2.61 V 2 VDS = 5 − ( −2.61) ⇒ VDS = 7.61 V (b) (i) Same as (a) VGS = VDS = 1.516 V (ii) VGS = VDS = 2.61 V 3.32 I D = K n (VGS − VTN ) 2 0.25 = ( 0.2 )(VGS − 0.6 ) 2 0.25 + 0.6 ⇒ VGS = 1.72 V ⇒ VS = −1.72 V 0.2 VD = 9 − ( 0.25 )( 24 ) ⇒ VD = 3 V VGS = 3.33 (a) 77. ID(mA) 1.0 0.808 Q-pt 0.5 3.81 RD = 10 V (V) DS 5 −1 ⇒ RD = 8 K 0.5 I DQ = 0.5 = 0.25 (VGS − 1.4 ) ⇒ VGS = 2.81 V 2 RS = −2.81 − ( −5 ) ⇒ RS = 4.38 K 0.5 Let RD = 8.2 K, RS = 4.3 K (b) Now −VGS − ( −5 ) 5 − VGS = I D = 0.25 (VGS − 1.4 ) 4.3 2 = 1.075 (VGS − 2.8 VGS + 1.96 ) 2 2 1.075 VGS − 2.01 VGS − 2.89 = 0 VGS = 2.01 ± 4.04 + 12.427 ⇒ VGS = 2.82 V 2 (1.075 ) I D = 0.25 ( 2.82 − 1.4 ) ⇒ I D = 0.504 mA 2 VDS = 10 − ( 0.504 )( 8.2 + 4.3) ⇒ VDS = 3.70 V (c) If RS = 4.3 + 10% = 4.73 K 2 5 − VGS = 1.18 (VGS − 2.8VGS + 1.96 ) 2 1.18 VGS − 2.31 VGS − 2.68 = 0 VGS = 2.31 ± 5.336 + 12.65 = 2.78 V 2 (1.18 ) I D = ( 0.25 )( 2.78 − 1.4 ) ⇒ I D = 0.476 mA 2 If Rs = 4.3 − 10% = 3.87 K 2 5 − VGS = ( 0.9675 ) (VGS − 2.8VGS + 1.96 ) 2 0.9675VGS − 1.71VGS − 3.10 = 0 VGS = 1.71 ± 2.924 + 12.0 = 2.88 V 2 ( 0.9675 ) I D = ( 0.25 )( 2.88 − 1.4 ) = 0.548 mA 2 3.34 VDD = VSD + I DQ R 9 = 2.5 + ( 0.1) R ⇒ R = 65 k Ω ⎛ k′ ⎞⎛ W ⎞ 2 p I DQ = ⎜ ⎟ ⎜ ⎟ (VSG + VTP ) ⎝ 2 ⎠⎝ L ⎠ W 2 ⎛ 0.025 ⎞ ⎛ W ⎞ =8 ( 0.1) = ⎜ ⎟ ⎜ ⎟ ( 2.5 − 1.5 ) ⇒ L⎠ L ⎝ 2 ⎠⎝ Then for L = 4 μ m, W = 32 μ m 78. 3.35 5 = I DQ RS + VSDQ = I DQ ( 2 ) + 2.5 I DQ = 1.25 mA IR = 10 = (1.25 )( 0.1) ⇒ R1 + R2 = 80 k Ω R1 + R2 I DQ = K p (VSG + VTP ) 2 1.25 = 0.5 (VSG + 1.5 ) ⇒ 2 1.25 − 1.5 = VSG 0.5 VSG = 0.0811 V VG = VS − VSG = 2.5 − 0.0811 = 2.42 V ⎛ R2 ⎞ VG = ⎜ ⎟ (10 ) − 5 ⎝ R1 + R2 ⎠ ⎛R ⎞ 2.42 = ⎜ 2 ⎟ (10 ) − 5 ⇒ R2 = 59.4 k Ω, R1 = 20.6 k Ω ⎝ 80 ⎠ 3.36 (a) ID(mA) 0.429 Q-pt RD = VD − ( −5 ) I DQ 5 V (V) SD = 5−2 ⇒ RD = 12 K 0.25 2 ⎞⎛ k′ ⎞ p ⎟ ⎜ ⎟ (VSG + VTP ) ⎠⎝ 2 ⎠ 2 ⎛ 0.035 ⎞ 0.25 = (15 ) ⎜ ⎟ (VSG − 1.2 ) ⇒ VSG = 2.18 V ⎝ 2 ⎠ 5 − 2.18 RS = ⇒ RS = 11.3 K 0.25 VSD = 2.18 − ( −2 ) = 4.18 V (b) k ′ = 35 + 5% = 36.75 μ A/V 2 p ⎛W ID = ⎜ ⎝L 5 − VSG 2 ⎛ 0.03675 ⎞ I D = (15 ) ⎜ ⎟ (VSG − 1.2 ) = 2 11.3 ⎝ ⎠ 2 3.11(VSG − 2.4VSG + 1.44 ) = 5 − VSG 2 3.11VSG − 6.46VSG − 0.522 = 0 79. VSG = 6.46 ± 41.73 + 6.49 = 2.155 V 2 ( 3.11) 5 − 2.155 = 0.252 mA 11.3 = 10 − ( 0.252 )(12 + 11.3) = 4.13 V ID = VSD k ′ = 35 − 5% = 33.25 μ A/V 2 p 5 − VSG 2 ⎛ 0.03325 ⎞ I D = (15 ) ⎜ ⎟ (VSG − 1.2 ) = 2 11.3 ⎝ ⎠ 2 2.82 (VSG − 2.4VSG + 1.44 ) = 5 − VSG 2 2.82VSG − 5.77VSG − 0.939 = 0 VSG = 5.77 ± 33.29 + 10.59 = 2.198 V 2 ( 2.82 ) 5 − 2.198 = 0.248 mA 11.3 = 10 − ( 0.248 )(12 + 11.3) = 4.22 V ID = VSD 3.37 ID = −VSD − ( −10 ) RD ⇒5= −6 + 10 ⇒ RD = 0.8 kΩ RD I D = K P (VSG + VTP ) ⇒ 5 = 3 (VSG − 1.75 ) 2 VSG = 2 5 + 1.75 = 3.04 V ⇒ VG = −3.04 3 ⎛ R2 ⎞ VG = ⎜ ⎟ (10 ) − 5 = −3.04 ⎝ R1 + R2 ⎠ Rin = R1 || R2 = 80 kΩ 1 ⋅ ( 80 )(10 ) = 5 − 3.04 ⇒ R1 = 408 kΩ R1 408 R2 = 80 ⇒ R2 = 99.5 kΩ 408 + R2 3.38 ⎛ 60 ⎞ K n1 = ⎜ ⎟ ( 4 ) = 120 μ A/V 2 ⎝ 2 ⎠ ⎛ 60 ⎞ K n 2 = ⎜ ⎟ (1) = 30 μ A/V 2 ⎝ 2 ⎠ For vI = 1 V , M1 Sat. region, M2 Non-sat region. (a) I D 2 = I D1 30 ⎡ 2 ( −VTNL )( 5 − vO ) − ( 5 − vO ) ⎤ = 120 (1 − 0.8 ) ⎣ ⎦ 2 We find vO − 6.4vO + 7.16 = 0 ⇒ vO = 4.955 V 2 (b) 2 For vI = 3 V , M1 Non-sat region, M2 Sat. region. I D 2 = I D1 2 30 ⎡ − ( −1.8 ) ⎤ = 120 ⎡ 2 ( 3 − 0.8 ) vO − vO ⎤ ⎣ ⎦ ⎣ ⎦ 2 2 We find 4vO − 17.6vO + 3.24 = 0 ⇒ vO = 0.193 V (c) For vI = 5 V , biasing same as (b) 2 30 ⎡ − ( −1.8 ) ⎤ = 120 ⎡ 2 ( 5 − 0.8 ) vO − vO ⎤ ⎣ ⎦ ⎣ ⎦ 2 80. 2 We find 4vO − 33.6vO + 3.24 = 0 ⇒ vO = 0.0976 V 3.39 For vI = 5 V , M1 Non-sat region, M2 Sat. region. I D1 = I D 2 ′ ′ 2 ⎛ kn ⎞ ⎛ W ⎞ ⎛ kn ⎞ ⎛ W ⎞ 2 ⎜ 2 ⎟ ⎜ L ⎟ ⎡ 2 (VGS1 − VTN 1 ) VDS 1 − VDS 1 ⎤ = ⎜ 2 ⎟ ⎜ L ⎟ (VGS 2 − VTN 2 ) ⎣ ⎦ ⎝ ⎠ ⎝ ⎠1 ⎝ ⎠ ⎝ ⎠2 2 W⎞ ⎡ 2 ⎛ ⎜ ⎟ ⎣ 2 ( 5 − 0.8 )( 0.15 ) − ( 0.15 ) ⎤ = (1) ⎡0 − ( −2 ) ⎤ ⎣ ⎦ ⎦ ⎝ L ⎠1 ⎛W ⎞ which yields ⎜ ⎟ = 3.23 ⎝ L ⎠1 3.40 a. M1 and M2 in saturation K n1 (VGS 1 − VTN 1 ) = K n 2 (VGS 2 − VTN 2 ) K n1 = K n 2 , VTN 1 = VTN 2 ⇒ VGS1 = VGS 2 = 2.5 V, V0 = 2.5 V 2 2 I D = (15 )( 40 )( 2.5 − 0.8 ) ⇒ I D = 1.73 mA 2 b. ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ > ⎜ ⎟ ⇒ VGS1 < VGS 2 ⎝ L ⎠1 ⎝ L ⎠ 2 40 (VGS1 − 0.8 ) = (15 )(VGS 2 − 0.8 ) 2 2 VGS 2 = 5 − VGS 1 1.633 (VGS 1 − 0.8 ) = ( 5 − VGS1 − 0.8 ) 2.633VGS 1 = 5.506 ⇒ VGS 1 = 2.09 V VGS 2 = 2.91 V, V0 = VGS1 = 2.91 V I D = (15 )(15 )( 2.91 − 0.8 ) ⇒ I D = 1.0 mA 2 3.41 (a) V1 = VGS 3 = 2.5 V 2 ⎛ W ⎞ ⎛ 0.06 ⎞ I D = 0.5 = ⎜ ⎟ ⎜ ⎟ ( 2.5 − 1.2 ) ⎝ L ⎠3 ⎝ 2 ⎠ ⎛W ⎞ ⎜ ⎟ = 9.86 ⎝ L ⎠3 V2 = 6 V ⇒ VGS 2 = V2 − V1 = 6 − 2.5 = 3.5 V 2 ⎛ W ⎞ ⎛ 0.06 ⎞ ⎛W ⎞ 0.5 = ⎜ ⎟ ⎜ ⎟ ( 3.5 − 1.2 ) ⇒ ⎜ ⎟ = 3.15 ⎝ L ⎠2 ⎝ 2 ⎠ ⎝ L ⎠2 VGS1 = 10 − V2 = 10 − 6 = 4 V 2 ⎛ W ⎞ ⎛ 0.06 ⎞ ⎛W ⎞ 0.5 = ⎜ ⎟ ⎜ ⎟ ( 4 − 1.2 ) ⇒ ⎜ ⎟ = 2.13 ⎝ L ⎠1 ⎝ 2 ⎠ ⎝ L ⎠1 (b) ′ kn1 = 0.06 + 5% = 0.063 mA/V 2 ′ ′ kn 2 = k n3 = 0.6 − 5% = 0.057 mA/V 2 2 ⎛ 0.057 ⎞ For M3: I D = ( 9.86 ) ⎜ ⎟ (V1 − 1.2 ) ⎝ 2 ⎠ 2 ⎛ 0.057 ⎞ For M2: I D = ( 3.15 ) ⎜ ⎟ (V2 − V1 − 1.2 ) 2 ⎠ ⎝ 81. 2 ⎛ 0.063 ⎞ For M1: I D = ( 2.13) ⎜ ⎟ (10 − V2 − 1.2 ) ⎝ 2 ⎠ 0.281(V1 − 1.2 ) = 0.0898 (V2 − V1 − 1.2 ) = 0.0671( 8.8 − V2 ) 2 2 Take square root. 0.530 (V1 − 1.2 ) = 0.300 (V2 − V1 − 1.2 ) = 0.259 ( 8.8 − V2 ) (1) 0.830V1 = 0.300V2 + 0.276 (2) 0.559V2 = 0.300V1 + 2.64 From (2) ⇒ V2 = 0.537V1 + 4.72 Substitute into (1) 0.830V1 = 0.300 [ 0.537V1 + 4.72] + 0.276 = 0.161V1 + 1.69 V1 = 2.53 V Then V2 = 0.537 ( 2.53) + 4.72 V2 = 6.08 V 3.42 ML in saturation MD in nonsaturation 2 ⎛W ⎞ ⎛W ⎞ 2 ⎡ ⎤ ⎜ ⎟ (VGSL − VTNL ) = ⎜ ⎟ ⎣ 2 (VGSD − VTND )VDSD − VDSD ⎦ ⎝ L ⎠L ⎝ L ⎠D W 2 2 ⎤ (1)( 5 − 0.1 − 0.8) = ⎛ ⎞ ⎣ 2 ( 5 − 0.8)( 0.1) − ( 0.1) ⎦ ⎜ ⎟ ⎡ L ⎠D ⎝ ⎛W ⎞ 16.81 = ⎜ ⎟ [ 0.83] ⎝ L ⎠D ⎛W ⎞ ⎜ ⎟ = 20.3 ⎝ L ⎠D 3.43 ML in saturation MD in nonsaturation 2 ⎛W ⎞ ⎛W ⎞ 2 ⎡ ⎤ ⎜ ⎟ (VGSL − VTNL ) = ⎜ ⎟ ⎣ 2 (VGSD − VTND )VDSD − VDSD ⎦ ⎝ L ⎠L ⎝ L ⎠D W 2 2 (1)(1.8 ) = ⎛ ⎞ ⎡ 2 ( 5 − 0.8 )( 0.05) − ( 0.05) ⎤ ⎜ ⎟ ⎣ ⎦ ⎝ L ⎠D ⎛W ⎞ 3.24 = ⎜ ⎟ [ 0.4175] ⎝ L ⎠D ⎛W ⎞ ⎜ ⎟ = 7.76 ⎝ L ⎠D 3.44 VDD − V0 5 − 0.1 = = 0.49 mA 10 RD Transistor biased in nonsaturation I D = 0.49 ID = 2 ⎛W ⎞ = ( 0.015 ) ⎜ ⎟ ⎡ 2 ( 4.2 − 0.8 )( 0.1) − ( 0.1) ⎤ ⎣ ⎦ ⎝L⎠ W ⎛W ⎞ 0.49 = ⎜ ⎟ 0.01005 ⇒ = 48.8 L ⎝L⎠ 3.45 2 82. 5 = I D RD + Vγ + VDS 5 = (12 ) RD + 1.6 + 0.2 ⇒ RD = 267 Ω 2 ⎛ k′ ⎞⎛ W ⎞ I D = ⎜ n ⎟ ⎜ ⎟ (VGS − VTN ) ⎝ 2 ⎠⎝ L ⎠ W 2 ⎛ 0.040 ⎞ ⎛ W ⎞ = 34 12 = ⎜ ⎟ ⎜ ⎟ ( 5 − 0.8 ) ⇒ 2 ⎠⎝ L ⎠ L ⎝ 3.46 5 = VSD + I D RD + Vγ 5 = 0.15 + (15 ) RD + 1.6 ⇒ RD = 217 Ω ⎛ k′ ⎞⎛ W ⎞ 2 p I D = ⎜ ⎟ ⎜ ⎟ (VSG + VTP ) ⎝ 2 ⎠⎝ L ⎠ W 2 ⎛ 0.020 ⎞⎛ W ⎞ = 85 15 = ⎜ ⎟⎜ ⎟ ( 5 − 0.8 ) ⇒ L ⎝ 2 ⎠⎝ L ⎠ 3.47 (a) VDD − VO ⎛W = 2⎜ RD ⎝L 5 − 0.2 ⎛W = 2⎜ 20 ⎝L ⎞⎛ 0.060 ⎞ 2 ⎟⎜ ⎟ ⎡( 2 )(VGS − VTN ) VO − VO ⎤ ⎣ ⎦ ⎠⎝ 2 ⎠ 2 ⎞ ⎟ ( 0.030 ) ⎡ 2 ( 5 − 0.8 )( 0.2 ) − ( 0.2 ) ⎤ ⎣ ⎦ ⎠ ⎛W ⎞ ⎛W ⎞ ⎛W ⎞ 0.24 = 0.0984 ⎜ ⎟ ⇒ ⎜ ⎟ = ⎜ ⎟ = 2.44 ⎝ L ⎠ ⎝ L ⎠1 ⎝ L ⎠ 2 (b) 5 − VO ⎛ 0.06 ⎞ 2 = ( 2.44 ) ⎜ ⎟ ⎡ 2 ( 5 − 0.8 ) VO − VO ⎤ ⎦ 20 2 ⎠⎣ ⎝ 5 − VO = 12.30VO − 1.464VO2 1.464VO2 − 13.30VO + 5 = 0 VO = 13.30 ± 176.89 − 29.28 2 (1.464 ) VO = 0.393 V 3.48 ′ 2 ⎞ ⎛ kn ⎞ ⎟ ⎜ ⎟ (VGS 1 − VTN ) 2⎠ ⎠1 ⎝ ′ 2 ⎞ ⎛ kn ⎞ ⎟ ⎜ ⎟ (VDS 2 ( sat ) ) ⎠2 ⎝ 2 ⎠ 2 ⎛ W ⎞ ⎛ 0.08 ⎞ ⎛W ⎞ ⎛W ⎞ 0.1 = ⎜ ⎟ ⎜ ⎟ ( 0.5 ) ⇒ ⎜ ⎟ = 10 = ⎜ ⎟ L ⎠2 ⎝ 2 ⎠ L ⎠2 ⎝ ⎝ ⎝ L ⎠1 ⎛ W ⎞ ⎛ 200 ⎞ ⎛ W ⎞ ⎜ ⎟ =⎜ ⎟ ⎜ ⎟ = 20 ⎝ L ⎠3 ⎝ 100 ⎠ ⎝ L ⎠ 2 M1 & M2 matched. 2 ⎛ 0.08 ⎞ Then 0.1 = (10 ) ⎜ ⎟ (VGS 1 − 0.25 ) 2 ⎠ ⎝ VGS1 = 0.75 V ⎛W I Q1 = ⎜ ⎝L ⎛W I Q1 = ⎜ ⎝L VD1 = −0.75 + 2 = 1.25 V RD = 2.5 − 1.25 ⇒ RD = 12.5 K 0.1 83. 3.49 (a) 2 ⎛ W ⎞ ⎛ k′ ⎞ p I Q 2 = ⎜ ⎟ ⎜ ⎟ (VSDB ( sat ) ) ⎝ L ⎠B ⎝ 2 ⎠ 2 ⎛ W ⎞ ⎛ 0.04 ⎞ 0.25 = ⎜ ⎟ ⎜ ⎟ ( 0.8 ) ⇒ ⎝ L ⎠B ⎝ 2 ⎠ ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = 19.5 = ⎜ ⎟ ⎝ L ⎠B ⎝ L ⎠A I KQ 2 ⎛ W ⎞ ⎛W ⎞ ⎜ ⎟ = ⎜ ⎟ IQ 2 ⎝ L ⎠B ⎝ L ⎠C ⎛ 100 ⎞ =⎜ ⎟ (19.5 ) = 7.81 ⎝ 250 ⎠ ′ 2 ⎛W ⎞ ⎛ kp ⎞ I Q 2 = ⎜ ⎟ ⎜ ⎟ (VSGA + VTP ) ⎝ L ⎠A ⎝ 2 ⎠ 2 ⎛ 0.04 ⎞ 0.25 = (19.5 ) ⎜ ⎟ (VSGA − 0.5 ) ⎝ 2 ⎠ VSGA = 1.30 V (b) VDA = 1.3 − 4 = −2.7 V RD = −2.7 − ( −5 ) 0.25 ⇒ RD = 9.2 K 3.50 ′ 2 ⎞ ⎛ kn ⎞ ⎟ ⎜ ⎟ (VDS 2 ( sat ) ) 2⎠ ⎠2 ⎝ 2 ⎞ ⎛ 0.06 ⎞ ⎛W ⎞ ⎛W ⎞ ⎟ ⎜ ⎟ ( 0.5 ) ⇒ ⎜ ⎟ = 53.3 = ⎜ ⎟ ⎠2 ⎝ 2 ⎠ ⎝ L ⎠2 ⎝ L ⎠1 2 ⎛ W ⎞ ⎛ k′ ⎞ I Q = ⎜ ⎟ ⎜ n ⎟ (VGS 1 − VTN ) L ⎠1 ⎝ 2 ⎠ ⎝ 2 ⎛ 0.06 ⎞ 0.4 = ( 53.3) ⎜ ⎟ (VGS 1 − 0.75 ) ⎝ 2 ⎠ VGS1 = 1.25 V VD1 = −1.25 + 4 = 2.75 V 5 − 2.75 RD = ⇒ RD = 5.625 K 0.4 ⎛W IQ = ⎜ ⎝L ⎛W 0.4 = ⎜ ⎝L 3.51 VDS ( sat ) = VGS − VP So VDS > VDS ( sat ) = −VP , I D = I DSS 3.52 VDS ( sat ) = VGS − VP = VGS + 3 = VDS ( sat ) a. VGS = 0 ⇒ I D = I DSS = 6 mA b. ⎛ V ⎞ ⎛ −1 ⎞ I D = I DSS ⎜ 1 − GS ⎟ = 6 ⎜ 1 − ⎟ ⇒ I D = 2.67 mA VP ⎠ ⎝ −3 ⎠ ⎝ 2 2 c. d. 2 ⎛ −2 ⎞ I D = 6 ⎜ 1 − ⎟ ⇒ I D = 0.667 mA ⎝ −3 ⎠ ID = 0 84. 3.53 ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ ⎛ 1 ⎞ 2.8 = I DSS ⎜1 − ⎟ ⎝ VP ⎠ 2 2 ⎛ 3 ⎞ 0.30 = I DSS ⎜1 − ⎟ ⎝ VP ⎠ ⎛ 1 ⎜1 − 2.8 ⎝ VP = 0.30 ⎛ 3 ⎜1 − VP ⎝ 2 2 ⎞ ⎟ ⎠ = 9.33 2 ⎞ ⎟ ⎠ ⎛ 1 ⎜1 − VP ⎝ ⎛ 3 ⎜1 − ⎝ VP ⎞ ⎟ ⎠ = 3.055 ⎞ ⎟ ⎠ 1 9.165 = 3.055 − 1− VP VP 8.165 = 2.055 ⇒ VP = 3.97 V VP 2 1 ⎞ ⎛ 2.8 = I DSS ⎜1 − ⎟ = I DSS ( 0.560 ) ⇒ I DSS = 5.0 mA ⎝ 3.97 ⎠ 3.54 VS = −VGS , VSD = VS − VDD Want VSD ≥ VSD ( sat ) = VP − VGS VS − VDD ≥ VP − VGS − VGS − VDD ≥ VP − VGS ⇒ VDD ≤ −VP So VDD ≤ −2.5 V ⎛ V ⎞ I D = 2 = I DSS ⎜1 − GS ⎟ VP ⎠ ⎝ 2 2 ⎛ V ⎞ 2 = 6 ⎜ 1 − GS ⎟ ⇒ VGS = 1.06 V ⇒ VS = −1.06 V ⎝ 2.5 ⎠ 3.55 I D = K n (VGS − VTN ) 2 18.5 = K n ( 0.35 − VTN ) 86.2 = K n ( 0.5 − VTN ) 2 2 Then ( 0.35 − VTN ) 18.5 = 0.2146 = ⇒ VTN = 0.221 V 2 86.2 ( 0.50 − VTN ) 2 18.5 = K n ( 0.35 − 0.221) ⇒ K n = 1.11 mA / V 2 2 3.56 I D = K (VGS − VTN ) 2 250 = K ( 0.75 − 0.24 ) ⇒ K = 0.961 mA / V 2 2 85. 3.57 2 ⎛ V ⎞ V V I D = I DSS ⎜ 1 − GS ⎟ = S = − GS VP ⎠ RS RS ⎝ 2 V ⎛ V ⎞ 10 ⎜ 1 − GS ⎟ = − GS 0.2 −5 ⎠ ⎝ 2 ⎛ 2V V ⎞ 2 ⎜1 + GS + GS ⎟ = −VGS 5 25 ⎠ ⎝ 2 2 9 VGS + VGS + 2 = 0 25 5 2 2VGS + 45VGS + 50 = 0 VGS = −45 ± ID = − ( 45 ) − 4 ( 2 )( 50 ) ⇒ VGS 2 ( 2) 2 = −1.17 V VGS 1.17 = ⇒ I D = 5.85 mA RS 0.2 VD = 20 − ( 5.85 )( 2 ) = 8.3 V VDS = VD − VS = 8.3 − 1.17 ⇒ VDS = 7.13 V 3.58 VDS = VDD − VS 8 = 10 − VS ⇒ VS = 2 V = I D RS = ( 5 ) RS ⇒ RS = 0.4 kΩ ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ ⎝ VP ⎠ 2 2 ⎛ −1 ⎞ 5 = I DSS ⎜1 − ⎟ Let I DSS = 10 mA ⎝ VP ⎠ 2 ⎛ −1 ⎞ 5 = 10 ⎜ 1 − ⎟ ⇒ VP = −3.41 V ⎝ VP ⎠ VG = VGS + VS = −1 + 2 = 1 V ⎛ R2 ⎞ 1 VG = ⎜ ⎟ VDD = ⋅ Rin ⋅ VDD R1 ⎝ R1 + R2 ⎠ 1 1 = ( 500 )(10 ) ⇒ R1 = 5 MΩ R1 5R2 = 0.5 ⇒ R2 = 0.556 MΩ 5 + R2 3.59 ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ ⎝ VP ⎠ 2 2 ⎛ V ⎞ 5 = 8 ⎜ 1 − GS ⎟ ⇒ VGS = 0.838 V 4 ⎠ ⎝ VSD = VDD − I D ( RS + RD ) = 20 − ( 5 )( 0.5 + 2 ) ⇒ VSD = 7.5 V 86. VS = 20 − ( 5 )( 0.5 ) = 17.5 V VG = VS + VGS = 17.5 + 0.838 = 18.3 V ⎛ R2 ⎞ 1 VG = ⎜ ⎟ VDD = ⋅ Rin ⋅ VDD R1 ⎝ R1 + R2 ⎠ 1 18.3 = (100 ) ( 20 ) ⇒ R1 = 109 kΩ R1 109 R2 = 100 ⇒ R2 = 1.21 MΩ 109 + R2 3.60 ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ 2 2 ⎛ V ⎞ 5 = 7 ⎜ 1 − GS ⎟ ⇒ VGS = 0.465 V 3 ⎠ ⎝ VSD = VDD − I D ( RS + RD ) 6 = 12 − ( 5 )( 0.3 + RD ) ⇒ RD = 0.9 kΩ VS = 12 − ( 5 )( 0.3) = 10.5 V VG = VS + VGS = 10.5 + 0.465 = 10.965 V ⎛ R2 ⎞ VG = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠ ⎛ R ⎞ 10.965 = ⎜ 2 ⎟ (12 ) ⇒ R2 = 91.4 kΩ ⇒ R1 = 8.6 kΩ ⎝ 100 ⎠ 3.61 ⎛ R2 ⎞ ⎛ 60 ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ ( 20 ) ⇒ VG = 6 V ⎝ 140 + 60 ⎠ ⎝ R1 + R2 ⎠ 2 ⎛ V ⎞ V V − VGS I D = I DSS ⎜ 1 − GS ⎟ = S = G VP ⎠ RS RS ⎝ ⎛ (8 )( 2 ) ⎜1 − ⎜ ⎝ 2 VGS ⎞ ⎟ = 6 − VGS ( −4 ) ⎟ ⎠ ⎛ V V2 ⎞ 16 ⎜ 1 + GS + GS ⎟ = 6 − VGS 2 16 ⎠ ⎝ 2 VGS + 9VGS + 10 = 0 VGS = −9 ± (9) 2 − 4 (10 ) 2 ⇒ VGS = −1.30 ⎛ ( −1.30 ) ⎞ I D = 8 ⎜1 − ⎟ ⇒ I D = 3.65 mA ⎜ ( −4 ) ⎟ ⎝ ⎠ VDS = VDD − I D ( RS + RD ) 2 = 20 − ( 3.65 )( 2 + 2.7 ) VDS = 2.85 V VDS > VDS ( sat ) = VGS − VP = −1.30 − ( −4 ) = 2.7 V (Yes) 3.62 87. VDS = VDD − I D ( RS + RD ) 5 = 12 − I D ( 0.5 + 1) ⇒ I D = 4.67 mA VS = I D RS = ( 4.67 ) ( 0.5 ) ⇒ VS = 2.33 V ⎛ R2 ⎞ ⎛ 20 ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ (12 ) ⇒ VG = 0.511 V R1 + R2 ⎠ ⎝ 450 + 20 ⎠ ⎝ VGS = VG − VS = 0.511 − 2.33 ⇒ VGS = −1.82 V ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ 2 ⎛ ( −1.82 ) ⎞ 4.67 = 10 ⎜ 1 − ⎟ ⇒ VP = −5.75 V ⎜ VP ⎟ ⎝ ⎠ 2 3.63 2 ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ , VGS = 0 VP ⎠ ⎝ I D = I DSS = 4 mA RD = VDD − VDS 10 − 3 ⇒ RD = 1.75 kΩ = 4 ID 3.64 VSD = VDD − I D RS 10 = 20 − (1) RS ⇒ RS = 10 kΩ R1 + R2 = VDD 20 = = 200 kΩ 0.1 I ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ 2 2 ⎛ V ⎞ 1 = 2 ⎜1 − GS ⎟ ⇒ VGS = 0.586 V 2 ⎠ ⎝ VG = VS + VGS = 10 + 0.586 = 10.586 ⎛ R2 ⎞ VG = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠ ⎛ R ⎞ 10.586 = ⎜ 2 ⎟ ( 20 ) ⇒ R2 = 106 kΩ ⎝ 200 ⎠ R1 = 94 kΩ 3.65 88. VDS = VDD − I D ( RS + RD ) 2 = 3 − ( 0.040 )(10 + RD ) ⇒ RD = 15 kΩ I D = K (VGS − VTN ) 2 40 = 250 (VGS − 0.20 ) ⇒ VGS = 0.60 V 2 VG = VGS + VS = 0.60 + ( 0.040 )(10 ) = 1.0 V ⎛ R2 ⎞ VG = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠ ⎛ R ⎞ 1 = ⎜ 2 ⎟ ( 3) ⇒ R2 = 50 kΩ ⎝ 150 ⎠ R1 = 100 kΩ 3.66 For VO = 0.70 V ⇒ VDS = 0.70 > VDS ( sat ) = VGS − VTN 0.75 − 0.15 = 0.6 Biased in the saturation region V − VDS 3 − 0.7 ⇒ I D = 46 μ A I D = DD = 50 RD I D = K (VGS − VTN ) ⇒ 46 = K ( 0.75 − 0.15 ) ⇒ K = 128 μ A / V 2 2 2 89. Chapter 4 Exercise Solutions EX4.1 g m = 2 K n (VGS − VTN ) and I D = K n (VGS − VTN ) 2 0.75 = 0.5 (VGS − 0.8 ) ⇒ VGS = 2.025 V g m = 2 ( 0.5 )( 2.025 − 0.8 ) ⇒ g m = 1.22 mA / V 2 ro = 1 λ I DQ 1 (0.01)(0.75) = 133 k Ω ro = 133 k Ω = EX4.2 Av = − g m RD g m = 2 K n I DQ = 2 ( 0.5)( 0.4 ) = 0.8944 mA/V Av = − ( 0.8944 )(10 ) = −8.94 EX4.3 (a) ⎛ R2 ⎞ ⎛ 320 ⎞ VGS = ⎜ ⎟ VDD = ⎜ ⎟ ( 5 ) = 1.905 V R1 + R2 ⎠ ⎝ 520 + 320 ⎠ ⎝ I DQ = 0.20 (1.905 − 0.8 ) = 0.244 mA 2 g m = 2 K n I DQ = 2 ( 0.2 )( 0.244 ) = 0.442 mA/V ro = ∞ (b) Av = − g m RD = − ( 0.422 )(10 ) = −4.22 (c) (d) Ri = R1 R2 = 520 320 = 198 K RO = RD = 10 K EX4.4 At transition point, I D = 1 mA I D = K n (VGSt − VTN ) = K n (VDS ( sat ) ) 2 2 1 = 0.2 (VDS ( sat ) ) ⇒ VDS ( sat ) = 2.236 V 2 5 − 2.236 + 2.236 = 3.62 V 2 5 − 3.62 RD = = 2.76 K 0.5 Want VDSQ = 0.5 = 0.2 (VGSQ − 0.8 ) ⇒ VGSQ = 2.38 V 2 ⎛ R2 ⎞ 1 VGSQ = ⎜ ⎟ VDD = ( R1 R2 ) VDD R1 ⎝ R1 + R2 ⎠ 1 So 2.38 = ( 200 )( 5 ) ⇒ R1 = 420 K and R2 = 382 K R1 90. Av = − g m RD ( 0.2 )( 0.5 ) = 0.6325 mA/V Av = − ( 0.6325 )( 2.76 ) g m = 2 K n I DQ = 2 = −1.75 EX4.5 (a) ⎛ R2 ⎞ 250 ⎛ ⎞ VG = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 = −3 V R1 + R2 ⎠ ⎝ 250 + 1000 ⎠ ⎝ (V − VGS ) − ( −5 ) 2 = K n (VGS − VTN ) ID = G 2 −3 − VGS + 5 = 2 ( 0.5 )(VGS − 0.6 ) 2 2 − VGS = VGS − 1.2VGS + 0.36 2 VGS − 0.2VGS − 1.64 = 0 VGS = 0.2 ± ( 0.04 ) + 4 (1.64 ) 2 2 = 1.385 V I DQ = ( 0.5 )(1.385 − 0.6 ) ⇒ I DQ = 0.308 mA VDSQ = 10 − ( 0.308 )(10 + 2 ) ⇒ VDSQ = 6.30 V 2 (b) Av = − g m RD 1 + g m RS g m = 2 K n I DQ = 2 ( 0.5 )( 0.308 ) g m = 0.7849 mA/V Av = − ( 0.7849 )(10 ) 1 + ( 0.7849 )( 2 ) ⇒ Av = −3.05 EX4.6 VSDQ = 3 V and I DQ = 0.5 mA ⇒ RD = I DQ = K P (VSG − VTP ) 5−3 ⇒ RD = 4 kΩ 0.5 2 0.5 = 1(VSG − 1) ⇒ VSG = 1.71 V ⇒ VGG = 5 − 1.71 ⇒ VGG = 3.29 V 2 Av = − g m RD g m = 2 K P I DQ = 2 (1)( 0.5 ) g m = 1.414 mA/V Av = − (1.414 )( 4 ) ⇒ Av = −5.66 Av = v0 −vsd 0.46sin ω t = =− = −5.66 ⇒ vi = 0.0813sin ω t vi vi vi EX4.7 a. VSG = 9 − I DQ RS , I DQ = K P (VSG − VTP VSG = 9 − ( 2 )(1.2 )(VSG − 2 ) 2 2 = 9 − 2.4 (VSG − 4VSG + 4 ) 2 2.4VSG − 8.6VSG + 0.6 = 0 ) 2 91. (8.6 ) − 4 ( 2.4 )( 0.6 ) 2 ( 2.4 ) 2 VSG = 3.51 V, I DQ = 2 ( 3.51 − 2 ) ⇒ I DQ = 4.57 mA VSDQ = 9 + 9 − I DQ (1.2 + 1) = 18 − ( 4.57 )( 2.2 ) ⇒ VSDQ = 7.95 V VSG = 2 8.6 ± b. V0 Ϫ VSG Vi gmVSG ϩ ϩ Ϫ RD RS CS ( 2 )( 4.57 ) = 6.046 mA/V g m = 2 K P I DQ = 2 V0 = g mVSG RD Av = − g m RD = − ( 6.046 )(1) ⇒ Av = −6.05 EX4.8 VDSQ = VDD − I DQ RS 5 = 10 − (1.5 ) RS ⇒ RS = 3.33 kΩ I DQ = K n (VGS − VTN ) ⇒ 1.5 = (1)(VGS − 0.8 ) 2 2 ⎛ R2 VGS = 2.025 V = VG − VS = VG − 5 ⇒ VG = 7.025 V = ⎜ ⎝ R1 + R2 So R2 = 281 kΩ, R1 = 119 kΩ Neglecting RSi , Av = g m ( RS 1 + g m ( RS −1 r0 ) r0 ) −1 r0 = ⎡ λ I DQ ⎤ = ⎡( 0.015 )(1.5 ) ⎤ = 44.4 kΩ ⎣ ⎦ ⎣ ⎦ RS r0 = 3.33 44.4 = 3.1 kΩ g m = 2 K n I DQ = 2 (1)(1.5 ) = 2.45 mA / V Av = ( 2.45)( 3.1) ⇒ Av = 0.884 1 + ( 2.45 )( 3.1) EX4.9 I DQ = K P (VSG − VTP ) 2 3 = 2 (VSG − 2 ) ⇒ VSG = 3.22 V 5 − VSG 5 − 3.22 I DQ = ⇒3= ⇒ RS = 0.593 kΩ RS RS 2 −1 −1 r0 = ⎡ λ I DQ ⎤ = ⎡( 0.02 )( 3) ⎤ = 16.7 kΩ ⎣ ⎦ ⎣ ⎦ ( 2 )( 3) = 4.9 mA / V g m ( r0 RS ) RL = ∞, Av = 1 + g m ( r0 RS ) g m = 2 K P I DQ = 2 For r0 RS = 16.7 0.593 = 0.573 kΩ ⎞ R2 ⋅10 ⎟ VDD = 400 ⎠ 92. Av = ( 4.9 )( 0.573) ⇒ Av = 0.737 1 + ( 4.9 )( 0.573) If Av is reduced by 10% ⇒ Av = 0.737 − 0.0737 = 0.663 g m ( r0 RS RL ) Av = Let r0 RS 0.663 = 1 + g m ( r0 RS RL ) RL = x ( 4.9 ) x ⇒ 0.663 = 4.9 x (1 − 0.663) 1 + ( 4.9 ) x x = 0.402 = 0.573 RL 0.573RL = 0.402 ⇒ ( 0.573 − 0.402 ) RL = ( 0.402 )( 0.573) ⇒ RL = 1.35 kΩ RL + 0.573 EX4.10 ⎛ R2 ⎞ 9.3 ⎞ ⎛ VG = ⎜ ⎟ VDD = ⎜ ⎟ (5) ⎝ 70.7 + 9.3 ⎠ ⎝ R1 + R2 ⎠ = 0.581 V I DQ = K p (VSG − VTP ) 2 = K P (VS − VG − VTP = ) 2 5 − VS RS Then ( 0.4 )( 5 )(VS − 0.581 − 0.8 ) = 5 − VS 2 2 (VS − 1.381) = 5 − VS 2 2 (VS2 − 2.762VS + 1.907 ) = 5 − VS 2VS2 − 4.52VS − 1.19 = 0 VS = 4.52 ± ( 4.52 ) + 4 ( 2 )(1.19 ) 2 ( 2) 2 VS = 2.50 V ⇒ I DQ = g m = 2 K P I DQ = 2 Av = = 5 − 2.5 = 0.5 mA 5 ( 0.4 )( 0.5 ) = 0.894 mA / V g m RS R1 R2 ⋅ 1 + g m RS R1 R2 + RSi ( 0.894 )( 5) 70.7 9.3 ⋅ ⇒ Av = 0.770 1 + ( 0.894 )( 5 ) 70.7 9.3 + 0.5 Neglecting RSi , Av = 0.817 R0 = RS 1 1 =5 = 5 1.12 ⇒ R0 = 0.915 kΩ 0.894 gm EX4.11 gmVsg V0 ϩ Vi ϩ Ϫ RS Vsg Ϫ RD RL 93. V0 = g mVsg ( RD RL ) and Vsg = Vi Av = g m ( RD RL ) I DQ = 5 − VSG = K p (VSG − VTP RS 5 − VSG = (1)( 4 )(VSG − 0.8 ) ) 2 2 2 5 − VSG = 4 (VSG − 1.6VSG + 0.64 ) 2 4VSG − 5.4VSG − 2.44 = 0 VSG = 5.4 ± (5.4) 2 + ( 4 )( 4 )( 2.44 ) 2 ( 4) VSG = 1.71 V 5 − 1.71 I DQ = = 0.822 mA 4 g m = 2 K p I DQ = 2 (1)( 0.822 ) = 1.81 mA / V Av = (1.81)( 2 4 ) = (1.81)(1.33) ⇒ Av = 2.41 Rin = RS 1 1 =4 = 4 0.552 ⇒ Rin = 0.485 kΩ gm 1.81 EX4.12 Kn2 = μ n Cox ⎛ W ⎞ Av = − 2 ⋅ ⎜ ⎟ = ( 0.015 )( 2 ) = 0.030 mA / V 2 ⎝ L ⎠2 K n1 K = −6 ⇒ n1 = 36 Kn2 Kn2 K n1 = ( 36 )( 0.030 ) = 1.08 mA / V 2 ⎛W ⎞ ⎛W ⎞ 1.08 = ( 0.015 ) ⎜ ⎟ ⇒ ⎜ ⎟ = 72 ⎝ L ⎠1 ⎝ L ⎠1 The transition point is found from vGSt − 1 = (10 − 1) − ( 6 )( vGSt − 1) 10 − 1 + 6 + 1 = 2.29 V 1+ 6 2.29 − 1 For Q-point in middle of saturation region VGS = + 1 ⇒ VGS = 1.645 V 2 vGSt = EX4.13 (a) Transition points: For M 2 : vOtB = VDD − VTNL = 5 − 1.2 = 3.8 V 2 2 For M 1 : K n1 ⎡( vOtA ) (1 + λ vOtA ) ⎤ = K n 2 ⎡(VTNL ) (1 + λ2 [VDD − vOtA ]) ⎤ ⎣ ⎦ ⎣ ⎦ 2 2 3 250 ⎡ vOtA + ( 0.01) vOtA ⎤ = 25 ⎡(1.2 ) (1 + ( 0.01)( 5 ) − ( 0.01) vOtA ) ⎤ ⎣ ⎦ ⎣ ⎦ 2 3 3 2 10 ⎡vOtA + ( 0.01) vOtA ⎤ = 1.512 − 0.0144vOtA ( 0.01) vOtA + vOtA + 0.00144vOtA − 0.512 = 0 ⎣ ⎦ which yields vOtA ≅ 0.388 V 94. 3.8 − 0.388 + 0.388 ⇒ VDSQ1 = 2.094 V 2 2 2 K n1 ⎡(VGS 1 − VTND ) (1 + λ1vO ) ⎤ = K n 2 ⎡(VTNL ) (1 + λ2 [VDD − vO ]) ⎤ ⎣ ⎦ ⎣ ⎦ Then middle of saturation region V0Q = 250 ⎡(VGS 1 − 0.8 ) (1 + [ 0.01][ 2.094]) ⎤ = 25 ⎡(1.2 ) (1 + [ 0.01][5 − 2.094]) ⎤ ⎣ ⎦ ⎣ ⎦ 2 2 2 10 ⎡(VGS1 − 0.8 ) = (1.0209 ) ⎤ = 1.482 ⎣ ⎦ (VGS1 − 0.8 ) 2 = 0.145 ⇒ VGS1 = 1.18 V I DQ = K n1 ⎡(VGS1 − 0.8 ) (1 + ( 0.01)( 2.094 ) ) ⎤ ⎣ ⎦ 2 b. 2 I DQ = ( 0.25 ) ⎡( 0.145 ) (1.02094 ) ⎤ ⇒ I DQ = 37.0 μ A ⎣ ⎦ − g m1 Av = = − g m1 ( r01 r02 ) I DQ ( λ1 + λ2 ) c. g m1 = 2 K n1 (VGS 1 − VTND ) = 2 ( 0.25 )(1.18 − 0.8 ) = 0.19 mA/V Av = −0.19 ⇒ Av = −257 ( 0.037 )( 0.01 + 0.01) EX4.14 Av = − g m ( ron rop ) ron = rop = 1 = 666.7 K ( 0.015)( 0.1) −250 = − g m ( 666.7 666.7 ) g m = 0.75 mA/V = 2 K n I DQ = 2 K n ( 0.1) K n = 1.406 mA/V 2 = ′ kn ⎛ W ⎜ 2⎝L ⎞ ⎛ 0.080 ⎞ ⎛ W ⎞ ⎟=⎜ ⎟⎜ ⎟ ⎠ ⎝ 2 ⎠⎝ L ⎠ ⎛W ⎞ ⎜ ⎟ = 35.2 ⎝ L ⎠1 EX4.15 (a) RO = So g m1 = 1 1 ro 2 ro1 ≈ g m1 g m1 1 1 = = 0.5 mA/V R0 2 g m1 = 2 K n I D ( 0.2 ) I D 0.5 = 2 (b) Av = g m1 ( ro1 ro 2 ) 1 + g m1 ( ro1 ro 2 ) ro1 = ro 2 = Av = ⇒ I D = 0.3125 mA 1 = 320 K ( 0.01)( 0.3125 ) 0.5 ( 320 320 ) 1 + ( 0.5 )( 320 320 ) Av = 0.988 EX4.16 95. (a) 1 ro1 2 K n I D + λ1 I D Av = = 1 1 λ2 I D + λ1 I D + ro 2 ro1 g m1 + 120 = 2 0.2 I D + 0.01I D 0.01I D + 0.01I D 2.4 I D − 0.01I D = 2 0.2 I D 2.39 I D = 2 0.2 ⇒ I D = 0.140 mA g m1 = 2 ( 0.2 )( 0.14 ) ⇒ g m1 = 0.335 mA/V (b) Ro = ro1 ro 2 ro1 = ro 2 = 1 = 714 K 0.01)( 0.14 ) ( Ro = 357 K EX4.17 R0 = RS 2 1 gm2 g m 2 = 0.632 mA/V, RS 2 = 8 kΩ R0 = 8 1 = 8 1.58 ⇒ R0 = 1.32 kΩ 0.632 EX4.18 a. I DQ1 = K n1 (VGS 1 − VTN 1 ) 2 1 = 1.2 (VGS 1 − 2 ) ⇒ VGS 1 = VGS 2 = 2.91 V 2 RS = 10 kΩ ⇒ VS1 = I DQ RS − 10 = (1)(10 ) − 10 = 0 ⎛ ⎞ R3 VG1 = 2.91 = ⎜ ⎟ (10 ) ⎝ R1 + R2 + R3 ⎠ ⎛ R ⎞ = ⎜ 3 ⎟ (10 ) ⇒ R3 = 145.5 kΩ ⎝ 500 ⎠ VDSQ1 = 3.5 ⇒ VS 2 = 3.5 V ⇒ VG 2 = 3.5 + 2.91 ⇒ VG 2 = 6.41 ⎛ R2 + R3 ⎞ VG 2 = ⎜ ⎟ (10 ) = 6.41 ⎝ R1 + R2 + R3 ⎠ ⎛ R + R3 ⎞ =⎜ 2 ⎟ (10 ) ⎝ 500 ⎠ R2 + R3 = 320.5 = R2 + 145.5 ⇒ R2 = 175 kΩ Then R1 + R2 + R3 = 500 = R1 + 175 + 145.5 ⇒ R1 = 179.5 kΩ Now VS 2 = 3.5 ⇒ VD 2 = VS 2 + VSDQ 2 = 3.5 + 3.5 = 7 V So RD = b. 10 − 7 ⇒ RD = 3 kΩ 1 From Example 6-18: 96. Av = − g m1 RD g m1 = 2 K n1 I DQ = 2 (1.2 )(1) = 2.19 mA / V Av = − ( 2.19 )( 3) ⇒ Av = −6.57 EX4.19 VS = I DQ RS = (1.2 )( 2.7 ) = 3.24 V VD = VS + VDSQ = 3.24 + 12 = 15.24 20 − 15.24 RD = ⇒ RD = 3.97 kΩ 1.2 ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ 2 2 ⎛ V ⎞ V 1.2 = 4 ⎜1 − GS ⎟ ⇒ GS = 0.4523 VP ⎠ VP ⎝ VGS = ( 0.4523)( −3) = −1.357 VG = VS + VGS = 3.24 − 1.357 = 1.883 V ⎛ R2 ⎞ ⎛ R2 ⎞ VG = ⎜ ⎟ ( 20 ) = ⎜ ⎟ ( 20 ) = 1.88 ⇒ R2 = 47 kΩ. R1 = 453 kΩ ⎝ 500 ⎠ ⎝ R1 + R2 ⎠ 1 1 = = 167 kΩ r0 = λ I DQ ( 0.005 )(1.2 ) 2 I DSS ⎛ VGS ⎞ 2 ( 4 ) ⎛ 1.357 ⎞ = 1− 1− ⎟ = 1.46 mA/V 3 ⎠ ( −VP ) ⎜ VP ⎟ 3 ⎜ ⎝ ⎝ ⎠ Av = − g m ( r0 RD RL ) = − (1.46 )(167 3.97 4 ) ⇒ Av = −2.87 gm = EX4.20 a. 2 2 ⎛ V ⎞ ⎛ V ⎞ V I DQ = I DSS ⎜ 1 − GS ⎟ 2 = 8 ⎜1 − GS ⎟ ⇒ GS = 0.5 VP ⎠ VP ⎠ VP ⎝ ⎝ VGS = ( 0.5 )( −3.5 ) ⇒ VGS = −1.75 Also I DQ = −VGS − ( −10 ) RS 2= 1.75 + 10 ⇒ RS = 5.88 kΩ RS b. ϩ Vi ϩ Ϫ gmVgs Vgs Ϫ V0 RS 97. gm = r0 = 2 I DSS VP ⎛ VGS ⎜1 − ⎝ VP ⎞ 2 ( 8 ) ⎛ 1.75 ⎞ ⎟= ⎜1 − ⎟ = 2.29 mA/V 3.5 ⎠ ⎠ 3.5 ⎝ 1 = 50 kΩ ( 0.01)( 2 ) Vi = Vgs + g m RS Vgs ⇒ Vgs = Av = c. Av = Vi 1 + g m RS ( 2.29 ) [5.88 50] V0 g m RS r0 = = ⇒ Av = 0.9234 V1 1 + g m RS r0 1 + ( 2.29 ) [5.88 50] g m ( RS 1 + g m ( RS ( 2.29 )( RS 1 + ( 2.29 )( RS RL ) ro RL ro ) RL ro ) RL ro ) = ( 0.80 )( 0.9234 ) = 0.7387 = 0.7387 ⇒ RS RL ro = 1.235 kΩ RS ro = 5.261 kΩ 5.261RL = 1.235 kΩ ⇒ RL = 1.61 kΩ 5.261 + RL TYU4.1 g m = 2 K n (VGS − VTN ) and I D = K n (VGS − VTN ) ⇒ VGS − VTN = 2 and g m = 2 K n I DQ Kn I DQ Kn = 2 K n I DQ ( 3.4 ) g2 = 1.445 mA / V Kn = m = 4 I DQ 4 ( 2) 2 K n = μ n Cox W 2 ⋅ L ⎛W 1.445 = ( 0.018 ) ⎜ ⎝L ⎞ W = 80.3 ⎟⇒ L ⎠ −1 2 ro = ⎡λ K n (VGS − VTN ) ⎤ = ⎡λ I DQ ⎤ ⎣ ⎦ ⎣ ⎦ −1 −1 ro = ⎡( 0.015 )( 2 ) ⎤ ⇒ ro = 33.3 k Ω ⎣ ⎦ TYU4.2 a. I DQ = K n (VGS − VTN ) 2 0.4 = 0.5 (VGS − 2 ) ⇒ VGS = 2.894 V 2 VDSQ = VDD − I DQ RD = 10 − ( 0.4 )(10 ) ⇒ VDSQ = 6 V b. g m = 2 K n (VGS − VTN ) = 2 ( 0.5 )( 2.894 − 2 ) ⇒ g m = 0.894 mA/V −1 r0 = ⎡ λ I DQ ⎤ , λ = 0 ⇒ r0 = ∞ ⎣ ⎦ v0 Av = = − g m RD = − ( 0.894 )(10 ) ⇒ Av = −8.94 vi c. vi = 0.4sin ω t ⇒ vds = − ( 8.94 )( 0.4 ) sin ω t vds = −3.58sin ω t 98. At VDS 1 = 6 − 3.58 = 2.42 VGS1 = 2.89 + 0.4 = 3.29 VGS1 − VTN = 3.29 − 2 = 1.29 = VDS ( sat ) So VDS 1 > VGS 1 − VTN ⇒ Biased in saturation region TYU4.3 I DQ = K n (VGS − VTN ) = ( 0.25 )( 2 − 0.8 ) ⇒ I DQ = 0.36 mA 2 a. 2 VDSQ = VDD − I DQ RD = 5 − ( 0.36 )( 5 ) ⇒ VDSQ = 3.2 V b. g m = 2 K n (VGS − VTN ) = 2 ( 0.25 )( 2 − 0.8 ) ⇒ g m = 0.6 mA/V, c. Av = r0 = ∞ v0 = − g m RD = − ( 0.6 )( 5 ) ⇒ Av = −3.0 vi TYU4.4 vi = vgs = 0.1sin ω t id = g m vgs = ( 0.6 )( 0.1) sin ω t id = 0.06sin ω t mA vds = ( −3)( 0.1) sin ω t = −0.3sin ω t ( V ) Then iD = I DQ + id = 0.36 + 0.06sin ω t = iD mA vDS = VDSQ + vds = 3.2 − 0.3sin ω t = vDS TYU4.5 a. VSDQ = VDD − I DQ RD 7 = 12 − I DQ ( 6 ) ⇒ I DQ = 0.833 mA I DQ = K P (VSG − VTP ) 2 0.833 = 2 (VSG − 1) ⇒ VSG = 1.65 V 2 g m = 2 K P (VSG − VTP ) = 2 ( 2 )(1.65 − 1) ⇒ g m = 2.58 mA/V, r0 = ∞ b. Av = v0 = − g m RD = − ( 2.58 )( 6 ) ⇒ Av = −15.5 vi V0 Ϫ Vi ϩ Ϫ gmVsg VSG RD ϩ TYU4.6 I DQ = K n (VGS − VTN ) ⇒ VGS − VTN = 2 g m = 2 K n (VGS − VTN ) = 2 K n So g m = 2 K n I DQ TYU4.7 η= γ 2 2φ f + vSB I DQ Kn I DQ Kn 99. η= (a) 0.40 2 2 ( 0.35 ) + 1 η= (b) ⇒ η = 0.153 0.40 2 2 ( 0.35 ) + 3 ⇒ η = 0.104 ( 0.5 )( 0.75) = 1.22 mA / V g mb = g m η = (1.22 )( 0.153) ⇒ g mb = 0.187 mA / V g mb = (1.22 )( 0.104 ) ⇒ g mb = 0.127 mA / V g m = 2 K n I DQ = 2 For (a), For (b), TYU4.8 I DQ = I Q = 0.5 mA W Let = 25 L K n = ( 20 )( 25 ) = 500 μ A / V 2 0.5 + 1.5 = 2.5 V ⇒ VS = −2.5 V 0.5 Av = − g m RD VGS = g m = 2 ( 0.5 )( 2.5 − 1.5 ) = 1 mA/V For Av = −4.0 ⇒ RD = 4 kΩ VD = 5 − ( 0.5 )( 4 ) = 3 V ⇒ VDSQ = 3 − ( −2.5 ) = 5.5 V TYU4.9 a. With RG ⇒ VGS = VDS ⇒ transistor biased in sat. region I D = K n (VGS − VTN ) = K n (VDS − VTN ) 2 2 VDS = VDD − I D RD = VDD − K n RD (VDS − VTN ) VDS = 15 − ( 0.15 )(10 )(VDS − 1.8 ) 2 2 2 = 15 − 1.5 (VDS − 3.6VDS + 3.24 ) 2 1.5VDS − 4.4VDS − 10.14 = 0 VDS = 4.4 ± ( 4.4 ) 2 + ( 4 )(1.5 )(10.14 ) 2 (1.5 ) ⇒ VDSQ = 4.45 V I DQ = ( 0.15 )( 4.45 − 1.8 ) ⇒ I DQ = 1.05 mA 2 b. Neglecting effect of RG: Av = − g m ( RD RL ) g m = 2 K n (VGS − VTN ) = 2 ( 0.15 )( 4.45 − 1.8 ) ⇒ g m = 0.795 mA/V Av = − ( 0.795 )(10 5 ) ⇒ Av = −2.65 c. RG ⇒ establishes VGS = VDS ⇒ essentially no effect on small-signal voltage gain. TYU4.10 a. 2 I DQ = K n (VGS − VTN ) I DQ = 0.8 ( 2 − VSG ) = 2 VSG VSG = 4 RS 2 3.2 ( 4 − 4VSG + VSG ) = VSG 2 3.2VSG − 13.8VSG + 12.8 = 0 100. (13.8) − 4 ( 3.2 )(12.8 ) 2 ( 3.2 ) 2 = 1.35 V ⇒ I DQ = 0.8 ( 2 − 1.35 ) ⇒ I DQ = 0.338 mA VSG = VSG 2 13.8 ± b. VDSQ = VDD − I DQ ( RD + RS ) 6 = 10 − ( 0.338 )( RD + 4 ) RD = 10 − ( 0.338 )( 4 ) − 6 0.338 ⇒ RD = 7.83 kΩ c. V0 ϩ Vgs Vi ϩ Ϫ gmVgs RD Ϫ RS Vi = Vgs + g mVgs RS ⇒ Vgs = Vi 1 + g m RS V0 = − g mVgs RD g m = 2 K n (VGS − VTN ) = 2 ( 0.8 )( −1.35 + 2 ) = 1.04 mA/V Av = − (1.04 )( 7.83) V0 − g m RD = = ⇒ Av = −1.58 1 + (1.04 )( 4 ) Vi 1 + g m RS TYU4.11 a. 5 = I DQ RS + VSG and I DQ = K p (VSG + VTP ) 2 0.8 = 0.5(VSG + 0.8) 2 ⇒ VSG = 0.465 V 5 = ( 0.8 ) RS + 0.465 ⇒ RS = 5.67 kΩ VSDQ = 10 − I DQ ( RS + RD ) 3 = 10 − ( 0.8 )( 5.67 + RD ) RD = 10 − ( 0.8 )( 5.67 ) − 3 0.8 ⇒ RD = 3.08 kΩ b. V0 Ϫ Vi ϩ Ϫ VSG gmVsg r0 RD ϩ V0 = g mVsg ( RD r0 ) = − g mVi ( RD r0 ) V Av = 0 = − g m ( RD r0 ) Vi g m = 2 K p (VSG + VTP ) = 2(0.5)(0.465 + 0.8) = 1.265 mA/V 1 1 = = 62.5 kΩ r0 = λ I 0 ( 0.02 )( 0.8 ) Av = − (1.265 )( 3.08 62.5 ) ⇒ Av = −3.71 101. TYU4.12 (a) V0 = g mVgs r0 Vi = Vgs + V0 ⇒ Vgs = Vi − V0 So V0 = g m r0 (Vi − V0 ) Av = ( 4 )( 50 ) V0 g m r0 = = ⇒ Av = 0.995 Vi 1 + g m r0 1 + ( 4 )( 50 ) ϩ Vgs Ϫ Ix I x + g mVgs = I x = g mVx + Vx Vx and Vgs = −Vx r0 Vx 1 1 ⇒ R0 = r0 = 50 ⇒ R0 ≅ 0.25 kΩ 4 r0 gm With RS = 4 kΩ ⇒ Av = (b) ϩ Ϫ r0 gmVgs r0 || Rs = 50 || 4 = 3.7 kΩ ⇒ Av = g m ( r0 RS ) 1 + g m ( r0 RS ) ( 4 )( 3.7 ) 1 + ( 4 )( 3.7 ) ⇒ Av = 0.937 TYU4.13 (a) g m = 2 K n I DQ ⇒ 2 = 2 K n ( 0.8 ) ⇒ K n = 1.25 mA / V 2 Kn = So μ n Cox W 2 ⋅ ⎛W ⎞ ⇒ 1.25 = ( 0.020 ) ⎜ ⎟ L ⎝L⎠ W = 62.5 L I DQ = K n (VGS − VTN ) ⇒ 0.8 = 1.25 (VGS − 2 ) ⇒ VGS = 2.8 V 2 2 b. −1 −1 ⎡ ⎤ r0 = ⎣ λ I DQ ⎦ = ⎡( 0.01)( 0.8 ) ⎤ = 125 kΩ ⎣ ⎦ g m ( r0 RL ) Av = 1 + g m ( r0 RL ) r0 RL = 125 4 = 3.88 Av = ( 2 )( 3.88 ) ⇒ Av = 0.886 1 + ( 2 )( 3.88 ) R0 = 1 1 r0 = 125 ⇒ R0 ≈ 0.5 kΩ gm 2 TYU4.14 102. Rin = 1 = 0.35 kΩ ⇒ g m = 2.86 mA/V gm V0 4 RD = RD RL = 2.4 = RD 4 = Ii 4 + RD ( 4 − 2.4 ) RD = ( 2.4 )( 4 ) ⇒ RD = 6 kΩ g m = 2 K n I DQ 2.86 = 2 K n ( 0.5 ) ⇒ K n = 4.09 mA / V 2 I DQ = K n (VGS − VTN ) 2 0.5 = 4.09 (VGS − 1) ⇒ VGS = 1.35 V ⇒ VS = −1.35 V, VD = 5 − ( 0.5 )( 6 ) = 2 V 2 VDS = VD − VS = 2 − ( −1.35 ) = 3.35 V We have VDS = 3.35 > VGS − VTN = 1.35 − 1 = 0.35 V ⇒ Biased in the saturation region TYU4.15 μC K n1 = n ox 2 Kn2 ⎛W ⋅⎜ ⎝L μ C ⎛W = n ox ⋅ ⎜ 2 ⎝L Av = − ⎞ 2 ⎟ = ( 0.020 )( 80 ) = 1.6 mA / V ⎠1 ⎞ 2 ⎟ = ( 0.020 )(1) = 0.020 mA / V ⎠2 K n1 1.6 =− ⇒ Av = −8.94 Kn2 0.020 The transition point is determined from vGSt − VTND = VDD − VTNL − K n1 ( vGSt − VTND ) Kn2 vGSt − 0.8 = ( 5 − 0.8 ) − ( 8.94 )( vGSt − 0.8 ) vGSt = ( 5 − 0.8) + ( 8.94 )( 0.8) + 0.8 1 + 8.94 vGSt = 1.22 V For Q-point in middle of saturation region VGS = 1.22 − 0.8 + 0.8 ⇒ VGS = 1.01 V 2 TYU4.16 a. I DQ 2 = 2mA, VDSQ 2 = 10 V I DQ 2 ⋅ RS 2 = 10 = 2 RS 2 ⇒ RS 2 = 5 kΩ I DQ 2 = K n 2 (VGS 2 − VTN 2 ) 2 2 = 1(VGS 2 − 2 ) ⇒ VSG 2 = 3.41 V ⇒ VG 2 = 3.41 V 2 10 − 3.41 ⇒ RD1 = 3.29 kΩ 2 = 10 V ⇒ VS 1 = 3.41 − 10 = −6.59 V Then RD1 = For VDSQ1 Then RS 1 = −6.59 − ( −10 ) 2 I D1 = K n1 (VGS 1 − VTN 1 ) ⇒ RS1 = 1.71 kΩ 2 2 = 1(VGS 1 − 2 ) ⇒ VGS 1 = 3.41 V 2 ⎛ R2 ⎞ R2 1 VGS1 = ⎜ = ⋅ Rin ⎟ ( 20 ) − I DQ1 RS 1 R1 + R2 ⎠ R1 + R2 R1 ⎝ 1 3.41 = ( 200 )( 20 ) − ( 2 )(1.71) ⇒ R1 = 586 kΩ R1 103. 586 R2 = 200 ⇒ ( 586 − 200 ) R2 = ( 200 )( 586 ) ⇒ R2 = 304 kΩ 586 + R2 g m1 = 2 K n1 I DQ1 = 2 (1)( 2 ) ⇒ g m1 = g m 2 = 2.828 mA/V b. From Example 6.17 −g g R ( R RL ) Av = m1 m 2 D1 S 2 1 + g m 2 ( RS 2 RL ) RS 2 RL = 5 4 = 2.222 kΩ − ( 2.828 )( 2.828 )( 3.29 )( 2.222 ) Av = 1 + ( 2.828 )( 2.222 ) ⇒ Av = −8.03 1 1 RS 2 = 5 = 0.3536 5 ⇒ R0 = 0.330 kΩ gm2 2.828 R0 = TYU4.17 From Example 6.19 g m = 3.0 mA/V, r0 = 41.7 kΩ R1 R2 = 420 180 = 126 kΩ ⎛ R1 R2 ⎞ Vgs = ⎜ ⎟ Vi ⎝ R1 R2 + Ri ⎠ ⎛ 126 ⎞ =⎜ ⎟ Vi = 0.863Vi ⎝ 126 + 20 ⎠ − g mVgs ( r0 RD RL ) Av = Vi = − ( 3.0 )( 0.863)( 41.7 2.7 4 ) = − ( 2.589 )( 41.7 1.61) = − ( 2.589 )(1.55 ) ⇒ Av = −4.01 TYU4.18 a. ⎛ R2 ⎞ VG1 = ⎜ ⎟ (VDD ) ⎝ R1 + R2 ⎠ ⎛ 430 ⎞ VG1 = ⎜ ⎟ ( 20 ) = 17.2 V ⎝ 430 + 70 ⎠ 2 ⎛ V ⎞ ⎛ V −V ⎞ I DQ1 = I DSS ⎜ 1 − GS ⎟ = 6 ⎜ 1 − G1 S1 ⎟ VP ⎠ 2 ⎝ ⎠ ⎝ 2 2 2 20 − VS1 ⎛ 17.2 VS 1 ⎞ ⎛V ⎞ = 6 ⎜1 − + = 6 ⎜ S 1 − 7.6 ⎟ and I DQ1 = 2 2 ⎟ 2 4 ⎝ ⎠ ⎝ ⎠ 2 20 − VS 1 ⎛V ⎞ = 6 ⎜ S1 − 7.6 ⎟ 4 2 ⎝ ⎠ 2 ⎛V ⎞ 20 − VS1 = 24 ⎜ S 1 − 7.6VS 1 + 57.76 ⎟ 4 ⎝ ⎠ Then = 6VS21 − 182.4VS 1 + 1386.24 6VS21 − 181.4VS 1 + 1366.24 = 0 VS 1 = 181.4 ± (181.4 ) − 4 ( 6 )(1366.24 ) 2 ( 6) 2 VS 1 = 14.2 V ⇒ VGS 1 = 17.2 − 14.2 = 3 V > VP 104. So VS1 = 16.0 ⇒ VGS1 = 17.2 − 16 = 1.2 < VP − Q 20 − 16 ⇒ I DQ1 = 1 mA 4 = 20 − I DQ1 ( RS 1 + RD1 ) = 20 − (1)( 8 ) ⇒ VSDQ1 = 12 V on I DQ1 = VSDQ1 VD1 = I DQ1 RD1 = (1)( 4 ) = 4 V = VG 2 2 ⎛ V − VS 2 ⎞ ⎛ V ⎞ I DQ 2 = I DSS ⎜ 1 − GS ⎟ = 6 ⎜ 1 − G 2 ⎟ ⎜ ⎟ VP ⎠ ( −2 ) ⎠ ⎝ ⎝ 2 2 2 V V ⎛ 4 V ⎞ ⎛ V ⎞ = 6 ⎜ 1 + − S 2 ⎟ = 6 ⎜ 3 − S 2 ⎟ and I DQ 2 = S 2 = S 2 RS 2 2 ⎠ 4 ⎝ 2 2 ⎠ ⎝ Then VS 2 ⎛ V ⎞ = 6⎜3 − S2 ⎟ 4 2 ⎠ ⎝ 2 ⎛ V2 ⎞ VS 2 = 24 ⎜ 9 − 3VS 2 + S 2 ⎟ 4 ⎠ ⎝ 2 6VS 2 − 73VS 2 + 216 = 0 VS 2 = ( 73) 73 ± 2 − 4 ( 6 )( 216 ) 2 ( 6) ⇒ VS 2 = 7.09 V or = 5.08 V For VS 2 = 5.08 V ⇒ VGS 2 = 4 − 5.08 = −1.08 transistor biased ON 5.08 ⇒ I DQ 2 = 1.27 mA 4 = 20 − VS 2 = 20 − 5.08 ⇒ VDS 2 = 14.9 V I DQ 2 = VDS 2 b. Vg2 Ϫ ϩ gm2Vgs2 Vgs2 Vi ϩ Ϫ RD1 V0 Vsg1 gm1Vsg1 RS2 ϩ Vg 2 = g m1Vsg1 RD1 = − g m1V1 RD1 V0 = g m 2Vgs 2 ( RS 2 RL ) Vg 2 = Vgs 2 + V0 ⇒ Vgs 2 = Vg 2 1 + g m 2 ( RS 2 RL ) Av = V0 − g m1 RD1 = Vi 1 + g m 2 ( RS 2 RL ) g m1 = 2 I DSS VP ⎛ VGS ⎞ ⎜1 − ⎟ VP ⎠ ⎝ 2 ( 6 ) ⎛ 1.2 ⎞ ⎜1 − ⎟ = 2.4 mA/V 2 ⎝ 2 ⎠ 2 ( 6 ) ⎛ 1.08 ⎞ gm2 = ⎜1 − ⎟ = 2.76 mA/V 2 ⎝ 2 ⎠ − ( 2.4 )( 4 ) = −2.05 = Av Then Av = 1 + ( 2.76 )( 4 )( 2 ) = Ϫ RL 105. Chapter 4 Problem Solutions 4.1 (a) ⎛ μ C ⎞⎛W ⎞ g m = 2 K n I D = 2 ⎜ n ox ⎟ ⎜ ⎟ I D ⎝ 2 ⎠⎝ L ⎠ 0.5 = 2 ⎛ ( 0.040 ) ⎜ ω⎞ W = 3.125 ⎟ ( 0.5 ) ⇒ L ⎝L⎠ (b) 2 ⎛ μ C ⎞⎛W ⎞ I D = ⎜ n ox ⎟ ⎜ ⎟ (VGS − VTN ) 2 ⎠⎝ L ⎠ ⎝ 0.5 = ( 0.04 )( 3.125 )(VGS − 0.8 ) ⇒ VGS = 2.80 V 2 4.2 ⎛ μ p Cox ⎞ ⎛ W ⎞ gm = 2 K p I D = 2 ⎜ ⎟ ⎜ ⎟ ID ⎝ 2 ⎠⎝ L ⎠ (a) 2 W ⎛ 50 ⎞ ⎛W ⎞ = 0.3125 ⎜ ⎟ = ( 20 ) ⎜ ⎟ (100 ) ⇒ L ⎝ 2⎠ ⎝L⎠ ⎛ μ p Cox ⎞ ⎛ W ⎞ 2 ID = ⎜ ⎟ ⎜ ⎟ (VSG + VTP ) 2 ⎠⎝ L ⎠ ⎝ (b) 100 = ( 20 )( 0.3125 )(VSG − 1.2 ) ⇒ VSG = 5.2 V 2 4.3 I D = K n (VGS − VTN ) (1 + λVDS ) 2 I D1 1 + λVDS1 3.4 1 + λ (10 ) = ⇒ = I D 2 1 + λVDS 2 3.0 1 + λ ( 5 ) 3.4 [1 + 5λ ] = 3.0 [1 + 10λ ] 3.4 − 3.0 = λ ( 3 ⋅10 − ( 3.4 ) ⋅ 5 ) ⇒ λ = 0.0308 ro = ΔVDS 5 = = 12.5 kΩ ΔI D 0.4 4.4 r0 = ID = 1 λ ID 1 λ r0 = 1 ( 0.012 )(100 ) ⇒ I D = 0.833 mA 4.5 2 I D = K n (VGS − VTN ) (1 + λVDS ) I D1 1 + λVDS 1 = I D 2 1 + λVDS 2 0.20 1 + λ ( 2 ) = 0.22 1 + λ ( 4 ) 0.20 (1 + 4λ ) = 0.22 (1 + 2λ ) ( 0.8 − 0.44 ) λ = 0.22 − 0.20 λ = 0.0556 V −1 106. ro = ΔVDS 2 = ⇒ ro = 100 K 0.02 ΔI D 4.6 (a) (i) ro = 1 1 = = 1000 K λ I D ( 0.02 )( 0.05 ) (ii) ro = 1 = 100 K ( 0.02 )( 0.5) (b) ΔI D = (i) ΔVDS 1 = = 0.001 mA = 1.0 μ A 1000 ro ΔI D 1.0 ⇒ 2% = 50 ID ΔI D = (ii) ΔVDS 1 = = 0.01 ⇒ 10 μ A ro 100 ΔI D 10 = ⇒ 2% ID 500 4.7 I D = 1.0 mA 1 1 ro = = = 100 K λ I D ( 0.01)(1) 4.8 Av = − g m ( r0 || RD ) = − (1)( 50 ||10 ) ⇒ Av = −8.33 4.9 RD = a. VDD − VD SQ I DQ = 10 − 6 ⇒ RD = 8 kΩ 0.5 For VGSQ = 2 V, for example, 2 ⎛ μ C ⎞⎛W ⎞ I DQ = ⎜ n ox ⎟ ⎜ ⎟ (VGSQ − VTN ) 2 ⎠⎝ L ⎠ ⎝ W 2 ⎛W ⎞ 0.5 = ( 0.030 ) ⎜ ⎟ ( 2 − 0.8 ) ⇒ = 11.6 L ⎝L⎠ b. g m = 2 K n I DQ = 2 K n (VGSQ − VTN ) g m = 2 ( 0.030 )(11.6 )( 2 − 0.8 ) ⇒ g m = 0.835 mA/V ro = 1 c. = 1 ⇒ r = 133 kΩ ( 0.015)( 0.50 ) o Av = − g m ( r0 RD ) = − ( 0.835 )(133 8 ) ⇒ Av = −6.30 λ I DQ 4.10 2 2 2 K n vgs = K n ⎡Vgs sinω t ⎤ = K nVgs sin 2ω t ⎣ ⎦ 1 [1 − cos 2ω t ] 2 2 K nVgs 2 So K n vgs = [1 − cos 2ω t ] 2 sin 2 ω t = 107. 2 K nVgs Ratio of signal at 2ω to that at ω : 2 ⋅ cos 2ω t 2 K n (VGSQ − VTN ) Vgs ⋅ sin ω t The coefficient of this expression is then: Vgs 4 (VGSQ − VTN ) 4.11 0.01 = Vgs 4 (VGSQ − VTN ) So Vgs = ( 0.01)( 4 )( 3 − 1) ⇒ Vgs = 0.08 V 4.12 Vo = − g mVgs ( ro RD ) ⎛ 50 ⎞ ⋅ Vi = ⎜ ⎟ ⋅ Vi = ( 0.9615 ) Vi R1 R2 + RSi ⎝ 50 + 2 ⎠ Av = − g m ( 0.9615 ) ( ro RD ) = − (1)( 0.9615 ) ( 50 10 ) ⇒ Av = −8.01 R1 R2 Vgs = 4.13 Av = − g m ( r0 || RD ) −10 = − g m (100 || 5 ) ⇒ g m = 2.1 mA/V 4.14 a. ⎛ R2 ⎞ VG = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠ ⎛ 200 ⎞ VG = ⎜ ⎟ (12 ) = 4.8 V ⎝ 200 + 300 ⎠ V − VGS 2 = K n (VGS − VTN ) ID = G RS 2 4.8 − VGS = (1)( 2 ) (VGS − 4VGS + 4 ) 2 2VGS − 7VGS + 3.2 = 0 VGS = 7± (7) 2 − 4 ( 2 )( 3.2 ) 2 ( 2) = 2.96 V I D = (1)( 2.96 − 2 ) ⇒ I D = 0.920 mA 2 VDS = VDD − I D ( RD + RS ) = 12 − ( 0.92 )( 3 + 2 ) ⇒ VDS = 7.4 V (b) Vo = − g mVg ( RD RL ) 1 + g m RS where Vg = Then R1 R2 R1 R2 + RSi ⋅ Vi = 300 200 300 200 + 2 ⋅ Vi = 120 ⋅ Vi = ( 0.9836 ) Vi 120 + 2 108. Av = − g m ( 0.9836 )( RD || RL ) 1 + g m RS g m = 2 (1)( 2.96 − 2 ) = 1.92 mA / V So Av = − (1.92 )( 0.9836 )( 3 || 3) 1 + (1.92 )( 2 ) ⇒ Av = −0.585 c. AC load line Ϫ1 Ϫ1 ϭ Slope ϭ 3.5 K 3͉͉3 ϩ 2 0.92 7.4 12 −1 Δi D = ⋅ Δvds 3.5 kΩ Δi D = 0.92 mA ⇒ Δvds = ( 0.92 )( 3.5 ) = 3.22 ⇒ 6.44 V peak-to-peak 4.15 a. I D Q = 3 mA ⇒ VS = I DQ RS = ( 3)( 0.5 ) = 1.5 V I DQ = K n (VGS − VTN ) 2 3 = ( 2 )(VGS − 2 ) ⇒ VGS = 3.225 V 2 VG = VGS + VS = 3.225 + 1.5 = 4.725 V ⎛ R2 ⎞ 1 VG = ⎜ ⎟ VDD = ⋅ Rin ⋅ VDD R1 ⎝ R1 + R2 ⎠ 1 4.725 = ( 200 )(15 ) ⇒ R1 = 635 kΩ R1 635R2 = 200 ⇒ R2 = 292 kΩ 635 + R2 b. − g m ( RD || RL ) Av = 1 + g m RS g m = 2 ( 2 )( 3.225 − 2 ) = 4.90 mA / V Av = − ( 4.90 )( 2 || 5 ) 1 + ( 4.90 )( 0.5 ) ⇒ Av = −2.03 4.16 From Problem 4.14: I D = 0.920 mA VDS = 7.4 V g m = 1.92 mA/V ⎛ R1 || R2 ⎞ Av = − g m ( RD || RL ) ⎜ ⎟ ⎝ R1 || R2 + RSi ⎠ ⎛ 200 || 300 ⎞ = − (1.92 )( 3 || 3) ⎜ ⎟ = − ( 2.88 )( 0.9836 ) ⎝ 200 || 300 + 2 ⎠ Av = −2.83 109. AC load line Ϫ1 Ϫ1 Slope ϭ ϭ 3͉͉3 1.5 K 0.92 7.4 12 −1 ⋅ ΔvDS , ΔiD = 0.92 mA ⇒ ΔvDS = ( 0.92 ) (1.5 ) = 1.38 1.5 kΩ ⇒ 2.76 V peak-to-peak output voltage swing ΔiD = 4.17 (a) Av = − g m RD −15 = −2 RD ⇒ RD = 7.5 K (b) Av = −5 = − g m RD 1 + g m RS − ( 2 )( 7.5 ) 1 + ( 2 ) RS ⇒ RS = 1 K 4.18 (a) Av = − g m RD 1 + g m RS (1) −8 = − g m RD 1 + g m (1) (2) −16 = − g m RD Then 8 = 16 ⇒ 1 + g m (1) g m = 1 mA/V RD = 16 K (b) Av = −10 = − (1)(16 ) 1 + (1) RS RS = 0.6 K 4.19 a. AC load line Ϫ1 Slope ϭ 5K IDQ VDS(sat) VDSQ 110. VDSQ = V + − I DQ RD − (−VGSQ ) Output Voltage Swing = VDSQ − VDS ( sat ) = ⎡V + − I DQ RD + VGSQ ⎤ − (VGSQ − VTN ) ⎣ ⎦ = V + − I DQ RD + VTN So ΔI D = I DQ = 1 1 ⎡V + − I DQ RD + VTN ⎤ ⋅ ΔVDS = ⎦ 5 kΩ 5 kΩ ⎣ 1 ⎡5 − I DQ (10) + 1⎤ ⎦ 5 kΩ ⎣ = 1.2 − 2 I DQ = I DQ ⇒ I DQ = 0.4 mA = I Q I DQ = b. ( 0.5 )( 0.4 ) = 0.8944 mA / V g m = 2 K n I DQ = 2 r0 = 1 1 = = 250 kΩ λ I DQ ( 0.01)( 0.4 ) Av = − g m ( RD || RL || r0 ) = − ( 0.8944 )(10 ||10 || 250 ) ⇒ Av = −4.38 4.20 (a) I DQ = K n (VGSQ − VTN ) 2 0.1 = 0.85 (VGSQ − 0.8 ) 2 VGSQ = 1.143 V RS = −1.143 − ( −5 ) ⇒ RS = 38.6 K 0.1 ΔV = Δ I ( RD RL ) ro ro = 1 = 0.1( RD RL ) ro RD RL ro = 10 K = 1 = 500 K ( 0.02 )( 0.1) 40 500 RD 40 500 + RD ⇒ 37.04 RD = 10 37.04 + RD RD = 13.7 K (b) gm = 2 Kn I D = 2 ( 0.85)( 0.1) g m = 0.583 mA/V ro = 500 K (c) Av = − g m ( RD RL ro ) = − ( 0.583) (13.7 40 500 ) = − ( 0.583)(10 ) Av = −5.83 4.21 a. 2 I D = K n (VGS − VTN ) 2 = 4 (VGS − ( −1) ) 2 VGS = −0.293 V ⇒ VS = 0.293 V = I D RS = (2) RS ⇒ RS = 0.146 kΩ 111. VD = VDS + VS = 6 + 0.293 = 6.293 RD = b. Av = 10 − 6.293 ⇒ RD = 1.85 kΩ 2 − g m ( RD RL ) 1 + g m RS g m = 2 K n (VGS − VTN ) g m = 2 ( 4 )( −0.293 + 1) = 5.66 mA/V Av = − ( 5.66 ) (1.85 2 ) 1 + ( 5.66 )( 0.146 ) ⇒ Av = −2.98 c. AC load line Ϫ1 1.85͉͉2 ϩ 0.146 Ϫ1 ϭ 1.107 K Slope ϭ 2 mA 10 6 Δv0 = ( ΔiD )(1.85 || 2 ) = ( 2 )(1.85 || 2 ) = 1.92 V Δvi = 1.92 = 0.645 ⇒ Vi = 0.645 V 2.98 4.22 a. VDSQ = VDD − I DQ ( RD + RS ) 2.5 = 5 − I DQ ( 2 + RS ) 2.5 I DQ = 2 + RS I DQ = K n (VGS − VTN ) = 2 −VGS −2.5 RS ⇒ VGS = − I DQ RS = 2 + RS RS 2 ⎡ −2.5RS ⎤ 2.5 Kn ⎢ − VTN ⎥ = 2 + RS 2 + RS ⎣ ⎦ 2 ⎡ 2.5 RS ⎤ 2.5 4 ⎢1 − ⎥ = 2 + RS ⎦ 2 + RS ⎣ 2 ⎡ 2 + RS − 2.5 RS ⎤ 2.5 4⎢ ⎥ = 2 + RS 2 + RS ⎣ ⎦ 4 ( 2 − 1.5RS ) 2 + RS 2 = 2.5 2 4 ( 4 − 6 RS + 2.25 RS ) = 2.5 ( 2 + RS ) 2 9 RS − 26.5RS + 11 = 0 RS = 26.5 ± ( 26.5) − 4 ( 9 )(11) 2 (9) 2 RS = 0.5 kΩ or 2.44 kΩ But RS = 2.44 kΩ ⇒ VGS = −1.37 Cut off. ⇒ RS = 0.5 kΩ, I DQ = 1.0 mA 112. b. Av = − g m ( RD || RL ) 1 + g m RS ( 4 )(1) = 4 mA / V g m = 2 K n I DQ = 2 Av = − ( 4 )( 2 || 2 ) 1 + ( 4 )( 0.5 ) ⇒ Av = −1.33 4.23 a. 5 = I DQ RS + VSDQ + I DQ RD − 5 5 = I DQ RS + 6 + I DQ (10 ) − 5 1. I DQ = 4 RS + 10 VS = VSDQ + I DQ RD − 5 = VSGQ 2. 1 + I DQ (10 ) = VSGQ 3. I DQ = K p (VSGQ − 2 ) 2 Choose RS = 10 kΩ ⇒ I DQ = 4 = 0.20 mA 20 VSGQ = 1 + (0.2)(10) = 3 V 0.20 = K P (3 − 2) 2 ⇒ K P = 0.20 mA / V 2 b. I DQ = ( 0.20 )( 3 − 2 ) = 0.20 mA 2 ( 0.2 )( 0.2 ) = 0.4 mA / V Av = − g m ( RD || RL ) = − ( 0.4 )(10 ||10 ) ⇒ Av = −2.0 g m = 2 K P I DQ = 2 c. 4 = 0.133 mA 30 = 1 + (0.133)(10) = 2.33 V Choose RS = 20 kΩ ⇒ I DQ = VSGQ 0.133 = K p (2.33 − 2) 2 ⇒ K p = 1.22 mA / V 2 g m = 2 (1.22)(0.133) = 0.806 mA/V Av = −(0.806)(10 10) ⇒ Av = −4.03 A larger gain can be achieved. 4.24 (a) I DQ = K p (VSGQ + VTP ) 2 0.25 = 0.8 (VSGQ − 0.5 ) 2 VSGQ = 1.059 V 3 − 1.059 RS = ⇒ RS = 7.76 K 0.25 VD = VS − VSDQ = 1.059 − 1.5 = −0.441 V RD = (b) −0.441 − ( −3) 0.25 ⇒ RD = 10.2 K 113. Av = − g m ( RD RL ) ( 0.8)( 0.25 ) = 0.8944 mA/V Av = − ( 0.8944 )(10.2 || 2 ) g m = 2 K p I DQ = 2 Av = −1.50 (c) ΔVO = ΔI ( RD || RL ) = 0.25 (10.2 || 2 ) = 0.418 So ΔVO = 0.836 peak-to-peak 4.25 I DQ = K n (VGSQ − VTN ) 2 g m = 2 K n I DQ 2.2 = 2 K n ( 6 ) ⇒ K n = 0.202 mA / V 2 6 = 0.202 ( 2.8 − VTN ) ⇒ VTN = −2.65 V 2 VDSQ = 18 − I DQ ( RS + RD ) 18 − 10 = 1.33 kΩ ⇒ RS = 1.33 − RD RS + RD = 6 g m ( RD RL ) Av = − 1 + g m RS ⎛ R ⋅1 ⎞ −2.2 ⎜ D ⎟ ⎝ RD + 1 ⎠ −1 = 1 + ( 2.2 )(1.33 − RD ) 1 + 2.93 − 2.2 RD = 2.2 RD 1 + RD ( 3.93 − 2.2 RD )(1 + RD ) = 2.2 RD 2 3.93 + 1.73RD − 2.2 RD = 2.2 RD 2 2.2 RD + 0.47 RD − 3.93 = 0 RD = −0.47 + ( 0.47 ) + 4 ( 2.2 )( 3.93) ⇒ RD = 1.23 kΩ, 2 ( 2.2 ) = 2.8 + ( 6 )( 0.1) = 3.4 V 2 VG = VGS + VS 1 1 VG = ⋅ Rin ⋅ VDD = (100 )(18 ) = 3.4 ⇒ R1 = 529 kΩ R1 R1 529 R2 = 100 ⇒ R2 = 123 kΩ 529 + R2 4.26 a. VSD(sat) VSDQ V1 RS = 0.10 kΩ 114. V1 = 9 + VSG , VSD ( sat ) = VSG + VTP VSDQ = = V1 − VSD ( sat ) 2 + VSD ( sat ) ( 9 + VSG ) − (VSG + VTP ) 2 + (VSG + VTP ) 9 + 1.5 + VSG − 1.5 2 = 3.75 + VSG = 9 + VSG − I DQ RD = VSDQ I DQ = K p (VSG + VTP ) Set RD = 0.1RL = 2 kΩ 3.75 = 9 − I DQ ( 2 ) ⇒ I DQ = 2.625 mA 2 b. g m = 2 K p I DQ = 2 r0 = ( 2 )( 2.625) = 4.58 mA / V 1 1 = = 38.1 kΩ λ I DQ ( 0.01)( 2.625 ) Open circuit. Av = − g m ( RD || r0 ) Av = −4.58 ( 2 || 38.1) ⇒ Av = −8.70 c. With RL Av = −4.58 ( 2 || 20 || 38.1) ⇒ Av = −7.947 ⇒ Change = 8.66% 4.27 (a) I DQ = K p (VSGQ + VTP ) 2 0.5 = 0.25 (VSGQ + 0.8 ) 2 VSGQ = 0.614 V = VS 10 − 0.614 RS = ⇒ RS = 18.8 K 0.5 VD = VS − VSDQ = 0.614 − 3 = −2.386 V RD = −2.386 − ( −10 ) 0.5 ⇒ RD = 15.2 K (b) Av = − g m RD ( 0.25)( 0.5 ) = 0.7071 mA/V Av = − ( 0.7071)(15.2 ) g m = 2 K p I DQ = 2 Av = −10.7 4.28 Av = − g m ( RD RL ) VDSQ = VDD − I DQ ( RS + RD ) 10 = 20 − (1)( RS + RD ) ⇒ RS + RD = 10 kΩ Let RD = 8 kΩ, RS = 2 kΩ Av = −10 = − g m ( 8 20 ) g m = 1.75 mA / V = 2 K n I DQ = 2 K n (1) ⇒ K n = 0.766 mA / V 2 115. VS = I DQ RS = (1)( 2 ) = 2 V I DQ = K n (VGS − VTN ) ⇒ 1 = 0.766 (VGS − 2 ) ⇒ VGS = 3.14 V 2 2 VG = VGS + VS = 3.14 + 2 = 5.14 VG = 1 1 ⋅ Rin ⋅ VDD ⇒ ( 200 )( 20 ) = 5.14 ⇒ R1 = 778 kΩ R1 R1 778R2 = 200 ⇒ R2 = 269 kΩ 778 + R2 4.29 Av = ( 4 )( 50 ) g m r0 = ⇒ Av = 0.995 1 + g m r0 1 + ( 4 )( 50 ) Ro = 1 1 r0 = 50 ⇒ R0 = 0.249 kΩ 4 gm Av = = R0 = g m ( r0 RS ) 1 + g m ( r0 RS ) 4 ( 2.38 ) 1 + 4 ( 2.38 ) = 4 ( 50 2.5 ) 1 + 4 ( 50 2.5 ) ⇒ Av = 0.905 1 1 r0 RS = 50 2.5 ⇒ R0 = 0.226 kΩ 4 gm 4.30 Av = 0.98 = g m ( RL ro ) 1 + g m ( RL ro ) g m ro ⇒ g m ro = 49 1 + g m ro Also 0.49 = 0.49 = 0.49 = g m ( RL ro ) 1 + g m ( RL ro ) ⎛ Rr ⎞ gm ⎜ L o ⎟ ⎝ RL + ro ⎠ = ⎛ Rr ⎞ 1 + gm ⎜ L o ⎟ ⎝ RL + ro ⎠ g m ( RL ro ) RL + ro + g m ( RL ro ) ( 49 )(1) 49 = 1 + ro + ( 49 )(1) 50 + ro ro = 50 K g m = 0.98 mA/V 4.31 (a) Av = ( 2 )( 25) g m ro = 1 + g m ro 1 + ( 2 )( 25 ) Av = 0.98 Ro = 1 1 ro = 25 = 0.5 || 25 gm 2 Ro = 0.49 K (b) 116. Av = g m ( ro || RL ) 1 + g m ( ro || RL ) = 2 ( 25 || 2 ) 1 + 2 ( 25 || 2 ) = 2 (1.852 ) 1 + 2 (1.852 ) Av = 0.787 4.32 Av = g m ( RL || ro ) 1 + g m ( RL || ro ) = 5 (10 || 100 ) 1 + ( 5 )(10 || 100 ) = 5 ( 9.09 ) 1 + 5 ( 9.09 ) Av = 0.978 Ro = 1 1 RL ro = 10 || 100 = 0.2 || 9.0909 gm 5 Ro = 0.196 K 4.33 a. ⎛ R2 ⎞ 396 ⎛ ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ (10 ) ⇒ VG = 2.42 V ⎝ 396 + 1240 ⎠ ⎝ R1 + R2 ⎠ 10 − (VSG + 2.42 ) 2 = K p (VSG + VTP ) I DQ = RS 2 7.58 − VSG = ( 2 )( 4 ) (VSG − 4VSG + 4 ) 2 8VSG − 31VSG + 24.4 = 0 VSG = 31 ± ( 31) 2 − 4 ( 8 )( 24.4 ) 2 (8) ⇒ VSG = 2.78 V I DQ = ( 2 )( 2.78 − 2 ) ⇒ I DQ = 1.21 mA 2 VSDQ = 10 − I DQ RS = 10 − (1.21)( 4 ) ⇒ VSDQ = 5.16 V b. g m = 2 K p I DQ = 2 r0 = Av = = ( 2 )(1.21) = 3.11 mA / V 1 1 = = 41.3 kΩ λ I DQ ( 0.02 )(1.21) g m ( RS RL r0 ) 1 + g m ( RS RL r0 ) 3.11( 4 4 41.3) 1 + ( 3.11) ( 4 4 41.3) ⇒ Av = 0.856 Ϫ Ii Vsg Vi ϩ Ϫ R1͉͉R2 r0 gmVsg ϩ RS RL I0 117. ⎛ RS r0 I 0 = − ( g mVsg ) ⎜ ⎜R r +R L ⎝ S 0 −Vi Vsg = 1 + g m ( RS RL r0 ) ⎞ ⎟ ⎟ ⎠ Vi = I i ( R1 R2 ) Ai = g m ( R1 R2 ) RS r0 I0 = ⋅ I i 1 + g m ( RS RL r0 ) RS r0 + RL ( 3.11) ( 396 1240 ) 4 41.3 ⋅ 1 + ( 3.11) ( 4 4 41.3) 4 41.3 + 4 ( 3.11)( 300 ) 3.647 = ⋅ ⇒ Av = 64.2 1 + ( 3.11)(1.908 ) 3.647 + 4 = R0 = 1 1 4 41.3 ⇒ R0 = 0.295 kΩ RS r0 = 3.11 gm 4.34 g m = 2 K n I Q = 2 (1)(1) = 2 mA / V 1 1 = = 100 kΩ λ I Q ( 0.01)(1) r0 = g m ( r0 RL ) ⇒ Av = 0.885 1 + g m ( r0 RL ) R0 = 1 1 r0 = 100 ⇒ R0 = 0.498 kΩ 2 gm 4.35 a. Av = = 2 (100 4 ) Av = 1 + 2 (100 4 ) gm ( 4) g m RL ⇒ 0.95 = 1 + g m RL 1 + gm ( 4) 0.95 = 4 (1 − 0.95 ) g m ⇒ g m = 4.75 mA/V ⎛1 ⎞⎛W g m = 2 ⎜ μ n Cox ⎟ ⎜ ⎝2 ⎠⎝ L ⎞ ⎟ IQ ⎠ 4.75 = 2 ⎛ ( 0.030 ) ⎜ W⎞ W = 47.0 ⎟ ( 4) ⇒ L ⎝L⎠ b. ⎛1 ⎞⎛ W g m = 2 ⎜ μ n Cox ⎟⎜ 2 ⎝ ⎠⎝ L 4.75 = 2 ⎞ ⎟ IQ ⎠ ( 0.030 )( 60 ) I Q ⇒ I Q = 3.13 mA 4.36 2 I DQ = K n (VGS − VTN ) 5 = 5 (VGS + 2 ) ⇒ VGS = −1 V ⇒ VS = −VGS = 1 V 2 I DQ = VS − ( −5 ) RS ⇒ RS = g m = 2 K n I DQ = 2 r0 = 1 λ I DQ = 1+ 5 ⇒ RS = 1.2 kΩ 5 ( 5 )( 5) = 10 mA / V 1 = 20 kΩ 0.01)( 5 ) ( 118. Av = = R0 = g m ( r0 RS RL ) 1 + g m ( r0 RS RL ) (10 ) ( 20 1.2 2 ) ⇒ Av = 0.878 1 + (10 ) ( 20 1.2 2 ) 1 1 || r0 || RS = || 20 ||1.2 ⇒ Ro = 91.9 Ω 10 gm 4.37 Av = g m (10 ) g m RS ⇒ 0.90 = 1 + g m RS 1 + g m (10 ) 0.90 = 10 (1 − 0.90 ) g m ⇒ g m = 0.90 mA/V 1 1 RS = 10 ⇒ R0 = 1 kΩ gm 0.90 With RL : R0 = Av = g m ( RS || RL ) ( 0.90 )(10 || 2 ) ⇒ Av = 0.60 1 + g m ( RS || RL ) 1 + ( 0.90 )(10 || 2 ) = 4.38 R0 = 1 RS gm Output resistance determined primarily by gm 1 = 0.2 kΩ ⇒ g m = 5 mA/V Set gm g m = 2 K n I DQ ⇒ 5 = 2 I DQ = K n (VGS − VTN ) ( 4 ) I DQ ⇒ I DQ = 1.56 mA 2 1.56 = 4 (VGS + 2 ) VGS = −1.38 V, VS = −VGS = 1.38 V 2 1.38 − ( −5 ) ⇒ RS = 4.09 kΩ 1.56 5 ( 4.09 2 ) g m ( RS RL ) Av = = ⇒ Av = 0.870 1 + g m ( RS RL ) 1 + 5 ( 4.09 2 ) RS = 4.39 a. g m = 2 K p I DQ = 2 ( 5)( 5 ) = 10 mA / V 1 1 = ⇒ Ro = 100 Ω g m 10 b. Open-circuit gain g m r0 But r0 = ∞ so Av = 1.0 Av = 1 + g m r0 With RL: g m RL Av = 1 + g m RL Ro = 0.50 = 4.40 10 RL ⇒ 0.50 = 10 (1 − 0.5 ) RL ⇒ RL = 0.1 kΩ 1 + 10 RL 119. −1 ⋅ Δv0 RS RL ΔiD = I DQ = Δv0 = − I DQ ⋅ RS RL = − v0 ( min ) = − Av = vi = I DQ ⋅ RS RL RS + RL I DQ RS R 1+ S RL g m ( RS RL ) 1 + g m ( RS RL ) = v0 vi − I DQ ( RS RL ) ⎡1 + g m ( RS RL ) ⎤ ⎣ ⎦ g m ( RS RL ) vi ( min ) = − I DQ ⎡1 + g m ( RS RL ) ⎤ ⎦ gm ⎣ 4.41 (a) VDD = VDSQ + I DQ RS 3 = 1.5 + ( 0.25 ) RS ⇒ RS = 6 K VS = 1.5 V I DQ = K n (VGSQ − VTN ) 0.25 = 0.5 (VGSQ − 0.4 ) 2 2 VGSQ = 1.107 V VG = VGSQ + VS + 1.107 + 1.5 = 2.607 V ⎛ R2 ⎞ 1 VG = ⎜ ⎟ VDD = − RL − VDD R1 + R2 ⎠ R1 ⎝ 1 2.607 = ( 300 )( 3) ⇒ R1 = 345.2 K ⇒ R2 = 2291 K R1 (b) g m RS Av = g m = 2 K n I DQ = 2 ( 0.5 )( 0.25 ) = 0.7071 mA/V 1 + g m RS Av = ( 0.7071)( 6 ) ⇒ Av = 0.809 1 + ( 0.7071)( 6 ) Ro = 1 1 RS = 6 = 1.414 || 6 gm ( 0.7071) Ro = 1.14 K 4.42 Av = + g m RD = ( 5 )( 4 ) Av = 20 Ri = 1 1 10 = 10 = 0.2 ||10 5 gm Ri = 0.196 K 4.43 a. 120. VGS + I DQ RS = 5 5 − VGS 2 = K n (VGS − VTN ) I DQ = RS 2 5 − VGS = (10 )( 3) (VGS − 2VGS + 1) 2 30VGS − 59VGS + 25 = 0 VGS = 59 ± ( 59 ) 2 − 4 ( 30 )( 25 ) 2 ( 30 ) ⇒ VGS = 1.35 V I DQ = ( 3)(1.35 − 1) ⇒ I DQ = 0.365 mA 2 VDSQ = 10 − ( 0.365 )( 5 + 10 ) ⇒ VDSQ = 4.53 V b. g m = 2 K n I DQ = 2 r0 = c. 1 ( 3)( 0.365 ) ⇒ g m = 2.093 mA / V 1 = ⇒r =∞ ( 0 )( 0.365 ) 0 Av = g m ( RD RL ) = ( 2.093) ( 5 4 ) ⇒ Av = 4.65 λ I DQ 4.44 a. I DQ = K p (VSG + VTP ) 2 0.75 = ( 0.5 )(VSG − 1) ⇒ VSG = 2.225 V 5 − 2.225 5 = I DQ RS + VSG ⇒ RS = ⇒ RS = 3.70 kΩ 0.75 VSDQ = 10 − I DQ ( RS + RD ) 6 = 10 − ( 0.75 )( 3.70 + RD ) ⇒ RD = 1.63 kΩ 2 b. Ri = 1 gm g m = 2 K p I DQ = 2 ( 0.5 )( 0.75) = 1.225 mA / V 1 ⇒ Ri = 0.816 kΩ 1.225 Ro = RD ⇒ Ro = 1.63 kΩ Ri = c. ⎞ ⎞⎛ RS ⋅i ⎟⎜ ⎜ R + [1/ g ] ⎟ i ⎟ ⎠⎝ S m ⎠ 3.70 ⎛ 1.63 ⎞ ⎛ ⎞ i0 = ⎜ ⎟⎜ ⎟ ii ⎝ 1.63 + 2 ⎠ ⎝ 3.70 + 0.816 ⎠ i0 = 0.368ii = i0 = 1.84sin ω t ( μ A ) ⎛ RD i0 = ⎜ ⎝ RD + RL v0 = i0 RL = (1.84 )( 2 ) sin ω t ⇒ v0 = 3.68sin ω t ( mV ) 4.45 a. 2 I DQ = K n (VGS − VTN ) 5 = 4 (VGS − 2 ) ⇒ VGS = 3.12 V V + = I DQ RD + VDSQ − VGS 2 10 = ( 5 ) RD + 12 − 3.12 ⇒ RD = 0.224 kΩ b. 121. g m = 2 K n I DQ = 2 Ri = ( 4 )( 5 ) ⇒ g m = 8.944 mA / V 1 1 = ⇒ Ri = 0.112 kΩ g m 8.94 Av = g m ( RD RL ) = ( 8.944) ( 0.224 2 ) ⇒ Av = 1.80 c. 4.46 a. 2 I DQ = K p (VSG + VTP ) 3 = 2 (VSG − 2 ) ⇒ VSG = 3.225 V V + = I DQ RS + VSG 10 − 3.225 RS = ⇒ RS = 2.26 kΩ 3 VSDQ = 20 − I DQ ( RS + RD ) 10 = 20 − ( 3)( 2.26 + RD ) ⇒ RD = 1.07 kΩ 2 b. Av = g m ( RD RL ) ( 2 )( 3) = 4.90 mA / V Av = ( 4.90 )(1.07 ||10 ) ⇒ Av = 4.74 g m = 2 K p I DQ = 2 4.47 a. Av = (W / L )D (W / L )L = 5 ⇒ (W / L ) D = 25 ( 0.5 ) ⇒ (W / L ) D = 12.5 1 ⎛W ⎞ μ n Cox ⎜ ⎟ = ( 30 )(12.5 ) = 375 μ A/V 2 2 ⎝ L ⎠D K L = ( 30 )( 0.5 ) = 15 μ A / V 2 KD = Transition point: vGSD − VTND = (VDD − VTNL ) − KD ( vGSD − VTND ) KL 375 ( vGSD − 2 ) 15 vGSD (1 + 5 ) = (10 − 2 ) + 2 + 5 ( 2 ) vGSD = 3.33 V and vDSD = 1.33 V 3.33 − 2 + 2 ⇒ VGSQ = 2.67 V VGSQ = 2 vGSD − 2 = (10 − 2 ) − ␯O 8 Q-point VDSQ 1.33 3.33 2 VGSQ b. ␯GSD 122. I DQ = K D (VGSQ − VTND ) 2 I DQ = 0.375 ( 2.67 − 2 ) ⇒ I DQ = 0.166 mA 2 VDSQ = 4.48 (a) 8 − 1.33 + 1.33 ⇒ VDSQ = 4.67 V 2 Transition point: Load: VOtB = VDD − VTNL = 10 − 2 = 8 V Driver: 2 2 K D ⎡( vOt A ) (1 + λD vOt A ) ⎤ = K L ⎡( −VTNL ) 1 + λL (VDD − vOt A ) ⎤ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ 2 3 0.5 ⎡ v0t A + ( 0.02 ) v0t A ⎤ = 0.1 ⎡( 4 )(1 + 0.02 ) (10 − v0t A ) ⎤ ⎣ ⎦ ⎣ ⎦ 3 2 We have 0.01v0t A + 0.5v0t A + 0.008v0t A − 0.48 = 0 ( ) Therefore v0t A = 0.963 V 8 − 0.963 + 0.963 = 4.48 V = VDSQ 2 2 2 Then K D ⎡(VGSD − VTND ) (1 + λD vOQ ) ⎤ = K L ⎡( −VTNL ) 1 + λL (VDD − vOQ ) ⎤ ⎣ ⎦ ⎣ ⎦ ⎡(VGSD − 1.2 )2 (1 + ( 0.02 )( 4.48 ) ) ⎤ = 0.1 ⎡( 4 ) (1 + 0.02 (10 − 4.48 ) ) ⎤ which yields VGSD = 2.103 V 0.5 ⎣ ⎦ ⎣ ⎦ b. 2 I DQ = K D ⎡(VGSD − VTND ) (1 + λD vOQ ) ⎤ ⎣ ⎦ 2 I DQ = 0.5 ⎡( 2.103 − 1.2 ) (1 + ( 0.02 )( 4.48 ) ) ⎤ ⎣ ⎦ So I DQ = 0.444 mA Now v0 Q = ( ) c. r0 D = r0 L = 1 1 = = 112.6 kΩ λ I DQ ( 0.02 )( 0.444 ) g mD = 2 K D (VGSD − VTND ) = 2 ( 0.5 )( 2.103 − 1.2 ) ⇒ g mD = 0.903 mA / V Then Av = − g mD ( r0 D r0 L ) = − ( 0.903) (112.6 112.6 ) or Av = −50.8 4.49 2 I D = K n (VGS − VTN ) , VDS = VGS I D = 0 when VDS = 1.5 V ⇒ VTN = 1.5 V 0.8 = K n ( 3 − 1.5 ) ⇒ K n = 0.356 mA / V 2 dI I go = D = = 2 K n (VDS − VTN ) = 2 ( 0.356 )( 3 − 1.5 ) ⇒ Ro = 0.936 k Ω dVDS Ro 2 4.50 a. I DQ = K nD (VGS − VTND ) = ( 0.5 ) ( 0 − ( −1) ) 2 2 I DQ = 0.5 mA I DQ = K nL (VGSL − VTNL ) = K nL (VDD − VO − VTNL ) 2 0.5 = 0.030 (10 − V0 − 1) 2 0.5 = 9 − V0 ⇒ V0 = 4.92 V 0.030 b. 2 123. I DD = I DL K nD (Vi − VTND ) = K nL (VDD − Vo − VTNL ) 2 2 K nD (Vi − VTND ) = VDD − Vo − VTNL K nL K nD (Vi − VTND ) K nL Vo = VDD − VTNL − Av = (W / L )D (W / L )L dVo K nD =− =− dVi K nL Av = − 500 ⇒ Av = −4.08 30 4.51 (a) I DQ = K L (VGSL − VTNL ) = K L (VDSL − VTNL ) 2 2 I D = ( 0.1)( 4 − 1) = 0.9 mA 2 I DQ = K D (VGSD − VTND ) 2 0.9 = (1)(VGSD − 1) ⇒ VGSD = 1.95 V VGG = VGSD + VDSL = 1.95 + 4 ⇒ VGG = 5.95 V 2 b. I DD = I DL K D (VGSD − VTND ) = K L (VGSL − VTNL ) 2 2 KD (VGG + Vi − Vo − VTND ) = Vo − VTNL KL ⎛ KD ⎞ Vo ⎜ 1 + ⎟= ⎜ KL ⎟ ⎝ ⎠ Av = (c) RLD KD (VGG + Vi − VTND ) + VTNL KL KD / KL dVo = ⇒ dVi 1 + K D / K L Av = 1 1+ KL / KD From Problem 4.49. 1 = 2 K L (VDSL − VTNL ) = 1 = 1.67 k Ω 2 ( 0.1)( 4 − 1) g m = 2 K D I DQ = 2 (1)( 0.9 ) = 1.90 mA / V Av = g m ( RLD || RL ) 1 + g m ( RLD || RL ) 4.52 a. = (1.90 )(1.67 || 4 ) 1 + (1.90 )(1.67 || 4 ) From Problem 4.51. g m ( RLD || RL ) (1.90 )(1.67 ||10 ) Av = = 1 + g m ( RLD || RL ) 1 + (1.90 )(1.67 || 10 ) Av = 0.731 b. ⇒ Av = 0.691 124. R0 = 1 1 RLD = 1.67 = 0.526 ||1.67 gm 1.90 R0 = 0.40 kΩ 4.53 ⎛ 85 ⎞ K n1 = ⎜ ⎟ ( 50 ) ⇒ 2.125 mA/V 2 ⎝ 2⎠ g m1 = 2 K n1 I D1 = 2 ro1 = ro 2 = 1 λ1 I D1 = ( 2.125 )( 0.1) = 0.9220 1 = 200 K 0.05 )( 0.1) ( 1 1 = = 133.3 K λ2 I D 2 ( 0.075 )( 0.1) Av = − g m1 ( ro1 || ro 2 ) = − ( 0.922 )( 200 ||133.3) Av = −73.7 4.54 K p1 = k ′ ⎛ w ⎞ ⎛ 40 ⎞ p 2 ⎜ ⎟ = ⎜ ⎟ ( 50 ) ⇒ 1.0 mA/V 2 ⎝L⎠ ⎝ 2 ⎠ g m1 = 2 K p1 I D1 = 2 (1)( 0.1) = 0.6325 mA/V ro1 = ro 2 = 1 λ1 I D1 = 1 = 133.3 K 0.075 )( 0.1) ( 1 1 = = 200 K λ2 I D 2 ( 0.05 )( 0.1) Av = − g m1 ( ro1 ro 2 ) = − ( 0.6325 )(133.3 || 200 ) Av = −50.6 4.55 (a) ⎛ 85 ⎞ K n1 = ⎜ ⎟ ( 50 ) ⇒ 2.125 mA/V 2 ⎝ 2⎠ g m1 = 2 K n1 I D1 = 2 ro1 = ro 2 = 1 λ1 I D1 = ( 2.125 )( 0.1) = 0.922 mA/V 1 ( 0.05 )( 0.1) = 200 K 1 1 = = 133.3 K λ2 I D 2 ( 0.075 )( 0.1) (b) Ri1 = 1 1 = = 1.085 K g m1 0.922 ⎛ ⎞ Ri1 1.085 ⎛ ⎞ Vgs1 = − ⎜ ⎟ Vi = − ⎜ ⎟ Vi = −0.956Vi Ri1 + 0.050 ⎠ ⎝ 1.085 + 0.050 ⎠ ⎝ Vgs1 = + ( 0.956 )( 0.922 )( 200 )(133.3) Av = − g m1 ( ro1 ro 2 ) ⋅ Vi Av = 70.5 1 1 = 0.05 + ⇒ Ri = 1.135 K 0.922 g m1 (c) Ri = 0.05 + (d) Ro ≈ ro1 ro 2 = 200 133.7 ⇒ Ro ≈ 80 K 125. 4.56 (a) g m1 = 2 K n I D1 = 2 ( 2 )( 0.1) = 0.8944 mA/V gm2 = 2 K p I D 2 = 2 ( 2 )( 0.1) = 0.8944 mA/V ro1 = ro 2 = 1 = 1 = 100 K λ I D ( 0.1)( 0.1) (b) The small-signal equivalent circuit gm2Vsg2 Vo ϩ ϩ rO2 gm1Vi Vi Vsg2 rO1 Ϫ Ϫ (1) g m1Vi + Vsg 2 ro1 + g m 2Vsg 2 + Vsg 2 − Vo ro 2 =0 (2) Vo Vo − Vsg 2 + = g m 2Vsg 2 ro ro 2 ⎛1 1 ⎞ ⎛ 1 ⎞ + gm2 ⎟ Vo ⎜ + ⎟ = Vsg 2 ⎜ ⎝ ro ro 2 ⎠ ⎝ ro 2 ⎠ 1 1 ⎞ 1 ⎛ ⎛ ⎞ Vo ⎜ + + 0.8944 ⎟ ⇒ Vsg 2 = Vo ( 0.03317 ) ⎟ = Vsg 2 ⎜ ⎝ 50 100 ⎠ ⎝ 100 ⎠ (1) ⎛ 1 1 ⎞ V g m1Vi + Vsg 2 ⎜ + g m 2 + ⎟ = o ro1 ro 2 ⎠ ro 2 ⎝ 1 ⎞ Vo ⎛ 1 0.8944 Vi + Vo ( 0.03317 ) ⎜ + 0.8944 + ⎟= 100 ⎠ 100 ⎝ 100 0.8944 Vi = Vo ( 0.01 − 0.03033) Vo = −44 Vi (c) For output resistance, set Vi = 0. gm2Vsg2 RO Ix rO2 ϩ Vsg2 Ϫ rO1 rO ϩ Ϫ Vx 126. (1) (2) g m 2Vsg 2 + I x = Vsg 2 ro1 Vx Vx − Vsg 2 + ro ro 2 + g m 2Vsg 2 + Vsg 2 − Vx ro 2 =0 (2) ⎛ 1 1 ⎞ V Vsg 2 ⎜ + g m 2 + ⎟ = x ro 2 ⎠ ro 2 ⎝ ro1 1 ⎞ Vx ⎛ 1 + 0.8944 + Vsg 2 ⎜ ⎟= 100 100 ⎠ 100 ⎝ Vsg 2 = Vx ( 0.010936 ) (1) ⎛1 1 ⎞ ⎛ 1 ⎞ I x = Vx ⎜ + ⎟ − Vsg 2 ⎜ + gm2 ⎟ ⎝ ro ro 2 ⎠ ⎝ ro 2 ⎠ 1 1 ⎞ ⎛ ⎛ 1 ⎞ + 0.8944 ⎟ I x = Vx ⎜ + ⎟ − Vx ( 0.010936 ) ⎜ ⎝ 50 100 ⎠ ⎝ 100 ⎠ I x = Vx ( 0.03 − 0.0098905 ) V Ro = x = 49.7 K Ix 4.57 a. 2 I D1 = K1 (VGS 1 − VTN 1 ) 0.4 = 0.1(VGS 1 − 2 ) ⇒ VGS1 = 4 V VS 1 = I D1 RS 1 = ( 0.4 )( 4 ) = 1.6 V VG1 = VS 1 + VGS 1 = 1.6 + 4 = 5.6 V 2 ⎛ R2 ⎞ 1 VG1 = ⎜ ⎟ VDD = ⋅ Rin ⋅ VDD R1 ⎝ R1 + R2 ⎠ 1 5.6 = ( 200 )(10 ) ⇒ R1 = 357 kΩ R1 357 R2 = 200 ⇒ R2 = 455 kΩ 357 + R2 VDS1 = VDD − I D1 ( RS1 + RD1 ) 4 = 10 − ( 0.4 )( 4 + RD1 ) ⇒ RD1 = 11 kΩ VD1 = 10 − ( 0.4 )(11) = 5.6 V I D 2 = K 2 (VSG 2 + VTP 2 ) 2 2 = 1(VSG 2 − 2 ) ⇒ VSG 2 = 3.41 V VS 2 = VD1 + VSG 2 = 5.6 + 3.41 = 9.01 10 − 9.01 RS 2 = ⇒ RS 2 = 0.495 kΩ 2 VSD 2 = VDD − I D 2 ( RS 2 + RD 2 ) 5 = 10 − ( 2 )( 0.495 + RD 2 ) ⇒ RD 2 = 2.01 kΩ 2 b. 127. Vi Vgs1 RD1 R1͉͉R2 ϩ Ϫ V0 Ϫ ϩ gm1Vgs1 Vgs2 RD2 gm2Vsg2 ϩ Ϫ V0 = g m 2Vsg 2 RD 2 = ( g m 2 RD 2 ) ( g m1Vgs1 RD1 ) Vgs1 = Vi V Av = 0 = g m1 g m 2 RD1 RD 2 Vi g m1 = 2 K1 I D1 = 2 ( 0.1)( 0.4 ) = 0.4 mA / V g m 2 = 2 K 2 I D 2 = 2 (1)( 2 ) = 2.828 mA / V Av = ( 0.4 )( 2.828 )(11)( 2.01) ⇒ Av = 25.0 c. 4 2 VSD(sat) 5 10 VSD ( sat ) = VSG + VTh = VDD − I D 2 ( RD 2 + RS 2 ) 2 = VDD − K p 2 ( RD 2 + RS 2 )VSD ( sat ) So 2 (1)( 2.01 + 0.495 )VSD ( sat ) + VSD ( sat ) − VDD = 0 2 2.50VSD ( sat ) + VSD ( sat ) − 10 = 0 VSD ( sat ) = −1 ± 1 + 4 ( 2.501)(10 ) 2 ( 2.501) VSD ( sat ) = 1.81 V 5 − 1.81 = 3.19 ⇒ Max. output swing = 6.38 V peak-to-peak 4.58 a. 10 ϭ 4 mA 2.5 VSD2(sat) 10 128. 2 VSD 2 ( sat ) = VDD − I D 2 ( RD 2 + RS 2 ) = VDD − K p 2 ( RD 2 + RS 2 ) VSD 2 ( sat ) 2 (1)( 2 + 0.5)VSD 2 ( sat ) + VSD 2 ( sat ) − 10 = 0 2 2.5VSD 2 ( sat ) + VSD 2 ( sat ) − 10 = 0 VSD 2 ( sat ) = −1 ± 1 + 4 ( 2.5 )(10 ) 2 ( 2.5 ) = 1.81 V 10 − 1.81 + 1.81 ⇒ VSDQ 2 = 5.91 V 2 VSDQ 2 = VDD − I DQ 2 ( RD 2 + RS 2 ) 5.91 = 10 − I DQ 2 ( 2 + 0.5 ) ⇒ I DQ 2 = 1.64 mA VS 2 = 10 − (1.64 )( 0.5 ) = 9.18 V VSDQ 2 = I DQ 2 = K p 2 (VSG 2 + VTP 2 ) 2 1.64 = (1)(VSG 2 − 2 ) ⇒ VSG 2 = 3.28 V VD1 = VS 2 − VSG 2 = 9.18 − 3.28 = 5.90 V 10 − 5.90 RD1 = ⇒ RD1 = 10.25 kΩ 0.4 2 I DQ1 = K n1 (VGS1 − VTN 1 ) 2 0.4 = ( 0.1)(VGS 1 − 2 ) ⇒ VGS1 = 4 V VS 1 = ( 0.4 )(1) = 0.4 V VG1 = VS 1 + VGS 1 = 0.4 + 4 = 4.4 V 2 ⎛ R2 ⎞ 1 VG1 = ⎜ ⎟ VDD = ⋅ Rm ⋅ VDD R1 + R2 ⎠ R1 ⎝ 1 4.4 = ⋅ ( 200 )(10 ) ⇒ R1 = 455 kΩ R1 455 R2 = 200 ⇒ R2 = 357 kΩ 455 + R2 b. I DQ 2 = 1.64 mA VSDQ 2 = 10 − (1.64 )( 2 + 0.5 ) ⇒ VSDQ 2 = 5.90 V VDSQ1 = 10 − ( 0.4 )(10.25 + 1) ⇒ VDSQ1 = 5.50 V (c) g m1 = 2 K n1 I DQ1 = 2 ( 0.1)( 0.4 ) = 0.4 mA / V g m 2 = 2 K p 2 I DQ 2 = 2 (1)(1.64 ) = 2.56 mA / V Av = g m1 g m 2 RD1 RD 2 = ( 0.4 )( 2.56 )(10.25 )( 2 ) ⇒ Av = 21.0 4.59 a. VDSQ 2 = 7 = VDD − I DQ 2 RS 2 = 10 − I DQ 2 ( 6 ) I DQ 2 = 0.5 mA I DQ 2 = K 2 (VGS 2 − VTN 2 ) 2 0.5 = ( 0.2 )(VGS 2 − 0.8 ) ⇒ VGS 2 = 2.38 V VS 1 = VS 2 + VGS 2 = 3 + 2.38 = 5.38 2 129. I DQ1 = VS 1 5.38 = = 0.269 mA 20 RS1 I DQ1 = K1 (VGS 1 − VTN 1 ) 2 0.269 = ( 0.2 )(VGS 1 − 0.8 ) ⇒ VGS 1 = 1.96 V VG1 = VS 1 + VGS 1 = 5.38 + 1.96 = 7.34 V 2 ⎛ R2 ⎞ 1 VG1 = ⎜ ⎟ VDD = ⋅ Rin ⋅ VDD R1 + R2 ⎠ R1 ⎝ 1 7.34 = ( 400 )(10 ) ⇒ R1 = 545 kΩ R1 545 R2 = 400 ⇒ R2 = 1.50 MΩ 545 + R2 b. I DQ1 = 0.269 mA, I DQ 2 = 0.5 mA VDSQ1 = 10 − ( 0.269 )( 20 ) ⇒ VDSQ1 = 4.62 V c. Av = g m1 RS 1 g R ⋅ m2 S 2 1 + g m1 RS1 1 + g m 2 RS 2 g m1 = 2 K1 I DQ1 = 2 ( 0.2 )( 0.269 ) = 0.464 mA / V ( 0.2 )( 0.5) = 0.6325 mA / V ( 0.464 )( 20 ) ( 0.6325)( 6 ) ⋅ Av = 1 + ( 0.464 )( 20 ) 1 + ( 0.6325 )( 6 ) = ( 0.9027 )( 0.7915 ) g m 2 = 2 K 2 I DQ 2 = 2 = Av = 0.714 gm1Vgs1 ϩ Vgs1 ϩ Vgs2 Ϫ Ϫ Ix RS2 RS1 R0 = gm2Vgs2 ϩ Ϫ Vx 1 1 RS 2 = 6 = 1.581 6 ⇒ Ro = 1.25 kΩ gm2 0.6325 4.60 (a) I DQ1 = 10 − VGS 1 2 = K n1 (VGS 1 − VTN 1 ) RS 2 2 10 − VGS 1 = ( 4 )(10 ) (VGS 1 − 4VGS 1 + 4 ) 2 40VGS 1 − 159VGS 1 + 150 = 0 VGS1 = 159 ± (159 ) − 4 ( 40 )(150 ) ⇒ VGS 1 = 2.435 V 2 ( 40 ) 2 130. I DQ1 = ( 4 )( 2.435 − 2 ) ⇒ I DQ1 = 0.757 mA 2 VDSQ1 = 20 − ( 0.757 )(10 ) ⇒ VDSQ1 = 12.4 V Also I DQ 2 = 0.757 mA VDSQ 2 = 20 − ( 0.757 )(10 + 5 ) ⇒ VDSQ 2 = 8.65 V g m1 = g m 2 = 2 KI DQ = 2 (b) ( 4 )( 0.757 ) ⇒ g m1 = g m 2 = 3.48 mA / V c. ϩ Vgs1 Vi ϩ Ϫ gm1Vgs1 gm2Vgs2 Ϫ RG V0 Ϫ RS1 Vgs2 RS2 RD ϩ V0 = − ( g m 2Vgs 2 ) ( RD RL ) Vgs 2 = ( − g m1Vgs1 − g m 2Vgs 2 ) ( RS1 RS 2 ) Vi = Vgs1 − Vgs 2 ⇒ Vgs1 = Vi + Vgs 2 Vgs 2 + g m 2Vgs 2 ( RS 1 RS 2 ) = − g m1 (Vi + Vgs 2 ) ( RS 1 RS 2 ) Vgs 2 + g m 2Vgs 2 ( RS 1 RS 2 ) + g m1Vgs 2 ( RS 1 Rs 2 ) = − g m1Vi ( RS1 RS 2 ) Vgs 2 = Av = − g m1Vi ( RS 1 RS 2 ) 1 + g m 2 ( RS 1 RS 2 ) + g m1 ( RS 1 RS 2 ) V0 g m1 g m 2 ( RS 1 RS 2 )( RD RL ) = Vi 1 + ( g m1 + g m 2 ) ( RS1 RS 2 ) ( 3.48 ) (10 10 )( 5 2 ) Av = ⇒ Av = 2.42 1 + ( 3.48 + 3.48 ) (10 10 ) 2 4.61 a. I DQ = 3 mA VS 1 = I DQ RS − 5 = ( 3)(1.2 ) − 5 = −1.4 V I DQ = K1 (VGS − VTN ) 2 3 = 2 (VGS − 1) ⇒ VGS = 2.225 V 2 VG1 = VGS + VS 1 = 2.225 − 1.4 = 0.825 V ⎛ ⎞ R3 ⎛ R3 ⎞ VG1 = ⎜ ⎟ ( 5 ) ⇒ 0.825 = ⎜ ⎟ ( 5 ) ⇒ R3 = 82.5 kΩ ⎝ 500 ⎠ ⎝ R1 + R2 + R3 ⎠ VD1 = VS1 + VDSQ1 = −1.4 + 2.5 = 1.1 V VG 2 = VD1 + VGS = 1.1 + 2.225 = 3.325 V ⎛ R2 + R3 ⎞ ⎛ R2 + R3 ⎞ VG 2 = ⎜ ⎟ ( 5 ) ⇒ 3.325 = ⎜ ⎟ ( 5) ⎝ 500 ⎠ ⎝ R1 + R2 + R3 ⎠ R2 + R3 = 332.5 ⇒ R2 = 250 kΩ RL 131. R1 = 500 − 250 − 82.5 ⇒ R1 = 167.5 kΩ VD 2 = VD1 + VDSQ 2 = 1.1 + 2.5 = 3.6 V RD = 5 − 3.6 ⇒ RD = 0.467 kΩ 3 b. Av = − g m1 RD ( 2 )( 3) = 4.90 mA / V g m1 = 2 K n I DQ = 2 Av = − ( 4.90 )( 0.467 ) ⇒ Av = −2.29 4.62 a. VS 1 = I DQ RS − 10 = ( 5 )( 2 ) − 10 ⇒ VS 1 = 0 I DQ = K1 (VGS 1 − VTN ) 2 5 = 4 (VGS1 − 1.5 ) ⇒ VGS 1 = 2.618 V VG1 = VGS 1 + VS1 = 2.618 V = IR3 = ( 0.1) R3 ⇒ R3 = 26.2 kΩ 2 VD1 = VS1 + VDSQ1 = 0 + 3.5 = 3.5 V VG 2 = VD1 + VGS = 3.5 + 2.62 = 6.12 V = ( 0.1)( R2 + R3 ) R2 + R3 = 61.2 kΩ ⇒ R2 = 35 kΩ VD 2 = VD1 + VDSQ 2 = 3.5 + 3.5 = 7.0 V 10 − 7 RD = ⇒ RD = 0.6 kΩ 5 10 − 6.12 R1 = ⇒ R1 = 38.8 kΩ 0.1 b. Av = − g m1 RD ( 4 )( 5 ) = 8.944 mA / V g m1 = 2 K n I DQ = 2 Av = − ( 8.944 )( 0.6 ) ⇒ Av = −5.37 4.63 a. ⎛ V ⎞ I DQ = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ ⎛ V ⎞ 4 = 6 ⎜ 1 − GS ⎟ ⎜ ( −3) ⎟ ⎝ ⎠ 2 2 ⎡ 4⎤ VGS = ( −3) ⎢1 − ⎥ ⇒ VGS = −0.551 V 6⎦ ⎣ VDSQ = VDD − I DQ RD 6 = 10 − ( 4 ) RD ⇒ RD = 1 kΩ b. 2 I DSS ⎛ VGS ⎞ 2 ( 6 ) ⎛ −0.551 ⎞ = 1− 1− ⎟ ⇒ g m = 3.265 mA/V −3 ⎠ ( −VP ) ⎜ VP ⎟ 3 ⎜ ⎝ ⎝ ⎠ 1 1 = ⇒ r0 = 25 kΩ r0 = λ I DQ ( 0.01)( 4 ) gm = c. Av = − g m ( r0 RD ) = − ( 3.265 )( 25 1) ⇒ Av = −3.14 132. 4.64 VGS + I DQ ( RS1 + RS 2 ) = 0 ⎛ V ⎞ I DQ = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ 2 2 VGS ⎛ V ⎞ + I DSS ( RS1 + RS 2 ) ⎜ 1 − GS ⎟ = 0 VP ⎠ ⎝ 2 ⎛ V ⎞ VGS + ( 2 )( 0.1 + 0.25 ) ⎜1 − GS ⎟ = 0 VP ⎠ ⎝ ⎛ 2 V2 ⎞ VGS + 0.7 ⎜ 1 − VGS + GS 2 ⎟ = 0 ⎜ ⎟ ⎝ ( −2 ) ( −2 ) ⎠ 2 0.175VGS + 1.7VGS + 0.7 = 0 VGS = −1.7 ± (1.7 ) − 4 ( 0.175)( 0.7 ) ⇒ VGS 2 ( 0.175 ) VGS ⎞ 2 ( 2 ) ⎛ −0.431 ⎞ 2 = −0.4314 V 2 I DSS ⎛ ⎜1 − ⎟= ⎜1 − ⎟ ⇒ g m = 1.569 mA/V −VP ⎝ −2 ⎠ VP ⎠ 2 ⎝ − g m ( RD RL ) − (1.569 ) ( 8 4 ) = ⇒ Av = −3.62 Av = 1 + g m RS1 1 + (1.569 )( 0.1) gm = Ai = i0 ( v0 / RL ) v0 RG ⎛ 50 ⎞ = = ⋅ = ( −3.62 ) ⎜ ⎟ ⇒ Ai = −45.2 ii ( vi / RG ) vi RL ⎝ 4⎠ 4.65 I DSS = 4 mA 2 V = DD = 10 V 2 = VDD − I DQ ( RS + RD ) I DQ = VDSQ VDSQ 10 = 20 − ( 4 )( RS + RD ) ⇒ RS + RD = 2.5 kΩ VS = 2 V = I DQ RS = 4 RS ⇒ RS = 0.5 kΩ, RD = 2.0 kΩ I DQ ⎛ V ⎞ = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ 2 2 ⎛ ⎞ ⎛ V 4⎞ 4 = 8 ⎜1 − GS ⎟ ⇒ VGS = ( −4.2 ) ⎜1 − ⎟ ⇒ VGS = −1.23 V ⎜ ( −4.2 ) ⎟ ⎜ ⎟ 8⎠ ⎝ ⎝ ⎠ VG = VS + VGS = 2 − 1.23 ⎛ R2 ⎞ ⎛ R2 ⎞ VG = 0.77 V = ⎜ ⎟ ( 20 ) = ⎜ ⎟ ( 20 ) ⇒ R2 = 3.85 kΩ, R1 = 96.2 KΩ ⎝ 100 ⎠ ⎝ R1 + R2 ⎠ 4.66 a. 133. I DSS = 5 mA 2 V 12 VDSQ = DD = =6V 2 2 12 − 6 RS = ⇒ RS = 1.2 kΩ 5 I DQ = I DQ ⎛ V ⎞ = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ 2 2 ⎛ ⎛ V ⎞ 5 ⎞ 5 = 10 ⎜ 1 − GS ⎟ ⇒ VGS = ( −5 ) ⎜ 1 − ⎟ ⇒ VGS = −1.464 V ⎜ ( −5 ) ⎟ ⎜ 10 ⎟ ⎝ ⎠ ⎝ ⎠ VG = VS + VGS = 6 − 1.464 = 4.536 V ⎛ R2 ⎞ 1 VG = ⎜ ⎟ VDD = ⋅ Rin ⋅ VDD R1 ⎝ R1 + R2 ⎠ 1 4.536 = (100 )(12 ) ⇒ R1 = 265 kΩ R1 265 R2 = 100 ⇒ R2 = 161 kΩ 265 + R2 b. ⎛ V ⎞ 2 (10 ) ⎛ −1.46 ⎞ 2I g m = DSS ⎜ 1 − GS ⎟ = ⎜1 − ⎟ ⇒ g m = 2.83 mA/V −VP ) ⎝ −5 ⎠ 5 ⎝ VP ⎠ ( r0 = 1 1 = = 20 kΩ λ I DQ ( 0.01)( 5 ) Av = ( g m r0 Rs RL ( ) 1 + g m r0 RS RL ) Av = ( 2.83) ( 20 1.2 0.5 ) ⇒ Av = 0.495 1 + ( 2.83) ( 20 1.2 0.5 ) R0 = 1 1 1.2 = 0.353 1.2 ⇒ R0 = 0.273 kΩ RS = 2.83 gm 4.67 a. ⎛ R2 ⎞ ⎛ 110 ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ (10 ) = 5.5 V ⎝ 110 + 90 ⎠ ⎝ R1 + R2 ⎠ I DQ = 10 − (VG − VGS ) RS ⎛ V ⎞ = I DSS ⎜1 − GS ⎟ VP ⎠ ⎝ ⎛ V ⎞ 10 − 5.5 + VGS = ( 2 )( 5 ) ⎜1 − GS ⎟ ⎝ 1.75 ⎠ 2 2 2 4.5 + VGS = 10 (1 − 1.143VGS + 0.3265VGS ) 2 3.265VGs − 12.43VGS + 5.5 = 0 VGS = 12.43 ± (12.43) − 4 ( 3.265 )( 5.5) ⇒ VGS 2 ( 3.265 ) 2 2 ⎛ 0.511 ⎞ I DQ = ( 2 ) ⎜1 − ⎟ ⇒ I DQ = 1.00 mA 1.75 ⎠ ⎝ VSDQ = 10 − (1.00 )( 5 ) ⇒ VSDQ = 5.0 V = 0.511 V 134. b. 2 I DSS ⎛ VGS ⎞ 2 ( 2 ) ⎛ 0.511 ⎞ ⎜1 − ⎟= ⎜1 − ⎟ ⇒ g m = 1.618 mA/V VP ⎝ VP ⎠ 1.75 ⎝ 1.75 ⎠ g m ( RS RL ) (1.618 ) ( 5 10 ) Av = = ⇒ Av = 0.844 1 + g m ( RS RL ) 1 + (1.618 ) ( 5 10 ) gm = ⎛R ⎞ i0 ( v0 / RL ) = = Av ⋅ ⎜ i ⎟ ii ( vi / Ri ) ⎝ RL ⎠ Ri = R1 R2 = 90 110 = 49.5 kΩ Ai = ⎛ 49.5 ⎞ Ai = ( 0.844 ) ⎜ ⎟ ⇒ Ai = 4.18 ⎝ 10 ⎠ c. AC load line Ϫ1 Slope ϭ 5͉͉10 Ϫ1 ϭ 3.33 K 1.0 5.0 10 Δid = 1.0 mA vsd = ( 3.33)(1.0 ) = 3.33 V Maximum swing in output voltage = 6.66 V peak-to-peak 4.68 ⎛ V ⎞ I DQ = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ 2 2 ⎛ 4⎞ ⎛ V ⎞ 4 = 8 ⎜1 − GS ⎟ ⇒ VGS = 4 ⎜ 1 − ⎟ ⇒ VGS = 1.17 V ⎜ 4 ⎠ 8⎟ ⎝ ⎝ ⎠ VSDQ = VDD − I DQ ( RS + RD ) 7.5 = 20 − 4 ( RS + RD ) ⇒ RS + RD = 3.125 kΩ ⎛ VGS ⎜1 − VP ⎝ RS = 3.125 − RD − g m RD Av = 1 + g m RS gm = 2 I DSS VP ⎞ 2 ( 8 ) ⎛ 1.17 ⎞ ⎟= ⎜1 − ⎟ ⇒ g m = 2.83 mA/V 4 ⎝ 4 ⎠ ⎠ −3 (1 + g m RS ) = − g m RD 3 ⎡1 + ( 2.83)( 3.125 − RD ) ⎤ = ( 2.83) RD ⎣ ⎦ 9.844 − 2.83RD = 0.9433RD ⇒ RD = 2.61 kΩ RS = 0.516 kΩ VS = 20 − ( 4 )( 0.516 ) ⇒ VS = 17.94 V VG = VS − VGS = 17.94 − 1.17 = 16.77 V ⎛ R2 ⎞ ⎛ R2 ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ ( 20 ) ⇒ R2 = 335 kΩ. R1 = 65 kΩ R1 + R2 ⎠ ⎝ 400 ⎠ ⎝ 135. Chapter 5 Exercise Problems EX5.1 β= α 1−α 0.980 = 49 1 − 0.980 0.995 = 199 For α = 0.995, β = 1 − 0.995 So 49 ≤ β ≤ 199 For α = 0.980, β = EX5.2 BVCEO = BVCBO n = β 200 3 120 or BVCEO = 40.5 V EX5.3 V − VBE ( on ) 2 − 0.7 I B = BB = ⇒ 6.5 μ A RB 200 I C = β I = (120 )( 6.5 μ A ) ⇒ I C = 0.78 mA VCE = VCC − I C RC = 5 − ( 0.78 )( 4 ) or VCE = 1.88 V P = I BVBE ( on ) + I CVCE = ( 0.0065 )( 0.7 ) + ( 0.78 )(1.88 ) ≅ 1.47 mW EX5.4 V + − VEB ( on ) − VBB 5 − 0.7 − 2.8 IB = = or I B = 4.62 μ A RB 325 I C = β I B = ( 80 )( 4.615 ) ⇒ I C = 0.369 mA VEC = 2 = 5 − ( 0.369 ) RC which yields RC = 8.13 k Ω EX5.5 (a) IB = VBB − VBE ( on ) RB = 2 − 0.7 ⇒ 5.91 μ A 220 I C = β I B = (100 )( 5.91 μ A ) ⇒ 0.591 mA I E = (1 + β ) I B = 0.597 mA VCE = VCC − I C RC = 10 − ( 0.591)( 4 ) or VCE = 7.64 V 6.5 − 0.7 ⇒ 26.4 μ A 220 Transistor is biased in saturation mode, so (b) IB = VCE = VCE ( sat ) = 0.2 V IC = VCC − VCE ( sat ) RC = 10 − 0.2 or I C = 2.45 mA 4 I E = I C + I B = 2.45 + 0.0264 ≅ 2.48 mA EX5.6 For 0 ≤ VI < 0.7 V , Qn is cutoff, VO = 9 V When Qn is biased in saturation, we have 0.2 = 9 − So, for VI ≥ 5.1 V , VO = 0.2 V EX5.7 (100 )(VI − 0.7 )( 4 ) 200 ⇒ VI = 5.1 V 136. IB = VBB − VBE ( on ) RB + (1 + β ) RE = 8 − 0.7 30 + ( 76 )(1.2 ) or I B = 60.2 μ A I C = β I B = ( 75 )( 60.23 μ A ) ⇒ 4.52 mA I E = (1 + β ) I B = 4.58 mA VCE = VCC − I C RC − I E RE = 12 − ( 4.517 )( 0.4 ) − ( 4.577 )(1.2 ) or VCE = 4.70 V EX5.8 For VC = 4 V and ICQ = 1.5 mA, V + − VC 10 − 4 RC = = ⇒ RC = 4 k Ω I CQ 1.5 ⎛ 1+ β ⎞ ⎛ 101 ⎞ IE = ⎜ ⎟ IC = ⎜ ⎟ (1.5 ) = 1.515 mA β ⎠ ⎝ 100 ⎠ ⎝ −V ( on ) − V − also I E = BE RE Then RE = −0.7 − ( −10 ) 1.515 ⇒ RE = 6.14 k Ω EX5.9 3 − 0.7 = 0.25 I EQ = RE RE = 9.2 kΩ ⎛ 75 ⎞ I CQ = ⎜ ⎟ ( 0.25 ) = 0.2467 mA ⎝ 76 ⎠ −0.7 + VCEQ + I CQ RC − 3 = 0 RC = 3 + 0.7 − 2 ⇒ RC = 6.89 kΩ 0.2467 EX5.10 5 = I E RE + VEB ( on ) + I B RB − 2 (a) (b) (c) (d) EX5.11 180 ⎞ ⎛ 5 + 2 − 0.7 = I E ⎜ 2 + ⎟ I E = 0.9859 mA 41 ⎠ ⎝ I C = 0.962 mA 180 ⎞ ⎛ 6.3 = I E ⎜ 2 + ⎟ I E = 1.2725 mA 61 ⎠ ⎝ I C = 1.25 mA 180 ⎞ ⎛ 6.3 = I E ⎜ 2 + ⎟ I E = 1.6657 mA 101 ⎠ ⎝ I C = 1.64 mA 180 ⎞ ⎛ 6.3 = I E ⎜ 2 + ⎟ I E = 1.97365 mA 151 ⎠ ⎝ I C = 1.94 mA 137. IE = VBB − VEB ( on ) RE ⇒ RE = 4 − 0.7 1.0 or RE = 3.3 k Ω I C = α I E = ( 0.992 )(1) = 0.992 mA I B = I E − I C = 1.0 − 0.992 or I B = 8 μ A VCB = I C RC − VCC = ( 0.992 )(1) − 5 or VCB = −4.01 V EX5.12 V + − (Vγ + VCE ( sat ) ) 5 − (1.5 + 0.2 ) R= = IC 15 or R = 220 Ω 15 IB = = 1 mA 15 v − VBE ( on ) 5 − 0.7 = = 4.3 k Ω RB = I 1 IB P = I BVBE (on) + I CVCE = (1)( 0.7 ) + (15 )( 0.2 ) = 3.7 mW EX5.13 (a) For V1 = V2 = 0, All currents are zero and VO = 5 V. (b) For V1 = 5 V, V2 = 0; IB2 = IC2 = 0, 5 − 0.7 I B1 = = 4.53 mA 0.95 5 − 0.2 I C1 = ⇒ I C1 = I R = 8 mA 0.6 VO = 0.2 V (c) For V1 = V2 = 5 V, IB1 = IB2 = 4.53 mA; IR = 8 mA, IC1 = IC2 = IR/2 = 4 mA, VO = 0.2 V EX5.14 vO = 5 − iC RC = 5 − β iB RC ΔvO = − βΔiB RC iB = VBB + ΔvI − VBE ( on ) RB ΔiB = ΔvI RB Then ΔvO − β RC = ΔvI RB Let β = 100, RC = 5 k Ω, RB = 100 k Ω So Δvo − (100 )( 5 ) = = −5 ΔvI 100 Want Q-point to be vo ( Q − pt ) = 2.5 = 5 − (100 ) I BQ ( 5 ) Then I BQ = 0.005 mA, I BQ = 0.005 = VBB − 0.7 100 so VBB = 1.2 V Also, I CQ = β I BQ = (100 )( 0.005 ) = 0.5 mA EX5.15 138. VCEQ = 2.5 = 5 − I CQ RC 5 − 2.5 = 10 k Ω 0.25 I CQ 0.25 = = = 0.002083 mA 120 β or RC = I BQ Then RB = 5 − 0.7 ⇒ RB = 2.06 M Ω 0.002083 EX5.16 (a) RTH = R1 R2 = 9 2.25 = 1.8 k Ω or VTH ⎛ R2 ⎞ ⎛ 2.25 ⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ ( 5) R1 + R2 ⎠ ⎝ 9 + 2.25 ⎠ ⎝ = 1.0 V I BQ = (b) VTH − VBE ( on ) RTH + (1 + β ) RE = 1 − 0.7 1.8 + (151)( 0.2 ) or I BQ = 9.375 μ A I CQ = β I BQ = (150 )( 9.375 μ A ) or I CQ = 1.41 mA I EQ = (1 + β ) I BQ ⇒ I EQ = 1.42 mA VCEQ = 5 − I CQ RC − I EQ RE = 5 − (1.41)(1) − (1.42 )( 0.2 ) or VCEQ = 3.31 V For β = 75 (c) I BQ = 1 − 0.7 = 17.6 μ A 1.8 + ( 76 )( 0.2 ) I CQ = β I BQ = ( 75 )(17.6 μ A ) or I CQ = 1.32 mA I EQ = (1 + β ) I BQ = ( 76 )(17.6 μ A ) or I EQ = 1.34 mA VCEQ = 5 − (1.32 )(1) − (1.34 )( 0.2 ) or VCEQ = 3.41 V EX5.17 VCEQ ≅ VCC − I CQ ( RC + RE ) or 2.5 ≅ 5 − I CQ (1 + 0.2 ) which yields I CQ = 2.08 mA, I CQ 2.08 I BQ = = = 0.0139 mA 150 β RTH = ( 0.1)(1 + β ) RE = ( 0.1)(151)( 0.2 ) or RTH = 3.02 k Ω ⎛ R2 ⎞ 1 Now VTH = ⎜ ⎟ ⋅ VCC = ⋅ RTH ⋅ VCC R1 + R2 ⎠ R1 ⎝ 1 so VTH = ( 3.02 )( 5 ) R1 We can write VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE 139. or 1 ( 3.02 )( 5 ) = ( 0.0139 )( 3.02 ) + 0.7 + (151)( 0.0139 )( 0.2 ) R1 We obtain R1 = 13 k Ω and then R2 = 3.93 k Ω EX5.18 β = 150, RTH = R1 || R2 5−0 I CQ = = 5 mA 1 I CQ 5 = = 0.0333 mA I BQ = β 150 I BQ = VTH − VBE ( on ) − ( −5 ) RTH + (1 + β ) RE Set RTH = ( 0.1)(1 + β ) RE ⎛ R2 ⎞ We have VTH = ⎜ ⎟ (10 ) − 5 ⎝ R1 + R2 ⎠ Then I BQ ⎛ R2 ⎞ ⎜ ⎟ (10 ) − 0.7 R + R2 ⎠ = 0.0333 = ⎝ 1 (1.1)(151)( 0.2 ) ⎛ R2 ⎞ which yields ⎜ ⎟ = 0.1806 ⎝ R1 + R2 ⎠ ⎛ RR ⎞ Now RTH = ⎜ 1 2 ⎟ = ( 0.1)(151)( 0.2 ) = 3.02 k Ω ⎝ R1 + R2 ⎠ so R1 ( 0.1806 ) = 3.02 k Ω We obtain R1 = 16.7 k Ω and R2 = 3.69 k Ω EX5.19 V + − V − − VECQ 5 − ( −5 ) − 5 I CQ ≅ = 4.5 + 0.5 RC + RE so I CQ = 1 mA, and I BQ = I CQ β = 1 = 0.00833 mA 120 RTH = ( 0.1)(1 + β ) RE = ( 0.1)(121)( 0.5 ) or RTH = 6.05 k Ω We can write V + = I EQ RE + VEB ( on ) + I BQ RTH + VTH We have VTH = 1 ⋅ RTH (10 ) − 5 and if we let I EQ ≅ I CQ = 1 mA, R1 then we have 5 = (1)( 0.5 ) + 0.7 + ( 0.00833)( 6.05 ) + which yields R1 = 6.91 k Ω and R2 = 48.6 k Ω EX5.20 ⎛ 2⎞ 2 ⎞ ⎛ I1 = I Q ⎜ 1 + ⎟ = ( 0.25 ) ⎜1 + ⎟ = 0.2625 mA ⎝ 40 ⎠ ⎝ β⎠ − 0 − VBE ( on ) − V 0 − 0.7 − ( −5 ) = R1 = 0.2625 I1 or R1 = 16.38 k Ω 1 ( 6.05 )(10 ) − 5 R1 140. For VCEO = 3 V , then VCO = 2.3 V ⎛ β ⎞ ⎛ 40 ⎞ I CO = ⎜ ⎟ ⋅ I Q = ⎜ ⎟ ( 0.25 ) = 0.2439 mA ⎝ 41 ⎠ ⎝ 1+ β ⎠ + V − VCO 5 − 2.3 RC = = = 11.07 k Ω I CO 0.2439 EX5.21 RTH 50 ||100 = 33.3 k Ω ⎛ 50 ⎞ VTH = VTH = ⎜ ⎟ (10 ) − 5 = −1.67 V ⎝ 50 + 100 ⎠ −1.67 − 0.7 − ( −5 ) I B1 = ⇒ 11.2 μ A 33.3 + (101)( 2 ) I C1 = 1.12 mA, I E1 = 1.13 mA VE1 = I E1 RE1 − 5 = (1.13)( 2 ) − 5 = −2.74 V VCE1 = 3.25 V ⇒ VC1 = 0.51 V Now VE 2 = 0.51 + 0.7 = 1.21 V 5 − 1.21 = 1.90 mA ⇒ I B 2 = 18.8 μ A 2 = 1.88 mA IE2 = IC 2 I R1 = I C1 − I B 2 = 1.12 − 0.0188 = 1.10 mA 5 − 0.51 = 4.08 k Ω 1.10 = 2.5 ⇒ VC 2 = VE 2 − VEC 2 RC1 = VEC 2 = 1.21 − 2.5 = −1.29 V RC 2 = −1.29 − ( −5 ) 1.88 = 1.97 k Ω EX5.22 12 = 240 k Ω = R1 + R2 + R3 0.05 Then VB1 = ( 0.5)( 2 ) + 0.7 = 1.7 V We find 1.7 = 34 k Ω 0.05 = ( 0.5 )( 2 ) + 4 + 0.7 = 5.7 V R3 = Also VB 2 ΔVR 2 = 5.7 − 1.7 = 4 V 4 = 80 k Ω 0.05 and R1 = 240 − 80 − 34 = 126 k Ω so R2 = VC 2 = 1 + 4 + 4 = 9 V Then RC = V + − VC 2 12 − 9 = = 6 kΩ I CQ 0.5 Test Your Understanding Exercises TYU5.1 α= β 1+ β For β = 75, α = 75 = 0.9868 76 141. For β = 125, α = 125 = 0.9921 126 TYU5.2 I E = (1 + β ) I B IE 0.780 = = 81.25 then β = 80.3 I B 0.00960 so 1 + β = Now α= β = 1+ β 80.25 = 0.9877 81.25 I C = α I E = ( 0.9877 )( 0.78 ) = 0.770 mA TYU5.3 β= α 0.990 = = 99 1 − α 1 − 0.990 Now I B = IE 2.15 = ⇒ 21.5 μ A and I C = α I E = ( 0.990 )( 2.15 ) = 2.13 mA 1 + β 100 TYU5.4 V 150 ro = A = IC IC For I C = 0.1 mA ⇒ ro = 1.5 M Ω For I C = 1.0 mA ⇒ ro = 150 k Ω For I C = 10 mA ⇒ ro = 15 k Ω TYU5.5 ⎛ V ⎞ I C = I O ⎜1 + CE ⎟ VA ⎠ ⎝ At VCE = 1 V , I C = 1 mA 1 ⎞ ⎛ For VA = 75 V , I C = 1 = I O ⎜1 + ⎟ ⇒ I O = 0.9868 mA 75 ⎠ ⎝ Then, at VCE = 10 V (a) ⎛ 10 ⎞ I C = ( 0.9868 ) ⎜ 1 + ⎟ = 1.12 mA ⎝ 75 ⎠ (b) At VCE 1 ⎞ ⎛ For VA = 150 V , I C = 1 = I O ⎜ 1 + ⎟ ⇒ I O = 0.9934 mA ⎝ 150 ⎠ 10 ⎞ ⎛ = 10 V , I C = ( 0.9934 ) ⎜ 1 + ⎟ = 1.06 mA ⎝ 150 ⎠ TYU5.6 BVCEO = BVCBO n β so BVCBO = 3 100 ( 30 ) = 139 V TYU5.7 (a) For VI = 0.2 V < VBE ( on ) ⇒ I B = I C = 0, VO = 5 V and P = 0 (b) IB = For VI = 3.6 V , transistor is driven into saturation, so VI − VBE ( on ) RB = V + − VCE ( sat ) 5 − 0.2 3.6 − 0.7 = 4.53 mA and I C = = = 10.9 mA RC 0.440 0.64 142. I C 10.9 = = 2.41 < β which shows that the transistor is indeed driven into saturation. Now, I B 4.53 Note that P = I BVBE ( on ) + I CVCE ( sat ) = ( 4.53)( 0.7 ) + (10.9 )( 0.2 ) = 5.35 mW TYU5.8 For VBC = 0 ⇒ VO = 0.7 V Then I C = I 5 − 0.7 9.77 = 9.77 mA and I B = C = = 0.195 mA 0.44 β 50 Now VI = I B RB + VBE ( on ) = ( 0.195 )( 0.64 ) + 0.7 or VI = 0.825 V Also P = I BVBE ( on ) + I CVCE = ( 0.195 )( 0.7 ) + ( 9.77 )( 0.7 ) = 6.98 mW TYU5.9 V + − VC 10 − 6.34 IC = = = 0.915 mA RC 4 −VBE ( on ) − V − And I E = Now α = RE = −0.7 − ( −10 ) 10 or IE = 0.930 I C 0.915 α 0.9839 = = 0.9839 and β = = = 61 I E 0.930 1 − α 1 − 0.9839 Also I B = I E − I C = 0.930 − 0.915 ⇒ 15 μ A and VCE = VC − VE = 6.34 − ( −0.7 ) = 7.04 V TYU5.10 10 − 0.7 IE = = 1.16 mA 8 1.1625 IB = = 22.8 μ A 51 I C = ( 50 )( 22.8 μ A ) = 1.14 mA VE = V + − I E RE = 10 − (1.1625 )( 8 ) = 0.7 V VC = I C RC − 10 = (1.1397 )( 4 ) − 10 = −5.44 VEC = 0.7 − ( −5.44 ) VEC = 6.14 V TYU5.11 VBB = I B RB + VBE ( on ) + I E RE or VBB = I B RB + VBE ( on ) + (1 + β ) I B RE Then I B = VBB − VBE ( on ) RB + (1 + β ) RE = 2 − 0.7 10 + ( 76 )(1) or I B = 15.1 μ A Also I C = ( 75 )(15.1 μ A ) = 1.13 mA and I E = ( 76 )(15.1 μ A ) = 1.15 mA Now VCE = VCC + VBB − I C RC − I E RE = 8 + 2 − (1.13)( 2.5 ) − (1.15 )(1) = 6.03 V TYU5.12 VCE = 2.5 V ⇒ VE = 2.5 V = I E RE We have VBB = I B RB + VBE ( on ) + VE 143. so I B = VBB − VBE ( on ) − VE RB = 5 − 0.7 − 2.5 10 or I B = 0.18 mA Then I E = (101)( 0.18 ) = 18.2 mA And RE = 2.5 = 0.138 k Ω ⇒ 138 Ω 18.18 TYU5.13 VBB = I E RE + VEB ( on ) + I B RB 2.2 = 0.0431 mA 51 ⎛ β ⎞ ⎛ 50 ⎞ and I C = ⎜ ⎟ ⋅ I E = ⎜ ⎟ ( 2.2 ) = 2.16 mA ⎝ 51 ⎠ ⎝ 1+ β ⎠ Then VBB = ( 2.2 )(1) + 0.7 + ( 0.0431)( 50 ) I E = 2.2 mA ⇒ I B = or VBB = 5.06 V Now VEC = 5 − I E RE = 5 − ( 2.2 )(1) = 2.8 V TYU5.14 (a) For vI = 0, iB = iC = 0, vO = 12 V , P = 0 For vI = 12 V , iB = (b) iC = VCC − VCE ( sat ) RC = vI − VBE ( on ) RB = 12 − 0.7 = 47.1 mA 0.24 12 − 0.1 = 2.38 A 5 vO = 0.1 V and P = iBVBE ( on ) + iCVCE ( sat ) = ( 0.0471)( 0.7 ) + ( 2.38 )( 0.1) = 0.271 W TYU5.15 For VCEQ = 2.5 V , I CQ = (a) I BQ = I CQ β = 5 − 2.5 = 1.25 mA 2 1.25 ⇒ I BQ = 12.5 μ A 100 5 − 0.7 = 344 k Ω 0.0125 IBQ is independent of β . Then RB = (b) For VCEQ = 1 V , I C = β= 5 −1 = 2 mA 2 IC 2 = ⇒ β = 160 I B 0.0125 For VCEQ = 4 V , I C = 5−4 = 0.5 mA 2 IC 0.5 = ⇒ β = 40 I B 0.0125 So 40 ≤ β ≤ 160 β= TYU5.16 5 − 0.7 I BQ = = 0.005375 mA 800 144. For β = 75, I CQ = β I BQ = ( 75 )( 0.005375 ) Or I CQ = 0.403 mA For β = 150, I CQ = (150 )( 0.005375 ) Or I CQ = 0.806 mA Largest I CQ ⇒ Smallest VCEQ 5 −1 = 4.96 k Ω 0.806 5−4 For β = 75, RC = = 2.48 k Ω 0.403 For β = 150, RC = 5 − 2.5 = 4.14 k Ω 0.604 = 5 − ( 0.403)( 4.14 ) = 3.33 V For a nominal I CQ = 0.604 mA and VCEQ = 2.5 V , RC = Now for I CQ = 0.403 mA, VCEQ For I CQ = 0.806 mA, VCEQ = 5 − ( 0.806 )( 4.14 ) = 1.66 V So, for RC = 4.14 k Ω, 1.66 ≤ VCEQ ≤ 3.33 V TYU5.17 (a) ⎛ β ⎞ ⎛ 100 ⎞ I CQ = ⎜ ⎟ ⋅ I EQ = ⎜ ⎟ (1) = 0.99 mA ⎝ 101 ⎠ ⎝1+ β ⎠ I EQ 1 = ⇒ 9.90 μ A I BQ = 1 + β 101 VB = − I BQ RB = − ( 0.0099 )( 50 ) or VB = −0.495 V ⎛ I CQ ⎞ ⎛ 0.99 × 10−3 ⎞ VBE = VT ln ⎜ ⎟ = ( 0.026 ) ln ⎜ −14 ⎟ ⎝ 3 × 10 ⎠ ⎝ IS ⎠ or VBE = 0.630 V Then VE = VB − VBE = −0.495 − 0.630 = −1.13 V VC = 10 − ( 0.99 )( 5 ) = 5.05 V Then VCEQ = VC − VE = 5.05 − ( −1.13) = 6.18 V (b) I EQ 1 = 0.0196 mA 1 + β 51 VB = − ( 0.0196 )( 50 ) = −0.98 V I EQ = 1 mA, I BQ = ⎛ β I CQ = ⎜ ⎝1+ β = ⎞ ⎛ 50 ⎞ ⎟ ⋅ I EQ = ⎜ ⎟ (1) = 0.98 mA ⎝ 51 ⎠ ⎠ ⎛ 0.98 × 10−3 ⎞ VBE = ( 0.026 ) ln ⎜ = 0.629 V −14 ⎟ ⎝ 3 × 10 ⎠ VE = −0.98 − 0.629 = −1.61 V VC = 10 − ( 0.98 )( 5 ) = 5.1 V VCEQ = VC − VE = 5.1 − ( −1.61) = 6.71 V TYU5.18 IQ IQ IB = = 1 + β 121 and 145. ⎛ IQ ⎞ VB = ⎜ ⎟ ( 20 ) = I Q ( 0.165 ) ⎝ 121 ⎠ VE = I Q ( 0.165 ) + 0.7 ⎛ β ⎞ ⎛ 120 ⎞ I CQ = ⎜ ⎟ ⋅ I EQ = ⎜ ⎟ ⋅ I Q = ( 0.992 ) I Q ⎝ 121 ⎠ ⎝ 1+ β ⎠ VC = I CQ RC − 5 = ( 0.992 ) I Q ( 4 ) − 5 = 3.97 I Q − 5 VECQ = VE − VC = ⎡ I Q ( 0.165 ) + 0.7 ⎤ − ⎡ I Q ( 3.97 ) − 5⎤ ⎣ ⎦ ⎣ ⎦ = −3.805 I Q + 5.7 Then 3 = 5.7 − 3.805I Q which yields I Q = 0.710 mA 146. Chapter 5 Problem Solutions 5.1 (a) β= iC 510 = ⇒ β = 85 6 iB β 85 ⇒ α = 0.9884 1 + β 86 iE = (1 + β ) iB = ( 86 )( 6 ) ⇒ iE = 516 μ A α= (b) = 2.65 ⇒ β = 53 0.050 53 α = ⇒ α = 0.9815 54 iE = (1 + β ) iB = ( 54 )( 0.050 ) ⇒ iE = 2.70 mA β= 5.2 (a) For β = 110: α = β 1+ β = 110 = 0.99099 111 180 = 0.99448 181 0.99099 ≤ α ≤ 0.99448 For β = 180: α = (b) I C = β I B = 110 ( 50 μ A ) ⇒ I C = 5.50 mA or I C = 180 ( 50 μ A ) ⇒ I C = 9.00 mA so 5.50 ≤ I C ≤ 9.0 mA 5.3 (a) iE α (b) 1.12 ⇒ 9.33 μ A 120 ⎛ 121 ⎞ = (1.12 ) ⎜ ⎟ = 1.13 mA ⎝ 120 ⎠ 120 = = 0.9917 121 50 = = 2.5 mA 20 ⎛ 21 ⎞ = ⎜ ⎟ ( 50 ) = 52.5 mA ⎝ 20 ⎠ 20 = = 0.9524 21 iB = iB iE α 5.4 (a) α β= 0.9 0.95 0.98 0.99 0.995 0.999 9 19 49 99 199 999 (b) α 1−α 147. α= β β 1+ β 0.9524 0.9804 0.9901 0.9934 0.9955 0.9975 20 50 100 150 220 400 5.5 1.2 ⇒ 14.8 μ A 81 ⎛ 80 ⎞ I C = (1.2 ) ⎜ ⎟ = 1.185 mA ⎝ 81 ⎠ 80 α= = 0.9877 81 VC = 5 − (1.185 )( 2 ) = 2.63 V IB = (a) 0.80 ⇒ 9.88 μ A 81 ⎛ 80 ⎞ I C = ( 0.80 ) ⎜ ⎟ = 0.790 mA ⎝ 81 ⎠ 80 α= = 0.9877 81 VC = 5 − ( 0.790 )( 2 ) = 3.42 V IB = (b) Yes, VC > VB so B-C junction is reverse biased in both areas. (c) 5.6 For VC = 0, I C = IE = IC α = 5 = 2.5 mA 2 2.5 ⇒ I E = 2.546 mA 0.982 5.7 (a) IC α VC VC (b) (c) 0.75 ⇒ 12.3 μ A 61 ⎛ 60 ⎞ = ( 0.75 ) ⎜ ⎟ = 0.738 mA ⎝ 61 ⎠ 60 = = 0.9836 61 = I C RC − 10 = ( 0.738 )( 5 ) − 10 = −6.31 V IB = 1.5 ⇒ 24.6 μ A 61 ⎛ 60 ⎞ I C = (1.5 ) ⎜ ⎟ = 1.475 mA ⎝ 61 ⎠ ⎛ 60 ⎞ α = ⎜ ⎟ = 0.9836 ⎝ 61 ⎠ VC = (1.475 )( 5 ) − 10 ⇒ VC = −2.625 V IB = Yes, VC < 0 in both cases so that B-C junction is reverse biased. 148. 5.8 IC = IE = VC − ( −10 ) RC IC α = = 10 − 1.2 = 1.76 mA 5 1.76 ⇒ I E = 1.774 mA 0.992 5.9 I C = I S eVRE / VT ⎛ 0.685 ⎞ = 10−13 exp ⎜ ⎟ ⇒ I C = 27.67 mA ⎝ 0.026 ⎠ ⎛1+ β ⎞ ⎛ 91 ⎞ IE = ⎜ ⎟ I C = ⎜ ⎟ ( 27.67 ) ⎝ 90 ⎠ ⎝ β ⎠ I D = 27.98 mA I 27.67 IB = C = ⇒ I B = 0.307 mA β 90 5.10 Device 1: iE = I Eo1evEB / VT ⇒ 0.5 × 10−3 = I Eo1e0.650 / 0.026 So that I EO1 = 6.94 × 10−15 A Device 2: 12.2 × 10−3 = I Eo 2 e0.650 / 0.026 Or I Eo 2 = 1.69 × 10−13 A Ratio of areas = I Eo 2 1.69 × 10−13 = ⇒ Ratio = 24.4 I Eo1 6.94 × 10−15 5.11 (a) ro = VA 250 = ⇒ ro = 250 k Ω 1 IC (b) ro = VA 250 = ⇒ ro = 2.50 M Ω IC 0.1 5.12 BVC E 0 = BVC B 0 3 β = 60 3 100 BVC E 0 = 12.9 V 5.13 BVC E 0 = 56 = 220 3 β β = 60.6 5.14 BVC B 0 3 β ⇒3β = 220 = 3.93 56 149. BVC E 0 = BVC B 0 3 β BVC B 0 = ( BVC E 0 ) 3 β = ( 50 ) 3 50 BVC B 0 = 184 V 5.15 (a) IE = −0.7 − ( −10 ) = 1.86 mA 5 ⎛ 75 ⎞ I C = (1.86 ) ⎜ ⎟ = 1.836 mA ⎝ 76 ⎠ VC = −0.7 + 4 = 3.3 V 10 − 3.3 ⇒ RC = 3.65 K 1.836 0.5 IB = = 0.00658 mA 76 VB = I B RB = ( 0.00658 )( 25 ) ⇒ VB = 0.164 V RC = (b) (c) ⎛ 75 ⎞ I C = ( 0.5 ) ⎜ ⎟ = 0.493 mA ⎝ 76 ⎠ −1 − ( −5 ) RC = ⇒ RC = 8.11 K 0.493 I O = E (10 ) + 0.7 + I E ( 4 ) − 8 76 7.3 = I E ( 4 + 0.132 ) ⇒ I E = 1.767 mA ⎛ 75 ⎞ I C = (1.767 ) ⎜ ⎟ = 1.744 mA ⎝ 76 ⎠ VCE = 8 − (1.744 )( 4 ) − ⎡(1.767 )( 4 ) − 8⎤ ⎣ ⎦ = 16 − 6.972 − 7.068 ⇒ VCE = 1.96 V (d) ⎛I ⎞ = I E (10 + 0.263 + 2 ) + 0.7 5 = I E (10 ) + ⎜ E ⎟ ( 20 ) + 0.7 + I E ( 2 ) ⎝ 76 ⎠ I E = 0.3506 mA ⇒ I B = 4.61 μ A VC = 5 − ( 0.3506 )(10 ) VC = 1.49 V 5.16 For Fig. 5.15 (a) RE = 5 + 5% = 5.25 K IE = −0.7 − ( −10 ) = 1.77 mA 5.25 I C = 1.75 mA 10 − 3.3 = 3.83 K RC = 1.75 RE = 5 − 5% = 4.75 K IE = −0.7 − ( −10 ) = 1.96 mA 4.75 I C = 1.93 mA 10 − 3.3 = 3.47 K RC = 1.93 So 1.75 ≤ I C ≤ 1.93 mA 3.47 ≤ RC ≤ 3.83 K For Fig. 5.15(c) RE = 4 + 5% = 4.2 K 150. IB = 8 − 0.7 = 0.0222 mA 10 + ( 76 )( 4.2 ) I C = 1.66 mA I E = 1.69 mA VCE = 16 − (1.66 )( 4 ) − (1.69 )( 4.2 ) = 16 − 6.64 − 7.098 ⇒ VCE = 2.26 V RE = 4 − 5% = 3.8 K 8 − 0.7 = 0.0244 I C = 1.83 mA IB = 10 + ( 76 )( 3.8 ) I E = 1.86 mA VCE = 16 − (1.83)( 4 ) − (1.86 )( 3.8 ) = 16 − 7.32 − 7.068 VCE = 1.61 V So 1.66 ≤ I C ≤ 1.83 mA 1.61 ≤ VCE ≤ 2.26 V 5.17 RB = VBB − VEB 2.5 − 0.7 = ⇒ RB = 120 k Ω IB 0.015 I CQ = ( 70 )(15μ A ) ⇒ 1.05 mA RC = VCC − VECQ I CQ = 5 − 2.5 ⇒ 1.05 5.18 (a) VB = − I B RB ⇒ I B = RC = 2.38 k Ω −VB − ( −1) = 500 RB I B = 2.0 μ A VE = −1 − 0.7 = −1.7 V IE = VE − ( −3) RE = −1.7 + 3 = 0.2708 mA 4.8 IE 0.2708 = (1 + β ) = = 135.4 ⇒ β = 134.4 0.002 IB β ⇒ α = 0.9926 1+ β I C = β I B ⇒ I C = 0.269 mA α= VCE = 3 − VE = 3 − ( −1.7 ) ⇒ VCE = 4.7 V (b) 5−4 ⇒ I E = 0.5 mA 2 4 = 0.7 + I B RB + ( I B + I C ) RC − 5 IE = I B + IC = I E I B + IC = I E 4 = 0.7 + I B (100 ) + ( 0.5 )( 8 ) − 5 I B = 0.043 ⇒ IE 0.5 = (1 + β ) = = 11.63 IB 0.043 β = 10.63, α = 5.19 β 1+ β ⇒ α = 0.9140 151. VB − 0.7 − ( −5 ) VB + 4.3 3 3 VB + 4.3 ⎞ ⎛ 50 ⎞ ⎛ ⎛ 50 ⎞ IC = ⎜ ⎟ I E = ⎜ ⎜ ⎟ 3 ⎟ ⎝ 51 ⎠ ⎝ 51 ⎠ ⎝ ⎠ ⎛ V + 4.3 ⎞ ⎛ 50 ⎞ VC = 5 − I C RC = 5 − ⎜ B ⎜ ⎟ (10 ) 3 ⎟ ⎝ 51 ⎠ ⎝ ⎠ IE = = Now VB = VC , so VB [1 + 3.27 ] = 5 − 14.1 = −9.05 VB = −2.12 V IE = −2.12 + 4.3 ⇒ I E = 0.727 mA 3 5.20 10 − VE 10 − 2 = ⇒ I E = 0.80 mA 10 10 VB = VE − 0.7 = 2 − 0.7 = 1.3 V IE = IB = VB 1.3 = ⇒ I B = 0.026 mA RB 50 I C = I E − I B = 0.80 − 0.026 ⇒ I C = 0.774 mA β= α= I C 0.774 = ⇒ β = 29.77 I B 0.026 β 1+ β = 29.77 ⇒ α = 0.9675 30.77 VEC = VE − VC = VE − ( I C RC − 10 ) = 2 − ⎡( 0.774 )(10 ) − 10 ⎤ ⎣ ⎦ VEC = 4.26 V Load line developed assuming the VB voltage can change and the RB resistor is removed. 1 mA 0.774 Q-point IC 4.26 5.21 5 − 0.7 ⇒ 17.2 μ A 250 I C = (120 )( 0.0172 ) = 2.064 mA IB = VC = ( 2.064 )(1.5 ) − 5 = −1.90 V VEC = 5 − ( −1.90 ) ⇒ VEC = 6.90 V 20 VEC 152. IC (mA) 6.67 Q-point 2.06 6.9 10 V EC (V) 5.22 ⎛ 50 ⎞ I C = ⎜ ⎟ (1) = 0.98 mA ⎝ 51 ⎠ VC = I C RC − 9 = ( 0.98 )( 4.7 ) − 9 or VC = −4.39 V 1 = 0.0196 mA 51 VE = I B RB + VEB ( on ) = ( 0.0196 )( 50 ) + 0.7 or VE = 1.68 V IB = 5.23 0.5 ⎛ 50 ⎞ I C = ⎜ ⎟ ( 0.5 ) = 0.49 mA, I B = = 0.0098 mA 51 ⎝ 51 ⎠ VE = I B RB + VEB ( on ) = ( 0.0098 )( 50 ) + 0.7 or VE = 1.19 V VC = I C RC − 9 = ( 0.49 )( 4.7 ) − 9 = −6.70 V Then VEC = VE − VC = 1.19 − ( −6.7 ) == 7.89 V PQ = I CVEC + I BVEB = ( 0.49 )( 7.89 ) + ( 0.0098 )( 0.7 ) or PQ = 3.87 mW Power Dissipated = PS = I Q ( 9 − VE ) = ( 0.5 )( 9 − 1.19 ) Or PS = 3.91 mW 5.24 I ⇒ I E1 = I E 2 = 0.5 mA 2 ≈ 0.5 mA = 5 − ( 0.5 )( 4 ) ⇒ VC1 = VC 2 = 3 V I E1 = I E 2 = I C1 = I C 2 VC1 = VC 2 5.25 (a) RE = 0 I B = 2 − 0.7 1.3 = RB RB ⎛ 1.3 ⎞ 5 − 2 = 0.8 ⇒ RC = 3.75 K I C = ( 80 ) ⎜ ⎟ = RC ⎝ RB ⎠ RB = 130 K (b) 0.8 ⎛ 81 ⎞ = 0.010 mA I E = 0.8 ⎜ ⎟ = 0.81 mA 80 ⎝ 80 ⎠ 2 = ( 0.010 )( RB ) + 0.7 + ( 0.81)(1) ⇒ RB = 49 K RE = 1 K IB = 5 = ( 0.8 ) RC + 2 + ( 0.81)(1) ⇒ RC = 2.74 K (c) For part (a) IB = 2 − 0.7 = 0.01 mA 130 153. I C = (120 )( 0.01) ⇒ I C = 1.20 mA VCE = 5 − (1.2 )( 3.75 ) ⇒ VCE = 0.5 V 2 = I B ( 49 ) + 0.7 + (121) I B (1) For part (b) I B = 0.00765 mA, I E = 0.925 mA, I C = 0.918 mA VCE = 5 − ( 0.918 )( 2.74 ) − ( 0.925 )(1) ⇒ VCE = 1.56 V Including RE result in smaller changes in Q-point values. 5.26 I BQ = a. VCC − VBE ( on ) RB 2 = 0.0333 mA 60 β 24 − 0.7 RB = ⇒ RB = 699 kΩ 0.0333 VCC − VCEQ 24 − 12 I CQ = ⇒ RC = ⇒ RC = 6 kΩ RC 2 I BQ = I CQ I BQ = b. = VCC − VBE ( on ) RB 24 − 0.7 699 = = 0.0333 mA ( Unchanged ) I CQ = β I BQ = (100 )( 0.0333) ⇒ I CQ = 3.33 mA VCEQ = VCC − I CQ RC = 24 − ( 3.33)( 6 ) ⇒ VCEQ = 4.02 V VCE = VCC − I C RC = 24 − I C ( 6 ) (c) IC (mA) 4 3.33 Q-pt (␤ ϭ 100) Q-pt (␤ ϭ 60) 2 4.02 12 5.27 a. VB = 0 ⇒ Cutoff ⇒ I E = 0, VC = 6 V b. VB = 1 V, I E = c. VB = 2 V. Assume active-mode 1 − 0.7 ⇒ I E = 0.3 mA 1 I C ≈ I E ⇒ VC = 6 − ( 0.3)(10 ) ⇒ VC = 3 V 2 − 0.7 = I E = 1.3 mA ≈ I C 1 VC = 6 − (1.3)(10 ) = −7 V! IE = 24 VCE 154. Transistor in saturation 2 − 0.7 ⇒ I E = 1.3 mA 1 VE = 1.3 V, VCE ( sat ) = 0.2 V IE = VC = VE + VCE ( sat ) = 1.3 + 0.2 ⇒ VC = 1.5 V 5.28 a. VBB = 0. ⎛ RL Cutoff V0 = ⎜ ⎝ RC + RL V0 = 3.33 V ⎞ ⎛ 10 ⎞ ⎟ VCC = ⎜ ⎟ ( 5) ⎝ 10 + 5 ⎠ ⎠ VBB = 1 V b. 1 − 0.7 ⇒ 6 μA 50 I C = β I B = ( 75 )( 6 ) ⇒ I C = 0.45 mA IB = 5 − V0 V = IC + 0 5 10 ⎛1 1 ⎞ 1 − 0.45 = V0 ⎜ + ⎟ ⇒ V0 = 1.83 V ⎝ 5 10 ⎠ Transistor in saturation V0 = VCE ( sat ) = 0.2 V c. 5.29 (a) β = 100 ⎛ 100 ⎞ (i) I Q = 0.1 mA I C = ⎜ ⎟ ( 0.1) = 0.0990 mA ⎝ 101 ⎠ VO = 5 − ( 0.099 )( 5 ) ⇒ VO = 4.505 V ⎛ 100 ⎞ (ii) I Q = 0.5 mA I C = ⎜ ⎟ ( 0.5 ) = 0.495 mA ⎝ 101 ⎠ VO = 5 − ( 0.495 )( 5 ) ⇒ VO = 2.525 V (iii) I Q = 2 mA Transistor is in saturation VO = −VBE ( sat ) + VCE ( sat ) = −0.7 + 0.2 ⇒ VO = −0.5 V (b) β = 150 ⎛ 150 ⎞ (i) I Q = 0.1 mA I C = ⎜ ⎟ ( 0.1) = 0.09934 mA ⎝ 151 ⎠ VO = 5 − ( 0.09934 )( 5 ) ⇒ VO = 4.503 V 4.503 − 4.505 × 100% = −0.044% 4.503 ⎛ 150 ⎞ (ii) I Q = 0.5 mA I C = ⎜ ⎟ ( 0.5 ) = 0.4967 mA ⎝ 151 ⎠ VO = 5 − ( 0.4967 )( 5 ) ⇒ VO = 2.517 V % change = 2.517 − 2.525 × 100% = −0.32% 2.525 (iii) I Q = 2 mA Transistor in saturation Vo = −8.5 V No change % change = 5.30 155. VCB = 0.5 V ⇒ VO = 0.5 V , I C = 5 − 0.5 = 0.90 mA 5 ⎛ 101 ⎞ IQ = ⎜ ⎟ ( 0.90 ) ⇒ I Q = 0.909 mA ⎝ 100 ⎠ 5.31 For I Q = 0, then PQ = 0 ⎛ 50 ⎞ For I Q = 0.5 mA, I C = ⎜ ⎟ ( 0.5 ) = 0.49 mA ⎝ 51 ⎠ 0.5 IB = = 0.0098 mA, VB = 0.490 V , VE = 1.19 V 51 VC = ( 0.49 )( 4.7 ) − 9 = −6.70 V ⇒ VEC = 7.89 V P ≅ I CVEC = ( 0.49 )( 7.89 ) ⇒ P = 3.87 mW For I Q = 1.0 mA, Using the same calculations as above, we find P = 5.95 mW For I Q = 1.5 mA, P = 6.26 mW For I Q = 2 mA, P = 4.80 mW For I Q = 2.5 mA, P = 1.57 mW For I Q = 3 mA, Transistor is in saturation. 0.7 + I B ( 50 ) = 0.2 + I C ( 4.7 ) − 9 I E = IQ = I B + IC ⇒ I B = 3 − IC Then, 0.7 + ( 3 − I C ) ( 50 ) = 0.2 + I C ( 4.7 ) − 9 Which yields I C = 2.916 mA and I B = 0.084 mA P = I BVEB + I CVEC = ( 0.084 )( 0.7 ) + ( 2.916 )( 0.2 ) or P = 0.642 mW 5.32 IE = VEE − VEB ( on ) RE = 9 − 0.7 ⇒ I E = 2.075 mA 4 I C = α I E = ( 0.9920 ) ( 2.075 ) ⇒ I C = 2.06 mA VBC + I C RC = VCC VBC = 9 − ( 2.06 ) ( 2.2 ) ⇒ VBC = 4.47 V 5.33 I CQ = I BQ = IR2 = VCC − VCEQ RC I CQ β = = 12 − 6 = 2.73 mA 2.2 2.73 ⇒ I BQ = 0.091 mA 30 0.7 − ( −12 ) = 0.127 mA 100 I R1 = I R 2 + I BQ = 0.127 + 0.091 = 0.218 mA V1 = I R1 R1 + 0.7 = ( 0.218 )(15 ) + 0.7 ⇒ V1 = 3.97 V 5.34 156. For VCE = 4.5 5 − 4.5 = 0.5 mA I CQ = 1 0.5 = 0.02 mA I BQ = 25 0.7 − ( −5 ) = 0.057 mA I R2 = 100 I R1 = I R 2 + I BQ = 0.057 + 0.02 = 0.077 mA V1 = I R1 R1 + VBE ( on ) = ( 0.077 )(15 ) + 0.7 = 1.86 V For VCE = 1.0 5 −1 = 4 mA 1 4 = = 0.16 mA 25 = 0.057 mA I CQ = I BQ I R2 I R1 = I R 2 + I BQ = 0.057 + 0.16 = 0.217 mA V1 = ( 0.217 )(15 ) + 0.7 ⇒ 3.96 V So 1.86 ≤ V1 ≤ 3.96 V IC 5 4 Range of Q-pt values 0.5 0 1 4.5 5 5.35 5 − 2.5 =5K 0.5 0.5 IB = = 0.00417 mA 120 5 − 0.7 RB = = 1032 K 0.00417 RC = (a) IC (mA) 1.0 Q-point 0.5 2.5 (b) Choose RC = 5.1 K RB = 1 MΩ 5 V (V) CE 157. For RB = 1 MΩ + 10% = 1.1 M, RC = 5.1 k + 10% = 5.61 K 5 − 0.7 I BQ = = 3.91 μ A ⇒ I CQ = 0.469 mA 1.1 VCEQ = 2.37 V RB = 1 MΩ + 10% = 1.1M, RC = 5.1 K − 10% = 4.59 K I BQ = 3.91 μ A ⇒ I CQ = 0.469 mA VCEQ = 2.85 V RB = 1 MΩ − 10% = 0.90 MΩ RC = 5.1 k + 10% = 5.61 K 5 − 0.7 I BQ = = 4.78 μ A ⇒ I C = 0.573 mA 0.90 VCEQ = 1.78 V RB = 1 MΩ − 10% = 0.90 MΩ RC = 5.1 k − 10% = 4.59 K I BQ = 4.78 μ A ⇒ I C = 0.573 mA VCEQ = 2.37 V IC (mA) 1.09 0.891 0.573 0.469 1.78 2.37 2.85 5 V (V) CE 5.36 VE 2 = 5 − VBE 2 VE1 = 5 − VBE1 VO = VE 2 − VE1 = ( 5 − VBE 2 ) − ( 5 − VBE1 ) VO = VBE1 − VBE 2 ⎛I ⎞ We have VBE1 = VE ln ⎜ E1 ⎟ ⎝ I EO ⎠ ⎛I ⎞ VBE 2 = VT ln ⎜ E 2 ⎟ ⎝ I EO ⎠ ⎡ ⎛I ⎞ ⎛I VO = VT ⎢ln ⎜ E1 ⎟ − ln ⎜ E 2 ⎝ I EO ⎣ ⎝ I EO ⎠ ⎞⎤ ⎟⎥ ⎠⎦ ⎛I ⎞ ⎛ 10 ⎞ VO = VT ln ⎜ E1 ⎟ = VT ln ⎜ I ⎟ ⎝ I ⎠ ⎝ IE2 ⎠ VO = 5.37 (a) kT ln (10 ) e RE = 0 (120 )( 4 ) VI − 0.7 IC = β I B VO = 5 − I C ( 4 ) = 5 − (VI − 0.7 ) 200 200 When VO = 0.2, 0.2 = 5 − 2.4 VI + 1.68 ⇒ VI = 2.7 IB = 158. VO(V) 5 0.2 0.7 2.7 5 VI (V) RE = 1 K VI − 0.7 V − 0.7 IB = = I I C = β IB 200 + (121)(1) 321 (b) (120 )( 4 ) (VI − 0.7 ) 321 When VO = 0.2 = 5 − 1.495VI + 1.047 VO = 5 − VI = 3.91 V VO(V) 5 0.2 0.7 3.91 5.38 For 4.3 ≤ VI ≤ 5 Q is cutoff I C = 0 VO = 0 If Q reaches saturation, VO = 4.8 4.8 IC = = 1.2 mA 4 5 − 0.7 − VI 1.2 IB = = 0.015 = ⇒ VI = 1.6 80 180 So VI ≤ 1.6, VO = 4.8 5 VI (V) 159. VO(V) 4.8 1.6 5.39 (a) 4.3 5 VI (V) For VI ≥ 4.3, Q is off and VO = 0 ⎛ 101 ⎞ When transistor enters saturation, 5 = ⎜ ⎟ I C (1) + 0.2 + I C ( 4 ) ⇒ I C = 0.958 mA ⎝ 100 ⎠ VO = 3.832 V I B = 0.00958 mA ⎛ 101 ⎞ 5=⎜ ⎟ ( 0.958 )(1) + 0.7 + ( 0.00958 )(180 ) + VI ⎝ 100 ⎠ VI = 5 − 0.7 − 0.9676 − 1.7244 ⇒ VI = 1.61 V For VI = 0, transistor in saturation 5 = I E (1) + 0.2 + I C ( 4 ) ⇒ 5 = I C (1) + I B (1) + 0.2 + I C ( 4 ) 5 = I E (1) + 0.7 + I B (180 ) 5 = I C (1) + I B (1) + 0.7 + I B (180 ) I E = IC + I B 4.8 = 5 I C + I B (1) 4.3 = 1I C + 181I B I B = 4.8 − 5 I C 4.3 = I C + (181)( 4.8 − 5 I C ) 904 I C = 864.5 I C = 0.956 mA VO = 3.825 V 160. VO(V) 3.832 3.825 1.61 4.3 5 VI (V) 5.40 RTH = R1 R2 = 33 10 = 7.67 kΩ ⎛ R2 ⎞ ⎛ 10 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (18 ) = 4.186 V R1 + R2 ⎠ ⎝ 10 + 33 ⎠ ⎝ V − VBE ( on ) 4.186 − 0.7 I BQ = TH = RTH + (1 + β ) RE 7.67 + ( 51)(1) I BQ = 0.0594 mA I CQ = β I BQ ⇒ I CQ = 2.97 mA I EQ = 3.03 mA VCEQ = VCC − I CQ RC − I EQ RE = 18 − ( 2.97 )( 2.2 ) − ( 3.03)(1) ⇒ VCEQ = 8.44 V 5.41 I CQ = 1.2 mA, VCEQ = 9 V , RTH = 50 k Ω 1.2 = 0.015 mA Also I B = 80 VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE ⎛ R2 ⎞ 1 1 VTH = ⎜ ⎟ (VCC ) = ⋅ RTH ⋅ VCC = ( 50 )(18 ) R1 R1 ⎝ R1 + R2 ⎠ 1 Then ( 50 )(18) = ( 0.015)( 50 ) + 0.7 + ( 81)( 0.015)(1) or R1 = 338 k Ω. R1 Then 338R2 = 50 ⇒ R2 = 58.7 k Ω 338 + R2 ⎛ 81 ⎞ I EQ = ⎜ ⎟ (1.2 ) = 1.215 mA ⎝ 80 ⎠ 18 = I CQ RC + VCEQ + I EQ RE 18 = (1.2 ) RC + 9 + (1.215 )(1) ⇒ RC = 6.49 k Ω 5.42 161. RTH = R1 R2 = 20 15 = 8.57 k Ω ⎛ R2 ⎞ ⎛ 15 ⎞ VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ (10 ) = 4.29 V ⎝ 15 + 20 ⎠ ⎝ R1 + R2 ⎠ I EQ VCC = I EQ RE + VEB ( on ) + ⋅ RTH + VTH 1+ β ⎛ 8.57 ⎞ 10 = I EQ (1) + 0.7 + I EQ ⎜ ⎟ + 4.29 ⎝ 101 ⎠ 10 − 0.7 − 4.29 5.01 Then I EQ = = ⇒ I EQ = 4.62 mA 8.57 1.085 1+ 101 I EQ ⎛ 4.62 ⎞ VB = ⋅ RTH + VTH = ⎜ ⎟ ( 8.57 ) + 4.29 or VB = 4.68 V 1+ β ⎝ 101 ⎠ 5.43 (a) RTH = 42 58 = 24.36 K ⎛ 42 ⎞ VTH = ⎜ ⎟ ( 24 ) = 10.08 V ⎝ 100 ⎠ 10.08 − 0.7 9.38 I BQ = = ⇒ 7.30 μ A 24.36 + (126 )(10 ) 1284.36 I CQ = 0.913 mA VCEQ = 14.8 V I EQ = 0.9202 IC (mA) 2.38 Q-point 0.913 14.8 24 V (b) R1 + 5% = 60.9, R2 + 5% = 44.1 RTH = 25.58 K 10.08 − 0.7 9.38 I BQ = = ⇒ 7.30 μ A 25.58 + 126 (10 ) 1285.58 I CQ = 0.912 mA VCEQ = 14.81 I EQ = 0.919 CE (V) VTH = 10.08 162. R1 + 5% = 60.9, R2 − 5% = 39.90 RTH = 24.11 K 9.50 − 0.7 8.8 I BQ = = = 6.85 μ A 24.11 + (126 )(10 ) 1284.11 I CQ = 0.857 mA VCEQ = 15.37 V I EQ = 0.8635 mA R1 − 5% = 55.1 K R2 + 5% = 44.1 K 10.67 − 0.7 9.97 I BQ = = = 7.76 μ A 24.50 + 1260 1284.5 I CQ = 0.970 mA I EQ = 0.978 mA VCEQ = 14.22 V R1 − 5% = 55.1 K R2 − 5% = 39.90 10.08 − 0.7 9.38 I BQ = = = 7.31 μ A 23.14 + 1260 1283.14 I CQ = 0.914 mA I EQ = 0.9211 mA VCEQ = 14.79 V So we have 0.857 ≤ I CQ ≤ 0.970 mA 14.22 ≤ VCEQ ≤ 15.37 V 5.44 a. RTH = R1 R2 = 25 8 = 6.06 kΩ ⎛ R2 ⎞ ⎛ 8 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ ( 24 ) ⎝ 8 + 25 ⎠ ⎝ R1 + R2 ⎠ = 5.82 V V − VBE (on) 5.82 − 0.7 I BQ = TH = RTH + (1 + β ) RE 6.06 + ( 76 )(1) I BQ = 0.0624 mA, I CQ = 4.68 mA I EQ = 4.74 VCEQ = VCC − I CQ RC − I EQ RE = 24 − ( 4.68 )( 3) − ( 4.74 )(1) VCEQ = 5.22 V b. I BQ = 5.82 − 0.7 ⇒ I BQ = 0.0326 mA 6.06 + (151)(1) I CQ = 4.89 mA I EQ = 4.92 VCEQ = 24 − ( 4.89 )( 3) − ( 4.92 )(1) VCEQ = 4.41 V 5.45 (a) VTH = 9.50 RTH = 24.50 K RTH = 23.14 K VTH = 10.67 V VTH = 10.08 163. I CQ ≅ I EQ = 0.4 mA 3 3 ⇒ RC = 7.5 k Ω; RE = ⇒ RE = 7.5 k Ω 0.4 0.4 9 R1 + R2 ≅ = 112.5 k Ω ( 0.2 )( 0.4 ) RC = ⎛ R2 ⎞ VTH = ⎜ ⎟ (VCC ) = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE ⎝ R1 + R2 ⎠ (112.5 − R2 ) R2 RR 0.4 , I BQ = = 0.004 mA RTH = 1 2 = 112.5 100 R1 + R2 ⎡ (112.5 − R2 ) R2 ⎤ ⎛ 9 ⎞ R2 ⎜ ⎥ + 0.7 + (101)( 0.004 )( 7.5 ) ⎟ = ( 0.004 ) ⎢ 112.5 ⎝ 112.5 ⎠ ⎣ ⎦ 2 We obtain R2 ( 0.08 ) = 0.004 R2 − 3.56 × 10−5 R2 + 3.73 From this quadratic, we find R2 = 48 k Ω ⇒ R1 = 64.5 k Ω (b) Standard resistor values: Set RE = RC = 7.5 k Ω and R1 = 62 k Ω, R2 = 47 k Ω Now RTH = R1 R2 = 62 47 = 26.7 k Ω ⎛ R2 ⎞ ⎛ 47 ⎞ VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ ( 9 ) = 3.88 V ⎝ 47 + 62 ⎠ ⎝ R1 + R2 ⎠ VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE So I BQ = 3.88 − 0.7 = 0.00406 mA 26.7 + (101)( 7.5 ) Then I CQ = 0.406 mA VRC = VRE = ( 0.406 )( 7.5 ) = 3.05 V 5.46 (a) RTH = R1 R2 = 12 2 = 1.714 K ⎛ R2 ⎞ ⎛ 2⎞ VTH = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 ⇒ VTH = −3.571 V R1 + R2 ⎠ ⎝ 14 ⎠ ⎝ (b) VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 5 −3.57 = I BQ (1.714 ) + 0.7 + (101) I BQ ( 0.5 ) − 5 I BQ = 5 − 0.7 − 3.571 0.729 = ⇒ 13.96 μ A 1.714 + (101)( 0.5 ) 52.21 I CQ = 1.396 mA, I EQ = 1.410 mA VCEQ = 10 − (1.396 )( 5 ) − (1.41)( 0.5 ) ⇒ VCEQ = 2.32 V (d) 164. RE = 0.5 + 5% = 0.525 K I BQ = RC = 5 + 5% = 5.25 K 0.729 ⇒ 13.32 μ A 1.714 + (101)( 0.525 ) I CQ = 1.332 mA I EQ = 1.345 mA VCEQ = 10 − (1.332 )( 5.25 ) − (1.345 )( 0.525 ) = 10 − 6.993 − 0.7061 ⇒ VCEQ = 2.30 V RE = 0.5 + 5% = 0.525 K I CQ = 1.332 mA RC = 5 − 5% = 4.75 K I EQ = 1.345 mA VCEQ = 10 − (1.332 )( 4.75 ) − (1.345 )( 0.525 ) = 10 − 6.327 − 0.7061 ⇒ VCEQ = 2.97 V RE = 0.5 − 5% = 0.475 K I BQ = RC = 5 + 5% = 5.25 K 0.729 ⇒ 14.67 μ A 1.714 + (101)( 0.475 ) I CQ = 1.467 mA I EQ = 1.482 mA VCEQ = 10 − (1.467 )( 5.25 ) − (1.482 )( 0.475 ) = 10 − 7.70175 − 0.70395 ⇒ VCEQ = 1.59 V RE = 0.5 − 5% = 0.475 K I CQ = 1.467 mA RC = 5 − 5% = 4.75 K I EQ = 1.482 mA VCEQ = 10 − (1.467 )( 4.75 ) − (1.482 )( 0.475 ) = 10 − 6.96825 − 0.70395 ⇒ VCEQ = 2.33 V 5.47 RTH = R1 R2 = 9 1 = 0.91 kΩ ⎛ R2 ⎞ ⎛ 1 ⎞ VTH = ⎜ ⎟ ( −12 ) = ⎜ ⎟ ( −12 ) = −1.2 V R1 + R2 ⎠ ⎝ 1+ 9 ⎠ ⎝ I EQ RE + VEB ( on ) + I BQ RTH + VTH = 0 I BQ = −VTH − VEB (on) 1.2 − 0.7 = RTH + (1 + β ) RE 0.90 + ( 76 )( 0.1) I BQ = 0.0588, I CQ = 4.41 mA I EQ = 4.47 mA Center of load line ⇒ VECQ = 6 V I EQ RE + VECQ + I CQ RC − 12 = 0 ( 4.47 ) ( 0.1) + 6 + ( 4.41) RC = 12 ⇒ RC = 1.26 kΩ 5.48 (a) RTH = 36 68 = 23.5 K ⎛ 36 ⎞ VTH = ⎜ ⎟ (10 ) = 3.46 V ⎝ 36 + 68 ⎠ 3.46 − 0.7 I BQ = = 0.00178 mA 23.5 + ( 51)( 30 ) I CQ = 0.0888 mA I EQ = 0.0906 mA VCE = 10 − ( 0.0888 )( 42 ) − ( 0.0906 )( 30 ) = 10 − 3.73 − 2.72 ⇒ VCE = 3.55 V 165. (b) R1 = 22.7, R2 = 12 K, RC = 14 K, RE = 10 K RTH = 7.85 k VTH = 3.46 3.46 − 0.7 I BQ = = 0.00533 mA 7.85 + ( 51)(10 ) I CQ = 0.266 mA I EQ = 0.272 mA VCE = 10 − ( 0.266 )(14 ) − ( 0.272 )(10 ) VCE = 3.56 V 5.49 (a) ⎛ R2 ⎞ ⎛ 68 ⎞ RTH = 36 68 = 23.5 K VTH = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 = 1.54 V R1 + R2 ⎠ ⎝ 36 + 68 ⎠ ⎝ 5 = ( 51) I BQ ( 30 ) + 0.7 + I B ( 23.5 ) + 1.54 I BQ = 2.76 = 1.78 μA ⇒ I CQ = 0.0888 mA 1553.5 I EQ = 0.0906 mA VECQ = 10 − ( 0.0906 )( 30 ) − ( 0.0888 )( 42 ) = 10 − 2.718 − 3.7296 ⇒ VECQ = 3.55 V (b) RTH = 12 22.7 = 7.85 K VTH = 1.54 RE = 10 K RC = 14 K 5 = ( 51) I BQ (10 ) + 0.7 + I B ( 7.85 ) + 1.54 I BQ = 2.76 = 5.33 μ A I CQ = 0.266 mA 517.85 I EQ = 0.272 mA VECQ = 10 − ( 0.272 )(10 ) − ( 0.266 )(14 ) = 10 − 2.72 − 3.724 ⇒ VECQ = 3.56 V 5.50 (a) RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.5 ) = 5.05 k Ω 1 VTH = ⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE R1 I BQ = I CQ β = 0.8 = 0.008 mA 100 1 Then ( 5.05)(10 ) = ( 0.008)( 5.05) + 0.7 + (101)( 0.008)( 0.5) R1 or R1 = 44.1 k Ω, 44.1R2 = 5.05 ⇒ R2 = 5.70 k Ω 44.1 + R2 ⎛ 101 ⎞ Now I EQ = ⎜ ⎟ ( 0.8 ) = 0.808 mA ⎝ 100 ⎠ VCC = I CQ RC + VCEQ + I EQ RE 10 = ( 0.8 ) RC + 5 + ( 0.808 )( 0.5 ) RC = 5.75 k Ω (b) For 75 ≤ β ≤ 150 166. ⎛ R2 ⎞ ⎛ 5.7 ⎞ VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ (10 ) = 1.145 V R1 + R2 ⎠ ⎝ 5.7 + 44.1 ⎠ ⎝ VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE For β = 75, I BQ = 1.145 − 0.7 = 0.0103 mA 5.05 + ( 76 )( 0.5 ) Then I CQ = ( 75 )( 0.0103) = 0.775 mA For β = 150, I BQ = 1.145 − 0.7 = 0.00552 mA 5.05 + (151)( 0.5 ) Then I CQ = 0.829 mA % Change = ΔI CQ = I CQ 0.829 − 0.775 × 100% ⇒ % Change = 6.75% 0.80 For RE = 1 k Ω (c) RTH = ( 0.1)(101)(1) = 10.1 k Ω VTH = 1 1 ⋅ RTH ⋅ VCC = (10.1)(10 ) = ( 0.008 )(10.1) + 0.7 + (101)( 0.008 )(1) R1 R1 which yields R1 = 63.6 k Ω And 63.6 R2 = 10.1 ⇒ R2 = 12.0 k Ω 63.6 + R2 ⎛ R2 ⎞ ⎛ 12 ⎞ Now VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ (10 ) = 1.587 V R1 + R2 ⎠ ⎝ 12 + 63.6 ⎠ ⎝ 1.587 − 0.7 = 0.0103 mA For β = 75, I BQ = 10.1 + ( 76 )(1) So I CQ = 0.773 mA For β = 150, I BQ = 1.587 − 0.7 = 0.00551 mA 10.1 + (151)(1) Then I CQ = 0.826 mA % Change = ΔI CQ I CQ = 0.826 − 0.773 × 100% ⇒ % Change = 6.63% 0.8 5.51 VCC ≅ I CQ ( RC + RE ) + VCEQ 10 = ( 0.8 )( RC + RE ) + 5 ⇒ RC + RE = 6.25 k Ω Let RE = 0.875 k Ω Then, for bias stable RTH = ( 0.1)(121)( 0.875 ) = 10.6 k Ω I BQ = 0.8 = 0.00667 mA 120 1 (10.6 )(10 ) = ( 0.00667 )(10.6 ) + 0.7 + (121)( 0.00667 )( 0.875 ) R1 So R1 = 71.8 k Ω and 71.8R2 = 10.6 ⇒ R2 = 12.4 k Ω 71.8 + R2 10 = 0.119 mA 71.8 + 12.4 This is close to the design specification. Then I R ≅ 5.52 167. = VCC − I CQ ( RC + RE ) I CQ ≈ I EQ ⇒ VCEQ 6 = 12 − I CQ ( 2 + 0.2 ) I CQ = 2.73 mA, I BQ = 0.0218 mA VCEQ = 6 V VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 6 ⎛ R2 ⎞ VTH = ⎜ ⎟ (12 ) − 6, ⎝ R1 + R2 ⎠ RTH = R 1 R2 Bias stable ⇒ RTH = ( 0.1)(1 + β ) RE = ( 0.1)(126 )( 0.2 ) = 2.52 kΩ ⎛ 1⎞ VTH = ⎜ ⎟ ( RTH )(12 ) − 6 ⎝ R1 ⎠ 1 ( 2.52 )(12 ) − 6 = ( 0.0218)( 2.52 ) + 0.7 + (126 )( 0.0218)( 0.2 ) − 6 R1 1 ( 30.24 ) = 0.7549 + 0.5494 R1 R1 = 23.2 kΩ, 23.2R 2 = 2.52 23.2 + R 2 R2 = 2.83 kΩ 5.53 a. ⎛ 81 ⎞ I CQ = 1 mA. I EQ = ⎜ ⎟ (1) = 1.01 mA ⎝ 80 ⎠ VCEQ = 12 − (1)( 2 ) − (1.01)( 0.2 ) ⇒ VCEQ = 9.80 V 1 = 0.0125 mA 80 = + ( 0.1) (1 + β ) RE = ( 0.1)( 81)( 0.2 ) = 1.62 kΩ I BQ = RTH ⎛ R2 ⎞ 1 1 VTH = ⎜ ( RTH )(12 ) − 6 = (19.44 ) − 6 ⎟ (12 ) − 6 = R1 + R2 ⎠ R1 R1 ⎝ VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 6 1 (19.44 ) − 6 = ( 0.0125 )(1.62 ) + 0.7 + ( 81)( 0.0125)( 0.2 ) − 6 R1 1 (19.44 ) = 0.923 R1 R1 = 21.1 kΩ, 21.1R2 = 1.62 21.1 + R2 R2 = 1.75 kΩ b. R1 = 22.2 kΩ or R1 = 20.0 kΩ R2 = 1.84 kΩ or R2 = 1.66 kΩ R2 ( max ) , R1 ( min ) RTH = (1.84 ) ( 20.0 ) = 1.685 kΩ ⎛ 1.84 ⎞ VTH = ⎜ ⎟ (12 ) − 6 = −4.99 V ⎝ 1.84 + 20.0 ⎠ −4.99 − 0.7 − ( −6 ) 0.31 = = 0.0173 mA I BQ = 1.685 + ( 81)( 0.2 ) 17.89 I CQ = 1.39 mA, I EQ = 1.40 mA 168. For max, RC ⇒ VCE = 12 − (1.39 )( 2 ) − (1.40 )( 0.2 ) VCE = 8.94 V R2 ( min ) , R1 ( max ) RTH = (1.66 ) ( 22.2 ) = 1.545 kΩ ⎛ 1.66 ⎞ VTH = ⎜ ⎟ (12 ) − 6 = −5.165 V ⎝ 1.66 + 22.2 ⎠ −5.165 − 0.7 + 6 0.135 = = 0.00761 mA ⇒ I CQ = 0.609 mA, I E = 0.616 I BQ = 1.545 + ( 81)( 0.20 ) 17.745 VCEQ = 12 − ( 0.609 )( 2 ) − ( 0.616 )( 0.2 ) VCEQ = V 10.7 V So 0.609 ≤ I C ≤ 1.39 mA 8.94 ≤ VCEQ ≤ 10.7 V 5.54 VCEQ ≅ VCC − I CQ ( RC + RE ) 5 = 12 − 3 ( RC + RE ) ⇒ RC + RE = 2.33 k Ω Let RE = 0.333 k Ω and RC = 2 k Ω Nominal value of β = 100 RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.333) = 3.36 k Ω 3 = 0.03 mA 100 1 1 = ⋅ RTH ⋅ (12 ) − 6 = ( 3.36 )(12 ) − 6 R1 R1 I BQ = VTH Then VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 6 1 ( 3.36 )(12 ) − 6 = ( 0.03)( 3.36 ) + 0.7 + (101)( 0.03)( 0.333) − 6 R1 which yields R1 = 22.3 k Ω and R2 = 3.96 k Ω ⎛ R2 ⎞ ⎛ 3.96 ⎞ Now VTH = ⎜ ⎟ (12 ) − 6 = ⎜ ⎟ (12 ) − 6 or VTH = −4.19 V ⎝ 3.96 + 22.3 ⎠ ⎝ R1 + R2 ⎠ For β = 75, VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 6 I BQ = VTH + 6 − 0.7 −4.19 + 6 − 0.7 = = 0.0387 mA ⇒ I C = 2.90 mA RTH + (1 + β ) RE 3.36 + ( 76 )( 0.333) For β = 150, I BQ = −4.19 + 6 − 0.7 = 0.0207 mA 3.36 + (151)( 0.333) Then I C = 3.10 mA Specifications are met. 5.55 RTH = R1 R2 = 3 12 = 2.4 k Ω ⎛ R2 ⎞ ⎛ 12 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ ( 20 ) = 16 V ⎝ 12 + 3 ⎠ ⎝ R1 + R2 ⎠ (a) For β = 75 20 = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH 20 − 0.7 − 16 = I BQ ⎡( 76 )( 2 ) + 2.4 ⎤ ⎣ ⎦ So I BQ = 0.0214 mA, I CQ = 1.60 mA, I EQ = 1.62 E VECQ = 20 − (1.6 )(1) − (1.62 )( 2 ) or VECQ = 15.16 V 169. (b) For β = 100, we find I BQ = 0.0161 mA, I CQ = 1.61 mA, VECQ = 15.13 V , I EQ = 1.63 mA 5.56 I CQ = 4.8 mA → I EQ = 4.84 mA VCEQ = VCC − I CQ RC − I EQ RE 6 = 18 − ( 4.8 )( 2 ) − ( 4.84 ) RE ⇒ RE = 0.496 kΩ RTH = ( 0.1)(1 + β ) RE = ( 0.1)(121)( 0.496 ) = 6.0 kΩ VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE I BQ = 0.040 mA VTH = 1 1 ⋅ RTH ⋅VCC = ( 6.0 )(18 ) R1 R1 1 ( 6.0 )(18) = ( 0.04 )( 6.0 ) + 0.70 + (121)( 0.04 )( 0.496 ) R1 1 (108 ) = 3.34 R1 R1 = 32.3 kΩ, 32.3 R2 = 6.0 32.3 + R2 R2 = 7.37 kΩ 5.57 For nominal β = 70 2 = 0.0286 mA → I EQ = 2.03 mA 70 = VCC − I CQ RC − I EQ RE I BQ = VCEQ 10 = 20 − ( 2 )( 4 ) − ( 2.03) RE ⇒ RE = 0.985 K RTH = ( 0.1)(1 + β ) RE = ( 0.1)( 71)( 0.985 ) = 6.99 K VTH = I BQ RTH + VBE ( on ) + I EQ RE 1 ⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + I EQ RE R1 1 ( 6.99 )( 20 ) = ( 0.0286 )( 6.99 ) + 0.70 + ( 2.03)( 0.985 ) R1 1 (139.8 ) = 2.90 R1 R1 = 48.2 K, 48.2 R2 = 6.99 48.2 + R2 R2 = 8.18 K Check: For β = 50 ⎛ 8.18 ⎞ VTH = ⎜ ⎟ ( 20 ) = 2.90 ⎝ 8.18 + 48.2 ⎠ V − VBE ( on ) 2.90 − 0.7 I BQ = TH = = 0.0384 mA RTH + (1 + β ) RE 6.99 + ( 51)( 0.985 ) I CQ = 1.92 mA For β = 90 I BQ = 2.90 − 0.7 = 0.0228 mA 6.99 + ( 91)( 0.985 ) I CQ = 2.05 mA Design criterion is satisfied. 170. 5.58 I CQ = 1 mA → I EQ = 1.017 mA VCEQ = VCC − I CQ RC − I EQ RE 5 = 15 − (1)( 5 ) − (1.017 ) RE ⇒ RE = 4.92 kΩ Bias stable: RTH = ( 0.1)(1 + β ) RE = ( 0.1)( 61)( 4.92 ) = 30.0 kΩ 1 = 0.0167 mA I BQ = 60 1 VTH = ⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + I EQ RE R1 1 ( 30.0 )(15 ) = ( 0.0167 )( 30.0 ) + 0.70 + (1.017 )( 4.92 ) R1 1 ( 448.5) = 6.197 R1 R1 = 72.5 kΩ, 72.5 R2 = 30.0 72.5 + R2 R2 = 51.2 kΩ Check: For β = 45 ⎛ 51.2 ⎞ VTH = ⎜ ⎟ (15 ) = 6.21V ⎝ 51.2 + 72.5 ⎠ V − VBE ( on ) 6.21 − 0.7 I BQ = TH = = 0.0215 mA RTH + (1 + β ) RE 30 + ( 46 )( 4.92 ) I CQ = 0.967 mA, ΔI C = 3.27% IC Check: For β = 75 6.21 − 0.7 I BQ = = 0.0136 mA 30.0 + ( 76 )( 4.92 ) ΔI C = 2.31% IC Design criterion is satisfied. I CQ = 1.023 mA, 5.59 (a) VCC ≅ I CQ ( RC + RE ) + VCEQ 3 = ( 0.1)( 5 RE + RE ) + 1.4 ⇒ RE = 2.67 k Ω 100 = 0.833 μ A 120 = ( 0.1)(1 + β ) RE = ( 0.1)(121)( 2.67 ) = 32.3 k Ω RC = 13.3 k Ω, I BQ = RTH VTH = 1 1 ⋅ RTH ⋅ VCC = ( 32.3)( 3) R1 R1 = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE = ( 0.000833)( 32.3) + 0.7 + (121)( 0.000833)( 2.67 ) which gives R1 = 97.3 k Ω, and R2 = 48.4 k Ω (b) 171. IR ≅ 3 3 = ⇒ 20.6 μ A R1 + R2 97.3 + 48.4 I CQ = 100 μ A P = ( I CQ + I R )VCC = (100 + 20.6 )( 3) or P = 362 μW 5.60 IE = 5 − VE 5 = = 1.67 mA RE 3 RTH = R1 || R2 = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 3) = 30.3 kΩ ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ ( 4 ) − 2 = ⋅ RTH ⋅ ( 4 ) − 2 R1 + R2 ⎠ R1 ⎝ I EQ = 0.0165 mA I BQ = 1+ β 5 = I EQ RE + VEB ( on ) + I B RTH + VTH 1 5 = (1.67 )( 3) + 0.7 + ( 0.0165 )( 30.3) + ( 30.3)( 4 ) − 2 R1 0.80 = 1 ( 30.3)( 4 ) ⇒ R1 = 152 kΩ R1 152 R2 = 30.3 ⇒ R2 = 37.8 kΩ 152 + R2 5.61 a. RTH = R1 R2 = 10 20 ⇒ RTH = 6.67 kΩ ⎛ R2 ⎞ ⎛ 20 ⎞ VTH = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 ⇒ VTH = 1.67 V ⎝ 20 + 10 ⎠ ⎝ R1 + R2 ⎠ b. 10 = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH 10 − 0.7 − 1.67 7.63 I BQ = = ⇒ I BQ = 0.0593 mA 6.67 + ( 61)( 2 ) 128.7 I CQ = 3.56 mA, I EQ =3.62 mA VE = 10 − I EQ RE = 10 − ( 3.62 )( 2 ) VE = 2.76 V VC = I CQ RC − 10 = ( 3.56 )( 2.2 ) − 10 VC = −2.17 V 5.62 V + − V − ≅ I CQ ( RC + RE ) + VECQ 20 = ( 0.5 )( RC + RE ) + 8 ⇒ ( RC + RE ) = 24 k Ω Let RE = 10 k Ω then RC = 14 k Ω Let β = 60 from previous problem. RTH = ( 0.1)(1 + β ) RE = ( 0.1)( 61)(10 ) Or RTH = 61 k Ω 172. 0.5 = 0.00833 mA 60 ⎛ R2 ⎞ 1 =⎜ ⎟ (10 ) − 5 = ⋅ RTH ⋅10 − 5 R1 ⎝ R1 + R2 ⎠ I BQ = VTH Now 10 = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH 10 = ( 61)( 0.00833)(10 ) + 0.7 + ( 0.00833) ( 61) + 1 ( 61) (10 ) − 5 R1 Then R1 = 70.0 k Ω and R2 = 474 k Ω 10 10 = ⇒ 18.4 μ A R1 + R2 70 + 474 So the 40 μ A current limit is met. IR ≅ 5.63 a. RTH = R1 R2 = 35 20 ⇒ RTH = 12.7 kΩ ⎛ R2 ⎞ ⎛ 20 ⎞ VTH = ⎜ ⎟ (7) − 5 = ⎜ ⎟ ( 7 ) − 5 ⇒ VTH = −2.45 V R1 + R2 ⎠ ⎝ 20 + 35 ⎠ ⎝ b. I BQ = = VTH − VBE ( on ) − ( −10 ) RTH + (1 + β ) RE −2.45 − 0.7 + 10 ⇒ I BQ = 0.136 mA 12.7 + ( 76 )( 0.5 ) I CQ = 10.2 mA, I EQ = 10.4 mA VCEQ = 20 − I CQ RC − I EQ RE = 20 − (10.2 )( 0.8 ) − (10.4 )( 0.5 ) VCEQ = 6.64 V c. R2 = 20 + 5% = 21 kΩ R1 = 35 − 5% = 33.25 kΩ RE = 0.5 − 5% = 0.475 kΩ RTH = R1 R2 = 21 33.25 = 12.9 kΩ ⎛ R2 ⎞ VTH = ⎜ ⎟ (7) − 5 ⎝ R1 + R2 ⎠ 21 ⎛ ⎞ =⎜ ⎟ ( 7 ) − 5 = −2.29 V ⎝ 21 + 33.25 ⎠ I BQ = −2.29 − 0.7 − ( −10 ) 12.9 + ( 76 )( 0.475 ) = 0.143 mA I CQ = 10.7 mA, I EQ = 10.9 mA For RC = 0.8 + 5% = 0.84 kΩ VCEQ = 20 − (10.7 )( 0.84 ) − (10.9 )( 0.475 ) ⇒ VCEQ = 5.83 V For RC = 0.8 − 5% = 0.76 kΩ 173. VCEQ = 20 − (10.7 )( 0.76 ) − (10.9 )( 0.475 ) ⇒ VCEQ = 6.69 V R2 = 20 − 5% = 19 kΩ R1 = 35 + 5% = 36.75 kΩ RE = 0.5 + 5% = 0.525 kΩ RTH = R1 R2 = 19 36.75 = 12.5 kΩ 19 ⎛ ⎞ VTH = ⎜ ⎟ ( 7 ) − 5 = −2.61 V ⎝ 19 + 36.75 ⎠ −2.61 − 0.7 − ( −10 ) = 0.128 mA I BQ = 12.5 + ( 76 )( 0.525 ) I CQ = 9.58 mA, I EQ = 9.70 mA For RC = 0.84 kΩ VCEQ = 20 − ( 9.58 )( 0.84 ) − ( 9.70 )( 0.525 ) ⇒ VCEQ = 6.86 V For RC = 0.76 kΩ VCEQ = 20 − ( 9.58 )( 0.76 ) − ( 9.70 )( 0.525 ) ⇒ VCEQ = 7.63 V So 9.58 ≤ I CQ ≤ 10.7 mA and 5.83 ≤ VCEQ ≤ 7.63 V 5.64 a. RTH = 500 kΩ 500 kΩ 70 kΩ = 250 kΩ 70 kΩ ⇒ RTH = 54.7 kΩ 5 − VTH 3 − VTH VTH − ( −5 ) + = 500 500 70 5 3 5 1 1 ⎞ ⎛ 1 + − = VTH ⎜ + + ⎟ − 0.0554 = VTH ( 0.0183) 500 500 70 ⎝ 500 500 70 ⎠ VTH = −3.03 V b. I BQ = = VTH − VBE ( on ) − ( −5 ) RTH + (1 + β ) RE −3.03 − 0.7 + 5 54.7 + (101)( 5 ) I BQ = 0.00227 mA I CQ = 0.227 mA, I EQ = 0.229 VCEQ = 20 − ( 0.227 )( 50 ) − ( 0.229 )( 5 ) VCEQ = 7.51 V 5.65 RTH = 30 || 60 || 20 ⇒ RTH = 10 kΩ 5 − VTH 5 − VTH VTH + = 30 60 20 5 ⎞ 1 1 ⎞ ⎛ 5 ⎛ 1 ⎜ + ⎟ = VTH ⎜ + + ⎟ ⎝ 30 60 ⎠ ⎝ 30 60 20 ⎠ VTH = 2.5 V For β = 100 174. I BQ = = VTH − VBE ( on ) − ( −5 ) RTH + (1 + β ) RE 2.5 − 0.7 + 5 10 + (101)( 0.2 ) I BQ = 0.225 mA I CQ = 22.5 mA, I EQ = 22.7 mA VCEQ = 15 − ( 22.5 )( 0.5 ) − ( 22.7 ) ( 0.2 ) VCEQ = −0.79! In saturation ⇒ VCEQ = 0.2 V VTH = I BQ RTH + VBE + I EQ RE − 5 2.5 + 5 − 0.7 = I BQ (10 ) + I EQ ( 0.2 ) 6.8 = I BQ (10 ) + I EQ ( 0.2 ) 14.8 = I CQ ( 0.5 ) + I EQ ( 0.2 ) Transistor in saturation, I EQ = I BQ + I CQ 6.8 = I BQ (10 ) + I BQ ( 0.2 ) + I CQ ( 0.2 ) 6.8 = I BQ (10.2 ) + I CQ ( 0.2 ) 51× 14.8 = I BQ ( 0.2 ) + I CQ ( 0.7 ) 754.8 = I BQ (10.2 ) + I CQ ( 35.7 ) 748 = I CQ ( 35.5 ) I CQ = 21.1 mA VCEQ = 0.2 V 5.66 I CQ = 50 μ A, I BQ = 0.625 μ A, I EQ = 50.6 μ A (a) 1 = 19.8 K 0.0506 5 = ( 0.050 ) RC + 5 + ( 0.0506 )(19.8 ) − 5 RE = RC = 80 K RTH = R1 R2 Design bias stable circuit. RTH = ( 0.1)( 51)(19.8 ) = 101 K ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ (10 ) − 5 = ⋅ RTH ⋅ (10 ) − 5 R1 ⎝ R1 + R2 ⎠ 1 So (101)(10 ) − 5 = I BQ (101) + 0.7 + ( 0.0506 )(19.8 ) − 5 R1 1 (1010 ) = 0.0631 + 0.7 + 1 R1 R1 = 573 K R2 = 123 K (b) 573 R2 = 101 573 + R2 175. RTH = 101 K, VTH = −3.23 V VTH = I BQ RTH + 0.7 + (121)(19.8 ) I BQ − 5 1.07 = I BQ (101 + 2395.8 ) ⇒ I BQ = 0.429 μ A I CQ = 0.0514 mA, I EQ = 0.0519 mA VCEQ = 10 − ( 0.0514 )( 80 ) − ( 0.0519 )(19.8 ) = 10 − 4.11 − 1.03 ⇒ VCEQ = 4.86 V 5.67 (a) 2 = 2.5 K 0.8 + I EQ RE I EQ = 0.80 mA, RE = 12 = I CQ RC + VCEQ 12 = ( 0.8 ) RC + 7 + 2 ⇒ RC = 3.75 K VTH = I BQ RTH + VBE + I EQ RE For a bias stable circuit RTH = ( 0.1)(12.1)( 2.5 ) = 30.25 K 1 1 ⋅ RTH ⋅ VCC = ( 30.25 )(12 ) = ( 0.00667 )( 30.25 ) + 0.7 + 2 R1 R1 1 ( 363) = 2.90 ⇒ R1 = 125 K R1 125 R2 = 30.25 ⇒ R2 = 39.9 K 125 + R2 (b) Let RE = 2.4 K, RC = 3.9 K R1 = 120 K R2 = 39 K Then RTH = R1 R2 = 120 39 = 29.4 K ⎛ 39 ⎞ VTH = ⎜ ⎟ (12 ) = 2.94 V ⎝ 120 + 39 ⎠ 2.94 − 0.7 2.24 = I BQ = ⇒ 7.00 μ A 29.4 + (121)( 2.4 ) 319.8 I CQ = 0.841 mΑ I EQ = 0.848 mA VCEQ = 12 − ( 0.841)( 3.9 ) − ( 0.848 )( 2.4 ) = 12 − 3.28 − 2.04 VCEQ = 6.68 V 5.68 (a) I CQ = 100 μ A, I BQ = 1.18 μ A, I EQ = 101 μ A 2 ⇒ RE = 19.8 K 0.101 9 = ( 0.101)(19.8 ) + 6 + ( 0.1) RC − 9 RE = RC = 100 K 176. Design a bias stable circuit. BTH = ( 0.1)( 86 )(19.8 ) = 170 K ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ (18 ) − 9 = (170 )(18 ) − 9 R1 + R2 ⎠ R1 ⎝ 9 = ( 0.101)(19.8 ) + 0.7 + ( 0.00118 )(170 ) + 1 (170 )(18 ) − 9 R1 203R2 = 170 203 + R2 R1 = 203 K R2 = 1046 K (b) β = 125 9 = (126 ) I BQ (19.8 ) + 0.7 + I BQ (170 ) + (15.07 − 9 ) 2.23 = 0.837 μ A 2664.8 = 0.1054 mA I BQ = I EQ I CQ = 0.1046 mA VECQ = 18 − ( 0.1046 )(100 ) − ( 0.1054 )(19.8 ) VECQ = 5.45 V 5.69 (a) 3 ⎛ 51 ⎞ I EQ ≈ ⎜ ⎟ ( 20 ) = 20.4 mA ⇒ RE = = 0.147 K 20.4 ⎝ 50 ⎠ 6 = 0.3 K VRC = 18 − 9 − 3 = 6 V RC = 20 For a bias stable circuit. RTH = ( 0.1)( 51)( 0.147 ) = 0.750 K V + = I EQ RE + VEB + I BQ RTH + VTH 18 = 3 + 0.7 + ( 0.4 )( 0.75 ) + 14 = 1 ( 0.75 )(18) R1 1 (13.5) ⇒ R1 = 0.964 K R1 ( 0.964 ) R2 0.964 + R2 = 0.75 ⇒ R2 = 3.38 K (b) Let RE = 0.15 K, RC = 0.3 K R1 = 1.0 K, R2 = 3.3 K ⎛ 3.38 ⎞ RTH = 1/13.3 = 0.767 K, VTH = ⎜ ⎟ (18 ) = 13.8 V ⎝ 1 + 3.38 ⎠ 18 − 13.8 − 0.7 3.5 = = 0.416 mA I BQ = 0.767 + ( 51)( 0.15 ) 8.417 I CQ = 20.8 mA, I EQ = 21.2 mA VECQ = 18 − ( 20.8 )( 0.3) − ( 21.2 )( 0.15 ) = 18 − 6.24 − 3.18 VECQ = 8.58 V 5.70 177. RTH = R1 R2 = 100 40 = 28.6 kΩ ⎛ R2 ⎞ ⎛ 40 ⎞ VTH = ⎜ ⎟ (10 ) = ⎜ ⎟ (10 ) = 2.86 V ⎝ 40 + 100 ⎠ ⎝ R1 + R2 ⎠ VTH − VBE ( on ) 2.86 − 0.7 I B1 = = RTH + (1 + β ) RE1 28.6 + (121) (1) I B1 = 0.0144 mA I C1 = 1.73 mA, I E1 = 1.75 mA 10 − VB 2 = I C1 + I B 2 3 VB 2 − VBE ( on ) − ( −10 ) IE2 = 5 10 − VB 2 VB 2 − 0.7 + 10 = I C1 + 3 (121) ( 5 ) ⎛1 ⎞ 10 9.3 1 − 1.73 − = VB 2 ⎜ + ⎜ 3 (121) ( 5 ) ⎟ ⎟ 3 605 ⎝ ⎠ 1.588 = VB 2 ( 0.335 ) ⇒ VB 2 = 4.74 V IE2 = I B2 4.74 − 0.7 − ( −10 ) 5 = 0.0232 mA ⇒ I E 2 = 2.808 mA I C 2 = 2.785 mA VCEQ1 = 4.74 − (1.75 ) (1) ⇒ VCEQ1 = 2.99 V VCEQ 2 = 10 − ( 4.74 − 0.7 ) ⇒ VCEQ 2 = 5.96 V 5.71 VE1 = −0.7 I R1 = VE 2 −0.7 − ( −5 ) = 0.215 mA 20 = −0.7 − 0.7 = −1.4 IE2 = −1.4 − ( −5 ) 1 ⇒ I E 2 = 3.6 mA I B 2 = 0.0444 mA I C 2 = 3.56 mA I E1 = I R1 + I B 2 = 0.215 + 0.0444 I E1 = 0.259 mA I B1 = 0.00320 mA I C1 = 0.256 mA 5.72 Current through V − source = I E1 + I E 2 and I E1 = I E 2 = (1 + β ) I B1 = ( 51)( 8.26 ) μ A So total current = 2 ( 51)( 8.26 ) μ A=843 μ A P − = I ⋅ V − = ( 0.843)( 5 ) ⇒ P − = 4.22 mW (From V − source) From Example 5.19, I Q = 0.413 mA ⎛ 50 ⎞ So I C 0 = ⎜ ⎟ ( 0.413) = 0.405 mA ⎝ 51 ⎠ 178. P + = I ⋅ V + = ( 0.405 )( 5 ) ⇒ P + = 2.03 mW (From V + source) 5.73 RTH = R1 R2 = 50 100 = 33.3 kΩ ⎛ R2 ⎞ VTH = ⎜ ⎟ (10 ) − 5 ⎝ R1 + R2 ⎠ ⎛ 100 ⎞ =⎜ ⎟ (10 ) − 5 = 1.67 V ⎝ 100 + 50 ⎠ 5 = I E1 RE1 + VEB ( on ) + I B1 RTH + VTH ⎛ 101 ⎞ I E1 = ⎜ ⎟ ( 0.8 ) = 0.808 mA ⎝ 100 ⎠ I B1 = 0.008 mA 5 = ( 0.808 ) RE1 + 0.7 + ( 0.008 )( 33.3) + 1.67 RE1 = 2.93 kΩ VE1 = 5 − ( 0.808 ) ( 2.93) = 2.63 V VC1 = VE1 − VECQ1 = 2.63 − 3.5 = −0.87 V VE 2 = −0.87 − 0.70 = −1.57 V IE2 = −1.57 − ( −5 ) RE 2 = 0.808 ⇒ RE 2 = 4.25 kΩ VCEQ 2 = 4 ⇒ VC 2 = −1.57 + 4 = 2.43 V 5 − 2.43 RC 2 = ⇒ RC 2 = 3.21 kΩ 0.8 I RC1 = I C1 − I B 2 = 0.8 − 0.008 = 0.792 mA RC1 = −0.87 − ( −5 ) 0.792 ⇒ RC1 = 5.21 kΩ 179. Chapter 6 Exercise Problems EX6.1 I BQ = (a) VBB − VBE ( on ) RB = 2 − 0.7 ⇒ 2 μA 650 Then I CQ = β I BQ = (100 )( 2 μ A ) = 0.20 mA VCEQ = VCC − I CQ RC = 5 − ( 0.2 )(15 ) = 2 V (b) Now I CQ 0.20 gm = = = 7.69 mA / V VT 0.026 and β VT (100 )( 0.026 ) rπ = = = 13 k Ω 0.20 I CQ (c) We find ⎛ r ⎞ Av = − g m RC ⎜ π ⎟ ⎝ rπ + RB ⎠ ⎛ 13 ⎞ = − ( 7.69 )(15 ) ⎜ ⎟ = −2.26 ⎝ 13 + 650 ⎠ EX6.2 V − VBE ( on ) 0.92 − 0.7 I BQ = BB = 100 RB or I BQ = 0.0022 mA and I CQ = β I BQ = (150 )( 0.0022 ) = 0.33 mA (a) gm = rπ = ro = I CQ VT β VT I CQ = 0.33 = 12.7 mA / V 0.026 = (150 )( 0.026 ) 0.33 = 11.8 k Ω VA 200 = = 606 k Ω I CQ 0.33 (b) vo = − g m vπ ( ro ⎛ rπ ⎞ RC ) and vπ = ⎜ ⎟ ⋅ vs ⎝ rπ + RB ⎠ so ⎛ rπ ⎞ vo = − gm ⎜ ⎟ ( ro RC ) vπ ⎝ rπ + RB ⎠ ⎛ 11.8 ⎞ = − (12.7 ) ⎜ ⎟ ( 606 15 ) ⎝ 11.8 + 100 ⎠ which yields Av = −19.6 EX6.3 (a) V − VEB ( on ) 1.145 − 0.7 I BQ = BB = 50 RB Av = or I BQ = 0.0089 mA Then I CQ = β I BQ = ( 90 )( 0.0089 ) = 0.801 mA 180. Now gm = rπ = ro = I CQ VT β VT I CQ = 0.801 = 30.8 mA / V 0.026 = ( 90 )( 0.026 ) 0.801 = 2.92 k Ω VA 120 = = 150 k Ω I CQ 0.801 We have Vo = g mVπ ( ro RC ) (b) ⎛ r ⎞ and Vπ = − ⎜ π ⎟ Vs ⎝ rπ + RB ⎠ so ⎛ rπ ⎞ V Av = o = − g m ⎜ ⎟ ( ro RC ) Vs ⎝ rπ + RB ⎠ ⎛ 2.92 ⎞ = − ( 30.8 ) ⎜ ⎟ (150 2.5 ) ⎝ 2.92 + 50 ⎠ which yields Av = −4.18 EX6.4 Using Figure 6.23 (a) For I CQ = 0.2 mA, 7.8 < hie < 15 k Ω, 60 < h fe < 125, 6.2 × 10−4 < hre < 50 × 10−4 , 5 < hoe < 13 μ mhos For I CQ = 5 mA, 0.7 < hie < 1.1 k Ω, 140 < h fe < 210, 1.05 × 10−4 < hre < 1.6 ×10−4 , (b) 22 < hoe < 35 μ mhos EX6.5 RTH = R1 R2 = 35.2 || 5.83 = 5 k Ω ⎛ R2 ⎞ ⎛ 5.83 ⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ (5) R1 + R2 ⎠ ⎝ 5.83 + 35.2 ⎠ ⎝ or VTH = 0.7105 V Then V − VBE ( on ) 0.7105 − 0.7 I BQ = TH = RTH 5 or I BQ = 2.1 μ A and I CQ = β I BQ = (100 )( 2.1 μ A ) = 0.21 mA VCEQ = VCC − I CQ RC = 5 − ( 0.21)(10 ) and VCEQ = 2.9 V Now gm = ro = I CQ VT = 0.21 = 8.08 mA 0.026 VA 100 = = 476 k Ω I CQ 0.21 And Av = − g m ( ro so Av = −79.1 EX6.6 RC ) = − ( 8.08 ) ( 476 10 ) 181. RTH = R1 R2 = 250 75 = 57.7 k Ω ⎛ R2 ⎞ ⎛ 75 ⎞ VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ (5) ⎝ 75 + 250 ⎠ ⎝ R1 + R2 ⎠ or VTH = 1.154 V I BQ = VTH − VBE ( on ) RTH + (1 + β ) RE = 1.154 − 0.7 57.7 + (121)( 0.6 ) or I BQ = 3.48 μ A I CQ = β I BQ = (120 )( 3.38 μ A ) = 0.418 mA (a) Now gm = rπ = I CQ VT = 0.418 = 16.08 mA / V 0.026 βVT (120 )( 0.026 ) = = 7.46 k Ω 0.418 I CQ We have Vo = − g mVπ RC We find Rib = rπ + (1 + β ) RE = 7.46 + (121)( 0.6 ) or Rib = 80.1 k Ω Also R1 R2 = 250 75 = 57.7 k Ω R1 R2 Rib = 57.7 80.1 = 33.54 k Ω We find ⎛ R1 R2 Rib ⎞ ⎛ 33.54 ⎞ Vs′ = ⎜ ⎟ ⋅ Vs = ⎜ ⎟ ⋅ Vs ⎝ 33.54 + 0.5 ⎠ ⎝ R1 R2 Rib + RS ⎠ or Vs′ = ( 0.985 ) Vs Now ⎡ ⎛1+ β ⎞ ⎤ ⎡ ⎛ 121 ⎞ ⎤ Vs′ = Vπ ⎢1 + ⎜ ⎟ RE ⎥ = Vπ ⎢1 + ⎜ ⎟ ( 0.6 ) ⎥ rπ ⎠ ⎥ ⎝ 7.46 ⎠ ⎢ ⎝ ⎣ ⎦ ⎣ ⎦ or Vπ = ( 0.0932 ) Vs′ = ( 0.0932 )( 0.985 ) Vs So Av = Vo = − (16.08 )( 0.0932 )( 0.985 )( 5.6 ) Vs or Av = −8.27 EX6.7 RTH = R1 R2 = 15 85 = 12.75 k Ω ⎛ R2 ⎞ ⎛ 85 ⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ (12 ) R1 + R2 ⎠ ⎝ 15 + 85 ⎠ ⎝ or VTH = 10.2 V Now VCC = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH 182. so 12 = (101) I BQ ( 0.5 ) + 0.7 + I BQ (12.75 ) + 10.2 which yields I BQ = 0.0174 mA and I CQ = β I BQ = (100 )( 0.0174 ) = 1.74 mA I EQ = (101)( 0.0174 ) = 1.76 mA VECQ = VCC − I EQ RE − I CQ RC = 12 − (1.76 )( 0.5 ) − (1.74 )( 4 ) or VECQ = 4.16 V Now rπ = β VT I CQ = (100 )( 0.026 ) 1.74 = 1.49 k Ω We have Vo = g mVπ ( RC || RL ) ⎛ V ⎞ Vs = −Vπ − ⎜ g mVπ + π ⎟ RE rπ ⎠ ⎝ Solving for Vπ and noting that β = g m rπ , we find Av = Vo Vs = = − β ( RC RL ) rπ + (1 + β ) RE − (100 )( 4 2 ) 1.49 + (101)( 0.5 ) or Av = −2.56 EX6.8 Dc analysis 10 − 0.7 I BQ = = 0.00439 mA 100 + (101)( 20 ) I CQ = 0.439 mA, I EQ = 0.443 mA Now rπ = (100 )( 0.026 ) = 5.92 k Ω 0.439 0.439 gm = = 16.88 mA / V 0.026 100 ro = = 228 k Ω 0.439 (a) Now Vo = − g mVπ ( ro RC ) ⎛ RB rπ ⎞ Vπ = ⎜ ⎟ ⋅ Vs ⎝ RB rπ + RS ⎠ RB rπ = 100 5.92 = 5.59 k Ω ⎛ 5.59 ⎞ Then Vπ = ⎜ ⎟ ⋅ Vs = ( 0.918 ) Vs ⎝ 5.59 + 0.5 ⎠ V Then Av = o = − (16.88 ) ( 0.918 ) ( 228 10 ) Vs or Av = −148 (b) Rin = RS + RB rπ = 0.5 + 100 5.92 = 6.09 k Ω 183. and Ro = RC ro = 10 228 = 9.58 k Ω EX6.9 (a) I CQ 0.25 gm = = = 9.615 mA / V 0.026 VT ro = VA 100 = = 400 k Ω I CQ 0.25 Av = − g m ( ro rc ) = − ( 9.615 )( 400 100 ) = −769 (b) Av = − g m ( ro rc rL ) = − ( 9.615 )( 400 100 100 ) Av = −427 EX6.10 5 − 0.7 I BQ = = 0.00672 mA 10 + (126 )( 5 ) I CQ = 0.84 mA, I EQ = 0.847 mA VCEQ = 10 − ( 0.84 )( 2.3) − ( 0.847 )( 5 ) or VCEQ = 3.83 V dc load line VCE ≅ (V + − V − ) − I C ( RC + RE ) or VCE = 10 − I C ( 7.3) ac load line (neglecting ro) vce = −ic ( RC RL ) = −ic ( 2.3 5 ) = −ic (1.58 ) IC (mA) 1.37 0.84 AC load line Q-point DC load line 3.83 10 V CE (V) EX6.11 (a) dc load line VEC ≅ VCC − I C ( RC + RE ) = 12 − I C ( 4.5 ) ac load line vec ≅ −ic ( RE + RC RL ) or vec = −ic ( 0.5 + 4 2 ) = −ic (1.83) Q-point values 12 − 0.7 − 10.2 = 0.0150 mA I BQ = 12.75 + (121)( 0.5 ) I CQ = 1.80 mA, I EQ = 1.82 mA VECQ = 3.89 V 184. IC (mA) AC load line 2.67 Q-point 1.80 DC load line 3.89 12 V (V) EC ΔiC = I CQ = 1.80 mA ΔvEC = (1.8 )(1.83) = 3.29 V (b) vEC ( min ) = 3.89 − 3.29 = 0.6 V So maximum symmetrical swing = 2 × 3.29 = 6.58 V peak-to-peak EX6.12 ⎛ R2 ⎞ 1 VTH = ⎜ (a) ⎟ ⋅ VCC = ⋅ RTH ⋅ VCC R1 ⎝ R1 + R2 ⎠ RTH = ( 0.1)(1 + β ) RE = ( 0.1)(121)(1) so RTH = 12.1 k Ω, VTH = 1 (12.1)(12 ) R1 We can write VCC = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH We have I CQ = 1.6 mA, I BQ = 1.6 = 0.0133 mA 120 Then 1 (12.1)(12 ) R1 12.1 + (121)(1) 12 − 0.7 − I BQ = 0.0133 = which yields R1 = 15.24 k Ω Since RTH = R1 R2 = 12.1 k Ω, we find R2 = 58.7 k Ω Also VECQ = 12 − (1.6 )( 4 ) − (1.61)(1) = 3.99 V (b) ac load line vec = −ic ( RC RL ) Want Δic = I CQ − 0.1 = 1.6 − 0.1 = 1.5 mA Also Δvec = 3.99 − 0.5 = 3.49 V Now Δvec 3.49 = = 2.327 k Ω = RC Δic 1.5 RL So 4 RL = 2.327 k Ω which yields RL = 5.56 k Ω EX6.13 185. RTH = R1 R2 = 25 50 = 16.7 k Ω ⎛ R2 ⎞ ⎛ 50 ⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ (5) ⎝ 25 + 50 ⎠ ⎝ R1 + R2 ⎠ or VTH = 3.33 V I BQ = VTH − VBE ( on ) RTH + (1 + β ) RE = 3.33 − 0.7 16.7 + (121)(1) or I BQ = 0.0191 mA Also I CQ = (120 )( 0.0191) = 2.29 mA Now gm = rπ = ro = I CQ VT = β VT I CQ 2.29 = 88.1 mA / V 0.026 = (120 )( 0.026 ) 2.29 = 1.36 k Ω VA 100 = = 43.7 k Ω I CQ 2.29 (a) ⎛ R1 R2 Rib ⎞ Vs′ = ⎜ ⎟ ⋅ Vs ⎝ R1 R2 Rib + RS ⎠ Rib = rπ + (1 + β ) ( RE ro ) = 1.36 + (121)(1 43.7 ) or Rib = 120 k Ω and R1 R2 = 16.7 k Ω Then R1 R2 Rib = 16.7 120 = 14.7 k Ω Now ⎛ 14.7 ⎞ Vs′ = ⎜ ⎟ ⋅ Vs = ( 0.967 ) Vs ⎝ 14.7 + 0.5 ⎠ and ⎛V ⎞ ⎛ 1+ β Vo = ⎜ π + g mVπ ⎟ ( RE ro ) = Vπ ⎜ ⎝ rπ ⎠ ⎝ rπ We have Vs′ = Vπ + Vo then ⎞ ⎟ RE ro ⎠ Vs′ ( 0.967 )Vs = ⎛ 1+ β ⎞ ⎛1+ β ⎞ 1+ ⎜ ⎟ RE ro 1 + ⎜ ⎟ RE ro rπ ⎠ ⎝ ⎝ rπ ⎠ We then obtain Vπ = V Av = o = Vs = ⎛1+ β ⎝ rπ ( 0.967 ) ⎜ ⎞ ⎟ RE ⎠ ⎛1+ β ⎞ 1+ ⎜ ⎟ RE ro ⎝ rπ ⎠ ( 0.967 )(1 + β ) RE ro rπ + (1 + β ) RE ro Now RE ro = 1 43.7 = 0.978 k Ω ro 186. Then Av = ( 0.967 )(121)( 0.978 ) = 0.956 1.36 + (121)( 0.978 ) (b) Rib = rπ + (1 + β )( RE ro ) or Rib = 1.36 + (121)( 0.978 ) = 120 k Ω EX6.14 For RS = 0, then r Ro = RE ro π 1+ β Using the parameters from Exercise 4.13, we obtain 1.36 Ro = 1 43.7 121 or Ro = 11.1 Ω EX6.15 (a) For I CQ = 1.25 mA and β = 100, we find I EQ = 1.26 mA and I BQ = 0.0125 mA Now VCEQ = 10 − I EQ RE Or 4 = 10 − (1.26 ) RE which yields RE = 4.76 k Ω Then RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 4.76 ) or RTH = 48.1 k Ω We have ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ (10 ) − 5 = ⋅ RTH (10 ) − 5 R1 ⎝ R1 + R2 ⎠ or VTH = 1 ( 481) − 5 R1 We can write I BQ = VTH − 0.7 − ( −5 ) RTH + (1 + β ) RE Or 1 ( 481) − 5 − 0.7 + 5 R1 0.0125 = 48.1 + (101)( 4.76 ) which yields R1 = 65.8 k Ω Since R1 R2 = 48.1 k Ω, we obtain R2 = 178.8 k Ω (b) 187. rπ = ro = β VT I CQ = (100 )( 0.026 ) 1.25 = 2.08 k Ω VA 125 = = 100 k Ω I CQ 1.25 We may note that g mVπ = g m ( I b rπ ) = β I b Also Rib = rπ + (1 + β )( RE RL ro ) = 2.08 + (101) ( 4.76 1 100 ) or Rib = 84.9 k Ω Now ⎛ RE ro ⎞ Io = ⎜ 1 + β ) Ib ⎜ R r + R ⎟( ⎟ L ⎠ ⎝ E o where ⎛ R1 R2 ⎞ Ib = ⎜ ⋅I ⎜R R +R ⎟ s ⎟ ib ⎠ ⎝ 1 2 We can then write ⎛ RE ro ⎞ ⎛ R1 R2 I AI = o = ⎜ ⎟ (1 + β ) ⎜ ⎜R R +R I s ⎜ RE ro + RL ⎟ ib ⎝ ⎠ ⎝ 1 2 We have RE ro = 4.76 100 = 4.54 k Ω ⎞ ⎟ ⎟ ⎠ so 48.1 ⎞ ⎛ 4.54 ⎞ ⎛ AI = ⎜ ⎟ (101) ⎜ ⎟ 4.54 + 1 ⎠ 48.1 + 84.9 ⎠ ⎝ ⎝ or AI = 29.9 r 2.08 (c) Ro = RE ro π = 4.76 100 1+ β 101 or Ro = 20.5 Ω EX6.16 (a) RTH = R1 R2 = 70 6 = 5.53 k Ω ⎛ R2 ⎞ ⎛ 6 ⎞ VTH = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 ⎝ 70 + 6 ⎠ ⎝ R1 + R2 ⎠ or VTH = −4.2105 V We find −4.2105 − 0.7 − ( −5 ) I BQ1 = ⇒ 2.91 μ A 5.53 + (126 )( 0.2 ) and I CQ1 = β I BQ1 = (125 )( 2.91 μ A ) = 0.364 mA I EQ1 = (1 + β ) I BQ1 = 0.368 mA At the collector of Q1, V − 0.7 − ( −5 ) 5 − VC1 = I CQ1 + C1 RC1 (1 + β )( RE 2 ) or 188. V − 0.7 − ( −5 ) 5 − VC1 = 0.364 + C1 5 (126 )(1.5) which yields VC1 = 2.99 V also VE1 = I EQ1 RE1 − 5 = ( 0.368 )( 0.2 ) − 5 or VE1 = −4.93 V Then VCEQ1 = VC1 − VE1 = 2.99 − ( −4.93) = 7.92 V We find V − 0.7 − ( −5 ) I EQ 2 = C1 = 4.86 mA 1.5 and ⎛ β ⎞ ⎛ 125 ⎞ I CQ 2 = ⎜ ⎟ ⋅ I EQ1 = ⎜ ⎟ ( 4.86 ) = 4.82 mA 1+ β ⎠ ⎝ 126 ⎠ ⎝ We find VE 2 = VC1 − 0.7 = 2.99 − 0.7 = 2.29 V and VCEQ 2 = 5 − VE 2 = 5 − 2.29 = 2.71 V (b) The small-signal transistor parameters are: rπ 1 = β VT I CQ1 g m1 = rπ 2 = gm2 = = I CQ1 VT β VT I CQ 2 I CQ 2 VT (125 )( 0.026 ) 0.364 0.364 = 14.0 mA / V 0.026 = = = 8.93 k Ω (125 )( 0.026 ) = 4.82 = 0.674 k Ω 4.82 = 185 mA / V 0.026 We find Rib1 = rπ 1 + (1 + β ) RE1 = 8.93 + (126 )( 0.2 ) Or Rib1 = 34.1 k Ω and Rib 2 = rπ 2 + (1 + β )( RE 2 RL ) = 0.674 + (126 )(1.5 10 ) = 165 k Ω The small-signal equivalent circuit is: 189. ϩ ϩ V␲1 r␲1 gm1V␲1 V␲2 ϩ Ϫ R1͉͉R2 RC1 RE1 We can write Vo = (1 + β ) I b 2 ( RE 2 RL ) where ⎛ RC1 ⎞ Ib2 = ⎜ ⎟ ( − g m1Vπ 1 ) ⎝ RC1 + Rib 2 ⎠ V Vπ 1 = s ⋅ rπ 1 Rib1 Then V Av = o Vs ⎛ RC1 ⎞ ⎛ − g m1rπ 1 ⎞ = (1 + β )( RE 2 RL ) ⎜ ⎟⎜ ⎟ ⎝ RC1 + Rib 2 ⎠ ⎝ Rib1 ⎠ so ⎛ 5 ⎞ ⎛ 125 ⎞ Av = − (126 )(1.5 10 ) ⎜ ⎟⎜ ⎟ ⎝ 5 + 165 ⎠ ⎝ 34.1 ⎠ or Av = −17.7 (c) Ri = R1 R2 Rib1 = 70 6 34.1 = 4.76 k Ω ⎛ r + RC1 ⎞ ⎛ 0.676 + 5 ⎞ and Ro = RE 2 ⎜ π 2 ⎟ = 1.5 ⎜ ⎟ ⎝ 126 ⎠ ⎝ 1+ β ⎠ or Ro = 43.7 Ω Test Your Understanding Exercises TYU6.1 I CQ 0.25 = = 9.62 mA / V gm = 0.026 VT rπ = ro = β VT I CQ = (120 )( 0.026 ) 0.25 = 12.5 k Ω VA 150 = = 600 k Ω I CQ 0.25 TYU6.2 V V 75 ro = A ⇒ I CQ = A = I CQ ro 200 k Ω or ICQ = 0.375 mA gm2V␲2 Ϫ Ϫ Vs r␲2 Vo RE2 RL 190. TYU6.3 RC RE Resulting gain is always smaller than this value. The effect of RS is very small. Set RC = 10 RE Now 5 ≅ I C ( RC + RE ) + VCEQ As a first approximation, Av ≅ − or 5 = ( 0.5 )( RC + RE ) + 2.5 which yields RC + RE = 5 k Ω So 10 RE + RE = 5 or RE = 0.454 k Ω and RC = 4.54 k Ω We have I CQ 0.5 I BQ = = = 0.005 mA β 100 and RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.454 ) or RTH = 4.59 k Ω also ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ ⋅ VCC = ⋅ RTH ⋅ VCC R1 + R2 ⎠ R1 ⎝ or 1 23 ( 4.59 )( 5 ) = R1 R1 We can write VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE VTH = or 23 = ( 0.005 )( 4.59 ) + 0.7 + (101)( 0.005 )( 0.454 ) R1 which yields R1 = 24.1 k Ω and since R1 R2 = 4.59 k Ω we find R2 = 5.67 k Ω TYU6.4 dc analysis RTH = R1 R2 = 15 85 = 12.75 k Ω ⎛ R2 ⎞ ⎛ 85 ⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ (12 ) ⎝ 15 + 85 ⎠ ⎝ R1 + R2 ⎠ or VTH = 10.2 V We can write 12 − 0.7 − VTH 12 − 0.7 − 10.2 I BQ = = RTH + (1 + β ) RE 12.75 + (101)( 0.5 ) or IBQ = 0.0174 mA 191. and I CQ = β I BQ = 1.74 mA ac analysis Vo = h fe I b ( RC RL ) and Ib = so Av = −Vs hie + (1 + h fe ) RE −h fe ( RC RL ) Vo = Vs hie + (1 + h fe ) RE For ICQ = 1.74 mA, we find h fe ( max ) = 110, h fe ( min ) = 70 hie ( max ) = 2 k Ω, hie ( min ) = 1.1 k Ω We obtain −110 ( 4 2 ) Av ( max ) = = −2.59 1.1 + (111)( 0.5 ) and Av ( min ) = −70 ( 4 2 ) 2 + ( 71)( 0.5 ) = −2.49 TYU6.5 First approximation, Av ≅ − RC RE This predicts a low value, so set RC =9 RE Now VCC ≅ I CQ ( RC + RE ) + VECQ or 7.5 = ( 0.6 )( 9 RE + RE ) + 3.75 which yields RE = 0.625 k Ω and RC = 5.62 k Ω We have RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.625 ) or RTH = 6.31 k Ω Also 1 1 VTH = ⋅ RTH ⋅ VCC = ( 6.31)( 7.5 ) R1 R1 We have I CQ 0.6 I BQ = = = 0.006 mA β 100 The KVL equation around the E-B loop gives VCC = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH or 7.5 = (101)( 0.006 )( 0.625 ) + 0.7 + ( 0.006 )( 6.31) + 1 ( 6.31)( 7.5) R1 which yields R1 = 7.41 k Ω Since RTH = R1 R2 = 6.31 k Ω, then R2 = 42.5 k Ω TYU6.6 192. ⎛R ⎞ − β RC = − ( 0.95 ) ⎜ C ⎟ rπ + (1 + β ) RE ⎝ RE ⎠ ⎛ 2 ⎞ or Av = − ( 0.95 ) ⎜ ⎟ = −4.75 ⎝ 0.4 ⎠ Assume rπ = 1.2 k Ω from Example 4.6. Then −β ( 2) = −4.75 1.2 + (1 + β )( 0.4 ) We have Av = or β = 76 TYU6.7 dc analysis: By symmetry, VTH = 0 RTH = R1 R2 = 20 20 = 10 k Ω We can write 0 − 0.7 − ( −5 ) I BQ = = 0.00672 mA 10 + (126 )( 5 ) I CQ = β I BQ = (125 )( 0.00672 ) = 0.84 mA Small-signal transistor parameters: β VT (125 )( 0.026 ) rπ = = = 3.87 k Ω I CQ 0.84 gm = ro = I CQ VT = 0.84 = 32.3 mA / V 0.026 VA 200 = = 238 k Ω I CQ 0.84 (a) We have Vo = − g mVπ ( ro RC RL ) and Vπ = Vs so Av = Vo Vs = − g m ( ro RC RL ) = − ( 32.3)( 238 2.3 5 ) or Av = −50.5 (b) Ro = ro RC = 238 2.3 = 2.28 k Ω TYU6.8 We find I CQ = 0.418 mA, then ⎛ 121 ⎞ VCEQ = 5 − ( 0.418 )( 5.6 ) − ⎜ ⎟ ( 0.418 )( 0.6 ) ⎝ 120 ⎠ or VCEQ = 2.41 V So ΔvCE = ( 2.41 − 0.5 ) × 2 or ΔvCE = 3.82 V peak-to-peak TYU6.9 dc load line VCE ≅ (10 + 10 ) − I CQ ( RC + RE ) or 193. VCE = 20 − I C (10 + RE ) ac load line vce = −ic RC = −ic (10 ) Now ΔvCE = VCEQ − 0.7 = ΔiC (10 ) = I CQ (10 ) so VCEQ − 0.7 = I CQ (10 ) We have that VCEQ = 20 − I CQ (10 + RE ) then I CQ (10 ) + 0.7 = 20 − I CQ (10 + RE ) or I CQ ( 20 + RE ) = 19.3 (1) The base current is found from 10 − 0.7 I BQ = 100 + (101) RE so I CQ = (100 )( 9.3) 100 + (101) RE Substituting into Equation (1), ⎡ (100 )( 9.3) ⎤ ⎢ ⎥ ( 20 + RE ) = 19.3 ⎢100 + (101) RE ⎥ ⎣ ⎦ which yields RE = 16.35 k Ω Then (100 )( 9.3) I CQ = = 0.531 mA 100 + (101)(16.35 ) and VCEQ ≅ 20 − ( 0.531)(10 + 16.35 ) = 6.0 V Now ΔvCE = VCEQ − 0.7 = 6 − 0.7 = 5.3 V or, maximum symmetrical swing = 2 × 5.3 = 10.6 V peak-to-peak TYU6.10 We can write 0 − 0.7 − ( −10 ) I BQ = ⇒ 6.60 μ A 100 + (131)(10 ) and I CQ = (130 )( 6.60 μ A ) = 0.857 mA Assume nominal small-signal parameters of: hie = 4 k Ω, h fe = 134 hre = 0, hoe = 12 μ S ⇒ 1 = 83.3 k Ω hoe We find ⎛ 1 ⎞ Rib = hie + (1 + h fe ) ⎜ RE RL ⎟ hoe ⎠ ⎝ = 4 + (135 )(10 ||10 || 83.3) = 641 k Ω To find the voltage gain: 194. RB Rib 100 641 ⋅ Vs = ⋅ Vs 100 641 + 10 RB Rib + RS Vs′ = or Vs′ = ( 0.896 ) Vs Also (1 + h fe ) R′ Vo = Vs′ hie + (1 + h fe ) R′ where R ′ = RE Then Av = RL 1 = 10 10 83.3 = 4.72 k Ω hoe Vo ( 0.896 )(135 )( 4.72 ) = = 0.891 Vs 4 + (135 )( 4.72 ) To find the current gain: ⎛ ⎞ 1 ⎜ RE ⎟ hoe ⎟ ⎛ RB ⎞ Io ⎜ 1 + h fe ) ⎜ Ai = = ⎟ ⎟( Ii ⎜ 1 ⎝ RB + Rib ⎠ + RL ⎟ ⎜ RE ⎜ ⎟ hoe ⎝ ⎠ ⎛ 10 83.3 ⎞ ⎛ 100 ⎞ =⎜ (135 ) ⎜ ⎟ 10 83.3 + 10 ⎟ ⎝ 100 + 641 ⎠ ⎝ ⎠ or Ai = 8.59 To find the output resistance: 1 hie + RS RB Ro = RE 1 + h fe hoe = 10 83.3 4 + 10 100 ⇒ 96.0 Ω 135 TYU6.11 We find RTH = R1 R2 = 50 50 = 25 k Ω ⎛ R2 ⎞ ⎛1⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ ( 5 ) = 2.5 V ⎝ 2⎠ ⎝ R1 + R2 ⎠ Now V − VEB ( on ) − VTH I BQ = CC RTH + (1 + β ) RE = 5 − 0.7 − 2.5 = 0.00793 mA 25 + (101)( 2 ) and I CQ = (100 )( 0.00793) = 0.793 mA The small-signal transistor parameters: I CQ 0.793 = = 30.5 mA / V gm = 0.026 VT rπ = ro = (a) β VT I CQ = (100 )( 0.026 ) 0.793 = 3.28 k Ω VA 125 = = 158 k Ω I CQ 0.793 195. Define R ′ = RE RL ro = 2 0.5 158 ≅ 0.40 k Ω Now Av = (1 + β ) R′ (101)( 0.4 ) = rπ + (1 + β ) R ′ 3.28 + (101)( 0.4 ) or Av = 0.925 (b) Rib = rπ + (1 + β ) R ′ = 3.28 + (101)( 0.4 ) or Rib = 43.7 k Ω Also Ro = RE ro rπ 3.28 = 2 158 1+ β 101 or Ro = 32.0 Ω TYU6.12 For VECQ = 2.5, I EQ = VCC − VECQ RE = 5 − 2.5 = 5 mA 0.5 then ⎛ 75 ⎞ I CQ = ⎜ ⎟ ( 5 ) = 4.93 mA ⎝ 76 ⎠ 5 I BQ = = 0.0658 mA 76 Small-signal transistor parmaters: β VT ( 75 )( 0.026 ) = = 0.396 k Ω rπ = I CQ 4.93 ro = VA 75 = = 15.2 k Ω I CQ 4.93 Define the small-signal base current into the base, then g mVπ = − β I b Now, ⎛ RE ro ⎞ Io = ⎜ ⎟ (1 + β ) I b ⎝ RE ro + RL ⎠ and ⎛ R1 R2 ⎞ Ib = ⎜ ⎟ ⋅ Ii ⎝ R1 R2 + Rib ⎠ The current gain is ⎛ RE ro ⎞ ⎛ R1 R2 ⎞ I AI = o = ⎜ ⎟ (1 + β ) ⎜ ⎟ I i ⎝ RE ro + RL ⎠ ⎝ R1 R2 + Rib ⎠ We have RE = RL = 0.5 k Ω Rib = rπ + (1 + β )( RE RL ro ) = 0.396 + ( 76 )( 0.5 0.5 15.2 ) = 19.1 k Ω and RE ro = 0.5 15.2 = 0.484 k Ω Then ⎛ R1 R2 ⎞ ⎛ 0.484 ⎞ AI = 10 = ⎜ ⎟ ⎟ ( 76 ) ⎜ ⎝ 0.484 + 0.5 ⎠ ⎝ R1 R2 + 19.1 ⎠ which yields R1 R2 = 6.975 k Ω Now 196. ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ ⋅ VCC = ⋅ RTH ⋅ VCC R1 + R2 ⎠ R1 ⎝ or 1 VTH = ( 6.975 )( 5 ) R1 We can write V − VEB ( on ) − VTH I BQ = CC RTH + (1 + β ) RE or 0.0658 = 5 − 0.7 − VTH 6.975 + ( 76 )( 0.5 ) which yields VTH = 1.34 = 1 ( 6.975)( 5 ) R1 or R1 = 26.0 k Ω and R2 = 9.53 k Ω TYU6.13 (a) dc analysis: V − VEB ( on ) 10 − 0.7 I EQ = EE = = 0.93 mA 10 RE ⎛ β ⎞ ⎛ 100 ⎞ I CQ = ⎜ ⎟ I EQ = ⎜ ⎟ ( 0.93) = 0.921 mA ⎝ 101 ⎠ ⎝ 1+ β ⎠ VECQ = VEE − I EQ RE − I CQ RC − ( −VCC ) = 10 − ( 0.93)(10 ) − ( 0.921)( 5 ) − ( −10 ) or VECQ = 6.1 V (b) Small-signal transistor parameters: β VT (100 )( 0.026 ) rπ = = = 2.82 k Ω I CQ 0.921 I CQ 0.921 = 35.42 mA / V 0.026 VT Small-signal current gain: I o = g mVπ and Vπ = Vs also ⎛ 1 ⎞ Vs + g mVπ = Vs ⎜ Ii = ⎜ R r + gm ⎟ ⎟ RE rπ ⎝ E π ⎠ Then g m ( RE rπ ) I g mVπ AI = o = = Ii ⎛ 1 ⎞ 1 + g m ( RE rπ ) + gm ⎟ Vπ ⎜ ⎝ RE rπ ⎠ gm = = = ( 35.42 ) (10 2.82 ) 1 + ( 35.42 ) (10 2.82 ) or AI = 0.987 (c) Small-signal voltage gain: 197. Vo = g mVπ RC = g mVs RC or V Av = o = g m RC = ( 35.42 )( 5 ) Vs or Av = 177 TYU6.14 (a) V − VBE ( on ) 10 − 0.7 I BQ = EE = RB + (1 + β ) RE 100 + (101)(10 ) or I BQ = 8.38 μ A and I CQ = 0.838 mA rπ = gm = ro = β VT = I CQ I CQ VT = (100 )( 0.026 ) 0.838 = 3.10 k Ω 0.838 = 32.23 mA / V 0.026 VA ∞ = =∞ I CQ 0.838 (b) Summing currents, we have ⎛ −V ⎞ ⎛ −V − Vs ⎞ V g mVπ + π = ⎜ π ⎟ + ⎜ π ⎟ rπ ⎝ RE ⎠ ⎝ RS ⎠ or ⎡⎛ 1 + β ⎞ 1 V 1 ⎤ Vπ ⎢⎜ + ⎥=− s ⎟+ rπ ⎠ RE RS ⎥ RS ⎢⎝ ⎣ ⎦ We can then write ⎤ V ⎡⎛ r ⎞ Vπ = − s ⎢⎜ π ⎟ RE RS ⎥ RS ⎣⎝ 1 + β ⎠ ⎦ Now Vo = − g mVπ ( RC RL ) So ( RC RL ) ⎡⎛ rπ ⎞ R R ⎤ Vo = gm ⎢⎜ ⎟ E S⎥ Vs RS ⎣⎝ 1 + β ⎠ ⎦ 32.23) (10 1) ⎡ 3.10 ( ⎤ = ⎢ 101 10 1⎥ 1) ( ⎣ ⎦ Av = or Av = 0.870 Now ⎛ RE Ie = ⎜ ⎝ RE + rπ ⎛ β Ic = ⎜ ⎝ 1+ β ⎞ ⎛ 10 ⎞ ⎟ ⋅ Ii = ⎜ ⎟ ⋅ I i = ( 0.763) I i ⎝ 10 + 3.10 ⎠ ⎠ ⎞ ⎛ 100 ⎞ ⎟ ⋅ Ie = ⎜ ⎟ ⋅ Ie ⎝ 101 ⎠ ⎠ ⎛ RC ⎞ ⎛ 10 ⎞ Io = ⎜ ⎟ ⋅ Ic = ⎜ ⎟ ⋅ I c = ( 0.909 ) I c RC + RL ⎠ ⎝ 10 + 1 ⎠ ⎝ So we have I ⎛ 100 ⎞ AI = o = ( 0.909 ) ⎜ ⎟ ( 0.763) Ii ⎝ 101 ⎠ 198. or AI = 0.687 (c) We have r 3.10 Ri = RE π = 10 1+ β 101 or Ri = 30.6 Ω Also Ro = RC = 10 k Ω TYU6.15 dc analysis: We can write 5 = I BQ RB + VBE ( on ) + I EQ RE or I BQ = I CQ = 5 − 0.7 4.3 = RB + (101) RE RB + (101) RE (100 )( 4.3) RB + (101) RE Also 5 = I CQ RC + VCEQ + I EQ RE − 5 or ⎡ ⎛ 101 ⎞ ⎤ VCEQ = 10 − I CQ ⎢ RC + ⎜ ⎟ RE ⎥ ⎝ 100 ⎠ ⎦ ⎣ ac analysis: Vo = − g mVπ ( RC RL ) and Vs = −Vπ − Vπ ⋅ RB = −Vπ rπ ⎛ RB ⎞ ⎜1 + ⎟ rπ ⎠ ⎝ or ⎛ r ⎞ Vπ = − ⎜ π ⎟ ⋅ Vs rπ + RB ⎠ ⎝ Then V β Av = o = ( RC RL ) Vs rπ + RB where β = g m rπ For I CQ = 1 mA, rπ = Now Av = 20 = β VT I CQ (100 ) ( 2 2 ) 2.6 + RB which yields RB = 2.4 k Ω Also (100 )( 4.3) I CQ = 1 = 2.4 + (101) RE which yields RE = 4.23 k Ω TYU6.16 (a) = (100 )( 0.026 ) 1 = 2.6 k Ω 199. dc analysis: For I EQ 2 = 1 mA, ⎛ 100 ⎞ I CQ 2 = ⎜ ⎟ (1) = 0.990 mA ⎝ 101 ⎠ I EQ 2 1 I EQ1 = = = 0.0099 mA 1 + β 101 and I EQ1 0.0099 I BQ1 = = = 0.000098 mA 1+ β 101 and I CQ1 = (100 )( 0.000098 ) = 0.0098 mA and VB1 = − I BQ1 RB = − ( 0.000098 )(10 ) or VB1 = −0.00098 ≅ 0 so VE1 = −0.7 V and VE 2 = −1.4 V Now I1 = I CQ1 + I CQ 2 = 0.0098 + 0.990 ≅ 1 mA then VO = 5 − (1)( 4 ) = 1 V and VCEQ 2 = 1 − ( −1.4 ) = 2.4 V VCEQ1 = 1 − ( −0.7 ) = 1.7 V (b) Small-signal transistor parameters: β VT (100 )( 0.026 ) = = 265 k Ω rπ 1 = 0.0098 I CQ1 g m1 = rπ 2 = gm2 = I CQ1 VT β VT I CQ 2 I CQ 2 VT 0.0098 = 0.377 mA 0.026 = = (100 )( 0.026 ) = 0.990 = 2.63 k Ω 0.990 = 38.1 mA / V 0.026 ro1 = ro 2 = ∞ (c) Small-signal voltage gain: From Figure 4.73 (b) Vo = − ( g m1Vπ 1 + g m 2Vπ 2 ) RC Vs = Vπ 1 + Vπ 2 and ⎛V ⎞ ⎛1+ β ⎞ Vπ 2 = ⎜ π 1 + g m1Vπ 1 ⎟ ⋅ rπ 2 = ⎜ ⎟ ⋅ Vπ 1rπ 2 ⎝ rπ 1 ⎠ ⎝ rπ 1 ⎠ Then ⎡ ⎛1+ β Vo = − ⎢ g m1Vπ 1 + g m 2 ⎜ ⎝ rπ 1 ⎣ Also ⎤ ⎞ ⎟ ⋅ rπ 2Vπ 1 ⎥ ⋅ RC ⎠ ⎦ 200. ⎛1+ β Vs = Vπ 1 + ⎜ ⎝ rπ 1 ⎞ ⎟ ⋅ rπ 2Vπ 1 ⎠ ⎡ ⎛ r ⎞⎤ = Vπ 1 ⎢1 + (1 + β ) ⎜ π 2 ⎟ ⎥ ⎝ rπ 1 ⎠ ⎦ ⎣ or Vπ 1 = Vs ⎛r ⎞ 1 + (1 + β ) ⎜ π 2 ⎟ ⎝ rπ 1 ⎠ Now ⎡ ⎛ rπ 2 ⎞ ⎤ ⎢ g m1 + g m 2 (1 + β ) ⎜ ⎟ ⎥ ⋅ RC V ⎝ rπ 1 ⎠ ⎦ Av = o = − ⎣ Vs ⎛r ⎞ 1 + (1 + β ) ⎜ π 2 ⎟ ⎝ rπ 1 ⎠ so ⎡ ⎛ 2.63 ⎞ ⎤ ⎢ 0.377 + ( 38.1)(101) ⎜ 265 ⎟ ⎥ ( 4 ) ⎝ ⎠⎦ Av = − ⎣ ⎛ 2.63 ⎞ 1 + (101) ⎜ ⎟ ⎝ 265 ⎠ or Av = −77.0 (d) Ri = rπ 1 + (1 + β ) rπ 2 = 265 + (101)( 2.63) or Ri = 531 k Ω TYU6.17 (a) dc analysis: For R1 + R2 + R3 = 100 k Ω I1 = VCC 12 = = 0.12 mA R1 + R2 + R3 100 Now VE1 = I CQ 2 RE = ( 0.5 )( 0.5 ) = 0.25 V VC1 = VE1 + VCEQ1 = 0.25 + 4 = 4.25 V and VC 2 = VC1 + VCEQ 2 = 4.25 + 4 = 8.25 V So RC = VCC − VC 2 12 − 8.25 = = 7.5 k Ω 0.5 I CQ Also VB1 = VE1 + VBE ( on ) = 0.25 + 0.7 = 0.95 V And ⎛ ⎞ R3 R3 VB1 = ⎜ (12 ) ⎟ ⋅ VCC ⇒ 0.95 = 100 ⎝ R1 + R2 + R3 ⎠ which yields R3 = 7.92 k Ω We have VB 2 = VC1 + VBE ( on ) = 4.25 + 0.7 = 4.95 V and 201. ⎛ R2 + R3 ⎞ VB 2 = ⎜ ⎟ ⋅ VCC ⎝ R1 + R2 + R3 ⎠ or ⎛ R + 7.92 ⎞ 4.95 = ⎜ 2 ⎟ (12 ) ⎝ 100 ⎠ which yields R2 = 33.3 k Ω Then R1 = 100 − 33.3 − 7.92 = 58.8 k Ω (b) Small-signal transistor parameters: (100 )( 0.026 ) βV rπ 1 = rπ 2 = T = I CQ 0.5 or rπ 1 = rπ 2 = 5.2 k Ω and I CQ 0.5 g m1 = g m 2 = = 0.026 VT or g m1 = g m 2 = 19.23 mA/V and ro1 = ro 2 = ∞ (c) Small-signal voltage gain: We have Vo = − g m 2Vπ 2 ( RC RL ) Also Vπ 2 + g m 2Vπ 2 = g m1Vπ 1 rπ 2 or ⎛ r ⎞ Vπ 2 = g m1Vπ 1 ⎜ π 2 ⎟ and Vπ 1 = Vs ⎝ 1+ β ⎠ We find ⎛ r ⎞ V Av = o = − g m 2 ( RC RL ) g m1 ⎜ π 2 ⎟ Vs ⎝1+ β ⎠ ⎛ β ⎞ = − g m 2 ( RC RL ) ⎜ ⎟ ⎝ 1+ β ⎠ We obtain ⎛ 100 ⎞ Av = − (19.23) ( 7.5 2 ) ⎜ ⎟ ⎝ 101 ⎠ or Av = −30.1 TYU6.18 (a) dc analysis: with vs = 0 RTH = R1 R2 = 125 30 = 24.2 k Ω ⎛ R2 ⎞ ⎛ 30 ⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ (12 ) ⎝ 125 + 30 ⎠ ⎝ R1 + R2 ⎠ or VTH = 2.32 V 202. Now VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE or I BQ = 2.32 − 0.7 = 0.0250 mA 24.2 + ( 81)( 0.5 ) and I CQ = ( 80 )( 0.025 ) = 2.00 mA Also ⎡ ⎛1+ β ⎞ ⎤ VCEQ = VCC − I CQ ⎢ RC + ⎜ ⎟ RE ⎥ ⎝ β ⎠ ⎦ ⎣ ⎡ ⎛ 81 ⎞ ⎤ = 12 − ( 2 ) ⎢ 2 + ⎜ ⎟ ( 0.5 ) ⎥ ⎣ ⎝ 80 ⎠ ⎦ or VCEQ = 6.99 V Power dissipated in RC: 2 PRC = I CQ RC = ( 2.0 ) ( 2 ) = 8.0 mW 2 Power dissipated in RL: I LQ = 0 ⇒ PRL = 0 Power dissipated in transistor: PQ = I BQVBEQ + I CQVCEQ = ( 0.025 )( 0.7 ) + ( 2.0 )( 6.99 ) = 14.0 mW (b) With vs = 18cos ω t ( mV ) β VT rπ = I CQ = (80 )( 0.026 ) 2.0 = 1.04 k Ω We can write β ( RC RL )VP cos ω t rπ Power dissipated in RL: vce = pRL = = vce ( rms ) RL 1 1 ⋅ 2 2 × 103 2 ⎤ 1 1 ⎡β = ⋅ ⋅ ⎢ ( RC RL ) VP ⎥ 2 RL ⎣ rπ ⎦ ⎡ 80 ⋅⎢ ( 2 2 ) ( 0.018)⎤ ⎥ ⎣1.04 ⎦ 2 2 or pRL = 0.479 mW Power dissipated in RC: Since RC = RL = 2 k Ω, we have pRC = 8.0 + 0.479 = 8.48 mW Power dissipated in transistor: From the text, we have 2 ⎛β ⎞ ⎛V ⎞ pQ ≅ I CQVCEQ − ⎜ ⎟ ⎜ P ⎟ ( RC RL ) ⎝ rπ ⎠ ⎝ 2 ⎠ 2 2 2 80 ⎛ ⎞ ⎛ 0.018 ⎞ 3 = ( 2 × 10−3 ) ( 6.99 ) − ⎜ ⎟ ( 2 × 10 2 × 3 ⎟ ⎜ ⎝ 1.04 × 10 ⎠ ⎝ 2 ⎠ or pQ = 13.0 mW TYU6.19 (a) dc analysis: 3 ) 203. RTH = R1 R2 = 53.8 10 = 8.43 k Ω ⎛ R2 ⎞ ⎛ 10 ⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ (5) R1 + R2 ⎠ ⎝ 53.8 + 10 ⎠ ⎝ or VTH = 0.7837 V Now 0.7837 − 0.7 I BQ = = 0.00993 mA 8.43 and I CQ = (100 )( 0.00993) = 0.993 mA We have VCEQ = VCC − I CQ RC or 2.5 = 5 − ( 0.993) RC which yields RC = 2.52 k Ω (b) Power dissipated in RC: 2 PRC = I CQ RC = ( 0.993) ( 2.52 ) 2 or PRC = 2.48 mW Power dissipated in transistor: PQ ≅ I CQVCEQ = ( 0.993)( 2.5 ) or PQ = 2.48 mW (c) ac analysis: Maximum ac collector current: ic = ( 0.993) cos ω t ( mA ) average ac power dissipated in RC: 1 1 2 2 pRC = ( 0.993) RC = ( 0.993) ( 2.52 ) 2 2 or pRC = 1.24 mW Now pRC 1.24 Fraction = = = 0.25 PRC + PQ 2.48 + 2.48 204. Chapter 6 Problem Solutions 6.1 a. gm = rπ = r0 = I CQ VT β VT I CQ = 2 ⇒ g m = 76.9 mA/V 0.026 = (180 )( 0.026 ) 2 ⇒ rπ = 2.34 kΩ VA 150 = ⇒ r0 = 75 kΩ I CQ 2 b. 0.5 ⇒ g m = 19.2 mA/V 0.026 (180 )( 0.026 ) rπ = ⇒ rπ = 9.36 kΩ 0.5 150 r0 = ⇒ r0 = 300 kΩ 0.5 gm = 6.2 (a) gm = rπ = ro = I CQ VT β VT I CQ = 0.8 = 30.8 mA/V 0.026 = (120 )( 0.026 ) 0.8 = 3.9 K VA 120 = = 150 K I CQ 0.8 (b) 0.08 = 3.08 mA/V 0.026 (120 )( 0.026 ) rπ = = 39 K 0.08 120 ro = = 1500 K 0.08 gm = 6.3 gm = rπ = r0 = 6.4 I CQ VT β VT I CQ ⇒ 200 = = I CQ 0.026 ⇒ I CQ = 5.2 mA (125)( 0.026 ) 5.2 ⇒ rπ = 0.625 kΩ VA 200 = ⇒ r0 = 38.5 kΩ I CQ 5.2 205. gm = rπ = I CQ ⇒ 80 = VT β VT I CQ 0.026 ⇒ 1.20 = I CQ ⇒ I CQ = 2.08 mA β ( 0.026 ) 2.08 ⇒ β = 96 6.5 (a) 2 − 0.7 = 0.0052 mA 250 I C = (120 )( 0.0052 ) = 0.624 mA I BQ = 0.624 ⇒ g m = 24 mA / V 0.026 (120 )( 0.026 ) rπ = ⇒ rπ = 5 k Ω 0.624 ro = ∞ gm = ⎛ r ⎞ ⎛ 5 ⎞ Av = − g m RC ⎜ π ⎟ = − ( 24 )( 4 ) ⎜ ⎟ ⇒ Av = −1.88 rπ + RB ⎠ ⎝ 5 + 250 ⎠ ⎝ v v vS = O = O ⇒ vS = −0.426sin100t V Av −1.88 (b) (c) 6.6 gm = I CQ VT , 1.08 ≤ I CQ ≤ 1.32 mA 1.08 1.32 ≤ gm ≤ ⇒ 41.5 ≤ g m ≤ 50.8 mA/V 0.026 0.026 (120 )( 0.026 ) β VT rπ = ; rπ ( max ) = = 2.89 kΩ I CQ 1.08 rπ ( min ) = (80 )( 0.026 ) 1.32 = 1.58 kΩ 1.58 ≤ rπ ≤ 2.89 kΩ 6.7 a. rπ = 5.4 = β VT I CQ = (120 )( 0.026 ) I CQ ⇒ I CQ = 0.578 mA 1 1 VCEQ = VCC = ( 5 ) = 2.5 V 2 2 VCEQ = VCC − I CQ RC ⇒ 2.5 = 5.0 − ( 0.578 ) RC ⇒ RC = 4.33 kΩ I CQ 0.578 = 0.00482 mA β 120 VBB = I BQ RB + VBE ( on ) = ( 0.00482 )( 25 ) + 0.70 ⇒ VBB = 0.820 V I BQ = b. = 206. rπ = gm = r0 = β VT I CQ I CQ = 0.578 = 5.40 kΩ 0.578 = 22.2 mA/V 0.026 = VT (120 )( 0.026 ) VA 100 = = 173 kΩ I CQ 0.578 ⎛ r ⎞ V0 = − g m ( r0 RC ) Vπ , Vπ = ⎜ π ⎟ VS ⎝ rπ + RB ⎠ β ( r0 RC ) ⎛ r ⎞ Av = − g m ⎜ π ⎟ ( r0 RC ) = − rπ + RB ⎝ rπ + RB ⎠ Av = − (120 ) ⎡173 ⎣ 4.33⎤ ⎦ 5.40 + 25 =− (120 )( 4.22 ) 30.4 ⇒ Av = −16.7 6.8 a. 1 VECQ = VCC = 5 V 2 VECQ = 10 − I CQ RC ⇒ 5 = 10 − ( 0.5 ) RC ⇒ RC = 10 kΩ I CQ 0.5 = 0.005 100 VEB ( on ) + I BQ RB = VBB = ( 0.70 ) + ( 0.005 )( 50 ) ⇒ VBB = 0.95 V I BQ = = β b. gm = rπ = r0 = c. I CQ = VT β VT I CQ 0.5 ⇒ g m = 19.2 mA/V 0.026 = (100 )( 0.026 ) 0.5 ⇒ rπ = 5.2 kΩ VA ∞ = ⇒ r0 = ∞ I CQ 0.5 Av = − (100 )(10 ) ⇒ A = −18.1 β RC =− v 5.2 + 50 rπ + RB 6.9 10 − 4 = 1.5 mA 4 1.5 I BQ = = 0.015 mA 100 (100 )( 0.026 ) rπ = = 1.73 K 1.5 5sin ω t ( mV ) v ib = be = = 2.89sin ω t ( μ A ) rπ 1.73 kΩ So I CQ = 207. iB ( t ) = I BQ + iEb = 15 + 2.89sin ω t ( μ A ) iC1 ( t ) = β iB ⇒ iC1 ( t ) = 1.5 + 0.289sin ω t ( mA ) vC ( t ) = 10 − iC1 ( t ) RC = 10 − [1.5 + 0.289sin ω t ] (γ ) vC1 ( t ) = 4 − 1.156sin ω t ( v ) Av = vC ( t ) vbe ( t ) = −1.156 ⇒ Av = −231 0.005 6.10 vo = 1.2sin ω t ( V ) iC ( t ) RC + vo = 0 ⇒ iC ( t ) = iC ( t ) = −0.60sin ω t ( mA ) ib ( t ) = iC ( t ) β −1.2sin ω t 2 = −6sin ω t ( μ A ) vbe ( t ) = ib ( t ) ⋅ rπ g m rπ = β 100 =2K 50 vbe ( t ) = −12sin ω t ( mV ) rπ = 6.11 a. I CQ ≈ I EQ VCEQ = 5 = 10 − I CQ ( RC + RE ) = 10 − I CQ (1.2 + 0.2) I CQ = 3.57 mA 3.57 = 0.0238 mA 150 R2 = RTH = ( 0.1)(1 + β ) RE I BQ = R1 VTH = = ( 0.1)(151)( 0.2 ) = 3.02 kΩ 1 ⋅ RTH ⋅ (10) − 5 R1 VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 5 1 (3.02)(10) − 5 = ( 0.0238)(3.02) + 0.7 + (151)( 0.0238)( 0.2) − 5 R1 1 ( 30.2 ) = 1.50 ⇒ R1 = 20.1 k Ω R1 20.1R2 = 3.02 ⇒ R2 = 3.55 kΩ 20.1 + R2 b. (150 )( 0.026 ) rp = = 1.09 kΩ 3.57 3.57 gm = = 137 mA/V 0.026 208. V0 ϩ V␲ ϩ Ϫ VS r␲ gmV␲ Ϫ R1͉͉R2 RC RE Av = (150 )(1.2 ) 2 β RC =2 ⇒ Av = 2 5.75 rp + (1 + β ) RE 1.09 + (151)( 0.2 ) 6.12 a. ⎛ R2 ⎞ ⎛ 50 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (12 ) = 10 V R1 + R2 ⎠ ⎝ 50 + 10 ⎠ ⎝ RTH = R1 R2 = 50 10 = 8.33 kΩ I BQ = 12 − 0.7 − 10 = 0.0119 mA 8.33 + (101)(1) I CQ = 1.19 mA, I EQ = 1.20 mA VECQ = 12 − (1.20 )(1) − (1.19 )( 2 ) VECQ = 8.42 V iC 4 1.19 8.42 12 ␯EC b. V0 Ϫ V␲ VS ϩ Ϫ r␲ gmV␲ ϩ R1͉͉R2 RC RE 209. rp = (100 )( 0.026 ) 1.19 V0 = g mVp RC = 2.18 kΩ ⎛V ⎞ VS = 2 Vp − ⎜ p + g mVp ⎟ RE ⎝ rp ⎠ ⎡ rπ + (1 + β ) RE ⎤ = −Vp ⎢ ⎥ rp ⎣ ⎦ 2 (100 )( 2 ) 2 β RC = ⇒ Av = 2 1.94 Av = rp + (1 + β ) RE 2.18 + (101)(1) c. Approximation: Assume rp does not vary significantly. RC = 2 kΩ ± 5% = 2.1 kΩ or 1.9 kΩ RE = 1 kΩ ± 5% = 1.05 kΩ or 0.95 kΩ For RC ( max ) = 2.1 kΩ and RE ( min ) Av = − (100 )( 2.1) 2.18 + (101)( 0.95 ) = −2.14 For RC ( min ) = 1.9 kΩ and RE ( max ) = 1.05 kΩ Av = − (100 )(1.9 ) 2.18 + (101)(1.05 ) = −1.76 So 1.76 ≤ Av ≤ 2.14 6.13 (a) VCC = ⎜ 1+ β ⎟ I CQ RE + VECQ + I CQ RC ⎜ ⎟ ⎛ ⎝ ⎞ β ⎠ ⎛ 101 ⎞ 12 = ⎜ ⎟ I CQ (1) + 6 + I CQ ( 2 ) ⎝ 100 ⎠ so that I CQ = 1.99 mA 1.99 = 0.0199 mA 100 = ( 0.1)(1 + β ) RE = ( 0.1)(101)(1) = 10.1 k Ω I BQ = RTH ⎛ R2 ⎞ 1 1 VTH = ⎜ ⎟ V = ⋅ R ⋅ V = (10.1)(12 ) R1 + R2 ⎠ CC R1 TH CC R1 ⎝ VCC = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH 121.2 12 = (101)( 0.0199)(1) + 0.7 + ( 0.0199)(10.1) + R1 which yields R1 = 13.3 k Ω and R2 = 41.6 k Ω (b) 6.14 Av = 2 (100 )( 2 ) 2 β RC = ⇒ Av = 2 1.95 rp + (1 + β ) RE 1.31 + (101)(1) 210. I CQ = 0.25 mA, I EQ = 0.2525 mA I BQ = 0.0025 mA I BQ RB + VBE ( on ) + I EQ ( RS + RE ) − 5 = 0 ( 0.0025)( 50 ) + 0.7 + ( 0.2525)( 0.1 + RE ) = 5 RE = 16.4 kΩ VE = − ( 0.0025 )( 50 ) − 0.7 = −0.825 V VC = VCEQ + VE = 3 − 0.825 = 2.175 V 5 − 2.175 RC = ⇒ RC = 11.3 kΩ 0.25 − β RC Av = rπ + (1 + β ) RS rπ = (100 )( 0.026 ) = 10.4 kΩ 0.25 − (100 )(11.3) Av = ⇒ Av = −55.1 10.4 + (101)( 0.1) Ri = RB ⎡ rπ + (1 + β ) RS ⎤ ⎣ ⎦ = 50 ⎡10.4 + (101)( 0.1) ⎤ ⎣ ⎦ Ri = 50 20.5 ⇒ Ri = 14.5 kΩ 6.15 (a) VCC > I CQ ( RC + RE ) + VCEQ 9 = I CQ ( 2.2 + 2 ) + 3.75 So that I CQ = 1.25 mA Assume circuit is to be designed to be bias stable. RTH = R1 R2 = ( 0.1)(1 + β ) RE = ( 0.1)(121)( 2 ) = 24.2 Ω 1.25 I BQ = = 0.01042 mA 120 1 VTH = ⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + I BQ (121)( RE ) R1 1 ( 24.2 )( 9 ) = ( 0.01042 )( 24.2 ) + 0.7 + ( 0.01042 )(121)( 2 ) R1 = 0.2522 + 0.7 + 2.5216 = 3.474 R1 = 62.7 K R2 = 39.4 K (b) 62.7 R2 = 24.2 62.7 + R2 211. 1.25 = 48.08 mA / V 0.026 (120 )( 0.026 ) rp = = 2.50 k Ω 1.25 100 ro = = 80 k Ω 1.25 gm = Vo ϩ IS V␲ r␲ R1͉͉R2Ϫ ro RC RL gmV␲ Vo = 2 g mVp ( ro RC RL ) Vp = I S ( R1 R2 rp ) Then Rm = Vo = 2 g m ( R1 R2 rp )( ro RC RL ) Is Rm = 2 48.08 ( 24.2 2.5 )( 80 2.2 1) = 2 48.08 ( 2.266 )( 0.6816 ) or Rm = Vo = −74.3 k Ω = −74.3 V / mA Is 6.16 a. I EQ = 0.80 mA, I BQ = 0.80 = 0.0121 mA 66 I CQ = 0.788 mA 0.3 ⇒ RB = 24.8 kΩ 0.0121 V − ( −5 ) 5 − 3 = ⇒ RC = 2.54 kΩ RC = C I CQ 0.788 VB = I BQ RB ⇒ RB = b. 0.788 = 30.3 mA / V 0.026 ( 65)( 0.026) rπ = = 2.14 kΩ 0.788 75 r0 = = 95.2 kΩ 0.788 ⎛ RC r0 ⎞ i0 = ⎜ g V , V = − vS ⎜R r +R ⎟ m π π ⎟ L ⎠ ⎝ C 0 gm = Gf = ⎛ RC r0 i0 = − gm ⎜ ⎜R r +R vS L ⎝ C 0 ⎞ ⎟ ⎟ ⎠ ⎛ 2.54 95.2 ⎞ = − ( 30.3) ⎜ ⎜ 2.54 95.2 + 4 ⎟ ⎟ ⎝ ⎠ G f = −11.6 mA/V 212. 6.17 (a) I CQ = 0.8 mA ⇒ I BQ = 0.00667 mA I BQ RS + 0.7 + (121) I BQ RE − 15 = 0 ( 0.01667 )( 2.5) + 0.7 + (121)( 0.00667 ) RE = 15 RE = 17.7 K VE = − ( 0.00667 )( 2.5 ) − 0.7 = −0.717 V VC = −0.717 + 7 = 6.283 V 15 − 6.283 RC = = 10.9 K 0.8 (120 )( 0.026 ) 0.8 gm = = 30.77 mA/V rπ = = 3.9 K 0.026 0.8 ⎛ r ⎞ vo = − g m ( RC RL ) ⋅ vπ vπ = ⎜ π ⎟ vS ⎝ rπ + RS ⎠ Av = − β ( RC RL ) rπ + RS = − (120 ) (10.9 5 ) 3.9 + 2.5 Av = −64.3 (b) For RS = 0 0.7 + (121)( 0.00667 ) RE = 15 RE = 17.7 K VE = −0.7 ⇒ VC = −0.7 + 7 = 6.3 15 − 6.3 ⇒ RC = 10.9 K 0.8 − β ( RC RL ) Av = = −30.77 (10.9 5 ) rπ RC = Av = −105 6.18 (a) 15 = ( 81) I BQ (10 ) + 0.7 + I BQ ( 2.5 ) 15 − 0.7 I BQ = = 0.0176 mA 2.5 + ( 81)(10 ) I CQ = 1.408 mA (80 )( 0.026 ) 1.408 = 54.15 mA/V rπ = 0.026 1.408 rπ = 1.48 K gm = RS V0 IS ϩ VS ϩ Ϫ V␲ Ϫ r␲ gmV␲ RC Io RL 213. Vo = − g mVσ ( RC RL ) ⇒ Av = i AI = o = iS − β ( RC RL ) rπ + RS = − ( 80 ) ( 5 5 ) 1.48 + 2.5 ⇒ Av = −50.3 ⎛ RC ⎞ − g mVπ ⎜ ⎟ ⎝ RC + RL ⎠ = − β ⎛ RC ⎞ ⎜ ⎟ V ⎝ RC + RL ⎠ π r AI = −40 π vo ( t ) = ( −50.3)( 4sin ω t ) vo ( t ) = −0.201sin ω t ( V ) 4 sin ω t ( mV ) = 1.005sin ω t ( μA ) 2.5 + 1.48 io = −40.2 sin ω t ( μA ) is = (b) 15 − 0.7 = 1.43 mA 10 ⎛ 80 ⎞ = ⎜ ⎟ (1.43) = 1.412 mA ⎝ 81 ⎠ I EQ = I CQ gm = (80 )( 0.026 ) 1.412 = 54.3 mA/V rπ = = 1.47 K 0.026 1.412 Av = − g m ( RC RL ) = − ( 54.3) ( 5 5 ) ⇒ Av = −136 ⎛ RC ⎞ ⎛ 5 ⎞ AI = − β ⎜ ⎟ = −80 ⎜ ⎟ ⇒ AI = −40 ⎝5+5⎠ ⎝ RC + RL ⎠ vo ( t ) = ( −136 )( 4sin ω t ) ⇒ vo ( t ) = −544sin ω t ⇒ vo ( t ) = −0.544sin ω t ( V ) is ( t ) = 4sin ω t ( mV ) = 2.72sin ω t ( μA ) 1.47 k io ( t ) = ( −40 )( 2.72sin ω t ) io ( t ) = −109sin ω t ( μA ) 6.19 RTH = R1 R2 = 27 15 = 9.64 K ⎛ R2 ⎞ ⎛ 15 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ ( 9 ) = 3.214 V R1 + R2 ⎠ ⎝ 15 + 27 ⎠ ⎝ V − VBE ( on ) 3.214 − 0.7 2.514 I BQ = TH = = RTH + (1 + β ) RE 9.64 + (101)(1.2 ) 130.84 I BQ = 0.0192 mA I CQ = 1.9214 mA gm = (100 )( 0.026 ) 1.92 = 73.9 mA/V rπ = = 1.35 K 0.026 1.92 214. RS V0 IS ϩ ϩ Ϫ VS ro = V␲ RTH r␲ gmV␲ Ϫ 100 = 52.1 K 1.92 ⎛ r R Vπ = ⎜ π TH ⎜r R +R S ⎝ π TH = 1.35 9.64 = 1.184 K ( Vo = − g mVπ r0 RC RL rπ RTH ⎛ 1.184 ⎞ Vπ = ⎜ ⎟ VS ⎝ 1.184 + 10 ⎠ = 0.1059VS ) ( Av = − ( 73.9 ) ( 0.1059 ) 52.1 2.2 2 ⎞ ⎟ VS ⎟ ⎠ ) = − ( 73.9 ) ( 0.1059 ) ( 52.1 1.0476 ) = − ( 73.9 ) ( 0.1059 ) (1.027 ) Av = −8.04 ⎛ ro RC − g mVπ ⎜ ⎜r R +R I L ⎝ o C AI = o = Vπ IS RTH rπ ⎞ ⎟ ⎟ ⎠ ⎛ ro RC ⎞ AI = − g m ( RTH rπ ) ⎜ ⎜r R +R ⎟ ⎟ L ⎠ ⎝ o C ro RC = 52.1 2.2 = 2.11 K RTH rπ = 9.64 1.35 = 1.184 K ⎛ 2.11 ⎞ AI = − ( 73.9 ) (1.184 ) ⎜ ⎟ ⎝ 2.11 + 2 ⎠ AI = −44.9 Ri = RTH rπ = 9.64 1.35 Ri = 1.184 K 6.20 a. 0.35 = 0.00347 mA 101 VB = 2 I B RB = 2 ( 0.00347 )(10 ) ⇒ VB = 2 0.0347 V I E = 0.35 mA, I B = VE = VB − VBE ( on ) ⇒ VE = 2 0.735 V b. r0 RC I0 RL 215. VC = VCEQ + VE = 3.5 − 0.735 = 2.77 V ⎛ b ⎞ ⎛ 100 ⎞ IC = ⎜ ⎟ IE = ⎜ ⎟ ( 0.35 ) = 0.347 mA ⎝ 101 ⎠ ⎝ 1+ b ⎠ V 1 − VC 5 − 2.77 RC = = ⇒ RC = 6.43 kΩ IC 0.347 (c) ⎛ RB rp ⎞ Av = 2 g m ⎜ R r ⎜ R r + R ⎟( C o ) ⎟ S ⎠ ⎝ B π 0.347 100 gm = = 13.3 mA/V , ro = = 288 k Ω 0.026 0.347 (100 )( 0.026 ) = 7.49 k Ω rp = 0.347 RB rp = 10 7.49 = 4.28 k Ω ⎛ 4.28 ⎞ Av = 2 (13.3) ⎜ ⎟ ( 6.43 288 ) ⇒ Av = 2 81.7 ⎝ 4.28 + 0.1 ⎠ d. ⎛ RB rp ⎞ Av = 2 g m ⎜ R r ⎜ R r + R ⎟( C 0 ) ⎟ S ⎠ ⎝ B p RB rp = 10 7.49 = 4.28 kΩ ⎛ 4.28 ⎞ Av = 2 (13.3) ⎜ ⎟ ( 6.43 288 ) ⇒ Av = 2 74.9 ⎝ 4.28 + 0.5 ⎠ 6.21 a. RTH = R1 R2 = 6 1.5 = 1.2 kΩ ⎛ R2 ⎞ + ⎛ 1.5 ⎞ VTH = ⎜ ⎟V = ⎜ ⎟ ( 5 ) = 1.0 V ⎝ 1.5 + 6 ⎠ ⎝ R1 + R2 ⎠ V − VBE ( on ) 1.0 − 0.7 I BQ = TH = = 0.0155 mA RTH + (1 + β ) RE 1.2 + (181)( 0.1) I CQ = 2.80 mA, I EQ = 2.81 VCEQ = V + − I CQ RC − I EQ RE = 5 − ( 2.8 )(1) − ( 2.81)( 0.1) ⇒ VCEQ = 1.92 V b. (180 )( 0.026 ) ⇒ rp = 1.67 kΩ 2.80 2.80 gm = ⇒ g m = 108 mA/V, r0 =` 0.026 (c) R1 R2 rp ⎛ ⎞ Av = 2 g m ⎜ ⎟ ( RC RL ) ⎝ R1 R2 rp + RS ⎠ R1 R2 rp = 6 1.5 1.67 = 0.698 k V rp = ⎛ 0.698 ⎞ Av = 2 (108 ) ⎜ ⎟ (1 1.2 ) ⇒ Av = 2 45.8 ⎝ 0.698 + 0.2 ⎠ 216. 6.22 a. 9 = I EQ RE + VEB ( on ) + I BQ RS 0.75 = 0.00926 mA I EQ = 0.75 mA, I BQ = 81 I CQ = 0.741 mA 9 = ( 0.75 ) RE + 0.7 + ( 0.00926 )( 2 ) ⇒ RE = 11.0 kΩ b. VE = 9 − ( 0.75 )(11) = 0.75 V VC = VE − VECQ = 0.75 − 7 = −6.25 V RC = VC − ( −9 ) I CQ = 9 − 6.25 ⇒ RC = 3.71 kΩ 0.741 c. ⎛ rp ⎞ Av = 2 g m ⎜ ⎟ ( RC || RL || r0 ) ⎝ rp + RS ⎠ (80 )( 0.026 ) rp = = 2.81 kΩ 0.741 80 r0 = = 108 kV 0.741 2 80 Av = ( 3.71||10 ||108 ) 2.81 + 2 Av = 2 43.9 d. Ri = RS + rp = 2 + 2.81 ⇒ Ri = 4.81 kΩ 6.23 I BQ = 4 − 0.7 = 0.00647 5 + (101)( 5 ) I CQ = 0.647 mA 80 ≤ h fe ≤ 120, 10 ≤ h0e ≤ 20 mS a. 2.45 kΩ ≤ hie ≤ 3.7 kΩ low gain high gain RS V0 IS ϩ VS ϩ Ϫ V␲ Ϫ r␲ gmV␲ RC Io RL 217. ⎛ 1 ⎞ V0 = 2 h fe I b ⎜ RC RL ⎟ ⎝ hoe ⎠ RB ?S V R + RS Ib = B RTH + hie RTH = RB RS = 5 1 = 0.833 kΩ High-gain ⎛ 5 ⎞ ⎜ ⎟ VS 5 +1⎠ Ib = ⎝ = 0.1838VS 0.833 + 3.7 Low-gain ⎛ 5 ⎞ ⎜ ⎟ VS ⎝ 5 +1⎠ Ib = = 0.2538VS 0.833 + 2.45 1 1 For hoe = 10 ⇒ || Rc || RL = || 4 || 4 0.010 hoe = 100 || 2 = 1.96 kΩ 1 || 4 || 4 = 50 || 2 = 1.92 kΩ 0.020 = (120 )( 0.1838 )(1.96 ) = 43.2 = ( 80 )( 0.2538 )(1.92 ) = 39.0 For hoe = 20 ⇒ Av Av max min 39.0 ≤ Av ≤ 43.2 b. Ri = RB hie = 5 3.7 = 2.13 kV or Ri = 5 2.45 = 1.64 kΩ 1.64 ≤ Ri ≤ 2.13 kΩ R0 = 1 1 4 = 100 || 4 = 3.85 kΩ RC = 0.010 hoe 1 || 4 = 50 || 4 = 3.70 kΩ 0.020 3.70 ≤ R0 ≤ 3.85 kΩ or R0 = 6.24 VCC ϭ 10 V RC R1 ␯o RS ϭ 1 k⍀ CC ␯s ϩ Ϫ R2 RE CE 218. Assume an npn transistor with b = 100 and VA = ∞. Let VCC = 10 V . 0.5 = 50 0.01 Bias at I CQ = 1 mA and let RE = 1 k Ω Av = For a bias stable circuit RTH = ( 0.1)(1 + b ) RE = ( 0.1)(101)(1) = 10.1 k Ω 1 1 101 VTH = ⋅ RTH ⋅ VCC = (10.1)(10 ) = R1 R1 R1 1 I BQ = = 0.01 mA 100 VTH = I BQ RTH + VBE ( on ) + (1 + b ) I BQ RE 101 = ( 0.01)(10.1) + 0.7 + (101)( 0.01)(1) R1 which yields R1 = 55.8 k Ω and R2 = 12.3 k Ω Now rp = (100 )( 0.026 ) 1 = 2.6 k Ω 1 = 38.46 mA/V 0.026 Vo = − g mVp RC gm = ⎛ R1} R2 } rp ⎞ ⎛ 10.1} 2.6 ⎞ where Vp = ⎜ ⎟ ⋅ Vs = ⎜ ⎟ .Vs ⎝ 10.1} 2.6 + 1 ⎠ ⎝ R1} R2 } rp + RS ⎠ or Vp = 0.674 Vs V Then Av = o = − ( 0.674 ) g m RC = − ( 0.674 )( 38.46 ) RC = −50 Vs which yields RC = 1.93 k Ω With this RC, the dc bias is OK. Finish Design, Set RC = 2 K RE = 1 K R1 = 56 K R2 = 12 K RTH = R1 R2 = 9.88 K ⎛ R2 ⎞ ⎛ 12 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (10 ) = 1.765 V ⎝ 12 + 56 ⎠ ⎝ R1 + R2 ⎠ 1.765 − 0.7 I BQ = = 9.60 μ A 9.88 + (101)(1) I CQ = 0.9605 mA rπ = RTH (100 )( 0.026 ) 0.9605 rπ = 2.125 K = 2.707 K gm = 0.9605 = 36.94 0.026 ⎛ RTH rπ ⎞ ⎛ 2.125 ⎞ Vπ = ⎜ ⎟ Vi = ⎜ ⎟ Vi = ( 0.680 ) Vi ⎝ 2.125 + 1 ⎠ ⎝ RTH rπ + RS ⎠ Av = − ( 0.680 ) g m RC = − ( 0.680 )( 36.94 )( 2 ) = −50.2 219. Design specification met. 6.25 a. I BQ = 6 − 0.7 = 0.0169 mA 10 + (101)( 3) I CQ = 1.69 mA, I EQ = 1.71 mA VCEQ = (16 + 6 ) − (1.69 )( 6.8 ) − (1.71)( 3) VCEQ = 5.38 V b. 1.69 ⇒ g m = 65 mA/V 0.026 (100 )( 0.026 ) rp = ⇒ rp = 1.54 kV , 1.69 (c) − β ( RC RL ) RB Rib Av = ⋅ rπ + (1 + β ) RE RB Rib + RS gm = r0 = ∞ Rib = rπ + (1 + β ) RE = 1.54 + (101)(3) = 304.5 k Ω RB Rib = 10 304.5 = 9.68 k Ω Then Av = − (100 )( 6.8 6.8 ) ⎛ 9.68 ⎞ ⋅⎜ ⎟ ⇒ Av = −1.06 1.54 + (101)( 3) ⎝ 9.68 + 0.5 ⎠ ⎛ RC ⎞ i0 = ⎜ ⎟ ( − β ib ) ⎝ RC + RL ⎠ ⎛ RB ib = ⎜ RB + rπ + (1 + β ) RE ⎝ ⎞ ⎟ iS ⎠ ⎞ ⎞⎛ RB ⎟⎜ ⎜ R + r + (1 + β ) R ⎟ ⎟ E ⎠ ⎠⎝ B π ⎞ 10 ⎛ 6.8 ⎞ ⎛ = − (100 ) ⎜ ⎟⎜ ⎜ 10 + 1.54 + (101)( 3) ⎟ ⇒ Ai = −1.59 ⎟ ⎝ 6.8 + 6.8 ⎠ ⎝ ⎠ (d) Ris = RS + RB Rib = 0.5 + 10 304.5 = 10.2 k Ω ⎛ RC Ai = − ( β ) ⎜ ⎝ RC + RL (e) Av = Av = 2 b ( RC RL ) rp + (1 + b ) RE 2 (100 ) ( 6.8} 6.8 ) 1.54 + (101)( 3) ⇒ Av = 2 1.12 Ai = same as ( c ) ⇒ Ai = 2 1.59 6.26 220. ie ϩ ϩ vCE ϩ vbe Ϫ Ϫ vCE gmv␲ ris ␲o Ϫ ϩ vCE gmv␲ Ϫ r= vCe 1 = g m vCe g m ⎛ 1 ⎞ So re = rp ⎜ ⎟ r0 ⎝ gm ⎠ 6.27 Let b = 100, VA = ∞ VCC RC R1 ␯o RS ϭ 100 ⍀ CC ␯s ϩ Ϫ R2 RE Let VCC = 2.5 V P = ( I R + I C ) VCC ⇒ 0.12 = ( I R + I C )( 2.5 ) ⇒ I R + I C = 48 mA, Let I R = 8mA, I C = 40 mA R1 + R2 > I BQ = VCC 2.5 ⇒ 312.5 k Ω = IR 8 40 = 0.4 mA 100 221. Let RE = 2 k Ω. For a bias stable circuit RTH = ( 0.1)(1 + b ) RE = ( 0.1)(101)( 2 ) = 20.2 k Ω 1 VTH = ⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + (1 + b ) I BQ RE R1 1 ( 20.2 )( 2.5 ) = ( 0.0004 )( 20.2 ) + 0.7 + (101)( 0.0004 )( 2 ) R1 which yields R1 = 64 k V and R2 = 29.5 k Ω (100 )( 0.026 ) rπ = 0.04 Av = = 65 k Ω Neglect RS Vo 2 b RC > Vs rπ + (1 + b ) RE −10 = 2 100 RC ⇒ RC = 26.7 k Ω 65 + (101)( 2 ) With this RC , dc biasing is OK. 6.28 100 = 20. 5 Assume a sign inversion from a common-emitter is not important. Use the configuration for Figure 6.31. Let RS = 0. Need an input resistance of Need a voltage gain of 5 × 102 3 = 25 × 103 = 25 k Ω 0.2 × 102 6 Ri = RTH Rib . Let RTH = 50 k Ω, Rib = 50 k Ω Ri = Rib = rp + (1 + b ) RE > (1 + b ) RE Rib 50 = = 0.495 k Ω 1 + b 101 Let RE = 0.5 k V , VCC = 10 V , I CQ = 0.2 mA For b = 100, RE = 0.2 = 0.002 mA 100 = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE Then I BQ = VTH 1 1 ⋅ RTH ⋅ VCC = ( 50)(10) = ( 0.002)(50) + 0.7 + (101)( 0.002 )( 0.5) R1 R1 which yields R1 = 555 k Ω and R2 = 55 k Ω Now Av = (100)( 0.026) − β RC = 13 k Ω , rπ = rπ + (1 + β ) RE 0.2 So −20 = − (100 ) RC 13 + (101)( 0.5) ⇒ RC = 12.7 k Ω [Note: I CQ RC = ( 0.2 )(12.7 ) = 2.54 V. So dc biasing is OK.] 6.29 222. VCC ϭ 10 V R1 RE CC ␯s ␯o ϩ Ϫ R2 RC b = 80, Av = 2 b RC rp + (1 + b ) RE First approximation: R ( Av ) ≈ C = 10 ⇒ RC = 10 RE RE Set RC = 12 RE VEC ≈ VCC − I C ( RC + RE ) = 10 − I C (13RE ) 1 For VEC = VCC = 5 2 5 = 10 − I C (13RE ) For I C = 0.7 mA I E = 0.709, I B = 0.00875 mA ⇒ RE = 0.55 kΩ − RC = 6.6 kΩ Bias stable ⇒ R1 R2 = RTH = ( 0.1)(1 + β ) RE = ( 0.1)(81)( 0.55 ) = 4.46 kΩ 1 10 = ( 0.709 )( 0.55 ) + 0.7 + ( 0.00875 )( 4.46 ) + ( 4.46 )(10 ) R1 8.87 = 1 ( 4.46 ) ⇒ R1 = 5.03 kΩ R1 5.03R2 = 4.46 ⇒ R2 = 39.4 kΩ 5.03 + R2 10 10 = = 0.225 mA R1 + R2 5.03 + 39.4 0.7 + 0.225 ≅ 0.925 mA from VCC source. Now rπ = Av = 6.30 (80 ) ( 0.026 ) 0.7 (80 )( 6.6 ) = 2.97 kΩ 2.97 + ( 81)( 0.55 ) = 11.1 223. ϩ5V RC R1 CC2 ␯o CC1 RL ϭ 10 K ␯s ϩ Ϫ R2 CE RE Ϫ5V β = 120 Let I CQ = 0.35 mA, I EQ = 0.353 mA I BQ = 0.00292 mA Let RE = 2 kΩ. For VCEQ = 4 V ⇒ 10 = 4 + ( 0.35) RC + ( 0.353)( 2) RC = 15.1 kΩ, rπ = Av = − β ( RC RL ) rπ (120 )( 0.026 ) =− = 8.91 kΩ 0.35 (120 ) (15.1 10 ) 8.91 Av = −81.0 For bias stable circuit: R1 R2 = RTH = ( 0.1)(1 + β ) RE = ( 0.1)(121)( 2 ) = 24.2 kΩ ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ (10) − 5 = ⋅ RTH ⋅ (10 ) − 5 R1 ⎝ R1 + R2 ⎠ VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 5 1 ( 24.2 )(10 ) − 5 = ( 0.00292 )( 24.2 ) + 0.7 + (121)( 0.00292 )( 2 ) − 5 R1 1 ( 242 ) = 1.477, R1 = 164 kΩ R1 164 R2 = 24.2 ⇒ R2 = 28.4 kΩ 164 + R2 10 = 0.052, 0.35 + 0.052 = 0.402 mA 164 + 28.4 So bias current specification is met. 6.31 From Prob. 6.12, 224. RTH = R1} R2 = 10}50 = 8.33 kΩ ⎛ R2 ⎞ ⎛ 50 ⎞ VTH = ⎜ ⎟ (12 ) = ⎜ ⎟ (12 ) = 10 V ⎝ 50 + 10 ⎠ ⎝ R1 + R2 ⎠ 12 − 0.7 − 10 = 0.0119 mA I BQ = 8.33 + (101)(1) I CQ = 1.19 mA, I EQ = 1.20 mA VECQ = 12 − (1.19 )( 2 ) − (120 )(1) = 8.42 V 1.19 1 8.42 11 12 For 1 ≤ vEC ≤ 11 DvEC = 11 − 8.42 = 2.58 ⇒ Output voltage swing = 5.16 V (peak-to-peak) 6.32 I BQ = 5 − 0.7 = 0.00315 mA 50 + (101)( 0.1 + 12.9) I CQ = 0.315 mA, I EQ = 0.319 mA VCEQ = ( 5 + 5 ) − ( 0.315 )( 6 ) − ( 0.319 )(13) VCEQ = 3.96 V AC load line Ϫ1 Slope ϭ 6.1 K 0.315 3.96 10 1 Δv 6.1 eC For ΔiC = 0.315 − 0.05 = 0.265 ⇒ ΔvEC = 1.62 ΔiC = − vEC ( min ) = 3.96 − 1.62 = 2.34 Output signal swing determined by current: 225. Max. output swing = 3.24 V peak-to-peak 6.33 From Problem 4.18, I CQ = 1.408 mA, I EQ = 1.426 mA (a) VECQ = 30 − (1.408 )( 5 ) − (1.426 )(10 ) = 8.7 V IC (mA) AC load line Ϫ1 Slope ϭ RC ͉͉RL Ϫ1 ϭ 2.5 kΩ 1.408 8.7 ␯EC (V) vEC ( max ) = 8.7 + ΔI C ⋅ ( 2.5 ) = 8.7 + (1.408 )( 2.5 ) = 12.22 Set vEC ( max ) = 12 = 8.7 + ΔI C ( 2.5 ) ⇒ ΔI C = 1.32 mA So ΔvEC (peak-to-peak) = 2(12 − 8.7) = 6.6 V (b) ΔiC (peak-to-peak) = 2(1.32) = 2.64 mA 6.34 I EQ I BQ VE VC VECQ = 0.80 mA, I CQ = 0.792 mA = 0.00792 mA = 0.7 + ( 0.00792 )(10 ) = 0.779 V = I CQ RC − 5 = ( 0.792 )( 4 ) − 5 = 2 1.83 V = 0.779 − ( −1.83) = 2.61 V Load line: Assume VE remains constant at ≈ 0.78 V IC (mA) AC load line Ϫ1 Slope ϭ RC ͉͉RL Ϫ1 ϭ 2.5 kΩ 1.408 8.7 ␯EC (V) 21 ? ec v 2 kV Collector current swing = 0.792 − 0.08 = 0.712 mA Dvec = ( 0.712 )( 2 ) = 1.424 V DiC = Output swing determined by current. Max. output swing = 2.85 V peak-to-peak 2.85 4 = 0.712 mA peak-to-peak Swing in i0 current = 226. 6.35 I BQ = 6 − 0.7 = 0.0169 mA 10 + (101)( 3) I CQ = 1.69 mA, I EQ = 1.71 mA VCEQ = (16 + 6 ) − (1.69 )( 6.8 ) − (1.71)( 3) VCEQ = 5.38 V AC load line Ϫ1 Slope ϭ 3.4 ϩ 3 Ϫ1 ϭ 6.4 K 1.69 5.38 22 1 DiC = 2 Dvce 6.4 For vce ( min ) = 1 V, Dvce = 5.38 − 1 = 4.38 V ⇒ DiC = 4.38 = 0.684 mA 6.4 Output swing limited by voltage: Δvce = Max. swing in output voltage = 8.76 V peak-to-peak 1 ΔiC ⇒ Δi0 = 0.342 mA 2 or Δi0 = 0.684 mA (peak-to-peak) Δi0 = 6.36 AC load line Ϫ1 Slope ϭ 1.05 K 2.65 Q-point ICQ VCEQ ro = 9 100 I CQ Neglect ro as (E) approx. dc load line VCE = 9 − I C ( 3.4 ) 227. ΔI C = I CQ − 0.1 ΔVCE = VCEQ − 1 Also ΔVCE = ΔI C ( RC RL ) = ΔI C (1.05 ) Or VCEQ − 1 = ( I CQ − 0.1) (1.05 ) Substituting the expression for the dc load line. ⎡9 − I CQ ( 3.4 ) − 1⎤ = ( I CQ − 0.1) (1.05 ) ⎣ ⎦ 8.105 = I CQ ( 4.45 ) ⇒ I CQ = 1.821 mA VCEQ = 2.81 V 1.821 I BQ = = 0.01821 100 RTH = ( 0.1)(101)(1.2 ) = 12.12 K 1 1 VTH = ⋅ RTH ⋅ VCC = (12.12 ) ( 9 ) = ( 0.01821) (12.12 ) + 0.7 + (101)( 0.01821)(1.2 ) R1 R1 = 0.2207 + 0.7 + 2.20705 R1 = 34.9 K R2 = 18.6 K 34.9 R2 = 12.12 34.9 + R2 6.37 dc load line 5 ϭ 4.55 mA 1 ϩ 0.1 AC load line Ϫ1 Slope ϭ 1͉͉1.2 Ϫ1 ϭ 0.545 K ICQ VCEQ 5 For maximum symmetrical swing ΔiC = I CQ − 0.25 ΔvCE = VCEQ − 0.5 and ΔiC = I CQ − 0.25 = VCEQ − 0.5 1 ⋅ | ΔvCE | 0.545 kΩ 0.545 VCEQ = 5 − I CQ (1.1) 0.545 ( I CQ − 0.25 ) = ⎡5 − I CQ (1.1) ⎤ − 0.5 ⎣ ⎦ ( 0.545 + 1.1) I CQ = 5 − 0.5 + 0.136 I CQ = 2.82 mA, I BQ = 0.0157 mA RTH = R1 R2 = ( 0.1)(1 + β ) RE = ( 0.1)(181)( 0.1) = 1.81 kΩ VTH = 1 ⋅ RTH ⋅ V + = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE R1 228. 1 (1.81)( 5 ) = ( 0.0157 )(1.81) + 0.7 + (181)( 0.0157 )( 0.1) R1 1 ( 9.05 ) = 1.013 ⇒ R1 = 8.93 kΩ R1 8.93R2 = 1.81 ⇒ R2 =2.27 kΩ 8.93 + R2 6.38 I CQ = 0.647 mA , VCEQ > 10 − ( 0.647 )( 9 ) = 4.18 V DiC = I CQ = 0.647 mA So DvCE = DiC ( 4} 4 ) = ( 0.647 )( 2 ) = 1.294 V Voltage swing is well within the voltage specification. Then DvCE = 2 (1.294 ) = 2.59 V peak-to-peak 6.39 a. RTH = R1} R2 = 10}10 = 5 kΩ ⎛ R2 ⎞ ⎛ 10 ⎞ VTH = ⎜ ⎟ (18 ) − 9 = ⎜ ⎟ (18 ) − 9 = 0 ⎝ 10 + 10 ⎠ ⎝ R1 + R2 ⎠ 0 − 0.7 − ( −9 ) = 0.0869 mA I BQ = 5 + (181)( 0.5 ) I CQ = 15.6 mA, I EQ = 15.7 mA VCEQ = 18 − (15.7 )( 0.5 ) ⇒ VCEQ = 10.1 V b. AC load line Ϫ1 Slope ϭ 0.5͉͉0.3 Ϫ1 ϭ 0.188 K 15.6 10.1 c. rπ = (180 )( 0.026 ) 15.6 (1 + β )( RE 18 = 0.30 kΩ RL ) ⎛ R1 R2 Rib ⋅⎜ rπ + (1 + β ) ( RE RL ) ⎝ R1 R2 Rib + RS Rib = rπ + (1 + β )( RE RL ) = 0.30 + (181)( 0.5 Av = ⎞ ⎟ ⎠ 0.3) or Rib = 34.2 k Ω R1 R2 Rib = 5 34.2 = 4.36 k Ω Av = d. (181)( 0.5 0.3) ⎛ 4.36 ⎞ ⋅⎜ ⎟ ⇒ Av 0.3 + (181)( 0.5 0.3) ⎝ 4.36 + 1 ⎠ = 0.806 229. Rib = rp + (1 + b ) ( RE } RL ) Rib = 0.30 + (181)( 0.188 ) ⇒ Rib = 34.3 kΩ Ro = RE rp + R1} R2 } RS 0.3 + 5}1 ⇒ Ro = 6.18 Ω = 0.5 1+ b 181 6.40 a. RTH = R1} R2 = 10}10 = 5 kΩ ⎛ R2 ⎞ VTH = ⎜ ⎟ ( −10 ) = 2 5 V ⎝ R1 + R2 ⎠ VTH = I BQ RTH + VBE ( on ) + (1 + b ) I BQ RE − 10 I BQ = 2 5 − 0.7 − ( −10 ) 5 + (121)( 2 ) = 0.0174 mA I CQ = 2.09 mA, I EQ = 2.11 mA VCEQ = 10 − ( 2.09 )(1) − ( 2.11)( 2 ) ⇒ VCEQ = 3.69 V b. AC load line Ϫ1 Slope ϭ 2͉͉2 Ϫ1 ϭ 1K 2.09 3.69 c. rπ = 10 (120 )( 0.026 ) = 1.49 kΩ 2.09 (1 + β ) ( RE RL ) ⎛ R1 R2 Rib ⎞ Av = ⋅⎜ ⎟ rπ + (1 + β ) ( RE RL ) ⎜ R1 R2 Rib + RS ⎟ ⎝ ⎠ Rib = rπ + (1 + β ) ( RE RL ) = 1.49 + (121) ( 2 2) Rib = 122.5 k Ω, Av = R1 R2 Rib = 5 122.5 = 4.80 k Ω (121) ( 2 2) ⎛ 4.80 ⎞ ⋅⎜ ⎟ ⇒ Av 1.49 + (121) ( 2 2) ⎝ 4.80 + 5 ⎠ = 0.484 d. Rib = rπ + (1 + b ) ( RE } RL ) Rib = 1.49 + (121) ( 2} 2 ) 1 Rib = 122 kΩ Ro = RE 6.41 a. rπ + R1} R2 } RS 1.49 + 5}5 1 Ro = 32.4 Ω =2 1+ b 121 230. RTH = R1 R2 = 60 40 = 24 kΩ ⎛ R2 ⎞ ⎛ 40 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ ( 5) = 2 V ⎝ 40 + 60 ⎠ ⎝ R1 + R2 ⎠ 5 − 0.7 − 2 I BQ = = 0.0130 mA 24 + ( 51)( 3) I CQ = 0.650 mA, I EQ = 0.663 mA VECQ = 5 − I EQ RE = 5 − ( 0.663)(3) ⇒ VECQ = 3.01 V b. 1.63 AC load line Ϫ1 Slope ϭ 51 Θ 50 Ι Θ3͉͉4Ι Ϫ1 ϭ 1.75 K 0.65 5 3.01 c. ( 50 )( 0.026 ) 80 = 2 kΩ, r0 = = 123 kΩ 0.650 0.65 ′ Define RL = RE RL r0 = 3 4 123 = 1.69 kΩ rπ = Av = ′ (1 + β ) RL ′ rπ (1 + β ) RL = ( 51)(1.69 ) 2 + ( 51)(1.69 ) ⎛ RE r0 Ai = (1 + β ) I b ⎜ ⎜R r +R L ⎝ E 0 ⇒ Av = 0.977 ⎞ ⎟ ⎟ ⎠ ⎛ RTH ⎞ Ib = I S ⎜ ⎟ ⎝ RTH + Rib ⎠ ′ Rib = rπ + (1 + β ) RL = 2 + ( 51)(1.69 ) = 88.2 RE r0 = 3 r0 = 3 123 = 2.93 24 ⎞ ⎛ 2.93 ⎞ ⎛ Ai = ( 51) ⎜ ⎟⎜ ⎟ ⇒ Ai = 4.61 2.93 + 4 ⎠ ⎝ 24 + 88.2 ⎠ ⎝ d. Rib = rπ + (1 + β ) RE RL r0 = 2 + ( 51)(1.69 ) ⇒ Rib = 88.2 kΩ rπ ⎛ 2⎞ RE = ⎜ ⎟ 3 = 0.0392 3 1+ β ⎝ 51 ⎠ R0 = 38.7 Ω e. Assume variations in rπ and r0 have negligible effects R1 = 60 ± 5% R1 = 63 kΩ, R1 = 57 kΩ R0 = R2 = 40 ± 5% R2 = 42 kΩ, R2 = 38 kΩ RE = 3 ± 5% RE = 3.15 kΩ, RE = 2.85 kΩ RL = 4 ± 5% RL = 4.2 kΩ, RL = 3.8 kΩ 231. ⎛ RE r0 ⎞ ⎛ RTH ⎞ Ai = (1 + β ) ⎜ ⎜ R r + R ⎟⎜ R + R ⎟ ⎟ L ⎠ ⎝ TH ib ⎠ ⎝ E 0 Rib = rπ + (1 + β ) ( RE RL r0 ) RTH ( max ) = 25.2 kΩ, RTH ( min ) = 22.8 kΩ Rib ( max ) = 92.5 kΩ, Rib ( min ) = 84.0 kΩ RE ( max ) , RL ( min ) , Rib = 88.6 kΩ RE ( min ) , RL ( max ) , Rib = 87.4 kΩ RE ( max ) r0 = 3.07 kΩ RE ( min ) r0 = 2.79 kΩ For RE ( min ) , RL ( max ) , RTH ( min ) 22.8 ⎞ ⎛ 2.79 ⎞ ⎛ Ai = ( 51) ⎜ ⎟⎜ ⎟ ⇒ Ai = 4.21 ⎝ 2.79 + 4.2 ⎠ ⎝ 22.8 + 87.4 ⎠ For RE ( max ) , RL ( min ) , RTH ( max ) 25.2 ⎞ ⎛ 3.07 ⎞⎛ Ai = ( 51) ⎜ ⎟⎜ ⎟ ⇒ Ai = 5.05 3.07 + 3.8 ⎠⎝ 25.2 + 88.6 ⎠ ⎝ 6.42 (a) 0.5 = 0.00617 mA 81 VB = I BQ RB = ( 0.00617 )(10 ) ⇒ VB = 0.0617 V I BQ = VE = VB + 0.7 ⇒ VE = 0.7617 V (b) ⎛ 80 ⎞ I CQ = ( 0.5 ) ⎜ ⎟ = 0.494 mA ⎝ 81 ⎠ I CQ 0.494 gm = = ⇒ g m = 19 mA / V VT 0.026 rπ = ro = β VT I CQ = (80 )( 0.026 ) 0.494 ⇒ rπ = 4.21 k Ω VA 150 ⇒ ro = 304 k Ω = I CQ 0.494 (c) RS Vs ϩ Ϫ IS VЈ S Ϫ RB V␲ r␲ ro gmV␲ ϩ Vo RL Io For RS = 0 ⎛V ⎞ Vo = − ⎜ π + g mVπ ⎟ ( RL ro ) ⎝ rπ ⎠ 232. −Vo ⎛1+ β ⎞ ⎜ ⎟ ( RL ro ) ⎝ rπ ⎠ Now Vs + Vπ = Vo so that Vπ = or Vs = Vo − Vπ = Vo + Vo ⎛ 1+ β ⎞ ⎜ ⎟ ( RL ro ) ⎝ rπ ⎠ We find (1 + β )( RL ro ) (81)( 0.5 304 ) = rπ + (1 + β )( RL ro ) 4.21 + ( 81)( 0.5 304 ) (81)( 0.5 ) ≅ ⇒ Av = 0.906 4.21 + ( 81)( 0.5 ) Rib = rπ + (1 + β )( RL ro ) ≅ 4.21 + ( 81)( 0.5 ) = 44.7 k Ω Av = Vo Vs = ⎛ RB ⎞ ⎛ ro ⎞ Ib = ⎜ ⎟ ⋅ I s and I o = ⎜ ⎟ (1 + β ) I b ⎝ RB + Rib ⎠ ⎝ ro + RL ⎠ Then ⎛ RB ⎞⎛ ro ⎞ Io = (1 + β ) ⎜ ⎟⎜ ⎟ Is ⎝ RB + Rib ⎠⎝ ro + RL ⎠ ⎛ 10 ⎞ Ai ≅ ( 81) ⎜ ⎟ (1) ⇒ Ai = 14.8 ⎝ 10 + 44.7 ⎠ (d) ⎛ RB + Rib ⎞ ⎛ 10 44.7 ⎞ Vs′ = ⎜ ⋅V = ⋅ V = 0.803) Vs ⎜ R R + R ⎟ s ⎜ 10 44.7 + 2 ⎟ s ( ⎟ ⎜ ⎟ s ⎠ ⎝ B ib ⎝ ⎠ Then Av = ( 0.803)( 0.906 ) ⇒ Av = 0.728 Ai = Ai = 14.8 (Unchanged) 6.43 (a) I CQ = 1.98 mA ro = rπ = (100 )( 0.026 ) 1.98 = 1.313 K VA 100 = I CQ 1.98 = 50.5 K rπ + RS 1.31 + 10 ro = 50.5 ⇒ Ro = 112 Ω 1+ β 101 0.112 50.5 ⇒ Ro ≅ 112 Ω (b) From Equation 4.68 (1 + β ) ( ro RL ) 100 Av = ro = = 50.5 K 1.98 rπ + (1 + β ) ( ro RL ) Ro = (i) 233. RL = 0.5 K (101) ( 50.5 0.5) 1.31 + (101) ( 50.5 0.5 ) (101)( 0.4951) ⇒ Av = 0.974 Av = 1.31 + (101)( 0.4951) Av = (ii) RL = 5 K Av = ro RL = 50.5 5 = 4.5495 (101)( 4.55) ⇒ Av = 0.997 1.31 + (101)( 4.55 ) 6.44 5 − 0.7 I CQ = 1.293 mA = 1.303 3.3 (125 )( 0.026 ) rπ = = 2.51 K 1.293 1.293 gm = = 49.73 mA/V 0.026 (a) Rib = rπ + (1 + β ) ( RE RL ) = 2.51 + (126 ) ( 3.3 1) I EQ = Rib = 99.2 K Ro = RE rπ 2.51 = 3.3 = 3.3 0.01992 1+ β 126 Ro = 19.8 Ω (b) v 2sin ω t ⇒ is ( t ) = 20.2sin ω t ( μ A ) is = s = Rib 99.2 veb ( t ) = −is ( t ) rπ = ( −20.2 )( 2.51) sin ω t veb ( t ) = −50.6sin ω t ( mV ) (126 ) ( 3.3 1) (1 + β ) ( RE RL ) (126 )( 0.7674 ) = = rπ + (1 + β ) ( RE RL ) 2.51 + (126 ) ( 3.3 1) 2.51 + (126 )( 0.7674 ) Av = 0.9747 ⇒ vo ( t ) = 1.95sin ω t ( V ) v (t ) ⇒ io ( t ) = 1.95sin ω t ( mA ) io ( t ) = o Av = RL 6.45 a. I EQ = 1 mA , VCEQ = VCC − I EQ RE 5 = 10 − (1)( RE ) ⇒ RE = 5 kΩ 1 = 0.0099 mA 101 10 = I BQ RB + VBE ( on ) + I EQ RE 10 = ( 0.0099 ) RB + 0.7 + (1)( 5 ) ⇒ RB = 434 kΩ I BQ = b. 234. ␯b ␯0 RB rπ = RE (100 )( 0.026 ) 0.99 (1 + β ) RE = 2.63 kΩ (101)( 5 ) v0 = = = 0.995 vb rπ + (1 + β ) RE 2.63 + (101)( 5 ) ⇒ vb = v0 4 = ⇒ vb = 4.02 V peak-to-peak at base 0.995 0.995 RS ␯b ␯S ϩ Ϫ RB͉͉Rib Rib = rπ + (1 + β ) RE = 508 kΩ RB Rib = 434 508 = 234 kΩ vb = RB Rib RB Rib + RS ⋅ vS = vb = 0.997vS ⇒ vS = 234vS 234 = vS 234 + 0.7 234.7 4.02 ⇒ vS = 4.03 V peak-to-peak 0.997 c. Rib = rπ + (1 + β ) ( RE RL ) Rib = 2.63 + (101) ( 5 1) = 86.8 kΩ RB Rib = 434 86.8 = 72.3 kΩ ⎛ 72.3 ⎞ vb = ⎜ ⎟ vS = 0.99vS = ( 0.99 )( 4.03) ⎝ 72.3 + 0.7 ⎠ vb = 3.99 V peak-to-peak (1 + β )( RE RL ) ⋅ vb rπ + (1 + β )( RE RL ) (101)( 0.833) = ( 3.99 ) 2.63 + (101)( 0.833) v0 = v0 = 3.87 V peak-to-peak 6.46 235. RTH = R1 R2 = 40 60 = 24 kΩ ⎛ 60 ⎞ VTH = ⎜ ⎟ (10 ) = 6 V ⎝ 60 + 40 ⎠ 6 − 0.7 β = 75 I BQ = = 0.0131 mA 24 + ( 76 )( 5 ) I CQ = 0.984 mA β = 150 I BQ = 6 − 0.7 = 0.00680 mA 24 + (151)( 5 ) I CQ = 1.02 mA β = 75 rπ = ( 75 )( 0.026 ) 0.984 β = 150 rπ = 3.82 kΩ = 1.98 kΩ β = 75 Rib = rπ + (1 + β )( RE RL ) = 65.3 kΩ β = 150 Rib = 130 kΩ (1 + β )( RE RL ) R1 R2 Rib Av = ⋅ rπ + (1 + β )( RE RL ) R1 R2 Rib + RS For β = 75, R1 R2 Rib = 40 60 65.3 = 17.5 k Ω Av = ( 76 )( 0.833) 17.5 ⋅ ⇒ Av = 0.789 1.98 + ( 76 )( 0.833) 17.5 + 4 For β = 150, R1 R2 Rib = 40 60 130 = 20.3 k Ω Av = (151)( 0.833) 20.3 ⋅ ⇒ Av = 0.811 3.82 + (151)( 0.833) 20.3 + 4 So 0.789 ≤ Av ≤ 0.811 ⎛ RE ⎞ ⎛ RTH ⎞ Ai = (1 + β ) ⎜ ⎟ ⎟⎜ ⎝ RE + RL ⎠ ⎝ RTH + Rib ⎠ β = 75 24 ⎞ ⎛ 5 ⎞⎛ Ai = ( 76 ) ⎜ ⎟⎜ ⎟ ⇒ Ai = 17.0 ⎝ 5 + 1 ⎠ ⎝ 24 + 65.3 ⎠ β = 150 ⎛ 5 ⎞ ⎛ 24 ⎞ Ai = (151) ⎜ ⎟ ⎜ ⎟ ⇒ Ai = 19.6 ⎝ 6 ⎠ ⎝ 24 + 130 ⎠ 17.0 ≤ Ai ≤ 19.6 6.47 (a) 236. ⎛ I ⎞ 9 = ⎜ E ⎟ (100 ) + VBE ( on ) + I E RE ⎝ 1+ β ⎠ 9 − 0.7 IE = ⎛ 100 ⎞ ⎜ ⎟ + RE ⎝ 1+ β ⎠ 8.3 = 2.803 mA ⎛ 100 ⎞ +1 ⎜ ⎟ ⎝ 51 ⎠ 8.3 β = 200 I E = = 5.543 mA ⎛ 100 ⎞ ⎜ ⎟ +1 ⎝ 201 ⎠ 2.80 ≤ I E ≤ 5.54 mA β = 50 I E = VE = I E RE , β = 50, VE = 2.80 V β = 200, VE = 5.54 V β = 50, I CQ = 2.748 mA, rπ = 0.473 K (b) β = 200, I CQ = 5.515 mA, rπ = 0.943 K Ri = RB ⎡ rπ + (1 + β ) RE ⎣ RL ⎤ ⎦ β = 50 ⇒ Ri = 100 ⎡ 0.473 + ( 51)(1 1) ⎤ = 100 25.97 = 20.6 K ⎣ ⎦ β = 200 ⇒ Ri = 100 ⎡0.943 + ( 201)(1 1) ⎤ = 100 101.4 = 50.3 K ⎣ ⎦ From Fig. (4.68) (1 + β ) ( RE RL ) ⎛ Ri ⎞ Av = ⋅⎜ ⎟ rπ + (1 + β ) ( RE RL ) ⎝ Ri + RS ⎠ = ( 51) (1 1) ⎛ 20.6 ⎞ ⋅⎜ ⎟ 0.473 + ( 51) (1 1) ⎝ 20.6 + 10 ⎠ β = 50 ⇒ Av = 0.661 β = 200 ⇒ Av = ( 201) (1 1) ⎛ 50.3 ⎞ ⎜ ⎟ 0.943 + ( 201) (1 1) ⎝ 50.3 + 10 ⎠ Av = 0.826 6.48 Vo = (1 + β ) I b RL Vs Ib = rπ + (1 + β ) RL so Av = (1 + β ) RL rπ + (1 + β ) RL For β = 100, RL = 0.5 k Ω rπ = (100 )( 0.026 ) 0.5 = 5.2 k Ω 237. (101)( 0.5 ) = 0.9066 5.2 + (101)( 0.5 ) Then Av ( min ) = Then β = 180, RL = 500 k Ω rπ = (180 )( 0.026 ) 0.5 Then Av ( max ) = = 9.36 k Ω (181)( 500 ) = 0.9999 9.36 + (181)( 500 ) 6.49 Rib IS ϩ Ib V␲ ␯S ϩ Ϫ gmV␲ ϭ ␤Ib r␲ Ϫ R1͉͉R2 RE RL I0 ⎛ RE ⎞ I 0 = (1+ β ) I b ⎜ ⎟ ⎝ RE + RL ⎠ ⎛ R1 R2 ⎞ Ib = I S ⎜ ⎟ ⎝ R1 R2 + Rib ⎠ Rib = rπ + (1 + β )( RE RL ) VCC = 10 V, For VCEQ = 5 V ⎛1+ β ⎞ 5 = 10 − ⎜ ⎟ I CQ RE ⎝ β ⎠ β = 80, For RE = 0.5 kΩ I CQ = 9.88 mA, I EQ = 10 mA, I BQ = 0.123 mA rπ = (80 )( 0.026 ) = 0.211 kΩ 9.88 Rib = 0.211 + ( 81)( 0.5 0.5 ) ⇒ Rib = 20.46 kΩ Ai = ⎛ RE ⎞ ⎛ R1 R2 ⎞ I0 = (1 + β ) ⎜ ⎟ ⎟⎜ IS ⎝ RE + RL ⎠⎝ R1 R2 + Rib ⎠ ⎞ R1 R2 ⎛ 1 ⎞⎛ 8 = ( 81) ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ R1 R2 + 20.46 ⎠ 0.1975 ⎡ R1} R2 + 20.46 ⎤ = R1} R2 ⎣ ⎦ R1} R2 ⇒ 5.04 kΩ 238. VTH = I BQ RTH + VBE ( on ) + (1 + b ) I BQ RE 1 ( 5.04 )(10 ) = ( 0.123)( 5.04 ) + 0.7 + (10 )( 0.5 ) ⇒ R1 = 7.97 kΩ R1 7.97 R2 = 5.04 ⇒ R2 = 13.7 kΩ 7.97 + R2 Now Ro = RE rπ 0.211 = 0.5 or Ro = 2.59 Ω 1+ b 81 (b) Rib = 0.211 + (81) ( 0.5 2) = 32.6 k Ω ⎛ 0.5 ⎞ ⎛ 5.04 ⎞ Ai = ( 81) ⎜ ⎟⎜ ⎟ = ( 81)( 0.2 )( 0.134 ) ⎝ 0.5 + 2 ⎠ ⎝ 5.04 + 32.6 ⎠ Ai = 2.17 6.50 Ri = RTH Rib where Rib = rπ + (1 + β ) RE 5 − 3.5 VCEQ = 3.5, I CQ = 0.75 mA 2 (120 )( 0.026 ) rπ = = 4.16 k Ω 0.75 Rib = 4.16 + (121) ( 2 ) = 246 k Ω Then Ri = 120 = RTH 246 ⇒ RTH = 234 k Ω 0.75 = 0.00625 mA I BQ = 120 VTH = I BQ RTH + VBE ( on ) + (1+ β ) I BQ RE 1 1 ⋅ RTH ⋅ VCC = ( 234)(5) = ( 0.00625) ( 234) + 0.7 + (121)( 0.00625 )( 2 ) R1 R1 which yields R1 = 318 k Ω and R2 = 886 k Ω 6.51 a. Let RE = 24 Ω and VCEQ = 1 VCC = 12 V ⇒ I EQ = 2 12 = 0.5 A 24 I CQ = 0.493 A, I BQ = 6.58 mA rπ = ( 75)( 0.026 ) 0.493 = 3.96 Ω Reb Is Ib ϩ V␲ VS ϩ Ϫ gmV␲ ϭ ␤Ib r␲ Ϫ R1 ͉͉ R2 ϭ Rrn RE Io RL 239. ⎛ RE ⎞ I 0 = (1 + β ) I b ⎜ ⎟ ⎝ RE + RL ⎠ ⎛ RTH ⎞ Ib = I S ⎜ ⎟ ⎝ RTH + Rib ⎠ Rib = rπ + (1 + β ) ( RE RL ) = 3.96 + ( 76 )( 24 8 ) ⇒ Rib = 460 Ω Ai = ⎛ RE ⎞ ⎛ RTH ⎞ I0 = (1 + β ) ⎜ ⎟ ⎟⎜ IS ⎝ RE + RL ⎠ ⎝ RTH + Rib ⎠ ⎞ ⎛ 24 ⎞ ⎛ RTH 8 = ( 76) ⎜ ⎟⎜ ⎟ ⎝ 24 + 8 ⎠ ⎝ RTH + 460 ⎠ RTH 0.140 = ⇒ RTH = 74.9 Ω (Minimum value) RTH + 460 dc analysis: 1 VTH = ⋅ RTH ⋅ VCC R1 = I BQ RTH + VBE ( on ) + I EQ RE 1 ( 74.9 )( 24 ) = ( 0.00658)( 74.9 ) + 0.70 + ( 0.5 )( 24 ) R1 = 13.19 R1 = 136 Ω, 136 R2 = 74.9 ⇒ R2 = 167 Ω 136 + R2 b. 0.493 AC load line Ϫ1 Slope ϭ 24͉͉8 Ϫ1 ϭ 6⍀ 12 24 1 ΔiC = − Δvce 6 For ΔiC = 0.493 ⇒ Δvce = ( 0.493)( 6 ) ⇒ Max. swing in output voltage for this design = 5.92 V peak-to-peak c. R0 = rπ 3.96 RE = 24 = 0.0521 24 ⇒ R0 = 52 mΩ 1+ β 76 6.52 The output of the emitter follower is ⎛ RL ⎞ vo = ⎜ ⎟ ⋅ vTH ⎝ RL + Ro ⎠ 240. Ro ϩ ϩ Ϫ ␯TH ␯O RL Ϫ For vO to be within 5% for a range of RL , we have RL ( min ) RL ( min ) + Ro = ( 0.95 ) RL ( max ) RL ( max ) + Ro 4 10 = ( 0.95 ) which yields Ro = 0.364 k Ω 4 + Ro 10 + Ro ⎛ r + R1 R2 RS ⎞ We have Ro = ⎜ π ⎟ RE ro 1+ β ⎝ ⎠ The first term dominates Let R1 R2 RS ≅ RS , then rπ + RS r +4 ⇒ 0.364 = π 1+ β 1+ β Ro ≅ or 0.364 = 0.364 ≅ rπ β VT 4 4 + = + 1 + β 1 + β I CQ (1 + β ) 1 + β VT 4 + I CQ 1 + β The factor V 4 4 4 = 0.044 to = 0.0305. We can set Ro ≅ 0.32 = T is in the range of I CQ 91 131 1+ β Or I CQ = 0.08125 mA. To take into account other factors, set I CQ = 0.15 mA, I BQ = 0.15 = 0.00136 mA 110 5 = 33.3 k Ω 0.15 Design a bias stable circuit. ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ (10) − 5 = ( RTH )(10) − 5 R1 ⎝ R1 + R2 ⎠ For VCEQ ≅ 5 V , set RE = RTH = ( 0.1)(1 + β ) RE = ( 0.1)(111)(33.3) = 370 k Ω VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 5 1 ( 370 )(10 ) − 5 = ( 0.00136 )( 370 ) + 0.7 + (111)( 0.00136 )( 33.3) − 5 R 1 which yields R1 = 594 k Ω and R2 = 981 k Ω So Now Av = (1 + β ) ( RE RL ) ⎛ RTH Rib ⎞ ⋅⎜ ⎟ rπ + (1 + β ) ( RE RL ) ⎝ RTH Rib + RS ⎠ Rib = rπ + (1 + β ) ( RE RL ) and rπ = β VT I CQ For β = 90, RL = 4 k Ω, rπ = 15.6 k Ω, Rib = 340.6 k Ω Av = ( 91)( 33.3 4 ) 370 340.6 ⋅ ⇒ Av = 0.9332 15.6 + ( 91)( 33.3 4 ) 370 340.6 + 4 241. For β = 90, RL = 10 k Ω Rib = 715.4 k Ω Av = ( 91)( 33.3 10 ) 370 715.4 ⋅ ⇒ Av = 0.9625 15.6 + ( 91)( 33.3 10 ) 370 715.4 + 4 For β = 130, RL = 4 k Ω rπ = 22.5 k Ω, Rib = 490 k Ω Av = (131)( 33.3 4 ) 370 490 ⋅ ⇒ Av = 0.9360 22.5 + (131)( 33.3 4 ) 370 490 + 4 For β = 130, RL = 10 k Ω Rib = 1030 k Ω (131)( 33.3 10 ) 370 1030 ⇒ Av = 0.9645 ⋅ 22.5 + (131)( 33.3 10 ) 370 1030 + 4 Now vO ( min ) = Av ( min ) .vS = 3.73sin ω t vO ( max ) = Av ( max ) .vS = 3.86sin ω t Av = ΔvO = 3.5% vO 6.53 2 2 PAVG = iL ( rms ) RL ⇒ 1 = iL ( rms )(12 ) so iL ( rms ) = 0.289 A ⇒ iL ( peak ) = 2 ( 0.289 ) iL ( peak ) = 0.409 A vL ( peak ) = iL ( peak ) ⋅ RL = ( 0.409 )(12 ) = 4.91 V 4.91 = 0.982 5 With RS = 10 k Ω, we will not be able to meet this voltage gain requirement. Need to insert a buffer or an Need a gain of op-amp voltage follower (see Chapter 9) between RS and CC1 . 1 Set I EQ = 0.5 A, VCEQ = (12 − ( −12 ) ) = 8 V 3 24 = I EQ RE + VCEQ = ( 0.5 ) RE + 8 ⇒ RE = 32 Ω 50 ( 0.5 ) = 0.49 A 51 β VT ( 50 )( 0.026 ) rπ = = = 2.65 Ω I CQ 0.49 Let β = 50, I CQ = Rib = rπ + (1 + β ) ( RE RL ) = 2.65 + ( 51) ( 32 12 ) Rib = 448 Ω Av = (1 + β ) ( RE RL ) ( 51) ( 32 12 ) = = 0.994 rπ + (1 + β ) ( RE RL ) 2.65 + ( 51) ( 32 12 ) So gain requirement has been met. 242. 0.49 = 0.0098 A = 9.8 mA 50 24 ≅ 10 I B = 98 mA Let I R ≅ R1 + R2 I BQ = So that R1 + R2 = 245 Ω VTH = R2 ( 24 ) − 12 = I BQ RTH + VBE ( on ) + I EQ RE − 12 R1 + R2 ( 0.0098) R1 R2 ⎛ R2 ⎞ + 0.7 + ( 0.5 )( 32 ) ⎜ 245 ⎟ ( 24 ) = 245 ⎝ ⎠ Now R1 = 245 − R2 So we obtain 2 4 × 10−5 R2 + 0.0882 R2 − 16.7 = 0 which yields R2 = 175 Ω and R1 = 70 Ω 6.54 (a) RTH = R1 R2 = 25.6 10.4 = 7.40 k Ω ⎛ R2 ⎞ ⎛ 10.4 ⎞ VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ (18 ) = 5.2 V ⎝ 10.4 + 25.6 ⎠ ⎝ R1 + R2 ⎠ VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE I BQ = 5.2 − 0.7 = 0.0117 mA 7.40 + (126 )( 3) Then I CQ = 1.46 mA and I EQ = 1.47 mA VCEQ = VCC − I CQ RC − I EQ RE VCEQ = 18 − (1.46 )( 4 ) − (1.47 )( 3) ⇒ VCEQ = 7.75 V (b) rπ = (125) ( 0.026 ) = 2.23 k Ω 1.46 1.46 gm = = 56.2 mA / V 0.026 Re Vo Ie Is RS RE r␲ Ib RTH ␤Ib RC RL 243. rπ + RTH 2.23 + 7.40 = = 0.0764 k Ω 1+ β 126 Re = Ie = − ( RS RE ) (R S RE ) + Re ⋅ Is = − (100 3) (100 3) + 0.0764 ⋅ Is or I e = − ( 0.974 ) I s ⎛ β Vo = − I c ( RC RL ) = − ⎜ ⎝ 1+ β ⎞ ⎟ I e ( RC RL ) ⎠ ⎛ β ⎞ Vo ⎛ 125 ⎞ = −⎜ ⎟ ( −0.974 )( RC RL ) = ⎜ ⎟ ( 0.974 )( 4 4 ) Is 1+ β ⎠ ⎝ 126 ⎠ ⎝ V Then Rm = o = 1.93 k Ω = 1.93 V / mA Is Then (c) ( ) ( ) Vs = I s RS R E Re = I s 100 3 0.0764 = I s ( 0.0744 ) Vs 0.0744 V V which yields o = o ( 0.0744 ) = 1.93 I s Vs or I s = or Av = 6.55 (a) Vo = 25.9 Vs β ( RC RL ) Av = , RL = 12 k Ω, β = 100 rπ + R1 R2 Let R1 R2 = 50 k Ω, I CQ = 0.5 mA VTH = I BQ RTH + VBE ( on ) + (1+ β ) I BQ RE (100 )( 0.026 ) 0.5 = 0.005 mA, rπ = = 5.2 k Ω 100 0.5 1 1 ⋅ RTH ⋅ VCC = ( 50 )(12 ) = ( 0.005 )( 50 ) + 0.7 + (101)( 0.005 )( 0.5 ) R1 R1 I BQ = which yields R1 = 500 k Ω and R2 = 55.6 k Ω Av = (b) I CQ (100 )(12 12 ) 5.2 + 50 = 10.9, Design criterion is met. = 0.5 mA, I EQ = 0.505 mA VCEQ = 12 − ( 0.5)(12) − ( 0.505)( 0.5) ⇒ VCEQ = 5.75 V 0.5 = 19.23 mA / V 0.026 Av = (19.23) (12 12 ) ⇒ Av = 115 Av = g m ( RC RL ) , g m = 6.56 a. Emitter current 244. I EQ = I CC = 0.5 mA 0.5 = 0.00495 mA 101 VE = I EQ RE = ( 0.5 )(1) ⇒ VE = 0.5 V I BQ = VB = VE + VBE ( on ) = 0.5 + 0.7 ⇒ VB = 1.20 V VC = VB + I BQ RB = 1.20 + ( 0.00495 )(100 ) ⇒ VC = 1.7 V b. rπ = (100 )( 0.026 ) = 5.25 kΩ (100 )( 0.00495 ) (100 )( 0.00495 ) gm = = 19.0 mA/V 0.026 Ri gmV␲ VS RS ϩ Ϫ Ϫ RE V␲ r␲ RB ϩ gmV␲ ΂ ΃ RE V RE ϩ RS S ϩ Ϫ RS ͉͉RE Ϫ V␲ Vπ = − RE RE r␲ ϩ Vo = − g mVπ ( RB RL ) Rie ⋅ VS = − ( 0.4971) VS Rie + RS Vo = (19 )( 0.4971) VS (100 1) Av = 9.37 c. VX ϩ Ϫ IX gmV␲ Ϫ RE V␲ r␲ ϩ IX = VX VX + − g mVπ , Vπ = −VX RE rπ IX 1 1 1 = = + + gm VX Ri RE rπ or Ri = RE rπ 1 1 = 1 5.253 gm 19 Ri = 0.84 0.05252 ⇒ Ri = 49.4 Ω 6.57 (a) V0 I EQ = 1 mA, I CQ = 0.9917 mA RL 245. VC = 5 − ( 0.9917 )( 2 ) = 3.017 V VE = −0.7 V VCEQ = 3.72 V (b) Av = g m ( RC RL ) 0.9917 = 38.14 mA/V 0.026 Av = ( 38.14 ) ( 2 10 ) ⇒ Av = 63.6 gm = 6.58 (a) 10 − 0.7 = 0.93 mA 10 = 0.921 mA I EQ = I CQ VECQ = 20 − ( 0.93)(10 ) − ( 0.921)( 5 ) VECQ = 6.10 V (b) 0.921 = 35.42 mA/V 0.026 Av = g m ( RC RL ) = ( 35.42 ) ( 5 50 ) gm = Av = 161 6.59 (a) I EQ = 0.93 mA, I CQ = 0.921 mA VECQ = 6.10 V (b) gm = 0.921 = 35.42 mA/V rπ = 2.82 K 0.026 From Eq. 6.90 ( RC RL ) ⎡ rπ R R ⎤ Av = g m ⎢1 + β E S ⎥ RS ⎣ ⎦ ( 35.42 ) ( 50 5 ) ⎡ 2.82 ⎤ = ⎢ 101 10 0.1⎥ 0.1 ⎣ ⎦ ( 35.42 )( 4.545 ) Av = [0.0218] 0.1 Av = 35.1 6.60 (a) ⎛ 60 ⎞ I CQ = ⎜ ⎟ (1) ⇒ I CQ = 0.984 mA ⎝ 61 ⎠ ⎛ 1⎞ VCEQ = I BQ RB + VBE ( on ) = ⎜ ⎟ (100 ) + 0.7 ⎝ 61 ⎠ VCEQ = 2.34 V (b) 246. Av = g m (R RL ) ⎡ rπ ⎤ RS ⎥ ⎢ RS ⎣1 + β ⎦ B 0.984 = 37.85 mA/V 0.026 rπ = 1.59 K gm = ( 37.85) (100 2 ) ⎡1.59 ⎤ ⎢ 61 0.05⎥ 0.05 ⎣ ⎦ = 1484 ⎡ 0.0261 0.05⎤ ⎣ ⎦ Av = 25.4 Av = 6.61 is ( peak ) = 2.5 mA, Vo ( peak ) = 5 mV vo 5 × 102 3 = = 2 × 103 = 2 k Ω is 2.5 × 102 6 From Problem 4.54 ⎛ RS RE ⎞ Vo ⎛ β ⎞ =⎜ ⎟ ( RC RL ) ⎜ ⎜R R +R ⎟ ⎟ Is ⎝ 1+ β ⎠ ie ⎠ ⎝ S E Let RC = 4 k Ω, RL = 5 k Ω, RE = 2 k Ω So we need Rm = Now β = 120, so we have ⎛ RS RE ⎛ 120 ⎞ 2=⎜ ⎟ ( 4 5) ⎜ ⎜R R +R ⎝ 121 ⎠ ie ⎝ S E RS RE = 0.9075 Then RS RE + Re ⎞ ⎛ RS RE ⎞ ⎟ = 2.204 ⎜ ⎟ ⎜R R +R ⎟ ⎟ ie ⎠ ⎠ ⎝ S E RS RE = 50 2 = 1.923 k Ω, so that Rie = 0.196 k Ω Assume VCEQ = 3 V VCC ≅ I CQ ( RC + RE ) + VCEQ 5 = I CQ ( 4 + 2 ) + 3 ⇒ I CQ = 0.333 mA rπ = (120 )( 0.026 ) = 9.37 k Ω 0.333 r + RTH 9.37 + RTH ⇒ 0.196 = Rie = π 1+ β 121 which yields RTH = 14.35 k Ω Now VTH = I BQ RTH + VBE ( on ) + I EQ RE 1 ⎛ 121 ⎞ = 0.00833 mA, I EQ = ⎜ ⎟ (1) = 1.008 mA 120 ⎝ 120 ⎠ 1 1 = ⋅ RTH ⋅ VCC = (14.35 )( 5 ) = ( 0.00833)(14.35 ) + 0.7 + (1.008 )( 2 ) R1 R1 I BQ = VTH which yields R1 = 25.3 k Ω and R2 = 33.2 k Ω 6.62 a. 247. 20 − 0.7 = 1.93 mA 10 = 1.91 mA I EQ = I CQ VECQ = VCC + VEB ( on ) − I C RC = 25 + 0.7 − (1.91)( 6.5 ) ⇒ VECQ = 13.3 V b. RS VS ϩ Ϫ Rie IS h fe Ib V0 Ie RE RC hie RL Ib Neglect effect hoe From Problem 6-16, assume 2.45 ≤ hie ≤ 3.7 kΩ 80 ≤ h fe ≤ 120 Vo = ( h fe I b ) ( RC Rie = RL ) ⎛ RE ⎞ hie , Ie = ⎜ ⎟ IS 1 + h fe ⎝ RE + Rie ⎠ ⎛ I ⎞ VS Ib = ⎜ e ⎟ , I S = ⎜ 1+ h ⎟ RS + RE fe ⎠ ⎝ Rie ⎛ h fe ⎞ ⎛ RE ⎞ ⎛ 1 Av = ⎜ R RL ) ⎜ ⎟×⎜ ⎜ 1+ h ⎟( C ⎟ fe ⎠ ⎝ RE + Rie ⎠ ⎝ RS + RE ⎝ High gain device: hie = 3.7 kΩ, h fe = 120 Rie = RE ⎞ ⎟ Rie ⎠ 3.7 = 0.0306 kΩ 121 Rie = 10 0.0306 = 0.0305 10 1 ⎛ 120 ⎞ ⎛ ⎞⎛ ⎞ Av = ⎜ ⎟ ( 6.5 5 ) ⎜ ⎟⎜ ⎟ ⇒ Av = 2.711 121 ⎠ 10 + 0.0306 ⎠ ⎝ 1 + 0.0305 ⎠ ⎝ ⎝ Low gain device: hie = 2.45 kΩ, h fe = 80 Rie = RE 2.45 = 0.03025 kΩ 81 Rie = 10 0.03025 = 0.0302 10 1 ⎛ 80 ⎞ ⎛ ⎞⎛ ⎞ Av = ⎜ ⎟ ( 6.5 5 ) ⎜ ⎟⎜ ⎟ ⇒ Av = 2.70 So Av ≈ constant ⎝ 81 ⎠ ⎝ 10 + 0.03025 ⎠ ⎝ 1 + 0.0302 ⎠ 2.70 ≤ Av ≤ 2.71 c. Ri = RE Rie We found 0.0302 ≤ Ri ≤ 0.0305 kΩ Neglecting hoe , Ro = RC = 6.5 kΩ 6.63 a. Small-signal voltage gain 248. Av = g m ( RC RL ) ⇒ 25 = g m ( RC 1) For VECQ = 3 V ⇒ VC = −VECQ + VEB ( on ) = −3 + 0.7 ⇒ VC = −2.3 VCC − I CQ RC + VC = 0 ⇒ I CQ = 5 − 2.3 2.7 = = I CQ RC RC For I CQ = 1 mA, RC = 2.7 kΩ 1 = 38.5 mA/V 0.026 Av = ( 38.5 )( 2.7 1) = 28.1 gm = Design criterion satisfied and VECQ satisfied. ⎛ 101 ⎞ IE = ⎜ ⎟ (1) = 1.01 mA ⎝ 100 ⎠ VEE = I E RE + VEB ( on ) ⇒ RE = b. rπ = β VT I CQ = (100)( 0.026) 1 5 − 0.7 ⇒ RE = 4.26 kΩ 1.01 ⇒ rπ = 2.6 kΩ, g m = 38.5 mA/V, ro = ∞ 6.64 a. ⎛ R2 ⎞ ⎛ 20 ⎞ VTH 1 = ⎜ ⎟ VCC = ⎜ ⎟ (10 ) ⇒ VTH 1 = 2.0 V ⎝ 20 + 80 ⎠ ⎝ R1 + R2 ⎠ RTH 1 = R1 R2 = 20 80 = 16 kΩ I B1 = 2 − 0.7 = 0.0111 mA 16 + (101)(1) I C1 = 1.11 mA ⇒ g m1 = rπ 1 = (100)( 0.026) 1.11 1.11 ⇒ g m1 = 42.74 mA/V 0.026 ⇒ rπ 1 = 2.34 kΩ ∞ ⇒ r01 = ∞ 1.11 ⎛ R4 ⎞ ⎛ 15 ⎞ VTH 2 = ⎜ ⎟ VCC = ⎜ ⎟ (10 ) = 1.50 V ⎝ 15 + 85 ⎠ ⎝ R3 + R4 ⎠ RTH 2 = R3 R4 = 15 85 = 12.75 kΩ r01 = IB2 = 1.50 − 0.70 = 0.01265 mA 12.75 + (101)( 0.5 ) I C 2 = 1.265 mA ⇒ g m 2 = rπ 2 = (100 )( 0.026 ) 1.26 1.265 ⇒ g m2 = 48.65 mA/V 0.026 ⇒ rπ 2 = 2.06 kΩ r02 = ∞ b. Av1 = − g m1 RC1 = − ( 42.7 )( 2 ) ⇒ Av1 = −85.48 Av 2 = − g m 2 ( RC 2 RL ) = − ( 48.5) ( 4 4) ⇒ Av 2 = −97.3 c. Input resistance of 2nd stage 249. Ri 2 = R3 R4 rπ 2 = 15 85 2.06 = 12.75 2.06 ⇒ Ri 2 = 1.773 kΩ ′1 = − g m1 ( RC1 Ri 2 ) = − ( 42.7 ) ( 2 1.77B) Av Av′1 = −40.17 Overall gain: Av = ( −40.17 )( −97.3) ⇒ Av = 3909 If we had Av1 ⋅ Av 2 = ( −85.48)( −97.3) = 8317 Loading effect reduces overall gain 6.65 a. ⎛ R2 ⎞ ⎛ 12.7 ⎞ VTH 1 = ⎜ ⎟ VCC = ⎜ ⎟ (12) ⇒ VTH 1 = 1.905 V ⎝ 12.7 + 67.3 ⎠ ⎝ R1 + R2 ⎠ RTH 1 = R1 R2 = 12.7 67.3 = 10.68 kΩ 1.905 − 0.70 I B1 = = 0.00477 mA 10.68 + (121)( 2 ) I C1 = 0.572 mA 0.572 g m1 = ⇒ g m1 = 22 mA/V 0.026 (120 )( 0.026 ) rπ 1 = ⇒ rπ 1 = 5.45 kΩ 0.572 ∞ r01 = ⇒ r01 = ∞ 0.572 ⎛ R4 ⎞ ⎛ 45 ⎞ VTH 2 = ⎜ ⎟ VCC = ⎜ ⎟ (12) ⇒ VTH 2 = 9.0 V ⎝ 45 + 15 ⎠ ⎝ R3 + R4 ⎠ RTH 2 = R3 R4 = 15 45 = 11.25 kΩ 9.0 − 0.70 I B2 = = 0.0405 mA 11.25 + (121)(1.6) I C2 = 4.86 mA 4.86 gm2 = ⇒ g m 2 = 187 mA/V 0.026 (120 )( 0.026 ) rπ 2 = ⇒ rπ 2 = 0.642 kΩ 4.86 r02 = ∞ b. I E1 = 0.577 mA VCEQ1 = 12 − ( 0.572 ) (10 ) − ( 0.577 ) ( 2 ) ⇒ VCEQ1 = 5.13 V I E 2 = 4.90 VCEQ 2 = 12 − ( 4.90 )(1.6 ) ⇒ VCEQ 2 = 4.16 V 250. Q1 AC load line Ϫ1 10͉͉7.92 Ϫ1 ϭ 4.42 K Slope ϭ 0.572 5.13 12 Q2 4.86 AC load line Ϫ1 Slope ϭ 1.6͉͉0.25 Ϫ1 ϭ 0.216 K 4.16 12 Ri 2 = R3 R4 Rib Rib = rπ 2 + (1 + β ) ( RE 2 RL ) = 0.642 + (121) (1.6 0.25 ) Rib = 26.8 Ri 2 = 15 45 26.8 Ri 2 = 7.92 kΩ c. Av1 = − g m1 ( RC1 Ri 2 ) = − ( 22 )(10 7.92 ) ⇒ Av 2 = −97.2 (1 + β )( RE 2 RL ) rπ 2 + (1 + β )( RE 2 RL ) (121)( 0.216 ) = = 0.976 0.642 + (121)( 0.216 ) Overall gain = ( −97.2 )( 0.976 ) = −94.9 Av 2 = d. RiS = R1 R2 rπ 1 = 67.3 12.7 5.45 ⇒ RiS = 3.61 kΩ Ro = rπ 2 + RS RE 2 where 1+ β RS = R3 R4 RC1 = 15 45 10 ⇒ RS = 5.29 kΩ Ro = 0.642 + 5.29 1.6 ⇒ 0.049 1.6 ⇒ Ro = 47.6 Ω 121 e. −1 ⋅ Δvce , ΔiC = 4.86 0.216 kΩ Δvce = ( 4.86 )( 0.216 ) = 1.05 V Max. output voltage swing = 2.10 V peak-to-peak ΔiC = 6.66 (a) 251. I R1 = IR2 IC 2 5 − 2 ( 0.7 ) = 72 mA 0.050 0.7 = = 1.4 mA 0.5 ⎛ β ⎞ =⎜ ⎟ ( 72 − 1.4 ) ⇒ I C 2 = 69.9 mA ⎝ 1+ β ⎠ 69.9 = 0.699 mA 100 ⎛ β ⎞ =⎜ ⎟ (1.4 + 0.699 ) ⇒ I C1 = 2.08 mA ⎝ 1+ β ⎠ IB2 = I C1 (b) Vs ϩ Ϫ ϩ V␲1 r␲1 Ϫ gm1V␲1 r␲2 0.5 k⍀ gm2V␲2 ϩ V␲2 Ϫ Vo 50 Ω Vs = Vπ 1 + Vπ 2 + Vo (1) ⎛V ⎞ V Vo = ⎜ π 2 + π 2 + g m 2Vπ 2 ⎟ ( 0.05 ) ⎝ 0.5 rπ 2 ⎠ rπ 2 = (100 )( 0.026 ) = 0.0372 k Ω 69.9 69.9 gm2 = = 2688 mA / V 0.026 V 1 ⎛ 1 ⎞ Vo = Vπ 2 ⎜ + + 2688 ⎟ ( 0.05 ) so that (1) Vπ 2 = o 135.8 ⎝ 0.5 0.0372 ⎠ (2) 252. Vπ 1 V V + g m1Vπ 1 = π 2 + π 2 rπ 1 0.5 rπ 2 rπ 1 = (100 )( 0.026 ) = 1.25 k Ω 2.08 2.08 g m1 = = 80 mA / V 0.026 1 ⎞ ⎛ 1 ⎞ ⎛ 1 Vπ 1 ⎜ + 80 ⎟ = Vπ 2 ⎜ + ⎟ 1.25 0.5 0.0372 ⎠ ⎝ ⎠ ⎝ ⎛ V ⎞ Vπ 1 ( 80.8 ) = Vπ 2 ( 28.88 ) = ⎜ o ⎟ ( 28.88 ) or (2) Vπ 1 = Vo ( 0.00261) ⎝ 136.7 ⎠ V V Then Vs = Vo ( 0.00261) + o + Vo = Vo (1.00993) or Av = o = 0.990 Vs 136.7 (c) Rib = rπ 1 (1 + β ) [ Rx ] Ix Vx ϩ Ϫ ϩ 0.5 k⍀ V␲2 r␲2 Ϫ gm2V␲2 Vo 50 ⍀ ⎛ 1 Vπ 2 Vπ 2 1 ⎞ + = Vπ 2 ⎜ + ⎟ 0.5 rπ 2 0.5 rπ 2 ⎠ ⎝ Vo V − Vπ 2 = x = I x + g m 2Vπ 2 0.05 0.05 Ix = ⎛ 1 ⎞ Ix ⎜ + gm2 ⎟ Vx ⎛ 1 ⎞ ⎝ 0.05 ⎠ − I x = Vπ 2 ⎜ + gm2 ⎟ = 0.05 0.05 ⎛ 1 ⎝ ⎠ 1 ⎞ + ⎜ ⎟ 0.5 rπ 2 ⎠ ⎝ We find Vx = Rx = 4.74 k Ω Ix Then Rib = 1.25 + (101) ( 2.89 ) ⇒ Rib = 480 k Ω 253. ϩ V␲1 r␲1 gm1V␲1 Ϫ r␲2 0.5 k⍀ gm2V␲2 ϩ V␲2 Ϫ Ix 50 ⍀ ϩ Ϫ Vx To find Ro: (1) (2) (3) Ix = Vx V − g m 2Vπ 2 − π 2 0.05 0.5 rπ 2 ⎛V ⎞ ⎛ 1 ⎞ Vπ 2 = ⎜ π 1 + g m1Vπ 1 ⎟ ( 0.5 rπ 2 ) = Vπ 1 ⎜ + 80 ⎟ ( 0.5 0.0372 ) or Vπ 2 = ( 2.77 ) Vπ 1 ⎝ 1.25 ⎠ ⎝ rπ 1 ⎠ Vπ 1 + Vπ 2 + Vx = 0 ⇒ Vπ 1 + ( 2.77 ) Vπ 1 + Vx = 0 so that Vπ 1 = − ( 0.2653) Vx and Vπ 2 = ( 2.77 ) ⎡ − ( 0.2653) Vx ⎤ = − ( 0.735 ) Vx ⎣ ⎦ Now I x = ⎛ Vx 1 ⎞ − Vπ 2 ⎜ g m 2 + ⎟ ⎜ 0.05 0.5 rπ 2 ⎟ ⎝ ⎠ So that I x = ⎡ ⎤ Vx Vx 1 + ( 0.735 ) Vx ⎢ 2688 + = 0.496 Ω ⎥ which yields Ro = Ix 0.05 0.5 0.0372 ⎥ ⎢ ⎣ ⎦ 6.67 a. RTH = R1 R2 = 335 125 = 91.0 kΩ ⎛ R2 ⎞ VTH = ⎜ ⎟ VCC ⎝ R1 + R2 ⎠ ⎛ 125 ⎞ =⎜ ⎟ (10 ) = 2.717 V ⎝ 125 + 335 ⎠ VTH = I B1 RTH + VBE1 + VBE 2 + I E 2 RE 2 I E 2 = (1 + β ) I E1 = (1 + β ) I B1 2 I B1 = 2.717 − 1.40 91.0 + (101) (1) 2 ⇒ I B1 = 0.128 μΑ I C1 = 12.8 μΑ I C 2 = β I E1 = β (1 + β ) I B1 = (100 )(101)( 0.128 μΑ ) I C 2 = 1.29 mΑ, I E 2 = 1.31 mΑ I RC = I C 2 + I C1 = 1.29 + 0.0128 = 1.30 mΑ 254. VC = 10 − I RC RC = 10 − (1.30 )( 2.2 ) = 7.14 V VE = I E 2 RE 2 = (1.30 )(1) = 1.30 V VCE 2 = 7.14 − 1.30 = 5.84 V VCE1 = VCE 2 − VBE 2 = 5.84 − 0.7 VCE1 = 5.14 V Summary: I C1 = 12.8 μΑ I C 2 = 1.29 mΑ VCE1 = 5.14 V VCE 2 = 5.84 V b. 0.0128 g m1 = = 0.492 mΑ / V 0.026 1.292 gm2 = = 49.7 mΑ / V 0.026 Rib Ib VS V0 ϩ V␲1 Ϫ ϩ Ϫ R1͉͉ R2 r␲1 gm1V␲1 RC ϩ V␲2 r␲2 gm2V␲2 Ϫ V0 = − ( g m1Vπ 1 + g m 2Vπ 2 ) RC VS = Vπ 1 + Vπ 2 , Vπ 1 = VS − Vπ 2 ⎛V ⎞ Vπ 2 = ⎜ π 1 + g m1Vπ 1 ⎟ rπ 2 rπ 1 ⎝ ⎠ ⎛1+ β ⎞ Vπ 2 = Vπ 1 ⎜ ⎟ rπ 2 ⎝ rπ 1 ⎠ V0 = − ⎡ g m1 (VS − Vπ 2 ) + g m 2Vπ 2 ⎤ RC ⎣ ⎦ V0 = − ⎡ g m1VS + ( g m 2 − g m1 ) Vπ 2 ⎤ RC ⎣ ⎦ ⎛r ⎞ Vπ 2 = (VS − Vπ 2 )(1 + β ) ⎜ π 2 ⎟ ⎝ rπ 1 ⎠ ⎡ ⎛ r ⎞⎤ ⎛r ⎞ Vπ 2 ⎢1 + (1 + β ) ⎜ π 2 ⎟ ⎥ = VS (1 + β ) ⎜ π 2 ⎟ ⎝ rπ 1 ⎠ ⎦ ⎝ rπ 1 ⎠ ⎣ 255. ⎧ ⎛r ⎞⎫ VS (1 + β ) ⎜ π 2 ⎟ ⎪ ⎪ ⎪ ⎝ rπ 1 ⎠ ⎪ R V0 = − ⎨ g m1VS + ( g m 2 − g m1 ) ⋅ ⎬ C ⎛r ⎞ ⎪ 1 + (1 + β ) ⎜ π 2 ⎟ ⎪ ⎪ ⎝ rπ 1 ⎠ ⎪ ⎩ ⎭ V0 Av = VS 2.01 ⎞ ⎫ ⎧ ( 49.7 − 0.492 )(101) ⎛ ⎜ ⎟⎪ ⎪ ⎪ ⎝ 203 ⎠ ⎪ 2.2 = − ⎨( 0.492 ) + ⎬ ⎛ 2.01 ⎞ ⎪ ⎪ 1 + (101) ⎜ ⎟ ⎪ ⎪ ⎝ 203 ⎠ ⎩ ⎭ Av = −55.2 c. Ris = R1 R2 Rib Rib = rπ 1 + (1 + β ) rπ 2 = 203 + (101)( 2.01) = 406 kΩ Ris = 91 406 = 74.3 kΩ = Ris R0 = RC = 2.2 kΩ 6.68 R0 Ix ϩ V␲1 r␲1 ϩ Ϫ ro1 gm1V␲1 Ϫ Vx VA ϩ V␲2 r␲2 Ϫ gm2V␲2 ro2 Vx Vx − VA + + g m1Vπ 1 ro 2 ro1 (1) I x = g m 2Vπ 2 + (2) Vx − VA VA + g m1Vπ 1 = ro1 rπ 1 rπ 2 Vπ 2 = VA = −Vπ 1 (3) Then from (2) ⎛ 1 Vx 1 ⎞ = VA ⎜ + g m1 + ⎟ ⎜r ro1 rπ 1 rπ 2 ⎟ ⎝ o1 ⎠ ⎛ 1 ⎛ ⎞ Vx Vx VA 1 ⎞ 1 + − − g m1VA or I x = Vx ⎜ + ⎟ + VA ⎜ g m 2 − − g m1 ⎟ ro 2 ro1 ro1 ro1 ⎝ ro1 ro 2 ⎠ ⎝ ⎠ Solving for VA from Equation (2) and substituting into Equation (1), we find (1) I x = g m 2VA + 256. 1 1 + g m1 + V ro1 rπ 1 rπ 2 Ro = x = Ix ⎞ 1 ⎛ 1 1 ⎞ 1 ⎛ 1 + gm2 ⎟ ⎜ + g m1 + ⎟+ ⎜ ro 2 ⎝ ro1 rπ 1 rπ 2 ⎠ ro1 ⎝ rπ 1 rπ 2 ⎠ For β = 100, VA = 100 V , I C1 = I Bias = 1 mA ro1 = ro 2 = rπ 1 = rπ 2 = 100 = 100 k Ω 1 (100 )( 0.026 ) 1 g m1 = g m 2 = Then Ro = = 2.6 k Ω 1 = 38.46 mA/V 0.026 1 1 + 38.46 + 100 2.6 2.6 ⎞ 1 ⎛ 1 1 ⎞ 1 ⎛ 1 + 38.46 + + 38.46 ⎟ ⎜ ⎟+ ⎜ ⎟ 100 ⎜ 100 2.6 2.6 ⎟ 100 ⎜ 2.6 2.6 ⎝ ⎠ ⎝ ⎠ or Ro = 50.0 k Ω Now I C 2 = 1 mA, I Bias = 0 Replace I Bias by I β ⋅ = C 2 , I C1 ≅ 0.01 mA β 1+ β 1+ β IC 2 100 100 = 100 k Ω, ro1 = = 10, 000 k Ω 1 0.01 1 = 38.46 mA/V , g m1 = 0.3846 mA/V gm2 = 0.026 (100 )( 0.026 ) = 2.6 k Ω, rπ 1 = 260 k Ω rπ 2 = 1 Then Ro = 66.4 k Ω ro 2 = 6.69 a. RTH = R1 R2 = 93.7 6.3 = 5.90 k Ω ⎛ R2 ⎞ VTH = ⎜ ⎟ VCC ⎝ R1 + R2 ⎠ 6.3 ⎞ ⎛ =⎜ ⎟ (12 ) = 0.756 V ⎝ 6.3 + 93.7 ⎠ 0.756 − 0.70 I BQ = = 0.00949 mA 5.90 I CQ = 0.949 mA VCEQ = 12 − ( 0.949 )( 6 ) ⇒ VCEQ = 6.305 V Transistor: PQ ≈ I CQVCEQ = ( 0.949 )( 6.305 ) ⇒ PQ = 5.98 mW 2 RC : PR = I CQ RC = ( 0.949 ) ( 6 ) ⇒ PR = 5.40 mW 2 b. 257. 2 AC load line Ϫ1 Slope ϭ 6͉͉105 Ϫ1 ϭ 5.68 K 0.949 6.31 12 100 r0 = = 105 kΩ 0.949 Peak signal current = 0.949 mA V0 ( max ) = ( 5.68 )( 0.949 ) = 5.39 V PRC = 2 2 1 V0 ( max ) 1 ⎡ ( 5.39 ) ⎤ ⋅ = ⎢ ⎥ ⇒ PRC = 2.42 mW RC 2 2⎢ 6 ⎥ ⎣ ⎦ 6.70 (a) 10 = I BQ RB + VBE ( on ) + (1 + β ) I BQ RE 10 − 0.7 I BQ = = 0.00369 mA 100 + (121)( 20 ) I CQ = 0.443 mA, I EQ = 0.447 mA For RC : PRC = ( 0.443) (10 ) ⇒ PRC = 1.96 mW 2 For RE : PRE = ( 0.447 ) ( 20 ) ⇒ PRE = 4.0 mW 2 (b) ΔiC = 0.667 − 0.443 = 0.224 mA 1 1 2 2 ( ΔiC ) RC = ( 0.224 ) (10 ) 2 2 = 0.251 mW Then P RC = P RC 6.71 a. I BQ = 10 − 0.7 = 0.00596 mA 50 + (151)(10 ) I CQ = 0.894 mA, I EQ = 0.90 mA VECQ = 20 − ( 0.894 )( 5 ) − ( 0.90 )(10 ) ⇒ VECQ = 6.53 V PQ ≅ I CQVECQ = ( 0.894 )( 6.53) ⇒ PQ = 5.84 mW 2 PRC ≅ I CQ RC = ( 0.894 ) ( 5 ) ⇒ PRC = 4.0 mW 2 2 PRE ≅ I EQ RE = ( 0.90 ) (10 ) ⇒ PRE = 8.1 mW 2 b. 258. AC load line Ϫ1 5͉͉2 Ϫ1 ϭ 1.43 K Slope ϭ 0.894 6.53 20 −1 ΔiC = ⋅ Δvec 1.43 kΩ ΔiC = 0.894 ⇒ Δvec = ( 0.894 )(1.43) = 1.28 V ⎛ 5 ⎞ Δi0 = ⎜ ⎟ ΔiC = 0.639 mA ⎝5+2⎠ 1 2 PRL = ( 0.639 ) ( 2 ) ⇒ PRL = 0.408 mW 2 1 2 PRC = ⋅ ( 0.894 − 0.639 ) ( 5 ) ⇒ PRC = 0.163 mW 2 PRE = 0 PQ = 5.84 − 0.408 − 0.163 ⇒ PQ = 5.27 mW 6.72 I BQ = 10 − 0.70 = 0.00838 mA 100 + (101)(10 ) I CQ = 0.838 mA, I EQ = 0.846 mA VCEQ = 20 − ( 0.838 )(10 ) − ( 0.846 )(10 ) ⇒ VCEQ = 3.16 V AC load line Ϫ1 RE ͉͉RL͉͉r0 Slope ϭ 0.838 3.16 20 100 r0 = = 119 kΩ 0.838 Neglecting base currents: a. RL = 1 kΩ slope = −1 −1 = 10 1 119 0.902 kΩ −1 ⋅ ΔVce 0.902 kΩ ΔiC = 0.838 ⇒ ΔVce = ( 0.902 )( 0.838 ) = 0.756 V ΔiC = 1 ( 0.756 ) ⇒ PRL = 0.286 mW 2 1 2 PRL = b. 259. RL = 10 kΩ slope = −1 −1 = 10 10 119 4.80 For ΔiC = 0.838 ⇒ Δvce = ( 0.838 )( 4.80 ) = 4.02 1 ( 3.16 ) ⇒ PRL = 0.499 mW 2 10 2 Max. swing determined by voltage PRL = 6.73 a. I BQ = 10 − 0.7 = 0.00838 mA 100 + (101)(10 ) I CQ = 0.838 mA, I EQ = 0.846 mA VCEQ = 20 − ( 0.838 )(10 ) − ( 0.846 )(10 ) ⇒ VCEQ = 3.16 V PQ ≅ I CQVCEQ = ( 0.838 )( 3.16 ) ⇒ PQ = 2.65 mW 2 PRC ≅ I CQ RC = ( 0.838 ) (10 ) ⇒ PRC = 7.02 mW 2 b. AC load line Ϫ1 RC ͉͉RL Ϫ1 Ϫ1 ϭ ϭ 10͉͉1 0.909 K Slope ϭ 0.838 3.16 20 −1 ΔiC = ⋅ Δvce 0.909 kΩ For ΔiC = 0.838 ⇒ Δvce = ( 0.909 )( 0.838 ) = 0.762 V ⎛ RC ⎞ ⎛ 10 ⎞ Δi0 = ⎜ ⎟ ΔiC = ⎜ ⎟ ΔiC = 0.762 mA ⎝ 10 + 1 ⎠ ⎝ RC + RL ⎠ 1 2 PRL = ( 0.762 ) (1) ⇒ PRL = 0.290 mW 2 1 2 PRC = ⋅ ( 0.838 − 0.762 ) (10 ) ⇒ PRC = 0.0289 mW 2 PQ = 2.65 − 0.290 − 0.0289 ⇒ PQ = 2.33 mW 260. Chapter 7 Exercise Solutions EX7.1 (a) (i) RS = RP = 4 kΩ 1 1 ω= = rS ( RS + RP ) CS CS = 1 2π f ( RS + RP ) = 1 2π ( 20 )( 4 + 4 ) × 10 3 CS = 0.995 μ F (ii) ⎛ RP ⎞ ω rS T ( jω ) = ⎜ ⎟ ⎝ RS + RP ⎠ 1 + ω 2 rS2 rS = ( RS + RP ) CS = 7.96 × 10 −3 RP 4 = = 0.5 RS + RP 4 + 4 f = 40 Hz T ( jω ) = ( 0.5)( 2π )( 40 ) ( 7.96 × 10 −3 ) ( ) 1 + ⎡ 2π ( 40 ) 7.96 × 10 −3 ⎤ ⎣ ⎦ T ( jω ) = 0.447 f = 80 Hz T ( jω ) = 2 ( 0.5)( 2π )(80 ) ( 7.96 × 10 −3 ) ( ) 1 + ⎡ 2π ( 80 ) 7.96 × 10 −3 ⎤ ⎣ ⎦ T ( jω ) = 0.485 2 f = 200 Hz T ( jω ) = ( 0.5 )( 2π )( 200 ) ( 7.96 × 10 −3 ) ( ) 1 + ⎡ 2π ( 200 ) 7.96 × 10 −3 ⎤ ⎣ ⎦ T ( jω ) = 0.498 (b) ω= 1 1 = rP ( RS RP ) CP CP = = 1 2π f ( RS ( RP ) 2π 500 × 10 CP = 63.7 pF EX7.2 a. 1 3 ) (10 10 ) × 10 3 2 261. ⎛ RP ⎞ RP RP = 0.891 = ⇒ (1 − 0.891) RP = 0.891 ⇒ RP = 8.20 kΩ 20 log10 ⎜ ⎟ = −1 ⇒ RP + RS ⎠ RP + RS RP + 1 ⎝ 1 1 ⇒ CS = fL = 2π ( RS + RP ) CS 2π (100 )(1 + 8.20 ) × 10 3 CS = 0.173 μ F fH = 1 2π ( RS RP ) CP 1 ⇒ CP = ( ) (1 || 8.20 ) × 10 2π 10 6 3 CP = 179 pF b. rS = ( RS + RP ) CS ( )( rS = 1 × 10 3 + 8.20 × 10 3 0.173 × 10 −6 ) rS = 1.59 ms open-circuit time-constant rP = ( RS RP ) CP ( rP = (1 8.20 ) × 10 3 179 × 10 −12 ) rP = 0.160 μ s short-circuit time-constant EX7.3 a. rS = ( Ri + RS i ) CC b. f = 1 2π rS RTH = R1 R2 = 2.2 20 = 1.98 kΩ ⎛ R2 ⎞ ⎛ 2.2 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (10 ) = 0.991 V ⎝ 2.2 + 20 ⎠ ⎝ R1 + R2 ⎠ VTH − VBE ( on ) 0.991 − 0.7 I BQ = = = 0.0132 mA RTH + (1 + β ) RE 1.98 + ( 201)( 0.1) I CQ = 2.636 mA rπ = gm = β VT I CQ ICQ VT = = ( 200 )( 0.026 ) 2.636 = 1.97 kΩ 2.636 = 101.4 mA/V 0.026 Rib = τ π + (1 + β ) RE = 1.97 + ( 201)( 0.1) = 22.1 kΩ RB = R1 R2 = 1.98 kΩ Ri = RB Rib = 1.98 22.1 = 1.817 kΩ rS = ( Ri + RSi ) CC ( = (1.817 + 0.1) ×10 3 )( 47 × 10 ) −6 = 90.1 ms 1 f = ⇒ f = 1.77 Hz 2π 90.1 × 10 −3 ( Midband Gain ) 262. Aν = = − β RC Ri ⋅ rπ + (1 + β ) RE Ri + RSi − ( 200 )( 2 ) 1.82 1.97 + ( 201)( 0.1) 1.82 + 0.1 ⋅ Aν = −17.2 EX7.4 I DQ = K n (VGS − VTN ) 2 a. 0.8 + 2 = VGS ⇒ VGS = 3.265 V 0.5 VS − ( −5 ) VS = −3.265 ⇒ I DQ = RS −3.265 + 5 ⇒ RS = 2.17 kΩ 0.8 5 ⇒ RD = 6.25 kΩ VD = 0 ⇒ RD = 0.8 b. rS = ( RD + RL ) CC = (10 + 6.25 ) × 10 3 × CC 1 1 f = ⇒ CC = 2π rS 2π f 16.25 × 10 3 RS = ( CC = ( 1 2π ( 20 ) 16.25 × 10 3 ) ) ⇒ CC = 0.49 μ F EX7.5 rS = ( RL + Ro ) CC 2 1 1 f = ⇒ CC 2 = 2π rS 2π f ( RL + Ro ) ⎧ rπ + ( RS RB ) ⎫ ⎪ ⎪ R0 = RE r0 ⎨ ⎬ 1+ β ⎪ ⎪ ⎩ ⎭ From Example 7-5, R0 = 35.5 Ω CC 2 = 1 2π (10 ) ⎡10 × 10 3 + 35.5⎤ ⎣ ⎦ CC 2 = 1.59 μ F EX7.6 a. I DQ = K P (VSG + VTP ) 2 1 − ( −2 ) = VSG ⇒ VSG = 3.41 V 0.5 VS = 3.41 5 − 3.41 ⇒ RS = 1.59 kΩ RS = 1 5 For VSDG = VSGQ ⇒ VD = 0 ⇒ RD = ⇒ RD = 5 kΩ 1 263. rP = ( RD RL ) CL b. f = 1 1 ⇒ CL = 2π rP 2π f ( RD RL ) CL = ( ) 2π 10 6 1 ( 5 10 ) × 10 3 ⇒ CL = 47.7 pF EX7.7 (a) RTH = 5 K VTH = −3.7527 I BQ = −3.7527 − 0.7 − ( −5) 0.54726 = 5 + (101)( 0.5) 55.5 = 0.00986 I CQ = 0.986 mA gm = 37.925 rπ = 2.637 K 5 − Vo Vo − = 1 − Vo ( 0.4 ) 5 5 Vo = 0.035 (b) 0.986 = Rib RS ϭ 0.1 k⍀ V0 ϩ Vi ϩ Ϫ RTH ϭ 5 k⍀ V␲ r␲ gmV␲ Ϫ RE ϭ 0.5 k⍀ Vo = − gmVπ ( RC RL ) Rib = rπ + (1 + β ) RE = 2.64 + (101)( 0.5) = 53.14 K Vπ = Vb Vb Vb = = 101 ⎞ 14.885 ⎛1+ β ⎞ ⎛ 1+ ⎜ ⎟ RE 1 + ⎜ 2.637 ⎟ ( 0.5) ⎝ ⎠ rπ ⎠ ⎝ ⎛ RTH Rib ⎞ ⎛ 5 53.14 ⎞ Vb = ⎜ ⎟ Vi = ⎜ ⎟ Vi RTH Rib + RS ⎠ ⎝ 5 53.14 + 0.1 ⎠ ⎝ ⎛ 4.57 ⎞ =⎜ ⎟ Vi = 0.9786 ⎝ 4.57 + 0.1 ⎠ Av = − (c) EX7.8 a. (37.925)( 2.5) 14.885 ( 0.9786 ) = Av = −6.23 RC ͉͉RL ϭ 2.5 k⍀ 264. I BQ = 0 − 0.7 − ( −10 ) = 0.0230 mA 0.5 + (101)( 4 ) I CQ = 2.30 mA β VT rπ = I CQ I CQ gm = rB = VT = = (100 )( 0.026 ) 2.30 = 1.13 kΩ 2.30 = 88.46 mA / V 0.026 RE ( RS + rπ ) CE RS + rπ + (1 + β ) RE ( 4 × 10 ) ( 0.5 + 1.13) C = 3 E 0.5 + 1.13 + (101)( 4 ) rB = 1 1 = = 0.7958 ms 2π f B 2π ( 200 ) rB = 16.07CE ⇒ CE = b. rA = RE CE = 4 × 10 3 ( fA = fβ = = )( 49.5 × 10 ) ⇒ r −6 A = 0.198 s 1 1 = ⇒ f A = 0.804 H Z 2π rA 2π ( 0.198 ) EX7.9 rπ = 0.796 × 10 −3 ⇒ CE = 49.5 μ F 16.07 β 0VT I CQ = (150 )( 0.026 ) 0.5 = 7.8 kΩ 1 2π rπ ( Cπ + Cμ ) 1 ( 2π 7.8 × 10 3 ) ( 2 + 0.3) × 10 −12 ⇒ f β = 8.87 MH Z EX7.10 rπ = fβ = = β 0VT I CQ = (150)(0.026) ⇒ rπ = 3.9 kΩ 1 1 2π rπ ( Cπ + Cμ ) ( 2π 3.9 × 10 1 3 ) ( 4 + 0.5) (10 ) −12 ⇒ f β = 9.07 MHz fT = β 0 f β = (150 )( 9.07 ) ⇒ fT = 1.36 GHz EX7.11 RTH = R1 R2 = 200 220 = 104.8 kΩ ⎛ R2 ⎞ ⎛ 220 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (5) = 2.619 V ⎝ 200 + 220 ⎠ ⎝ R1 + R2 ⎠ 2.62 − 0.7 = 0.009316 mA I BQ = 105 + (101)(1) I CQ = 0.9316 mA 265. gm = rπ = I CQ = VT β VT I CQ 0.9316 ⇒ gm = 35.83 mA/V 0.026 = (100)(0.026) ⇒ rπ = 2.79 kΩ 0.932 a. C M = Cμ ⎡1 + gm ( RC RL ) ⎤ ⎣ ⎦ C M = ( 2 ) ⎡1 + ( 35.83 )( 2.2 4.7 ) ⎤ ⇒ C M = 109 pF ⎣ ⎦ b. RB = rS f3 dB = = R2 = 100 200 220 = 51.17 kΩ R1 1 2π ( RB rπ ) (Cπ + Cμ ) 1 ⇒ f3dB = 0.506 MHz 2π [ 51.17 2.79] × 10 3 × (10 + 109 ) × 10 −12 EX7.12 fT = gm 2π ( Cgs + Cgd ) gm = 2 K n I DQ =2 ( 0.2 )( 0.4 ) = 0.5657 mA/V fT = 0.5657 × 10 −3 ⇒ fT = 333 MHz 2π ( 0.25 + 0.02 ) × 10 −12 EX7.13 dc analysis ⎛ R2 ⎞ ⎛ 166 ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ (10 ) = 4.15 V R1 + R2 ⎠ ⎝ 166 + 234 ⎠ ⎝ V I D = S and VS = VG − VGS RS K n (VGS − VTN ) = 2 VG − VGS RS 2 ( 0.5)( 0.5) (VGS − 4VGS + 4 ) = 4.15 − VGS 2 0.25VGS − 3.15 = 0 ⇒ VGS = 3.55 V gm = 2K n (VGS − VTN ) = 2 ( 0.5 )( 3.55 − 2 ) = 1.55 mA/V Small-signal equivalent circuit. Ri ϭ 10 k⍀ Vi ϩ Ϫ RG ϭ R1͉͉R2 Cgs V0 ϩ Cgd Vgs RD gmVgs Ϫ a. C M = Cgd (1 + gm ( RD RL ) ) C M = ( 0.1) ⎡1 + (1.55) ( 4 20 )⎤ ⇒ C M = 0.617 pF ⎣ ⎦ b. RL 266. 1 fH = 2πτ P τ P = ( RG Ri ) ( Cgs + C M ) RG = R1 R2 = 234 166 = 97.1 kΩ RG Ri = 97.1 10 = 9.07 kΩ ( ) rP = 9.07 × 10 3 (1 + 0.617 ) × 10 −12 = 14.7 ns fH = 1 ( 2π 14.7 × 10 −9 ) ⇒ f H = 10.9 MHz EX7.14 dc analysis VTH = 0, RTH = 10 kΩ I BQ = 0 − 0.7 − ( −5 ) 10 + (126 )( 5 ) = 0.00672 mA I CQ = 0.840 mA rπ = gm = r0 = β VT I CQ I CQ VT = = (125)( 0.026 ) 0.840 = 3.87 kΩ 0.840 = 32.3 mA/V 0.026 VA 200 = = 238 kΩ I CQ 0.84 High-frequency equivalent circuit C␮ RS ϭ 1 k⍀ V0 ϩ Vi RB ϭ V R1͉͉R2 ␲ Ϫ ϩ Ϫ r␲ r0 C␲ a. Miller Capacitance ′ C M = Cμ (1 + gm RL ) ′ RL = r0 RC RL ′ RL = 238 2.3 5 = 1.565 kΩ C M = ( 3 ) ⎡1 + ( 32.3 )(1.57 ) ⎤ ⇒ Cμ = 155 pF ⎣ ⎦ b. Req = RS RB rπ = RS R1 R2 rπ Req = 1 20 20 3.87 = 0.736 kΩ rP = Re q ( Cπ + C M ) ( ) = 0.736 × 10 3 ( 24 + 155 ) × 10 −12 = 1.314 × 10 fH = c. ( −7 1 2π 1.314 × 10 −7 ) ⇒ f H = 1.21 MHz gmV␲ RC RL 267. ( Av )M ( Av )M ⎡ RB τ π ⎤ ′ = − gm RL ⎢ ⎥ ⎢ RB τ π + RS ⎥ ⎣ ⎦ ⎡ 10 3.87 ⎤ = − ( 32.3 )(1.565 ) ⎢ ⎥ ⇒ ( Av ) M = −37.2 ⎣ 10 3.87 + 1 ⎦ EX7.15 The dc analysis 10 − 0.7 = 0.00838 mA I BQ = 100 + (101)(10 ) I CQ = 0.838 mA rπ = gm = β VT I CQ = (100 )( 0.026 ) 0.838 = 3.10 kΩ I CQ = 32.22 mA/V VT For the input ⎡⎛ r ⎞ ⎤ rPπ = ⎢⎜ π ⎟ RE RS ⎥ Cπ ⎣⎝ 1 + β ⎠ ⎦ ⎡ 3.10 ⎤ 10 1⎥ × 10 3 × 24 × 10 −12 =⎢ ⎣ 101 ⎦ = 7.13 × 10 −10 s f Hπ = 1 1 = ⇒ f Hπ = 223 MHz 2π rPπ 2π 7.13 × 10 −10 ( ) For the output rP μ = [ RC RL ] Cμ = (10 1) × 10 3 × 3 × 10 −12 = 2.73 × 10 −9 1 1 fHμ = = ⇒ f H μ = 58.4 MHz 2π rP μ 2π 2.73 × 10 −9 ( ( Aν )M = gm ( RC ) ⎡ ⎛ rπ ⎞ ⎤ ⎢ RE ⎜ ⎟ ⎥ ⎝1+ β ⎠ ⎥ RL ) ⎢ ⎢ ⎥ ⎛ rπ ⎞ ⎢ RE ⎜ ⎟ + RS ⎥ ⎢ ⎥ ⎝1+ β ⎠ ⎣ ⎦ ⎡ ⎛ 3.1 ⎞ ⎤ ⎢ 10 ⎜ 101 ⎟ ⎥ ⎝ ⎠ ⎥⇒ A = ( 32.22 )(10 1) ⎢ ( ν )M = 0.870 ⎛ 3.1 ⎞ ⎥ ⎢ ⎢ 10 ⎜ 101 ⎟ + 1 ⎥ ⎝ ⎠ ⎦ ⎣ EX7.16 ⎛ ⎞ R3 7.92 ⎛ ⎞ VB1 = ⎜ (12 ) = 0.9502 V ⎟ (12 ) = ⎜ R1 + R2 + R3 ⎠ 58.8 + 33.3 + 7.92 ⎟ ⎝ ⎠ ⎝ Neglecting lose currents 0.9502 − 0.7 IC = = 0.50 mA 0.5 β VT (100 )( 0.026 ) rπ = = = 5.2 K IC 0.5 IC 0.5 = = 19.23 mA/V VT 0.026 From Eq (7.127(a)), gm = 268. τ Pπ = [ RS RB1 rπ ][Cπ 1 + C M 1 ] RB1 = R2 R3 = 33.3 7.92 = 6.398 K C M 1 = 2Cμ 1 = 6 pF Then 3 −12 τ Pπ = [1 6.398 5.2] × 10 × [24 + 6] × 10 ⇒ 22.24 ns 1 1 = ⇒ 7.15 MHz f Hπ = 2πτ Pπ 2π ( 22.24 × 10 −9 ) From Eq (7.128(a)) 3 −12 τ P μ = [ RC RL ] C μ 2 = ( 7.5 2 ) × 10 × 3 × 10 ⇒ 4.737 ns 1 1 fHμ = = = 33.6 MHz 2πτ Pμ 2π ( 4.737 × 10 −9 ) From Eq. 7.133 Av Av M M ⎡ RB1 rπ 1 ⎤ RC ) ⎢ ⎥ ⎣ RB1 rπ 1 + RS ⎦ ⎡ 6.40 5.2 ⎤ = (19.23)( 7.5 2 ) ⎢ ⎥ ⎣ 6.40 5.2 + 1 ⎦ = gm 2 ( RC ⎡ 2.869 ⎤ = (19.23)(1.579 ) ⎢ ⎣ 2.869 + 1 ⎥ ⎦ = 22.5 TYU7.1 a. V0 = − ( gmVπ ) RL rπ Vπ = × Vi 1 rπ + + RS sCC T (s) = = V0 ( s ) Vi ( s ) = − gm rπ RL rπ + RS + (1 / sCC ) − gm rπ RL ( sCC ) 1 + s ( rπ + RS ) CC gm rπ = β T (s) = − β RL rπ + RS ⎛ s ( rπ + RS ) CC ×⎜ ⎜ 1 + s (r + R ) C π S C ⎝ ⎞ ⎟ ⎟ ⎠ Then τ = ( rπ + RS ) CC b. 1 f3− dB = 2π ( rπ + RS ) CC f3− dB = 1 2π ⎡ 2 × 10 + 1 × 10 3 ⎤ ⎡10 −6 ⎤ ⎣ ⎦⎣ ⎦ ( 2 )( 50 )( 4 ) rg R T ( jω ) max = π m L = 2 +1 rπ + RS T ( jω ) max = 133 c. 3 ⇒ f3 dB = 53.1 Hz 269. ͉T( j␻)͉ 133 f 53.1 Hz TYU7.2 a. ⎛ V0 = − g mVπ ⎜ RL ⎝ 1 ⎞ ⎟ sCL ⎠ ⎛ r ⎞ Vπ = ⎜ π ⎟ × Vi ⎝ rπ + RS ⎠ 1 ⎛ ⎜ RL × sC V0 ( s ) rπ L ⎜ T (s) = = − gm 1 Vi ( s ) rπ + RS ⎜ ⎜ RL + sC L ⎝ T (s) = ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ − β RL ⎛ 1 ×⎜ ⎟ rπ + RS ⎝ 1 + sRL CL ⎠ Then τ = RL CL b. 1 1 f3− dB = = ⇒ f3dB = 3.18 MH Z 3 2π RL CL 2π 5 × 10 10 × 10 −12 ( T ( jω ) = )( gm rπ RL ( 75)(1.5)( 5) = 1.5 + 0.5 rπ + RS ) T ( jω ) max = 281 c. 281 ͉T( j␻)͉ 3.18 MHz f TYU7.3 a. Open-circuit time constant ( CL → open ) rS = ( RS + rπ ) CC ( ) = ( 0.25 + 2 ) × 10 3 2 × 10 −6 = 4.5 ms Short-circuit time constant ( CC → short ) ( )( rP = RL CL = 4 × 10 3 50 × 10 −12 rP = 0.2 μ s b. Midband gain ) 270. ⎛ rπ ⎞ V0 = − gmVπ RL , Vπ = ⎜ ⎟ Vi ⎝ τ π + RS ⎠ V −g r R Av = 0 = m π L Vi rπ + RS = − ( 65 )( 2 )( 4 ) 2 + 0.25 Av = −231 c. fL = 1 1 = ⇒ f L = 35.4 Hz 2π rS 2π 4.5 × 10 −3 fH = 1 1 ⇒ f H = 0.796 MHz = 2π rP 2π 0.2 × 10 −6 ( ) ( ) TYU7.4 Computer Analysis TYU7.5 Computer Analysis TYU7.6 (100 )( 0.026 ) βV = 10.4 kΩ rπ = 0 T = 0.25 I CQ fβ = 1 2π rπ ( Cπ + C μ ) Cπ + Cμ = 1 1 = 2π f β rπ 2π 11.5 × 10 6 10.4 × 10 3 ( )( Cπ + Cμ = 1.33 pF Cμ = 0.1 pF ⇒ Cπ = 1.23 pF TYU7.7 h fe = β0 1 + j ( f / fβ ) f β = 5 MH Z , β 0 = 100 At f = 50 MH Z h fe = 100 ⎛ 50 ⎞ 1+ ⎜ ⎟ ⎝ 5 ⎠ 2 ⇒ h fe = 9.95 ⎛ 50 ⎞ Phase = − tan −1 ⎜ ⎟ ⇒ Phase = −84.3° ⎝ 5 ⎠ TYU7.8 f 500 ⇒ f β = 4.17 MHz fβ = T = β 0 120 fβ = 1 2π rπ (Cπ + C μ ) Cπ + Cμ = 1 1 = 2π f β rπ 2π (4.167 × 10 6 )(5 × 10 3 ) Cπ + Cμ = 7.639 pF Cμ = 0.2 pF ⇒ Cπ = 7.44 pF ) 271. TYU7.9 (a) gm = 2K n (VGS − VTN ) = 2 ( 0.4 )( 3 − 1) ⇒ gm = 1.6 mA/V ′ gm = 80% of gm = 1.28 mA/V gm ′ gm = 1 + gm rS 1 + gm rS = gm ′ gm ⎛ gm ⎞ 1 ⎛ 1.6 ⎞ − 1⎟ = − 1⎟ ⎜ ⎜ ′ gm 1.6 ⎝ 1.28 ⎠ ⎝ ⎠ rS = 0.156 kΩ ⇒ rS = 156 ohms 1 gm rS = (b) gm = 2K n (VGS − VTN ) = 2 ( 0.4 )( 5 − 1) ⇒ gm = 3.2 mA/V gm 3.2 ′ gm = = = 2.134 1 + gm rS 1 + ( 3.2 )( 0.156 ) Δgm 3.2 − 2.134 = ⇒ A 33.3% reduction 3.2 gm TYU7.10 fT = = Cgs = = gm 2π ( CgsT + C gdT ) gm 2π ( Cgs + Cgsp + Cgdp ) gm − Cgsp − C gdp 2π fT 0.5 × 10 −3 ( 2π 500 × 10 6 ) − ( 0.01 + 0.01) × 10 −12 ⇒ Cgs = 0.139 pF TYU7.11 fT = gm 2π (Cgs + Cgsp + Cgdp ) Cgsp = Cgdp 2Cgsp = gm 1× 10−3 − C gs = − 0.4 × 10−12 2π fT 2π (350 × 106 ) 2Cgsp = 0.0547 pF ⇒ Cgsp = Cgdp ≅ 0.0274 pF TYU7.12 dc analysis ⎛ 50 ⎞ VG = ⎜ ⎟ (10 ) − 5 = −2.5 ⎝ 50 + 150 ⎠ VS = VG − VGS . I D = K n (VGS − VTN ) = 2 VS − ( −5 ) RS VG − VGS + 5 RS 272. 2 (1)( 2 ) ⎡VGS − 1.6VGS + 0.64 ⎤ = −2.5 − VGS + 5 ⎣ ⎦ 2 2VGS − 2.2VGS − 1.22 = 0 2.2 ± VGS = ( 2.2 ) 2 + 4 ( 2 )(1.22 ) 2 (2) ⇒ VGS = 1.505 V gm = 2 K n (VGS − VTN ) = 2 (1)(1.505 − 0.8 ) = 1.41 mA/V Equivalent circuit Cgd Ri V0 ϩ Vi RG ϭ Vgs R1͉͉R2 ϩ Ϫ Cgs gmVgs RD Ϫ C M = Cgd (1 + gm RD ) = ( 0.2 ) ⎡1 + (1.42 )( 5 ) ⎤ ⇒ C M = 1.61 pF ⎣ ⎦ (a) (b) τ P = ( Ri RG ) ( Cgs + CM ) rP = [ 20 50 150 ] × 10 3 × ( 2 + 1.61) × 10 −12 ( )( ) = 13 × 10 3 3.62 × 10 −12 = 4.71 × 10 −8 s fH = c. 1 1 = ⇒ f H = 3.38 MHz 2π rP 2π 4.71 × 10 −8 ( ( Av )M ( Av )M ) ⎛ RG ⎞ = − gm RD ⎜ ⎟ ⎝ RG + RS ⎠ ⎛ 37.5 ⎞ = − (1.41)( 5 ) ⎜ ⎟ ⇒ ( Av ) M = −4.60 ⎝ 37.5 + 20 ⎠ TYU7.13 Computer Analysis 273. Chapter 7 Problem Solutions 7.1 a. T (s) = T (s) = V0 ( s ) Vi ( s ) = 1/ ( sC1 ) ⎡1/ ( sC1 ) ⎤ + R1 ⎣ ⎦ 1 1 + sR1C1 b. 1 ͉T ͉ fH ϭ 159 Hz f fH = 1 1 = ⇒ f H = 159 Hz 2π R1C1 2π (103 )(10−6 ) c. V0 ( s ) = Vi ( s ) ⋅ 1 1 + sR1C1 For a step function Vi ( s ) = 1 s K K2 1 1 = 1+ V0 ( s ) = ⋅ s 1 + sR1C1 s 1 + sR1C1 = = K1 (1 + sR1C1 ) + K 2 s s (1 + sR1C1 ) K1 + s ( K1 R1C1 + K 2 ) s (1 + sR1C1 ) K 2 = − K1 R1C1 and K1 = 1 V0 ( s ) = − R1C1 1 + s 1 + sR1C1 = 1 1 − 1 s +s R1C1 v0 ( t ) = 1 − e− t / R1C1 7.2 a. T (s) = T (s) = b. V0 ( s ) Vi ( s ) = R2 R2 + ⎡1/ ( sC2 ) ⎤ ⎣ ⎦ sR2 C2 1 + sR2 C2 274. 1 ͉T ͉ fL ϭ 1.59 Hz 1 1 fL = = ⇒ f L = 1.59 Hz 4 2π R2 C2 2π (10 )(10 × 10−6 ) c. V0 ( s ) = Vi ( s ) ⋅ sR2 C2 1 + sR2 C2 Vi ( s ) = 1 s V0 ( s ) = R2 C2 1 = 1 + sR2 C2 s + 1 R2 C2 v0 ( t ) = e− t / R2C2 7.3 a. T (s) = RP V0 ( s ) Vi ( s ) 1 sCP ⎛ 1 1 ⎞ + ⎜ RS + ⎟ sCP ⎝ sCS ⎠ RP = RP 1 RP ⋅ sCP RP 1 = = sCP R + 1 1 + sRP CP P sCP Then RP ⎛ 1 ⎞ RP + ⎜ RS + ⎟ (1 + sRP CP ) sCS ⎠ ⎝ RP = RP CP 1 RP + RS + + + sRS RP CP CS sCS T (s) = ⎛ RP T (s) = ⎜ ⎝ RP + RS b. ⎤⎞ ⎞ ⎛ ⎡ sRP RS RP C 1 ⋅ P+ + ⋅ CP ⎥ ⎟ ⎟ × ⎜ 1/ ⎢1 + ⎥⎟ ⎠ ⎜ ⎢ RP + RS CS s ( RS + RP ) CS RS + RP ⎦⎠ ⎝ ⎣ 275. −11 ⎛ ⎤⎞ 1 ⎛ 10 ⎞ ⎜ ⎡ 10 10 T (s) = ⎜ × 1/ ⎢1 + ⋅ −6 + + s ( 5 × 103 ) ⋅10−11 ⎥ ⎟ ⎟ s ( 2 × 104 ) ⋅10−6 ⎝ 10 + 10 ⎠ ⎜ ⎢ 20 10 ⎥⎟ ⎦⎠ ⎝ ⎣ 1 1 ≅ ⋅ 1 2 1+ + s ( 5 × 10−8 ) s ( 0.02 ) s = jω T ( jω ) = 1 ⋅ 2 T ( jω ) = 1 1 ⋅ 2 ⎡ ⎤ 1 1 + j ⎢ω ( 5 × 10−8 ) − ⎥ ω ( 0.02 ) ⎦ ⎣ 1 1 For ω L = = = 50 4 ( RS + RR ) CS ( 2 ×10 )(10−6 ) 1 ⎡ ⎤ 1 1 + j ⎢( 50 ) ( 5 × 10−8 ) − ( 50 )( 0.02 ) ⎥ ⎣ ⎦ 1 1 1 1 ≈ ⋅ ⇒ T ( jω ) = ⋅ 2 1− j 2 2 For ωH = ( RS 1 1 = = 2 × 107 RP ) CP ( 5 × 103 )(10−11 ) 1 ⎡ ⎤ 1 ⎥ 1 + j ⎢( 2 × 107 )( 5 × 10−8 ) − ⎢ ( 2 ×107 ) ( 0.02 ) ⎥ ⎣ ⎦ 1 1 1 1 ⇒ T ( jω ) = ⋅ T ( jω ) ≈ ⋅ 2 1+ j 2 2 T ( jω ) = 1 ⋅ 2 In each case, T ( jω ) = c. RP 2 RP + RS 1 ⋅ RS = RP = 10 kΩ, CS = CP = 0.1 μ F T (s) = ⎛ ⎡ ⎤⎞ 1 ⎜ ⎢ 1 1 1 ⋅ 1/ 1 + ⋅ + + s ( 5 × 103 )(10−7 ) ⎥ ⎟ ⎥⎟ 2 ⎜ ⎢ 2 1 s 2 × 104 (10−7 ) ⎜ ⎟ ⎦⎠ ⎝ ⎣ ( ) s = jω T ( jω ) = 1 ⋅ 2 1 ⎡ ⎤ 1 1 ⎥ 1 + + j ⎢ω ( 5 × 10−4 ) − 2 ω ( 2 × 10−3 ) ⎥ ⎢ ⎣ ⎦ 1 = 500 For ω = ( 2 ×104 )(10−7 ) T ( jω ) = 1 ⋅ 2 1 ⎡ ⎤ 1 ⎥ 1.5 + j ⎢( 500 ) ( 5 × 10−4 ) − ⎢ ( 500 ) ( 2 ×10−3 ) ⎥ ⎣ ⎦ 1 1 = ⋅ ⇒ T ( jω ) = 0.298 2 1.5 − j ( 0.75 ) 276. For ω = 1 ( 5 ×10 )(10 ) −7 3 = 2 × 103 ⎧ ⎛ ⎫ ⎡ ⎤ ⎞⎪ 1 ⎪ ⎜ 1 ⎥ ⎟⎬ ⋅ ⎨1/ 1.5 + j ⎢( 2 × 103 )( 5 × 10−4 ) − 2 ⎪ ⎜ ⎢ ( 2 ×103 )( 2 ×10−3 ) ⎥ ⎟⎪ ⎣ ⎦ ⎠⎭ ⎩ ⎝ 1 1 = ⋅ ⇒ T ( jω ) = 0.298 2 1.5 + j ( 0.75 ) T ( jω ) = In each case, T ( jω ) < 7.4 Circuit (a): V R2 T= 0 = Vi 1 R2 + sC1 = = RP 2 RP + RS 1 ⋅ R2 R1 (1/ sC1 ) R2 + R1 + (1/ sC1 ) R2 R2 + R1 1 + sR1C1 = R2 (1 + sR1C1 ) R2 + sR1 R2 C1 + R1 or (1 + sR1C1 ) Vo R2 = ⋅ Vi R1 + R2 1 + sR1 R2 C1 Low frequency: Vo R2 20 2 = = = Vi R1 + R2 10 + 20 3 High frequency: Vo =1 Vi τ 1 = R1C1 = (104 )(10 × 10−6 ) = 0.10 ⇒ f1 = 1 = 1.59 Hz 2πτ 1 τ 2 = ( R1 R 2 ) C1 = (10 20 ) × 103 × (10 × 10−6 ) ⇒ τ 2 = 0.0667 ⇒ f 2 = ͉T ͉ 1.0 0.67 1.59 2.39 f Circuit (b): 1 sC2 R2 V 1 + sR2 C2 T= o = = R2 Vi 1 + R1 R2 + R1 1 + sR2 C2 sC2 R2 ⎞ ⎛ R2 ⎞ ⎛ 1 =⎜ ⎟ ⎟⎜ ⎜ ⎟ ⎝ R1 + R2 ⎠ ⎝ 1 + s ( R1 R2 ) C2 ⎠ Low frequency: Vo R2 20 2 = = = Vi R1 + R2 20 + 10 3 τ = ( R1 R2 ) C2 = (10 20 ) × 103 × 10 × 10−6 = 0.0667 f = 1 2πτ = 2.39 Hz 1 = 2.39 Hz 2πτ 2 277. 0.67 ͉T ͉ 2.39 f 7.5 a. rS = ( Ri + RP ) CS = [30 + 10] × 103 × 10 × 10−6 ⇒ rS = 0.40 s rP = ( Ri RP ) CP = ⎡30 10 ⎤ × 103 × 50 × 10−12 ⇒ rP = 0.375 μ s ⎣ ⎦ b. 1 1 fL = = ⇒ f L = 0.398 Hz 2π rS 2π ( 0.4 ) fH = 1 1 = ⇒ f H = 424 kHz 2π rP 2π ( 0.375 × 10−6 ) At midband. CS → short, CP → open Vo = I i ( Ri RP ) T ( s ) = Ri R P = 30 10 ⇒ T ( s ) = 7.5 k Ω c. 7.5 k⍀ ͉T ͉ fH fL 7.6 (a) T= T 1 (1 + j 2π f τ ) max =1 At f = 2 ⇒T = ( 1 1 + ( 2π f τ ) 2 ) 2 = 1 1 + ( 2π f τ ) 2 1 1 1 ⇒T = = 2 2πτ 2 1 + (1) ⎛1⎞ = 20 log10 ⎜ ⎟ ⇒ T dB ≅ −6 dB ⎝ 2⎠ −1 Phase = 2 tan ( 2π f τ ) = −2 tan −1 (1) = −2 ( 45° ) ⇒ Phase = −90° (b) Slope = −2 ( 6dB / oct ) = −12dB / oct = −40dB / decade T dB Phase = −2 ( 90° ) ⇒ Phase = −180° 7.7 (a) 278. −10 ( jω ) T ( jω ) = T = jω ⎞ jω ⎞ ⎛ ⎛ 20 ⎜ 1 + ⎟ ( 2000 ) ⎜1 + ⎟ 20 ⎠ 2000 ⎠ ⎝ ⎝ ⎛ jω ⎞ −5 × 10−3 ⎜ −4 ⎟ ⎝ ω ⎠ = 2.5 × 10 ( jω ) = jω ⎞ ⎛ jω ⎞ ⎛ jω ⎞⎛ jω ⎞ ⎛ 1+ 1+ 1+ ⎜1 + ω ⎟ ⎜ 2000 ⎟ ⎜ 20 ⎟⎜ 2000 ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ 2.5 × 10−4 (ω ) ⎛ω ⎞ ⎛ ω ⎞ 1+ ⎜ ⎟ ⋅ 1+ ⎜ ⎟ ⎝ 20 ⎠ ⎝ 2000 ⎠ 2 2 ͉T ͉ 5 ϫ 10Ϫ3 2.5 ϫ 10Ϫ4 1 20 2000 ␻ (b) jω ⎞ ⎟ 10 ⎠ ⎝ T ( jω ) = jω ⎞ ⎛ 1000 ⎜ 1 + ⎟ ⎝ 1000 ⎠ (10 )(10 ) ⎛1 + ⎜ T = ( 0.10 ) ⎛ω ⎞ 1+ ⎜ ⎟ ⎝ 10 ⎠ ⎛ ω ⎞ 1+ ⎜ ⎟ ⎝ 1000 ⎠ 2 2 ͉T ͉ 10 0.10 10 ␻ 1000 7.8 (a) T (s) = 105 = 10 ( 5 + 10 )( 5 + 500 ) (10 )( 500 ) ⎛ jω ⎞ ⎜ ⎟ 10 ⎝ 10 ⎠ = ⋅ 500 ⎛ jω ⎞⎛ jω ⎞ + 1⎟⎜ + 1⎟ ⎜ ⎝ 10 ⎠⎝ 500 ⎠ ⋅ 5 ⎛ 5 ⎞⎛ 5 ⎞ + 1⎟ ⎜ + 1⎟⎜ ⎝ 10 ⎠⎝ 500 ⎠ 279. ͉T ͉ 0.02 10 (b) (c) (d) 500 ␻(rod/s) Midband gain = 0.02 ω = 500 rad/s ω = 10 rad/s 7.9 (a) T ( s) = = T ( jω ) = 2 × 104 ( S + 10 )( S + 10 ) 3 5 2 × 104 (10 )(10 ) 3 5 ⋅ 1 ⎛ S ⎞⎛ S ⎞ ⎜ 3 + 1 ⎟ ⎜ 5 + 1⎟ 10 10 ⎝ ⎠⎝ ⎠ 2 × 10−4 ⎛ jω ⎞ ⎛ jω ⎞ ⎜ 3 + 1 ⎟ ⎜ 5 + 1⎟ ⎝ 10 ⎠ ⎝ 10 ⎠ ͉T ͉ 2 ϫ 10Ϫ4 10 3 (b) (c) (d) 2 × 10 10 5 ␻(rod/s) −4 ω = 103 rad/s no low freq −3dB freq. 7.10 a. ⎛ r ⎞ V0 = − g mVπ RL Vπ = ⎜ π ⎟ Vi rπ + RS ⎠ ⎝ ⎛ r ⎞ ⎛ 5.2 ⎞ T = g m RL ⎜ π ⎟ = ( 29 )( 6 ) ⎜ ⎟ ⎝ 5.2 + 0.5 ⎠ ⎝ rπ + RS ⎠ Tmidband = 159 b. rS = ( RS + rπ ) CC 1 1 1 fL = ⇒ rS = = ⇒ rS = 5.31 ms Open-circuit 2π rS 2π f L 2π ( 30 ) rP = 1 1 = ⇒ τ P = 0.332 μ s Short-circuit 2π f H 2π ( 480 × 103 ) 280. c. rS 5.31× 10−3 = ⇒ CC = 0.932 μ F ( RS + τ π ) ( 0.5 + 5.2 ) ×103 CC = rP = RL CL rP 0.332 × 10−6 = ⇒ CL = 55.3 pF RL 6 × 103 CL = 7.11 Computer Analysis 7.12 Computer Analysis 7.13 a. RTH = R1 R2 = 10 1.5 = 1.304 kΩ ⎛ R2 ⎞ ⎛ 1.5 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (12 ) = 1.565 V ⎝ 1.5 + 10 ⎠ ⎝ R1 + R2 ⎠ 1.565 − 0.7 I BQ = = 0.0759 mA 1.30 + (101)( 0.1) I CQ = 7.585 mA rπ = (100 )( 0.026 ) = 0.343 kΩ 7.59 7.59 = 292 mA/V gm = 0.026 Ri = R1 R2 ⎡ rπ + (1 + β ) RE ⎤ ⎣ ⎦ = 10 1.5 ⎡0.343 + (101)( 0.1) ⎤ ⎣ ⎦ = 1.30 10.44 ⇒ Ri = 1.159 kΩ r = ( RS + Ri ) CC = [ 0.5 + 1.16] × 103 × 0.1× 10−6 r = 1.659 × 10−4 s fL = 1 1 = ⇒ f L = 959 Hz 2π r 2π (1.66 × 10−4 ) b. RS Rib V␲ Vi ϩ Ϫ V0 ϩ Ib R1͉͉R2 r␲ gmV␲ Ϫ RC RE 281. V0 = − ( β I b ) RC R1b = rπ + (1 + β ) RE = 0.343 + (101)( 0.1) = 10.44 kΩ ⎛ R1 R2 ⎞ Ib = ⎜ I ⎜R R +R ⎟ i ⎟ ib ⎠ ⎝ 1 2 ⎛ 1.30 ⎞ =⎜ ⎟ I i = ( 0.111) I i ⎝ 1.30 + 10.4 ⎠ Vi Ii = RS + R1 R2 Rib = Ii = V0 = Vi Vi 0.5 + (1.3) (10.44 ) Vi 1.659 β RC ( 0.111) 1.659 ⇒ V0 Vi = (100 )(1)( 0.111) midband 1.659 ⇒ V0 Vi = 6.69 midband c. 6.69 ͉V ͉ V 0 i fL ϭ 959 Hz f 7.14 I DQ = 0.5 mA ⇒ VS = ( 0.5 )( 0.5 ) = 0.25 V 0.5 + 1.5 = 3.08 V 0.2 + VS = 3.08 + 0.25 ⇒ VG = 3.33 V I DQ = K n (VGS − VTN ) ⇒ VGS = 2 VG = VGS ⎛ R2 ⎞ 1 VG = ⎜ ⎟ VDD ⇒ 3.33 = ⋅ Rin ⋅ VDD R1 + R2 ⎠ R1 ⎝ 1 3.33 = ( 200 )( 9 ) ⇒ R1 = 541 kΩ R1 541R2 = 200 ⇒ R2 = 317 kΩ 541 + R2 VD = VDSQ + VS = 4.5 + 0.25 = 4.75 9 − 4.75 ⇒ RD = 8.5 kΩ 0.5 1 1 1 fL = ⇒ rL = = = 7.96 ms 2π rL 2π f L 2π ( 20 ) RD = rL = Rin ⋅ CC ⇒ CC = rL 7.96 × 10−3 = ⇒ CC = 0.0398 μ F 200 × 103 Rin g m = 2 ( 0.2 )( 3.08 − 1.5 ) = 0.632 mA/V Av midband = ( 0.632 )(8.5) g m RD = ⇒ Av = 4.08 1 + g m RS 1 + ( 0.632 )( 0.5 ) 282. 4.08 ͉A␯͉ fL ϭ 20 Hz Phase f fL Ϫ90 Ϫ135 Ϫ180 7.15 I DQ I DQ = K n (VGS − VTN ) ⇒ VGS = 2 Kn + VTN = 1 + 1 = 2.414 V 0.5 VS = −2.414 V −2.414 − ( −5 ) RS = VD = VDSQ ⇒ RS = 2.59 kΩ 1 + VS = 3 − 2.414 = 0.586 V 5 − 0.59 ⇒ RD = 4.41 kΩ 1 RD = b. CC ϩ Vi ϩ Ϫ RG Vgs gmVgs RD Ϫ RS ⎛ ⎜ RD I 0 = − ( g mVgs ) ⎜ ⎜R +R + 1 L ⎜ D sCC ⎝ Vi Vgs = 1 + g m RS I0 ( s ) Vi ( s ) = T (s) = = c. ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ⎡ ⎤ − gm sCC ⋅ RD ⎢ ⎥ 1 + g m RS ⎢1 + s ( RD + RL ) CC ⎥ ⎣ ⎦ I0 ( s ) Vi ( s ) s ( RD + RL ) CC − g m RD 1 ⋅ ⋅ 1 + g m RS RD + RL 1 + s ( RD + RL ) CC I0 RL 283. fL = 1 1 1 → rL = = = 15.92 ms 2π rL 2π f L 2π (10 ) rL = ( RD + RL ) CC ⇒ CC = rL 15.9 × 10−3 = ⇒ CC = 1.89 μ F RD + RL ( 4.41 + 4 ) × 103 7.16 a. 9 − VSG 2 = I D = K P (VSG + VTP ) RS 2 9 − VSG = ( 0.5 )(12 ) (VSG − 4VSG + 4 ) 2 6VSG − 23VSG + 15 = 0 VSG = ( 23) 23 ± 2 − 4 ( 6 )(15 ) 2 ( 6) ⇒ VSG = 3 V g m = 2 K P (VSG + VTP ) = 2 ( 0.5 )( 3 − 2 ) ⇒ g m = 1 mA/V Ro = 1 RS = 1 12 ⇒ Ro = 0.923 kΩ gm b. r = ( R0 + RL ) CC c. fL = CC = 1 2πτ ⇒r= 1 2π f L = 1 = 7.96 ms 2π ( 20 ) 7.96 × 10−3 r = ⇒ CC = 0.729 μ F Ro + RL ( 0.923 + 10 ) × 103 7.17 a. 1 = 0.00833 mA 120 R1 || R2 = ( 0.1)(1 + β )( RE ) = ( 0.1)(121)( 4 ) = 48.4 kΩ I CQ = 1 mA, I BQ = VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE 1 ⋅ RTH ⋅ VCC = ( 0.00833)( 48.4 ) + 0.7 + (121)( 0.00833)( 4 ) R1 1 ( 48.4 )(12 ) = 5.135 R1 R1 = 113 kΩ 113R2 = 48.4 ⇒ R2 = 84.7 k Ω 113 + R2 b. r R0 = π RE r0 1+ β rπ = (120 )( 0.026 ) 1 = 3.12 kΩ 80 = 80 kΩ 1 3.12 R0 = 4 80 = 0.02579 4 80 ⇒ R0 = 25.6 Ω 121 c. r0 = 284. r = ( R0 + RL ) CC 2 r = ( 0.0256 + 4 ) × 103 × 2 × 10−6 = 8.05 × 10−3 s f = 1 1 = ⇒ f = 19.8 Hz 2π r 2π ( 8.05 × 10−3 ) 7.18 (a) 5 − 0.7 = 1.075 mA I CQ = 1.064 mA 4 = 10 − (1.064 )( 2 ) − (1.075 )( 4 ) = 3.57 V I EQ = VCEQ VCEQ gm = rπ = I CQ = VT β VT 1.064 = 40.92 mA/V 0.026 = I CQ (100 )( 0.026 ) 1.064 = 2.44 K (b) For CC1 , Req1 = RS + RE rπ 2440 = 200 + 4000 1+ β 101 Req1 = 224.0r , τ 1 = Req1CC1 = 1.053 ms For CC 2 , Req 2 = RC + RL = 2 + 47 = 49 K τ 2 = Req 2 ⋅ Cc 2 = 49 ms f1 = (c) 1 2πτ 1 = 1 2π (1.053 × 10−3 ) ⇒ f1 = 151 Hz 7.19 (a) τ H = ( RC RL ) CL = ( 2 47 ) × 103 × 10 × 10−12 = 1.918 × 10−8 s fH = 1 2πτ H = 1 2π (1.918 × 10−8 ) (b) 1 1 + ( f .2πτ H ) 2 = 0.1 2 2 ⎛ 1 ⎞ ⎜ ⎟ = 100 = 1 + ( f .2πτ H ) ⎝ 0.1 ⎠ f = 99 2πτ H = 2π (1.918 × 10−8 ) f = 82.6 MHz 7.20 (a) 99 ⇒ f H = 8.30 MHz 285. 5 − VSG 2 = K P (VSG + VTP ) R1 2 5 − VSG = (1)(1.2 )(VSG − 1.5 ) = (1.2 ) (VSG − 3VSG + 2.25 ) 2 2 1.2VSG − 2.6VSG − 2.3 = 0⇒ VSG = 2.84 V I DQ = 1.8 mA VSDQ = 10 − (1.8 )(1.2 + 1.2 ) ⇒ VSDQ = 5.68 V g m = 2 K P I DQ = 2 (1)(1.8 ) = 2.683 mA / V ro = ∞ (b) Ris = 1 1 = = 0.3727 k Ω g m 2.68 Ri = 1.2 0.373 = 0.284 k Ω For CC1 , τ s1 = ( 284 + 200 ) ( 4.7 × 10−6 ) = 2.27 ms For CC 2 , τ s 2 = (1.2 x103 + 50 × 103 )(10−6 ) = 51.2 ms (c) CC2 dominates, 1 1 f 3− dB = = = 3.1 Hz 2πτ s 2 2π ( 51.2 × 10−3 ) 7.21 ′ Assume VTN = 1V , kn = 80μ A / V 2 , λ = 0 Neglecting RSi = 200Ω , Midband gain is: Av = g m RD Let I DQ = 0.2 mA, VDSQ = 5V Then RD = 9−5 ⇒ RD = 20 k Ω 0.2 We need g m = Av RD = 10 ⎛ k′ ⎞⎛ W ⎞ = 0.5 mA / V 2 and g m = 2 K n I DQ = 2 ⎜ n ⎟ ⎜ ⎟ I DQ 20 ⎝ 2 ⎠⎝ L ⎠ W ⎛ 0.080 ⎞ ⎛ W ⎞ or 0.5 = 2 ⎜ = 7.81 ⎟ ⎜ ⎟ ( 0.2 ) ⇒ L ⎝ 2 ⎠⎝ L ⎠ Let 9 9 R1 + R2 = = = 225 k Ω ( 0.2 ) I DQ ( 0.2 )( 0.2 ) ⎛ R2 ⎞ 2 ⎛ R2 ⎞ ⎛ 0.080 ⎞ I DQ = 0.2 = ⎜ ⎟ (9 ) = ⎜ ⎟ ( 7.81)(VGS − 1) ⇒ VGS = 1.80 = ⎜ ⎟ (9) ⇒ ⎝ 2 ⎠ ⎝ 225 ⎠ ⎝ R1 + R2 ⎠ R2 = 45 k Ω, R1 = 180 k Ω RTH = R1 R2 = 180 45 = 36 k Ω τ1 = 1 2π f1 = 1 7.96 × 10−4 = 7.958 × 10−4 s = ( RSi + RTH ) CC or CC = ⇒ 2π ( 200 ) ( 200 + 36 ×103 ) CC = 0.022 μ F τ2 = 7.22 a. 1 2π f 2 = 1 2π ( 3 x103 ) = 5.305 × 10−5 s = RD CL or CL = 5.31× 10−5 ⇒ CL = 2.65 nF 20 × 103 286. T (s) = T (s) = R2 + (1/ sC ) R2 + (1/ sC ) + R1 1 + sR2 C 1 + s ( R1 + R2 ) C rA = R2 C , rB = ( R1 + R2 ) C b. ͉T ͉ 1 0.0909 fA fB f c. rA = R2 C = (103 )(100 × 10−12 ) = 10−7 s = rA rB = ( R1 + R2 ) C = [10 + 1] × 103 × 100 × 10−12 = 1.1× 10−6 s = rB fA ≈ 1 1 = ⇒ f A = 1.59 MHz 2π rA 2π (10−7 ) fB ≈ 1 1 = ⇒ f B = 0.145 MHz 2π rB 2π (1.1× 10−6 ) 7.23 I BQ = 10 − 0.7 = 0.00997 mA 430 + ( 201)( 2.5 ) I CQ = ( 200 ) I BQ = 1.995 mA rπ = ( 200 )( 0.026 ) = 2.61 k Ω 1.99 Rib = 2.61 + ( 201)( 2.5 ) = 505 k Ω τs = 1 = 2π f L 1 = 0.0106 s 2π (15 ) = Req CC = ( 0.5 + 505 430 ) × 103 CC = 232.7 × 103 CC Or CC = 4.55 × 10−8 F ⇒ 45.5 nF 7.24 10 = I BQ (300) + 0.7 + ( 201) I BQ (1) ⇒ I BQ = 0.0186 mA ⇒ I CQ = 3.7126 mA rπ = β VT I CQ = ( 200 )( 0.026 ) 3.71 = 1.40 K Ri = rπ + (1 + β ) RE = 1.40 + ( 201)(1) = 202.4 K Req = RS + RB Ri = 0.1 + 300 202.4 = 121 K τ L = Req ⋅ CC fL = CC = 7.25 1 2πτ L τL Req ⇒τL = = 1 2π f = 1 = 7.958 × 10−3 s 2π ( 20 ) 7.958 × 10−3 ⇒ CC = 0.0658 μ F 121× 103 287. RTH = R1 R2 = 1.2 1.2 = 0.6 k Ω ⎛ R2 ⎞ ⎛ 1.2 ⎞ VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ ( 5 ) = 2.5 V ⎝ 1.2 + 1.2 ⎠ ⎝ R1 + R2 ⎠ 2.5 − 0.7 = 0.319 mA I BQ = 0.6 + (101)( 0.05 ) I CQ = 31.9 mA rπ = (100 )( 0.026 ) 31.9 = 0.0815 k Ω 1 so that f 3− dB ( CC1 ) > τ C and f = C1 C2 f 3− dB ( CC 2 ) = 25 Hz = 1 2πτ 2 , so that τ 2 = 1 = 0.006366 s = Req CC 2 2π ( 25 ) where ⎛ r + R1 R2 RS ⎞ Req = RL + RE ⎜ π ⎟ 1+ β ⎝ ⎠ ⎛ 81.5 + 600 300 ⎞ = 10 + 50 ⎜ ⎟ = 10 + 50 2.787 ⇒ 101 ⎝ ⎠ 0.00637 Req = 12.64 Ω ⇒ CC 2 = ⇒ CC 2 = 504 μ F 12.6 Rib = rπ + (1 + β ) RE Assume CC 2 an open Rib = 81.5 + (101)( 50 ) = 5132 Ω τ 1 = (100 )τ 2 = (100 )( 0.006366 ) = 0.6366 s = Req1CC1 Req1 = RS + RTH Rib = 300 + 600 5132 = 837.2 Ω So CC1 = 0.6366 ⇒ CC1 = 760 μ F 837.2 7.26 From Problem 7.25 RTH = 0.6 K, I CQ = 31.9 mA, rπ = 81.5 Ω 1 so f 3− dB ( CC 2 ) f 3− dB ( CC1 ) 2πτ Then f 3− dB ( CC 2 ) ⇒ CC1 acts as an open circuit and for f 3− dB ( CC1 ) ⇒ CC 2 acts as a short circuit. τC2 τ C1 and f = f 3− dB ( CC1 ) = 20 Hz = 1 2πτ C1 ⇒ τ C1 = 0.007958 s Rib = rπ + (1 + β ) ( RE RL ) = 81.5 + (101) ( 50 10 ) = 923.2 Ω τ C1 ⇒ Req1 = RS + RTH Rib = 300 + 600 923.2 = 663.7 Ω 0.007958 ⇒ CC1 = 12 μ F 663.7 = 100τ C1 = 0.7958 s CC1 = τC2 ⎛ r + RTH ⎞ Req 2 = RL + RE ⎜ π ⎟ = 10 + 50 ⎝ 1+ β ⎠ Req 2 = 10 + 50 6.748 = 15.95 Ω 0.7958 CC 2 = ⇒ CC 2 = 0.050 F 15.95 7.27 a. ⎛ 81.5 + 600 ⎞ ⎜ ⎟ ⎝ 101 ⎠ 288. I D = K n (VGS − VTN ) VGS = 2 ID 0.5 + VTN = + 0.8 = 1.8 V Kn 0.5 −VGS − ( −5 ) 5 − 1.8 ⇒ RS = 6.4 kΩ 0.5 0.5 VD = VDSQ + VS = 4 − 1.8 = 2.2 V 5 − 2.2 RD = ⇒ RD = 5.6 kΩ 0.5 (b) RS = gm = 2 Kn I D = 2 = ( 0.5 )( 0.5 ) = 1 mA / V rA = RS CS = ( 6.4 × 103 )( 5 × 10−6 ) = 3.2 × 10−2 s 1 1 = ⇒ f A = 4.97 Hz fA = 2π rA 2π ( 3.2 × 10−2 ) ⎡ 6.4 × 103 ⎤ ⎞ −6 ⎥ ( 5 × 10 ) ⎟ CS = ⎢ 1 + (1)( 6.4 ) ⎥ ⎢ ⎠ ⎣ ⎦ = 4.32 × 10−3 s 1 1 = ⇒ f B = 36.8 Hz fB = 2π rB 2π ( 4.32 × 10−3 ) ⎛ RS rB = ⎜ ⎝ 1 + g m RS c. g m RD (1 + sRS CS ) Av = ⎡ ⎛ RS 1 + g m RS ⎝ (1 + g m RS ) ⎢1 + s ⎜ ⎞ ⎤ ⎟ CS ⎥ ⎠ ⎦ ⎣ As RS becomes large g m RD ( sRS CS ) Av → ⎡ ⎛ R ⎞ ⎤ ( g m RS ) ⎢1 + s ⎜ S ⎟ CS ⎥ ⎝ g m RS ⎠ ⎦ ⎣ ⎡ ⎛ 1 ⎞ ⎤ ⎟ CS ⎥ ⎣ ⎝ gm ⎠ ⎦ ( g m RD ) ⎢ s ⎜ Av = ⎛ 1 1+ s ⎜ ⎝ gm ⎞ ⎟ CS ⎠ The corner frequency f B = gm = 2 Kn I D = 2 fB = ( 0.5 )( 0.5 ) = 1 mA / V 1 ⇒ f B = 31.8 Hz ⎛ 1 ⎞ −6 2π ⎜ −3 ⎟ ( 5 × 10 ) ⎝ 10 ⎠ 7.28 a. 1 and the corresponding f A → 0 2π (1/ g m ) CS fB = RE ( RS + rπ ) CE 1 and rB = 2π rB RS + rπ + (1 + β ) RE 289. For RS = 0 rB = I EQ = RE rπ CE rπ + (1 + β ) RE −0.7 − ( −10 ) 5 = 1.86 mA β = 75 ⇒ I CQ = 1.84 mA β = 125 ⇒ I CQ = 1.85 mA For f B ≤ 200 Hz ⇒ rB ≥ 1 = 0.796 ms 2π ( 200 ) rπ αβ so smallest rB will occur for smallest β . β = 75; rπ = ( 75)( 0.026 ) = 1.06 kΩ 1.84 ( 5 ×103 ) (1.06 ) CE ⇒ C = 57.2 μ F 0.796 × 10−3 = E 1.06 + ( 76 )( 5 ) b. (125 )( 0.026 ) For β = 125; rπ = = 1.76 kΩ 1.85 ( 5 ×103 ) (1.76 ) ( 57.2 ×10−6 ) = 0.797 ms rB = 1.76 + (126 )( 5 ) fB = 1 1 = ⇒ f B = 199.7 Hz Essentially independent of β . 2π rB 2π ( 0.797 × 10−3 ) rA = RE CE = ( 5 × 103 )( 57.2 × 10−6 ) = 0.286 sec fA = 1 1 = ⇒ f A = 0.556 Hz Independent of β . 2π rA 2π ( 0.286 ) 7.29 a. Expression for the voltage gain is the same as Equation (7.66) with RS = 0. b. rA = RE CE RE rπ CE rB = rπ + (1 + β ) RE 7.30 5 − 0.7 I C = 0.99 mA = 1 mA = I E 4.3 β VT (100 )( 0.026 ) rπ = = = 2.626 K 0.99 I CQ rA = RE CE = ( 4.3 × 103 )( 5 × 10−6 ) = 2.15 ×10 −2 s rB = ( 4.3 ×103 )( 2.626 ×103 )( 5 ×10−6 ) = 1.292 ×10−4 s RE rπ CE = rπ + (1 + β ) RE 2.626 × 103 + (101) ( 4.3 × 103 ) fA = 1 1 = = 7.40 Hz 2π rA 2π ( 2.15 × 10−2 ) fB = 1 1 = = 1.23 × 103 = 1.23 kHz 2π rB 2π (1.292 × 10−4 ) 290. Av w→0 = Av w →∞ = (100 )( 2 ) β RC = = 0.458 rπ + (1 + β ) RE 2.626 + (101)( 4.3) β RC rπ = (100 )( 2 ) 2.626 = 76.2 ͉A␯͉ 76.2 0.458 7.40 Hz 1.23 kHz f 7.31 rH = ( RL RC ) CL = (10 5 ) × 103 × 15 × 10−12 rH = 5 × 10−8 s 1 1 = ⇒ f H = 3.18 MHz 2π rH 2π ( 5 × 10−8 ) fH = 10 − 0.7 = 0.93 mA, I CQ = 0.921 mA 10 0.921 gm = = 35.4 mA/V 0.026 Av = g m ( RC RL ) = 35.4 ( 5 10 ) ⇒ Av = 118 I EQ = 7.32 ⎛ R2 ⎞ VG = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠ ID = ⎛ 166 ⎞ =⎜ ⎟ (10 ) ⎝ 166 + 234 ⎠ = 4.15 V VG − VGS 2 = K n (VGS − VTN ) RS 2 4.15 − VGS = ( 0.5 )( 0.5 ) (VGS − 4VGS + 4 ) 2 0.25VGS − 3.15 = 0 ⇒ VGS = 3.55 V g m = 2 K n (VGS − VTN ) = 2 ( 0.5 )( 3.55 − 2 ) g m = 1.55 mA / V 1 1 = 0.5 = 0.5 0.645 1.55 gm R0 = RS R0 = 0.282 kΩ r = ( R0 RL ) CL and f H = βω ≈ f H = 5 MHz ⇒ r = CL = 7.33 r R0 RL = 1 2π r 1 2π ( 5 × 106 ) = 3.18 × 10−8 s 3.18 × 10−8 ⇒ CL = 121 pF ( 0.282 4 ) ×103 291. (a) Low-frequency RS CC Vo ϩ ϩ Ϫ Vs RB V␲ RC r␲ Ϫ RL gmV␲ Mid-Band RS Vo ϩ ϩ Ϫ Vs RB V␲ r␲ Ϫ RC RL gmV␲ High-frequency RS Vo ϩ ϩ Ϫ Vs RB V␲ Ϫ r␲ RC gmV␲ (b) ͉Am͉ fL fH f (c) 12 − 0.7 = 11.3 μ A 1 MΩ = 1.13 mA I BQ = I CQ rπ = (100 )( 0.026 ) = 2.3 k Ω 1.13 1.13 gm = = 43.46 mA / V 0.026 Am = ⎛ R R ⎞ V0 ( midband ) = − g m ( RC RL ) ⎜ B π ⎟ ⎜R r +R ⎟ Vs S ⎠ ⎝ B π ⎛ 1000 2.3 ⎞ = − ( 43.46 ) ( 5.1 500 ) ⎜ ⎜ 1000 2.3 + 1 ⎟ ⎟ ⎝ ⎠ ⎛ 2.29 ⎞ = ( 43.46 )( 5.05 ) ⎜ ⎟ ⇒ Am = 153 ⎝ 2.29 + 1 ⎠ Am dB = 43.7 dB RL CL 292. fL = 1 2πτ L , τ L = ( RS + RB rπ ) CC or τ L = (1 + 1000 2.3) × 103 (10 × 10−6 ) ⇒ τ L = 3.29 × 10−2 s ⇒ f L = 4.83 Hz fH = 1 , τ H = ( RC RL ) CL ⇒ τ L = ( 5.1 500 ) × 103 (10 × 10−12 ) 2πτ H = 5.05 × 10−8 s ⇒ f H = 3.15 MHz 7.34 a. V0 Ϫ Vi ϩ Ϫ Vsg RD gmVsg RL CL ϩ ⎛ 1 ⎞ V0 = ( g mVsg ) ⎜ R0 RL ⎟ sCL ⎠ ⎝ Vsg = −Vi Av ( s ) = V0 ( s ) ⎛ 1 ⎞ = − g m ⎜ RD RL ⎟ Vi ( s ) sCL ⎠ ⎝ ⎡ 1 ⎢ RD RL ⋅ sCL = − gm ⎢ ⎢ 1 ⎢ RD RL + sCL ⎢ ⎣ Av ( s ) = − g m ( RD RL ) ⋅ b. c. ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ 1 1 + s ( RD RL ) CL r = ( RD RL ) CL r = (10 20 ) × 103 × 10 × 10−12 ⇒ r = 6.67 × 10 −8 s fH = 1 1 = ⇒ f H = 2.39 MHz 2π r 2π ( 6.67 × 10−8 ) From Example 7.6, gm = 0.705 mA/V Av = g m ( RD RL ) = ( 0.705 ) (10 20 ) ⇒ Av = 4.7 7.35 Computer Analysis 7.36 Computer Analysis 7.37 Computer Analysis 7.38 293. fT = gm 2π ( Cπ + Cμ ) I CQ gm = fT = 1 = 38.46 mA/V 0.026 = VT 38.46 × 10−3 2π (10 + 2 ) × 10−12 fT = 510 MHz fT fβ = = = β 510 ⇒ f β = 4.25 MHz 120 5000 MHz ⇒ f β = 33.3 MHz 150 gm 7.39 fT fβ = fT = β 2π ( Cπ + Cμ ) 0.5 = 19.23 mA/V 0.026 19.2 × 10 −3 5 × 109 = 2π ( Cπ + 0.15 ) × 10−12 gm = Cπ + 0.15 = 19.2 × 10−3 2π (10−12 )( 5 × 109 ) = 0.612 pF Cπ = 0.462 pF 7.40 fβ = a. fT β = 2000 MHz = 13.3 MHz = f β 150 b. h fe = 150 1 + j ( f / fβ ) h fe = 150 1 + ( f / fβ ) 2 = 10 2 ⎛ f ⎞ ⎛ 150 ⎞ 2 1+ ⎜ ⎟ = ⎜ = 225 ⎜ f ⎟ ⎝ 10 ⎟ ⎠ ⎝ β⎠ f = f β ⋅ 224 = (13.33) 224 ⇒ f = 199.6 MHz 7.41 (a) V0 = − g mVπ RL where 1 sC1 rπ 1 + srπ C1 Vπ = ⋅ Vi = ⋅ Vi rπ 1 + rb rπ + rb 1 + srπ C1 sC1 rπ = ⎞ ⎛ r ⎞⎛ rπ 1 ⋅ Vi = ⎜ π ⎟ ⎜ ⎟ ⋅ Vi rπ + rb + srb rπ C1 ⎝ rπ + rb ⎠ ⎜ 1 + s ( rb rπ ) C1 ⎟ ⎝ ⎠ 294. So Av ( s ) = V0 ( s ) ⎞ ⎛ r ⎞⎛ 1 = − g m RL ⎜ π ⎟ ⎜ ⎟ ⎜ 1 + s ( rb rπ ) C1 ⎟ Vi ( s ) ⎝ rπ + rb ⎠ ⎝ ⎠ (100 )( 0.026 ) (b) Midband gain: rπ = 1 = 2.6 k Ω, g m = 1 = 38.46 mA / V 0.026 (i) For rb = 100 Ω (ii) ⎛ 2.6 ⎞ Av1 = − ( 38.46 )( 4 ) ⎜ ⎟ ⇒ Av1 = −148.1 ⎝ 2.6 + 0.1 ⎠ For rb = 500 Ω (c) (i) ⎛ 2.6 ⎞ Av 2 = − ( 38.46 )( 4 ) ⎜ ⎟ ⇒ Av 2 = −129.0 ⎝ 2.6 + 0.5 ⎠ 1 , τ = ( rb rπ ) C1 f 3− dB = 2πτ For rb = 100 Ω τ 1 = ( 0.1 2.6 ) × 103 ( 2.2 ×10−12 ) = 2.12 × 10−10 s ⇒ f 3− db = 751 MHz (ii) For rb = 500 Ω τ 2 = ( 0.5 2.6 ) × 103 ( 2.2 ×10−12 ) = 9.23 ×10−10 s f 3− dB = 173 MHz 7.42 (b) f = 10 kHz = 104 Z i = 200 + ( 2500 1 − j (104 )(1.333 × 10−6 ) 1 + (10 ) (1.333 ×10 ) 4 2 ) −6 2 = 200 + 2500 − j 33.3 = 2700 − j 33.3 f = 100 kHz = 105 (c) Z i = 200 + ( 2500 1 − j (105 )(1.333 × 10−6 ) 1 + (10 ) (1.333 ×10 ) 5 2 ) −6 2 Z i = 200 + 2456 − j 327 = 2656 − j 327 f = 1 MHz = 106 (d) Z i = 200 + ( 2500 1 − j (106 )(1.333 × 10−6 ) 1 + (10 ) (1.333 ×10 ) 6 2 ) −6 2 Z i = 200 + 900 − j1200 = 1100 − j1200 7.43 a. CM = Cμ (1 + g m RL ) b. RB͉͉RS rb V0 ϩ RB • Vi RB ϩ RS ϩ Ϫ V␲ Ϫ r␲ C␲ CM gmV␲ RL 295. V0 = − g mVπ RL rπ Vπ = Let Cπ + CM = Ci 1 sCi ⎛ RB ⎞ ⋅⎜ ⎟ Vi R + RS ⎠ 1 rπ + RB RS + rb ⎝ B sC1 Av ( s ) = Vo ( s ) Vi ( s ) 1 ⎡ ⎤ rπ ⋅ ⎢ ⎥ sCi ⎢ ⎥ 1 ⎢ ⎥ rπ + ⎥ ⎛ RB ⎞ ⎢ sCi = − g m RL ⎜ ⎥ ⎟⎢ ⎝ RB + RS ⎠ ⎢ rπ ⋅ 1 ⎥ ⎢ ⎥ sCi + RB RS + rb ⎥ ⎢ 1 ⎢ rπ + ⎥ sCi ⎢ ⎥ ⎣ ⎦ ⎡ ⎤ ⎛ RB ⎞ rπ = − g m RL ⎜ ⎥ ⎟× ⎢ ⎝ RB + RS ⎠ ⎢ rπ + (1 + srπ Ci ) ( RB RS + rb ) ⎥ ⎣ ⎦ Let Req = ( RB RS + rb ) ⎛ RB Av ( s ) = − β RL ⎜ ⎝ RB + RS Av ( s ) = c. ⎡ ⎤ ⎞ ⎢ 1 ⎥ × ⎟ ⎢ ⎡1 + s rπ Req Ci ⎤ ⎥ ⎠ ⎢ ( rπ + Req ) ⎣ ⎦⎥ ⎣ ⎦ ( ) − β RL ⎛ RB ⎞ 1 ⋅⎜ ⎟⋅ rπ + Req ⎝ RB + RS ⎠ 1 + s rπ Req Ci ( fH = ( 1 ) ) 2π rπ Req Ci 7.44 High Freq. ⇒ CC1 , CC 2 , CE → short circuits C␮ V0 ϩ IS R1͉͉R2 V␲ Ϫ r␲ C␲ gmV␲ RC RL I0 296. gm = fT = I CQ VT = 5 = 192.3 mA/V 0.026 gm 2π ( Cπ + Cμ ) ⇒ 250 × 106 = 192 × 10−3 2π ( Cπ + Cμ ) Cπ + Cμ = 122.4 pF ⇒ Cμ = 5 pF, Cπ = 117.4 pF ( CM = Cμ 1 + g m ( RC RL ) ) = 5 ⎡1 + (192.3) (1 1) ⎤ ⇒ CM = 485.8 pF ⎣ ⎦ Ci = Cπ + CM = 117 + 485 = 603 pF rπ = ( 200 )( 0.026 ) = 1.04 kΩ 5 Req = R1 R2 rπ = 5 1.04 = 0.861 kΩ r = Req ⋅ Ci = ( 0.861× 103 )( 603 × 10−12 ) = 5.19 × 10 −7 s 1 f = 2π r = 1 2π ( 5.19 × 10−7 ) ⇒ f = 307 kHz 7.45 RTH = R1 R2 = 60 5.5 = 5.04 kΩ ⎛ R2 ⎞ ⎛ 5.5 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (15 ) = 1.26 V ⎝ 5.5 + 60 ⎠ ⎝ R1 + R2 ⎠ 1.26 − 0.7 I BQ = = 0.0222 mA 5.04 + (101)( 0.2 ) I CQ = 2.22 mA rπ = (100 )( 0.026 ) = 1.17 kΩ 2.22 2.22 gm = = 85.4 mA/V 0.026 Lower 3 – dB frequency: rL = Req ⋅ CC1 Req = RS + R1 R2 rπ = 2 + 60 5.5 1.17 = 2.95 k Ω rL = ( 2.95 ×103 )( 0.1× 10−6 ) = 2.95 × 10−4 s fL = 1 1 = ⇒ f L = 540 Hz 2π rL 2π ( 2.95 × 10−4 ) Upper 3 – dB frequency: 297. gm fT = 2π ( Cπ + Cμ ) ⇒ 400 × 106 = 85.4 × 10−3 2π ( Cπ + Cμ ) Cπ + Cμ = 34 pF; Cμ = 2 pF ⇒ Cπ = 32 pF CM = Cμ (1 + g m RC ) = 2 (1 + ( 85.4 )( 4 ) ) ⇒ CM = 685 pF Ci = Cπ + CM = 32 + 685 = 717 pF Req = RS R1 R2 rπ = 2 60 5.5 1.17 = 0.644 kΩ = ( 0.644 × 103 )( 717 × 10−12 ) r = Req ⋅ Ci = 4.62 × 10−7 s fH = 1 2π r ⇒ f H = 344 kHz 7.46 RTH = R1 R2 = 600 55 = 50.38 K ⎛ R2 ⎞ ⎛ 55 ⎞ VTH = ⎜ ⎟ (15 ) = ⎜ ⎟ (15 ) = 1.2595 V ⎝ 600 + 55 ⎠ ⎝ R1 + R2 ⎠ 1.26 − 0.7 I BQ = = 0.00222 mA 50.4 + (101)( 2 ) I CQ = 0.2217 mA rπ = (100 )( 0.026 ) = 11.73 K 0.222 0.2217 gm = = 8.527 mA/V 0.026 Lower – 3dB Fig. τ L = R e q1 Cc1 ; Req1 = RS + RTH r π = 0.50 + 50.38 11.73 = 10.0 K τ L = (10 × 10 3 )( 0.1×10 ) = 10 −6 −3 s τ L = R e q1 Cc1 ; Req1 = RS + RTH r π = 0.50 + 50.38 11.73 = 10.0 K τ L = (10 × 103 )( 0.1×10−6 ) = 10−3 s 1 fL = 2πτ 2 = 1 2π (10−3 ) ⇒ f L = 159 Hz Upper – 3dB Fig. gm 8.527 × 10−3 = = 400 × 106 fT = −12 2π ( Cπ + Cμ ) 2π ( Cπ + 2 ) × 10 Cπ + Cμ = 3.393 pF ⇒ Cπ = 1.393 pF CM = Cμ (1 + g m RC ) = 2 ⎡1 + ( 8.527 )( 40 ) ⎤ = 684 pF ⎣ ⎦ CT = Cπ + CM = 1.393 + 684 = 685.4 pF Req 2 = RS RTH rπ = 0.5 50.38 11.73 = 50.38 0.480 = 0.4750 K τ H = R eq 2 .CT fH = = ( 0.4750 × 103 ) ( 685.4 × 10−12 ) = 3.256 × 10−7 s 1 1 = ⇒ f H = 489 KHz 2πτ H 2π ( 3.256 × 10−7 ) 298. 7.47 fT = gm 2π ( Cgs + Cgd ) ⎛ 40 ⎞ g m = 2 K n I D , K n = (15 ) ⎜ ⎟ = 60 μ A / V 2 ⎝ 10 ⎠ gm = 2 fT = ( 60 )(100 ) = 154.9 μ A / V 154.9 × 10−6 ⇒ fT = 44.8 MHz 2π ( 0.5 + 0.05 ) × 10−12 7.48 fT = gm 2π ( Cgs + Cgd ) g m = 2 K n (VGs − VT ) ID then g m = 2 K n I D Kn I D = K n (VGs − VTN ) or VGs − VTN = 2 So fT = 2 Kn I D 2π ( Cgs + Cgd ) = Kn I D π ( Cgs + Cgd ) ( 0.2 ×10 )( 20 ×10 ) −3 (a) fT = (b) fT = (c) 10 = −6 π ( 0.5 + 0.1) × 10−12 ⇒ fT = 33.6 MHz ( 0.2 ×10 )( 250 ×10 ) −3 −6 π ( 0.5 + 0.1) × 10−12 ( 0.2 ×10 ) I ⇒ fT = 118.6 MHz −3 9 D π ( 0.5 + 0.1) × 10−12 ⇒ I D = 17.8 mA 7.49 fT = gm 2π ( Cgs + Cgd ) Cgs + Cgd = WLCox ⎛W ⎞⎛ μ C ⎞ g m = 2 K n (VGS − VTN ) = 2 ⎜ ⎟ ⎜ n ox ⎟ (VGS − VTN ) ⎝ L ⎠⎝ 2 ⎠ ⎛W ⎞ ⎜ ⎟ ( μ n Cox )(VGS − VTN ) L Then fT = ⎝ ⎠ 2π WLCox fT = μ n (VGS − VTN ) 2π L2 (a) fT = (b) fT = 7.50 a. 450 ( 0.5 ) 2π (1.2 × 10−4 ) 450 ( 0.5 ) 2 2π ( 0.18 × 10−4 ) ⇒ fT = 2.49 GHz ⇒ fT = 111 GHz 299. ( CM = Cgd ′ 1 + g m ( ro RD ) ) CM = 5 ⎡1 + ( 3) (15 10 ) ⎤ ⇒ CM = 95 pF ⎣ ⎦ b. r = ( ri ) ( Cgs + CM ) r = (10 × 103 ) ( 50 + 95 ) × 10−12 = 1.45 × 10−6 s f = 1 1 = ⇒ f = 110 kHz 2π r 2π (1.45 × 10−6 ) 7.51 fT = gm 2π ( CgsT + CgdT ) ( Eq. 7.104 ) ⎛ 2⎞ Let CgdT = 0 and CgsT = ⎜ ⎟ (WLCox ) ⎝ 3⎠ ⎛μ C g m = 2 K n I D = 2 ⎜ n ox ⎝ 2 ⎞ ⎡W ⎤ ⎟ ⎢ L ⎥ ID ⎠⎣ ⎦ ⎛1 ⎞⎛ W ⎞ 2 ⎜ μ n Cox ⎟⎜ ⎟ I D ⎝2 ⎠⎝ L ⎠ So fT = ⎛ 2⎞ 2π ⎜ ⎟ (W LCox ) ⎝ 3⎠ ⎛1 ⎞⎛ W ⎞ ⎜ μ n Cox ⎟⎜ ⎟ I D 2 3 ⎝ ⎠⎝ L ⎠ = ⋅ 2π L W Cox fT = 3 2π L ⋅ μn I D 2W Cox L 7.52 (a) ′ gm = gm 1 + g m rS g m = 2 K n (VGS − VTN ) −8 ⎛ μ C ⎞ ⎛ W ⎞ ⎛ ( 400 ) ( 7.25 × 10 ) ⎞ ⎟ (10 ) K n = ⎜ n ox ⎟ ⎜ ⎟ = ⎜ ⎜ ⎟ 2 ⎝ 2 ⎠⎝ L ⎠ ⎝ ⎠ K n = 1.45 × 10−4 A / V 2 For VGS = 5 V g m = 2 (1.45 × 10 −4 ) ( 5 − 0.65 ) = 1.26 × 10−3 A/V ′ g m = ( 0.80 ) g m = 1.01× 10−3 A/V 1.01× 10−3 = 1.26 × 10−3 1 + (1.26 × 10−3 ) rS 1 + (1.26 × 10−3 ) rS = 1.25 ⇒ rS = 196 Ω b. 300. For VGS = 3 V g m = 2 (1.45 × 10−4 ) ( 3 − 0.65 ) = 0.6815 × 10−3 A/V ′ gm = 0.6815 × 10−3 1 + ( 0.6815 × 10−3 ) (196 ) ′ g m = 0.60 × 10−3 A/V Re duced by ≈ 12% 7.53 a. ϩ Vgs Ii Ri gmVgs Ϫ RL I0 rS I 0 = g mVgs and Vgs = I i Ri − g mVgs rS so Vgs = Then Ai = b. I i Ri 1 + g m rS I0 g R = m i I i 1 + g m rS As an approximation, consider ϩ Ii Vgs Ri CgsT I0 CM RL gЈ Vgs m Ϫ In this case I gm 1 ′ ′ ′ Ai = 0 = g m Ri ⋅ where CM = C gdT (1 + g m RL ) and g m = Ii 1 + g m rs 1 + sRi ( C gsT + CM ) c. As rS increases, CM decreases, so the bandwidth increases, but the current gain magnitude decreases. 7.54 ⎛ R2 ⎞ ⎛ 225 ⎞ VGS = ⎜ ⎟ VDD = ⎜ ⎟ (10 ) ⎝ 225 + 500 ⎠ ⎝ R1 + R2 ⎠ VGS = 3.10 V g m = 2 K n (VGS − VTN ) = 2 (1)( 3.10 − 2 ) g m = 2.207 mA / V a. CgdT Ri V0 Vi b. ϩ Ϫ ϩ R1͉͉R2 Vgs Ϫ CgsT gmVgs RD 301. CM = C gdT (1 + g m RD ) = (1) ⎡1 + ( 2.207 )( 5 ) ⎤ ⎣ ⎦ CM = 12 pF c. r = ( Ri R1 R2 ) ( CgsT + CM ) Ri R1 R2 = 1 500 225 = 1 155 = 0.9936 kΩ r = ( 0.9936 × 103 ) ( 5 + 12 ) × 10−12 = 1.69 × 10 −8 s fH = Av = 1 2π r = 1 2π (1.69 × 10−8 ) − g mVgs RD Vi and Vgs = ⇒ f H = 9.42 MHz R1 R2 R1 R2 + Ri ⋅ Vi = 155 ⋅ Vi = 0.994 Vi 155 + 1 Av = − ( 2.2 )( 5 )( 0.994 ) ⇒ Av = −10.9 7.55 RTH = R1 R2 = 33 22 = 13.2 kΩ ⎛ R2 ⎞ ⎛ 22 ⎞ VTH = ⎜ ⎟ ( 5) = ⎜ ⎟ ( 5) = 2 V ⎝ 22 + 33 ⎠ ⎝ R1 + R2 ⎠ 2 − 0.7 I BQ = = 0.00261 mA 13.2 + (121)( 4 ) I CQ = 0.3138 rπ = (120 )( 0.026 ) = 9.94 kΩ 0.3138 0.3138 gm = = 12.07 mA/V 0.026 100 r0 = = 318 kΩ 0.3138 a. gm fT = 2π ( Cπ + Cμ ) Cπ + Cμ = gm 12.07 × 10−3 = 2π fT 2π ( 600 × 106 ) Cπ + Cμ = 3.20 pF; Cμ = 1 pF ⇒ Cπ = 2.20 pF ( ) CM = Cμ ⎡1 + g m ro RC RL ⎤ ⎣ ⎦ = (1) ⎡1 + (12.07 ) 318 4 5 ⎤ ⎣ ⎦ CM = 27.6 pF b. ( ) 302. r = Req ( Cπ + CM ) Req = R1 R2 Rs rπ = 33 22 2 rπ = 1.74 9.94 kΩ ⇒ Req = 1.48 kΩ r = (1.48 × 103 ) ( 2.20 + 27.6 ) × 10−12 r = 4.41× 10 −8 s fH = 1 2π r 1 = ( 2π ( 4.41× 10−8 ) V0 = − g mVπ ro RC RL ⇒ f H = 3.61 MHz ) ⎛ R1 R2 rπ ⎞ Vπ = ⎜ ⎟V ⎜ R1 R2 rπ + RS ⎟ i ⎝ ⎠ R1 R2 rπ = 33 22 9.94 = 5.67 kΩ ⎛ 5.67 ⎞ vπ = ⎜ ⎟ Vi = 0.739Vi ⎝ 5.67 + 2 ⎠ r0 RC RL = 318 4 5 = 2.18 kΩ Av = − (12.07 )( 0.739 )( 2.18 ) Av = −19.7 7.56 RTH = R1 R2 = 40 5 = 4.44 kΩ ⎛ R2 ⎞ ⎛ 5 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (10 ) = 1.111 V ⎝ 5 + 40 ⎠ ⎝ R1 + R2 ⎠ 1.111 − 0.7 = 0.00633 mA I BQ = 4.44 + (121)( 0.5 ) I CQ = 0.760 mA rπ = (120 )( 0.026 ) = 4.11 kΩ 0.760 0.760 = 29.23 mA/V gm = 0.026 r0 = ∞ fT = gm 2π ( Cπ + Cμ ) Cπ + Cμ = gm 29.23 × 10−3 = 2π fT 2π ( 250 × 106 ) Cπ + Cμ = 18.6 pF; Cμ = 3 pF ⇒ Cπ = 15.6 pF a. CM = Cμ ⎡1 + g m ( RC RL ) ⎤ ⎣ ⎦ ⎡1 + ( 29.2 ) ( 5 2.5 ) ⎤ ⇒ CM = 149 pF CM = 3 ⎣ ⎦ For upper frequency: x 303. rH = Req ( Cπ + CM ) Req = rπ R1 R2 RS = 4.11 40 5 0.5 Req = 0.405 kΩ rH = ( 0.405 × 103 ) (15.6 + 149 ) × 10−12 = 6.67 ×10 −8 s 1 fH = ⇒ f H = 2.39 MHz 2π rH For lower frequency: rL = Req CC1 Req = RS + R1 R2 rπ = 0.5 + 40 5 4.11 Req = 2.64 kΩ rL = ( 2.64 × 103 )( 4.7 × 10−6 ) = 1.24 × 10−2 s fL = 1 ⇒ f L = 12.8 Hz 2π rL b. ͉A␯͉ 39.5 fL fH V0 = − g mVπ ( RC RL ) ⎛ R1 R2 rπ ⎞ Vπ = ⎜ ⎟V ⎜ R1 R2 rπ + RS ⎟ i ⎝ ⎠ ⎛ 2.135 ⎞ Vπ = ⎜ ⎟ Vi = 0.8102Vi ⎝ 2.135 + 0.5 ⎠ AV = ( 29.23)( 0.8102 ) ( 5 2.5 ) AV = 39.5 7.57 I D = K P (VSG + VTP ) = 2 9 − VSG RS 2 ( 2 )(1.2 ) (VSG − 4VSG + 4 ) = 9 − VSG 2 2.4VSG − 8.6VSG + 0.6 = 0 VSG = 8.6 ± (8.6 ) − 4 ( 2.4 )( 0.6 ) 2 ( 2.4 ) 2 VSG = 3.512 V g m = 2 K P (VSG + VTP ) = 2 ( 2 )( 3.512 − 2 ) g m = 6.049 mA / V I D = ( 2 )( 3.512 − 2 ) = 4.572 mA 1 1 ⇒ r0 = 21.9 kΩ r0 = = λ I o ( 0.01)( 4.56 ) 2 f 304. ( CM = CgdT 1 + g m ( ro RD ) a. ) CM = (1) ⎡1 + ( 6.04 ) ( 21.9 1) ⎤ ⇒ CM = 6.785 pF ⎣ ⎦ b. rH = ( Ri RG ) ( CgsT + CM ) rH = ( 2 100 ) × 103 (10 + 6.78 ) × 10−12 rH = 3.29 × 10−8 s 1 fH = → f H = 4.84 MHz 2πτ H V0 = − g m ( ro RD ) ⋅ Vgs ⎛ RG ⎞ ⎛ 100 ⎞ Vgs = ⎜ ⎟ Vi = ⎜ ⎟ Vi ⎝ 102 ⎠ ⎝ RG + Ri ⎠ ⎛ 100 ⎞ Av = − ( 6.04 ) ⎜ ⎟ ( 21.9 1) ⎝ 102 ⎠ Av = −5.67 7.58 ⎛ R2 ⎞ ⎛ 22 ⎞ VG = ⎜ ⎟ ( 20 ) − 10 = ⎜ ⎟ ( 20 ) − 10 R1 + R2 ⎠ ⎝ 22 + 8 ⎠ ⎝ VG = 4.67 V ID = 10 − VSG − 4.67 2 = K P (VSG + VTP ) RS 2 5.33 − VSG = (1)( 0.5 ) (VSG − 4VSG + 4 ) 2 0.5VSG − VSG − 3.33 = 0 VSG = 1 ± 1 + 4 ( 0.5 )( 3.33) 2 ( 0.5 ) ⇒ VSG = 3.77 V g m = 2 K p (VSG + VTP ) = 2 (1)( 3.77 − 2 ) g m = 3.54 mA / V b. ( CM = CgdT 1 + g m ( RD RL ) ) CM = ( 3) ⎡1 + ( 3.54 ) ( 2 5 ) ⎤ ⇒ CM = 18.2 pF ⎣ ⎦ a. r = Req ( CgsT + CM ) Req = Ri R1 R2 = 0.5 8 22 = 0.461 kΩ r = ( 0.461× 103 ) (15 + 18.2 ) × 10−12 = 1.53 × 10−8 s fH = 1 2π r ⇒ f H = 10.4 MHz c. V0 = − g mVgs ( RD RL ) ⎛ R R ⎞ ⎛ 5.87 ⎞ Vgs = ⎜ 1 2 ⎟ Vi = ⎜ ⎟ Vi ⇒ Vgs = ( 0.9215 ) Vi ⎜ R1 R2 Ri ⎟ ⎝ 5.87 + 0.5 ⎠ ⎝ ⎠ Av = − ( 3.54 )( 0.9215 ) ( 2 5 ) ⇒ Av = −4.66 305. 7.59 ⎛ 100 ⎞ I E = 0.5 mA ⇒ I CQ = ⎜ ⎟ ( 0.5 ) = 0.495 mA ⎝ 101 ⎠ 0.495 = 19.0 mA/V gm = 0.026 (100 )( 0.026 ) = 5.25 kΩ rπ = 0.495 a. Input: From Eq. 7.107b ⎡ rπ ⎤ rPπ = ⎢ RE RS ⎥ Cπ ⎣1 + β ⎦ ⎡ 5.25 ⎤ 0.5 0.05⎥ × 103 × 10 × 10−12 =⎢ ⎣ 101 ⎦ = 2.43 × 10−10 s 1 ⇒ f H π = 656 MHz 2π rpπ Output: From Eq. 7.108b rPμ = ( RB RL ) Cμ = (100 1) × 103 × 10−12 f Hπ = = 9.90 × 10−10 s fH μ = 1 ⇒ f H μ = 161 MHz 2π rPμ b. RS Vi gmV␲ Ϫ ϩ Ϫ RE V␲ r␲ ϩ V0 = − g m Vπ ( RB RL ) g mVπ + Vπ Vπ Vi − ( −Vπ ) + + =0 rπ RE RS ⎡ 1 1 1 ⎤ −V + ⎥= i Vπ ⎢ g m + + rπ RE RS ⎦ RS ⎣ 1 1 1 ⎤ −Vi ⎡ + + = Vπ ⎢19 + 5.25 0.5 0.05 ⎥ 0.05 ⎣ ⎦ Vπ ( 41.19 ) = −Vi ( 20 ) Vπ = − ( 0.4856 ) Vi V0 = − (19 )( −0.4856 )(100 1) Vi Av = 9.14 c. r = CL ( RL RB ) = (15 × 10−12 ) (1 100 ) × 103 r = 1.485 × 10−8 s 1 → f = 10.7 MHz 2π r Since f < f H μ ⇒ 3d B freq. dominated by CL . f = 7.60 V0 RB RL 306. 20 − 0.7 = 1.93 mA 10 ⎛ 100 ⎞ I CQ = ⎜ ⎟ (1.93) = 1.91 mA ⎝ 101 ⎠ 1.91 gm = = 73.5 mA/V 0.026 (100 )( 0.026 ) rπ = = 1.36 kΩ 1.91 a. Input: ⎡ rπ ⎤ rPπ = ⎢ RE RS ⎥ ⋅ Cπ ⎣1 + β ⎦ I EQ = rPπ ⎡1.36 ⎤ 10 1⎥ × 103 × 10 × 10−12 =⎢ ⎣ 101 ⎦ = 1.327 × 10−10 s 1 ⇒ f Pπ = 1.20 GHz 2π rPπ Output: rPπ = ( RC RL ) Cμ = ( 6.5 5 ) × 103 × 10−12 f Pπ = rPπ = 2.826 × 10−9 s 1 f Pμ = → f Pμ = 56.3 MHz 2π rPπ b. V0 = − g m Vπ ( RC RL ) g mVπ + Vπ Vπ Vi − ( −Vπ ) + + =0 rπ RE RS ⎛ Vi 1 1 1 ⎞ + Vπ ⎜ g m + + ⎟=− rπ RE RS ⎠ RS ⎝ 1 1 1 ⎤ −V ⎡ + + = i Vπ ⎢ 73.5 + 1.36 10 1 ⎥ (1) ⎣ ⎦ Vπ ( 75.34 ) = −Vi ⇒ Vπ = − ( 0.01327 ) Vi V0 = − ( 73.5 )( −0.01327 ) ( 6.5 5 ) Vi Av = 2.76 c. r = CL ( RL Rc ) = (15 × 10−12 ) ( 6.5 5 ) × 103 r = 4.24 × 10−8 s f = 1 → f = 3.75 MHz 2π r Since f < fp μ , 3dB frequency is dominated by CL. 7.61 307. VGS + I D RS = 5 5 − VGS 2 ID = = K n (VGS − VTN ) RS 5 − VGS = ( 3)(10 ) (V 2GS − 2VGS + 1) 30V 2GS − 59VGS + 25 = 0 VGS = 59 ± ( 59 ) − 4 ( 30 )( 25 ) 2 2 ( 30 ) ⇒ VGS = 1.349 V g m = 2 K n (VGS − VTN ) = 2 ( 3)(1.35 − 1) g m = 2.093 mA / V On the output: rPμ = ( RD RL ) Cgd T = ( 5 4 ) × 103 × 4 × 10−12 rPμ = 8.89 × 10−9 s f Pμ = 1 → f Pμ = 17.9 MHz 2π rPμ Ri Vi gmVgs Ϫ ϩ Ϫ RS Vgs RD ϩ V0 = − g mVgs ( RD RL ) g mVgs + Vgs RS + Vi − ( −Vgs ) RS =0 ⎛ V 1 1⎞ + ⎟=− i Vgs ⎜ g m + RS Ri ⎠ Ri ⎝ V 1 1⎞ ⎛ Vgs ⎜ 2.093 + + ⎟ = − i 10 2 ⎠ 2 ⎝ Vgs = ( 0.1857 )Vi Av = V0 = ( 2.093)( 0.1857 ) ( 5 4 ) Vi Av = 0.864 7.62 dc analysis ID = V + − VSG 2 = K P (VSG + VTP ) RS 5 − VSG = (1)( 4 )(VSG − 0.8 ) 2 2 = 4 (VSG − 1.6VSG + 0.64 ) 2 4VSG − 5.4VSG − 2.44 = 0 VSG = V0 5.4 ± ( 5.4 ) 2 + 4 ( 4 )( 2.44 ) 2 ( 4) = 1.707 g m = 2 K P (VSG + VTP ) = 2 (1)(1.707 − 0.8 ) g m = 1.81 mA / V RL 308. Ri gmVgs Ϫ RS Vgs CgsT CgdT RD ϩ 3 ⋅ dB frequency due to CgsT : Req = 1 RS Ri gm 1 2π Req ⋅ CgsT fA = 1 4 0.5 = 0.246 kΩ 1.81 1 fA = = 162 MHz 2π ( 246 ) ( 4 × 10−12 ) Req = 3 − dB frequency due to CgdT 1 2π ( RD RL ) CgdT fB = = 1 2π ( 2 4 ) × 103 × 10−12 f = 119 MHz Midband gain Ri Vi gmVgs V0 Ϫ ϩ Ϫ Vgs RS RD ϩ 1 RS gm 1 4 1.81 Vgs = ⋅ Vi = ⋅ Vi 1 1 RS + R i 4 + 0.5 gm 1.81 − − = −0.492Vi V0 = − g mVgs ( RD RL ) Av = ( 0.492 )(1.81) ( 4 2 ) ⇒ Av = 1.19 7.63 rπ = (120 )( 0.026 ) 1.02 g m = 39.23 mA/V a. = 3.059 kΩ RL RL 309. Input: f H π = 1 2π rπ rπ = ⎡ Rs R2 R3 rπ ⎤ ( Cπ + 2Cμ ) ⎣ ⎦ Req = 0.1 20.5 28.3 3.06 = 0.096 kΩ rπ = ( 96 ) (12 + 2 ( 2 ) ) × 10−12 = 1.537 × 10−9 s f Hπ = 1 2π (1.536 × 10−9 ) Output: f H μ = = 103.6 MHz 1 2π rμ rμ = ( RC RL ) Cμ = (15 10 ) × 103 × 2 × 10−12 = 6.67 × 10−9 fH μ = 1 2π ( 6.67 × 10−9 ) = 23.9 MHz b. A = g m ( RC ⎡ R2 R3 rπ ⎤ RL ) ⎢ ⎥ ⎣ R2 R3 rπ + RS ⎦ R2 R3 rπ = 20.5 28.3 3.059 = 2.433 kΩ ⎡ 2.433 ⎤ A = ( 39.23) ( 5 10 ) ⎢ ⎥ ⇒ A = 125.6 ⎣ 2.433 + 0.1 ⎦ c. CL = 15 pF > Cμ ⇒ CL dominates frequency response. 310. Chapter 8 Exercise Solutions EX8.1 VCC = 30 V, VCE = 30 − I C RC , I CVCE = 10 1 Maximum power at VCE = VCC = 15 2 10 10 2 IC = = = VCE 15 3 a. 2 So 15 = 30 − RL ⇒ RL = 22.5 Ω ⇒ Maximum Power = 10 W 3 b. VCC = 15 V, I C ,max = 2 A VCE = 15 − I C RL 0 = 15 − 2 RL ⇒ RL = 7.5 Ω ⇒ Maximum Power = (1)( 7.5 ) = 7.5 W EX8.2 (a) (b) ΔT = Pθ = ( 8 )( 2.4 ) ΔT = 19.2° C ΔT = Pθ ΔT 85 = P= θ 3.7 P = 23.0 W EX8.3 Power = iD ⋅ vDS = (1)(12 ) = 12 watts c. Tsink = Tamb + P ⋅θ snk − amb Tsink = 25 + (12 )( 4 ) ⇒ Tsink = 73°C b. Tcase = Tsink + P ⋅θ case − snk Tcase = 73 + (12 )(1) ⇒ Tcase = 85°C a. Tdev = Tcase + P ⋅θ dev − case Tdev = 85 + (12 )( 3) ⇒ Tdev = 121°C EX8.4 311. θ dev − case = PD ,max = TJ ,max − Tamb PD,rated = 200 − 25 = 3.5°C/W 50 TJ ,max − Tamb θ dev − case + θ case −snk + θ snk − amb 200 − 25 ⇒ PD ,max = 29.2 W 3.5 + 0.5 + 2 = Tamb + PD ,max (θ case −snk + θ snk − amb ) = 25 + ( 29.2 )( 0.5 + 2 ) ⇒ Tcase = 98°C = Tcase EX8.5 I DQ = a. 10 − 4 ⇒ I DQ = 60 mA 0.1 b. ⎛9⎞ vds = − ⎜ ⎟ ( 60 )( 0.050 ) = −2.7 V ⇒ vDS ( min ) = 4 − 2.7 = 1.3 V ⎝ 10 ⎠ So maximum swing is determined by drain-to-source voltage. VPP = 2 × ( 2.5 ) = 5.0 V c. 1 VP2 1 ( 2.5 ) ⋅ = ⋅ ⇒ PL = 31.25 mW 2 RL 2 0.1 2 PL = PS = VDD ⋅ I DQ = (10 )( 60 ) = 600 mW η= PL PS = 31.25 ⇒ η = 5.2% 600 EX8.6 Computer Analysis EX8.7 No Exercise Problem EX8.8 a. vI = v0 + vGSn − VBB 2 dv dvI = 1 + GSn dv0 dv0 iDn = K n ( vGSn − VTN ) vGSn = iDn + VTN Kn dvGSn dvGSn diDn = ⋅ dvo diDn dvo 2 312. So dvGSn = diDn 1 1 1 ⋅ ⋅ K n 2 iDn At v0 = 0, iDn = 0.050 A So iDn dvGSn 1 1 1 = ⋅ ⋅ =5 diDn 2 0.050 0.2 = iL + iDp For a small change in v0 → Δ iL = Δ iDn − ( −Δ iDp ) 1 Δ iL 2 di 1 di 1 1 1 1 = ⋅ = 0.025 or Dn = ⋅ L = ⋅ dv0 2 dv0 2 RL 2 20 So Δ iDn = Then dvGSn = ( 5 )( 0.025 ) = 0.125 dv0 Then dv dvI 1 = 1 + 0.125 = 1.125 and Av = 0 = ⇒ Av = 0.889 dv0 dvI 1.125 For v0 = 5 V, iL = 0.25 A = iDn , and iDp = 0 b. dvGSn dvGSn diDn = ⋅ dv0 diDn dv0 dvGSn = diDn 1 1 1 1 1 1 ⋅ ⋅ = ⋅ ⋅ = 2.24 0.2 2 0.25 K n 2 iDn diDn diL 1 = = = 0.05 dv0 dv0 20 dvGSn = ( 2.24 )( 0.05 ) = 0.112 dv0 dvI = 1 + 0.112 = 1.112 dv0 Av = dv0 1 = ⇒ Av = 0.899 dvI 1.112 EX8.9 a. 2 ′ ′ Rb = rπ + (1 + β ) RE and RE = a 2 RL = (10 ) ( 8 ) = 800 Ω Ri = 1.5 kΩ = RTH Rb V 18 I Q = 2CC = = 22.5 mA a RL (10 )2 ( 8 ) rπ = (100 )( 0.026 ) = 0.116 kΩ 22.5 Rb = 0.116 + (101)( 0.8 ) = 80.9 kΩ 1.5 = RTH 80.9 = RTH ( 80.9 ) RTH + ( 80.9 ) ⇒ ( 80.9 − 1.5 ) RTH = (1.5 )( 80.9 ) ⇒ RTH = 1.53 kΩ ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ VCC = ⋅ RTH ⋅ VCC R1 + R2 ⎠ R1 ⎝ 313. 22.5 = 0.225 mA β 100 V − 0.7 1 = TH ⇒ (1.53)(18 ) = ( 0.225 )(1.53) + 0.7 ⇒ R1 = 26.4 kΩ RTH R1 I BQ = I BQ IQ = 26.4 R2 = 1.53 26.4 + R2 ( 26.4 − 1.53) R2 = (1.53)( 26.4 ) ⇒ R2 = 1.62 kΩ b. vE = 0.9VCC = ( 0.9 )(18 ) = 16.2 V iE = 0.9 I CQ = ( 0.9 )( 22.5 ) = 20.25 mA v 16.2 v0 = E = ⇒ VP = 1.62 V a 10 i0 = aiE = (10 )( 20.25 ) ⇒ I P = 203 mA PL = 1 (1.62 )( 0.203) ⇒ PL = 0.164 W 2 EX8.10 a. ⎛ IC ⎞ ⎞ ⎜ ⎟ ⎟ ⇒ VBE = VT ln ⎜ ⎟ ⎠ ⎝ I SQ ⎠ ⎛ 5 × 10−3 ⎞ = ( 0.026 ) ln ⎜ = 0.6225 V ⇒ VD1 = VD 2 = 0.6225 −13 ⎟ ⎝ 2 × 10 ⎠ ⎛V I C = I SQ exp ⎜ BE ⎝ VT VBE ⎛ 0.6225 ⎞ I Bias = I D = I SD exp ⎜ ⎟ ⎝ 0.026 ⎠ ⎛ 0.6225 ⎞ = 5 × 10−13 exp ⎜ ⎟ ⎝ 0.026 ⎠ I Bias = 12.5 mA b. V0 = 2 V, iL = 2 = 26.7 mA 0.075 1st approximation: iCn ≅ 26.7 mA, iBn = 0.444 mA ⎛ 26.7 × 10−3 ⎞ VBE = ( 0.026 ) ln ⎜ = 0.6661 −13 ⎟ ⎝ 2 × 10 ⎠ I D = 12.5 − 0.444 = 12.056 mA ⎛ 12.056 × 10−3 ⎞ VD = ( 0.026 ) ln ⎜ ⎟ = 0.6216 −13 ⎝ 5 × 10 ⎠ 2VD = 1.243 V VEB = 2VD − VBE = 0.5769 ⎛ 0.5769 ⎞ icP = 2 × 10−13 exp ⎜ ⎟ = 0.866 mA ⎝ 0.026 ⎠ 2nd approximation: iEn = iL + iCP = 26.7 + 0.866 ≅ 27.6 mA = iEn ⎛ 60 ⎞ iCn = ⎜ ⎟ ( 27.6 ) ⇒ iCn = 27.1 mA ⎝ 61 ⎠ iBn = 0.452 mA I D = 12.5 − 0.452 ⇒ I D = 12.05 mA 314. ⎛ 27.1× 10−3 ⎞ ⇒ VBEn = 0.6664 V VBEn = ( 0.026 ) ln ⎜ −13 ⎟ ⎝ 2 × 10 ⎠ ⎛ 12.05 × 10−3 ⎞ VD = ( 0.026 ) ln ⎜ ⎟ = 0.6215 V −13 ⎝ 5 × 10 ⎠ 2VDD = 1.243 V VEB = 1.243 − 0.6664 ⇒ VEBp = 0.5766 V ⎛ 0.5766 ⎞ iCP = 2 × 10−13 exp ⎜ ⎟ ⇒ iCP = 0.856 mA ⎝ 0.026 ⎠ c. 10 = 133 mA 0.075 ≅ iL = 133 mA ⇒ iCN = 131 mA V0 = 10 V, iL = iEn iBn = 2.18 mA ⇒ I D = 12.5 − 2.18 ⇒ I D = 10.3 mA ⎛ 10.3 × 10−3 ⎞ VD = ( 0.026 ) ln ⎜ = 0.6175 −13 ⎟ ⎝ 5 × 10 ⎠ 2VDD = 1.235 V ⎛ 131× 10−3 ⎞ ⇒ VBEn = 0.7074 V VBEn = ( 0.026 ) ln ⎜ −13 ⎟ ⎝ 2 × 10 ⎠ VEBp = 1.235 − 0.7074 ⇒ VEBp = 0.5276 V ⎛ 0.5276 ⎞ iCP = 2 × 10−13 exp ⎜ ⎟ ⇒ iCP = 0.130 mA ⎝ 0.026 ⎠ EX8.11 No Exercise Problem EX8.12 a. vI = 0 = v0 , vB 3 = 0.7 V 12 − 0.7 11.3 I R1 = = ⇒ I R1 = 45.2 mA R1 0.25 If transistors are matched, then iE1 = iE 3 i iR1 = iE1 + iB 3 = iE1 + E 3 1+ β ⎛ 1 ⎞ 1⎞ ⎛ iR1 = iE1 ⎜ 1 + ⎟ = iE1 ⎜ 1 + ⎟ 1+ β ⎠ 41 ⎠ ⎝ ⎝ 45.2 iE1 = ⇒ iE1 = iE 2 = 44.1 mA 1.024 i 44.1 iB1 = iB 2 = E1 = ⇒ iB1 = iB 2 = 1.08 mA 1+ β 41 b. 315. For vI = 5 V ⇒ v0 = 5 V 5 ⇒ i0 = 0.625 A 8 0.625 iE 3 ≅ 0.625 A, iB 3 = ⇒ iB 3 = 15.2 mA 41 12 − 5.7 = 25.2 mA vB 3 = 5.7 V ⇒ iR1 = 0.25 10 = 0.244 mA iE1 = 25.2 − 15.2 ⇒ iE1 = 10.0 mA ⇒ iB1 = 41 vB 4 = 5 − 0.7 = 4.3 V i0 = I R2 = 4.3 − ( −12 ) 0.25 = 65.2 mA ≅ iE 2 65.2 = 1.59 mA 41 iI = iB 2 − iB1 = 1.59 − 0.244 ⇒ iI = 1.35 mA iB 2 = i0 625 = ⇒ AI = 463 iI 1.35 From Equation (8.54) (1 + β ) R ( 41)( 250 ) AI = = = 641 2 RL 2 (8) AI = c. TYU8.1 24 = 1.2 A = I D ( max ) 20 For I D = 0 ⇒ VDS ( max ) = 24 V For VDS = 0, I D ( max ) = Maximum power when VDS = ID = VDS ( max ) 2 I D ( max ) 2 = 12 V and = 0.6 A ⇒ PD ( max ) = (12 )( 0.6 ) = 7.2 Watts TYU8.2 Maximum power at center of load line Pmax = ( 0.05 )(10 ) ⇒ Pmax = 0.5 W TYU8.3 316. a. PQ = VCEQ ⋅ I CQ = ( 7.5 )( 7.5 ) PQ = 56.3 mW 1 VP2 1 ( 6.5 ) ⋅ = ⋅ ⇒ PL = 21.1 mW 2 RL 2 1 2 b. PL = PS = (15 )( 7.5 ) ⇒ PS = 113 mW η= PL PS = 21.1 ⇒ η = 18.7% 113 PQ = 56.3 − 21.1 = 35.2 mW TYU8.4 a. PL = 1 VP2 20 ⋅ ⇒ VP = 2 RL PL = 2 ( 8 )( 25 ) ⇒ VP = 20 V ⇒ VCC = ⇒ VCC = 25 V 2 RL 0.8 b. IP = VP 20 = ⇒ I P = 2.5 A 8 RL c. PQ = VCCVP VP2 − π RL 4 RL ( 25 )( 20 ) ( 20 ) PQ = − π (8) 4 (8) 2 d. η= = 19.9 − 12.5 ⇒ PQ = 7.4 W π VP π 20 = ⋅ ⇒ η = 62.8% 4VCC 4 25 TYU8.5 a. PL = ( 4) 1 VP2 ⋅ = ⇒ PL = 80 mW 2 RL 2 ( 0.1) b. IP = VP 4 ⇒ I P = 40 mA = RL 0.1 c. PQ = VCCVP VP2 − π RL 4 RL PQ = ( 5 )( 4 ) ( 4 ) − = 63.7 − 40 ⇒ PQ = 23.7 mW π ( 0.1) 4 ( 0.1) 2 2 d. η= π VP 4VCC TYU8.6 a. 1 ⎛ 2V I CQ ≅ ⋅ ⎜ CC 2 ⎝ RL = π 4 ⋅ ⇒ η = 62.8% 4 5 ⎞ VCC 12 = = 8 mA ⎟= ⎠ RL 1.5 RTH = R1 R2 ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ VCC = ⋅ RTH ⋅ VCC R1 ⎝ R1 + R2 ⎠ I CQ VTH − VBE 8 = = 0.107 mA = I BQ = β 75 RTH + (1 + β ) RE 317. Let RTH = (1 + β ) RE = ( 76 )( 0.1) = 7.6 kΩ 1 ⋅ ( 7.6 )(12 ) − 0.7 R1 0.107 = 7.6 + 7.6 1 ⋅ ( 91.2 ) = 2.33 ⇒ R1 = 39.1 kΩ R1 39.1R2 = 7.6 ⇒ ( 39.1 − 7.6 ) R2 = ( 7.6 )( 39.1) ⇒ R2 = 9.43 kΩ 39.1 + R2 b. 2 2 1 1 PL = ⋅ ( 0.9 I CQ ) RL = ⎡( 0.9 )( 8 ) ⎤ (1.5 ) ⇒ PL = 38.9 mW ⎦ 2 2⎣ PS = VCC I CQ = (12 )( 8 ) = 96 mW PQ = PS − PL = 96 − 38.9 ⇒ PQ = 57.1 mW η= PL PS = 38.9 ⇒ η = 40.5% 96 TYU8.7 I E = I E 3 + IC 4 + IC 5 = I E 3 + IC 4 + β5 I B5 = I E 3 + β 4 I B 4 + β 5 (1 + β 4 ) I B 4 I E = (1 + β 3 ) I B 3 + β 4 β 3 I B 3 + β 5 (1 + β 4 ) β 3 I B 3 If β 4 and β 5 are large, then I E ≅ β 3 β 4 β 5 I B 3 So that composite current gain is β ≅ β 3 β 4 β 5 318. Chapter 8 Problem Solutions 8.1 a. b. i. VD D = 80 V Maximum power at VDS = ID = RD = ii. VD D 2 PT 25 = = 0.625 A VD S 40 80 − 40 ⇒ RD = 64 Ω 0.625 VD D = 50 V Maximum power at VD S = VD D ID = PT 25 = =1 A VDS 25 RD = 50 − 25 ⇒ RD = 25 Ω 1 8.2 a. = 40 V 2 = 25 V 319. PQ (max) = I C Q ⋅ So I C Q = RL = VC C 2 2 PQ (max) VCC = 2(20) = 1.67 A 24 VCC − (VCC / 2) 24 − 12 = ⇒ RL = 7.2 Ω I CQ 1.67 1.67 ⇒ 20.8 mA 80 24 − 0.7 RB = ⇒ RB = 1.12 kΩ 20.8 b. I CQ ⋅ RL (1.67 )( 7.2 ) = = 462 Av = g m RL = VT 0.026 IB = I CQ β = V0 (max) = 12 V ⇒ VP = V0 (max) 12 = ⇒ VP ≅ 26 mV Av 462 8.3 a. For maximum power delivered to the load, set VC EQ = VC C 2 Set VC C = 25 V = VCE ( sus ) Then I Cm = VCC 25 = RL 0.1 I Cm = 250 mA < I C ,max 25 − 12.5 = 125 mA 0.1 V PQ ( max ) = I CQ ⋅ CC = ( 0.125 )(12.5 ) 2 = 1.56 W < PD,max I CQ = 125 = 1.25 mA 100 25 − 0.7 RB = ⇒ RB = 19.4 kΩ 1.25 1 2 1 2 b. PL ( max ) = ⋅ I CQ ⋅ RL = ( 0.125 ) (100 ) ⇒ PL ( max ) = 0.781 W ( rms ) 2 2 I BQ = 8.4 Point (b): Maximum power delivered to load. Point (a): Will obtain maximum signal current output. Point (c): Will obtain maximum signal voltage output. 320. 8.5 a. b. VGG = 5 V, I D = 0.25 ( 5 − 4 ) = 0.25 A, VD S = 37.5 V, P = 9.375 W 2 VGG = 6 V, I D = 0.25 ( 6 − 4 ) = 1.0 A, VD S = 30 V, P = 30 W 2 VGG = 7 V, I D = 0.25 ( 7 − 4 ) = 2.25 A, VD S = 17.5 V, P = 39.375 W 2 VGG = 8 V, I D 2 = 0.25 ⎡ 2 ( 8 − 4 )VD S − VD S ⎤ ⎣ ⎦ = 40 − VD S 10 I D = 3.71 A, P = 10.8 W VGG = 9 V, I D ⇒ VD S = 2.92 2 = 0.25 ⎡ 2 ( 9 − 4 ) VD S − VD S ⎤ ⎣ ⎦ 40 − VD S ⇒ VD S = 1.88 V 10 I D = 3.81 A, P = 7.16 W c. Yes, at VGG = 7 V, P = 39.375 W > PD ,max = 35 W = 8.6 a. VDD = 25 V 2 50 − 25 = = 1.25 A 20 Set VDSQ = I DQ I DQ = K n (VGS − VTN ) 2 1.25 + 4 = VGS = 6.5 V 0.2 ⎛ R2 ⎞ VGS = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠ Let R1 + R2 = 100 kΩ ⎛ R ⎞ 6.5 = ⎜ 2 ⎟ ( 50 ) ⇒ R2 = 13 kΩ ⎝ 100 ⎠ R1 = 87 kΩ b. PD = I DQVDSQ = (1.25 )( 25 ) ⇒ PD = 31.25 W c. I D ,max = 2 I DQ ⇒ I D ,max = 2.5 A VDS ,max = VDD ⇒ VDS ,max = 50 V PD ,max = 31.25 W d. 321. V0 = g m RL Vi g m = 2 K n I DQ = 2 ( 0.2 )(1.25) = 1 A / V V0 = (1)( 20 )( 0.5 ) = 10 V 1 V02 1 (10 ) ⋅ = ⋅ ⇒ PL = 2.5 W 2 RL 2 20 2 PL = PQ = 31.25 − 2.5 ⇒ PQ = 28.75 W 8.7 (a) PD = PD ,max − ( Slope ) (T j − 25 ) (b) 60 + 25 ⇒ T j ,max = 145°C 0.5 T j ,max − Tcase 145 − 25 or θ dev − amb = ⇒ θ dev − amb = 2°C/W = 60 θ dev − amb At PD = 0, T j ,max = (c) PD ,max 8.8 PD ,rated = T j ,max − Tamb θ dev − case or θ dev − case = T j ,max − Tamb PD ,rated 150 − 25 = 2.5°C/W 50 − Tamb = PD (θ dev − case + θ case − amb ) = Then Tdev 150 − 25 = PD ( 2.5 + θ case − amb ) ⇒ 125 = PD ( 2.5 + θ case − amb ) 8.9 PD = I D ⋅ VDS = ( 4 )( 5 ) = 20 W Tdev − Tamb = PD (θ dev − case + θ case − snk + θ snk − amb ) Tdev − 25 = 20 (1.75 + 0.8 + 3) = 111 ⇒ Tdev = 136°C Tdev − Tcase = PD ⋅θ dev − case = ( 20 )(1.75 ) = 35 Tcase = Tdev − 35 = 136 − 35 ⇒ Tcase = 101°C Tcase − Tsink = PD ⋅θ case − snk = ( 20 )( 0.8 ) = 16°C Tsink = Tcase − 16 = 101 − 16 ⇒ Tsink = 85°C 8.10 322. Tdev − Tamb = PD (θ dev − case + θ case − amb ) 200 − 25 = 25 ( 3 + θ case − amb ) ⇒ θ case − amb = 4°C/W 8.11 θ dev − case = PD = = T j ,max − Tamb PD ,rated = 175 − 25 = 10°C/W 15 T j ,max − Tamb θ dev − case + θ case −snk + θ snk − amb 175 − 25 ⇒ PD = 10 W 10 + 1 + 4 8.12 η= PL PS PS = VCC ⋅ I Q ⎛V ⎞ PL = VP ⋅ I P = ⎜ CC ⎟ ( I Q ) ⎝ 2 ⎠ 1 ⋅ VCC ⋅ I Q ⇒ η = 50% η= 2 VCC ⋅ I Q 8.13 vo ( max ) = 4.8 V iC 3 = iC 2 = vI = vo + 0.7 −0.7 − ( −5 ) 1 iL ( max ) = −4.3 mA = so − 3.6 ≤ vI ≤ 5.5 V vo ( min ) = −4.3 V 8.14 = 4.3 mA vS ( min ) 1 323. I D 3 = K (VGS 3 − VTN ) = 2 0 − VGS 3 − ( −5 ) R 12 (VGS 3 − 0.5 ) = 5 − VGS 3 2 2VGS 3 − 11VGS − 2 = 0 2 VGS 3 = 11 ± (11) 2 + 4 (12 )( 2 ) 2 (12 ) VGS 3 = VGS 2 = 1.072 V I D 3 = I D 2 = 12 (1.072 − 0.5 ) = 3.93 mA VDS 2 ( sat ) = VGS 2 − VTN = 1.072 − 0.5 = 0.572 V 2 vo ( min ) : i2 ( max ) = −3.93 = vI ( min ) = vo ( min ) + VTN vI ( min ) = −3.43 V V0 ( min ) 1 = −3.93 + 0.5 ⇒ V0 ( min ) = −3.93 V vo ( max ) = 5 − VDS ( sat ) = 5 − 0.572 vo ( max ) = 4.43 V 4.43 I D1 ( max ) = 3.93 + = 8.36 mA 1 I D1 = 8.36 = 12 (VGS 1 − 0.5 ) ⇒ VGS 1 = 1.33 V vI ( max ) = vo + VGS1 = 4.43 + 1.33 ⇒ vI ( max ) = 5.76 V 2 8.15 a. Neglect base currents. v0 ( max ) = V + − VCE (sat) = 10 − 0.2 = 9.8 V 9.8 9.8 iL (max) = I Q = = ⇒ I Q = 9.8 mA RL 1 0 − 0.7 − ( −10 ) ⇒ R = 949 Ω 9.8 iE1 ( max ) = 2 I Q ⇒ iE1 ( max ) = 19.6 mA R= iE1 ( min ) = 0 iL ( max ) = I Q = 9.8 mA iL ( min ) = − I Q = −9.8 mA b. 2 1 1 ( iL ( max ) ) RL = 2 ( 9.8)2 (1) ⇒ PL = 48.02 mW 2 PS = I Q (V + − V − ) + I Q ( 0 − V − ) PL = = 9.8 ( 20 ) + 9.8 (10 ) ⇒ PS = 294 mW η= PL PS = 8.16 a. I Q ( min ) = R= b. 48.02 ⇒ η = 16.3% 294 v0 ( max ) RL 0 − 0.7 − ( −12 ) 100 = 10 ⇒ I Q ( min ) = 100 mA 0.1 ⇒ R = 113 Ω 324. PQ1 = I Q ⋅ VCE1 = (100 )(12 ) ⇒ PQ1 = 1.2 W P (source) = 2 I Q (12 ) = 2.4 W c. (10 ) 1 VP2 ⋅ = = 0.5 W 2 RL 2 (100 ) 2 PL = PS = 1.2 + 2.4 = 3.6 W η= PL PS = 0.5 ⇒ η = 13.9% 3.6 8.17 I D1 = K n (VGS − VTN ) = 12 ( 0 − ( −1.8 ) ) 2 2 I D1 = 38.9 mA (a) For RL = ∞ vo ( max ) = 4.8 V VDS ( sat ) = VGS − VTN = 1.8 V vo ( min ) = −5 + 1.8 = −3.2 V vI = vo + 0.7 ⇒ −2.5 ≤ vI ≤ 5.5 V For RL = 500 Ω vo ( max ) = 4.8 V (b) For vo < 0, vo ( min ) = −3.2 V ′ I2 = vo −3.2 = = −6.4 mA RL 0.5 −2.5 ≤ vI ≤ 5.5 V (c) ′ For vo = −2V , I 2 ( max ) = −38.9 mA −2 R2 ( min ) = ⇒ RL ( min ) = 51.4 Ω −38.9 1 v2 1 ( 2) PL = ⋅ o = ⋅ ⇒ PL = 38.9 mW 2 RL 2 51.4 2 PL = 10 ( 38.9 ) = 389 mW % = 38.9 = 10% 389 8.18 + V 2 (V ) PL = P = RL RL 2 1 (V ) 1 (V ) + ⋅ , V − = −V + PS = ⋅ 2 RL 2 RL + 2 (V ) = + 2 So PS η= 8.19 (a) PL PS RL ⇒ η = 100% − 2 325. As maximum conversion efficiency π V η = , P = 0.785 4 VCC ⎛4⎞ So V p ( max ) = ( 0.785 )( 5 ) ⎜ ⎟ ⎝π ⎠ V p ( max ) = 5 V (b) Maximum power dissipation occurs when V p = (c) P ( max ) = θ 2= 2VCC π 2 VCC 2 π RL ( 5) 2 π 2 RL ⇒ RL = 1.27 Ω 8.20 P= (a) 2 1 Vp ⋅ 2 RL 2 1 Vp ⋅ ⇒ V p = 49 V ⇒ V + = 52 V, V − = −52 V 2 24 V 49 IP = P = = 2.04 A RL 24 50 = (b) η= (c) π VP ⋅ = 4 VCC π ⎛ 49 ⎞ ⎜ ⎟ 4 ⎝ 52 ⎠ η = 74.0% 8.21 (a) VDS ≥ VDS ( sat ) = VGS − VTN = VGS VDS = 10 − Vo ( max ) and I D = I L = K n (VGS ) Vo ( max ) RL VGS = = K n (VGS ) 2 Vo ( max ) RL ⋅ K n So 10 − Vo ( max ) = 2 ⎡10 − V0 ( max ) ⎤ = ⎣ ⎦ Vo ( max ) RL ⋅ K n = 2 VGS b. 20.5 ± ( 5 )( 0.4 ) V0 ( max ) V02 ( max ) − 20.5V0 ( max ) + 100 = 0 V0 ( max ) = Vo ( max ) V0 ( max ) 100 − 20V0 ( max ) + V02 ( max ) = iL = 2 ( 20.5 ) 2 2 2 − 4 (100 ) ⇒ V0 ( max ) = 8 V 8 ⇒ iL = 1.6 mA 5 i 1.6 = L = = 2 V ⇒ VI = 10 V Kn 0.4 = 2 ( 5) π = 3.183 V 326. 1 (8) PL = ⋅ = 6.4 mW 2 5 20 (1.6 ) PS = = 10.2 mW 2 π η= PL PS = 6.4 ⇒ η = 62.7% 10.2 8.22 vO = iL RL and iL = iD = K n ( vGS − VTN ) or iL = K n ( vGS ) and vGS = vI − vO Then 2 vO = K n RL ( vI − vO ) or vO = 2 ( vI − vO ) 2 2 2 ⎛ dv ⎞ dv0 = ( 2 )( 2 )( vI − v0 ) ⎜ 1 − 0 ⎟ dvI ⎝ dvI ⎠ dv0 ⎡1 + 4 ( vI − v0 ) ⎤ = 4 ( vI − v0 ) ⎦ dvI ⎣ or 4 ( vI − v0 ) dv0 = dvI 1 + 4 ( vI − v0 ) For vI = 10 V, v0 = 8 V ⇒ At vI = 0, v0 = 0 ⇒ 4 (10 − 8 ) dv0 dv = ⇒ 0 = 0.889 dvI 1 + 4 (10 − 8 ) dvI dv0 =0 dvI At vI = 1, v0 = 0.5 ⇒ dv0 = 0.667 dvI 8.23 a. ⎛i ⎞ ⎛ 5 × 10−3 ⎞ VBE = VT ln ⎜ C ⎟ = ( 0.026 ) ln ⎜ −13 ⎟ ⎝ 5 × 10 ⎠ ⎝ IS ⎠ V VBE = BB = 0.5987 V ⇒ VBB = 1.1973 V 2 PQ = iC ⋅ vCE = ( 5 )(10 ) ⇒ PQ = 50 mW b. 327. v0 = −8 V iL = −8 ⇒ iL = −80 mA 0.1 iCp ≈ 80 mA ⎛ iCp ⎞ ⎛ 80 × 10−3 ⎞ vEB = VT ln ⎜ ⎟ = ( 0.026 ) ln ⎜ −13 ⎟ ⎝ 5 × 10 ⎠ ⎝ IS ⎠ vEB = 0.6708 V VBB − vEB + v0 = 0.5987 − 0.6708 − 8⇒ vI = −8.072 V 2 = VBB − vEB = 1.1973 − 0.6708 = 0.5265 V vI = VBE ⎛v ⎞ ⎛ 0.5265 ⎞ iCn = I S exp ⎜ BE ⎟ = 5 × 10−13 exp ⎜ ⎟ ⇒ iCn = 0.311 mA ⎝ 0.026 ⎠ ⎝ VT ⎠ 2 PL = iL RL = ( 80 ) ( 0.1) ⇒ PL = 640 mW 2 PQn = iCn ⋅ vCE = ( 0.311) (10 − ( −8 ) ) ⇒ PQn = 5.60 mW PQp = iCp ⋅ vEC = ( 80 )( 2 ) ⇒ PQp = 160 mW 8.24 (a) iDn = K n ( vGSn − VTN ) 2 V 0.5 + 2 = vGSn = 2.5 V = BB ⇒ VBB = 5.0 V 2 2 Pn = ( 0.5 )(10 ) ⇒ Pn = Pp = 5 mW (b) VDS = VGS − VTN ⇒ VDS = VGS − 2 VDS = 10 − vo ( max ) and v ( max ) v ( max ) iL + VTN = O +2 = O +2 Kn RL K n ( 2 )(1) VGS = so 10 − v0 ( max ) = v0 ( max ) 2 +2−2 = v0 ( max ) 2 so v0 ( max ) = 8 V iDn = iL = 8 ⇒ iDn = iL = 8 mA 1 8 + 2 ⇒ VGS = 4 V 2 V Then vI = vo + VGS − BB = 8 + 4 − 2.5 ⇒ vI = 9.5 V 2 VBB ⎞ ⎛ vSGp = vo − ⎜ vI − ⎟ = 8 − ( 9.5 − 2.5 ) 2 ⎠ ⎝ vSGp = 1 V ⇒ M p cutoff ⇒ iDp = 0 VGS = 2 PL = iL RL = ( 8 ) (1) ⇒ PL = 64 mW 2 PMn = iDn ⋅ vDS = ( 8 )(10 − 8 ) ⇒ PMn = 16 mW PMp = iDp ⋅ vSD ⇒ PMp = 0 8.25 a. 328. v0 = 24 V ⇒ iL = 24 ⇒ iL ≈ iN = 3 A 8 3 ⇒ iBn = 73.2 mA 41 For iD = 25 mA ⇒ iR1 = 25 + 73.2 = 98.2 mA iBn = ⎛i VBE = VT ln ⎜ N ⎝ IS Then 98.2 = ⎞ 3 ⎛ ⎞ ⎟ = ( 0.026 ) ln ⎜ −12 ⎟ ⎝ 6 × 10 ⎠ ⎠ = 0.7004 V 30 − ( 24 + 0.7 ) R1 ⇒ R1 = 5.3 ⇒ R1 = 53.97 Ω 98.2 ⎛ 25 × 10−3 ⎞ = 0.5759 V VD = ( 0.026 ) ln ⎜ −12 ⎟ ⎝ 6 × 10 ⎠ VEB = 2VD − VBE = 2 ( 0.5759 ) − 0.7004 = 0.4514 V ⎛V ⎞ ⎛ 0.4514 ⎞ iP = I S exp ⎜ EB ⎟ = ( 6 × 10−12 ) exp ⎜ ⎟ ⇒ iP = 0.208 mA VT ⎠ ⎝ 0.026 ⎠ ⎝ b. Neglecting base current 30 − 0.6 30 − 0.6 iD ≈ = ⇒ iD ≈ 545 mA R1 53.97 ⎛ 0.545 ⎞ VD = ( 0.026 ) ln ⎜ = 0.656 V −12 ⎟ ⎝ 6 × 10 ⎠ Approximation for iD is okay. Diodes and transistors matched ⇒ iN = iP = 545 mA 8.26 (a) I D1 = K1 (VGS 1 − VTN ) VGS1 = 2 5 +1 = 2 V 5 I D 3 = K 3 (VGS 3 − VTN ) 2 200 = K 3 ( 2 − 1) ⇒ K n3 = K p 4 = 200 μ A / V 2 2 (b) vI + VSG 4 + VGS 3 − VGS1 = vO For vo large, iL = i1 = K n1 (VGS1 − VTN ) VGS1 = iL + VTN = K n1 ⎛ So vI + 2 + 2 − ⎜ ⎜ ⎝ vo + VTN RL K n1 ⎞ vo + 1⎟ = v0 ( 0.5)( 5) ⎟ ⎠ v0 −3 2.5 dv 1 dv dvI 1 =1= 0 + ⋅ ⋅ 0 dvI dvI 2 2.5v0 dvI vI = v0 + ⎡ ⎤ 1 ⎢1 + ⎥ ⎢ 2 2.5v0 ⎥ ⎣ ⎦ For vO = 5 V : 1= dv0 dvI 2 329. 1= ⎤ dv dv0 ⎡ dv 1 ⎢1 + ⎥ = 0 (1.1414 ) ⇒ 0 = 0.876 dvI ⎢ 2 2.5 ( 5 ) ⎥ dvI dvI ⎣ ⎦ 8.27 vO = vI + I Dn + VTN Kn VBB − VGS and VGS = 2 For vO ≈ 0, I Dn = I DQ + iL = I DQ + Then vO = vI + I DQ + ( vO / RL ) I DQ v VBB V or vO = vI + BB − VTN − ⋅ 1+ O − VTN − 2 Kn I DQ RL 2 Kn For vO small, vO ≅ vI + ⎡ 1 I DQ vO ⎢1 + ⋅ ⎢ 2 Kn ⎣ Now dvO = dvI ⎡ 1 ⎢1 + ⋅ ⎢ 2 ⎣ So ⋅ I DQ VBB 1 v − VTN − ⋅ 1+ ⋅ O Kn 2 2 I DQ RL I DQ V 1 ⎤ ⎥ = vI + BB − VTN − 2 I DQ RL ⎥ Kn ⎦ 1 1 ⎤ ⋅ ⎥ K n I DQ RL ⎥ ⎦ I DQ = 0.95 1 I DQ 1 1 ⋅ ⋅ = − 1 = 0.0526 2 K n I DQ RL 0.95 For RL = 0.1 k Ω, then Or vO RL 1 K n I DQ = 0.01052 K n I DQ = 95.1 We can write g m = 2 K n I DQ = 190 mA/V This is the required transconductance for the output transistor. This implies a very large transistor. 8.28 Av = − g m RL So −12 = − g m ( 2 ) ⇒ g m = 6 mA/V= I CQ VT I CQ = ( 6 )( 0.026 ) ⇒ I CQ = 0.156 mA But for maximum symmetrical swing, set I CQ = VCC 10 = = 5 mA ⇒ Av > 12 2 RL Maximum power to the load: 2 1 VCC (10 ) ⋅ = ⇒ PL ( max ) = 25 mW 2 RL 2 ( 2) 2 PL ( max ) = PS = VCC ⋅ I CQ = (10 )( 5 ) = 50 mW So η = 50% 330. 5 = 0.0278 mA β 180 R1 = RTH = 6 kΩ I CQ I BQ = = VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE Set RE = 20 Ω VTH = ( 0.0278 )( 6 ) + 0.7 + (181)( 0.0278 )( 0.020 ) VTH = 0.967 V 1 ⋅ RTH ⋅ VCC R1 VTH = 0.967 = 1 ( 6 )(10 ) ⇒ R1 = 62.0 kΩ R1 R2 = 6.64 kΩ 8.29 I CQ = VCC 15 = = 15 mA 1 RL I BQ = 15 = 0.15 mA 100 (15) 1 V2 ⇒ PL ( max ) = 112.5 mW PL ( max ) = ⋅ CC = 2 RL 2 (1) 2 Let RTH = 10 kΩ VTH = I BQ RTH + VBE + (1 + β ) I BQ RE = ( 0.15 )(10 ) + 0.7 + (101)( 0.15 )( 0.1) VTH = 3.715 = 1 1 ⋅ RTH ⋅ VCC = ⋅ (10 )(15 ) R1 R1 R1 = 40.4 kΩ R2 = 13.3 kΩ 8.30 ⎛ R2 ⎞ ⎛ 1.55 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (10 ) R1 + R2 ⎠ ⎝ 1.55 + 0.73 ⎠ ⎝ = 6.80 V RTH = R1 R2 = 0.73 1.55 = 0.496 kΩ I BQ = VTH − VBE 6.80 − 0.70 = RTH + (1 + β ) RE 0.496 + ( 26 )( 0.02 ) I BQ = 6.0 mA, I CQ = 150 mA ′ ′ Av = − g m RL and RL = a 2 RL = ( 3) ( 8 ) = 72 Ω 2 gm = I CQ VT = 150 ⇒ 5.77 A/V 0.026 331. Av = − ( 5.77 )( 72 ) = −415 Vo ′ = Av ⋅ Vi = ( 415 )( 0.017 ) = 7.06 V 7.06 = 2.35 V 3 7.06 Vo = = 2.35 V 3 7.06 Vo = = 2.35 V 3 PS = I CQ ⋅ VCC = ( 0.15 )(10 ) = 1.5 W Vo = η= PL PS = 0.345 ⇒ η = 23% 1.5 8.31 a. Assuming the maximum power is being delivered, then 36 9 Vo′ ( peak ) = 36 V ⇒ Vo = = 9 V ⇒ Vrms = ⇒ Vrms = 6.36 V 4 2 36 Vo = ⇒ Vo = 25.5 V b. 2 c. Secondary I rms = Primary I P = PL 2 = ⇒ I rms = 0.314 A Vrms 6.36 0.314 ⇒ I P = 78.6 mA 4 d. PS = I CQ .VCC = ( 0.15 )( 36 ) = 5.4 W 2 ⇒ η = 37% η= 5.4 8.32 a. 332. ⎛V ve = ⎜ π + g mVπ ⎝ rπ ⎞ ′ ⎟ RE = Vπ ⎠ ⎛1 ⎞ ′ ⎜ + g m ⎟ RE rπ ⎝ ⎠ ⎛1+ β = Vπ ⎜ ⎝ rπ vi = Vπ + ve ⇒ Vπ = vi − ve ⎞ ′ ⎟ RE ⎠ ⎛ 1+ β ⎞ ′ ve = (vi − ve ) ⎜ ⎟ RE ⎝ rπ ⎠ 1+ β ′ ⋅ RE 2 ′ ⎛n ⎞ (1 + β ) RE ve rπ v ′ = = = e where RE = ⎜ 1 ⎟ RL ′ vi 1 + 1 + β ⋅ R ′ rπ + (1 + β ) RE vi ⎝ n2 ⎠ E rπ ⎛n ⎞ ve so ve − v0 ⎜ 1 ⎟ ⎛ n1 ⎞ ⎝ n2 ⎠ ⎜ ⎟ ⎝ n2 ⎠ ′ (1 + β ) RE v 1 ⋅ so 0 = ′ vi ⎛ n1 ⎞ rπ + (1 + β ) RE ⎜ ⎟ ⎝ n2 ⎠ v0 = b. I n1 1 2 1 2 ⋅ I P RL , a = , I CQ = P so PL = .a 2 I CQ RL n2 a 2 2 PS = I CQ .VCC For η = 50% : PL = 1 2 2 ⋅ a I CQ RL a 2 I R VCC VCC V PL CQ L so a 2 = = 0.5 = 2 = = ⇒ a 2 = CC I CQ ⋅ VCC I CQ ⋅ RL ( 0.1)( 50 ) 2VCC 5 PS c. 49 ( 0.026 ) r β VT R0 = π = = ⇒ R0 = 0.255 Ω 1 + β (1 + β ) I CQ ( 50 )( 0.1) 8.33 a. With a 10:1 transformer ratio, we need a current gain of 8 through the transistor. ⎛ R1 R2 ⎞ ⎛ R1 R2 ie ie = (1 + β ) ib and ib = ⎜ ⎟ ii so we need = 8 = (1 + β ) ⎜ ⎜R R +R ⎟ ⎜R R +R ii ib ⎠ ib ⎝ 1 2 ⎝ 1 2 ′ ′ Rib = rπ + (1 + β ) RL ≈ (1 + β ) RL = (101)( 0.8 ) = 80.8 ⎞ ⎟ where ⎟ ⎠ 333. ⎛ ⎞ R1 R2 Then 8 = (101) ⎜ ⎜ R R + 80.8 ⎟ ⎟ ⎝ 1 2 ⎠ R1 R2 = 0.0792 or R1 R2 = 6.95 kΩ R1 R2 + 80.8 2VCC V 12 ′ = RL ⇒ I CQ = CC = = 15 mA ′ 2 I CQ RL 0.8 Set 15 = 0.15 mA 100 = I BQ RTH + VBE I BQ = VTH 1 ⋅ RTH ⋅ VCC = I BQ RTH + VBE R1 1 ( 6.95)(12 ) = ( 0.15)( 6.95) + 0.7 ⇒ R1 = 47.9 kΩ then R2 = 8.13 kΩ R1 b. I I e = 0.9 I CQ = 13.5 mA = L ⇒ I L = 135 mA a 1 2 PL = ( 0.135 ) ( 8 ) ⇒ PL = 72.9 mW 2 PS = VCC I CQ = (12 )(15 ) ⇒ PS = 180 mW η= PL PS ⇒ η = 40.5% 8.34 a. VP = 2 RL PL VP = 2 ( 8 )( 2 ) = 5.66 V = peak output voltage IP = VP 5.66 = = 0.708 A = peak output current RL 8 Set Ve = 0.9VCC = aVP to minimize distortion Then a = ( 0.9 )(18) 5.66 ⇒ a = 2.86 b. 1 ⎛ I P ⎞ 1 ⎛ 0.708 ⎞ = ⎜ ⎟ ⇒ I CQ = 0.275 A 0.9 ⎜ a ⎟ 0.9 ⎝ 2.86 ⎠ ⎝ ⎠ Then PQ = VCC I Q = (18 )( 0.275 ) ⇒ PQ = 4.95 W Power rating of transistor Now I CQ = 8.35 a. Need a current gain of 8 through the transistor. 334. ⎛ R1 R2 ib = 8 = (1 + β ) ⎜ ⎜R R +R ii ib ⎝ 1 2 ⎞ ⎟ where Rib ≈ (1 + β )( 0.9 ) = 90.9 kΩ ⎟ ⎠ ⎛ ⎞ R1 R2 8 =⎜ = 0.0792 or R1 R2 = 7.82 kΩ ⎜ R R + 90.9 ⎟ ⎟ 101 ⎝ 1 2 ⎠ 2VCC 12 = 0.9 kΩ ⇒ I CQ = = 13.3 mA 2 I CQ 0.9 Set 13.3 = 0.133 mA 100 1 Then ( 7.82 )(12 ) = ( 0.133)( 7.82 ) + 0.7 ⇒ R1 = 53.9 kΩ and R 2 = 9.15 kΩ R1 b. I I e = ( 0.9 ) I CQ = 12 mA = L ⇒ I L = 120 mA a 1 2 PL = ( 0.12 ) ( 8 ) ⇒ PL = 57.6 mW 2 PS = VCC I CQ = (12 )(13.3) ⇒ PS = 159.6 mW I BQ = η= PL 57.6 = ⇒ η = 36.1% PS 159.6 8.36 a. All transistors are matched. ⎛1+ β ⎞ iC 3 mA = iE1 + iB 3 = ⎜ ⎟ iC + β ⎝ β ⎠ ⎛ 61 1 ⎞ 3 = ⎜ + ⎟ iC ⇒ iC = 2.90 mA ⎝ 60 60 ⎠ b. For vo = 6 V , let RL = 200 Ω. 6 = 0.03 A = 30 mA ≅ iE 3 200 30 iB 3 = = 0.492 mA 61 iE1 = 3 − 0.492 = 2.508 mA io = 2.508 ⇒ iB1 = 41.11 μ A 61 3 iE 2 ≅ 3 mA ⇒ iB 2 = ⇒ 49.18 μ A 61 iI = iB 2 − iB1 = 49.18 − 41.11 ⇒ iI = 8.07 μ A iB1 = Current gain Ai = VBE 3 30 × 10−3 ⇒ Ai = 3.72 × 103 8.07 × 10−6 ⎛i ⎞ ⎛ 30 × 10−3 ⎞ = VT ln ⎜ E 3 ⎟ = ( 0.026 ) ln ⎜ −13 ⎟ ⎝ 5 × 10 ⎠ ⎝ IS ⎠ VBE 3 = 0.6453 V ⎛i ⎞ ⎛ 2.508 × 10−3 ⎞ VEB1 = VT ln ⎜ E1 ⎟ = ( 0.026 ) ln ⎜ ⎟ −13 ⎝ 5 × 10 ⎠ ⎝ IS ⎠ VEB1 = 0.5807 V 335. vI = v0 + VBE 3 − VEB1 = 6 + 0.6453 − 0.5807 vI = 6.0646 V Voltage gain Av = v0 6 = ⇒ Av = 0.989 vI 6.0646 8.37 a. For i0 = 1 A, I B 3 ≅ 1 ⇒ 20 mA 50 ⎡10 − ( v0,max + VBE 3 ) ⎤ 10 − VEB1 = 2⎢ − 20 ⎥ R1 R1 ⎢ ⎥ ⎣ ⎦ 10 − VBE 2vo,max = + 40 If, for simplicity, we assume VEB1 = VBE 3 = 0.7 V, then R1 R1 We can then write If we assume v0,max = 4 V, then b. 9.3 2 ( 4 ) = + 40 which yields R1 = R2 = 32.5 Ω R1 R1 9.3 ⇒ I E1 = 0.286 A = I E 2 32.5 = 10 I S1,2 , then I E 3 = I E 4 = 2.86 A For vI = 0, I E1 = Since I S 3,4 c. We can write ⎧ rπ 1 ⎫ ⎪ rπ 3 + R1 ⎪ 1 + β1 ⎪ 1⎪ R0 = ⎨ ⎬ 2⎪ 1 + β3 ⎪ ⎪ ⎪ ⎩ ⎭ ( 50 )( 0.026 ) βV = 0.4545 Ω Now rπ 3 = 3 T = 2.86 IC 3 rπ 1 = β1VT I C1 = (120 )( 0.026 ) 0.286 = 10.91 Ω So ⎧ 10.91 ⎫ 0.4545 + 32.5 1⎪ 121 ⎪ ⎪ ⎪ R0 = ⎨ ⎬ 2⎪ 51 ⎪ ⎪ ⎪ ⎩ ⎭ 10.91 = 32.5 0.0902 = 0.0900 32.5 121 Then R0 = 8.38 1 ⎧ 0.4545 + 0.0900 ⎫ ⎨ ⎬ or R 0 = 0.00534 Ω 2⎩ 51 ⎭ 336. { } 1 rπ 1 + (1 + β ) ⎡ R1 ( rπ 3 + (1 + β ) 2 RL ) ⎤ ⎣ ⎦ 2 iC1 ≈ 7.2 mA and iC 3 ≈ 7.2 mA Ri = Then rπ = ( 60 )( 0.026 ) 7.2 = 0.217 kΩ { } 1 0.217 + ( 61) ⎡ 2 ( 0.217 + ( 61)( 0.2 ) ) ⎤ ⎣ ⎦ 2 1 = 0.217 + 61 ⎡ 2 12.4 ⎤ or Ri = 52.6 kΩ ⎣ ⎦ 2 So Ri = { } 8.39 a. b. I1 = K1 (VSG + VTP ) = 2 V + − VSG R1 5 = 10 (VSG − 2 ) ⇒ VSG = 2.707 V 2 10 − 2.707 ⇒ R1 = R2 = 1.46 kΩ R1 RL = 100 Ω For a sinusoidal output signal: c. 5= 337. 1 (v ) 1 ( 5) PL = ⋅ o = ⋅ ⇒ PL = 125 mW 2 RL 2 0.1 2 iD 3 ≈ 2 ( vo ) ( 5 ) RL = 0.1 ⇒ iD 3 = 50 mA 50 + 2 = 4.236 V 10 10 − ( 4.236 + 5 ) I1 = ⇒ I D1 = 0.523 mA 1.46 0.523 VSG1 = + 2 = 2.229 V 10 vI = 5 + 4.236 − 2.229 ⇒ vI = 7.007 V VGS 3 = I D2 = (VI − VGS ) − ( −10 ) = 10 (VGS − 2 ) 2 1.46 17.007 − VGS 2 = 10 (VGS − 4VGS + 4 ) 1.46 2 14.6VGS − 57.4VGS + 41.4 = 0 VGS = 57.4 ± ( 57.4 ) − 4 (14.6 )( 41.4 ) 2 (14.6 ) 2 VGS 2 = 2.98 V I D 2 = 10 ( 2.98 − 2 ) ⇒ I D 2 = 9.60 mA 2 VG 4 = vI − VGS 2 = 7 − 2.98 = 4.02 V VSG 4 = 5 − 4.02 = 0.98 V ⇒ I D 4 = 0 8.40 For v0 = 0 I Q = I C 3 + I C 2 + I E1 ⎛ 1+ βn ⎞ IC 3 I B3 = I E 2 = ⎜ ⎟ IC 2 = βn ⎠ βn ⎝ I C 3 = (1 + β n ) I C 2 ⎛ β ⎞ I I B 2 = I C 1 = ⎜ P ⎟ I E1 = C 2 βn ⎝ 1+ βP ⎠ ⎛ β ⎞ I C 2 = β n ⎜ P ⎟ I E1 ⎝ 1+ βP ⎠ ⎛ β ⎞ I C 3 = (1 + β n ) β n ⎜ P ⎟ I E1 ⎜1+ β ⎟ p ⎠ ⎝ ⎛ β ⎞ ⎛ β ⎞ I Q = (1 + β n ) β n ⎜ P ⎟ I E1 + β n ⎜ P ⎟ I E1 + I E1 ⎝1+ βP ⎠ ⎝ 1+ βP ⎠ ⎛ 10 ⎞ ⎛ 10 ⎞ = ( 51)( 50 ) ⎜ ⎟ I E1 + ( 50 ) ⎜ ⎟ I E1 + I E1 ⎝ 11 ⎠ ⎝ 11 ⎠ I Q = 2318.18I E1 + 45.45 I E1 + I E1 I E1 = 1.692 μ A ⇒ I C1 = 1.534 μ A ⎛ 10 ⎞ I C 2 = ( 50 ) ⎜ ⎟ (1.692 ) ⇒ I C 2 = 76.9 μ A ⎝ 11 ⎠ ⎛ 10 ⎞ I C 3 = ( 51)( 50 ) ⎜ ⎟ (1.692 ) ⇒ I C 3 = 3.92 mA ⎝ 11 ⎠ 338. Because of rπ 1 and Z, neglect effect of r0. Then neglecting r01, r02 and r03, we find I X = g m 3Vπ 3 + g m 2Vπ 2 + g m1Vπ 1 + VX rπ 1 + Z Now ⎛ r ⎞ Vπ 1 = ⎜ π 1 ⎟ VX , Vπ 2 ≅ g m1Vπ 1rπ 2 ⎝ rπ 1 + Z ⎠ and Vπ 3 = ( g m1Vπ 1 + g m 2Vπ 2 ) rπ 3 = ⎡ g m1Vπ 1 + g m 2 ( g m1Vπ 1rπ 2 ) ⎤ rπ 3 ⎣ ⎦ ⎛ r ⎞ Vπ 3 = ⎜ π 1 ⎟ [ g m1 + g m1 g m 2 rπ 2 ] rπ 3 ⋅ VX ⎝ rπ 1 + Z ⎠ ( β + β1 β 2 ) rπ 3 ⋅ VX Vπ 3 = 1 rπ 1 + Z ⎛ r ⎞ ⎛ βr ⎞ and Vπ 2 = g m1 ⎜ π 1 ⎟ rπ 2VX = ⎜ 1 π 2 ⎟ VX ⎝ rπ 1 + Z ⎠ ⎝ rπ 1 + Z ⎠ ( β + β1 β 2 ) β 3 VX ββ β1 Then I X = 1 ⋅ V X + 1 2 ⋅ VX + ⋅ VX + rπ 1 + Z rπ 1 + Z rπ 1 + Z rπ 1 + Z Then R0 = rπ 1 = rπ 1 + Z VX = I X 1 + β1 + β1 β 2 + ( β1 + β1 β 2 ) β 3 (10 )( 0.026 ) 1.534 Z = 25 kΩ Then = 0.169 MΩ R0 = 169 + 25 1 + (10 ) + (10 )( 50 ) + ⎡10 + (10 )( 50 ) ⎤ ( 50 ) ⎣ ⎦ R0 = 194 = 0.00746 kΩ or Ro = 7.46 Ω 26, 011 8.41 a VBB Neglect base currents. ⎛I ⎞ = 2VD = 2VT ln ⎜ Bias ⎟ ⎝ IS ⎠ ⎛ 5 × 10−3 ⎞ = 2 ( 0.026 ) ln ⎜ ⇒ VBB = 1.281 V −13 ⎟ ⎝ 10 ⎠ 339. VBE1 + VEB 3 = VBB I E1 = I E 3 + I C 2 ⎛ β ⎞ I B2 = IC 3 = ⎜ P ⎟ I E 3 ⎝ 1+ βP ⎠ ⎛ β ⎞ IC 2 = β n I B 2 = β n ⎜ P ⎟ I E 3 ⎝ 1+ βP ⎠ ⎛ β I E1 = I E 3 + β n ⎜ P ⎝ 1+ βP ⎞ ⎟ IE3 ⎠ ⎡ ⎛ β I E1 = I E 3 ⎢1 + β n ⎜ P ⎝ 1+ βP ⎣ ⎞⎤ ⎟⎥ ⎠⎦ ⎡ ⎛ 1+ βn ⎞ ⎛ 1+ βP ⎞ ⎛ βP ⎜ ⎟ I C1 = ⎜ ⎟ I C 3 ⎢1 + β n ⎜ ⎝ βP ⎠ ⎝ 1+ βP ⎝ βn ⎠ ⎣ ⎞⎤ ⎟⎥ ⎠⎦ ⎡I ⎤ ⎡I ⎤ VBE1 = VT ln ⎢ C1 ⎥ , VEB 3 = VT ln ⎢ C 3 ⎥ IS ⎦ ⎣ ⎣ IS ⎦ (1.01) I C1 = ⎛ ⎜ 21 ⎞ ⎟ IC 3 ⎝ 20 ⎠ I C1 ⎡ ⎛ 20 ⎞ ⎤ ⎢1 + (100 ) ⎜ 21 ⎟ ⎥ ⎝ ⎠⎦ ⎣ ⎡ 21 ⎤ = I C 3 ⎢ + 100 ⎥ = 101.05 I C 3 ⎣ 20 ⎦ = 100.05 I C 3 ⎛ 100.05 I C 3 ⎞ ⎛ IC 3 ⎞ VT ln ⎜ ⎟ + VT ln ⎜ ⎟ = VBB IS ⎝ ⎠ ⎝ IS ⎠ 2 ⎛ 100.05 I C 3 ⎞ VT ln ⎜ ⎟ = VBB 2 IS ⎝ ⎠ 2 100.05 I C 3 I 2 S ⎛V ⎞ = exp ⎜ BB ⎟ ⎝ VT ⎠ ⎛V ⎞ exp ⎜ BB ⎟ = 0.4995 mA = I C3 100.05 ⎝ VT ⎠ Then I E 3 = 0.5245 mA Now I C1 = 100.05 I C 3 = 49.97 mA = I C1 IC 3 = IS ⎛ 20 ⎞ I C 2 = (100 ) ⎜ ⎟ ( 0.5245 ) = 49.95 mA = I C 2 ⎝ 21 ⎠ ⎛I ⎞ ⎛ 49.97 × 10−3 ⎞ VBE1 = VT ln ⎜ C1 ⎟ = 0.026 ln ⎜ ⎟ −13 ⎝ 10 ⎠ ⎝ IS ⎠ = 0.70037 ⎛I ⎞ ⎛ 0.4995 × 10−3 ⎞ VEB 3 = VT ln ⎜ C 3 ⎟ = 0.026 ln ⎜ ⎟ 10−13 ⎝ ⎠ ⎝ IS ⎠ = 0.58062 Note: VBE1 + VEB 3 = 0.70037 + 0.58062 = 1.28099 = VBB b. 340. v0 = 10 V ⇒ iE1 ≈ 10 = 0.10 A = iC1 100 100 = 1 mA 100 ⎛ 4 × 10−3 ⎞ = 2 ( 0.026 ) ln ⎜ = 1.2694 V −13 ⎟ ⎝ 10 ⎠ iB1 = VBB ⎛ 0.1 ⎞ VBE1 = ( 0.026 ) ln ⎜ −13 ⎟ = 0.7184 ⎝ 10 ⎠ VEB 3 = 1.2694 − 0.7184 = 0.55099 V ⎛ 0.55099 ⎞ I C 3 = 10−13 exp ⎜ ⎟ = 0.1598 mA ⎝ 0.026 ⎠ V 2 (10 ) PL = 0 = ⇒ PL = 1 W RL 100 2 PQ1 = iC1 ⋅ vCE1 = ( 0.1)(12 − 10 ) ⇒ PQ1 = 0.2 W PQ 3 = iC 3 ⋅ vEC 3 = ( 0.1598 ) (10 − [ 0.7 − 12]) ⇒ PQ 3 = 3.40 mW iC 2 = (100 )( iC 3 ) = (100 )( 0.1598 ) = 15.98 mA PQ 2 = iC 2 ⋅ vCE 2 = (15.98 ) (10 − [ −12]) ⇒ PQ 2 = 0.352 W 8.42 a. ⎛ 10 × 10−3 ⎞ ⇒ VBB = 1.74195 V VBB = 3 ( 0.026 ) ln ⎜ −12 ⎟ ⎝ 2 × 10 ⎠ VBE1 + VBE 2 + VEB 3 = VBB I C1 ≈ IC 2 βn , IC 3 ≈ IC 2 β n2 ⎛I ⎞ ⎛I ⎞ ⎛I ⎞ VT ln ⎜ C1 ⎟ + VT ln ⎜ C 2 ⎟ + VT ln ⎜ C 3 ⎟ = VBB ⎝ IS ⎠ ⎝ IS ⎠ ⎝ IS ⎠ 3 ⎡ IC ⎤ VT ln ⎢ 3 2 3 ⎥ = VBB ⎣ βn IS ⎦ ⎛V ⎞ I C 2 = β n I S 3 exp ⎜ BB ⎟ ⎝ VT ⎠ ⎛ 1.74195 ⎞ = ( 20 ) ( 20 × 10−12 ) 3 exp ⎜ ⎟ ⎝ 0.026 ⎠ I C 2 = 0.20 A, I C1 ≈ 10 mA, I C 3 ≈ 0.5 mA ⎛ 10 × 10−3 ⎞ VBE1 = ( 0.026 ) ln ⎜ ⇒ VBE1 = 0.58065 V −12 ⎟ ⎝ 2 × 10 ⎠ ⎛ 0.2 ⎞ VBE 2 = ( 0.026 ) ln ⎜ ⇒ VBE 2 = 0.6585 V −12 ⎟ ⎝ 2 × 10 ⎠ ⎛ 0.5 × 10−3 ⎞ VEB 3 = ( 0.026 ) ln ⎜ ⇒ VEB 3 = 0.50276 V −12 ⎟ ⎝ 2 × 10 ⎠ b. 1 V2 1 V2 PL = 10 W= ⋅ 0 = ⋅ 0 ⇒ V0 ( max ) = 20 V 2 RL 2 20 For v0 ( max ) : 341. 2 v0 ( 20 ) = ⇒ PL = 20 W 20 RL 2 PL = 20 = −1 A 20 iC 5 + iC 4 + iE 3 = −io ( max ) = 1 A i0 ( max ) = − ⎞ ⎟ =1 ⎟ ⎠ i ⎛ β ⎞ i ⎛ β ⎞ ⎡ 1 ⎛ 1+ β p iC 5 + C 5 ⎜ n ⎟ + C 5 ⎜ n ⎟ ⎢ ⎜ βn ⎝ 1 + βn ⎠ βn ⎝ 1 + βn ⎠ ⎣ βn ⎜ β p ⎢ ⎝ iC 5 + iC 5 ⎛ β n ⎞ iC 4 ⎛ 1 + β p ⋅⎜ ⎜ ⎟+ βn ⎝ 1 + βn ⎠ βn ⎜ β p ⎝ ⎞⎤ ⎟⎥ = 1 ⎟⎥ ⎠⎦ 1 ⎛ 20 ⎞ ⎤ ⎛ 1 ⎞ ⎡ 1 ⎛ 6 ⎞ ⎤ ⎡ iC 5 ⎢1 + ⎜ ⎟ ⎥ + ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ = 1 ⎣ 20 ⎝ 21 ⎠ ⎦ ⎝ 21 ⎠ ⎣ 20 ⎝ 5 ⎠ ⎦ iC 5 (1.05048 ) = 1 iC 5 = 0.952 A iC 4 = 0.0453 A iE 3 = 0.00272 A ⎛5⎞ iC 3 = 0.00272 ⎜ ⎟ ⎝6⎠ = 0.002267 A ⎛ 2.267 × 10−3 ⎞ VEB 3 = ( 0.026 ) ln ⎜ ⎟ = 0.54206 V −12 ⎝ 2 × 10 ⎠ VBE1 + VBE 2 = 1.74195 − 0.54206 = 1.19989 ⎛ I ⎞ ⎛I ⎞ VT ln ⎜ C 2 ⎟ + VT ln ⎜ C 2 ⎟ = 1.19989 ⎝ βn IS ⎠ ⎝ IS ⎠ ⎛ 1.19989 ⎞ iC 2 = β n ⋅ I S exp ⎜ ⎟ ⎝ 0.026 ⎠ = 20 (18.83) mA iC 2 = 93.9 mA iC 2 ⎛ β n ⎞ 93.9 = 4.47 mA ⎜ ⎟= β n ⎝ 1 + β n ⎠ 21 PQ 2 = I C 2 ( 24 − ( −20 ) ) = ( 0.0939 ) ( 44 ) = 4.13 W iC1 = PQ 5 = ( 0.952 ) ( −10 − ( −24 ) ) = 13.3 W 342. Chapter 9 Exercise Solutions EX9.1 Av = −15 = − R2 R1 R1 = Ri = 20 K ⇒ R2 = 15 R1 = 15 ( 20 ) = 300 K EX9.2 We can write ACL = − R2 ⎛ R3 ⎞ R3 ⎜1 + ⎟ − R1 ⎝ R4 ⎠ R1 R1 = R1 = 10 kΩ Want ACL = −50 Set R2 = R3 = 50 kΩ ⎛ R ⎞ ACL = −50 = −5 ⎜ 1 + 3 ⎟ − 5 ⎝ R4 ⎠ R R 50 1+ 3 = 9 ⇒ 3 = 8 = R4 R4 R4 R4 = 6.25 kΩ EX9.3 R2 1 ⋅ R1 ⎡ ⎛ R2 ⎞ ⎤ 1 ⎢1 + ⎜1 + ⎟ ⎥ Ad ⎝ R1 ⎠ ⎦ ⎣ R Ri = R1 = 25 kΩ Let 2 = x R1 We have ACL = − x 1 1+ (1 + x ) 5 × 103 −x = x 1.0002 + 5 × 103 x ⎞ ⎛ 12 ⎜ 1.0002 + ⎟=x 5 × 103 ⎠ ⎝ −12 = − 12.0024 = x − ( 2.4 × 10 −3 ) x R2 12.0024 = 12.0313 = 0.9976 25 kΩ R2 = 300.78 kΩ x= EX9.4 ⎛R ⎞ R R R v0 = − ⎜ F vI 1 + F vI 2 + F vI 3 + F VI 4 ⎟ R2 R3 R4 ⎝ R1 ⎠ 343. We need RF R R R = 7, F = 14, F = 3.5, F = 10 R1 R2 R3 R4 Set RF = 280 kΩ 280 7 280 R2 = 14 280 R3 = 3.5 280 R4 = 10 Then R1 = = 40 kΩ = 20 kΩ = 80 kΩ = 28 kΩ EX9.5 We may note that R3 R R R 3 20 = = 2 and F = = 2 so that 3 = F R2 1.5 R1 10 R2 R1 Then iL = − ( −3) − vI ⇒ iL = 2 mA = R2 1.5 kΩ vL = iL Z L = ( 2 × 10−3 ) ( 200 ) = 0.4 V i4 = vL 0.4 = = 0.267 mA R2 1.5 kΩ i3 = i4 + iL = 0.267 + 2 = 2.267 mA v0 = i3 R3 + vL = ( 2.267 × 10−3 )( 3 × 103 ) − 0.4 ⇒ v0 = 7.2 V EX9.6 Refer to Fig. 9.24 R1 = 2 R1 = 5 kΩ Let R1 = R3 = 2.5 kΩ Set R2 = R4 Differential Gain = EX9.7 v0 R2 R2 = = 100 = ⇒ R2 = R4 = 250 kΩ v1 R1 2.5 kΩ 344. We have the general relation that ⎛ R ⎞ ⎛ [ R4 / R3 ] ⎞ R v0 = ⎜ 1 + 2 ⎟ ⎜ v − 2v ⎜ 1 + [ R / R ] ⎟ I 2 R I1 ⎟ R1 ⎠ ⎝ ⎝ 4 3 ⎠ 1 R1 = R3 = 10 kΩ, R2 = 20 kΩ, R4 = 21 kΩ ⎛ 20 ⎞ ⎛ [ 21/10] ⎞ ⎛ 20 ⎞ v0 = ⎜ 1 + ⎟ ⎜ ⎟ vI 2 − ⎜ ⎟ vI 1 10 ⎠ ⎜ 1 + [ 21/10] ⎟ ⎝ ⎝ 10 ⎠ ⎝ ⎠ v0 = 2.0323vI 2 − 2.0vI a. vI 1 = 1, vI 2 = −1 v0 = −2.0323 − 2.0 ⇒ v0 = −4.032 V b. vI 1 = vI 2 = 1 V v0 = 2.0323 − 2.0 ⇒ v0 = 0.0323 V c. vcm = vI 1 = vI 2 so common-mode gain v0 = 0.0323 vcm Acm = d. ⎛ A ⎞ C M R RdB = 20 log10 ⎜ d ⎟ ⎝ Acm ⎠ 2.0323 ⎛ 1⎞ Ad = − ( 2.0 ) ⎜ − ⎟ = 2.016 2 ⎝ 2⎠ ⎛ 2.016 ⎞ C M R RdB = 20 log10 ⎜ ⎟ = 35.9 d B ⎝ 0.0323 ⎠ EX9.8 v0 = − R4 ⎛ 2 R2 ⎞ ⎜1 + ⎟ ( vI 1 − vI 2 ) R3 ⎝ R1 ⎠ Differential gain (magnitude) = R4 ⎛ 2 R2 ⎞ ⎜1 + ⎟ R3 ⎝ R1 ⎠ Minimum Gain ⇒ Maximum R1 = 1 + 50 = 51 kΩ So Ad = 20 ⎛ 2 (100 ) ⎞ ⎜1 + ⎟ ⇒ Ad = 4.92 20 ⎝ 51 ⎠ Maximum Gain ⇒ Minimum R1 = 1 kΩ Ad = 20 ⎛ 2 (100 ) ⎞ ⎜1 + ⎟ ⇒ Ad = 201 20 ⎝ 1 ⎠ Range of Differential Gain = 4.92 − 201 EX9.9 Time constant = r = R1C2 = (104 )( 0.1×10−6 ) = 1 m sec −1 ×t 0 ≤ t ≤ 1 ⇒ v0 = R1 C2 At t = 1 m sec ⇒ v0 = −1 V 0 ≤ t ≤ 2 ⇒ v0 = −1 + 1 × ( t − 1) R1 C2 At t = 2 m sec ⇒ v0 = −1 + ( 2 − 1) 1 =0 345. EX9.10 v0 = vI 1 + 10vI 2 − 25vI 3 − 80vI 4 From Figure 9.40, v13 input to R1, vI4 input to R2, vI1 input to RA, and vI2 input to RB. From Equation (9.94) RF R = 25 and F = 80 R1 R2 Set RF = 500 kΩ, then R1 = 20 kΩ, and R2 = 6.25 kΩ. ⎛ R ⎞⎛ R ⎞ ⎛ R ⎞⎛ R ⎞ Also ⎜1 + F ⎟⎜ P ⎟ = 1 and ⎜1 + F ⎟ ⎜ P ⎟ = 10 ⎝ RN ⎠ ⎝ RA ⎠ ⎝ RN ⎠ ⎝ RB ⎠ where RN = R1 R2 = 20 6.25 = 4.76 kΩ and RP = RA RB RC We find that RA = 10 RB Let RA = 200 kΩ, RB = 20 kΩ 500 ⎞ ⎛ RP ⎞ ⎛ RP ⎞ ⎛ Now ⎜1 + ⎟ = 1 = (106 ) ⎜ ⎟⎜ ⎟ ⎝ 4.76 ⎠ ⎝ RA ⎠ ⎝ 200 ⎠ Then RP = 1.89 kΩ RA RB = 200 20 = 18.2 kΩ So RP = 1.89 = 18.2 RC ⇒ RC = 2.11 kΩ 18.2 + RC EX9.11 Computer Analysis TYU9.1 ACL = − R2 −100 kΩ = ⇒ ACL = −10 10 kΩ R1 vI = 0.25 V ⇒ vo = −2.5 V i1 = vI 0.25 = = 0.025 mA ⇒ i1 = 25 μ A R1 10 kΩ i2 = i1 = 25 μ A Ri = R1 = 10 kΩ TYU9.2 (a) − R2 Av = R1 + RS −100 = −4.926 19 + 1.3 −100 = −5.076 Av ( max ) = 19 + 0.7 so 4.926 ≤ Av ≤ 5.076 Av ( min ) = (b) 346. 0.1 = 5.076 μ A 19 + 0.7 0.1 = 4.926 μ A i1 ( min ) = 19 + 1.3 so 4.926 ≤ i1 ≤ 5.076 μ A i1 ( max ) = (c) Maximum current specification is violated. TYU9.3 v0 = Ad ( v2 − v1 ) Ad = 103 a. v2 = 0, v0 = 5 v 5 v1 = − 0 = − 3 ⇒ v1 = −5 mV Ad 10 b. v1 = 5, v0 = −10 v0 = v2 − v1 Ad −10 = v2 − 5 ⇒ v2 = 4.99 V 103 c. v1 = 0.001, v2 = −0.001 v0 = 103 ( −0.001 − 0.001) v0 = −2 V d. v2 = 3, v0 = 3 v0 = Ad ( v2 − v1 ) v0 = v2 − v1 Ad 3 = 3 − v1 ⇒ v1 = 2.997 V 103 TYU9.4 ⎛R ⎞ R R v0 = − ⎜ 4 vI 1 + 4 vI 2 + 4 vI 3 ⎟ R2 R3 ⎝ R1 ⎠ ⎡⎛ 40 ⎞ ⎤ ⎛ 40 ⎞ ⎛ 40 ⎞ v0 = − ⎢⎜ ⎟ ( 250 ) + ⎜ ⎟ ( 200 ) + ⎜ ⎟ ( 75 ) ⎥ ⎝ 20 ⎠ ⎝ 30 ⎠ ⎣⎝ 10 ⎠ ⎦ v0 = − [1000 + 400 + 100] v0 = −1500 μ V = −1.5 mV TYU9.5 347. vI 1 + vI 2 + vI 3 RF = ( vI 1 + vI 2 + vI 3 ) R 3 RF 1 = ⇒ R1 = R2 = R3 ≡ R = 1 M Ω R 3 1 Then RF = M Ω = 333 k Ω 3 vO = TYU9.6 v ⎛ R ⎞ R Av = 0 = ⎜ 1 + 2 ⎟ = 5 so that 2 = 4 vI ⎝ R1 ⎠ R1 For v0 = 10 V, vI = 2 V Then i1 = 2 = 50 μ A ⇒ R1 = 40 kΩ R1 Then R2 = 160 kΩ we find i2 = v0 − vI 10 − 2 = = 50 μ A 160 R2 TYU9.7 v0 = Aod ( v2 − v1 ) = Aod ( vI − v1 ) v0 v − vI = −v1 or v1 = vI − 0 Aod Aod i1 = v −v v1 = i2 and i2 = 0 1 R1 R2 ⎛ 1 ⎞ v −v Then v1 ⎜ ⎟ = 0 1 R2 ⎝ R1 ⎠ ⎛ 1 1 ⎞ v v1 ⎜ + ⎟ = 0 ⎝ R1 R2 ⎠ R2 ⎛ R ⎞ ⎛ R ⎞⎛ v ⎞ v0 ⎜ 1 + 2 ⎟ v1 = ⎜ 1 + 2 ⎟ ⎜ vI − 0 ⎟ R1 ⎠ R1 ⎠⎝ Aod ⎠ ⎝ ⎝ ⎛ R2 ⎞ ⎜1 + ⎟ R1 ⎠ v0 ⎝ So Av = = vI 1 ⎛ R2 ⎞ 1+ ⎜1 + ⎟ Aod ⎝ R1 ⎠ TYU9.8 348. ⎛ Rb For vI 2 = 0, v2 = ⎜ ⎝ Rb + Ra ⎞ ⎟ vI 1 and ⎠ ⎛ R ⎞ ⎛ Rb ⎞ v0 ( vI 1 ) = ⎜ 1 + 2 ⎟ ⎜ ⎟ vI 1 R1 ⎠⎝ Rb + Ra ⎠ ⎝ ⎛ 70 ⎞⎛ 50 ⎞ = ⎜ 1 + ⎟⎜ ⎟ vI 1 5 ⎠⎝ 50 + 25 ⎠ ⎝ = 10vI 1 ⎛ Ra ⎞ For vI 1 = 0, v2 = ⎜ ⎟ vI 2 ⎝ Rb + Ra ⎠ ⎛ R ⎞ ⎛ Ra ⎞ v0 ( vI 2 ) = ⎜ 1 + 2 ⎟ ⎜ ⎟ vI 2 R1 ⎠⎝ Rb + Ra ⎠ ⎝ ⎛ 70 ⎞⎛ 25 ⎞ = ⎜ 1 + ⎟⎜ ⎟ vI 2 5 ⎠⎝ 25 + 50 ⎠ ⎝ = 5vI 2 Then v0 = v0 ( vI 1 ) + v0 ( vI 2 ) v0 = 10vI 1 + 5vI 2 TYU9.9 Ri so i1 = i2 = iS = 100 μ A RS v0 = −iS RF We want −10 = − (100 × 10−6 ) RF ⇒ RF = 100 kΩ TYU9.10 We want iL = 1 mA when vI = −5 V iL = − ( −5 ) −VI −v ⇒ R2 = I = ⇒ R2 = 5 kΩ R2 i2 10−3 vL = iL Z L = (10−3 ) ( 500 ) ⇒ vL = 0.5 V i4 = vL 0.5 = ⇒ i4 = 0.1 mA R2 5 kΩ i3 = i4 + iL = 0.1 + 1 = 1.1 mA If op-amp is biased at ±10 V, output must be limited to ≈ 8 V. So v0 = i3 R3 + vL 8 = (1.1× 10−3 ) R3 + 0.5 ⇒ R3 = 6.82 kΩ Let R3 = 7.0 kΩ 349. Then we want R3 RF 7 = = = 1.4 R2 R1 5 Can choose R1 = 10 kΩ and RF = 14 kΩ TYU9.11 a. v −v i1 = I 1 I 2 R1 ′ v01 = vI 1 + i1 R2 , v02 = vI 2 − i1 R2 and v0 = v0 = R4 [vI 2 − i1 R2 − vI 1 − i1 R2′ ] R3 v0 = R4 ( v02 − v01 ) R3 R4 ′ ⎦ ⎡( vI 2 − vI 1 ) − i1 ( R2 + R2 ) ⎤ R3 ⎣ ⎤ ⎛ vI 2 − vI 1 ⎞ R4 ⎡ ′ ⎢ ( vI 2 − vI 1 ) − ⎜ ⎟ ( R2 + R2 ) ⎥ R3 ⎣ ⎝ R1 ⎠ ⎦ For common-mode input vI 2 = vI 1 ⇒ v0 = 0 ⇒ Common Gain = 0, C M R R = ∞ v0 = b. ′ Ad ( min ) ⇒ R2 min, R1 max ⎛ 20 ⎞ ⎡ 100 + 95 ⎤ Ad = ⎜ ⎟ ⎢1 + ⎥ = 4.82 51 ⎦ ⎝ 20 ⎠ ⎣ ⎛ 20 ⎞ ⎡ 100 + 105 ⎤ Ad ( max ) = ⎜ ⎟ ⎢1 + ⎥ = 206 1 ⎝ 20 ⎠ ⎣ ⎦ c. A CM RR = d Acm Acm = 0 ⇒ C M R R = ∞ TYU9.12 Differential Gain = R4 ⎛ 2 R2 ⎞ ⎜1 + ⎟ R3 ⎝ R1 ⎠ Let R3 = R4 so the difference amplifier gain is unity. Minimum Gain ⇒ Maximum R1 ⎛ 2 R2 ⎞ So ⎜1 + ⎜ R ( max ) ⎟ = 2 ⎟ 1 ⎝ ⎠ We want 2 R2 = R1 ( max ) Maximum Gain ⇒ Minimum R1 ⎛ 2 R2 ⎞ So ⎜1 + ⎜ R ( min ) ⎟ = 1000 or 2 R2 = 999 R1 ( min ) ⎟ 1 ⎝ ⎠ If R2 = 50 kΩ, let R1 ( min ) = 100 Ω fixed resistor and let R1 ( max ) = 100 kΩ + 100 Ω = 100.1 pot Then actual differential gain is in the range of 1.999 − 1001 TYU9.13 −1 10 μ sec −10 × 10−6 ×t 0 = r r − (10 ) (10 × 10−6 ) After 10 pulses: v0 = −5 = r End of 1st pulse: v0 = 350. 100 × 10−6 = 20 μ sec = r 5 r = R1C2 = 20 μ sec = 20 × 10−6 So r = For example, C2 = 0.01× 10−6 = 0.01 μ F ⇒ R1 = 2 kΩ TYU9.14 ⎡ ⎤ + R − ΔR R + ΔR v01 = ⎢ − ⎥V ⎢ ( R − ΔR ) + ( R + ΔR ) ( R + ΔR ) + ( R − ΔR ) ⎥ ⎣ ⎦ ⎡ R − ΔR R + ΔR ⎤ + V =⎢ − 2R ⎥ ⎣ 2R ⎦ ⎛ R − ΔR − R − ΔR ⎞ + =⎜ ⎟V 2R ⎝ ⎠ ⎛ ΔR ⎞ + v01 = − ⎜ ⎟V ⎝ R ⎠ For V + = 3.5 V, ΔR = 50, R = 10 × 103 ⎛ 50 ⎞ v01 = − ⎜ 4 ⎟ ( 3.5 ) = −1.75 × 10−2 ⎝ 10 ⎠ Need an amplifier with a gain of Ad = v0 5 = ⇒ Ad = −285.7 vi −1.75 × 10−2 Use instrumentation amplifier, Fig. 9-25. Connect v01 to vI1 and ( −v01 ) to vI2. ⎛ R ⎞ ⎛ 2R ⎞ Ad = ⎜ 4 ⎟ ⎜ 1 + 2 ⎟ = 285.7 R1 ⎠ ⎝ R3 ⎠ ⎝ Let R4 = 150 kΩ, R3 = 10 kΩ Then R2 = 9.02 R1 Let R2 = 100 kΩ, R1 = 11.1 kΩ TYU9.15 ⎡1 ⎤ + R v01 = ⎢ − ⎥V ⎢ 2 R + R (1 + δ ) ⎥ ⎣ ⎦ ⎡ R + R (1 + δ ) − 2 R ⎤ + =⎢ ⎥V ⎢ 2 ( R + R (1 + δ ) ) ⎥ ⎣ ⎦ = Rδ ×V + 2R ( 2 + δ ) ⎛δ ⎞ v01 ≈ ⎜ ⎟ V + ⎝4⎠ + V = 5 For δ = 0.01 ⎛ 0.01 ⎞ v01 = ⎜ ⎟ ( 5 ) = 0.0125 ⎝ 4 ⎠ v 5 = 400 Need a gain Ad = 0 = v01 0.0125 ⎛ R ⎞ ⎛ 2R ⎞ Use an instrumentation amplifier Ad = 400 = ⎜ 4 ⎟ ⎜1 + 2 ⎟ R1 ⎠ ⎝ R3 ⎠ ⎝ R Let R4 = 150 kΩ, R3 = 10 kΩ then 2 = 12.8 R1 Let R2 = 150 kΩ, R1 = 11.7 kΩ 351. Chapter 9 Problem Solutions 9.1 (a) vO = Ad ( v2 − v1 ) ( ) 1 = Ad 10−3 − ( −10−3 ) ⇒ Ad = 500 (b) 1 = 500 ( v2 − 10−3 ) = 1 + 0.5 = 500v2 v2 = 3 mV (c) 5 = 500 (1 − v1 ) ⇒ 500v1 = 495 v1 = 0.990 V (d) (e) vO = 0 − 3 = 500 ( v2 − ( −0.5 ) ) −250 − 3 = 500v2 v2 = −0.506 V 9.2 (a) 1 ⎛ ⎞ −3 v2 = ⎜ ⎟ vI = ( 0.49975 × 10 ) ( 3) 1 + 2000 ⎠ ⎝ v2 = 1.49925 × 10−3 vO = Aod ( v2 − v1 ) = ( 5 × 103 )(1.49925 × 10−3 − 0 ) vO = 7.49625 V (b) vO = Aod ( v2 − v1 ) 3 = Aod (1.49925 × 10−3 − 0 ) Aod = 2 × 103 9.3 Av = − R2 = −12 ⇒ R2 = 12 R1 R1 Ri = R1 = 25 kΩ ⇒ R2 = (12 )( 25 ) = 300 kΩ 9.3 (a) (b) v2 = 3.00 V vO = Aod ( v2 − v1 ) 2.500 = Aod ( 3.010 − 3.00 ) Aod = 250 9.4 352. ⎛ Ri ⎞ vid = ⎜ ⎟ vI ⎝ Ri + 25 ⎠ ⎛ Ri ⎞ 0.790 = ⎜ ⎟ ( 0.80 ) ⎝ Ri + 25 ⎠ 0.9875 ( Ri + 25 ) = Ri 24.6875 = 0.0125 Ri Ri = 1975 K 9.5 200 ⎫ = −10 ⎪ 20 ⎪ and ⎬ for each case ⎪ Ri = 20 kΩ ⎪ ⎭ Av = − 9.6 a. 100 = −10 10 Ri = R1 = 10 kΩ b. 100 100 Av = − = −5 10 Ri = R1 = 10 kΩ c. 100 Av = − = −5 10 + 10 Ri = 10 + 10 = 20 K Av = − 9.7 353. I1 = vI 0.5 ⇒ R1 = ⇒ R1 = 5 K R1 0.1 R2 = 15 ⇒ R2 = 75 K R1 9.8 Av = − R2 R1 Av = −10 (a) (b) (c) (d) (e) (f) Av = −1 Av = −0.20 Av = −10 Av = −2 Av = −1 9.9 Av = − R2 R1 R1 = 20 K, R2 = 40 K (a) (b) (c) (d) R1 = 20 K, R2 = 200 K R1 = 20 K, R2 = 1000 K R1 = 80 K, R2 = 20 K 9.10 Av = − R2 = −8 ⇒ R2 = 8 R1 R1 For vI = −1, i1 = 1 = 15 μ A ⇒ R1 = 66.7 kΩ ⇒ R2 = 533.3 kΩ R1 9.11 Av = − R2 = −30 ⇒ R2 = 30 R1 R1 Set R2 = 1 MΩ ⇒ R1 = 33.3 kΩ 9.12 a. Av = ⎛R ⎞ R2 1.05R2 ⇒ = 1.105 ⎜ 2 ⎟ R1 0.95 R1 ⎝ R1 ⎠ ⎛R ⎞ 0.95R2 = 0.905 ⎜ 2 ⎟ 1.05R1 ⎝ R1 ⎠ Deviation in gain is +10.5% and − 9.5% b. ⎛R ⎞ ⎛R ⎞ 1.01R2 0.99 R2 = 1.02 ⎜ 2 ⎟ ⇒ = 0.98 ⎜ 2 ⎟ Av ⇒ 0.99 R1 1.01R1 ⎝ R1 ⎠ ⎝ R1 ⎠ Deviation in gain = ±2% 9.13 (a) 354. Av = vO −15 = = −15 vl 1 vO = −15vl ⇒ vO = −150sin ω t ( mV ) (b) i2 = i1 = vI = 10sin ω t ( μ A ) R1 vO ⇒ iL = −37.5sin ω t ( μ A ) RL iL = iO = iL − i2 iO = −47.5sin ω t ( μ A ) 9.14 Av = − R2 R1 + R5 Av = −30 ± 2.5% ⇒ 29.25 ≤ Av ≤ 30.75 So R2 R2 = 29.25 and = 30.75 R1 + 2 R1 + 1 We have 29.25 ( R1 + 2 ) = 30.75 ( R1 + 1) Which yields R1 = 18.5 k Ω and R2 = 599.6 k Ω For vI = 25 mV , then 0.731 ≤ vO ≤ 0.769 V 9.15 vO1 = − R2 120 , vI = − ( 0.2 ) ⇒ vO1 = −1.2 V 20 R1 R4 ⎛ −75 ⎞ , vO1 = ⎜ ⎟ ( −1.2 ) ⇒ vO = +6 V R3 ⎝ 15 ⎠ 0.2 ⇒ i1 = i2 = 10 μ A i1 = i2 = 20 v −1.2 ⇒ i3 = i4 = −80 μ A i3 = i4 = O1 = 15 R3 1st op-amp: 90 μ A into output terminal 2nd op-amp: 80 μ A out of output terminal. vO = − 9.16 (a) R2 22 =− ⇒ Av = −22 1 R1 (b) From Eq. (9.23) R2 1 1 Av = − ⋅ = −22 ⋅ 1 R1 ⎡ ⎡ ⎤ 1 ⎛ R2 ⎞ ⎤ ⎢1 + ⎢1 + 104 ( 23) ⎥ ⎜1 + ⎟ ⎥ ⎣ ⎦ R1 ⎠ ⎦ ⎣ Aod ⎝ Av = − Av = −21.95 (c) 355. Want Av = −22 ( 0.98 ) = −21.56 So − 21.56 = 1+ −22 1 1+ ( 23) Aod 1 22 ( 23) = Aod 21.56 1 ( 23) = 0.020408 ⇒ Aod = 1127 Aod 9.17 (a) R2 1 ⋅ R1 ⎡ 1 ⎛ R2 ⎞ ⎤ ⎢1 + ⎜1 + ⎟ ⎥ R1 ⎠ ⎦ ⎣ Aod ⎝ 100 1 =− ⋅ 1 25 ⎡ ⎤ ⎢1 + 5 × 103 ( 5 ) ⎥ ⎣ ⎦ Av = −3.9960 Av = − (b) vO = −3.9960 (1.00 ) ⇒ vO = −3.9960 V (c) 4 − 3.9960 × 100% = 0.10% 4 (d) vO = Aod ( v2 − v1 ) = − Aod v1 − ( −3.9960 ) vO = Aod 5 × 10+3 v1 = − v1 = 0.7992 mV 9.18 vO = Aod ( v2 − v1 ) = − Aod v1 v −5 v1 = − O = Aod 5 × 10+3 v1 = −1 mV 9.19 Av = − R2 ⎛ R3 R3 ⎞ + ⎟ ⎜1 + R1 ⎝ R4 R2 ⎠ a. −10 = − 10 = b. 9.20 a. R2 ⎛ 100 100 ⎞ + ⎜1 + ⎟ 100 ⎝ 100 R2 ⎠ 2 R2 + 1 ⇒ R2 = 450 kΩ 100 2R 100 = 2 + 1 ⇒ R2 = 4.95 MΩ 100 356. R2 ⎛ R3 R3 ⎞ + ⎟ ⎜1 + R1 ⎝ R4 R2 ⎠ R1 = 500 kΩ Av = − R2 ⎛ R3 R3 ⎞ + ⎟ ⎜1 + 500 ⎝ R4 R2 ⎠ Set R2 = R3 = 500 kΩ 80 = ⎛ 500 ⎞ 500 80 = 1⎜ 1 + + 1⎟ = 2 + ⇒ R4 = 6.41 kΩ R4 R4 ⎝ ⎠ b. For vI = −0.05 V −0.05 ⇒ i1 = i2 = −0.1 μ A i1 = i2 = 500 kΩ v X = −i2 R2 = − ( −0.1× 10−6 )( 500 × 103 ) = 0.05 i4 = − vX 0.05 =− ⇒ i4 = −7.80 μ A 6.41 R4 i3 = i2 + i4 = −0.1 − 7.80 ⇒ i3 = −7.90 μ A 9.21 (a) Av = −1000 = − R2 −500 = R1 R1 R1 = 0.5 K (b) − R2 ⎛ R3 R3 ⎞ + ⎟ ⎜1 + R1 ⎝ R4 R2 ⎠ −250 ⎛ 500 500 ⎞ −1250 −1000 = + ⎜1 + ⎟= R1 ⎝ 250 250 ⎠ R1 R1 = 1.25 K Av = 9.22 i1 = vI = i2 R ⎛v ⎞ v A = −i2 R = − ⎜ I ⎟ R = −vI ⎝R⎠ v v i3 = − A = I R R 357. vA vA 2v 2v − =− A = I R R R R ⎛ 2vI ⎞ vB = v A − i4 R = −vI − ⎜ ⎟ ( R ) = −3vI ⎝ R ⎠ i4 = i2 + i3 = − ( −3vI ) 3vI vB =− = R R R 2vI 3vI 5vI i6 = i4 + i5 = + = R R R v0 5vI ⎞ ⎛ v0 = vB − i6 R = −3vI − ⎜ ⎟ R ⇒ v = −8 ⎝ R ⎠ I i5 = − From Figure 9.12 ⇒ Av = −3 9.23 (a) R2 1 ⋅ R1 ⎡ 1 ⎛ R2 ⎞ ⎤ ⎢1 + ⎜1 + ⎟ ⎥ R1 ⎠ ⎦ ⎣ Aod ⎝ 50 1 =− ⋅ ⇒ Av = −4.99985 10 ⎡ 1 ⎛ 50 ⎞ ⎤ 1+ 1 + ⎟⎥ ⎢ 2 × 105 ⎜ 10 ⎝ ⎠⎦ ⎣ Av = − (b) vO = − ( 4.99985 ) (100 × 10−3 ) ⇒ vO = −499.985 mV (c) Error = 0.5 − 0.499985 × 100% ⇒ 0.003% 0.5 9.24 a. From Equation (9.23) R2 1 Av = − ⋅ R1 ⎡ 1 ⎛ R2 ⎞ ⎤ ⎢1 + ⎜1 + ⎟ ⎥ R1 ⎠ ⎦ ⎣ Aod ⎝ 100 1 =− ⋅ = −0.9980 100 ⎡ 1 ⎛ 100 ⎞ ⎤ 1 + 3 ⎜1 + ⎟⎥ ⎢ 10 ⎝ 100 ⎠ ⎦ ⎣ Then v0 = Av ⋅ vI = ( −0.9980 )( 2 ) ⇒ v0 = −1.9960 V b. v0 = Aod ( v A − vB ) ⎛ 1 vB v0 − vB 1 ⎞ v = ⇒ vB ⎜ + ⎟ = 0 R1 R2 R1 R2 ⎠ R2 ⎝ v0 vB = ⎛ R2 ⎞ ⎜1 + ⎟ R1 ⎠ ⎝ 358. Then v0 = Aod v A − Aod v0 ⎛ R2 ⎞ ⎜1 + ⎟ R1 ⎠ ⎝ ⎡ ⎤ ⎢ ⎥ Aod ⎥ v0 ⎢1 + = Aod v A ⎢ ⎛ R ⎞⎥ 2 ⎢ ⎜1 + ⎟ ⎥ R1 ⎠ ⎥ ⎢ ⎝ ⎣ ⎦ ⎡ ⎛ R2 ⎞ ⎤ ⎢ ⎜ 1 + ⎟ + Aod ⎥ R1 ⎠ ⎥=A v v0 ⎢ ⎝ od A ⎢ ⎛ R ⎞ ⎥ ⎢ ⎜1 + 2 ⎟ ⎥ R1 ⎠ ⎥ ⎢ ⎝ ⎣ ⎦ ⎛ R2 ⎞ Aod ⎜ 1 + ⎟ v A R1 ⎠ ⎝ v0 = ⎛ R ⎞ Aod + ⎜ 1 + 2 ⎟ R1 ⎠ ⎝ ⎛ R2 ⎞ ⎜1 + ⎟ vA R1 ⎠ v0 = ⎝ 1 ⎛ R2 ⎞ 1+ ⎜1 + ⎟ Aod ⎝ R1 ⎠ ⎛ 10 ⎞ ⎛ vI ⎞ ⎜1 + ⎟ ⎜ ⎟ 10 ⎠ ⎝ 2 ⎠ So v0 = ⎝ = 0.9980vI 1 ⎛ 10 ⎞ 1 + 3 ⎜1 + ⎟ 10 ⎝ 10 ⎠ For vI = 2 V v0 = 1.9960 V 9.25 ii = (a) vl v v R = i2 = − O ⇒ O = − 2 R1 R2 vl R1 (b) ⎞ vl v 1 ⎛ R2 = i3 + O = i3 + ⎜ − ⋅ vl ⎟ R1 RL RL ⎝ R1 ⎠ v ⎛ R ⎞ Then i3 = l ⎜ 1 + 2 ⎟ R1 ⎝ RL ⎠ i2 = i1 = 9.26 ⎛ R3 R1 ⎞ + ⎛ 0.1 1 ⎞ VX .max = ⎜ 10 ⇒ VX .max = 0.09008 V ⋅V = ⎜ ⎜R R +R ⎟ ⎟ ⎜ 0.1 1 + 10 ⎟ ( ) ⎟ 4 ⎠ ⎝ 3 1 ⎝ ⎠ R vO = 2 ⋅ VX .max R1 10 = R2 R ( 0.09008 ) ⇒ 2 = 111 R1 R1 So R2 = 111 k Ω 9.27 (a) 359. ⎛R ⎞ R R vO = − ⎜ F ⋅ vI 1 + F ⋅ vI 2 + F ⋅ vI 3 ⎟ R2 R3 ⎝ R1 ⎠ ⎡⎛ 100 ⎞ ⎤ ⎛ 100 ⎞ ⎛ 100 ⎞ = − ⎢⎜ ⎟ ( 0.5 ) + ⎜ ⎟ ( 0.75 ) + ⎜ ⎟ ( 2.5 ) ⎥ ⎝ 50 ⎠ ⎝ 20 ⎠ ⎝ 100 ⎠ ⎣ ⎦ = − [1 + 3.75 + 2.5] vO = −7.25 V (b) ⎡⎛ 100 ⎞ ⎛ 100 ⎞ ⎛ 100 ⎞ ⎤ −2 = − ⎢⎜ ⎟ (1) + ⎜ ⎟ ( 0.8 ) + ⎜ ⎟ vI 3 ⎥ ⎝ 20 ⎠ ⎝ 100 ⎠ ⎦ ⎣⎝ 50 ⎠ 2 = 2 + 4 + vI 3 vI 3 = −4 V 9.28 vo = R R − RF ⋅ vI 1 − F ⋅ vI 2 − F ⋅ vI 3 R1 R2 R3 = −4vI 1 − 8vI 2 − 2vI 3 RF =4 R1 RF =8 R2 RF =2 R3 Largest resistance = RF = 250 K ⇒ R1 = 62.5 K R2 = 31.25 K 9.29 v0 = −4vI 1 − 0.5vI 2 = − RF =4 R1 RF R vI 1 − F vI 2 R1 R2 RF = 0.5 ⇒ R1 is the smallest resistor R2 i = 100 μ A = vI 2 = R1 R1 ⇒ R1 = 20 kΩ ⇒ RF = 80 kΩ ⇒ R2 = 160 kΩ 9.30 vI 1 = ( 0.05 ) 2 sin ( 2π ft ) = 0.0707 sin ( 2π ft ) 1 1 ⇒ 10 ms f = 1 kHz ⇒ T = 3 ⇒ 1 ms vI 2 ⇒ T2 = 100 10 R R 10 10 vO = − F ⋅ vI 1 − F ⋅ vI 2 = − ⋅ vI 1 − ⋅ vI 2 R1 R2 1 5 vO = − (10 ) ( 0.0707 sin ( 2π ft ) ) − ( 2 )( ±1 V ) vO = −0.707 sin ( 2π ft ) − ( ±2 V ) R3 = 125 K 360. 9.31 vO = − RF R R ⋅ vI 1 − F ⋅ vI 2 − F ⋅ vI 3 R1 R2 R3 20 20 20 ⋅ vI 1 − ⋅ vI 2 − ⋅ vI 3 10 5 2 K sin ω t = −2vI 1 − 4 [ 2 + 100sin ω t ] − 0 vO = − Set vI 1 = −4 mV 9.32 Only two inputs. ⎡R ⎤ R vO = − ⎢ F ⋅ vI 1 + F ⋅ vI 2 ⎥ R2 ⎣ R1 ⎦ 1 ⎡ ⎤ = − ⎢3vI 1 + ⋅ vI 2 ⎥ 4 ⎣ ⎦ RF RF 1 =3 = R1 R2 4 Smallest resistor = 10 K = R1 RF = 30 K R2 = 120 K 9.33 ⎡R ⎤ R vO = − ⎢ F ⋅ vI 1 + F ⋅ vI 2 ⎥ R1 R2 ⎣ ⎦ R − RF −R −5 − 5sin ω t = ( 2.5sin ω t ) F ⋅ ( 2 ) ⇒ F = 2 R1 R2 R1 RF = 2.5 R2 RF = largest resistor ⇒ RF = 200 K R1 = 100 K R2 = 80 K 9.34 a. v0 = − RF R R R ⋅ a3 ( −5 ) − F ⋅ a2 ( −5 ) − F ⋅ a1 ( −5 ) − F ⋅ a0 ( −5 ) R3 R2 R1 R0 So v0 = b. RF ⎡ a3 a2 a1 a0 ⎤ + + + ( 5) 10 ⎢ 2 4 8 16 ⎥ ⎣ ⎦ R 1 v0 = 2.5 = F ⋅ ⋅ 5 ⇒ RF = 10 kΩ 10 2 c. i. v0 = 10 1 ⋅ ⋅ 5 ⇒ v0 = 0.3125 V 10 16 361. v0 = ii. 10 ⎡ 1 1 1 1 ⎤ + + + ( 5 ) ⇒ v0 = 4.6875 V 10 ⎢ 2 4 8 16 ⎥ ⎣ ⎦ 9.35 (a) 10 ⋅ vI 1 1 20 20 vO = − ⋅ vO1 − ⋅ vI 2 = − ( 20 )( −10 ) vI 1 − ( 20 ) vI 2 1 1 vO = 200vI 1 − 20vI 2 vO1 = − (b) vI 1 = 1 + 2sin ω t ( mV ) vI 2 = −10 mV Then vO = 200 (1 + 2sin ω t ) − 20 ( −10 ) So vO = 0.4 + 0.4sin ω t (V ) 9.36 For one-input v1 = − v0 Aod v −v vI 1 − v1 v1 = + 1 0 R1 R2 R3 RF ⎡1 VI 1 1 1 ⎤ v0 = v1 ⎢ + + ⎥− R1 R1 R2 R3 RF ⎦ RF ⎣ =− v0 Aod ⎡1 1 1 ⎤ v0 + ⎢ + ⎥− R1 R2 R3 RF ⎦ RF ⎣ ⎧ 1 1 1 ⎛ 1 1 ⎞⎫ ⎪ ⎪ = −v0 ⎨ + + ⎜ + ⎟⎬ ⎪ Aod RF RF Aod ⎝ R1 R2 R3 ⎠ ⎪ ⎩ ⎭ =− ⎫ v0 ⎧ 1 RF 1 +1+ ⋅ ⎨ ⎬ RF ⎩ Aod Aod R1 R2 R3 ⎭ ⎧ ⎫ ⎪ ⎪ R 1 ⎪ ⎪ v0 = − F ⋅ vI 1 ⋅ ⎨ ⎬ where RP = R1 R2 R3 R1 1 ⎛ RF ⎞ ⎪ ⎪1 + 1+ ⎪ Aod ⎜ RP ⎟ ⎪ ⎝ ⎠⎭ ⎩ Therefore, for three-inputs v0 = 9.37 ⎛R ⎞ R R −1 × ⎜ F ⋅ vI 1 + F ⋅ vI 2 + F ⋅ vI 3 ⎟ R2 R3 1 ⎛ RF ⎞ ⎝ R1 ⎠ 1+ ⎜1 + ⎟ Aod ⎝ RP ⎠ 362. ⎛ R ⎞ R Av = 12 = ⎜ 1 + 2 ⎟ ⇒ 2 = 11 R1 ⎠ R1 ⎝ v v 0.5 i1 = I ⇒ R1 = I = R1 i1 0.15 R1 = 3.33 K R2 = 36.7 K 9.38 ⎛ 1 ⎞ vB = ⎜ ⎟ vI ⎝ 1 + 500 ⎠ a. b. ⎛ 1 ⎞ v0 = Aod ⎜ ⎟ vi ⎝ 501 ⎠ ⎛ 1 ⎞ 2.5 = Aod ⎜ ⎟ ( 5 ) ⇒ Aod = 250.5 ⎝ 501 ⎠ ⎛ 1 ⎞ v0 = 5000 ⎜ ⎟ ( 5 ) ⇒ v0 = 49.9 V ⎝ 501 ⎠ 9.39 ⎛ R ⎞ Av = ⎜ 1 + 2 ⎟ R1 ⎠ ⎝ Av = 11 (a) (b) (c) (d) (e) (f) Av = 2 Av = 1.2 Av = 11 Av = 3 Av = 2 9.40 (a) R2 = 1 ⇒ R1 = R2 = 20 K R1 (b) R2 = 9 ⇒ R1 = 20 K, R2 = 180 K R1 (c) R2 = 49 ⇒ R1 = 20 K, R2 = 980 K R1 (d) R2 = 0 can set R2 = 20 K, R1 = ∞ (open circuit) R1 9.41 ⎛ 50 ⎞ ⎡⎛ 20 ⎞ ⎛ 40 ⎞ ⎤ v0 = ⎜ 1 + ⎟ ⎢⎜ ⎟ vI 2 + ⎜ ⎟ vI 1 ⎥ ⎝ 50 ⎠ ⎣⎝ 20 + 40 ⎠ ⎝ 20 + 40 ⎠ ⎦ v0 = 1.33vI 1 + 0.667vI 2 9.42 (a) 363. vI 1 − v2 vI 2 − v2 v2 + = 20 40 10 ⎛ 100 ⎞ vO = ⎜ 1 + ⎟ v2 = 3v2 50 ⎠ ⎝ Now 2vI 1 − 2v2 + vI 2 − v2 = 4v2 ⎛v ⎞ 2vI 1 + vI 2 = 7v2 = 7 ⎜ o ⎟ ⎝3⎠ 6 3 So vO = ⋅ vI 1 + ⋅ vI 2 7 7 (b) (c) 6 ⎛ 3⎞ ( 0.2 ) + ⎜ ⎟ ( 0.3) ⇒ vO = 0.3 V 7 ⎝7⎠ ⎛6⎞ ⎛ 3⎞ vO = ⎜ ⎟ ( 0.25 ) + ⎜ ⎟ ( −0.4 ) ⇒ vO = 42.86 mV ⎝7⎠ ⎝7⎠ vO = 9.43 ⎛ R4 ⎞ v2 = ⎜ ⎟ vI ⎝ R3 + R4 ⎠ ⎛ R ⎞ ⎛ R ⎞ ⎛ R4 ⎞ vO = ⎜1 + 2 ⎟ v2 = ⎜ 1 + 2 ⎟ ⎜ ⎟ vI R1 ⎠ R1 ⎠⎝ R3 + R4 ⎠ ⎝ ⎝ Av = vO ⎛ R2 ⎞ ⎛ R4 ⎞ = ⎜1 + ⎟ ⎜ ⎟ vI ⎝ R1 ⎠ ⎝ R3 + R4 ⎠ 9.44 (a) vO ⎛ 50 x ⎞ = ⎜1 + ⎜ (1 − x ) 50 ⎟ ⎟ vI ⎝ ⎠ vO ⎛ x ⎞ 1− x + x = ⎜1 + ⎟= vI ⎝ 1 − x ⎠ 1− x v 1 Av = O = vI 1 − x (b) 1 ≤ Av ≤ ∞ (c) If x = 1, gain goes to infinity. 9.45 Change resister values as shown. 364. i1 = vI = i2 R ⎛v ⎞ vx = i2 2 R + vI = ⎜ I ⎟ 2 R + vI = 3vI ⎝R⎠ v x 3I i3 = = R R 4v v 3v i4 = i2 + i3 = I + I = I R R R ⎛ 4vI ⎞ v0 = i4 2 R + vx = ⎜ ⎟ 2 R + 3vI ⎝ R ⎠ v0 = 11 vI 9.46 (a) (b) vO =1 vI From Exercise TYU9.7 ⎛ R2 ⎞ ⎜1 + ⎟ R1 ⎠ ⎝ vO = vI ⎡ 1 ⎛ R2 ⎞ ⎤ ⎢1 + ⎜1 + ⎟ ⎥ R1 ⎠ ⎦ ⎣ Aod ⎝ But R2 = 0, R1 = ∞ vO v 1 1 = = ⇒ O = 0.999993 1 1 vI 1 + vI 1+ Aod 1.5 × 105 (b) 9.47 Want vO 1 ⇒ Aod = 99 = 0.990 = 1 vI 1+ Aod 365. v0 = Aod ( vI − v0 ) ⎛ 1 ⎞ + 1⎟ v0 = vI ⎜ ⎝ Aod ⎠ v0 1 = vI ⎛ 1 ⎞ ⎜1 + ⎟ Aod ⎠ ⎝ v Aod = 104 ; 0 = 0.99990 vI Aod = 103 ; v0 = 0.9990 vI Aod = 102 ; v0 = 0.990 vI Aod = 10; v0 = 0.909 vI 9.48 ⎛ R ⎞ v0 A = ⎜ 1 + 2 ⎟ vI R1 ⎠ ⎝ ⎛ R ⎞ ⎛ R ⎞ v01 = ⎜1 + 2 ⎟ vI , v02 = − ⎜ 1 + 2 ⎟ vI R1 ⎠ R1 ⎠ ⎝ ⎝ So v01 = −v02 9.49 (a) iL = vI R1 (b) vO1 = iL RL + vI = iL RL + iL R1 vOI ( max ) ≅ 10 V = iL (1 + 9 ) = 10iL So iL ( max ) ≅ 1 mA Then vI ( max ) ≅ iL R1 = (1)( 9 ) ⇒ vI ( max ) ≅ 9 V 9.50 (a) ⎛ 20 ⎞ ⎛ 20 ⎞ vX = ⎜ ⎟ ⋅ vI = ⎜ ⎟ ( 6 ) = 2 ⎝ 20 + 40 ⎠ ⎝ 60 ⎠ vO = 2 V (b) (c) Same as (a) ⎛ 6 ⎞ vX = ⎜ ⎟ ( 6 ) = 0.666 V ⎝ 6 + 48 ⎠ ⎛ 10 ⎞ vO = ⎜ 1 + ⎟ ⋅ v X ⇒ vO = 1.33 V ⎝ 10 ⎠ 9.51 a. 366. Rin = v −v v1 and 1 0 = i1 and v0 = − Aod v1 i1 RF So i1 = v1 − ( − Aod v1 ) Then Rin = RF = v1 (1 + Aod ) RF v1 RF = i1 1 + Aod b. ⎛ RS ⎞ RF ⋅ i1 i1 = ⎜ ⎟ iS and v0 = − Aod ⋅ 1 + Aod ⎝ RS + Rin ⎠ ⎛ A ⎞⎛ RS ⎞ So v0 = − RF ⎜ od ⎟⎜ ⎟ iS ⎝ 1 + Aod ⎠⎝ RS + Rin ⎠ Rin = RF 10 = = 0.009990 1 + Aod 1001 ⎞ RS ⎛ 1000 ⎞ ⎛ v0 = − RF ⎜ ⎟ iS ⎟⎜ ⎝ 1001 ⎠ ⎝ RS + 0.009990 ⎠ ⎞ RS ⎛ 1000 ⎞ ⎛ Want ⎜ ⎟ ≤ 0.990 ⎟⎜ ⎝ 1001 ⎠ ⎝ RS + 0.009990 ⎠ which yields RS ≥ 1.099 kΩ 9.52 vO = iC RF , 0 ≤ iC ≤ 8 mA For vO ( max ) = 8 V, Then RF = 1 k Ω 9.53 v 10 i = I so 1 = ⇒ R = 10 kΩ R R In the ideal op-amp, R1 has no influence. ⎛ R ⎞ Output voltage: v0 = ⎜1 + 2 ⎟ vI R⎠ ⎝ v0 must remain within the bias voltages of the op-amp; the larger the R2, the smaller the range of input voltage vI in which the output is valid. 9.54 (a) 367. iL = − vI R2 10mA = − ( −10V ) R2 R2 = 1 K R 1 = F R2 R1 R3 Also vL = (10mA )( 0.05k ) = 0.5 V 0.5 = 0.5 mA i2 = 1 iR 3 = 10 + 0.5 = 10.5 mA v −v 13 − 0.5 Limit vo to 13V ⇒ R3 = O L = R3 = 1.19 K 10.5 iR 3 Then RF R3 1.19 R = = = 1.19 = F 1 R1 R2 R1 For example, RF = 119 K, R1 = 100 K (b) From part (a), vO = 13 V when vI = −10 V 9.55 (a) i1 = i2 and i2 = vx + iD , vx = −i2 RF R2 ⎛R Then i1 = −i1 ⎜ F ⎝ R2 ⎛ R Or iD = i1 ⎜ 1 + F R2 ⎝ ⎞ ⎟ + iD ⎠ ⎞ ⎟ ⎠ (b) R1 = vI 5 = ⇒ R1 = 5 k Ω i1 1 ⎛ R ⎞ R 12 = (1) ⎜ 1 + F ⎟ ⇒ F = 11 R2 ⎠ R2 ⎝ For example, R2 = 5 k Ω, RF = 55 k Ω 9.56 VX VX − vO + R2 R3 (1) IX = (2) VX VX − vO + =0 R1 RF 368. ⎛ R ⎞ From (2) vO = VX ⎜ 1 + F ⎟ R1 ⎠ ⎝ ⎛ 1 ⎛ R ⎞ 1 ⎞ 1 Then (1) I X = VX ⎜ + ⎟ − ⋅ VX ⎜1 + F ⎟ R1 ⎠ ⎝ ⎝ R2 R3 ⎠ R3 IX R R 1 1 1 1 1 = = + − − F = − F VX R0 R2 R3 R3 R1 R3 R2 R1 R3 = R1 R3 − R2 RF R1 R2 R3 or Ro = R1 R2 R3 R1 R3 − R2 RF Note: If RF 1 = ⇒ R2 RF = R1 R3 then Ro = ∞, which corresponds to an ideal current source. R1 R3 R2 9.57 R2 R4 = =5 R1 R3 Minimum resistance seen by vI1 is R1. Set R1 = R3 = 25 kΩ Then R2 = R4 = 125 kΩ Ad = iL = v0 ⇒ v0 = iL RL = ( 0.5 )( 5 ) = 2.5 V RL v0 = 5 ( vI 2 − vI 1 ) 2.5 = 5 ( vI 2 − 2 ) ⇒ vI 2 = 2.5 V 9.58 vO = R2 ( vI 2 − vI 1 ) R1 R2 R R and 2 = 4 with R2 = R4 and R1 = R3 R1 R1 R3 Differential input resistance R 20 Ri = 2 R1 ⇒ R1 = i = = 10 K 2 2 R2 (a) = 50 ⇒ R2 = R4 = 500 K R1 Ad = R1 = R3 = 10 K (b) R2 = 20 ⇒ R2 = R4 = 200 K R1 R1 = R3 = 10 K (c) R2 = 2 ⇒ R2 = R4 = 20 K R1 R1 = R3 = 10 K (d) R2 = 0.5 ⇒ R2 = R4 = 5 K R1 R1 = R3 = 10 K 9.59 369. We have ⎞ ⎛ R ⎞⎛ R / R ⎞ ⎛R ⎞ ⎛ R ⎞⎛ ⎛ R2 ⎞ 1 vO = ⎜ 1 + 2 ⎟ ⎜ 4 3 ⎟ vI 2 − ⎜ 2 ⎟ vI 1 or vO = ⎜ 1 + 2 ⎟ ⎜ ⎟ vI 2 − ⎜ ⎟ vI 1 R1 ⎠ ⎝ 1 + R4 / R3 ⎠ R1 ⎠ ⎝ 1 + R3 / R4 ⎠ ⎝ ⎝ R1 ⎠ ⎝ ⎝ R1 ⎠ Set R2 = 50 (1 + x ) , R1 = 50 (1 − x ) R3 = 50 (1 − x ) , R4 = 50 (1 + x ) ⎡ ⎤ ⎢ ⎥ ⎡ ⎛ 1 + x ⎞⎤ 1+ x ⎞ 1 ⎥ vI 2 − ⎛ vO = ⎢1 + ⎜ ⎟⎥ ⎢ ⎜ ⎟ vI 1 ⎝ 1 − x ⎠ ⎦ ⎢1 + ⎛ 1 − x ⎞ ⎥ ⎝ 1− x ⎠ ⎣ ⎢ ⎜ 1+ x ⎟ ⎥ ⎠⎦ ⎣ ⎝ ⎤ ⎡1 − x + (1 + x ) ⎤ ⎡ 1+ x ⎛ 1+ x ⎞ vO = ⎢ ⎥ vI 2 − ⎜ ⎥⋅⎢ ⎟ vI 1 1− x ⎝ 1− x ⎠ ⎢ ⎣ ⎦ ⎣1 + x + (1 − x ) ⎥ ⎦ ⎛ 1+ x ⎞ ⎛1+ x ⎞ =⎜ ⎟ vI 2 − ⎜ ⎟ vI 1 ⎝1− x ⎠ ⎝ 1− x ⎠ For vI 1 = vI 2 ⇒ vO = 0 Set R2 = 50 (1 + x ) R1 = 50 (1 − x ) R3 = 50 (1 + x ) R4 = 50 (1 − x ) ⎛ ⎞ 1 ⎟ ⎛ 1+ x ⎞⎜ ⎛ 1+ x ⎞ vO = ⎜1 + ⎟ ⎜ 1 + x ⎟ vI 2 − ⎜ ⎟ vI 1 ⎝ 1− x ⎠⎜1+ ⎝ 1− x ⎠ ⎟ ⎜ ⎟ ⎝ 1− x ⎠ ⎛ 1+ x ⎞ = vI 2 − ⎜ ⎟ vI 1 ⎝ 1− x ⎠ vI 1 = vI 2 = vcm vO 1 + x 1 − x − (1 + x ) −2 x = 1− = = vcm 1− x 1− x 1− x Set R2 = 50 (1 − x ) R1 = 50 (1 + x ) R3 = 50 (1 − x ) R4 = 50 (1 + x ) ⎛ ⎞ 1 ⎟ ⎛ 1− x ⎞⎜ ⎛ 1− x ⎞ vO = ⎜ 1 + ⎟ ⎜ 1 − x ⎟ vI 2 − ⎜ ⎟ vI 1 ⎝ 1+ x ⎠⎜ 1+ ⎝ 1+ x ⎠ ⎟ ⎜ ⎟ ⎝ 1+ x ⎠ ⎛ 1− x ⎞ = ⎜1 − ⎟ vcm ⎝ 1+ x ⎠ 1 + x − (1 − x ) 2x 1+ x 1+ x Worst common-mode gain Acm = Acm = (b) −2 x 1− x = 370. −2 x −2 ( 0.01) = = −0.0202 1 − x 1 − 0.01 −2 ( 0.02 ) For x = 0.02, Acm = = −0.04082 1 − 0.02 −2 ( 0.05 ) For x = 0.05, Acm = = −0.1053 1 − 0.05 1 1 For this condition, set vI 2 = + , vI 1 = − ⇒ vd = 1 V 2 2 1 1 ⎡ ⎛ 1 + x ⎞ ⎤ 1 ⎡1 − x + (1 + x ) ⎤ 1 2 = Ad = ⎢1 + ⎜ ⎥= ⋅ ⎟⎥ = ⎢ 2 ⎣ ⎝ 1 − x ⎠⎦ 2 ⎣ 1− x ⎦ 2 1− x 1− x For x = 0.01, Acm = For x = 0.01 Ad = 1.010 C M R RdB = 20 log10 1.010 = 33.98 dB 0.0202 C M R RdB = 20 log10 1.020 = 27.96 dB 0.04082 For x = 0.02, Ad = 1 = 1.020 0.98 For x = 0.05 Ad = 1 1.0526 = 1.0526 C M R RdB = 20 log10 ≅ 20 dB 0.95 0.1053 9.60 ⎛ 10R ⎞ ⎛ 10 ⎞ vy = ⎜ ⎟ v2 = ⎜ ⎟ ( 2.65 ) ⇒ v y = vx = 2.40909 V 10R+R ⎠ ⎝ ⎝ 11 ⎠ v2 − v y 2.65 − 2.40909 i3 = i4 = = = 0.0120 mA 20 R v −v 2.50 − 2.40909 i1 = i2 = 1 x = = 0.0045455 mA 20 R vO = vx − i2 (10R ) = ( 2.40909 ) − ( 0.0045455 )( 200 ) vO = 1.50 V 9.61 iE = (1 + β )( iB ) = ( 81)( 2 ) = 162 mA = 10 R R = 61.73 Ω 9.62 a. From superposition: R2 v01 = − ⋅ vI 1 R1 ⎛ R ⎞ ⎛ R1 ⎞ v02 = ⎜ 1 + 2 ⎟ ⎜ ⎟ vI 2 R1 ⎠ ⎝ R3 + R4 ⎠ ⎝ Setting vI 1 = vI 2 = vcm ⎡ ⎛ ⎢⎛ R ⎞ ⎜ 1 v0 = v01 + v02 = ⎢⎜ 1 + 2 ⎟ ⎜ R3 R1 ⎠ ⎜ ⎢⎝ ⎜ 1+ R ⎢ 4 ⎝ ⎣ ⎤ ⎞ ⎟ R ⎥ ⎟ − 2 ⎥ vcm ⎟ R1 ⎥ ⎟ ⎥ ⎠ ⎦ 371. ⎛ v R ⎛ R ⎞⎜ 1 Acm = 0 = 4 ⋅ ⎜ 1 + 2 ⎟ ⎜ R4 vcm R3 ⎝ R1 ⎠ ⎜ ⎜ 1+ R 3 ⎝ R4 ⎛ R2 ⎞ R2 ⎛ R4 ⎞ ⎜1 + ⎟ − ⎜1 + ⎟ R R1 ⎠ R1 ⎝ R3 ⎠ = 3⎝ ⎛ R4 ⎞ ⎜1 + ⎟ ⎝ R3 ⎠ ⎞ ⎟ R ⎟− 2 ⎟ R1 ⎟ ⎠ R4 R2 − R R1 Acm = 3 ⎛ R4 ⎞ ⎜1 + ⎟ ⎝ R3 ⎠ b. Max. Acm Max. Acm ⇒ Min. R4 R and Max. 2 R3 R1 47.5 52.5 − 10.5 9.5 = 4.5238 − 5.5263 ⇒ A = cm 47.5 1 + 4.5238 1+ 10.5 9.63 vI 1 − v A v A − vB vA − v0 = + R1 + R2 Rv R2 (1) vI 2 − vB vB − v A vB = + R1 + R2 Rv R2 (2) ⎛ R1 ⎞ ⎛ R2 ⎞ v− = ⎜ ⎟ vA + ⎜ ⎟ vI 1 ⎝ R1 + R2 ⎠ ⎝ R1 + R2 ⎠ ⎛ R1 ⎞ ⎛ R2 ⎞ v+ = ⎜ ⎟ vB + ⎜ ⎟ vI 2 ⎝ R1 + R2 ⎠ ⎝ R1 + R2 ⎠ (3) (4) max = 0.1815 372. Now v− = v+ ⇒ R1vA + R2 vI 1 = R1vB + R2 vI 2 R So that v A = vB + 2 ( vI 2 − vI 1 ) R1 ⎛ 1 v vI 1 1 1 ⎞ v = vA ⎜ + + ⎟− B − 0 R1 + R2 R1 + R2 RV R2 ⎠ RV R2 ⎝ ⎛ 1 vI 2 1 1 ⎞ v = vB ⎜ + + ⎟− A R1 + R2 R1 + R2 RV R2 ⎠ RV ⎝ (1) ( 2) Then ⎛ 1 v ⎛ R ⎞⎛ 1 vI 1 1 1 ⎞ v 1 1 ⎞ + ⎟ ( vI 2 − vI 1 ) = vB ⎜ + + ⎟ − B − 0 + ⎜ 2 ⎟⎜ + R1 + R2 R1 + R2 RV R2 ⎠ RV R2 ⎝ R1 ⎠ ⎝ R1 + R2 RV R2 ⎠ ⎝ ⎛ 1 ⎤ vI 2 R2 1 1 ⎞ 1 ⎡ = vB ⎜ + + ⎟− ⎢ vB + ( vI 2 − vI 1 ) ⎥ R1 + R2 R1 + R2 RV R2 ⎠ RV ⎣ R1 ⎦ ⎝ Subtract (2) from (1) ⎛ R ⎞⎛ 1 v 1 1 1 ⎞ 1 R2 + + ⎟ ( vI 2 − vI 1 ) − 0 + ⋅ ( vI 2 − vI 1 ) ( vI 1 − vI 2 ) = ⎜ 2 ⎟ ⎜ R1 + R2 R1 ⎠ ⎝ R1 + R2 RV R2 ⎠ R2 RV R1 ⎝ ⎧ ⎫ v0 1 1 ⎞ 1 1 R2 ⎪ ⎪⎛ R ⎞ ⎛ 1 = ( vI 2 − vI 1 ) ⎨⎜ 2 ⎟ ⎜ + + ⎟+ + ⋅ ⎬ R2 ⎪⎝ R1 ⎠ ⎝ R1 + R2 RV R2 ⎠ R1 + R2 RV R1 ⎪ ⎩ ⎭ ⎛ R ⎞ ⎧ R2 R R1 R ⎫ v0 = ( vI 2 − vI 1 ) ⎜ 2 ⎟ ⎨ + 2 +1+ + 2⎬ R1 + R2 RV ⎭ ⎝ R1 ⎠ ⎩ R1 + R2 RV v0 = 9.64 R2 ⎞ 2 R2 ⎛ ⎜1 + ⎟ ( vI 2 − vI 1 ) R1 ⎝ RV ⎠ (1) (2) 373. i1 = vI 1 − vI 2 ( 0.50 − 0.030sin ω t ) − ( 0.50 + 0.030sin ω t ) = 20 R1 −0.060sin ω t 20 i1 = −3sin ω t ( μ A ) vO1 = i1 R2 + vI 1 = ( −0.0030sin ω t )(115 ) + 0.50 − 0.030sin ω t vO1 = 0.50 − 0.375sin ω t = vO 2 = vI 2 − i1 R2 = 0.50 + 0.030sin ω t − ( −0.003sin ω t )(115 ) vO 2 = 0.50 + 0.375sin ω t R4 200 ⎡ 0.50 + 0.375sin ω t − ( 0.50 − 0.375sin ω t ) ⎤ ( vO 2 − vO1 ) = ⎦ 50 ⎣ R3 vO = vO = 3sin ω t ( V ) i3 = vO 2 0.50 + 0.375sin ω t = 50 + 200 R3 + R4 i3 = 2 + 1.5sin ω t ( μ A ) i2 = vO1 − vO ( 0.5 − 0.375sin ω t ) − ( 3sin ω t ) = 250 R3 + R4 i2 = 2 − 13.5sin ω t ( μ A ) 9.65 ⎛ 40 ⎞ vOB = ⎜1 + ⎟ vI = 2.1667 sin ω t ⎝ 12 ⎠ 30 vOC = − vI = −1.25sin ω t 12 (a) (b) (c) vO = vOB − vOC = 2.1667 sin ω t − ( −1.25sin ω t ) vO = 3.417 sin ω t vO 3.417 = = 6.83 0.5 vI (d) 9.66 iO = vI R 9.67 vO R ⎛ 2R ⎞ = 4 ⎜1 + 2 ⎟ vI 2 − vI 1 R3 ⎝ R1 ⎠ 200 ⎛ 2 (115 ) ⎞ vO = ⎜1 + ⎟ ( 0.06sin ω t ) R1 ⎠ 50 ⎝ 230 For vO = 0.5 = 1.0833 ⇒ R1 = 212.3 K R1 Ad = vO = 8 V 9.68 230 = 32.33 ⇒ R1 = 7.11 K ⇒ R1 f = 7.11 K, R1 (potentiometer) = 205.2 K R1 374. ⎛ 2 R2 ⎞ ⎜1 + ⎟ ( vI 2 − vI 1 ) R1 ⎠ ⎝ Set R2 = 15 K, Set R1 = 2 K + 100 k ( Rot ) R4 R3 vO = Want R4 ≈8 R3 Set R3 = 10 K R 4 = 75 K Now Gain (min) = 75 ⎛ 2 (15 ) ⎞ ⎜1 + ⎟ = 9.71 10 ⎝ 102 ⎠ Gain ( max ) = 75 ⎛ 2 (15 ) ⎞ ⎜1 + ⎟ = 120 10 ⎝ 2 ⎠ 9.69 For a common-mode gain, vcm = vI 1 = vI 2 Then ⎛ R ⎞ R v01 = ⎜ 1 + 2 ⎟ vcm − 2 vcm = vcm R1 ⎠ R1 ⎝ ⎛ R ⎞ R v02 = ⎜ 1 + 2 ⎟ vcm − 2 vcm = vcm R1 ⎠ R1 ⎝ From Problem 9.62 we can write R4 R4 − ′ R3 R3 Acm = ⎛ R4 ⎞ ⎜1 + ⎟ R3 ⎠ ⎝ ′ R3 = R4 = 20 kΩ, R3 = 20 kΩ ± 5% 20 1− ′ R3 1 ⎛ 20 ⎞ Acm = = ⎜1 − ⎟ ′ R3 ⎠ 1 + 1) 2 ⎝ ( ′ For R3 = 20 kΩ − 5% = 19 kΩ 1 ⎛ 20 ⎞ ⎜ 1 − ⎟ = −0.0263 2 ⎝ 19 ⎠ ′ For R3 = 20 kΩ + 5% = 21 kΩ Acm = 1 ⎛ 20 ⎞ ⎜ 1 − ⎟ = 0.0238 2⎝ 21 ⎠ So Acm max = 0.0263 Acm = 9.70 a. v0 = 1 ⋅ vI ( t ′ ) dt ′ R1C2 ∫ 0.5 ∫ 0.5sin ω t dt = − ω v0 = 0.5 = f = cos ω t 1 ( 0.5 ) 0.5 ⋅ = R1C2 ω 2π R1C2 f 1 1 = ⇒ f = 31.8 Hz 2π R1C2 2π ( 50 × 103 )( 0.1× 10−6 ) Output signal lags input signal by 90° 375. b. i. f = ii. f = 0.5 2π ( 50 × 103 )( 0.1× 10−6 ) ⇒ f = 15.9 Hz 0.5 ( 0.1)( 2π ) ( 50 ×103 )( 0.1×10−6 ) 9.71 − vI ⋅ t 1 ∫ vI ( t ) dt = RC RC vI = −0.2 vO = − Now 8= − ( −0.2 )( 2 ) (a) RC RC = 0.05 s (b) 14 = ( 0.2 ) t 0.05 ⇒ t = 3.5 s 9.72 a. v0 = vI − R2 1 jω C2 R1 R2 ⋅ =− 1 jω C2 ⎛ 1 ⎞ R1 ⎜ R2 + ⎟ jω C2 ⎠ ⎝ v0 R 1 =− 2⋅ vI R1 1 + jω R2 C2 b. v0 R =− 2 vI R1 c. f = 1 2π R2 C2 9.73 a. R ( jω C1 ) v0 − R2 = =− 2 vI R + 1 1 + jω R1C1 1 jω C1 v0 R jω R1C1 =− 2⋅ vI R1 1 + jω R1C1 b. v0 R =− 2 vI R1 c. f = 1 2π R1C1 9.74 Assuming the Zener diode is in breakdown, ⇒ f = 159 Hz 376. vO = − i2 = R2 1 ⋅ Vz = − ( 6.8 ) ⇒ vO = −6.8 V R1 1 0 − vO 0 − ( −6.8 ) = ⇒ i2 = 6.8 mA R2 1 10 − Vz 10 − 6.8 − i2 = − 6.8 ⇒ iz = −6.2 mA!!! Rs 5.6 Circuit is not in breakdown. Now 10 − 0 10 = i2 = ⇒ i2 = 1.52 mA 5.6 + 1 Rs + R1 iz = vO = −i2 R2 = − (1.52 )(1) ⇒ vO = −1.52 V iz = 0 9.75 ⎡ ⎤ ⎛ v ⎞ v ⎛ vI ⎞ vO = −VT ln ⎜ I ⎟ = − ( 0.026 ) ln ⎢ −14 I 4 ⎥ ⇒ vO = −0.026 ln ⎜ −10 ⎟ I s R1 ⎠ ⎢ (10 )(10 ) ⎥ ⎝ 10 ⎠ ⎝ ⎣ ⎦ For vI = 20 mV , vO = 0.497 V For vI = 2 V , vO = 0.617 V 9.76 377. ⎛ 333 ⎞ v0 = ⎜ ⎟ ( v01 − v02 ) = 16.65 ( v01 − v02 ) ⎝ 20 ⎠ ⎛i ⎞ v01 = −vBE1 = −VT ln ⎜ C1 ⎟ ⎝ IS ⎠ ⎛i ⎞ v02 = −vBE 2 = −VT ln ⎜ C 2 ⎟ ⎝ IS ⎠ ⎛i ⎞ ⎛i ⎞ v01 − v02 = −VT ln ⎜ C1 ⎟ = VT ln ⎜ C 2 ⎟ ⎝ iC 2 ⎠ ⎝ iC1 ⎠ v v iC 2 = 2 , iC1 = 1 R2 R1 ⎛v R ⎞ So v01 − v02 = VT ln ⎜ 2 ⋅ 1 ⎟ ⎝ R2 v1 ⎠ Then ⎛v R ⎞ v0 = (16.65 )( 0.026 ) ln ⎜ 2 ⋅ 1 ⎟ ⎝ v1 R2 ⎠ ⎛v R ⎞ v0 = 0.4329 ln ⎜ 2 ⋅ 1 ⎟ ⎝ v1 R2 ⎠ ln ( x ) = log e ( x ) = ⎡ log10 ( x ) ⎤ ⋅ ⎡log e (10 ) ⎤ ⎣ ⎦ ⎣ ⎦ = 2.3026 log10 ( x ) ⎛v R ⎞ Then v0 ≅ (1.0 ) log10 ⎜ 2 ⋅ 1 ⎟ ⎝ v1 R2 ⎠ 9.77 vO = − I s R evI / VT = − (10−14 )(104 ) evI / VT vO = (10 ( −10 )e ) vI / 0.026 For vI = 0.30 V , vo = 1.03 × 10−5 V For vI = 0.60 V , vo = 1.05 V 378. Chapter 10 Exercise Solutions EX10.1 V + − VBE ( on ) 10 − 0.7 = I REF = 15 R1 I REF = 0.62 mA I REF 0.62 = 2 2 1+ 1+ β 75 I 0 = 0.604 mA I0 = EX10.2 V + − VBE ( on ) − V − 5 − 0.7 − ( −5 ) I REF = = R1 12 I REF = 0.775 mA I 0.775 I 0 = REF = = 0.7549 mA 2 2 1+ 1+ 75 β ΔI 0 = ( 0.02 )( 0.7549 ) = 0.0151 mA and ΔI 0 = r0 = ΔV 1 ΔVCE 2 ⇒ r0 = CE 2 ΔI 0 r0 V 4 = 265 kΩ = A ⇒ VA = ( 265 )( 0.7549 ) ⇒ VA ≅ 200 V 0.0151 I0 EX10.3 V + − 2VBE ( on ) 9 − 2 ( 0.7 ) I REF = = R1 12 I REF = 0.6333 mA I0 = I REF 0.6333 = = 0.6331 mA 2 2 1+ 1+ β (1 + β ) 75 ( 76 ) I 0 = 0.6331 mA = I C1 I B1 = I B 2 = I0 β ⇒ I B1 = I B 2 = 8.44 μ A I E 3 = I B1 + I B 2 ⇒ I E 3 = 16.88 μ A I B3 = I E3 ⇒ I B 3 = 0.222 μ A 1+ β EX10.4 ⎛I ⎞ I 0 RE = VT ln ⎜ REF ⎟ ⎝ I0 ⎠ ⎛ I ⎞ 0.026 ⎛ 0.75 ⎞ V ln ⎜ RE = T ln ⎜ REF ⎟ = ⎟ ⇒ RE = 3.54 kΩ I 0 ⎝ I 0 ⎠ 0.025 ⎝ 0.025 ⎠ 5 − 0.7 R1 = ⇒ R1 = 5.73 kΩ 0.75 VBE1 − VBE 2 = I 0 RE = ( 0.025 )( 3.54 ) ⇒ VBE1 − VBE 2 = 88.5 mV EX10.5 379. 5 − 0.7 − ( −5 ) ⇒ I REF = 0.775 mA 12 ⎛I ⎞ I 0 RE = VT ln ⎜ REF ⎟ ⎝ I0 ⎠ ⎛ 0.775 ⎞ I 0 ( 6 ) = ( 0.026 ) ln ⎜ ⎟ ⇒ I 0 ≅ 16.6 μ A ⎝ I0 ⎠ I REF = EX10.6 ⎛I ⎞ I 0 RE = VT ln ⎜ REF ⎟ ⎝ I0 ⎠ 0.026 ⎛ 0.70 ⎞ RE = ln ⎜ ⎟ ⇒ RE = 3.465 kΩ 0.025 ⎝ 0.025 ⎠ I 0.025 gm2 = 0 = ⇒ g m 2 = 0.9615 mA/V VT 0.026 β VT rπ 2 = r02 = I0 = (150 )( 0.026 ) 0.025 = 156 kΩ VA 100 = = 4000 kΩ I 0 0.025 ′ RE = RE || rπ 2 = 3.47 ||156 = 3.39 kΩ ′ R0 = r02 (1 + g m 2 RE ) = 4000 ⎡1 + ( 0.962 )( 3.39 ) ⎤ ⎣ ⎦ R0 = 17.04 MΩ 1 3 dI 0 = ⋅ dVC 2 = ⇒ dI 0 = 0.176 μ A R0 17, 040 EX10.7 I REF = I R + I BR + I B1 + ... + I BN I R = I 01 = I 02 = ... = I 0 N and I BR = I B1 = I B 2 = ... = I BN = ⎛I ⎞ ⎛ N +1⎞ I REF = I 01 + ( N + 1) ⎜ 01 ⎟ = I 01 ⎜ 1 + ⎟ β ⎠ β ⎠ ⎝ ⎝ I REF So I 01 = I 02 = ... = I 0 N = N +1 1+ β I 01 1 = 0.90 = N +1 I REF 1+ 50 N +1 1 1+ = 50 0.9 1 ⎛ ⎞ − 1⎟ ( 50 ) N +1 = ⎜ ⎝ 0.9 ⎠ ⎛ 1 ⎞ − 1⎟ ( 50 ) − 1 N =⎜ 0.9 ⎠ ⎝ N = 4.55 ⇒ N = 4 EX10.8 I 01 β 380. VDS ( sat ) = 1 V = VGS 2 − VTN = VGS 2 − 2 ⇒ VGS 2 = 3 V 2 2 ⎛ μ C ⎞⎛W ⎞ I O = K n 2 (VGS 2 − VTN ) = ⎜ n ox ⎟ ⎜ ⎟ (VGS 2 − VTN ) ⎝ 2 ⎠ ⎝ L ⎠2 2 ⎛W ⎞ ⎛W ⎞ 0.20 = ( 0.020 ) ⎜ ⎟ ( 3 − 2 ) ⇒ ⎜ ⎟ = 10 L ⎠2 ⎝ ⎝ L ⎠2 2 ⎛ μ C ⎞⎛W ⎞ I REF = ⎜ n ox ⎟ ⎜ ⎟ (VGS 1 − VTN ) 2 ⎠ ⎝ L ⎠1 ⎝ VGS 1 = VGS 2 2 ⎛W ⎞ ⎛W ⎞ 0.5 = ( 0.020 ) ⎜ ⎟ ( 3 − 2 ) ⇒ ⎜ ⎟ = 25 ⎝ L ⎠1 ⎝ L ⎠1 VGS 3 = V + − VGS 1 = 10 − 3 = 7 V ⎛ μ C ⎞⎛W ⎞ 2 I REF = ⎜ n ox ⎟ ⎜ ⎟ (VGS 3 − VTN ) ⎝ 2 ⎠ ⎝ L ⎠3 2 ⎛W ⎞ ⎛W ⎞ 0.5 = ( 0.020 ) ⎜ ⎟ ( 7 − 2 ) ⇒ ⎜ ⎟ = 1 L ⎠3 ⎝ ⎝ L ⎠3 EX10.9 I REF = K n (VGS − VTN ) 2 0.020 = 0.080 (VGS − 1) a. 2 VGS = 1.5 V all transistors b. VG 4 = VGS 3 + VGS1 + V − = 1.5 + 1.5 − 5 = −2 V VS 4 = VG 4 − VGS 4 = −2 − 1.5 = −3.5 V VD 4 ( min ) = VS 4 + VDS 4 ( sat ) and VDS 4 ( sat ) = VGS 4 − VTN = 1.5 − 1 = 0.5 V So VD 4 ( min ) = −3.5 + 0.5 ⇒ VD 4 ( min ) = −3.0 V c. R0 = r04 + r02 (1 + g m r04 ) 1 1 r02 = r04 = = = 2500 kΩ λ I 0 ( 0.02 )( 0.020 ) g m = 2 K n (VGS − VTN ) = 2 ( 0.080 )(1.5 − 1) ⇒ g m = 0.080 mA / V R0 = 2500 + 2500 (1 + ( 0.080 )( 2500 ) ) ⇒ R0 = 505 MΩ EX10.10 For Q2 : vDS ( min ) = VP = 2 V ⇒ VS ( min ) = vDS ( min ) − 5 = 2 − 5 ⇒ VS ( min ) = −3 V I 0 = I DSS 2 (1 + λ vDS 2 ) = 0.5 (1 + ( 0.15 )( 2 ) ) ⇒ I 0 = 0.65 mA ⎛ v ⎞ I 0 = I DSS1 ⎜ 1 − GS 1 ⎟ ⎝ VP1 ⎠ 2 2 ⎛ v ⎞ 0.65 = 0.80 ⎜ 1 − GS1 ⎟ −2 ⎠ ⎝ vGS 1 = 0.0986 ⇒ vGS 1 = −0.197 V −2 vGS 1 = VI − VS − 0.197 = VI − ( −3) ⇒ VI ( min ) = −3.2 V 381. Vgs 2 = 0, Vgs1 = −VX IX = VX VX − V1 + + g m1VX r02 r01 V1 V1 − VX + = g m1VX RD r01 (1) (2) ⎛ 1 ⎞ VX ⎜ + g m1 ⎟ r01 ⎝ ⎠ V1 = 1 1 + RD r01 ⎞ 1 ⎛ 1 ⎜ + g m1 ⎟ r01 ⎝ r01 IX 1 1 1 ⎠ = = + + g m1 − 1 1 VX R0 r02 r01 + RD r01 1 ⎡ ⎤ ⎢ ⎞ r01 ⎥ 1 ⎛ 1 ⎥ = + ⎜ + g m1 ⎟ ⎢1 − 1 1 ⎥ r02 ⎝ r01 ⎠⎢ + ⎢ RD r01 ⎥ ⎣ ⎦ 1 ⎛ ⎞ ⎞ ⎜ RD ⎟ 1 ⎛ 1 ⎟ = + ⎜ + g m1 ⎟ ⎜ r02 ⎝ r01 ⎠⎜ 1 + 1 ⎟ ⎜R ⎟ ⎝ D r01 ⎠ 382. r01 ⇒ For RD ⎞ 1 1 ⎛ 1 ≅ + ⎜ + g m1 ⎟ R0 r02 ⎝ r01 ⎠ For Q1: 2I g m1 = DSS 1 VP ⎛ VGS 1 ⎞ 2 ( 0.8 ) ⎛ −0.197 ⎞ ⎜1 − ⎟= ⎜1 − ⎟ −2 ⎠ 2 ⎝ VP ⎠ ⎝ g m1 = 0.721 mA/V 1 1 = = 10.3 kΩ r0 = λ I 0 ( 0.15 )( 0.65 ) 1 1 1 = + + 0.721 = 0.915 ⇒ R0 = 1.09 kΩ R0 10.3 10.3 EX10.11 a. ⎛V ⎞ I REF = I S exp ⎜ EB 2 ⎟ ⎝ VT ⎠ ⎛I ⎞ ⎛ 0.5 × 10−3 ⎞ VEB 2 = VT ln ⎜ REF ⎟ = ( 0.026 ) ln ⎜ ⎟ ⇒ VEB 2 = 0.521 V −12 ⎝ 10 ⎠ ⎝ IS ⎠ 5 − 0.521 R1 = ⇒ R1 = 8.96 kΩ b. 0.5 c. Combining Equations (10.79), (10.80), and (10.81), we find ⎛ VEC 2 ⎞ ⎜1 + ⎟ ⎡ VAP ⎠ ⎛ V ⎞⎤ ⎛ V ⎞ I S 0 ⎢ exp ⎜ I ⎟ ⎥ ⎜ 1 + CEo ⎟ = I REF × ⎝ ⎛ VEB 2 ⎞ ⎝ VT ⎠ ⎦ ⎝ VAN ⎠ ⎣ ⎜1 + ⎟ ⎝ VAP ⎠ 2.5 ⎞ ⎛ ⎜1 + ⎟ ⎡ ⎤⎛ ⎛V ⎞ 2.5 ⎞ 100 ⎠ = ( 0.5 × 10−3 ) ⎝ 10−12 ⎢ exp ⎜ I ⎟ ⎥ ⎜ 1 + ⎟ ⎛ 0.521 ⎞ ⎝ VT ⎠ ⎦ ⎝ 100 ⎠ ⎣ ⎜1 + ⎟ 100 ⎠ ⎝ ⎛V 1.025 ×10 −12 exp ⎜ I ⎝ VT d. ⎞ ⎛ VI −4 ⎟ = 5.098 × 10 exp ⎜ ⎠ ⎝ VT ⎞ 8 ⎟ = 4.974 × 10 ⇒ VI = 0.521 V ⎠ ⎛ 1 ⎞ 1 −⎜ ⎟ − ⎝ VT ⎠ = 0.026 = −38.46 ⇒ A = −1923 Av = v 1 1 1 1 0.01 + 0.01 + + VAN VAP 100 100 EX10.12 I CQ 0.8 gm = = = 30.77 mA/V 0.026 VT r0 = r02 = VA 80 = = 100 kΩ I CQ 0.8 a. V0 = − g mVπ 1 ( r0 || r02 ) , Vπ 1 = Vi Av = − g m ( r0 || r02 ) = − ( 30.77 ) [100 ||100] ⇒ Av = −1538 b. Av = − g m ( r0 || r02 || RL ) Av = − 1540 = −770 − 770 = − ( 30.77 )( 50 || RL ) ⇒ ( 50 || RL ) = 25 ⇒ RL = 50 kΩ 2 EX10.13 (a) Neglecting effect of λ and RL 383. I O = I REF = K n (VIQ − VTN ) 0.40 = 0.25 (VIQ − 1) 2 2 Then VIQ = 2.265 V r0 = r02 = b. 1 1 = = 125 kΩ λ I 0 ( 0.02 )( 0.4 ) g m = 2 K n (VIQ − VTN ) = 2 ( 0.25 )( 2.26 − 1) = 0.632 mA/V Av = − g m ( r0 || r02 ) = − ( 0.632 )(125 ||125 ) ⇒ Av = −39.5 Av = − g m ( r0 || r02 || RL ) − c. 39.4 = − ( 0.632 )( 62.5 || RL ) ⇒ 62.5 || RL = 31.25 ⇒ RL = 62.5 kΩ 2 TYU10.1 For I 0 = 0.75 mA ⎛ 2⎞ 2 ⎞ ⎛ I REF = I 0 ⎜ 1 + ⎟ = ( 0.75 ) ⎜1 + ⎟ ⎝ 100 ⎠ ⎝ β⎠ I REF = 0.765 mA I REF = R1 = V + − VBE ( on ) − V − R1 5 − 0.7 − ( −5 ) 0.765 R1 = 12.2 kΩ TYU10.2 10 − ( 0.7 )( 2 ) I REF = = 0.717 mA 12 I 0 ≈ I REF = 0.717 V 100 r0 = A = ⇒ r0 = 139.5 kΩ I 0 0.717 ΔI 0 = 1 4 ΔVCE 2 = ⇒ ΔI 0 = 0.0287 mA r0 139.5 TYU10.3 I 0 = I REF ⋅ I B3 = I0 β 1 0.50 = ⇒ I 0 = 0.4996 mA ⎛ ⎞ ⎛ 2 2 ⎞ ⎜1 + ⎜ β (1 + β ) ⎟ ⎜ 1 + 50 ( 51) ⎟ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⇒ I B 3 = 9.99 μ A ⎛1+ β ⎞ I E3 = ⎜ ⎟ I C 3 = I E 3 = 0.5096 mA ⎝ β ⎠ IE3 0.5096 IC 2 = = ⇒ I C 2 = 0.490 mA = I C1 2⎞ ⎛ 2⎞ ⎛ ⎜ 1 + ⎟ ⎜ 1 + 50 ⎟ ⎠ ⎝ β⎠ ⎝ I B1 = I B 2 = TYU10.4 IC 2 β ⇒ I B1 = I B 2 = 9.80 μ A 384. I REF = ′ ′ k n1 ⎛ W ⎞ kn 3 ⎛ W ⎞ 2 2 ⎜ ⎟ (VGS 1 − VTN 1 ) = ⎜ ⎟ ( 5 − VGS 1 − VTN 3 ) 2 ⎝ L ⎠1 2 ⎝ L ⎠3 1 ⎡⎛ 40 ⎞ ⎛ 17.3 ⎞ ⎤ ⎢⎜ 38 ⎟ ⎜ 2.70 ⎟ ⎥ (VGS1 − 0.98 ) = ( 3.98 − VGS 1 ) ⇒ VGS 1 = 1.814 V ⎠⎦ ⎣⎝ ⎠ ⎝ 2 ⎛ 0.040 ⎞ I REF = ⎜ ⎟ (17.3)(1.814 − 0.98 ) ⎝ 2 ⎠ I REF = 0.241 mA 2 ′ kn 2 ⎛ W ⎞ 2 ⎜ ⎟ (VGS 1 − VTN 2 ) 2 ⎝ L ⎠2 2 ⎛ 0.042 ⎞ =⎜ ⎟ ( 6.92 )(1.814 − 1.0 ) ⎝ 2 ⎠ I O = 0.0963 mA IO = TYU10.5 a. From Equation (10.52), ⎛ 3 3 ⎞ ⎜ 1− ⎟ 12 × 10 + ⎜ 12 ⎟ × 1.8 VGS 1 = ( ) ⎜ 3 3 ⎟ 1+ ⎜1+ ⎟ 12 12 ⎠ ⎝ ⎛ 0.5 ⎞ ⎛ 1 − 0.5 ⎞ VGS 1 = ⎜ ⎟ (10 ) + ⎜ ⎟ × (1.8 ) ⎝ 1 + 0.5 ⎠ ⎝ 1 + 0.5 ⎠ VGS 1 = 3.93 V also VDS1 = 3.93 V I REF = (12 )( 0.020 ) [3.93 − 1.8] ⎡1 + ( 0.01)( 3.93) ⎤ ⇒ I REF = 1.13 mA ⎣ ⎦ W / L )2 (1 + λVDS 2 ) ( b. I 0 = I REF × × (W / L )1 (1 + λVDS1 ) 2 ⎛ 6 ⎞ ⎡1 + ( 0.01)( 2 ) ⎤ ⎦ ⇒ I = 0.555 mA I 0 = (1.13) × ⎜ ⎟ × ⎣ 0 ⎝ 12 ⎠ ⎡1 + ( 0.01)( 3.93) ⎤ ⎣ ⎦ c. For VDS 2 = 6 V ⇒ I o = 0.576 mA TYU10.6 K n1 (VGS 1 − VTN ) = K n3 (VGS 3 − VTN ) 2 2 ⎛ 0.10 ⎞ VGS 1 − 2 = ⎜ ⎜ 0.25 ⎟ (VGS 3 − 2 ) ⎟ ⎝ ⎠ − 2 = ( 0.6325 )(VGS 3 − 2 ) VGS 1 VGS 3 = 10 − VGS 1 VGS 1 − 2 = ( 0.6325 )(10 − VGS 1 ) − ( 0.6325 )( 2 ) 1.6325VGS1 = 7.06 ⇒ VGS 1 = 4.325 V I REF = K n1 (VGS 1 − VTN ) = ( 0.25 ) ( 4.325 − 2 ) ⇒ I REF = 1.35 mA 2 2 I O = 3K n 2 (VGS 1 − VTN ) = 3 ( 0.25 )( 4.325 − 2 ) I O = 3I REF ⇒ I O = 4.05 mA 2 TYU10.7 2 385. I REF = 0.20 = K n1 (VGS 1 − VTN ) = 0.15 (VGS 1 − 1) ⇒ VGS 1 = VGS 2 = 2.15 V 2 I O = K n 2 (VGS 2 − VTN ) = 2 I O = K n 3 (VGS 3 − VTN ) 2 0.15 2 ( 2.15 − 1) ⇒ I O = 0.10 mA 2 2 0.10 = 0.15 (VGS 3 − 1) ⇒ VGS 3 = 1.82 V 2 TYU10.8 All transistors are identical ⇒ I 0 = I REF = 250 μ A I REF = K n (VGS − VTN ) 2 0.25 = 0.20 (VGS − 1) ⇒ VGS = 2.12 V 2 TYU10.9 For Q1 : iD = I DSS 1 (1 + λ vDS1 ) 2 ⎛ v ⎞ For Q2 : iD = I DSS 2 ⎜ 1 − GS 2 ⎟ (1 + λ vDS 2 ) VP ⎠ ⎝ vGS 2 = −vDS1 and vDS 2 = VDS − vDS 1 So 2 ⎡ −v ⎤ I DSS 1 (1 + λ vDS1 ) = I DSS 2 ⎢1 − DS 1 ⎥ ⎡1 + λ (VDS − vDS1 ) ⎤ ⎦ VP ⎦ ⎣ ⎣ I DSS 1 = I DSS 2 2 ⎡ v ⎤ ⎡1 + ( 0.1) vDS1 ⎤ = ⎢1 − DS1 ⎥ ⎡1 + ( 0.1) ( 3) − ( 0.1) vDS 1 ⎤ ⎣ ⎦ ⎦ 2 ⎦ ⎣ ⎣ 2 1 + 0.1vDS 1 = (1 − vDS 1 + 0.25vDS 1 ) (1.3 − 0.1vDS 1 ) 3 2 This becomes 0.025vDS 1 − 0.425vDS 1 + 1.5vDS 1 − 0.3 = 0 We find vDS 1 = 0.2127 V, vDS 2 = 2.787 V, vGS 2 = −0.2127 V iD = I DSS 1 (1 + λ vDS 1 ) = 2 ⎡1 + ( 0.1)( 0.2127 ) ⎤ ⎣ ⎦ iD = 2.04 mA R0 = r02 + r01 (1 + g m 2 r02 ) 2 I DSS ⎛ vGS 2 ⎞ 2 ( 2 ) ⎛ −0.2127 ⎞ ⎜1 − ⎟= ⎜1 − ⎟ VP ⎠ −VP ⎝ 2 ⎝ −2 ⎠ g m = 1.787 mA/V 1 1 r02 = r04 = = = 5 kΩ λ I DSS ( 0.1)( 2 ) gm = R0 = 5 + 5 ⎡1 + (1.787 )( 5 ) ⎤ ⇒ R0 = 54.7 kΩ ⎣ ⎦ TYU10.10 a. b. c. ⎛ 0.1× 10−3 ⎞ ⇒ VEB 2 = 0.557 V VEB 2 = ( 0.026 ) ln ⎜ −14 ⎟ ⎝ 5 × 10 ⎠ 5 − 0.557 R1 = ⇒ R1 = 44.4 kΩ 0.1 386. ⎛ VEC 2 ⎞ ⎜ 1+ V ⎟ ⎤ ⎛ VCE 0 ⎞ ⎞ AP ⎟ = I REF × ⎜ 1+ ⎟ ⎟⎥ ⎜ VAN ⎠ ⎜ 1 + VEB 2 ⎟ ⎠⎦ ⎝ ⎜ VAP ⎟ ⎝ ⎠ 2.5 ⎞ ⎛ ⎜ 1 + 100 ⎟ ⎡ ⎛ VI ⎞ ⎤ ⎛ 2.5 ⎞ −14 −3 5 × 10 ⎢ exp ⎜ ⎟ ⎥ ⎜ 1 + ⎟ = ( 0.1× 10 ) ⎜ 0.557 ⎟ ⎝ VT ⎠ ⎦ ⎝ 100 ⎠ ⎜ ⎟ ⎣ ⎜1+ ⎟ 100 ⎠ ⎝ ⎛V ⎞ ( 5.125 ×10−14 ) exp ⎜ VI ⎟ = 1.019 ×10−4 ⎝ T⎠ ⎡ ⎛V I S 0 ⎢ exp ⎜ I ⎝ VT ⎣ ⎛V exp ⎜ I ⎝ VT 1 − 0.026 ⇒ Av = 1 1 + 100 100 d. ⎞ 9 ⎟ = 1.988 × 10 ⇒ VI = 0.557 V ⎠ Av = −1923 TYU10.11 I REF = K p1 (VSG + VTP ) a. 2 0.25 = 0.20 (VSG − 1) ⇒ VSG = 2.12 V 2 b. VDSO From Equation (10.89) ⎡1 + λP (V + − VSG ) ⎤ K (V − V )2 TN ⎦− n I = Vo = ⎣ I REF ( λn + λP ) λn + λP 5= 1 + ( 0.015 )(10 − 2.12 ) 0.030 ( 0.2 )(VI − 1) 0.25 ( 0.030 ) 2 − 0.15 = 1.12 − 0.8 (VI − 1) ⇒ VI = 2.10 V 2 c. Av = −2 K n (VI − VTN ) I REF ( λn + λP ) Av = − 2 ( 0.2 )( 2.10 − 1.0 ) 0.25 ( 0.030 ) ⇒ Av = −58.7 TYU10.12 (a) I REF = K p1 (VSG + VTP ) 2 80 = 50 (VSG − 1) ⇒ VSG = 2.26 V 2 (b) VDSo ⎡1 + λ p (V + − VSG ) ⎤ K (V − V )2 TN ⎦− n I = Vo = ⎣ λn + λ p I REF ( λn + λ p ) ⎡1 + ( 0.015 )(10 − 2.26 ) ⎤ ( 50 )(VI − 1) ⎦− 5= ⎣ 0.030 (80 )( 0.030 ) 2 20.83 (VI − 1) = 32.2 ⇒ VI = 2.243 V 2 (c) Av = TYU10.13 a. −2 K n (VI − VTN ) I REF ( λn + λ p ) = −2 ( 50 )( 2.243 − 1) (80 )( 0.030 ) ⇒ Av = −51.8 387. gm = IC 0 0.5 = ⇒ g m = 19.2 mA/V 0.026 VT r0 = VAN 120 = ⇒ r0 = 240 kΩ I CQ 0.5 r02 = VAP 80 = ⇒ r02 = 160 kΩ I CQ 0.5 Av = − g m ( r0 || r02 || RL ) = − (19.2 ) [ 240 ||160 || 50] ⇒ Av = −631 b. TYU10.14 1 = 38.46 mA/V 0.026 (100 )( 0.026 ) rπ 1 = rπ 2 = = 2.6 K 1 80 rO1 = rO 2 = = 80 K 1 120 rO = = 120 K 1 1 80 = 0.0257 K RO1 = 2.6 38.46 For R1 = 9.3 K I C = 1mA, g m = ( RO1 + RE ) = 9.3 ( 0.0257 + 1) = 0.924 K ′′ RE = 1 [ 2.6 + 0.924] = 0.779 K RO 2 = 80 ⎡1 + ( 38.46 )( 0.779 ) ⎤ = 2476.7 K ⎣ ⎦ Av = − g m ( rO || RO 2 ) = − ( 38.46 )(120 || 2476.7 ) = − ( 38.46 )(114.5 ) R ′ = R1 Av = −4404 For RL = 100 K Av = −38.46 ⎡114.5 100 ⎤ = −2053 ⎣ ⎦ For RL = 10 K Av = −38.46 [114.5 ||10] = −354 TYU10.15 M 1 and M 2 identical ⇒ I o = I REF a. I O = K n (VI − VYN ) 2 0.25 = 0.2 (VI − 1) VI = 2.12 V g m = 2 K n (VI − VTN ) = 2 ( 0.2 )( 2.12 − 1) ⇒ g m = 0.447 mA/V 2 r0 n = r0 p = b. 1 λn I 0 = 1 ( 0.01)( 0.25 ) ⇒ r0 n = 400 kΩ 1 1 = ⇒ r0 p = 200 kΩ λ p I 0 ( 0.02 )( 0.25 ) Av = − g m ( r0 || r02 || RL ) Av = − ( 0.447 ) [ 400 || 200 ||100] ⇒ Av = −25.5 388. Chapter 10 Problem Solutions 10.1 a. I1 = I 2 = 0 − 2Vγ − V − R1 + R2 2Vγ + I 2 R2 = VBE + I C R3 R2 2Vγ + ( −2Vγ − V − ) = VBE + IC R3 R1 + R2 ⎧ ⎫ R2 ⎞ ⎪ ⎪ − ⎛ ⎨2Vγ − ( 2Vγ + V ) ⎜ ⎟ − VBE ⎬ R1 + R2 ⎠ ⎪ ⎪ ⎝ ⎩ ⎭ Vγ = VBE and R1 = R2 IC = IC = b. 1 R3 1 ⎧ 1 ⎫ − ⎨2Vγ − ( 2Vγ + V ) − VBE ⎬ R3 ⎩ 2 ⎭ or I C = c. −V − 2 R3 I C = 2 mA = − ( −10 ) 2 R3 I1 = I 2 = 2 mA = ⇒ R3 = 2.5 kΩ −2 ( 0.7 ) − ( −10 ) R1 + R2 ⇒ R1 + R2 = 4.3 kΩ ⇒ R1 = R2 = 2.15 kΩ 10.2 (a) ⎛I ⎞ VBE1 = VT ln ⎜ C1 ⎟ ⎝ IS ⎠ (i) I REF = I C1 = 10 μ A, (ii) I REF = I C1 = 100 μ A, (iii) (b) ⎛ 10 × 10−6 ⎞ VBE1 = ( 0.026 ) ln ⎜ ⎟ = 0.5388 V −14 ⎝ 10 ⎠ I O = 10 μ A ⎛ 100 × 10−6 ⎞ VBE1 = ( 0.026 ) ln ⎜ ⎟ = 0.5987 V −14 ⎝ 10 ⎠ I O = 100 μ A ⎛ 10−3 ⎞ I REF = I C1 = 1 mA, VBE1 = ( 0.026 ) ln ⎜ −14 ⎟ = 0.6585 V ⎝ 10 ⎠ I O = 1 mA IO = I REF 2 1+ β (i) IO = 10 ⇒ I O = 9.615 μ A VBE1 = VBE 2 2 1+ 50 ⎛I ⎞ = VT ln ⎜ O ⎟ ⎝ IS ⎠ ⎛ 9.615 × 10−6 ⎞ = ( 0.026 ) ln ⎜ ⎟ −14 ⎝ 10 ⎠ = 0.5378 V (ii) IO = ⎛ 96.15 × 10−6 ⎞ 100 ⇒ I O = 96.15 μ A VBE1 = ( 0.026 ) ln ⎜ ⎟ −14 2 ⎝ 10 ⎠ 1+ 50 = 0.5977 V 389. IO = (iii) 10.3 I REF = ⎛ 0.9615 × 10−3 ⎞ ⇒ I O = 0.9615 mA VBE1 = ( 0.026 ) ln ⎜ ⎟ 2 10−14 ⎝ ⎠ 1+ 50 = 0.6575 V 1 V + − VBE ( on ) − V − R1 ⇒ 0.250 = 3 − 0.7 − ( −3) R1 R1 = 21.2 K I REF 0.250 = ⇒ I C1 = I C 2 = 0.2419 mA 2 2 1+ 1+ 60 β = 4.03 μ A I C1 = I C 2 = I B1 = I B 2 10.4 I REF = V + − VBE ( on ) − V − R1 = 5 − 0.7 − ( −5 ) 18.3 I REF = 0.5082 mA I REF 0.5082 = ⇒ I C1 = I C 2 = 0.4958 mA 2 2 1+ 1+ 80 β = ( 6.198 μ A ) I C1 = I C 2 = I B1 = I B 2 10.5 (a) I REF = (b) R1 = V + − VBE ( on ) − V − R1 V + − VBE ( on ) − V − = or R1 = 15 − 0.7 − ( −15 ) 0 − 0.7 − ( −15 ) I REF 0.5 Advantage: Requires smaller resistance. (c) For part (a): 29.3 = 0.526 mA I O ( max ) = ( 58.6 )( 0.95 ) I O ( min ) = 0.5 ⇒ R1 = 28.6 k Ω 29.3 = 0.476 mA ( 58.6 )(1.05 ) ΔI O = 0.526 − 0.476 = 0.05 mA ⇒ ±5% For part (b): 14.3 = 0.526 mA I O ( max ) = ( 28.6 )( 0.95) I O ( min ) = 14.3 = 0.476 mA ( 28.6 )(1.05 ) ΔI O = 0.05 mA ⇒ ±5% 10.6 a. ⇒ R1 = 58.6 k Ω ⎛ 2⎞ 2 ⎞ ⎛ I REF = I 0 ⎜ 1 + ⎟ = 2 ⎜ 1 + ⎟ or I REF = 2.04 mA ⎝ 100 ⎠ ⎝ β⎠ 15 − 0.7 R1 = ⇒ R1 = 7.01 kΩ 2.04 390. r0 = b. VA 80 = = 40 kΩ 2 I0 ΔI 0 1 ⎛ 1 ⎞ = ⇒ ΔI 0 = ⎜ ⎟ ( 9.3) = 0.2325 mA ΔVCE r0 ⎝ 40 ⎠ ΔI 0 0.2325 ΔI = ⇒ 0 = 11.6% 2 I0 I0 10.7 I 0 = nI C1 I REF = I C1 + I B1 + I B 2 = I C1 + I C1 β + I0 β ⎛ ⎛ 1+ n ⎞ 1 n⎞ I REF = I C1 ⎜ 1 + + ⎟ = I C1 ⎜ 1 + ⎟ β ⎠ ⎝ β β⎠ ⎝ = I0 ⎛ 1 + n ⎞ nI REF ⎜1 + ⎟ or I 0 = n⎝ β ⎠ ⎛ 1+ n ⎞ ⎜1 + β ⎟ ⎝ ⎠ 10.8 IO = I REF 2 ⎞ ⎛ ⇒ I REF = ( 0.20 ) ⎜ 1 + ⎟ = 0.210 mA 2 40 ⎠ ⎝ 1+ β 5 − 0.7 4.3 = ⇒ R1 = 20.5 K R1 = I REF 0.21 10.9 a. b. 5 − 0.7 = 0.239 mA 18 0.239 I0 = ⇒ I 0 = 0.230 mA 2 1+ 50 VA 50 r0 = = = 218 kΩ I 0 0.230 I REF = 1 ⎛ 1 ⎞ ⋅ ΔVEC = ⎜ ⎟ (1.3) = 0.00597 mA ⇒ I 0 = 0.236 mA r0 ⎝ 217 ⎠ ⎛ 1 ⎞ ΔI 0 = ⎜ ⎟ ( 3.3) = 0.01516 mA ⇒ I 0 = 0.245 mA ⎝ 217 ⎠ ΔI 0 = c. 10.10 5 − 0.7 − ( −5 ) ⇒ R1 = 9.3 kΩ a. I REF = 1 = b. I 0 = 2 I REF ⇒ I 0 = 2 mA c. For VEC 2 ( min ) = 0.7 ⇒ RC 2 = 10.11 R1 5 − 0.7 ⇒ RC 2 = 2.15 kΩ 2 391. I O = 0.50 mA ⇒ I OA = I OB = 0.25 mA ⎛ 3⎞ 3 ⎞ ⎛ I REF = I OA ⎜ 1 + ⎟ = 0.25 ⎜ 1 + ⎟ ⎝ 60 ⎠ ⎝ β⎠ I REF = 0.2625 mA R1 = 2.5 − 0.7 ⇒ R1 = 6.86 K 0.2625 10.12 R1 = 10 − 0.7 = 37.2 K 0.25 10.13 I 2 = 2 I1 and I3 = 3I1 (a) I 2 = 1.0 mA, I 3 = 1.5 mA (b) I1 = 0.25 mA, I 3 = 0.75 mA (c) I1 = 0.167 mA, I 2 = 0.333 mA 10.14 a. 392. I 0 = I C1 and I REF = I C1 + I B 3 = I C1 + VBE 2 I C1 VBE = + R2 R2 β I E 3 = I B1 + I B 2 + I REF = I C1 + I REF − 2 I C1 β (1 + β ) + VBE 1 + β ) R2 ( ⎛ ⎞ VBE 2 = I 0 ⎜1 + ⎜ β (1 + β ) ⎟ ⎟ (1 + β ) R2 ⎝ ⎠ I REF − I0 = IE3 1+ β VBE 1 + β ) R2 ( ⎛ ⎞ 2 ⎜1 + ⎜ β (1 + β ) ⎟ ⎟ ⎝ ⎠ ⎛ ⎞ 2 0.7 I REF = ( 0.70 ) ⎜ 1 + ⎜ ( 80 )( 81) ⎟ + ( 81)(10 ) ⎟ ⎝ ⎠ I REF = 0.700216 + 0.000864 b. I REF = 0.7011 mA = 10 − 2 ( 0.7 ) R1 ⇒ R1 = 12.27 kΩ 10.15 a. I ES 1+ β = (1 + N ) I BR I 0i = I CR and I REF = I CR + I BS = I CR + I ES = I BR + I B1 + I B 2 + ... + I BN = (1 + N ) I CR β Then I REF = I CR + or I 0i = b. I REF ⎛ (1 + N ) ⎞ ⎜1 + ⎜ β (1 + β ) ⎟ ⎟ ⎝ ⎠ ⎡ ⎤ 6 I REF = ( 0.5 ) ⎢1 + ⎥ = 0.5012 mA ⎢ ( 50 )( 51) ⎥ ⎣ ⎦ R1 = 10.16 (1 + N ) I CR β (1 + β ) 5 − 2 ( 0.7 ) − ( −5 ) 0.5012 ⇒ R1 = 17.16 kΩ 393. ⎛ ⎞ ⎡ ⎤ 2 2 I REF = I 0 ⎜ 1 + ⎜ β (1 + β ) ⎟ = ( 0.5 ) ⎢1 + ( 50 )( 51) ⎥ ⇒ I REF = 0.5004 mA ⎟ ⎢ ⎥ ⎝ ⎠ ⎣ ⎦ 5 − 2 ( 0.7 ) − ( −5 ) R1 = ⇒ R1 = 17.19 kΩ 0.5004 10.17 I 0 = I REF ⋅ 1 ⎛ ⎞ 2 ⎜1 + ⎜ β (2 + β ) ⎟ ⎟ ⎝ ⎠ For I 0 = 0.8 mA ⎛ 2 ⎞ I REF = ( 0.8 ) ⎜ 1 + ⎜ 25 ( 27 ) ⎟ ⇒ I REF = 0.8024 mA ⎟ ⎝ ⎠ 18 − 2 ( 0.7 ) ⇒ R1 = 20.69 kΩ R1 = 0.8024 10.18 394. The analysis is exactly the same as in the text. We have 1 I 0 = I REF ⋅ ⎛ ⎞ 2 ⎜1 + ⎜ β (2 + β ) ⎟ ⎟ ⎝ ⎠ 10.19 2 = 0.0267 mA 75 1 I C1 = 1 mA, I B1 = = 0.0133 mA 75 I E 3 = I B1 + I B 2 = 0.0133 + 0.0267 = 0.04 mA I 0 = 2 mA, I B 2 = I E3 0.04 = = 0.000526 mA 1+ β 76 = I C1 + I B 3 ⇒ I REF = 1.000526 ≈ 1 mA I B3 = I REF R1 = 10 − 2 ( 0.7 ) I REF = 8.6 ⇒ R1 = 8.6 kΩ 1 10.20 (a) Assuming RO ≈ rO 3 = RO = VA V = A I O I REF β ro3 2 100 = = 400 K 0.25 (100 )( 400 ) 2 ⇒ RO = 20 MΩ (b) RO = ΔV ΔV 5 ⇒ ΔI O = = ΔI O 20 MΩ 20 MΩ ΔI O = 0.25 μ A 10.21 I REF = V + − VBE1 − V − 5 − 0.7 = 9.3 R1 I REF = 0.4624 mA VT ⎛ I REF ⎞ 0.026 ⎛ 0.4624 ⎞ ln ⎜ ln ⎜ ⎟= ⎟ 1.5 RE ⎝ I O ⎠ ⎝ IO ⎠ ⎛ 0.4624 ⎞ I O = 0.01733ln ⎜ ⎟ ⎝ IO ⎠ IO = By trial and error I O 41.7 μ A VBE 2 = 0.7 − I O RE VBE 2 = 0.7 − ( 0.0417 )(1.5 ) VBE 2 = 0.6375 V 10.22 (a) 395. I REF = V + − VBE1 − V − 5 − 0.7 − ( −5 ) = ⇒ I REF = 93 μ A 100 R1 ⎛I ⎞ ⎛ 93 × 10−3 mA ⎞ I O RE = VT ln ⎜ REF ⎟ ⇒ I O (10 ) = 0.026 ln ⎜ ⎟ IO ⎝ IO ⎠ ⎝ ⎠ By trial and error, I O ≅ 6.8 μ A ′ Ro = ro 2 (1 + g m 2 RE ) Now 30 = 4.41 M Ω 6.8 0.0068 = 0.2615 mA / V gm2 = 0.026 (100 )( 0.026 ) = 382.4 k Ω rπ 2 = 0.0068 So ′ RE = rπ 2 || RE = 382 || 10 = 9.74 k Ω Then Ro = 4.41 ⎡1 + ( 0.262 )( 9.74 ) ⎤ ⇒ Ro = 15.6 M Ω ⎣ ⎦ ro 2 = VBE1 − VBE 2 = I o RE = ( 0.0068 )(10 ) ⇒ VBE1 − VBE 2 = 0.068 V (b) 10.23 I ⋅ ΔVC R0 ΔI 0 = ′ R0 = r02 (1 + g m 2 RE ) V 80 r02 = A = = 11.76 MΩ I 0 6.8 gm2 = rπ 2 = I 0 0.0068 = = 0.2615 mA/V VT 0.026 (80 )( 0.026 ) = 306 kΩ 0.0068 ′ RE = RE rπ 2 = 10 306 = 9.68 K R0 = (11.76 ) ⎡1 + ( 0.2615 )( 9.68 ) ⎤ = 41.54 MΩ ⎣ ⎦ Now ⎛ 1 ⎞ ΔI 0 = ⎜ ⎟ ( 5 ) ⇒ ΔI 0 = 0.120 μ A ⎝ 41.54 ⎠ 10.24 (a) I REF = 5 − 0.7 − ( −5 ) R1 = 0.50 R1 = 18.6 K ⎛I ⎞ I O RE = VT ln ⎜ REF ⎟ ⎝ IO ⎠ 0.026 ⎛ 0.50 ⎞ RE = ln ⎜ ⎟ 0.050 ⎝ 0.050 ⎠ RE = 1.20 K 396. (b) ′ RO = rc 2 [1 + RE g m 2 ] ′ RE = RE rπ 2 rπ 2 = ( 75 )( 0.026 ) = 39 K 0.050 VA 100 ro 2 = = ⇒ 2 MΩ I O 0.05 (c) gm2 = 0.050 = 1.923 mA/V 0.026 ′ RE = 1.20 39 = 1.164 K RO = 2 ⎡1 + (1.164 )(1.923) ⎤ ⇒ RO = ( 6.477 ) MΩ ⎣ ⎦ ΔV 5 ΔI O = = = 0.772 μ A RO 6.477 ΔI O 0.772 × 100% = × 100 = 1.54% IO 50 10.25 Let R1 = 5 k Ω, Then I REF = 12 − 0.7 − ( −12 ) 5 ⇒ I REF = 4.66 mA Now ⎛I ⎞ 0.026 ⎛ 4.66 ⎞ I O RE = VT ln ⎜ REF ⎟ ⇒ RE = ln ⎜ ⎟ ⇒ RE ≅ 1 k Ω 0.10 ⎝ 0.10 ⎠ ⎝ IO ⎠ 10.26 ⎛I ⎞ VBE = VT ln ⎜ REF ⎟ ⎝ IS ⎠ ⎛ 10−3 ⎞ −15 0.7 = ( 0.026 ) ln ⎜ ⎟ ⇒ I S = 2.03 × 10 A IS ⎠ ⎝ ⎛ 2 × 10−3 ⎞ At 2 mA, VBE = ( 0.026 ) ln ⎜ −15 ⎟ ⎝ 2.03 × 10 ⎠ = 0.718 V 15 − 0.718 ⇒ R1 = 7.14 kΩ 2 ⎛ I ⎞ 0.026 ⎛ 2 ⎞ V RE = T ln ⎜ REF ⎟ = ⋅ ln ⎜ ⎟ ⇒ RE = 1.92 kΩ I 0 ⎝ I 0 ⎠ 0.050 ⎝ 0.050 ⎠ R1 = 10.27 a. 10 − 0.7 = 0.465 mA 20 Let V − = 0 I REF ≈ ⎛I ⎞ VBE ≅ VT ln ⎜ REF ⎟ ⎝ IS ⎠ ⎛ 10−3 ⎞ −15 0.7 = ( 0.026 ) ln ⎜ ⎟ ⇒ I S = 2.03 ×10 A IS ⎠ ⎝ Then ⎛ 0.465 × 10−3 ⎞ = 0.680 V VBE ≅ ( 0.026 ) ln ⎜ −15 ⎟ ⎝ 2.03 × 10 ⎠ Then 397. I REF ≅ b. 10.28 I REF ≈ 10 − 0.680 ⇒ I REF = 0.466 mA 20 RE = VT ⎛ I REF ln ⎜ I0 ⎝ I0 10 − 0.7 − ( −10 ) 40 ⎞ 0.026 ⎛ 0.466 ⎞ ⋅ ln ⎜ ⎟= ⎟ ⇒ RE = 400Ω 0.10 ⎝ 0.10 ⎠ ⎠ = 0.4825 mA ⎛I ⎞ VBE ≅ VT ln ⎜ REF ⎟ ⎝ IS ⎠ ⎛ 10−3 ⎞ −15 0.7 = ( 0.026 ) ln ⎜ ⎟ ⇒ I S = 2.03 × 10 A ⎝ IS ⎠ Now ⎛ 0.4825 × 10−3 ⎞ VBE = ( 0.026 ) ln ⎜ = 0.681 V −15 ⎟ ⎝ 2.03 × 10 ⎠ VBE1 = 0.681 V So 10 − 0.681 − ( −10 ) ⇒ I REF = 0.483 mA 40 ⎛I ⎞ I 0 RE = VT ln ⎜ REF ⎟ ⎝ I0 ⎠ ⎛ 0.483 ⎞ I 0 (12 ) = ( 0.026 ) ln ⎜ ⎟ ⎝ I0 ⎠ I REF ≅ By trial and error. ⇒ I 0 ≅ 8.7 μ A VBE 2 = VBE1 − I 0 RE = 0.681 − ( 0.0087 )(12 ) ⇒ VBE 2 = 0.5766 V 10.29 VBE1 + I REF RE1 = VBE 2 + I 0 RE 2 VBE1 − VBE 2 = I 0 RE 2 − I REF RE1 For matched transistors ⎛I ⎞ VBE1 = VT ln ⎜ REF ⎟ ⎝ IS ⎠ ⎛I ⎞ VBE 2 = VT ln ⎜ 0 ⎟ ⎝ IS ⎠ ⎛I ⎞ Then VT ln ⎜ REF ⎟ = I 0 RE 2 − I REF RE1 ⎝ I0 ⎠ Output resistance looking into the collector of Q2 is increased. 10.30 V + − VBE1 − V − 5 − 0.7 − ( −5 ) = = 0.3174 mA R1 + RE1 27.3 + 2 (a) I REF = (b) I O = I REF = 0.3174 mA Using the same relation as for the widlar current source. 398. RO = ro 2 ⎡1 + g m 2 ( RE rπ 2 ) ⎤ ⎣ ⎦ ro 2 = rπ 2 = VA 80 = = 252 K I O 0.3174 gm2 = 0.3174 = 12.21 mA/V 0.026 (100 )( 0.026 ) = 8.192 K RE || rπ 2 = 2 || 8.192 = 1.608 K 0.3174 RO = 252 ⎡1 + (12.21)(1.608 ) ⎤ ⇒ RO = 5.2 MΩ ⎣ ⎦ (c) I O = I REF = 5 − 0.7 − ( −5 ) = 0.3407 mA 27.3 V 80 RO = ro 2 = A = ⇒ RO = 235 K I O 0.3407 10.31 Assume all transistors are matched. a. 2VBE1 = VBE 3 + I 0 RE ⎛I ⎞ VBE1 = VT ln ⎜ REF ⎟ ⎝ IS ⎠ ⎛I ⎞ VBE 3 = VT ln ⎜ 0 ⎟ ⎝ IS ⎠ ⎛I ⎞ ⎛I ⎞ 2VT ln ⎜ REF ⎟ − VT ln ⎜ 0 ⎟ = I 0 RE IS ⎠ ⎝ ⎝ IS ⎠ ⎡ ⎛I ⎞ ⎛I VT ⎢ ln ⎜ REF ⎟ − ln ⎜ 0 ⎢ ⎝ IS ⎠ ⎝ IS ⎣ 2 ⎞⎤ ⎟ ⎥ = I 0 RE ⎠⎥ ⎦ ⎛ I 2 REF ⎞ VT ln ⎜ ⎟ = I 0 RE ⎝ I0 I S ⎠ b. ⎛ 0.7 ⎞ −15 VBE = 0.7 V at 1 mA ⇒ 10−3 = I S exp ⎜ ⎟ or I S = 2.03 × 10 A ⎝ 0.026 ⎠ ⎛ 0.1× 10−3 ⎞ VBE at 0.1 mA ⇒ VBE = ( 0.026 ) ln ⎜ = 0.640 V −15 ⎟ ⎝ 2.03 × 10 ⎠ 0.640 Since I 0 = I REF , then VBE = I 0 RE ⇒ RE = or RE = 6.4 kΩ 0.1 10.32 (a) I REF = 5 − 0.7 − ( −5 ) R1 = 0.80 mA R1 = 11.6 K RE 2 = 0.026 ⎛ 0.80 ⎞ ln ⎜ ⎟ ⇒ RE 2 = 1.44 K 0.050 ⎝ 0.050 ⎠ RE 3 = 0.026 ⎛ 0.80 ⎞ ln ⎜ ⎟ ⇒ RE 2 = 4.80 K 0.020 ⎝ 0.020 ⎠ (b) VBE 2 = 0.7 − ( 0.05 )(1.44 ) ⇒ VBE 2 = 0.628 V VBE 3 = 0.7 − ( 0.02 )( 4.80 ) ⇒ VBE 3 = 0.604 V 399. 10.33 (a) VBE1 = VBE 2 I REF = V + − 2VBE1 − V − R1 + R2 Now 2VBE1 + I REF R2 = VBE 3 + I O RE or I O RE = 2VBE1 − VBE 3 + I REF R2 We have ⎛I ⎞ ⎛I ⎞ VBE1 = VT ln ⎜ REF ⎟ and VBE 3 = VT ln ⎜ O ⎟ ⎝ IS ⎠ ⎝ IS ⎠ (b) Let R1 = R2 and I O = I REF ⇒ VBE1 = VBE 3 ≡ VBE Then VBE = I O RE − I REF R2 = I O ( RE − R2 ) so I REF = I O = = V + − V − − 2 I O ( RE − R2 ) 2 R2 ⎛R ⎞ V −V − IO ⎜ E ⎟ + IO 2 R2 ⎝ R2 ⎠ + − Then IO = V + −V − 2 Rε (c) Want I O = 0.5 mA So RE = 2 R2 = 5 − ( −5 ) 2 ( 0.5 ) ⇒ RE = 10 k Ω 5 − 2 ( 0.7 ) − ( −5 ) 0.5 Then R1 = R2 = 8.6 k Ω = 17.2 k Ω 10.34 a. 20 − 0.7 − 0.7 = 1.55 mA 12 I 01 = 2 I REF = 3.1 mA I 02 = I REF = 1.55 mA I 03 = 3I REF = 4.65 mA b. VCE1 = − I 01 RC1 − ( −10 ) = − ( 3.1)( 2 ) + 10 ⇒ VCE1 = 3.8 V I REF = VEC 2 = 10 − I 02 RC 2 = 10 − (1.55 )( 3) ⇒ VEC 2 = 5.35 V VEC 3 = 10 − I 03 RC 3 = 10 − ( 4.65 )(1) ⇒ VEC 3 = 5.35 V 10.35 a. Ist approximation 400. 20 − 1.4 = 2.325 mA 8 ⎛ 2.32 ⎞ Now VBE − 0.7 = ( 0.026 ) ln ⎜ ⎟ ⇒ VBE = VEB = 0.722 V ⎝ 1 ⎠ Then 2nd approximation 20 − 2 ( 0.722 ) I REF ≅ = 2.32 mA 8 I 01 = 2 I REF = 4.64 mA I 02 = I REF = 2.32 mA I 03 = 3I REF = 6.96 mA b. At the edge of saturation, VCE = VBE = 0.722 V I REF ≅ RC 2 RC 3 0 − 0.722 − ( −10 ) ⇒ RC1 = 2.0 kΩ 4.64 10 − 0.722 ⇒ RC 2 = 4.0 kΩ = 2.32 10 − 0.722 ⇒ RC 3 = 1.33 kΩ = 6.96 RC1 = 10.36 I C1 = I C 2 = 10 − 0.7 − 0.7 − ( −10 ) 10 = 1.86 mA I C 3 = I C 4 = 1.86 mA ⎛ 1.86 ⎞ I C 5 ( 0.5 ) = 0.026 ln ⎜ ⎟ ⎝ IC 5 ⎠ By Trial and error. ⇒ I C 5 = 0.136 mA = I C 6 = I C 7 2 I C 3 ( 0.8 ) + VCE 3 = 10 ⇒ VCE 3 = 10 − 2 (1.86 )( 0.8 ) VCE 3 = 7.02 V 5 = VEB 6 + VCE 5 + I C 5 ( 0.5 ) − 10 VCE 5 = 5 + 10 − 0.7 − ( 0.136 )( 0.5 ) VCE 5 = 14.2 V 5 = VEC 7 + I C 7 ( 0.8 ) VEC 7 = 5 − ( 0.136 )( 0.8 ) VEC 7 = 4.89 V 10.37 I C1 = I C 2 = 10 − 0.7 − 0.7 − ( −10 ) 10 ⇒ I C1 = I C 2 = 1.86 mA I C 4 = I C 5 = 1.86 mA ⎛I ⎞ ⎛ 1.86 ⎞ I C 3 RE1 = VT ln ⎜ C1 ⎟ ⇒ I C 3 ( 0.3) = 0.026 ln ⎜ ⎟ ⎝ IC 3 ⎠ ⎝ IC 3 ⎠ By trial and error I C 3 = 0.195 mA ⎛I ⎞ ⎛ 1.86 ⎞ I C 6 RE 2 = VT ln ⎜ C 5 ⎟ ⇒ I C 6 ( 0.5 ) = 0.026 ln ⎜ ⎟ ⎝ IC 6 ⎠ ⎝ IC 6 ⎠ By trial and error I C 6 = 0.136 mA 401. 10.38 10 − 0.7 = 1 mA 6.3 + 3 VBE ( QR ) = 0.7 V as assumed VRER = I REF ⋅ RER = (1)( 3) = 3 V I REF = VRE1 = 3 V ⇒ RE1 = VRE1 3 = ⇒ RE1 = 3 kΩ I 01 1 VRE 2 = 3 V ⇒ RE 2 = VRE 2 3 = ⇒ RE 2 = 1.5 kΩ I 02 2 VRE 3 = 3 V ⇒ RE 3 = VRE 3 3 = ⇒ RE 3 = 0.75 kΩ I 03 4 I 01 = 1 mA I 02 = 2 mA I 03 = 4 mA 10.38 VDS 2 ( sat ) = 2 V = VGS 2 − VTN 2 = VGS 2 − 1.5 ⇒ VGS 2 = 3.5 V 2 ⎛1 ⎞⎛W ⎞ I O = ⎜ μ n Cox ⎟ ⎜ ⎟ (VGS 2 − VTN 2 ) 2 L ⎠2 ⎝ ⎠⎝ 2 ⎛W ⎞ ⎛W ⎞ 250 = ( 20 ) ⎜ ⎟ ( 3.5 − 1.5 ) ⇒ ⎜ ⎟ = 3.125 ⎝ L ⎠2 ⎝ L ⎠2 ⎛1 ⎞⎛W ⎞ I REF = ⎜ μ n Cox ⎟ ⎜ ⎟ (VGS 2 − VTN 1 ) ⎝2 ⎠ ⎝ L ⎠1 2 ⎛W ⎞ ⎛W ⎞ 100 = ( 20 ) ⎜ ⎟ ( 3.5 − 1.5 ) ⇒ ⎜ ⎟ = 1.25 L ⎠2 ⎝ ⎝ L ⎠1 Now VGS 3 = 10 − VGS 2 = 10 − 3.5 = 6.5 V 2 ⎛W ⎞ ⎛W ⎞ So 100 = ( 20 ) ⎜ ⎟ ( 6.5 − 1.5 ) ⇒ ⎜ ⎟ = 0.2 ⎝ L ⎠3 ⎝ L ⎠3 10.39 I REF = 2.5 − VGS ⎛ 0.08 ⎞ 2 =⎜ ⎟ ( 6 )(VGS − 0.5 ) 15 ⎝ 2 ⎠ 2 2.5 − VGS = 3.6 (VGS − VGS + 0.25 ) 2 3.6VGS − 2.6VGS − 1.6 = 0 VGS = 2.6 ± 6.76 + 23.04 2 ( 3.6 ) VGS = 1.12 V (1.1193) 2.5 − 1.1193 ⇒ I REF = 92.0 μ A ( 92.05 ) I REF = 15 I o = 92.0 μ A VDS 2 ( sat ) = VGS − VTN = 1.1193 − 0.5 VDS 2 ( sat ) = 0.619 V 10.39 a. From Equation (10.50), 402. VGS1 = VGS 2 VGS 1 = VGS 2 ⎛ ⎛ 5 ⎞ 5 ⎞ ⎜ ⎟ ⎜ 1− ⎟ 25 ⎟ 5 + ⎜ 25 ⎟ 0.5 =⎜ ( ) ( ) ⎜ ⎜ 5 ⎟ 5 ⎟ 1+ 1+ ⎜ ⎟ ⎜ ⎟ 25 ⎠ 25 ⎠ ⎝ ⎝ ⎛ 0.447 ⎞ ⎛ 1 − 0.447 ⎞ =⎜ ⎟ (5) + ⎜ ⎟ ( 0.5 ) 1 + 0.447 ⎠ ⎝ ⎝ 1 + 0.447 ⎠ = 1.74 V I REF ≅ K n1 (VGS 1 − VTN ) = (18 )( 25 )(1.74 − 0.5 ) ⇒ I REF = 0.692 mA 2 2 2 ⎛1 ⎞⎛W ⎞ I O = ⎜ μ n Cox ⎟ ⎜ ⎟ (VGS 2 − VTN ) (1 + λVDS 2 ) 2 L ⎠2 ⎝ ⎠⎝ b. I 0 = (18 )(15 )(1.74 − 0.5 ) ⎡1 + ( 0.02 )( 2 ) ⎤ ⎣ ⎦ = ( 415 )(104 ) ⇒ I 0 = 0.432 mA 2 I 0 = ( 415 ) ⎡1 + ( 0.02 )( 4 ) ⎤ ⇒ I 0 = 0.448 mA ⎣ ⎦ c. 10.40 (a) 2 ⎛ 80 ⎞ ⎛ W ⎞ I REF = 50 = ⎜ ⎟ ⎜ ⎟ (VGS − 0.5 ) ⎝ 2 ⎠ ⎝ L ⎠1 2.0 − VGS I REF = 0.050 = R Design such that VDS 2 ( sat ) = 0.25 = VGS − 0.5 VGS = 0.75 V 2 − 0.75 ⇒ R = 25 K R 2 ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ 50 = ⎜ ⎟ ⎜ ⎟ ( 0.75 − 0.5 ) ⇒ ⎜ ⎟ = 20 ⎝ 2 ⎠ ⎝ L ⎠1 ⎝ L ⎠1 So 0.050 = ⎛W ⎞ ⎜ ⎟ 20 50 ⎛W ⎞ ⎝ L ⎠1 I REF = ⇒ = ⇒ ⎜ ⎟ = 40 100 ⎝ L ⎠ 2 IO ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ L ⎠2 ⎝ L ⎠2 1 1 RO = = ⇒ RO = 667 K (b) λ I O ( 0.015 )( 0.1) (c) ΔI O = ΔV 1 = ⇒ 1.5 μ A RO 666 ΔI O ⎛ 1.5 ⎞ × 100% = ⎜ ⎟ × 100% ⇒ 1.5% IO ⎝ 100 ⎠ 10.41 (a) 2 ⎛ 80 ⎞ I REF = 250 = ⎜ ⎟ ( 3)(VGS − 1) 2⎠ ⎝ VGS = 2.44 V I O = 250 μ A at VDS 2 = VGS = 2.44 V 1 1 = = 200 K RO = λ I O ( 0.02 )( 0.25 ) 403. ΔI O = (i) ΔV 3 − 2.44 = ⇒ 2.8 μ A RO 200 I O = 252.8 μ A ΔI O = (ii) ΔV 4.5 − 2.44 = ⇒ 10.3 μ A RO 200 I O = 260.3 μ A ΔI O = (iii) ΔV 6 − 2.44 = ⇒ 17.8 μ A RO 200 I O = 267.8 μ A 4.5 ( 250 ) = 375 μ A at VDS = 2.44 V 3 1 1 RO = = = 133.3 K λ I O ( 0.02 )( 0.375 ) IO = (b) ΔI O = (i) ΔV 3 − 2.44 = ⇒ 4.20 μ A RO 133.3 I O = 379.2 μ A ΔI O = (ii) ΔV 4.5 − 2.44 = ⇒ 15.5 μ A RO 133.3 I O = 390.5 μ A ΔI O = (iii) ΔV 6 − 2.44 = ⇒ 26.7 μ A RO 133.3 I O = 401.7 μ A 10.41 VSD 2 ( sat ) = 0.25 = VSG + VTP = VSG − 0.4 ⇒ VSG 2 = 0.65 V k′ ⎛ W ⎞ 2 p I O = ⎜ ⎟ (VSG 2 + VTP ) 2 ⎝ L ⎠2 40 ⎛ W ⎞ 2 ⎛W ⎞ 25 = ⎜ ⎟ ( 0.65 − 0.4 ) ⇒ ⎜ ⎟ = 20 2 ⎝ L ⎠2 ⎝ L ⎠2 I REF = 75 μ A = (W /L )1 ⎛W ⎞ ⋅ I O ⇒ ⎜ ⎟ = 60 (W /L )2 ⎝ L ⎠1 k′ ⎛W ⎞ 2 p ⎜ ⎟ (VSG 3 + VTP ) 2 ⎝ L ⎠3 = 3 − 0.65 = 2.35 V I REF = VSG 3 Then 75 = 40 ⎛ W ⎞ 2 ⎛W ⎞ ⎜ ⎟ ( 2.35 − 0.4 ) ⇒ ⎜ ⎟ = 0.986 2 ⎝ L ⎠3 ⎝ L ⎠3 10.42 (a) VGS = VTN 1 + I REF 0.5 = 1+ =2V K n1 0.5 ⎛ I I O = K n 2 ⎜ REF ⎜ K n1 ⎝ 2 ⎞ ⎛ I REF ⎞ ⎟ = Kn2 ⎜ ⎟ ⎟ ⎝ K n1 ⎠ ⎠ 404. ⎛ 0.5 ⎞ I 0 ( max ) = ( 0.5 )(1.05 ) ⎜ ⎟ ⇒ I 0 ( max ) = 0.525 mA ⎝ 0.5 ⎠ ⎛ 0.5 ⎞ I 0 ( min ) = ( 0.5 )( 0.95 ) ⎜ ⎟ ⇒ I 0 ( min ) = 0.475 mA ⎝ 0.5 ⎠ So 0.475 ≤ I 0 ≤ 0.525 mA (b) ⎡ I ⎤ I O = K n 2 ⎢ REF + VTN 1 − VTN 2 ⎥ ⎢ K n1 ⎥ ⎣ ⎦ 2 2 ⎡ 0.5 ⎤ I 0 ( min ) = ( 0.5 ) ⎢ + 1 − 1.05⎥ ⇒ I 0 (min) = 0.451 mA ⎣ 0.5 ⎦ 2 ⎡ 0.5 ⎤ I 0 ( max ) = ( 0.5 ) ⎢ + 1 − 0.95⎥ ⇒ I 0 (max) = 0.551 mA ⎣ 0.5 ⎦ So 0.451 ≤ I 0 ≤ 0.551 mA 10.43 (1) (2) Ix = Vx − VA + g mVgs 2 ro Ix = VA + g mVgs1 ro Vgs1 = Vx , Vgs 2 = −VA So (1) Ix = ⎛1 ⎞ Vx − VA ⎜ + g m ⎟ ro ⎝ ro ⎠ (2) Ix = VA + g mVx ⇒ VA = ro [ I x − g mVx ] ro 405. Then Ix = ⎛1 ⎞ Vx − ro ( I x − g mVx ) ⎜ + g m ⎟ ro ⎝ ro ⎠ Ix = ⎡I ⎤ Vx g 2 − ro ⎢ x + g m I x − m ⋅ Vx − g mVx ⎥ ro ro ⎣ ro ⎦ Ix = Vx 2 − I x − g m ro I x + g mVx + g m roVx ro ⎡1 ⎤ 2 I x [ 2 + g m ro ] = Vx ⎢ + g m + g m ro ⎥ ro ⎣ ⎦ 1 Since g m >> ro I x [ 2 + g m ro ] ≅ Vx ( g m )(1 + g m ro ) Then Vx 2 + g m ro = Ro = Ix g m (1 + g m ro ) Usually, g m ro >> 2, so that Ro ≅ 1 gm 10.44 VDS 2 (sat) = 2 = VGS 2 − 0.8 ⇒ VGS 2 = 2.8 V I O = 200 = 60 ⎛ W ⎞ 2 ⎛W ⎞ ⎜ ⎟ ( 2.8 − 0.8 ) ⇒ ⎜ ⎟ = 1.67 2 ⎝ L ⎠2 ⎝ L ⎠2 ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ I REF ⎜ L ⎟1 0.4 ⎝ L ⎠1 ⎛W ⎞ ⎝ ⎠ = ⇒ = ⇒ ⎜ ⎟ = 3.33 IO 0.2 (1.67 ) ⎝ L ⎠1 ⎛W ⎞ ⎜ ⎟ L ⎠2 ⎝ VGS 3 = 6 − 2.8 = 3.2 V 2 ⎛ 60 ⎞ ⎛ W ⎞ ⎛W ⎞ I REF = 400 = ⎜ ⎟ ⎜ ⎟ ( 3.2 − 0.8 ) ⇒ ⎜ ⎟ = 2.31 ⎝ 2 ⎠ ⎝ L ⎠3 ⎝ L ⎠3 10.45 (a) 2 2 ⎛ 60 ⎞ ⎛ 60 ⎞ I REF = ⎜ ⎟ ( 20 )(VGS 1 − 0.7 ) = ⎜ ⎟ ( 3)(VGS 3 − 0.7 ) ⎝ 2 ⎠ ⎝ 2 ⎠ VGS1 + VGS 3 = 5 20 (VGS1 − 0.7 ) = 5 − VGS1 − 0.7 3 3.582VGS 1 = 6.107 ⇒ VGS1 = VGS 2 = 1.705 V 2 ⎛ 60 ⎞ I O = ⎜ ⎟ (12 )(1.705 − 0.7 ) = 363.6 μ A at VDS 2 = 1.705 V ⎝ 2 ⎠ 2 ⎛ 60 ⎞ I REF = ⎜ ⎟ ( 20 )(1.705 − 0.7 ) = 606 μ A ⎝ 2 ⎠ (b) RO = ΔI O = 1 λ IO = 1 = 183.4 K 0.015 )( 0.3636 ) ( ΔV 1.5 − 1.705 = ⇒ −1.12 μ A RO 183.4 I O = 362.5 μ A 406. (c) ΔI O = ΔV 3 − 1.705 = ⇒ 7.06 μ A RO 183.4 I O = 370.7 μ A 10.46 2 2 ⎛ 50 ⎞ ⎛ 50 ⎞ I REF = ⎜ ⎟ (15 )(VSG1 − 0.5 ) = ⎜ ⎟ ( 3)(VSG 3 − 0.5 ) ⎝ 2⎠ ⎝ 2⎠ VSG1 + VSG 3 = 10 ⇒ VSG 3 = 10 − VSG1 15 (VSG1 − 0.5) = 10 − VSG1 − 0.5 3 3.236VSG1 = 10.618 ⇒ VSG1 = 3.28 V 2 ⎛ 50 ⎞ I REF = ⎜ ⎟ (15 )( 3.28 − 0.5 ) ⇒ I REF = 2.90 mA ⎝ 2⎠ I O = I REF = 2.90 mA VSD 2 (sat) = VSG 2 + VTP = 3.28 − 0.5 ⇒ VSD 2 (sat) = 2.78 V 10.47 VSD 2 (sat) = 1.2 = VSG 2 − 0.35 ⇒ VSG 2 = 1.55 V 2 ⎛ 50 ⎞⎛ W ⎞ ⎛W ⎞ I O = 100 = ⎜ ⎟⎜ ⎟ (1.55 − 0.35 ) ⇒ ⎜ ⎟ = 2.78 ⎝ 2 ⎠⎝ L ⎠ 2 ⎝ L ⎠2 W W I REF 200 ⎛W ⎞ L1 L1 = ⇒ = ⇒ ⎜ ⎟ = 5.56 W IO 100 2.78 ⎝ L ⎠1 L 2 VSG1 + VSG 3 = 4 ⇒ VSG 3 = 2.45 V ( ) ( ) ( ) 2 ⎛ 50 ⎞⎛ W ⎞ ⎛W ⎞ I REF = 200 = ⎜ ⎟⎜ ⎟ ( 2.45 − 0.35 ) ⇒ ⎜ ⎟ = 1.81 ⎝ 2 ⎠⎝ L ⎠3 ⎝ L ⎠3 10.48 2 2 ⎛ 80 ⎞ ⎛ 80 ⎞ I REF = ⎜ ⎟ ( 25 )(VSG1 − 1.2 ) = ⎜ ⎟ ( 4 )(VSG 3 − 1.2 ) 2⎠ 2⎠ ⎝ ⎝ 10 − VSG1 VSG1 + 2VSG 3 = 10 ⇒ VSG 3 = 2 10 − VSG1 25 − 1.2 Then (VSG1 − 1.2 ) = 4 2 3VSG1 = 6.8 ⇒ VSG1 = 2.27 V 2 ⎛ 80 ⎞ I REF = ⎜ ⎟ ( 25 )( 2.267 − 1.2 ) ⇒ I REF = I O = 1.14 mA ⎝ 2⎠ VSD 2 (sat) = VSG 2 + VTP = 2.27 − 1.2 ⇒ VSD 2 ( sat ) = 1.07 V 10.49 VSD 2 (sat) = 1.8 = VSG 2 − 1.4 ⇒ VSG 2 = 3.2 V 2 ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ I O = ⎜ ⎟ ⎜ ⎟ ( 3.2 − 1.4 ) = 100 ⇒ ⎜ ⎟ = 0.772 ⎝ 2 ⎠ ⎝ L ⎠2 ⎝ L ⎠2 W I REF L1 = W IO L 2 W 200 ⎛W ⎞ L1 = ⇒ ⎜ ⎟ = 1.54 100 0.772 ⎝ L ⎠1 ( ) ( ) ( ) 407. Assume M3 and M4 are matched. 10 − 3.2 = 3.4 V 2VSG 3 + VSG1 = 10 ⇒ VSG 3 = 2 2 ⎛ 80 ⎞ ⎛ W ⎞ I REF = 200 = ⎜ ⎟ ⎜ ⎟ ( 3.4 − 1.4 ) 2 ⎠ ⎝ L ⎠3,4 ⎝ ⎛W ⎞ ⎜ ⎟ = 1.25 ⎝ L ⎠3,4 10.50 (a) ⎛ k′ ⎞⎛ W ⎞ 2 p I REF = ⎜ ⎟ ⎜ ⎟ (VSG1 + VTP ) ⎝ 2 ⎠ ⎝ L ⎠1 ⎛ k′ ⎞⎛ W ⎞ 2 p = ⎜ ⎟ ⎜ ⎟ (VSG 3 + VTP ) 2 ⎠ ⎝ L ⎠3 ⎝ But VSG 3 = 3 − VSG1 So 25 (VSG1 − 0.4 ) = 5 ( 3 − VSG1 − 0.4 ) 2 2 which yields VSG1 = 1.08 V and VSG 3 = 1.92 V I REF = 20 ( 25 )(1.08 − 0.4 ) ⇒ I REF = 231 μ A 2 (W / L )2 15 = = 0.6 I REF (W / L )1 25 Then I O = ( 0.6 )( 231) = 139 μ A IO = (b) VDS 2 ( sat ) = 1.08 − 0.4 = 0.68 V VR = 3 − 0.68 = 2.32 = I O R then R= 2.32 ⇒ R = 16.7 k Ω 0.139 10.51 VSD 2 (sat) = 0.35 = VSG 2 − 0.4 ⇒ VSG 2 = 0.75 V 2 ⎛W ⎞ ⎛W ⎞ I O = 80 = ( 20 ) ⎜ ⎟ ( 0.75 − 0.4 ) ⇒ ⎜ ⎟ = 32.7 L ⎠2 ⎝ ⎝ L ⎠2 ( ) ( ) ( ) W W I REF 50 L1 L1 = ⇒ = ⇒ W = 20.4 L1 W 80 32.7 IO L 2 VSG 3 = 3 − 0.75 = 2.25 ( ) 2 ⎛W ⎞ ⎛W ⎞ I REF = 50 = ( 20 ) ⎜ ⎟ ( 2.25 − 0.4 ) ⇒ ⎜ ⎟ = 0.730 ⎝ L ⎠3 ⎝ L ⎠3 10.52 a. I REF = K n (VGS − VTN ) 2 100 = 100 (VGS − 2 ) ⇒ VGS = 3 V 2 For VD 4 = −3 V, I 0 = 100 μ A 408. R0 = r04 + r02 (1 + g m r04 ) b. r02 = r04 = 1 1 = = 500 kΩ λ I 0 ( 0.02 )( 0.1) g m = 2 K n (VGS − VTN ) = 2 ( 0.1)( 3 − 2 ) = 0.2 mA / V R0 = 500 + 500 ⎡1 + ( 0.2 )( 500 ) ⎤ ⎣ ⎦ R0 = 51 MΩ ΔI 0 = 1 6 ⋅ ΔVD 4 = ⇒ ΔI 0 = 0.118 μ A 51 R0 10.53 Vgs 4 = − I X r02 VS 6 = ( I X − g mVgs 4 ) r04 + I X r02 = ( I X + g m I X r02 ) r04 + I X r02 VS 6 = I X ⎡ r02 + (1 + g m r02 ) r04 ⎤ = −Vgs 6 ⎣ ⎦ ⎛ V − VS 6 VX 1 ⎞ I X = g mVgs 6 + X = − VS 6 ⎜ g m + ⎟ r06 r06 r06 ⎠ ⎝ ⎛ 1 ⎞ ⎡ ⎤ ⎜ g m + ⎟ ⎣ r02 + (1 + g m r02 ) r04 ⎦ r06 ⎠ ⎝ ⎧ ⎛ ⎫ V 1 ⎞ ⎪ ⎪ I X ⎨1 + ⎜ g m + ⎟ ⎡ r02 + (1 + g m r02 ) r04 ⎤ ⎬ = X ⎣ ⎦ r06 ⎠ ⎪ ⎝ ⎪ r06 ⎩ ⎭ VX = R0 = r06 + (1 + g m r06 ) ⎡ r02 + (1 + g m r02 ) r04 ⎤ ⎣ ⎦ IX IX = VX − IX r06 I 0 ≈ I REF = 0.2 mA = 0.2 (VGS − 1) 2 VGS = 2 V g m = 2 K n (VGS − VTN ) = 2 ( 0.2 )( 2 − 1) = 0.4 mA / V r02 = r04 = r06 = 1 = 1 = 250 kΩ ( 0.02 )( 0.2 ) R0 = 250 + ⎡1 + ( 0.4 )( 250 ) ⎤ × {250 + ⎡1 + ( 0.4 )( 250 ) ⎤ ( 250 )} ⎣ ⎦ ⎣ ⎦ λ I0 R0 = 2575750 kΩ ⇒ R0 = 2.58 × 109 Ω 409. 10.54 ′ ′ kn ⎛ W ⎞ kn ⎛ W ⎞ 2 2 ⎜ ⎟ (VGS1 − VTN ) = ⎜ ⎟ (VGS 3 − VTN ) 2 ⎝ L ⎠1 2 ⎝ L ⎠3 k′ ⎛ W ⎞ 2 p = ⎜ ⎟ (VGS 4 + VTP ) 2 ⎝ L ⎠4 (1) 50 ( 20 )(VGS 1 − 0.5 ) = 50 ( 5 )(VGS 3 − 0.5 ) 2 2 (2) 50 ( 20 )(VGS 1 − 0.5 ) = 20 (10 )(VGS 4 − 0.5 ) 2 2 (3) VSG 4 + VGS 3 + VGS1 = 6 From (1) 4 (VGS 1 − 0.5 ) = (VGS 3 − 0.5 ) ⇒ VGS 3 = 2 (VGS 1 − 0.5 ) + 0.5 2 2 From (2) 5 (VGS1 − 0.5 ) = (VGS 4 − 0.5 ) ⇒ VSG 4 = 5 (VGS1 − 0.5 ) + 0.5 2 2 Then (3) becomes 5 (VGS1 − 0.5 ) + 0.5 + 2 (VGS 1 − 0.5 ) + 0.5 + VGS1 = 6 which yields VGS 1 = 1.36 V and VGS 3 = 2.22 V , VSG 4 = 2.42 V k′ ⎛ W ⎞ 2 2 Then I REF = n ⎜ ⎟ (VGS1 − VTN ) = 50 ( 20 )(1.36 − 0.5 ) or I REF = I O = 0.740 mA 2 ⎝ L ⎠1 VGS 1 = VGS 2 = 1.36 V VDS 2 ( sat ) = VGS 2 − VTN = 1.36 − 0.5 ⇒ VDS 2 ( sat ) = 0.86 V 10.55 VDS 2 ( sat ) = 0.5 V = VGS 2 − VTN = VGS 2 − 0.5 ⇒ VGS 2 = 1 V ′ kn ⎛ W ⎞ 2 ⎜ ⎟ (VGS 2 − VTN ) 2 ⎝ L ⎠2 2 ⎛W ⎞ ⎛W ⎞ = 50 ⎜ ⎟ (1 − 0.5 ) ⇒ ⎜ ⎟ = 4 ⎝ L ⎠2 ⎝ L ⎠2 I O = 50 μ A = VGS1 = VGS 2 = 1 V ⇒ I REF = 150 = ′ kn ⎛ W ⎞ 2 2 ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ (VGS 1 − VTN ) = 50 ⎜ ⎟ (1 − 0.5 ) ⇒ ⎜ ⎟ = 12 2 ⎝ L ⎠1 ⎝ L ⎠1 ⎝ L ⎠1 VGS 3 + VSG 4 + VGS 1 = 6 2VGS 3 = 6 − 1 = 5 V ⇒ VGS 3 = 2.5 V 2 ⎛W ⎞ ⎛W ⎞ I REF = 150 = 50 ⎜ ⎟ ( 2.5 − 0.5 ) ⇒ ⎜ ⎟ = 0.75 ⎝ L ⎠3 ⎝ L ⎠3 k′ ⎛ W ⎞ 2 p ⎜ ⎟ (VSG 4 + VTP ) 2 ⎝ L ⎠4 2 ⎛W ⎞ ⎛W ⎞ 150 = 20 ⎜ ⎟ ( 2.5 − 0.5 ) ⇒ ⎜ ⎟ 1.88 L ⎠4 ⎝ ⎝ L ⎠4 I REF = 10.56 a. As a first approximation 2 I REF = 80 = 80 (VGS 1 − 1) ⇒ VGS 1 = 2 V Then VDS 1 ≅ 2 ( 2 ) = 4 V The second approximation 80 = 80 (VGS1 − 1) ⎡1 + ( 0.02 )( 4 ) ⎤ ⎣ ⎦ 80 2 Or = (VGS 1 − 1) ⇒ VGS 1 = 1.962 86.4 2 410. Then I O = K n (VGS1 − VTN ) (1 + λnVGS 1 ) 2 = 80 (1.962 − 1) ⎡1 + ( 0.02 )(1.962 ) ⎤ ⎣ ⎦ Or I 0 = 76.94 μ A 2 b. From a PSpice analysis, I 0 = 77.09 μ A for VD 3 = −1 V and I 0 = 77.14 μ A for VD 3 = 3 V. The change is ΔI 0 ≈ 0.05 μ A or 0.065%. 10.57 a. For a first approximation, I REF = 80 = 80 (VGS 4 − 1) ⇒ VGS 4 = 2 V 2 As a second approximation I REF = 80 = 80 (VGS 4 − 1) ⎡1 + ( 0.02 )( 2 ) ⎤ ⎣ ⎦ Or VGS 4 = 1.98 V = VGS 1 2 I O = K n (VGS 2 − VTN ) (1 + λVGS 2 ) 2 To a very good approximation I 0 = 80 μ A b. From a PSpice analysis, I 0 = 80.00 μ A for VD 3 = −1 V and the output resistance is R0 = 76.9 MΩ. Then For VD = +3 V 1 4 ΔI 0 = ⋅ VD 3 = = 0.052 μ A R0 76.9 I 0 = 80.05 μ A 10.58 (a) VDS 3 ( sat ) = VGS 3 − VTN or VGS 3 = VDS 3 ( sat ) + VTN = 0.2 + 0.8 = 1.0 k′ ⎛ W ⎞ 2 I D = n ⎜ ⎟ (VGS 3 − VTN ) 2⎝L⎠ 2 ⎛W ⎞ ⎛W ⎞ 50 = 48 ⎜ ⎟ ( 0.2 ) ⇒ ⎜ ⎟ = 26 ⎝L⎠ ⎝ L ⎠3 (b) VGS 5 − VTN = 2 (VGS 3 − VTN ) VGS 5 = 0.8 + 2 ( 0.2 ) ⇒ VGS 5 = 1.2 V (c) VD1 ( min ) = 2VDS ( sat ) = 2 ( 0.2 ) ⇒ VD1 ( min ) = 0.4 V 10.59 (a) K n1 = R= = ′ kn ⎛ W ⎞ 2 ⎜ ⎟ = 50 ( 5 ) = 250 μ A / V 2 ⎝ L ⎠1 1 K n1 I D1 ⎛ ⎜1 − ⎜ ⎝ (W / L )1 ⎞ ⎟ (W / L )2 ⎟ ⎠ ⎛ 5 ⎞ ⎜1 − ⎟ = ( 8.944 )( 0.6838 ) ⎜ 50 ⎟ ( 0.25 )( 0.05 ) ⎝ ⎠ 1 R = 6.12 k Ω (b) 411. V + − V − = VSD 3 ( sat ) + VGS 1 VSD 3 ( sat ) = VSG 3 + VTP I D1 = 50 = 20 ( 5 )(VSG 3 − 0.5 ) ⇒ VSG 3 = 1.207 V Then VSD 3 ( sat ) = 1.21 − 0.5 = 0.707 V 2 Also I D1 = 50 = 50 ( 5 )(VGS1 − 0.5 ) ⇒ VGS 1 = 0.9472 V 2 Then (V + − V − ) min = 0.71 + 0.947 = 1.66 V (c) 2 ⎛W ⎞ ⎛W ⎞ I O1 = 25 = 50 ⎜ ⎟ ( 0.947 − 0.5 ) ⇒ ⎜ ⎟ = 2.5 ⎝ L ⎠5 ⎝ L ⎠5 2 ⎛W ⎞ ⎛W ⎞ I O 2 = 75 = 20 ⎜ ⎟ (1.207 − 0.5 ) ⇒ ⎜ ⎟ = 7.5 ⎝ L ⎠6 ⎝ L ⎠6 10.60 1 ( 5 ) = 1.667 V 3 2 ⎛1 ⎞⎛W ⎞ I REF = ⎜ μ n Cox ⎟ ⎜ ⎟ (VGS 3 − VTN ) ⎝2 ⎠ ⎝ L ⎠3 2 ⎛W ⎞ ⎛W ⎞ ⎛W ⎞ ⎛W ⎞ 100 = ( 20 ) ⎜ ⎟ (1.667 − 1) ⇒ ⎜ ⎟ = ⎜ ⎟ = ⎜ ⎟ = 11.25 ⎝ L ⎠3 ⎝ L ⎠3 ⎝ L ⎠ 4 ⎝ L ⎠5 VGS 3 = 2 ⎛1 ⎞⎛W ⎞ I O1 = ⎜ μ n Cox ⎟ ⎜ ⎟ (VGS 3 − VTN ) ⎝2 ⎠ ⎝ L ⎠1 ⎛W ⎞ ⎜ ⎟ I REF ⎝ L ⎠3 Or = I 01 ⎛W ⎞ ⎜ ⎟ ⎝ L ⎠1 ⎛ W ⎞ ⎛ I 01 ⎜ ⎟ =⎜ ⎝ L ⎠1 ⎝ I REF ⎞ ⎛ W ⎞ ⎛ 0.2 ⎞ ⎛W ⎞ ⎟⎜ ⎟ = ⎜ ⎟ (11.25 ) ⇒ ⎜ ⎟ = 22.5 ⎝ L ⎠1 ⎠ ⎝ L ⎠3 ⎝ 0.1 ⎠ ⎛ W ⎞ ⎛ I ⎞ ⎛ W ⎞ ⎛ 0.3 ⎞ ⎛W ⎞ And ⎜ ⎟ = ⎜ 02 ⎟ ⎜ ⎟ = ⎜ ⎟ (11.25 ) ⇒ ⎜ ⎟ = 33.75 L ⎠ 2 ⎝ I REF ⎠ ⎝ L ⎠3 ⎝ 0.1 ⎠ ⎝ ⎝ L ⎠2 10.61 I REF = 24 − VSGP − VGSN R Also I REF = 40 (1)(VGSN − 1.2 ) I REF = 18 (1)(VSGP − 1.2 ) 2 2 Then 40 (VGSN − 1.2 ) = 18 (VSGP − 1.2 ) which yields VSGP = 6.325 (VGSN − 1.2 ) + 1.2 4.243 412. 2 Then ⎡ 0.040 (VGSN − 1.2 ) ⎤ ⋅ R = 24 − VGSN − 1.49 (VGSN − 1.2 ) − 1.2 ⎣ ⎦ which yields VGSN = 2.69 V and VSGP = 3.43 V 24 − 3.42 − 2.69 ⇒ I REF = 89.4 μ A 200 Now I REF = 89.4 = 17.9 μ A 5 I 2 = (1.25 )( 89.4 ) = 112 μ A I 3 = ( 0.8 )( 89.4 ) = 71.5 μ A I 4 = 4 ( 89.4 ) = 358 μ A I1 = 10.61 a. g m ( M 0 ) = 2 K n I REF gm ( M 0 ) = 2 ( 0.25)( 0.2 ) ⇒ g m ( M 0 ) = 0.447 mA/V r0 n = 1 1 = ⇒ r0 n = 250 kΩ λn I REF ( 0.02 )( 0.2 ) r0 p = 1 1 = ⇒ r0 p = 167 kΩ λ p I REF ( 0.03)( 0.2 ) b. Av = − g m ( r0 n || r0 p ) = − ( 0.447 )( 250 ||167 ) ⇒ Av = −44.8 c. RL = 205 ||167 = r0 n || r0 p or RL = 100 kΩ 10.62 We have VGSN = 2.69 V and VSGP = 3.43 V 10 − 2.69 − 3.43 3.88 So I REF = = ⇒ I REF = 19.4 μ A R 200 Then I1 = ( 0.2 )(19.4 ) = 3.88 μ A I 2 = (1.25 )(19.4 ) = 24.3 μ A I 3 = ( 0.8 )(19.4 ) = 15.5 μ A I 4 = 4 (19.4 ) = 77.6 μ A 10.63 I D2 (W L ) = (W L ) (W L ) = (W L ) 2 ⋅ I REF = 9 ( 200 ) ⇒ I D 2 = 120 μ A 15 1 IO 4 ⎛ 20 ⎞ ⋅ I D 2 = ⎜ ⎟ (120 ) ⇒ I O = 267 μ A ⎝ 9 ⎠ 3 2 ⎛ 40 ⎞ I O = 266.7 = ⎜ ⎟ ( 20 )(VSG 4 − 0.6 ) 2 ⎠ ⎝ VSG 4 = 1.416 V VSD 4 (sat) = 1.416 − 0.6 ⇒ VSD 4 ( sat ) = 0.816 V 10.64 2 ⎛ 40 ⎞ ⎛ W ⎞ I REF = 50 = ⎜ ⎟ ⎜ ⎟ (VSG1 − 0.6 ) ⎝ 2 ⎠ ⎝ L ⎠1 1.75 − VSG1 I REF = = 50 R 413. VSD 2 (sat) = 0.35 = VSG 2 − 0.6 ⇒ VSG 2 = 0.95 V 1.75 − 0.95 R= ⇒ R = 16 K 0.05 2 ⎛ 40 ⎞⎛ W ⎞ ⎛W ⎞ 50 = ⎜ ⎟⎜ ⎟ ( 0.95 − 0.6 ) ⇒ ⎜ ⎟ = 20.4 2 ⎠⎝ L ⎠1 ⎝ ⎝ L ⎠1 ( ) W I O1 120 ⎛W ⎞ L 2 = = ⇒ = 49 50 I REF ( 20.4 ) ⎜ L ⎟2 ⎝ ⎠ ( ) W I D 3 25 ⎛W ⎞ L 3 = = ⇒ ⎜ ⎟ = 10.2 I REF 50 ( 20.4 ) ⎝ L ⎠3 VDS 5 (sat) = 0.35 = VGS 5 − 0.4 ⇒ VGS 5 = 0.75 V 2 ⎛ 100 ⎞⎛ W ⎞ ⎛W ⎞ IO 2 = ⎜ ⎟⎜ ⎟ ( 0.75 − 0.4 ) = 150 ⇒ ⎜ ⎟ = 24.5 2 ⎠⎝ L ⎠5 ⎝ ⎝ L ⎠5 W I D4 I D3 25 ⎛W ⎞ L 4 = = = ⇒ ⎜ ⎟ = 4.08 I O 2 I O 2 150 24.5 ⎝ L ⎠4 ( ) 10.65 For vGS = 0, iD = I DSS (1 + λ vDS ) a. b. c. VD = −5 V, vDS = 5 iD = ( 2 ) ⎡1 + ( 0.05 )( 5 ) ⎤ ⇒ iD = 2.5 mA ⎣ ⎦ VD = 0, vDS = 10 iD = ( 2 ) ⎡1 + ( 0.05 )(10 ) ⎤ ⇒ iD = 3 mA ⎣ ⎦ VD = 5 V, vDS = 15 V iD = ( 2 ) ⎡1 + ( 0.05 )(15 ) ⎤ ⇒ iD = 3.5 mA ⎣ ⎦ 10.66 ⎛ V ⎞ I 0 = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ ⎛ V ⎞ 2 = 4 ⎜ 1 − GS ⎟ VP ⎠ ⎝ 2 2 VGS 2 = 1− = 0.293 4 VP So VGS = ( 0.293)( −4 ) = −1.17 V VS and VS = −VGS R ( −1.17 ) −V ⇒ R = 0.586 kΩ R = GS = − 2 I0 Finish solution: See solution Then I 0 = 10.66 Completion of solution Need vDS ≥ vDS ( sat ) = vGS − VP = −1.17 − ( −4 ) vDS ≥ 2.83 V So VD ≥ vDS ( sat ) + VS = 2.83 + 1.17 ⇒ VD ≥ 4 V 414. 10.67 ⎛V ⎞ I REF = I S 1 exp ⎜ EB1 ⎟ ⎝ VT ⎠ a. ⎛I ⎞ ⎛ 1× 10−3 ⎞ or VEB1 = VT ln ⎜ REF ⎟ = ( 0.026 ) ln ⎜ ⇒ VEB1 = 0.5568 −13 ⎟ ⎝ 5 × 10 ⎠ ⎝ I S1 ⎠ 5 − 0.5568 R1 = ⇒ R1 = 4.44 kΩ 1 From Equations (10.79) and (10.80) and letting VCE 0 = VEC 2 = 2.5 V b. c. 10 −12 2.5 ⎞ ⎛ ⎜ 1 + 80 ⎟ ⎛ VI ⎞ ⎡ 2.5 ⎤ −3 exp ⎜ ⎟ ⎢1 + ⎥ = 10 ⎜ 0.5568 ⎟ ⎝ VT ⎠ ⎣ 120 ⎦ ⎜1+ ⎟ ⎜ ⎟ 80 ⎠ ⎝ ⎛V ⎞ ⎛ 1.03125 ⎞ 1.0208333 × 10−12 exp ⎜ I ⎟ = (10−3 ) ⎜ ⎟ ⎝ 1.00696 ⎠ ⎝ VT ⎠ Then VI = 0.026 ln (1.003222 × 109 ) So VI = 0.5389 V d. Av = − (1/ VT ) (1/ VAN ) + (1/ VAP ) 1 −38.46 Av = 0.026 = 1 1 0.00833 + 0.0125 + 120 80 Av = −1846 − 10.68 a. b. c. other. ⎛I ⎞ ⎛ 0.5 × 10−3 ⎞ VBE = VT ln ⎜ REF ⎟ = ( 0.026 ) ln ⎜ ⎟ ⇒ VBE = 0.5208 −12 ⎝ 10 ⎠ ⎝ I S1 ⎠ 5 − 0.5208 R1 = ⇒ R1 = 8.96 kΩ 0.5 Modify Eqs. 10.79 and 10.80 to apply to pnp and npn, and set the two equation equal to each ⎛ V ⎞⎛ V I CO = I SO exp ⎜ EBO ⎟⎜ 1 + ECO VAP ⎝ VT ⎠⎝ ⎞ ⎛ VBE ⎞ ⎛ VCE 2 ⎞ ⎟ ⎟ = I C 2 = I S 2 exp ⎜ ⎟ ⎜1 + ⎠ ⎝ VT ⎠ ⎝ VAN ⎠ ⎛ V ⎞ ⎛ 2.5 ⎞ ⎛ VBE ⎞ ⎛ 2.5 ⎞ −12 5 × 10−13 exp ⎜ EBO ⎟ ⎜ 1 + ⎟ ⎜1 + ⎟ = 10 exp ⎜ ⎟ 80 ⎠ VT ⎠ ⎝ VT ⎠ ⎝ 120 ⎠ ⎝ ⎝ ⎛V ⎞ ⎛V ⎞ 5.15625 × 10−13 exp ⎜ EBO ⎟ = 1.020833 × 10−12 exp ⎜ BE ⎟ VT ⎠ ⎝ ⎝ VT ⎠ ⎛V ⎞ exp ⎜ EBO ⎟ ⎝ VT ⎠ = 1.9798 = exp ⎛ VEBO − VBE ⎞ ⎜ ⎟ VT ⎛V ⎞ ⎝ ⎠ exp ⎜ BE ⎟ VT ⎠ ⎝ VEBO = VBE + VT ln (1.9798 ) = 0.5208 + ( 0.026 ) ln (1.9798 ) VEBO = 0.5386 ⇒ VI = 5 − 0.5386 ⇒ VI = 4.461 V 415. Av = d. − (1/ VT ) (1/ VAN ) + (1/ VAP ) 1 −38.46 0.026 = Av = 1 1 0.00833 + 0.0125 + 120 80 Av = −1846 − 10.69 a. M1 and M2 matched. For I REF = I 0 , we have VSD 2 = VSG = VSG 3 = VDS 0 = 2.5 V For M1 and M3: 2 ⎛1 ⎞⎛W ⎞ I REF = ⎜ μ p Cox ⎟ ⎜ ⎟ (VSG + VTP ) (1 + λPVSD ) ⎝2 ⎠ ⎝ L ⎠1 2 ⎛W ⎞ ⎛W ⎞ ⎛W ⎞ ⎛W ⎞ 100 = 10 ⎜ ⎟ ( 2.5 − 1) ⎡1 + ( 0.02 )( 2.5 ) ⎤ ⇒ ⎜ ⎟ = 4.23 = ⎜ ⎟ = ⎜ ⎟ ⎣ ⎦ ⎝ L ⎠1 ⎝ L ⎠1 ⎝ L ⎠3 ⎝ L ⎠ 2 For M0: 2 ⎛1 ⎞⎛W ⎞ I O = ⎜ μ n Cox ⎟ ⎜ ⎟ (VGS − VTN ) (1 + λnVDS ) ⎝2 ⎠ ⎝ L ⎠0 2 ⎛W ⎞ ⎛W ⎞ 100 = 20 ⎜ ⎟ ( 2 − 1) ⎡1 + ( 0.02 )( 2.5 ) ⎤ ⇒ ⎜ ⎟ = 4.76 ⎣ ⎦ L ⎠0 ⎝ ⎝ L ⎠0 b. r0 n = r0 p = 1 λ I0 = 1 ( 0.02 )( 0.1) = 500 kΩ ⎛1 ⎞⎛W ⎞ g m = 2 K n I O = 2 ⎜ μ n Cox ⎟ ⎜ ⎟ I O 2 ⎝ ⎠ ⎝ L ⎠o ( 0.02 )( 4.76 )( 0.1) =2 g m = 0.195 mA/V Av = − g m ( r0 n || r0 p ) = − ( 0.195 )( 500 || 500 ) ⇒ Av = −48.8 10.70 I REF = K p1 (VSG + VTP ) a. 2 100 = 100 (VSG − 1) ⇒ VSG = 2 V 2 b. From Eq. 10.89 ⎡1 + λ p (V + − VSG ) ⎤ K (V − V )2 TN ⎦− n I VO = ⎣ λn + λ p I REF ( λn + λ p ) ⎡1 + ( 0.02 )(10 − 2 ) ⎤ ⎦ − 100 (VI − 1) 5= ⎣ 0.02 + 0.02 100 ( 0.02 + 0.02 ) 2 5 = 29 − (VI − 1) 2 (VI − 1) 0.04 = 0.96 VI = 1.98 V 2 416. Av = − g m ( r0 n || r0 p ) c. r0 n = r0 p = 1 λ I REF = 1 = 500 kΩ 0.02 )( 0.1) ( ( 0.1)( 0.1) = 0.2 mA / V Av = − ( 0.2 )( 500 || 500 ) ⇒ Av = −50 g m = 2 K n I REF = 2 10.71 5 − 0.6 5 − 0.6 = = 0.22 mA R1 20 From Eq. 10.96 ⎛I ⎞ 0.22 −⎜ C ⎟ − ⎝ VT ⎠ 0.026 Av = = ⎛ IC I C ⎞ 0.22 + 1 + 0.22 1 + + ⎜ ⎟ 90 ⎝ VAN RL VAP ⎠ 140 RL I REF = = −8.4615 −8.4615 = 1 1 0.0015714 + + 0.002444 0.004016 + RL RL RL = ∞, Av = −2107 (a) (b) (c) RL = 250 K, Av = −1056 RL = 100 K, Av = −604 10.72 I REF = 5 − 0.6 = 0.1257 mA 35 Then I CO = 2 I REF = 0.2514 mA From Eq. 10.96 −0.2514 −9.6692 0.026 Av = = 0.2514 0.2514 1 1 0.002095 + 0.0031425 + + + 120 80 RL RL Av = (a) (b) −9.6692 1 RL RL = ∞ Av = −1846 0.0052375 + RL = 250 K, Av = −1047 10.73 (a) To a good approximation, output resistance is the same as the widlar current source. R0 = r02 ⎡1 + g m 2 ( rπ 2 || RE ) ⎤ ⎣ ⎦ (b) Av = − g m 0 ( r0 || RL || R0 ) 10.74 Output resistance of Wilson source 417. β r03 R0 ≅ 2 Then Av = − g m ( r0 || R0 ) V 80 r03 = AP = = 400 kΩ I REF 0.2 r0 = VAN 120 = = 600 kΩ I REF 0.2 gm = I REF 0.2 = = 7.692 mA/V VT 0.026 ⎡ (80 )( 400 ) ⎤ Av = −7.69 ⎢600 ⎥ = −7.69 [ 600 ||16, 000] ⇒ Av = −4448 2 ⎢ ⎥ ⎣ ⎦ 10.75 (a) I D 2 = I D 0 = I REF = 200 μ A For M 2 ; ro 2 = 1 λP I D 2 = 1 = 250 K 0.02 )( 0.2 ) ( ⎛ 0.04 ⎞ gm2 = 2 K P I D2 = 2 ⎜ ⎟ ( 35 )( 0.2 ) ⎝ 2 ⎠ g m 2 = 0.748 mA/V 1 1 For M 0 ; r∞ = = = 333 K λn I Do ( 0.015 )( 0.2 ) ⎛ 0.08 ⎞ g mo = 2 ⎜ ⎟ ( 20 )( 0.2 ) ⇒ g mo = 0.80 mA/V ⎝ 2 ⎠ Av = − g mo ( ro 2 || roo ) = − ( 0.80 )( 250 || 333) (b) Av = −114.3 Want Av = −57.15 = −0.80 (142.8 || RL ) 142.8 RL 142.8 || RL = 71.375 = ⇒ RL = 143 K 142.8 + RL (c) 10.76 Assume M1, M2 matched I REF = I D 2 = I Do = 200 μ A 1 1 ro 2 = = = 250 K λ p I D 2 ( 0.02 )( 0.2 ) roo = 1 1 = = 333 K λn I D 0 ( 0.015 )( 0.2 ) Av = − g mo ( ro 2 roo ) −100 = − g mo ( 250 333) ⇒ g mo = 0.70 mA/V ⎛ 0.08 ⎞⎛ W ⎞ g mo = 2 ⎜ ⎟⎜ ⎟ ( 0.2 ) = 0.70 ⎝ 2 ⎠⎝ L ⎠0 ⎛W ⎞ ⎜ ⎟ = 15.3 ⎝ L ⎠0 418. ⎛ k ′ ⎞⎛ W ⎞ ⎛ k ′ ⎞ ⎛ W ⎞ p Now ⎜ n ⎟ ⎜ ⎟ = ⎜ ⎟ ⎜ ⎟ 2 ⎠ ⎝ L ⎠0 ⎝ 2 ⎠ ⎝ L ⎠2 ⎝ ⎛ 80 ⎞ ⎛ 40 ⎞⎛ W ⎞ ⎜ ⎟ (15.3) = ⎜ ⎟⎜ ⎟ ⎝ 2⎠ ⎝ 2 ⎠⎝ L ⎠ 2 ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = ⎜ ⎟ = 30.6 ⎝ L ⎠ 2 ⎝ L ⎠1 10.77 Since Vsg 3 = 0, the circuit becomes 419. ′ I x = − g mVsg 2 + Vx − Vsg 2 r02 and Vsg 2 = I x ro 3 Then ⎛ r ⎞ V ′ I x ⎜ 1 + g m ro3 + o3 ⎟ = x ro 2 ⎠ ro 2 ⎝ so that ⎛ Vx r ⎞ ′ = Ro = ro 2 ⎜1 + g m ro 3 + o 3 ⎟ Ix ro 2 ⎠ ⎝ or ′ Ro = ro 2 + ro 3 (1 + g m ro 2 ) vo = − g m1 ( ro1 Ro ) vi Av = Now g m1 = 2 ro1 = ( 0.050 )( 20 )( 0.10 ) = 0.632 mA / V 1 1 = = 500 k Ω λn I DQ ( 0.02 )( 0.10 ) ′ g m = 2 K p I DQ = 2 ro 2 = ro 3 = ( 0.020 )(80 )( 0.1) = 0.80 mA / V 1 1 = = 500 k Ω λ p I DQ ( 0.020 )( 0.1) Then Ro = 500 + 500 ⎡1 + ( 0.8 )( 500 ) ⎤ ⇒ 201 M Ω ⎣ ⎦ Av = − ( 0.632 ) ( 500 201000 ) ⇒ Av = −315 10.78 From Eq. 10.105 2 − gm Av = 1 1 + ro3 ro 4 ro1 ro 2 ⎛ k′ ⎞⎛W ⎞ g m = 2 ⎜ n ⎟ ⎜ ⎟ I D1 ⎝ 2 ⎠⎝ L ⎠ =2 ( 0.050 ) ( 20 ) ( 0.08 ) g m = 0.5657 mA / V ro = 1 1 = = 625 K λ I D ( 0.02 )( 0.08 ) − ( 0.5657 ) −0.3200 Av = = 1 1 2 ( 0.00000256 ) + 2 2 ( 625) ( 625) 2 Av = −62,500 10.79 420. V − ( −Vπ 2 ) Vπ 2 Vπ 2 + + g m 2Vπ 2 + O rπ 2 ro1 ro 2 (1) g m1Vi = (2) VO VO − ( −Vπ 2 ) + + g m 2Vπ 2 = 0 RO 3 ro 2 (1) ⎛ 1 1 1 ⎞ V + + gm2 + ⎟ + O g m1Vi = Vπ 2 ⎜ ro 2 ⎠ ro 2 ⎝ rπ 2 ro1 (2) ⎛ 1 ⎛ 1 ⎞ 1 ⎞ VO ⎜ + ⎟ + Vπ 2 ⎜ + gm2 ⎟ = 0 ⎝ RO 3 ro 2 ⎠ ⎝ ro 2 ⎠ g m >> 1 ro (1) ⎛ 1 + β ⎞ VO g m1Vi = Vπ 2 ⎜ ⎟+ ⎝ rπ 2 ⎠ ro 2 (2) ⎛ 1 1 ⎞ VO ⎜ + ⎟ + Vπ 2 ⋅ g m 2 = 0 ⎝ RO 3 ro 2 ⎠ (3) Vπ 2 = − VO gm2 ⎛ 1 1 ⎞ + ⎟ ⎜ ⎝ RO 3 ro 2 ⎠ Then ⎛ 1 1 ⎞⎛ 1 + β ⎞ VO + ⎟⎜ ⎜ ⎟+ ⎝ RO 3 ro 2 ⎠⎝ rπ 2 ⎠ ro 2 ⎛ 1 1 ⎞ ⎛ 1 + β ⎞ VO = −VO ⎜ + ⎟⎜ ⎟+ RO 3 ro 2 ⎠ ⎝ β ⎠ ro 2 ⎝ (1) g m1Vi = − VO gm2 VO ⎛ 1 + β ⎞ ⎜ ⎟ RO 3 ⎝ β ⎠ ⎛ β ⎞ VO = − g m1 RO 3 ⎜ ⎟ Vi ⎝1+ β ⎠ From Equation (10.20) RO 3 ≈ β rO 3 So ≈− 421. VO − g m1ro3 β 2 = 1+ β Vi Av = gm = 0.25 = 9.615 mA/V 0.026 ro3 = Av = 80 = 320 K 0.25 − ( 9.615 )( 320 )(120 ) 121 2 = −366,165 422. Chapter 11 Exercise Solutions EX11.1 vE = −VBE ( on ) ⇒ vE = −0.7 V I C1 = I C 2 = 0.5 mA vC1 = vC 2 = 10 − ( 0.5 )(10 ) ⇒ vC1 = vC 2 = 5 V EX11.2 iC 2 1 = = 0.99 IQ ⎛ vd ⎞ 1 + exp ⎜ ⎟ ⎝ VT ⎠ ⎛v ⎞ 1 1 + exp ⎜ d ⎟ = VT ⎠ 0.99 ⎝ ⎛v exp ⎜ d ⎝ VT ⎞ 1 −1 ⎟= 0.99 ⎠ ⎡ 1 ⎤ − 1⎥ ⇒ vd = −119.5 mV vd = VT ln ⎢ ⎣ 0.99 ⎦ EX11.3 a. v1 = v2 = 0 ⇒ vE = 0.7 V ΔVRC = ( 0.25 )( 8 ) = 2 V ⇒ vC1 = vC 2 = −3 V ⇒ vEC1 = 3.7 V b. v1 = v2 = 2.5 V ⇒ vE = 3.2 V ⇒ vEC1 = 6.2 V c. v1 = v2 = −2.5 V ⇒ vE = −1.8 V ⇒ vEC1 = 1.2 V EX11.4 Let I Q = 1 mA, then I CQ1 = I CQ 2 = 0.5 mA 0.5 = 19.23 mA / V 0.026 v 1 At vC 2 , Ad = c 2 = g m RC 2 vd 2 g m1 = g m 2 = 1 (19.23) RC 2 ⇒ RC 2 = 15.6 k Ω 2 v 1 At vC1 , Ad = c1 = − g m RC1 vd 2 So, 150 = 1 (19.23) RC1 ⇒ RC1 = 10.4 k Ω 2 If V + = +10 V and V − = −10 V , dc biasing is OK. So, −100 = − EX11.5 Acm = 1+ ⎛ I Q RC ⎞ −⎜ ⎟ ⎝ 2VT ⎠ = (1 + β ) IQ RO VT β ⎡ ( 0.8 )(12 ) ⎤ −⎢ ⎥ ⎣ 2 ( 0.026 ) ⎦ = −0.0594 (101)( 0.8 )(100 ) 1+ ( 0.026 )(100 ) vo = Acm vcm = ( −0.0594 )(1sin ω t )( mV ) = −59.4sin ω t ( μV ) EX11.6 423. vd = 40 μV , vcm = 0 vo = ( 50 )( 40 ) ⇒ 2.0 mV (a) (b) vd = 40 μV vcm = 200 μV Assuming Acm > 0 50 Acm 60 = 20 log10 Acm = 0.05 vo = ( 50 )( 40 ) + ( 0.05 )( 200 ) vo = 2.010 mV EX11.7 Diff. Gain Ad = a. I Q RC 4VT For v1 = v2 = 5 V ⇒ Minimum collector voltage vC 2 = 5 V ⇒ IQ ⋅ RC = 15 − 5 = 10 V or I Q RC = 20 V for max. Ad 2 20 Then Ad = ⇒ Ad ( max ) = 192 2 ( 0.026 ) If I Q = 0.5 mA, RC = 40 kΩ b. ⎛ I Q RC ⎞ −⎜ ⎟ ⎝ 2VT ⎠ Acm = ⎡ (1 + β ) I Q R0 ⎤ ⎢1 + ⎥ VT β ⎣ ⎦ Then ⎛ 20 ⎞ −⎜ ⎜ 2 ( 0.026 ) ⎟ ⎟ ⎝ ⎠ Acm = ⇒ Acm = −0.199 ⎡ ( 201)( 0.5 )(100 ) ⎤ ⎢1 + ( 0.026 )( 200 ) ⎥ ⎣ ⎦ ⎛ 192 ⎞ and C M RRdB = 20 log10 ⎜ ⎟ ⇒ C M RRdB = 59.7 dB ⎝ 0.199 ⎠ EX11.8 Rid = 2 ⎡ rπ + (1 + β ) RE ⎤ ⎣ ⎦ rπ = β VT I CQ = (100 )( 0.026 ) 0.25 = 10.4 K Rid = 2 ⎡10.4 + (101)( 0.5 ) ⎤ = 122 K ⎣ ⎦ EX11.9 Ad = g m RC 2 (1 + g m RE ) ( 9.62 )(10 ) 2 ⎡1 + ( 9.62 ) RE ⎤ ⎣ ⎦ 1 + ( 9.62 ) RE = 4.81 ⇒ RE = 0.396 K Rid = 2 ⎡ rπ + (1 + β ) RE ⎤ = 2 ⎡10.4 + (101)( 0.396 ) ⎤ ⎣ ⎦ ⎣ ⎦ 10 = Rid = 100.8 K 424. EX11.10 10 − VGS 4 2 = K n 3 (VGS 4 − VTN ) I1 = R1 10 − VGS 4 = ( 0.1)( 80 )(VGS 4 − 0.8 ) 2 2 10 − VGS 4 = 8 (VGS 4 − 1.6VGS 4 + 0.64 ) 2 8VGS 4 − 11.8VGS 4 − 4.88 = 0 VGS 4 = 11.8 ± (11.8) + 4 ( 8)( 4.88) 2 (8) 2 = 1.81 V 10 − 1.81 = 0.102 mA 80 0.102 = ID2 = = 0.0512 mA 2 I1 = I Q = I D1 = K n1 (VGS 1 − VTN ) 2 0.0512 = 0.050 (VGS 1 − 0.8 ) ⇒ VGS 1 = 1.81 V 2 v01 = v02 = 5 − ( 0.0512 )( 40 ) = 2.95 V Max vcm : VDS 1 ( sat ) = VGS1 − VTN = 1.81 − 0.8 = 1.01 V vcm ( max ) = v01 − VDS 1 ( sat ) + VGS 1 = 2.95 − 1.01 + 1.81 vcm ( max ) = 3.75 V Min vcm : VDS 4 ( sat ) = VGS 4 − VTN = 1.81 − 0.8 = 1.01 V vcm ( min ) = VGS 1 + VDS 4 ( sat ) − 5 = 1.81 + 1.01 − 5 vcm ( min ) = −2.18 V −2.18 ≤ vcm ≤ 3.75 V EX11.11 (1)( 2 ) ⇒ g f ( max ) = 1 mA / V 2 ( 2) = g f RD = (1)( 5 ) ⇒ Ad = 5 g f ( max ) = Ad K n IQ = EX11.12 ⎛ K Kn iD1 1 = + ⋅ vd 1 − ⎜ n ⎜ 2 IQ IQ 2 2IQ ⎝ ⎞ 2 ⎟ vd ⎟ ⎠ Using the parameters in Example 11.11, K n = 0.5 mA / V 2 , I Q = 1 mA, then ⎛ 0.5 ⎞ 2 iD1 1 0.5 = 0.90 = + ⋅ vd 1 − ⎜ ⎜ 2 (1) ⎟ vd ⎟ 2 2 (1) IQ ⎝ ⎠ By trial and error, vd = 0.894 V EX11.13 a. 425. gm = IQ 2VT = 0.5 = 9.615 mA/V 2 ( 0.026 ) r02 = VA 2 125 = = 500 kΩ I C 2 0.25 r04 = VA 4 85 = = 340 kΩ I C 4 0.25 Ad = g m ( r02 r04 ) = ( 9.615 ) ( 500 340 ) ⇒ Ad = 1946 b. c. ( Ad = g m r02 r04 RL ) Ad = ( 9.615 ) ⎡500 340 100 ⎤ ⇒ Ad = 644 ⎣ ⎦ β VT (150 )( 0.026 ) rπ = = = 15.6 kΩ I CQ 0.25 Rid = 2rπ ⇒ Rid = 31.2 kΩ d. R0 = r02 r04 = 500 340 ⇒ R0 = 202 kΩ EX11.14 80 80 2 2 I REF = (10 )(VGS 4 − 0.5 ) = ( 0.33)(VGS 5 − 0.5 ) 2 2 VGS 4 + 2VGS 5 = 6 1 VGS 5 = ( 6 − VGS 4 ) 2 10 1 (VGS 4 − 0.5) = ( 6 − VGS 4 ) − 0.5 0.33 2 6.0VGS 4 = 5.252 ⇒ VGS 4 = 0.8754 V 80 2 I REF = (10 )( 0.8754 − 0.5 ) = 53.37 μ A 2 Also I Q = I REF = 53.37 μ A ⎛ 80 ⎞ g m = 2kn I Q = 2 ⎜ ⎟ (10 )( 53.37 ) ⇒ 0.2066 mA/V ⎝ 2⎠ 1 1 = = 1873.7 K ro1 = ro8 = λ ID ⎛ 0.05337 ⎞ ( 0.02 ) ⎜ ⎟ 2 ⎝ ⎠ Ad = g m ( ro1 ro8 ) = ( 0.2066 )(1873.7 1873.7 ) Ad = 194 EX11.15 Ad = g m ( ro 2 RO ) 400 = g m ( 500 101000 ) ⇒ g m = 0.8 mA / V g m = 2 K n I DQ ⎛ 0.08 ⎞ ⎛ W ⎞ ⎛W ⎞ ⎛W ⎞ 0.8 = 2 ⎜ ⎟ ⎜ ⎟ ( 0.1) ⇒ ⎜ ⎟ = ⎜ ⎟ = 40 ⎝ 2 ⎠ ⎝ L ⎠1 ⎝ L ⎠1 ⎝ L ⎠ 2 EX11.16 426. I e 6 = (1 + β ) I b 6 = I b 7 I c 7 = β I b 7 = β (1 + β ) I b 6 Ic7 = β (1 + β ) = (100 )(101) = 1.01× 104 Ib6 EX11.17 r + 300 ROA = π A 1+ β ⎛ β I CA = ⎜ ⎝1+ β ⎛ β =⎜ ⎝1+ β ⎞ ⎟ I EA ⎠ ⎞⎛ 1 ⎞ ⎟⎜ ⎟ (1) ⇒ 9.803 μ A ⎠⎝ 1 + β ⎠ (100 )( 0.026 ) rπ A = = 265.2 K 0.009803 265.2 + 300 ROA = = 5.596 K 101 r + ROA RO = π B 10 1+ β rπ B = (100 )( 0.026 ) = 2.626 K 0.99 2.626 + 5.596 10 RO = 101 RO = 81.4 104 = 80.7 Ω EX11.18 10 − 0.7 − ( −10 ) I1 = = 0.6 ⇒ R1 = 32.2 K R1 ⎛I I Q R2 = VT ln ⎜ 1 ⎜I ⎝ Q ⎞ ⎟ ⎟ ⎠ . ( 0.2 ) R2 = ( 0.026 ) ln ⎛ ⎜ 0.6 ⎞ ⎟ ⇒ R2 = 143 Ω ⎝ 0.2 ⎠ I R 6 = I1 ⇒ R3 = 0 10 = I C1 RC + VCE1 − 0.7 10.7 = ( 0.1) RC + 4 ⇒ RC = 67 K vo 2 = −0.7 + 4 = 3.3 V vE 4 = 3.3 − 1.4 = 1.9 V I R4 = vE 4 1.9 ⇒ R4 = ⇒ R4 = 3.17 K 0.6 R4 vC 3 = vO 2 − 1.4 + vCE 4 = 3.3 − 1.4 + 3 = 4.9 V I R 5 = I R 4 = 0.6 = 10 − 4.9 ⇒ R5 = 8.5 K R5 vE 5 = 4.9 − 0.7 = 4.2 V ⇒ R6 = R7 = 0 − ( −10 ) 5 ⇒ R7 = 2 K 4.2 − 0.7 = 5.83 K 0.6 427. EX11.19 Ri 2 = rπ 3 + (1 + β ) rπ 4 (100 )( 0.026 ) rπ 4 = 0.6 β 2VT rπ 3 ≈ I R4 = 4.333 K (100 ) ( 0.026 ) 2 = 0.6 = 433.3 K Ri 2 = 433.3 + (101)( 4.333) ⇒ Ri 2 = 871 K Ri 3 = rπ 5 + (1 + β ⎡ R6 + rπ 6 + (1 + β ) R7 ⎤ ⎣ ⎦ (100 )( 0.026 ) rπ 5 = 0.6 (100 )( 0.026 ) rπ 6 = = 4.333 K = 0.52 K 5 Ri 3 = 4.333 + (101) ⎡5.83 + 0.52 + (101)( 2 ) ⎤ ⎣ ⎦ Ri 3 = 21.0 MΩ gm 0.1 ( RC Ri 2 ) g m = 0.026 = 3.846 mA/V 2 3.846 Ad 1 = ( 67 871) = 119.64 2 ⎛I ⎞ 0.6 A2 = ⎜ R 4 ⎟ ( R5 Ri 3 ) = (8.5 21000 ) = 98.037 2 ( 0.026 ) ⎝ 2VT ⎠ Ad 1 = A = Ad 1 ⋅ A2 = (119.64 )( 98.037 ) = 11, 729 EX11.20 fz = 1 1 = 2π RO CO 2π (10 × 106 )( 0.2 × 10−12 ) f z = 79.6 kHz fp = 1 2π Req CO Req = 51.98 Ω from Example 11.20 1 fp = 2π ( 51.98 ) ( 0.2 × 10−12 ) f p = 15.3 GHz TYU11.1 Vd = V1 − V2 = 2 + 0.005sin ω t − ( 0.5 − 0.005sin ω t ) ⇒ Vd = 1.5 + 0.010sin ω t ( V ) V1 + V2 2 2 + 0.005sin ω t + 0.5 − 0.005sin ω t = ⇒ Vcm = 1.25 V 2 Vcm = TYU11.2 For v1 = v2 = +4 V ⇒ Minimum vC1 = vC 2 = 4 V IQ I C1 = I C 2 = = 1 mA 2 10 − 4 RC = ⇒ RC = 6 kΩ 1 428. TYU11.3 From Equation (11.41) g R CMRR = m O ⎛ ΔRC ⎞ ⎜ ⎟ ⎝ RC ⎠ For CMRR |dB = 75 dB ⇒ CMRR = 5623.4 Then 5623.4 = ( 3.86 )(100 ) ΔRC ⋅ (10 ) Or ΔRC = 0.686 K TYU11.4 From Equation (11.49) 1 + 2 RO g m CMRR = ⎛ Δg ⎞ 2⎜ m ⎟ ⎝ gm ⎠ For CMRR |dB = 90 dB ⇒ CMRR = 31622.8 ⎛ 1 + 2 (100 )( 3.86 ) ⎞ Then 31622.8 = ⎜ ⎟ ( 3.86 ) 2Δg m ⎝ ⎠ Δg m 0.0472 = = 0.0122 ⇒ 1.22% Or Δg m = 0.0472 mA/V or 3.86 gm TYU11.5 For v1 = v2 = 5 V ⇒ min vC1 = vC 2 = 5 V So I C1 RC = 10 − 5 = 0.25 RC ⇒ RC = 20 kΩ Ad = I Q RC = 4VT ( 0.5 )( 20 ) ⇒ Ad 4 ( 0.026 ) = 96.2 Let I Q = 0.5 mA C M RRdB = 95 db ⇒ C M RR = 5.62 × 104 ⇒ Acm = Acm ⎛ I Q RC ⎞ ⎜ ⎟ ⎝ 2VT ⎠ = = 1.71× 10−3 ⎡ (1 + β ) I Q R0 ⎤ ⎢1 + ⎥ VT β ⎣ ⎦ ⎡ ( 0.5 )( 20 ) ⎤ ⎢ ⎥ ⎣ 2 ( 0.026 ) ⎦ = 1.71× 10−3 ⎡ ( 201)( 0.5 ) R0 ⎤ ⎢1 + ⎥ ⎣ ( 0.026 )( 200 ) ⎦ 1 + 19.33R0 = 1.125 × 105 ⇒ R0 = 5.82 × 103 kΩ = 5.82 MΩ We have R0 = r04 ⎡1 + g m 4 ( R2 rπ 4 ) ⎤ ⎣ ⎦ r04 = VA 125 = = 250 kΩ I Q 0.5 gm4 = rπ 2 = IQ VT β VT IQ = 0.5 = 19.23 mA/V 0.026 = ( 200 )( 0.026 ) 0.5 = 10.4 kΩ 96.2 ⇒ Acm = 1.71× 10−3 5.62 × 104 429. 5.820 = 250 ⎡1 + g m 4 ( R2 rπ 4 ) ⎤ ⎣ ⎦ 19.23 ( R2 rπ 2 ) = 22.28 (R 2 rπ 2 ) = 1.159 kΩ R2 (10.4 ) R2 + 10.4 = 1.159 R2 (10.4 − 1.16 ) = (1.16 )(10.4 ) ⇒ R2 = 1.30 kΩ Let I1 = 1 mA ⎛I I Q R2 − I1 R3 = VT ln ⎜ 1 ⎜I ⎝ Q ⎞ ⎟ ⎟ ⎠ ( 0.5)(1.304 ) − (1) R3 = ( 0.026 ) ln ⎛ ⎜ 1 ⎞ ⎟ ⇒ R3 = 0.634 kΩ ⎝ 0.5 ⎠ If VBE ( Q3 ) ≅ 0.7 V 10 − 0.7 − ( −10 ) R1 + R3 = 1 = 19.3 ⇒ R1 ≅ 18.7 kΩ TYU11.6 a. v0 = Ad vd + Acm vcm vd = v1 − v2 = 0.505sin ω t − 0.495sin ω t = 0.01sin ω t v +v 0.505sin ω t + 0.495sin ω t vcm = 1 2 = 2 2 = 0.50sin ω t v0 = ( 60 )( 0.01sin ω t ) + ( 0.5 )( 0.5sin ω t ) ⇒ v0 = 0.85sin ω t ( V ) b. vd = v1 − v2 = 0.5 + 0.005sin ω t − ( 0.5 − 0.005sin ω t ) = 0.01sin ω t v +v vcm = 1 2 2 0.5 + 0.005sin ω t + 0.5 − 0.005sin ω t = 2 = 0.5 v0 = ( 60 )( 0.01sin ω t ) + ( 0.5 )( 0.5 ) ⇒ v0 = 0.25 + 0.6sin ω t ( V ) TYU11.7 a. b. I B1 = I B 2 = rπ = β VT I CQ IQ / 2 = 1 ⇒ I B1 = I B 2 = 6.62 μ A 151 (1 + β ) (150 )( 0.026 ) = 1 = 3.9 kΩ Rid = 2rπ = 2 ( 3.9 ) = 7.8 kΩ Ib = c. Vd 10sin ω t ( mV ) = ⇒ I b = 1.28sin ω t ( μ A ) Rid 7.8 kΩ Ricm ≅ 2 (1 + β ) R0 = 2 (151)( 50 ) ⇒ 15.1 MΩ Ib = Vcm 3sin ω t = ⇒ I b = 0.199sin ω t ( μ A ) Ricm 15.1 MΩ 430. TYU11.8 Ad = g f RD 8 = g f ( 4 ) ⇒ g f ( max ) = 2 mA / V g f ( max ) = ( 2) 2 Kn IQ 2 ⎛ 4⎞ = K n ⎜ ⎟ ⇒ K n = 2 mA / V 2 ⎝ 2⎠ TYU11.9 From Example 11-10, I Q = 0.587 mA Kn2 IQ ( 0.1)( 0.587 ) ⋅ (16) ⇒ Ad = 2.74 2 1 1 = ⇒ R0 = 85.2 kΩ For M 4 , R0 = λ4 I Q ( 0.02)( 0.587 ) Ad = 2 ⋅ RD = g m = 2 K n (VGS 2 − VTN ) = 2 ( 0.1)( 2.71 − 1) = 0.342 mA/V Acm = − ( 0.342)(16) − g m RD = ⇒ Acm = −0.0923 1 + 2 g m R0 1 + 2 ( 0.342)(85.2) ⎛ 2.74 ⎞ C M RRdB = 20 log10 ⎜ ⎟ ⇒ C M RRdB = 29.4 dB ⎝ 0.0923 ⎠ TYU11.10 1 CMRR = ⎡1 + 2 2 K n I Q ⋅ Ro ⎤ ⎦ 2⎣ CMRRdB = 60 dB ⇒ C M RR = 1000 1 1000 = ⎡1 + 2 2 ( 0.1)( 0.2 ) ⋅ R0 ⎤ ⎦ 2⎣ 2000 = 1 + 0.4 R0 ⇒ R0 ≅ 5 MΩ TYU11.11 Ro = ro 4 + ro 2 (1 + g m 4 ro 4 ) Assume I REF = I O = 100 μ A and λ = 0.01 V −1 ro 2 = ro 4 = 1 1 = ⇒1 MΩ λ I D ( 0.01)( 0.1) Let K n ( all devices ) = 0.1 mA / V 2 ( 0.1)( 0.1) = 0.2 mA / V Ro = 1000 + 1000 (1 + ( 0.2 )(1000 ) ) ⇒ 202 M Ω Then g m 4 = 2 K n I D = 2 Now VGS1 = VGS 2 = ID 0.05 + VTN = + 1 = 1.707 V 0.1 Kn VDS1 ( sat ) = VGS 1 − VTN = 1.707 − 1 = 0.707 V So vo1 ( min ) = +4 − VGS 1 + VDS 1 ( sat ) = 4 − 1.707 + 0.707 vo1 ( min ) = 3 V = 10 − I D RD = 10 − ( 0.05 ) RD ⇒ RD = 140 kΩ For a one-sided output, the differential gain is: 431. Ad = 1 g m1 RD where g m1 = 2 K n I D 2 =2 ( 0.1)( 0.05) = 0.1414 mA / V 1 ( 0.1414 )(140 ) ⇒ Ad = 9.90 2 The common-mode gain is: 2 K n I Q ⋅ RD 2 ( 0.1)( 0.1) ⋅ (140 ) Acm = = ⇒ Acm = 0.0003465 1 + 2 2 K n I Q ⋅ Ro 1 + 2 2 ( 0.1)( 0.1) ⋅ ( 202000 ) Ad = Ad ⇒ CMRRdB = 89.1 dB Acm Then CMRRdB = 20 log10 TYU11.12 I B5 = a. IQ β (1 + β ) = 0.5 ⇒ 15.3 nA (180 )(181) So I 0 = 15.3 nA b. For a balanced condition VEC 4 = VEC 3 = VEB 3 + VEB 5 ⇒ VEC 4 = 1.4 V VCE 2 = VC 2 − VE 2 = (10 − 1.4 ) − ( −0.7 ) ⇒ VCE 2 = 9.3 V TYU11.13 Ad = 2 g f ( r02 r04 ) gf = IQ 4VT = 0.2 = 1.923 mA/V 4 ( 0.026 ) r02 = VA 2 120 = = 1200 kΩ I C 2 0.1 r04 = VA 4 80 = = 800 kΩ I C 4 0.1 Ad = 2 (1.923) (1200 800 ) ⇒ Ad = 1846 TYU11.14 P = ( I Q + I REF ) ( 5 − ( −5 ) ) 10 = ( 0.1 + I REF )(10 ) ⇒ I REF = 0.9 mA R1 = 5 − 0.7 − ( −5 ) I REF = 9.3 ⇒ R1 = 10.3 k Ω 0.9 ⎛I ⎞ I Q RE = VT ln ⎜ REF ⎟ ⎜ I ⎟ ⎝ Q ⎠ 0.026 ⎛ 0.9 ⎞ ln ⎜ RE = ⎟ ⇒ RE = 0.571 kΩ 0.1 ⎝ 0.1 ⎠ ro 2 = VA 2 120 = ⇒ 2.4 M Ω I C 2 0.05 ro 4 = VA 4 80 = ⇒ 1.6 M Ω I C 4 0.05 gm = 0.05 = 1.923 mA / V 0.026 ( ) ( ) Ad = g m ro 2 ro 4 RL = (1.923) 2400 1600 90 ⇒ Ad = 158 432. TYU11.15 a. R0 = r02 r04 120 r02 = = 1.2 MΩ 0.1 80 r04 = = 0.8 MΩ 0.1 R0 = 1.2 0.8 ⇒ R0 = 0.48 MΩ b. Ad ( open circuit ) = 2 g f ( r02 r04 ) ( Ad ( with load ) = 2 g f r02 r04 RL For Ad ( with load ) = ) 1 Ad ( open circuit ) ⇒ RL = ( r02 r04 ) ⇒ RL = 0.48 MΩ 2 TYU11.16 2Kn 1 Ad = 2 ⋅ I Q ( λ2 + λ4 ) =2 2 ( 0.1) 0.1 ⋅ 1 ( 0.01 + 0.015) ⇒ Ad = 113 TYU11.17 For the MOSFET, I D = 25 − g m1 = 2 K n I D = 2 75 = 24.26 μ A 101 ( 20 )( 24.26 ) = 44.05 μ A/V For the Bipolar, I E = 100 − 25 = 75 μ A ⎛ 100 ⎞ IC = ⎜ ⎟ ( 75 ) = 74.26 μ A ⎝ 101 ⎠ (100 )( 0.026 ) rπ 2 = = 35.0 K 0.07426 0.07426 gm2 = = 2.856 mA/V 0.026 g (1 + g m 2 rπ ) ( 44.05 ) ⎡1 + ( 2.856 )( 35 ) ⎤ ⎣ ⎦ C g m = m1 = 1 + g m1rπ 1 + ( 0.04405 )( 35 ) C g m = 1.75 mA/V TYU11.18 From Figure 11.43 80 r04 = = 160 kΩ 0.5 R0 ≅ β r04 = (150 )(160 ) kΩ ⇒ R0 = 24 MΩ From Figure 11.44 1 1 r06 = = ⇒ r06 = 160 kΩ λ I D ( 0.0125 )( 0.5 ) 0.5 = 0.5 (VGS − 1) ⇒ VGS = 2 V 2 g m 6 = 2 K n (VGS − VTN ) = 2 ( 0.5 )( 2 − 1) = 1 mA / V r04 = 160 kΩ R0 = ( g m 6 )( r06 )( β r04 ) = (1)(160 )(150 )(160 ) ⇒ R0 = 3.840 MΩ 433. TYU11.19 From Equation (11.126) 2 (1 + β ) β VT 2 (121)(120 )( 0.026 ) ⇒ Ri = 1.51 MΩ Ri = = IQ 0.5 rπ 11 = β VT = IQ (120 )( 0.026 ) 0.5 = 6.24 kΩ ′ RE = rπ 11 R3 = 6.24 0.1 = 0.0984 kΩ g m11 = r011 = IQ VT = 0.5 = 19.23 mA/V 0.026 VA 120 = = 240 kΩ I Q 0.5 ′ Then RC11 = r011 (1 + g m11 RE ) = 240 ⎡1 + (19.23)( 0.0984 ) ⎤ ⎣ ⎦ = 694 kΩ rπ 8 = β VT IC 8 = (120 )( 0.026 ) 2 = 1.56 kΩ Rb8 = rπ 8 + (1 + β ) R4 = 1.56 + (121)( 5 ) = 607 kΩ Then RL 7 = RC11 Rb8 = 694 607 = 324 kΩ ⎡ 0.5 ⎤ ⎛ IQ ⎞ Then Av = ⎜ ⎥ ( 324 ) ⇒ Av = 3115 ⎟ RL 7 = ⎢ ⎢ 2 ( 0.026 ) ⎥ ⎝ 2VT ⎠ ⎣ ⎦ ⎛r +Z ⎞ R0 = R4 || ⎜ π 8 ⎟ ⎝ 1+ β ⎠ Z = RC11 RC 7 V 120 RC 7 = A = = 240 kΩ I Q 0.5 Z = 694 240 = 178 kΩ ⎛ 1.56 + 178 ⎞ R0 = 5 || ⎜ ⎟ = 5 1.48 ⇒ R0 = 1.14 kΩ ⎝ 121 ⎠ TYU11.20 ⎛ IQ Av = ⎜ ⎝ 2VT ⎞ ⎟ RL 7 ⎠ ⎛ 0.5 ⎞ 103 = ⎜ ⎜ 2 ( 0.026 ) ⎟ RL 7 ⇒ RL 7 = 104 kΩ ⎟ ⎝ ⎠ 434. Chapter 11 Problem Solutions 11.1 (a) −0.7 − ( −3) RE = 0.1 ⇒ RE = 23 K 3 − 1.5 = 0.05 ⇒ RC = 30 K RC (b) vCE 2 = 6 − iC 2 ( RC + 2 RE ) = 6 − iC 2 ( 76 ) (c) vcm ( max ) ⇒ vCB 2 = 0 ⇒ vCE 2 = 0.7 V So 0.7 = 6 − iC 2 ( 76 ) ⇒ iC 2 = 69.74 μ A ( v ( max ) − 0.7 ) − ( −3) = 2 ( 0.06974 ) ⇒ vCM ( max ) = 0.908 V 23 ( min ) ⇒ VS = −3 V ⇒ vCM ( min ) = −2.3 V CM vCM 11.2 Ad = 180, C M RRdB = 85 dB Ad 180 = ⇒ Acm = 0.01012 Acm Acm Assume the common-mode gain is negative. v0 = Ad vd + Acm vcm C M RR = 17, 783 = = 180vd − 0.01012vcm v0 = 180 ( 2sin ω t ) mV − ( 0.01012 )( 2sin ω t ) V v0 = 0.36sin ω t − 0.02024sin ω t Ideal Output: v0 = 0.360sin ω t ( V ) Actual Output: v0 = 0.340sin ω t ( V ) 11.3 a. 435. I1 = 10 − 2 ( 0.7 ) IC 2 = ⇒ I1 = 1.01 mA 8.5 I1 2 1+ β (1 + β ) = 1.01 ⇒ I C 2 ≅ 1.01 mA 2 1+ (100 )(101) ⎛ 100 ⎞ ⎛ 1.01 ⎞ IC 4 = ⎜ ⎟⎜ ⎟ ⇒ I C 4 ≅ 0.50 mA ⎝ 101 ⎠ ⎝ 2 ⎠ VCE 2 = ( 0 − 0.7 ) − ( −5 ) ⇒ VCE 2 = 4.3 V VCE 4 = ⎡5 − ( 0.5 )( 2 ) ⎤ − ( −0.7 ) ⇒ VCE 4 = 4.7 V ⎣ ⎦ b. For VCE 4 = 2.5 V ⇒ VC 4 = −0.7 + 2.5 = 1.8 V 5 − 1.8 IC 4 = ⇒ I C 4 = 1.6 mA 2 ⎛ 1+ β ⎞ ⎛ 101 ⎞ IC 2 + ⎜ ⎟ ( 2IC 4 ) = ⎜ ⎟ ( 2 )(1.6 ) ⇒ I C 2 = 3.23 mA β ⎠ ⎝ 100 ⎠ ⎝ I1 ≈ I C 2 = 3.23 mA R1 = 10 − 2 ( 0.7 ) 3.23 ⇒ R1 = 2.66 kΩ 11.4 a. Neglecting base currents 30 − 0.7 ⇒ R1 = 73.25 kΩ I1 = I 3 = 400 μ A ⇒ R1 = 0.4 VCE1 = 10 V ⇒ VC1 = 9.3 V 15 − 9.3 ⇒ RC = 28.5 kΩ RC = 0.2 b. (100 )( 0.026 ) rπ = = 13 kΩ 0.2 50 r0 ( Q3 ) = = 125 kΩ 0.4 We have Ad = (100 )( 28.5) ⇒ Ad 2 ( rπ + RB ) 2 (13 + 10 ) β RC = = 62 ⎧ ⎫ ⎪ ⎪ β RC ⎪ 1 ⎪ Acm = − ⎨ ⎬ 2r0 (1 + β ) ⎪ rπ + RB ⎪ 1+ ⎪ rπ + RB ⎪ ⎩ ⎭ ⎧ ⎫ ⎪ (100 )( 28.5 ) ⎪ 1 ⎪ ⎪ =− ⎨ ⎬ ⇒ Acm = −0.113 13 + 10 ⎪ 2 (125 )(101) ⎪ 1+ ⎪ ⎪ 13 + 10 ⎭ ⎩ ⎛ 62 ⎞ C M RRdB = 20 log10 ⎜ ⎟ ⇒ C M RRdB = 54.8 dB ⎝ 0.113 ⎠ c. 436. Rid = 2 ( rπ + RB ) = 2 (13 + 10 ) ⇒ Rid = 46 kΩ 1 ⎡ rπ + RB + 2 (1 + β ) r0 ⎤ ⎦ 2⎣ 1 = ⎡13 + 10 + 2 (101)(125 ) ⎤ ⇒ Ricm = 12.6 MΩ ⎦ 2⎣ Ricm = 11.5 vCM ( max ) ⇒ VCB = 0 so that vCM ( max ) = 5 − (a) vCM ( max ) = 3 V IQ 2 ( RC ) = 5 − (b) Vd ⎛ I CQ ⎞ Vd ⎛ 0.25 ⎞ ⎛ 0.018 ⎞ =⎜ =⎜ ⎟⋅ ⎟⎜ ⎟ = 0.08654 mA 2 ⎝ VT ⎠ 2 ⎝ 0.026 ⎠ ⎝ 2 ⎠ = ΔI ⋅ RC = ( 0.08654 ) ( 8 ) = 0.692 V ΔI = g m ⋅ ΔVC 2 (c) ⎛ 0.25 ⎞ ⎛ 0.010 ⎞ ΔI = ⎜ ⎟⎜ ⎟ = 0.04808 mA ⎝ 0.026 ⎠ ⎝ 2 ⎠ ΔVC 2 = ( 0.04808 )( 8 ) = 0.385 V 11.6 P = ( I1 + I C 4 ) (V + − V − ) I1 ≅ I C 4 so 1.2 = 2 I1 ( 6 ) ⇒ I1 = I C 4 = 0.1 mA R1 = 3 − 0.7 − ( −3) 0.1 ⇒ R1 = 53 k Ω For vCM = +1V ⇒ VC1 = VC 2 = 1 V ⇒ RC = 3 −1 ⇒ RC = 40 k Ω 0.05 One-sided output 1 0.05 Ad = g m RC where g m = = 1.923 mA / V 2 0.026 Then 1 Ad = (1.923)( 40 ) ⇒ Ad = 38.5 2 11.7 a. IE ( 2 ) + I E (85) − 5 2 5 − 0.7 IE = ⇒ I E = 0.050 mA 85 + 1 ⎛ β ⎞ ⎛ I E ⎞ ⎛ 100 ⎞⎛ 0.050 ⎞ I C1 = I C 2 = ⎜ ⎟⎜ ⎟ = ⎜ ⎟⎜ ⎟ ⎝ 1 + β ⎠ ⎝ 2 ⎠ ⎝ 101 ⎠⎝ 2 ⎠ 0 = 0.7 + Or I C1 = I C 2 = 0.0248 mA VCE1 = VCE 2 = ⎡5 − I C1 (100 ) ⎤ − ( −0.7 ) ⎣ ⎦ So VCE1 = VCE 2 = 3.22 V b. vcm ( max ) for VCB = 0 and VC = 5 − I C1 (100 ) = 2.52 V So vcm ( max ) = 2.52 V vcm ( min ) for Q1 and Q2 at the edge of cutoff ⇒ vcm ( min ) = −4.3 V (c) Differential-mode half circuits ( 0.5) 2 (8) 437. ⎛V ⎞ vd ′ = Vπ + ⎜ π + g mVπ ⎟ .RE 2 ⎝ rπ ⎠ ⎡ (1 + β ) ⎤ ′ = Vπ ⎢1 + RE ⎥ rπ ⎣ ⎦ Then − Vπ = − ( vd / 2 ) ⎡ (1 + β ) ⎤ ′ RE ⎥ ⎢1 + rπ ⎣ ⎦ vo = − g mVπ RC ⇒ Ad = rπ = β VT = I CQ β RC 1 ⋅ ′ 2 rπ + (1 + β ) RE (100 )( 0.026 ) ′ = 105 k Ω RE = 2 k Ω 0.0248 Then Ad = 11.8 a. RE = 1 (100 )(100 ) ⇒ Ad = 16.3 ⋅ 2 105 + (101)( 2 ) For v1 = v2 = 0 and neglecting base currents −0.7 − ( −10 ) 0.15 ⇒ RE = 62 kΩ b. Ad = rπ = Ad = v02 β RC = vd 2 ( rπ + RB ) β VT I CQ = (100 )( 0.026 ) 0.075 (100 )( 50 ) ⇒ Ad 2 ( 34.7 + 0.5 ) = 34.7 kΩ = 71.0 ⎡ ⎤ ⎢ ⎥ β RC ⎢ 1 ⎥ Acm = − rπ + RB ⎢ 2 RE (1 + β ) ⎥ ⎢1 + ⎥ rπ + RB ⎥ ⎢ ⎣ ⎦ ⎡ ⎤ ⎥ 1 =− ⎢ ⎥ ⇒ Acm = −0.398 34.7 + 0.5 ⎢ 2 ( 62 )(101) ⎥ ⎢1 + 34.7 + 0.5 ⎥ ⎣ ⎦ 71.0 ⇒ C M RRdB = 45.0 dB C M RRdB = 20 log10 0.398 c. Rid = 2 ( rπ + RB ) (100 )( 50 ) ⎢ Rid = 2 ( 34.7 + 0.5 ) ⇒ Rid = 70.4 kΩ Common-mode input resistance 1 Ricm = ⎡ rπ + RB + 2 (1 + β ) RE ⎤ ⎦ 2⎣ 1 = ⎡34.7 + 0.5 + 2 (101)( 62 ) ⎤ ⇒ Ricm = 6.28 MΩ ⎦ 2⎣ 11.9 438. (a) v1 = v2 = 1 V ⇒ VE = 1.6 IE = 9 − 1.6 ⇒ 18.97 μ A 390 IE = 9.49 μ A I C1 = I C 2 = 9.39 μ A 2 vC1 = vC 2 = ( 9.39 )( 0.51) − 9 = −4.21 V (b) 9.39 gm = ⇒ 0.361 mA/V 0.026 V ΔI = g m d = ( 0.361× 10−3 ) ( 0.005 ) = 1.805 μ A 2 ΔvC = (1.805 × 10−6 )( 510 × 103 ) = 0.921 V ⇒ vC 2 = −4.21 + 0.921 ⇒ −3.29 V vC1 = −4.21 − 0.921 ⇒ −5.13 V 11.10 (a) v1 = v2 = 0 I E1 = I E 2 ≅ 6 μ A β = 60 I C1 = I C 2 = 5.90 μ A vC1 = vC 2 = ( 5.90 )( 0.360 ) − 3 = −0.875 V VEC1 = VEC 2 = +0.6 − ( −0.875 ) = 1.475 V (b) (i) 5.90 gm = ⇒ 0.227 mA/V 0.026 Ad = g m RC = ( 0.227 )( 360 ) = 81.7 Acm = 0 (ii) ( 60 )( 0.026 ) g R Ad = m C = 40.8 rπ = 2 0.0059 = 264 K Acm = 11.11 − ( 0.227 )( 360 ) = −0.0442 2 ( 61)( 4000 ) 1+ 264 439. For v1 = v2 = 0.20 V I C1 = I C 2 = 0.1 mA vC1 = vC 2 = ( 0.1)( 30 ) − 10 = −7 V 0.1 gm = = 3.846 mA/V 0.026 v ΔI = g m d = ( 3.846 )( 0.008 ) ⇒ 30.77 μ A 2 ΔvC = ΔI ⋅ RC = ( 30.77 × 10−6 )( 30 × 103 ) = 0.923 V v2 ↑⇒ I C 2 ↓⇒ vC 2 ↓⇒ vC1 = −7 + 0.923 = −6.077 V vC 2 = −7 − 0.923 = −7.923 V 11.12 RC = 50 K For v1 = v2 = 0 IE = −0.7 − ( −10 ) 75 = 0.124 mA I C1 = I C 2 = 0.0615 mA 0.0615 gm = = 2.365 mA/V 0.026 (120 )( 0.026 ) rπ = = 50.7 K 0.0615 Differential Input v V v1 = d v2 = − d 2 2 Half-circuit. V ΔR ⎞ ⎛ ΔI = + g m d ⇒ ΔvC1 = −ΔI ⎜ RC + ⎟ 2 2 ⎠ ⎝ vo = ΔvC1 − ΔvC 2 ΔR ⎞ ⎛ ΔvC 2 = +ΔI ⎜ RC − ⎟ 2 ⎠ ⎝ ΔR ⎞ ΔR ⎞ ⎛ ⎛ = −ΔI ⎜ RC + ⎟ − ΔI ⎜ RC − ⎟ 2 ⎠ 2 ⎠ ⎝ ⎝ = −2ΔIRC ⎛ V ⎞ = −2 ⎜ g m d ⎟ RC 2⎠ ⎝ Ad = − g m RC = − ( 2.365 )( 50 ) = −118.25 Common-mode input. 440. ⎛V ⎞ vcm = Vπ + ⎜ π + g mVπ ⎟ ( 2 RE ) ⎝ rπ ⎠ vcm Vπ = ⎛ β⎞ 1 + ⎜ 1 + ⎟ ( 2 RE ) ⎝ rπ ⎠ ΔI = g mVπ = g m vcm β vcm = rπ + (1 + β )( 2 RE ) ⎛1+ β ⎞ 1+ ⎜ ⎟ ( 2 RE ) ⎝ rπ ⎠ ΔR ⎞ ⎛ − β ⎜ RC + ⎟ ⋅ vcm 2 ⎠ ⎝ ΔvC1 = −ΔIR1 = rπ + (1 + β )( 2 RE ) ΔvC 2 ΔR ⎞ ⎛ − β ⎜ RC − ⎟ vcm 2 ⎠ ⎝ = −ΔIR2 = rπ + (1 + β )( 2 RE ) vo = ΔvC1 − ΔvC 2 ΔR ⎞ ΔR ⎞ ⎛ ⎛ − β ⎜ RC + ⎟ vcm ⎟ vcm + β ⎜ RC − 2 ⎠ 2 ⎠ ⎝ ⎝ = [ ] ⎛ ΔR ⎞ −2 β ⎜ ⎟ vcm ⎝ 2 ⎠ = rπ + (1 + β )( 2 RE ) Acm = − (120 )( 0.5 ) − βΔR = rπ + (1 + β )( 2 RE ) 50.7 + (121)( 2 )( 75 ) = −0.0032966 118.25 C M RR = = 35,870.5 0.0032966 C M R R ∫ = 91.1 dB dB 11.13 v1 = v2 = 0 IE = −0.7 − ( −10 ) 75 = 0.124 mA I C1 = I C 2 = 0.0615 mA 0.0615 gm = = 2.365 mA/V 0.026 Δg m = 0.01 gm g m1 = 2.377 mA/V g m 2 = 2.353 mA/V rπ = [ ] (120 )( 0.026 ) 0.0615 = 50.7 K 441. Vd 2 V = − g m1 d Rc 2 Vd = + gm2 Rc 2 ΔI = g m ΔvC1 ΔvC 2 vo = ΔvC1 − ΔvC 2 = − g m1 Vd V RC − g m 2 d RC 2 2 Vd RC ( g m1 + g m 2 ) 2 R −50 Ad = − C ( g m1 + g m 2 ) = ( 2.377 + 2.353) ⇒ Ad = −118.25 2 2 Common-Mode − g m1 RC vcm − g m 2 RC vcm ΔvC1 = ΔvC 2 = ⎛ 1+ β ⎞ ⎛ 1+ β ⎞ 1+ ⎜ 1+ ⎜ ⎟ ( 2 RE ) ⎟ ( 2 RE ) ⎝ rπ ⎠ ⎝ rπ ⎠ =− − ( g m1 − g m 2 ) RC − ( 2.377 − 2.353) ( 50 ) vo = Acm = = vcm ⎛ 1+ β ⎞ ⎛ 121 ⎞ 1+ ⎜ 1+ ⎜ ⎟ ( 2 )( 75 ) ⎟ ( 2 RE ) ⎝ 50.7 ⎠ ⎝ rπ ⎠ −1.2 = ⇒ Acm = −0.003343 358.99 C M R R ∫ = 91 dB dB 11.14 (a) v1 = v2 = 0 vE = +0.7 V 5 − 0.7 IE = = 4.3 mA 1 I C1 = I C 2 = 2.132 mA vC1 = vC 2 = ( 2.132 )(1) − 5 = −2.87 V (b) v1 = 0.5, v2 = 0 Q2 on Q1 off ⎛ 120 ⎞ I C1 = 0, I C 2 = 4.3 ⎜ ⎟ mA = 4.264 mA ⎝ 121 ⎠ vC1 = −5 V vC 2 = ( 4.264 ) (1) − 5 vC 2 = −0.736 V 2.132 = 82.0 mA/V 0.026 (82.0 ) v V ΔI = g m d ΔvC = ΔI ⋅ RC = g m d ⋅ RC = ⋅ Vd (1) = 41.0Vd 2 2 2 Vd = 0.015 ⇒ Δvc = 0.615 V (c) vE ≈ 0.7 V gm = vC 2 ↓ vC1 ↑ vC1 = −2.87 + 0.615 = −2.255 V vC 2 = −2.87 − 0.615 = −3.485 V 11.15 442. (a) gm = IC 1 = = 38.46 mA/V VT 0.026 Ad = vo 1 = = 100 vd 0.01 Ad = g m RC 100 = 38.46 RC Rc = 2.6 K (b) With v1 = v2 = 0 vC1 = vC 2 = 10 − (1)( 2.6 ) = 7.4 V ⇒ vcm ( max ) = 7.4 V 11.16 a. i. ( v01 − v02 ) = 0 ii. I C1 = I C 2 = 1 mA v01 − v02 = ⎡V + − I C1 RC1 ⎤ − ⎡V + − I C 2 RC 2 ⎤ ⎣ ⎦ ⎣ ⎦ = I C ( RC 2 − RC1 ) = (1)( 7.9 − 8 ) ⇒ v01 − v02 = −0.1 V b. ⎛v ⎞ I 0 = ( I S 1 + I S 2 ) exp ⎜ BE ⎟ ⎝ VT ⎠ ⎛v ⎞ 2 × 10−3 So exp ⎜ BE ⎟ = −13 −13 ⎝ VT ⎠ 10 + 1.1× 10 = 9.524 × 109 ⎛v I C1 = I S 1 exp ⎜ BE ⎝ VT ⎞ −13 9 ⎟ = (10 )( 9.524 × 10 ) ⇒ I C1 = 0.952 mA ⎠ I C 2 = (1.1× 10−13 )( 9.524 × 109 ) ⇒ I C 2 = 1.048 mA i. v01 − v02 = I C 2 RC 2 − I C1 RC1 ⇒ v01 − v02 = (1.048 − 0.952 )( 8 ) ⇒ v01 − v02 = 0.768 V ii. v01 − v02 = (1.048 )( 7.9 ) − ( 0.952 )( 8 ) v01 − v02 = 8.279 − 7.616 ⇒ v01 − v02 = 0.663 V 11.17 From Equation (11.12(b)) IQ iC 2 = 1 + evd / VT 1 0.90 = 1 + evd / VT 1 So evd / VT = − 1 = 0.111 0.90 vd = VT ln ( 0.111) = ( 0.026 ) ln ( 0.111) ⇒ vd = −0.0571 V 11.18 From Example 11.2, we have 443. vd ( max ) 1 − − vd ( max ) / 0.026 4 ( 0.026 ) 1 + e = 0.02 v ( max ) 0.5 + d 4 ( 0.026 ) 0.5 + ⎡ v ( max ) ⎤ 1 0.98 ⎢ 0.5 + d ⎥= − vd ( max ) / 0.026 4 ( 0.026 ) ⎥ 1 + e ⎢ ⎣ ⎦ 0.490 + 9.423vd ( max ) = By trial and error vd ( max ) = 23.7 mV 1 1+ e − vd ( max ) / 0.026 11.19 a. For I1 = 1 mA, VBE3 = 0.7 V 20 − 0.7 R1 = ⇒ R1 = 19.3 kΩ 1 ⎛ I ⎞ 0.026 ⎛ 1 ⎞ V R2 = T ⋅ ln ⎜ 1 ⎟ = ⋅ ln ⎜ ⎟ ⇒ R2 = 0.599 kΩ ⎜I ⎟ 0.1 IQ ⎝ 0.1 ⎠ ⎝ Q⎠ b. (180 )( 0.026 ) rπ 4 = = 46.8 kΩ 0.1 0.1 gm = = 3.846 mA/V 0.026 100 r04 = ⇒ 1 MΩ 0.1 From Chapter 10 R0 = r04 ⎡1 + g m ( RE rπ 4 ) ⎤ ⎣ ⎦ RE rπ 4 = 0.599 46.8 = 0.591 R0 = (1) ⎡1 + ( 3.846 )( 0.591) ⎤ = 3.27 MΩ ⎣ ⎦ r01 = 100 ⇒ 2 MΩ 0.05 ⎡ ⎛ r ⎞⎤ Ricm ≅ ⎡(1 + β ) R0 ⎤ ⎢(1 + β ) ⎜ 01 ⎟⎥ ⎣ ⎦ ⎝ 2 ⎠⎦ ⎣ = ⎣(181)( 3.27 ) ⎦ ⎣(181)(1) ⎤ ⎡ ⎤ ⎡ ⎦ = 592 181 ⇒ Ricm = 139 MΩ (c) From Eq. (11.32(b)) − g m RC Acm = 2 (1 + β ) Ro 1+ rπ + RB 0.05 = 1.923 mA / V 0.026 (180 )( 0.026 ) rπ = = 93.6 k Ω 0.05 RB = 0 gm = Then Acm = − (1.923)( 50 ) ⇒ Acm = −0.00760 2 (181)( 3270 ) 1+ 93.6 444. 11.19 For vCM = 3.5 V and a maximum peak-to-peak swing in the output voltage of 2 V, we need the quiescent collector voltage to be VC = 3.5 + 1 = 4.5 V Assume the bias is ±10 V , and I Q = 0.5 mA. Then I C = 0.25 mA 10 − 4.5 ⇒ RC = 22 k Ω 0.25 (100 )( 0.026 ) = 10.4 k Ω In this case, rπ = 0.25 Then (100 )( 22 ) Ad = = 101 So gain specification is met. 2 (10.4 + 0.5 ) Now RC = For CMRRdB = 80 dB ⇒ 1 ⎡ (1 + β ) I Q Ro ⎤ 1 ⎡ (101)( 0.5 ) Ro ⎤ ⎥ ⇒ Ro = 1.03 M Ω ⎢1 + ⎥ = ⎢1 + 2⎣ VT β ⎦ 2 ⎢ ( 0.026 )(100 ) ⎥ ⎣ ⎦ Need to use a Modified Widlar current source. Ro = ro ⎡1 + g m ( RE1 rπ ) ⎤ ⎣ ⎦ CMRR = 104 = If VA = 100V , then ro = rπ = (100 )( 0.026 ) 100 = 200 k Ω 0.5 = 5.2 k Ω 0.5 0.5 gm = = 19.23 mA / V 0.026 Then 1030 = 200 ⎡1 + (19.23)( RE1 rπ ) ⎤ ⇒ RE1 rπ = 0.216 k Ω = RE1 5.2 ⇒ RE1 = 225 Ω ⎣ ⎦ Also let RE 2 = 225 Ω and I REF ≅ 0.5 mA 11.20 (a) (b) RE = −0.7 − ( −10 ) 0.25 ⇒ RE = 37.2 k Ω 445. ⎛1+ β ⎞ Vπ 1 V V Ve + g mVπ 1 + π 2 + g mVπ 2 = e or (1) ⎜ ⎟ (Vπ 1 + Vπ 2 ) = rπ rπ RE rπ ⎠ RE ⎝ ⎛ r ⎞ Vπ 1 V1 − Ve = ⇒ Vπ 1 = ⎜ π ⎟ (V1 − Ve ) rπ RB + rπ rπ + RB ⎠ ⎝ Vπ 2 = V2 − Ve Then ⎛1+ β ⎝ rπ ⎞ ⎡ rπ ⎤ V (V1 − Ve ) + (V2 − Ve )⎥ = e ⎟⎢ ⎠ ⎣ rπ + RB ⎦ RE From this, we find (1) ⎜ V1 + Ve = rπ + RB ⋅ V2 rπ ⎡ rπ + RB r + RB ⎤ +1+ π ⎢ ⎥ rπ ⎦ ⎣ RE (1 + β ) Now Vo = − g mVπ 2 RC = − g m RC (V2 − Ve ) We have rπ = (120 )( 0.026 ) 0.125 ≅ 25 k Ω, gm = 0.125 = 4.81 mA / V 0.026 (i) Set V1 = Vd V and V2 = − d 2 2 Then ⎛ ⎛ 25 + 0.5 ⎞ ⎞ Vd ⎜ 1 − ⎜ 25 ⎟ ⎟ ( −0.02 ) ⎝ ⎠⎠ ⎝ Ve = = 2 2.026 ⎡ 25 + 0.5 25 + 0.5 ⎤ +1+ ⎢ ⎥ 25 ⎦ ⎣ ( 37.2 )(121) Vd 2 So Ve = −0.00494Vd Now V ⎛ V ⎞ Vo = − ( 4.81)( 50 ) ⎜ − d − ( −0.00494 )Vd ⎟ ⇒ Ad = o = 119 Vd ⎝ 2 ⎠ (ii) Set V1 = V2 = Vcm Then ⎛ 25 + 0.5 ⎞ Vcm ⎜ 1 + ⎟ V ( −2.02 ) 25 ⎠ ⎝ Ve = = cm 2.02567 ⎡ 25 + 0.5 25 + 0.5 ⎤ +1+ ⎢ ⎥ 25 ⎦ ⎣ ( 37.2 )(121) Ve = Vcm ( 0.9972 ) Then Vo = − ( 4.81)( 50 ) ⎡Vcm − Vcm ( 0.9972 ) ⎤ ⎣ ⎦ or Acm = Vo = −0.673 Vcm 11.21 From Equation (11.18) 446. v0 = vC 2 − vC1 = g m RC vd gm = I CQ VT For I Q = 2 mA, I CQ = 1 mA 1 = 38.46 mA/V 0.026 Now 2 = ( 38.46 ) RC ( 0.015 ) Then g m = So RC = 3.47 kΩ Now VC = V + − I C RC = 10 − (1)( 3.47 ) = 6.53 V For VCB = 0 ⇒ vcm ( max ) = 6.53 V 11.22 The small-signal equivalent circuit is A KVL equation: v1 = Vπ 1 − Vπ 2 + v2 v1 − v2 = Vπ 1 − Vπ 2 A KCL equation Vπ 1 V + g mVπ 1 + π 2 + g mVπ 2 = 0 rπ rπ ⎛1 ⎞ + g m ⎟ = 0 ⇒ Vπ 1 = −Vπ 2 ⎝ rπ ⎠ (Vπ 1 + Vπ 2 ) ⎜ Then v1 − v2 = 2Vπ 1 ⇒ Vπ 1 = 1 1 ( v1 − v2 ) and Vπ 2 = − ( v1 − v2 ) 2 2 At the v01 node: v01 v01 − v02 + + g mVπ 1 = 0 RC RL ⎛ 1 ⎛ 1 1 ⎞ v01 ⎜ + ⎟ − v02 ⎜ ⎝ RL ⎝ RC RL ⎠ At the v02 node: ⎞ 1 ⎟ = g m ( v2 − v1 ) ⎠ 2 (1) v02 v02 − v01 + + g mVπ 2 = 0 RC RL ⎛ 1 ⎛ 1 ⎞ 1 1 ⎞ v02 ⎜ + ⎟ − v01 ⎜ ⎟ = g m ( v1 − v2 ) RC RL ⎠ ⎝ RL ⎠ 2 ⎝ From (1): (2) 447. ⎛ R ⎞ 1 v02 = v01 ⎜ 1 + L ⎟ − g m RL ( v2 − v1 ) ⎝ RC ⎠ 2 Substituting into (2) ⎛ R ⎞⎛ 1 ⎛ 1 ⎛ 1 1 ⎞ 1 1 ⎞ + + v01 ⎜1 + L ⎟ ⎜ ⎟ − g m RL ( v2 − v1/ ) ⎜ ⎟ − v01 ⎜ RC ⎠ ⎝ RC RL ⎠ 2 RC RL ⎠ ⎝ RL ⎝ ⎝ ⎡ ⎛ RL ⎛ 1 RL ⎞⎤ 1 ⎞ 1 + 2 + + 1⎟ ⎥ v01 ⎜ ⎟ = g m ( v1 − v2 ) ⎢1 − ⎜ ⎝ RC RC RC ⎠ 2 ⎠⎦ ⎣ ⎝ RC v01 ⎛ RL ⎞ 1 ⎛ RL ⎞ ⎜2+ ⎟ = − gm ⎜ ⎟ ( v1 − v2 ) RC ⎝ RC ⎠ 2 ⎝ RC ⎠ For v1 − v2 = vd 1 − g m RL v01 Av1 = = 2 vd ⎛ RL ⎞ ⎜2+ ⎟ RC ⎠ ⎝ 1 g m RL v02 From symmetry: Av 2 = = 2 vd ⎛ RL ⎞ ⎜2+ ⎟ RC ⎠ ⎝ Then Av = v02 − v01 g m RL = vd ⎛ RL ⎞ ⎜2+ ⎟ RC ⎠ ⎝ 11.23 The small-signal equivalent circuit is KVL equation: v1 = Vπ 1 − Vπ 2 + v2 or v1 − v2 = Vπ 1 − Vπ 2 KCL equation: ⎞ 1 ⎟ = g m ( v1 − v2 ) ⎠ 2 448. Vπ 1 V + g mVπ 1 + g mVπ 2 + π 2 = 0 rπ rπ ⎛1 ⎞ + g m ⎟ = 0 ⇒ Vπ 1 = −Vπ 2 ⎝ rπ ⎠ (Vπ 1 + Vπ 2 ) ⎜ Then v1 − v2 = −2Vπ 2 or Vπ 2 = − Now v0 = − g mVπ 2 ( RC 1 = g m ( RC 2 For v1 − v2 ≡ vd ⇒ Ad = RL ) 1 ( v1 − v2 ) 2 RL )( v1 − v2 ) v0 1 = g m ( RC vd 2 RL ) 11.23 a. 10 − 7 ⇒ RD = 6 kΩ 0.5 I Q = I D1 + I D 2 ⇒ I Q = 1 mA RD = b. 10 = I D ( 6 ) + VDS − VGS and VGS = ID + VTN Kn For I D = 0.5 mA, VGS = 0.5 + 2 = 3.12 V 0.4 and VDS = 10.12 Load line is actually nonlinear. c. Maximum common-mode voltage when M 1 and M 2 reach the transition point, or VDS ( sat ) = VGS − VTN = 3.12 = 2 = 1.12V Then vcm = v02 − vDS ( sat ) + VGS = 7 − 1.12 + 3.12 Or vcm ( max ) = 9 V Minimum common-mode voltage, voltage across I Q becomes zero. So vcm ( min ) = −10 + 3.12 ⇒ vcm ( min ) = −6.88 V 11.24 449. We have VC 2 = − g mVπ 2 RC = − g m (Vb 2 − Ve ) RC and VC1 = − g mVπ 1 RC = − g m (Vb1 − Ve ) RC Then V0 = VC 2 − VC1 = − g m (Vb 2 − Ve ) RC − ⎡ − g m (Vb1 − Ve ) RC ⎤ ⎣ ⎦ = g m RC (Vb1 − Vb 2 ) Differential gain Ad = V0 = g m RC Vb1 − Vb 2 Common-mode gain Acm = 0 11.25 (a) vcm = 3 V ⇒ VC1 = VC 2 = 3 V 10 − 3 Then RC = ⇒ RC = 70 k Ω 0.1 (b) CMRRdB = 75 dB ⇒ CMRR = 5623 Now CMRR = 5623 = 1 ⎡ (1 + β ) I Q Ro ⎤ ⎢1 + ⎥ β VT 2⎣ ⎦ 1 ⎡ (151)( 0.2 ) Ro ⎤ ⎢1 + ⎥ ⇒ Ro = 1.45 M Ω 2 ⎢ (150 )( 0.026 ) ⎥ ⎣ ⎦ Use a Widlar current source. ′ Ro = ro [1 + g m RE ] Let VA of current source transistor be 100 V. 100 0.2 = 500 k Ω, g m = = 7.69 mA / V Then ro = 0.2 0.026 (150 )( 0.026 ) rπ = = 19.5 k Ω 0.2 ′⎦ ′ So 1450 = 500 ⎡1 + ( 7.69 ) RE ⎤ ⇒ RE = 0.247 k Ω ⎣ ′ Now RE = RE rπ ⇒ 0.247 = RE 19.5 ⇒ RE = 250Ω ⎛I Then I Q RE = VT ln ⎜ REF ⎜ I ⎝ Q ⎞ ⎟ ⎟ ⎠ ⎛ I REF ⎞ ⎟ ⇒ I REF = 1.37 mA ⎟ ⎝ ( 0.2 ) ⎠ ( 0.2 )( 0.250 ) = ( 0.026 ) ln ⎜ ⎜ Then R1 = 10 − 0.7 − ( −10 ) 11.26 At terminal A. RTHA = RA R = 1.37 ⇒ R1 = 14.1 k Ω R (1 + δ ) ⋅ R R (1 + δ ) + R = R (1 + δ ) 2+δ ≅ R = 5 kΩ 2 Variation in RTH is not significant ⎛ RA ⎞ + R (1 + δ )( 5 ) 5 (1 + δ ) VTHA = ⎜ = ⎟V = R (1 + δ ) + R 2+δ ⎝ RA + R ⎠ 450. At terminal B. R = 5 kΩ 2 ⎛ R ⎞ + VTHB = ⎜ ⎟ V = 2.5 V ⎝R+R⎠ From Eq. (11.27) − β RC (V2 − V1 ) VO = where V2 = VTHB and V1 = VTHA 2 ( rπ + RB ) RTHB = R R = RB = 5 k Ω, rπ = So VO = (120 )( 0.026 ) 0.25 − (120 )( 3)(V2 − V1 ) 2 (12.5 + 5 ) = 12.5 k Ω = −10.3 (V2 − V1 ) We can find V2 − V1 = VTHB − VTHA ⎡ 5 (1 + δ ) ⎤ VTHB − VTHA = 2.5 − ⎢ ⎥ ⎣ 2+δ ⎦ 2.5 ( 2 + δ ) − 5 (1 + δ ) 2.5δ − 5δ = = 2+δ 2+δ −2.5δ ≅ = −1.25δ 2 Then VO = − (10.3)( −1.25 ) δ = 12.9δ So for −0.01 ≤ δ ≤ 0.01 We have −0.129 ≤ VO 2 ≤ 0.129 V 11.27 a. Rid = 2rπ rπ = (180 )( 0.026 ) 0.2 So Rid = 46.8 kΩ = 23.4 kΩ Assuming rμ → ∞, then b. Ricm ≅ ⎡(1 + β ) R0 ⎤ ⎣ ⎦ Ricm = ⎡(181)(1) ⎤ ⎣ ⎦ = 181 ⇒ Ricm = 181 MΩ 11.28 (a) 10 − 0.7 − ( −10 ) I1 = = 0.5 ⇒ R1 = 38.6 K R1 R2 = (b) 0.026 ⎛ 0.5 ⎞ ln ⎜ ⎟ ⇒ R2 = 236 Ω 0.14 ⎝ 0.14 ⎠ 451. Ricm ≈ (1 + β ) Ro 0.14 = 5.385 mA/V 0.026 (180 )( 0.026 ) rπ 4 = = 33.4 K 0.14 ′ RE = 33.4 0.236 = 0.234 K ′ Ro = ro 4 (1 + g m 4 RE ) g m 4 = 100 = 714 K 0.14 Ro = 714 ⎡1 + ( 5.385 )( 0.234 ) ⎤ ⎣ ⎦ ro 4 = = 1614 K Ricm = (181)(1614 ) ≈ 292 MΩ (c) Acm = − g m1 RC 2 (1 + β ) Ro 1+ rπ 1 g m1 = rπ 1 = 0.07 = 2.692 mA/V 0.026 (180 )( 0.026 ) 0.07 − ( 2.692 )( 40 ) 2 (181)(1614 ) 1+ 66.86 Acm = −0.0123 = 66.86 K Acm = 11.29 Ad 1 = g m1 ( R1 rπ 3 ) g m1 = rπ 3 = Ad 2 = I Q1 / 2 VT β VT IQ2 / 2 = 19.23I Q1 = 2 (100 )( 0.026 ) IQ 2 = 5.2 IQ 2 IQ 2 / 2 g m 3 R2 , g m3 = = 19.23I Q 2 2 VT (19.23) I Q 2 ⋅ R2 ⇒ I Q 2 R2 = 3.12 V 2 Maximum vo 2 − vo1 = ±18 mV for linearity Then 30 = vo3 ( max ) = ( ±18 )( 30 ) mV ⇒ ±0.54 V so I Q 2 R2 = 3.12 V is OK. From Ad 1 : 452. ⎛ ⎛ 5.2 ⎞ ⎞ ⎜ R1 ⎜ ⎜I ⎟ ⎟ ⎟ ⎜ ⎝ Q2 ⎠ ⎟ 20 = 19.23I Q1 ( R1 rπ 3 ) = 19.23I Q1 ⎜ ⎟ ⎜ R + ⎛ 5.2 ⎞ ⎟ ⎜ ⎟ ⎜ 1 ⎜ IQ 2 ⎟ ⎟ ⎝ ⎠⎠ ⎝ 19.23I Q1 R1 ( 5.2 ) 20 = I Q 2 R1 + 5.2 I Q1 Let 2 ⋅ R1 = 5V ⇒ I Q1 R1 = 10 V Then 20 = 19.23 (10 )( 5.2 ) I Q 2 R1 + 5.2 Now I Q1 R1 = 10 ⇒ R1 = ⇒ I Q 2 R1 = 44.8 V 10 I Q1 ⎛ 10 ⎞ ⎛ IQ2 ⎞ = 4.48 So I Q 2 ⎜ ⎟ = 44.8 ⇒ ⎜ ⎜I ⎟ ⎟ ⎜I ⎟ ⎝ Q1 ⎠ ⎝ Q1 ⎠ Let I Q1 = 100 μ A, I Q 2 = 448 μ A Then I Q 2 R2 = 3.12 ⇒ R2 = 6.96 k Ω I Q1 R1 = 10 ⇒ R1 = 100 k Ω 11.30 a. 20 − VGS 3 2 I1 = = 0.25 (VGS 3 − 2 ) 50 2 20 − VGS 3 = 12.5 (VGS 3 − 4VGS 3 + 4 ) 2 12.5VGS 3 − 49VGS 3 + 30 = 0 VGS 3 = 49 ± ( 49 ) 2 − 4 (12.5 )( 30 ) 2 (12.5 ) ⇒ VGS 3 = 3.16 V 20 − 3.16 ⇒ I1 = I Q = 0.337 mA 50 IQ = ⇒ I D1 = 0.168 mA 2 I1 = I D1 0.168 = 0.25 (VGS 1 − 2 ) ⇒ VGS1 = 2.82 V VDS 4 = −2.82 − ( −10 ) ⇒ VDS 4 = 7.18 V 2 VD1 = 10 − ( 0.168 )( 24 ) = 5.97 V VDS1 = 5.97 − ( −2.82 ) ⇒ VDS 1 = 8.79 V (b) (c) 453. Max vCM ⇒ VDS 1 = VDS 2 = VDS ( sat ) = VGS1 − VTN 2.82 − 2 = 0.82 V Now VD1 = 10 − ( 0.168 )( 24 ) = 5.97 V VS ( max ) = 5.97 − VDS1 ( sat ) = 5.97 − 0.82 VS ( max ) = 5.15 V vCM ( max ) = VS ( max ) + VGS1 = 5.15 + 2.82 vCM ( max ) = 7.97 V vCM ( min ) = V − + VDS 4 ( sat ) + VGS 1 VDS 4 ( sat ) = VGS 4 − VTN = 3.16 − 2 = 1.16 V Then vCM ( min ) = −10 + 1.16 + 2.82 ⇒ vCM ( min ) = −6.02 V 11.31 a. I D1 = I D 2 = 120 μ A = 100 ( VGS1 − 1.2 ) ⇒ VGS 1 = VGS 2 = 2.30 V 2 For v1 = v2 = −5.4 V and VDS1 = VDS 2 = 12 V ⇒ −5.4 − 2.30 + 12 = 4.3 V = VD 10 − 4.3 ⇒ RD = 47.5 kΩ 0.12 I Q = I D1 + I D 2 ⇒ I Q = I1 = 240 μ A RD = I1 = 240 = 200 (VGS 3 − 1.2 ) ⇒ VGS 3 = 2.30 V 2 R1 = 20 − 2.3 ⇒ R1 = 73.75 kΩ 0.24 b. r04 = 1 λ IQ ΔI Q = = 1 ( 0.01)( 0.24 ) = 416.7 kΩ 1 5.4 ⋅ ΔVDS = ⇒ ΔI Q ≅ 13 μ A r04 416.7 11.32 (a) I Q = 160 μ A k′ ⎛ W ⎞ 2 I D = n ⎜ ⎟ (VGS − VTN ) 2⎝L⎠ 80 2 80 = ( 4 )(VOS − 0.5 ) 2 80 = 160 (Vo5 − 0.5 ) 2 80 + 0.5 = 1.207 V 160 5−2 = 37.5 K VDS = 2 − ( −1.207 ) = 3.21 V RD = 0.08 (c) VDS ( sat ) = VGS − VTN = 1.207 − 0.5 = 0.707 V VGS = Then VS = VO 2 − VDS ( sat ) = 2 − 0.707 = +1.29 V And v1 = v2 = vcm = VGS + VS = 1.207 + 1.29 vcm = 2.50 V (b) 454. 11.33 vD = 5 − ( 0.2 )( 8 ) = 3.4 V VGS = ID + VTN Kn 0.2 + 0.8 = 1.694 V 0.25 VDS ( sat ) = VGS − VTN = 1.694 − 0.8 = 0.894 V VS = VD − VDS ( sat ) = 3.4 − 0.894 = 2.506 = vCM = VS + VGS = 2.506 + 1.694 ⇒ vCM = 4.2 V (b) ΔvD = ΔI D ⋅ RD ΔI D = g m ⋅ Vd 2 gm = 2 Kn I D =2 ( 0.25)( 0.2 ) = 0.4472 mA/V ΔI D = ( 0.4472 )( 0.05 ) ⇒ 22.36 μ A ΔvD = ( 22.36 × 10−6 )( 8 × 103 ) = 0.179 V vD 2 = 3.4 + ΔvD vD 2 = 3.4 + 0.179 ⇒ vD 2 = 3.58 V (c) vd = −50 mV ΔI D = − ( 0.4472 )( 0.025 ) ⇒ −11.18 μ A ΔvD = − (11.18 × 10−6 )( 8 × 103 ) = −0.0894 V vD 2 = 3.4 − 0.0894 ⇒ vD 2 = 3.31 V 11.34 a. I D1 = I D 2 = 0.5 mA v01 − v02 = ⎡V + − I D1 RD1 ⎤ − ⎡V + − I D 2 RD 2 ⎤ ⎣ ⎦ ⎣ ⎦ v01 − v02 = I D 2 RD 2 − I D1 RD1 = I D ( RD 2 − RD1 ) i. RD1 − RD 2 = 6 kΩ, v01 − v02 = 0 ii. RD1 = 6 kΩ, RD 2 = 5.9 kΩ v01 − v02 = ( 0.5 )( 5.9 − 6 ) ⇒ v01 − v02 = −0.05 V b. 455. K n1 = 0.4 mA / V 2 , K n 2 = 0.44 mA / V 2 VGS1 = VGS 2 I Q = ( K n1 + K n 2 )(VGS − VTN ) 2 1 = ( 0.4 + 0.44 )(VGS − VTN ) ⇒ (VGS − VTN ) = 1.19 2 2 I D1 = ( 0.4 )(1.19 ) = 0.476 mA I D 2 = ( 0.44 )(1.19 ) = 0.524 mA i. RD1 = RD 2 = 6 kΩ v01 − v02 = ( 0.524 − 0.476 )( 6 ) ⇒ v01 − v02 = 0.288 V ii. RD1 = 6 kΩ, RD 2 = 5.9 kΩ v01 − v02 = ( 0.524 )( 5.9 ) − ( 0.476 )( 6 ) = 3.0916 − 2.856 ⇒ v01 − v02 = 0.236 V 11.35 (a) From Equation (11.69) ⎛ K Kn iD 2 1 = − ⋅ vd 1 − ⎜ n ⎜ 2IQ IQ 2 2IQ ⎝ 0.90 = 0.50 − ⎞ 2 ⎟ vd ⎟ ⎠ ⎡ 0.1 ⎤ 2 0.1 ⋅ vd 1 − ⎢ ⎥ vd 2 ( 0.25 ) ⎢ 2 ( 0.25 ) ⎥ ⎣ ⎦ 2 +0.40 = − ( 0.4472 ) vd 1 − ( 0.2 ) vd 2 0.8945 = −vd 1 − ( 0.2 ) vd Square both sides 2 2 0.80 = vd (1 − [ 0.2] vd ) ( 0.2 ) ( vd2 ) 2 vd = 2 2 − vd + 0.80 = 0 1 ± 1 − 4 ( 0.2 )( 0.80 ) 2 ( 0.2 ) = 4V 2 or 1V 2 Then vd = ± 2 V or ± 1 V But vd max = IQ kn = 0.25 = 1.58 0.1 So vd = ±1V, ⇒ vd = −1V b. 11.36 From part (a), vd ,max = 1.58 V 456. ⎛i ⎞ d ⎜ D1 ⎟ ⎜I ⎟ ⎝ Q⎠= dvd ⎛ K Kn ⋅ 1− ⎜ n ⎜ 2I 2IQ ⎝ Q ) vd vd =0 Kn 2IQ = So linear ⎞ 2 ⎟ vd + ( ⎟ ⎠ Kn iD1 1 = + ⋅ vd IQ 2 2 IQ ⎡1 ⎤ ⎛K ⎞ 2 Kn Kn 1 + ⋅ vd ( max ) − ⎢ + ⋅ vd ( max ) ⋅ 1 − ⎜ n ⎟vd ( max ) ⎥ 2 2IQ 2 IQ ⎢2 ⎥ ⎝ 2I n ⎠ ⎣ ⎦ = 0.02 Then Kn 1 + ⋅v 2 2 I Q d ( max ) ⎡1 ⎤ ⎡1 ⎛K Kn Kn 0.98 ⎢ + ⋅ vd ( max ) ⎥ = ⎢ + ⋅ vd ( max ) ⋅ 1 − ⎜ n ⎜ 2I 2IQ 2IQ ⎢2 ⎥ ⎢2 ⎝ Q ⎣ ⎦ ⎣ 0.49 + 0.98 ⎤ ⎞ 2 ⎟ vd ( max ) ⎥ ⎟ ⎥ ⎠ ⎦ ⎡1 ⎤ ⎛ 0.15 ⎞ 2 0.15 0.15 ⋅ vd ( max ) = ⎢ + ⋅ vd ( max ) ⋅ 1 − ⎜ ⎟ vd ( max ) ⎥ ⎜ 2 ( 0.2 ) ⎟ 2 ( 0.2 ) 2 ( 0.2 ) ⎢2 ⎥ ⎝ ⎠ ⎣ ⎦ 2 0.49 + 0.600 vd ( max ) = 0.50 + 0.6124 vd ( max ) ⋅ 1 − ( 0.6124 ) vd ( max ) 2 0.600 vd ( max ) = 0.010 + 0.6124 vd ( max ) ⋅ 1 − ( 0.6124 ) vd ( max ) By trial and error vd ( max ) ≈ 0.429 V 11.37 (b) gm = 2 K p I D = 2 ( 0.05 )( 0.008696 ) = 0.0417 mA/V Vd = ( 0.0417 )( 0.05 ) = 0.002085 mA 2 ΔvD = ( 0.002085 )( 510 ) = 1.063 ΔI = g m vD 2 ↑⇒ vD 2 = 1.063 − 4.565 = −3.502 V vD1 = −1.063 − 4.565 = −5.628 V 9 = I S RS + VSG + 1 I S = 2I D 8 = 2 K P RS (VSG + VTP ) + VSG 2 8 = ( 2 )( 0.05 )( 390 )(VSG − 0.8 ) + VSG 2 2 8 = 39 (VSG − 1.6VSG + 0.64 ) + VSG 2 39VSG − 61.4VSG + 16.96 = 0 VSG = 61.4 ± 3769.96 − 4 ( 39 )(16.96 ) 2 ( 39 ) = 1.217 V VS = 2.217 I S = 0.01739 mA I D1 = I D 2 ⇒ 8.696 μ A vD1 = vD 2 = ( 8.696 )( 0.510 ) − 9 = −4.565 V (b) 457. g m = 2 K P I DQ = 2 ( 0.05 )( 0.008696 ) = 0.0417 mA/V Vd = ( 0.0417 )( 0.05 ) = 0.002085 mA 2 ΔvD = ( 0.002085 )( 510 ) = 1.063 V ΔvD = ΔI D ⋅ RD ΔI D = g m ⋅ v1 ↑, I D1 ↓, vD1 ↓ vD1 = −4.565 − 1.063 = −5.628 V vD 2 = −4.565 + 1.063 = −3.502 V 11.38 (a) v1 = v2 = 0 I D = K n (VSG + VTP ) ID = 6 μA 2 6 + 0.4 = VSG 30 VSG = 0.847 V VS = +0.847 V vD = I D RD − 3 = ( 6 )( 0.36 ) − 3 = −0.84 V VSD = VS − vD = 0.847 − ( −0.84 ) vSD = 1.69 V (b) (i) Ad = g m RD g m = 2 K n I D =2 ( 30 )( 6 ) = 26.83 μ A/V Ad = ( 26.83)( 0.36 ) ⇒ Ad = 9.66 Acm = 0 (ii) ( 26.83)( 0.36 ) g R Ad = m D = ⇒ Ad = 4.83 2 2 − ( 26.83)( 0.36 ) − g m RD Acm = = = −0.0448 1 + 2 g m RO 1 + 2 ( 26.83)( 4 ) 11.39 458. For v1 = v2 = −0.30 V I D1 = I D 2 = 0.1 mA VSG = ID − VTP KP 0.1 +1 = 2 V 0.1 = vD 2 = ( 0.1)( 30 ) − 10 = −7 V = vD1 gm = 2 K p I D = 2 ( 0.1)( 0.1) = 0.2 mA/V ⎛V ⎞ ΔI D = g m ⎜ d ⎟ = ( 0.2 )( 0.1) = 0.02 mA ⎝ 2⎠ ΔvD = ( ΔI D ) RD = ( 0.02 )( 30 ) = 0.6 V vD 2 ↑⇒ vD 2 = −7 + 0.6 ⇒ vD 2 = −6.4 V vD1 = −7 − 0.6 ⇒ vD1 = −7.6 V 11.40 For v1 = v2 = 0 0 = VGS + 2 I D RS − 10 10 = VGS + 2 K n RS (VGS − VTN ) 2 = VGS + 2 ( 0.15 )( 75 )(VGS − 1) 2 2 22.5VGS − 44VGS + 12.5 = 0 So VGS = 1.61 V and I D = ( 0.15 )(1.61 − 1) ⇒ 55.9 μ A 2 gm = 2 Kn I D = 2 ( 0.15 )( 0.0559 ) g m = 0.1831 mA/V Use Half-circuits – Differential gain ΔR ⎞ ⎛V ⎞⎛ vD1 = − g m ⎜ d ⎟ ⎜ RD + ⎟ 2 ⎠ ⎝ 2 ⎠⎝ ΔR ⎞ ⎛V ⎞⎛ vo 2 = g m ⎜ d ⎟ ⎜ RD − ⎟ 2 ⎠ ⎝ 2 ⎠⎝ vo = vD1 − vD 2 = − g mVd RD v Ad = o = − g m RD Vd Now – Common-Mode Gain 459. Vi = Vgs + g mVgs ( 2 RS ) = Vcm Vcm Vgs = 1 + g m ( 2 RS ) vD1 vD 2 ΔR ⎞ ⎛ − g m ⎜ RD + D ⎟ Vcm 2 ⎠ ⎝ = 1 + g m ( 2 RS ) ΔR ⎞ ⎛ − gm ⎜ RD − D ⎟ Vcm 2 ⎠ ⎝ = 1 + g m ( 2 RS ) vO = vD1 − vD 2 So vo = Acm = − g m ( ΔRD ) Vcm 1 + g m ( 2 RD ) − g m ( ΔRD ) vo = Vcm 1 + g m ( 2 RS ) Then Ad = − ( 0.1831)( 50 ) = −9.16 Acm = − ( 0.1831)( 0.5 ) 1 + ( 0.1831)( 2 )( 75 ) = −0.003216 C M R R ∫ = 69.1 dB bB 11.41 a. Ad = g m ( r02 r04 ) r02 = VA 2 150 = = 375 kΩ I C 2 0.4 r04 = VA 4 100 = = 250 kΩ I C 4 0.4 gm = IC 2 0.4 = = 15.38 mA/V VT 0.026 Ad = (15.38 ) ( 375 250 ) ⇒ Ad = 2307 b. RL = r02 r04 = 375 250 ⇒ RL = 150 kΩ 11.41 From 11.40 I D1 = I D 2 = 55.9 μ A g m = 0.183 mA/V 460. ⎛ +V ⎞ ΔvD 2 = + g m 2 ⎜ d ⎟ RD ⎝ 2 ⎠ V V vO = ΔvD1 − ΔvD 2 = − g m1 d RD − g m 2 d RD 2 2 −V −V ⎛ Δg Δg ⎞ ⎞ ⎛ vO = d ⋅ RD ( g m 2 + g m1 ) = d ⋅ RD ⎜ g m − m + ⎜ g m − m ⎟ ⎟ 2 2 2 2 ⎠⎠ ⎝ ⎝ Ad = − g m RD = − ( 0.183) ( 50 ) = −9.15 Ad : ΔvD1 = − g m1 Vd ⋅ RD 2 Δg ⎞ Δg ⎞ ⎛ ⎛ − ⎜ g m + M ⎟ RD vcm ⎜ g m − M ⎟ RD vCM 2 ⎠ 2 ⎠ ⎝ ⎝ = + 1 + g m ( 2 RS ) 1 + g m ( 2 RS ) ACM : vO = ΔvD1 − ΔvD 2 −Δg m RD vO = vcm 1 + g m ( 2 RS ) Acm = Acm = Δg m = ( 0.01) ( 0.183) = 0.00183 − ( 0.00183) ( 50 ) 1 + ( 0.183)( 2 ) ( 75 ) = −0.003216 C M R R ∫ = 69.1 dB dB 11.42 (a) v1 = v2 = 0 5 = 2 I D RS + VSG 5 = 2 K p RS (VSG + VTP ) + VSG 2 2 5 = 2 ( 0.5 )( 2 ) (VSG − 1.6VSG + 0.64 ) + VSG 2 5 = 2VSG − 2.2VSG + 1.28 2 2VSG − 2.2VSG − 3.72 = 0 VSG = 2.2 ± 4.84 + 4 ( 2 )( 3.72 ) 2 ( 2) VSG = 2.02 V vS = 2.02 V, 5 − 2.02 = 1.49 mA 2 = I D 2 = 0.745 mA IS = I D1 vD1 = vD 2 = ( 0.745 (1) − 5 ) ⇒ vD1 = vD 2 = −4.26 V (b) 5 = I S RS + VSG 2 5 = ( I D1 + I D 2 ) RS + VSG 2 2 2 5 = ⎡ K p (VSG1 + VTP ) + K p (VSG 2 + VTP ) ⎤ RS + VSG 2 ⎣ ⎦ VSG1 = VSG 2 − 1 5 = ( 0.5 )( 2 ) ⎡(VSG 2 − 1.8 ) + (VSG 2 − 0.8 ) ⎤ + VSG 2 ⎣ ⎦ 2 2 5 = ⎡VSG 2 − 3.6VSG 2 + 3.24 + VSG 2 − 1.6VSG 2 + 0.64 ⎤ + VSG 2 ⎣ ⎦ 2 2 5 = 2VSG 2 − 4.2VSG 2 + 3.88 2 2VSG 2 − 4.2VSG 2 − 1.12 = 0 VSG 2 = 4.2 ± 17.64 + 4 ( 2 ) (1.12 ) 2 ( 2) 2 461. VSG 2 = 2.339 V VSG1 = 1.339 V vS = 2.339 V I D1 I D1 vD1 vD1 = 0.5 (1.339 − 0.8 ) = 0.1453 mA = ( 0.1453)(1) − 5 = −4.855 V 2 I D2 I D2 vD 2 vD 2 = 0.5 ( 2.339 − 0.8 ) = 1.184 mA = (1.184 ) (1) − 5 = −3.816 V 2 (c) ΔI = g m Vd 2 gm = 2 K p I D vS ≈ 2.02 V = 2 ( 0.5 )( 0.745 ) g m = 1.22 mA/V ΔI = (1.22 )( 0.1) = 0.122 mA ΔvD = ( ΔI ) RD = ( 0.122 )(1) = 0.122 V vD 2 ↓ vD1 ↑ vD1 = −4.26 + 0.122 vD 2 = −4.26 − 0.122 vD1 = −4.138 V vD 2 = −4.382 V 11.43 a. gf = IQ 4VT ⇒ I Q = g f ( 4VT ) = ( 8 )( 4 )( 0.026 ) ⇒ I Q = 0.832 mA Neglecting base currents. 30 − 0.7 R1 = ⇒ R1 = 35.2 kΩ 0.832 V 100 r04 = r02 = A = = 240 kΩ b. I CQ 0.416 462. gm = I CQ VT = 0.416 = 16 mA / V 0.026 Ad = g m ( r02 || r04 ) = 16 ( 240 || 240 ) ⇒ Ad = 1920 Rid = 2rπ , rπ = (180 )( 0.026 ) 0.416 = 11.25 kΩ ⇒ Rid = 22.5 kΩ R0 = r02 || r04 ⇒ R0 = 120 kΩ c. Max. common-mode voltage when VCB = 0 for Q1 and Q2 . Therefore vcm ( max ) = V + − VEB ( Q3 ) = 15 − 0.7 vcm ( max ) = 14.3 V Min. common-mode voltage when VCB = 0 for Q5 . Therefore vcm ( min ) = 0.7 + 0.7 + ( −15 ) = −13.6 V So −13.6 ≤ vcm ≤ 14.3 V 1 (1 + β )( 2 R0 ) 2 V 100 R0 = A = = 120 kΩ I Q 0.832 Ricm ≅ Ricm = (181)(120 ) ⇒ Ricm = 21.7 MΩ 11.43 (a) gm = 2 Kn I D =2 ( 0.4 )(1) g m = 1.265 mA/V v 1 Ad = o = = 10 vd 0.1 Ad = g m RD 10 = (1.265 ) RD RD = 7.91 K (b) Quiescent v1 = v2 = 0 vD1 = vD 2 = 10 − (1)( 7.91) = 2.09 V VGS = ID 1 + VTN = + 0.8 = 2.38 V Kn 0.4 VDS ( sat ) = 2.38 − 0.8 = 1.58 So vcm = vD − VDS ( sat ) + VGS = 2.09 − 1.58 + 2.38 vcm = 2.89 V 11.44 463. g m RD 2 For vCM = 2.5 V IQ = 0.25 mA I D1 = I D 2 = 2 Ad = Let VD1 = VD 2 = 3 V , then RD = Then 100 = g m ( 28 ) 2 k′ ⎛ W And g m = 2 n ⎜ 2⎝L 10 − 3 ⇒ RD = 28 k Ω 0.25 ⇒ g m = 7.14 mA / V ⎞ ⎟ ID ⎠ ⎛ 0.080 ⎞ ⎛ W ⎞ 7.14 = 2 ⎜ ⎟ ⎜ ⎟ ( 0.25 ) ⇒ ⎝ 2 ⎠⎝ L ⎠ ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = ⎜ ⎟ = 1274 (Extremely large transistors to meet the gain requirement.) ⎝ L ⎠1 ⎝ L ⎠ 2 Need ACM = 0.10 From Eq. (11.64(b)) g m RD ACM = 1 + 2 g m Ro ( 7.14 )( 28) ⇒ Ro = 140 k Ω 1 + 2 ( 7.14 ) Ro For the basic 2-transistor current source 1 1 Ro = ro = = = 200 k Ω λ I Q ( 0.01)( 0.5 ) So 0.10 = This current source is adequate to meet common-mode gain requirement. 11.45 Not in detail, Approximation looks good. a. −V − ( −5 ) 2 I S = GS 1 and I S = 2 I D = 2 K n (VGS 1 − VTN ) RS 5 − VGS 1 2 = 2 ( 0.050 )(VGS 1 − 1) 20 2 5 − VGS 1 = 2 (VGS1 − 2VGS1 + 1) 2 2VGS1 − 3VGS 1 − 3 = 0 VGS1 = 3± ( 3) 2 + 4 ( 2 )( 3) 2 ( 2) ⇒ VGS1 = 2.186 V 5 − 2.186 ⇒ I S = 0.141 mA 20 I I D1 = I D 2 = S ⇒ I D1 = I D 2 = 0.0704 mA 2 v02 = 5 − ( 0.0704 )( 25 ) ⇒ v02 = 3.24 V IS = b. g m = 2 K n (VGS − VTN ) = 2 ( 0.05 )( 2.186 − 1) g m = 0.119 mA/V 1 1 r0 = = = 710 kΩ λ I DQ ( 0.02 )( 0.0704 ) 464. Vgs1 = v1 − VS , Vgs 2 = v2 − VS v01 v −V + g mVgs1 + 01 S = 0 RD r0 ⎛ 1 V 1⎞ v01 ⎜ + ⎟ + g m ( v1 − VS ) − S = 0 r0 ⎝ RD r0 ⎠ v02 v − VS + g mVgs 2 + 02 =0 RD r0 ⎛ 1 V 1⎞ v02 ⎜ + ⎟ + g m ( v2 − VS ) − S = 0 RD r0 ⎠ r0 ⎝ v − V v − VS V + g mVgs 2 = S g mVgs1 + 01 S + 02 r0 r0 RS (1) (2) g m ( v1 − VS ) + v01 v02 2VS V + − + g m ( v2 − VS ) = S r0 r0 r0 RS g m ( v1 + v2 ) + v01 v02 + = VS r0 r0 ⎧ 2 1 ⎫ ⎨2 g m + + ⎬ r0 RS ⎭ ⎩ (3) From (1) ⎛ 1⎞ VS ⎜ g m + ⎟ − g m v1 r0 ⎠ v01 = ⎝ ⎛ 1 1⎞ + ⎟ ⎜ ⎝ RD r0 ⎠ Then ⎛ 1⎞ VS ⎜ g m + ⎟ − g m v1 r0 ⎠ ⎧ v 2 1 ⎫ g m ( v1 + v2 ) + ⎝ + 02 = VS ⎨2 g m + + ⎬ (3) r0 r0 RS ⎭ ⎛ 1 1⎞ ⎩ r0 ⎜ + ⎟ RD r0 ⎠ ⎝ 465. ⎛ 1 ⎛ ⎛ 1 ⎧ 1⎞ 1⎞ 1⎞ 2 1 ⎫ ⎛ 1 1⎞ g m ( v1 + v2 ) r0 ⎜ + ⎟ + VS ⎜ g m + ⎟ − g m v1 + v02 ⎜ + ⎟ = VS ⎨2 g m + + ⎬ ⋅ r0 ⎜ + ⎟ r0 ⎠ r0 RS ⎭ ⎝ RD r0 ⎠ ⎝ RD r0 ⎠ ⎝ ⎝ RD r0 ⎠ ⎩ ⎧ ⎫ ⎛ 1 ⎛ r ⎞ r0 ⎞ ⎛ 1⎞ 2 1 ⎞⎛ 1 ⎞⎪ ⎪⎛ g m ( v1 + v2 ) ⎜ 1 + 0 ⎟ − g m v1 + v02 ⎜ + ⎟ = VS ⎨⎜ 2 g m + + ⎟ ⎜1 + ⎟ − ⎜ gm + ⎟⎬ RD ⎠ RD r0 ⎠ r0 RS ⎠ ⎝ RD ⎠ ⎝ r0 ⎠ ⎪ ⎪⎝ ⎝ ⎝ ⎩ ⎭ ⎛ 1 ⎧ ⎛ r r ⎞ r r 1⎞ 2 1 2 1⎫ g m ⎜ v1 ⋅ 0 + v2 + v2 ⋅ 0 ⎟ + v02 ⎜ + ⎟ = VS ⎨2 g m + + + 2gm ⋅ 0 + + 0 − gm − ⎬ RD ⎠ r0 RS RD RD RS RD r0 ⎭ ⎝ RD ⎝ RD r0 ⎠ ⎩ ⎧ ⎫ ⎛ 1 ⎛ r r ⎞ r0 ⎞ 2 1⎞ 1 1 ⎛ ⎪ g m ⎜ v1 ⋅ 0 + v2 + v2 ⋅ 0 ⎟ + v02 ⎜ + ⎟ = VS ⎨2 g m + + (1 + g m r0 )⎪ (4) ⎬ ⎜1 + ⎟+ RD ⎠ r0 RS ⎝ RD ⎠ RD ⎪ ⎪ ⎝ RD ⎝ RD r0 ⎠ ⎩ ⎭ ⎛ 1 ⎛ 1⎞ 1⎞ Then substituting into (2), v02 ⎜ + ⎟ + g m v2 = VS ⎜ g m + ⎟ r0 ⎠ ⎝ RD r0 ⎠ ⎝ 710 ⎤ 1 ⎤ ⎡ 710 ⎡1 + v2 + v2 (4) Substitute numbers: ( 0.119 ) ⎢ v1 ⎥ + v02 ⎢ 25 + 710 ⎥ 25 ⎦ ⎣ 25 ⎣ ⎦ ⎧ ⎫ 1 1 ⎛ 710 ⎞ 2 = VS ⎨0.119 + + ⎜1 + ⎟ + ⎡1 + ( 0.119 )( 710 ) ⎤ ⎬ ⎣ ⎦ 710 20 ⎝ 25 ⎠ 25 ⎩ ⎭ ( 0.119 ) [ 28.4v1 + 29.4v2 ] + ( 0.0414 ) v02 = VS {0.1204 + 1.470 + 6.8392} = VS ( 8.4296 ) or VS = 0.4010v1 + 0.4150v2 + 0.00491v02 1 ⎞ 1 ⎞ ⎛ 1 ⎛ (2) Then v02 ⎜ + ⎟ + ( 0.119 ) v2 = VS ⎜ 0.119 + ⎟ 710 ⎠ ⎝ 25 710 ⎠ ⎝ v02 ( 0.0414 ) + v2 ( 0.119 ) = ( 0.1204 ) [ 0.401v1 + 0.4150v2 + 0.00491v02 ] v02 ( 0.0408 ) = ( 0.04828 ) v1 − ( 0.0690 ) v2 v02 = (1.183) v1 − (1.691) v2 vd 2 vd v2 = vcm − 2 v ⎞ v ⎞ ⎛ ⎛ So v02 = (1.183) ⎜ vcm + d ⎟ − (1.691) ⎜ vcm − d ⎟ 2⎠ 2⎠ ⎝ ⎝ Or v02 = 1.437vd − 0.508vcm ⇒ Ad = 1.437, Acm = −0.508 Now v1 = vcm + ⎛ 1.437 ⎞ C M R RdB = 20 log10 ⎜ ⎟ ⇒ C M R RdB = 9.03 dB ⎝ 0.508 ⎠ 11.46 KVL: 466. v1 = Vgs1 − Vgs 2 + v2 So v1 − v2 = Vgs1 − Vgs 2 KCL: g mVgs1 + g mVgs 2 = 0 ⇒ Vgs1 = −Vgs 2 1 1 So Vgs1 = ( v1 − v2 ) , Vgs 2 = − ( v1 − v2 ) 2 2 Now v02 v02 − v01 + = − g mVgs 2 RD RL ⎛ 1 1 ⎞ v01 = v02 ⎜ + ⎟− ⎝ RD RL ⎠ RL v01 v01 − v02 + = − g mVgs1 RD RL ⎛ 1 1 ⎞ v02 = v01 ⎜ + ⎟− RD RL ⎠ RL ⎝ ⎛ R ⎞ From (1): v01 = v02 ⎜ 1 + L ⎟ + g m RLVgs 2 RD ⎠ ⎝ (1) (2) Substitute into (2): ⎛ ⎛ 1 v02 R ⎞⎛ 1 1 ⎞ 1 ⎞ − g mVgs1 = v02 ⎜1 + L ⎟ ⎜ + + ⎟ + g m RL ⎜ ⎟ Vgs 2 − RD ⎠ ⎝ RD RL ⎠ RD RL ⎠ RL ⎝ ⎝ ⎛ ⎛ 1 R ⎞⎛ 1 ⎞ R 1 ⎞ − g m ⋅ ( v1 − v2 ) + g m ⎜ 1 + L ⎟ ⎜ ⎟ ( v1 − v2 ) = v02 ⎜ + L + ⎟ 2 ⎝ RD ⎠ ⎝ 2 ⎠ ⎝ RD RD RD ⎠ 1 ⋅ g m RL v02 ⎛ v02 RL ⎞ 1 ⎛ RL ⎞ gm ⎜ = 2 ( v1 − v2 ) = ⎜ 2 + ⎟ ⇒ Ad 2 = ⎟ 2 ⎝ RD ⎠ RD ⎝ RD ⎠ v1 − v2 ⎛ RL ⎞ ⎜2+ ⎟ RD ⎠ ⎝ 1 − ⋅ g m RL v01 From symmetry Ad 1 = = 2 v1 − v2 ⎛ RL ⎞ ⎜2+ ⎟ RD ⎠ ⎝ Then Av = 11.47 v02 − v01 g m RL = v1 − v2 ⎛ RL ⎞ ⎜2+ ⎟ RD ⎠ ⎝ 467. v1 − v2 = Vgs1 − Vgs 2 and g mVgs1 + g mVgs 2 = 0 ⇒ Vgs1 = −Vgs 2 Then v1 − v2 = −2Vgs 2 Or Vgs 2 = − 1 ( v1 − v2 ) 2 v0 = − g mVgs 2 ( RD RL ) = Or Ad = gm ( RD RL ) ( v1 − v2 ) 2 gm ( RD RL ) 2 11.48 From Equation (11.64(a)), Ad = We need Ad = Then 10 = Kn IQ 2 ⋅ RD 2 = 10 0.2 K n ( 0.5 ) ⋅ RD or K n ⋅ RD = 20 2 If we set RD = 20 k Ω, then K n = 1 mA / V 2 For this case VD = 10 − ( 0.25 )( 20 ) = 5 V 0.25 + 1 = 1.5 V 1 VDS ( sat ) = VGS − VTN = 1.5 − 1 = 0.5 V VGS = Then vcm ( max ) = VD − VDS ( sat ) + VGS = 5 − 0.5 + 1.5 Or vcm ( max ) = 6 V 11.49 Vd 1 = − g mVgs1 RD = − g m RD (V1 − Vs ) Vd 2 = − g mVgs 2 RD = − g m RD (V2 − Vs ) Now Vo = Vd 2 − Vd 1 = − g m RD (V2 − Vs ) − ( − g m RD (V1 − Vs ) ) Vo = g m RD (V1 − V2 ) Define V1 − V2 ≡ Vd V Then Ad = o = g m RD and Acm = 0 Vd 11.49 Ad = g m ( r02 r04 ) g m = 2 kn I DQ =2 ( 0.12 )( 0.075 ) = 0.1897 mA/V 1 1 r02 = = = 889 kΩ λn I DQ ( 0.015 )( 0.075 ) r04 = 1 λ p I DQ = 1 = 667 kΩ 0.02 )( 0.075 ) ( Ad = ( 0.1897 ) ( 889 667 ) ⇒ Ad = 72.3 11.50 (a) 468. ⎛ K′ ⎞⎛W K n1 = K n 2 = ⎜ n ⎟ ⎜ ⎝ 2 ⎠⎝ L VGS1 = VGS 2 = ⎞ ⎛ 0.080 ⎞ 2 ⎟=⎜ ⎟ (10 ) = 0.40 mA / V ⎠ ⎝ 2 ⎠ ID 0.1 + VTN = + 1 = 1.5 V Kn 0.4 VDS1 ( sat ) = 1.5 − 1 = 0.5 V For vCM = +3 V ⇒ VD1 = VD 2 = vCM − VGS 1 + VDS 1 ( sat ) = 3 − 1.5 + 0.5 ⇒ VD1 = VD 2 = 2 V RD = 10 − 2 ⇒ RD = 80 k Ω 0.1 (b) 1 g m RD and g m = 2 ( 0.4 )( 0.1) = 0.4 mA / V 2 1 Then Ad = ( 0.4 )( 80 ) = 16 2 16 C M R RdB = 45 ⇒ C M R R = 177.8 = Acm Ad = So Acm = 0.090 Acm = g m RD 1 + 2 g m Ro 0.090 = ( 0.4 )(80 ) 1 + 2 ( 0.4 ) Ro ⇒ Ro = 443 k Ω If we assume λ = 0.01 V −1 for the current source transistor, then 1 1 ro = = = 500 k Ω λ I Q ( 0.01)( 0.2 ) So the CMRR specification can be met by a 2-transistor current source. ⎛W ⎞ ⎛W ⎞ Let ⎜ ⎟ = ⎜ ⎟ = 1 ⎝ L ⎠3 ⎝ L ⎠ 4 IQ 0.2 ⎛ 0.080 ⎞ 2 Then K n 3 = K n 4 = ⎜ + VTN = + 1 = 3.24 V ⎟ (1) = 0.040 mA / V and VGS 3 = 2 ⎠ 0.04 K n3 ⎝ For vCM = −3 V , VD 3 = −3 − VGS1 = −3 − 1.5 = −4.5 V ⇒ VDS 3 ( min ) = −4.5 − ( −10 ) = 5.5 V > VDS 3 ( sat ) So design is OK. ⎛W ⎞ On reference side: For ⎜ ⎟ ≥ 1, VGS ( max ) = 3.24 V ⎝L⎠ 20 − VGS 3 = 20 − 3.24 = 16.76 V Then 16.67 = 5.17 ⇒ We need six transistors in series. 3.24 469. 20 − 3.24 = 2.793 V 6 2 ⎛ K′ ⎞⎛W ⎞ = ⎜ n ⎟ ⎜ ⎟ (VGS − VTN ) ⎝ 2 ⎠⎝ L ⎠ VGS = I REF 2 ⎛ 0.080 ⎞⎛ W ⎞ ⎛W ⎞ 0.2 = ⎜ ⎟⎜ ⎟ ( 2.793 − 1) ⇒ ⎜ ⎟ = 1.56 for each of the 6 transistors. ⎝ 2 ⎠⎝ L ⎠ ⎝L⎠ 11.51 Ad = 1 g m RD 2 gm = 2 Kn I D = 2 ( 0.25 )( 0.25) = 0.50 mA / V 1 ( 0.50 )( 3) = 0.75 2 From Problem 11.26 Ad = 470. V1 = VA = 5 (1 + δ ) 2+δ , V2 = VB = 2.5 V and V1 − V2 = 1.25δ Then Vo 2 = Ad ⋅ (V1 − V2 ) = ( 0.75 )(1.25δ ) = 0.9375δ So for −0.01 ≤ δ ≤ 0.01 −9.375 ≤ Vo 2 ≤ 9.375 mV 11.52 From previous results v −v Ad 1 = o 2 o1 = g m1 R1 = 2 K n1 I Q1 ⋅ R1 = 20 v1 − v2 and Ad 2 = Set I Q1 R1 2 vo3 1 1 2 K n3 I Q 2 ⋅ R2 = 30 = g m 3 R2 = 2 vo 2 − vo1 2 I Q 2 R2 = 5 V and 2 = 2.5 V Let I Q1 = I Q 2 = 0.1 mA Then R1 = 100 k Ω, R2 = 50 k Ω 2 ⎛ 0.06 ⎞ ⎛ W ⎞ ⎛ 20 ⎞ ⎛W ⎞ ⎛W ⎞ Then 2 ⎜ ⎟ ⎜ ⎟ ( 0.1) = ⎜ ⎟ ⇒ ⎜ ⎟ = ⎜ ⎟ = 6.67 ⎝ 2 ⎠ ⎝ L ⎠1 ⎝ 100 ⎠ ⎝ L ⎠1 ⎝ L ⎠ 2 2 ⎛ 2 ( 30 ) ⎞ ⎛ 0.060 ⎞ ⎛ W ⎞ ⎛W ⎞ ⎛W ⎞ and 2 ⎜ ⎟ ⇒ ⎜ ⎟ = ⎜ ⎟ = 240 ⎟ ⎜ ⎟ ( 0.1) = ⎜ ⎝ 2 ⎠ ⎝ L ⎠3 ⎝ L ⎠3 ⎝ L ⎠ 4 ⎝ 50 ⎠ 11.53 a. iD1 ⎛ v ⎞ = I DSS ⎜ 1 − GS 1 ⎟ VP ⎠ ⎝ ⎛ v ⎞ iD 2 = I DSS ⎜ 1 − GS 2 ⎟ VP ⎠ ⎝ iD1 − iD 2 2 2 ⎛ v ⎞ ⎛ v ⎞ = I DSS ⎜1 − GS 1 ⎟ − I DSS ⎜ 1 − GS 2 ⎟ VP ⎠ VP ⎠ ⎝ ⎝ I DSS = ( vGS 2 − vGS1 ) VP I DSS =− VP ⋅ vd = I DSS ( −VP ) ⋅ vd iD1 + iD 2 = I Q ⇒ iD 2 = I Q − iD1 ( iD1 − I Q − iD1 ) 2 = I DSS ( −VP ) 2 2 ⋅ vd iD1 − 2 iD1 ( I Q − iD1 ) + ( I Q − iD1 ) = Then iD1 ( I Q − iD1 ) = Square both sides I DSS ( −VP ) 2 2 ⋅ vd ⎤ I 1⎡ 2 ⎢ I Q − DSS 2 ⋅ vd ⎥ 2⎢ ⎥ ( −VP ) ⎦ ⎣ 471. 2 ⎤ I 1⎡ 2 i − iD1 I Q + ⎢ I Q − DSS 2 ⋅ vd ⎥ = 0 4⎢ ( −VP ) ⎥ ⎣ ⎦ 2 D1 ⎤ I ⎛ 1⎞⎡ 2 I Q ± I − 4 ⎜ ⎟ ⎢ I Q − DSS 2 ⋅ vd ⎥ ⎝ 4⎠⎢ ( −VP ) ⎥ ⎣ ⎦ 2 2 Q iD1 = 2 2 2 ⎡ ⎤ 2 1 2 ⎢ 2 2 I Q I DSS vd ⎛ I DSS vd ⎞ ⎥ ⎟ iD1 = IQ − IQ − ± +⎜ 2 2 ⎢ 2 2 ( −VP ) ⎜ ( −VP ) ⎟ ⎥ ⎝ ⎠ ⎦ ⎣ Use + sign IQ iD1 1 2 I Q I DSS 2 = + ⋅ vd 2 2 ( −VP )2 iD1 = IQ IQ 2 + ⎛ I ⎞ 2 − ⎜ DSS 2 ⋅ vd ⎟ ⎜ ( −V ) ⎟ P ⎝ ⎠ 2 2 2 2 2 2 2 2 I DSS ⎛ I DSS 1 IQ vd −⎜ ⎜ I 2 ( −VP ) IQ ⎝ Q ⎞ ⎛ vd ⎞ ⎟ ⎜ ⎟ ⎟ V ⎠ ⎝ P⎠ ⎞ 2 I DSS ⎛ I DSS −⎜ ⎟ ⋅ vd ⎜ I IQ ⎠ ⎝ Q ⎞ ⎛ vd ⎞ ⎟ ⎜ ⎟ ⎟ V ⎠ ⎝ P⎠ ⎞ 2 I DSS ⎛ I DSS −⎜ ⎟ ⋅ vd ⎜ I IQ ⎠ ⎝ Q ⎞ ⎛ vd ⎞ ⎟ ⎜ ⎟ ⎟ V ⎠ ⎝ P⎠ Or iD1 1 ⎛ 1 = +⎜ I Q 2 ⎝ −2VP We had iD 2 = I Q − iD1 Then iD 2 1 ⎛ 1 = −⎜ I Q 2 ⎝ −2VP b. If iD1 = I Q , then 1 ⎛ 1 1= +⎜ 2 ⎝ −2VP ⎞ 2 I DSS ⎛ I DSS −⎜ ⎟ ⋅ vd ⎜ I IQ ⎠ ⎝ Q 2 I DSS ⎛ I DSS VP = vd −⎜ ⎜ I IQ ⎝ Q Square both sides 2 ⎞ ⎛ vd ⎞ ⎟ ⎜ ⎟ ⎟ V ⎠ ⎝ P⎠ 2 2 ⎞ ⎛ vd ⎞ ⎟ ⎜ ⎟ ⎟ V ⎠ ⎝ P⎠ 2 472. VP 2 ⎛ I DSS ⎜ ⎜ I ⎝ Q 2 vd = ⎡ 2I ⎛I = v ⎢ DSS − ⎜ DSS ⎜ I ⎢ IQ ⎝ Q ⎣ 2 d 2 2 2 ⎞ ⎛ vd ⎞ ⎤ ⎟ ⎜ ⎟ ⎥ ⎟ V ⎠ ⎝ P⎠ ⎥ ⎦ 2 ⎞ ⎛ 1 ⎞ 2 2 2 I DSS 2 ⋅ vd + VP ⎟ ⎜ ⎟ ( vd ) − ⎟ V IQ ⎠ ⎝ P⎠ ⎛ 2I 2 I DSS ± ⎜ DSS ⎜ I IQ ⎝ Q 2 ⎞ ⎛I ⎟ − 4 ⎜ DSS ⎟ ⎜ I ⎠ ⎝ Q ⎛ 2I 2 ⎜ DSS ⎜ IQ ⎝ 2 2 =0 2 ⎞ ⎛ 1 ⎞ 2 ⎟ ⎜ ⎟ (VP ) ⎟ V ⎠ ⎝ P⎠ 2 ⎞ ⎛ 1 ⎞ ⎟ ⎜ ⎟ ⎟ ⎝ VP ⎠ ⎠ 2 2 ⎛ IQ ⎞ 2 vd = (VP ) ⎜ ⎟ ⎝ I DSS ⎠ 1/ 2 ⎛ IQ ⎞ Or vd = VP ⎜ ⎟ ⎝ I DSS ⎠ c. For vd small, IQ 1 IQ 2 I DSS + ⋅ ⋅ vd iD1 ≈ IQ 2 2 ( −VP ) gf = diD1 d vd vd → 0 = 2 I DSS 1 IQ ⋅ ⋅ IQ 2 ( −VP ) ⎛ 1 ⎞ I Q I DSS Or ⇒ g f ( max ) = ⎜ ⎟ 2 ⎝ −VP ⎠ 11.53 Ad = g m ( ro 2 Ro ) Want Ad = 400 From Example 11.15, ro 2 = 1 M Ω Assuming that g m = 0.283 mA / V for the PMOS from Example 11.15, then Ro = 285 M Ω. ⎛ k ′ ⎞⎛ W ⎞ So 400 = g m (1000 285000 ) ⇒ g m = 0.4014 mA / V = 2 ⎜ n ⎟ ⎜ ⎟ I DQ ⎝ 2 ⎠ ⎝ L ⎠1 ⎛ 0.080 ⎞ ⎛ W ⎞ ⎛W ⎞ ⎛W ⎞ 0.04028 = ⎜ ⎟ ⎜ ⎟ ( 0.1) ⇒ ⎜ ⎟ = ⎜ ⎟ = 10.1 ⎝ 2 ⎠ ⎝ L ⎠1 ⎝ L ⎠1 ⎝ L ⎠ 2 11.54 a. I Q = I D1 + I D 2 ⇒ I Q = 1 mA v0 = 7 = 10 − ( 0.5 ) RD ⇒ RD = 6 kΩ b. ⎛ 1 ⎞ I Q ⋅ I DSS g f ( max ) = ⎜ ⎟ 2 ⎝ −VP ⎠ ⎛ 1 ⎞ (1)( 2 ) g f ( max ) = ⎜ ⎟ ⇒ g f ( max ) = 0.25 mA/V 2 ⎝ 4⎠ c. g R Ad = m D = g f ( max ) ⋅ RD 2 Ad = ( 0.25 )( 6 ) ⇒ Ad = 1.5 473. 11.55 a. IS = −VGS − ( −5 ) RS ⎛ V ⎞ = ( 2 ) I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ ⎛ V ⎞ 5 − VGS = ( 2 )( 0.8 )( 20 ) ⎜ 1 − GS ⎟ ⎜ ( −2 ) ⎟ ⎝ ⎠ 1 2 ⎞ ⎛ 5 − VGS = ( 2 )16 ⎜1 + VGS + VGS ⎟ 4 ⎝ ⎠ 2 8VGS + 33VGS + 27 = 0 VGS = 2 2 −33 ± 1089 − 4 ( 8 )( 27 ) 2 (8) = −1.125 V IS = 5 − ( −1.125 ) 20 = 0.306 mA I D1 = I D 2 = 0.153 mA vo 2 = 1.17 V (b) 11.56 Equivalent circuit and analysis is identical to that in problem 11.36. 1 ⋅ g m RL Ad 2 = 2 ⎛ RL ⎞ ⎜2+ ⎟ RD ⎠ ⎝ 1 − ⋅ g m RL Ad 1 = 2 ⎛ RL ⎞ ⎜2+ ⎟ RD ⎠ ⎝ Av = v02 − v01 g m RL = vd ⎛ RL ⎞ ⎜2+ ⎟ RD ⎠ ⎝ 11.57 (a) Ad = g m ( ro 2 ro 4 ) 0.1 = 3.846 mA/V 0.026 120 ro 2 = = 1200 K 0.1 80 ro 4 = = 800 K 0.1 Ad = ( 3.846 ) (1200 800 ) gm = Ad = 1846 (b) 474. For Ad = 923 = ( 3.846 ) (1200 800 RL ) 240 = 480 RL = 480 RL ⇒ RL = 480 K 480 + RL 11.58 (a) ⎛ 2⎞ I Q = 250 μ A I REF = I Q ⎜ 1 + ⎟ ⎝ β⎠ 2 ⎞ ⎛ = 250 ⎜1 + ⎟ = 252.8 μ A ⎝ 180 ⎠ 5 − ( 0.7 ) − ( −5 ) R1 = ⇒ R1 = 36.8 K 0.2528 (b) 0.125 = 4.808 mA/V Ad = g m ( ro 2 ro 4 ) gm = 0.026 150 = 1200 K ro 2 = 0.125 100 = 800 K Ad = ( 4.808 ) (1200 800 ) ro 4 = 0.125 Ad = 2308 (c) Rid = 2rπ = 2 (180 )( 0.026 ) 0.125 ⇒ Rid = 74.9 K Ro = ro 2 ro 4 = 1200 800 = 480 K = Ro (d) vcm ( max ) = 5 − 0.7 = 4.3 V vcm ( min ) = 0.7 + 0.7 − 5 = −3.6 V 11.59 a. ⎛ IQ ⎞ ⎛ 1 ⎞ I 0 = I B3 + I B 4 ≈ 2 ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠⎝ β ⎠ I Q 0.2 I0 = = ⇒ I0 = 2 μ A β 100 b. V 100 = 1000 kΩ r02 = r04 = A = I CQ 0.1 gm = I CQ VT = 0.1 = 3.846 mA/V 0.026 Ad = g m ( r02 r04 ) = ( 3.846 ) (1000 1000 ) ⇒ Ad = 1923 c. Ad = g m r02 r04 RL ( ) Ad = ( 3.846 ) (1000 1000 250 ) ⇒ Ad = 641 11.60 a. 475. Ad = g m ( r02 r04 RL ) gm = I CQ VT = IQ 2VT V 125 r02 = A 2 = I CQ I CQ r04 = VA 4 80 = I CQ I CQ If I Q = 2 mA, then g m = 38.46 mA/V r02 = 125 kΩ, r04 = 80 kΩ So Ad = 38.46 ⎡125 80 200 ⎤ ⎣ ⎦ Or Ad = 1508 For each gain of 1000. lower the current level For I Q = 0.60 mA, I CQ = 0.30 mA 0.3 gm = = 11.54 mA/V 0.026 125 r02 = = 417 kΩ 0.3 80 r04 = = 267 kΩ 0.3 Ad = 11.54 ⎡ 417 267 200 ⎤ = 1036 ⎣ ⎦ So I Q = 0.60 mA is adequate b. For V + = 10 V, VBE = VEB = 0.6 V For VCB = 0, vcm ( max ) = V + − 2VEB = 10 − 2 ( 0.6 ) Or vcm ( max ) = 8.8 V 11.61 a. From symmetry. VGS 3 = VGS 4 = VDS 3 = VDS 4 = 0.1 +1 0.1 Or VDS 3 = VDS 4 = 2 V 0.1 +1 = 2 V 0.1 = VSD 2 = VSG1 − (VDS 3 − 10 ) = 2 − ( 2 − 10 ) VSG1 = VSG 2 = VSD1 Or VSD1 = VSD 2 = 10 V b. r0 n = r0 p = 1 λn I DQ 1 = = 1 ⇒ 1 MΩ 0.01)( 0.1) ( 1 ⇒ 0.667 MΩ ( 0.015 )( 0.1) g m = 2 K p (VSG + VTP ) = 2 ( 0.1)( 2 − 1) = 0.2 mA / V Ad = g m ( ron rop ) = ( 0.2 ) (1000 667 ) ⇒ Ad (c) λP I DQ = 80 476. IQ I D 2 = I D1 = 2 = 0.1 mA 1 1 = = 1000 k Ω λn I D 4 ( 0.01)( 0.1) ro 4 = 1 ro 2 = λP I D 2 = 1 ( 0.015)( 0.1) = 667 k Ω Ro = ro 2 ro 4 = 667 1000 = 400 k Ω 11.62 Ad = g m ( ro 4 ro 2 ) ⎛ 0.08 ⎞ gm = 2 ⎜ ⎟ ( 2.5 )( 0.05 ) ⎝ 2 ⎠ = 0.1414 mA/V 1 ro 4 = = 1000 K ( 0.02 )( 0.05 ) ro 2 = 1 = 1333 K ( 0.015)( 0.05 ) Ad = ( 0.1414 ) (1000 1333) Ad = 80.8 11.63 R04 = r04 ⎡1 + g m 4 ( R rπ 4 ) ⎤ ⎣ ⎦ 80 = 800 K 0.1 0.1 gm4 = = 3.846 0.026 (100 )( 0.026 ) rπ 4 = 0.1 = 26 K r04 = R rπ 4 = 1 26 = 0.963 K Assume β = 100 rπ 3 = (100 )( 0.026 ) 0.1 = 26 kΩ 0.1 = 3.846 mA/V 0.026 R04 = 800 ⎡1 + ( 3.846 )( 0.963) ⎤ ⇒ 3.763 MΩ ⎣ ⎦ g m3 = ⇒ R0 = 3.763MΩ Then Av = − g m ( r02 R0 ) 120 = 1200 kΩ 0.1 0.1 gm = = 3.846 mA/V 0.026 Av = − ( 3.846 ) ⎡1200 3763⎤ ⇒ Av = −3499 ⎣ ⎦ r02 = b. For 477. R = 0, r04 = Av = − g m ( r02 80 = 800 kΩ 0.1 r04 ) = − ( 3.846 ) ⎡1200 800 ⎤ ⇒ Av = −1846 ⎣ ⎦ (c) For part (a), Ro = ( 3.763 1.2 ) = 0.910 M Ω For part (b), Ro = (1.2 0.8 ) = 0.48 M Ω 11.64 I B5 = IE5 I +I I +I = B3 B4 = C 3 C 4 1+ β 1+ β β (1 + β ) Now I C 3 + I C 4 ≈ I Q IQ So I B 5 ≈ β (1 + β ) I B6 = I Q1 IE6 = 1 + β β (1 + β ) For balance, we want I B 6 = I B 5 So that I Q1 = I Q 11.65 Resistance looking into drain of M4. Vsg 4 ≅ I X R1 I X ± g m 4Vsg 4 = VX − Vsg 4 r04 ⎡ R ⎤ V I X ⎢1 + g m 4 R1 + 1 ⎥ = X r04 ⎦ r04 ⎣ ⎡ R ⎤ Or R0 = r04 ⎢1 + g m 4 R1 + 1 ⎥ r04 ⎦ ⎣ a. 478. Ad = g m 2 ( ro 2 Ro ) g m 2 = 2 K n I DQ = 2 ( 0.080 )( 0.1) = 0.179 mA / V 1 1 ro 2 = = = 667 k Ω λn I DQ ( 0.015 )( 0.1) g m 4 = 2 K P I DQ = 2 ro 4 = 1 λ p I DQ ( 0.080 )( 0.1) = 0.179 mA / V 1 = = 500 k Ω 0.02 )( 0.1) ( 1 ⎤ ⎡ R0 = 500 ⎢1 + ( 0.179 )(1) + = 590.5 kΩ 500 ⎥ ⎣ ⎦ Ad = ( 0.179 ) ⎡667 590.5⎤ ⇒ Ad = 56.06 ⎣ ⎦ b. When R1 = 0, R0 = r04 = 500 kΩ Ad = ( 0.179 ) ⎡667 500 ⎤ ⇒ Ad = 51.15 ⎣ ⎦ (c) For part (a), Ro = ro 2 Ro = 667 590.5 ⇒ Ro = 313 k Ω For part (b), Ro = ro 2 ro 4 = 667 500 ⇒ Ro = 286 kΩ 11.66 Let β = 100, VA = 100 V 479. ro 2 = VA 100 = = 1000 k Ω I CQ 0.1 ′ ′ Ro 4 = ro 4 [1 + g m RE ] where RE = rπ RE Now rπ = (100 )( 0.026 ) = 26 k Ω 0.1 0.1 gm = = 3.846 mA / V 0.026 ′ RE = 26 1 = 0.963 k Ω Then Ro 4 = 1000 ⎡1 + ( 3.846 )( 0.963) ⎤ = 4704 k Ω ⎣ ⎦ Ad = g m ( ro 2 Ro 4 ) = 3.846 (1000 4704 ) ⇒ Ad = 3172 11.67 (a) For Q2, Q4 Vx − Vπ 4 V + g m 2Vπ 2 + g m 4Vπ 4 + x ro 2 ro 4 (1) Ix = (2) g m 2Vπ 2 + (3) Vπ 4 = −Vπ 2 From (2) Vx − Vπ 4 V = π4 ro 2 rπ 4 rπ 2 ⎡ 1 ⎤ Vx 1 = Vπ 4 ⎢ + + gm2 ⎥ ro 2 rπ 4 rπ 2 ro 2 ⎢ ⎥ ⎣ ⎦ 480. Now ⎞ ⎛ 120 ⎞ ⎟=⎜ ⎟ ( 0.5 ) = 0.496 mA ⎠ ⎝ 121 ⎠ ⎛ 120 ⎞ ⎛ IQ ⎞ ⎛ 1 ⎞ ⎛ β ⎞ ⎟ ⇒ I C 2 = 0.0041 mA = ⎜ ⎟⎜ ⎟⎜ ⎟ = ( 0.5 ) ⎜ ⎜ (121)2 ⎟ ⎝ 2 ⎠⎝ 1+ β ⎠⎝ 1+ β ⎠ ⎝ ⎠ ⎛ β IC 4 = ⎜ ⎝ 1+ β IC 2 ⎞ ⎛ IQ ⎟⎜ ⎠⎝ 2 So rπ 2 = (120 )( 0.026 ) = 761 k Ω 0.0041 0.0041 gm2 = = 0.158 mA/V 0.026 100 ro 2 = ⇒ 24.4 M Ω 0.0041 (120 ) ( 0.026 ) rπ 4 = = 6.29 k Ω 0.496 0.496 gm4 = = 19.08 mA / V 0.026 100 ro 4 = = 202 k Ω 0.496 Now ⎡ ⎤ Vx Vx 1 1 = Vπ 4 ⎢ + + 0.158⎥ ⇒ which yields Vπ 4 = ro 2 ( 0.318) ro 2 ⎢ 6.29 761 24400 ⎥ ⎣ ⎦ From (1), ⎛ V V 1 ⎞ I x = x + x + Vπ 4 ⎜ g m 4 − g m 2 − ⎟ ro 2 ro 4 ro 2 ⎠ ⎝ 1 ⎞⎤ ⎡ ⎛ ⎜ 19.08 − 0.158 − ⎟ Ix ⎢ 1 V 1 24400 ⎠ ⎥ ⎥ which yields Ro 2 = x = 135 k Ω =⎢ + +⎝ Vx ⎢ 24400 202 Ix ( 0.318)( 24400 ) ⎥ ⎢ ⎥ ⎣ ⎦ 80 Now ro 6 = = 160 k Ω 0.5 Then Ro = Ro 2 ro 6 = 135 160 ⇒ Ro = 73.2 k Ω (b) c c Ad = g m Ro where g m = Δi vd / 2 481. Δi = g m1Vπ 1 + g m 3Vπ 3 and Vπ 1 + Vπ 3 = vd 2 ⎛V ⎞ Also ⎜ π 1 + g m1Vπ 1 ⎟ rπ 3 = Vπ 3 ⎝ rπ 1 ⎠ ⎛1+ β ⎞ So Vπ 1 ⎜ ⎟ rπ 3 = Vπ 3 ⎝ rπ 1 ⎠ ⎛ 121 ⎞ Or Vπ 1 ⎜ ⎟ ( 6.29 ) = Vπ 3 ≅ Vπ 1 ⎝ 761 ⎠ v v Then 2Vπ 1 = d ⇒ Vπ 1 = d 2 4 ⎛v ⎞ ⎛v ⎞ So Δi = ( g m1 + g m 3 ) Vπ 1 = ( 0.158 + 19.08 ) ⎜ d ⎟ = 9.62 ⎜ d ⎟ ⎝ 4⎠ ⎝ 2⎠ Δi c = 9.62 ⇒ Ad = ( 9.62 )( 73.2 ) ⇒ Ad = 704 So g m = vd / 2 Now Rid = 2 Ri where Ri = rπ 1 + (1 + β ) rπ 3 Ri = 761 + (121)( 6.29 ) = 1522 k Ω Then Rid = 3.044 M Ω 11.69 (a) Ad = 100 = g m ( ro 2 ro 4 ) Let I Q = 0.5 mA 1 1 ro 2 = = = 200 k Ω λn I D ( 0.02 )( 0.25 ) ro 4 = 1 1 = = 160 k Ω λP I D ( 0.025 )( 0.25 ) Then 100 = g m ( 200 160 ) ⇒ g m = 1.125 mA / V ⎛ K′ ⎞⎛W gm = 2 ⎜ n ⎟ ⎜ ⎝ 2 ⎠⎝ L ⎞ ⎟ ID ⎠ ⎛ 0.080 ⎞ ⎛ W ⎞ ⎛W ⎞ 1.125 = 2 ⎜ ⎟ ⎜ ⎟ ( 0.25 ) ⇒ ⎜ ⎟ = 31.6 ⎝ 2 ⎠⎝ L ⎠ ⎝ L ⎠n ⎛W ⎞ ⎛W ⎞ Now ⎜ ⎟ somewhat arbitrary. Let ⎜ ⎟ = 31.6 L ⎠P ⎝ ⎝ L ⎠P 11.70 482. Ad = g m ( ro 2 ro 4 ) P = ( I Q + I REF ) (V + − V − ) Let I Q = I REF Then 0.5 = 2 I Q ( 3 − ( −3) ) ⇒ I Q = I REF = 0.0417 mA ro 2 = 1 1 = = 3205 k Ω λn I D ( 0.015 )( 0.0208 ) ro 4 = 1 1 = = 2404 k Ω λP I D ( 0.02 )( 0.0208 ) Then Ad = 80 = g m ( 3205 2404 ) ⇒ g m = 0.0582 mA/V ⎛ k ′ ⎞⎛ W ⎞ gm = 2 ⎜ n ⎟ ⎜ ⎟ I D ⎝ 2 ⎠ ⎝ L ⎠n ⎛ 0.080 ⎞⎛ W ⎞ ⎛W ⎞ 0.0582 = 2 ⎜ ⎟⎜ ⎟ ( 0.0208 ) ⇒ ⎜ ⎟ = 1.02 ⎝ 2 ⎠⎝ L ⎠ n ⎝ L ⎠n 11.71 Ad = g m ( ro 2 Ro ) ≈ g m ro 2 1 ro 2 = λn I D = 1 = 666.7 K ( 0.015)( 0.1) Ad = 400 = g m ( 666.7 ) g m = 0.60 mA/V ⎛ k′ ⎞⎛ W = 2 ⎜ n ⎟⎜ ⎝ 2 ⎠⎝ L ⎞ ⎟ ID ⎠ ⎛ 0.08 ⎞ ⎛ W ⎞ 0.60 = 2 ⎜ ⎟ ⎜ ⎟ ( 0.1) ⎝ 2 ⎠⎝ L ⎠ ⎛W ⎞ 0.090 = 0.004 ⎜ ⎟ ⎝L⎠ ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = ⎜ ⎟ = 22.5 ⎝ L ⎠1 ⎝ L ⎠ 2 11.72 483. Ad = g m ( Ro 4 Ro 6 ) where Ro 4 = ro 4 + ro 2 [1 + g m 4 ro 4 ] Ro 6 = ro 6 + ro8 [1 + g m 6 ro 6 ] We have 1 = 1667 k Ω ro 2 = ro 4 = ( 0.015 )( 0.040 ) ro 6 = ro8 = 1 = 1250 k Ω 0.02 )( 0.040 ) ( ⎛ 0.060 ⎞ gm4 = 2 ⎜ ⎟ (15 )( 0.040 ) = 0.268 mA/V ⎝ 2 ⎠ ⎛ 0.025 ⎞ gm6 = 2 ⎜ ⎟ (10 )( 0.040 ) = 0.141 mA/V ⎝ 2 ⎠ Then Ro 4 = 1667 + 1667 ⎡1 + ( 0.268 )(1667 ) ⎤ ⇒ 748 M Ω ⎣ ⎦ Ro 6 = 1250 + 1250 ⎡1 + ( 0.141)(1250 ) ⎤ ⇒ 222.8 M Ω ⎣ ⎦ (a) Ro = Ro 4 Ro 6 = 748 222.8 ⇒ Ro = 172 M Ω (b) Ad = g m 4 ( Ro 4 Ro 6 ) = ( 0.268 )(172000 ) ⇒ Ad = 46096 11.73 Ad = g m ( ro 2 ro 4 ) ro 2 = ro 4 = = 1 λ ID 1 ( 0.02 )( 0.1) gm = 2 Kn I D = 2 = 500 K ( 0.5)( 0.1) = 0.4472 mA/V Ad = ( 0.4472 ) ( 500 500 ) ⇒ Ad = 112 Ro = ro 2 ro 4 = 500 500 ⇒ Ro = 250 K 11.74 (a) I DP = K p (VSG + VTP ) 2 0.4 + 1 = VSG 3 = 1.894 V 0.5 I DN = K n (VGS − VTN ) 2 0.4 + 1 = VGS 1 = 1.894 V 0.5 VDS1 ( sat ) = VGS1 − VTN = 1.894 − 1 = 0.894 V V + = VSG 3 + VDS1 ( sat ) − VGS 1 + vCM V + = 1.894 + 0.894 − 1.894 + 4 ⇒ V + = 4.89 V = −V − (b) 484. Ad = g m ( ro 2 ro 4 ) ro 2 = ro 4 = 1 λ ID gm = 2 Kn I D 1 = = 166.7 K ( 0.015 )( 0.4 ) = 2 ( 0.5 )( 0.4 ) = 0.8944 mA/V Ad = ( 0.8944 ) (166.7 166.7 ) ⇒ Ad = 74.5 11.75 (a) For vcm = +2V ⇒ V + = 2.7 V If I Q is a 2-transistor current source, V − = vcm − 0.7 − 0.7 V − = −3.4 V ⇒ V + = −V − = 3.4 V (b) 100 = 1000 K Ad = g m ( ro 2 ro 4 ) ro 2 = 0.1 60 = 600 K ro 4 = 0.1 0.1 = 3.846 mA/V gm = 0.026 Ad = ( 3.846 ) (1000 600 ) ⇒ Ad = 1442 11.76 (a) (b) V + = −V − = 3.4 V 75 = 1250 K 0.06 40 = 666.7 K ro 4 = 0.06 0.06 = 2.308 mA/V gm = 0.026 Ad = ( 2.308 ) (1250 666.7 ) ro 2 = Ad = 1004 11.77 g m1 = 2 K n I Bias1 = 2 gm2 = rπ 2 = I CQ VT β VT I CQ = = ( 0.2 )( 0.25 ) = 0.447 mA/V 0.75 = 28.85 mA/V 0.026 (120 )( 0.026 ) 0.75 = 4.16 kΩ 485. i0 = g m1Vgs1 + g m 2Vπ 2 Vπ 2 = g m1Vgs1rπ 2 and vi = Vgs1 + Vπ 2 i0 = Vgs1 ( g m1 + g m 2 ⋅ g m1rπ 2 ) vi = Vgs1 + g m1Vgs1rπ 2 and Vgs1 = i0 = vi ⋅ C gm = g m1 (1 + β ) vi 1 + g m1rπ 2 1 + g m1rπ 2 ( 0.447 )(121) i0 g m1 (1 + β ) = = vi 1 + g m1rπ 2 1 + ( 0.447 )( 4.16 ) C ⇒ g m = 18.9 mA/V 11.78 r0 ( M 2 ) = r0 ( Q2 ) = 1 λn I DQ = 1 ( 0.01)( 0.2 ) = 500 kΩ VA 80 = = 400 kΩ I CQ 0.2 g m ( M 2 ) = 2 K n I DQ = 2 ( 0.2 )( 0.2 ) = 0.4 mA/V Ad = g m ( M 2 ) ⎡ r0 ( M 2 ) r0 ( Q2 ) ⎤ ⎣ ⎦ = 0.4 ⎡500 400 ⎤ ⇒ Ad = 88.9 ⎣ ⎦ If the IQ current source is ideal, Acm = 0 and C M RRdB = ∞ 11.79 a. b. Assume RL is capacitively coupled. Then 486. I CQ + I DQ = I Q I DQ = VBE 0.7 = = 0.0875 mA R1 8 I CQ = 0.9 − 0.0875 = 0.8125 mA g m1 = 2 K P I DQ = 2 (1)( 0.0875 ) ⇒ g m1 = 0.592 mA/V gm2 = rπ 2 = I CQ VT β VT I CQ = 0.8125 ⇒ g m 2 = 31.25 mA/V 0.026 = (100 )( 0.026 ) 0.8125 ⇒ rπ 2 = 3.2 kΩ c. V0 = ( − g m1Vsg − g m 2Vπ 2 ) RL Vi + Vsg = V0 ⇒ Vsg = V0 − Vi Vπ 2 = ( g m1Vsg ) ( R1 rπ 2 ) V0 = − ⎡ g m1Vsg + g m 2 g m1Vsg ( R1 rπ 2 ) ⎤ RL ⎣ ⎦ ⎡ ⎤ V0 = − (V0 − Vi ) ⎣ g m1 + g m 2 g m1 ( R1 rπ 2 ) ⎦ RL ⎡ g m1 + g m 2 g m1 ( R1 rπ 2 ) ⎦ RL ⎤ V0 = ⎣ Vi 1 + ⎡ g m1 + g m 2 g m1 ( R1 rπ 2 ) ⎤ RL ⎣ ⎦ We find Av = g m1 + g m 2 g m1 ( R1 rπ 2 ) = 0.592 + ( 31.25 )( 0.592 ) ( 8 3.2 ) = 42.88 Then Av = ( 42.88 )( RL ) 1 + ( 42.88 )( RL ) 11.80 a. I DQ I CQ Assume RL is capacitively coupled. 0.7 = = 0.0875 mA 8 = 1.2 − 0.0875 = 1.11 mA g m1 = 2 K p I DQ = 2 (1)( 0.0875 ) ⇒ g m1 = 0.592 mA/V gm2 = rπ 2 = b. I CQ VT β VT I CQ = = 1.11 ⇒ g m 2 = 42.7 mA/V 0.026 (100 )( 0.026 ) 1.11 ⇒ rπ 2 = 2.34 kΩ 487. Vsg = VX I X = g m 2Vπ 2 + g m1Vsg (g V m1 sg )(R 1 rπ 2 ) = Vπ 2 I X = VX ⎡ g m1 + g m 2 g m1 ( R1 rπ 2 ) ⎤ ⎣ ⎦ VX 1 R0 = = IX g m1 + g m 2 g m1 ( R1 rπ 2 ) = 1 0.592 + ( 0.592 )( 42.7 ) ( 8 2.34 ) ⇒ R0 = 21.6 Ω 11.81 (a) (1) g m 2Vπ + (2) g m 2Vπ + Then Vπ = From (1) Vo − ( −Vπ ) ro 2 Vo − ( −Vπ ) g m1Vi ⎛ 1 1⎞ ⎜ + ⎟ ⎝ ro1 rπ ⎠ ro 2 =0 = g m1Vi + ⎛ 1 1⎞ −Vπ −Vπ + or 0 = g m1Vi − Vπ ⎜ + ⎟ ro1 rπ ⎝ ro1 rπ ⎠ 488. ⎛ Vo 1 ⎞ =0 ⎜ g m 2 + ⎟ Vπ + ro 2 ⎠ ro 2 ⎝ ⎛ 1 ⎞ ⎜ gm2 + ⎟ ro 2 ⎠ ⎛ 1 ⎞ Vo = −ro 2 ⎜ g m 2 + ⎟ Vπ = −ro 2 g m1Vi ⎝ ro 2 ⎠ ⎛ 1 1⎞ ⎝ ⎜ + ⎟ ⎝ ro1 rπ ⎠ ⎛ 1 ⎞ − g m1ro 2 ⎜ g m 2 + ⎟ ro 2 ⎠ V ⎝ Av = o = Vi ⎛ 1 1⎞ ⎜ + ⎟ ⎝ ro1 rπ ⎠ Now ( 0.25)( 0.025 ) = 0.158 mA / V g m1 = 2 K n I Q = 2 gm2 = ro1 = VT 1 λ IQ = = 0.025 = 0.9615 mA / V 0.026 1 ( 0.02 )( 0.025) = 2000 k Ω VA 50 = = 2000 k Ω I Q 0.025 ro 2 = rπ = IQ β VT IQ = (100 )( 0.026 ) 0.025 = 104 k Ω Then 1 ⎞ ⎛ − ( 0.158 )( 2000 ) ⎜ 0.9615 + ⎟ 2000 ⎠ ⎝ ⇒ Av = −30039 Av = 1 ⎞ ⎛ 1 + ⎜ ⎟ ⎝ 2000 104 ⎠ To find Ro; set Vi = 0 ⇒ g m1Vi = 0 489. I x = g m 2Vπ + Vx − ( −Vπ ) Vπ = − I x ( ro1 rπ ) ro 2 Then ⎛ V 1 ⎞ I x = ⎜ g m 2 + ⎟ ( − I x ) ( ro1 rπ ) + x ro 2 ⎠ ro 2 ⎝ Combining terms, Ro = ⎡ ⎛ Vx 1 ⎞⎤ = ro 2 ⎢1 + ( ro1 rπ ) ⎜ g m 2 + ⎟ ⎥ Ix ro 2 ⎠ ⎦ ⎝ ⎣ ⎡ 1 ⎞⎤ ⎛ = 2000 ⎢1 + ( 2000 104 ) ⎜ 0.9615 + ⎟ ⇒ Ro = 192.2 M Ω 2000 ⎠ ⎥ ⎝ ⎣ ⎦ (b) Vo − ( −Vgs 3 ) (1) g m 3Vgs 3 + (2) g m 3Vgs 3 + (3) −Vgs 3 − ( −Vπ 2 ) ( −Vπ 2 ) Vπ 2 + g m 2Vπ 2 + = g m1Vi + rπ 2 ro 2 ro1 From (2), Vπ 2 = ro3 Vo − ( −Vgs 3 ) ro3 =0 = g m 2Vπ 2 + −Vgs 3 − ( −Vπ 2 ) ro 2 Vgs 3 ⎛ 1 ⎞ ro 2 ⎜ g m 2 + ⎟ ro 2 ⎠ ⎝ Then (3) or Vgs 3 ⎛ 1 1 1⎞ + gm2 + + ⎟ = g m1Vi + Vπ 2 ⎜ ro 2 ro1 ⎠ ro 2 ⎝ rπ 2 ⎛ 1 ⎞ Vgs 3 or 0 = Vπ 2 ⎜ g m 2 + ⎟ − ro 2 ⎠ ro 2 ⎝ 490. Vgs 3 ⎡ 1 1 1⎤ + ⎥ = g m1Vi + ⎢ + gm2 + ro 2 ro1 ⎦ ro 2 ⎛ 1 ⎞ r ro 2 ⎜ g m 2 + ⎟ ⎣ π 2 ro 2 ⎠ ⎝ Vgs 3 Vgs 3 1 1 ⎤ ⎡ 1 ⎢104 + 0.9615 + 2000 + 2000 ⎥ = 0.9615Vi + 2000 1 ⎞⎣ ⎛ ⎦ 2000 ⎜ 0.9615 + ⎟ 2000 ⎠ ⎝ Then Vgs 3 = 1.83 × 105 Vi Vgs 3 ⎛ −V 1 ⎞ 1 ⎞ ⎛ 5 From (1), ⎜ g m 3 + ⎟ Vgs 3 = o or Vo = −2000 ⎜ 0.158 + ⎟ (1.83 ×10 ) Vi ro 3 ⎠ ro3 2000 ⎠ ⎝ ⎝ V Av = o = −5.80 × 107 Vi To find Ro Vx − ( −Vgs 3 ) (1) I x = g m 3Vgs 3 + (2) g m 3Vgs 3 + (3) Vπ 2 = − I x ( ro1 rπ 2 ) ro3 Vx − ( −Vgs 3 ) ro 3 = g m 2Vπ 2 + ⎛ 1 ⎞ V From (1) I x = Vgs 3 ⎜ g m 3 + ⎟ + x ro 3 ⎠ ro3 ⎝ 1 ⎞ Vx ⎛ I x = Vgs 3 ⎜ 0.158 + ⎟+ 2000 ⎠ 2000 ⎝ V Ix − x 2000 So Vgs 3 = 0.1585 −Vgs 3 − ( −Vπ 2 ) ro 2 491. From (2), ⎡ ⎛ 1 1 ⎤ V 1 ⎞ + ⎥ + x = Vπ 2 ⎜ g m 2 + ⎟ Vgs 3 ⎢ g m 3 + ro 3 ro 2 ⎦ ro 3 ro 2 ⎠ ⎣ ⎝ 1 1 ⎤ Vx 1 ⎞ ⎡ ⎛ + Vgs 3 ⎢ 0.158 + ⎥ + 2000 = Vπ 2 ⎜ 0.9615 + 2000 ⎟ 2000 2000 ⎦ ⎣ ⎝ ⎠ Vx ⎡ I − Vx / 2000 ⎤ Then ⎢ x ⎥ ( 0.159 ) + 2000 = − I x ( 2000 104 ) ( 0.962 ) ⎣ 0.1585 ⎦ V We find Ro = x = 6.09 × 1010 Ω Ix 11.82 Assume emitter of Q1 is capacitively coupled to signal ground. ⎛ 80 ⎞ I CQ = 0.2 ⎜ ⎟ = 0.1975 mA ⎝ 81 ⎠ 0.2 I DQ = = 0.00247 mA 81 (80 )( 0.026 ) rπ = = 10.5 k Ω 0.1975 0.1975 g m ( Q1 ) = = 7.60 mA / V 0.026 gm ( M1 ) = 2 K n I D = 2 g m ( M 1 ) = 0.0445 mA / V ( 0.2 )( 0.00247 ) Vi = Vgs + Vπ and Vπ = g m ( M 1 ) Vgs rπ or Vgs = ⎛ 1 Then Vi = Vπ ⎜ 1 + ⎜ g (M )r m 1 π ⎝ Vπ g m ( M 1 ) rπ ⎞ Vi ⎟ or Vπ = ⎟ ⎛ ⎞ 1 ⎠ ⎜1 + ⎜ g (M ) r ⎟ ⎟ m 1 π ⎠ ⎝ − g m ( Q1 ) RC V Vo = − g m ( Q1 ) Vπ RC ⇒ Av = o = Vi ⎛ ⎞ 1 ⎜1 + ⎜ g (M )r ⎟ ⎟ m 1 π ⎠ ⎝ − ( 7.60 )( 20 ) Then Av = ⇒ Av = −48.4 ⎛ ⎞ 1 ⎜1 + ⎜ ( 0.0445 )(10.5 ) ⎟ ⎟ ⎝ ⎠ 492. 11.83 Using the results from Chapter 4 for the emitter-follower: ⎡ rπ 9 + r07 R011 ⎤ ⎢ rπ 8 + ⎥ 1+ β ⎥ R0 = R4 || ⎢ ⎢ ⎥ 1+ β ⎢ ⎥ ⎣ ⎦ β VT (100 )( 0.026 ) rπ 8 = = = 2.6 kΩ 1 IC8 IC 8 IC 9 ≈ rπ 9 = β = 1 = 0.01 mA 100 (100 )( 0.026 ) = 260 kΩ 0.01 V 100 r07 = A = = 500 kΩ I Q 0.2 r011 = VA 100 = = 500 kΩ I Q 0.2 ′ R011 = r011 [1 + g m RE ] , g m = rπ 11 = (100 ) ( 0.026 ) 0.2 = 7.69 0.026 = 13 kΩ 0.2 ′ RE = 0.2 13 = 0.197 kΩ R011 = 500 ⎡1 + ( 7.69 )( 0.197 ) ⎤ = 1257 kΩ ⎣ ⎦ Then 260 + 500 1257 ⎤ ⎡ ⎢ 2.6 + ⎥ 101 ⎥ R0 = 5 || ⎢ 101 ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ = 5 0.0863 ⇒ R0 = 0.0848 K ⇒ 84.8 Ω 11.84 Ri = rπ 1 + (1 + β ) rπ 2 rπ 2 = (100 )( 0.026 ) 0.5 = 5.2 kΩ (100 )( 0.026 ) (100 ) ( 0.026 ) = = 520 kΩ 0.5 ( 0.5 /100 ) Ri = 520 + (101)( 5.2 ) ⇒ Ri ≅ 1.05 MΩ (100 )( 0.026 ) rπ 3 + 50 2 rπ 1 = R0 = 5 R0 = 5 101 , rπ 3 = 1 = 2.6 kΩ 2.6 + 50 = 5 0.521 ⇒ R0 = 0.472 kΩ 101 493. ⎛V ⎞ V0 = − ⎜ π 3 + g m 3Vπ 3 ⎟ ( 5 ) ⎝ rπ 3 ⎠ ⎛1+ β ⎞ V0 = −Vπ 3 ⎜ ⎟ ( 5) ⎝ rπ 3 ⎠ (1) (V − V ) Vπ 3 = g m 2Vπ 2 + 0 π 3 50 rπ 3 ⎛ 1 1 ⎞ V g m 2Vπ 2 = Vπ 3 ⎜ + ⎟− 0 ⎝ rπ 3 50 ⎠ 50 ⎛V ⎞ Vπ 2 = ⎜ π 1 + g m1Vπ 1 ⎟ rπ 2 ⎝ rπ 1 ⎠ ⎛ 1+ β = Vπ 1 ⎜ ⎝ rπ 1 ⎞ ⎟ rπ 2 ⎠ and Vin = Vπ 1 + Vπ 2 gm2 = (2) (3) (4) 0.5 = 19.23 mA/V 0.026 Then ⎛ 101 ⎞ V0 = −Vπ 3 ⎜ ⎟ ( 5 ) ⇒ Vπ 3 = −V0 ( 0.005149 ) ⎝ 2.6 ⎠ And 1 ⎞ V ⎛ 1 + ⎟− 0 19.23Vπ 2 = −V0 ( 0.005149 ) ⎜ ⎝ 2.6 50 ⎠ 50 = −V0 ( 0.02208 ) (1) (2) Or Vπ 2 = −V0 ( 0.001148 ) And Vπ 1 = Vin − Vπ 2 = Vin + V0 ( 0.001148 ) (4) So ⎛ 101 ⎞ −V0 ( 0.001148 ) = ⎡Vin + V0 ( 0.001148 ) ⎤ ⎜ ⎣ ⎦ 520 ⎟ ( 5.2 ) ⎝ ⎠ −V0 ( 0.001148 ) − V0 ( 0.001159 ) = Vin (1.01) ⇒ Av = 11.85 (3) V0 = −438 Vin 494. I2 = 5 = 1 mA 5 1 + 0.8 = 2.21 V 0.5 2.21 − ( −5 ) I1 = = 0.206 mA 35 VGS 2 = V0 = ( g m 2Vgs 2 ) ( R2 r02 ) Vgs 2 = ( g m1Vsg1 ) ( r01 R1 ) − V0 and Vsg1 = −Vin So Vgs 2 = − ( g m1Vin ) ( r01 R1 ) − V0 Then V0 = g m 2 ( R2 r02 ) ⎡ − ( g m1Vin ) ( r01 R1 ) − V0 ⎤ ⎣ ⎦ Av = V0 − g m 2 ( R2 r02 ) g m1 ( r01 R1 ) = Vin 1 + g m 2 ( R2 r02 ) gm2 = 2 Kn2 I D 2 = 2 ( 0.5 )(1) = 1.414 mA / V g m1 = 2 K p1 I D1 = 2 ( 0.2 )( 0.206 ) = 0.406 mA / V r01 = 1 λ1 I D1 r02 = = 1 ( 0.01)( 0.206 ) = 485 kΩ 1 1 = = 100 kΩ λ2 I D 2 ( 0.01)(1) R2 r02 = 5 100 = 4.76 kΩ R1 r01 = 35 485 = 32.6 kΩ Then Av = − (1.414 )( 4.76 )( 0.406 )( 32.6 ) 1 + (1.414 )( 4.76 ) So ⇒ Av = −11.5 Output Resistance—From the results for a source follower in Chapter 6. 1 1 R0 = R2 r02 = 5 100 gm2 1.414 = 0.707 4.76 So R0 = 0.616 kΩ 11.86 a. 495. R2 = 5 ⇒ R2 = 10 kΩ 0.5 I D2 0.5 − VTP 2 = + 1 = 2.41 V K p2 0.25 VSG 2 = R1 = 5 − ( −2.41) 0.1 ⇒ R1 = 74.1 kΩ b. V0 = − ( g m 2Vsg 2 ) ( r02 R2 ) Vsg 2 = V0 − ⎡ − ( g m1Vgs1 ) ( r01 R1 ) ⎤ and Vgs1 = Vin ⎣ ⎦ Av = V0 − ( g m 2 ) ( r02 R2 ) ( g m1 ) ( r01 R1 ) = Vin 1 + ( g m 2 ) ( r02 R2 ) g m1 = 2 K n1 I D1 = 2 gm2 = 2 K p 2 I D 2 = 2 r01 = r02 = 1 λ1 I D1 = ( 0.1)( 0.1) = 0.2 mA / V ( 0.25)( 0.5) = 0.707 mA / V 1 ( 0.01)( 0.1) = 1000 kΩ 1 1 = = 200 kΩ λ2 I D 2 ( 0.01)( 0.5 ) r02 R2 = 200 10 = 9.52 kΩ r01 R1 = 1000 74.1 = 69.0 kΩ Then Av = − ( 0.707 )( 9.52 )( 0.2 )( 69 ) 1 + ( 0.707 )( 9.52 ) So ⇒ Av = −12.0 R0 = 1 1 10 200 R2 r02 = 0.707 gm2 = 1.414 9.52 Or R0 = 1.23 kΩ 11.87 a. I C 2 = 0.25 mA 5−2 R= ⇒ R = 12 kΩ 0.25 v − VBE ( on ) 2 − 0.7 I C 3 = 02 ⇒ RE1 = ⇒ RE1 = 2.6 kΩ RE1 0.5 RC = 5 − v03 5 − 3 = ⇒ RC = 4 kΩ IC 3 0.5 ⎡ v03 − VBE ( on ) ⎤ − ( −5 ) ⎦ IC 4 = ⎣ RE 2 RE 2 = 3 − 0.7 + 5 ⇒ RE 2 = 2.43 kΩ 3 496. b. Input resistance to base of Q3, Ri 3 = rπ 3 + (1 + β ) RE1 rπ 3 = (100 )( 0.026 ) = 5.2 kΩ 0.5 Ri 3 = 5.2 + (101)( 2.6 ) = 267.8 kΩ v 1 Ad 1 = 02 = g m 2 ( R Ri 3 ) vd 2 0.25 = 9.62 mA/V 0.026 1 Ad 1 = ( 9.62 ) (12 267.8 ) ⇒ Ad 1 = 55.2 2 − β ( RC Ri 4 ) v Now 03 = v02 rπ 3 + (1 + β ) RE1 gm2 = where Ri 4 = rπ 4 + (1 + β ) RE 2 and (1 + β ) RE 2 v0 = v03 rπ 4 + (1 + β ) RE 2 (100 )( 0.026 ) rπ 4 = = 0.867 kΩ 3 (101)( 2.43) v0 = = 0.9965 v03 0.867 + (101)( 2.43) Ri 4 = 0.867 + (101)( 2.43) = 246.3 kΩ rπ 3 = 5.2 kΩ So v03 − (100 ) ( 4 246.3) = = −1.47 5.2 + (101)( 2.6 ) v02 v0 = ( 55.2 )( 0.9965 )( −1.47 ) ⇒ Ad = −80.9 vd Using Equation (11.32b) − g m 2 ( R Ri 3 ) So Ad = c. Acm1 = 1+ rπ 2 = 2 (1 + β ) R0 rπ 2 (100 )( 0.026 ) = 10.4 kΩ 0.25 − ( 9.62 ) (12 267.8 ) = −0.0569 = Acm1 Acm1 = 2 (101)(100 ) 1+ 10.4 ⎛ v0 ⎞⎛ v03 ⎞ Then Acm = ⎜ ⎟⎜ ⎟ ⋅ Acm1 ⎝ v03 ⎠⎝ v02 ⎠ = ( 0.9965 )( −1.47 )( −0.0569 ) ⇒ Acm = 0.08335 ⎛ 80.9 ⎞ C M RRdB = 20 log10 ⎜ ⎟ ⇒ C M RRdB = 59.7 dB ⎝ 0.08335 ⎠ 11.88 a. RC1 = 10 − v01 10 − 2 = ⇒ RC1 = 80 kΩ I C1 0.1 RC 2 = 10 − v04 10 − 6 = ⇒ RC 2 = 20 kΩ IC 4 0.2 497. b. Ad 1 = v01 − v02 = − g m1 ( RC1 rπ 3 ) vd 0.1 = 3.846 mA/V 0.026 (180 )( 0.026 ) rπ 3 = = 23.4 kΩ 0.2 Ad 1 = − ( 3.846 ) ( 80 23.4 ) ⇒ Ad 1 = −69.6 g m1 = Ad 2 = v04 1 = g m 4 RC 2 v01 − v02 2 0.2 = 7.692 mA/V 0.026 1 Ad 2 = ( 7.692 )( 20 ) = 76.9 2 Then Ad = ( 76.9 )( −69.6 ) ⇒ Ad = −5352 gm4 = 11.89 a. Neglect the effect of r0 in determining the differential-mode gain. v02 1 Ad 1 = = g m 2 ( RC Ri 3 ) where Ri 3 = rπ 3 + (1 + β ) RE vd 2 A2 = I1 = − β RC 2 rπ 3 + (1 + β ) RE 12 − 0.7 − ( −12 ) R1 = 23.3 = 1.94 mA ≈ I C 5 12 1 ⋅ (1.94 ) gm2 = 2 = 37.3 mA/V 0.026 ( 200 )( 0.026 ) rπ 3 = IC 3 1 (1.94 )(8) = 4.24 V 2 4.24 − 0.7 = 1.07 mA IC 3 = 3.3 ( 200 )( 0.026 ) = 4.86 kΩ rπ 3 = 1.07 Ri 3 = 4.86 + ( 201)( 3.3) = 668 kΩ v02 = 12 − Ad 1 = 1 ( 37.3) ⎡8 668⎤ = 147.4 ⎣ ⎦ 2 Then Ad = Ad 1 ⋅ A2 = (147.4 )( −1.197 ) ⇒ Ad = −176 R0 = r05 = Acm1 = − g m 2 ( RC Ri 3 ) 1+ rπ 2 = VA 80 = = 41.2 kΩ I C 5 1.94 2 (1 + β ) R0 rπ 2 ( 200 )( 0.026 ) 1 ⋅ (1.94 ) 2 = 5.36 kΩ 498. Acm1 = − ( 37.3) ( 8 668 ) 2 ( 201)( 41.2 ) 1+ 5.36 A2 = −1.197 = −0.09539 Acm = ( −0.09539 )( −1.197 ) ⇒ Acm = 0.114 b. vd = v1 − v2 = 2.015sin ω t − 1.985sin ω t vd = 0.03sin ω t ( V ) v +v vcm = 1 2 = 2.0sin ω t 2 v03 = Ad vd + Acm vcm = ( −176 )( 0.03) + ( 0.114 )( 2 ) Or v03 = −5.052sin ω t Ideal, Acm = 0 So v03 = Ad vd = ( −176 )( 0.03) v03 = −5.28sin ω t c. Rid = 2rπ 2 = 2 ( 5.36 ) ⇒ Rid = 10.72 kΩ 2 Ricm ≅ 2 (1 + β ) R0 (1 + β ) r0 VA 80 = = 82.5 kΩ 1 IC 2 ⋅ (1.94 ) 2 2 Ricm = ⎡ 2 ( 201)( 41.2 ) ⎤ ⎡( 201)( 82.5 ) ⎤ ⎣ ⎦ ⎣ ⎦ r0 = = 16.6 MΩ 16.6 MΩ So ⇒ Ricm = 4.15 MΩ 11.90 a. 24 − VGS 4 2 = kn (VGS 4 − VTh ) I1 = R1 24 − VGS 4 = ( 55 )( 0.2 )(VGS 4 − 2 ) 2 2 24 − VGS 4 = 11 (VGS 4 − 4VGS 4 + 4 ) 2 11VGS 4 − 43VGS 4 + 20 = 0 VGS 4 = 43 ± ( 43) 2 − 4 (11)( 20 ) 2 (11) = 3.37 V 24 − 3.37 = 0.375 mA = I Q 55 ⎛ 0.375 ⎞ v02 = 12 − ⎜ ⎟ ( 40 ) = 4.5 V ⎝ 2 ⎠ v02 − VGS 3 2 = I D 3 = kn (VGS 3 − VTh ) R5 I1 = 499. 2 4.5 − VGS 3 = ( 0.2 ) ( 6 ) (VGS 3 − 4VGS 3 + 4 ) 2 1.2VGS 3 − 3.8VGS 3 + 0.3 = 0 VGS 3 = I D3 = 3.8 ± ( 3.8) 2 − 4 (1.2 ) ( 0.3) 2 (1.2 ) = 3.09 V 4.5 − 3.09 = 0.235 mA 6 gm2 = 2 Kn I D2 = 2 ( 0.2 ) ⎛ ⎜ ⎝ 0.375 ⎞ ⎟ 2 ⎠ = 0.387 mA/V 1 1 g m 2 RD = ( 0.387 )( 40 ) ⇒ Ad 1 = 7.74 2 2 − g m 3 RD 2 A2 = 1 + g m 3 R5 Ad 1 = g m3 = 2 K n I D3 = 2 ( 0.2 )( 0.235 ) = 0.434 mA/V A2 = − ( 0.434 ) ( 4 ) 1 + ( 0.434 )( 6 ) = −0.482 So Ad = Ad 1 ⋅ A2 = ( 7.74 ) ( −0.482 ) ⇒ Ad = −3.73 R0 = r05 = Acm1 = 1 1 = = 133 kΩ λ I Q ( 0.02 )( 0.375 ) − ( 0.387 ) ( 40 ) − g m 2 RD = 1 + 2 g m 2 R0 1 + 2 ( 0.387 ) (133) = −0.149 Acm = ( −0.149 )( −0.482 ) ⇒ Acm = 0.0718 b. vd = v1 − v2 = 0.3sin ω t v +v vcm = 1 2 = 2sin ω t 2 v03 = Ad vd + Acm vcm = ( −3.73)( 0.3) + ( 0.0718 )( 2 ) ⇒ v03 = −0.975sin ω t ( V ) Ideal, Acm = 0 v03 = Ad vd = ( −3.73)( 0.3) Or ⇒ v03 = −1.12sin ω t ( V ) 11.91 The low-frequency, one-sided differential gain is 500. ⎛ r ⎞ v02 1 = g m RC ⎜ π ⎟ vd 2 ⎝ rπ + RB ⎠ 1 ⋅ β RC = 2 rπ + RB Av 2 = rπ = (100 )( 0.026 ) 0.5 = 5.2 kΩ 1 ⋅ (100 )(10 ) ⇒ Av 2 = 87.7 Av 2 = 2 5.2 + 0.5 CM = Cμ (1 + g m RC ) 0.5 = 19.23 mA/V 0.026 CM = 2 ⎡1 + (19.23)(10 ) ⎤ ⇒ CM = 387 pF ⎣ ⎦ gm = 1 2π ⎡ rπ RB ⎤ ( Cπ + CM ) ⎣ ⎦ 1 = So ⇒ f H = 883 kHz 3 ⎡ ⎤ 2π ⎣5.2 0.5⎦ × 10 × ( 8 + 387 ) × 10−12 fH = 11.92 From Equation (11.117), f Z = a. 1 1 = 2π R0 C0 2π ( 5 × 106 )( 0.8 × 10−12 ) Or f Z = 39.8 kHz b. From Problem 11.69, f H = 883 kHz. From Equation (11.116(b)), the low-frequency common- mode gain is − g m RC Acm = ⎡⎛ RB ⎞ 2 (1 + β ) R0 ⎤ ⎢⎜ 1 + ⎥ ⎟+ rπ ⎠ rπ ⎣⎝ ⎦ rπ = 5.2 kΩ, g m = 19.23 mA/V So Acm = − (19.23)(10 ) ⎡⎛ 0.5 ⎞ 2 (101) ( 5 × 106 ) ⎤ ⎢⎜ 1 + ⎥ ⎟+ 5.2 × 103 ⎢⎝ 5.2 ⎠ ⎥ ⎣ ⎦ −4 = −9.9 × 10 ⎛ 87.7 ⎞ C M RRdB = 20 log10 ⎜ = 98.9 dB −4 ⎟ ⎝ 9.9 × 10 ⎠ 501. 11.93 a. From Equation (7.72), fT = gm 2π ( Cπ + Cμ ) 1 = 38.46 mA/V 0.026 38.46 × 10−3 Then 800 × 106 = 2π ( Cπ + Cμ ) gm = Or Cπ + Cμ = 7.65 × 10−12 F = 7.65 pF And Cπ = 6.65 pF CM = Cμ (1 + g m RC ) = 1 ⎡1 + ( 38.46 )(10 ) ⎤ ⎣ ⎦ = 386 pF 1 fH = 2π ⎡ rπ RB ⎤ ( Cπ + CM ) ⎣ ⎦ rπ = (120 )( 0.026 ) 1 = 3.12 kΩ 1 2π ⎡3.12 1⎤ × 10 × ( 6.65 + 386 ) × 10−12 ⎣ ⎦ Or f H = 535 kHz fH = b. 3 From Equation (11.140), f Z = 1 1 = 2π R0 C0 2π (10 × 106 )(10−12 ) Or f Z = 15.9 kHz 11.94 The differential-mode half circuit is: ⎛v ⎞ g m ⎜ d ⎟ RC ⎝ 2⎠ or Av = v02 = rπ ⎛1+ β ⎞ 1+ ⎜ RE ⎟ rπ ⎠ ⎝ rπ = (100 )( 0.026 ) 0.5 ⎛1⎞ ⎜ ⎟ β RC ⎝ 2⎠ + (1 + β ) RE = 5.2 kΩ ⎛1⎞ ⎜ ⎟ (100 )(10 ) 500 2 = Av = ⎝ ⎠ 5.2 + (101) RE 5.2 + (101) RE 502. a. For RE = 0.1 kΩ : Av = 32.7 b. For RE = 0.25 kΩ : Av = 16.4 503. Chapter 12 Exercise Solutions EX12.1 A 1 + Aβ A A ⇒ Aβ = −1 1 + Aβ = Af Af Af = a. β= 1 1 1 1 − = − = 0.05 − 0.0001 ⇒ β = 0.0499 Af A 20 104 Af b. = 20 (1/ β ) (1/ 0.0499 ) = 20 = 0.998 20.040 EX12.2 A 1 + Aβ 1 1 1 1 − = − 6 = 0.01 − 10−6 β= Af A 100 10 Af = β = 0.009999 dA ⎛ Af ⎞ dA = ⋅ Af (1 + β A) A ⎜ A ⎟ A ⎝ ⎠ dAf ⎛ 100 ⎞ dAf = ⎜ 6 ⎟ ( 20 ) % ⇒ = 0.002% Af Af ⎝ 10 ⎠ dAf = 1 ⋅ EX12.3 Bandwidth = ω H (1 + β A0 ) ⎛A ⎞ ⎛ 105 ⎞ = ω H ⎜ 0 ⎟ = ( 2π )(10 ) ⎜ ⎟ ⎝ 100 ⎠ ⎝ At ⎠ ω = ( 2π ) (104 ) rad / sec ⇒ f = 10 kHz EX12.4 (a) vOA = A1 A2 vi + A2 vn = (100 )(10 ) vi + (10 ) vn So 1000vi S = = 100 i No 10vn Ni (b) A1 A2 A2 vOC = vi + vn 1 + β Α1 Α2 1 + β Α1 Α2 = 105 10 v + vn 5 i 1 + ( 0.001)10 1 + ( 0.001) (105 ) So 103 vi S = = 104 i N o 0.1vn Ni 504. EX12.5 a. Vε = VS − V fb = 100 − 99 = 1 m V 5 ⇒ Av = 5000 V/V 0.001 V fb 0.099 V fb = β V0 ⇒ β = = ⇒ β = 0.0198 V/V V0 5 V0 = AvVε ⇒ Av = Avf = Av 1 + β Av = 5000 ⇒ Avf = 50 V/V 1 + ( 0.0198 )( 5000 ) Rif = Ri (1 + β Av ) = ( 5 ) ⎡1 + ( 0.0198 )( 5000 ) ⎤ ⇒ Rif = 500 kΩ ⎣ ⎦ b. R0 f = R0 4 = ⇒ R0 f ⇒ 40 Ω 1 + β Av 1 + ( 0.0198 )( 5000 ) EX12.6 a. I ε = I S − I fb = 100 − 99 = 1 μ A Ai = β= Aif = I0 5 = ⇒ Ai = 5000 A/A I ε 0.001 I fb I0 = 0.099 ⇒ β = 0.0198 A/A 5 Ai 5000 ⇒ Aif = 50 A/A = 1 + Ai β 1 + ( 5000 )( 0.0198 ) Rif = b. Ri 5 = ⇒ Rif ⇒ 50 Ω 1 + β Ai 1 + ( 0.0198 )( 5000 ) R0 f = (1 + β Ai ) R0 = ⎡1 + ( 0.0198 )( 5000 ) ⎤ ( 4 ) ⇒ R0 f = 400 kΩ ⎣ ⎦ EX12.7 Avf = Avf = Av 104 = ⇒ Avf = 3.9984 Av 104 1+ 1+ 1 + ( R2 / R1 ) 1 + ( 30 /10 ) 105 = 3.99984 105 1+ 1 + ( 30 /10 ) 3.99984 − 3.9984 × 100% ⇒ 0.0360% 3.9984 EX12.8 Use a non inverting op-amp. R R 1 + 2 = 15 ⇒ 2 = 14 R1 R1 Let β= R2 = 140 K R1 = 10 K 1 = 0.066667 R2 1+ R1 505. Input resistance. Rif = 5 ( 0.06667 ) ( 5 × 103 ) ≅ 1.67 MΩ Rof = 50 ( 0.066667 ) ( 5 × 103 ) ≡ 0.15Ω EX12.9 ⎛ ⎞ ⎟ ⎞⎜ RE ⎜ ⎟ ⋅ ii ⎟ ⎠ ⎜ R + rk ⎟ ⎜ E 1+ h ⎟ FE ⎠ ⎝ 80 )( 0.026 ) ( rk = = 4.16 k Ω 0.5 Then rk 4.16 = = 0.0514 k Ω 1 + hFE 81 Then we want ⎞ io RE ⎛ 80 ⎞ ⎛ = 0.95 = ⎜ ⎟ ⎜ ⎟ ii ⎝ 81 ⎠ ⎝ RE + 0.0514 ⎠ ⎛ h io = ⎜ FE ⎝ 1 + hFE or ⎛ ⎞ RE ⎜ ⎟ = 0.9619 ⎝ RE + 0.0514 ⎠ which yields RE ( min ) = 1.30 k Ω and ⎛ 81 ⎞ V + = I E RE + 0.7 = ⎜ ⎟ ( 0.5 )(1.3) + 0.7 ⇒ V + ( min ) = 1.36 V ⎝ 80 ⎠ EX12.10 Use the configuration shown in figure 12.20. RS = 500 Ω, RL = 200 Ω Let 1 + RF = 15 R1 For example, let R1 = 2 K RF = 28 K EX12.11 a. ⎛ R2 ⎞ ⎛ 20 ⎞ VG = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 ⎝ 20 + 30 ⎠ ⎝ R1 + R2 ⎠ VG = −1 V VS = −1 − VGS ID = VS − ( −5 ) RS = K n (VGS − VTN ) 2 ( −1) − VGS + 5 = (1.5)( 0.4 )(VGS − 2 ) 2 506. 4 − VGS = 0.6 (V 2GS − 4VGS + 4 ) 0.6V 2GS − 1.4VGS − 1.6 = 0 VGS = 1.4 ± (1.4 ) 2 + 4 ( 0.6 )(1.6 ) 2 ( 0.6 ) VGS = 3.174 V g m = 2 K n (VGS − VTN ) = 2 (1.5 )( 3.17 − 2 ) g m = 3.52 mA / V ⎛ RD ⎞ ⎛ 2 ⎞ I0 = − ⎜ ⎟ ( g mVgs ) = − ⎜ ⎟ ( 3.52 ) Vgs ⎝ 2+2⎠ ⎝ RD + RL ⎠ I 0 = −1.76Vgs Vi = Vgs + g mVgs RS ⇒ Vgs = Vgs = Vi 1 + g m RS Vi = ( 0.4153) Vi 1 + ( 3.52 )( 0.4 ) I 0 = − (1.76 )( 0.4153)Vi ⇒ Agf = I0 = −0.731 mA / V Vi b. For K n = 1 mA / V 2 From dc analysis: 4 − VGS = (1)( 0.4 )(VGS − 2 ) 2 4 − VGS = 0.4 (V 2GS − 4VGS + 4 ) 0.4V 2GS − 0.6VGS − 2.4 = 0 VGS = 0.6 ± ( 0.6 ) + 4 ( 0.4 )( 2.4 ) 2 ( 0.4 ) 2 VGS = 3.31V g m = 2 (1)( 3.31 − 2 ) = 2.623 mA / V ⎛ 2 ⎞ I0 = − ⎜ ⎟ ( 2.623) Vgs = −1.311Vgs ⎝ 2+2⎠ Vi Vgs = = Vi ( 0.488 ) 1 + ( 2.623)( 0.4 ) I 0 = − (1.31)( 0.488 )Vi 507. Agf = I0 = −0.6398 mA / V Vi % change = 0.731 − 0.6398 ⇒ 12.5% 0.731 EX12.12 Use the circuit with the configuration shown in Figure 12.27. The LED replaces RL . 1 ⇒ RE = 100 Ω RE Agf = 10 mS = 10 × 10−3 = EX12.13 dc analysis: 10 − V0 V0 = ID + 4.7 47 + 20 I D = K n (VGS − VTN ) (1) 2 (2) ⎛ 20 ⎞ VGS = ⎜ (3) ⎟ V0 = 0.2985V0 ⎝ 20 + 47 ⎠ 2.13 − V0 ( 0.213) = I D + ( 0.0149 ) V0 (1) I D = 2.13 − V0 ( 0.2279 ) From (2): 2.13 − V0 ( 0.2279 ) = 1 ⎡( 0.2985V0 ) − 1.5⎤ ⎣ ⎦ 2 2.13 − V0 ( 0.2279 ) = 0.0891V0 − 0.8955V0 + 2.25 2 0.0891V0 − 0.6676V0 + 0.12 = 0 2 V0 = 0.6676 ± ( 0.6676 ) − 4 ( 0.0891)( 0.12 ) 2 ( 0.0891) 2 V0 = 7.31 V 10 − 7.31 7.31 − = 0.572 − 0.109 4.7 67 I D = 0.463 mA ID = VGS = a. 0.463 + 1.5 = 2.18 1 g m = 2 K n (VGS − VTN ) = 2 (1)( 2.18 − 1.5 ) ⇒ g m = 1.36 mA/V 508. V0 − Vgs V0 + g mVgs + = 0 (1) RD RF Vgs − Vi RS + Vgs − V0 RF =0 (2) ⎛ 1 1 ⎞ V0 Vi + + Vgs ⎜ ⎟= ⎝ RS RF ⎠ RF RS V 1 ⎞ V ⎛ 1 Vgs ⎜ + ⎟ = 0 + i ⎝ 20 47 ⎠ 47 20 Vgs ( 0.0713) = V0 ( 0.0213) + Vi ( 0.050 ) Vgs = V0 ( 0.299 ) + Vi ( 0.701) From (1): V0 V Vgs + (1.36 ) Vgs + 0 − =0 4.7 47 47 V0 ( 0.213) + (1.36 ) Vgs + V0 ( 0.213) − Vgs ( 0.0213) = 0 V0 ( 0.234 ) + Vgs (1.34 ) = 0 V0 ( 0.234 ) + (1.34 ) ⎡(V0 )( 0.299 ) + (Vi )( 0.701) ⎤ = 0 ⎣ ⎦ V0 ( 0.635 ) + Vi ( 0.939 ) = 0 ⇒ Avf = V0 = −1.48 Vi b. For K n = 1.5 mA / V 2 From dc analysis: 2.13 − V0 ( 0.2279 ) = 1.5 ⎡( 0.2985V0 ) − 1.5⎤ ⎣ ⎦ 2 = 1.5 ⎡0.0891V0 − 0.8955V0 + 2.25⎤ ⎣ ⎦ 2 = 0.1337V0 − 1.343V0 + 3.375 2 0.1337V0 − 1.115V0 + 1.245 = 0 2 V0 = V0 = 1.115 ± (1.115) − 4 ( 0.1337 )(1.245 ) 2 ( 0.1337 ) 2 1.115 ± 0.7597 ⇒ V0 = 7.01 V 2 ( 0.1337 ) 10 − 7.01 7.01 − = 0.636 − 0.105 4.7 67 I D = 0.531 mA ID = 0.531 + 1.5 = 2.09 1.5 g m = 2 K n (VGS − VTN ) = 2 (1.5 )( 2.09 − 1.5 ) VGS = = 1.77 mA/V From ac analysis: V0 V Vgs + (1.77 )Vgs + 0 − =0 4.7 47 47 509. V0 ( 0.213) + (1.77 ) Vgs + V0 ( 0.0213) − Vgs ( 0.0213) = 0 V0 ( 0.234 ) + Vgs (1.75 ) = 0 V0 ( 0.234 ) + (1.75 ) ⎡V0 ( 0.299 ) + Vi ( 0.701) ⎤ = 0 ⎣ ⎦ V0 ( 0.757 ) + Vi (1.23) = 0 ⇒ Avf = % change = V0 = −1.62 Vi 1.62 − 1.48 ⇒ 9.46% 1.48 EX12.14 a. Input resistance. V0 − Vgs V0 + g mVgs + = 0 (1) RD RF Ii = Vgs − V0 RF (2) So V0 = Vgs − I i RF ⎛ 1 ⎛ 1 ⎞ 1 ⎞ V0 ⎜ + ⎟ + Vgs ⎜ g m − ⎟ = 0 (1) RF ⎠ ⎝ RD RF ⎠ ⎝ 1 ⎞ 1 ⎞ ⎛ 1 ⎛ ⎡Vgs − I i ( 47 ) ⎤ ⎜ ⎣ ⎦ 4.7 + 47 ⎟ + Vgs ⎜1.36 − 47 ⎟ = 0 ⎝ ⎠ ⎝ ⎠ ⎡Vgs − I i ( 47 ) ⎤ ( 0.234 ) + Vgs (1.34 ) = 0 ⎣ ⎦ Vgs Vgs (1.57 ) = I i (11.0 ) ⇒ Rif = = 7.0 kΩ Ii Output Resistance. 510. IX = VX VX + g mVgs + RD RS + RF ⎛ RS ⎞ ⎛ 20 ⎞ Vgs = ⎜ ⎟ VX = ⎜ ⎟ VX = 0.2985VX ⎝ 20 + 47 ⎠ ⎝ RS + RF ⎠ V VX I X = X + (1.36)(0.2985)VX + 4.7 20 + 47 I X = VX [ 0.213 + 0.406 + 0.0149] R0 f = b. VX = 1.58 kΩ IX From part (a) 1 ⎞ ⎛ ⎡Vgs − I i ( 47 ) ⎤ ( 0.234 ) + Vgs ⎜ 1.77 − ⎟ = 0 ⎣ ⎦ 47 ⎠ ⎝ Vgs Vgs (1.98 ) = I i (11) ⇒ Rif = = 5.56 kΩ Ii IX = VX VX + (1.77 )( 0.2985 ) VX + 4.7 20 + 47 I X = VX [ 0.213 + 0.528 + 0.0149] ⇒ R0 f = VX = 1.32 kΩ IX EX12.15 Use the circuit with the configuration shown in Figure 12.41 Let RF = 10 K EX12.16 ⎛ 5.5 ⎞ VTH = ⎜ ⎟ (10 ) = 0.9735 V ⎝ 5.5 + 51 ⎠ RTH = 5.5 || 51 = 4.965 kΩ I BQ = 0.973 − 0.7 = 0.00217 mA 4.96 + (121)(1) I CQ = 0.2605 mA rπ = 11.98 kΩ, g m = 10.02 mA / V Req = RS || R1 || R2 || rπ = (10 ) || 51|| 5.5 || 12 = 2.598 kΩ From Equation (12.99(b)): ⎛ ⎞ Req T = ( g m RC ) ⎜ ⎜R +R +R ⎟ ⎟ F eq ⎠ ⎝ C 2.598 ⎛ ⎞ = (10 )(10 ) ⎜ ⎟ ⇒ T = 2.75 10 + 82 + 2.598 ⎠ ⎝ EX12.17 Computer Analysis EX12.18 511. Vε = −Vt , V0 = AvVε = − AvVt ⎛ R1 || Ri ⎞ ⎛ R1 || Ri ⎞ Vr = ⎜ ⎟ V0 = − ⎜ ⎟ ( AvVt ) R1 || Ri + R2 ⎠ ⎝ ⎝ R1 || Ri + R2 ⎠ T =− ⎛ R1 || Ri ⎞ Vr = Av ⎜ ⎟ Vt ⎝ R1 || Ri + R2 ⎠ or T= Av R 1+ 2 R1 Ri EX12.19 ⎛ f′ ⎞ Phase = −180° = −3 tan −1 ⎜ 5 ⎟ ⎝ 10 ⎠ ⎛ f′ ⎞ or tan −1 ⎜ 5 ⎟ = 60° ⇒ f ′ = 1.732 × 105 Hz ⎝ 10 ⎠ β (100 ) T ( f ′) = 1 = 3 2 ⎡ ⎛ f′ ⎞ ⎤ ⎢ 1+ ⎜ 5 ⎟ ⎥ ⎢ ⎝ 10 ⎠ ⎥ ⎣ ⎦ β (100 ) ⇒ β = 0.08 = 3 ⎡ 1 + (1.732 )2 ⎤ ⎢ ⎥ ⎣ ⎦ EX12.20 Afo = 1000 = 7.092 1 + ( 0.140 )(1000 ) 512. ( 0.140 )(1000 ) T =1= ⎛ f ⎞ ⎛ f ⎞ 1 + ⎜ 3 ⎟. 1 + ⎜ 4 ⎟ 10 ⎠ ⎝ ⎝ 5 × 10 ⎠ 2 ⎛ f ⎞ 1+ ⎜ 6 ⎟ ⎝ 10 ⎠ 2 By trial and error, at f = 7.65 × 104 Hz ( 0.140 )(1000 ) 2 2 1 + ( 76.5 ) 1 + (1.53) 1 + ( 0.0765 ) ( 0.140 )(1000 ) = = 1.00 76.5 )(1.828 )(1.0 ) ( T = 2 Then φ = − ⎡ tan −1 (76.5) + tan −1 (1.53) + tan −1 (0.0765) ⎤ ⎣ ⎦ = − [89.25 + 56.83 + 4.37 ] = −150.45 Phase Marge = 180 − 150.45 = 29.5° EX12.21 The new loop gain function is 105 1 T ′( f ) = × ⎛ f ⎞⎛ f ⎞ ⎛1 + j ⋅ f ⎞ ⎛1 + j ⋅ f ⎞ ⎟⎜ ⎟ ⎜1 + j ⋅ ⎟⎜ 1 + j ⋅ ⎟ ⎜ 107 ⎠ ⎝ 5 × 108 ⎠ f PD ⎠ ⎝ 5 × 105 ⎠ ⎝ ⎝ ⎧ ⎛ f ⎞ f ⎞ ⎪ ⎪ ⎛ f ⎞ ⎛ f ⎞⎫ −1 ⎛ Phase = − ⎨ tan −1 ⎜ + tan −1 ⎜ 7 ⎟ + tan −1 ⎜ ⎟ + tan ⎜ 5 ⎟ 8 ⎟⎬ ⎝ 5 × 10 ⎠ ⎝ 10 ⎠ ⎝ 5 × 10 ⎠ ⎪ ⎪ ⎝ f PD ⎠ ⎩ ⎭ For a phase margin 45° ⇒ Phase = −135°, the poles are far apart so this will occur at approximately f ′ = 5 × 105 Hz. Then 105 T ′( f ) = 1 = 2 ⎛ 5 × 105 ⎞ 1+ ⎜ ⎟ × 1+1× 1× 1 ⎝ f PD ⎠ 105 1= (1.414 ) ⎛ 5 × 105 ⎞ 1+ ⎜ ⎟ ⎝ f PD ⎠ 2 2 ⎛ 5 × 105 ⎞ 9 1+ ⎜ ⎟ = 5 × 10 ⎝ f PD ⎠ f PD ≅ 5 × 105 5 × 109 ⇒ f PD = 7.07 Hz EX12.22 Af ( 0 ) = A0 1 + β A0 = 2 × 105 1 + ( 0.05 ) ( 2 × 105 ) ⇒ Af (0) ≅ 20 f C = f PD (1 + β A0 ) = 100 ⎡1 + ( 0.05 ) ( 2 × 105 ) ⎤ ⇒ fC ≅ 1 MHz ⎣ ⎦ EX12.23 Phase Margin = 45° ⇒ Phase = −135° This will occur at approximately f ′ = 107 Hz 513. 105 T ′( f ) = 1 = 2 ⎛ 107 ⎞ 1+ ⎜ ⎟ × 2× 1 ⎝ f PD ⎠ 2 ⎛ 107 ⎞ 9 1+ ⎜ ⎟ = 5 × 10 f PD ⎠ ⎝ 107 ⇒ f PD = 141 Hz f PD ≅ 5 × 109 TYU12.1 A Af = 1 + Aβ A f + A f Aβ = A Af = A (1 − Af β ) A= Af 1 − Af β = 80 1 − ( 80 )( 0.0120 ) A = 2000 TYU12.2 dAf dA ⎛ Af ⎞ dA 1 = ⋅ =⎜ ⎟. Af (1 + β A) A ⎝ A ⎠ A dA dAf = A Af ⎛ A ⋅⎜ ⎜A ⎝ f ⎞ ⎛ 5 × 105 ⎞ dA = ±5% ⎟ = ( 0.001) ⎜ ⎟⇒ ⎟ A ⎝ 100 ⎠ ⎠ TYU12.3 Af ⋅ f H = A0 ⋅ f1 Af = 6 A0 ⋅ f1 (10 ) ( 8 ) ⇒ Af ( 0 ) = 32 = fH 250 × 103 TYU12.4 Vε = VS − V fb = 100 − 99 = 1 mV Ag = β= Agf = I 0 5 mA = ⇒ Ag = 5 A/V Vε 1 mV V fb = I0 99 mV ⇒ β = 19.8 V/A 5 mA Ag 1 + β Ag TYU12.5 = 5 ⇒ Agf = 0.05 A/V = 50 mA/V 1 + (19.8 )( 5 ) 514. I ε = I S − I fb = 100 − 99 = 1 μ A Az = β= Azf = V0 5V = ⇒ Az = 5 × 106 V/A Iε 1 μ A I fb = V0 99 μ A ⇒ β = 1.98 × 10−5 A/V 5V Az 1 + β Az TYU12.6 rπ = a. gm = I CQ = VT = 5 × 106 1 + (1.98 × 10−5 )( 5 × 106 ) ⇒ Azf = 5 × 104 V/A = 50 V/mA hFEVT (100 )( 0.026 ) = = 5.2 kΩ I CQ 0.5 0.5 = 19.23 mA / V 0.026 ⎛1 ⎞ ⎛ 1 ⎞ + 19.23 ⎟ ( 2 ) ⎜ + g m ⎟ RE ⎜ rπ 5.2 ⎠ ⎠ = ⎝ Avf = ⎝ ⎛1 ⎞ ⎛ 1 ⎞ + 19.23 ⎟ ( 2 ) 1 + ⎜ + g m ⎟ RE 1 + ⎜ ⎝ 5.2 ⎠ ⎝ rπ ⎠ (19.42 )( 2 ) ⇒ Avf = 0.97490 1 + (19.42 )( 2 ) = rπ + (1 + hFE ) RE = 5.2 + (101)( 2 ) ⇒ Rif = Rif rπ 5.2 =2 ⇒ R0 f = 0.0502 kΩ ⇒ 50.2 Ω 1 + hFE 101 R0 f = RE b. = 207.2 kΩ hFE = 150 ⇒ rπ = 7.8 kΩ, g m = 19.23mA/V ⎛ 1 ⎞ + 19.23 ⎟ ( 2 ) ⎜ (19.36 )( 2 ) 7.8 ⎠ Avf = ⎝ = 1 + (19.36 )( 2 ) ⎛ 1 ⎞ 1+ ⎜ + 19.23 ⎟ ( 2 ) 7.8 ⎝ ⎠ Avf = 0.97482 ⇒ 0.0082% change in Avf Rif = 7.8 + (101)( 2 ) = 209.8 kΩ ⇒ 1.25% change in Rif R0 f = RE rπ 7.8 =2 = 2 0.0517 1 + hFE 151 R0 f = 50.36 Ω ⇒ 0.319% change in R0 f TYU12.7 515. V0 = ( g mVgs ) RS Vi = Vgs + g m RSVgs Vi Vgs = 1 + g m RS g m = 2 K n I DQ = 2 ( 0.2 )( 0.25 ) = 0.447 mA / V Avf = V0 g m RS = Vi 1 + g m RS Avf = ( 0.447 )( 5 ) ⇒ Avf 1 + ( 0.447 )( 5 ) = 0.691 Rif = ∞ I X = g mVgs = VX RS Vgs = −VX ⎛ 1 ⎞ I X = VX ⎜ g m + ⎟ RS ⎠ ⎝ 1 1 R0 f = RS = 5 gm 0.447 R0 f = 1.55 kΩ TYU12.8 Computer Analysis TYU12.9 Computer Analysis TYU12.10 a. Agf = b. Agf = hFE ⋅ Ag 1 + ( hFE Ag ) RE = ( 200 ) (103 ) ⇒ Agf 1 + ( 200 ) (103 )(103 ) ( 200 ) (104 ) ⇒ Agf 1 + ( 200 ) (104 )(103 ) ≅ 1 mA/V = 10−3 A/V=1 mA/V 516. Percent change is negligible. ( 4.5 × 10−7 % ) ( 200 ) (104 ) ( 200 ) (103 ) − 1 + ( 200 ) (104 )(103 ) 1 + ( 200 ) (103 )(103 ) 10−3 = − = ( 200 ) (104 ) ⎡1 + ( 200 ) (104 )(103 )⎤ ⎣ ⎦ (10 )( 2 ×10 )( 2 ×10 ) −3 9 8 (200)(103 )[1 + (200)(104 )(103 )] (10−3 )(2 × 109 )(2 × 108 ) 200 × 104 − 200 × 103 (10−3 )( 2 ×109 )( 2 ×108 ) TYU12.11 T = Ai β = T ( f1 ) Ai 0 β = 4.5 × 10−9 (10 ) ( 0.01) 5 = ⎛ ⎛ f ⎞ f ⎞ ⎜ 1 + j ⋅ ⎟ 1 + j ⋅ ⎜ 10 ⎟ ⎝ ⎠ f1 ⎠ ⎝ 3 10 =1= 2 ⎛ f′⎞ 1+ ⎜ E ⎟ ⎝ 10 ⎠ ⎛ f′⎞ 1 + ⎜ E ⎟ = 106 ⎝ 10 ⎠ 2 f E′ = 10 106 − 1 ⇒ f E′ ≈ 104 Hz ⎛ 104 ⎞ ⎛ f′⎞ Phase = φ = − tan −1 ⎜ E ⎟ = − tan −1 ⎜ ⎟ ⎝ 10 ⎠ ⎝ 10 ⎠ = − tan −1 (103 ) φ ≈ 90° Phase Margin = 180 − 90 ⇒ Phase Margin = 90° TYU12.12 Αι 0 β ⎛ f ⎞⎛ f ⎞ ⎜ 1+ j ⋅ ⎟ ⎜1 + j ⋅ ⎟ f1 ⎠ ⎝ f2 ⎠ ⎝ ⎡ ⎛ f ⎞ ⎛ f ⎞⎤ Phase = − ⎢ tan −1 ⎜ ⎟ + tan −1 ⎜ ⎟ ⎥ ⎝ f1 ⎠ ⎝ f2 ⎠⎦ ⎣ T = Ai β = Phase Margin = 60° ⇒ Phase = −120° ⎡ ⎛ f ⎞ ⎛ f ⎞⎤ −120° = − ⎢ tan −1 ⎜ 4 ⎟ + tan −1 ⎜ 5 ⎟ ⎥ ⎝ 10 ⎠ ⎝ 10 ⎠ ⎦ ⎣ At f ′ = 7.66 × 104 Hz, Phase = − ⎡ tan −1 ( 7.66 ) + tan −1 ( 0.766 ) ⎤ ⎣ ⎦ = − [82.56 + 37.45] = −120° 517. (10 ) β 5 T ( f ′) = 1 = 1+ ( 7.66 ) × 1 + ( 0.766 ) 2 (10 ) β 2 5 1= ( 7.725 )(1.26 ) ⇒ β = 9.73× 10−5 TYU12.13 Phase Margin = 60° ⇒ Phase = −120° ⎛ f′ ⎞ Phase = −120° = −3 tan −1 ⎜ 5 ⎟ ⎝ 10 ⎠ ⎛ f′ ⎞ tan −1 ⎜ 5 ⎟ = 40° ⇒ f ′ = 0.839 × 105 Hz ⎝ 10 ⎠ β (100 ) T ( f ′) = 1 = 3 2 ⎡ ⎛ f′ ⎞ ⎤ ⎢ 1+ ⎜ 5 ⎟ ⎥ ⎢ ⎝ 10 ⎠ ⎥ ⎣ ⎦ β (100 ) = ⇒ β = 0.0222 3 ⎡ 1 + ( 0.839 )2 ⎤ ⎢ ⎥ ⎣ ⎦ 518. Chapter 12 Problem Solutions 12.1 (a) Af = A 1 + Aβ 120 = 5 × 105 1 + (5 × 105 ) β 120 + (120)(5 × 105 ) β = 5 × 105 β = 0.008331 (b) 5 × 103 120 = 1 + (5 × 103 ) β 120 + (120)(5 × 103 ) β = 5 × 103 β = 0.008133 12.2 Af = (a) A 1 + Aβ β = 0.15 T = Aβ T =∞ (i) A = 80 dB ⇒ A = 104 ⇒ T = 1.5 × 103 (ii) (iii) T = 15 1 (i) Af = = 6.667 β (ii) Af = 6.662 (iii) Af = 6.25 (b) (i) (ii) (iii) (i) T =∞ T = 2.5 × 103 T = 25 1 Af = = 4.00 (ii) Af = 3.9984 (iii) Af = 3.846 β 12.3 (a) A 1 ≅ = 125 1 + Aβ β β = 0.0080 (b) Af = 519. Af = (125)(0.9975) = 124.6875 124.6875 = A 1 + (0.008) A 124.6875 [1 + (0.008) A] = A 124.6875 = A [1 − 0.9975] A = 49,875 12.4 (a) 100 = 104 1 + (104 ) β β = 9.9 × 10−3 (b) ⎛ Af ⎞ dA =⎜ ⎟ ⎝ A⎠ A Af dAf dAf ⎛ 100 ⎞ = ⎜ 4 ⎟ (−0.10) = −0.001 ⇒ = −0.10% Af ⎝ 10 ⎠ 12.5 ⎛ Af ⎞ dA =⎜ ⎟ ⎝ A⎠ A Af dAf −0.001 = Af = 500 500 = Af 5 × 104 (−0.10) 5 × 104 1 + (5 × 104 ) β β = 1.98 ×10−3 12.6 (a) For Fig. P12.6(a) vo1 vo 200 10 = = vi 1 + 200 β1 vo1 1 + 10 β1 ⎛ 200 ⎞ ⎛ 10 ⎞ Af = ⎜ ⎟⎜ ⎟ = 40 ⎝ 1 + 200 β1 ⎠ ⎝ 1 + 10 β1 ⎠ 50 = (1 + 200 β1 )(1 + 10 β1 ) = 1 + 210 β1 + 2000 β12 2000 β12 + 210 β1 − 49 = 0 −210 ± 44100 + 4(2000)(49) 2(2000) β1 = 0.1126 For Fig P12.6(b) (200)(10) 40 = 1 + (200)(10) β 2 β1 = β 2 = 0.0245 Fig. P12.6(a) (b) 520. A1 = 180 180 10 ⎛ ⎞⎛ ⎞ Af = ⎜ ⎟ ⎜ 1 + (10)(0.1126) ⎟ = (8.4634)(4.704) ⎝ 1 + (180)(0.1126) ⎠ ⎝ ⎠ Af = 39.81 dAf Af = 39.81 − 40 ⇒ −0.475% 40 Fig. P12.6(b) (180)(10) = 39.91 Af = 1 + (180)(10)(0.0245) dAf 39.91 − 40 = ⇒ −0.225% 40 Af (c) Fig. P12.6(b) is a better feedback circuit. 12.7 (a) VO = (−10)(−15)(−20)Vε = −3000Vε Vε = β VO + VS So VO = −3000( β VO + VS ) We find V −3000 Avf = O = VS 1 + 3000 β For Avf = −120 = (b) Then Avf = −3000 ⇒ β = 0.008 1 + 3000 β Now VO = (−9)(−13.5)(−18)Vε = −2187Vε −2187 −2187 = = −118.24 1 + 2187 β 1 + 2187(0.008) % change = 120 − 118.24 × 100 ⇒ 1.47% change 120 12.8 (105 )(4) = (50) f B ⇒ f B = 8 kHz 12.9 (a) (50) f3− dB = (105 )(4) ⇒ f 3− dB = 8 kHz (b) (10) f3− dB = (105 )(4) ⇒ f3− dB = 40 kHz 12.10 (50)(20 × 103 ) = 5A0 so A0 = 2 × 105 12.11 Low freq. Af = A0 1 + A0 β 5000 ⇒ β = 0.0098 1 + (5000) β Freq. response 100 = 521. Af = A 1 + Aβ 5000 f ⎞⎛ f ⎞ ⎛ ⎜1 + j f ⎟⎜ 1 + j f ⎟ 1 ⎠⎝ 2 ⎠ = ⎝ (5000)(0.0098) 1+ f ⎞⎛ f ⎞ ⎛ ⎜ 1 + j f ⎟⎜ 1 + j f ⎟ 1 ⎠⎝ 2 ⎠ ⎝ 5000 f ⎞⎛ f ⎞ ⎛ ⎜ 1 + j f ⎟ ⎜ 1 + j f ⎟ + 49 1 ⎠⎝ 2 ⎠ ⎝ 5000 = f f ⎛ jf ⎞ ⎛ jf 1+ j + j + ⎜ ⎟⎜ f1 f 2 ⎝ f1 ⎠ ⎝ f 2 = = ⎞ ⎟ + 49 ⎠ 5000 f f ⎛ jf ⎞ ⎛ jf ⎞ 50 + j + j + ⎜ ⎟ ⎜ ⎟ f1 f 2 ⎝ f1 ⎠ ⎝ f 2 ⎠ Also Af = Af 0 f ⎞⎛ f ⎞ ⎛ ⎜1 + j f ⎟ ⎜ 1 + j f ⎟ A ⎠⎝ B ⎠ ⎝ = 100 f f ⎛ f ⎞⎛ f ⎞ 1+ j j +j + j fA fB ⎜ f A ⎟ ⎜ fB ⎟ ⎝ ⎠⎝ ⎠ So 100 100 = 1 ⎛ jf ⎞ ⎛ jf ⎞ f f ⎛ f ⎞⎛ f ⎞ f f +j + ⎜ j ⎟⎜ j ⎟ 1 + j +j + ⎜ ⎟⎜ ⎟ 1+ j fA f B ⎝ f A ⎠⎝ f B ⎠ 50 f1 50 f 2 50 ⎝ f1 ⎠ ⎝ f 2 ⎠ Then 1 1 1 1 + = + f A f B 50 f1 50 f 2 and 1 1 = f A f B 50 f1 f 2 f1 = 10 and f 2 = 2000 1 1 1 1 + = + = 0.002 + 0.000010 = 0.002010 f A f B 50(10) 50(2000) and f 1 1 1 = ⇒ = B f A f B (50)(10)(2000) f A 106 Then fB 1 + = 0.002010 6 fB 10 10−6 f B2 + 1 = 2.01 + 10−3 f B 10−6 f B2 − 2.01× 10−3 f B + 1 = 0 fB = 2.01× 10−3 ± 4.0401× 10 −6 − 4(10−6 )(1) 2(10−6 ) fB = 2.01× 10−3 ± 2.0025 × 10−4 2(10−6 ) + sign f B = 1.105 × 103 Hz + sign f A = 9.05 × 102 Hz 522. 12.12 (a) Fig. P12.6(a) ⎡ ⎤ ⎛ 200 ⎞ ⎢ ⎥ ⎜ f ⎟ ⎢ ⎥ ⎜1+ j ⎟ ⎜ ⎟ f1 ⎠ 10 ⎤ ⎢ ⎥⎡ ⎝ Af = ⎢ ⎥ ⎢1 + (10)(0.1126) ⎥ 200 ⎞ ⎛ ⎣ ⎦ ⎢1 + ⎜ ⎟ (0.1126) ⎥ ⎢ ⎜ 1+ j f ⎟ ⎥ ⎢ ⎜ ⎥ f1 ⎟ ⎠ ⎣ ⎝ ⎦ 200 ⎡ ⎤ =⎢ ⎥ (4.704) f ⎞ ⎛ ⎢ ⎜1 + j ⎟ + 22.52 ⎥ f1 ⎠ ⎢ ⎥ ⎣⎝ ⎦ = 940.73 23.52 + j f f1 = 940.73 1 ⋅ f 23.52 1 + j (23.52) f1 40 f −3dB = (23.52)(100) ⇒ 2.352 kHz jf 1+ (23.52) f1 Fig P12.6(b) (200)(10) f 1+ j f1 2000 = Af = f (0.0245)(200)(10) 1+ 1 + j + 49 f f1 1+ j f1 = = (b) 2000 1 ⋅ f −3dB = (50)(100) ⇒ 5 KHz 50 1 + j f (50) f1 Overall feedback ⇒ wider bandwidth. 12.13 v0 = A1 A2 vi + A1vn v0 = (100)vi + (1)vn = (100)(10) + (1)(1) ⇒ 12.14 (a) S0 1000 = = 1000 N0 1 523. (b) Circuit (b) – less distortion 12.15 (a) Low input R ⇒ Shunt input Low output R ⇒ Shunt output Or a Shunt-Shunt circuit (b) High input R ⇒ Series input High output R ⇒ Series output Or a series-Series circuit (c) Shunt-Series circuit (d) Series-Shunt circuit 12.16 (a) Ri (max) = Ri (1 + T ) = 10(1 + 104 ) ⇒ Ri (max) ≅ 105 k Ω 524. Ri 10 = ≅ 10−3 k Ω 1 + T 1 + 104 Or Ri (min) = 1Ω Ri (min) = Ro (max) = Ro (1 + T ) = 1(1 + 104 ) ⇒ Ro (max) ≅ 104 k Ω (b) Ro 1 = ≅ 10−4 k Ω 1 + T 1 + 104 Or Ro (min) = 0.1Ω Ro (min) = 12.17 io Series output = current signal and Shunt input = current vi signal. Also, Shunt output = voltage signal and Series input = voltage signal. Two possible solutions are shown. Overall Transconductance Amplifier, Ag = 12.18 ⎛ R1 || Ri ⎞ Vε = − ⎜ ⎟ ⋅ Vx ⎝ R1 || Ri + R2 ⎠ 12.19 Vε = Vi − V fb = 50 − 48 = 2mV Av = β= Vo 5 = = 2.5 × 103 V/V Vε 0.002 V fb Vo = 0.048 = 0.0096 V/V 5 V 5 Arf = o = = 100 V/V Vi 0.05 12.20 525. ⎛ R ⎞ R Avf ≈ ⎜1 + 2 ⎟ = 20 ⇒ 2 = 19 R1 ⎠ R1 ⎝ vd = iS Ri vS − vd (vs − vd ) − v0 + R1 R2 (1) v0 − A0 L vd v0 − (vs − vd ) + =0 R0 R2 (2) is = ⎛ 1 (v − v ) 1 ⎞ A v v0 ⎜ + ⎟ = 0 L d + S d R0 R2 ⎝ R0 R2 ⎠ A0 L vd (vS − vd ) + R0 R2 v0 = ⎛ 1 1 ⎞ ⎜ + ⎟ ⎝ R0 R2 ⎠ From (1): 1 ⎡ A0 L vd (vS − vd ) ⎤ ⋅⎢ + R2 ⎥ vS − vd vS − vd R2 ⎣ R0 ⎦ + − iS = R1 R2 ⎛ 1 1 ⎞ ⎜ + ⎟ ⎝ R0 R2 ⎠ A0 L 1 1 ⎞ ⎛ ⎛ − ⎜ 1 ⎜ 1 R0 R2 R2 ⎟ 1 1 ⎟ − vd ⎜ + − + iS = vS ⎜ + R ⎜ R1 R2 ⎜ R1 R2 1 + R2 ⎟ 1+ 2 ⎜ ⎜ R0 ⎟ R0 ⎝ ⎠ ⎝ vd = iS Ri ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ⎧ ⎡⎛ 1 ⎡⎛ 1 1 ⎞ ⎛ R ⎞ A0 L 1 ⎤ ⎫ 1 ⎞ ⎛ R2 ⎞ 1 − ⎥⎪ ⎪ Ri ⎢⎜ + ⎟ ⎜ 1 + 2 ⎟ + ⎢ ⎜ + ⎟ ⎜1 + ⎟ − R1 R2 ⎠ ⎝ R0 ⎠ R0 R2 ⎦ ⎪ ⎪ ⎝ ⎢ ⎝ R1 R2 ⎠ ⎝ R0 ⎠ R2 iS ⎨1 + ⎣ ⎬ = vS ⎢ R2 R ⎪ ⎪ 1+ 1+ 2 ⎢ R0 R0 ⎪ ⎪ ⎢ ⎣ ⎩ ⎭ ⎧ R ⎡1 R 1 ⎡ 1 R2 1 1 A0 L ⎤ ⎫ 1⎤ ⎪ ⎪ iS ⎨1 + 2 + Ri ⎢ + 2 ⋅ + + ⎥ ⎬ = vS ⎢ + ⋅ + ⎥ ⎪ R0 ⎣ R1 R1 R0 R0 R0 ⎦ ⎪ ⎣ R1 R1 R0 R0 ⎦ ⎩ ⎭ ⎧ ⎡R ⎛ R ⎞ ⎤⎫ ⎡ R ⎛ R ⎞⎤ ⎪ ⎪ iS ⎨ R0 + R2 + Ri ⎢ 0 + ⎜1 + 2 ⎟ + A0 L ⎥ ⎬ = vS ⎢ 0 + ⎜ 1 + 2 ⎟ ⎥ (1) R1 ⎠ R1 ⎠ ⎦ ⎪ ⎣ R1 ⎝ ⎦⎪ ⎣ R1 ⎝ ⎩ ⎭ Let R2 = 190 kΩ , R1 = 10 kΩ ⎧ ⎡ 0.1 ⎤⎫ ⎡ 0.1 ⎤ iS ⎨0.1 + 190 + 100 ⋅ ⎢ + 20 + 105 ⎥ ⎬ = vS ⎢ + 20 ⎥ ⎣ 10 ⎦⎭ ⎣ 10 ⎦ ⎩ 7 iS (1.000219 × 10 ) = vS (20.01) vS ≅ 5 × 105 kΩ ⇒ Rif ≅ 500 MΩ iS Output Resistance Rif = ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ 526. IX = VX − A0 L vd VX + R0 R2 + R1 || Ri vd = − R1 || Ri ⋅ VX R1 || Ri + R2 A0 L ⋅ R1 || Ri IX 1 1 1 = = + + VX R0 f R0 R0 ( R1 || Ri + R2 ) R2 + R1 || Ri R1 || Ri = 10 ||100 = 9.09 1 1 105 ⎛ 9.09 ⎞ 1 = + ⋅⎜ ⎟+ R0 f 0.1 0.1 ⎝ 9.09 + 190 ⎠ 190 + 9.09 = 10 + 4.566 × 104 + 0.00502 R0 f = 2.19 × 10−5 kΩ ⇒ R0 f = 0.0219 Ω 12.21 a. vS − vd v0 − (vS − vd ) v = and vd = 0 R1 R2 A 527. ⎛ 1 vS vS v 1 ⎞ + = 0 + vd ⎜ + ⎟ R1 R2 R2 R1 R2 ⎠ ⎝ v0 v0 ⎛ 1 1 ⎞ + ⎜ + ⎟ R2 A ⎝ R1 R2 ⎠ ⎛ 1 1 ⎞ v ⎡ 1 ⎛ R ⎞⎤ vS ⎜ + ⎟ = 0 ⎢1 + ⎜ 1 + 2 ⎟ ⎥ R1 ⎠ ⎦ ⎝ R1 R2 ⎠ R2 ⎣ A ⎝ = ⎛ R2 ⎞ ⎜1 + ⎟ R1 ⎠ v0 = ⎝ vS 1⎛ R ⎞ 1 + ⎜1 + 2 ⎟ A⎝ R1 ⎠ which can be written as v A Avf = 0 = vS ⎡ ⎛ R ⎞⎤ 1 + ⎢ A / ⎜1 + 2 ⎟ ⎥ R1 ⎠ ⎦ ⎣ ⎝ 1 b. β= R 1+ 2 R1 20 = c. 105 1 + (105 ) β 105 −1 So β = 20 5 ⇒ β = 0.04999 10 R2 1 R 1 Then = −1 = − 1 ⇒ 2 = 19.004 R1 β 0.04999 R1 A → 9 × 104 9 × 104 Af = = 19.99956 1 + (9 × 104 )(0.04999) d. ΔAf Af = ΔAf −4.444 × 10−4 = −2.222 × 10 −3 % ⇒ = −0.005% 20 Af 12.22 Aif = Rif = Ai 1000 = = 90.9 A/A 1 + β i Ai 1 + (1000)(0.01) Ri 1 + β i Ai = 1 ⇒ Rif = 90.9 Ω 1 + (0.01)(1000) Rof = Ro (1 + β i Ai ) = 10 [1 + (0.01)(1000) ] ⇒ Rof = 110 kΩ 12.23 528. I ε = I i − I fb = 50 − 47.5 = 2.5 μ A Ai = βi = Io 5 = = 2000 A/A I ε 0.0025 I fb Io = 0.0475 = 0.0095 A/A 5 I 5 Aif = o = = 100 A/A I i 0.05 12.24 a. Assume that V1 is at virtual ground. V0 = − I fb RF Now I fb = I 0 + I fb RF V0 = I0 − R3 R3 I fb = I S − I ε and I 0 = Ai I ε = I0 Ai so I0 Ai From above ⎛ R ⎞ I fb ⎜ 1 + F ⎟ = I 0 R3 ⎠ ⎝ I fb = I S − ⎛ I 0 ⎞ ⎛ RF ⎞ ⎜ I S − ⎟ ⎜1 + ⎟ = I0 Ai ⎠ ⎝ R3 ⎠ ⎝ ⎡ ⎛ R ⎞ 1 ⎛ R ⎞⎤ I S ⎜ 1 + F ⎟ = I 0 ⎢1 + ⎜ 1 + F ⎟ ⎥ R3 ⎠ R3 ⎠ ⎦ ⎝ ⎣ Ai ⎝ or 529. Aif = ⎛ RF ⎞ ⎜1 + ⎟ R3 ⎠ ⎝ = ⎡ 1 ⎛ RF ⎞ ⎤ ⎢1 + ⎜ 1 + ⎟⎥ R3 ⎠ ⎦ ⎣ Ai ⎝ Ai = = Aif Ai 1+ ⎛ RF ⎞ ⎜1 + ⎟ R3 ⎠ ⎝ I0 IS b. βi = c. 25 = 1 ⎛ RF ⎞ ⎜1 + ⎟ R3 ⎠ ⎝ 105 1 + (105 ) β i 105 −1 so β i = 25 5 ⇒ β i = 0.03999 10 RF R 1 1 = −1 = − 1 ⇒ F = 24.0 so R3 β i R3 0.03999 Ai = 105 − (0.15)(105 ) = 8.5 × 104 d. so Aif = so ΔAif Aif 8.5 × 104 = 24.9989 1 + (8.5 × 104 )(0.03999) =− 1.10 × 10−3 = −4.41× 10 −5 ⇒ −4.41× 10 −3 % 25 12.24 b. V0 = ( g m1 Vgs + g m 2 Vπ ) RL (1) Vi = Vgs + V0 (2) ⎛ 1 + hFE ⎞ V1 Vπ V + g m 2 Vπ + 1 = 0 ⇒ Vπ ⎜ =0 ⎟+ rπ RE 2 ⎝ rπ ⎠ RE 2 Vπ V −V = g m1 Vgs + 1 π = 0 rπ RD1 or ⎡V ⎤ V V1 = RD1 ⎢ π + π − g m1 Vgs ⎥ (4) ⎣ rπ RD1 ⎦ Then ⎛ 1 + hFE Vπ ⎜ ⎝ rπ ⎞ RD1 ⎟+ ⎠ RE 2 ⎧⎛ 1 + hFE ⎪ Vπ ⎨⎜ ⎪⎝ rπ ⎩ Let ⎡ Vπ Vπ ⎤ − g m1 Vgs ⎥ = 0 (3) ⎢ + ⎣ rπ RD1 ⎦ ⎫ ⎞ RD1 ⎛ RD1 ⎞ 1 ⎪ + ⎬ = g m1 ⎜ ⎟+ ⎟ Vgs RE 2 rπ RE 2 ⎪ ⎝ RE 2 ⎠ ⎠ ⎭ (3) 530. Vπ ⋅ ⎛R ⎞ 1 = g m1 ⎜ D1 ⎟ Vgs Req ⎝ RE 2 ⎠ ⎛R ⎞ so Vπ = g m1 Req ⎜ D1 ⎟ Vgs ⎝ RE 2 ⎠ Then ⎡ ⎛R ⎞ ⎤ V0 = ⎢ g m1 Vgs + g m1 g m 2 Req ⎜ D1 ⎟ Vgs ⎥ RL ⎝ RE 2 ⎠ ⎦ ⎣ so ⎡ ⎛ R ⎞⎤ V0 = g m1 RL ⎢1 + g m 2 Req ⎜ D1 ⎟ ⎥ (Vi − V0 ) ⎝ RE 2 ⎠ ⎦ ⎣ so ⎡ ⎛ R ⎞⎤ g m1 RL ⎢1 + g m 2 Req ⎜ D1 ⎟ ⎥ V ⎝ RE 2 ⎠ ⎦ ⎣ Av = 0 = Vi ⎡ ⎛ R ⎞⎤ 1 + g m1 RL ⎢1 + g m 2 Req ⎜ D1 ⎟ ⎥ ⎝ RE 2 ⎠ ⎦ ⎣ c. Set Vi = 0 I X + g m1 Vgs + g m 2Vπ = VX RL Vgs = −VX From part (b), we have ⎛R ⎞ ⎛R ⎞ Vπ = g m1 Req ⎜ D1 ⎟ Vgs = − g m1 Req ⎜ D1 ⎟ VX RE 2 ⎠ ⎝ ⎝ RE 2 ⎠ Then ⎛R ⎞ IX 1 1 = = + g m1 g m 2 Req ⎜ D1 ⎟ VX R0 RL ⎝ RE 2 ⎠ or R0 = RL 1 g m1 1 ⎛R ⎞ g m1 g m 2 Req ⎜ D1 ⎟ ⎝ RE 2 ⎠ 12.25 I S = I ε + I fb , V1 = I ε Ri I fb = I 0 + V0 and V0 = V1 − I fb RF R3 I 0 = Ai I ε ⇒ I ε = I0 Ai (1) 531. Now I fb = Ai I ε + 1 (V1 − I fb RF ) R3 ⎡ R ⎤ V I fb ⎢1 + F ⎥ = Ai Iε + 1 R3 ⎦ R3 ⎣ I fb = I S − I e ⎡ R ⎤ V ( I S − I e ) ⎢1 + F ⎥ = Ai I e + 1 R3 ⎣ R3 ⎦ ⎡⎛ R ⎞ ⎤ V ⎡ R ⎤ I S ⎢1 + F ⎥ = I e ⎢⎜ 1 + F ⎟ + Ai ⎥ + 1 R3 ⎠ ⎣ R3 ⎦ ⎣⎝ ⎦ R3 V Iε = 1 Ri ⎧ ⎫ ⎤ 1⎪ ⎤ ⎪ 1 ⎡⎛ RF ⎞ ⎟ + Ai ⎥ + ⎬ ⎥ = V1 ⎨ ⋅ ⎢⎜ 1 + R3 ⎠ ⎦ ⎪ Ri ⎣⎝ ⎦ R3 ⎪ ⎩ ⎭ ⎛ RF ⎞ ⎜1 + ⎟ R3 ⎠ V1 ⎝ Rif = = I S ⎧ 1 ⎡⎛ RF ⎞ ⎤ 1⎫ ⎪ ⎪ ⎨ ⋅ ⎢⎜ 1 + ⎟ + Ai ⎥ + ⎬ Ri ⎣⎝ R3 ⎠ R3 ⎪ ⎪ ⎦ ⎩ ⎭ ⎡ R I S ⎢1 + F ⎣ R3 The 1/ R3 term in the denominator will be negligible. Using the results of Problem 12.15: 25 Rif = ⎧1 ⎡ 5 ⎫ ⎨ ⎣ (25) + 10 ⎤ ⎬ ⎦⎭ 2 ⎩ Rif ≅ 5 × 10−4 kΩ ⇒ Rif = 0.5 Ω Output Resistance (Let Z L = 0) IX = VX VX + Ai I ε + R3 RF + Ri Iε = VX RF + Ri so A + 1 RF IX 1 1 , = = + i = 24 VX R0 f R3 RF + Ri R3 Let RF = 240 kΩ, R3 = 10 k Ω 1 1 105 + 1 = + Rof 10 240 + 2 532. so R0 f ≈ RF + Ri 240 + 2 ⇒ R0 f ≈ 2.42 × 10−3 kΩ or R0 f ≈ 2.42 Ω = 5 10 + 1 Ai + 1 12.26 Agf = Ag 5 = = 0.2 A/V 1 + β z Ag 1 + (4.8)(5) Rif = Ri [1 + β z Ag ] = (10) [1 + (4.8)(5) ] = 250 kΩ Rof = Ro [1 + β z Ag ] = (10) [1 + (4.8)(5) ] = 250 kΩ 12.27 Vε = Vi − V fb = 40 − 38 = 2.0 mV I o 8 mA = = 4 A/V Vε 2 mV Ag = V fb βz = I0 Agf = = 38 mV = 4.75 V/A 8 mA I0 8 mA = = 0.2 A/V Vi 40 mV 12.28 IE = V − Vε (1 + hFE ) ⋅ I0 = S hFE RE Also I 0 = hFE ( AgVε ) so Vε = I0 hFE Ag Then V I0 1 + hFE ⋅ I0 = S − hFE RE hFE Ag RE ⎡1 + hFE ⎤ VS 1 + ⎢ ⎥ I0 = hFE hFE Ag RE ⎥ RE ⎢ ⎣ ⎦ ⎡ Ag (1 + hFE ) RE + 1 ⎤ VS ⎢ ⎥ I0 = hFE Ag RE RE ⎢ ⎥ ⎣ ⎦ I0 1 = VS RE ⎡ ⎤ hFE Ag RE hFE Ag I0 ⋅⎢ ≈ ⎥⇒ ⎢1 + Ag (1 + hFE ) RE ⎥ VS 1 + (hFE Ag ) RE ⎣ ⎦ 533. b. Bz = RE c. 10 = 5 × 105 1 + (5 × 105 ) β z 5 × 105 −1 β z = 10 5 ⇒ β z = RE = 0.099998 kΩ 5 × 10 d. If Ag → 5.5 × 105 then Agf = ΔAgf Agf 5.5 × 105 = 10.0000182 1 + (5.5 × 105 )(0.099998) = 1.82 × 10−5 ⇒ 1.82 × 10−4 % 10 12.29 I E = (1 + hFE ) AgVε , I E = Now (1 + hFE ) Ag I S Ri = Vε − I S and Vε = I S Ri , Vε = VS − Vε = VS − I S Ri RE 1 ⋅ (VS − I S Ri ) − I S RE ⎡ ⎤ V Ri + 1⎥ I S = S ⎢(1 + hFE ) Ag Ri + RE ⎦ RE ⎣ ⎡ ⎤ VS R = RE ⎢(1 + hFE ) Ag Ri + i + 1⎥ IS RE ⎦ ⎣ From Problem 12.16: (1 + hFE ) Ag ≈ hFE Ag = 5 × 105 mS Rif = RE ≈ 0.1 kΩ 20 ⎤ ⎡ +1 so Rif = (0.1) ⎢(5 × 105 )(20) + 0.1 ⎥ ⎣ ⎦ or Rif = 106 kΩ 534. Vπ = AgVε rπ I X = g mVπ + VX − (−Vε ) R0 (1) Vε = −( I X + AgVε )( RE || Ri ) (2) or Vε ⎡1 + Ag ( Rε || Ri ) ⎤ = − I X ( RE || Ri ) ⎣ ⎦ Now: V V I X = g m Ag rπ Vε + X + ε (1) R0 R0 ⎛ 1 ⎞ ⎡ − I X ( RE || Ri ) ⎤ VX I X = ⎜ g m Ag rπ + ⎟ ⎢ ⎥+ R0 ⎠ ⎢1 + Ag ( RE || Ri ) ⎥ R0 ⎝ ⎣ ⎦ VX R0 f = IX ⎧ ⎛ 1 ⎞ ⎡ ( RE || Ri ) ⎤ ⎫ ⎪ ⎪ = R0 ⎨1 + ⎜ g m Ag rπ + ⎟ ⎢ ⎥⎬ R0 ⎠ ⎢1 + Ag ( RE || Ri ) ⎥ ⎪ ⎪ ⎣ ⎦⎭ ⎩ ⎝ g m rπ Ag = hFE Ag = 5 × 105 mS Let hFE = 100 so Ag = 5× 103 mS RE || Ri = 0.1 || 20 ≈ 0.1 kΩ Then ⎧ ⎤⎫ 1 ⎞⎡ 0.1 ⎪ ⎛ ⎪ R0 f = 50 ⎨1 + ⎜ 5 × 105 + ⎟ ⎢ ⎥⎬ 3 50 ⎠ ⎣1 + (5 × 10 )(0.1) ⎦ ⎪ ⎪ ⎝ ⎩ ⎭ or R0 f = 5.04 MΩ 12.30 Azf = Az 1 + β g Az = 5 = 0.2 V/μA 1 + (4.8)(5) Rif = Ri 1 = ⇒ Rif = 40 Ω 1 + β g Az 1 + (4.8)(5) Rof = Ro 1 = ⇒ Rof = 40 Ω 1 + β g Az 1 + (4.8)(5) 12.31 535. I ε = I i − I fb = 40 − 38 = 2 μA Az = Vo 8 V = =4 Iε 2 μA βg = Azf = I fb Vo = 38 μA = 4.75 8 V Vo 8 V = = 0.2 I i 40 μA 12.32 a. Assuming V1 is at virtual ground (−V0 ) = − I fb RF and (−V0 ) = − Az Iε ⇒ I ε = I fb = I S − I ε ⎛V ⎞ So V0 = ( I S − I ε ) RF = I S RF − ⎜ 0 ⎟ RF ⎝ Az ⎠ ⎡ R ⎤ V0 ⎢1 + F ⎥ = I S RF ⎣ Az ⎦ V RF AR so Azf = 0 = = z F I S ⎡ RF ⎤ Az + RF ⎢1 + A ⎥ z ⎦ ⎣ Az Az = or Azf = ⎛ 1 ⎞ 1 + Az β g 1 + Az ⎜ ⎟ ⎝ RF ⎠ 1 RF b. βg = c. 5 × 104 = 5 × 106 1 + (5 × 106 ) β g 5 × 106 −1 4 β g = 5 × 10 6 ⇒ β g = 1.98 × 10−5 5 × 10 1 RF = ⇒ RF = 50.5 kΩ βg d. Az = (0.9)(5 × 106 ) = 4.5 × 106 V0 Az 536. Azf = ΔAzf Azf 4.5 × 106 = 4.994 × 104 1 + (4.5 × 106 )(1.98 × 10−5 ) =− 55.4939 = −1.11× 10−3 ⇒ −0.111% 5 × 104 12.33 V1 = I ε Ri , − V0 = − Az I ε ⇒ V0 = Ax I ε I fb = I S − I ε and −V0 = V1 − I fb RF − Az I ε = V1 − ( I S − I ε ) RF ⎛V ⎞ ⎛V ⎞ − Az ⎜ 1 ⎟ = V1 − I S RF + ⎜ 1 ⎟ RF ⎝ Ri ⎠ ⎝ Ri ⎠ ⎡ A R ⎤ I S RF = V1 ⎢1 + z + F ⎥ ⎣ Ri Ri ⎦ V RF Rif = 1 = I S ⎡ Az RF ⎤ ⎢1 + R + R ⎥ i i ⎦ ⎣ Or, using the results from problem 12.18. 50.5 × 103 Rif = ⎡ 5 × 106 50.5 × 103 ⎤ ⎢1 + 10 × 103 + 10 × 103 ⎥ ⎣ ⎦ = 50.5 × 103 ⇒ Rif = 99.79 Ω [1 + 500 + 5.05] 12.34 Assume I CQ = 0.2 mA Then rπ = (100)(0.026) = 13 k Ω 0.2 537. (1) ⎛ 1 1 Vi − VA VA VA − Vo V V 1 ⎞ = + ⇒ i + o = VA ⎜ + + ⎟ Ri R1 R2 Ri R2 ⎝ Ri R1 R2 ⎠ Now Vi Vo ⎛1 1 1⎞ + = VA ⎜ + + ⎟ ⇒ Vi + Vo = VA (12) 10 10 ⎝ 10 1 10 ⎠ 1 or VA = (Vi + Vo ) 12 ⎛ AvVε − Vo ⎞ Vo − VA (2) ⎜ ⎟ (1 + hFE ) = R2 ⎝ Ro + rπ ⎠ where Vε = Vi − VA Then ⎛ Av (Vi − VA ) − Vo ⎞ Vo − VA ⎜ ⎟ (1 + hFE ) = R0 + rπ R2 ⎝ ⎠ we find AvVi (1 + hFE ) Vo (1 + hFE ) Vo AvVA (1 + hFE ) VA − − = − Ro + rπ Ro + rπ R2 Ro + rπ R2 Then (5 × 103 )(101)Vi Vo (101) Vo ⎛ (5 × 103 )(101) 1 ⎞ − − =⎜ − ⎟ VA 14 14 10 ⎝ 14 10 ⎠ Rearranging terms, we find V Avf = o =10.97 Vi ⎛ Vi ⎞ Vi Vi = =⎜ ⎟ Ri I i ⎛ Vi − VA ⎞ ⎝ Vi − VA ⎠ ⎜ ⎟ ⎝ Ri ⎠ 1 1 VA = (Vi + Vo ) = (Vi + 10.97Vi ) = 0.9975Vi 12 12 Then 1 ⎛ ⎞ Rif = ⎜ ⎟ (10 k Ω) ⇒ Rif = 4 M Ω ⎝ 1 − 0.9975 ⎠ To find the output resistance: Rif = 538. Ix + ( AvVε )(1 + hFE ) Vx = Ro + rπ R2 + R1 || Ri ⎛ R1 || Ri ⎞ Vε = − ⎜ ⎟ ⋅ Vx ⎝ R1 || Ri + R2 ⎠ Now R1 || Ri = 1 || 10 = 0.909 Then Vε = −0.0833Vx Now 3 ⎧ ⎫ 1 ⎪ ⎡ (5 × 10 )(0.0833) + 1 ⎤ ⎪ I x = Vx ⎨ ⎢ ⎬ ⎥ (101) + 1 + 13 10 + 0.909 ⎪ ⎪⎣ ⎦ ⎩ ⎭ = Vx {3.012 × 103 + 0.0917} Or Vx = Rof = 3.32 × 10−4 k Ω ⇒ Rof = 0.332 Ω Ix 12.35 a. Neglecting base currents I C 2 = 0.5 mA, VC 2 = 12 − (0.5)(22.6) = 0.7 V I C1 = 0.5 mA ⇒ v0 = 0 Then I C 3 = 2 mA b. rπ 1 = rπ 2 = hFE ⋅ VT (100)(0.026) = = 5.2 kΩ I C1 0.5 0.5 = 19.23 mA / V 0.026 (100)(0.026) rπ 3 = = 1.3 kΩ 2 2 g m3 = = 76.92 mA / V 0.026 g m1 = g m 2 = 539. Vπ 1 V + g m1Vπ 1 + g m1Vπ 2 + π 2 = 0 rπ 1 rπ 1 ⎛ 1 ⎞ (Vπ 1 + Vπ 2 ) ⎜ + g m1 ⎟ = 0 ⇒ Vπ 1 = −Vπ 2 ⎝ rπ 1 ⎠ Vπ 1 ( RS + rπ 1 ) − Vπ 2 + Vb 2 Vi = rπ 1 (1) ⎛ R ⎞ or Vi = Vπ 1 ⎜ 1 + S ⎟ − Vπ 2 + Vb 2 ⎝ rπ 1 ⎠ But Vπ 2 = −Vπ 1 so ⎛ R ⎞ Vi = Vπ 1 ⎜ 2 + S ⎟ + Vb 2 (2) rπ 1 ⎠ ⎝ V02 V −V + g m1Vπ 2 + 02 0 = 0 RC rπ 3 (3) Vπ 3 V V −V + g m 3Vπ 3 = 0 + 0 b 2 rπ 3 RL R2 Vπ 3 = V02 − V0 so ⎛ 1 + hFE (V02 − V0 ) ⎜ ⎝ rπ 3 ⎞ ⎛ 1 1 ⎞ V + ⎟ − b2 ⎟ = V0 ⎜ RL R2 ⎠ R2 ⎝ ⎠ (4) Vb 2 − V0 Vb 2 Vπ 2 + + = 0 (5) R2 R1 rπ 1 Substitute numbers into (2), (3), (4) and (5): 1 ⎞ ⎛ Vi = −Vπ 2 ⎜ 2 + ⎟ + Vb 2 5.2 ⎠ ⎝ Vi = −Vπ 2 (2.192) + Vb 2 (2) 1 ⎞ ⎛ 1 ⎛ 1 ⎞ + V02 ⎜ ⎟ + (19.23)Vπ 2 − V0 ⎜ ⎟ = 0 ⎝ 22.6 1.3 ⎠ ⎝ 1.3 ⎠ V02 (0.8135) + (19.23)Vπ 2 − (0.7692)V0 = 0 (3) ⎛ 101 ⎞ ⎛ 101 1 1 ⎞ ⎛ 1 ⎞ V02 ⎜ + + ⎟ − Vb 2 ⎜ ⎟ ⎟ = V0 ⎜ ⎝ 1.3 ⎠ ⎝ 1.3 4 50 ⎠ ⎝ 50 ⎠ V02 (77.69) = V0 (77.96) − Vb 2 (0.02) (4) ⎛ 1 1⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ Vb 2 ⎜ + ⎟ − V0 ⎜ ⎟ + Vπ 2 ⎜ ⎟=0 ⎝ 50 10 ⎠ ⎝ 50 ⎠ ⎝ 5.2 ⎠ Vb 2 (0.120) − V0 (0.020) + Vπ 2 (0.1923) = 0 (5) 540. From (2): Vb 2 = Vi + Vπ 2 (2.192). Substitute in (4) and (5) to obtain: V02 (77.69) = V0 (77.96) − [Vi + Vπ 2 (2.192)](0.02) (4′) [Vi + Vπ 2 (2.192)](0.120) − V0 (0.020) + Vπ 2 (0.1923) = 0 (5′) So we now have the following three equations: V02 (0.8135) + (19.23)Vπ 2 − (0.7692)V0 = 0 (3) V02 (77.69) = V0 (77.96) − Vi (0.02) − Vπ 2 (0.04384) (4′) (0.120)Vi + Vπ 2 (0.4553) − V0 (0.020) = 0 (5′) From (3): V02 = V0 (0.9455) − Vπ 2 (23.64). Substitute for V02 in (4′) to obtain: (77.69)[Vo (0.9455) − Vπ 2 (23.64)] = V0 (77.96) − Vi (0.02) − Vπ 2 (0.04384) or 0 = V0 (4.504) − Vi (0.02) + Vπ 2 (1836.5) Next, solve (5′) for Vπ 2 : (0.120)Vi + Vπ 2 (0.4553) − V0 (0.020) = 0 Vπ 2 = V0 (0.04393) − Vi (0.2636) Finally, 0 = V0 (4.504) − Vi (0.02) + (1836.5)[V0 (0.04393) − Vi (0.2636)] 0 = V0 (85.18) − Vi (484.12) So V 484.12 ⇒ Avf = 5.68 Avf = 0 = Vi 85.18 12.36 a. RTH = R1 || R2 = 400 || 75 = 63.2 kΩ ⎛ R2 ⎞ ⎛ 75 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (10) = 1.579 V R1 + R2 ⎠ ⎝ 75 + 400 ⎠ ⎝ 1.579 − 0.7 I BQ1 = = 0.007106 mA 63.2 + (121)(0.5) I CQ1 = 0.853 mA VC1 = 10 − (0.853)(8.8) = 2.49 V 2.49 − 0.7 = 0.497 mA 3.6 = 10 − (0.497)(13) = 3.54 V IC 2 ≈ VC 2 IC 3 ≈ Then 3.54 − 0.7 = 2.03 mA 1.4 541. (120)(0.026) = 3.66 kΩ 0.853 0.853 g m1 = = 32.81 mA / V 0.026 (120)(0.026) rπ 2 = = 6.28 kΩ 0.497 0.497 gm2 = = 19.12 mA / V 0.026 (120)(0.026) rπ 3 = = 1.54 kΩ 2.03 2.03 g m3 = = 78.08 mA / V 0.026 b. rπ 1 = Vi = Vπ 1 + Vε 1 ⇒ Vε 1 = Vi − Vπ 1 (1) Vπ 1 V V −V + g m1Vπ 1 = ε 1 + ε 1 0 rπ 1 RE1 RF (2) Vπ 2 = −( g m1Vπ 1 )( RC1 || rπ 2 ) (3) g m 2Vπ 2 + Vπ 3 + V0 Vπ 3 + =0 RC 2 rπ 3 (4) Vπ 3 V V −V + g m 3Vπ 3 = 0 + 0 ε 1 (5) rπ 3 RE 3 RF Substitute numbers in (2), (3), (4) and (5): 1⎞ V ⎛ 1 ⎞ ⎛ 1 Vπ 1 ⎜ + 32.81⎟ = (Vi − Vπ 1 ) ⎜ + ⎟− 0 ⎝ 3.66 ⎠ ⎝ 0.5 10 ⎠ 10 (2) or Vπ 1 (35.18) = Vi (2.10) − V0 (0.10) Vπ 2 = −(32.81)Vπ 1 (88 || 6.28) or Vπ 2 = −Vπ 1 (120.2) (3) (19.12)Vπ 2 + Vπ 3 V0 Vπ 3 + + =0 13 13 1.54 or Vπ 2 (19.12) + Vπ 3 (0.7263) + V0 (0.07692) = 0 (4) 1 ⎞ V −V ⎛ 1 ⎞ ⎛ 1 Vπ 3 ⎜ + 78.08 ⎟ = V0 ⎜ + ⎟ − i π1 10 ⎝ 1.54 ⎠ ⎝ 1.4 10 ⎠ or Vπ 3 (78.73) = V0 (0.8143) − Vi (0.10) + Vπ 1 (0.10) (5) Now substituting Vπ 2 = −Vπ 1 (120.2) in (4): (19.12)[−Vπ 1 (120.2)] + Vπ 3 (0.7263) + V0 (0.07692) = 0 542. or −Vπ 1 (2298.2) + Vπ 3 (0.7263) + V0 (0.07692) = 0 Then Vπ 3 = Vπ 1 (3164.3) − V0 (0.1059) Substituting Vπ 3 = Vπ 1 (3164.3) − V0 (0.1059) in (5): (78.73)[Vπ 1 (3164.3) − V0 (0.1059)] = V0 (0.8143) − Vi (0.10) + Vπ 1 (0.10) or Vπ 1 (2.49 × 105 ) − V0 (9.152) = −Vi (0.10) Then Vπ 1 = V0 (3.674 × 10−5 ) − Vi (4.014 × 10−7 ) Now substituting Vπ 1 = V0 (3.674 × 10−5 ) −Vi (4.014 × 10−7 ) in (2): (35.18)[V0 (3.674) ×10−5 ) − Vi (4.014 ×10 −7 )] = Vi (2.10) − V0 (0.10) or V0 (0.1013) = Vi (2.10) So V0 = 20.7 Vi Rif = c. I RB1 = I b1 = Vi and I i = I RB1 + I b1 Ii Vi RB1 Vπ 1 rπ 1 Now Vπ 1 = (20.7Vi )(3.674 × 10−5 ) − Vi (4.014 × 10−7 ) Vπ 1 = Vi (7.60 × 10−4 ) Then Vi Rif = Vi V (7.60 × 10−4 ) + i 63.2 3.66 1 = 0.01582 + 2.077 × 10−4 or Rif = 62.4 kΩ d. To determine R0 f : Equation (1) is modified to Vπ 1 + Ve1 = 0 (Vi = 0) Equation (5) is modified to: Vπ 3 (78.73) + I X = V0 (0.8143) + Vπ 1 (0.10) (5) Now Vπ 1 (35.18) = −V0 (0.10) (2) Vπ 2 = −Vπ 1 (120.2) (3) Vπ 2 (19.12) + Vπ 3 (0.7263) + V0 (0.07692) = 0 (4) Now Vπ 1 = −V0 (0.002843) so 543. Vπ 2 = −(−V0 )(0.002843)(120.2) Vπ 2 = V0 (0.3417) Then V0 (0.3417)(19.12) + Vπ 3 + (0.7263) + V0 (0.07692) = 0 or Vπ 3 = −V0 (9.101) (4) So then −V0 (9.101)(78.73) + I X = V0 (0.8143) + (0.10)(−V0 )(0.002843) or I X = V0 (717.3) (5) or V R0 f = 0 = 0.00139 kΩ ⇒ R0 f = 1.39 Ω IX 12.37 (a) (1) Vi − VA V −V V + g m1Vπ 1 = A + A O rπ 1 RE RF (2) VB V + g m1Vπ 1 + B = 0 RC1 rπ 2 (3) VC V − Vo + g m 2Vπ 2 + C =0 RC 2 rπ 3 (4) g m 3Vπ 3 + VC − VO VO − VA = rπ 3 RF Vπ 1 = Vi − VA Vπ 2 = VB Vπ 3 = VC − VO 544. ⎛ 1 ⎞ V V −V + g m1 ⎟ = A + A O rπ 1 RE RF ⎝ ⎠ (1) (Vi − VA ) ⎜ (2) ⎛ 1 1 ⎞ VB ⎜ + ⎟ + g m1 (Vi − VA ) = 0 RC1 rπ 2 ⎠ ⎝ (3) ⎛ 1 VO 1 ⎞ VC ⎜ + =0 ⎟ + g m 2VB − rπ 3 ⎝ RC 2 rπ 3 ⎠ (4) (VC − VO ) ⎜ g m3 + (1) (2) (3) (4) (Vi − VA )(555.5) = VA (20) + (VA − VO )(0.8333) VB (5.109) + 550(Vi − VA ) = 0 VC (3.257) + 178VB − VO (1.718) = 0 (VC − VO )(173.7) = (VO − VA )(0.8333) (1) (2) (3) (4) Vi (555.5) + VO (0.8333) = VA (576.3) VB (5.109) + 550Vi − VA (550) = 0 VC (3.257) + 178VB − VO (1.718) = 0 VC (173.7) + VA (0.8333) = VO (174.5) ⎛ 1 ⎞ VO − VA ⎟= rπ 3 ⎠ RF ⎝ (100)(0.026) 14.3 rπ 1 = = 0.182 K g m1 = = 550 mA/V 14.3 0.026 (100)(0.026) 4.62 rπ 2 = = 0.563 K g m 2 = = 178 mA/V 4.62 0.026 (100)(0.026) 4.47 rπ 3 = = 0.582 K g m 3 = = 172 mA/V 4.47 0.026 V −V 1 ⎞ V + 550 ⎟ = A + A O (1) (Vi − VA ) ⎛ ⎜ 1.2 ⎝ 0.182 ⎠ 0.05 1 1 ⎞ ⎛ (2) VB ⎜ + ⎟ + (550)(Vi − VA ) = 0 ⎝ 0.3 0.563 ⎠ VO 1 ⎞ ⎛ 1 VC ⎜ + =0 (3) ⎟ + 178 VB − 0.65 0.582 ⎠ 0.582 ⎝ 1 ⎞ VO − VA (4) (VC − VO ) ⎛172 + ⎜ ⎟= 0.582 ⎠ 1.2 ⎝ VB = VA (107.7) − Vi (107.7) From (2) From (4) VC = VO (1.0046) − VA (0.004797) Substitute into (3) (3.257) [VO (1.0046) − VA (0.004797) ] +(178)[VA (107.7) − Vi (107.7)] − Vo (1.718) = 0 VO (3.272) − VA (0.01562) + VA (19170.6) − Vi (19170.6) − VO (1.718) = 0 VA (19170.6) = Vi (19170.6) − VO (1.554) VA = Vi (1.00) − VO (0.00008106) Substitute into (1) 545. Vi (555.5) + Vo (0.8333) = (576.3) [Vi (1.00) − Vo (0.00008106) ] = Vi (576.3) − Vo (0.0467) Vo (0.880) = Vi (20.8) Vo = Avf = 23.6 Vi Ideal R + RE 1.2 + 0.05 = = 25.0 Avf = F 0.05 RE Rif = (b) Vi V V − VA and I i = π 1 = i rπ 1 rπ 1 Ii We have VA = Vi (1.00) − Vo (0.00008106) = Vi (1.00) − (23.6)Vi (0.00008106) VA = Vi (0.99809) Vi (1 − 0.99809) = Vi (0.01051) 0.182 Vi Rif = ⇒ Rif = 95.1 K Vi (0.01051) Then I i = To find Rof , set Vi = 0 I X + g m 3Vπ 3 + Vπ 3 Vx − VA = rπ 3 RF Vπ 3 = VC − VX I X + (VC − VX )( g m 3 + V − VA 1 )= X rπ 3 RF For Vi = 0, we have VC = VX (1.0046) − VA (0.004797) VA (576.3) = VX (0.8333) VA = VX (0.001446) VC = VX (1.0046) − VX (0.001446)(0.004797) VC = VX (1.0046) 1 ⎞ VX (1 − 0.004797) ⎛ I X + VX (1.0046 − 1.0) ⎜172 + ⎟= 0.582 ⎠ 1.2 ⎝ I X + VX (0.7991) = VX (0.8293) I X = VX (0.03024) Rof = VX = 33.1 K IX 12.38 RD1 = RE1 = 8 K VGG = VGS + ( I D + I C ) RL ID = 10 − VD 8 IC = 10 − (VD + 0.7) 8 546. VGG = 4.5 = 36 = ID + VTN + ( I D + I C ) RL kn (10 − VD ) ⎛ 10 − VD ⎞ ⎛ 9.3 − VD ⎞ +1+ ⎜ ⎟ (1.8) + ⎜ ⎟ (1.8) 8(0.4) 8 ⎠ ⎝ 8 ⎠ ⎝ 8 3.2 10 − VD + 8 + 18 − 1.8VD + 16.74 − 1.8VD 36 = 4.472 10 − VD + 42.74 − 3.6VD 3.6VD − 6.74 = 4.472 10 − VD 2 12.96VD − 48.528VD + 45.4276 = 19.999(10 − VD ) 2 12.96VD − 28.528VD − 154.6 = 0 28.528 ± 813.847 + 4(12.96)(154.6) 2(12.96) VD = 4.726 V VD = 10 − 4.726 = 0.6593 mA 8 9.3 − 4.726 IC = 8 = 0.5718 mA 0.6593 VGS = + 1 = 2.284 V 0.4 VO = 4.5 − 2.284 = 2.216 V ID = VDS = VD − VO = 4.726 − 2.216 = 2.51V (V ( sat ) = 1.28 V ) DS Also VO = ( I D + I C ) RL = (0.6593 + 0.5718)(1.8) = 2.216 V OK VECQ 2 = (4.726 + 0.7) − 2.216 = 3.21 V (b) 547. (1) Vi = Vgs + Vo (2) V VA + g m1Vgs = π RD1 rπ 2 (3) VB Vπ + + g m 2Vπ = 0 RE 2 rπ 2 Vπ = VB − VA Vgs = Vi − Vo (2) VA V − VA + g m1 (Vi − Vo ) = B RD1 rπ 2 (3) VB VB − VA + + g m 2 (VB − VA ) = 0 RE 2 rπ 2 (4) g m1Vgs + g m 2Vπ = VO RL g m1 (Vi − Vo ) + g m 2 (VB − VA ) = VO RL g m1 = 2 K n I D1 = 2 (0.4)(0.6593) = 1.027 mA/V gm2 = rπ 2 = I C 2 0.5718 = = 21.99 mA/V VT 0.026 (100)(0.026) = 4.547 K 0.5718 548. (2) (3) VA V − VA + (1.027)(Vi − Vo ) = B 8 4.547 VB VB − VA + + (21.99)(VB − VA ) = 0 8 4.547 0.125VA + 1.027Vi − 1.027Vo = 0.2199VB − 0.2199VA (2) (3) 0.125VB + 0.2199VB − 0.2199VA + 21.99VB − 21.99VA = 0 (2) 0.3449VA − 0.2199VB = −1.027Vi + 1.027Vo (3) 22.21VA = 22.335VB VA = 1.0056VB Then (0.3449)(1.0056)VB − 0.2199VB = −1.027Vi + 1.027Vo 0.1269VB = −1.027Vi + 1.027Vo VB = −8.093Vi + 8.093Vo VA = −8.138Vi + 8.138Vo Then (4) (1.027)(Vi − Vo ) + (21.99) [ −8.093Vi + 8.093Vo + 8.138Vi − 8.138Vo ] = Vo 1.8 1.027Vi − 1.027Vo − 177.965Vi + 177.965Vo + 178.955Vi − 178.955Vo = 0.5556Vo 2.017Vi − 2.5726Vo = 0 Vo = 0.784 = AV Vi Set Vi = 0 I X + g m 2Vπ + g m1Vgs = VX RL I X + g m 2 (VB − VA ) + g m1 (−VX ) = VX RL VB = 8.093VX VA = 8.138VX I X + (21.99) [8.093VX − 8.138VX ] + 1.027( −VX ) = VX (0.5556) I X = 2.572VX VX = Rof = 0.389 K IX 12.39 a. g m = 2 K n I DQ = 2 (0.5)(0.5) = 1 mA / V 549. V0 = ( g mVgs ) RS Vi = Vgs + V0 so Vgs = Vi − V0 Then V0 = g m RS (Vi − V0 ) Av = g m RS 1(2) = ⇒ Av = 0.667 1 + g m RS 1 + (1)(2) To determine R0 f I X + g mVgs = VX and Vgs = −VX RS IX 1 1 = = gm + VX R0 f RS so R0 f = 1 1 RS = 2 ⇒ R0 f = 0.667 kΩ 1 gm For K n = 0.8 mA / V 2 b. g m = 2 (0.8)(0.5) = 1.265 mA / V Av = (1.265)(2) = 0.7167 1 + (1.265)(2) ΔAf = Af 0.7167 − 0.667 ⇒ 7.45% increase 0.667 1 || 2 = 0.7905 || 2 1.265 = 0.5666 kΩ R0 f = R0 f ΔR0 f R0 f = 0.5666 − 0.667 ⇒ 15.05% decrease 0.667 12.40 dc analysis: RTH 1 = 150 || 47 = 35.8 kΩ , ⎛ 47 ⎞ VTH 1 = ⎜ ⎟ (25) = 5.96 V ⎝ 47 + 150 ⎠ RTH 2 = 33 || 47 = 19.4 kΩ , ⎛ 33 ⎞ VTH 2 = ⎜ ⎟ (25) = 10.3 V ⎝ 33 + 47 ⎠ 5.96 − 0.7 I B1 = = 0.0187 mA 35.8 + (51)(4.8) I C1 = (50)(0.0187) = 0.935 mA 10.3 − 0.7 = 0.03705 mA 19.4 + (51)(4.7) = (50)(0.03705) = 1.85 mA I B2 = IC 2 (50)(0.026) = 1.39 kΩ; 0.935 (50)(0.026) = 0.703 kΩ rπ 2 = 1.85 rπ 1 = 550. 0.935 = 35.96 mA / V 0.026 1.85 = = 71.15 mA / V 0.026 g m1 = gm2 VS = Vπ 1 + Ve (1) Vπ 1 V V − V0 + g m1Vπ 1 = e + e rπ 1 R1 RF (2) g m1Vπ 1 + Vπ 2 Vπ 2 Vπ 2 + + =0 RC1 RB 2 rπ 2 (3) V0 V0 − Ve (4) + =0 RC 2 RF Substitute numerical values in (2), (3) and (4): Ve = VS − Vπ 1 (1) g m 2Vπ 2 + Vπ 1 1 ⎞ ⎛ 1 ⎛ 1 ⎞ + (35.96)Vπ 1 = (VS − Vπ 1 ) ⎜ + ⎟ − V0 ⎜ ⎟ 1.39 ⎝ 0.1 4.7 ⎠ ⎝ 4.7 ⎠ or Vπ 1 (46.89) = VS (10.213) − V0 (0.2128) (2) 1 1 ⎞ ⎛1 + (35.96)Vπ 1 + Vπ 2 ⎜ + ⎟=0 ⎝ 10 19.4 0.703 ⎠ or (35.96)Vπ 1 + Vπ 2 (1.574) = 0 (3) 1 ⎞ ⎛ 1 ⎛ 1 ⎞ + (71.15)Vπ 2 + V0 ⎜ ⎟ − (VS − Vπ 1 ) ⎜ ⎟=0 ⎝ 4.7 4.7 ⎠ ⎝ 4.7 ⎠ or (71.15)Vπ 2 + V0 (0.4255) − VS (0.2128) + Vπ 1 (0.2128) = 0 (4) From (3): Vπ 2 = −Vπ 1 (22.85) Then substitute in (4): −(71.15)Vπ 1 (22.85) + V0 (0.4255) − VS (0.2128) + Vπ 1 (0.2128) = 0 or −Vπ 1 (1625.6) + V0 (0.4255) − VS (0.2128) = 0 From (2): Vπ 1 = VS (0.2178) − V0 (0.004538) Then −(1625.6)[VS (0.2178) − V0 (0.004538)] + V0 (0.4255) − VS (0.2128) = 0 or −VS (354.3) + V0 (7.802) = 0 Finally V ⇒ 0 = 45.4 VS 551. 12.41 For example, use a z-stage amplifier. Each stage shown in Fig. 12.29. 12.42 For M 3 : K n 3 = ′ kn 2 ⎛W ⎞ ⎛W ⎞ ⋅ ⎜ ⎟ Let ⎜ ⎟ = 25 L ⎠3 ⎝ ⎝ L ⎠3 ⎛ 0.080 ⎞ 2 Then K n 3 = ⎜ ⎟ (25) = 1 mA / V ⎝ 2 ⎠ Want vo = 0 for vi = 0, so that I D 3 = I Q 2 = 0.4 = 1 ⋅ (VGS 3 − VTN ) 2 0.4 + 2 = 2.63 V 1 For VGS 3 = 2.63V ⇒ VG 3 = 12 − I D 2 RD Or 2.63 = 12 − (0.1) RD ⇒ RD = 93.7 k Ω Then VGS 3 = g m 3 = 2 K n 3 I D 3 = 2 (1)(0.4) = 1.26 mA / V ⎛ R1 ⎞ ⎛ 10 ⎞ VA = ⎜ ⎟ (Vo ) = ⎜ ⎟ (Vo ) = 0.0769Vo R1 + R2 ⎠ ⎝ 120 + 10 ⎠ ⎝ (Small amount of feedback) (1) Vi = Vgs1 − Vgs 2 + VA 552. g mVgs1 + g mVgs 2 = 0 ⇒ Vgs1 = −Vgs 2 (2) Then Vi = −2Vgs 2 + VA ⇒ Vgs 2 = 1 (VA − Vi ) 2 Vgs 2 = 0.0384 Vo − 0.5Vi (3) VB = − g mVgs 2 RD = − g m RD [0.03846Vo − 0.5Vi ] (4) Vgs 3 = VB − Vo and Vo = g m 3Vgs 3 [ RL || ( R1 + R2 )] So Vo = g m 3 [ RL || ( R1 + R2 ) ] (VB − Vo ) Then Vo = g m 3 ⎡ RL || ( R1 + R2 ) ⎤ ⎡ − g m RD ( 0.03846Vo − 0.5Vi ) − Vo ⎤ ⎣ ⎦⎣ ⎦ Or Vo ⎡1 + g m 3 [ RL || ( R1 + R2 )][ g m RD (0.03846) + 1]⎤ = g m 3 [ RL || ( R1 + R2 ) ][ 0.5 g m RD ] ⋅ Vi ⎣ ⎦ Now RL || ( R1 + R2 ) = 4 ||130 = 3.88 So Vo ⎡1 + (1.26)(3.88) [ g m (93.7)(0.03846) + 1]⎤ = (1.26)(3.88)(0.5) g m (93.7)Vi ⎣ ⎦ Rearranging terms, we find Vo 229 g m = = 10 ⇒ g m = 1.11 mA / V Vi 5.89 + 17.6 g m We have ⎛ k ′ ⎞⎛ W gm = 2 ⎜ n ⎟ ⎜ ⎝ 2 ⎠⎝ L ⎞ ⎛ 0.080 ⎞ ⎛ W ⎟ I D ⇒ 1.11 = 2 ⎜ ⎟⎜ ⎠ ⎝ 2 ⎠⎝ L ⎞ ⎛W ⎞ ⎛W ⎞ ⎟ (0.1) ⇒ ⎜ ⎟ = ⎜ ⎟ = 77 ⎠ ⎝ L ⎠1 ⎝ L ⎠ 2 12.43 Assuming an ideal op-amp, then from Equation (12.58) I0 R 20 = 1+ 1 = = 100 IS R2 0.2 Then R1 = 99 R2 For example, set R2 = 5 kΩ and R1 = 495 kΩ 12.44 ⎛ h ⎞ ⎛ 100 ⎞ I C1 = ⎜ FE ⎟ I E1 = ⎜ ⎟ (0.2) = 0.198 mA ⎝ 101 ⎠ ⎝ 1 + hFE ⎠ VC1 = 10 − (0.198)(40) = 2.08 V (a) 2.08 − 0.7 = 1.38 mA 1 ⎛ 100 ⎞ =⎜ ⎟ (1.38) = 1.37 mA ⎝ 101 ⎠ IE2 = IC 2 For Q1 : (100)(0.026) = 13.1 k Ω 0.198 0.198 = = 7.62 mA / V 0.026 rπ 1 = g m1 553. For Q2 : (100)(0.026) = 1.90 k Ω 1.37 1.37 = = 52.7 mA / V 0.026 rπ 2 = gm2 (b) IS = Vπ 1 Vπ 1 Vπ 1 − Ve + + RS rπ 1 RF g m1Vπ 1 + Vπ 2 + Ve Vπ 2 + =0 RC1 rπ 2 (1) (2) Vπ 2 V V −V (3) + g m 2Vπ 2 = e + e π 1 rπ 2 RE RF Substitute numerical values in (1), (2), and (3): 1 1⎞ ⎛1 ⎛1⎞ + ⎟ − Ve ⎜ ⎟ I S = Vπ 1 ⎜ + 10 13.1 10 ⎠ ⎝ ⎝ 10 ⎠ I S = Vπ 1 (0.2763) − Ve (0.10) (1) 1 ⎞ ⎛ 1 ⎛ 1 ⎞ (7.62)Vπ 1 + Vπ 2 ⎜ + ⎟ + Ve ⎜ ⎟ = 0 ⎝ 40 1.90 ⎠ ⎝ 40 ⎠ (7.62)Vπ 1 + Vπ 2 (0.5513) + Ve (0.025) = 0 (2) ⎛ 1 ⎞ ⎛1 1 ⎞ ⎛ 1⎞ + 52.7 ⎟ = Ve ⎜ + ⎟ − Vπ 1 ⎜ ⎟ Vπ 2 ⎜ ⎝ 1.90 ⎠ ⎝ 1 10 ⎠ ⎝ 10 ⎠ (3) Vπ 2 (53.23) = Ve (1.10) − Vπ 1 (0.10) From (3), Ve = Vπ 2 (48.39) + Vπ 1 (0.0909) Substituting into (1), I S = Vπ 1 (0.2763) − (0.10) [Vπ 2 (48.39) + Vπ 1 (0.0909)] or I S = Vπ 1 (0.2672) − Vπ 2 (4.839) (1′) and substituting into (2), (7.62)Vπ 1 + Vπ 2 (0.5513) + ( 0.025 ) ⎡Vπ 2 ( 48.39 ) + Vπ 1 ( 0.0909 ) ⎤ = 0 ⎣ ⎦ or (7.622)Vπ 1 + Vπ 2 (1.761) = 0 ⇒ Vπ 1 = −Vπ 2 (0.2310) Then substituting (2′) into (1′), we obtain I S = (0.2672)(−Vπ 2 )(0.2310) − Vπ 2 (4.839) or I S = −Vπ 2 (4.901) Now (2′) 554. ⎛ RC 2 ⎞ I O = − g m 2Vπ 2 ⎜ ⎟ ⎝ RC 2 + RL ⎠ ⎛ 2 ⎞ = −(52.7) ⎜ ⎟ Vπ 2 = −(42.16)Vπ 2 ⎝ 2 + 0.5 ⎠ Then ⎛ −IS ⎞ I O = −(42.16) ⎜ ⎟ ⎝ 4.901 ⎠ or I Aif = O = 8.60 Is (c) Ri = Vπ 1 and Ri = RS || Rif IS We had Vπ 1 = −Vπ 2 (0.2310) and I S = −Vπ 2 (4.901) so ⎛ −Vπ 1 ⎞ IS = − ⎜ ⎟ (4.901) = Vπ 1 (21.22) ⎝ 0.2310 ⎠ Then V 1 = 0.04713 Ri = π 1 = IS 21.22 Finally 10 Rif 0.04713 = ⇒ Rif = 47.4 Ω 10 + Rif 12.45 (a) Vπ 1 V −V + π 1 e2 RF RS RB1 rπ 1 (1) Ii = (2) g m1Vπ 1 + (3) Vπ 2 V V −V + g m 2Vπ 2 = e 2 + e 2 π 1 rπ 2 RE 2 RF (4) ⎛ RC 2 ⎞ I o = −( g m 2Vπ 2 ) ⎜ ⎟ ⎝ RC 2 + RL ⎠ VC1 V + π2 = 0 RC1 RB 2 rπ 2 Now (1)′ Ii = Vπ 1 V − e2 RS RB1 rπ 1 RF RF ⎛ ⎞ RF ⎟V − I R So Ve 2 = ⎜ ⎜ RS RB1 rπ 1 RF ⎟ π 1 i F ⎝ ⎠ Now, from (2) V + Ve 2 Vπ 2 g m1Vπ 1 + π 2 + =0 Rc1 RB 2 rπ 2 (2)′ ⎛ Vπ 2 Ve 2 1 ⎞ + =0 ⎜ g m1 + ⎟ Vπ 1 + rπ 2 ⎠ RC1 RB 2 rπ 2 RC1 RB 2 ⎝ 555. Also (3)′ ⎛ ⎛ 1 Vπ 1 1 ⎞ 1 ⎞ = Ve 2 ⎜ + ⎜ gm2 + ⎟ Vπ 2 + ⎟ rπ 2 ⎠ RF ⎝ RE 2 RF ⎠ ⎝ And ⎛ I ⎞ ⎛ R + RL ⎞ Vπ 2 = − ⎜ O ⎟ ⎜ C 2 ⎟ ⎝ g m 2 ⎠ ⎝ RC 2 ⎠ Substitute (1)′ into (2)′ and (3)′ (4)′ ⎡ ⎤ RF ⎢ − Vπ 1 − I i RF ⎥ = 0 ⎢ RS RB1 rπ 1 RF ⎥ ⎣ ⎦ ⎡ ⎤ ⎛ Vπ 1 ⎛ 1 RF 1 ⎞ 1 ⎞ =⎜ + − Vπ 1 − I i RF ⎥ (3)″ ⎜ gm2 + ⎟ Vπ 2 + ⎟⎢ rπ 2 ⎠ RF ⎝ RE 2 RF ⎠ ⎢ RS RB1 rπ 1 RF ⎥ ⎝ ⎣ ⎦ Solve for Vπ 1 from (2)″ and substitute into (3)″. Also use Equation (4)′. (2)″ ⎛ Vπ 2 1 ⎞ 1 + ⎜ g m1 + ⎟ Vπ 1 + rπ 2 ⎠ RC1 RB 2 rπ 2 RC1 RB 2 ⎝ (b) RB1 = R1 R2 = 20 80 = 16 K ⎛ 20 ⎞ VTH 1 = ⎜ ⎟ (10) = 2 V ⎝ 100 ⎠ 2 − 0.7 I BQ1 = = 0.0111 mA 16 + (101)(1) I CQ1 = 1.11 mA RTH 2 = 15 85 = 12.75 K ⎛ 15 ⎞ VTH 2 = ⎜ ⎟ (10) = 1.5 V ⎝ 100 ⎠ 1.5 − 0.7 I BQ 2 = = 0.01265 mA 12.75 + (101)(0.5) I CQ 2 = 1.265 mA 1.11 = 42.69 mA/V 0.026 1.265 gm2 = = 48.65 mA/V 0.026 (100)(0.026) rπ 1 = = 2.34 K 1.11 (100)(0.026) rπ 2 = = 2.06 K 1.265 Now RC1 RB 2 = 2 12.75 = 1.729 K g m1 = RS RB1 rπ 1 RF ≅ RB1 rπ 1 RF = 16 2.34 10 = 1.695 K RC1 RB 2 rπ 2 = 1.729 2.06 = 0.940 K Now (2)″ (3)″ Vπ 2 1 ⎞ 1 ⎡ 10 ⎛ ⎤ + ⋅ Vπ 1 − I i (10) ⎥ ⎜ 42.69 + ⎟ Vπ 1 + ⎢ 2.06 ⎠ 0.940 1.729 ⎣1.695 ⎝ ⎦ 46.587Vπ 1 + 1.064Vπ 2 − 5.784 I i = 0 Vπ 1 ⎛ 1 1 ⎞ ⎡ 10 ⎛ 101 ⎞ ⎤ =⎜ + ⎟⎢ ⋅ Vπ 1 − I i (10) ⎥ ⎜ ⎟ Vπ 2 + 2.06 ⎠ 10 ⎝ 0.5 10 ⎠ ⎣1.695 ⎝ ⎦ 49.03Vπ 2 = 12.29Vπ 1 − 21I i 556. From (2)″ Vπ 1 = (0.1242) I i − (0.02284)Vπ 2 Then 49.03Vπ 2 = 12.29 [ (0.1242) I i − (0.02284)Vπ 2 ] − 21I i (3)″ 49.31Vπ 2 = −19.47 I i From (4)′ ⎛ I ⎞⎛ 4 + 4 ⎞ Vπ 2 = − ⎜ o ⎟ ⎜ ⎟ = −(0.0411) I o ⎝ 48.65 ⎠ ⎝ 4 ⎠ Then (49.31) [ −(0.0411) I o ] = −19.47 I i Io = Ai = 9.61 Ii 12.46 a. RTH = 13.5 || 38.3 = 9.98 kΩ ⎛ 13.5 ⎞ VTH = ⎜ ⎟ (10) = 2.606 V ⎝ 13.5 + 38.3 ⎠ (120)(2.606 − 0.7) I C1 = = 1.75 mA 9.98 + (121)(1) VC1 = 10 − (1.75 )( 3) = 4.75 V 4.75 − 0.7 = 0.50 mA 8.1 (120)(0.026) rπ 1 = = 1.78 kΩ 1.75 1.75 g m1 = = 67.31 mA / V 0.026 (120)(0.026) rπ 2 = = 6.24 kΩ 0.50 0.50 gm2 = = 19.23 mA / V 0.026 b. IC 2 ≈ VS − Vπ 1 Vπ 1 V −V = + π 1 e2 RS RB || rπ 1 RF g m1Vπ 1 + Vπ 2 + Ve 2 Vπ 2 + =0 RC1 rπ 2 Vπ 2 V V −V + g m 2Vπ 2 = ε 2 + ε 2 π 1 rπ 2 RE 2 RF and (1) (2) (3) 557. V0 = −( g m 2Vπ 2 ) RC 2 (4) Substitute numerical values in (1), (2), and (3) VS Vπ 1 ⎡ 1 1 ⎤ Ve 2 = Vπ 1 ⎢ + + ⎥− 0.6 ⎣ 0.6 9.98 ||1.78 1.2 ⎦ 1.2 VS (1.67) = Vπ 1 (4.011) − Ve 2 (0.8333) (1) 1 ⎞ Ve 2 ⎛1 (67.31)Vπ 1 + Vπ 2 ⎜ + =0 ⎟+ ⎝ 3 6.24 ⎠ 3 or Vπ 1 (67.31) + Vπ 2 (0.4936) + Ve 2 (0.3333) = 0 (2) V V ⎛ 1 ⎞ V Vπ 1 ⎜ + 19.23 ⎟ = e 2 + e 2 − π 2 ⎝ 6.24 ⎠ 8.1 1.2 1.2 or Vπ 2 (19.39) = Ve 2 (0.9568) − Vπ 1 (0.8333) (3) From (1) Ve 2 = Vπ 1 (4.813) − VS (2.00) Then Vπ 1 (67.31) + Vπ 2 (0.4936) + (0.3333)[Vπ 1 (4.813) − VS (2.00)] = 0 or Vπ 1 (68.91) + Vπ 2 (0.4936) − VS (0.6666) = 0 (2′) and Vπ 2 (19.39) = (0.9568) [Vπ 1 (4.813) − VS (2.00) ] − Vπ 1 (0.8333) or Vπ 2 (19.39) = Vπ 1 (3.772) − VS (1.914) (3′) We find Vπ 1 = VS (0.009673) − Vπ 2 (0.007163) Then Vπ 2 (19.39) = (3.772)[VS (0.009673) − Vπ 2 (0.007163)] − VS (1.914) Vπ 2 (19.42) = VS ( −1.878) or Vπ 2 = −VS (0.09670) so that V0 = −(19.23)(4)(−VS )(0.09670) Then V0 = 7.44 VS 12.47 Using the circuit from Problem 12.32, we have Rif = VS − Vπ 1 RS From Problem 12.32 Vπ 1 = VS (0.009673) − Vπ 2 (0.007163) Where I S = = VS (0.009673) − (0.007163)(−VS )(0.09670) = VS (0.01037) So Rif = VS (0.01037) ⋅ (0.6) = 0.00629 kΩ VS − VS (0.01037) Vπ 1 . IS 558. or Rif = 6.29 Ω 12.48 RTH = 1.4 ||17.9 = 1.298 kΩ ⎛ 1.4 ⎞ VTH = ⎜ ⎟ (10) = 0.7254 V ⎝ 1.4 + 17.9 ⎠ 0.7254 − 0.7 = 0.0196 mA I B1 = 1.298 I C1 = (50)(0.0196) = 0.98 mA Neglecting dc base currents, VB 2 = 10 − (0.98)(7) = 3.14 V 3.14 − 0.7 = 3.25 mA 0.25 + 0.5 ⎛ 50 ⎞ I C 2 = ⎜ ⎟ (3.25) = 3.19 mA ⎝ 51 ⎠ (50)(0.026) rπ 1 = = 1.33 kΩ 0.98 0.98 g m1 = = 37.7 mA / V 0.026 (50)(0.026) rπ 2 = = 0.408 kΩ 3.19 3.19 gm2 = = 123 mA / V 0.026 IE2 = IS = Vπ 1 V −V + π1 1 R1 || R2 || rπ 1 RF g m1Vπ 1 + Vπ 2 Vπ 2 + Ve 2 + =0 rπ 2 RC1 Vπ 2 V −V + g m 2Vπ 2 = e 2 1 rπ 2 RE1 (1) (2) (3) Ve 2 − Vπ 1 V −V V (4) = 1 + 1 π1 RE1 RE 2 RF Enter numerical values in (1), (2), (3) and (4): Vπ 1 V −V IS = + π1 1 17.9 ||1.4 ||1.33 5 or 559. I S = Vπ 1 (1.722) − V1 (0.20) (1) (37.7)Vπ 1 + Vπ 2 V + Ve 2 + π2 =0 0.408 7 or Vπ 1 (37.7) + Vπ 2 (2.594) + Ve 2 (0.1429) = 0 Vπ 2 V −V + (123)Vπ 2 = e 2 1 0.408 0.25 or Vπ 2 (125.5) = Ve 2 (4) − V1 (4) (2) (3) Ve 2 − V1 V −V V = 1 + 1 π1 0.25 0.50 5 or Ve 2 (4) = V1 (6.20) − Vπ 1 (0.20) (4) From (4): Ve 2 = V1 (1.55) − Vπ 1 (0.05) Then substituting in (3): Vπ 2 (125.5) = (4)[V1 (1.55) − Vπ 1 (0.05)] − V1 (4) or Vπ 2 (125.5) = V1 (2.20) − Vπ 1 (0.20) (3′) and substituting in (2): Vπ 1 ( 37.7 ) + Vπ 2 ( 2.594 ) + ( 0.1429 ) ⎡V1 (1.55 ) − Vπ 1 ( 0.05 ) ⎤ = 0 ⎣ ⎦ or Vπ 1 (37.69) + Vπ 2 (2.594) + V1 (0.2215) = 0 Now V1 = −Vπ 1 (170.16) − Vπ 2 (11.71) Then substituting in (1): I S = Vπ 1 (1.722) − (0.20)[ −Vπ 1 (170.16) − Vπ 2 (11.71)] or I S = Vπ 1 (35.75) + Vπ 2 (2.342) and substituting in (3′): Vπ 2 (125.5) = (2.20)[ −Vπ 1 (170.16) − Vπ 2 (11.71)] − Vπ 1 (0.20) or Vπ 2 (151.3) = −Vπ 1 (374.55) so that Vπ 1 = −Vπ 2 (0.4040) Then I S = (35.75)[−Vπ 2 (0.4040)] + Vπ 2 (2.342) I S = −Vπ 2 (12.10) ⎛ RC 2 ⎞ I 0 = −( g m 2Vπ 2 ) ⎜ ⎟ ⎝ RC 2 + RL ⎠ ⎛ 2.2 ⎞ = −(123) ⎜ ⎟ Vπ 2 = −(64.43)Vπ 2 ⎝ 2.2 + 2 ⎠ or Vπ 2 = −(0.01552) I 0 Then I0 I 1 = ⇒ 0 = 5.33 I S (0.01552)(12.10) IS 12.49 560. For example, use the circuit shown in Figure P12.30 12.50 rπ 1 = 6.24 kΩ, rπ 2 = 3.12 kΩ, rπ 3 = 1.56 kΩ g m1 = 19.23mA / V , g m 2 = 38.46 mA / V, g m 3 = 76.92 mA / V VS = Vπ 1 + Ve1 (1) Vπ 1 V V −V + g m1Vπ 1 = e1 + e1 e 3 rπ 1 RE1 RF Vπ 2 = − g m1Vπ 1 ( RC1 || rπ 2 ) g m 2Vπ 2 + (2) (3) Vπ 3 + Ve 3 Vπ 3 + =0 RC 2 rπ 3 (4) Vπ 3 V V −V (5) + g m 3Vπ 3 = e 3 + e 3 e1 rπ 3 RE 2 RF Enter numerical values in (2)-(5): Vπ 1 1 ⎞ ⎛ 1 ⎛ 1 ⎞ + (19.23)Vπ 1 = Ve1 ⎜ + ⎟ − Ve 3 ⎜ ⎟ 6.24 0.1 0.8 ⎠ ⎝ ⎝ 0.8 ⎠ or Vπ 1 (19.39) = Ve1 (11.25) − Ve 3 (1.25) (2) Vπ 2 = −(19.23)Vπ 1 (5 || 3.12) = −(36.94)Vπ 1 (3) 1 ⎞ ⎛1 ⎛1⎞ (38.46)Vπ 2 + Vπ 3 ⎜ + ⎟ + Ve 3 ⎜ ⎟ = 0 ⎝ 2 1.56 ⎠ ⎝ 2⎠ or Vπ 2 (38.46) + Vπ 3 (1.141) + Ve 3 (0.5) = 0 (4) 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎛ 1 ⎞ Vπ 3 ⎜ + 76.92 ⎟ = Ve 3 ⎜ + ⎟ − Ve 3 ⎜ ⎟ ⎝ 1.56 ⎠ ⎝ 0.1 0.8 ⎠ ⎝ 0.8 ⎠ or Vπ 3 (77.56) = Ve 3 (11.25) − Ve1 (1.25) (5) From (1) Vπ 1 = VS − Ve1 Then (VS − Ve1 )(19.39) = Ve1 (11.25) − Ve3 (1.25) or VS (19.39) = Ve1 (30.64) − Ve 3 (1.25) (2′) Vπ 2 = −VS (36.94) + Ve1 (36.94) (3′) (38.46)[−VS (36.94) + Ve1 (36.94)] + Vπ 3 (1.141) + Ve 3 (0.5) = 0 From (5): Ve 3 = Vπ 3 (6.894) + Ve1 (0.1111) Then VS (19.39) = Ve1 (30.64) − (1.25)[Vπ 3 (6.894) + Ve1 (0.1111)] (4′) 561. or VS (19.39) = Ve1 (30.50) − Vπ 3 (8.6175) (2″) and −VS (1420.7) + Ve1 (1420.7) + Vπ 3 (1.141) + (0.5)[Vπ 3 (6.894) + Ve1 (0.1111)] = 0 or −VS (1420.7) + Ve1 (1420.76) + Vπ 3 (4.588) = 0 (4″) From (2″): Ve1 = VS (0.6357) + Vπ 3 (0.2825) Then substituting in (4″): −VS (1420.7) + (1420.76)[VS (0.6357) + Vπ 3 (0.2825)] + Vπ 3 (4.588) = 0 −VS (517.5) + Vπ 3 (405.95) = 0 Now I 0 = g m 3Vπ 3 = 76.92Vπ 3 or Vπ 3 = I 0 (0.0130) Then −VS (517.5) + I 0 (0.0130)(405.95) = 0 or I0 = 98.06 mA / V VS 12.52 (100)(0.026) = 5.2 kΩ 0.5 0.5 g m1 = g m 2 = = 19.23 mA / V 0.026 (100)(0.026) rπ 3 = = 1.3 kΩ 2 2 g m3 = = 76.92 mA / V 0.026 rπ 1 = rπ 2 = Vπ 1 V + g m1Vπ 1 + g m 2Vπ 2 + π 2 = 0 rπ 1 rπ 2 (1) Since rπ 1 = rπ 2 and g m1 = g m 2 , then Vπ 1 = −Vπ 2 VS = Vπ 1 − Vπ 2 + Ve 3 = −2Vπ 2 + Ve 3 (2) g m 2Vπ 2 + Vπ 3 Vπ 3 + Ve 3 + =0 rπ 3 RC 2 Vπ 3 V V + g m 3Vπ 3 = e3 + π 2 rπ 3 RF rπ 2 (4) (3) 562. ⎛ RC 3 ⎞ I0 = − ⎜ ⎟ ( g m 3Vπ 3 ) ⎝ RC 3 + RL ⎠ From (2): Ve 3 = VS + 2Vπ 2 (19.23) Vπ 2 + (5) Vπ 3 Vπ 3 1 + + (VS + 2Vπ 2 ) = 0 1.3 18.6 18.6 or (19.23)Vπ 2 + (0.8230)Vπ 3 + (0.05376)VS = 0 (3′) V ⎛ 1 ⎞ ⎛1⎞ + 76.92 ⎟ = ⎜ ⎟ (VS + 2Vπ 2 ) + π 2 Vπ 3 ⎜ 1.3 10 ⎠ 5.2 ⎝ ⎠ ⎝ or (77.69)Vπ 3 = (0.3923)Vπ 2 + (0.1)VS (4′) ⎛ 2 ⎞ I0 = − ⎜ (5′) ⎟ (76.92)Vπ 3 = −(51.28)Vπ 3 ⎝ 2 +1⎠ From (3′): Vπ 2 = −(0.04255)Vπ 3 − (0.002780)VS Then (77.69)Vπ 3 = (0.3923)[ −(0.04255)Vπ 3 − (0.002780)VS ] + (0.1)VS (77.71)Vπ 3 = (0.0989)VS or Vπ 3 = (0.001273)VS so that I 0 = −(51.28)(0.001273)VS or I0 = −(0.0653) mA/V VS 12.52 A= 5 × 103 2 f ⎞⎛ f ⎞ ⎛ ⎜1 + j 4 ⎟ ⎜1 + j 5 ⎟ 10 ⎠ ⎝ 10 ⎠ ⎝ ⎛ f ⎞ ⎛ f ⎞ Phase = φ = − tan −1 ⎜ 4 ⎟ − 2 tan −1 ⎜ 5 ⎟ ⎝ 10 ⎠ ⎝ 10 ⎠ By trial and error, when f = 1.095 × 105 Hz , φ ≅ 180° For T = 1 at f = 1.095 × 105 Hz, 1= β ( 5 × 103 ) ⎛ f ⎞ 1+ ⎜ 4 ⎟ ⎝ 10 ⎠ 2 ⎡ ⎛ f ⎞ ⋅ ⎢1 + ⎜ 5 ⎟ ⎢ ⎝ 10 ⎠ ⎣ 2 ⎤ ⎥ ⎥ ⎦ ⇒1= β ( 5 × 103 ) (10.996 )( 2.199 ) 12.53 Use the basic circuit shown in Figure 12.27. ⇒ β = 4.84 × 10−3 563. For the ideal case I0 1 = Vi RE we want I0 = 10−3 A/V = 1 mA/V Vi Set RE = 1 kΩ Since the op-amp has a finite gain, finite input resistance, and finite output resistance, the closed-loop gain is slightly less than the ideal. RE will need to be slightly decreased to increase the gain. 12.54 dc analysis I E RE + VEB (on) + I B RB + VCC = 0 5 − 0.7 = 0.0343 100 + (51)(0.5) I C = (50)(0.0343) = 1.71 mA IB = (50)(0.026) = 0.760 kΩ 1.71 1.71 gm = = 65.77 mA/V 0.026 a. Then rπ = To determine Rif : IS + Vπ V − (−Vπ ) + 0 =0 RB || rπ RF (1) 564. V0 V0 − (−Vπ ) (2) + RC RF Now from (2): V ⎛1 1 ⎞ (65.77)Vπ − π = V0 ⎜ + ⎟ 10 ⎝ 1 10 ⎠ (65.67)Vπ = V0 (1.10) or V0 = (59.7)Vπ and from (1): Vπ V V + π + 0 =0 IS + 100 0.760 10 10 g mVπ = I S + Vπ (0.8543) + (0.1)(59.7)Vπ = 0 I S = −Vπ (6.824) Now (−Vπ ) ⇒ Rif = 147 Ω Rif = IS To determine R0 f : IX = VX VX + − g mVπ RC RF + RB rπ (3) ⎛ −( RB rπ ) ⎞ Vπ = ⎜ ⎜ ( R r ) + R ⎟ (VX ) (4) ⎟ F ⎠ ⎝ B π Now ⎛ −(100 0.760) ⎞ Vπ = ⎜ ⎜ (100 0.760) + 10 ⎟ (VX ) = −(0.07014)VX ⎟ ⎝ ⎠ so 1 ⎛1 ⎞ I X = VX ⎜ + + (65.77)(0.07014) ⎟ 1 10.754 ⎝ ⎠ VX ⇒ R0 f = 175 Ω R0 f = IX b. From part (a), we find I Vπ = − S 6.824 then ⎛ −IS ⎞ V0 = (59.7) ⎜ ⎟ ⎝ 6.824 ⎠ or 565. V0 = −8.75 kΩ VS c. If capacitance is finite, a phase shift will be introduced. 12.55 dc analysis: VGS = VDS ID = VDD − VGS = K n (VGS − VTN ) 2 RD 10 − VGS = (0.20)(8)(VGS − 2) 2 2 10 − VGS = 1.6(VGS − 4VGS + 4) 2 1.6VGS − 5.4VGS − 3.6 = 0 5.4 ± (5.4) 2 + 4(1.6)(3.6) = 3.95 V 2(1.6) 10 − 3.95 ID = = 0.756 mA 8 g m = 2 K n I D = 2 (0.2)(0.756) ⇒ g m = 0.778 mA/V VGS = a. Vgs − VS RS + Vgs − V0 RF =0 ⎛ 1 1 ⎞ VS V0 Vgs ⎜ + + ⎟= RS RF ⎠ RS RF ⎝ V0 − Vgs V0 + g mVgs + =0 RD RF (1) ⎛ 1 ⎛ 1 ⎞ 1 ⎞ V0 ⎜ + − g m ⎟ (2) ⎟ = Vgs ⎜ ⎝ RD RF ⎠ ⎝ RF ⎠ So from (1): 1 ⎞ VS V0 ⎛1 Vgs ⎜ + + ⎟= ⎝ 10 100 ⎠ 10 100 or Vgs (0.11) = VS (0.10) + V0 (0.010) Vgs = VS (0.909) + V0 (0.0909) Then from (2): 566. ⎛1 1 ⎞ ⎛ 1 ⎞ V0 ⎜ + − 0.778 ⎟ ⎟ = Vgs ⎜ ⎝ 8 100 ⎠ ⎝ 100 ⎠ V0 (0.135) = Vgs (−0.768) = (−0.768)[VS (0.909) + V0 (0.0909)] V0 (0.2048) = −VS (0.6981) so V Av = 0 = −3.41 VS b. We have Vgs = VS (0.909) + V0 (0.0909) = VS (0.909) + (0.0909) − (3.41VS ) = 0.599VS Now V V0 (−3.41VS ) RS Azf = 0 = = I S VS − Vgs VS − 0.599VS RS or (−3.41)(10) Azf = ⇒ Azf = −85.0 V/ma 0.401 Vgs Vgs 0.401VS 0.599VS c. Rif = = = ⋅ (10) ⇒ Rif = 14.9 kΩ 0.401VS IS RS d. IX = VX VX + g mVgs + RD RS + RF ⎛ RS ⎞ ⎛ 10 ⎞ Vgs = ⎜ ⎟ VX = ⎜ ⎟ VX RS + RF ⎠ ⎝ 10 + 100 ⎠ ⎝ = (0.0909)VX 1 ⎤ ⎡1 I X = VX ⎢ + (0.778)(0.0909) + 10 + 100 ⎥ ⎣8 ⎦ IX 1 = = 0.2048 ⇒ R0 f = 4.88 kΩ VX R0 f 12.56 As g m → ∞, V0 − RF −100 = = = −10 VS RS 10 567. To be within 10% of ideal, V0 = −10(0.9) = −9 VS From Problem 12.41, we had Vgs = VS (0.909) + V0 (0.0909) = VS (0.909) + ( −9VS )(0.0909) = 0.0909VS Also from Problem 12.41, we had V0 (0.135) = Vgs (0.010 − g m ) or (−9VS )(0.135) = (0.0909)VS (0.010 − g m ) −1.215 = 0.000909 − 0.0909 g m or g m = 13.36 mA/V 12.57 dc analysis RTH = 24 || 150 = 20.7 kΩ ⎛ 24 ⎞ VTH = ⎜ ⎟ (12) = 1.655 V ⎝ 24 + 150 ⎠ 1.655 − 0.7 I BQ = = 0.00556 mA 20.7 + (151)(1) so I CQ = 0.834 mA (150)(0.026) = 4.68 kΩ 0.834 0.834 gm = = 32.08 mA / V 0.026 rπ = VS − Vπ Vπ V − V0 = + π RS RB || rπ RF (1) V0 V0 − Vπ + = 0 (2) RC RF From (1): ⎡1 VS 1 1 ⎤ V0 = Vπ ⎢ + + ⎥− 5 5 20.7 || 4.68 RF ⎦ RF ⎣ g mVπ + or 568. ⎛ 1 ⎞ V0 VS (0.20) = Vπ ⎜ 0.4620 + ⎟− RF ⎠ RF ⎝ From (2): ⎛ ⎛1 1 ⎞ 1 ⎞ ⎜ 32.08 − ⎟ Vπ + V0 ⎜ + ⎟=0 RF ⎠ ⎝ ⎝ 6 RF ⎠ so ⎛ 1 ⎞ −V0 ⎜ 0.1667 + ⎟ RF ⎠ ⎝ (2) Vπ = ⎛ 1 ⎞ ⎜ 32.08 − ⎟ RF ⎠ ⎝ Then ⎡ ⎛ 1 ⎢ −V0 ⎜ 0.1667 + RF ⎛ 1 ⎞⎢ ⎝ VS (0.20) = ⎜ 0.4620 + ⎟ RF ⎠ ⎢ ⎛ 1 ⎞ ⎝ ⎢ ⎜ 32.08 − ⎟ RF ⎠ ⎢ ⎝ ⎣ ⎞⎤ ⎟⎥ ⎠ ⎥ − V0 ⎥ RF ⎥ ⎥ ⎦ Neglect the RF in the denominator term. Now V0 V = −5 ⇒ VS = − 0 = −V0 (0.20) 5 VS ⎡ −V ( 0.1667 RF + 1) ⎤ −V0 (0.20)(0.20) RF = (0.4620 RF + 1) ⎢ 0 ⎥ − V0 32.08RF ⎣ ⎦ 2 −1.283RF = −(0.4620 RF + 1)(0.1667 RF + 1) − 32.08 RF 2 1.206 RF − 32.71RF − 1 = 0 32.71 ± (32.71) 2 + 4(1.206)(1) 2(1.206) so that RF = 27.2 kΩ RF = 12.58 dc analysis RTH = 4 || 15 = 3.16 kΩ = RB ⎛ 4 ⎞ VTH = ⎜ ⎟12 = 2.526 V ⎝ 4 + 15 ⎠ 2.526 − 0.7 I BQ = = 0.00251 3.16 + (181)(4) I CQ = 0.452 mA (180)(0.026) = 10.4 kΩ 0.452 0.452 gm = = 17.4 mA/V 0.026 rπ = 569. Vi − Vπ 1 Vπ 1 V −V = + π1 0 RS RB || rπ RF (1) g mVπ 1 + Vπ 2 =0 RC || RB || rπ (2) g mVπ 2 + Vπ 3 =0 RC || RB || rπ (3) g mVπ 3 + V0 V0 V0 − Vπ 1 + + =0 RC RL RF (4) Now RC || RB || rπ = 8 || 3.16 ||10.4 = 1.86 kΩ RB || rπ = 3.16 ||10.4 = 2.42 kΩ Now substituting in (2): V (17.4)Vπ 1 + π 2 = 0 or Vπ 2 = −(32.36)Vπ 1 1.86 and substituting in (3): V (17.4)Vπ 2 + π 3 = 0 1.86 V (17.4)[−(32.36)Vπ 1 ] + π 3 = 0 1.86 or Vπ 3 = (1047.3)Vπ 1 Substitute numerical values in (1): ⎛1 Vi 1 1 ⎞ V0 = Vπ 1 ⎜ + + ⎟− 10 10 2.42 RF ⎠ RF ⎝ or ⎛ 1 ⎞ V0 Vi (0.10) = Vπ 1 ⎜ 0.513 + ⎟− RF ⎠ RF ⎝ Substitute numerical values in (4): ⎛ 1 1 1 ⎞ Vπ 1 =0 (17.4)(1047.3)Vπ 1 + V0 ⎜ + + ⎟− ⎝ 8 4 RF ⎠ RF ⎛ ⎛ 1 ⎞ 1 ⎞ Vπ 1 ⎜ 1.822 × 104 − ⎟ + V0 ⎜ 0.375 + ⎟=0 RF ⎠ RF ⎠ ⎝ ⎝ ⎛ 1 ⎞ −V0 ⎜ 0.375 + ⎟ RF ⎠ ⎝ Vπ 1 = 1 1.822 × 104 − RF so that 570. ⎡ ⎛ 1 ⎞⎤ ⎢ −V0 ⎜ 0.375 + ⎟⎥ RF ⎠ ⎥ V0 ⎛ 1 ⎞⎢ ⎝ − Vi (0.10) = ⎜ 0.513 + ⎟ RF ⎠ ⎢ 1.822 × 104 − 1 ⎥ RF ⎝ ⎢ ⎥ RF ⎥ ⎢ ⎣ ⎦ V V We have 0 = −80 or Vi = − 0 80 Vi ⎡ ⎛ 1 ⎞⎤ ⎢ − ⎜ 0.375 + ⎟⎥ RF ⎠ ⎥ 1 (0.10) ⎛ 1 ⎞⎢ ⎝ − = ⎜ 0.513 + − ⎟ 80 RF ⎠ ⎢1.822 × 104 − 1 ⎥ RF ⎝ ⎢ ⎥ RF ⎥ ⎢ ⎣ ⎦ Neglect that 1/ RF term in the denominator. (0.513RF + 1)(0.375 RF + 1) −(0.00125 RF ) = − −1 1.822 × 104 RF 2 22.775 RF = (0.513RF + 1)(0.375 RF + 1) + 1.822 × 10 4 RF We find 2 22.58 RF − 1.822 × 104 RF − 1 = 0 RF = 1.822 × 104 ± (1.822 ×10 4 ) 2 + 4(22.58)(1) 2(22.58) or RF = 0.807 MΩ 12.59 a. VS − (−Vd ) −Vd − V1 = RS RF or ⎛ 1 1 ⎞ VS V1 + + =0 Vd ⎜ ⎟+ ⎝ RS RF ⎠ RS RF V0 − V1 V1 V1 − (−Vd ) = + R1 R2 RF or ⎛ 1 V0 1 1 ⎞ Vd = V1 ⎜ + + ⎟+ R1 ⎝ R1 R2 RF ⎠ RF (1) (2) 571. V0 A0 L Substitute numerical values in (1) and (2): V0 ⎛ 1 1 ⎞ VS V1 ⋅⎜ + ⎟ + + = 0 104 ⎝ 5 10 ⎠ 5 10 or V0 (0.3 × 10−4 ) + VS (0.20) + V1 (0.10) = 0 and V0 = A0 LVd or Vd = (1) V0 ⎛ 1 1 1⎞ V ⎛1⎞ = V1 ⎜ + + ⎟ + 04 ⋅ ⎜ ⎟ 50 ⎝ 50 10 10 ⎠ 10 ⎝ 10 ⎠ or V0 (0.02 − 10−5 ) = V1 (0.22) (2) ⎛ 0.02 − 10−5 ⎞ Then V1 = V0 ⎜ ⎟ ⎝ 0.22 ⎠ and ⎡ ⎛ 0.02 − 10−5 ⎞ ⎤ V0 (0.3 × 10−4 ) + VS (0.20) + (0.10) ⎢V0 ⎜ ⎟⎥ = 0 ⎣ ⎝ 0.22 ⎠ ⎦ V0 ⎡0.3 × 10−4 − 0.4545 × 10−5 + 0.00909⎤ + VS (0.20) = 0 ⎣ ⎦ Then b. V0 V −0.20 = ⇒ 0 = −21.94 −3 VS 9.115 × 10 VS Rif = Now Vd = −Vd −V ⋅ R = d S VS − (−Vd ) VS + Vd RS V0 21.94VS =− A0 L 104 (21.94 × 10−4 )(5) 1 − 21.94 × 10−4 or Rif = 1.099 × 10−2 kΩ ⇒ Rif = 10.99 Ω Then Rif = c. Because of the A0LVd source, R0 f = 0 12.60 For example, use the circuit shown in Figure 12.41 12.61 Break the loop 572. It = Iε Now Ai I t + V0 V0 + =0 R1 RF + RS Ri ⎛ RS ⎞ V0 Ir = ⎜ ⎟⋅ ⎝ RS + Ri ⎠ RF + RS Ri ⎛ R + Ri ⎞ or V0 = I r ⎜ S ⎟ ⋅ ( RF + RS Ri ) ⎝ RS ⎠ Then ⎛ 1 ⎞ ⎡ ⎛ RS + Ri ⎞ ⎤ 1 Ai I t + ⎜ + ⎟ × ⎢Ir ⎜ ⎟ ( RF + RS Ri ) ⎥ = 0 ⎜R R +R R ⎟ F S i ⎠ ⎣ ⎝ RS ⎠ ⎦ ⎝ 1 Ai I T = − r ⇒T = It ⎡1 ⎤ ⎛ RS + Ri ⎞ 1 ⎢ + ⎥⎜ ⎟ ( RF + RS Ri ) R1 RF + RS Ri ⎦ ⎝ RS ⎠ ⎣ 12.62 Vπ 1 V V −V + g m1Vπ 1 = ε 1 + ε 1 0 rπ 1 RE1 RF g m1Vπ 1 + Vr RC1 rπ 2 (1) = 0 ⇒ Vr = −( g m1Vπ 1 )( RC1 rπ 2 ) Vπ 2 = Vt so that g m 2Vt + Vπ 3 + V0 Vπ 3 + = 0 (3) RC 2 rπ 3 Vπ 3 V V −V + g m 3Vπ 3 = 0 + 0 ε 1 (4) rπ 3 RE 3 RF From (4): ⎛ 1 ⎛ 1 ⎞ V 1 ⎞ V0 ⎜ + + g m3 ⎟ + ε 1 ⎟ = Vπ 3 ⎜ ⎝ RE 3 RF ⎠ ⎝ rπ 3 ⎠ RF But Vε 1 = −Vπ 1 ⎛ 1 ⎞ V + g m3 ⎟ − π 1 Vπ 3 ⎜ ⎝ rπ 3 ⎠ RF so V0 = ⎛ 1 1 ⎞ + ⎜ ⎟ RE 3 RF ⎠ ⎝ Then (2) 573. ⎡⎛ 1 ⎞ ⎛ 1 1 ⎞⎤ + Vπ 1 ⎢⎜ + g m1 ⎟ − ⎜ ⎟⎥ = ⎠ ⎝ RE1 RF ⎠ ⎦ ⎣⎝ rπ 1 ⎛ 1 ⎞ V −Vπ 3 ⎜ + g m3 ⎟ + π 1 rπ 3 ⎝ ⎠ RF ⎛ 1 1 ⎞ RF ⋅ ⎜ + ⎟ ⎝ RE 3 RF ⎠ (1′) and ⎛ 1 ⎞ V + g m3 ⎟ − π 1 Vπ 3 ⎜ ⎛ 1 1 ⎞ ⎝ rπ 3 ⎠ RF = 0 + g m 2Vt + Vπ 3 ⎜ ⎟+ ⎛ 1 1 ⎞ ⎝ RC 2 rπ 3 ⎠ + RC 2 ⋅ ⎜ ⎟ RE 3 RF ⎠ ⎝ (3′) From (3′), solve for Vπ 3 and substitute into (1′). Then from (1′), solve for Vπ 1 and substitute into (2). Then T =− Vr . Vt 12.63 Vr Vr Vr − Ve + + =0 RS rπ 1 RF g m1Vt + (1) Vπ 2 + Ve Vπ 2 + = 0 (2) RC1 rπ 2 Vπ 2 V V − Vr + g m 2Vπ 2 = e + e (3) rπ 2 RE RF Using the parameters from Problem 12.29, we obtain 1 1⎞ V ⎛1 Vr ⎜ + + ⎟− e = 0 ⎝ 10 15.8 10 ⎠ 10 or Vr (0.2633) = Ve (0.10) (1) 1 ⎞ Ve ⎛ 1 =0 (7.62)Vt + Vπ 2 ⎜ + ⎟+ ⎝ 40 2.28 ⎠ 40 or Vt (7.62) + Vπ 2 (0.4636) + Ve (0.025) = 0 (2) ⎛ 1 ⎞ ⎛1 1 ⎞ V Vπ 2 ⎜ + 52.7 ⎟ = Ve ⎜ + ⎟ − r 2.28 ⎝ ⎠ ⎝ 1 10 ⎠ 10 or Vπ 2 (53.14) = Ve (1.10) − Vr (0.10) Then Vπ 2 = Ve (0.0207) − Vr (0.001882) (3) Substituting in (2): Vt (7.62) + (0.4636)[Ve (0.0207) − Vr (0.001882)] + Ve (0.025) = 0 574. or Vt (7.62) + Ve (0.03460) − Vr (0.0008725) = 0 From (1) Ve = Vr (2.633) Then Vt (7.62) + Vr (2.633)(0.03460) − Vr (0.0008725) = 0 Vt (7.62) + Vr (0.09023) = 0 Vr = −84.45 Vt Now V T = − r ⇒ T = 84.45 Vt or 12.64 Vπ = −Vt g mVπ = V0 V0 + RC RF + RB rπ (1) and ⎛ RB rπ ⎞ (2) Vr = ⎜ V ⎜R r +R ⎟ 0 ⎟ F ⎠ ⎝ B π Now ⎛1 ⎞ 1 (65.77)Vπ = V0 ⎜ + ⎜ 1 10 + 100 0.760 ⎟ ⎟ ⎝ ⎠ or (65.77)Vπ = V0 (1.0930) and ⎛ 0.754 ⎞ Vr = ⎜ ⎟ V0 = (0.07011)V0 ⎝ 10 + 0.754 ⎠ so V0 = (14.26)Vr Then (65.77)(−Vt ) = (14.26)Vr (1.0930) Vr = −4.22 so that T = 4.22 Vt 12.64 Want f1 = 12 MHz for a phase margin of 45° TdB ( 0 ) = 80 dB ⇒ T ( 0 ) = 104 Then 575. T ( 0) T(f)= ⎛ f ⎜1 + j f PD ⎝ Set f = f1 and T ⎞⎛ f ⎞ ⎟⎜ 1 + j 6 ⎟ 12 × 10 ⎠ ⎠⎝ =1 So 104 T =1= 2 ⎛ 12 × 106 ⎞ 1+ ⎜ ⎟ ⋅ 2 ⎝ f PD ⎠ which yields 12 × 106 104 = ⇒ f PD = 1.70 kHz f PD 2 12.65 a. ⎛ f ⎞ −1 ⎛ f ⎞ φ = − tan −1 ⎜ ⎟ − 2 tan ⎜ 4 ⎟ 5 × 102 ⎠ ⎝ ⎝ 10 ⎠ or ⎛ f ⎞ ⎛ f ⎞ −180 = − tan −1 ⎜ 180 2 ⎟ − 2 tan −1 ⎜ 180 ⎟ ⇒ f180 ≈ 1.05 × 104 Hz 4 ⎝ 5 × 10 ⎠ ⎝ 10 ⎠ β (105 ) T ( f180 ) = 1 = b. 2 2 ⎛ 1.05 × 104 ⎞ ⎡ ⎛ 1.05 ×104 ⎞ ⎤ 1+ ⎜ ⎟ ⎢1 + ⎜ ⎟ ⎥ 2 4 ⎝ 5 × 10 ⎠ ⎢ ⎝ 10 ⎠ ⎥ ⎣ ⎦ 5 β (10 ) or 1= (21.02)(2.105) β = 4.42 × 10−4 12.65 A0 = 80 dB ⇒ A0 = 104 A0 Af ( 0 ) = 1 + β A0 or 5 = 104 1 + β (104 ) ⇒ β ≈ 0.2 Then T ( 0 ) = β A0 = 0.2 × 104 Inserting a dominate pole ⎛ f ⎞ −1 ⎛ f ⎞ −1 ⎛ f ⎞ φ = − tan −1 ⎜ ⎟ − tan ⎜ 6 ⎟ − tan ⎜ 7 ⎟ 10 ⎠ f PD ⎠ ⎝ ⎝ 10 ⎠ ⎝ If we want a phase margin of 45°, then ⎛ f ⎞ ⎛ f ⎞ −135° ≈ −90° − tan −1 ⎜ 6 ⎟ − tan −1 ⎜ 7 ⎟ 10 ⎠ ⎝ ⎝ 10 ⎠ By trial and error, f ≈ 0.845 MHz Then 0.2 × 104 1 T =1= × 2 2 2 ⎛ 0.845 × 106 ⎞ ⎛ 0.845 ⎞ ⎛ 0.845 ⎞ 1+ ⎜ 1+ ⎜ 1+ ⎜ ⎟ ⎟ ⎟ ⎝ 10 ⎠ f PD ⎝ 1 ⎠ ⎝ ⎠ 576. 0.845 × 106 0.2 ×104 ≈ f PD (1.309 )(1.0036 ) so f PD = 555 Hz 12.66 β (103 ) T = β Av = (a) f ⎞ ⎛ ⎜1 + j 4 ⎟ 10 ⎠ ⎝ f f φ = −2 tan −1 4 − tan −1 5 10 10 Set φ = −180° 2 f ⎞ ⎛ ⎜1 + j 5 ⎟ 10 ⎠ ⎝ By trial and error, f = 4.58 ×10 4 Hz Set T = 1 at f = 4.58 ×10 4 Hz (b) β (103 ) 1= 2 ⎛ 4 2 ⎞ 4 2 ⎛ ⎞ ⎛ ⎞ ⎜ 1 + ⎜ 4.58 ×10 ⎟ ⎟ 1 + ⎜ 4.58 ×10 ⎟ 4 5 ⎜ ⎝ 10 ⎠ ⎟ ⎝ 10 ⎠ ⎝ ⎠ β (103 ) ⇒ β = 0.02417 1= (21.976)(1.10) 103 = 39.7 1 + (10 )(0.02417) (c) Avfo = (d) Wait T < 1 at f = 4.58 × 104 Hz, so system is stable for smaller values of β . 3 12.67 ⎛ f ⎞ ⎛ f ⎞ ⎛ f ⎞ − tan −1 ⎜ − tan −1 ⎜ 5 ⎟ 4 ⎟ 4 ⎟ ⎝ 10 ⎠ ⎝ 5 × 10 ⎠ ⎝ 10 ⎠ 4 At f = 8.1× 10 Hz, φ = −180.28° φ = − tan −1 ⎜ Determine T ( f ) at this frequency. T = β (103 ) × = a. 1 ⎛ 8.1× 104 ⎞ 1+ ⎜ ⎟ 4 ⎝ 10 ⎠ 2 × 1 ⎛ 8.1× 104 ⎞ 1+ ⎜ 4 ⎟ ⎝ 5 × 10 ⎠ β (103 ) (8.161)(1.904)(1.287) For β = 0.005 T ( f ) = 0.250 < 1 ⇒ Stable b. For β = 0.05 T ( f ) = 2.50 > 1 ⇒ Unstable 12.68 (b)Phase margin = 80° ⇒ φ = −100° ⎛ f ⎞ ⎛ f ⎞ − tan −1 ⎜ 3 ⎟ 4 ⎟ ⎝ 10 ⎠ ⎝ 5 × 10 ⎠ 3 By trial and error, f = 1.16 × 10 Hz φ = −100 = −2 tan −1 ⎜ 2 × 1 ⎛ 8.1× 104 ⎞ 1+ ⎜ ⎟ 5 ⎝ 10 ⎠ 2 577. Then β (5 × 103 ) T =1= 2 ⎛ 3 2 ⎞ 3 2 ⎜ 1 + ⎛ 1.16 × 10 ⎞ ⎟ ⋅ 1 + ⎛ 1.16 × 10 ⎞ ⎜ ⎟ ⎜ 3 4 ⎟ ⎜ ⎝ 10 ⎠ ⎟ ⎝ 5 × 10 ⎠ ⎝ ⎠ β (5 × 103 ) = ⇒ β = 4.7 × 10−4 (2.35)(1.00) 12.69 For β = 0.005, c. T ( f ) = 1(0 dB) at f ≈ 2.10 × 104 Hz Then 4 4 ⎛ 2.10 × 104 ⎞ −1 ⎛ 2.10 × 10 ⎞ −1 ⎛ 2.10 × 10 ⎞ ⎟ − tan ⎜ ⎟ − tan ⎜ ⎟ 3 4 5 ⎝ 10 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠ φ = − tan −1 ⎜ or = −87.27 − 64.54 − 11.86 φ = −163.7 System is stable. Phase margin = 16.3° For β = 0.05, T ( f ) = 1 (0 dB) at f ≈ 6.44 × 104 Hz 578. Then ⎛ 6.44 × 104 ⎞ ⎛ 6.44 × 104 ⎞ ⎛ 6.44 × 104 ⎞ − tan −1 ⎜ − tan −1 ⎜ ⎟ ⎟ ⎟ 3 4 5 ⎝ 10 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠ φ = − tan −1 ⎜ or = −89.11 − 81.17 − 32.78 φ = −203.1° ⇒ System is unstable. 12.70 β (105 ) T = Aβ = f ⎞⎛ f ⎞⎛ f ⎞ ⎛ 1 + j 5 ⎟ ⎜1 + j ⎜1 + j ⎟ 4 ⎟⎜ 5 × 10 ⎠ ⎝ 10 ⎠ ⎝ 5 × 105 ⎠ ⎝ Phase Margin = 60° ⇒ φ = −120° So ⎛ f ⎞ ⎛ f ⎞ ⎛ f ⎞ −120 = − tan −1 ⎜ − tan −1 ⎜ 5 ⎟ − tan −1 ⎜ 4 ⎟ 5 ⎟ ⎝ 5 × 10 ⎠ ⎝ 10 ⎠ ⎝ 5 × 10 ⎠ By trial and error, at f = 105 Hz , φ ≅ −120° Then β (105 ) T = 1= 2 2 2 ⎛ 105 ⎞ ⎛ 105 ⎞ ⎛ 105 ⎞ ⋅ 1+ ⎜ 5 ⎟ ⋅ 1+ ⎜ 1+ ⎜ 4 ⎟ 5 ⎟ ⎝ 5 × 10 ⎠ ⎝ 10 ⎠ ⎝ 5 × 10 ⎠ 1= β (105 ) ( 2.236 )(1.414 )(1.02 ) ⇒ β = 3.22 × 10−5 12.71 (a) 100 = 105 ⇒ β = 9.99 × 10−3 1 + (105 ) β T = 1 = β Av = 1= (9.99 × 10−3 )(105 ) ⎛ f ⎞ 1+ ⎜ 3 ⎟ ⎝ 10 ⎠ 999 2 2 ⎛ f ⎞ 1+ ⎜ 5 ⎟ ⎝ 10 ⎠ ⎛ f ⎞ ⎛ f ⎞ 1+ ⎜ 3 ⎟ 1+ ⎜ 5 ⎟ ⎝ 10 ⎠ ⎝ 10 ⎠ 5 f = 3.08 × 10 Hz 2 2 Phase f f − tan −1 5 3 10 10 5 3.08 ×10 3.08 × 105 = − tan −1 − tan −1 103 105 = −89.81 − 72.01 φ = −161.8 Stable (b) Phase Margin = 180 − 161.8 = 18.2° φ = − tan −1 12.72 579. φ = − tan −1 (a) For φ = −180° f f f − tan −1 6 − tan −1 4 5 × 10 10 5 × 107 By trial and error, f180 = 7.25 × 106 Hz T = (b) = (0.10)(105 ) ⎛ 7.25 × 106 ⎞ 1+ ⎜ ⎟ 4 ⎝ 5 × 10 ⎠ 2 ⎛ 7.25 ×106 ⎞ 1+ ⎜ ⎟ 6 ⎝ 10 ⎠ 2 ⎛ 7.25 ×106 ⎞ 1+ ⎜ ⎟ 7 ⎝ 5 × 10 ⎠ 104 (145)(7.319)(1.01) T = 9.33 System is unstable T =1= 104 ⎛ f ⎞ 1+ ⎜ 4 ⎟ ⎝ 5 × 10 ⎠ 2 ⎛ f ⎞ 1+ ⎜ 6 ⎟ ⎝ 10 ⎠ 2 ⎛ f ⎞ 1+ ⎜ 7 ⎟ ⎝ 5 × 10 ⎠ 2 f = 2.14 ×107 Hz 7 7 ⎛ 2.14 × 107 ⎞ −1 ⎛ 2.14 × 10 ⎞ −1 ⎛ 2.14 × 10 ⎞ ⎟ − tan ⎜ ⎟ − tan ⎜ ⎟ 4 6 7 ⎝ 5 × 10 ⎠ ⎝ 10 ⎠ ⎝ 5 × 10 ⎠ φ = − tan −1 ⎜ = −89.87 − 87.32 − 23.17 φ = −200.4° For f = 7.25 × 106 Hz (c) (0.0010)(105 ) T = 2 ⎛ 7.25 × 106 ⎞ ⎛ 7.25 ×106 ⎞ 1+ ⎜ ⎟ 1+ ⎜ ⎟ 4 6 ⎝ 5 × 10 ⎠ ⎝ 10 ⎠ 100 = (145)(7.319)(1.01) 2 ⎛ 7.25 ×106 ⎞ 1+ ⎜ ⎟ 7 ⎝ 5 × 10 ⎠ T = 0.0933 System is stable. T =1= 100 ⎛ f ⎞ 1+ ⎜ 4 ⎟ ⎝ 5 × 10 ⎠ 2 ⎛ f ⎞ 1+ ⎜ 6 ⎟ ⎝ 10 ⎠ 2 ⎛ f ⎞ 1+ ⎜ 7 ⎟ ⎝ 5 × 10 ⎠ 2 f = 2.13 ×106 Hz 2.13 × 106 2.13 × 106 2.13 × 106 − tan −1 − tan −1 5 × 104 106 5 × 107 = −88.66 − 64.85 − 2.44 φ = −155.9° φ = − tan −1 12.73 (a) f180 f − 2 tan −1 180 5 × 103 105 = 1.05 × 105 Hz φ = −180 = − tan −1 f180 2 2 580. (0.0045)(2 × 103 ) T = (b) 2 2 ⎛ 1.05 × 105 ⎞ ⎡ ⎛ 1.05 × 105 ⎞ ⎤ 1+ ⎜ ⎢1 + ⎜ ⎟ ⎥ 3 ⎟ 5 ⎝ 5 × 10 ⎠ ⎢ ⎝ 10 ⎠ ⎥ ⎣ ⎦ 9 = (21.02)(2.1025) 0 T = f180 = 0.204 System is stable 9 T =1= ⎛ f ⎞ 1+ ⎜ 3 ⎟ ⎝ 5 × 10 ⎠ 2 ⎡ ⎛ f ⎞2 ⎤ ⎢1 + ⎜ 5 ⎟ ⎥ ⎢ ⎝ 10 ⎠ ⎥ ⎣ ⎦ f = 3.88 × 104 Hz 3.88 × 104 3.88 ×10 4 − 2 tan −1 3 5 × 10 105 = −82.66 − 42.41 φ = −125.1° φ = − tan −1 (c) T = = (0.15)(2 × 103 ) ⎛ 1.05 × 105 ⎞ 1+ ⎜ 3 ⎟ ⎝ 5 × 10 ⎠ 2 ⎡ ⎛ 1.05 × 105 ⎞ 2 ⎤ ⎢1 + ⎜ ⎟ ⎥ 5 ⎢ ⎝ 10 ⎠ ⎥ ⎣ ⎦ 300 (21.02)(2.1025) T = 6.79 System is unstable T = 1= 300 ⎛ f ⎞ 1+ ⎜ 3 ⎟ ⎝ 5 × 10 ⎠ 2 ⎡ ⎛ f ⎞2 ⎤ ⎢1 + ⎜ 5 ⎟ ⎥ ⎢ ⎝ 10 ⎠ ⎥ ⎣ ⎦ f = 2.33 × 105 Hz 2.33 × 105 2.33 × 105 − 2 tan −1 5 × 103 105 = −88.77 − 133.54 φ = − tan −1 φ = −222.3° 12.74 Phase Margin = 45° ⇒ φ = −135° φ = −135° ⎛ f ⎞ ⎛ f ⎞ ⎛ f ⎞ ⎛ f ⎞ = − tan −1 ⎜ 3 ⎟ − tan −1 ⎜ 4 ⎟ − tan −1 ⎜ 5 ⎟ − tan −1 ⎜ 6 ⎟ ⎝ 10 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠ 4 At f = 10 Hz, φ = −135.6° 581. T =1 1 = β (103 ) × 2 1 × ⎛ 104 ⎞ ⎛ 104 ⎞ 1+ ⎜ 3 ⎟ 1+ ⎜ 4 ⎟ ⎝ 10 ⎠ ⎝ 10 ⎠ 1 1 × × 2 4 2 ⎛ 10 ⎞ ⎛ 104 ⎞ 1+ ⎜ 5 ⎟ 1+ ⎜ 6 ⎟ ⎝ 10 ⎠ ⎝ 10 ⎠ 1= 2 × β (103 ) (10.05)(1.414)(1.005)(1.00) or β = 0.01428 12.75 1 1 × f ⎛ ⎞ f ⎞ ⎛1 + j 3 ⎟ ⎜1 + j ⎟ ⎜ 300 × 10 ⎠ f PD ⎠ ⎝ ⎝ 1 1 × × f ⎞ ⎛ f ⎛ ⎞ ⎜1 + j ⎜1 + j ⎟ 6 ⎟ 2 × 10 ⎠ ⎝ 25 × 106 ⎠ ⎝ Phase Margin = 45° ⇒ φ = −135° at f = 300 kHz T = 5000 × 3 ⎛ 300 × 103 ⎞ −1 ⎛ 300 × 10 ⎞ −135° = − tan −1 ⎜ −0−0 ⎟ − tan ⎜ 3 ⎟ ⎝ 300 × 10 ⎠ ⎝ f PD ⎠ = −90° − 45° Now T = 1@ f = 300 kHz T =1≈ 5000 2 ⎛ 300 × 103 ⎞ 1+ ⎜ ⎟ ⋅ 2 ⋅1 ⋅1 ⎝ f PD ⎠ 2 ⎛ 300 × 103 ⎞ ⎛ 5000 ⎞ 1+ ⎜ ⎟ =⎜ ⎟ ⎝ f PD ⎠ ⎝ 2 ⎠ f PD ≈ 12.76 (a) 2 300 × 103 2 ⇒ f PD = 84.8 Hz 5000 At f = 106 Hz, 106 106 − 2 tan −1 6 104 10 = −89.4 − 90 ≈ −180° φ = − tan −1 582. T = = 103 ⎛ 106 ⎞ 1+ ⎜ 4 ⎟ ⎝ 10 ⎠ 2 ⎡ ⎛ 106 ⎞ 2 ⎤ ⎢1 + ⎜ 6 ⎟ ⎥ ⎢ ⎝ 10 ⎠ ⎥ ⎣ ⎦ 103 =5 (100)(2) T > 1 at f = f180 = 106 Hz, System is unstable. 103 T= (b) ⎛ f ⎞⎛ f ⎞⎛ f ⎞ ⎜1 + j ⎟⎜ 1 + j 4 ⎟ ⎜ 1 + j 6 ⎟ f PD ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠ ⎝ Phase margin = 45° ⇒ φ = −135 φ = −135 = − tan −1 f f PD − tan −1 2 f f − 2 tan −1 6 4 10 10 4 f ≅ 10 Hz T =1= 1≅ 103 ⎛ 104 ⎞ 1+ ⎜ ⎟ ⎝ f PD ⎠ 2 ⎛ 104 ⎞ 1+ ⎜ 4 ⎟ ⎝ 10 ⎠ 2 ⎡ ⎛ 104 ⎞ 2 ⎤ ⎢1 + ⎜ 6 ⎟ ⎥ 10 ⎠ ⎥ ⎢ ⎣ ⎝ ⎦ 103 ⎛ 104 ⎞ ⎜ ⎟ (1.414)(1.0) ⎝ f PD ⎠ (104 )(1.414) f PD = 103 f PD = 14.14 Hz 12.77 (a) ⎛ f180 ⎞ ⎛ f ⎞ ⎛ f ⎞ − tan −1 ⎜ 180 4 ⎟ − tan −1 ⎜ 180 ⎟ 4 ⎟ 5 ⎝ 10 ⎠ ⎝ 5 × 10 ⎠ ⎝ 10 ⎠ φ = −180 = − tan −1 ⎜ f180 ≅ 8.06 × 104 Hz (b) 500 T = 2 ⎛ 8.06 × 10 ⎞ ⎛ 8.06 × 104 ⎞ 1+ ⎜ 1+ ⎜ ⎟ ⎟ 4 4 ⎝ 10 ⎠ ⎝ 5 × 10 ⎠ 500 = (8.122)(1.897)(1.284) 4 2 ⎛ 8.06 ×104 ⎞ 1+ ⎜ ⎟ 5 ⎝ 10 ⎠ T = 25.3 500 ⎛ ⎞⎛ f f ⎞⎛ f ⎞⎛ f ⎞ ⎜1 + j ⎟⎜ 1 + j 4 ⎟ ⎜ 1 + j ⎟ ⎜1 + j 5 ⎟ 10 ⎠ ⎝ 5 × 104 ⎠ ⎝ 10 ⎠ f PD ⎠ ⎝ ⎝ Phase Margin = 60° ⇒ φ = −120° (c) T= −120 = − tan −1 f f f f − tan −1 4 − tan −1 − tan −1 5 10 5 × 104 10 f PD Assume tan −1 f f PD ≅ 90° 2 583. Then f ≅ 4.2 × 103 Hz 500 T =1= 2 ⎛ 4.2 × 103 ⎞ ⎛ 4.2 × 103 ⎞ 1+ ⎜ ⎟ 1+ ⎜ ⎟ 4 ⎝ 10 ⎠ ⎝ f PD ⎠ 500 1= 2 2 ⎛ 4.2 × 103 ⎞ 1+ ⎜ ⎟ (1.085)(1.004)(1.0) ⎝ f PD ⎠ 4.2 × 103 500 ≅ (1.0846)(1.0035)(1.0) f PD f PD = 9.14 Hz 12.78 50 = (a) T= 104 ⇒ β = 0.0199 1 + (104 ) β (0.0199)(104 ) ⎛ f ⎞⎛ f ⎞ ⎜1 + j ⎟⎜ 1 + j 5 ⎟ 10 ⎠ f PD ⎠ ⎝ ⎝ Phase margin = 45° ⇒ φ = −135° −135 = − tan −1 f f PD − tan −1 f 105 5 f = 10 Hz T =1= (0.0199)(104 ) ⎛ 105 ⎞ 1+ ⎜ ⎟ ⎝ f PD ⎠ 2 ⎛ 105 ⎞ 1+ ⎜ 5 ⎟ ⎝ 10 ⎠ 2 105 (0.0199)(104 ) = f PD 1.414 f PD = 711 Hz 20 = (b) T= 104 ⇒ β = 0.0499 1 + (104 ) β (0.0499)(104 ) f ⎞⎛ f ⎞ ⎛ ⎜1 + j ⎟⎜ 1 + j 5 ⎟ 711 ⎠⎝ 10 ⎠ ⎝ T =1= (0.0499)(104 ) ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ 711 ⎠ 2 ⎛ f ⎞ 1+ ⎜ 5 ⎟ ⎝ 10 ⎠ 2 f = 1.76 × 105 Hz 5 ⎛ 1.76 × 105 ⎞ −1 ⎛ 1.76 × 10 ⎞ ⎟ − tan ⎜ ⎟ 5 ⎝ 711 ⎠ ⎝ 10 ⎠ φ = − tan −1 ⎜ = −89.77 − 60.40 φ = −150.2 Phase Margin = 180 − 150.2 = 29.8° ⎛ 4.2 × 103 ⎞ 1+ ⎜ 4 ⎟ ⎝ 5 × 10 ⎠ 2 ⎛ 4.2 × 103 ⎞ 1+ ⎜ ⎟ 5 ⎝ 10 ⎠ 2 584. 12.79 (a) 20 = AO = 100 dB ⇒ AO = 105 105 ⇒ β = 0.04999 1 + (105 ) β (0.04999)(105 ) ⎛ f ⎞⎛ f ⎞⎛ f ⎞ ⎜1 + j ⎟⎜ 1 + j 6 ⎟⎜ 1 + j 7 ⎟ f PD ⎠ ⎝ 10 ⎠⎝ 10 ⎠ ⎝ Phase Margin = 45° ⇒ φ = −135° T= −135 = − tan −1 f f f − tan −1 6 − tan −1 7 10 10 f PD f ≈ 106 Hz T =1= 1= (0.04999)(105 ) ⎛ 106 ⎞ 1+ ⎜ ⎟ ⎝ f PD ⎠ 2 ⎛ 106 ⎞ 1+ ⎜ 6 ⎟ ⎝ 10 ⎠ 2 ⎛ 106 ⎞ 1+ ⎜ 7 ⎟ ⎝ 10 ⎠ 2 (0.04999)(105 ) 2 ⎛ 106 ⎞ 1+ ⎜ ⎟ (1.414)(1.005) ⎝ f PD ⎠ 106 (0.04999)(105 ) = f PD (1.414)(1.005) f PD = 2.84 Hz (b) T =1= 5= 105 ⇒ β = 0.19999 1 + (105 ) β (0.19999)(105 ) ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ 284 ⎠ 2 ⎛ f ⎞ 1+ ⎜ 6 ⎟ ⎝ 10 ⎠ 2 ⎛ f ⎞ 1+ ⎜ 7 ⎟ ⎝ 10 ⎠ 2 f = 2.25 × 106 Hz 6 6 ⎛ 2.25 × 106 ⎞ −1 ⎛ 2.25 × 10 ⎞ −1 ⎛ 2.25 × 10 ⎞ ⎟ − tan ⎜ ⎟ − tan ⎜ ⎟ 6 7 ⎝ 284 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠ φ = − tan −1 ⎜ = −89.99 − 66.04 − 12.68 φ = −168.7 Phase Margin = 180 − 168.7 = 11.3° 12.80 a. T( f ) = T (0) = 100 dB ⇒ T (0) = 105 105 f ⎞⎛ f ⎞⎛ f ⎛ ⎞ 1+ j ⎜1 + j ⎟ ⎜1 + j ⎟ 6 ⎟⎜ 10 ⎠ ⎝ 5 × 10 ⎠ ⎝ 10 × 106 ⎠ ⎝ T =1= = 105 × 1 ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ 10 ⎠ 2 × 1 ⎛ f ⎞ 1+ ⎜ 6 ⎟ ⎝ 5 × 10 ⎠ 2 × 1 f ⎛ ⎞ 1+ ⎜ 6 ⎟ ⎝ 10 × 10 ⎠ 2 585. By trial and error f = 0.976 MHz ⎛ 0.976 × 106 ⎞ −1 ⎛ 0.976 ⎞ −1 ⎛ 0.976 ⎞ ⎟ − tan ⎜ ⎟ − tan ⎜ ⎟ 10 5 ⎠ ⎝ ⎝ 10 ⎠ ⎝ ⎠ φ = − tan −1 ⎜ = −90° − 11.05° − 5.574° = −106.6° Phase Margin = 180° − 106.6° = 73.4° b. f P′1 ∝ 1 10 75 so = CF f P′1 20 or f P′1 = 2.67 Hz Now T =1= = 105 × × 1 ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ 2.67 ⎠ 2 × 1 ⎛ f ⎞ 1+ ⎜ 6 ⎟ ⎝ 5 × 10 ⎠ 2 1 2 f ⎛ ⎞ 1+ ⎜ 6 ⎟ ⎝ 10 × 10 ⎠ By trial and error f ≈ 2.66 ×105 Hz then ⎛ 2.66 × 105 ⎞ −1 ⎛ 0.266 ⎞ −1 ⎛ 0.266 ⎞ φ = − tan −1 ⎜ ⎟ − tan ⎜ ⎟ − tan ⎜ ⎟ ⎝ 5 ⎠ ⎝ 10 ⎠ ⎝ 2.67 ⎠ = −90° − 3.045° − 1.524° = −94.57° Phase Margin = 180° − 94.57° = 85.4° 12.81 1 where τ = ( Ro1 Ri 2 ) Ci 2πτ = (500 1000) × 103 × 2 × 10−12 ⇒ τ = 6.67 × 10−7 s (a) f 3− dB = Then 1 ⇒ f 3− dB = 239 kHz 2π (6.67 × 10−7 ) (b) For 1 1 f PD = 10 Hz, τ = = = 0.0159 s 2π f PD 2π (10) f 3− dB = Then τ = ( Ro1 Ri 2 ) ( Ci + CM ) 0.0159 = ( 500 1000 ) × 103 × ( Ci + CM ) or ( Ci + CM ) = 4.77 ×10−8 = 2 x10−12 + CM ⇒ CM = 0.0477 μ F 12.82 Assuming a phase margin of 45°. 586. f ⎛ f ⎞ ⎛ ⎞ −135° ≈ −90° − tan −1 ⎜ − tan −1 ⎜ 6 ⎟ 6 ⎟ ⎝ 2 × 10 ⎠ ⎝ 25 × 10 ⎠ By trial and error, f ≈ 1.74 MHz Then T =1 1 = 5000 × 2 ⎛ 1.74 × 106 ⎞ 1+ ⎜ ⎟ f PD ⎠ ⎝ 1 1 × × 2 2 ⎛ 1.74 ⎞ ⎛ 1.74 ⎞ 1+ ⎜ 1+ ⎜ ⎟ ⎟ ⎝ 2 ⎠ ⎝ 25 ⎠ or 1.74 × 106 5000 ≈ f PD (1.325)(1.0024) so f PD = 462 Hz 12.83 20 = 105 ⇒ β = 0.04999 1 + (105 ) β Phase Margin = 60° ⇒ φ = −120° f −120 = − tan −1 Assume tan −1 f PD − tan −1 f f − tan −1 6 10 5 × 107 f = 90° f PD f ≅ 30° ⇒ f = 5.77 × 105 Hz 106 (0.04999)(105 ) T = 1= 2 2 2 ⎛ 5.77 × 105 ⎞ ⎛ 5.77 × 105 ⎞ ⎛ 5.77 ×105 ⎞ 1+ ⎜ 1+ ⎜ 1+ ⎜ ⎟ ⎟ ⎟ 6 7 f PD ⎝ 10 ⎠ ⎝ 5 × 10 ⎠ ⎝ ⎠ tan −1 1= 5.77 × 105 f PD 4.999 × 103 2 ⎛ 5.77 × 105 ⎞ 1+ ⎜ ⎟ (1.155)(1.0) f PD ⎝ ⎠ 4.999 ×103 (1.155)(1.0) f PD = 133.3 Hz 587. Chapter 13 Exercise Solutions EX13.1 I C1 = I C 2 ≅ 9.5 μ A 9.5 μ A I B1 = I B 2 = = 0.0475 μ A ⇒ I B1 = I B 2 = 47.5 nA 200 EX13.2 5 − 0.6 − 0.6 − (−5) = 0.22 mA I REF = 40 I C17 = I C13 B = 0.75I REF = (0.75)(0.22) = 0.165 mA 0.165 (0.165)(0.1) + 0.6 + I C16 = 200 50 = 0.000825 + 0.01233 I C16 = 13.2 μ A EX13.3 ⎛V ⎞ 0.18 ×10−3 = 10−14 exp ⎜ D ⎟ ⎝ VT ⎠ ⎛ 0.18 × 10−3 ⎞ VD = VT ln ⎜ ⎟ −14 ⎝ 10 ⎠ ⎛ 0.18 × 10−3 ⎞ = (0.026) ln ⎜ ⎟ −14 ⎝ 10 ⎠ VD = 0.6140 VBB = 2VDD ≅ 1.228 V ⎛ V /2 ⎞ I C14 = I C 20 = I S exp ⎜ BB ⎟ ⎝ VT ⎠ ⎛ 0.6140 ⎞ = 3 × 10−14 exp ⎜ ⎟ ⎝ 0.026 ⎠ I C14 = I C 20 = 0.541 mA EX13.4 100 ro 6 = ⇒ 10.5 MΩ 0.0095 Then, using results from Example 13.4 588. Ract1 = ro 6 [1 + g m 6 ( R2 || rπ 6 ) ] = 10.5 [1 + (0.365)(1|| 547) ] = 14.3 MΩ V 100 ro 4 = A = ⇒ 10.5 MΩ I C 4 0.0095 ⎛ 9.5 ⎞ Ad = − ⎜ ⎟ (10.5 ||14.3 || 4.07 ) = −889 ⎝ 0.026 ⎠ EX13.5 100 = 556 K 0.18 Ri 3 = rπ 22 + (1 + β P )[ R19 || R20 ] = 7.22 + (51)(556 ||111) ⇒ 4.73 MΩ 100 Ract 2 = = 185 K 0.54 100 Ro17 = = 185 K 0.54 −(200)(201)(50)(185 || 4730 ||185) Av 2 = 4070[50 + [9.63 + (201)(0.1)]] −182358786.9 = 32450.1 Av 2 = −562 R19 = ro13 A = EX13.6 100 100 ro17 = = 185 K ro13 B = = 185 K 0.54 0.54 RC17 = 185 [1 + (20.8)(0.1|| 9.63) ] RC17 = 566 K 7.22 + 566 ||185 = 2.88 K Re 22 = 51 100 = 556 K RC19 = 0.18 0.65 + 2.88 || 556 Re 20 = = 0.0689 51 = 68.9 Ω RO = 22 + 68.9 = 90.9 Ω EX13.7 Ci = C1 (1+ | A2 |) = 30(1 + 562) = 16890 pF Ri 2 = 4.07 MΩ Ro1 = Ract1 || ro 4 = 14.3 ||10.5 = 6.05 MΩ Then Req = Ro1 || Ri 2 = 6.05 || 4.07 = 2.43 MΩ Then f PD = 1 1 = 6 2π Req Ci 2π (2.43 × 10 )(16890 × 10−12 ) = 3.88 Hz 589. EX13.8 For M 5 , M 6 ⎛ 40 ⎞ K P = (12.5) ⎜ ⎟ = 250 μ A / V 2 ⎝ 2 ⎠ 5 + 5 − VSG 5 0.25(VSG 5 − 0.5) 2 = 225 2 56.25(VSG 5 − VSG 5 + 0.25) = 10 − VSG 5 2 56.25VSG 5 − 55.25VSG 5 + 4.0625 = 0 55.25 ± 3052.5625 − 914.0625 2(56.25) = 0.902 V VSG 5 = VSG 5 10 − 0.902 ⇒ 40.4 μ A 225 Current in M 1 − M 4 = 20.2 μ A I set = I Q = EX13.9 ⎛ 40 ⎞ K P1 = K P 2 = (12.5) ⎜ ⎟ = 250 μ A/V 2 ⎝ 2 ⎠ From Exercise Ex 13.8, I D1 = I D 2 = 20.2 μ A ro 4 = ro 2 = 1 1 = = 2.48 MΩ λ I D (0.02)(0.0202) Ad = 2 K P1 I Q (ro 2 || ro 4 ) = 2(0.25)(0.0404)(2480 || 2480) Ad = 176 ⎛ 80 ⎞ K n 7 = (12.5) ⎜ ⎟ = 500 μ A / V 2 ⎝ 2⎠ g m 7 = 2 K n 7 I Q = 2 (0.50)(0.0404) = 0.284 mA/V ro 7 = ro8 = 1 λ ID7 = 1 ⇒ 1.24 MΩ (0.02)(0.0404) Av 2 = g m 7 (ro 7 || ro8 ) = (0.284)(1240 || 1240) = 176 AV = Ad ⋅ AV 2 = (176)(176) = 30,976 EX13.10 (a) ⎛ k ′ ⎞⎛ W ⎞ ⎛ I Q1 ⎞ g m1 = 2 ⎜ n ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ L ⎠1 ⎝ 2 ⎠ ⎛ 0.08 ⎞ ⎛ 0.25 ⎞ =2 ⎜ ⎟ (20) ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ = 0.6325 mA/V 1 = 800 K (0.01)(0.25/ 2) 1 ro 4 = = 533.3 K (0.015)(0.25/ 2) ro 2 = Ad 1 = g m1 (ro 2 || ro 4 ) = (0.6325)(800 || 533.3) Ad 1 = 202.4 590. ⎛ k′ ⎞⎛ W ⎞ g m5 = 2 ⎜ n ⎟ ⎜ ⎟ IQ 2 ⎝ 2 ⎠ ⎝ L ⎠5 ⎛ 0.04 ⎞ =2 ⎜ ⎟ (80)(0.25) ⎝ 2 ⎠ = 1.265 mA/V 1 = 266.7 K (0.015)(0.25) 1 ro9 = = 400 K (0.01)(0.25) ro 5 = A2 = − g m 5 (ro 5 || ro 9 ) = −(1, 265)(266.7 || 400) A2 = −202.4 Overall gain (assuming A3 = 1) A = Ad 1 ⋅ A2 = (202.4)(−202.4) = −40,966 EX13.11 I D1 = I D 2 = 25 μ A g m1 = g m8 = 2 k′ ⎛ W p ⎜ 2 ⎝L ⎞ ⎛ 40 ⎞ ⎟ I DQ = 2 ⎜ ⎟ (25)(25) ⇒ g m1 = g m8 = 224 μ A / V ⎠ ⎝ 2 ⎠ ⎛ k′ ⎞⎛ W gm6 = 2 ⎜ n ⎟ ⎜ ⎝ 2 ⎠⎝ L ⎞ ⎛ 80 ⎞ ⎟ I DQ = 2 ⎜ ⎟ (25)(25) ⇒ g m 6 = 316 μ A / V ⎠ ⎝ 2⎠ 1 1 = = 2 MΩ ro1 = ro 6 = ro8 = ro10 = λ I D (0.02)(25) ro 4 = 1 λ ID4 = 1 =1 MΩ (0.02)(50) Ro8 = g m8 (ro8 ro10 ) = (224)(2)(2) = 896 M Ω Ro 6 = g m 6 (ro 6 ) ( ro 4 || ro1 ) = 316(2) (1|| 2 ) = 421 M Ω Then Ad = g m1 ( Ro 6 Ro8 ) = 224 ( 421 896 ) ⇒ Ad = 64,158 EX13.12 V + − V − = VEB1 + VEB 6 + VBE 7 + I1 R1 = 0.6 + 0.6 + 0.6 + (0.236)(8) = 3.69 V So V + = −V − = 1.845 V EX13.13 Ad = 2 K P I Q 5 ( Ri 2 ) = 2(1)(0.2)(26) Ad = 16.4 EX13.14 591. rπ 13 = β VT I C13 = (200)(0.026) 0.20 = 26 kΩ Ri 2 = rπ 13 + (1 + β ) RE13 = 26 + 201(1) = 227 kΩ r010 = r012 = g m12 = rπ 12 = 1 λ I D10 = 1 = 500 kΩ (0.02)(0.1) VA 50 = = 500 kΩ I C12 0.1 I C12 0.1 = = 3.85 mA / V VT 0.026 β VT I C12 = (200)(0.026) = 52 kΩ 0.1 Ract1 = r012 [1 + g m12 (rπ 12 || R5 )] = 500[1 + (3.85)(52 || 0.5)] = 1453 kΩ Ad = 2 K n I Q 5 ⋅ ( ro10 || Ract1 || R12 ) = 2(0.6)(0.2) ⋅ (500 ||1453 || 227) = (0.490)(141) ⇒ Ad = 69.1 EX13.15 From Example 13.15, f PD = 265 Hz Av = Adi ⋅ A2 = (16.4)(1923) = 31,537 fT = f PD ⋅ Av = (265)(31,537) = 8.36 MHz TYU13.1 Computer Analysis TYU13.2 Computer Analysis TYU13.3 Vi N (max) = V + − VEB (on) = 15 − 0.6 = 14.4 V Vi N (min) ≅ 4VBE (on) + V + = 4(0.6) − 15 = −12.6 V −12.6 ≤ Vi N (cm) ≤ 14.4 V TYU13.4 592. V0 (max) ≅ V + − 2VBE (on) = 15 − 2(0.6) a. V0 (max) = 13.8 V V0 (min) = 3VBE (on) + V − = 3(0.6) − 15 V0 (min) ≅ −13.2 V −13.2 ≤ V0 ≤ 13.8 V V0 (max) = 5 − 1.2 = 3.8 V b. V0 (min) ≅ 3VBE + V − = 3(0.6) − 5 = −3.2 V −3.2 ≤ V0 ≤ 3.8 V TYU13.5 15 − 2(0.6) − (−15) = 0.72 mA I REF ≅ 40 ⎛I ⎞ ⎛ 0.72 × 10−3 ⎞ VBE = VT ln ⎜ REF ⎟ = (0.026) ln ⎜ ⎟ −14 ⎝ 10 ⎠ ⎝ IS ⎠ = 0.650 V So I REF = 30 − 2(0.65) ⇒ I REF = 0.718 mA 40 VBE11 = 0.650 V ⎛I ⎞ I C10 R4 = VT ln ⎜ REF ⎟ ⎝ I C10 ⎠ ⎛ 0.718 ⎞ I C10 (5) = (0.026) ln ⎜ ⎟ ⎝ I C10 ⎠ By trial and error: I C10 = 18.9 μ A VBE10 = VBE11 − I C10 R4 = 0.650 − (0.0189)(5) ⇒ VBE10 ≅ 0.556 V I C10 18.9 = = 9.45 μ A 2 2 ⎛I ⎞ ⎛ 9.45 × 10−6 ⎞ = VT ln ⎜ C 6 ⎟ = (0.026)ln ⎜ ⎟ ⇒ VBE 6 = 0.537 V −14 ⎝ 10 ⎠ ⎝ IS ⎠ IC 6 ≅ VBE 6 TYU13.6 10 − 0.6 − 0.6 − (−10) I REF = ⇒ I REF = 0.47 mA 40 ⎛I ⎞ I C10 R4 = VT ln ⎜ REF ⎟ ⎝ I C10 ⎠ ⎛ 0.47 ⎞ I C10 (5) = (0.026) ln ⎜ ⎟ ⎝ I C10 ⎠ By trial and error: ⇒ I C10 ≅ 17.2 μ A IC 6 ≅ I C13 B I C10 ⇒ I C 6 = 8.6 μ A 2 = (0.75) I REF ⇒ I C13 B = 0.353 mA I C13 A = (0.25) I REF ⇒ I C13 A = 0.118 mA 593. TYU13.7 R0 = R6 + RE14 RE14 = rπ 14 + R0 d R013 A 1 + βn The diode resistance can be found as ⎛V ⎞ I D = I S exp ⎜ D ⎟ ⎝ VT ⎠ ⎛ 1 1 ∂I D = = IS ⎜ rd ∂VD ⎝ VT ⎞ ⎛ VD ⎟ ⋅ exp ⎜ ⎠ ⎝ VT ⎞ ID ⎟= ⎠ VT or rd = VT V 0.026 = T = ⇒ 144 Ω I D I C13 A 0.18 RE 22 = rπ 22 + R017 R013 B 1+ βP R013 B = r013 B = 92.6 kΩ R017 = r017 ⎡1 + g m17 ( R8 rπ 17 ) ⎦ = 283 kΩ ⎤ ⎣ From previous calculations RE 22 = 1.51 kΩ R0 d = 2rd + RE 22 = 2(0.144) + 1.51 = 1.80 kΩ R013 A = r013 A = 278 kΩ βV (200)(0.026) rπ 14 = n T = = 1.04 kΩ I C14 5 1.04 + 1.8 || 278 ⇒ 14.1 Ω 201 R0 = R6 + RE14 = 27 + 14.1 ⇒ R0 ≅ 41 Ω RE14 = 594. TYU13.8 For M 6 we have VSG 5 = VSG 6 = 1.06 V So VSD 6 (sat) = 1.06 − 0.5 = 0.56 V For M 1 and M 2 ID = IQ = K P (VSG1 + VTP ) 2 2 0.0397 = 0.125(VSG1 − 0.5) 2 ⇒ VSG1 = 0.898 V 2 So maximum input voltage = V + − VSD 6 (sat) − VSG1 = 5 − 0.56 − 0.898 ⇒ Vi N (max) = 3.54 V For M 3, K p = (6.25)(20) = 125 μ A / V 2 I D3 = IQ = 39.7 μA 2 2 39.7 = 125(VGS 3 − VTN ) 2 2 VTN = 0.5V ⇒ VGS 3 = 0.898 V VSD1 (sat) = 0.898 − 0.5 = 0.398 V Vi N (min) = V − + VGS 3 + VSD1 (sat) − VSG1 = −5 + 0.898 + 0.398 − 0.898 Vi N (min) = −4.60 V −4.60 ≤ Vi N (cm) ≤ 3.54 V TYU13.9 V0 (max) = V + − VSD 8 (sat) VSG 8 = VSG 5 = 1.06 V VSD 8 (sat) = 1.06 − 0.5 = 0.56 V V0 (max) = 5 − 0.56 = 4.44 V V0 (min) = V − + VDS 7 (sat) VGS 7 = 1.06 ⇒ VDS 7 (sat) = 1.06 − 0.5 = 0.56 V0 (min) = −5 + 0.56 = −4.44 −4.44 ≤ V0 ≤ 4.44 V TYU13.10 (a) For M 5, K p 5 = 125 μ A / V 2 595. V + − V − − VSG 5 Rset K p 5 (VSG 5 + VTP ) 2 = 5 + 5 − VSG 5 100 2 12.5(VSG 5 − VSG 5 + 0.25) = 10 − VSG 5 0.125(VSG 5 − 0.5) 2 = 2 12.5VSG 5 − 11.5VSG 5 − 6.875 = 0 11.5 ± (11.5) 2 + 4(12.5)(6.875) 2(12.5) = 1.33 V VSG 5 = VSG 5 Then 10 − 1.33 ⇒ 86.7 μ A 100 IQ = ID4 = = 43.35 μ A 2 I REF = I Q = I D 8 = I D 7 = I D1 = I D 2 = I D 3 K p1 = K p 2 = 125μ A / V 2 (b) ro 2 = ro 4 = I λ ID = 1 = 1153 k Ω (0.02)(0.04335) Input stage gain Ad = 2 K p1 I Q ⋅ (ro 2 ro 4 ) = 2(0.125)(0.0867) ⋅ (1153 1153) ⇒ Ad = 84.9 Transconductance of M 7 g m 7 = 2 K n 7 I D 7 = 2 (0.250)(0.0867) = 0.294 mA / V ro 7 = ro8 = 1 λ I D7 = 1 = 577 k Ω (0.02)(0.0867) Second stage gain Av 2 = g m 7 (ro 7 ro8 ) = (0.294)(577 577) ⇒ Av 2 = 84.8 Overall gain = Ad ⋅ Av 2 = (84.9)(84.8) ⇒ A = 7, 200 TYU13.11 (a) Ad = Bg m1 ( ro 6 ro8 ) ⎛ k ′ ⎞⎛ W g m1 = 2 ⎜ n ⎟ ⎜ ⎝ 2 ⎠⎝ L g m1 = 400 μ A / V ro 6 = ro8 = 1 λP I D 6 = ⎞ ⎛ 80 ⎞ ⎟ I D1 = 2 ⎜ ⎟ (20)(50) ⎠ ⎝ 2⎠ 1 = 0.333 M Ω (0.02)(150) 1 1 = = 0.333 M Ω λn I D 8 (0.02)(150) Ad = 3(400) ( 0.333 || 0.333) ⇒ Ad = 200 (b) f PD = 1 2π Ro (CL + CP ) where Ro = ro 6 || ro8 = 0.333 || 0.333 M Ω 596. f PD = 1 ⇒ f PD = 477 kHz 2π ( 0.333 || 0.333) × 106 × 2 × 10−12 f PD ⋅ Ad = (477 × 103 )(200) ⇒ 95.5 MHz TYU13.12 (a) From Exercise TYU 13.11, g m1 = 400 μ A / V ro 6 = ro8 = ro10 = ro12 = 0.333 M Ω ⎛ k′ g m10 = 2 ⎜ P ⎝ 2 ⎞⎛W ⎞ ⎛ 40 ⎞ ⎟ ⎜ L ⎟ I D10 = 2 ⎜ 2 ⎟ (20)(150) ⇒ g m10 = 490 μ A / V ⎝ ⎠ ⎠⎝ ⎠ ⎛ k ′ ⎞⎛ W ⎞ ⎛ 80 ⎞ g m12 = 2 ⎜ n ⎟ ⎜ ⎟ I D12 = 2 ⎜ ⎟ (20)(150) ⇒ g m12 = 693 μ A / V ⎝ 2⎠ ⎝ 2 ⎠⎝ L ⎠ Ro10 = g m10 (ro10 ro 6 ) = (490)(0.333)(0.333) = 54.4 M Ω Ro12 = g m12 (ro12 ro8 ) = (693)(0.333)(0.333) = 77.0 M Ω Ad = Bg m1 ( Ro10 || Ro12 ) = 3(400)(54.4 || 77.0) ⇒ Ad = 38, 254 Ro = Ro10 || Ro12 = 54.4 || 77.0 = 31.9 M Ω (b) 1 = 2.50 kHz 2π (31.9 × 106 )(2 × 10−12 ) f PD ⋅ Ad = (2.5 × 103 )(38, 254) ⇒ 95.6 MHz f PD = TYU13.13 (a) Ad = g m1 ( Ro 6 || Ro8 ) From Example 13.10, g m1 = 316 μ A / V , Ro8 = 316 M Ω Now Ro 6 = g m 6 (ro 6 )(ro 4 || ro1 ) ro1 = 1 M Ω, ro 4 = 0.5 M Ω I 50 gm6 = C 6 = ⇒ 1.923 mA / V VT 0.026 ro 6 = VA6 80 = = 1.6 M Ω I C 6 50 Then Ro 6 = (1.923)(1600)(0.5 || 1) = 1026 M Ω Ad = (316)(1026 || 316) ⇒ Ad = 76,343 (b) 1 ⇒ f PD = 329 Hz 2π (316 || 1026) × 106 × 2 × 10−12 f PD ⋅ Ad = (329)(76,343) ⇒ 25.1 MHz f PD = TYU13.14 For Q7 and R1 VSG = VBE 7 + I1 R1 = 0.6 + I1 (5) For M 8 : I 2 = K p (VSG + VTP ) 2 I 2 = 0.3(VSG − 1.4) 2 By trial and error: 597. VSG = 2.54 V I1 = I 2 = 0.388 mA TYU13.15 For J 6 biased in the saturation region ⇒ I C 3 = I DSS = 300 μ A Q1 , Q2 , Q3 are matched ⇒ I C1 = I C 2 = I C 3 = 300 μ A 598. Chapter 13 Problem Solutions 13.1 Computer Simulation 13.2 Computer Simulation 13.3 ( Ad = g m1 ro 2 ro 4 Ri 6 (a) ) g m1 = I C1 20 = ⇒ 0.769 mA / V VT 0.026 ro 2 = VA 2 80 = = 4 MΩ I C 2 20 ro 4 = VA 4 80 = = 4 MΩ I C 2 20 Ri 6 = rπ 6 + (1 + β n ) ⎡ R1 rπ 7 ⎤ ⎣ ⎦ (120)(0.026) rπ 7 = = 15.6 k Ω 0.2 V (on) 0.6 = = 0.030 mA I C 6 ≅ BE 20 R1 (120)(0.026) = 104 k Ω 0.030 rπ 6 = Then Ri 6 = 104 + (121) ⎡ 20 15.6 ⎤ ⇒ 1.16 M Ω ⎣ ⎦ Then ( ) Ad = 769 4 4 1.16 ⇒ Ad = 565 Now ⎛ R1 Vo = − I c 7 ro 7 = −( β n I b 7 )ro 7 = − β n ro 7 ⎜ ⎝ R1 + rπ 7 ⎞ ⎟ Ic6 ⎠ ⎛ R1 ⎞ Vo1 = − β n (1 + β n )ro 7 ⎜ ⎟ I b 6 and I b 6 = R1 + rπ 7 ⎠ Ri 6 ⎝ Then − β n (1 + β n )ro 7 ⎛ R1 ⎞ V Av 2 = o = ⎜ ⎟ Vo1 Ri 6 ⎝ R1 + rπ 7 ⎠ ro 7 = VA 80 = = 400 k Ω I C 7 0.2 So Av 2 = −(120)(121)(400) ⎛ 20 ⎞ ⎜ ⎟ ⇒ Av 2 = −2813 1160 ⎝ 20 + 15.6 ⎠ Overall gain = Ad ⋅ Av 2 = (565)(−2813) ⇒ A = −1.59 ×106 (b) Rid = 2rπ 1 and rπ 1 = (80)(0.026) = 104 k Ω 0.020 Rid = 208 k Ω (c) f PD = 1 and CM = (10)(1 + 2813) = 28,140 pF 2π Req CM Req = ro 2 ro 4 Ri 6 = 4 4 1.16 = 0.734 M Ω f PD = 1 = 7.71 Hz 2π (0.734 × 10 )(28,140 × 10−12 ) 6 599. Gain-Bandwidth Product = (7.71)(1.59 × 106 ) ⇒ 12.3 MHz 13.4 a. b. Q3 acts as the protection device. Same as part (a). 13.5 If we assume VBE (on) = 0.7 V, then Vin = 0.7 + 0.7 + 50 + 5 So breakdown voltage ≈ 56.4 V. 13.6 (a) I REF = 15 − 0.6 − 0.6 − (−15) = 0.50 ⇒ R5 = 57.6 k Ω R5 ⎛I ⎞ I C10 R4 = VT ln ⎜ REF ⎟ ⎝ I C10 ⎠ 0.026 ⎛ 0.50 ⎞ R4 = ln ⎜ ⎟ ⇒ R4 = 2.44 k Ω 0.030 ⎝ 0.030 ⎠ 5 − 0.6 − 0.6 − (−5) ⇒ I REF = 0.153 mA 57.6 ⎛ 0.153 ⎞ I C10 (2.44) = (0.026) ln ⎜ ⎟ ⎝ I C10 ⎠ (b) I REF = By trial and error, I C10 ≅ 21.1 μ A 13.7 (a) I REF ≅ 0.50 mA ⎛I ⎞ ⎛ 0.50 × 10−3 ⎞ VBE = VT ln ⎜ REF ⎟ = (0.026) ln ⎜ ⎟ ⇒ VBE11 = 0.641V = VEB12 −14 ⎝ 10 ⎠ ⎝ IS ⎠ Then 15 − 0.641 − 0.641 − (−15) ⇒ R5 = 57.4 k Ω R5 = 0.50 0.026 ⎛ 0.50 ⎞ ln ⎜ R4 = ⎟ ⇒ R4 = 2.44 k Ω 0.030 ⎝ 0.030 ⎠ ⎛ 0.030 × 10−3 ⎞ VBE10 = 0.026 ln ⎜ ⎟ ⇒ VBE10 = 0.567 V −14 ⎝ 10 ⎠ (b) From Problem 13.6, I REF ≅ 0.15 mA ⎛ 0.15 × 10−3 ⎞ VBE11 = VEB12 = 0.026 ln ⎜ ⎟ = 0.609 V −14 ⎝ 10 ⎠ 5 − 0.609 − 0.609 − (−5) ⇒ I REF = 0.153 mA Then I REF = 57.4 Then I C10 ≅ 21.1 μ A from Problem 13.6 13.8 5 − 0.6 − 0.6 − (−5) ⇒ I REF = 0.22 mA 40 ⎛I ⎞ I C10 R4 = VT ln ⎜ REF ⎟ ⎝ I C10 ⎠ a. I REF = ⎛ 0.22 ⎞ I C10 (5) = (0.026) ln ⎜ ⎟ ⎝ I C10 ⎠ 600. By trial and error; I C10 ≅ 14.2 μ A I C10 ⇒ I C 6 = 7.10 μ A 2 = 0.75 I REF ⇒ I C17 = 0.165 mA IC 6 ≅ I C17 I C13 A = 0.25I REF ⇒ I C13 A = 0.055 mA (b) Using Example 13.4 rπ 17 = 31.5 kΩ ′ RE = 50 [31.5 + (201)(0.1)] = 50 51.6 = 25.4 kΩ rπ 16 = β nVT I C16 and 0.165 (0.165)(0.1) + 0.6 + = 0.0132 mA 200 50 rπ 16 = 394 kΩ Then Ri 2 = 394 + (201)(25.4) ⇒ 5.5 MΩ rπ 6 = 732 kΩ 0.00710 gm6 = = 0.273 mA / V 0.026 50 r06 = = 7.04 MΩ 0.0071 Then Ract1 = 7.04[1 + (0.273)(1 732)] = 8.96 MΩ 50 r04 = = 7.04 MΩ 0.0071 Then ⎛ 7.1 ⎞ Ad = − ⎜ ⎟ (7.04 8.96 5.5) ⎝ 0.026 ⎠ or Ad = −627 Gain of differential amp stage I C16 = Using Example 13.5, and neglecting the input resistance to the output stage: V 50 Ract 2 = A = = 303 kΩ I C13 B 0.165 Av 2 = −(200)(201)(50)(303) (303) (5500)[50 + 31.5 + (201)(0.1)] or Av 2 = −545 Gain of second stage 13.9 I C10 = 19 μ A From Equation (13.6) ⎡ β 2 + 2β P + 2 ⎤ ⎡ (10) 2 + 2(10) + 2 ⎤ = 2I ⎢ I C10 = 2 I ⎢ P ⎥ ⎥ 2 2 ⎣ (10) + 3(10) + 2 ⎦ ⎣ β P + 3β P + 2 ⎦ ⎡122 ⎤ = 2I ⎢ ⎥ ⎣132 ⎦ So ⎛ 132 ⎞ 2 I = (19) ⎜ ⎟ = 20.56 μ A ⎝ 122 ⎠ I C 2 = I = 10.28 μ A 601. IC 9 = I B9 2I = 20.56 ⇒ I C 9 = 17.13 μ A 2⎞ ⎛ 1+ ⎟ ⎜ ⎝ 10 ⎠ ⎛ 2 ⎞ ⎜1 + ⎟ βP ⎠ ⎝ I 17.13 = C9 = ⇒ I B 9 = 1.713 μ A βP 10 IB4 = 10.28 I = ⇒ I B 4 = 0.9345 μ A (1 + β P ) 11 ⎛ β IC 4 = I ⎜ P ⎝1+ βP ⎞ ⎛ 10 ⎞ ⎟ = (10.28) ⎜ ⎟ ⇒ I C 4 = 9.345 μ A ⎝ 11 ⎠ ⎠ 13.10 VB 5 − V − = VBE (on) + I C 5 (1) = 0.6 + (0.0095)(1) = 0.6095 0.6095 IC 7 = ⇒ I C 7 = 12.2 μ A 50 I C 8 = I C 9 = 19 μ A I REF = 0.72 mA I E13 = I REF = 0.72 mA I C14 = 138 μ A Power = (V + − V − ) [ I C 7 + I C 8 + I C 9 + I REF + I E13 + I C14 ] = 30[0.0122 + 0.019 + 0.019 + 0.72 + 0.72 + 0.138] ⇒ Power = 48.8 mW Current supplied by V + and V − = I C 7 + I C 8 + I C 9 + I REF + I E13 + I C14 = 1.63 mA 13.11 (a) vcm (min) = −15 + 0.6 + 0.6 + 0.6 + 0.6 = −12.6 V vcm (max) = +15 − .6 = 14.4 V So − 12.6 ≤ vcm ≤ 14.4 V (b) vcm (min) = −5 + 4(0.6) = −2.6 V vcm (max) = 5 − 0.6 = 4.4 V So − 2.6 ≤ vcm ≤ 4.4 V 13.12 If v0 = V − = −15 V , the base voltage of Q14 is pulled low, and Q18 and Q19 are effectively cut off. As a first approximation 0.6 I C14 = = 22.2 mA 0.027 22.2 I B14 = = 0.111 mA 200 Then I C15 = I C13 A − I B14 = 0.18 − 0.111 = 0.069 mA Now ⎛I ⎞ VBE15 = VT ln ⎜ C15 ⎟ ⎝ 15 ⎠ ⎛ 0.069 × 10−3 ⎞ = (0.026) ln ⎜ ⎟ −14 ⎝ 10 ⎠ = 0.589 V 602. As a second approximation 0.589 I C14 = ⇒ I C14 = 21.8 mA 0.027 21.8 I B14 = = 0.109 mA 200 and I C15 = 0.18 − 0.109 ⇒ I C15 = 0.071 mA 13.13 a. Neglecting base currents: I D = I BIAS Then ⎛I ⎞ VBB = 2VD = 2VT ln ⎜ D ⎟ ⎝ IS ⎠ ⎛ 0.25 × 10−3 ⎞ = 2(0.026) ln ⎜ −14 ⎟ ⎝ 2 × 10 ⎠ or VBB = 1.2089 V ⎛V / 2⎞ I CN = I CP = I S exp ⎜ BB ⎟ ⎝ VT ⎠ ⎛ 1.2089 ⎞ = 5 × 10−14 exp ⎜ ⎟ ⎝ 2(0.026) ⎠ So I CN = I CP = 0.625 mA For vI = 5 V, v0 ≅ 5 V b. 5 = 1.25 mA 4 As a first approximation I CN ≈ iL = 1.25 mA iL ≅ ⎛ 1.25 × 10−3 ⎞ VBEN = (0.026) ln ⎜ = 0.6225 V −14 ⎟ ⎝ 5 × 10 ⎠ Neglecting base currents, VBB = 1.2089 V Then VEBP = 1.2089 − 0.6225 = 0.5864 V ⎛ 0.5864 ⎞ I CP = 5 × 10−14 exp ⎜ ⎟ ⇒ I CP = 0.312 mA ⎝ 0.026 ⎠ As a second approximation, I CN = iL + I CP = 1.25 + 0.31 ⇒ I CN ≅ 1.56 mA 13.14 R1 + R2 = VBB 1.157 = = 64.28 kΩ (0.1) I BIAS 0.018 ⎛I VBE = VT ln ⎜ C ⎝ IS ⎞ ⎛ (0.9) I BIAS ⎞ ⎟ = (0.026) ln ⎜ ⎟ IS ⎠ ⎝ ⎠ ⎛ 0.162 × 10−3 ⎞ = (0.026) ln ⎜ ⎟ −14 ⎝ 10 ⎠ 603. VBE = 0.6112 V ⎛ R2 ⎞ VBE = ⎜ ⎟ VBB ⎝ R1 + R2 ⎠ ⎛ R ⎞ 0.6112 = ⎜ 2 ⎟ (1.157) ⎝ 64.28 ⎠ So R2 = 33.96 kΩ Then R1 = 30.32 kΩ 13.15 ( Ad = − g m ro 4 ro 6 Ri 2 (a) ) From example 13.4 9.5 gm = = 365 μ A / V , ro 4 = 5.26 M Ω 0.026 Now ro 6 = ro 4 = 5.26 M Ω Assuming R8 = 0, we find ′ Ri 2 = rπ 16 + (1 + β n ) RE = 329 + (201) ( 50 9.63) ⇒ 1.95 M Ω Then ( ) Ad = −(365) 5.26 5.26 1.95 ⇒ Ad = −409 (b) Av 2 = From Equation (13.20), ( − β n (1 + β n ) R9 Ract 2 Ri 3 R017 { } ) Ri 2 R9 + ⎡ rπ 17 + (1 + β n ) Rg ⎤ ⎣ ⎦ For Rg = 0, Ri 2 = 1.95 M Ω Using the results of Example 13.5 Av 2 = ( −200(201)(50) 92.6 4050 92.6 (1950){50 + 9.63} )⇒A v2 = −792 13.16 Let I C10 = 40 μ A, then I C1 = I C 2 = 20 μ A. Using Example 13.5, Ri 2 = 4.07 MΩ (200)(0.026) = 260 kΩ 0.020 0.020 gm6 = = 0.769 mA/V 0.026 50 r06 = ⇒ 2.5 MΩ 0.02 Then Ract1 = 2.5[1 + (0.769)(1 260)] = 4.42 MΩ 50 ⇒ 2.5 MΩ r06 = 0.02 Then rπ 6 = 604. ⎛ I CQ ⎞ Ad = − ⎜ ⎟ (r04 Ract1 Ri 2 ) ⎝ VT ⎠ ⎛ 20 ⎞ = −⎜ ⎟ (2.5 4.42 4.07) ⎝ 0.026 ⎠ So Ad = −882 13.17 From Problem 13.8 I1 = I 2 = 7.10 μ A, I C17 = 0.165 mA, I C13 A = 0.055 mA I C16 ≈ I B17 + I E17 R8 + VBE17 R9 = 0.165 (0.165)(0.1) + 0.6 + 200 50 = 0.000825 + 0.01233 I C16 = 0.0132 mA (200)(0.026) = 31.5 K 0.165 1 RE = R9 [ rπ 17 + (1 + β ) R8 ] = 50 [31.5 + (201)(0.1)] rπ 17 = = 50 51.6 = 25.4 K rπ 16 = (200)(0.026) = 394 K 0.0132 Then 1 Ri 2 = rπ 16 + (1 + β ) RE = 394 + (201)(25.4) ⇒ 5.50 MΩ Now (200)(0.026) rπ 6 = = 732 K 0.0071 0.0071 gm6 = = 0.273 mA/V 0.026 50 ro 6 = ⇒ 7.04 MΩ 0.0071 Ract1 = ro 6 [1 + g m 6 ( R rπ 6 )] = 7.04[1 + (0.273)(1 732)] = 8.96 MΩ 50 ro 4 = ⇒ 7.04 MΩ 0.0071 Then Ad = − g m1 (ro 4 Ract1 Ri 2 ) ⎛ 7.10 ⎞ = −⎜ ⎟ (7.04 8.96 5.5) ⎝ 0.026 ⎠ Ad = −627 50 50 ⇒ 303 K Ro17 = = 303 K 0.165 0.165 From Eq. (13.20), assuming Ri 3 → ∞ Now Ract 2 = Av 2 ≅ − = β (1 + β ) R9 ( Ract 2 R017 ) Ri 2 { R9 + [rπ 17 + (1 + β ) R8 ]} −(200)(201)(50)(303 303) (5500)[50 + 31.5 + (201)(0.1)] Av 2 = −545 = −3.045 × 108 5.588 × 105 Overall gain Av = (−627)(−545) = 341, 715 605. 13.18 Using results from 13.17 ⎛ 100 ⎞ Ri 2 = 5.50 MΩ, Ract1 ⎜ ⎟ [1 + (0.273)(1 732)] ⇒ 17.93 MΩ ⎝ 0.0071 ⎠ 100 ⇒ 14.08 MΩ ro 4 = 0.0071 ⎛ 7.10 ⎞ Ad = − ⎜ ⎟ (14.08 17.93 5.50) ⎝ 0.026 ⎠ Ad = −885 Now 100 100 Ract 2 = = 606 K Ro17 = = 606 K 0.165 0.165 −(200)(201)(50)(606 606) −6.09 × 108 Av 2 = = (5500)[50 + 31.5 + (201)(0.1)] 5.588 × 105 Av 2 = −1090 Overall gain Av = (−885)(−1090) = 964, 650 13.19 Now Re14 = rπ 14 + R01 and R0 = R6 + Re14 1+ βP Assume series resistance of Q18 and Q19 is small. Then R01 = r013 A Re 22 where Re 22 = rπ 22 + R017 r013 B 1+ βP and R017 = r017 [1 + g m17 ( R8 rπ 17 )] Using results from Example 13.6, rπ 17 = 9.63 kΩ rπ 22 = 7.22 kΩ g m17 = 20.8 mA/V r017 = 92.6 kΩ Then R017 = 92.6[1 + (20.8)(0.1 9.63)] = 283 kΩ 50 r013 B = = 92.6 kΩ 0.54 Then 7.22 + 283 92.6 = 1.51 kΩ Re 22 = 51 R01 = r013 A Re 22 = 278 1.51 = 1.50 kΩ (50)(0.026) = 0.65 kΩ rπ 14 = 2 Then 0.65 + 1.50 Re14 = = 0.0422 kΩ 51 or Re14 = 42.2 Ω Then R0 = 42.2 + 27 ⇒ R0 = 69.2 Ω 13.20 606. ⎡ ⎛ r Rid = 2 ⎢ rπ 1 + (1 + β n ) ⎜ π 3 ⎝ 1+ βP ⎣ β n = 200, β P = 10 (a) I C1 = 9.5 μ A ⎞⎤ ⎟⎥ ⎠⎦ (200)(0.026) = 547 K 0.0095 (10)(0.026) rπ 3 = = 27.4 K 0.0095 Then (201)(27.4) ⎤ ⎡ Rid = 2 ⎢547 + ⎥ 11 ⎣ ⎦ Rid ⇒ 2.095 MΩ rπ 1 = (b) I C1 = 7.10 μ A (200)(0.026) rπ 1 = = 732 K 0.0071 (10)(0.026) rπ 3 = = 36.6 K 0.0071 (201)(36.6) ⎤ ⎡ Rid = 2 ⎢ 732 + ⎥ 11 ⎣ ⎦ Rid ⇒ 2.80 MΩ 13.21 We can write A0 ⎛ f ⎞⎛ f ⎞ ⎜1 + j ⎟⎜ 1 + j ⎟ f PD ⎠⎝ f1 ⎠ ⎝ 181, 260 = f ⎞⎛ f ⎞ ⎛ ⎜1 + j ⎟ ⎜1 + j ⎟ 10.7 ⎠ ⎝ f1 ⎠ ⎝ Phase: ⎛ f ⎞ −1 ⎛ f ⎞ φ = − tan −1 ⎜ ⎟ − tan ⎜ ⎟ ⎝ 10.7 ⎠ ⎝ f1 ⎠ A( f ) = For a Phase margin = 70°, φ = −110° So ⎛ f ⎞ −1 ⎛ f ⎞ −110° = − tan −1 ⎜ ⎟ − tan ⎜ ⎟ ⎝ 10.7 ⎠ ⎝ f1 ⎠ Assuming f 10.7, we have ⎛ f ⎞ f tan −1 ⎜ ⎟ = 20° ⇒ = 0.364 f1 ⎝ f1 ⎠ At this frequency, A( f ) = 1, so 607. 1= = 181, 260 2 ⎛ f ⎞ 2 1+ ⎜ ⎟ ⋅ 1 + (0.364) ⎝ 10.7 ⎠ 170,327 2 ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ 10.7 ⎠ f = 170,327 ⇒ f = 1.82 MHz or 10.7 Then, second pole at f f1 = ⇒ f1 = 5 MHz 0.364 13.22 a. Original g m1 and g m 2 ⎛W ⎞⎛ μ C ⎞ K p1 = K p 2 = ⎜ ⎟⎜ P ox ⎟ = (12.5)(10) ⎝ L ⎠⎝ 2 ⎠ = 125 μ A / V 2 So ⎛ IQ ⎞ g m1 = g m 2 = 2 K p1 ⎜ ⎟ ⎝ 2⎠ = 2 (0.125)(10) = 0.09975 mA/V ⎛W ⎞ If ⎜ ⎟ is increased to 50, then ⎝L⎠ K p1 = K p 2 = (50)(10) = 500 μ A / V 2 So g m1 = g m 2 = 2 (0.5)(0.0199) = 0.1995 mA/V b. Gain of first stage Ad = g m1 (r02 r04 ) = (0.1995)(5025 5025) or Ad = 501 Voltage gain of second stage remains the same, or Av 2 = 251 Then Av = Ad ⋅ Av 2 = (501)(251) or Ad = 125, 751 13.24 a. K p = (10)(20) = 200 μ A / V 2 = 0.2 mA / V 2 I REF = I SET 10 − VSG − (−10) 200 = k P (VSG − 1.5) 2 = 2 20 − VSG = (0.2)(200)(VSG − 3VSG + 2.25) 2 40VSG − 119VSG + 70 = 0 VSG = Then 119 ± (119) 2 − 4(40)(70) ⇒ VSG = 2.17 V 2(40) 608. 20 − 2.17 ⇒ I REF = 89.2 μ A 200 M 5 , M 6 , M 8 matched transistors so that I REF = I Q = I D 7 = I REF = 89.2 μ A b. Small-signal voltage gain of input stage: Ad = 2 K p1 I Q ⋅ ( ro 2 ro 4 ) 1 = 1.12 MΩ ⎛ 89.2 ⎞ (0.02) ⎜ ⎟ ⎝ 2 ⎠ 1 1 r04 = = = 2.24 MΩ λn I D ⎛ 89.2 ⎞ (0.01) ⎜ ⎟ ⎝ 2 ⎠ Then Ad = 2(200)(89.2) ⋅ (1.12 2.24) r02 = 1 = λP I D or Ad = 141 Small-signal voltage gain of second stage: Av 2 = g m 7 (r07 r08 ) K n 7 = (20)(20) = 400 μ A / V 2 So g m 7 = 2 K n 7 I D 7 = 2 (0.4)(0.0892) = 0.378 mA/V 1 1 r08 = = = 561 kΩ λP I D 7 (0.02)(0.0892) r07 = 1 λn I D 7 = 1 = 1121 kΩ (0.01)(0.0892) So Av 2 = (0.378)(1121 561) ⇒ Av 2 = 141 Then overall voltage gain Av = Ad ⋅ Av 2 = (141)(141) ⇒ Av = 19,881 13.25 Small-signal voltage gain of input stage: Ad = 2 K p1 I Q ⋅ ( ro 2 ro 4 ) K p1 = (10)(10) = 100 μ A / V 2 1 1 = = 1000 kΩ IQ ⎞ ⎛ ⎛ 0.2 ⎞ λP ⎜ ⎟ (0.01) ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2⎠ 1 1 r04 = = = 2000 kΩ ⎛ IQ ⎞ ⎛ 0.2 ⎞ (0.005) ⎜ λn ⎜ ⎟ ⎟ ⎝ 2 ⎠ ⎝ 2⎠ r02 = Then Ad = 2(0.1)(0.2) ⋅ (1000 2000) or Ad = 133 Small-signal voltage gain of second stage: Av 2 = g m 7 ( r07 r08 ) K n 7 = (20)(20) = 400 μ A / V 2 So 609. g m 7 = 2 K n 7 I D 7 = 2 (0.4)(0.2) = 0.566 mA/V 1 1 r08 = = = 500 kΩ λP I D 7 (0.01)(0.2) r07 = 1 λn I D 7 = 1 = 1000 kΩ (0.005)(0.2) So Av 2 = (0.566)(1000 500) ⇒ Av 2 = 189 Then overall voltage gain is Av = Ad ⋅ Av 2 = (133)(189) ⇒ Av = 25,137 13.26 f PD = 1 2π Req Ci where Req = r04 r02 and Ci = C1 (1 + Av 2 ) We can find that Av 2 = 251 and r04 = r02 = 5.025 MΩ Now Req = 5.025 5.025 = 2.51 MΩ and Ci = 12(1 + 251) = 3024 pF So 1 f PD = 2π (2.51× 106 )(3024 × 10−12 ) or f PD = 21.0 Hz 13.27 f PD = 1 2π Req Ci where Req = r04 r02 From Problem 13.22, r02 = 1.12 MΩ, r04 = 2.24 MΩ and Av 2 = 141 So 1 8= 2π (1.12 2.24) × 106 × Ci or Ci = 2.66 × 10−8 = C1 (1 + Av 2 ) = C1 (142) or C1 = 188 pF 13.28 R0 = r07 r08 We can find that r07 = r08 = 2.52 MΩ Then R0 = 2.52 2.52 or R0 = 1.26 MΩ 610. 13.29 a. V0 = ( g m1Vgs1 )(r01 r02 ) VI = Vgs1 + V0 Then V0 = g m1 (r01 r02 )(VI − V0 ) or Av = g m1 (r01 r02 ) 1 + g m1 (r01 r02 ) b. I X + g m1Vgs1 = VX VX and Vgs1 = −VX + r02 r01 R0 = 1 r r g m1 01 02 13.30 (a) (b) 2 ⎛ 80 ⎞ I Q 2 = ⎜ ⎟ (20) [1.1737 − 0.7 ] ⎝ 2⎠ I Q 2 = 180 μ A 2 ⎛ 80 ⎞ I D 6 = ⎜ ⎟ (25) (VGS 6 − 0.7 ) = 25 ⇒ VGS 6 = 0.8581 V ⎝ 2⎠ 2 ⎛ 40 ⎞ I D 7 = ⎜ ⎟ (50) (VSG 7 − 0.7 ) = 25 ⇒ VSG 7 = 0.8581 V 2 ⎠ ⎝ Set VSG 8 P = VGS 8 N = 0.8581 V ⎛ 40 ⎞ ⎛ W ⎞ ⎛W ⎞ 180 = ⎜ ⎟ ⎜ ⎟ (0.8581 − 0.7) 2 ⇒ ⎜ ⎟ = 360 ⎝ 2 ⎠ ⎝ L ⎠8 P ⎝ L ⎠8 P ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ 180 = ⎜ ⎟ ⎜ ⎟ (0.8581 − 0.7) 2 ⇒ ⎜ ⎟ = 180 ⎝ 2 ⎠ ⎝ L ⎠8 N ⎝ L ⎠8 N 13.31 611. VGS11 ⇒ 2 ⎛ 80 ⎞ 200 = ⎜ ⎟ (20) (VGS 11 − 0.7 ) ⎝ 2⎠ VGS 11 = 1.20 V Let M 12 = 2 transistors in series. Than 5 − 1.20 = 1.90 V 2 2 ⎛ 80 ⎞⎛ W ⎞ ⎛W ⎞ ⎛W ⎞ 200 = ⎜ ⎟⎜ ⎟ (1.90 − 0.7 ) ⇒ ⎜ ⎟ = ⎜ ⎟ = 3.47 L ⎠12 A ⎝ L ⎠12 B 2 ⎠⎝ L ⎠12 ⎝ ⎝ VGS12 = 13.32 (a) 2 ⎛ 80 ⎞ I Q 2 = 250μ A = ⎜ ⎟ (5) (VGS 8 − 0.7 ) ⎝ 2⎠ ⇒ VGS 8 = 1.818 V 1.818 ⇒ VGS 6 = VSG 7 = = 0.909 V 2 ⎛ 80 ⎞ I D 6 = I D 7 = ⎜ ⎟ (25)(0.909 − 0.7) 2 = 43.7 μ A ⎝ 2⎠ (b) ⎛ 80 ⎞ ⎛ 250 ⎞ g m1 = 2 ⎜ ⎟ (15) ⎜ ⎟ ⇒ 0.5477 mA/V ⎝ 2⎠ ⎝ 2 ⎠ 1 ro 2 = = 800 K ( 0.01)( 0.125) r04 = 1 = 533.3K ( 0.015)( 0.125) Ad 1 = g m1 ( ro 2 ro 4 ) = ( 0.5477 ) ( 800 Ad 1 = 175 Second stage: 533.3) 612. A2 = − g m 5 (ro 5 ro 9 ) ⎛ 40 ⎞ g m 5 = 2 ⎜ ⎟ (80)(250) ⇒ 1.265 mA/V ⎝ 2 ⎠ 1 r05 = = 266.7 K (0.015)(0.25) 1 r09 = = 400 K (0.01)(0.25) A2 = −(1.265)(266.7 400) A2 = −202 Assume the gain of the output stage ≈ 1, then Av = Ad 1 ⋅ A2 = (175)(−202) Av = −35,350 13.33 (a) Ad = g m1 ( Ro 6 Ro8 ) g m1 = 2 K n I DQ = 2 (0.5)(0.025) ⇒ 224 μ A / V g m1 = g m8 g m 6 = 2 (0.5)(0.025) ⇒ 224 μ A / V ro1 = ro 6 = ro8 = ro10 = ro 4 = 1 1 = = 2.67 M Ω λ I DQ (0.015)(25) 1 1 = ⇒ 1.33 M Ω λ I D 4 ( 0.015 )( 50 ) Now Ro8 = g m8 (ro8 ro10 ) = (224)(2.67)(2.67) = 1597 M Ω Ro 6 = g m 6 (ro 6 )(ro 4 ro1 ) = (224)(2.67)(2.67 1.33) ⇒ Ro 6 = 531 M Ω Then Ad = (224)(531 1597) ⇒ Ad = 89, 264 (b) Ro = Ro 6 Ro8 = 531 1597 ⇒ Ro = 398 M Ω (c) f PD = 1 1 = ⇒ f PD = 80 Hz 2π Ro CL 2π ( 398 × 106 )( 5 × 10−12 ) GBW = (89, 264)(80) ⇒ GBW = 7.14 MHz 13.34 (a) ro1 = ro8 = ro10 = ro 6 = ro 4 = 1 λn I D = 1 1 = = 2 MΩ λ p I D (0.02)(25) 1 = 2.67 M Ω (0.015)(25) 1 1 = = 1.33 M Ω λn I D 4 (0.015)(50) ⎛ 35 ⎞ ⎛ W ⎞ ⎛W ⎞ g m1 = 2 ⎜ ⎟ ⎜ ⎟ (25) = 41.8 ⎜ ⎟ = g m8 ⎝ 2 ⎠ ⎝ L ⎠1 ⎝ L ⎠1 ⎛ 80 ⎞⎛ W ⎞ ⎛W ⎞ g m 6 = 2 ⎜ ⎟⎜ ⎟ (25) = 63.2 ⎜ ⎟ ⎝ 2 ⎠⎝ L ⎠6 ⎝ L ⎠6 Ro = Ro 6 Ro8 = [ g m 6 (ro 6 )(ro 4 ro1 )] [ g m8 (ro8 ro10 )] 613. ⎛W ⎞ ⎛W ⎞ Define X 1 = ⎜ ⎟ and X 6 = ⎜ ⎟ L ⎠1 ⎝ ⎝ L ⎠6 Then ⎡ ⎤ ⎣ Ro = ⎣ 63.2 X 6 ( 2.67 ) (1.33 2 ) ⎦ ⎡ 41.8 X 1 ( 2 )( 2 ) ⎤ ⎦ 22,539 X 1 X 6 = 134.8 X 6 167.2 X 1 = 134.8 X 6 + 167.2 X 1 Ad = g m1 Ro ⎛ 22,539 X 1 X 6 ⎞ = (41.8 X 1 ) ⎜ ⎟ 134.8 X 6 + 167.2 X 1 ⎠ ⎝ = 10, 000 1 ⎛W ⎞ ⎛W ⎞ Now X 6 = ⎜ ⎟ = ⎜ ⎟ = 0.674 X 1 L ⎠6 2.2 ⎝ L ⎠1 ⎝ We then find ⎛W ⎞ ⎛W ⎞ X 12 = ⎜ ⎟ = 4.06 = ⎜ ⎟ ⎝ L ⎠1 ⎝ L ⎠p and ⎛W ⎞ ⎜ ⎟ = 1.85 ⎝ L ⎠n 13.35 Let V + = 5V , V − = −5V P = IT (10) = 3 ⇒ IT = 0.3 mA ⇒ I REF = 0.1 mA = 100 μ A 1 ro1 = ro8 = ro10 = = 1 MΩ (0.02)(50) 1 ro 6 = = 1.33 MΩ (0.015)(50) 1 ro 4 = = 0.667 M Ω (0.015)(100) ⎛ 35 ⎞ ⎛ W ⎞ g m1 = 2 ⎜ ⎟ ⎜ ⎟ (50) = 59.2 X 1 = g m8 ⎝ 2 ⎠ ⎝ L ⎠1 ⎛W ⎞ where X 1 = ⎜ ⎟ ⎝ L ⎠1 Assume all width-to-length ratios are the same. ⎛ 80 ⎞ ⎛ W ⎞ g m 6 = 2 ⎜ ⎟ ⎜ ⎟ (50) = 89.4 X 1 ⎝ 2 ⎠⎝ L ⎠ Now Ro = Ro 6 Ro8 = ⎡ g m 6 ( ro 6 ) ( ro 4 ro1 ) ⎤ ⎡ g m8 ( ro8 ro10 ) ⎤ ⎦ ⎣ ⎦ ⎣ = ⎡89.4 X 1 (1.33) ( 0.667 1) ⎤ ⎡59.2 X 1 (1)(1) ⎤ ⎦ ⎣ ⎦ ⎣ ( 47.6 X 1 )( 59.2 X 1 ) = [ 47.6 X 1 ] [59.2 X 1 ] = 47.6 X 1 + 59.2 X 1 So Ro = 26.4 X 1 Now Ad = g m1 Ro = ( 59.2 X 1 )( 26.4 X 1 ) = 25, 000 So that X 12 = 13.36 W = 16 for all transistors L 614. (a) Ad = Bg m1 (ro 6 ro8 ) ro 6 = ro8 = 1 1 = = 0.741 M Ω λ I DQ (0.015)(90) ⎛ k ′ ⎞⎛ W ⎞ g m1 = 2 ⎜ n ⎟ ⎜ ⎟ I D1 = 2 (500)(30) = 245 μ A / V ⎝ 2 ⎠⎝ L ⎠ Ad = (3)(245)(0.741 0.741) ⇒ Ad = 272 (b) Ro = ro 6 ro8 = 0.741 0.741 ⇒ Ro = 371 k Ω (c) f PD = 1 1 = ⇒ f PD = 85.8 kHz 2π Ro C 2π (371× 103 )(5 × 10−12 ) GBW = (272)(85.8 × 103 ) ⇒ GBW = 23.3 MHz 13.37 (a) 1 = 0.5 M Ω (0.02)(2.5)(40) 1 ro8 = = 0.667 M Ω (0.015)(2.5)(40) Ad = Bg m1 ( ro 6 ro8 ) ro 6 = 400 = (2.5) g m1 ( 0.5 0.667 ) ⇒ g m1 = 560 μ A / V ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ g m1 = 560 = 2 ⎜ ⎟ ⎜ ⎟ (40) ⇒ ⎜ ⎟ = 49 ⎝ 2 ⎠⎝ L ⎠ ⎝L⎠ Assume all (W/L) ratios are the same except for ⎛W ⎞ ⎛W ⎞ M 5 and M 6 . ⎜ ⎟ = ⎜ ⎟ = 122.5 ⎝ L ⎠5 ⎝ L ⎠ 6 (b) Assume the bias voltages are V + = 5V , V − = −5V . ⎛W ⎞ ⎛W ⎞ Assume ⎜ ⎟ = ⎜ ⎟ = 49 ⎝ L ⎠ A ⎝ L ⎠B ⎛ 80 ⎞ I Q = ⎜ ⎟ (49)(VGSA − 0.5) 2 = 80 ⇒ VGSA = 0.702 V ⎝ 2⎠ Then ⎛ 80 ⎞ ⎛ W ⎞ I REF = 80 = ⎜ ⎟ ⎜ ⎟ (VGSC − 0.5) 2 ⎝ 2 ⎠ ⎝ L ⎠C For four transistors 615. 10 − 0.702 = 2.325 V 4 ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ 80 = ⎜ ⎟ ⎜ ⎟ (2.325 − 0.5) 2 ⇒ ⎜ ⎟ = 0.60 ⎝ 2 ⎠ ⎝ L ⎠C ⎝ L ⎠C VGSC = (c) f 3− dB = 1 2π Ro C Ro = 0.5 0.667 = 0.286 M Ω 1 = 185 kHz 2π (286 × 103 )(3 × 10−12 ) GBW = (400)(185 × 103 ) ⇒ 74 MHz f 3− dB = 13.38 (a) From previous results, we can write Ro10 = g m10 (ro10 ro 6 ) Ro12 = g m12 (ro12 ro8 ) Ad = Bg m1 ( Ro10 Ro12 ) Now ro10 = ro 6 = 1 1 = = 0.5 M Ω (0.02)(2.5)(40) λP B ( I Q / 2 ) ro12 = ro8 = 1 1 = = 0.667 M Ω (0.015)(2.5)(40) λn B ( I Q / 2 ) Assume all transistors have the same width-to-length ratios except for M 5 and M 6 . ⎛W Let ⎜ ⎝L Then ⎞ 2 ⎟= X ⎠ ⎛ k′ ⎞⎛ W ⎞ ⎛ 35 ⎞ p g m10 = 2 ⎜ ⎟ ⎜ ⎟ ( I DQ10 ) = 2 ⎜ ⎟ X 2 (2.5)(40) ⎝ 2⎠ ⎝ 2 ⎠ ⎝ L ⎠10 = 83.67 X ⎛ k′ ⎞⎛ W ⎞ ⎛ 80 ⎞ g m12 = 2 ⎜ n ⎟ ⎜ ⎟ ( I DQ12 ) = 2 ⎜ ⎟ X 2 (2.5)(40) ⎝ 2⎠ ⎝ 2 ⎠ ⎝ L ⎠12 = 126.5 X ⎛ 80 ⎞ g m1 = 2 ⎜ ⎟ X 2 (40) = 80 X ⎝ 2⎠ Then Ro10 = (83.67 X )(0.5)(0.5) = 20.9 X M Ω Ro12 = (126.5 X )(0.667)(0.667) = 56.3 X M Ω We want 20, 000 = (2.5)(80 X )[20.9 X 56.3 X ] ⎡ (20.9 X )(56.3 X ) ⎤ 2 = 200 X ⎢ ⎥ = 3048 X ⎣ 20.9 X + 56.3 X ⎦ Then ⎛W ⎞ X 2 = 6.56 = ⎜ ⎟ ⎝L⎠ Then ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = ⎜ ⎟ = (2.5)(6.56) = 16.4 ⎝ L ⎠ 6 ⎝ L ⎠5 (b) Assume bias voltages are V + = 5V , V − = −5V 616. ⎛W ⎞ ⎛W ⎞ Assume ⎜ ⎟ = ⎜ ⎟ = 6.56 ⎝ L ⎠ A ⎝ L ⎠B ⎛ 80 ⎞ I Q = 80 = ⎜ ⎟ (6.56)(VGSA − 0.5) 2 ⇒ VGSA = 1.052 V ⎝ 2⎠ Need 5 transistors in series 10 − 1.052 VGSC = = 1.79 V 5 Then ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ I REF = 80 = ⎜ ⎟ ⎜ ⎟ (1.79 − 0.5) 2 ⇒ ⎜ ⎟ = 1.20 ⎝ 2 ⎠ ⎝ L ⎠C ⎝ L ⎠C (c) f 3− dB = 1 where Ro = Ro10 Ro12 2π Ro C Now Ro10 = 20.9 6.56 = 53.5 M Ω Ro12 = 56.3 6.56 = 144 M Ω Then Ro = 53.5 144 = 39 M Ω 1 = 1.36 kHz 2π (39 × 106 )(3 × 10−12 ) GBW = (20, 000)(1.36 x103 ) ⇒ GBW = 27.2 MHz f 3− dB = 13.39 Ad = g m ( M 2 ) ⋅ ⎡ ro 2 ( M 2 ) ro 2 (Q2 ) ⎤ ⎣ ⎦ ⎛ 40 ⎞ g m ( M 2 ) = 2 ⎜ ⎟ (25)(100) = 447 μ A / V ⎝ 2 ⎠ 1 1 ro 2 ( M 2 ) = = = 500 k Ω λ I DQ (0.02)(0.1) ro 2 (Q2 ) = VA 120 = = 1200 k Ω I CQ 0.1 Then Ad = 447(0.5 1.2) ⇒ Ad = 158 13.40 617. Ad = g m ( M 2 ) ⋅ ⎡ ro 2 ( M 2 ) ro 2 (Q2 ) ⎤ ⎣ ⎦ ⎛ 80 ⎞ g m ( M 2 ) = 2 ⎜ ⎟ (25)(100) = 632 μ A / V ⎝ 2⎠ 1 1 ro 2 ( M 2 ) = = = 667 k Ω λ I DQ (0.015)(0.1) VA 80 = = 800 k Ω I CQ 0.1 ro 2 (Q2 ) = Ad = (632) ( 0.667 0.80 ) ⇒ Ad = 230 13.41 (a) I REF = 200 μ A Ad = g m1 ( Ro 6 Ro8 ) K n = K p = 0.5 mA / V 2 λn = λ p = 0.015 V −1 where Ro8 = g m8 (ro8 ro10 ) Ro 6 = g m 6 (ro 6 ) ( ro 4 ro1 ) Now g m8 = 2 K P I D 8 = 2 (0.5)(0.1) = 0.447 mA/V 1 1 ro8 = = = 667 k Ω λP I D 8 (0.015)(0.1) ro10 = 1 = 667 k Ω λP I D 8 gm6 = IC 6 0.1 = = 3.846 mA/V 0.026 VT ro 6 = VA 80 = = 800 k Ω I C 6 0.1 ro 4 = 1 1 = = 333 k Ω λn I D 4 (0.015)(0.2) ro1 = 1 λ p I D1 = 1 = 667 k Ω (0.015)(0.1) g m1 = 2 K P I D1 = 2 (0.5)(0.1) = 0.447 mA/V So Ro8 = (0.447)(667)(667) ⇒ 198.9 M Ω Ro 6 = (3.846)(800)(333 667) ⇒ 683.4 M Ω Then Ad = 447(198.9 683.4) ⇒ Ad = 68,865 13.42 Assume biased at V + = 10V , V − = −10V . P = 3I REF (20) = 10 ⇒ I REF = 167 μ A Ad = g m1 ( Ro 6 Ro8 ) = 25, 000 ′ kn = 80 μ A / V 2 , k ′ = 35 μ A / V 2 p λn = 0.015V −1 , λ p = 0.02 V −1 ⎛W ⎞ ⎛W ⎞ Assume ⎜ ⎟ = 2.2 ⎜ ⎟ L ⎠p ⎝ ⎝ L ⎠n 618. Ro8 = g m8 ( ro8 ro10 ) Ro 6 = g m 6 (ro 6 )(ro 4 ro1 ) ro8 = ro10 = 1 λP I D 8 1 λP I D 8 = 1 = 0.60 M Ω (0.02)(83.3) = 0.60 M Ω ⎛ k′ ⎞⎛ W ⎞ ⎛ 35 ⎞ p g m8 = 2 ⎜ ⎟ ⎜ ⎟ I D 8 = 2 ⎜ ⎟ (2.2) X 2 (83.3) ⎝ 2⎠ ⎝ 2 ⎠ ⎝ L ⎠8 = 113.3 X ⎛W ⎞ where X 2 = ⎜ ⎟ ⎝ L ⎠n VA 80 ro 6 = = = 0.960 M Ω I C 6 83.3 ro 4 = ro1 = gm6 = 1 λn I D 4 = 1 = 0.40 M Ω (0.015)(167) 1 1 = = 0.60 M Ω λ p I D1 (0.02)(83.3) IC 6 83.3 = = 3204 μ A / V 0.026 VT ′ ⎛ kp ⎞⎛ W ⎞ ⎛ 35 ⎞ g m1 = 2 ⎜ ⎟ ⎜ ⎟ I D1 = 2 ⎜ ⎟ (2.2) X 2 (83.3) ⎝ 2⎠ ⎝ 2 ⎠ ⎝ L ⎠1 = 113.3 X Now Ro 6 = (3204)(0.960) ⎡0.40 0.60 ⎤ = 738 M Ω ⎣ ⎦ Ro8 = (113.3 X )(0.60)(0.60) = 40.8 X M Ω Then Ad = 25, 000 = (113.3 X ) ⎡ 738 40.8 X ⎤ ⎣ ⎦ ⎡ 30,110 X ⎤ = (113.3 X ) ⎢ ⎥ ⎣ 738 + 40.8 X ⎦ which yields X = 2.48 or ⎛W ⎞ X 2 = 6.16 = ⎜ ⎟ ⎝ L ⎠n and ⎛W ⎞ ⎜ ⎟ + (2.2)(6.16) = 12.3 ⎝ L ⎠P 13.43 For vcm (max), assume VCB (Q5 ) = 0. Then VS = 15 − 0.6 − 0.6 = 13.8 V 0.236 I D 9 = I D10 = = 0.118 mA 2 Using parameters given in Example 13.11 I 0.118 VSG = D 9 − VTP = + 1.4 = 2.17 V 0.20 KP Then vcm (max) = 13.8 − 2.17 ⇒ vcm (max) = 11.6 V 619. For vcm (min) , assume VSD ( M 9 ) = VSD ( sat ) = VSG + VTP = 2.17 − 1.4 = 0.77 V Now VD10 = I D10 (0.5) + 0.6 + I D10 (0.5) − 15 = 0.118 + 0.6 − 15 ⇒ VD10 = −14.28 V Then vcm (min) = −14.28 + VSD (sat) − VSG = −14.28 + 0.77 − 2.17 = −15.68 V Then, common-mode voltage range −15.68 ≤ vcm ≤ 11.6 Or, assuming the input is limited to ±15 V, then −15 ≤ vcm ≤ 11.6 V 13.44 For I1 = I 2 = 300 μ A, VSG = VBE + (0.3)(8) = 0.6 + 2.4 = 3.0 V Then I 2 = K P (VSG + VTP ) 2 0.3 = K P (3 − 1.4)2 ⇒ K P = 0.117 mA / V 2 13.45 For VCB = 0 for both Q6 and Q7 , then VS = 0.6 + 0.6 + VSG + (−VS ) So 2VS = 1.2 + VSG Now I1 0.6 + I 2 R1 = VSG = + VTP and I1 = I 2 KP Also I1 = I 2 = K P (VSG + VTP ) 2 so 0.6 + (0.25)(8)(VSG − 1.4) 2 = VSG 2 0.6 + 2(VSG − 2.8VSG + 1.96) = VSG 2 2VSG − 6.6VSG + 4.52 = 0 6.6 ± (6.6) 2 − 4(2)(4.52) = 2.33 V 2(2) Then 2VS = 1.2 + 2.33 = 3.53 and VS = 1.765 V VSG = 13.46 I C 5 = I C 4 = 300 μ A Using the parameters from Examples 13.12 and 13.13, we have βV (200)(0.026) Ri 2 = rπ 13 = n T = = 17.3 kΩ 0.3 I C13 Ad = 2 K n I Q 5 ⋅ ( Ri 2 ) = 2(0.6)(0.3) ⋅ (17.3) or Ad = 10.38 Now 620. I C13 0.3 = = 11.5 mA/V 0.026 VT g m13 = r013 = VA 50 = = 167 kΩ I C13 0.3 Then | Av 2 | = g m13 ⋅ r013 = (11.5)(167) or Av 2 = 1917 Overall gain: Av = (10.38)(1917) = 19,895 Assuming the resistances looking into Q4 and into the output stage are very large, we have 13.47 | Av 2 | = β R013 rπ 13 + (1 + β ) RE13 where R013 = r013 ⎡1 + g m13 ( RE13 rπ 13 ) ⎤ ⎣ ⎦ I C13 = 300 μ A, r013 = 50 = 167 kΩ 0.3 0.3 = 11.5 mA / V 0.026 (200)(0.026) = = 17.3 kΩ 0.3 g m13 = rπ 13 So R013 = (167) ⎡1 + (11.5) (1 17.3) ⎤ ⇒ 1.98 MΩ ⎣ ⎦ Then | Av 2 | = (200)(1980) = 1814 17.3 + (201)(1) Now Ci = C1 (1 + Av 2 ) = 12 [1 + 1814] ⇒ Ci = 21, 780 pF 1 f PD = 2π Req Ci Req = Ri 2 r012 r010 Neglecting R3 , r010 = 1 1 = = 333 kΩ λ I D10 (0.02)(0.15) Neglecting R5 , 50 r012 = = 333 kΩ 0.15 Ri 2 = rπ 13 + (1 + β ) RE13 = 17.3 + (201)(1) = 218 kΩ Then f PD = 1 2π ⎡ 218 333 333⎤ × 103 × ( 21, 780 ) × 10−12 ⎣ ⎦ or f PD = 77.4 Hz Unity-Gain Bandwidth Gain of first stage: 621. Ad = 2 K n I Qs ⋅ ( R12 ro12 ro10 ) = 2(0.6)(0.3) ⋅ (218 333 333) = (0.6)(218 333 333) or Ad = 56.6 Overall gain: Av = (56.6)(1814) = 102, 672 Then unity-gain bandwidth = (77.4)(102, 672) ⇒ 7.95 MHz 13.48 Since VGS = 0 in J 6 , I REF = I DSS ⇒ I DSS = 0.8 mA 13.49 a. Ri 2 = rπ 5 + (1 + β ) [ rπ 6 + (1 + β ) RE ] (100)(0.026) = 13 kΩ 0.2 I 200 μ A ≅ C6 = = 2 μA 100 β rπ 6 = IC 5 So rπ 5 = (100)(0.026) = 1300 kΩ 0.002 Then Ri 2 = 1300 + (101) [13 + (101)(0.3) ] or Ri 2 = 5.67 MΩ Av = g m 2 ( r02 r04 Ri 2 ) b. gm2 = 2 ⋅ I D ⋅ I DSS VP = 2 ⋅ (0.1)(0.2) 3 = 0.0943 mA / V r02 = 1 1 = = 500 kΩ λ I D (0.02)(0.1) r04 = VA 5.0 = = 500 kΩ I C 4 0.1 Then Av = (0.0943)[500 || 500 || 5670] or Av = 22.6 13.50 a. Need VSD (QE ) ≥ VSD ( sat ) = VP For minimum bias ±3 V Set VP = 3 V and VZK = 3 V I REF 2 = VZK − VD1 R3 3 − 0.6 ⇒ R3 = 24 kΩ 0.1 Set bias in QE = I REF 2 + I Z 2 = 0.1 + 0.1 = 0.2 mA Therefore, so that R3 = 622. I DSS = 0.2 mA b. Neglecting base currents 12 − 0.6 I 01 = I REF 1 = 0.5 mA = R4 so that R4 = 22.8 kΩ 13.51 a. gm2 We have 2 = ⋅ I D ⋅ I DSS | VP | = 2 ⋅ (0.5)(1) 4 = 0.354 mA/V r02 = r04 = 1 λ ID = 1 = 100 kΩ (0.02)(0.5) VA 100 = = 200 kΩ I D 0.5 0.5 = 19.23 mA/V 0.026 (200)(0.026) = 10.4 kΩ rπ 4 = 0.5 So R04 = r04 ⎡1 + g m 4 ( rπ 4 R2 ) ⎤ ⎣ ⎦ gm4 = = 200 ⎡1 + (19.23) (10.4 0.5 ) ⎤ ⎣ ⎦ = 2035 kΩ Ad = g m 2 ( r02 R04 RL ) For RL → ∞ Ad = 0.354 (100 || 2035 ) = 33.7 b. With these parameter values, gain can never reach 500. Similarly for this part, gain can never reach 700. 623. Chapter 14 Exercise Solutions EX14.1 a. ACL = b. dACL ACL ACL = −50 ⇒ ACL = −49.949 ⎡ ⎛ 1 ⎞ ⎤ ⎢1 + ⎜ 5 × 104 ⎟ ( 51) ⎥ ⎠ ⎣ ⎝ ⎦ dA 51 = 10 × ⇒ CL = 0.0102% ACL 5 × 104 −50 ⇒ ACL = −49.943 51 ⎤ ⎡ ⎢1 + 4.5 × 104 ⎥ ⎣ ⎦ EX14.2 a. For R0 = 0 1 1 1 = + (1 + 104 ) = 0.1 + 103 Ri f 10 10 ⇒ Ri f = 10−3 kΩ = 1 Ω b. For R0 = 10 kΩ 1 1 1 ⎡1 + 104 + 1 ⎤ 104 = + ×⎢ ⎥ ≅ 0.1 + Ri f 10 10 ⎣ 1 + 1 + 1 ⎦ 3 (10 ) Rif = 3 × 10−3 kΩ ⇒ Ri f = 3 Ω EX14.3 ⎛ 40 ⎞ 40 (1 + 104 ) + 99 ⎜ 1 + ⎟ 1 ⎠ ⎝ Ri f = 99 1+ 1 5 4 × 10 + 4.059 × 103 ≅ 100 Ri f = 4.04 × 103 kΩ ⇒ Ri f = 4.04 MΩ EX14.4 R 1 + 2 = 100 R1 a. 1 1 ⎡ 105 ⎤ = ⎢ ⎥ = 10 ⇒ R0 f = 0.1 Ω R0 f 100 ⎣100 ⎦ b. 1 1 ⎡ 105 ⎤ 2 = ⎢ ⎥ = 10 R0 f 10 ⎣100 ⎦ R0 f = 10−2 kΩ ⇒ R0 f = 10 Ω EX14.5 From Equation (14.43) 624. ACL ( f ) = = ACL 0 1+ j ⋅ f f PD ( A0 /ACL 0 ) 25 f 1+ j ⋅ ( 50 ) (104 / 25) = 25 1+ j ⋅ f 2 × 104 f = 2 kHz a. v0 = 25 ⇒ v0 ( peak ) = 1.25 mV vI f = 20 kHz b. v0 1 = ⋅ 25 ⇒ v0 ( peak ) = 0.884 mV vI 2 c. f = 100 kHz v0 25 25 = = = 4.90 2 vI 5.099 1 + (100 / 20 ) ⇒ v0 = 0.245 mV EX14.6 Full-scale response = 1× 5 = 5 V t= 5 ⇒ t = 2.5 μ s 2 EX14.7 a. FPBW = SR 0.63 × 106 = 2π V0 ( max ) 2π (1) FPBW = 1.0 × 105 ⇒ FPBW = 100 kHz b. FPBW = 0.63 × 106 = 1.0 × 104 2π (10 ) ⇒ FPBW = 10 kHz EX14.8 ⎛I ⎞ ⎛ 1.85 × 10−14 ⎞ V0 S = VT ln ⎜ S 2 ⎟ = (0.026) ln ⎜ −14 ⎟ ⎝ 2 × 10 ⎠ ⎝ I S1 ⎠ ⇒ V0 S = 2.03 mV EX14.9 We need iC1 = iC 2 , vEC 3 = vEC 4 = 0.6 V, and vCE1 = vCE 2 = 10 V By Equation (14.60(a)) ⎡ ⎛ v ⎞ ⎤ ⎛ 10 ⎞ iC1 = I S 1 ⎢exp ⎜ BE1 ⎟ ⎥ ⎜ 1 + ⎟ ⎝ VT ⎠ ⎦ ⎝ 50 ⎠ ⎣ ⎡ ⎛ v ⎞ ⎤ ⎛ 0.6 ⎞ = I S 3 ⎢ exp ⎜ EB 3 ⎟ ⎥ ⎜1 + ⎟ 50 ⎠ ⎝ VT ⎠ ⎦ ⎝ ⎣ By Equation (14.60(b)) 625. ⎡ ⎛ v ⎞ ⎤ ⎛ 10 ⎞ iC 2 = I S 2 ⎢ exp ⎜ BE 2 ⎟ ⎥ ⎜ 1 + ⎟ ⎝ VT ⎠ ⎦ ⎝ 50 ⎠ ⎣ ⎡ ⎛ v ⎞ ⎤ ⎛ 0.6 ⎞ = I S 4 ⎢ exp ⎜ EB 4 ⎟ ⎥ ⎜ 1 + ⎟ 50 ⎠ ⎝ VT ⎠ ⎦ ⎝ ⎣ I S1 = I S 2, take the ratio: ⎛v −v ⎞ I exp ⎜ BE1 BE 2 ⎟ = S 3 VT ⎝ ⎠ IS 4 vBE1 − vBE 2 = V0 S ⎛I ⎞ = VT ln ⎜ S 3 ⎟ ⎝ IS 4 ⎠ = 0.026 ⋅ ln (1.05 ) ⇒ V0 S = 1.27 m V EX14.10 VOS = 0.020 = I Q ⎛ ΔK n ⎞ 1 ⋅ ⋅⎜ ⎟ 2 2Kn ⎝ Kn ⎠ 1 150 ⎛ ΔK n ⎞ ⋅ ⋅ 2 2 ( 50 ) ⎜ 50 ⎟ ⎝ ⎠ ⇒ ΔK n = 1.63μ A / V 2 ⇒ ΔK n 1.63 = ⇒ 3.26% 50 Kn EX14.11 ⎛ R5 ⎞ + Want ⎜ ⎟V = 5 m V ⎝ R5 + R4 ⎠ R5 R5 = R4 so R5 × V + = 0.005 R4 ( 0.005)(100 ) = 0.05 kΩ 10 ⇒ R5 = 50 Ω EX14.12 R1′ = 25 1 = 0.9615 kΩ ′ R2 = 75 1 = 0.9868 kΩ For I Q = 100 μ A ⇒ iC1 = iC 2 = 50 μ A From Equation (14.75) ⎛ 50 × 10−6 ⎞ ( 0.026 ) ln ⎜ −14 ⎟ + ( 0.050 )( 0.9615 ) ⎝ 10 ⎠ ⎛i ⎞ = ( 0.026 ) ln ⎜ C 2 ⎟ + ( 0.050 )( 0.9868 ) ⎝ IS 4 ⎠ 0.58065 + 0.048075 ⎛i ⎞ = ( 0.026 ) ln ⎜ C 2 ⎟ + 0.04934 ⎝ IS 4 ⎠ 626. ⎛i ⎞ ln ⎜ C 2 ⎟ = 22.284 ⎝ IS 4 ⎠ 50 × 10−6 = 4.7625 × 109 IS 4 I S 4 ≅ 1.05 × 10−14 A EX14.13 From Equation (14.79) ⎛ R ⎞ v0 = I B1 R2 − I B 2 R3 ⎜ 1 + 2 ⎟ R1 ⎠ ⎝ For v0 = 0 ⎛ 100 ⎞ 0 = (1.1× 10−6 ) (100 kΩ ) − (1.0 × 10−6 ) R3 ⎜ 1 + ⎟ 10 ⎠ ⎝ R3 (11) = (1.1)(100 kΩ ) ⇒ R3 = 10 kΩ TYU14.1 v1CM ( max ) = V + − VSD1 ( sat ) − VSG1 v1CM ( min ) = V − + VDS 4 ( sat ) + VSD1 ( sat ) − VSG1 We have: ′ I REF = 100 μ A, kn = 80 μ A / V 2 , k ′ = 40 μ A / V 2 , p ⎛W ⎜ ⎝L For ⎞ ⎟ = 25 ⎠ M1 : 2 ⎛ 40 ⎞ I D = 50 = ⎜ ⎟ ( 25 )(VSG1 + VTP ) ⎝ 2 ⎠ So 50 = 500 (VSG1 − 0.5 ) ⇒ VSG1 = 0.816 V 2 VSD1 ( sat ) = 0.816 − 0.5 = 0.316 V Then vCM ( max ) = V + − 0.316 − 0.816 = V + − 1.13 V For M 4 : 2 ⎛ 80 ⎞ I D = 100 = ⎜ ⎟ ( 25 )(VGS 4 − VTN ) ⎝ 2⎠ So 100 = 1000 (VGS 4 − 0.5 ) ⇒ VGS 4 = 0.816 V 2 VDS 4 ( sat ) = 0.816 − 0.5 = 0.316 V VCM ( min ) = V − + 0.316 + 0.316 − 0.816 = V − − 0.184 So V − − 0.184 ≤ vCM ≤ V + − 1.13 V TYU14.2 vo ( max ) = V + − VSD 8 ( sat ) − VSG10 vo ( min ) = V − + VDS 4 ( sat ) + VDS 6 ( sat ) Now 627. 50 VSG 8 = VSG10 = + 0.5 = 0.816 V ( 40/ 2 )( 25) VSD 8 ( sat ) = VSD10 ( sat ) = 0.316 V So vo ( max ) = V + − 0.316 − 0.816 = V + − 1.13 Also VGS 6 = 50 + 0.5 = 0.724 V (80/ 2 )( 25) VGS 4 = 100 + 0.5 = 0.816 V (80/ 2 )( 25) VDS 6 ( sat ) = 0.724 − 0.5 = 0.224 V VDS 4 ( sat ) = 0.816 − 0.5 = 0.316 V So vo ( min ) = V − + 0.316 + 0.224 = V − + 0.54 Then V − + 0.54 ≤ vo ≤ V + − 1.13 V TYU14.3 500 = −25 20 Within 0.1% ⇒ −25 + ( 0.001)( 25 ) ⇒ ACL = −24.975 ACL ( ideal ) = − −24.975 = −25 ⎡ 26 ⎤ ⎢1 + A ⎥ 0L ⎦ ⎣ −25 26 = − 1 = 0.0010 A0 L −24.975 A0 L = 25.974 TYU14.4 ACL ( ∞ ) a. ACL = b. dACL 100 = 10 × 5 = 0.01% 10 ACL ⎡ A (∞) ⎤ 1 + ⎢ CL ⎥ ⎣ A0 L ⎦ R 495 ACL ( ∞ ) = 1 + 2 = 1 + = 100 R1 5 100 ACL = ⇒ ACL = 99.90 100 1+ 5 10 ACL ( ∞ ) = 100 ACL = 99.90 − ( 0.0001)( 99.90 ) ⇒ ACL = 99.89 TYU14.5 628. ACL ( ∞ ) − ACL = 1− ACL ( ∞ ) ACL = 1− ACL ( ∞ ) 1 A (∞) 1 + CL A0 L ACL ( ∞ ) ACL ( ∞ ) −1 A0 L A0 L So 0.001 = = A (∞) A (∞) 1 + CL 1 + CL A0 L A0 L 1+ 0.001 = 0.999 ⋅ ACL ( ∞ ) = ACL ( ∞ ) A0 L 0.001 0.001 ⋅ A0 L = ⋅ (104 ) ⇒ ACL ( ∞ ) = 10.010 0.999 0.999 or ACL = (1 − 0.001)(10.010 ) ⇒ ACL = 10.0 TYU14.6 ii ⎛ Rif ⎞ =⎜ ⎟ I1 ⎝ Ri ⎠ iI 0.1 a. = = 1× 10−5 i1 104 iI 10 = = 1× 10−3 i1 104 b. TYU14.7 Voltage follower R2 = 0, R1 = ∞ Rif = Ri (1 + A0 L ) = 10 (1 + 5 × 105 ) ≅ 5 × 106 kΩ ⇒ Rif = 5000 MΩ TYU14.8 f3dB = (105 ) (10 ) ⇒ 20 kHz fT = 50 ACL 0 f max = f 3dB = V0 ( max ) = SR 2π V0 ( max ) SR 0.8 × 106 = 2π f3dB 2π ( 20 × 103 ) ⇒ V0 ( max ) = 6.37 V TYU14.9 v0 = I B1 R3 = (10−6 )( 200 × 103 ) a. ⇒ v0 = 0.20 V b. R4 = R1 R2 R3 = 100 50 200 ⇒ R4 = 28.6 kΩ 629. Chapter 14 Problem Solutions 14.1 Ad = vo = −80 vi vo (max) = 4.5 ⇒ vi (max) = 56.25 mV So vi (max)rms = 56.25 2 = 39.77 mV 14.2 (a) 4.5 = 0.028125 mA 160 4.5 iL = = 4.5 mA 1 Output Circuit = 4.528 mA v −4.5 vi = − o = ⇒ vi = −0.05625 V A 80 (b) v 4.5 io ≈ 15 mA = o = RL RL i2 = ⇒ RL (min) = 300 Ω 14.3 (1) (2) v2 = 12.5 mV (3) (4) (5) AOL = 2 × 104 v1 = 8 μ V AOL = 1000 vo = 2 V 14.4 From Eq. (14.4) 630. ACL = −15 = − R2 / R1 1 ⎛ R2 ⎞ 1+ ⎜1 + ⎟ AoL ⎝ R1 ⎠ − R2 / R1 1 ⎛ R2 ⎞ 1+ ⎜1 + ⎟ R1 ⎠ 2 × 103 ⎝ ⎛R ⎞ 15 ⎜ 2 ⎟ 1 ⎞ ⎛ ⎝ R1 ⎠ = R2 15 ⎜ 1 + ⎟+ R1 2 × 103 ⎠ 2 × 103 ⎝ 15.0075 = R2 (0.9925) R1 R2 = 15.12 R1 14.5 vI − v1 v1 − v0 v1 = + and v0 = − A0 L v1 R1 R2 Ri so that v1 = − v0 A0 L ⎛ 1 vI v0 1 1⎞ + = v1 ⎜ + + ⎟ R1 R2 ⎝ R1 R2 Ri ⎠ So ⎡1 vI 1 ⎛ 1 1 1 ⎞⎤ = −v0 ⎢ + + ⎟⎥ ⎜ + R1 ⎣ R2 A0 L ⎝ R1 R2 Ri ⎠ ⎦ Then v0 −(1/ R1 ) = = ACL vI ⎡ 1 1 ⎛ 1 1 1 ⎞⎤ + ⎟⎥ ⎢ + ⎜ + ⎣ R2 A0 L ⎝ R1 R2 Ri ⎠ ⎦ From Equation (14.20) for RL = ∞ and R0 = 0 1 1 1 (1 + A0 L ) = + ⋅ Rif Ri R2 1 For Ri = 1 kΩ a. −(1/ 20) ⎡ 1 1 ⎛ 1 1 1 ⎞⎤ ⎢100 + 103 ⎜ 20 + 100 + 1 ⎟ ⎥ ⎝ ⎠⎦ ⎣ −0.05 = [0.01 + 1.06 × 10−3 ] ACL = or 631. ⇒ ACL = −4.52 1 1 1 + 103 = + ⇒ Rif = 90.8 Ω Rif 1 100 For Ri = 10 kΩ b. ACL = = −(1/ 20) ⎡ 1 1 ⎛ 1 1 1 ⎞⎤ ⎢100 + 103 ⎜ 20 + 100 + 10 ⎟ ⎥ ⎝ ⎠⎦ ⎣ −0.05 [0.01 + 1.6 × 10−4 ] or ⇒ ACL = −4.92 1 1 1 + 103 = + ⇒ Rif = 98.9 Ω Rif 10 100 For Ri = 100 kΩ c. ACL = = −(1/ 20) ⎡ 1 1 ⎛ 1 1 1 ⎞⎤ ⎢100 + 103 ⎜ 20 + 100 + 100 ⎟ ⎥ ⎝ ⎠⎦ ⎣ −0.05 [0.01 + 7 × 10−5 ] or ⇒ ACL = −4.965 1 1 1 + 103 = + ⇒ Rif = 99.8 Ω Rif 100 100 14.6 ⎛ R2 ⎞ ⎜1 + ⎟ R1 ⎠ v ⎝ ACL = o = vi ⎡ 1 ⎛ R2 ⎞ ⎤ ⎢1 + ⎜1 + ⎟ ⎥ R1 ⎠ ⎦ ⎣ AOL ⎝ For the ideal: ⎛ R2 ⎞ 0.10 = 50 ⎜1 + ⎟ = R1 ⎠ 0.002 ⎝ vo (actual ) = (0.10)(1 − 0.001) = 0.0999 So 632. 0.0999 50 = = 49.95 1 0.002 1 + (50) AOL which yields AOL = 1000 14.7 From Equation (14.18) ⎛A 1 ⎞ − ⎜ OL − ⎟ v ⎝ Ro R2 ⎠ Avf 1 = o1 = v1 ⎛ 1 1 1 ⎞ + + ⎟ ⎜ RL Ro R2 ⎠ ⎝ Or ⎛ 5 × 103 1 ⎞ −⎜ − ⎟ 1 100 ⎠ −(4.99999 × 103 ) vo1 = ⎝ ⋅ v1 = ⋅ v1 1.11 ⎛1 1 1 ⎞ + + ⎜ ⎟ ⎝ 10 1 100 ⎠ vo1 = −4.504495 × 103 ⋅ v1 Now i1 vi − v1 = ≡K v1 R1v1 Then vi − v1 = KR1v1 which yields vi v1 = KR1 + 1 Now, from Equation (14.20) 1 ⎤ ⎡ 1 + 5 × 103 + ⎥ 1 1 ⎢ 10 K= + ⎢ ⎥ 10 100 ⎢ 1 + 1 + 1 ⎥ ⎢ 10 100 ⎥ ⎣ ⎦ ⎡ 5.0011× 103 ⎤ = (0.1) + (0.01) ⎢ ⎥ = 45.15495 1.11 ⎣ ⎦ Then vi vi v1 = = ( 45.15495)(10 ) + 1 452.5495 We find vi ⎡ ⎤ vo1 = −4.504495 × 103 ⎢ ⎥ ⎣ 452.5495 ⎦ Or v Avf 1 = o1 = −9.9536 vi For the second stage, RL = ∞ 633. ⎛ 5 × 103 1 ⎞ −⎜ − ⎟ 1 100 ⎠ vo 2 = ⎝ ⋅ v1′ = −4.950485 × 103 ⋅ v1′ ⎛1 1 ⎞ ⎜ + ⎟ ⎝ 1 100 ⎠ ⎡ ⎤ 1 1 ⎢1 + 5 × 103 ⎥ K≡ + ⎢ ⎥ = 49.61485 10 100 ⎢ 1 + 1 ⎥ ⎢ 100 ⎥ ⎣ ⎦ vo1 vo1 vo1 v1′ = = = KR1 + 1 (49.61485)(10) + 1 497.1485 Then vo 2 −4.950485 × 103 = = −9.95776 497.1485 vo1 So v Avf = o 2 = (−9.9536)(−9.95776) ⇒ Avf = 99.12 vi 14.8 a. v1 − vI v v −v + 1 + 1 0 =0 R3 + Ri R1 R2 ⎡ 1 vI 1 1⎤ v v1 ⎢ + + ⎥= 0 + ⎣ R3 + Ri R1 R2 ⎦ R2 R3 + Ri v0 v0 − A0 L vd v0 − v1 + + =0 RL R0 R2 or ⎡1 A v 1 1⎤ v v0 ⎢ + + ⎥ = 1 + 0L d RL R0 R2 ⎦ R2 R0 ⎣ ⎛ v −v ⎞ vd = ⎜ I 1 ⎟ ⋅ Ri ⎝ R3 + Ri ⎠ So substituting numbers: vI 1⎤ v 1 ⎡ 1 v1 ⎢ + + ⎥= 0 + ⎣10 + 20 10 40 ⎦ 40 10 + 20 or v1[0.15833] = v0 [0.025] + vI [0.03333] 1 ⎤ v (104 )vd ⎡1 1 v0 ⎢ + + ⎥= 1 + 0.5 ⎣1 0.5 40 ⎦ 40 (1) (2) (3) (1) (2) 634. or v0 [3.025] = v1 [ 0.025] + ( 2 × 104 ) vd ⎛ v −v ⎞ vd = ⎜ I 1 ⎟ ⋅ 20 = 0.6667 ( vI − v1 ) ⎝ 10 + 20 ⎠ So v0 [3.025] = v1 [ 0.025] + ( 2 × 104 ) ( 0.6667 )( vI − v1 ) (3) (2) or v0 [3.025] = 1.333 ×10 4 vI − 1.333 × 104 v1 From (1): v1 = v0 ( 0.1579 ) + vI ( 0.2105 ) Then v0 [3.025] = 1.333 × 104 vI − 1.333 × 104 ⎡ v0 ( 0.1579 ) + vI ( 0.2105 ) ⎤ ⎣ ⎦ v0 ⎡ 2.1078 ×103 ⎤ = vI ⎡1.0524 ×104 ⎤ ⎣ ⎦ ⎣ ⎦ or v ACL = 0 = 4.993 vI To find Rif : Use Equation (14.27) ⎛ 0.5 0.5 ⎞ iI ⎜ 1 + + ⎟ 1 40 ⎠ ⎝ 3 ⎧⎛ 1 1 ⎞ ⎛ 0.5 0.5 ⎞ 0.5 ⎫ (10 ) vd = v1 ⎨⎜ + ⎟ ⎜ 1 + + − ⎬− ⎟ 1 40 ⎠ (40) 2 ⎭ 40 ⎩⎝ 10 40 ⎠ ⎝ iI (1.5125) = v1{(0.125)(1.5125) − 0.0003125} − 25vd or iI (1.5125) = vI {0.18875} − 25vd Now vd = iI Ri = iI (20) and v1 = vI − iI (20) So iI (1.5125) = [vI − iI (20)] ⋅ [0.18875] − 25iI (20) iI [505.3] = vI (0.18875) or vI = 2677 kΩ iI Now Rif = 10 + 2677 ⇒ Rif = 2.687 MΩ To determine R0 f : Using Equation (14.36) ⎡ 1 1 ⎢ A0 L = ⋅⎢ R2 ′ R0 f R0 ⎢ ⎢1 + R R 1 i ⎣ ′ or R0 f = 3.5 Ω ⎤ ⎡ ⎤ ⎥ 1 ⎢ 103 ⎥ ⎥= ⎥ ⋅⎢ ⎥ 0.5 ⎢1 + 40 ⎥ ⎥ ⎢ 10 20 ⎥ ⎣ ⎦ ⎦ Then R0 f = 1 kΩ 3.5 Ω ⇒ R0 f = 3.49 Ω b. Using Equation (14.16) dACL dA ⎛ 5 ⎞ = (−10) ⎜ 3 ⎟ ⇒ CL = −(0.05)% ACL ACL ⎝ 10 ⎠ 14.9 635. v0 − A0 L vd v0 − vI + = 0 and vd = vI − v0 R0 Ri So v0 A0 L v v − ⋅ (vI − v0 ) + 0 − I = 0 R0 R0 Ri Ri ⎡1 A ⎡1 A ⎤ 1⎤ v0 ⎢ + 0 L + ⎥ = vI ⎢ + 0 L ⎥ ⎣ R0 R0 Ri ⎦ ⎣ Ri R0 ⎦ ⎡ 1 (104 ) 1 ⎤ ⎡ 1 (104 ) ⎤ + + + v0 ⎢ ⎥ = vI ⎢ ⎥ ⎣ 0.2 0.2 100 ⎦ ⎣100 0.2 ⎦ v0 [5.000501× 104 ] = vI [5.000001× 10 4 ] So ACL = Set vI = 0 b. i0 = v0 = 0.9999 vI v0 − A0 L vd v0 + and vd = −v0 R0 Ri ⎡1 A 1⎤ i0 = v0 ⎢ + 0 L + ⎥ R0 R0 Ri ⎦ ⎣ Then 1 1 A0 L 1 = + + R0 f R0 R0 Ri or 1 1 (104 ) 1 = + + R0 f 0.2 0.2 100 which yields R0 f ≅ 0.02 Ω 14.10 636. vI 1 − v1 vI 2 − v1 v1 − v0 + = 20 10 40 vI 1 vI 2 v0 1 1⎤ ⎡1 + + = v1 ⎢ + + ⎥ 20 10 40 ⎣ 20 10 40 ⎦ v and v0 = − A0 L v1 so that v1 = − 0 A0L Then ⎧1 1 ⎛ 7 ⎞⎫ ⋅ vI 1 (0.05) + vI 2 (0.10) = −v0 ⎨ + ⎟⎬ 3 ⎜ ⎩ 40 2 × 10 ⎝ 40 ⎠ ⎭ = −v0 [2.50875 × 10−2 ] ⇒ v0 = −1.993vI 1 − 3.986vI 2 Δv0 2 − 1.993 Δv = ⇒ 0 = 0.35% 2 v0 v0 14.11 ⎛ 40 ⎞ ⎛ 4⎞ vB = ⎜ ⎟ v2 = ⎜ ⎟ v2 = 0.8v2 ⎝ 40 + 10 ⎠ ⎝ 5⎠ v1 − v A v A − v0 = 10 40 v1 v0 1 ⎞ ⎛1 + = vA ⎜ + ⎟ 10 40 ⎝ 10 40 ⎠ v1 (0.1) + v0 (0.025) = vA (0.125) v0 = A0 L vd = A0 L (vB − v A ) or v0 = A0 L [0.8v2 − v A ] v0 − 0.8v2 = −v A A0 L ⇒ v A = 0.8v2 − Then v0 A0 L (1) (2) (3) 637. ⎡ v ⎤ v1 (0.1) + v0 (0.025) = (0.125) ⎢0.8v2 − 0 ⎥ A0 L ⎦ ⎣ 0.125 ⎤ ⎡ v1 (0.1) − v2 (0.1) = −v0 ⎢0.025 + 103 ⎥ ⎣ ⎦ −2 = −v0 [2.5125 × 10 ] ⇒ Ad = ⇒ v0 = 3.9801 v2 − v1 ΔAd 0.0199 = ⇒ 0.4975% Ad 4 14.12 a. Considering the second op-amp and Equation (14.20), we have ⎡ ⎤ 1 1 1 ⎢1 + 100 ⎥ 101 = + ⋅⎢ ⎥ = 0.10 + (0.1)(11) Rif 2 10 0.1 ⎢ 1 + 1 ⎥ ⎢ 0.1 ⎥ ⎣ ⎦ So Rif 2 = 0.0109 kΩ The effective load on the first op-amp is then RL1 = 0.1 + Rif 2 = 0.1109 kΩ Again using Equation (14.20), we have 1 1 + 100 + 1 1 1 0.1109 = 0.10 + 110.017 = + ⋅ 11.017 Rif 10 1 1 + 1 + 1 0.1109 1 so that Rif = 99.1 Ω b. To determine R0 f : For the first op-amp, we can write, using Equation (14.36) ⎡ ⎤ ⎡ ⎤ ⎥ 1 ⎢ 100 ⎥ 1 1 ⎢ A0 L ⎥ = ⋅⎢ ⎥ = ⋅⎢ R2 ⎥ 1 ⎢ 40 ⎥ R0 f 1 R0 ⎢ 1+ 1+ ⎢ R1 || Ri ⎥ ⎢ 1 || 10 ⎥ ⎣ ⎦ ⎣ ⎦ which yields R0 f 1 = 0.021 kΩ For the second op-amp, then ⎡ ⎢ A0 L 1 1 ⎢ = ⋅ R2 R0 f R0 ⎢ ⎢1 + ( R + R ) || R 1 0f1 i ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ ⎡ ⎤ ⎥ 1 ⎢ 100 ⎥ = ⋅⎢ 0.10 1 ⎢1 + ⎥ ⎢ (0.121) ||10 ⎥ ⎣ ⎦ or R0 f = 18.4 Ω c. To find the gain, consider the second op-amp. 638. v01 − (−vd 2 ) vd 2 −vd 2 − v02 + = 0.1 Ri 0.1 (1) v01 v02 1 1 ⎞ ⎛ 1 + vd 2 ⎜ + + ⎟=− 0.1 0.1 ⎝ 0.1 10 0.1 ⎠ or v01 (10) + vd 2 (20.1) = −v02 (10) v02 − A0 L vd 2 v02 − (−vd 2 ) + =0 R0 0.1 (2) v02 ⎛ 100 1 ⎞ v02 − vd 2 ⎜ − =0 ⎟+ 1 ⎝ 1 0.1 ⎠ 0.1 v02 (11) − vd 2 (90) = 0 or vd 2 = v02 (0.1222) Then Equation (1) becomes v01 (10) + v02 (0.1222)(20.1) = −v02 (10) or v01 = −v02 (1.246) Now consider the first op-amp. vI − (−vd 1 ) vd 1 −vd 1 − v01 + = 1 Ri 1 (1) ⎛1 1 1⎞ vI (1) + vd 1 ⎜ + + ⎟ = −v01 (1) ⎝ 1 10 1 ⎠ or vI (1) + vd 1 (2.1) = −v01 (1) v01 v − A0 L vd 1 v01 − (−vd 1 ) + 01 + =0 0.1109 R0 1 1 1⎞ ⎛ 1 ⎛ 100 1 ⎞ v01 ⎜ + + ⎟ − vd 1 ⎜ − ⎟=0 ⎝ 0.1109 1 1 ⎠ ⎝ 1 1⎠ v01 (11.017) − vd 1 (99) = 0 or vd 1 = v01 (0.1113) Then Equation (1) becomes (2) 639. vI (1) + v01 (0.1113)(2.1) = −v01 or vI = −v01 (1.234) We had v01 = −v02 (1.246) So vI = v02 (1.246)(1.234) or v02 = 0.650 vI v02 =1 vI So ratio of actual to ideal = 0.650. d. Ideal 14.13 (a) For the op-amp. A0 L ⋅ f 3dB = 106 106 = 50 Hz 2 × 104 For the closed-loop amplifier. 106 f 3dB = = 40 kHz 25 (b) Open-loop amplifier. 2 × 104 2 × 104 ⇒| A | = A= 2 f ⎛ f ⎞ 1+ j 1+ ⎜ f3dB ⎟ ⎝ f3dB ⎠ f 3dB = f = 0.25 f 3dB ⇒ A = f = 5 f 3− dB ⇒ A = 2 × 104 1 + (0.25) 2 = 1.94 × 104 2 × 104 = 3.92 × 103 1 + (5) 2 Closed-loop amplifier 25 f = 0.25 f 3dB ⇒ A = = 24.25 1 + (0.25) 2 f = 5 f 3− dB ⇒ A = 25 1 + (5) 2 = 4.90 14.14 The open loop gain can be written as A0 A0 L ( f ) = ⎛ f ⎞⎛ f ⎞ ⎜1 + j ⋅ ⎟⎜ 1 + j ⋅ ⎟ f PD ⎠ ⎝ 5 × 106 ⎠ ⎝ where A0 = 2 × 105. The closed-loop response is A0 L ACL = 1 + β A0 L At low frequency, 2 × 105 100 = 1 + β (2 × 105 ) So that β = 9.995 × 10−3. Assuming the second pole is the same for both the open-loop and closed-loop, then ⎛ f ⎞ f ⎞ −1 ⎛ φ = − tan −1 ⎜ ⎟ − tan ⎜ 6 ⎟ f PD ⎠ ⎝ 5 × 10 ⎠ ⎝ 640. For a phase margin of 80°, φ = −100°. So ⎛ f ⎞ −100 = −90 − tan −1 ⎜ 6 ⎟ ⎝ 5 × 10 ⎠ or f = 8.816 × 105 Hz Then A0 L = 1 2 × 105 = ⎛ 8.816 × 105 ⎞ 1+ ⎜ ⎟ f PD ⎝ ⎠ 2 ⎛ 8.816 × 105 ⎞ 1+ ⎜ ⎟ 6 ⎝ 5 × 10 ⎠ 2 or 8.816 × 105 ≅ 1.9696 × 105 f PD or f PD = 4.48 Hz 14.15 (a) 1st stage (10) f3− dB = 1 MHz ⇒ f 3− dB = 100 kHz 2nd stage (50) f3− dB = 1 MHz ⇒ f 3− dB = 20 kHz Bandwidth of overall system ≅ 20 kHz (b) If each stage has the same gain, so 2 K = 500 ⇒ K = 22.36 Then bandwidth of each stage (22.36) f 3− dB = 1 MHz ⇒ f 3− dB = 44.7 kHz 14.16 A= Ao 1+ j A= f f3− dB Ao ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ f3− dB ⎠ 2 ⇒ 200 = 5 × 104 ⎛ 104 ⎞ 1+ ⎜ ⎟ ⎝ f3− dB ⎠ Then fT = (5 × 104 )(40) ⇒ fT = 2 MHz 14.17 2 ⇒ f3− dB = 40 Hz 641. (5 × 104 ) f PD = 106 ⇒ f PD = 20 Hz (25) f 3− dB 106 ⇒ f3− dB = 40 kHz Av = Avo 1+ j f ⇒ Av = f3− dB 25 f ⎛ ⎞ 1+ ⎜ 3 ⎟ ⎝ 40 × 10 ⎠ 2 At f = 0.5 f 3− dB = 20 kHz Av = 25 1 + (0.5) 2 = 22.36 At f = 2 f 3− dB = 80 kHz Av = 25 1 + (2) 2 14.18 (20 × 103 ) ⋅ Avf = 11.18 MAX = 106 ⇒ Avf MAX = 50 14.19 From Equation (14.55), SR 10 × 106 F P BW = = 2π VP 0 2π (10) or F P BW = f max = 159 kHz 14.20 a. VP 0 Using Equation (14.55), 8 × 106 = 2π (250 × 103 ) or VP 0 = 5.09 V b. 1 1 = = 4 × 10−6 s f 250 × 103 One-fourth period = 1 μ s Period T = Slope = VP 0 = SR = 8 V/μ s 1μ s ⇒ VP 0 = 8 V 14.21 For input (a), maximum output is 5 V. 642. S R = 1 V/μs so For input (b), maximum output is 2 V. For input (c), maximum output is 0.5 V so the output is 14.22 For input (a), max v01 = 3 V. Then v02 max = 3(3) = 9 V For input (b), max v01 = 1.5 V. 643. Then v02 max = 3 (1.5 ) = 4.5 V 14.23 f MAX = 20 kHz, SR = 0.8 V / μ s V po = SR 0.8 × 106 = ⇒ V po = 6.37 V 2π f MAX 2π (20 × 103 ) 14.24 ⎛V ⎞ ⎛V ⎞ I1 = I S 1 exp ⎜ BE1 ⎟ , I 2 = I S 2 exp ⎜ BE 2 ⎟ ⎝ VT ⎠ ⎝ VT ⎠ Want I1 = I 2 , so ⎛V ⎞ 5 × 10−14 (1 + x) exp ⎜ BE1 ⎟ I1 ⎝ VT ⎠ =1= I2 ⎛V ⎞ 5 × 10−14 (1 − x) exp ⎜ BE 2 ⎟ ⎝ VT ⎠ = ⎛ V −V ⎞ (1 + x) exp ⎜ BE1 BE 2 ⎟ VT (1 − x) ⎝ ⎠ Or ⎛ V − VBE1 ⎞ ⎛ VOS ⎞ 1+ x = exp ⎜ BE 2 ⎟ = exp ⎜ ⎟ 1− x VT ⎝ ⎠ ⎝ VT ⎠ ⎛ 0.0025 ⎞ = exp ⎜ ⎟ = 1.10 ⎝ 0.026 ⎠ Now 1 + x = (1 − x )(1.10) ⇒ x = 0.0476 ⇒ 4.76% 14.25 From Equation (14.62), vCE 2 ⎛ vCE1 ⎞ ⎛ ⎜1+ V ⎟ I ⎜ 1+ V AN ⎟ AN ⎜ = S3 ⋅⎜ ⎜ 1 + vEB ⎟ I S 4 ⎜ 1 + vEC 4 ⎜ V ⎟ ⎜ VAP AP ⎠ ⎝ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ For vCE 2 = 0.6 V, then vEC 4 = 5 V. We have vCE1 = 5 V so 644. 5 ⎞ ⎛ ⎛ 0.6 ⎞ ⎜ 1 + 80 ⎟ I S 3 ⎜ 1 + 80 ⎟ ⋅⎜ ⎜ ⎟= ⎟ ⎜ 1 + 0.6 ⎟ I S 4 ⎜ 1 + 5 ⎟ ⎜ ⎟ ⎜ ⎟ 80 ⎠ 80 ⎠ ⎝ ⎝ or I S 3 (1.0625) 2 = = 1.112 I S 4 (1.0075) 2 So I S 3 = (10−14 ) (1.112 ) or I S 3 = 1.112 × 10−14 A 14.26 By superposition: R vo (vi ) = − 2 ⋅ vi = −50vi R1 ⎛ R ⎞ vo (vos ) = ⎜ 1 + 2 ⎟ ⋅ vos = 51vos R1 ⎠ ⎝ So vo = vo ( vi ) + vo ( vos ) = −50vi + 51vos For vi = 20 mV and vos + 2.5 mV vo = −50(0.02) + 51(0.0025) = −0.8725 V For vi = 20 mV and vos = −2.5 mV vo = −50(0.02) + 51(−0.0025) = −1.1275 V So −1.1275 ≤ vo ≤ −0.8725 V 14.27 vo = −50vi = −50 ⎡ ±2.5 mV + sin ω t ( mV ) ⎤ ⎣ ⎦ vo = [ ±0.125 − 0.25sin ω t ] ( v ) 14.28 645. 0.5 × 10−3 = 5 × 10−8 A 104 Also i dV 1 I I = C o ⇒ Vo = ∫ Idt = ⋅ t dt C0 C Then 5 × 10−8 5= t ⇒ t = 103 s 10 × 10−6 I= 14.29 a. ⎛ 100 ⎞ | v01 | = 10 ⎜ 1 + ⎟ or | v01 | = 110 mV 10 ⎠ ⎝ Then ⎛ 50 ⎞ | v02 | = | v01 | (5) + 10 ⎜ 1 + ⎟ = (110)(5) + (10)(6) ⎝ 10 ⎠ or | v02 | = 610 mV 14.30 v0 due to vI 1 ⎞ ⎛ v0 = (0.5) ⎜1 + ⎟ = 0.9545 V ⎝ 1.1 ⎠ Wiper arm at V + = 10 V, (using superposition) ⎛ R1 || R5 ⎞ ⎛ 0.0909 ⎞ v1 = ⎜ ⎟ (10) = ⎜ ⎟ (10) R1 || R5 + R4 ⎠ ⎝ 0.0909 + 10 ⎠ ⎝ = 0.090 ⎛ 1⎞ Then v01 = − ⎜ ⎟ (0.090) = −0.090 ⎝ 1⎠ Wiper arm in center, v1 = 0 and v02 = 0 Wiper arm at V − = −10 V, v1 = −0.090 So v03 = 0.090 Finally, total output v0 : (from superposition) Wiper arm at V + , v0 = 0.8645 V Wiper arm in center, v0 = 0.9545 V Wiper arm at V − , 646. v0 = 1.0445 V 14.31 ′ a. R1′ = R2 = 0.5 || 25 = 0.490 kΩ or ′ R1′ = R2 = 490 Ω b. From Equation (14.75), ⎛ 125 × 10−6 ⎞ (0.026) ln ⎜ + (0.125) R1′ −14 ⎟ ⎝ 2 × 10 ⎠ ⎛ 125 × 10−6 ⎞ ′ = (0.026) ln ⎜ + (0.125) R2 −14 ⎟ ⎝ 2.2 × 10 ⎠ ′ 0.586452 + (0.125) R1′ = 0.583974 + (0.125) R2 ′ 0.002478 = (0.125)( R2 − R1′) ′ So R2 − R1′ = 0.0198 kΩ ⇒ 19.8 Ω Then R2 (1 − x) Rx R × Rx − 1 = 0.0198 R2 + (1 − x) Rx R1 + xRx (0.5)(1 − x)(50) (0.5)(50) x − = 0.0198 (0.5) + (1 − x)(50) (0.5) + x(50) 25(1 − x) 25 x − = 0.0198 50.5 − 50 x 0.5 + 50 x (0.5 + 50 x)(25 − 25 x) − (25 x)(50.5 − 50 x) = 0.0198 (50.5 − 50 x )(0.5 + 50 x) 25 {0.5 − 0.5 x + 50 x − 50 x 2 − 50.5 x + 50 x 2 } = 0.0198 {25.25 + 2525 x − 25 x − 2500 x 2 } 25 {0.5 − x} = 0.0198 {25.25 + 2500 x − 2500 x 2 } 0.5 − x = 0.019998 + 1.98 x − 1.98 x 2 1.98 x 2 − 2.98 x + 0.48 = 0 x= 2.98 ± (2.98) 2 − 4(1.98)(0.48) 2(1.98) So x = 0.183 and 1 − x = 0.817 14.32 R1′ = R1 ||15 = 0.5 ||15 = 0.4839 kΩ ′ R2 = R2 || 35 = 0.5 || 35 = 0.4930 kΩ From Equation (14.75), ⎛i ⎞ ⎛i ⎞ ′ (0.026) ln ⎜ C1 ⎟ + iC1 R1′ = (0.026) ln ⎜ C 2 ⎟ + iC 2 R2 ⎝ IS3 ⎠ ⎝ IS 4 ⎠ ⎛i ⎞ ′ (0.026) ln ⎜ C1 ⎟ = iC 2 R2 − iC1 R1′ ⎝ iC 2 ⎠ ⎛i ⎞ ⎡ i R′ ⎤ ′ (0.026) ln ⎜ C1 ⎟ = iC 2 R2 ⎢1 − C1 ⋅ 1 ⎥ ′ iC 2 ⎠ iC 2 R2 ⎦ ⎝ ⎣ ⎡ ⎛i ⎞ ⎛ i ⎞⎤ (0.026) ln ⎜ C1 ⎟ = iC 2 (0.4930) ⎢1 − (0.9815) ⎜ C1 ⎟ ⎥ ⎝ iC 2 ⎠ ⎝ iC 2 ⎠ ⎦ ⎣ By trial and error: 647. iC1 = 252 μ A and iC 2 = 248 μ A or iC1 = 1.0155 iC 2 14.33 From Eq. (14.79), we have ⎛ R ⎞ vo = I B1 R2 − I B 2 R3 ⎜ 1 + 2 ⎟ R1 ⎠ ⎝ I B1 = 1 μ A I B 2 = 2 μ A Setting vo = 0, we have ⎛ 200 ⎞ 0 = (10−6 )( 200 × 103 ) − ( 2 × 10−6 ) R3 ⎜ 1 + ⎟ 20 ⎠ ⎝ 200 × 10−3 R3 = ⇒ R = 9.09 K ( 2 ×10−6 ) (11) 3 14.34 1+ R2 = 80 R1 R1 = 6.329 K V f = vI = 5sin ω t ( mV ) I1 + I B = I 2 (a) VI = 0 ⇒ VX = 0 I 2 = I B VO = (10−6 )(500 × 103 ) ⇒ vo = 0.50 V (b) v − VX VX + IB = o R1 R2 ⎛ 1 ⎛ 1 v 1 ⎞ 1 ⎞ VX ⎜ + ⎟ + I B = 0 ⇒ vo = R2 ⎜ + ⎟ vI + I B R2 R1 ⎝ R1 R2 ⎠ ⎝ R1 R2 ⎠ vo = 80 ⎡5sin ω t ( mV ) ⎤ + (10−6 )( 500 × 103 ) ⎣ ⎦ vo = [ 0.5 + 0.4sin ω t ] ( v ) 14.35 a. 648. For I B 2 = 1 μ A, then v0 = − (10−6 )(104 ) or v0 = −0.010 V b. If a 10 kΩ resistor is included in the feedback loop Now v0 = − I B 2 (10) + I B1 (10) = 0 Circuit is compensated if I B1 = I B 2 . 14.36 From Equation (14.83), we have v0 = R2 I 0S where R2 = 40 kΩ and I 0 S = 3 μ A. Then v0 = ( 40 × 103 )( 3 × 10−6 ) or v0 = 0.12 V 14.37 a. Assume all bias currents are in the same direction and into each op-amp. v01 = I B1 (100 kΩ ) = (10−6 )(105 ) ⇒ v01 = 0.1 V Then v02 = v01 ( −5 ) + I B1 ( 50 kΩ ) = ( 0.1)( −5 ) + (10−6 )( 5 × 104 ) = −0.5 + 0.05 or v02 = −0.45 V b. Connect R3 = 10 ||100 = 9.09 kΩ resistor to noninverting terminal of first op-amp, and R3 = 10 || 50 = 8.33 kΩ resistor to noninverting terminal of second op-amp. 649. 14.38 a. For a constant current through a capacitor. 1 t v0 = ∫ I dt C 0 0.1× 10−6 or v0 = ⋅ t ⇒ v0 = (0.1)t 10−6 v0 = 1 V b. At t = 10 s, c. Then 100 × 10−12 v0 = ⋅ t ⇒ v0 = (10−4 )t −6 10 At t = 10 s, v0 = 1 mV 14.39 a. Assume all bias currents are into the op-amp. v01 = I B1 ( 50 kΩ ) = (10 ×10−6 )( 50 ×103 ) or v01 = v02 = 0.5 V v03 = ( −1)( v01 ) + (10 × 10−6 )( 20 × 103 ) or v03 = −0.3 V b. RA = 10 || 50 ⇒ RA = 8.33 kΩ RB = 20 || 20 ⇒ RB = 10 kΩ c. Assume the worst case offset current, that is, I 0 S = I B1 − I B 2 or I 0 S = I B 2 − I B1 . From Equation (14.83), v01 = R2 I 0 S = ( 50 × 103 )( 2 × 10−6 ) or v01 = v02 = 0.1 V v03 = ( −1) v01 − I 0 S R2 = ( −1)( 0.1) − ( 2 × 10−6 )( 20 × 103 ) or v03 = −0.14 V 14.40 a. Using Equation (14.79), Circuit (a), ⎛ 50 ⎞ v0 = ( 0.8 × 10−6 )( 50 × 103 ) − ( 0.8 ×10−6 )( 25 × 103 ) ⎜1 + ⎟ ⎝ 50 ⎠ or v0 = 0 Circuit (b), ⎛ 50 ⎞ v0 = ( 0.8 × 10−6 )( 50 × 103 ) − ( 0.8 × 10 −6 )(103 ) ⎜1 + ⎟ ⎝ 50 ⎠ −2 = 4 × 10 − 1.6 or v0 = −1.56 V b. Assume I B1 = 0.7 μ A and I B 2 = 0.9 μ A, then using Equation (14.79): Circuit (a), 650. ⎛ 50 ⎞ v0 = ( 0.7 × 10−6 )( 50 × 103 ) − ( 0.9 × 10−6 )( 25 × 103 ) ⎜ 1 + ⎟ ⎝ 50 ⎠ = 0.035 − 0.045 or v0 = −0.010 V Circuit (b), ⎛ 50 ⎞ v0 = ( 0.7 × 10−6 )( 50 × 103 ) − ( 0.9 × 10−6 )(106 ) ⎜1 + ⎟ ⎝ 50 ⎠ = 0.035 − 1.8 or v0 = −1.765 V 14.41 a. If R = 0, ⎛ 100 ⎞ v0,max = ⎜1 + ⎟ V0 S + I B (100 kΩ) 10 ⎠ ⎝ = (11)(10 × 10−3 ) + (2 × 10−6 )(100 × 103 ) v0,max = 0.110 + 0.20 ⇒ v0,max = 0.310 V b. R = 10.1|| 100 = 9.17 kΩ = R 14.42 a. ⎛ Ri ⎞ ⎜ ⎟ (15) = 0.010 V ⎝ Ri + R2 ⎠ 15 = 0.0006667 15 + R2 15(1 − 0.0006667) = 0.0006667 R2 Then R2 = 22.48 MΩ b. R1 = Ri || RF = 15 || 10 ⇒ R1 = 6 kΩ 14.43 a. Assume the offset voltage polarities are such as to produce the worst case values, but the bias currents are in the same direction. Use superposition: 651. Offset voltages ⎛ 100 ⎞ | v01 | = ⎜1 + ⎟ (10) = 110 mV =| v01 | 10 ⎠ ⎝ ⎛ 50 ⎞ | v02 | = (5)(110) + ⎜ 1 + ⎟ (10) ⎝ 10 ⎠ ⇒ | v02 | = 610 mV Bias Currents: v01 = I B (100 kΩ) = (2 ×10−6 )(100 × 103 ) = 0.2 V Then v02 = (−5)(0.2) + (2 × 10−6 )(50 × 103 ) = −0.9 V Worst case: v01 is positive and v02 is negative, then v01 = 0.31 V and v02 = −1.51 V b. Compensation network: If we want ⎛ RB ⎞ + + ⎜ ⎟ V = 20 mV and V = 10 V RB + RC ⎠ ⎝ ⎛ 8.33 ⎞ ⎜ ⎟ (10) = 0.020 ⎝ 8.33 + RC ⎠ or RC ≅ 4.15 MΩ 14.44 Assume bias currents are in same direction, but assume polarity of offset voltages are such as to produce the worst case output. a. Let I B1 = 5.5 μ A, I B 2 = 4.5 μ A Bias Current Effects: v01 = I B1 (50 kΩ) = 0.275 V ⇒ v02 = 0.275 V v03 = I B1 (20 kΩ) − v01 ⇒ v03 = −0.165 V Offset Voltage Effects: ⎛ 50 ⎞ v01 = (5) ⎜1 + ⎟ = 30 mV ⇒ v02 = 30 mV ⎝ 10 ⎠ ⎛ 20 ⎞ v03 = −v01 − 5 ⎜ 1 + ⎟ ⇒ v03 = −40 mV ⎝ 20 ⎠ Total Effect: v01 = 0.305 V and v02 = 0.305 V v03 = −0.205 V 14.45 For circuit (a), effect of bias current: v0 = (50 × 103 )(100 × 10−9 ) ⇒ 5 mV Effect of offset voltage 652. ⎛ 50 ⎞ v0 = (2) ⎜1 + ⎟ = 4 mV ⎝ 50 ⎠ So net output voltage is v0 = 9 mV For circuit (b), effect of bias current: Let I B 2 = 550 nA, I B1 = 450 nA, then from Equation (14.79), ⎛ 50 ⎞ v0 = (450 × 10−9 )(50 × 103 ) − (550 × 10−9 )(106 ) ⎜1 + ⎟ ⎝ 50 ⎠ −2 = 2.25 × 10 − 1.1 or v0 = −1.0775 V If the offset voltage is negative, then v0 = (−2)(2) = −4 mV So the net output voltage is v0 = −1.0815 V 14.46 a. At T = 25°C, V0 S = 2 mV so the output voltage for each circuit is v0 = 4 mV b. For T = 50°C, the offset voltage for is V0 S = 2 mV + (0.0067)(25) = 2.1675 mV so the output voltage for each circuit is v0 = 4.335 mV 14.47 a. At T = 25°C, V0 S = 1 mV, then ⎛ 50 ⎞ v01 = (1) ⎜ 1 + ⎟ ⇒ v01 = 6 mV ⎝ 10 ⎠ and ⎛ 60 ⎞ ⎛ 60 ⎞ v02 = v01 ⎜ 1 + ⎟ + (1) ⎜ 1 + ⎟ ⎝ 20 ⎠ ⎝ 20 ⎠ = 6(4) + (1)(4) ⇒ v02 = 28 mV b. At T = 50°C, V0 S = 1 + (0.0033)(25) = 1.0825 mV, then v01 = (1.0825)(6) ⇒ v01 = 6.495 mV and v02 = (6.495)(4) + (1.0825)(4) or v02 = 30.31 mV 14.48 25°C; I B = 500 nA, I 0 S = 200 nA 50°C, I B = 500 nA + (8 nA / °C)(25°C) = 700 nA I 0 S = 200 nA + (2 nA / °C)(25°C) = 250 nA a. Circuit (a): For I B , bias current cancellation, v0 = 0 Circuit (b): For I B , Equation (14.79), ⎛ 50 ⎞ v0 = (500 × 10−9 )(50 × 103 ) − (500 × 10 −9 )(106 ) ⎜ 1 + ⎟ ⎝ 50 ⎠ = 0.025 − 1.00 ⇒ v0 = −0.975 V b. Due to offset bias currents. Circuit (a): 653. v0 = (200 × 10−9 )(50 × 103 ) ⇒ v0 = 0.010 V Circuit (b): Let I B 2 = 600 nA I B1 = 400 nA Then ⎛ 50 ⎞ v0 = (400 × 10−9 )(50 × 103 ) − (600 × 10−9 )(106 ) ⎜ 1 + ⎟ ⎝ 50 ⎠ = 0.020 − 1.20 ⇒ v0 = −1.18 V c. Circuit (a): Due to I B , v = 0 Circuit (b): Due to I B , ⎛ 50 ⎞ v0 = (700 × 10−9 )(50 × 103 ) − (700 × 10−9 )(106 ) ⎜1 + ⎟ ⎝ 50 ⎠ = 0.035 − 1.40 ⇒ v0 = −1.365 V Circuit (a): Due to I 0 S , v0 = (250 × 10−9 )(50 × 103 ) ⇒ v0 = 0.0125 V Circuit (b): Due to I 0 S , Let I B 2 = 825 nA I B1 = 575 nA Then ⎛ 50 ⎞ v0 = (575 × 10−9 )(50 × 103 ) − (825 × 10−9 )(106 ) ⎜ 1 + ⎟ ⎝ 50 ⎠ = 0.02875 − 1.65 ⇒ v0 = −1.62 V 14.49 25°C; I B = 2 μ A, I 0 S = 0.2 μ A 50°C, I B = 2 μ A + (0.020 μ A / °C)(25°C ) = 2.5 μ A I 0 S = 0.2 μ A + (0.005 μ A / °C)(25°C) = 0.325 μ A a. Due to I B : (Assume bias currents into op-amp). v01 = I B (50 kΩ) = (2 × 10−6 )(50 × 103 ) ⇒ v01 = 0.10 V ⎛ 60 ⎞ ⎛ 60 ⎞ v02 = v01 ⎜ 1 + ⎟ + I B (60 kΩ) − I B (50 kΩ) ⎜ 1 + ⎟ 20 ⎠ ⎝ ⎝ 20 ⎠ 3 −6 −6 = (0.1)(4) + (2 × 10 )(60 × 10 ) − (2 × 10 )(60 × 103 )4 or v02 = 0.12 V b. Due to I 0 S : 1st op-amp. Let I B1 = 2.1 μ A 2nd op-amp. Let I B1 = 2.1 μ A I B 2 = 1.9 μ A v01 = I B1 (50 kΩ) = (2.1× 10−6 )(50 × 103 ) ⇒ v01 = 0.105 V ⎛ 60 ⎞ ⎛ 60 ⎞ v02 = v01 ⎜ 1 + ⎟ + I B1 (60 kΩ) − I B 2 (50 kΩ) ⎜1 + ⎟ 20 ⎠ ⎝ ⎝ 20 ⎠ 3 −6 = (0.105)(4) + (2.1× 10 )(60 × 10 ) − (1.9 × 10 −6 )(50 × 103 )(4) or v02 = 0.166 V c. Due to I B : 654. v01 = (2.5 × 10 −6 )(50 × 103 ) ⇒ v01 = 0.125 V ⎛ 60 ⎞ ⎛ 60 ⎞ v01 = v02 ⎜ 1 + ⎟ + I B (60 kΩ) − I B (50 kΩ) ⎜1 + ⎟ 20 ⎠ ⎝ ⎝ 20 ⎠ −6 3 = (0.125)(4) + (2.5 × 10 )(60 × 10 ) − (2.5 × 106 )(50 × 103 (4) or v02 = 0.15 V Due to I 0 S : Let I B1 = 2.625 μ A I B 2 = 2.3375 μ A v01 = I B1 (50 kΩ) = (2.6625 × 10−6 )(50 × 103 ) ⇒ v01 = 1.133 V ⎛ 60 ⎞ ⎛ 60 ⎞ v02 = v01 ⎜ 1 + ⎟ + I B1 (60 kΩ) − I B 2 (50 kΩ) ⎜ 1 + ⎟ ⎝ 20 ⎠ ⎝ 20 ⎠ = (0.133)(4) + (2.6625 × 10−6 )(60 × 103 ) − (2.3375 × 10−6 )(50 × 103 )(4) or v02 = 0.224 V 14.50 ⎛ R4 ⎞ ⎛ R2 ⎞ vB = ⎜ ⎟ vI 1 and v0 (vI 2 ) = vB ⎜ 1 + ⎟ R1 ⎠ ⎝ ⎝ R3 + R4 ⎠ or ⎛ R4 ⎞ ⎛ R2 ⎞ v0 (vI 2 ) = ⎜ ⎟ ⎜ 1 + ⎟ vI 2 R1 ⎠ ⎝ R3 + R4 ⎠ ⎝ or vI 1 . v0 (vI 1 ) = − R2 ⋅ vI R1 Then ⎛ R4 ⎞⎛ R2 ⎞ R2 v0 = ⎜ ⎟⎜ 1 + ⎟ vI 2 − ⋅ vI 1 R1 ⎠ R1 ⎝ R3 + R4 ⎠ ⎝ V V we can write vI 2 = vcm + d and vI 1 = Vcm − d ⋅ Then 2 2 ⎛ R4 ⎞⎛ R2 ⎞ ⎛ Vd ⎞ R2 ⎛ Vd ⎞ v0 = ⎜ ⎟⎜ 1 + ⎟ ⎜ Vcm + ⎟ − ⋅ ⎜ Vcm − ⎟ R1 ⎠ ⎝ 2 ⎠ R1 ⎝ 2⎠ ⎝ R3 + R4 ⎠ ⎝ Common-mode gain ⎛ R4 ⎞ ⎛ R2 ⎞ R2 v Acm = 0 = ⎜ ⎟ ⎜1 + ⎟ − Vcm ⎝ R3 + R4 ⎠ ⎝ R1 ⎠ R1 655. Differential mode gain v 1 ⎡⎛ R4 ⎞⎛ R2 ⎞ R2 ⎤ Ad = 0 = ⎢⎜ ⎟⎜ 1 + ⎟ + ⎥ Vd 2 ⎣⎝ R3 + R4 ⎠ ⎝ R1 ⎠ R1 ⎦ Then A CM RR = d Acm 1 ⎡⎛ R4 ⎞ ⎛ R2 ⎞ R2 ⎤ ⋅ ⎢⎜ ⎟ ⎜1 + ⎟ + ⎥ R1 ⎠ R1 ⎦ 2 ⎣⎝ R3 + R4 ⎠ ⎝ = ⎡⎛ R4 ⎞ ⎛ R2 ⎞ R2 ⎤ ⎢⎜ ⎟ ⎜1 + ⎟ − ⎥ R1 ⎠ R1 ⎦ ⎣⎝ R3 + R4 ⎠ ⎝ ⎡ ⎢ 1 ⎢ R4 ⋅ 2 ⎢ R3 ⎢ ⎢ ⎣ ⎤ ⎥ ⎛ R2 ⎞ R2 ⎥ 1 ⋅ ⋅ ⎜1 + ⎟ + R1 ⎠ R1 ⎥ ⎛ R4 ⎞ ⎝ 1+ ⎟ ⎥ ⎜ ⎥ ⎝ R3 ⎠ ⎦ CM RR = ⎛ R2 ⎞ R2 R4 1 ⋅ ⋅ ⎜1 + ⎟ − R3 ⎛ R4 ⎞ ⎝ R1 ⎠ R1 ⎜1 + ⎟ ⎝ R3 ⎠ Minimum CMRR ⇒ Maximum denominator R R ⇒ maximum 4 and minimum 2 . Then R3 R1 R4 (1.02)(50) = = 5.204 R3 (0.98)(10) R2 (0.98)(50) = = 4.804 R1 (1.02)(10) Then 1 ⎡ 5.204 ⎤ ⋅ (5.804) + (4.804) ⎥ 2 ⎢ 6.204 ⎦ CMRR = ⎣ ⎡ 5.204 ⎤ ⎢ 6.204 ⋅ (5.804) − (4.804) ⎥ ⎣ ⎦ 1 ⋅ (9.6725) = 2 (0.06447) CMRR = 75.0 ⇒ CMRRdB = 20 log10 (75.0) ⇒ CMRRdB = 37.5 dB 14.51 Use the results of Problem 14.50: R 1 + x ⎛ 50 ⎞ Let 4 = ⋅ ⎜ ⎟ ≈ (1 + 2 x)(5) R3 1 − x ⎝ 10 ⎠ R 1 − x ⎛ 50 ⎞ Let 2 = ⋅ ⎜ ⎟ ≈ (1 − 2 x)(5) R1 1 + x ⎝ 10 ⎠ Then 656. CMRR = ⎤ 1 ⎡ (1 + 2 x ) 5 ⋅ ( 6 − 10 x ) + (1 − 2 x )( 5 ) ⎥ ⎢ 2 ⎣ 6 + 10 x ⎦ ⎡ (1 + 2 x ) 5 ⎤ ⋅ ( 6 − 10 x ) − (1 − 2 x )( 5 ) ⎥ ⎢ ⎣ 6 + 10 x ⎦ 1 2 ⎡30 + 10 x − 100 x + 30 − 10 x − 100 x 2 ⎤ ⎣ ⎦ = 2 2 2 ⎡30 + 10 x − 100 x − ( 30 − 10 x − 100 x ) ⎤ ⎣ ⎦ a. 20 x = b. 20 x = 1 ⋅ [60 − 200 x 2 ] 30 − 100 x 2 2 = = 20 x 20 x For CMRRdB = 90 dB ⇒ CMRR = 31, 623 x will be small, neglect the x 2 term. Then 30 ⇒ x = 0.0000474 = 0.00474% 31, 623 For CMRRdB = 60 dB ⇒ CMRR = 1000. Then 30 ⇒ x = 0.0015 = 0.15% 1000 657. Chapter 15 Exercise Solutions EX15.1 For the circuit shown in Figure 15.7 1 f 3dB = 2π RC or 1 1 RC = = = 3.979 × 10−6 2π f3dB 2π ( 40 × 103 ) For R = 75 K Then C = 5.31×10−11 = 53.1 pF We have C3 = 1.414C = 75.1 pF C4 = 0.707C = 37.5 pF EX15.2 1 fC = CReq or C= 1 1 = f c Req (105 )( 20 × 106 ) C = 0.5 pF EX15.3 Low-frequency gain: T = − f 3dB = C1 30 =− = −6 C2 5 3 −12 fC C2 (100 × 10 )( 5 × 10 ) = ⇒ f 3dB = 6.63 kHz 2π CF 2π (12 × 10−12 ) EX15.4 1 f0 = 2π 3RC 1 1 RC = = = 6.13 × 10−6 3 2π f 0 3 2π (15 × 10 ) 3 Let C = 0.001 μF = 1 nF Then R = 6.13 kΩ so R2 = 8R = 49 kΩ EX15.5 f0 = C= 1 1 ⇒C = 2π RC 2π f 0 R 1 2π ( 800 ) (104 ) ⇒ C ≅ 0.02 μ F R2 = 2 R1 = 2 (10 ) ⇒ R2 = 20 kΩ EX15.6 658. ⎛ R1 ⎞ VTH = ⎜ ⎟ VH ⎝ R1 + R2 ⎠ ⎛ R1 ⎞ 2=⎜ ⎟ (12) ⎝ R1 + 20 ⎠ 2 ( R1 + 20 ) = 12 R1 40 = 10 R1 ⇒ R1 = 4 kΩ EX15.7 ⎛ R1 ⎞ VTH − VTL = ⎜ ⎟ (VH − VL ) ⎝ R1 + R2 ⎠ ⎛ R1 ⎞ 0.10 = ⎜ ⎟ (10 − [ −10]) ⎝ R1 + R2 ⎠ 1+ R2 R 20 = = 200 ⇒ 2 = 199 R1 0.10 R1 ⎛ R2 ⎞ VS = ⎜ ⎟ VREF ⎝ R1 + R2 ⎠ ⎛ R ⎞ 1 ⎞ ⎛ VREF = ⎜1 + 1 ⎟ VS = ⎜ 1 + ⎟ (1) ⇒ VREF = 1.005 V R2 ⎠ ⎝ 199 ⎠ ⎝ I= VH − VBE ( on ) − Vγ R + 0.1 10 − 0.7 − 0.7 R + 0.1 = = 43 kΩ 0.2 R = 42.9 kΩ EX15.8 At t = 0− , let v0 = −5 so v X = −2.5. For t > 0 ⎛ −t ⎞ v X = 10 + ( −2.5 − 10 ) exp ⎜ ⎟ ⎝ rX ⎠ When v X = 5.0, output switches ⎛ t ⎞ 5.0 = 10 − 12.5 exp ⎜ − 1 ⎟ ⎝ rX ⎠ ⎛ t ⎞ 10 − 5 5.0 = exp ⎜ − 1 ⎟ = ⎝ rX ⎠ 12.5 12.5 ⎛ t ⎞ 12.5 ⎛ 12.5 ⎞ exp ⎜ + 1 ⎟ = ⇒ t1 = rX ⋅ ln ⎜ ⎟ ⇒ t1 = rX ( 0.916 ) ⎝ 5.0 ⎠ ⎝ rX ⎠ 5.0 During the next part of the cycle ⎛ t ⎞ v X = −5 + ( 5 − [ −5]) exp ⎜ − ⎟ ⎝ rX ⎠ When v X = −2.5, output switches 659. ⎛ t ⎞ −2.5 = −5 + 10 exp ⎜ − 2 ⎟ ⎝ rX ⎠ ⎛ t ⎞ 5 − 2.5 2.5 exp ⎜ − 2 ⎟ = = 10 10 ⎝ rX ⎠ ⎛ t ⎞ 10 ⎛ 10 ⎞ exp ⎜ + 2 ⎟ = ⇒ t2 = rX ⋅ ln ⎜ ⎟ ⇒ t2 = rX (1.39 ) rX ⎠ 2.5 ⎝ 2.5 ⎠ ⎝ 1 Period = t1 + t2 = T = ⎡( 0.916 ) + (1.39 ) ⎤ rX = 2.31rX ⇒ Frequency = ⎣ ⎦ 2.31rX rX = ( 50 × 103 )( 0.01× 10 −6 ) = 5 × 10 −4 s ⇒ f = 866 Hz Duty cycle = ( 0.916 ) t1 × 100% = × 100% ⇒ Duty cycle = 39.7% t1 + t2 ( 0.916 ) + (1.39 ) EX15.9 a. rX = RX C X ⎛ R1 ⎞ ⎛ 10 ⎞ vY = ⎜ ⎟ v0 = ⎜ ⎟ (12 ) = 1.2 V ⎝ 10 + 90 ⎠ ⎝ R1 + R2 ⎠ R1 = 0.10 β= R1 + R2 0.7 ⎤ ⎡ ⎢ 1 + 12 ⎥ ⎡1 + Vγ /VP ⎤ T = rX ln ⎢ ⎥ ⎥ = rX ln ⎢ ⎣ 1− β ⎦ ⎢1 − (0.10) ⎥ ⎢ ⎥ ⎣ ⎦ T = 50 × 10−6 = rX ln [1.18] = (0.162) rX 50 × 10−6 ⇒ RX = 3.09 kΩ (0.1× 10−6 )(0.162) b. Recovery time ⎛ t ⎞ v X = VP + (−1.2 − VP ) exp ⎜ − ⎟ ⎝ rX ⎠ RX = When v X = Vγ , t = t2 660. ⎛ t ⎞ 0.7 = 12 + ( −1.2 − 12 ) exp ⎜ − 2 ⎟ ⎝ rX ⎠ ⎛ t ⎞ 12 − 0.7 = 0.856 exp ⎜ − 2 ⎟ = 13.2 ⎝ rX ⎠ ⎛ 1 ⎞ t2 = rX ln ⎜ ⎟ = ( 0.155 ) rX ⎝ 0.856 ⎠ rX = ( 3.09 × 103 )( 0.1× 10−6 ) = 3.09 × 10−4 ⇒ t2 = 48.0 μ s EX15.10 T = 1.1 RC T = 75 × 10−6 Let C = 10 nF Then 75 × 10−6 R= = 6.82 K (1.1) (10 ×10−9 ) EX15.11 1 1 = ⇒ f = 802 Hz 0.693 ( RA + 2 RB ) C ( 0.693) ⎡ 20 + 2 ( 80 ) ⎤ × 103 × ( 0.01× 10−6 ) ⎣ ⎦ R + RB 20 + 80 Duty cycle = A × 100% = × 100% ⇒ Duty cycle = 55.6% 20 + 2 ( 80 ) RA + 2 RB f = EX15.12 P= a. 1 VP2 ⋅ 2 RL VP = 2 RL P = 2 ( 8 )(1) ⇒ VP = 4 V IP = VP 4 = ⇒ I P = 0.5 A RL 8 VCE = 12 − 4 = 8 V b. I C ≈ 0.5 A So P = I C ⋅ VCE = ( 0.5 )( 8 ) ⇒ P = 4 W EX15.13 VP = 2 RL PL = 2 ( 8 )(10 ) = 12.65 V ⎛ V ⎞ PS = VS ⎜ P ⎟ ⎝ π RL ⎠ VS = PS π R2 (10 ) π ( 8 ) = VP 12.65 VS = 19.9 V EX15.14 Line regulation = Now dV0 dV dV = 0 ⋅ Z+ + dV dVZ dV 661. dV0 ⎛ 10 ⎞ = ⎜1 + ⎟ = 2 dVZ ⎝ 10 ⎠ dVZ ⎛ rZ ⎞ 10 =⎜ = 0.00227 ⎟= + dV ⎝ rZ + R1 ⎠ 10 + 4400 So Line regulation = ( 2 )( 0.00227 ) = 0.00454 0.454% EX15.15 V1 V0 − V1 ⎛1 1⎞ V = ⇒ V1 ⎜ + ⎟ = 0 10 10 ⎝ 10 10 ⎠ 10 V ⎛ 2⎞ V V1 ⎜ ⎟ = 0 ⇒ V0 = 2V1 ⇒ V1 = 0 10 ⎠ 10 2 ⎝ V0 − V1 V0 V0 − A0 L (VZ − V1 ) + + =0 RL R0 10 V0 V0 V0 A0 LVZ V1 A0 LV1 + + − = − R0 R0 10 RL R0 10 = V0 A V − 0L 0 2(10) 2 R0 V0 V 1000 ( 6.3) V0 (1000 ) V0 + I0 + 0 − = − 10 0.5 0.5 20 2 ( 0.5 ) V0 [0.10 + 2.0 − 0.05 + 1000] + I 0 = 12, 600 V0 (1002.05) + I 0 = 12, 600 For I 0 = 1 mA ⇒ V0 = 12.5732 For I 0 = 100 mA ⇒ V0 = 12.4744 Load reg = V0 ( NL ) − V0 ( FL ) V0 ( NL ) × 100% 12.5732 − 12.4744 × 100% 12.5732 Load reg = 0.786% = EX15.16 a. 662. IC 3 = VZ − 3VBE ( on ) IC 3 = R1 + R2 + R3 5.6 − 3 ( 0.6 ) 3.9 + 3.4 + 0.576 ⎛I ⎞ I C 4 R4 = VT ln ⎜ C 3 ⎟ ⎝ IC 4 ⎠ = 3.8 ⇒ I C 3 = 0.482 mA 7.88 ⎛ 0.482 ⎞ I C 4 (0.1) = (0.026) ln ⎜ ⎟ ⎝ IC 4 ⎠ By trial and error I C 4 = 0.213 mA VB 7 = 2(0.6) + (0.482)(3.9) ⇒ VB 7 = 3.08 V b. ⎛ R13 ⎞ ⎜ ⎟ V0 = VB 8 = VB 7 ⎝ R13 + R12 ⎠ ⎛ 2.23 ⎞ ⎜ ⎟ (5) = 3.08 ⎝ 2.23 + R12 ⎠ ( 2.23)( 5 ) = ( 3.08 )( 2.23) + ( 3.08) R12 11.15 = 6.868 = 3.08R12 ⇒ R12 = 1.39 kΩ TYU15.1 1 2π RC 1 1 RC = = = 1.59 × 10−5 2π f3dB 2π (104 ) f 3dB = Let C = 0.01 μ F ⇒ R = 1.59 kΩ Then C1 = 0.03546 μ F C2 = 0.01392 μ F C3 = 0.002024 μ F 1 T = 6 = 1 ⎛ f ⎞ ⎛ 20 ⎞ 1+ ⎜ ⎟ 1+ ⎜ ⎟ ⎝ 10 ⎠ ⎝ f3d ⎠ T = 0.124 or T = −18.1 dB 6 TYU15.2 f 3dB = RC = 1 1 ⇒ RC = 2π RC 2π f3dB 1 2π ( 50 × 103 ) = 3.18 × 10−6 Let C = 0.001 μ F = 1 nF ⇒ R = 3.18 kΩ Then 663. R1 = 2.94 kΩ R2 = 3.44 kΩ R3 = 1.22 kΩ R4 = 8.31 kΩ T = 0.01 = 1 ⎛ f ⎞ 1 + ⎜ 3− dB ⎟ ⎝ f ⎠ 8 8 2 ⎛ f ⎞ ⎛ 1 ⎞ 4 1 + ⎜ 3− dB ⎟ = ⎜ ⎟ = 10 f ⎠ ⎝ 0.01 ⎠ ⎝ 2 ⎛ f3− dB ⎞ f 3dB ⇒ f ≅ 15.8 kHz ⎜ ⎟ ≅ 10 ⇒ f = 10 ⎝ f ⎠ TYU15.3 1-pole 2-pole 3-pole 4-pole T = T = T = T = 1 ⎛ 12 ⎞ 1+ ⎜ ⎟ ⎝ 10 ⎠ 1 2 ⎛ 12 ⎞ 1+ ⎜ ⎟ ⎝ 10 ⎠ 1 4 ⎛ 12 ⎞ 1+ ⎜ ⎟ ⎝ 10 ⎠ 6 1 ⎛ 12 ⎞ 1+ ⎜ ⎟ ⎝ 10 ⎠ 8 ⇒ −3.87 dB ⇒ −4.88 dB ⇒ −6.0 dB ⇒ −7.24 dB TYU15.4 1 Req = fC C or f C C = 1 1 = = 2 × 10−7 Req 5 × 106 If C = 10 pF ⇒ fC = 20 kHz TYU15.5 1 1 f0 = = ⇒ f 0 ≅ 65 kHz 4 2π 6 RC 2π 6 (10 )(100 × 10−12 ) R2 = 29 R = 29 (104 ) ⇒ R2 = 290 kΩ TYU15.6 664. f0 = 1 ⎛ CC ⎞ 2π L ⋅ ⎜ 1 2 ⎟ ⎝ C1 + C2 ⎠ = 1 ⎡ (10−9 ) 2 2π (10 ) ⎢ −9 ⎣ 2 × 10 −6 ⎤ ⋅⎥ ⎦ ⇒ f 0 = 7.12 MHz C2 = gm R C1 C2 1 1 ⋅ = ⇒ g m = 0.25 mA / V C1 R 4 × 103 We have ⎛ k ′ ⎞⎛ W ⎞ g m = 2 ⎜ ⎟⎜ ⎟ (VGS − VTh ) ⎝ 2 ⎠⎝ L ⎠ k ′ ≅ 20 μ A / V 2 , VGS − VTh ≅ 1 V gm = So W 0.25 × 10−3 = = 12.5 L ( 20 × 10−6 ) (1) and a value of W / L = 12.5 is certainly reasonable. TYU15.7 ⎛R ⎞ VTH = − ⎜ 1 ⎟ VL ⎝ R2 ⎠ ⎛R ⎞ R 0.10 = − ⎜ 1 ⎟ ( −10 ) ⇒ 1 = 0.010 R2 ⎝ R2 ⎠ Let R1 = 0.10 kΩ then R2 = 10 kΩ TYU15.8 a. ⎛ R2 ⎞ ⎛ 10 ⎞ VS = ⎜ ⎟ VREF = ⎜ ⎟ ( 2) ⎝ 1 + 10 ⎠ ⎝ R1 + R2 ⎠ VS = 1.82 V ⎛ R1 ⎞ ⎛ 1 ⎞ VTH = VS + ⎜ ⎟ VH = 1.82 + ⎜ ⎟ (10 ) R1 + R2 ⎠ ⎝ 1 + 10 ⎠ ⎝ VTH = 2.73 V VTL = b. ⎛ R1 ⎞ ⎛ 1 ⎞ VS + ⎜ ⎟ VL = 1.82 + ⎜ ⎟ ( −10 ) R1 + R2 ⎠ ⎝ 1 + 10 ⎠ ⎝ VTL = 0.91 V 665. TYU15.9 ⎛ R ⎞ VS = ⎜ 1 + 1 ⎟ VREF ⎝ R2 ⎠ ⎛R ⎞ ⎛R ⎞ VTH = VS − ⎜ 1 ⎟ VL and VTL = VS − ⎜ 1 ⎟ VH ⎝ R2 ⎠ ⎝ R2 ⎠ ⎛R ⎞ Hysteresis Width = VTH − VTL = ⎜ 1 ⎟ (VH − VL ) ⎝ R2 ⎠ ⎛R ⎞ ⎛R ⎞ 2.5 = ⎜ 1 ⎟ ( 5 − [ −5]) = 10 ⎜ 1 ⎟ R2 ⎠ ⎝ ⎝ R2 ⎠ R So 1 = 0.25 R2 Then ⎛ R ⎞ VS = −1 = ⎜ 1 + 1 ⎟ VREF = (1 + 0.25)VREF ⇒ VREF = −0.8 V R2 ⎠ ⎝ Then VTH = −1 − ( 0.25 )( −5 ) ⇒ VTH = 0.25 V VTL = −1 − ( 0.25 )( 5 ) ⇒ VTL = −2.25 V TYU15.10 ⎛ R1 ⎞ 1 ⎛ 10 ⎞ vX = ⎜ ⎟ v0 = ⎜ ⎟ v0 = v0 3 ⎝ 10 + 20 ⎠ ⎝ R1 + R2 ⎠ 10 t = 0, vX = − 3 ⎛ t ⎞ ⎛ 10 ⎞ v X = 10 + ⎜ − − 10 ⎟ exp ⎜ − ⎟ ⎝ 3 ⎠ ⎝ rX ⎠ Output switches when v X = 10 3 666. 10 = 10 − 13.33 exp 3 ⎛ t1 ⎞ ⎜− ⎟ ⎝ rX ⎠ ⎛ t ⎞ 10 − 3.33 6.67 = exp ⎜ − 1 ⎟ = 13.33 13.33 ⎝ rX ⎠ ⎛ t ⎞ 13.33 ≅2 exp ⎜ + 1 ⎟ = ⎝ rX ⎠ 6.67 t1 = rX ln (2) = (0.693)rX T = 2(0.693)rX f = 1 2(0.693)rX rX = RX C X = (104 )( 0.1×10 −6 ) = 1×10 −3 ⇒ f = 722 Hz ⇒ Duty cycle = 50% TYU15.11 ⎛ R1 ⎞ 20 = 0.333 β =⎜ ⎟= R1 + R2 ⎠ 20 + 40 ⎝ rX = RX C X = (104 )( 0.01× 10−6 ) = 1× 10−4 ⎛ 1 + Vγ / VP T = rX ln ⎜ ⎝ 1− β Recovery time 0.7 ⎤ ⎡ ⎢ 1+ 8 ⎥ ⎞ −4 ⎥ ⇒ T = 48.9 μ s ⎟ = (1× 10 ) ln ⎢ ⎠ ⎢1 − 0.333 ⎥ ⎢ ⎥ ⎣ ⎦ 667. ⎛ R1 ⎞ ⎛ 20 ⎞ vY = ⎜ ⎟ v0 = ⎜ ⎟ (8) = 2.667 V R1 + R2 ⎠ ⎝ 20 + 40 ⎠ ⎝ ⎛ t ⎞ 0.7 = 8 + ( −2.667 − 8 ) exp ⎜ − 2 ⎟ ⎝ rX ⎠ ⎛ t ⎞ 8 − 0.7 exp ⎜ − 2 ⎟ = = 0.6844 ⎝ rX ⎠ 10.66 ⎛ 1 ⎞ t2 = rX ln ⎜ ⎟ ⇒ t2 = 37.9 μ s ⎝ 0.685 ⎠ TYU15.12 f = 1 ( 0.693)( RA + RB ) C RA + RB = 1 ( 0.693) fC Let C = 0.01 μ F, RA + RB = f = 1kHz 1 ( 0.693) (103 )( 0.01×10−6 ) Duty cycle = 55 = = 1.443 × 105 RA + RB × 100% RA + 2 RB (1.443 ×10 ) (100 ) (1.443 ×10 ) + R (1.443 ×10 ) (100 − 55) ⇒ R = 5 55 = 5 B 5 RB 55 B = 118 kΩ so RA = 26.2 kΩ TYU15.13 v01 ⎛ R2 ⎞ ⎛ 30 ⎞ a. = ⎜ 1 + ⎟ = ⎜1 + ⎟ = 2.5 vI ⎝ R1 ⎠ ⎝ 20 ⎠ v02 R 50 =− 4 =− = −2.5 20 vI R3 (b) P= 1 VL2 1 [12 − (−12)]2 ⋅ = ⋅ = 240 mW 2 RL 2 1.2 Or P = 0.24 W c. 12 = V pi = 4.8 V 2.5 668. Chapter 15 Problem Solutions 15.1 (a) For example: vo ⎛ R2 ⎞ R = ⎜1 + ⎟ = 10 ⇒ 2 = 9 vi ⎝ R1 ⎠ R1 Corner Frequency: 1 f = = 5 × 103 ⇒ RC = 3.18 × 10−5 2π RC (b) For Example: Low-Frequency: 1 − R2 jω C vo R 1 = =− 2⋅ vi R1 R1 1 + jω R2 C So, set R2 = 15 ⇒ For example, R1 = 10 k Ω, R2 = 150 k Ω R1 R2 C = 1 1 = = 1.06 × 10−5 2π f3− dB 2π (15 × 103 ) Then C = 70.7 pF 15.2 (a) Av = 1 ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ f3− dB ⎠ 2 = 1 1 + (2) 2 = 0.447 ⇒ Av = −7 dB 669. (b) (c) Av = Av = 1 ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ f3− dB ⎠ 1 2 ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ f3− dB ⎠ 2 = = 1 1 + (2) 4 1 1 + (2)6 = 0.2425 ⇒ Av = −12.3 dB = 0.1240 ⇒ Av = −18.1 dB 15.3 (a) Figure 15.6 1 f = 2π RC 1 1 RC = = = 7.958 × 10−6 2π f 2π (20 × 103 ) Let R1 = R2 = R = 10 K C = 795.8 pF So C4 = 0.707C = 562.6 pF C3 = 1.414C = 1.125 nF (b) 1 T = = 0.777 (i) 4 ⎛ 18 ⎞ 1+ ⎜ ⎟ ⎝ 20 ⎠ 1 T = = 0.707 (ii) 4 ⎛ 20 ⎞ 1+ ⎜ ⎟ ⎝ 20 ⎠ 1 T = = 0.637 (iii) 4 ⎛ 22 ⎞ 1+ ⎜ ⎟ ⎝ 20 ⎠ 15.4 Use Figure 15.10(b) 1 f3− dB = 2π RC or 1 RC = = 3.18 × 10−6 2π (50 × 103 ) For example, let C = 100 pF Then R = 31.8 kΩ And R1 = 8.97 kΩ R2 = 22.8 kΩ R3 = 157 kΩ From Equation (15.26) 1 T = 6 ⎛ f ⎞ 1 + ⎜ 3dB ⎟ ⎝ f ⎠ 670. We find f kHz 30 35 40 45 |T| 0.211 0.324 0.456 0.589 15.5 From Equation (15.7). Y1Y2 T (s) = Y1Y2 + Y4 (Y1 + Y2 + Y3 ) For a high-pass filter, let Y1 = Y2 = sC , Y3 = 1 1 , and Y4 = R3 R4 Then s 2C 2 1 ⎛ 1 ⎞ s 2 C 2 + ⎜ sC + sC + ⎟ R4 ⎝ R3 ⎠ 1 = 1 ⎛ 1 ⎞ 1+ ⎜2+ ⎟ sR4 C ⎝ sR3C ⎠ T (s) = Define r3 = R3C and r4 = R4 C 1 1 ⎛ 1 ⎞ 1+ ⎜2+ ⎟ sr4 ⎝ sr3 ⎠ Set s = jω T (s) = 1 1 ⎛ 1 ⎞ 1+ ⎜2+ ⎟ jω r4 ⎝ jω r3 ⎠ 1 = j ⎛ j ⎞ 1− ⎜2− ⎟ ω r4 ⎝ ω r3 ⎠ T ( jω ) = = 1 ⎛ 1 ⎞ 2j ⎜1 − 2 ⎟− ω r3 r4 ⎠ ω r4 ⎝ −1/ 2 2 ⎧⎛ 1 ⎞ 4 ⎫ ⎪ ⎪ ( jω ) = ⎨⎜ 1 − 2 T ⎟ + 2 2⎬ ω r3 r4 ⎠ ω r4 ⎪ ⎪⎝ ⎩ ⎭ For a maximally flat filter, we want dT =0 dω ω →∞ Taking the derivative, we find d T ( jω ) dω 2 1 ⎧⎛ 1 ⎞ 4 ⎫ ⎪ ⎪ = − ⎨⎜ 1 − 2 ⎟ + 2 2⎬ 2 ⎪⎝ ω r3 r4 ⎠ ω r4 ⎪ ⎩ ⎭ −3 / 2 ⎡ ⎛ 1 ⎞ ⎛ 2 ⎞ 4(−2) ⎤ × ⎢ 2 ⎜1 − 2 ⎟⎜ 3 ⎟+ 3 2 ⎥ ⎢ ⎝ ω r3 r4 ⎠ ⎝ ω r3 r4 ⎠ ω r4 ⎥ ⎣ ⎦ 671. or d T ( jω ) dω =0 ω →∞ ⎡⎛ 4 ⎞⎛ 1 ⎞ 8 ⎤ = ⎢⎜ 3 ⎟⎜ 1 − 2 ⎟− 3 2 ⎥ ⎢⎝ ω r3 r4 ⎠⎝ ω r3 r4 ⎠ ω r4 ⎥ ⎣ ⎦ = 4 ⎡ 1 ⎢ ω 3 ⎢ r3 r4 ⎣ Then ⎡ 1 ⎛ 1 ⎞ 2⎤ ⎢ ⎜1 − 2 ⎟− 2 ⎥ ⎢ r3 r4 ⎝ ω r3 r4 ⎠ r4 ⎥ ⎣ ⎦ ⎛ 1 ⎞ 2⎤ ⎜1 − 2 ⎟− 2 ⎥ ⎝ ω r3 r4 ⎠ r4 ⎥ ⎦ =0 ω →∞ 1 2 = ⇒ 2r3 = r4 r3 r4 Then the transfer function can be written as: So that 2 ⎧⎡ ⎤ 1 4 ⎫ ⎪ ⎪ T ( jω ) = ⎨ ⎢1 − 2 2 ⎥ + 2 2 ⎬ ω (2r3 ) ⎦ ω (4r3 ) ⎪ ⎪⎣ ⎩ ⎭ −1/ 2 ⎧ 1 1 1 ⎫ = ⎨1 − 2 2 + + ⎬ ω r3 4(ω 2 r32 ) 2 ω 2 r32 ⎭ ⎩ −1/ 2 −1/ 2 ⎧ ⎫ 1 = ⎨1 + 2 2 2 ⎬ ⎩ 4(ω r3 ) ⎭ 3 − dB frequency 1 2ω 2 r32 = 1 or ω = = 2r3 Define 1 ω= RC So that R R3 = 2 1 2 R3C We had 2r3 = r4 or 2( R3C ) = R4 C ⇒ R4 = 2 R3 So that R4 = 2 ⋅ R 15.6 From Equation (15.25) 1 T = − 25 dB ⇒ T = 0.0562 2N ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ f 3− dB ⎠ f f 3− dB So = 20 =2 10 672. 0.0562 = 1 1 + (2) 2 N 1 + (2) 2 N = 316.6 ⇒ (2)2 N = 315.6 2 N ⋅ ln (2) = ln (315.6) ⇒ N = 4.15 ⇒ N = 5 A 5-pole filter 15.7 T = 1 ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ f 3dB ⎠ 2N At f = 12 kHz, T = 0.9 1 0.9 = ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ f 3dB ⎠ ⎛ 12 ⎞ ⎜ ⎟ ⎝ f3dB ⎠ Also 2N = 2N = ⎛ 14 ⎞ ⎜ ⎟ ⎝ f3dB ⎠ ⎛ 12 ⎞ ⎜ ⎟ ⎝ f3dB ⎠ 2N 1 ⎛ 14 ⎞ 1+ ⎜ ⎟ ⎝ f 3dB ⎠ 2N 1 − 1 = 9999 (0.01) 2 2N 2N ⎛ 12 ⎞ 1+ ⎜ ⎟ ⎝ f 3dB ⎠ 1 − 1 = 0.2346 (0.9) 2 0.01 = ⎛ 14 ⎞ ⎜ ⎟ ⎝ f3dB ⎠ 1 = 2N ⎛ 14 ⎞ =⎜ ⎟ ⎝ 12 ⎠ 2N = 9999 = 4.262 × 104 0.2346 (1.16667) 2 N = 4.262 × 104 N = 35 Then 0.9 = ⎛ 12 ⎞ ⎜ ⎟ ⎝ f3dB ⎠ 1 ⎛ 12 ⎞ 1+ ⎜ ⎟ ⎝ f 3dB ⎠ 2N 2N = 0.2346 ⎛ ⎞ ⎜ ⎟ ⎛ 12 ⎞ ⎝ 2N ⎠ = (0.2346)0.014286 ⎜ ⎟ = (0.2346) ⎝ f3dB ⎠ = 0.9795 So f 3dB = 12.25 kHz 1 15.8 673. T = 1 2N ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ f 3dB ⎠ (a) For N = 3 1 T = = 0.2841 1 + (1.5)6 For N = 5 1 T = = 0.1306 1 + (1.5)10 (b) For N = 7 1 T = = 0.05843 1 + (1.5)14 (c) 15.9 Consider For low-frequency: vo R2 + R3 = vi R1 + R2 + R3 For high-frequency: vo R2 = vi R1 + R2 So we need ⎛ R2 ⎞ R2 + R3 = 25 ⎜ ⎟ R1 + R2 + R3 ⎝ R1 + R2 ⎠ Let R1 + R2 = 50 k Ω and R2 = 1.5 k Ω ⇒ R1 = 48.5 k Ω Then 674. 1.5 + R3 ⎛ 1.5 ⎞ = 25 ⎜ ⎟ ⇒ R3 = 144 k Ω 50 + R3 ⎝ 50 ⎠ Connect the output of this circuit to a non-inverting op-amp circuit. At low-frequency: vo1 = R2 + R3 1.5 + 144 ⋅ vi = ⋅ vi = 0.75vi R1 + R2 + R3 48.5 + 1.5 + 144 Need to have vo = 25. ⎛ R ⎞ ⎛ R ⎞ R vo = 25 = ⎜1 + 5 ⎟ ⋅ vo1 = ⎜ 1 + 5 ⎟ (0.75)vi ⇒ 5 = 32.3 R4 ⎝ R4 ⎠ ⎝ R4 ⎠ To check at high-frequency. R2 1.5 vo1 = vi = vi = 0.03vi R1 + R2 1.5 + 48.5 vo = (1 + 32.3)vo1 = (33.3)(0.03)vi = (1.0)vi which meets the design specification Consider the frequency response. 1 R2 + R3 vo1 sC = 1 vi R1 + R2 + R3 sC Now R3 1 = R3 sC 1 + sR3C Then, we find vo1 R3 + R2 (1 + sR3C ) = vi R3 + ( R1 + R2 )(1 + sR3C ) which can be rearranged as ( R2 + R3 ) (1 + s ( R2 || R3 )C ) vo1 = vi ( R1 + R2 + R3 ) 1 + s ( R3 || ( R1 + R2 ) ) C ( ) So fL ≅ 1 1 1 = = 3 2π ( R2 || R3 ) C 2π (1.5 || 144 ) × 10 C ( 9.33 × 103 ) C fH ≅ 1 1 = 2π ( R3 || ( R1 + R2 ) ) C 2π (144 || 50 ) × 103 C = Set 1 ( 2.33 ×10 ) C 5 675. ⎤ fL + fH 1 ⎡ 1 1 ⎥ = ⎢ + 3 5 2 2 ⎢ ( 9.33 × 10 ) C ( 2.33 × 10 ) C ⎥ ⎣ ⎦ Which yields C = 2.23 nF 25 kHz = 15.10 Av = 1 ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ f3− dB ⎠ − 100 dB ⇒ 10−5 2N So 10−5 = 1 ⎛ 770 ⎞ 1+ ⎜ ⎟ ⎝ 12 ⎠ 2N or 2 ⎛ 1 ⎞ 1 + (64.2) 2 N = ⎜ −5 ⎟ = 1010 ⎝ 10 ⎠ or (64.2) 2 N ≅ 1010 Now N 1 2 3 Left Side 4.112 × 103 1.7 × 107 7 × 1010 So, we need a 3rd order filter. 15.11 676. Low-pass: −50 dB ⇒ 3.16 × 10−3 Then 1 1 3.16 × 10−3 = = 4 4 ⎛ f ⎞ ⎛ 60 ⎞ 1+ ⎜ ⎟ 1+ ⎜ ⎟ ⎝ fL ⎠ ⎝ fL ⎠ We find f L = 3.37 Hz High Pass: 1 1 3.16 × 10−3 = = 4 4 ⎛ fH ⎞ ⎛ f ⎞ 1+ ⎜ H ⎟ 1+ ⎜ ⎟ ⎝ 60 ⎠ ⎝ f ⎠ We find f H = 1067 Hz Bandwidth: BW = f H − f L = 1067 − 3.37 ⇒ BW ≅ 1064 Hz 15.12 a. v vI = − 02 − R4 R3 v0 ⎛ 1 ⎞ R1 ⎜ ⎟ ⎝ sC ⎠ (1) v0 v = − 01 (2) R2 ⎛ 1 ⎞ ⎜ ⎟ ⎝ sC ⎠ v01 v (3) = − 02 ⇒ v01 = −v02 R5 R5 Then ⎛ 1 ⎞ v0 v = + 02 or v02 = v0 ⎜ ⎟ (2) R2 ⎛ 1 ⎞ ⎝ sR2 C ⎠ ⎜ ⎟ ⎝ sC ⎠ And 677. ⎛ 1 ⎞ v0 ⋅⎜ ⎟− sR2 C ⎠ ⎛ 1 ⎞ ⎝ R1 ⎜ ⎟ ⎝ sC ⎠ v vI =− 0 R4 R3 ⎡ ⎤ ⎢ ⎥ 1 1 ⎥ = −v0 ⎢ + ⎢ R3 ( sR2 C ) R1 ⋅ (1/ sC ) ⎥ ⎢ R1 + (1/ sC ) ⎥ ⎣ ⎦ (1) ⎡ 1 + sR1C ⎤ 1 = −v0 ⎢ + ⎥ R1 ⎦ ⎣ R3 ( sR2 C ) ⎡ R + (1 + sR1C )( sR2 R3C ) ⎤ = −v0 ⎢ 1 ⎥ ( sC ) R1 R2 R3 ⎣ ⎦ Then ⎤ v0 ( sC )( R1 R2 R3 ) 1 ⎡ =− ⎢ ⎥ vI R4 ⎣ R1 + sR2 R3C + s 2 R1 R2 R3C 2 ⎦ or − Av ( s ) = 1 R4 v0 = 1 1 vI + sC + R1 sCR2 R3 − Av ( jω ) = b. 1 R4 1 1 + jω C + R1 jω CR2 R3 or − Av ( jω ) = 1 R4 ⎡ ⎤ 1 1 + j ⎢ω C − ω CR2 R3 ⎥ R1 ⎣ ⎦ R1 1 =− ⋅ R4 ⎪ ⎧ ⎫ ⎡ R1 ⎤ ⎪ ⎨1 + j ⎢ω R1C − ⎥⎬ ω CR2 R3 ⎦ ⎪ ⎪ ⎣ ⎩ ⎭ R1 1 ⋅ Av ( jω ) = 2 −1/ 2 R4 ⎧ R1 ⎤ ⎫ ⎪ ⎡ ⎪ ⎨1 + ⎢ω R1C − ⎥ ⎬ ω CR2 R3 ⎦ ⎪ ⎪ ⎣ ⎩ ⎭ Av ⎡ R1 ⎤ when ⎢ω R1C − ⎥=0 ω CR2 R3 ⎦ ⎣ max Then Av max = R1 85 = ⇒ Av 3 R4 max = 28.3 Now ⎡ ω R1C ⎢1 − ⎣ Then ⎤ 1 1 ⎥ = 0 or ω = 2 ω C R2 R3 ⎦ C R2 R3 2 678. f = 1 2π C R2 R3 = 1 2π (0.1× 10−6 ) (300) 2 So f = 5.305 kHz To find the two 3 − dB frequencies, ⎡ R1 ⎤ ⎢ω R1C − ⎥ = ±1 ω CR2 R3 ⎦ ⎣ ω 2 R1 R2 R3 C 2 − R1 = ±ω R2 R3 C ω 2 (85 × 103 )(300) 2 (0.1× 10−6 ) 2 − 85 × 103 = ±ω (300) 2 (0.1× 10−6 ) ω 2 (7.65× 10−5 ) − 85 × 103 = ±ω (9 × 10−3 ) ω 2 (7.65× 10−5 ) ± ω (9 × 10−3 ) − 85 × 103 = 0 (9 × 10−3 ) 2 + 4(7.65 × 10−5 )(85 × 10−3 ) ± (9 × 10−3 ) ± 2(7.65 × 10−5 ) 2(7.65 × 10−5 ) We find f = 5.315 kHz and f = 5.296 kHz ω= 15.13 a. vI − v A v = A (1) R ⎛ 1 ⎞ ⎜ ⎟ ⎝ sC ⎠ vI − vB vB − v0 = (2) R R and v A = vB So vI ⎛1 ⎞ ⎛ 1 + sRC ⎞ = v A ⎜ + sC ⎟ = v A ⎜ ⎟ (1) R R ⎝ ⎠ ⎝ R ⎠ or vI vA = 1 + sRC Then 2vI vI + v0 = 2vB = 2vA = (2) 1 + sRC ⎡ 2 ⎤ ⎡1 − sRC ⎤ v0 = vI ⎢ − 1 = vI ⎢ ⎥ 1 + sRC ⎥ ⎣ ⎦ ⎣1 + sRC ⎦ Now v0 1 − jω RC = A( jω ) = 1 + jω RC vI A= 1 + ω 2 R2C 2 1 + ω 2 R2C 2 ⇒ A =1 Phase: φ = −2 tan −1 (ω RC ) b. f 0 RC = (104 )(15.9 × 10−9 ) = 1.59 × 10−4 φ 0 679. −11.4 −53.1 −90° −157 −169 10 2 5 × 103 1/ 2π RC = 103 Hz 5 × 103 10 4 15.14 a. Vi Vi − V0 + =0 R1 R2 || (1/ sC ) Vi Vi − V0 + =0 R1 ⎡ R2 ⎤ ⎢1 + sR C ⎥ 2 ⎣ ⎦ R2 1 (Vi ) + Vi = V0 ⋅ R1 1 + sR2 C V0 R2 + R1 (1 + sR2 C ) ( R2 + R1 ) [1 + s ( R1 || R2 )C ] = = Vi R1 (1 + sR2 C ) R1 (1 + sR2 C ) ⇒ V0 ⎛ R2 ⎞ ⎡1 + s ( R1 || R2 )C ⎤ = ⎜1 + ⎟ ⎢ ⎥ Vi ⎝ R1 ⎠ ⎣ (1 + sR2 C ) ⎦ ⇒ f 3dB1 = 1 2π R2 C ⇒ f 3dB2 = 1 2π ( R1 || R2 )C b. Vi V − V0 + i =0 R1 || (1/ sC ) R2 Vi V V + i = 0 ⎛ R1 ⎞ R2 R2 ⎜ ⎟ ⎝ 1 + sR1C ⎠ ⎡R ⎤ Vi ⎢ 2 ⋅ (1 + sR1C ) + 1⎥ = V0 ⎣ R1 ⎦ Vi ⋅ [ R2 + R1 + sR1 R2 C ] = V0 R1 V0 R2 + R1 V ⎛ R ⎞ 1 = ⋅ [1 + s ( R1 || R2 )C ] ⇒ 0 = ⎜1 + 2 ⎟ [1 + s ( R1 || R2 )C ] ⇒ f3dB = 2π ( R1 || R2 )C Vi R1 Vi ⎝ R1 ⎠ 15.15 a. 680. −V0 Vi = R1 + (1/ sC1 ) R2 || (1/ sC2 ) ⎛ sC1 ⎞ ⎛ 1 + sR2 C2 ⎞ Vi ⎜ ⎟ = −V0 ⎜ ⎟ ⎝ 1 + sR1C1 ⎠ ⎝ sC2 ⎠ V0 − sR2 C1 − sR2 C1 = = Vi (1 + sR1C1 )(1 + sR2 C2 ) 1 + sR1C1 + sR2 C2 + s 2 R1 R2 C1C2 ⎡ ⎤ ⎢ ⎥ V0 R sC1 ⎥ = − 2 ×⎢ ⎥ Vi R1 ⎢ 1 ⎛ R2 C2 ⎞ 2 ⎢ + sC1 ⎜ 1 + ⋅ ⎟ + s R2 C1C2 ⎥ R1 C1 ⎠ ⎢ R1 ⎥ ⎝ ⎣ ⎦ or ⎡ ⎤ ⎢ ⎥ V0 R2 ⎢ 1 ⎥ =− ⋅ T (s) = ⎥ Vi R1 ⎢ 1 ⎛ R2 C2 ⎞ + ⎜ 1 + ⋅ ⎟ + sR2 C2 ⎥ ⎢ R1 C1 ⎠ ⎢ sR1C1 ⎝ ⎥ ⎣ ⎦ b. T ( jω ) = − R2 1 × 2 2 1/ 2 R1 ⎧ 1 ⎞ ⎫ ⎪⎛ R2 C2 ⎞ ⎛ ⎪ ⎨⎜ 1 + . ⎟ + ⎜ ω R2 C2 − ⎟ ⎬ R1 C1 ⎠ ⎝ ω R1C1 ⎠ ⎪ ⎪⎝ ⎩ ⎭ ⎛ 1 ⎞ when ⎜ ω R2 C2 − ⎟ = 0, we want ω R1C1 ⎠ ⎝ R 1 T ( jω ) = 50 = 2 ⋅ R1 ⎛ R2 C2 ⎞ ⎜1 + ⋅ ⎟ R1 C1 ⎠ ⎝ At the 3 − dB frequencies, we want ⎛ ⎛ R2 C2 ⎞ 1 ⎞ ⎜ ω R2 C2 − ⎟ = ± ⎜1 + ⋅ ⎟ ω R1C1 ⎠ R1 C1 ⎠ ⎝ ⎝ For f = 5 kHz, use + sign and for f = 200 Hz, use − sign. ω1 = 2π (200) = 1257 ω 2 = 2π (5 × 103 ) = 3.142 × 104 Define r2 = R2 C2 and r1 = R1C1 Then 50 = R2 ⋅ R1 1 r 1+ 2 r1 ⎛ ⎛ r2 ⎞ 1 ⎞ ⎜ ω 2 r2 − ⎟ = + ⎜1 + ⎟ ω 2 r1 ⎠ ⎝ ⎝ r1 ⎠ ⎛ ⎛ r2 ⎞ 1 ⎞ ⎜ ω1 r2 − ⎟ = − ⎜1 + ⎟ ω1 r1 ⎠ ⎝ ⎝ r1 ⎠ From (2) 2 ω 2 r1r2 − 1 r1 + r2 = ω 2 r1 r1 (1) (2) (3) 681. or ω 2 r1 r2 − 1 ω2 = r1 + r2 r1 (ω 2 r2 − 1) = 1 ω2 + r2 So 1 +r 2 ω r1 = 2 ω 2 r2 − 1 Substituting into (3), we find ⎡ ⎤ ⎢ r (ω r − 1) ⎥ 1 ⎥ ω1r2 − = − ⎢1 + 2 2 2 1 ⎡ 1 ⎤ ⎢ + r2 ⎥ + r2 ⎥ ⎢ω ⎢ ⎥ ω2 ⎣ ⎦ ⎥ ω1 ⎢ 2 ⎢ ω 2 r2 − 1 ⎥ ⎢ ⎥ ⎣ ⎦ ⎡⎛ 1 ⎤ ⎡1 ⎤ 1 ⎞ ω1r2 ⎢ + r2 ⎥ − (ω 2 r2 − 1) = − ⎢⎜ + r2 ⎟ + r2 (ω 2 r2 − 1) ⎥ ⎣ω2 ⎦ ω1 ⎠ ⎣⎝ ω 2 ⎦ ω1 ω2 1 1 ⋅ r2 + ω1r22 − ⋅ r2 + =− − r2 − ω 2 r22 + r2 ω2 ω1 ω1 ω2 ⎛ 1 ⎛ω ω ⎞ 1 ⎞ (ω1 + ω 2 )r22 + ⎜ 1 − 2 ⎟ r2 + ⎜ + ⎟=0 ⎝ ω1 ω 2 ⎠ ⎝ ω 2 ω1 ⎠ (3.2677 × 104 )r22 − 24.96r2 + 8.273 × 10−4 = 0 (24.96) 2 − 4(3.2677 ×10 4 )(8.273 × 10−4 ) 24.96 ± 2(3.2677 ×10 4 ) 2(3.2677 × 104 ) Since ω 2 is large, r2 should be small so use minus sign: r2 = r2 = 3.47 × 10−5 Then 3.18 × 10−5 + 3.47 × 10−5 r1 = ⇒ r1 = 7.32 × 10−4 9.09 × 10−2 Now R 1 50 = 2 ⋅ R1 3.47 × 10−5 1+ 7.32 × 10−4 Then R2 = 52.37 or R2 = 524 kΩ R1 Also r1 = R1C1 so that C1 = 0.0732 μ F r2 = R2 C2 so that C2 = 66.3 pF 15.16 Gain = 10 dB ⇒ Gain = 3.162 For example, we may have 682. Want R4 = 2.162 R3 For example, let R3 = 50 kΩ, R4 = 108 kΩ f1 = 1 = 200 2π R1C1 So R1C1 = 1 = 0.796 × 10−3 2π (200) For example, let R1 = 200 kΩ ⇒ A large input resistance C1 = 0.00398 μ F 1 1 f2 = = 50 × 103 ⇒ R2 C2 = = 3.18 × 10−6 2π R2 C2 2π (50 × 103 ) For example, let R2 = 10 kΩ and C2 = 318 pF 15.17 f C = 100 kHz 1 Req = fC C a. For C = 1 pF, Req = 10 MΩ b. For C = 10 pF, Req = 1 MΩ c. For C = 30 pF, Req = 333 kΩ 15.18 a. From Equation (15.28), V1 − V2 Q= ⋅ TC Req and f C = 100 kHz so that TC = 1 ⇒ 10 μ s 100 × 103 Now Req = So 1 1 = ⇒ 1 MΩ 3 f C C (100 × 10 )(10 × 10−12 ) 683. Q= (2 − 1)(10 × 10−6 ) = 10 × 10−12 C 106 or Q = 10 pC Q 10 × 10−12 or I eq = 1 μ A = TC 10 × 10−6 b. I eq = c. Q = CV so find the time that V0 reaches 99% of its full value. V0 = V1 (1 − e − t / r ) where r = RC Then 0.99 = 1 − e − t / r or e − t / r = 0.01 or t = r ln (100) r = RC = (103 )(10 × 10−12 ) = 10−8 s Then t = 4.61×10−8 s 15.19 Low frequency gain = −10 ⇒ f 3dB = 10 × 103 Hz = C1 = 10 C2 fC C2 2π CF Set f C = 10 f3dB = 100 kHz Then C2 2π (10 × 103 ) = = 0.628 CF 100 × 103 The largest capacitor is C1 , so let C1 = 30 pF Then C2 = 3 pF and CF = 4.78 pF 15.20 a. Req = Time constant = Req ⋅ CF = r where 1 1 = = 2 × 106 Ω −12 3 f C C1 (100 × 10 )(5 × 10 ) Then r = (2 × 106 )(30 × 10−12 ) or r = 60 μ s b. v0 = − 1 vI ⋅ dt r∫ or Δv0 = (1)TC 1 ,TC = r fC 684. So Δv0 = 1 (60 × 10 )(100 × 103 ) −6 or Δv0 = 0.167 V c. Now Δv0 = 13 = N (0.167) or N = 78 clock pulses 15.21 1 1 = 2π 3RC 2π 3(104 )(0.10 × 10−12 ) f O = 91.9 MHz fO = R2 = 8R = 80 kΩ 15.22 a. ⎛ sRCV ⎞ R ⋅ v0 = ⎜ ⎟ ⋅ v0 R + (1/ sCV ) ⎝ 1 + sRCV ⎠ R ⎛ sRC ⎞ ⋅v = ⋅v v2 = 1 1 ⎜ 1 + sRC ⎟ 1 ⎝ ⎠ R+ sC R ⎛ sRC ⎞ v3 = ⋅v = ⋅v 1 2 ⎜ 1 + sRC ⎟ 2 ⎝ ⎠ R+ sC R2 v0 = − ⋅ v3 R Then 2 R ⎛ sRC ⎞ ⎛ sRCV ⎞ v0 = − 2 ⎜ ⎟ v0 ⎟ ⎜ R ⎝ 1 + sRC ⎠ ⎝ 1 + sRCV ⎠ v1 = Set s = jω ⎛ ⎞ ⎛ jω RCV ⎞ −ω 2 R 2 C 2 ⎟ ⎜ ⎟⎜ 1 + 2 jω RC − ω 2 R 2 C 2 ⎠ ⎝ 1 + jω RCV ⎠ ⎝ The real part of the denominator must be zero. 1 − ω 2 R 2 C 2 − 2ω 2 R 2 CCV = 0 so 1 ω0 = R C (C + 2CV ) 1= − R2 R b. f 0,max = 1 2π (10 ) (10 4 −11 )(10−11 + 2[10−11 ]) f 0,max = 919 kHz f 0,min = 1 2π (10 ) (10 4 f 0,min = 480 kHz −11 )(10−11 + 2[50 × 10−12 ]) 685. 15.23 1 fO = RC = 2π 6 RC 1 2π 6 fO RC = 2.32 × 10 = 1 2π 6(28 × 103 ) −6 2.32 × 10−6 ⇒ R = 18.56 kΩ 125 × 10−12 R2 = 29 R ⇒ R2 = 538.4 kΩ R= 15.24 v0 − v1 v1 v1 − v2 (1) = + 1 1 R sC sC v or (v0 − v1 ) sC = 1 + (v1 − v2 ) sC R v1 − v2 v2 v2 (2) = + 1 1 R +R sC sC v v ( sC ) or (v1 − v2 ) sC = 2 + 2 R 1 + sRC v0 v2 (3) =− 1 R2 +R sC v v sC or 2 =− 0 1 + sRC R2 so −v0 v2 = (1 + sRC ) sR2 C From (2) 1 sC ⎤ ⎡ v1 ( sC ) = v2 ⎢ sC + + R 1 + sRC ⎥ ⎣ ⎦ or v (1 + sRC ) ⎡ 1 1 ⎤ ⋅ ⎢1 + + v1 = − 0 ⎥ sR2 C ⎣ sRC 1 + sRC ⎦ From (1) 1 ⎡ ⎤ v0 ( sC ) = v1 ⎢ sC + + sC ⎥ − v2 ( sC ) R ⎣ ⎦ Then v0 1 ⎤ ⎡ −v0 (1 + sRC ) ⎤ ⎡1 + sRC 1 ⎤ ⎡ v0 = ⎢ 2 + + ⎥×⎢ ⎥⎢ ⎥ + sR C ⋅ (1 + sRC ) sRC ⎦ ⎣ sR2 C 1 + sRC ⎦ ⎣ ⎦ ⎣ sRC 2 2 ⎡1 + 2sRC ⎤ ⎡1 + sRC ⎤ ⎡ (1 + sRC ) + sRC ⎤ 1 + sRC −1 = ⎢ ⎥⎢ ⎥− ⎥⎢ ⎣ sRC ⎦ ⎣ sR2 C ⎦ ⎣ ( sRC )(1 + sRC ) ⎦ sR2 C −1 = (1 + 2 sRC )(1 + 2sRC + s 2 R 2 C 2 + sRC ) (1 + sRC )( sRC ) 2 − ( sRC ) 2 ( sR2 C ) ( sRC ) 2 ( sR2 C ) 686. Set s = (1 + 2 jω RC ) (1 + 3 jω RC + ω 2 R 2 C 2 ) (1 + jω RC ) ( −ω 2 R 2C 2 ) jω − 1 = − ( −ω 2 R 2C 2 ) ( jω R2C ) ( −ω 2 R 2C 2 ) ( jω R2C ) The real part of the numerator must be zero. 1 − ω 2 R 2 C 2 − 6ω 2 R 2 C 2 + ω 2 R 2 C 2 = 0 6ω 2 R 2 C 2 = 1 so that 1 ω0 = 6 RC Condition for oscillation: 2 jω RC + 3 jω RC − 2 jω 3 R 3 C 3 + jω 3 R 3 C 3 −1 = (−ω 3 R 2 C 2 )( jω R2 C ) 1= 5 − ω 2 R2 C 2 (ω RC )(ω R2 C ) But ω = ω0 = 1 6RC Then 1⎞ ⎛ 2 2 1 ⎜ 5 − ⎟ (6 R C ) 6⎠ 6 1= =⎝ ( RC )( R2 C ) RR2 C 2 2 2 6R C ⎛ 29 ⎞ ⎜ ⎟ (6 R ) R 6 1= ⎝ ⎠ or 2 = 29 R2 R 5− 15.25 Let RF 1 = RF 2 = RF 3 ≡ RF ⎞⎛ 1 ⎞ ⎟ ⎜ 1 + 5R ⎟ vo ⎠⎝ C ⎠ ⎛ R ⎞⎛ 1 ⎞ Vo 2 = ⎜ 1 + F ⎟ ⎜ ⎟ vo1 R ⎠ ⎝ 1 + 5 RC ⎠ ⎝ vo3 + vo 2 v v + o3 + o3 = 0 R 1/ sC R 2 ⎛ ⎞ v vo3 ⎜ + sC ⎟ = o 2 ⎝R ⎠ R ⎛ R Vo1 = ⎜1 + F R ⎝ ⎛ 1 ⎞ vo3 = ⎜ ⎟ vo 2 ⎝ 2 + 5RC ⎠ R vo = − F ⋅ vo 3 R R ⎛ 1 ⎞ ⎛ RF ⎞ ⎛ 1 ⎞ ⎛ RF ⎞ ⎛ 1 ⎞ vo = − F ⋅ ⎜ ⎜ ⎟ vo ⎟ ⎜1 + ⎟ ⎜1 + ⎟⎜ R ⎝ 2 + 5 RC ⎠ ⎝ R ⎠ ⎝ 1 + 5 RC ⎠ ⎝ R ⎟ ⎝ 1 + 5 RC ⎠ ⎠ 1= − RF R ⎛ RF ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎜1 + R ⎟ ⎜ 2 + 5R ⎟ ⎜ 1 + 5R ⎟ ⎜ 1 + 5R ⎟ ⎝ ⎠ ⎝ C ⎠⎝ C ⎠⎝ C ⎠ 2 687. Let S = jω 1= − RF R ⎞⎛ ⎞⎛ ⎞ 1 1 1 ⎛ RF ⎞ ⎛ ⎜ 1 + R ⎟ ⎜ 2 + jω R ⎟ ⎜ 1 + jω R ⎟ ⎜ 1 + jω R ⎟ ⎝ ⎠ ⎝ C ⎠⎝ C ⎠⎝ C ⎠ =− RF R ⎤ ⎞⎡ 1 1 ⎛ RF ⎞ ⎛ ⎜ 1 + R ⎟ ⎜ 2 + jω R ⎟ ⎢ 2 2 2 ⎥ ⎝ ⎠ ⎝ C ⎠ ⎣1 + 2 jω RC − ω R C ⎦ =− RF R ⎤ 1 ⎛ RF ⎞ ⎡ ⎜1 + R ⎟ ⎢ 2 2 2 2 2 2 3 3 3 ⎥ ⎝ ⎠ ⎣ 2 + 4 jω RC − 2ω R C + jω RC − 2ω R C − jω R C ⎦ 2 2 2 ⎤ RF ⎛ RF ⎞ ⎡ 1 ⎜ 1 + R ⎟ ⎢ 2 − 4ω 2 R 2 C 2 + 5 jω RC − jω 3 R 3C 3 ⎥ R ⎝ ⎠ ⎣ ⎦ Imaginary Term must be zero 3 5 jω 0 RC − jω 0 R 3 C 3 = 0 2 1= − 2 5 − jω 0 R 2 C 2 = 0 ω0 = 5 RC Then R 1= − F R 1= − RF R ⎡ ⎤ 2 ⎥ 1 ⎛ RF ⎞ ⎢ ⎢ ⎥ 1+ ⎜ ⎟ R ⎠ ⎢ 4R2 C 2 − 5 ⎥ ⎝ ⎢ 2 − R2C 2 ⎥ ⎣ ⎦ 2 ⎛ RF ⎞ ⎡ 1 ⎤ 1 RF ⎜ 1 + R ⎟ ⎢ 2 − 20 ⎥ = 18 ⋅ R ⎦ ⎝ ⎠ ⎣ ⎛ RF ⎞ ⎜1 + R ⎟ ⎝ ⎠ 2 2 R ⎛ R ⎞ R 18 = F ⎜ 1 + F ⎟ ⇒ F = 2 R ⎝ R ⎠ R 15.26 a. v0 − v01 v v −v = 01 + 01 02 R R ⎛ 1 ⎞ ⎜ ⎟ sC ⎠ ⎝ v01 − v02 v v −v = 02 + 02 03 R R ⎛ 1 ⎞ ⎜ ⎟ ⎝ sC ⎠ v02 − v03 v v = 03 + 03 R ⎛ 1 ⎞ R ⎜ ⎟ ⎝ sC ⎠ R v0 = − F ⋅ v03 R We can write the equations as v0 − v01 = v01 ( sRC ) + v01 − v02 v01 − v02 = v02 ( sRC ) + v02 − v03 v02 − v03 = v03 ( sRC ) + v03 and R v0 = − F ⋅ v03 R (1) (2) (3) (4) (1) (2) (3) (4) 688. Combining terms, we find v0 = v01 (2 + sRC ) − v02 (1) v01 = v02 (2 + sRC ) − v03 (2) v02 = v03 (2 + sRC ) (3) and R v0 = − F ⋅ v03 (4) R Combining Equations (3) and (2) v01 = v03 (2 + sRC ) 2 − v03 = v03 ⎡ (2 + sRC ) 2 − 1⎤ ⎣ ⎦ (2) Then Equation (1) is v0 = v03 ⎡(2 + sRC ) 2 − 1⎤ (2 + sRC ) − v03 (2 + sRC ) ⎣ ⎦ Using Equation (4), we find R − F ⋅ v03 = V03 ⎡(2 + sRC ) 2 − 1⎤ (2 + sRC ) − (2 + sRC ) ⎣ ⎦ R To find the frequency of oscillation, set s = jω and set the imaginary part of the right side of the equation to zero. We will have R − F = (2 + jω RC )[4 + 4 jω RC − ω 2 R 2 C 2 − 1 − 1] R Then jω RC (2 − ω 2 R 2 C 2 ) + 8 jω RC = 0 or jω RC ⎡ 2 − ω 2 R 2 C 2 + 8⎤ = 0 ⎣ ⎦ { } Then the frequency of oscillation is 1 10 f0 = ⋅ 2π RC The condition to sustain oscillations is determined from R − F = 2[2 − ω 2 R 2 C 2 ] − 4ω 2 R 2 C 2 R or R − F = 4 − 6ω 2 R 2 C 2 R 10 Setting ω 2 = 2 2 , we have RC RF − = 4 − 6(10) R or RF = 56 R b. For R = 5 kΩ and f 0 = 5 kHz, we find 10 ⇒ C = 0.02 μ F 2π (5 × 103 )(5 × 103 ) and RF = 56(5) ⇒ RF = 280 kΩ C= 15.27 a. We can write 689. ⎛ R1 vA = ⎜ ⎝ R1 + R2 ⎛ Zp ⎞ ⎟ v0 and vB = ⎜ ⎜Z +Z ⎠ s ⎝ p where Z p = RB and Z s = RA + ⎞ ⎟ v0 ⎟ ⎠ RB 1 = sCB 1 + sRB CB 1 + sRA C A 1 = sC A sC A Setting v A = vB , we have RB R1 1 + sRB CB = RB 1 + sRAC A R1 + R2 + sC A 1 + sRB CB R1 sRB C A = (1) R1 + R2 sRB C A + (1 + sRAC A )(1 + sRB CB ) To find the frequency of oscillation, set s = jω and set the real part of the denominator on the right side of Equation (1) equal to zero. The denominator term is jω RB C A + (1 + jω RAC A )(1 + jω RB CB ) or jω RB C A + 1 + jω RAC A + jω RB CB − ω 2 RA RB C ACB (2) Then from (2), we must have 2 1 − ω 0 RA RB C ACB = 0 or f0 = 1 2π RA RB C ACB b. To find the condition for sustained oscillation, combine Equations (1) and (2). Then R1 jω RB C A = R1 + R2 jω RB C A + jω RAC A + jω RB CB ) or R2 R C = 1+ A + B R1 RB C A Then R2 RA CB = + R1 RB C A 1+ 15.28 a. We can write ⎛ R1 ⎞ vA = ⎜ ⎟ v0 ⎝ R1 + R2 ⎠ and ⎛ ⎞ R || sL vB = ⎜ ⎟ v0 ⎝ R || sL + R + sL ⎠ Setting v A = vB , we have 690. sRL ⎡ ⎤ ⎢ ⎥ R1 R + sL =⎢ ⎥ ⋅ v0 sRL R1 + R2 ⎢ + R + sL ⎥ ⎢ R + sL ⎥ ⎣ ⎦ R1 sRL = (1) R1 + R2 sRL + ( R + sL) 2 To find the frequency of oscillation, set s = jω and se the real part of the denominator on the right side of Equation (1) equal to zero. The denominator term is: jω RL + ( R + jω L) 2 or jω RL + R 2 + 2 jω RL − ω 2 L2 (2) Then 2 R 2 − ω 0 L2 = 0 or R 1 f0 = ⋅ L 2π b. To find the condition for sustained oscillations, combine Equations (1) and (2). R1 jω RL 1 = = R1 + R2 jω RL + 2 jω RL 3 Then R 1+ 2 = 3 R1 so that R2 =2 R1 15.29 1 2π RC 1 1 RC = = 2π fO 2π (28 × 103 ) fO = RC = 5.684 × 10−6 For example, Let R = 20 K Then C = 284.2 pF Also R2 =2 R1 15.30 From Equation (15.59) 1 f0 = ⎛ CC ⎞ 2π L ⎜ 1 2 ⎟ ⎝ C1 + C2 ⎠ and from Equation (15.61) C2 = gm R C1 691. Now, g m = 2 kn I DQ = 2 (0.5)(1) = 1.414 mA / V We have C1 = 0.01 μ F, R = 4 kΩ, f 0 = 400 kHz So C2 = g m RC1 = (1.414)(4)(0.01) or C2 = 0.0566 μ F and 400 × 103 = 1 ⎡ (0.01)(0.0566) ⎤ 2π L ⎢ × 10−6 0.01 + 0.0566 ⎥ ⎣ ⎦ 2 ⎡ ⎤ 1 L(8.5 × 10−9 ) = ⎢ = 1.58 × 10−13 3 ⎥ ⎣ 2π (400 × 10 ) ⎦ Then L = 18.6 μ H 15.31 Vπ = −V0 V0 ⎛ 1 ⎞ ⎜ ⎟ ⎝ sC2 ⎠ + V0 V0 − V1 + = g mVπ = − g mV0 RL ⎛ 1 ⎞ ⎜ ⎟ ⎝ sC1 ⎠ ⎡ ⎤ 1 V0 ⎢ sC2 + sC1 + + g m ⎥ = V1 ( sC1 ) RL ⎣ ⎦ V1 V0 − V1 + + g mVπ = 0 sL ⎛ 1 ⎞ ⎜ ⎟ ⎝ sC1 ⎠ (1) (2) ⎛ 1 ⎞ V1 ⎜ + sC1 ⎟ = V0 ( sC1 + g m ) ⎝ sL ⎠ V0 ( sC1 + g m ) V1 = ⎛ 1 ⎞ ⎜ + sC1 ⎟ sL ⎝ ⎠ Then ⎡ ⎤ V ( sC1 )( sC1 + g m ) 1 V0 ⎢ s (C1 + C2 ) + + gm ⎥ = 0 RL ⎛ 1 ⎞ ⎣ ⎦ ⎜ + sC1 ⎟ ⎝ sL ⎠ ⎡ ⎤⎛ 1 1 ⎞ + g m ⎥ ⎜ + sC1 ⎟ = sC1 ( sC1 + g m ) ⎢ s (C1 + C2 ) + RL ⎠ ⎣ ⎦ ⎝ sL g C1 + C2 sC 1 + s 2 C1 (C1 + C2 ) + + 1 + sg m C1 + m = s 2 C12 + sg m C1 L sRL L RL sL C1 + C2 sC g 1 + s 2 C1C2 + + 1 + m =0 L sRL L RL sL Set s = jω g C1 + C2 jω C1 1 − ω 2 C1C2 + + + m =0 L jω RL L RL jω L 692. Then ω2 = C1 + C2 C1 + C2 ⇒ ω0 = C1C2 L C1C2 L and gm ω C1 1 + = ω L ω RL L RL Then gm (C + C2 )C1 1 + = 1 L RL L C1C2 LRL gm + 1 C1 + C2 = RL C2 RL g m RL + 1 = C1 C + 1 or 1 = g m RL C2 C2 15.32 a. V0 V0 V0 + + g mVπ + =0 (1) 1 sL1 R + sL2 sC ⎛ ⎞ ⎜ sL2 ⎟ Vπ = ⎜ (2) ⎟ V0 ⎜ 1 + sL ⎟ ⎜ 2 ⎟ ⎝ sC ⎠ Then ⎧ g (s2 L C ) ⎫ 1 sC ⎪ 1 ⎪ V0 ⎨ + + + m 2 2 ⎬=0 2 ⎪ sL1 R 1 + s L2 C 1 + s L2 C ⎪ ⎩ ⎭ ⎧ R (1 + s 2 L2 C ) + ( sL1 )(1 + s 2 L2 C ) s 2 RL1C + g m ( sRL1 )( s 2 L2 C ) ⎫ ⎪ ⎪ + ⎨ ⎬=0 ( sRL1 )(1 + s 2 L2 C ) ( sRL1 )(1 + s 2 L2 C ) ⎪ ⎪ ⎩ ⎭ Set s = jω . Both real and imaginary parts of the numerator must be zero. R(1 − ω 2 L2 C ) + jω L1 (1 − ω 2 L2 C ) − ω 2 RL1C + ( jω g m RL1 )(−ω 2 L2 C ) = 0 Real part: R (1 − ω 2 L2 C ) − ω 2 RL1C = 0 R = ω 2 RC ( L1 + L2 ) or 1 ω0 = C ( L1 + L2 ) b. Imaginary part: jω L1 1 − ω 2 L2C − jω gm RL1 ω 2 L2C = 0 ( ) ( ( L1 = ω L1 L2C + gm RL1 ω L2C 2 2 ) ) 693. Now ω 2 = 1 ( L1 + L2 ) 1= 1 [ L2 C + g m RL2 C ] C ( L1 + L2 ) 1= L2 L (1 + g m R ) ⇒ 1 = (1 + g m R ) L1 + L2 L2 or L1 = gm R L2 15.33 ω 0 = 2π (800 × 103 ) = 1 C ( L1 + L2 ) or C ( L1 + L2 ) = 3.96 × 10−14 Also L1 = gm R L2 For example, if R = 1 kΩ, then L1 = (20)(1) = 20 L2 So L1 = 20 L2 Then C (21L2 ) = 3.96 × 10−14 or CL2 = 1.89 × 10−15 C = 0.01 μ F If then L2 = 0.189 μ H and L1 = 3.78 μ H 15.34 v0 − v1 v1 v1 − vB = + R ⎛ 1 ⎞ R ⎜ ⎟ sC ⎠ ⎝ and vB v −v + B 1 =0 R ⎛ 1 ⎞ ⎜ ⎟ ⎝ sC ⎠ or (1) (2) 1⎞ v ⎛ vB ⎜ sC + ⎟ = 1 ⇒ v1 = vB (1 + sRC ) R⎠ R ⎝ From (1) 2⎞ v ⎛ v0 ( sC ) = v1 ⎜ sC + ⎟ − B R⎠ R ⎝ or v0 ( sRC ) = vB (1 + sRC )(2 + sRC ) − vB = vB [ (1 + sRC )(2 + sRC ) − 1] Now 694. ⎛ R ⎞⎡ ⎤ ⎛ R2 ⎞ ⎡ sRC sRC ⎤ T ( s ) = ⎜1 + 2 ⎟ ⎢ ⎥ = ⎜1 + R ⎟ ⎢ 2 2 2 ⎥ R1 ⎠ ⎣ (1 + sRC )(2 + sRC ) − 1 ⎦ ⎝ ⎣ 2 + 3sRC + s R C − 1 ⎦ ⎝ 1 ⎠ or ⎛ R ⎞⎡ sRC ⎤ T ( s ) = ⎜1 + 2 ⎟ ⎢ 2 2 2 ⎥ R1 ⎠ ⎣ s R C + 3sRC + 1 ⎦ ⎝ ⎛ R ⎞⎡ ⎤ jω RC T ( jω ) = ⎜ 1 + 2 ⎟ ⎢ ⎥ R1 ⎠ ⎣1 − ω 2 R 2 C 2 + 3 jω RC ⎦ ⎝ Frequency of oscillation: 1 f0 = 2π RC Condition for oscillation: ⎛ R ⎞ ⎡ jω RC ⎤ 1 = ⎜1 + 2 ⎟ ⎢ R1 ⎠ ⎣ 3 jω RC ⎥ ⎦ ⎝ or R2 =2 R1 15.35 vb − vo vb vb − va + + =0 1 1 R sC sC vb − vo (1) + 2vb ⋅ sC − va ⋅ sC = 0 R Va − Vb Va (2) + =0 1 R sC 1⎞ ⎛ ⎛ 1 + sRC ⎞ Va ⎜ sC + ⎟ = vb ⋅ sC ⇒ vb = va ⎜ ⎟ R⎠ ⎝ ⎝ sRC ⎠ From (1) ⎛1 ⎞ v vb ⎜ + 2 sC ⎟ = o + va ⋅ sC ⎝R ⎠ R Substitute (2) into (1) 695. ⎛ 1 + sRC ⎞ ⎛ 1 + 2 sRC ⎞ vo va ⎜ ⎟⎜ ⎟ = + va ⋅ sC R ⎝ sRC ⎠ ⎝ ⎠ R ⎡ (1 + sRC )(1 + 2 sRC ) ⎤ v va ⎢ − sC ⎥ = o ( sRC ) ⋅ R ⎣ ⎦ R ⎡ (1 + sRC )(1 + 2 sRC ) ⎤ va ⎢ − sRC ⎥ = vo sRC ⎣ ⎦ 2 2 2 vo (1 + sRC )(1 + 2 sRC ) − s R C = va sRC va sRC = vo 1 + 3sRC + 2( sRC ) 2 − s 2 R 2 C 2 va sRC = vo 1 + 3sRC + ( sRC ) 2 ⎛ R ⎞ sRC T ( s ) = ⎜1 + 2 ⎟ ⋅ R1 ⎠ 1 + 3sRC + ( sRC ) 2 ⎝ ⎛ R ⎞⎡ ⎤ jω RC T ( jω ) = ⎜ 1 + 2 ⎟ ⎢ ⎥ R1 ⎠ ⎣1 − ω 2 R 2 C 2 + 3 jω RC ⎦ ⎝ 2 So 1 − ω 0 R 2 C 2 = 0 So f O = 1 2π RC ⎛ R ⎞⎛ 1 ⎞ Also 1 = ⎜ 1 + 2 ⎟ ⎜ ⎟ R1 ⎠ ⎝ 3 ⎠ ⎝ R So 2 = 2 R1 15.36 v0 − v1 v1 v1 − vB = + sL R R ⎛ sL ⎞ vB = ⎜ ⎟ v1 ⎝ R + sL ⎠ or (1) (2) ⎛ R + sL ⎞ v1 = ⎜ ⎟ vB ⎝ sL ⎠ Then v0 ⎛ 1 2⎞ v = v1 ⎜ + ⎟ − B sL ⎝ sL R ⎠ R or v0 ⎛ R + sL ⎞ ⎛ 1 2 ⎞ vB =⎜ ⎟ ⎜ + ⎟ vB − sL ⎝ sL ⎠ ⎝ sL R ⎠ R ⎧⎛ R + sL ⎞⎛ R + 2 sL ⎞ 1 ⎫ = vB ⎨⎜ ⎟⎜ ⎟− ⎬ ⎩⎝ sL ⎠⎝ sRL ⎠ R ⎭ Then (1) 696. vB = v0 1 ⋅ sL ⎧ ( R + sL)( R + 2 sL) − ( sL) 2 ⎨ ( sL)( sRL) ⎩ ⎫ ⋅⎬ ⎭ Now ⎛ R ⎞⎛ sRL ⎞ T ( s ) = ⎜1 + 2 ⎟ ⎜ 2 ⎟ R1 ⎠ ⎝ R + 3sRL + 2 s 2 L2 − s 2 L2 ⎠ ⎝ or ⎛ R ⎞⎛ sRL ⎞ T ( s ) = ⎜1 + 2 ⎟ ⎜ 2 2 ⎟ R1 ⎠ ⎝ s L + 3sRL + R 2 ⎠ ⎝ And ⎛ R ⎞⎛ ⎞ jω RL T ( jω ) = ⎜ 1 + 2 ⎟ ⎜ 2 ⎟ R1 ⎠ ⎝ R − ω 2 L2 + 3 jω RL ⎠ ⎝ R Frequency of oscillation: f 0 = 2π L Condition for oscillation: ⎛ R ⎞⎛ 1 ⎞ 1 = ⎜1 + 2 ⎟ ⎜ ⎟ R1 ⎠ ⎝ 3 ⎠ ⎝ or R2 =2 R1 15.37 From Equation (15.65(b)), the crossover voltage is R vI = − 2 ⋅ VREF R1 Let R2 = RVAR + RF where RVAR is the potentiometer and RF is the fixed resistor. Let VREF = −5 V, RF = 10 kΩ, and RVAR = 40 kΩ Then we have R ⎛ 10 ⎞ vI = − F ⋅ VREF = − ⎜ ⎟ (−5) = 1 V R1 ⎝ 50 ⎠ and ⎛ 50 ⎞ vI = − ⎜ ⎟ (−5) = 5 V ⎝ 50 ⎠ 15.38 ⎛ R1 ⎞ VTH − VTL = ⎜ ⎟ (VH − VL ) ⎝ R1 + R2 ⎠ ⎛ R1 ⎞ ⎛ R1 ⎞ 0.2 = ⎜ ⎟ (13 − (−13) ) = ⎜ ⎟ 26 ⎝ R1 + R2 ⎠ ⎝ R1 + R2 ⎠ R1 So = 0.007692 R1 + R2 13 = 0.25 ⇒ R1 + R2 = 52 R1 + R2 Then I= 697. R1 = (0.007692)(52) = 0.4 kΩ = R1 So R2 = 51.6 kΩ 15.39 a. ⎛ R1 ⎞ ⎛ 10 ⎞ VTH = ⎜ ⎟ VH = ⎜ ⎟ (10) R1 + R2 ⎠ ⎝ 10 + 40 ⎠ ⎝ so VTH = 2 V ⎛ R1 ⎞ ⎛ 10 ⎞ VTL = ⎜ ⎟ VL = ⎜ ⎟ (−10) ⎝ 10 + 40 ⎠ ⎝ R1 + R2 ⎠ so VTL = −2 V b. vI = 5sin ω t 15.40 a. Upper crossover voltage when v0 = +VP , Now ⎛ R1 ⎞ vB = ⎜ ⎟ (+VP ) ⎝ R1 + R2 ⎠ and ⎛ RA ⎞ ⎛ RB ⎞ vA = ⎜ ⎟ VREF + ⎜ ⎟ VTH RA + RB ⎠ ⎝ ⎝ RA + RB ⎠ v A = vB so that ⎛ R1 ⎞ ⎛ RA ⎜ ⎟ VP = ⎜ ⎝ R1 + R2 ⎠ ⎝ RA + RB or ⎞ ⎛ RB ⎟ VREF + ⎜ ⎠ ⎝ RA + RB ⎞ ⎟ VTH ⎠ ⎛ R + RB ⎞ ⎛ R1 ⎞ ⎛ RA ⎞ VTH = ⎜ A ⎟⎜ ⎟ VP − ⎜ ⎟ VREF ⎝ R1 + R2 ⎠ ⎝ RB ⎠ ⎝ RB ⎠ Lower crossover voltage when v0 = −VP So ⎛ R + RB ⎞ ⎛ R1 ⎞ ⎛ RA ⎞ VTL = − ⎜ A ⎟ ⎜ ⎟ VP − ⎜ ⎟ VREF ⎝ R1 + R2 ⎠ ⎝ RB ⎠ ⎝ RB ⎠ b. ⎛ 10 + 20 ⎞ ⎛ 5 ⎞ ⎛ 10 ⎞ VTH = ⎜ ⎟ ⎜ ⎟ (10) − ⎜ ⎟ (2) 5 + 20 ⎠ ⎝ 20 ⎠ ⎝ ⎝ 20 ⎠ or VTH = 2 V and ⎛ 10 + 20 ⎞ ⎛ 5 ⎞ VTL = − ⎜ ⎟ ⎜ ⎟ (10) − 1 ⇒ VTL = −4 V ⎝ 5 + 20 ⎠ ⎝ 20 ⎠ 15.41 a. 698. vB VREF − vB v0 − vB = + R1 R3 R2 ⎛ 1 v 1 1 ⎞ V + ⎟ = REF + 0 vB ⎜ + R3 R2 ⎝ R1 R2 R3 ⎠ VTH = vB when v0 = +VP and VTL = vB when v0 = −VP So VREF VP + R3 R2 VTH = ⎛ 1 1 1 ⎞ + ⎟ ⎜ + ⎝ R1 R2 R3 ⎠ and VTL = VREF VP − R3 R2 ⎛ 1 1 1 ⎞ + ⎟ ⎜ + ⎝ R1 R2 R3 ⎠ b. VREF ⎛ 1 1 1 ⎞ + ⎟ R3 ⎜ + ⎝ R1 R2 R3 ⎠ −10 −5 = ⎛ 1 1 1⎞ + ⎟ 10 ⎜ + R1 R2 10 ⎠ ⎝ VS = 1 1 1 1 + = − = 0.10 R1 R2 5 10 ΔVT = VTH − VTL = 0.2 = 2VP R2 ⎛ 1 1 1 ⎞ + ⎟ ⎜ + R1 R2 R3 ⎠ ⎝ 2(12) R2 (0.10 + 0.10) So R2 = 600 kΩ Then 1 1 + = 0.10 R1 R2 1 1 + = 0.10 ⇒ R1 = 10.17 kΩ R1 600 c. VTH = −5 + 0.1 = −4.9 VTL = −5 − 0.1 = −5.1 15.42 a. If the saturated output voltage is VP < 6.2 V, then the circuit behaves as a comparator where v0 < 6.2 V. 699. If the saturated output voltage is VP > 6.2 V, the output will flip to either +VP or −VP and the input has no control. b. Same as part (a) except the curve at vI ≈ 0 will have a finite slope. c. Circuit works as a comparator as long as v01 < 8.7 V and v02 > −3.7 V. Otherwise the input has no control. 15.43 a. Switching point is when v0 = 0. Then ⎛ R2 ⎞ v+ = vI ≡ VS = ⎜ ⎟ VREF ⎝ R1 + R2 ⎠ VTH occurs when vo = VH , then by superposition ⎛ R1 ⎞ ⎛ R2 ⎞ v+ = VTH = ⎜ ⎟ VH + ⎜ ⎟ VREF ⎝ R1 + R2 ⎠ ⎝ R1 + R2 ⎠ or ⎛ R1 ⎞ VTH = VS + ⎜ ⎟ VH ⎝ R1 + R2 ⎠ VTL occurs when v0 = VL , then by superposition ⎛ R1 ⎞ ⎛ R2 ⎞ v+ = VTL = ⎜ ⎟ VL + ⎜ ⎟ VREF ⎝ R1 + R2 ⎠ ⎝ R1 + R2 ⎠ or ⎛ R1 ⎞ VTL = VS + ⎜ ⎟ VL ⎝ R1 + R2 ⎠ b. Now For VTH = 2 V and VTL = 1 V, then VS = 1.5 V ⎛ 10 ⎞ 2 = 1.5 + ⎜ ⎟ (10) ⎝ 10 + R2 ⎠ 0.5 10 = ⇒ R2 = 190 kΩ 10 10 + R2 ⎛ 190 ⎞ Now VS = 1.5 = ⎜ ⎟ VREF ⎝ 10 + 190 ⎠ so VREF = 1.579 V 15.44 a. Now Switching point when v0 = 0. ⎛ R2 ⎞ v+ = VREF = ⎜ ⎟ vI where vI = VS . ⎝ R1 + R2 ⎠ Then ⎛ R + R2 ⎞ ⎛ R1 ⎞ VS = ⎜ 1 ⎟ VREF = ⎜ 1 + ⎟ VREF ⎝ R2 ⎠ ⎝ R2 ⎠ Now upper crossover voltage for v1 occurs when v0 = VL and v+ = VREF . Then 700. VTH − VREF VREF − VL = R1 R2 or VTH = − ⎛ R1 R ⎞ ⋅ VL + VREF ⎜1 + 1 ⎟ R2 R2 ⎠ ⎝ R1 ⋅ VL R2 Lower crossover voltage for vI occurs when v0 = VH and vI = VREF . Then VH − VREF VREF − VTL = R2 R1 or VTH = VS − or VTL = − ⎛ R1 R ⎞ ⋅ VH + VREF ⎜1 + 1 ⎟ R2 ⎝ R2 ⎠ or VTL = VS − R1 ⋅ VH R2 For VTH = −1 and VTL = −2, VS = −1.5 V. Then VTL = VS − b. so that R1 = 0.833 kΩ Now ⎛ R ⎞ VS = ⎜ 1 + 1 ⎟ VREF ⎝ R2 ⎠ ⎛ 0.833 ⎞ −1.5 = ⎜ 1 + ⎟ VREF 20 ⎠ ⎝ which gives VREF = −1.44 V 15.45 (a) vo (max) = 4.7 + 0.7 = 5.4 V vo (min) = −5.4 V ⎛ R1 ⎞ VTH − TTL = ⎜ ⎟ ( 5.4 − (−5.4) ) ⎝ R1 + R2 ⎠ ⎛ 2 ⎞ 0.8 = ⎜ ⎟ (10.8) ⎝ 2 + R2 ⎠ 2 + R2 = 27 ⇒ R2 = 25 K (b) R= Neglecting current in R1 and R2 , 13.0 − 5.4 ⇒ R = 15.2 K 0.5 15.46 a. v0 = VREF + 2Vγ 5 = VREF + 2(0.7) or VREF = 3.6 V b. R1 R ⋅ VH ⇒ −2 = −1.5 − 1 (12) R2 20 701. ⎛ R1 ⎞ VTH = ⎜ ⎟ (VREF + 2Vγ ) ⎝ R1 + R2 ⎠ ⎛ R1 ⎞ 0.5 = ⎜ ⎟ (5) ⎝ R1 + R2 ⎠ R2 R = 10 ⇒ 2 = 9 R1 R1 For example, let R2 = 90 kΩ and R1 = 10 kΩ c. For vI = 10 V, and v0 is in its low state. D1 is on and D2 is off. or 1 + v1 − (v1 + 0.7) VREF − v1 v1 − v0 + = 100 1 1 For v1 = −0.7, then 10 − 0 3.6 − (−0.7) −0.7 − v0 + = 100 1 1 or v0 = −5.1 V 15.47 For v0 = High = (VREF + 2Vγ ). Then switching point is when. ⎛ R1 ⎞ vI = vB = ⎜ ⎟ v0 ⎝ R1 + R2 ⎠ ⎛ R1 ⎞ or VTH = ⎜ ⎟ (VREF + 2Vγ ) ⎝ R1 + R2 ⎠ Lower switching point is when ⎛ R1 ⎞ v1 = vB = ⎜ ⎟ v0 and v0 = −(VREF + 2Vγ ) ⎝ R1 + R2 ⎠ so ⎛ R1 ⎞ VTL = − ⎜ ⎟ (VREF + 2Vγ ) ⎝ R1 + R2 ⎠ 15.48 By symmetry, inverting terminal switches about zero. Now, for v0 low, upper diode is on. VREF − v1 = v1 − v0 v0 = 2v1 − VREF where v1 = −Vγ so v0 = −(VREF + 2Vγ ) Similarly, in the high state v0 = (VREF + 2Vγ ) Switching occurs when non-inverting terminal is zero. So for v0 low. VTH − 0 0 − ⎡ − (VREF + 2Vγ ) ⎤ ⎣ ⎦ = R1 R2 or VTH = R1 ⋅ (VREF + 2Vγ R2 ) 702. By symmetry R VTL = − 1 ⋅ (VREF + 2vγ ) R2 15.49 f = 1 2.2 RX C X 1 1 = 2.2 f (2.2)(12 × 103 ) RX C X = 3, 788 × 10−5 RX C X = For example, Let RX = 56 K C X = 680 pF Within 1 of 1% of design specification. 2 15.50 (a) ⎛ R1 ⎞ ⎛ 20 ⎞ vx1 = ⎜ ⎟ vo = ⎜ ⎟ vo ⎝ 20 + 5 ⎠ ⎝ R1 + R2 ⎠ So vH = ±9.6 Also vx2 = 12 + (−9.6 − 12)e − t / τ x = 12 − 21.6e − t / τ x Set vx1 = vx2 9.6 = 12 − 21.6e − t / τ x T = τ x ln (9) 2 1 1 f = = T 2τ x ln (9) e+ t / τ x = 9 ⇒ t1 = 1 2(22 × 10 )(0.2 × 10−6 ) ln (9) f = 51.7 Hz Duty cycle = 50% f = 3 15.51 (a) ⎛ R1 ⎞ ⎛ 20 ⎞ + vx1 = ⎜ ⎟ VH = ⎜ ⎟ (15) = 12 V ⎝ 20 + 5 ⎠ ⎝ R1 + R2 ⎠ ⎛ 20 ⎞ − vx1 = ⎜ ⎟ (−10) = −8 V ⎝ 20 + 5 ⎠ + vx 2 = 15 + (−8 − 15)e − t / τ x = 15 − 23e− t / τ x Then 12 = 15 − 23e − t1 / τ x e − t1 / τ x = 7.667 ⇒ t1 − τ x ln (7.667) ⎛ − t2 − t1 ⎞ ⎜ ⎟ ⎜ τ ⎟ x ⎠ − vx 2 = −10 + (12 + 10)e⎝ Then 703. −8 = −10 + 22e ⎛ − t2 − t1 ⎞ ⎜ ⎟ ⎜ τ ⎟ x ⎝ ⎠ − ( t2 − t1 ) e τx = 11 ⇒ t2 − t1 = τ x ln (11) Period = t2 = T = τ x [ ln (7.667) + ln (11) ] = τ x (4.435) τ x = (22 × 103 )(0.2 × 10−6 ) = 4.4 × 10−3 T = 1.95 × 10−2 1 f = = 51.2 Hz T t1 = τ x ln(7.667) = (4.4 ×10 −3 ) ln (7.667) t1 = 8.962 × 10−3 t1 8.962 × 10−3 = T 1.951× 10−2 Duty Cycle = 45.9% Duty cycle = 15.52 t1 = 1.1RX C X = (1.1)(10 4 )(0.1× 10−6 ) ⇒ t1 = 1.1 ms 0 < t < t1 , vY = 10(1 − e− t / rY ) rY = RY CY Now = (2 × 103 )(0.02 × 10−6 ) = 4 × 10−5 s t1 = 2.75 rY ⇒ CY completely charges during each cycle. 15.53 a. Switching voltage ⎛ R1 + R3 ⎞ ⎛ 10 + 10 ⎞ vX = ⎜ ⎟ ⋅ VP = ⎜ ⎟ (±10) ⎝ 10 + 10 + 10 ⎠ ⎝ R1 + R3 + R2 ⎠ So v X = ±6.667 V Using Equation (15.83(b)) 2 ⎛ 2 ⎞ v X = VP + ⎜ − VP − VP ⎟ e − t1 / rX = VP 3 ⎝ 3 ⎠ 5 2 Then 1 − ⋅ e− t1 / rX = 3 3 1 5 − t1 / rX or t1 = rX ln (5) = ⋅e 3 3 T 1 1 t1 = = ⇒ t1 = 0.001 s = 2 2 f 2(500) 10−3 = rX ln (5) ⇒ rX = 6.21× 10−4 = RX (0.01× 10−6 ) So RX = 62.1 kΩ b. Switching voltage ⎛ ⎞ R1 vX = ⎜ ⎟ (±VP ) ⎝ R1 + R3 + R2 ⎠ 10 1 ⎛ ⎞ =⎜ ⎟ (±VP ) = ⋅ (±VP ) 3 ⎝ 10 + 10 + 10 ⎠ 704. Using Equation (15.83(b)) 1 ⎛ 1 ⎞ v X = VP + ⎜ − VP − VP ⎟ e − t1 / rX = VP 3 ⎝ 3 ⎠ 4 − t1 / rX 1 = Then 1 − ⋅ e 3 3 2 4 − t1 / rX = ⋅e 3 3 t1 = rX ln (2) = (6.21× 10−4 ) ln (2) = 4.30 × 10−4 s T = 2t1 = 8.6 × 10−4 s 1 f = ⇒ f = 1.16 kHz T 15.54 From Equation (15.92) ⎛ ⎛ Vγ ⎞ ⎞ ⎜1+ ⎜ ⎟ ⎟ V T = rX ln ⎜ ⎝ P ⎠ ⎟ ⎜ 1− β ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ R1 10 where β = = = 0.2857 R1 + R2 10 + 25 so 0.7 ⎤ ⎡ ⎢ 1+ 5 ⎥ 100 = rX ln ⎢ ⎥ ⎢1 − 0.2857 ⎥ ⎢ ⎥ ⎣ ⎦ so τ X = 213.9 μ s = RX C X For example, RX = 10 kΩ, C X = 0.0214 μ F ⎛ R1 ⎞ ⎛ 10 ⎞ vY = ⎜ ⎟ VP = ⎜ ⎟ (5) = 1.43 V ⎝ 10 + 25 ⎠ ⎝ R1 + R2 ⎠ and v X = 0.7 V To trigger the circuit, vY must be brought to a voltage less than v X . Therefore minimum triggering pulse is −0.73 V. Using Equation (15.82) for T < t < T ′ v X = VP + (−0.2857VP − VP )e − t ′ / rX Recovery period is when v X = Vγ = 0.7 V. 0.7 = 5 + (−6.43)e − t ′ / rX 6.43e− t ′ / rX = 4.3 or t ′ = rX ln (1.495) rX = 213.9 μ s so t ′ = T ′ − T = 86.1 μ s 15.55 (a) 705. ⎡1 + (Vr / VP ) ⎤ T =τx ⎢ ⎥ ⎣ 1− β ⎦ τ x = Rx Cx β = 1 2 T 0.69τ x = (0.69)(47 × 103 )(0.2 × 10−6 ) T = 6.49 ms Recovery Time 0.4τ x (b) = 3.76 ms 15.56 a. From Equation (15.95) T = 1.1 RC For T = 60 s = 1.1 RC then RC = 54.55 s For example, let C = 50 μ F and R = 1.09 MΩ b. Recovery time: capacitor is discharged by current through the discharge transistor. 5 − 0.7 If V + = 5 V, then I B ≅ = 0.043 mA 100 If β = 100, I C = 4.3 mA VC = 1 Ic ∫ IC dt = C ⋅ t C 2 + ⋅ V = 3.33 V 3 V ⋅ C (3.33)(50 × 10−6 ) So that t = C = IC ⋅ 4.3 × 10−3 Capacitor has charged to So recovery time t ≈ 38.7 ms 15.57 T = 1.1 RC 5 × 10−6 = 1.1 RC so RC = 4.545 × 10−6 s For example, let C = 100 pF and R = 45.5 kΩ From Problem (15.53), recovery time V ⋅ C (3.33)(100 × 10−12 ) t≅ C = IC 4.3 × 10−3 or t = 77.4 ns 15.58 From Equation (15.102), 1 f = (0.693)(20 + 2(20)) × 103 × (0.1× 10−6 ) or f = 240.5 Hz Duty cycle = 20 + 20 × 100% = 66.7% 20 + 2(20) 706. 15.59 f = 1 (0.693)( RA + 2 RB )C RA = R1 = 10 kΩ, RB = R2 + xR3 So 10 kΩ ≤ RB ≤ 110 kΩ 1 = 627 kHz (0.693)(10 + 2(110)) × 103 × (0.01× 10−6 ) 1 = = 4.81 kHz (0.693)(10 + 2(10)) × 103 × (0.01× 10−6 ) f min = f max So 627 Hz ≤ f ≤ 4.81 kHz Duty cycle = RA + RB × 100% RA + 2 RB Now 10 + 10 × 100% = 66.7% 10 + 2(10) and 10 + 110 × 100% = 52.2% 10 + 2(110) So 52.2 ≤ Duty cycle ≤ 66.7% 15.60 1 kΩ ≤ RA ≤ 51 kΩ 1 kΩ ≤ RB ≤ 51 kΩ 1 = 1.40 Hz (0.693)(1 + 2(51)) × 103 × (0.01× 10−6 ) 1 f max = = 2.72 kHz (0.693)(51 + 2(1)) × 103 × (0.01× 10−6 ) or 1.40 kHz ≤ f ≤ 2.72 kHz f min = Duty cycle = RA + RB × 100% RA + 2 RB 1 + 51 × 100% = 50.5% 1 + 2(51) or 51 + 1 × 100% = 98.1% 51 + 2(1) or 50.5% ≤ Duty cycle ≤ 98.1% 15.61 a. I E3 = IE 4 = V + − 3VEB R1 A + R1B Assume VEB = 0.7 707. I E3 = I E 4 = 22 − 3(0.7) = 0.398 mA 25 + 25 Now ⎛ 20 ⎞ I C 3 = I C 4 = I C 5 = I C 6 = ⎜ ⎟ (0.398) ⎝ 21 ⎠ I C 3 = I C 4 = I C 5 = I C 6 = 0.379 mA 0.398 ⎛ 20 ⎞ ⎜ ⎟ ⇒ I C1 = I C 2 = 0.018 mA 21 ⎝ 21 ⎠ I D = 0.398 mA, current in D1 and D2 I C1 = I C 2 = b. ⎛I ⎞ ⎛ 0.398 × 10−3 ⎞ VBB = 2VD = 2VT ln ⎜ D ⎟ = 2(0.026) ln ⎜ ⎟ −13 ⎝ 10 ⎠ ⎝ IS ⎠ or VBB = 1.149 V = VBE 7 + VEB 8 Now IC 7 ≈ IC 4 + IC 9 + I E8 I C 4 = 0.379 mA ⎛ 20 ⎞ I B9 = IC 8 = ⎜ ⎟ I E 8 ⎝ 21 ⎠ So ⎛I ⎞ I E 8 = 1.05 I B 9 = 1.05 ⎜ C 9 ⎟ ⎝ 100 ⎠ ⎛ 100 ⎞ ⎛ 21 ⎞ IC 7 = IC 4 + ⎜ ⎟ I E 8 + I E 8 = I C 4 + (96.24) ⎜ ⎟ I C 8 ⎝ 1.05 ⎠ ⎝ 20 ⎠ So I C 7 = 0.379 mA + 101I C 8 and ⎛I ⎞ ⎛I ⎞ VBE 7 = VT ln ⎜ C 7 ⎟ ; VEB 8 = VT ln ⎜ C 8 ⎟ IS ⎠ ⎝ ⎝ IS ⎠ Then ⎡ ⎛I ⎞ ⎛ I ⎞⎤ 1.149 = 0.026 ⎢ ln ⎜ C 7 ⎟ + ln ⎜ C 8 ⎟ ⎥ ⎝ I S ⎠⎦ ⎣ ⎝ IS ⎠ ⎡ I (0.379 × 10−3 ) + 101I C 8 ⎤ 44.19 = ln ⎢ C 8 ⎥ (10−13 ) 2 ⎣ ⎦ 2 (10−13 ) 2 exp (44.19) = 101I C 8 + 3.79 × 10 −4 I C 8 2 −7 1.554 × 10 = 101I C 8 + 3.79 × 10−4 I C 8 IC 8 = (3.79 × 10−4 ) 2 + 4(101)(1.554 × 10−7 ) −3.79 × 10−4 ± 2(101) 2(101) I C 8 = 37.4 μ A I C 7 = 0.379 + 101(0.0374) ⇒ I C 7 = 4.16 mA ⎛ 21 ⎞ I C 9 = 4.16 − 0.379 − 0.0374 ⎜ ⎟ ⎝ 20 ⎠ I C 9 = 3.74 mA c. P = (0.398 + 0.398 + 4.16)(22) ⇒ P = 109 mW 708. 15.62 a. b. From Figure 15.47, 3.7 W to the load V + ≈ 19 V c. P= 1 VP2 2 RL or VP = 2 RL P = 2(10)(3.7) ⇒ VP = 8.6 V 15.63 1 VP2 P= 2 RL so VP = 2 RP = 2(10)(20) ⇒ 20 V peak-to-peak output voltage Maximum output voltage of each op-amp = ±10 V. Current is (20 /10) = 2 A. Bias op-amps at ±12 V. For A1 , v01 ⎛ R2 ⎞ R = ⎜1 + ⎟ = 15 ⇒ 2 = 14 vI ⎝ R1 ⎠ R1 For A2 , v02 R = 4 = 15 vI R3 For example, let R1 = R3 = 10 kΩ, and R2 = 140 kΩ and R4 = 150 kΩ. 15.64 a. v01 = iR2 + vI where i = vI R1 Then ⎛ R ⎞ v01 = vI ⎜ 1 + 2 ⎟ R1 ⎠ ⎝ Now ⎛R ⎞ v02 = −iR3 = −vI ⎜ 3 ⎟ ⎝ R1 ⎠ So ⎛ R ⎞ ⎡ ⎛ R ⎞⎤ vL = v01 − v02 = vI ⎜1 + 2 ⎟ − ⎢ −vI ⎜ 3 ⎟ ⎥ R1 ⎠ ⎣ ⎝ ⎝ R1 ⎠ ⎦ Av = b. vL R R = 1+ 2 + 3 vI R1 R1 Want Av = 10 ⇒ R2 R3 + =9 R1 R1 ⎛ R ⎞ R Also want ⎜1 + 2 ⎟ = 3 R1 ⎠ R1 ⎝ R2 ⎛ R2 ⎞ R + ⎜1 + ⎟ = 9 so 2 = 4 R1 ⎝ R1 ⎠ R1 For R1 = 50 kΩ, R2 = 200 kΩ Then and 709. R3 = 5 so R3 = 250 kΩ R1 P= c. 1 VP2 2 RL or VP = 2 RL P = 2(20)(10) = 20 V So peak values of output voltages are v01 = v02 = 10 V Peak load current = 20 =1 A 20 15.65 (a) ⎛ R ⎞ vo1 = ⎜1 + 2 ⎟ vI R1 ⎠ ⎝ ⎛ R2 ⎞ ⎜ 1 + ⎟ vI R1 ⎠ ⎝ vL = vo1 − vo 2 vo 2 = − Av = R4 R3 vL ⎛ R2 ⎞ ⎛ R4 ⎞ = ⎜1 + ⎟ ⎜1 + ⎟ vI ⎝ R1 ⎠ ⎝ R3 ⎠ For v01 − 12 V, vo 2 = −12 V when R3 = R4 (b) So vL = 24sin ω + (V ) ⎛ R ⎞⎛ R ⎞ 10 = ⎜ 1 + 2 ⎟ ⎜1 + 4 ⎟ R1 ⎠ ⎝ R3 ⎠ ⎝ Let R3 = R4 (c) Then R2 =4 R1 15.66 (a) From Problem 15.65 vO ⎛ R2 ⎞ ⎛ R4 ⎞ = ⎜1 + ⎟ ⎜1 + ⎟ vI ⎝ R1 ⎠⎝ R3 ⎠ For vo1 = vo 2 ⇒ R3 = R4 Then ⎛ R ⎞ vO = 2 ⎜1 + 2 ⎟ vI R1 ⎠ ⎝ ⎛ R ⎞ ⎛R ⎞ 15 = 2 ⎜1 + 2 ⎟ ⇒ ⎜ 2 ⎟ = 6.5 R1 ⎠ ⎝ R1 ⎠ ⎝ (b) Peak output voltage VP = 2 RL PL = 2(16)(20) = 25.3 V Then Vo1 = Vo 2 = 12.65 V iL (peak) = 25.3 = 1.58 A 16 710. 15.67 Line regulation = ΔV0 ΔV + Now ΔI = ΔV + and ΔVZ = rZ ⋅ ΔI and ΔV0 = 10ΔVZ R1 So ΔV0 = 10 ⋅ rZ ⋅ ΔV + R1 So Line regulation = ΔV0 10(15) = ⇒ 1.61% ΔV + 9300 15.68 R0 f = − ΔV0 ΔI 0 So R0 f = −(−10 × 10−3 ) 1 or R0 f = 10 mΩ 15.69 For V0 = 8 V V + (min) = V0 + I 0 (max) R11 + VBE11 + VBE10 + VEB 5 This assumes VBC 5 = 0. Then V + (min) = 8 + (0.1)(1.9) + 0.6 + 0.6 + 0.6 V + (min) = 9.99 V 15.70 a. IC 3 = IC 5 = VZ − 3VBE (npn) R1 + R2 + R3 6.3 − 3(0.6) = 0.571 mA 0.576 + 3.4 + 3.9 1 ⎛ 0.6 ⎞ IC 8 = ⎜ ⎟ = 0.106 mA 2 ⎝ 2.84 ⎠ Neglecting current in Q9 , total collector current and emitter current in Q5 is 0.571 + 0.106 = 0.677 Now I Z 2 R4 + VEB 4 = VEB 5 IC 3 = IC 5 = ⎛I ⎞ VEB 4 = VT ln ⎜ Z 2 ⎟ ⎝ I5 ⎠ ⎛I ⎞ VEB 5 = VT ln ⎜ C 5 ⎟ ⎝ 2I S ⎠ 711. ⎛ I ⎞ Then I Z 2 R4 = VT ln ⎜ C 5 ⎟ ⎝ 2I Z 2 ⎠ ⎛ 0.677 ⎞ 0.026 R4 = ⋅ ln ⎜ ⎟ 0.25 ⎝ 2(0.25) ⎠ or R4 = 31.5 Ω b. From Example 15.16, VB 7 = 3.43 V . Then ⎛ R13 ⎞ ⎜ ⎟ V0 = VB 8 = VB 7 ⎝ R12 + R13 ⎠ or ⎛ 2.23 ⎞ ⎜ ⎟ (12) = 3.43 ⎝ 2.23 + R12 ⎠ 3.43(2.23 + R12 ) = (2.23)(12) which yields R12 = 5.57 kΩ 15.71 Line regulation = ΔV0 ΔV + Now ΔVB 7 = ΔI C 3 ⋅ R1 ⎛ R13 ⎞ and ⎜ ⎟ (ΔV0 ) = ΔVB 7 = ΔI C 3 R1 ⎝ R12 + R13 ⎠ ΔVZ ΔI Z ⋅ rZ and ΔI C 3 = = R1 + R2 + R3 R1 + R2 + R3 and ΔI Z = V ΔV + where r0 = A IZ r0 Then ⎛ 0.015 ⎞ (0.4288)( ΔV0 ) = ΔI C 3 (3.9) = (3.9) ΔI Z ⎜ ⎟ ⎝ 7.876 ⎠ 50 = 87.6 kΩ r0 = 0.571 Then ⎛ ΔV + ⎞ (0.4288)(ΔV0 ) = (0.00743) ⎜ ⎟ ⎝ 87.6 ⎠ So ΔV0 = 0.0198% ΔV + 15.72 a. 712. IZ = 25 − 5 = 10 R1 + rZ 20 = 2 kΩ = R1 + 0.01 ⇒ R1 = 1.99 kΩ 10 b. In the ideal case; ⎛ R3 + R4 ⎞ ⎜ ⎟ V0 = VZ ⎝ R2 + R3 + R4 ⎠ So R1 + rZ = ⎛ 2 +1 ⎞ ⎜ ⎟ V0 = 5 ⇒ V0 = 6.67 V ⎝ 2 +1+1⎠ and ⎛ ⎞ R4 ⎜ ⎟ V0 = VZ ⎝ R2 + R3 + R4 ⎠ ⎛ 1 ⎞ ⎜ ⎟ V0 = 5 ⇒ V0 = 20 V ⎝ 2 +1+1⎠ So 6.67 ≤ V0 ≤ 20 V c. 3 3 ⋅ V0 so vd = 5 − ⋅ V0 4 4 and V0 = A0 L vd − VBE V1 = ⎛I ⎞ and VBE = VT ln ⎜ 0 ⎟ ⎝ IS ⎠ Now 3 ⎛ ⎞ V0 = A0 L ⎜ 5 − ⋅ V0 ⎟ − VBE 4 ⎝ ⎠ ⎛ 3 ⎞ V0 ⎜ 1 + ⋅ A0 L ⎟ = 5 A0 L − VBE 4 ⎝ ⎠ 5 A0 L − VBE V0 = 3 1 + ⋅ A0 L 4 713. Load regulation = V0 (NL) − V0 (FL) V0 (NL) 5 A0 L − VBE (NL) 5 A0 L − VBE (FL) − (1 + 3 A0 L ) (1 + 3 A0 L ) 4 4 = 5 A0 L − VBE (NL) (1 + 3 A0 L ) 4 ⎛ I (FL) ⎞ VT ln ⎜ 0 ⎟ VBE (FL) − VBE (NL) ⎝ I 0 (NL) ⎠ = = 5 A0 L − VBE (NL) 5 A0 L − VBE (NL) V0 6.67 = = 1.67 mA 4 k Ω 4 kΩ 1 ⎛ ⎞ (0.026) ln ⎜ −3 ⎟ ⎝ 1.67 × 10 ⎠ ⇒ 3.33 × 10−4 % Load regulation = 4 5(10 ) − 0.7 I 0 (FL) = 1 A, I 0 (NL) = 15.73 V 5.6 IE = Z = = 1.12 mA 5 R2 I0 = β 1+ β ⎛ 100 ⎞ ⋅ IE = ⎜ ⎟ (1.12) ⇒ I 0 = 1.109 mA Load current ⎝ 101 ⎠ For VBC = 0 ⇒ V0 = 20 − VZ − 0.6 = 20 − 5.6 − 0.6 or V0 = 13.8 V Then V 13.8 ⇒ RL = 12.4 kΩ RL = 0 = I 0 1.109 So 0 ≤ RL ≤ 12.4 kΩ 714. Chapter 16 Exercise Solutions EX16.1 COX = γ = ( 3.9 ) (8.85 ×10−14 ) 200 × 10−8 = 1.726 ×10−7 F / cm 2 2 (1.6 × 10−19 ) (11.7 ) ( 8.85 ×10 −14 )(1015 ) 1.726 × 10−7 1 = 0.1055 V 2 ⎡ ⎤ ΔVTN = r ⎣ 0.576 + VSB − 0.576 ⎦ = ( 0.1055 ) ⎡ 0.576 + 5 + 0.576 ⎤ ⎣ ⎦ ΔVTN = 0.169 V EX16.2 (a) vo = VDD − I D RD ⎛ k′ ⎞⎛ W vo = 3 − ⎜ n ⎟ ⎜ ⎝ 2 ⎠⎝ L vo = 0.1 ⎞ 2 ⎡ ⎤ ⎟ ⎣ 2 ( 3 − 0.5 ) vo − vo ⎦ RD ⎠ 2 ⎛ 0.06 ⎞ ⎡ 0.1 = 3 − ⎜ ⎟ ( 5 ) ⎣( 5 )( 0.1) − ( 0.1) ⎤ RD ⎦ ⎝ 2 ⎠ 0.1 = 3 − 0.0735RD RD = 39.5 K (b) 2 ⎛ 0.06 ⎞ ⎜ ⎟ ( 5 )( 39.5 )(VIt − 0.5 ) + (VIt − 0.5 ) − 3 = 0 2 ⎠ ⎝ 5.925 (VIt − 0.5 ) + (VIt − 0.5 ) − 3 = 0 2 (VIt − 0.5 ) = VOt = −1 ± 1 + 4 ( 5.925 )( 3) 2 ( 5.925 ) VOt = 0.632 V VIt = 1.132 V EX16.3 (a) (i) vo = VDD − VTNL = 3 − 0.4 vo = 2.6 V (ii) 2 ⎛W ⎞ ⎛W ⎞ 2 ⎜ ⎟ ⎡ 2 ( vI − 0.4 ) vo − vo ⎤ = ⎜ ⎟ [VDD − vo − 0.4] ⎣ ⎦ ⎝ L ⎠D ⎝ L ⎠L 2 16 ⎡ 2 ( 2.6 − 0.4 ) vo − vo ⎤ = 2 [3 − vo − 0.4] ⎣ ⎦ 2 2 2 35.2vo − 8vo = 6.76 − 5.2vo + vo 2 9vo − 40.4vo + 6.76 = 0 vo = 40.4 ± 1632.16 − 4 ( 9 )( 6.76 ) vo = 0.174 V 2 (9) 715. (b) 2 ⎛ 60 ⎞ iD = ⎜ ⎟ (2) [3 − 0.174 − 0.4] ⎝ 2 ⎠ iD = 353.1 μ A P = iD ⋅ VDD = 1.06 mW EX16.4 (a) 2 ⎛W ⎞ ⎛W ⎞ 2 ⎜ ⎟ ⎡ 2 ( vI − 0.4 ) vo − vo ⎤ = ⎜ ⎟ ( − ( −0.8 ) ) ⎦ L ⎠D ⎣ L ⎠L ⎝ ⎝ 2 6 ⎡ 2 ( 3 − 0.4 ) vo − vo ⎤ = 2 ( 0.64 ) ⎣ ⎦ 2 6vo − 31.2vo + 1.28 = 0 vo = 31.2 ± 973.44 − 4 ( 6 )(1.28 ) 2(6) vo = 41.4 mV (b) 2 2 ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ ( vIt − 0.4 ) = ⎜ ⎟ ( −(−0.8) ) L ⎠D L ⎠L ⎝ ⎝ 6 ( vIt − 0.4 ) = 2 ( 0.64 ) 2 ⇒ vIt = 0.862 V ⎫ ⎬ Driver vOt = 0.862 − 0.4 = 0.462 V ⎭ vIt = 0.862 V ⎫ ⎬ Load vOt = VDD + VTNL = 3 − 0.8 = 2.2 V ⎭ (c) 2 ⎛ 60 ⎞ iD = ⎜ ⎟ (2) ( − ( −0.8 ) ) = 38.4 μ A ⎝ 2 ⎠ P = iD ⋅ VDD = 115.2 μ W EX16.5 We have { } 0.73 ⎤} ⎦ VOH = VDD − VTNLO + r ⎡ 2φ fP + VSB − 2φ fP ⎤ ⎣ ⎦ { VOH = 5 − 0.8 + 0.35 ⎡ 0.73 + VOH − ⎣ VOH − 4.499 = −0.35 0.73 + VOH Squaring both sides 2 VOH − 8.998VOH + 20.241 = 0.1225(0.73 + VOH ) 2 VOH − 9.1205VOH + 20.15 = 0 9.1205 ± 83.1835 − 4(20.15) 2 = 3.76 V VOH = VOH EX16.6 a. i. A = logic 1 = 10 V, B = logic 0 “A” driver in nonsaturation. “B” driver off 716. ′ 2 ⎛ kn ⎞ ⎛ W ⎞ ⎛ k′ ⎞⎛ W ⎞ 2 ⎜ 2 ⎟ ⎜ L ⎟ ( −VTNL ) = ⎜ 2 ⎟ ⎜ L ⎟ ⎡ 2 ( vI − vTND ) VOL − VOL ⎤ ⎣ ⎦ ⎝ ⎠ ⎝ ⎠D ⎝ ⎠ ⎝ ⎠L 2 ( 3) = (10 ) ⎡ 2 (10 − 1.5 ) V0 L − V02L ⎤ ⎣ ⎦ 2 9 = 5 (17V0 L − V02L ) 5V02L − 85V0 L + 9 = 0 (85 ) 85 ± 2 − 4 ( 5 )( 9 ) ⇒ V0 L = 0.107 V 2(5) ii. A = B = logic 1 ′ 2 ⎛ kn ⎞ ⎛ W ⎞ ⎛ k′ ⎞⎛ W ⎞ 2 ⎡ ⎜ 2 ⎟ ⎜ L ⎟ ( −VTNL ) = 2 ⎜ 2 ⎟ ⎜ L ⎟ ⎣ 2 ( vI − vTND )VOL − VOL ⎤ ⎦ ⎝ ⎠ ⎝ ⎠D ⎝ ⎠ ⎝ ⎠L V0 L = 2 ( 3) = ( 2 )(10 ) ⎡ 2 (10 − 1.5 ) V0 L − V02L ⎤ ⎣ ⎦ 2 9 = 10 (17V0 L − V02L ) 10V02L − 170V0 L + 9 = 0 V0 L = 170 ± (170 ) − 4 (10 )( 9 ) ⇒ V0 L = 0.0531 V 2 (10 ) 2 b. Both cases. 35 2 iD = ⋅ ( 2 )( 3) = 315 μ A ⇒ P = iD ⋅ VDD ⇒ P = 3.15 mW 2 EX16.7 800 = 160 μ A 5 35 ⎛ W ⎞ 2 ⎛W ⎞ ⎛W ⎞ iD = 160 = ⋅ ⎜ ⎟ (1.4 ) = 34.3 ⎜ ⎟ ⇒ ⎜ ⎟ = 4.66 L ⎠L 2 ⎝ L ⎠L ⎝ ⎝ L ⎠L P = iD ⋅ VDD ⇒ iD = iD = 160 = 35 ⎛ W ⎞ ⎡ 2 ⎛W ⎞ ⋅ ⎜ ⎟ 2 ( 5 − 0.8 )( 0.12 ) − ( 0.12 ) ⎤ ⇒ ⎜ ⎟ = 9.20 ⎦ ⎝ L ⎠D 2 ⎝ L ⎠D ⎣ EX16.8 (a) VOPt VONt (b) VDD 2.1 = = 1.05 V 2 2 = VIt − VTD = 1.05 − (−0.4) = 1.45 V = VIt − VTN = 1.05 − 0.4 = 0.65 V VIt = VIt = VOPt 2.1 + (−0.4) + 0.5(0.4) 1 + 0.5 = 1.16 + 0.4 = 1.56 V = 1.16 V VONt = 1.16 − 0.4 = 0.76 V (c) VIt = VOPt VONt EX16.9 2.1 + (−0.4) + 2(0.4) = 0.938 V 1+ 2 = 0.938 + 0.4 = 1.338 V = 0.538 V 717. 2 P = f ⋅ CL ⋅ VDD ( 0.10 ×10 ) = f ( 0.5 ×10 ) ( 3) −6 −12 2 f = 2.22 × 104 Hz ⇒ f = 22.2 kHz EX16.10 a. K n / K p = 200 / 80 = 2.5 ⇒ VIt = 10 − 2 + 2.5(2) 1 + 2.5 ⇒ VIt = 4.32 V V0 Pt = 6.32 V V0 Nt = 2.32 V b. VIL = 2 + ⎤ 10 − 2 − 2 ⎡ 2.5 ⋅ ⎢2 − 1⎥ ⇒ VIL = 3.39 V 2.5 − 1 ⎣ 2.5 + 3 ⎦ 1 {(1 + 2.5)(3.39) + 10 − (2.5)(2) + 2} 2 = 9.43 V V0 HU = V0 HU VIH = 2 + ⎤ 10 − 2 − 2 ⎡ 2(2.5) ⋅⎢ − 1⎥ ⇒ VIH = 4.86 V 2.5 − 1 ⎢ 3(2.5) + 1 ⎥ ⎣ ⎦ (4.86)(1 + 2.5) − 10 − (2.5)(2) + 2 2(2.5) = 0.802 V V0 LU = V0 LU c. NM L = VIL − V0 LU = 3.39 − 0.802 ⇒ NM L = 2.59 V NM H = V0 HU − VIH = 9.43 − 4.86 ⇒ NM H = 4.57 V EX16.11 3 PMOS in series and 3 NMOS in parallel. Worst Case: Only one NMOS is ON in Pull-down mode ⇒ same as the CMOS inverter ⇒ Wn = W . All 3 PMOS are on during pull-up mode ⇒ W p = 3(2W ) = 6W . EX16.12 NMOS: Worst Case, M NA , M NB on, Wn = 2(W ) or M NC , M ND or M NC , M NE on ⇒ Wn = 2(W ). PMOS: M PA and M PC on or M PA and M PB on ⇒ WP = 2(2W ) = 4W If M PD and M PE on, need WP = 2(4W ) = 8W EX16.13 a. vI = φ = 5 V ⇒ v0 = 4 V b. vI = 3 V, φ = 5 V ⇒ v0 = 3 V c. vI = 4.2 V, φ = 5 V ⇒ v0 = 4 V d. vI = 5 V, φ = 3 V ⇒ v0 = 2 V EX16.14 (a) vI = 8V , φ = 10V ⇒ vGSD = 8 V M D in nonsaturation 718. 2 K D ⎡ 2(vGSD − VTND )vO − vO ⎤ ⎣ ⎦ K L [VDD − vO − VTNL ] 2 KD ⎡ K 2 2 2 ( 8 − 2 )( 0.5 ) − ( 0.5 ) ⎤ = [10 − 0.5 − 2] ⇒ D = 9.78 ⎦ KL ⎣ KL (b) vI = φ = 8V ⇒ vGSD = 6 V KD ⎡ K 2 2 2 6 − 2 )( 0.5 ) − ( 0.5 ) ⎤ = [10 − 0.5 − 2] ⇒ D = 15 ⎣ ( ⎦ KL KL EX16.15 16 K ⇒ 16384 cells Total Power = 125 mW = (2.5) IT ⇒ IT = 50 mA 50 mA ⇒ I = 3.05 μ A 16384 V 2.5 ⇒ R = 0.82 M Ω or R = DD = 3.05 I Then, for each cell, I = Now, I ≅ VDD R TYU16.1 P = iD ⋅ VDD ⇒ iD = 750 = 150 μ A 5 35 ⎛ W ⎞ 2 ⎜ ⎟ (5 − 0.2 − 0.8) 2 ⎝ L ⎠L ⎛W ⎞ ⎛W ⎞ 150 = 280 ⎜ ⎟ ⇒ ⎜ ⎟ = 0.536 L ⎠L ⎝ ⎝ L ⎠L 150 = ⎛ k′ ⎞⎛ W ⎞ 2 iD = ⎜ n ⎟ ⎜ ⎟ ⎡ 2(vI − VTND )vO − vO ⎤ ⎦ 2 ⎠ ⎝ L ⎠D ⎣ ⎝ 35 ⎛ W ⎞ 150 = ⎜ ⎟ ⎡ 2(4.2 − 0.8)(0.2) − (0.2) 2 ⎤ ⎦ 2 ⎝ L ⎠D ⎣ ⎛W ⎞ ⎛W ⎞ 150 = 23.1 ⋅ ⎜ ⎟ ⇒ ⎜ ⎟ = 6.49 L ⎠D ⎝ ⎝ L ⎠D TYU16.2 P = iD ⋅ VDD ⇒ I D = 350 = 70 μ A 5 2 ⎛ k′ ⎞⎛ W ⎞ iD = ⎜ n ⎟ ⎜ ⎟ ( −VTNL ) ⎝ 2 ⎠ ⎝ L ⎠L 35 ⎛ W ⎞ 2 ⎛W ⎞ 70 = ⋅ ⎜ ⎟ ( 2 ) ⇒ ⎜ ⎟ = 1 2 ⎝ L ⎠L ⎝ L ⎠L 35 ⎛ W ⎞ ⎡ 2 ⋅ ⎜ ⎟ 2 ( 5 − 0.8 )( 0.05 ) − ( 0.05 ) ⎤ ⎦ 2 ⎝ L ⎠D ⎣ ⎛W ⎞ ⎛W ⎞ 70 = 7.31 ⋅ ⎜ ⎟ ⇒ ⎜ ⎟ = 9.58 ⎝ L ⎠D ⎝ L ⎠D iD = TYU16.3 719. 800 = 160 μ A 5 35 ⎛ W ⎞ 2 ⎛W ⎞ iD = 160 = ⋅ ⎜ ⎟ (1.4 ) ⇒ ⎜ ⎟ = 4.66 2 ⎝ L ⎠L ⎝ L ⎠L P = iD ⋅ VDD ⇒ iD = iD = 160 μ A = 35 1 ⎛ W ⎞ ⎡ 2 ⎛W ⎞ ⋅ ⋅ ⎜ ⎟ 2 ( 5 − 0.8 )( 0.12 ) − ( 0.12 ) ⎤ ⇒ ⎜ ⎟ = 27.6 ⎦ ⎝ L ⎠D 2 3 ⎝ L ⎠D ⎣ TYU16.4 a. From the load transistor: kn ⎞ ⎛ W ⎞ 35 2 2 ⎛ ′ I DL = ⎜ ⎟ ⎜ ⎟ (VGSL − VTNL ) = ( 0.5 )( 5 − 0.15 − 0.7 ) 2 ⎝ 2 ⎠ ⎝ L ⎠L or I DL = 150.7 μ A Maximum v0 occurs when either A or B is high and C is high. For the two NMOS is series, the effective k N is cut in half, so I DL = ′ 1 ⎡⎛ k n ⎞ ⎛ W ⎞ ⎤ 2 ⎢⎜ ⎟ ⎜ ⎟ ⎥ ⎡ 2 (VGSD − VTND ) VDS − VDS ⎤ ⎦ 2 ⎣⎝ 2 ⎠ ⎝ L ⎠ D ⎦ ⎣ or 1 ⎡ 35 ⎛ W ⎞ ⎤ ⎡ 2 ⎢ ⋅ ⎜ ⎟ ⎥ 2 ( 5 − 0.7 )( 0.15 ) − ( 0.15 ) ⎤ ⎦ 2 ⎣ 2 ⎝ L ⎠D ⎦ ⎣ which yields ⎛W ⎞ ⎜ ⎟ = 13.6 ⎝ L ⎠D 150.7 = b. P = iD ⋅ VDD = (150.7 )( 5 ) ⇒ P = 753 μ W TYU16.5 a. v0 (max) occurs when A = B = 1 and C = D = 0 or A = B = 0 and C = D = 1 1 ⎛W ⎞ ⎛W ⎞ 2 2 ⎜ ⎟ (−VTNL ) = ⋅ ⎜ ⎟ ⎡ 2(vI − VTND )vO − vO ⎤ ⎦ 2 ⎝ L ⎠D ⎣ ⎝ L ⎠L 1 ⎛W ⎞ ⎡ ⎤ (0.5)(1.2)2 = ⋅ ⎜ ⎟ ⎣ 2(5 − 0.7)(0.15) − (0.15) 2 ⎦ 2 ⎝ L ⎠D ⎛W ⎞ ⎛W ⎞ 0.72 = (0.634) ⎜ ⎟ ⇒ ⎜ ⎟ = 1.14 ⎝ L ⎠D ⎝ L ⎠D b. 2 ⎛ k′ ⎞⎛ W ⎞ ⎛ 35 ⎞ iD = ⎜ n ⎟ ⎜ ⎟ (−VTNL ) 2 = ⎜ ⎟ (0.5) [ −(−1.2) ] ⎝ 2⎠ ⎝ 2 ⎠ ⎝ L ⎠L iD = 12.6 μ A P = iD ⋅ VDD = (12.6)(5) ⇒ P = 63 μ W TYU16.6 a. K n = K p = 50 μ A / V 2 VIt = 2.5 V iD (max) = K n (VIt − VTN ) 2 = 50(2.5 − 0.8) 2 ⇒ iD (max) = 145 μ A 720. b. K n = K p = 200 μ A / V 2 VIt = 2.5 V iD (max) = (200)(2.5 − 0.8) 2 ⇒ iD (max) = 578 μ A TYU16.7 a. 5 − 2 + (1)(0.8) VIt = 1+1 VIt = 1.9 V V0 Pt = 3.9 V V0 Nt = 1.1 V b. 3 VIL = 0.8 + ⋅ [5 − 2 − 0.8] ⇒ VIL = 1.63 V 8 1 V0 HU = {2(1.63) + 5 − 0.8 + 2} 2 V0 HU = 4.73 V 5 VIH = 0.8 + (5 − 2 − 0.8) ⇒ VIH = 2.18 V 8 1 V0 LU = {2(2.18) − 5 − 0.8 + 2} 2 V0 LU = 0.275 V c. NM L = VIL − V0 LU = 1.63 − 0.275 ⇒ NM L = 1.35 V NM H = V0 HU − VIH = 4.73 − 2.18 ⇒ NM H = 2.55 V TYU16.8 721. TYU16.9 NMOS − 2 transistors in series Wn = 2 (W ) = 2W PMOS − 2 transistors in series W p = 2 ( 2W ) = 4W TYU16.10 The NMOS part of the circuit is: 722. TYU16.11 The NMOS part of the circuit is: TYU16.12 Exclusive-OR A 0 1 0 1 B 0 0 1 1 f 0 1 1 0 723. TYU16.13 NMOS conducting for 0 ≤ vI ≤ 4.2 V ⇒ NMOS Conducting: 0 ≤ t ≤ 8.4 s NMOS Cutoff: 8.4 ≤ t ≤ 10 s PMOS cutoff for 0 ≤ vI ≤ 1.2 V ⇒ PMOS Cutoff: 0 ≤ t ≤ 2.4 s PMOS Conducting: 2.4 ≤ t ≤ 10 s TYU16.14 (a) 1 K ⇒ 32 × 32 array Each row and column requires a 5-bit word ⇒ 6 transistors per row and column, ⇒ 32 × 6 + 32 × 6 = 384 transistors plus buffer transistors. (b) 4 K ⇒ 64 × 64 array Each row and column requires a 6-bit word ⇒ 7 transistors per row and column ⇒ 64 × 7 + 64 × 7 = 896 transistors plus buffer transistors. (c) 16 K ⇒ 128 × 128 array Each row and column requires a 7-bit word ⇒ 8 transistors per row and column ⇒ 128 × 8 + 128 × 8 = 2048 transistors plus buffer transistors. TYU16.15 From Equation (16.84) 2 (W / L )nA 2 (VDDVTN ) − 3VTN 2(2.5)(0.4) − 3(0.4) 2 = = = 0.526 2 2 (W / L )n1 (VDD − 2VTN ) ( 2.5 − 2(0.4) ) From Equation (16.86) 724. (W / L ) p (W / L )nB = 2 ′ ⎡ 2(2.5)(0.4) − 3(0.4) 2 ⎤ kn 2 (VDDVTN ) − 3VTN ⋅ = (2.5) ⎢ ⎥ = 0.862 2 k′ (2.5 − 0.4) 2 (VDD + VTP ) ⎣ ⎦ p ⎛W ⎞ ⎛W ⎞ So ⎜ ⎟ of transmission gate device must be < 0.526 times the ⎜ ⎟ of the NMOS transistors in the L⎠ ⎝ ⎝L⎠ ⎛W ⎞ ⎛W ⎞ inverter cell. The ⎜ ⎟ of the PMOS transistors must be < 0.862 times the ⎜ ⎟ of the transmission gate ⎝L⎠ ⎝L⎠ W⎞ W⎞ ⎛ ⎛ devices. Then the ⎜ ⎟ of the PMOS devices must be < 0.453 times ⎜ ⎟ of NMOS devices in cell. ⎝L⎠ ⎝L⎠ TYU16.16 Initial voltage across the storage capacitor = VDD − VTN = 3 − 0.5 = 2.5 V . Now dV I −I = C or V = − ⋅ t + K dt C 2.5 = 1.25 V , and C = 0.05 pF . Then where K = 2.5 V , t = 1.5 ms, V = 2 I (1.5 × 10−3 ) 1.25 = 2.5 − ⇒ (0.05 × 10−12 ) I = 4.17 × 10−11 A ⇒ I = 41.7 pA 725. Chapter 16 Problem Solutions 16.1 (a) ΔVTN = Cax = 2e ∈s N a Cax ⎡ 2φ fp + VSB − 2φ fp ⎤ ⎣ ⎦ ∈ax (3.9)(8.85 × 10 −14 ) = = 7.67 × 10−8 450 × 10−8 tax 2e ∈s N a = ⎡ 2 (1.6 × 10 −19 ) (11.7 ) ( 8.85 × 10−14 )( 8 × 1015 ) ⎤ ⎣ ⎦ Then 5.15 × 10−8 ⎡ ΔVTN = ⋅ 2(0.343) + VSB − 2(0.343) ⎤ ⎦ 7.67 × 10−8 ⎣ For VSB = 1 V : 1/ 2 = 5.15 × 10 −8 ΔVTN = 0.671 ⎡ 1.686 − 0.686 ⎤ ⇒ ΔVTN = 0.316 V ⎣ ⎦ For VSB = 1 V : ⎡ ⎤ ΔVTN = 0.671 ⎣ 2.686 − 0.686 ⎦ ⇒ ΔVTN = 0.544 V (b) For VGS = 2.5 V, VDS = 5 V, transistor biased in the saturation region. ID For VSB ID For VSB = K n (VGS − VTN ) 2 = 0, = 0.2(2.5 − 0.8) 2 = 0.578 mA = 1, I D = 0.2 ( 2.5 − [ 0.8 + 0.316]) = 0.383 mA 2 For VSB = 2, I D = 0.2 ( 2.5 − [ 0.8 + 0.544]) = 0.267 mA 2 16.2 (a) ID = VDD − vO 2 = K n ⎡ 2(VGS − VTN )vO − vO ⎤ ⎣ ⎦ RD 5 − (0.1) 2 = K n ⎡ 2 ( 5 − 0.8 )( 0.1) − ( 0.1) ⎤ 3 ⎣ ⎦ 40 × 10 −5 8 × 10 ⎛ W ⎞ or K n = 1.476 × 10−4 A / V 2 = ⎜ ⎟ 2 ⎝L⎠ ⎛W ⎞ So that ⎜ ⎟ = 3.69 ⎝L⎠ b. From Equation (16.10). K n RD [VIt − VTN ] + [VIt − VTN ] − VDD = 0 2 (0.1476)(40) [VIt − 0.8] + [VIt − 0.8] − 5 = 0 2 −1 ± (1) 2 + 4(0.1476)(40)(5) 2(0.1476)(40) or [VIt − 0.8] = 0.839 or [VIt − 0.8] = 726. So that VIt = 1.64 V P = I D (max) ⋅ VDD 5 − (0.1) = 0.1225 mA 40 or P = 0.6125 mW and I D (max) = 16.3 a. From Equation (16.10), the transistor point is found from K n RD (VIt − VTN ) 2 + (VIt − VTN ) − VDD = 0 K n = 50 μ A / V 2 , RD = 20 k Ω, VTN = 0.8 V (0.05)(20)(VIt − VTN ) 2 + (VIt − VTN ) − 5 = 0 −1 ± 1 + 4(0.05)(20)(5) = 1.79 V So VIt = 2.59 V 2(0.05)(20) V0t = 1.79 V Output voltage for vI = 5 V is determined from Equation (16.12): VIt − VTN = 2 v0 = 5 − (0.05)(20) ⎡ 2(5 − 0.8)v0 − v0 ⎤ ⎣ ⎦ 2 v0 − 9.4v0 + 5 = 0 So v0 = b. 9.4 ± (9.4) 2 − 4(1)(5) = 0.566 V 2(1) For RD = 200 kΩ, −1 ± 1 + 4(0.05)(200)(5) = 0.659 V So VIt = 1.46 V 2(0.05)(200) V0t = 0.659 V (VIt − VTN ) = 2 v0 = 5 − (0.05)(200) ⎡ 2 ( 5 − 0.8 ) v0 − v0 ⎤ ⎣ ⎦ 2 or 10v0 − 85v0 + 5 = 0 v0 = 16.4 (a) 85 ± (85 ) − 4 (10 )( 5 ) = 0.0592 V 2 (10 ) 2 P = IV 727. 0.25 = I (33) ⇒ I = 75.76 μA 3.3 − 0.15 R= ⇒ R = 41.6 K 0.07576 2 ⎛ k ′ ⎞⎛ W ⎞ I = ⎜ n ⎟ ⎜ ⎟ (VGS − VTN ) 2 ⎠⎝ L ⎠ ⎝ 2 ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ 75.76 = ⎜ ⎟ ⎜ ⎟ ( 3.3 − 0.8 ) ⇒ ⎜ ⎟ = 0.303 ⎝ 2 ⎠⎝ L ⎠ ⎝L⎠ (b) VDS (sat) = VGS − VTN V − VDS (sat) 2 I D = K n (VGS − VTN ) = DD R 3.3 − (VGS − 0.8 ) ⎛ 0.08 ⎞ 2 ⎜ ⎟ ( 0.303) (VGS − 1.6VGS + 0.64 ) = 41.6 ⎝ 2 ⎠ 2 0.504 (VGS − 1.6VGS + 0.64 ) = 4.1 − VGS 2 0.504VGS + 0.1936VGS − 3.777 = 0 −0.1936 ± 0.03748 + 4(0.504)(3.777) 2(0.504) = 2.55 V VGS = VGS For 0.8 ≤ VGS ≤ 2.55 V Transistor biased in saturation region 16.5 (a) P = I ⋅ VDD 0.25 = I (3.3) ⇒ I = 75.76 μA For Sat Load 2 ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ I = 75.76 = ⎜ ⎟ ⎜ ⎟ ( 3.3 − 0.15 − 0.8 ) ⇒ ⎜ ⎟ = 0.343 2 ⎠ ⎝ L ⎠L ⎝ ⎝ L ⎠L 2 2 ⎛ 80 ⎞ ⎛ 80 ⎞ ⎛ W ⎞ ⎜ ⎟ ( 0.343)( 3.3 − 0.15 − 0.8 ) = I = 75.76 = ⎜ ⎟ ⎜ ⎟ ⎡ 2 ( 2.5 − 0.8 )( 0.15 ) − ( 0.15 ) ⎤ ⎦ 2⎠ 2 ⎠ ⎝ L ⎠D ⎣ ⎝ ⎝ ⎛W ⎞ ⎜ ⎟ = 3.89 ⎝ L ⎠D Eq 16.21 ⎛ 3.89 ⎞ 3.3 − 0.8 + 0.8 ⎜1 + ⎟ ⎜ 0.343 ⎟ 5.994 ⎝ ⎠= VIt = 4.3677 3.89 1+ 0.343 0.8 ≤ VGS ≤ 1.372 V 16.6 (a) From Equation (16.23) KD ⎡ K 2 2 2 3 − 0.5 )( 0.25 ) − ( 0.25 ) ⎤ = ( 3 − 0.25 − 0.5 ) ⇒ D = 4.26 ⎣ ( ⎦ KL KL 728. KD ⎡ K 2 2 2 2.5 − 0.5 )( 0.25 ) − ( 0.25 ) ⎤ = ( 3 − 0.25 − 0.5 ) ⇒ D = 5.4 ⎣ ( ⎦ KL KL (b) iD = K L (VGSL − VTNL ) = K L (VDD − vO − VTNL ) 2 2 (c) ⎛ 0.080 ⎞ 2 =⎜ ⎟ (1)(3 − 0.25 − 0.5) ⇒ iD = 0.203 mA ⎝ 2 ⎠ P = iD ⋅ VDD = (0.203)(3) ⇒ P = 0.608 mW for both parts (a) and (b). 16.7 P = 0.4 mW = iD ⋅ VDD = iD (3) ⇒ iD = 0.1333 mA iD = K L (VDD − vO − VTNL ) 2 2 ⎛ 0.080 ⎞ ⎛ W ⎞ ⎛W ⎞ 0.1333 = ⎜ ⎟ ⎜ ⎟ ( 3 − 0.1 − 0.5 ) = ( 0.2304 ) ⎜ ⎟ 2 ⎠ ⎝ L ⎠L ⎝ ⎝ L ⎠L ⎛W ⎞ So ⎜ ⎟ = 0.579 ⎝ L ⎠L KD ⎡ 2 2 2 2.5 − 0.5 )( 0.1) − ( 0.1) ⎤ = ( 3 − 0.1 − 0.5 ) ⎣ ( ⎦ KL ⇒ KD ⎛W ⎞ = 14.8 so that ⎜ ⎟ = 8.55 KL ⎝ L ⎠D VIt = ( 3 − 0.5 + 0.5 1 + 14.8 ) 1 + 14.8 or VIt = 1.02 V , VOt = 0.52 V 16.8 We have KD 2 2 ⎡ 2 ( vI − VTND ) vO − vO ⎤ = (VDD − vO − VTNL ) ⎣ ⎦ KL (W / L )D ⎡ 2 2 2 V − V − VTN )( 0.08VDD ) − ( 0.08VDD ) ⎤ = (VDD − 0.08VDD − VTN ) ⎣ ( DD TN ⎦ (W / L )L (W / L )D ⎡ 2 2 2 ⎦ ⎣ 2 (VDD − 2 ( 0.2 )VDD ) ( 0.08VDD ) − 0.0064VDD ⎤ = ⎡( 0.92 − 0.2 )VDD ⎤ = 0.5184VDD ⎦ ⎣ (W / L )L (W / L )D (W / L )D = 5.4 [0.096] = 0.5184 ⇒ (W / L )L (W / L )L 16.9 VOH = VB − VTN = Logic 1 So (a) VB = 4 V ⇒ VOH (b) VB = 5 V ⇒ VOH (c) VB = 6 V ⇒ VOH (d) VB = 7 V ⇒ VOH For vI = VOH = 3V = 4V = 5V = 5 V ,since VDS = 0 729. 2 K D ⎡ 2 ( vI − VT ) vO − vO ⎤ = K L [VB − vO − VT ] ⎣ ⎦ 2 Then (a) 2 (1) ⎡ 2 ( 3 − 1)VOL − VOL ⎤ = ( 0.4 ) [ 4 − VOL − 1] ⎣ ⎦ (b) ⎡ (1) ⎣ 2 ( 4 − 1)VOL − V (c) 2 (1) ⎡ 2 ( 5 − 1)VOL − VOL ⎤ = ( 0.4 ) [6 − VOL − 1] ⎣ ⎦ 2 2 OL ⇒ VOL = 0.657 V ⎤ ⎦ = ( 0.4 ) [5 − VOL − 1] ⇒ VOL = 0.791 V 2 2 ⇒ VOL = 0.935 V (d) Load in non-sat region iDD = iOL 2 2 (1) ⎡ 2 ( 5 − 1) VOL − VOL ⎤ = ( 0.4 ) ⎡ 2 ( 7 − VOL − 1)( 5 − VOL ) − ( 5 − VOL ) ⎤ ⎣ ⎦ ⎣ ⎦ 2 2 8VOL − VOL = ( 0.4 ) ⎡ 2 ( 6 − VOL )( 5 − VOL ) − ( 25 − 10VOL + VOL ) ⎤ ⎣ ⎦ 2 2 = ( 0.4 ) ⎡ 2 ( 30 − 11VOL + VOL ) − 25 + 10VOL − VOL ⎤ ⎣ ⎦ 2 2 = ( 0.4 ) ⎡60 − 22VOL + 2VOL − 25 + 10VOL − VOL ⎤ ⎣ ⎦ 2 2 8VOL − VOL = 14 − 4.8VOL + 0.4VOL 2 1.4VOL − 12.8VOL + 14 = 0 VOL = 12.8 ± 163.84 − 4 (1.4 )(14 ) 2 (1.4 ) VOL = 1.27V For load VDS ( sat ) = 7 − 1.27 − 1 = 4.73V VDS = 5 − 1.27 = 3.73 non-sat 16.10 a. For load VOt = VDD + VTNL = 5 − 2 = 3 V KD ⋅ (VIt − VTND ) = −VTNL KL 500 (VIt − 0.8 ) = − ( −2 ) 100 ⇒ VIt = 1.69 V ⎫ ⎬ Load VOt = 3 V ⎭ Driver: VOt = VIt − VTND = 1.69 − 0.8 = 0.89 V VIt = 1.69 V ⎫ ⎬ Driver V0t = 0.89 V ⎭ From Equation (16.29(b)): b. 2 500 2 ⎡ ⎤ ⎣ ⋅ ⎣ 2(5 − 0.8)v0 − v0 ⎦ = ⎡ − ( −2 ) ⎤ ⎦ 100 2 5v0 − 42v0 + 4 = 0 v0 = c. 42 ± ( 42 ) − 4 ( 5)( 4 ) ⇒ v0 = 0.0963 V 2 (5) 2 2 iD = K L ( −VTNL ) = 100 ⎡ − ( −2 ) ⎤ ⇒ iD = 400 μ A ⎣ ⎦ 2 730. 16.11 2 2 ⎛ 500 ⎞ ⎡ ⎜ ⎟ ⎣ 2 ( 3 − 0.5 )( 0.1) − ( 0.1) ⎤ = ( −VTNL ) ⎦ ⎝ 50 ⎠ So ( −VTNL ) 2 = 4.9 ⇒ VTNL = −2.21 V 16.12 (a) P = iD ⋅ VDD 150 = iD ⋅ ( 3) ⇒ iD = 50 μ A iD = K L (−VTNL ) 2 2 ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ 50 = ⎜ ⎟ ⎜ ⎟ ⎡ − ( −1) ⎤ ⇒ ⎜ ⎟ = 1.25 ⎦ 2 ⎠ ⎝ L ⎠L ⎣ ⎝ ⎝ L ⎠L 2 KD ⎡ 2 2 ( 3 − 0.5 )( 0.1) − ( 0.1) ⎤ = ⎡ − ( −1) ⎤ ⎦ ⎦ ⎣ KL ⎣ K D (W / L ) D ⎛W ⎞ = = 2.04 ⇒ ⎜ ⎟ = 2.55 K L (W / L ) L ⎝ L ⎠D For the Load: VOt = VDD + VTNL = 3 − 1 ⇒ VOt = 2 V 2.04 (VIt − 0.5 ) = ⎡ − ( −1) ⎤ ⇒ VIt = 1.20 V ⎣ ⎦ For the Driver: VOt = VIt − VTND = 1.20 − 0.5 ⇒ VOt = 0.70 V VIt = 1.20 V (b) NM L = VIL − VOLU NM H = VOHU − VIH VIL = 0.5 + ⎡ − ( −1) ⎤ ⎣ ⎦ = 0.902 V ( 2.04 )(1 + 2.04 ) 2 ⎡ − ( −1) ⎤ ⎦ = 1.31 V VIH = 0.5 + ⎣ 3 ( 2.04 ) Then VOHU = ( 3 − 1) + ( 2.04 )( 0.902 − 0.5 ) = 2.82 V VOLU = (1.31 − 0.5) = 0.405 V 2 NM L = 0.902 − 0.405 ⇒ NM L = 0.497 V NM H = 2.82 − 1.31 ⇒ NM H = 1.51 V 16.13 a. From Equation (16.29(b)): 2 ⎛W ⎞ ⎡ ⎛W ⎞ 2 ⎜ ⎟ ⎣ 2 ( 2.5 − 0.5 )( 0.05 ) − ( 0.05 ) ⎤ = ⎜ ⎟ [− ( −1)] ⎦ ⎝ L ⎠L L ⎠D ⎝ ⎛W ⎞ ⎜ ⎟ =1 ⎝ L ⎠L 731. ⎛W ⎞ ⎜ ⎟ = 5.06 ⎝ L ⎠D Then 2 ⎛ 80 ⎞ iD = ⎜ ⎟ (1) ⎡ − ( −1) ⎤ ⎣ ⎦ ⎝ 2⎠ or iD = 40 μ A b. P = iD ⋅ VDD = ( 40 )( 2.5 ) ⇒ P = 100 μ W 16.14 a. vI = 0.5 V ⇒ iD = 0 ⇒ P = 0 vI = 5 V, From Equation (16.12), i. ii. 2 v0 = 5 − ( 0.1)( 20 ) ⎡ 2 ( 5 − 1.5 ) v0 − v0 ⎤ ⎣ ⎦ 2 2v0 − 15v0 + 5 = 0 v0 = 15 ± (15 ) − 4 ( 2 )( 5 ) ⇒ v0 = 0.35 V 2 ( 2) 2 5 − 0.35 = 0.2325 mA 20 P = iD ⋅ VDD = ( 0.2325 )( 5 ) ⇒ P = 1.16 mW iD = b. ii. vI = 0.25 V ⇒ iD = 0 ⇒ P = 0 i. vI = 4.3 V, From Equation (16.23), 2 100 ⎡ 2 ( 4.3 − 0.7 ) v0 − v0 ⎤ = 10 [5 − v0 − 0.7 ] ⎣ ⎦ 2 2 2 10 ⎡7.2v0 − v0 ⎤ = 18.49 − 8.6v0 + v0 ⎣ ⎦ Then 2 11v0 − 80.6v0 + 18.49 = 0 v0 = 80.6 ± (80.6 ) − 4 (11)(18.49 ) ⇒ v0 = 0.237 V 2 (11) 2 Then iD = 10 [5 − 0.237 − 0.7 ] = 165 μ A 2 P = iD ⋅ VDD = (165 )( 5 ) ⇒ P = 825 μ W c. i. ii. vI = 0.03 V ⇒ iD = 0 ⇒ P = 0 vI = 5 V 2 iD = K L ( −VTNL ) = (10 ) ⎡ − ( −2 ) ⎤ = 40 μ A ⎣ ⎦ 2 P = iD ⋅ VDD = ( 40 )( 5 ) ⇒ P = 200 μ W 16.15 vo1 = 3.8V Load & Driver in Sat, region, ML1, MD1 732. iDL = iDD 2 2 ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ ( vGSL − VTNL ) = ⎜ ⎟ ( vGSD − VTND ) ⎝ L ⎠L ⎝ L ⎠D (1) ( 5 − vo1 − 0.8 ) = (10 )( vI − 0.8 ) 2 0.16 = 10 ( vI − 0.8 ) 2 2 vI = 0.9265V Now MD2: Non Sat and ML2: Sat iDL = iDD 2 ⎛W ⎞ ⎛W ⎞ 2 ⎡ ⎤ ⎜ ⎟ ( vGSL − VTNL ) = ⎜ ⎟ ⎣ 2 ( vo1 − VTND ) vo 2 − vo 2 ⎦ ⎝ L ⎠L ⎝ L ⎠D (1)( 5 − vo 2 − 0.8 ) ( 4.2 − vo 2 ) 2 2 2 = (10 ) ⎡ 2 ( 3.8 − 0.8 ) vo 2 − vo 2 ⎤ ⎣ ⎦ 2 = 10 ⎡ 6vo 2 − vo 2 ⎤ ⎣ ⎦ 2 2 17.64 − 8.4vo 2 + vo 2 = 60vo 2 − 10vo 2 2 11vo 2 − 68.4vo 2 + 17.64 = 0 vo 2 = 68.4 ± 4678.56 − 4 (11)(17.64 ) 2 (11) vo 2 = 0.270 V 16.16 a. VIH From Equation (16.41), 2 ⎡ − ( −2 ) ⎤ ⎦ ⇒ V = 1.95 V = v = 0.8 + ⎣ IH 01 3( 4) M D 2 in non-saturation and M L 2 in saturation. 2 K D ⎡ 2 ( v01 − VTND ) v02 − v02 ⎤ = K L ( −VTNL ) ⎣ ⎦ 2 ⎡ ⎤ 4 ⎣ 2 (1.95 − 0.8 ) v02 − v02 ⎦ = (1) ⎡ − ( −2 ) ⎤ ⎣ ⎦ 2 2 2 4v02 − 9.2v02 + 4 = 0 v02 = 9.2 ± ( 9.2 ) − 4 ( 4 )( 4 ) ⇒ v02 = 0.582 V 2 ( 4) 2 Both M D1 and M L1 in saturation region. From Equation (16.28(b)). 4 ⋅ ( vI − 0.8 ) = − ( −2 ) or vI = 1.8 V b. VIL = 0.8 + ( +2 ) = 1.25 V = v01 4 (1 + 4 ) M D 2 in saturation, M L 2 in non-saturation 733. 2 2 K D [ vO1 − VTND ] = K L ⎡ 2 ( −VTNL )( 5 − vO 2 ) − ( 5 − vO 2 ) ⎤ ⎣ ⎦ 4 (1.25 − 0.8 ) = 2 ( 2 )( 5 − v02 ) − ( 5 − v02 ) 2 ( 5 − v02 ) 5 − v02 = 2 2 − 4 ( 5 − v02 ) + 0.81 = 0 ( 4) 4± 2 − 4 (1)( 0.81) 2 (1) = 0.214 V so v02 = 4.786 V To find vI : 4 ( v01 − 0.8 ) = (1) ( − ( −2 ) ) 2 2 v01 − 0.8 = 1 v01 − 1.8 = V VIH = 1.95 V, VIL = 1.25 V c. 16.17 a. i. Neglecting the body effect, v0 = VDD − VTN Assume VDD = 5 V, then v0 = 4.2 V ii. Taking the body effect into account: From Problem 16.1. VTN = VTN 0 + 0.671 ⎡ 0.686 + VSB − 0.686 ⎤ ⎣ ⎦ and VSB = v0 Then v0 = 5 − ⎡ 0.8 + 0.671 0.686 + v0 − 0.686 ⎤ ⎣ ⎦ ( ) v0 = 4.756 − 0.671 0.686 + v0 0.671 0.686 + v0 = 4.756 − v0 2 0.450 ( 0.686 + v0 ) = 22.62 − 9.51v0 + v0 2 v0 − 9.96v0 + 22.3 = 0 v0 = b. 9.96 ± ( 9.96 ) 2 − 4 ( 22.3) ⇒ v0 = 3.40 V 2 PSpice results similar to Figure 16.13(a). 16.18 Results similar to Figure 16.13(b). 16.19 a. M X on, M Y cutoff. From Equation (16.29(b)): 2 KD ⎡ 2 2 5 − 0.8 )( 0.2 ) − ( 0.2 ) ⎤ = ⎡ − ( −2 ) ⎤ ⎦ ⎣ ( ⎦ ⎣ KL or b. KD = 2.44 KL For v X = vY = .5 V 734. 2 2 ( 2.44 ) ⎡ 2 ( 5 − 0.8 ) v0 − v0 ⎤ = ⎡ − ( −2 ) ⎤ ⎦ ⎣ ⎦ ⎣ 2 2 4.88v0 − 41.0v0 + 4 = 0 v0 = 41 ± ( 41) − 4 ( 4.88 )( 4 ) 2 ( 4.88 ) 2 or v0 = 0.0987 V c. 2 ⎛ 80 ⎞ iD = ⎜ ⎟ (1) ⎡ − ( −2 ) ⎤ = 160 μ A ⎦ 2⎠ ⎣ ⎝ P = (160 )( 5 ) ⇒ P = 800 μ W for both parts (a) and (b). 16.20 (a) Maximum value of vO in low state- when only one input is high, then, 2 KD ⎡ 2 2 ( 3 − 0.5 )( 0.1) − ( 0.1) ⎤ = ⎡ − ( −1) ⎤ ⎦ ⎦ ⎣ KL ⎣ KD = 2.04 KL (b) P = iD ⋅ VDD 0.1 = iD (3) ⇒ iD = 33.3 μ A 2 ⎛ k′ ⎞⎛ W ⎞ iD = ⎜ n ⎟ ⎜ ⎟ ( −VTNL ) 2 ⎠ ⎝ L ⎠L ⎝ 2 ⎛ 80 ⎞⎛ W ⎞ ⎛W ⎞ 33.3 = ⎜ ⎟⎜ ⎟ ⎡ − ( −1) ⎤ ⇒ ⎜ ⎟ = 0.8325 ⎣ ⎦ ⎝ 2 ⎠⎝ L ⎠ L ⎝ L ⎠L ⎛W ⎞ Then ⎜ ⎟ = 1.70 ⎝ L ⎠D (c) 2 2 3 ( 2.04 ) ⎡ 2 ( 3 − 0.5 ) vO − vO ⎤ = ⎡ − ( −1) ⎤ ⇒ vO = 0.0329 V ⎦ ⎣ ⎦ ⎣ 16.21 (a) One driver in non-sat, 2 2 I D = K L ( −VTNL ) = K D ⎡ 2 (VGS − VTND ) VDSD − VDSD ⎤ ⎣ ⎦ 2 K 2 ⎡ − ( −1) ⎤ = D ⎡ 2 ( 3.3 − 0.5 )( 0.1) − ( 0.1) ⎤ ⎣ ⎦ ⎣ ⎦ KL KD = 1.82 KL (b) 735. P = ± VDD 0.1 = ± ( 3.3) ⇒ I = 30.3 μ A 2 ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ 30.3 = I = ⎜ ⎟ ⎜ ⎟ ⎡ − ( −1) ⎤ ⇒ ⎜ ⎟ = 0.7575 ⎣ ⎦ ⎝ 2 ⎠ ⎝ L ⎠L ⎝ L ⎠L ⎛W ⎞ ⎜ ⎟ = 1.38 ⎝ L ⎠D (c) 0.1 = 0.05 V 2 0.1 = 0.0333 V Three inputs High, vo ≈ 3 0.1 = 0.025 V Four inputs High, vo ≈ 4 Two inputs High, vo ≈ (i) (ii) (iii) 16.22 a. P = iD ⋅ VDD 250 = iD ( 5 ) ⇒ iD = 50 μΑ ⎛ k' ⎞⎛W ⎞ 2 iD = ⎜ n ⎟ ⎜ ⎟ [ −VTNL1 ] 2 ⎠ ⎝ L ⎠ ML1 ⎝ 2 ⎛ 60 ⎞ ⎛ W ⎞ 50 = ⎜ ⎟ ⎜ ⎟ ⎡ − ( −2 ) ⎤ ⎣ ⎦ ⎝ 2 ⎠ ⎝ L ⎠ ML1 ⎛W ⎞ So that ⎜ ⎟ = 0.417 ⎝ L ⎠ ML1 KD 2 2 ⎡ 2 ( vI − VTND ) vO − vO ⎤ = [ −VTNL ] ⎣ ⎦ KL 2 KD ⎡ 2 2 5 − 0.8 )( 0.15 ) − ( 0.15 ) ⎤ = ⎡ − ( −2 ) ⎤ ⎦ ⎣ ( ⎦ ⎣ KL or KD ⎛W ⎞ = 3.23 ⇒ ⎜ ⎟ = 1.35 KL ⎝ L ⎠ MD1 b. For v X = vY = 0 ⇒ v01 = 5 and v03 = 4.2 Then 2 2 2 K D 2 ⎡ 2 ( vO1 − VTND ) vO 2 − vO 2 ⎤ + K D 3 ⎡ 2 ( vO 3 − VTND ) vO 2 − vO 2 ⎤ = K L 2 [ −VTNL 2 ] ⎣ ⎦ ⎣ ⎦ K D 2 ∝ 8, K D 3 ∝ 8, K L 2 ∝ 1 2 2 8 ⎡ 2 ( 5 − 0.8 ) v02 − v02 ⎤ + 8 ⎡ 2 ( 4.2 − 0.8 ) v02 − v02 ⎤ = (1) ⎡ − ( −2 ) ⎤ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 2 2 67.2v02 − 8v02 + 54.4v02 − 8v02 = 4 Then 2 16v0 − 121.6v0 + 4 = 0 v02 = 121.6 ± (121.6 ) − 4 (16 )( 4 ) 2 (16 ) So v02 = 0.0330 V 16.23 2 2 736. a. We can write 2 2 K x ⎡ 2 ( v X − VTN ) vDSX − vDSX ⎤ = K y ⎡ 2 ( vY − vDSX − VTN ) vDSY − vDSY ⎤ = K L [VDD − vO − VTN ] ⎣ ⎦ ⎣ ⎦ 2 where v0 = vDSX + vDSY We have v X = vY = 9.2 V , VDD = 10 V , VTN = 0.8 V 2 2 As a good first approximation, neglect the vDSX and vDSY terms. Let v0 ≈ 2vDSX . Then from the first and third terms in the above equation. 9 ⎡ 2 ( 9.2 − 0.8 ) vDSX ⎤ ≅ (1)(10 − 2vDSX − 0.8 ) ⎣ ⎦ (151.2 ) vDSX 2 ≅ 84.64 − 36.8vDSX So that vDSX = 0.450 V From the first and second terms of the above equation. 9 ⎡ 2 ( 9.2 − 0.8 ) vDSX ⎤ ≅ 9 ⎡ 2 ( 9.2 − vDSX − 0.8 ) vDSY ⎤ ⎣ ⎦ ⎣ ⎦ or (16.8)( 0.45) = 2 ( 9.2 − 0.45 − 0.8) vDSY which yields vDSY = 0.475 V Then v0 = vDSX + vDSY = 0.450 + 0.475 or v0 = 0.925 V We have vGSX = 9.2 V and vGSY = 9.2 − vDSX = 9.2 − 0.45 or vGSY = 8.75 V b. Since v0 is close to ground potential, the body effect will have minimal effect on the results. From a PSpice analysis: For part (a): vDSX = 0.462 V, vDSY = 0.491 V, v0 = 0.9536 V, vGSX = 9.2 V, and vGSY = 8.738 V For part (b): vDSX = 0.441 V, vDSY = 0.475 V, v0 = 0.9154 V, vGSX = 9.2 V, and vGSY = 8.759 V 16.24 a. We can write 2 2 2 K x ⎡ 2 ( v X − VTNX ) vDSX − vDSX ⎤ = K y ⎡ 2 ( vY − vDSX − VTNY ) vDSY − vDSY ⎤ = K L [ −VTNL ] ⎣ ⎦ ⎣ ⎦ 2 From the first and third terms, (neglect vDSX ), 4 ⎡ 2 ( 5 − 0.8 ) vDSX ⎤ = (1) ⎡ − ( −1.5 ) ⎤ ⎣ ⎦ ⎣ ⎦ 2 or vDSX = 0.067 V 2 From the second and third terms, (neglect vDSY ), 4 ⎡ 2 ( 5 − 0.067 − 0.8 ) vDSY ⎤ = (1) ⎡ − ( −1.5 ) ⎤ ⎣ ⎦ ⎣ ⎦ 2 or vDSY = 0.068 V Now vGSX = 5, vGSY = 5 − 0.067 ⇒ vGSY = 4.933 V and v0 = vDSX + vDSY ⇒ v0 = 0.135 V Since v0 is close to ground potential, the body-effect has little effect on the results. 16.25 737. (a) 0.2 = 0.05 V 4 2 − VDSD ⎤ ⎦ We have VDS of each driver ≈ K L [ −VTNL ] = K D ⎡ 2 (VGSD − VTN ) VDSD ⎣ 2 2 K 2 ⎡ − ( −1) ⎤ = D ⎡ 2 ( 3.3 − 0.4 )( 0.05 ) − ( 0.05 ) ⎤ ⎣ ⎦ ⎦ KL ⎣ KD = 3.478 KL (b) P = I VDD 0.15 = I ( 3.3) ⇒ I = 45.45 μ A 2 ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ 45.45 = ⎜ ⎟ ⎜ ⎟ ⎡ − ( −1) ⎤ ⇒ ⎜ ⎟ = 1.14 ⎣ ⎦ ⎝ 2 ⎠ ⎝ L ⎠L ⎝ L ⎠L ⎛W ⎞ ⎜ ⎟ = 3.95 ⎝ L ⎠D 16.26 Complement of (B AND C) OR A ⇒ ( B ⋅ C ) + A 16.27 Considering a truth table, we find A B Y 0 0 0 0 1 1 1 0 1 1 1 0 which shows that the circuit performs the exclusive-OR function. 16.28 ( A + B )(C + D) 16.29 (a) Carry-out = A • ( B + C ) + B • C 738. (b) For vO1 = Low = 0.2 V 2 KD ⎡ 2 2 ( 5 − 0.8 )( 0.2 ) − ( 0.2 ) ⎤ = ⎡ − ( −1.5 ) ⎤ ⇒ ⎦ ⎦ ⎣ KL ⎣ ⎛W ⎞ ⎛W ⎞ For ⎜ ⎟ = 1, then ⎜ ⎟ = 1.37 ⎝ L ⎠L ⎝ L ⎠D ⎛W ⎞ So, for M 6 : ⎜ ⎟ = 1.37 ⎝ L ⎠6 To achieve the required composite conduction parameter, ⎛W ⎞ For M 1 − M 5 : ⎜ ⎟ = 2.74 ⎝ L ⎠1− 5 16.30 AE: VDS ≈ 0.075 2 2 ⎛W ⎞ (1) ⎡ − ( −1) ⎤ ≅ ⎜ ⎟ ⎡ 2 ( 3.3 − 0.4 )( 0.075 ) − ( 0.075 ) ⎤ ⎣ ⎦ ⎦ L ⎠D ⎣ ⎝ ⎛W ⎞ ⎛W ⎞ ⎛W ⎞ ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = ⎜ ⎟ = 2.33 ⇒ ⎜ ⎟ = ⎜ ⎟ = ⎜ ⎟ = 4.66 ⎝ L ⎠ A ⎝ L ⎠E ⎝ L ⎠ B ⎝ L ⎠C ⎝ L ⎠ D 16.31 Not given 739. 16.32 a. From Equation (16.43), 5 − 0.8 + 0.8 vI = VIt = = VIt = 2.5 V 1+1 p − channel, V0 Pt = 2.5 − ( −0.8 ) ⇒ V0 Pt = 3.3 V n − channel, V0 Nt = 2.5 − 0.8 ⇒ V0 Nt = 1.7 V c For vI = 2 V, NMOS in saturation and PMOS in nonsaturation. From Equation (16.49), 2 = ⎡ 2 ( 5 − 2 − 0.8 )( 5 − v0 ) − ( 5 − v0 ) ⎤ ⎣ ⎦ 1.44 = 4.4(5 − v0 ) − (5 − v0 ) 2 ( 2 − 0.8) 2 So ( 5 − v0 ) − 4.4 ( 5 − v0 ) + 1.44 = 0 2 ( 5 − v0 ) = 4.4 ± ( 4.4 ) 2 − 4 (1)(1.44 ) 2 or 5 − v0 = 0.356 ⇒ v0 = 4.64 V By symmetry, for vI = 3 V, v0 = 0.356 V 16.33 (a) ⎛ 80 ⎞ K n = ⎜ ⎟ ( 2 ) = 80 μ A / V 2 ⎝ 2⎠ ⎛ 40 ⎞ K p = ⎜ ⎟ ( 4 ) = 80 μ A / V 2 ⎝ 2 ⎠ VDD + VTP + (i) VIt = Kn ⋅ VTN Kp Kn Kp 1+ = 3.3 − 0.4 + (1)(0.4) 1+1 VIt = 1.65 V PMOS: VOt = VIt − VTP = 1.65 − ( −0.4 ) ⇒ VOt = 2.05 V NMOS: VOt = VIt − VTN = 1.65 − ( 0.4 ) ⇒ VOt = 1.25 V (iii) For vO = 0.4 V : NMOS: Non-sat: PMOS:Sat 2 K n ⎡ 2 (VGSN − VTN )VDS − VDS ⎤ = K p [VSGP + VTP ] ⎣ ⎦ 2 2 ( vI − 0.4 )( 0.4 ) − ( 0.4 ) = ( 3.3 − vI − 0.4 ) ⇒ vI = 1.89 V 2 For vO = 2.9 V , By symmetry vI = 1.65 − (1.89 − 1.65 ) ⇒ vI = 1.41 V (b) ⎛ 80 ⎞ K n = ⎜ ⎟ ( 2 ) = 80 μ A/V 2 ⎝ 2⎠ ⎛ 40 ⎞ K p = ⎜ ⎟ ( 2 ) = 40 μ A/V 2 ⎝ 2 ⎠ 2 740. 80 ⋅ ( 0.4 ) 40 ⇒ VIt = 1.44 V 80 1+ 40 3.3 − 0.4 + VIt = (i) PMOS: VOt = 1.44 − ( −0.4 ) ⇒ VOt = 1.84 V NMOS: VOt = 1.44 − 0.4 ⇒ VOt = 1.04 V For vO = 0.4 V (iii) ⎤ = 40 3.3 − vI − 0.4]2 ⇒ vI = 1.62 V ⎦ ( )[ For vO = 2.9 V : NMOS: Sat, PMOS: Non-sat ⎡ (80 ) ⎣ 2 ( vI − 0.4 )( 0.4 ) − ( 0.4 ) (80 ) [vI − 0.4] 2 16.34 (a) (i) 2 = ( 40 ) ⎡ 2 ( 3.3 − vI − 0.4 )( 0.4 ) − ( 0.4 ) ⎤ ⇒ vI = 1.18 V ⎣ ⎦ From Eq. (16.43), switching voltage VDD + VTP + vI t = (ii) 2 Kn ⋅ VTN Kp Kn 1+ Kp = 2 ( 4) ( 0.4 ) 3.2266 12 = ⇒ vIt = 1.776 V 1.8165 2 ( 4) 1+ 12 3.3 + ( −0.4 ) + v0 = 3.1 V , PMOS, non-sat; NMOS, sat ′ ′ ⎛ kp ⎞⎛ W ⎞ 2 ⎛ kn ⎞ ⎛ W ⎞ 2 ⎜ ⎟ ⎜ ⎟ ⎡ 2 (VSG + VTP ) VSD − VSD ⎤ = ⎜ ⎟ ⎜ ⎟ (VGS − VTN ) ⎦ 2 ⎠⎝ L ⎠p ⎣ 2 ⎠ ⎝ L ⎠n ⎝ ⎝ 2 2 ⎛ 40 ⎞ ⎛ 80 ⎞ ⎜ ⎟ (12 ) ⎡ 2 ( 3.3 − vI − 0.4 )( 3.3 − 3.1) − ( 3.3 − 3.1) ⎤ = ⎜ ⎟ ( 4 ) [ vI − 0.4] ⎣ ⎦ ⎝ 2⎠ ⎝ 2 ⎠ 12 [1.16 − 0.4vI − 0.04] = 8 ⎡ vI2 − 0.8vI + 0.16 ⎤ ⎣ ⎦ 8vI2 − 1.6vI − 12.16 = 0 vI = (iii) 1.6 ± 2.56 + 4 ( 8 ) (12.16 ) 2 (8) ⇒ vI = 1.337 V v0 = 0.2 V PMOS: sat, NMOS, non-sat. 2 2 ⎛ 40 ⎞ ⎛ 80 ⎞ ⎜ ⎟ (12 ) [3.3 − vI − 0.4] = ⎜ ⎟ ( 4 ) ⎡ 2 ( vI − 0.4 )( 0.2 ) − ( 0.2 ) ⎤ ⎣ ⎦ ⎝ 2 ⎠ ⎝ 2⎠ 12 ⎡8.41 − 5.8vI + vI2 ⎤ = 8 [ 0.4vI − 0.2] ⎣ ⎦ 12vI2 − 72.8vI + 102.52 = 0 vI = 72.8 ± 5299.84 − 4 (12 )(102.52 ) vI = 2.222 V (b) 2 (12 ) 741. vIt = (i) 2 ( 6) ( 0.4 ) 3.5928 4 = 2.732 2 (6) 1+ 4 3.3 + ( −0.4 ) + vIt = 1.315 V (ii) From (a), (ii) 4 [1.16 − 0.4vI − 0.04] = 12 ⎡vI2 − 0.8vI + 0.16 ⎤ ⎣ ⎦ 12vI2 − 8vI − 2.56 = 0 vI = 8 ± 64 + 4 (12 )( 2.56 ) 2 (12 ) ⇒ vI = 0.903 V (iii) From (a), (iii) ⎡8.41 − 5.8vI + vI2 ⎤ = 12 [ 0.4vI − 0.2] 4⎣ ⎦ 4vI2 − 28vI + 36.04 = 0 vI = 28 ± 784 − 4 ( 4 )( 36.04 ) 2 ( 4) 16.35 a. ⇒ vI = 1.70 V For vO1 = 0.6 < VTN ⇒ vO 2 = 5 V N1 in nonsaturation and P in saturation. From Equation (16.45), 1 ⎡ 2 ( vI − 0.8 )( 0.6 ) − ( 0.6 )2 ⎤ = [5 − vI − 0.8]2 ⎣ ⎦ 1.2vI − 1.32 = 17.64 − 8.4vI + vI2 or vI2 − 9.6vI + 18.96 = 0 vI = 9.6 ± ( 9.6 ) 2 − 4 (1)(18.96 ) 2 or vI = 2.78 V V0 Nt ≤ v02 ≤ V0 Pt b. From symmetry, VIt = 2.5 V V0 Pt = 2.5 + 0.8 = 3.3 V and V0 Nt = 2.5 − 0.8 = 1.7 V So 1.7 ≤ v02 ≤ 3.3 V 16.36 a. V0 Nt ≤ v01 ≤ V0 Pt By symmetry, VIt = 2.5 V V0 Pt = 2.5 + 0.8 = 3.3 V and V0 Nt = 2.5 − 0.8 = 1.7 V So 1.7 ≤ v01 ≤ 3.3 V b. For vO 2 = 0.6 < VTN ⇒ vO 3 = 5 V N 2 in nonsaturation and P2 in saturation. From Equation (16.57), 742. ⎡ 2 ( vI 2 − 0.8 )( 0.6 ) − ( 0.6 )2 ⎤ = [5 − vI 2 − 0.8]2 ⎣ ⎦ 1.2vI 2 − 1.32 = 17.64 − 8.4vI 2 + vI22 or vI22 − 9.6vI 2 + 18.96 = 0 So vI 2 = v01 = 2.78 V For v01 = 2.78, both N1 and P in saturation. Then 1 vI = 2.5 V 16.37 a. iPeak = K n ( vI − VTN ) iPeak = 0.1 ⋅ ( 2.5 − 0.8 ) = 0.538 ( mA ) 1/ 2 b. iPeak = 0.1 ⋅ (1.65 − 0.8 ) = 0.269 ( mA ) 1/ 2 16.38 (a) ⎛ 50 ⎞ K n = ⎜ ⎟ ( 2 ) = 50 μ A / V 2 ⎝ 2⎠ ⎛ 25 ⎞ K p = ⎜ ⎟ ( 4 ) = 50 μ A / V 2 ⎝ 2 ⎠ I D , peak = K n ( v1 − VTN ) = 50 ( 2.5 − 0.8 ) 2 or I D , peak = 144.5 μ A (b) K n = 50 μ A / V 2 , K p = 25 μ A/V 2 From Equation (16.55), 50 5 − 0.8 + (0.8) 25 vIt = = 2.21 V 50 1+ 25 Then I D , peak = K n (VIt − VTN ) 2 = 50 ( 2.21 − 0.8 ) or I D , peak = 99.4 μ A 16.39 2 2 743. (a) Switching Voltage, Eq. (16.43) 2 ( 4) ( 0.4 ) 8 vIt = = 1.65 V = vIt 2 ( 4) 1+ 8 80 ⎞ 2 ⎛ iD, peak = ⎜ ⎟ ( 4 )(1.65 − 0.4 ) ⇒ iD , peak = 250 μA ⎝ 2⎠ (b) 3.3 − 0.4 + 2 ( 4) ( 0.4 ) 4 = 1.436 V = vIt vIt = 2 ( 4) 1+ 4 2 ⎛ 80 ⎞ iD , peak = ⎜ ⎟ ( 4 )(1.436 − 0.4 ) ⇒ iD , peak = 172 μA ⎝ 2⎠ (c) 3.3 − 0.4 + 2 ( 4) ( 0.4 ) 12 vIt = ⇒ vIt = 1.776 V 2 ( 4) 1+ 12 2 ⎛ 80 ⎞ iD , peak = ⎜ ⎟ ( 4 )(1.776 − 0.4 ) ⇒ iD , peak = 303 μA ⎝ 2⎠ 3.3 − 0.4 + 16.40 2 a. P = fCLVDD For VDD = 5 V, P = (10 × 106 )(0.2 ×10−12 )(5) 2 or P = 50 μ W For VDD = 15 V, P = (10 × 106 )(0.2 × 10−12 )(15) 2 or P = 450 μ W b. For VDD = 5 V, P = (10 × 106 )(0.2 ×10−12 )(5) 2 or P = 50 μ W 16.41 (a) 2 P = fCLVDD = (150 × 106 )( 0.4 ×10−12 ) ( 5 ) = 1.5 × 10 −3 W / inverter 2 Total power: PT = ( 2 × 106 )(1.5 × 10−3 ) ⇒ PT = 3000 W !!!! (b) For f = 300 MHz 2 1.5 ×10 = ( 300 ×106 )( 0.4 × 10−12 ) VDD ⇒ VDD = 3.54 V −3 16.42 3 = 3 × 10−7 W 7 10 (a) P= (b) 2 P = fCLVDD ⇒ CL = P 2 fVDD 744. (i) CL = (ii) CL = (iii) CL = 3 × 10−7 ( 5 ×10 ) ( 5) 6 ⇒ CL = 0.0024 pF 2 3 × 10−7 2 ⇒ CL = 0.00551pF 2 ⇒ CL = 0.0267 pF ( 5 ×10 ) ( 3.3) 6 3 × 10−7 ( 5 ×10 ) (1.5) 6 16.43 10 = 2 × 10−6 W 5 × 106 P CL = 2 fVDD P= (a) (b) 2 × 10−6 ⇒ CL = 0.01 pF (i) CL = (ii) CL = (iii) CL = 16.44 (a) For vI ≅ VDD , NMOS in nonsaturation (8 ×10 ) ( 5) 6 2 2 × 10−6 2 ⇒ CL = 0.023 pF 2 ⇒ CL = 0.111 pF (8 ×10 ) ( 3.3) 6 2 × 10−6 (8 ×10 ) (1.5) 6 2 iD = K n ⎡ 2 ( vI − VTN ) vDS − vDS ⎤ and vDS ≅ 0 ⎣ ⎦ So di 1 = D ≅ K n ⎡ 2 (VDD − VTN ) ⎤ ⎣ ⎦ rds dvDS Or rds = 1 ′ ⎛ kn ⎞ ⎛ W ⎞ ⎜ 2 ⎟ ⎜ L ⎟ ⋅ 2 (VDD − VTN ) ⎝ ⎠ ⎝ ⎠n or 1 rds = W ′⎛ kn ⎜ ⎝L For vI ≅ 0, ⎞ ⎟ ⋅ (VDD − VTN ) ⎠n PMOS in nonsaturation 2 iD = K p ⎡ 2 (VDD − vI + VTP ) vSD − vSD ⎤ ⎣ ⎦ and vSD ≅ 0 for vI ≅ 0. So ⎛ k′ ⎞⎛ W ⎞ di 1 p = D ≅ ⎜ ⎟ ⎜ ⎟ ⋅ 2 (VDD + VTP ) rsd dvSD ⎝ 2 ⎠ ⎝ L ⎠ p or rsd = 1 1 k′ p ⎛W ⎜ ⎜ Lp ⎝ ⎞ ⎟ ⋅ (VDD + VTP ) ⎟ ⎠ 745. ⎛W ⎞ ⎛W ⎞ (b) For ⎜ ⎟ = 2, ⎜ ⎟ = 4 ⎝ L ⎠n ⎝ L ⎠p rds = rsd = 1 ⇒ rds = 2.38 k Ω 50 )( 2 )( 5 − 0.8 ) ( 1 ( 25 )( 4 )( 5 − 0.8 ) ⇒ rsd = 2.38 k Ω ⎛W ⎞ For ⎜ ⎟ = 2,. ⎝ L ⎠p rsd = 1 ⇒ rsd = 4.76 k Ω 25 )( 2 )( 5 − 0.8 ) ( Now, for NMOS: vds = id rds or id = vds 0.5 = ⇒ id = 0.21 mA rds 2.38 For PMOS: For rsd = 2.38 k Ω, id = vsd 0.5 = ⇒ id = 0.21 mA rsd 2.38 For rsd = 4.76 k Ω , id = vsd 0.5 = ⇒ id = 0.105 mA rsd 4.76 16.45 From Equation (16.63) 3 VIL = 1.5 + ⋅ (10 − 1.5 − 1.5 ) ⇒ VIL = 4.125 V 8 and Equation (16.62) 1 V0 HU = ⋅ ⎡ 2 ( 4.125 ) + 10 − 1.5 + 1.5⎤ ⎦ 2 ⎣ or V0 HU = 9.125 V From Equation (16.69) 5 VIH = 1.5 + ⋅ (10 − 1.5 − 1.5 ) ⇒ VIH = 5.875 V 8 and Equation (16.68) 1 V0 LU = ⋅ ⎡ 2 ( 5.875 ) − 10 − 1.5 + 1.5⎤ ⎦ 2 ⎣ or V0 LU = 0.875 V Now NM L = VIL − V0 LU = 4.125 − 0.875 ⇒ NM L = 3.25 V NM H = V0 HU − VTH = 9.125 − 5.875 ⇒ NM H = 3.25 V 16.46 From Equation (16.71) ⎡ VIL = 1.5 + ⎤ 100 ⎥ 50 − 1 ⎥ = 1.5 + 7 ⎡ 2 ( 0.632 ) − 1⎤ 2 ⎣ ⎦ ⎢ 100 + 3 ⎥ ⎢ ⎥ 50 ⎣ ⎦ (10 − 1.5 − 1.5 ) ⎢ ⎢ ⎛ 100 ⎞ − 1⎟ ⎜ ⎝ 50 ⎠ 746. or VIL = 3.348 V From Equation (16.70) 1 ⎧⎛ 100 ⎞ ⎫ ⎛ 100 ⎞ V0 HU = ⋅ ⎨⎜1 + ⎟ ( 3.348 ) + 10 − ⎜ ⎟ (1.5 ) + 1.5⎬ 2 ⎩⎝ 50 ⎠ ⎝ 50 ⎠ ⎭ or V0 HU = 9.272 V From Equation (16.77) VIH ⎡ ⎤ ⎛ 100 ⎞ ⎢ 2⎜ ⎥ ⎟ (10 − 1.5 − 1.5 ) ⎢ ⎝ 50 ⎠ = 1.5 + − 1⎥ = 1.5 + 7 [1.51 − 1] ⎥ ⎛ 100 ⎞ ⎢ ⎛ 100 ⎞ − 1⎟ ⎢ 3 ⎜ ⎜ ⎟ +1 ⎥ ⎝ 50 ⎠ ⎢ ⎝ 50 ⎠ ⎥ ⎣ ⎦ or VIH = 5.07 V From Equation (16.76) 100 100 ( 5.07 ) ⎛1 + ⎞ − 10 − ⎛ ⎞ (1.5 ) + 1.5 ⎜ ⎟ ⎜ ⎟ 50 ⎠ ⎝ ⎝ 50 ⎠ V0 LU = ⎛ 100 ⎞ 2⎜ ⎟ ⎝ 50 ⎠ or V0 LU = 0.9275 V Now NM L = VIL − V0 LU = 3.348 − 0.9275 or NM L = 2.42 V NM H = V0 HU − VIH = 9.272 − 5.07 or NM H = 4.20 V 16.47 (a) Kn = KP 3 (VDD + VTP − VTN ) 8 3 = 0.4 + ( 3.3 − 0.4 − 0.4 ) ⇒ VIL = 1.3375 V 8 1 VOHu = {2 (1.3375 ) + 3.3 − 0.4 + 0.4} 2 VOHu = 2.9875 V VIL = VTN + VIH = 0.4 + 5 ( 3.3 − 0.4 − 0.4 ) ⇒ VIH = 1.9625 V 8 1 {2 (1.9625) − 3.3 − 0.4 + 0.4} 2 = 0.3125 V VOLu = VOLu NM H = VOHu − VIH = 2.9875 − 1.9625 ⇒ NM H = 1.025 V NM L = VIL − VOLu = 1.3375 − 0.3125 ⇒ NM L = 1.025 V (b) 747. ⎡ ⎤ 2 ( 4) ⎢ ⎥ ( 3.3 − 0.4 − 0.4 ) ⎢ 2.5 12 VIL = 0.4 + 2 − 1⎥ = 0.4 + ⎡( −0.147 ) ⎤ ⎦ ⎢ 2 ( 4) ⎥ ⎛ ( 2 )( 4 ) ⎞ ( −0.333) ⎣ +3 ⎥ − 1⎟ ⎢ ⎜ 12 ⎦ ⎝ 12 ⎠ ⎣ VIL = 1.505 V VOHu = ⎫ ⎛ 2 ( 4) ⎞ 1 ⎧⎛ ( 2 )( 4 ) ⎞ ⎪ ⎪ 1 ⎟ (1.505 ) + 3.3 − ⎜ ⎟ ( 0.4 ) + 0.4 ⎬ = {2.5083 + 3.3 − 0.2667 + 0.4} ⎨⎜1 + 2 ⎪⎝ 12 ⎠ ⎪ 2 ⎝ 12 ⎠ ⎩ ⎭ VOHu = 2.9708 V ⎡ ⎤ ⎛ 2 ( 4) ⎞ ⎢ 2⎜ ⎥ ⎟ ( 3.3 − 0.4 − 0.4 ) ⎢ ⎝ 12 ⎠ 2.5 − 1⎥ = 0.4 + VIH = 0.4 + [ −0.2302] ⇒ VIH = 2.1282 V ⎢ ⎥ ⎛ 2 ( 4) ⎞ ( −0.333) ( 2 )( 4 ) + 1 ⎥ − 1⎟ ⎢ 3 ⎜ ⎥ 12 ⎝ 12 ⎠ ⎢ ⎣ ⎦ ⎛ 2 ( 4) ⎞ ⎛ 2 ( 4) ⎞ ( 2.1282 ) ⎜1 + ⎟ − 3.3 − ⎜ ⎟ ( 0.4 ) + 0.4 12 ⎠ 3.547 − 3.3 − 0.2667 + 0.4 ⎝ ⎝ 12 ⎠ VOLu = = ⇒ VOLu = 0.2853 V 1.333 ⎛ 2 ( 4) ⎞ 2⎜ ⎟ ⎝ 12 ⎠ NM H = VOHu − VIH = 2.9708 − 2.1282 ⇒ NM H = 0.8426 V NM L = VIL − VOLu = 1.505 − 0.2853 ⇒ NM L = 1.22 V 16.48 a. v A = vB = 5 V N1 and N 2 on, so vDS1 ≈ vDS 2 ≈ 0 V P and P2 off 1 So we have a P3 − N 3 CMOS inverter. By symmetry, vC = 2.5 V (Transition Point). b. For v A = vB = vC ≡ vI Want K n ,eff = K p ,eff ′ kn ⎛ W ⎞ k ′ ⎛ 3W ⎞ ⋅⎜ ⎟ = P ⋅⎜ ⎟ 2 ⎝ 3L ⎠ n 2 ⎝ L ⎠ P ′ ′ With kn = 2k P , then 2 1 ⎛W ⎞ 1 ⎛W ⎞ ⋅ ⋅⎜ ⎟ = ⋅3⋅⎜ ⎟ 2 3 ⎝ L ⎠n 2 ⎝ L ⎠ P 9 ⎛W ⎞ ⎛W ⎞ Or ⎜ ⎟ = ⋅ ⎜ ⎟ L ⎠n 2 ⎝ L ⎠ P ⎝ c. We have ⎛ k ′ ⎞ ⎛ W ⎞ ⎛ 2k ′ ⎞ ⎛ 9 ⎞ ⎛ W ⎞ p Kn = ⎜ n ⎟ ⎜ ⎟ = ⎜ ⎟⎜ ⎟⎜ ⎟ 2 ⎠ ⎝ L ⎠n ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ L ⎠ p ⎝ ⎛ k′ ⎞⎛ W ⎞ p Kp = ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ L ⎠p Then from Equation (16.55) 748. Kn ⋅ ( 0.8 ) Kp 5 + ( −0.8 ) + VIt = 1+ Kn Kp Now Kn ⎛9⎞ = ( 2) ⎜ ⎟ = 9 Kp ⎝ 2⎠ Then VIt = 5 + ( −0.8 ) + 3 ( 0.8 ) 1+ 3 ⇒ VIt = 1.65 V 16.49 By definition, NMOS is on if gate voltage is 5 V and is off if gate voltage is 0 V. State N1 N2 N3 N4 N5 v0 1 2 3 4 0 0 5 0 off off on on on off on on off on off off on on off on Logic function ( v X OR vY ) ⊗ ( v X AND vZ ) Exclusive OR of ( vX OR vY ) with ( vX AND vZ ) 16.50 ⎛W ⎞ NMOS in Parallel ⇒ ⎜ ⎟ = 2 ⎝ L ⎠n ⎛W ⎞ 4-PMOS in series ⇒ ⎜ ⎟ = 4 ( 4 ) = 16 ⎝ L ⎠p (b) CL doubles ⇒ current must double to maintain switching speed. ⎛W ⎞ ⇒⎜ ⎟ =4 ⎝ L ⎠n ⎛W ⎞ ⎜ ⎟ = 32 ⎝ L ⎠p 16.51 ⎛W ⎞ 4-NMOS in series ⎜ ⎟ = 4 ( 2 ) = 8 ⎝ L ⎠n ⎛W ⎞ 4-PMOS in parallel ⎜ ⎟ = 4 ⎝ L ⎠p (b) 16.52 ⎛W ⎜ ⎝L ⎛W ⎜ ⎝L ⎞ ⎟ = 16 ⎠n ⎞ ⎟ =8 ⎠p on off on on 749. ⎛W ⎞ NMOS in parallel ⇒ ⎜ ⎟ = 2 ⎝ L ⎠n ⎛W ⎞ 3-PMOS in series ⇒ ⎜ ⎟ = 3 ( 4 ) = 12 ⎝ L ⎠P (a) (b) ⎛W ⎜ ⎝L ⎛W ⎜ ⎝L ⎞ ⎟ =4 ⎠n ⎞ ⎟ = 24 ⎠p 16.53 ⎛W ⎞ 3-NMOS in series ⎜ ⎟ = 3 ( 2 ) = 6 ⎝ L ⎠n ⎛W ⎞ 3-PMOS in parallel ⎜ ⎟ = 4 ⎝ L ⎠p (a) (b) 16.54 (a) (b) ⎛W ⎜ ⎝L ⎛W ⎜ ⎝L ⎞ ⎟ = 12 ⎠n ⎞ ⎟ =8 ⎠p Y = A( B + C )( D + E ) ⎛W ⎞ For NMOS in pull down mode, 3 in series ⇒ ⎜ ⎟ = 3 ( 2 ) = 6 ⎝ L ⎠n For PMOS ⎛W ⎞ ⎜ ⎟ =4 ⎝ L ⎠P, A (c) ⎛W ⎞ = 2 ( 4) = 8 ⎜ ⎟ ⎝ L ⎠ P , B ,C , D , E 750. 16.55 (a) (b) Y = A( BD + CE ) (c) NMOS: 3 transistors in series for pull down mode. ⎛W ⎞ For twice the speed: ⎜ ⎟ = 2 ( 3)( 2 ) = 12 ⎝ L ⎠n ⎛W ⎞ PMOS: ⎜ ⎟ = 2 ( 4 ) = 8 ⎝ L ⎠P, A ⎛W ⎞ = 2 ( 2 )( 4 ) = 16 ⎜ ⎟ ⎝ L ⎠ P , B ,C , D , E 16.56 (a) (b) Y = A + BC + DE ⎛W ⎞ ⎛W ⎞ NMOS: ⎜ ⎟ = 2 =4 ⎜ ⎟ ⎝ L ⎠n, A ⎝ L ⎠ n , B ,C , D , E PMOS: 3 transistors in series for the pull-up mode (c) 751. ⎛W ⎞ ⎜ ⎟ = 3 ( 4 ) = 12 ⎝ L ⎠p 16.57 (a) (b) Y = A + ( B + D)(C + E ) (c) ⎛W ⎞ NMOS: ⎜ ⎟ = 2 ( 2 ) = 4 ⎝ L ⎠n, A PMOS: ⎛W ⎞ ⎜ ⎟ = ( 2 ) 3 ( 4 ) = 24 ⎝ L ⎠p 16.58 (a) A classic design is shown: ⎛W ⎞ = ( 2 )( 4 ) = 8 ⎜ ⎟ ⎝ L ⎠ n , B ,C , D , E 752. A, B, C signals supplied through inverters. ⎛W ⎞ ⎛W ⎞ (b) For Inverters, ⎜ ⎟ = 1 and ⎜ ⎟ = 2 ⎝ L ⎠p ⎝ L ⎠n ⎛W ⎞ ⎛W ⎞ For PMOS in Logic function, let ⎜ ⎟ = 1 , then for NMOS in Logic function, ⎜ ⎟ = 2.25 ⎝ L ⎠p ⎝ L ⎠n 16.59 (a) A classic design is shown: 753. (b) ⎛W ⎞ ⎛W ⎞ =2 ⎜ ⎟ = 1, ⎜ ⎟ L ⎠ ND ⎝ ⎝ L ⎠ NA, NB , NC ⎛W ⎞ ⎛W ⎞ = 8, ⎜ ⎟ =4 ⎜ ⎟ ⎝ L ⎠ PA, PB ⎝ L ⎠ PC , PD 16.60 ( A OR B ) AND C 16.61 ⎛W ⎞ 5-NMOS in series ⇒ ⎜ ⎟ = 5 ( 2 ) = 10 ⎝ L ⎠n ⎛W ⎞ 5-PMOS in parallel ⇒ ⎜ ⎟ = 4 ⎝ L ⎠p 16.62 By definition: NMOS off if gate voltage = 0 NMOS on if gate voltage = 5 V PMOS off if gate voltage = 5 V PMOS on if gate voltage = 0 State 1 2 N1 off on P 1 on off NA off on NB off off NC off off v01 5 5 N2 on on P2 off off v02 0 0 754. 3 4 5 6 off on off on on off on off off off off off off off off on off on off on 5 5 5 0 on on on off off off off on Logic function is v02 = ( v A OR vB ) AND vC 16.63 State 1 2 3 4 5 6 Logic function: v03 = ( vX OR vZ ) AND vY 16.64 v01 v02 v03 5 0 5 5 5 0 5 0 5 0 5 5 0 5 0 5 0 0 0 0 0 5 755. 16.65 756. 16.66 757. 16.67 2 I = −C dVC dt So 1 ( 2I ) ⋅ t C For ΔVC = −0.5 V ΔVC = − 758. −0.5 = − 16.68 (a) (i) (ii) (iii) (b) (i) (ii) (iii) 16.69 (a) (i) (ii) (iii) (b) (i) (ii) (iii) 2 ( 2 x 10−12 ) ⋅ t 25 x 10−15 ⇒ t = 3.125 ms vO = 0 vO = 4.2 V vO = 2.5 V vO = 0 vO = 3.2 V vO = 2.5 V vo = 0 vo = 2.9 V vo = 2.4 V vo = 0 vo = 2.0 V vo = 2.0 V 16.70 Neglect the body effect. v01 (logic 1) = 4.2 V , v02 (logic 1) = 5 V a. vI = 5 V ⇒ vGS 1 = 4.2 V b. M 1 in nonsaturation and M 2 in saturation. From Equation (16.23) ⎛W ⎜ ⎝L ⎛W ⎜ ⎝L Or ⎛W ⎜ ⎝L 2 ⎞ ⎛W ⎞ 2 ⎟ ⎡ 2 ( vGS 1 − VTND ) vO1 − vO1 ⎤ = ⎜ ⎟ (VDD − vO1 − VTNL ) ⎣ ⎦ L ⎠L ⎠D ⎝ 2 2 ⎞ ⎡ ⎟ ⎣ 2 ( 4.2 − 0.8 )( 0.1) − ( 0.1) ⎤ = (1) [5 − 0.1 − 0.8] ⎦ ⎠D ⎞ ⎛W ⎞ ⎟ ( 0.67 ) = 16.81 ⇒ ⎜ ⎟ = 25.1 ⎠D ⎝ L ⎠D Now v01 = 4.2 V ⇒ vGS 3 = 4.2 V M 3 in nonsaturation and M 4 in saturation. From Equation (16.29(b)). ⎛W ⎜ ⎝L ⎛W ⎜ ⎝L 2 ⎞ ⎛W ⎞ 2 ⎟ ⎡ 2 ( vGS 3 − VTND ) vO 2 − vO 2 ⎤ = ⎜ ⎟ [ −VTNL ] ⎣ ⎦ ⎠D ⎝ L ⎠L 2 2 ⎞ ⎟ ⎡ 2 ( 4.2 − 0.8 )( 0.1) − ( 0.1) ⎤ = ( 2 ) ⎡ − ( −1.5 ) ⎤ ⎣ ⎦ ⎣ ⎦ ⎠D ⎛W ⎞ ⎜ ⎟ (0.67) = 2.25 ⎝ L ⎠D 759. ⎛W ⎞ Or ⎜ ⎟ = 3.36 ⎝ L ⎠D 16.71 A B Y 0 0 1 0 1 0 1 0 0 1 1 ⇒ indeterminate 0.1 Without the top transistor, the circuit performs the exclusive-NOR function. 16.72 B A A 0 1 0 0 1 1 1 0 0 1 0 1 Y = A + AB = A + B Z = Y or Z = AB B 1 0 1 0 Y 0 1 1 1 Z 1 0 0 0 16.73 16.74 For φ = 1, φ = 0, then Y = B. And for φ = 0, φ = 1, then Y = A . A multiplexer. 16.75 Y = AC + BC 16.76 Y = AB + AB = A ⊗ B 16.77 760. A 0 1 0 1 B 0 0 1 1 Y 0 1 1 0 Exclusive-OR function. 16.78 This circuit is referred to as a two-phase ratioed circuit. The same width-to-length ratios between the driver and load transistors must be maintained as discussed previously with the enhancement load inverter. When φ1 is high, v01 becomes the complement of vI . When φ2 goes high, then v0 becomes the complement of v01 or is the same as vI . The circuit is a shift register. 16.79 Let Q = 0 and Q = 1 ; as S increases, Q decreases. When Q reaches the transition point of the M 5 − M 6 inverter, the flip-flop with change state. From Equation (16.28(b)), KL VIt = ⋅ ( −VTNL ) + VTND KD where K L = K 6 and K D = K 5 . Then 30 VIt = ⋅ ⎡ − ( −2 ) ⎤ + 1 ⇒ VIt = Q = 2.095 V ⎦ 100 ⎣ This is the region where both M 1 and M 3 are biased in the saturation region. Then S= K3 ⋅ ( −VTNL ) + VTND = K1 30 ⋅ ⎡ − ( −2 ) ⎤ + 1 ⎦ 200 ⎣ or S = 1.77 V This analysis neglects the effect of M 2 starting to turn on at the same time. 16.80 Let vY = R, v X = S , v02 = Q, and v01 = Q. Assume VThN = 0.5 V and VThP = −0.5 V. For S = 0, we have the following: 761. If we want the switching to occur for R = 2.5 V, then because of the nonsymmetry between the two circuits, we cannot have Q and Q both equal to 2.5 V. Set R = Q = 2.5 V and assume Q goes low. For the M 1 − M 5 inverter, M 1 in nonsaturation and M 5 in saturation. Then 2 2 K n ⎡ 2 ( 2.5 − 0.5 ) Q − Q ⎤ = K p [ 2.5 − 0.5] ⎢ ⎥ ⎣ ⎦ Or 2 ⎛ Kp ⎞ 4Q − Q = 4 ⎜ ⎟ ⎝ Kn ⎠ For the other circuit, M 2 − M 4 in saturation and M 6 in nonsaturation. Then ( ) 2 K n ( 2.5 − 0.5 ) + K n (Q − 0.5) 2 = K p ⎡ 2 5 − Q − 0.5 ( 2.5 ) − ( 2.52 ) ⎤ ⎣ ⎦ Combining these equations and neglecting the Q 3 term, we find Kp Q = 1.4 V and kn = 0.9 16.81 3.3 + ( −0.4 ) + 0.5 vIt = = 1.7 V 1+1 vI = 1.5 V NMOS Sat; PMOS Non Sat 2 = ⎡ 2 ( 3.3 − vI − 0.4 )( 3.3 − vo1 ) − ( 3.3 − vo1 ) ⎤ ⇒ vo1 = 2.88 V ⎣ ⎦ vI = 1.6 V vo1 = 2.693 V vI = 1.7 V vo1 = variable (switching region) ( vI − 0.5) 2 vI = 1.8 V NMOS Non Sat; PMOS Sat ( 3.3 − VI − 0.4 ) 2 2 = ⎡ 2 ( vI − 0.5 ) vo1 − vo1 ⎤ ⇒ vo1 = 0.607 V ⎣ ⎦ Now vI = 1.5 V, vo1 = 2.88 V ⇒ vo ≈ 0V vI = 1.6 V, vo1 = 2.693 V NMOS Non Sat; PMOS Sat 2 ( 3.3 − vo1 − 0.4 ) = ⎡ 2 ( vo1 − 0.5 ) vo − vo2 ⎤ ⎣ ⎦ vo = 0.00979 V vI = 1.7 V, v o1 = Switching Mode ⇒ v0 = Switching Mode. vI = 1.8 V, vo1 = 0.607 V NMOS Sat; PMOS Non Sat ( v01 − 0.5) 2 2 = ⎡ 2 ( 3.3 − v01 − 0.4 )( 3.3 − v0 ) − ( 3.3 − v0 ) ⎤ ⇒ v0 = 3.298 V ⎣ ⎦ 16.82 For R = φ = VDD and S = 0 ⇒ Q = 0, Q = 1 For S = φ = VDD and R = 0 ⇒ Q = 1, Q = 1 The signal φ is a clock signal. For φ = 0, The output signals will remain in their previous state. 16.83 a. Positive edge triggered flip-flop when CLK = 1, output of first inverter is D and then Q = D = D . 762. b. For example, put a CMOS transmission gate between the output and the gate of M 1 driven by a CLK pulse. 16.84 For J = 1, K = 0, and CLK = 1; this makes Q = 1 and Q = 0 . For J = 0, K = 1, and CLK = 1 , and if Q = 1, then the circuit is driven so that Q = 0 and Q = 1. If initially, Q = 0, then the circuit is driven so that there is no change and Q = 0 and Q = 1. J = 1, K = 1, and CLK = 1, and if Q = 1, then the circuit is driven so that Q = 0. If initially, Q = 0 , then the circuit is driven so that Q = 1. So if J = K = 1, the output changes state. 16.85 For J = v X = 1, K = vY = 0, and CLK = vZ = 1, then v0 = 0. For J = v X = 0, K = vY = 1, and CLK = vZ = 1, then v0 = 1. Now consider J = K = CLK = 1. With v X = vZ = 1, the output is always v0 = 0, So the output does not change state when J = K = CLK = 1. This is not actually a J − K flip-flop. 16.86 64 K ⇒ 65,536 transistors arranged in a 256 × 256 array. (a) Each column and row decoder required 8 inputs. (b) (i) Address = 01011110 so input = a7 a6 a5 a4 a3 a2 a1a0 (ii) Address = 11101111 so input = a7 a6 a5 a4 a3 a2 a1a0 (c) (i) (ii) Address = 00100111 so input = a7 a6 a5 a4 a3 a2 a1a0 Address = 01111011 so input = a7 a6 a5 a4 a3 a2 a1a0 16.87 (a) 1-Megabit memory ⇒ = 1, 048,576 ⇒ 1024 × 1024 Nuclear & input row and column decodes lines necessary = 10 (b) 250K × 4 bits ⇒ 262,144 × 4 bits ⇒ 512 × 512 For 512 lines ⇒ 9 row and column decoder lines necessary. 16.88 Put 128 words in a 8 × 16 array, which means 8 row (or column) address lines and 16 column (or row) address lines. 16.89 Assume the address line is initially uncharged, then dV 1 I I = C C or VC = ∫ Idt = ⋅ t C C dt −12 VC ⋅ C ( 2.7 ) ( 5.8 × 10 ) ⇒ = I 250 × 10−6 t = 6.26 × 10 −8 s ⇒ 62.6 ns Then t = 16.90 763. (a) ⎛W or ⎜ ⎝L (b) 5 − 0.1 ⎛ 35 ⎞⎛ W = ⎜ ⎟⎜ 1 ⎝ 2 ⎠⎝ L 2 ⎞⎡ ⎟ ⎣ 2 ( 5 − 0.7 )( 0.1) − ( 0.1) ⎤ ⎦ ⎠ ⎞ ⎟ = 0.329 ⎠ 16 K ⇒ 16,384 cells 2 = 2 μA 1 Power per cell = (2 μ A)(2 V ) = 4 μW Total Power = PT = (4 μW )(16,384) ⇒ PT = 65.5 mW iD ≅ Standby current = (2 μ A)(16,384) ⇒ IT = 32.8 mA 16.91 16 K ⇒ 16,384 cells 200 ⇒ 12.2 μW 16,384 V P 12.2 2.5 iD = = = 4.88 μ A ≅ DD = ⇒ R = 0.512 M Ω VDD R R 2.5 PT = 200 mW ⇒ Power per cell = If we want vO = 0.1 V for a logic 0, then ⎛ k′ ⎞⎛ W ⎞ 2 iD = ⎜ n ⎟ ⎜ ⎟ ⎡ 2 (VDD − VTN ) vO − vO ⎤ ⎦ 2 ⎠⎝ L ⎠ ⎣ ⎝ 2 ⎛ 35 ⎞ ⎛ W ⎞ 4.88 = ⎜ ⎟ ⎜ ⎟ ⎡ 2 ( 2.5 − 0.7 )( 0.1) − ( 0.1) ⎤ ⎣ ⎦ ⎝ 2 ⎠⎝ L ⎠ ⎛W ⎞ So ⎜ ⎟ = 0.797 ⎝L⎠ 16.92 Q = 0, Q = 1 So D = Logic 1 = 5 V A very short time after the row has been addressed, D remains charged at VDD = 5 V . Then M p 3, M A, and M N 1 begin to conduct and D decreases. In steady-state, all three transistors are biased in the nonsaturation region. Then 2 2 2 K p 3 ⎡ 2 (VSG 3 + VTP 3 ) VSD 3 − VSD 3 ⎤ = K nA ⎡ 2 (VGSA − VTNA ) VDSA − VDSA ⎤ = K n1 ⎡ 2 (VGS 1 − VTN 1 ) VDS 1 − VDS 1 ⎤ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ Or 2 2 K p 3 ⎡ 2 (VDD + VTP 3 )(VDD − D ) − (VDD − D ) ⎤ = K nA ⎡ 2 (VDD − Q − VTNA )( D − Q ) − ( D − Q ) ⎤ ⎣ ⎦ ⎣ ⎦ = K n1 ⎡ 2 (VDD − VTN 1 ) Q − Q 2 ⎤ (1) ⎣ ⎦ Equating the first and third terms: 2 ⎛ 20 ⎞ ⎡ 2 ⎤ ⎛ 40 ⎞ ⎡ ⎤ ⎜ ⎟ (1) ⎣ 2 ( 5 − 0.8 )( 5 − D ) − ( 5 − D ) ⎦ = ⎜ ⎟ ( 2 ) ⎣ 2 ( 5 − 0.8 ) Q − Q ⎦ (2) ⎝ 2 ⎠ ⎝ 2 ⎠ As a first approximation, neglect the ( 5 − D ) and Q 2 terms. We find 2 Q = 1.25 − 0.25 D (3) Then, equating the first and second terms of Equation (1): 2 2 ⎛ 20 ⎞ ⎡ ⎛ 40 ⎞ ⎜ ⎟ (1) ⎣ 2 ( 5 − 0.8 )( 5 − D ) − ( 5 − D ) ⎤ = ⎜ ⎟ (1) ⎡ 2 ( 5 − Q − 0.8 )( D − Q ) − ( D − Q ) ⎤ ⎦ ⎝ 2 ⎠ ⎣ ⎦ 2 ⎠ ⎝ 764. Substituting Equation (3), we find as a first approximation: D = 2.14 V Substituting this value of D into equation (2), we find 2 8.4 ( 5 − 2.14 ) − ( 5 − 2.14 ) = 4 ⎡8.4Q − Q 2 ⎤ ⎣ ⎦ We find Q = 0.50 V Using this value of Q, we can find a second approximation for D by equating the second and third terms of equation (1). We have 2 20 ⎡ 2 ( 4.2 − Q )( D − Q ) − ( D − Q ) ⎤ = 40 ⎡ 2 ( 4.2Q ) − Q 2 ⎤ ⎣ ⎦ ⎣ ⎦ Using Q = 0.50 V , we find D = 1.79 V 16.93 Initially M N 1 and M A turn on. M N 1, Nonsat; M A , sat. K nA [VDD − Q − VTN ] = K n1 ⎡ 2 (VDD − VTN 1 ) Q − Q 2 ⎤ ⎣ ⎦ 2 2 ⎛ 40 ⎞ ⎛ 40 ⎞ 2 ⎡ ⎤ ⎜ ⎟ (1) [5 − Q − 0.8] = ⎜ ⎟ ( 2 ) ⎣ 2 ( 5 − 0.8 ) Q − Q ⎦ 2 ⎠ 2 ⎠ ⎝ ⎝ which yields Q = 0.771 V Initially M P 2 and M B turn on Both biased in nonsaturation reagion 2 2 K P 2 ⎡ 2 (VDD + VTP 3 ) VDD − Q − VDD − Q ⎤ = K nB ⎡ 2 (VDD − VTNB ) Q − Q ⎤ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ 2 2 ⎛ 20 ⎞ ⎡ ⎤ ⎛ 40 ⎞ ⎡ ⎤ ⎜ ⎟ ( 4 ) ⎢ 2 ( 5 − 0.8 ) 5 − Q − 5 − Q ⎥ = ⎜ ⎟ (1) ⎢ 2 ( 5 − 0.8 ) Q − Q ⎥ ⎦ ⎣ ⎦ ⎝ 2 ⎠ ⎣ ⎝ 2 ⎠ ( ) ( ( ) ( ) ) which yields Q = 3.78 V Note: (W / L) ratios do not satisfy Equation (16.86) 16.94 For Logic 1, v1: ( 5 )( 0.05 ) + ( 4 )(1) = (1 + 0.05 ) v1 ⇒ v1 = 4.0476 V v2 : (5)(0.025) + (4)(1) = (1 + 1.025)v2 ⇒ v2 = 4.0244 V For Logic 0, v1: (0)(0.05) + (4)(1) = (1 + 0.05)v1 ⇒ v1 = 3.8095 V v2 : (0)(0.025) + (4)(1) = (1 + 0.025)v2 ⇒ v2 = 3.9024 V 16.95 Not given 16.96 Not given 16.97 Not given 765. 16.98 Quantization error = (a) Or LSB ≤ 0.10 V For a 6-bit word , LSB = 1 − LSB = (b) 1 LSB ≤ 1% ≤ 0.05 V 2 5 = 0.078125 V 64 5 = 0.078125 V 64 3.5424 × 64 = 45.34 ⇒ n = 45 5 Digital Output = 101101 45 × 5 = 3.515625 64 . 1 Δ = 3.5424 − 3.515625 = 0.026775 < LSB. 2 (c) 16.99 Quantization error = (a) 1 − LSB = 0.10 V 1 LSB ≤ 0.5% ≤ 0.05 V 2 10 = 0.078125 V 128 1 − LSB = 0.078125 V For a 7-beit word, LSB = (b) 3.5424 ×128 = 45.34272 ⇒ n = 45 10 Digital output = 0101101 (c) Now 45 × 10 = 3.515625 128 Δ = 3.5424 − 3.515625 = 0.026775 V < 16.100 (a) (b) ⎛0 1 0 1 ⎞ vo = ⎜ + + + ⎟ ( 5 ) ⎝ 2 4 8 16 ⎠ vo = 1.5625 V ⎛1 0 1 0 ⎞ vo = ⎜ + + + ⎟ ( 5 ) ⎝ 2 4 8 16 ⎠ vo = 3.125 V 16.101 ⎛1⎞ LSB = ⎜ ⎟ ( 5 ) = 0.3125 V ⎝ 16 ⎠ 1 LSB = 0.15625 V 2 ⎛ 10 ⎞ Now vo = ⎜ ⎟ (5) ⎝ 20 + ΔR1 ⎠ (a) 1 LSB 2 766. For vo = 2.5 + 0.15625 = 2.65625 V (10 )( 5) ⇒ ΔR1 = −1.176 K 2.65625 For vo = 2.5 − 0.15625 = 2.34375 V 20 + ΔR1 = 20 + ΔR1 = (10 )( 5 ) 2.34375 V ΔR1 = +1.333 K For ΔR1 = 1.176 K ⇒ ΔR1 = 5.88% ⎛ ⎞ 10 For R4 : vo = ⎜ ⎟ ( 5) ⎝ 160 + ΔR4 ⎠ vo = 0.3125 + 0.15625 = 0.46875 V (b) (10 )( 5 ) ⇒ ΔR4 = −53.33K 0.46875 Or vo = 0.3125 − 0.15625 = 0.15625 V 160 + ΔR4 = (10 )( 5 ) ⇒ ΔR4 = 160 K 0.15625 For ΔR4 = 53.33K ⇒ ΔR4 = 33.33% 160 + ΔR4 = 16.102 (a) R5 = 320 kΩ R6 = 640 kΩ R7 = 1280 kΩ R8 = 2560 kΩ (b) ⎛ 10 ⎞ vo = ⎜ ⎟ ( 5 ) = 0.01953125 V ⎝ 2560 ⎠ 16.103 (a) V −5 ⇒ I1 = −0.50 mA I1 = REF = 2 R 10 I I 2 = 1 = −0.25 mA 2 I2 I 3 = = −0.125 mA 2 I3 I 4 = = −0.0625 mA 2 I4 I 5 = = −0.03125 mA 2 I I 6 = 5 = −0.015625 mA 2 Δvo = I 6 RF = ( 0.015625 )( 5 ) (b) Δvo = 0.078125 V (c) vo = − [ I 2 + I 5 + I 6 ] RF = [ 0.25 + 0.03125 + 0.015625] ( 5 ) vo = 1.484375 V 767. (d) For 101010; vo = ( 0.50 + 0.125 + 0.03125 )( 5 ) = 3.28125 V For 010101; vo = ( 0.25 + 0.0625 + 0.015625 )( 5 ) = 1.640625 V Δvo = 1.640625 V 16.104 1 5 ⎛V ⎞⎛ R ⎞ V = 0.3125 V LSB = ⎜ REF ⎟ ⎜ ⎟ = REF = 2 8 R ⎠ ⎝ 2 ⎠ 16 16 ⎝ Ideal 3 ⎛ 3V ⎞ v A for 011 ⇒ ⎜ REF ⎟ ( R ) = VREF = 1.875 V 8R ⎠ 8 ⎝ 1 Range of v A = 1.875 ± LSB 2 or 1.5625 ≤ vA ≤ 2.1875 V 16.105 6-bits ⇒ 26 = 64 resistors 26 − 1 = 63 comparators 16.106 (a) 10- bit output ⇒ 1024 clock periods 1 1 1 clock period = = 6 = 1 μS f 10 May conversion time = 1024 μS = 1.024 mS (b) 1 1⎛ 5 ⎞ LSB = ⎜ ⎟ = 0.002441406 V 2 2 ⎝ 1024 ⎠ ⎛ 5 ⎞ v′ = (128 + 16 + 2 ) ⎜ A ⎟ = 0.712890625 V ⎝ 1024 ⎠ 1 So range of v A = v′ ± LSB A 2 0.710449219 ≤ v A ≤ 0.715332031 V (c) 0100100100 ⇒ 256 + 32 + 4 = 292 clock pulses 16.107 N ×5 ⇒ N = 640 ⇒ 512 + 128 1024 Output = 1010000000 (b) N ×5 ⇒ N = 381.19 ⇒ N = 381 ⇒ 256 + 64 + 32 + 16 + 8 + 4 + 1 1.8613 = 1024 Output = 0101111101 (a) 3.125 = 768. Chapter 17 Exercise Solutions EX17.1 a. −0.7 − (−5) iE = = 1 mA ⇒ RE = 4.3 kΩ RE iC1 = iC 2 = 0.5 mA = 5 − 3.5 ⇒ RC1 = RC 2 = 3 kΩ RC1 v1 = 1 V i. (1 − 0.7) − (−5) iE = ⇒ iE = 1.23 mA 4.3 iC1 = iE = v01 = 5 − (1.23)(3) ⇒ v01 = 1.31 V b. v02 = 5 V v1 = −1 V ii. iE = 1 mA ⇒ v02 = 5 − (1)(3) ⇒ v02 = 2 V v01 = 5 V EX17.2 P (iCXY + iCR + i3 + i4 )(5.2) a. v X = vY = logic 1 ⇒ iCXY = 3.22 mA iCR = 0 −0.7 + 5.2 i3 = = 3 mA 1.5 −1.4 + 5.2 i4 = = 2.53 mA 1.5 P = (3.22 + 0 + 3 + 2.53)(5.2) ⇒ P = 45.5 mW b. v X = vY = logic 0 ⇒ iCXY = 0 iCR = 2.92 mA i3 = 2.53 mA i4 = 3 mA P = (0 + 2.92 + 2.53 + 3)(5.2) ⇒ P = 43.9 mW EX17.3 3.5 − (2.45 + 0.7) = 1.4 K 0.25 3.5 − 2(0.7) R1 + R2 = = 8.4 k ⇒ R2 = 7 K 0.25 2.45 R5 = = 9.8 K 0.25 R1 = EX17.4 769. I MAX = 1 = −0.7 − (−5.2) ⇒ R3 = R4 = 4.5 K R3 −0.7 − 0.7 − (−5.2) = 3.22 mA 1.18 i5 = i1 = 1.40 mA i3 = 1.0 mA −1.4 − (−5.2) i4 = = 0.844 mA 4.5 P = (3.22 + 1.4 + 1.4 + 1.0 + 0.844)(5.2) = 40.8 mW iCXY = EX17.5 ⎛ v − 0.7 − (−5.2) ⎞ iL = ⎜ oR ⎟ (10) (1.18)(51) ⎝ ⎠ voR − (−5.2) i3 = 1.5 voR + 5.2 ⎛ voR + 4.5 ⎞ ⎡ 0 − (VoR + 0.7) ⎤ ⎢ ⎥ (51) = 1.5 + ⎜ (1.18)(51) ⎟ (10) 0.24 ⎣ ⎦ ⎝ ⎠ ⎡⎛ 51 ⎞ 1 1 − (6) ⎤ (0.7)(51) 5.2 4.5(10) −voR ⎢⎜ + ⎟+ ⎥ = 0.24 + 1.5 + (1.18)(51) ⎣⎝ 0.24 ⎠ 1.5 (1.18)(51) ⎦ −voR [212.50 + 0.6667 + 0.166168] = 148.75 + 3.4666 + 0.747757 −voR [213.3328] = 152.9644 voR = −0.7170 EX17.6 (100)(0.026) rπ 3 = = 2.6 K 1 1 g m3 = = 38.46 mA / V 0.026 Vn Vn = Ib3 = RC 2 + rπ 3 + (1 + β ) R3 0.24 + 2.6 + (101)(4.5) Ib3 = Vn 457.34 ⎛ Vn ⎞ VO = − I b 3 ( RC 2 + rπ 3 ) = − ⎜ ⎟ (0.24 + 2.6) ⎝ 457.34 ⎠ VO = −0.00621 Vn ⎛ Vn ⎞ ′ VO = (1 + β ) I b 3 R3 = (101) ⎜ ⎟ (4.5) ⎝ 457.34 ⎠ ′ VO = 0.9938 Vn EX17.7 P = I Q ⋅ VCC ⇒ 0.2 = I Q (1.7) ⇒ I Q = 117.6 μ A QR on ⇒ v0 = 1.7 − I Q RC = 1.7 − 0.4 ⇒ RC = VR = 1.7 + 1.3 ⇒ VR = 1.5 V 2 0.4 ⇒ RC = 3.40 kΩ 0.1176 770. EX17.8 (a) v X = vY = 5 V v1 = VBE ( sat ) + 2VY = 0.8 + 2(0.7) = 2.2 V 5 − 2.2 i1 = = 0.70 mA 4 V − VCE ( sat ) 5 − 0.1 iRC = CC = = 1.225 mA RC 4 P = (i1 + iRC )VCC = (0.70 + 1.225)(5) or P = 9.625 mW (b) v X = vY = 0 ⇒ v1 = 0.70 V V − v 5 − 0.70 i1 = CC 1 = = 1.075 mA 4 R1 P = i1 ⋅ VCC = (1.075)(5) ⇒ P = 5.375 mW EX17.9 vB1 = 0.1 + 0.8 = 0.9 V (a) 5 − 0.9 i1 = = 0.683 mA 6 All other currents ≈ 0 vB1 = 0.8 + 0.8 + 0.7 = 2.3 V (b) 5 − 2.3 = 0.45 mA i1 = 6 iB 2 = (1 + 0.2)(0.45) = 0.54 mA 5 − (0.8 + 0.1) i2 = = 2.733 mA 1.5 iE 2 = 2.733 + 0.54 = 3.273 mA 0.8 i4 = = 0.533 mA 1.5 iB 0 = 3.273 − 0.533 = 2.74 mA 5 − 0.1 i3 = = 2.23 mA 2.2 i 2.733 For Q2 : 2 = = 5.06 < β F 0.54 iB 2 For Q0 : i3 2.23 = = 0.81 < β F iB 0 2.74 EX17.10 v X = vY = 3.6 V, Output low: 5 − 2.3 = 0.45 mA 6 = (1 + 2 β R )i1 i1 = iB 2 iB 2 = 0.54 mA vB 3 = 0.8 + 0.1 = 0.9 771. 5 − 0.9 = 2.05 mA 2 0.8 ⇒ iB 0 = 2.06 mA iB 0 = 2.05 + 0.54 − 1.5 5 − 0.9 ′ iL = = 0.683 mA 6 ′ β F ⋅ iB 0 = N ⋅ iL ⇒ (20)(2.06) = N (0.683) i2 = ⇒ N = 60 EX17.11 iRC = a. 5 − 0.4 = 2.04 mA 2.25 2 + 2.04 ′ ⇒ iC = 3.67 mA 1 1+ 10 3.67 ′ iB = = 0.367 mA 10 iD = 2 − 0.367 ⇒ iD = 1.63 mA ′ iC = ′ iD = 0 ⇒ iB = iB = 2 mA b. ′ ′ iC = β iB = (10)(2) = 20 mA = iRC + iL iL = 20 − 2.04 ⇒ iL (max) ≈ 18 mA EX17.12 v1 = 0.4 + 0.3 = 0.7, i1 = (a) 5 − 0.7 = 0.1075 mA 40 All transistor currents are zero. P = (0.1075)(5 − 0.4) ⇒ 495 μ W 5 − 1.4 = 0.090 mA 40 5 − 1.1 vC 2 = 0.7 + 0.4 = 1.1 V, i2 = iC 2 = = 0.325 mA 12 iB 0 ≈ iB 2 + iC 2 = 0.09 + 0.325 = 0.415 mA v1 = 1.4 V, i1 = iB 2 = (b) iC 0 ≈ 0 P = (i1 + i2 )(5) = (0.09 + 0.325)(5) = 2.08 mW TYU17.1 QR on a. iE = 1.5 − 0.7 − ( −3.5) = 2 mA ⇒ RE = 2.15 kΩ RE iCR ≈ iE = 2 mA = v X = vY = 2 V ⇒ Q1 and Q2 on b. iE = 3.5 − 2 ⇒ RC 2 = 0.75 kΩ RC 2 2 − 0.7 − (−3.5) 4.8 = ⇒ iE = 2.233 mA 2.15 RE RC1 = 3.5 − 2 1.5 = ⇒ RC1 = 0.672 kΩ 2.233 iCXY 772. TYU17.2 logic 1 = −0.7 V Q1 and Q2 on when v X = vY = −0.7 V −0.7 − 0.7 − (−5.2) = 2.5 ⇒ RE = 1.52 kΩ RE iE = vNOR = −1.5 ⇒ RC1 = 0 − (−1.5 + 0.7) ⇒ RC1 = 320 Ω 2.5 −1.5 − 0.7 ⇒ VR = −1.1 V 2 −1.1 − 0.7 − (−5.2) = 2.237 mA QR on ⇒ iE = 1.52 0 − (−1.5 + 0.7) ⇒ RC 2 = 358 Ω RC 2 = 2.237 −0.7 − (−5.2) ⇒ R3 = R4 = 1.8 kΩ R3 = R4 = 2.5 VR = TYU17.3 N M H = −0.70 − (−0.93) ⇒ N M H = 0.23 V N M L = −1.17 − ( −1.40) ⇒ N M L = 0.23 V TYU17.4 State 1 2 3 4 5 6 7 8 A 0 1 0 0 1 1 0 1 B 0 0 1 0 1 0 1 1 C 0 0 0 1 0 1 1 1 ( A AND C ) OR ( B AND C ) true for true for states 6 and 8 states 3 and 5 Output goes high for these 4 states TYU17.5 A 0 B 0 C 0 v0 0 Q01 off “on” off off on on off on Q02 off off on off on off on on Q03 off off off on off on on on Q1 off off off on off on on on Q2 on on on off on off off off QR on on “off” on off “off” on off v0 0 0 1 0 1 1 0 1 773. 1 0 0 1 1 0 1 0 1 0 1 0 1 1 0 0 1 0 1 1 1 1 1 1 0 0 0 1 ⇒ ( A ⊕ B) ⊕ C TYU17.6 v X = vY = 0.1 ⇒ v1 = 0.8 a. 5 − 0.8 i1 = ⇒ i1 = 0.7 mA 6 i2 = iR = iB = iRC = 0 v0 = 5 V b. c. Same as part (a). v X = vY = 5 V ⇒ v1 = 0.8 + 0.7 + 0.7 = 2.2 V i1 = i2 = 5 − 2.2 ⇒ i1 = i2 = 0.467 mA 6 0.8 ⇒ iR = 0.053 mA 15 iB = i2 − iR ⇒ iB = 0.414 mA iR = iRC = 5 − 0.1 ⇒ iRC = 0.98 mA 5 vC = 0.1 V TYU17.7 (a) From Tyu17.6, iB = 0.414 mA, iRC = 0.98 mA ′ v1′ = 0.1 + 0.7 = 0.8, iL = 5 − 0.8 = 0.70 mA 6 ′ iC ,max = β iB = iRC + NiL (25)(0.414) = 0.98 + N (0.70) N = 13.4 → N = 13 (b) iC , max = β iB = (25)(0.414) = 10.35 mA < 15 mA So from part (a), N = 13 TYU17.8 From EX17.9, iB 0 = 2.74 mA, i3 = 2.23 mA ′ vo = 0.1 V, so vB1 = 0.1 + 0.8 = 0.9 V 5 − 0.9 = 0.6833 mA 6 iC 0, max = β iB 0 = i3 + Ni1′ (20)(2.74) = 2.23 + N (0.6833) N = 76.9 ⇒ N = 76 i1′ = TYU17.9 774. Q1 in saturation 5 − 0.9 iB1 = ⇒ iB1 = 0.683 mA 6 iC1 = iB 2 = iC 2 = 0 iB 0 = iC 0 = 0 vB 4 = 0.1 + 0.7 = 0.8 V 0.1 iB 4 = ⇒ iB 4 = 1.19 μ A (21)(4) iC 4 = 23.8 μ A iB 3 = iC 3 = 0 TYU17.10 (a) v X = vY = 0.4, vB1 = 0.4 + 0.7 = 1.1 V 5 − 1.1 = 1.393 mA 2.8 P = i1 (5 − 0.4) = (1.393)(5 − 0.4) = 6.41 mW i1 = (b) v X = vY = 3.6 V 5 − 2.1 = 1.036 mA 2.8 5 − 1.1 = 5.132 mA vC 2 = 0.7 + 0.4 = 1.1 V, i2 = 0.76 0.4 vE 4 = 1.1 − 0.7 = 0.4 V, iR 4 = = 0.1143 mA 3.5 ⎛ β ⎞ ⎛ 25 ⎞ iR 3 = ⎜ ⎟ iR 4 = ⎜ ⎟ (0.1143) = 0.1099 ≈ 0.11 mA ⎝ 26 ⎠ ⎝1+ β ⎠ vB1 = 2.1, i1 = P = (i1 + i2 + iR 3 )(5) = (1.036 + 5.132 + 0.11)(5) or P = 31.4 mW TYU17.11 (a) v X = 0.4 V, vC1 = 0.4 + 0.7 = 1.1 V 5 − 1.1 ⇒ 97.5 μ A 40 v X = 3.6 V, vC1 = 3(0.7) = 2.1 V (b) i1 = i1 = 5 − 2.1 ⇒ 72.5 μ A 40 775. Chapter 17 Problem Solutions 17.1 a. vI = −1.5 Vi Q1 off , Q2 on −0.7 − (−3.5) ⇒ iE = 0.56 mA iE = 5 iC1 = 0 ⇒ v01 = 3.5 V iC 2 = iE ⇒ v02 = 3.5 − iE RC 2 = 3.5 − ( 0.56 )( 2 ) or v02 = 2.38 V vI = 1.0 Vi Q1 on, Q2 off b. iE = iC 2 (1 − 0.7 ) − ( −3.5) 5 = 0 ⇒ v02 = 3.5 V ⇒ iE = 0.76 mA c. logic 0 at v02 (low level) = 2.38 V Then v01 = 2.38 = 3.5 − (0.76) RC1 or RC1 = 1.47 kΩ 17.2 (a) iC 2 = I Q = 0.5 = 3−0 ⇒ RC 2 = 6 K RC 2 (b) iC1 = I Q = 0.5 = 3 −1 ⇒ RC1 = 4 K RC1 (c) ⎛V ⎞ I S exp ⎜ BE1 ⎟ iC1 ⎝ VT ⎠ = IQ ⎡ ⎛V ⎞ ⎛ V ⎞⎤ I S ⎢ exp ⎜ BE1 ⎟ + exp ⎜ BE 2 ⎟ ⎥ VT ⎠ ⎝ ⎝ VT ⎠ ⎦ ⎣ 1 ⎛ VBE 2 − VBE1 ⎞ 1 + exp ⎜ ⎟ VT ⎝ ⎠ vI = VBE1 − VBE 2 = So iC1 = IQ 1 ⎛ −v ⎞ 1 + exp ⎜ I ⎟ ⎝ VT ⎠ 0.1 1 = = 0.2 0.5 ⎛ −v ⎞ 1 + exp ⎜ I ⎟ ⎝ VT ⎠ 776. ⎛ −v ⎞ 1 exp ⎜ I ⎟ = −1 = 4 ⎝ VT ⎠ 0.2 ( −vI ) = ( 0.026 ) ln (4) vI = −0.0360 V 17.3 (a) vI = 0.5 V, Q1 on, Q2 off = v02 = 3 V v01 = 3 − (1)(0.5) = 2.5 V (b) vI = −0.5 V Q1 off, Q2 on ⇒ v01 = 3 V v02 = 3 − (1)(0.5) = 2.5 V 17.4 (a) Q2 on, vE = −1.2 − 0.7 = −1.9V iE = iC 2 = −1.9 − ( −5.2 ) = 1.32 mA 2.5 v2 = −1V = −iC 2 RC 2 = − (1.32 )( RC 2 ) RC 2 = 0.758 k Ω Q1 on, vE = −0.7 − 0.7 = −1.40 V (b) iE = iC1 = −1.4 − ( −5.2 ) = 1.52 mA 2.5 v1 = −1V = −iC1 RC1 = − (1.52 )( RC1 ) RC1 = 0.658 k Ω (c) For vin = −0.7 V , Q1 on, Q2 off ⇒ vO1 = −0.70V vO 2 = −1 − 0.7 ⇒ vO 2 = −1.7 V For vin = −1.7 V , Q1 off , Q2 on ⇒ vO 2 = −0.7 V vO1 = −1 − 0.7 ⇒ vO1 = −1.7 V (d) (i) For vin = −0.7V , iE = 1.52 mA iC 4 = iC 3 = −1.7 − ( −5.2 ) 3 −0.7 − ( −5.2 ) = 1.17 mA = 1.5 mA 3 P = ( iE + iC 4 + iC 3 )( 5.2 ) = (1.52 + 1.17 + 1.5 )( 5.2 ) or P = 21.8 mW (ii) For vin = −1.7V , iE = 1.32 mA iC 4 = iC 3 = −0.7 − ( −5.2 ) 3 −1.7 − ( −5.2 ) = 1.5 mA = 1.17 mA 3 P = (1.32 + 1.5 + 1.17 )( 5.2 ) or P = 20.7 mW 777. 17.5 3.7 − 0.7 = 1.5 mA 0.67 + 1.33 VR = I 3 R4 + Vγ = (1.5 )(1.33) + 0.7 I3 = a. or VR = 2.70 V logic 1 level = 3.7 − 0.7 ⇒ 3.0 V b. For v X = vY = logic 1. 3 − 0.7 iE = = 2.875 mA = iRC1 0.8 vB 3 = 3.7 − ( 2.875 )( 0.21) = 3.10 V ⇒ v01 ( logic 0 ) = 2.4 V For v X = vY = logic 0, QR on 2.7 − 0.7 iE = = 2.5 mA = iRC 2 0.8 vB 4 = 3.7 − ( 2.5 )( 0.24 ) = 3.1 V ⇒ v02 ( logic 0 ) = 2.4 V 17.6 I REF = (a) 1.1 − 0.7 − ( −1.9 ) I Q = I REF 20 = 0.115 mA = 0.115 mA vo1 (max) = 1.1 − 0.7 = 0.4 V (b) i5 (max) = vo1 (max) − ( −1.9 ) R5 0.4 + 1.9 = 11.5 K 0.2 vo1 = −0.4 V ⇒ vB 5 = +0.3 V 1.1 − 0.3 RC1 = = 6.96 K 0.115 vo 2 = −0.4 V ⇒ vB 6 = +0.3V 1.1 − 0.3 RC 2 = = 6.96 K 0.115 R5 = R6 = (c) (d) 17.7 logic 1 + logic 0 1 + 0 = = 0.5 V 2 2 For i2 = 1 mA VR = R5 = 0.5 − ( −2.3) 1 For QR on, iE = or ⇒ R5 = 2.8 kΩ VR − VBE − ( −2.3) RE 778. 0.5 − 0.7 + 2.3 ⇒ RE = 2.1 kΩ 1 = VR + VBE = 0.5 + 0.7 = 1.2 V RE = VB 2 i1 = 1.2 − 1.4 − ( −2.3) R2 or 1.2 − 1.4 + 2.3 ⇒ R2 = 2.1 kΩ 1 1.7 − 1.2 R1 = ⇒ R1 = 0.5 kΩ 1 1 − (−2.3) i3 = = 3 ⇒ R3 = 1.1 kΩ R3 R2 = i4 = 0 − (−2.3) = 3 ⇒ R4 = 0.767 kΩ R4 For QR on, v0 R = logic 0 = 0 V ⇒ vB 3 = 0.7 V iE = iCR = 1 mA So 1.7 − 0.7 RC 2 = ⇒ RC 2 = 1 kΩ 1 For vI = logic 1 = 1 V, iE = 1 − 0.7 − ( −2.3) For vNOR = 1.238 mA 2.1 = logic 0 = 0 V, vB 4 = 0.7 V Then 1.7 − 0.7 RC1 = ⇒ RC1 = 0.808 kΩ 1.238 17.8 Maximum iE for vI = logic 1 = 3.3 V Then iE = 5 mA = 3.3 − 0.7 ⇒ RE = 0.52 kΩ RE For v02 = logic 1 = 3.3 V iE 3 = 3.3 − 0 = 5 mA R3 or R3 = 0.66 kΩ By symmetry, R2 = 0.66 kΩ For Q1 on, iE = iRC1 = 5 mA = So RC1 = 0.12 kΩ 4 − ( 2.7 + 0.7 ) RC1 779. For QR on, 3 − 0.7 = 4.423 mA = iRC 2 0.52 4 − ( 2.7 + 0.7 ) and iRC 2 = 4.423 = RC 2 iE = ⇒ RC 2 = 0.136 kΩ 17.9 Neglecting base currents: (a) I E1 = 0, I E 3 = 0 5 − 0.7 ⇒ I E 5 = 1.72 mA 2.5 Y = 0.7 V I E5 = (b) 5 − 0.7 ⇒ I E1 = 0.239 mA 18 =0 I E1 = I E3 5 − 0.7 ⇒ I E 5 = 1.72 mA 2.5 Y = 0.7 V I E5 = (c) I E1 = I E 3 = 5 − 0.7 ⇒ I E1 = I E 3 = 0.239 mA 18 I E 5 = 0, Y = 5 V (d) Same as (c). 17.10 (a) (b) VR = −(1)(1) − 0.7 ⇒ VR = −1.7 V QR off , then vO1 = Logic 1 = −0.7 V QR on, then vO1 = −(1)(2) − 0.7 ⇒ vO1 = Logic 0 = −2.7 V QA / QB − off , then vO 2 = Logic 1 = −0.7 V QA / QB − on, then vO 2 = −(1)(2) − 0.7 ⇒ vO 2 = Logic 0 = −2.7 V (c) A = B = Logic 0 = −2.7 V , QR on, VE = −1.7 − 0.7 ⇒ VE = −2.4 V A = B = Logic 1 = −0.7 V , QA / QB on, VE = −0.7 − 0.7 ⇒ VE = −1.4 V (d) A = B = Logic 1 = −0.7 V , QA / QB on, 780. −2.7 − (−5.2) = 1.67 mA 1.5 −0.7 − (−5.2) iC 2 = = 3 mA 1.5 P = (1.67 + 1 + 1 + 1 + 3)(5.2) ⇒ P = 39.9 mW iC 3 = A = B = Logic 0 = −2.7 V iC 3 = 3 mA, iC 2 = 1.67 mA P = 39.9 mW 17.11 a. b. AND logic function logic 0 = 0 V 5 − (1.6 + 0.7) = 2.25 mA 1.2 V2 = (2.25)(0.8) ⇒ logic 1 = 1.8 V Q3 on, i = c. 5 − 0.7 ⇒ iE1 = 1.65 mA 2.6 5 − (0.7 + 0.7) = ⇒ iE 2 = 3 mA 1.2 iC 3 = 0, iC 2 = iE 2 = 3 mA iE1 = iE 2 V2 = 0 d. 5 − (1.8 + 0.7) ⇒ iE1 = 0.962 mA 2.6 5 − (1.6 + 0.7) = ⇒ iE 2 = 2.25 mA 1.2 iC 2 = 0, iC 3 = iE 2 = 2.25 mA iE1 = iE 2 V2 = 1.8 V 17.12 a. b. 3.5 + 3.1 − 0.7 ⇒ VR = 2.6 V 2 For Q1 on, v X = vY = logic 1 = 3.5 V VR = 3.5 − (0.7 + 0.7) = 0.175 mA 12 0.175 0.4 Want iRC1 = = ⇒ RC1 = 4.57 kΩ 2 RC1 iE = c. For Q2 on, iE = Want iRC 2 = 2.6 − 0.7 = 0.158 mA 12 0.158 0.4 = ⇒ RC 2 = 5.06 kΩ 2 RC 2 For vY = logic 1 = 3.5 V 3.5 − 0.7 = 0.35 mA, i E = 0.175 mA iR1 = 8 P = ( iR1 + iE )(VCC ) = ( 0.35 + 0.175 )( 3.5 ) ⇒ P = 1.84 mW d. 781. 17.13 a. logic 1 = 0.2 V logic 0 = −0.2 V iE = b. ( 0 − 0.7 ) − ( −3.1) RE = 0.8 So RE = 3 kΩ 0.8 0.4 = ⇒ R1 = 1 kΩ 2 R1 c. Want iR1 = d. For v X = vY = logic 1 = 0.2 V iE = ( 0.2 − 0.7 ) − ( −3.1) iR 2 = e. 3 = 0.867 mA 0.4 ⇒ iR 2 = 0.4 mA 1 iD 2 = 0.467 mA iE = 0.867 mA 0.2 − (−3.1) i3 = = 1 mA 3.3 −0.2 − (−3.1) i4 = = 0.879 mA 3.3 P = (0.867 + 1 + 0.879)(0.9 − (−3.1)) or P = 11.0 mW 17.14 a. ( −0.9 − 0.7 ) − ( −3) i1 = ⇒ i1 = 1.4 mA 1 ( −0.2 − 0.7 ) − ( −3) ⇒ i3 = 0.14 mA i3 = 15 ( −0.2 − 0.7 ) − ( −3) ⇒ i4 = 0.14 mA i4 = 15 i2 + iD = i1 + i3 = 1.4 + 0.14 = 1.54 mA 0.4 i2 = ⇒ i2 = 0.8 mA 0.5 iD = 0.74 mA v0 = −0.4 V b. i1 = 1.4 mA (0 − 0.7) − (−3) ⇒ i3 = 0.153 mA 15 i4 = i3 ⇒ i4 = 0.153 mA i3 = i2 + iD = i4 ⇒ i2 = 0.153 mA iD = 0 v0 = −(0.153)(0.5) ⇒ v0 = −0.0765 V 782. c. i1 = ( 0 − 0.7 − 0.7 ) − ( −3) 1 ⇒ i1 = 1.6 mA (−0.2 − 0.7) − (−3) ⇒ i3 = 0.14 mA 15 i4 = i3 ⇒ i4 = 0.14 mA i3 = i2 + iD = i3 ⇒ i2 = 0.14 mA iD = 0.0 v0 = −(0.14)(0.5) ⇒ v0 = −0.07 V (0 − 0.7 − 0.7) − (−3) ⇒ i1 = 1.6 mA 1 (0 − 0.7) − (−3) ⇒ i3 = 0.153 mA i3 = 15 i4 = i3 ⇒ i4 = 0.153 mA d. i1 = i2 + iD = i1 + i4 = 1.6 + 0.153 = 1.753 mA 0.4 i2 = ⇒ i2 = 0.8 mA 0.5 iD = 0.953 mA v0 = −0.40 V 17.15 a. i. A = B = C = D = 0 ⇒ Q1 − Q4 cutoff So VDD = 2 I E R 1 + VEB + I B R2 and I I IB = E = E 1 + β P 51 Then I 2.5 = 2 I E (2) + 0.7 + E ⋅ (15) 51 15 ⎞ ⎛ 2.5 = 0.7 = I E ⋅ ⎜ 4 + ⎟ 51 ⎠ ⎝ So I E = 0.419 mA and Y = 2.5 − 2(0.4192)(2) ⇒ Y = 0.823 V ii. A = C = 0, B = D = 2.5 V Now vB 5 = vB 6 = 2.5 − 0.7 = 1.8 V and Y = vB 5 + 0.7 ⇒ Y = 2.5 V b. Y = ( A OR B ) AND (C OR D) 17.16 a. logic 1 = 0 V logic 0 = −0.4 V 783. v01 = A OR B b. v02 = C OR D v03 = v01 OR v02 or v03 = ( A OR B ) AND (C OR D) 17.17 a. For CLOCK = high, I DC flows through the left side of the circuit.. If D is high, I DC flows through the left R resistor pulling Q low. If D is low. I DC flows through the right R resistor pulling Q low. For CLOCK = low, I DC flows through the right side of the circuit maintaining Q and Q in their previous state. P = ( I DC + 0.5 I DC + 0.1I DC + 0.1 I DC )( 3) b. P = 1.7 I DC ( 3) = (1.7 )( 50 )( 3) ⇒ P = 255 μ W 17.18 (a) vI = 0 ⇒ V1 = 0.7 V 3.3 − 0.7 = 0.433 mA 6 iB = iC = 0 i1 = vo = 3.3 V (b) vI = 3.3 V v1 = 0.7 + 0.8 = 1.5 V 3.3 − 1.5 i1 = = 0.3 mA 6 0.8 iR = = 0.04 mA 20 iB = 0.3 − 0.04 = 0.26 mA 3.3 − 0.1 iC = = 0.8 mA 4 vo = 0.1 V 17.19 i. i1 = For v X = vY = 0.1 V ⇒ v ′ = 0.8 V 5 − 0.8 ⇒ i1 = 0.525 mA 8 i3 = i4 = 0 ii. For v X = vY = 5 V, v′ = 0.8 + 0.7 + 0.7 ⇒⇒ v′ = 2.2 V 5 − 2.2 ⇒ i1 = 0.35 mA 8 0.8 ⇒ i4 = 0.297 mA i4 = i1 − 15 5 − 0.1 i3 = ⇒ i3 = 2.04 mA 2.4 i1 = 784. 17.20 a. For v X = vY = 5 V , both Q1 and Q2 driven into saturation. v1 = 0.8 + 0.7 + 0.8 ⇒ v1 = 2.3 V 5 − 2.3 ⇒ i1 = iB1 = 0.675 mA 4 5 − (0.8 + 0.7 + 0.1) i2 = ⇒ i2 = 1.7 mA 2 i4 = iB1 + i2 ⇒ i4 = 2.375 mA i1 = 0.8 ⇒ i5 = 0.08 mA 10 = i4 − i5 ⇒ iB 2 = 2.295 mA i5 = iB 2 i3 = 5 − 0.1 ⇒ i3 = 1.225 mA 4 v0 = 0.1V b. ′ iL 5 − (0.1 + 0.7) = 1.05 mA 4 ′ iC (max) = β iB 2 = NiL + i3 = (20)(2.295) = N (1.05) + 1.225 So N = 42 17.21 DX and DY off, Q1 forward active mode v1 = 0.8 + 0.7 + 0.7 = 2.2 V 5 = i1 R1 + i2 R2 + v1 and i1 = (1 + β )i2 So 5 − 2.2 = i2 [ (1 + β ) R1 + R2 ] Assume β = 25 5 − 2.2 i2 = ⇒ i2 = 0.0589 mA (26)(1.75) + 2 i1 = (1 + β )i2 = (26)(0.05895) ⇒ i1 = 1.53 mA i3 = β i2 ⇒ i3 = 1.47 mA 0.8 = 0.0589 + 1.47 − 0.16 ⇒ 5 iBo = 1.37 mA Qo in saturation iBo = i2 + i3 − iCo = 5 − 0.1 ⇒ iCo = 0.817 mA 6 17.22 vI = 0 V, Q1 forward actions (a) 5 − 0.7 iB = = 0.717 mA 6 iC = (25)(0.71667) = 17.9 mA iE = (26)(0.71667) = 18.6 mA 785. VI = 0.8 V (b) 5 − (0.8 + 0.7) = 0.583 mA 6 Because of the relative doping levels of the Emitter and collector, and because of the difference in B-C and B-E areas, we have −iC ≈ iB = 0.583 mA and iE = small value. (c) vI = 3.6 Q1 inverse active. 5 − (0.8 + 0.7) iB = = 0.583 mA 6 iE = − β R iE = −(0.5)(0.583) = −0.292 mA iC = −iB − iE = −0.583 − 0.292 ⇒ iC = −0.875 mA iB = 17.23 a. i1 = i. v X = vY = 0.1 V, so Q1 in saturation. 5 − (0.1 + 0.8) ⇒ i1 = 0.683 mA 6 ⇒ iB 2 = i2 = i4 = iB 3 = i3 = 0 v X = vY = 5 V, so Q1 in inverse active mode. ii. Assume Q2 and Q3 in saturation. 5 − (0.8 + 0.8 + 0.7) ⇒ i1 = iB 2 = 0.45 mA 6 5 − (0.8 + 0.1) ⇒ i2 = 2.05 mA i2 = 2 0.8 ⇒ i4 = 0.533 mA i4 = 1.5 i1 = iB 3 = ( iB 2 + i2 ) − i4 = 0.45 + 2.05 − 0.533 or iB 3 = 1.97 mA i3 = b. 5 − 0.1 ⇒ i3 = 2.23 mA 2.2 For Q3 : i3 2.23 = = 1.13 < β iB 3 1.97 For Q2 : i2 2.05 = = 4.56 < β iB 2 0.45 Since ( I C / I B ) < β , then each transistor is in saturation. 17.24 (a) v X = vY = Logic 1 786. v ′ = 0.8 + 2 ( 0.7 ) = 2.2 V 5 − 2.2 i1 = = 0.35 mA 8 0.8 i4 = i1 − = 0.35 − 0.0533 = 0.2967 mA 15 5 − 0.1 i3 = = 2.04 mA 2.4 5 − ( 0.1 + 0.7 ) ′ iL = = 0.525 mA 8 Assume β = 25 Then ( 25 )( 0.2967 ) = 2.04 + N ( 0.525 ) So N = 10.2 ⇒ N = 10 (b) Now 5 = 2.04 + N ( 0.525 ) So N = 5.64 ⇒ N = 5 17.25 a. iB1 = iB 2 For v X = vY = 5 V, Q, in inverse active mode. 5 − ( 0.8 + 0.8 + 0.7 ) = 0.45 mA 6 = iB1 + 2 β R iB1 = 0.45(1 + 2 [ 0.1]) = 0.54 mA iC 2 = 5 − ( 0.8 + 0.1) 2 iB 3 = ( iB 2 + iC 2 ) − = 2.05 mA 0.8 = 0.54 + 2.05 − 0.533 1.5 or iB 3 = 2.06 mA Now 5 − (0.1 + 0.8) = 0.683 mA 6 Then ′ iC 3 (max) = β F iB 3 = NiL or (20)(2.06) = N (0.683) ⇒ N = 60 ′ iL = b. ′ From above, for v0 high, I L = (0.1)(0.45) = 0.045 mA. Now ⎛ 5 − 4.9 ⎞ (21)(0.1) ′ I L (max) = (1 + β F ) ⎜ ⎟ = R2 ⎠ 2 ⎝ = 1.05 mA So ′ I L (max) = NI L or 1.05 = N (0.045) ⇒ N = 23 17.26 787. Vin = 0.1V : QI , Sat : Qs , Qo , Cutoff (a) 5 − (0.1 + 0.8) = 1.025 mA 4 P = iI (5 − 0.1) = (1.025)(4.9) ⇒ iI = P = 5.02 mW Vin = 5V , QI , Inverse Active; QS , Qo , Saturation (b) vBI = 0.7 + 0.8 + 0.7 = 2.2V 5 − 2.2 = 0.7 mA 4 iEI = β R ⋅ iI = (0.1)(0.7) = 0.07 mA iI = Vout = 0.7 + 0.1 = 0.8 V 5 − 0.8 i2 = = 4.2 mA 1 P = (iI + iEI + i2 )(5) = (0.7 + 0.07 + 4.2)(5) ⇒ P = 24.9 mW 17.27 a. iB1 = v X = vY = vZ = 0.1 V 5 − (0.1 + 0.8) ⇒ iB1 = 1.05 mA 3.9 Then iC1 = iB 2 = iC 2 = iB 3 = iC 3 = 0 v X = vY = vZ = 5 V b. iB1 = 5 − (0.8 + 0.8 + 0.7) ⇒ iB1 = 0.692 mA 3.9 Then iC1 = iB 2 = iB1 (1 + 3β R ) = (0.692)(1 + 3[0.5]) ⇒ iC1 = iB 2 = 1.73 mA 5 − (0.1 + 0.8) ⇒ iC 2 = 2.05 mA 2 0.8 = iB 2 + iC 2 − = 1.73 + 2.05 − 1.0 0.8 ⇒ iB 3 = 2.78 mA iC 2 = iB 3 5 − 0.1 = 2.04 mA 2.4 5 − (0.1 + 0.8) ′ iL = = 1.05 mA 3.9 ′ iC 3 = iR 3 + 5iL = 2.04 + (5)(1.05) ⇒ iC 3 = 7.29 mA iR 3 = 17.28 a. v X = vY = vZ = 2.8 V, Q1 biased in the inverse active mode. 788. 2.8 − (0.8 + 0.8 + 0.7) ⇒ iB1 = 0.25 mA 2 = iB1 (1 + 3β R ) = 0.25(1 + 3 [0.3]) ⇒ iB 2 = 0.475 mA iB1 = iB 2 vC 2 = 0.8 + 0.1 = 0.9 V 0.9 − (0.7 + 0.1) 0.1 = (1 + β F )(0.5) (101)(0.5) iB 4 = iR 2 = 0.00198 mA (Negligible) 5 − 0.9 = = 4.56 mA 0.9 ⇒ iC 2 = 4.56 mA 0.8 = 0.475 + 4.56 − 0.8 1 = 4.235 mA iB 3 = iB 2 + iC 2 − ⇒ iB 3 v X = vY = vZ = 0.1 V b. 5 − (0.1 + 0.8) ⇒ iB1 = 2.05 mA 2 From part (a), ′ iL = β R ⋅ iB1 = (0.3)(0.25) = 0.075 mA Then ′ 5iL 5(0.075) iB 4 = = ⇒ iB 4 = 0.00371 mA 1+ βF 101 iB1 = 17.29 a. v X = vY = vZ = 0.1 V 2 − (0.1 + 0.8) + iB 3 RB1 where (2 − 0.7) − (0.9) 0.4 iB 3 = = RB 2 1 iB1 = ⇒ iB 3 = 0.4 mA Then iB1 = 1.1 + 0.4 ⇒ iB1 = 1.5 mA 1 iB 2 = 0 = iC 2 ′ Q3 in saturation iC 3 = 5iL For v0 high, ′ ′ vB1 = 0.8 + 0.7 = 1.5 V ⇒ Q3 off 2 − 1.5 = 0.5 mA 1 ′ ′ iL = β R iB1 = (0.2)(0.5) = 0.1 mA ′ iB1 = Then 789. iC 3 = 0.5 mA v X = vY = vZ = 2 V b. From part (a), ⇒ iB1 = 0.5 mA iB 3 = 0 = iC 3 iB 2 = iB1 (1 + 3β R ) = (0.5)(1 + 3 [0.2]) iB 2 = 0.8 mA ′ ′ iC 2 = 5iL , and from part (a), iL = 1.5 mA So iC 2 = 7.5 mA 17.30 5.8 − 0.7 = 0.51 mA 10 5 − (0.7 − 0.3) = 4.6 mA IC − I D = 1 Now I I I D = 0.51 − I B = 0.51 − C = 0.51 − C 50 β Then I ⎞ 1 ⎞ ⎛ ⎛ I C − I D = I C − ⎜ 0.51 − C ⎟ = I C ⎜ 1 + ⎟ − 0.51 = 4.6 50 ⎠ ⎝ 50 ⎠ ⎝ So I C = 5.01 mA IB + ID = (a) IC 5.01 ⇒ I B = 0.1002 mA 50 I D = 0.51 − 0.1002 ⇒ I D = 0.4098 mA IB = β = VCE = 0.4 V I D = 0, VCE = VCE ( sat ) = 0.1 V (b) 5.8 − 0.8 ⇒ I B = 0.5 mA 10 5 − 0.1 ⇒ I C = 4.9 mA IC = 1 IB = 17.31 a. v X = vY = 0.4 V vB1 = 0.4 + 0.7 ⇒ vB1 = 1.1 V 5 − 1.1 ⇒ iB1 = 1.39 mA 2.8 = 0.4 + 0.4 ⇒ vB 2 = 0.8 V iB1 = vB 2 iB 2 = iC 2 = iB 0 = iC 0 = iB 5 = iC 5 = iB 3 = iC 3 = 0 ( No load ) 790. 5 = iB 4 R2 + VBE + (1 + β )iB 4 R4 5 − 0.7 ⇒ iB 4 = 0.0394 mA 0.76 + (31)(3.5) = β F iB 4 ⇒ iC 4 = 1.18 mA iB 4 = iC 4 vB 4 = 5 − (0.0394)(0.76) ⇒ vB 4 = 4.97 V v X = vY = 3.6 V b. vB1 = 0.7 + 0.7 + 0.3 ⇒ vB1 = 1.7 V vB 2 = 1.4 V vB 0 = 0.7 V vC 2 = 1.1 V 5 − 1.7 ⇒ iB1 = 1.1786 mA 2.8 = iB1 (1 + 2 β R ) = 1.18(1 + 2 [0.1]) iB1 = iB 2 iB 2 = 1.41 mA 1.1 − 0.7 ⇒ iB 4 = 0.00369 mA (31)(3.5) 5 − 1.1 = = 5.13 mA ⇒ iC 2 ≈ 5.13 mA 0.76 ≈ iB 2 + iC 2 iB 4 = iR 2 iB 0 iB 0 = 6.54 mA 17.32 (a) vI = 0, v1 = 0.3 V 1.5 − 0.3 = 1.2 mA 1 iB = iC = 0, vO = 1.5 V i1 = (b) vI = 1.5 V v1 = 0.7 + 0.3 = 1 V 1.5 − 1 = 0.5 mA 1 0.7 iR = = 0.035 mA 20 iB = 0.5 − 0.035 = 0.465 mA i1 = 1.5 − 0.4 = 0.917 mA 1.2 vO = 0.4 V iC = 17.33 a. Assuming the output transistor Q2 is a Schottky transistor, then ′ v0 = 0.4 V, iL = Then 2.5 − (0.4 + 0.3) = 0.5 RB1 791. RB1 = 3.6 kΩ Then iB1 = 2.5 − (0.7 + 0.8) 1.0 = = 0.278 mA 3.6 RB1 0.7 = 1.50 mA 0.7 iE1 = iB1 + iC1 ⇒ iC1 = 1.50 − 0.278 = 1.222 mA iB 2 = 0.5 mA, iE1 = 0.5 + and iC1 = 2.5 − (0.7 + 0.1) = 1.222 mA RC1 ⇒ RC1 = 1.39 kΩ b. v X = vY = 0.4 V, vB1 = 0.7 V vC 2 = 2.5 − 0.7 ⇒ vC 2 = 1.8 V All transistor currents are zero. c. vB1 = 1.5 V, vC1 = 0.8 V Currents calculated in part (a). ′ iB 2 = 0.5 mA, iL = 0.5 mA d. ′ iC 2 (max) = β iB 2 = NiL or (50)(0.5) = N (0.5) So N = 50 17.34 a. For v X = vY = 3.6 V 5 − 2.1 = 0.29 mA 10 5 − 1.8 vC1 = 0.7 + 0.7 + 0.4 = 1.8 V ⇒ iC1 = = 0.32 mA 10 1.4 iB 2 = iB1 + iC1 − = 0.29 + 0.32 − 0.0933 15 So iB 2 = 0.517 mA vC 2 = 0.7 + 0.4 = 1.1 V 5 − 1.1 iC 2 = = 0.951 mA 4.1 vB1 = 3(0.7) = 2.1 ⇒ iB1 = 0.7 = 0.517 + 0.951 − 0.175 4 or iB 5 = 1.293 mA ′ For v0 = 0.4 V, vB1 = 0.4 + 0.7 = 1.1 V Then 1.1 − 0.7 ′ iB1 = = 0.00086 mA (31)(15) 5 − 1.1 ′ ′ iL = − 0.00086 or iL ≈ 0.39 mA 10 ′ So iC 5 (max) = β iΒ 5 = NiL iB 5 = iB 2 + iC 2 − (30)(1.293) = N (0.39) ⇒ N = 99 792. b. P = (0.29 + 0.32 + 0.951)(5) + (99)(0.39)(0.4) P = 7.805 + 15.444 or P = 23.2 mW (Assumming 99 load circuits which is unreasonably large.) 17.35 a. Assume no load. For v X = logic 0 = 0.4 V 5 − (0.4 + 0.7) iE1 = = 0.0975 mA 40 Essentially all of this current goes to ground from VCC . P = iE1 ⋅ VCC = (0.0975)(5) ⇒ P = 0.4875 mW 5 − (3)(0.7) = 0.0725 mA 40 5 − (0.7 + 0.7 + 0.4) = 0.064 mA iR 2 = 50 5 − (0.7 + 0.4) = 0.26 mA iR 3 = 15 P = (0.0725 + 0.064 + 0.26)(5) P = 1.98 mW b. iR1 = c. For v0 = 0, vC 7 = 0.7 + 0.4 = 1.1 V iR 7 = 17.36 (a) 5 − 1.1 ⇒ iR 7 = 78 mA ≈ iSC 0.050 vl = vO = 25 V ; A transient situation vDS ( M N ) = 2.5 − 0.7 = 1.8 V vGS ( M N ) = 2.5 − 0.7 = 1.8 V ⇒ M N in saturation vSD ( M P ) = 5 − (2.5 + 0.7) = 1.8 V vSG ( M P )5 − 2.5 = 2.5 V ⇒ M P in saturation iDN = K n (vGSN − VTN ) 2 = (0.1)(1.8 − 0.8) 2 ⇒ iDN = 0.1 mA iDP = K P (vSGP + VTP ) 2 = (0.1)(2.5 − 0.8) 2 ⇒ iDP = 0.289 mA iC1 = β iDP = (50)(0.289) ⇒ iC1 = 14.45 mA iC 2 = β iDN = (50)(0.1) ⇒ iC 2 = 5 mA Difference between iE1 and iDN + iC 2 is a load current. (b) Assume iC1 = 14.45 mA is a constant i ⋅t (V )(C ) 1 VC = ∫ iC1dt = C1 ⇒ t = C C C iC1 t= (c) (5)(15 × 10−12 ) ⇒ t = 5.19 ns 14.45 × 10−3 t= (5)(15 × 10−12 ) ⇒ t = 260 ns 0.289 × 10−3 793. 17.37 (a) Assume R1 = R2 = 10 kΩ; β = 50 0.7 = 0.07 mA 10 NMOS in saturation region; vGSN = 2.5 − 0.7 = 1.8 V Then iR1 = iR 2 = iDN = K n ( vGSN − VTN ) = ( 0.1)(1.8 − 0.8 ) iDN = 0.10 mA 2 2 Then iB 2 = 0.03 ⇒ iC 2 = (50)(0.03) = 1.5 mA iE1 = 1.53 mA ⇒ iB1 = 0.03 mA ⇒ iC1 = 1.5 mA So iDP = 0.10 mA Now, M P biased in non-saturation region vSGP = 2.5 V 2 iDP = 0.10 = 0.10 ⎡ 2(2.5 − 0.8)vSD − vSD ⎤ ⎣ ⎦ 2 0.10 vSD − 0.34 vSD + 0.10 = 0 vSD = 0.34 ± (0.34) 2 − 4(0.10)(0.10) 2(0.10) 794. Or vSD = 0.325 V Then vo = 5 − 0.325 − 0.7 vo = 3.975 V 1 i ⋅t ∫ idt = C C Cv (15 × 10−12 )(5) t= = i 1.53 × 10−3 t = 49 ns (c) Cv (15 × 10−12 )(5) t= = i 0.1× 10−3 t = 0.75 μ s (b) v= 795. Ramey Soft Innovation with Excellence. Date:- ______________ There’s more to a product than just code. 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