Microelectronic Circuit Design 3rd Soln Edition by R. Jaeger

April 4, 2018 | Author: Anonymous | Category: Documents
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1.1 Answering machine Alarm clock Automatic door Automatic lights ATM Automobile: Engine controller Temperature control ABS Electronic dash Navigation system Automotive tune-up equipment Baggage scanner Bar code scanner Battery charger Cable/DSL Modems and routers Calculator Camcorder Carbon monoxide detector Cash register CD and DVD players Ceiling fan (remote) Cellular phones Coffee maker Compass Copy machine Cordless phone Depth finder Digital Camera Digital watch Digital voice recorder Digital scale Digital thermometer Electronic dart board Electric guitar Electronic door bell Electronic gas pump Elevator Exercise machine Fax machine Fish finder Garage door opener GPS Hearing aid Invisible dog fences Laser pointer LCD projector Light dimmer Keyboard synthesizer Keyless entry system Laboratory instruments Metal detector Microwave oven Model airplanes MP3 player Musical greeting cards Musical tuner Pagers Personal computer Personal planner/organizer (PDA) Radar detector Broadcast Radio (AM/FM/Shortwave) Razor Satellite radio receiver Security systems Sewing machine Smoke detector Sprinkler system Stereo system Amplifier CD/DVD player Receiver Tape player Stud sensor Talking toys Telephone Telescope controller Thermostats Toy robots Traffic light controller TV receiver & remote control Variable speed appliances Blender Drill Mixer Food processor Fan Vending machines Video game controllers Wireless headphones & speakers Wireless thermometer Workstations Electromechanical Appliances* Air conditioning and heating systems Clothes washer and dryer Dish washer Electrical timer Iron, vacuum cleaner, toaster Oven, refrigerator, stove, etc. *These appliances are historically based only upon on-off (bang-bang) control. However, many of the high end versions of these appliances have now added sophisticated electronic control. 1-1 ©R. C. Jaeger & T. N. Blalock 6/9/06 1.2 B = 19.97 x 100.1997(2020−1960) = 14.5 x 1012 = 14.5 Tb/chip 1.3 (a) 0.1977(Y2 −1960) B2 19.97 x10 0.1977(Y2 −Y1 ) 0.1977(Y2 −Y1 ) = = 10 so 2 = 10 0.1977(Y1 −1960) B1 19.97 x10 Y2 − Y1 = log2 = 1.52 years 0.1977 (b) Y2 − Y1 = log10 = 5.06 years 0.1977 1.4 N = 1610 x10 1.5 0.1548(2020−1970) = 8.85 x 1010 transistors/μP (2 ) N 2 1610 x10 0.1548(Y2 −Y1 ) = = 10 0.1548(Y1 −1970) N1 1610 x10 log2 ( a ) Y2 − Y1 = = 1.95 years 0.1548 log10 ( b ) Y2 − Y1 = = 6.46 years 0.1548 0.1548 Y −1970 1.6 F = 8.00 x10 −0.05806(2020−1970) μm = 10 nm . No, this distance corresponds to the diameter of only a few atoms. Also, the wavelength of the radiation needed to expose such patterns during fabrication is represents a serious problem. 1.7 From Fig. 1.4, there are approximately 600 million transistors on a complex Pentium IV microprocessor in 2004. From Prob. 1.4, the number of transistors/μP will be 8.85 x 1010. in 2020. Thus there will be the equivalent of 8.85x1010/6x108 = 148 Pentium IV processors. 1-2 ©R. C. Jaeger & T. N. Blalock 6/9/06 1.8 1.5W tube)= 113 MW! P = 75 x106 tubes ( 1.9 1.10 ( ) I= 1.13 x 108W = 511 kA! 220V D, D, A, A, D, A, A, D, A, D, A 10.24V 10.24V 10.24V = = 2.500 mV VMSB = = 5.120V 12 2 2 bits 4096 bits 1001001001102 = 211 + 28 + 25 + 22 + 2 = 234210 VO = 2342(2.500 mV )= 5.855 V VLSB = 1.11 VLSB = 5V mV 5V = = 19.53 bit 2 bits 256 bits 8 and 128 + 8 + 4 + 2) = 100011102 14210 = ( 10 2.77V = 142 LSB mV 19.53 bit 1.12 VLSB = 2.5V 2.5V mV = = 2.44 bit 2 bits 1024 bits 10 01011011012 = 28 + 26 + 25 + 23 + 22 + 20 ( ) 10 = 36510 ⎛ 2.5V ⎞ VO = 365 ⎜ ⎟ = 0.891 V ⎝ 1024 ⎠ 1.13 mV 6.83V 14 10V = 0.6104 and 2 bits = 11191 bits 14 10V bit 2 bits 1119110 = (8192 + 2048 + 512 + 256 + 128 + 32 + 16 + 4 + 2 + 1) 10 VLSB = 1119110 = 101011101101112 1.14 ( ) A 4 digit readout ranges from 0000 to 9999 and has a resolution of 1 part in 10,000. The number of bits must satisfy 2B ≥ 10,000 where B is the number of bits. Here B = 14 bits. 1.15 VLSB = 5.12V mV V 5.12V = 1.25 and VO = ( 1011101110112 ) VLSB ± LSB = 12 bit 2 2 bits 4096 bits 11 9 8 7 5 4 3 VO = 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 1 1.25mV ± 0.0625V ( ) 10 VO = 3.754 ± 0.000625 or 3.753V ≤ VO ≤ 3.755V 1-3 6/9/06 1.16 IB = dc component = 0.002 A, ib = signal component = 0.002 cos (1000t) A 1.17 VGS = 4 V, vgs = 0.5u(t-1) + 0.2 cos 2000 t Volts 1.18 vCE = [5 + 2 cos (5000t)] V 1.19 vDS = [5 + 2 sin (2500t) + 4 sin (1000t)] V 1.20 V = 10 V, R1 = 22 kΩ, R2= 47 kΩ and R3 = 180 kΩ. + V 1 I2 R2 R V 1 + V I 3 2 R 3 - V1 = 10V V2 = 10V 22 kΩ + 47 kΩ 180 kΩ ( 22 kΩ ) = 10V 22 kΩ = 3.71 V 22 kΩ + 37.3kΩ 37.3kΩ = 6.29 V Checking : 6.29 + 3.71 = 10.0 V 22 kΩ + 37.3kΩ ⎛ ⎞ 180 kΩ 180 kΩ 10V I2 = I1 =⎜ = 134 μA ⎟ 47 kΩ + 180 kΩ ⎝ 22 kΩ + 37.3kΩ ⎠ 47 kΩ + 180kΩ ⎛ ⎞ 47kΩ 47kΩ 10V I3 = I1 =⎜ = 34.9 μA ⎟ 47 kΩ + 180kΩ ⎝ 22 kΩ + 37.3kΩ ⎠ 47 kΩ + 180kΩ 10V = 169μA and I1 = I2 + I3 22 kΩ + 37.3kΩ Checking : I1 = 1-4 ©R. C. Jaeger & T. N. Blalock 6/9/06 1.21 V = 18 V, R1 = 56 kΩ, R2= 33 kΩ and R3 = 11 kΩ. + V 1 I 2 R V 1 + V I3 R R 2 2 3 - V1 = 18V 56 kΩ 56 kΩ + 33kΩ 11kΩ 18V ( ) = 15.7 V V2 = 18V 33kΩ 11kΩ 56kΩ + 33kΩ 11kΩ ( ) = 2.31 V Checking : V1 + V2 = 15.7 + 2.31 = 18.0 V which is correct. I1 = 56kΩ + 33kΩ 11kΩ ( ) = 280 μA I2 = I1 11kΩ 11kΩ = (280 μA) = 70.0 μA 33kΩ + 11kΩ 33kΩ + 11kΩ Checking : I2 + I3 = 280 μA I3 = I1 33kΩ 33kΩ = (280 μA) = 210 μA 33kΩ + 11kΩ 33kΩ + 11kΩ 1.22 I1 = 5mA (5.6kΩ + 3.6kΩ) = 3.97 mA (5.6kΩ + 3.6kΩ)+ 2.4kΩ I2 = 5mA 2.4 kΩ = 1.03 mA 9.2 kΩ + 2.4 kΩ V3 = 5mA 2.4 kΩ 9.2 kΩ ( kΩ = 3.72V )5.6k3.6 Ω + 3.6 kΩ and Checking : I1 + I2 = 5.00 mA 1.23 I2 R2 = 1.03mA(3.6 kΩ)= 3.71 V 150 kΩ 150 kΩ = 125 μA I3 = 250μA = 125 μA 150 kΩ + 150 kΩ 150 kΩ + 150 kΩ 82 kΩ V3 = 250μA 150 kΩ 150 kΩ = 10.3V 68 kΩ + 82 kΩ Checking : I1 + I2 = 250 μA and I2 R2 = 125μA(82 kΩ)= 10.3 V I2 = 250μA ( ) 1-5 6/9/06 1.24 + v v R 1 s + g v m v th - Summing currents at the output node yields: v + .002v = 0 so v = 0 and v th = vs − v = vs 5x10 4 + v R 1 g v m ix vx Summing currents at the output node : ix = − v − 0.002v = 0 but v = −vx 5 x10 4 vx v 1 ix = + 0.002vx = 0 Rth = x = = 495 Ω 4 1 ix 5 x10 + gm R1 Thévenin equivalent circuit: 495 Ω v s 1-6 ©R. C. Jaeger & T. N. Blalock 6/9/06 1.25 The Thévenin equivalent resistance is found using the same approach as Problem 1.24, and ⎛ 1 ⎞−1 Rth = ⎜ + .025⎟ = 39.6 Ω ⎝ 4 kΩ ⎠ + v vs R 1 g v m in The short circuit current is : v in = + 0.025v and v = vs 4kΩ v i n = s + 0.025vs = 0.0253vs 4kΩ Norton equivalent circuit: 0.0253v s 39.6 Ω 1-7 6/9/06 1.26 (a) βi vs R 1 + R2 v th - i Vth = Voc = −β i R2 but i =− vs R1 and ix Vth = β vs R2 39 kΩ = 120 vs = 46.8 vs R1 100 kΩ βi R 1 R2 i Rth v x Rth = vx ; ix ix = vx + βi R2 but i = 0 since VR 1 = 0. Rth = R2 = 39 kΩ. Thévenin equivalent circuit: 39 k Ω 58.5v s (b) βi i s + R2 v th - R 1 i ⎛ i ⎞ Vth = Voc = −β i R2 where i + bi + is = 0 and Vth = −β ⎜ − s ⎟ R2 = 38700 is ⎝ β + 1⎠ 1-8 ©R. C. Jaeger & T. N. Blalock 6/9/06 βi R 1 R2 i Rth v x Rth = vx ; ix ix = vx + βi but R2 i + βi = 0 so i = 0 and Rth = R2 = 39 kΩ Thévenin equivalent circuit: 39 k Ω 38700i s 1.27 βi vs R 1 R2 i in in = −β i but i = − vs R1 and in = β R1 vs = 100 vs = 1.33 x 10−3 vs 75kΩ Norton equivalent circuit: From problem 1.26(a), Rth = R2 = 56 kΩ. 0.00133v s 56 k Ω 1-9 6/9/06 1.28 is v s βi R 1 R2 i is = vs v v β +1 − β i = s + β s = vs R1 R1 R1 R1 R= vs R 100kΩ = 1 = = 1.24 Ω is β + 1 81 1.29 The open circuit voltage is vth = −g mv R2 and v = +is R1. vth = − g m R1 R2is = −(0.0025) 105 106 i s= 2.5 x 108 is For is = 0, v = 0 and Rth = R2 = 1 MΩ ( )( ) 1.30 5V 3V f (Hz) 0 0 1.31 500 1000 2V f (kHz) 0 9 10 11 4 cos(20000πt + 2000πt )+ cos(20000πt − 2000πt ) 2 v = 2cos(22000πt )+ 2cos( 18000πt ) 1.32 v = 4sin (20000πt )sin (2000πt )= [ ] 2∠36 o A = −5 0 = 2 x105 ∠36 o 10 ∠0 A = 2 x105 ∠A = 36o 1-10 ©R. C. Jaeger & T. N. Blalock 6/9/06 1.33 (a) A = 1.34 10−1∠ − 12 o 10−2 ∠ − 45o o A = = 5 ∠ − 45 = 100∠ − 12 o (b) 2 x10−3 ∠0 o 10−3 ∠0 o R2 620 kΩ 180 kΩ =− = −44.3 (b) Av = − = −10.0 14kΩ R1 18 kΩ 62 kΩ = −38.8 1.6 kΩ (a) Av = − 1.35 vo (t ) = − (c) Av = − R2 v s (t )= (−90.1 sin 750πt ) mV R1 IS = VS 0.01V = = 11.0μA and R1 910Ω is = ( 11.0 sin 750πt ) μA 1.36 Since the voltage across the op amp input terminals must be zero, v- = v+ and vo = vs. Therefore Av = 1. 1.37 Since the voltage across the op amp input terminals must be zero, v- = v+ = vs. Also, i- = 0. v− − vo v + i− + − = 0 R2 R1 or vs − v o vs + =0 R2 R1 and A v = vo R = 1+ 2 vs R1 1.38 Writing a nodal equation at the inverting input terminal of the op amp gives v −v v1 − v− v2 − v− + = i− + − o R1 R2 R3 vo = − but v- = v+ = 0 and i- = 0 R3 R v1 − − 3 v2 = −0.255sin 3770t − 0.255sin10000t volts R1 R2 1-11 6/9/06 1.39 ⎛b b b ⎞ ⎛ 0 1 1⎞ ⎛ 1 0 0⎞ vO = −VREF ⎜ 1 + 2 + 3 ⎟ (a) vO = −5⎜ + + ⎟ = −1.875V (b) vO = −5⎜ + + ⎟ = −2.500V ⎝2 4 8⎠ ⎝ 2 4 8⎠ ⎝ 2 4 8⎠ b1b2b 3 vO (V) 0 -0.625 -1.250 -1.875 -2.500 -3.125 -3.750 -4.375 000 001 010 011 100 101 110 111 1.40 Low-pass amplifier Amplitude 10 f 6 kHz 1-12 ©R. C. Jaeger & T. N. Blalock 6/9/06 1.41 Band-pass amplifier Amplitude 20 f 1 kHz 5 kHz 1.42 High-pass amplifier Amplitude 16 f 10 kHz 1.43 15000πt ) vO (t ) = 10 x 5sin (2000πt )+ 10 x 3cos(8000πt )+ 0 x 3cos( vO (t ) = 50sin(2000πt )+ 30cos(8000πt ) volts [ ] 1.44 12000πt ) vO (t ) = 20 x 0.5sin (2500πt )+ 20 x 0.75cos(8000πt )+ 0 x 0.6cos( vO (t ) = 10.0sin (2500πt )+ 15.0cos(8000πt ) volts [ ] 1.45 The gain is zero at each frequency: vo(t) = 0. 1-13 6/9/06 1.46 t=linspace(0,.005,1000); w=2*pi*1000; v=(4/pi)*(sin(w*t)+sin(3*w*t)/3+sin(5*w*t)/5); v1=5*v; v2=5*(4/pi)*sin(w*t); v3=(4/pi)*(5*sin(w*t)+3*sin(3*w*t)/3+sin(5*w*t)/5); plot(t,v) plot(t,v1) plot(t,v2) plot(t,v3) 2 1 0 -1 -2 0 1 2 3 4 (a) 10 5 0 -5 -10 0 5 x10-3 1 2 3 4 (b) 5 x10-3 1-14 ©R. C. Jaeger & T. N. Blalock 6/9/06 10 5 0 -5 -10 0 1 2 3 4 (c) 5 x10-3 10 5 0 -5 -10 0 (d) 1.47 1 2 3 4 5 x10-3 ( c ) 3000( 1 − .10) ≤ R ≤ 3000( 1 + .10) or 2700Ω ≤ R ≤ 3300Ω Vnom = 2.5V ΔV ≤ 0.05V 0.05 = 0.0200 or 2.00% 2.50 ( b ) 3000( 1 − .05)≤ R ≤ 3000( 1 + .05) or 2850Ω ≤ R ≤ 3150Ω ( a ) 3000( 1 − .01)≤ R ≤ 3000( 1 + .01) or 2970Ω ≤ R ≤ 3030Ω 1.48 T= 1.49 1.50 1 − .5)≤ C ≤ 20000μF ( 1 + .2) or 10000μF ≤ R ≤ 24000μF 20000μF ( 8200( 1 − 0.1)≤ R ≤ 8200( 1 + 0.1) or 7380Ω ≤ R ≤ 9020Ω The resistor is within the allowable range of values. 1-15 6/9/06 1.51 (a) 5V ( 1 − .05)≤ V ≤ 5V ( 1 + .05) or 5.75V ≤ V ≤ 5.25V V = 5.30 V exceeds the maximum range, so it is out of the specification limits. (b) If the meter is reading 1.5% high, then the actual voltage would be 5.30 = 5.22V which is within specifications limits. Vmeter = 1.015Vact or Vact = 1.015 1.52 TCR = R nom ΔR 6562 − 6066 Ω = = 4.96 o ΔT 100 − 0 C = R o + TCR (ΔT)= 6066 + 4.96(27)= 6200Ω 0 C 1-16 ©R. C. Jaeger & T. N. Blalock 6/9/06 1.53 + R V 1 V1 I2 R2 + V2 I3 R 3 - Let RX = R2 R3 then V1 = V R1 = R1 + RX R min X = V1max = 47kΩ(0.9)+ 180 kΩ(0.9) 10( 1.05) 33.5kΩ 1+ 22 kΩ( 1.1) = 4.40V 180 kΩ)(0.9) 47 kΩ(0.9)( V1 R 1+ X R1 = 33.5kΩ R max = X V1min = 10(0.95) 47 kΩ( 1.1)+ 180 kΩ( 1.1) = 3.09V 180 kΩ)( 1.1) 47 kΩ( 1.1)( = 41.0 kΩ 41.0 kΩ 1+ 22 kΩ(0.9) V R1 R2 R3 I1 = V R1 + RX and I2 = I1 R3 = R2 + R3 R1 + R2 + I2max = 10( 1.05) 22000(0.9)+ 47000(0.9)+ 22000(0.9)(47000)(0.9) 180000( 1.1) = 158 μA I2min = 10(0.95) 22000( 1.1)+ 47000( 1.1)+ R2 = R2 + R3 V R1 + R3 + R1 R3 R2 1.1) 22000( 1.1)(47000)( 180000(0.9) = 114 μA I3 = I1 I3max = 10( 1.05) 22000(0.9)+ 180000(0.9)+ 180000)(0.9) 22000(0.9)( 47000( 1.1) = 43.1 μA I3min = 10(0.95) 22000( 1.1)+ 180000( 1.1)+ 180000)( 1.1) 22000( 1.1)( 47000(0.9) = 28.3 μA 1-17 6/9/06 1.54 I1 = I R2 + R3 =I R1 + R2 + R3 1 1+ R1 R2 + R3 and similarly I2 = I I1max = 1+ I2max = 1+ 5600( 1.05)+ 3600( 1.05) 5600(0.95)+ 3600(0.95) 2400( 1.05) I 1 1 R + + 2 R1 R3 R1 R3 5( 1.02) mA = 1.14 mA I2min = 2400(0.95) 5( 1.02) 1 R + R3 1+ 2 R1 mA = 4.12 mA I1min = 1+ 5600(0.95)+ 3600(0.95) 1.05) 5600( 1.05)+ 3600( 2400(0.95) 5(0.98) mA = 0.936 mA 2400( 1.05) 5(0.98) mA = 3.80 mA 1+ V3 = I2 R3 = V3max = 5600(0.95) 1 1 + + 2400( 1.05) 3600( 1.05) 2400( 1.05)(3600)( 1.05) 5(0.98) 5( 1.02) = 4.18 V V3min = 5600( 1.05) 1 1 + + 2400(0.95) 3600(0.95) 2400(0.95)(3600)(0.95) = 3.30 V 1.55 From Prob. 1.24 : Rth = 1 gm + 1 R1 = 619 Ω min Rth = max Rth = 1 1 0.002(0.8)+ 5 x10 4 ( 1.2) 1 1 0.002( 1.2)+ 5 x10 4 (0.8) = 412 Ω 1-18 ©R. C. Jaeger & T. N. Blalock 6/9/06 1.56 For one set of 200 cases using the equations in Prob. 1.53. R1 = 22000 * (0.9 + 0.2 * RAND()) V = 10 * (0.95 + 0.1* RAND()) R1 = 4700 * (0.9 + 0.2 * RAND()) R3 = 180000 * (0.9 + 0.2 * RAND()) V1 Min Max Average 3.23 V 3.71 V 3.71 V I2 116 μA 151 μA 133 μA I3 29.9 μA 40.9 μA 35.1 μA 1.57 For one set of 200 cases using the Equations in Prob. 1.54: R1 = 2400 * (0.95 + 0.1* RAND()) R1 = 5600 * (0.95 + 0.1* RAND()) R3 = 3600 * (0.95 + 0.1* RAND()) I1 Min Max Average 3.82 mA 4.09 mA 3.97 mA I2 0.96 mA 1.12 mA 1.04 mA V3 3.46 V 4.08 V 3.73 V I = 0.005* (0.98 + 0.04 * RAND()) 1.58 1.59 3.29, 0.995, -6.16; 3.295, 0.9952, -6.155 (a) (1.763 mA)(20.70 kΩ) = 36.5 V (b) 36 V (c) (0.1021 A)(97.80 kΩ) = 9.99 V; 10 V 1-19 6/9/06 CHAPTER 2 2.1 Based upon Table 2.1, a resistivity of 2.6 μΩ-cm < 1 mΩ-cm, and aluminum is a conductor. 2.2 Based upon Table 2.1, a resistivity of 1015 Ω-cm > 105 Ω-cm, and silicon dioxide is an insulator. 2.3 I max 2.4 ⎛ 10−8 cm2 ⎞ ⎛ 7 A ⎞ = ⎜10 1μm) ⎜ ⎟ = 500 mA ⎟(5μm)( 2 cm 2 ⎠ ⎝ ⎝ μm ⎠ EG ⎛ ⎞ ni = BT 3 exp⎜ − ⎟ −5 ⎝ 8.62 x10 T ⎠ 31 For silicon, B = 1.08 x 10 and EG = 1.12 eV: ni = 2.01 x10 /cm 30 3 -10 3 6.73 x10 /cm 2.27 x10 /cm 13 3 9 3 8.36 x 10 /cm . 8.04 x 10 /cm . 15 3 13 3 For germanium, B = 2.31 x 10 and EG = 0.66 eV: ni = 35.9/cm 2.5 Define an M-File: function f=temp(T) ni=1E14; f=ni^2-1.08e31*T^3*exp(-1.12/(8.62e-5*T)); ni = 10 /cm 2.6 14 3 for T = 506 K ni = 10 /cm3 for T = 739 K 16 ⎛ ⎞ EG ni = BT 3 exp⎜ − −5 ⎟ ⎝ 8.62 x10 T ⎠ T = 100 K: ni = 6.03 x 10-19/cm 3 with B = 1.27x1029 K −3cm−6 6 3 11 3 T = 300 K and EG = 1.42 eV: ni = 2.21 x10 /cm T = 500 K: ni = 2.79 x10 /cm 20 2.7 ⎛ cm2 ⎞⎛ V ⎞ 6 cm vn = −μn E = ⎜ −700 ⎟⎜ 2500 ⎟ = −1.75 x10 V − s ⎠⎝ cm ⎠ s ⎝ ⎛ V ⎞ cm2 ⎞⎛ 5 cm v p = +μ p E = ⎜ +250 ⎟⎜ 2500 ⎟ = +6.25 x10 cm ⎠ V − s ⎠⎝ s ⎝ ⎛ 1 ⎞⎛ cm ⎞ 4 A jn = − qnvn = −1.60 x10−19 C ⎜1017 3 ⎟⎜ −1.75 x106 ⎟ = 2.80 x10 s ⎠ cm ⎠⎝ cm2 ⎝ ⎛ 1 ⎞⎛ cm ⎞ −10 A j p = qnv p = 1.60 x10−19 C ⎜103 3 ⎟⎜ 6.25 x105 ⎟ = 1.00 x10 s ⎠ cm 2 ⎝ cm ⎠⎝ ( ) ( ) 2.8 ⎛ E ⎞ ni2 = BT 3 exp⎜ − G ⎟ ⎝ kT ⎠ 10 2 31 B = 1.08 x1031 (10 ) = 1.08 x10 ⎛ ⎞ 1.12 T 3 exp⎜ − ⎟ ⎝ 8.62 x10−5 T ⎠ Using a spreadsheet, solver, or MATLAB yields T = 305.22K Define an M-File: function f=temp(T) f=1e20-1.08e31*T^3*exp(-1.12/(8.62e-5*T)); Then: fzero('temp',300) | ans = 305.226 K 2.9 v= j − 1000 A / cm 2 cm = = − 105 2 Q s 0.01C / cm 2.10 C ⎞⎛ cm ⎞ MA ⎛ 6 A j = Qv = ⎜ 0.4 3 ⎟⎜10 7 =4 2 ⎟ = 4 x10 2 cm ⎠⎝ sec ⎠ cm cm ⎝ 21 2.11 ⎛ V ⎞ cm2 ⎞⎛ 6 cm vn = −μn E = ⎜−1000 ⎟⎜ −2000 ⎟ = +2.00 x10 V − s ⎠⎝ cm ⎠ s ⎝ ⎛ V ⎞ cm 2 ⎞⎛ 5 cm v p = +μ p E = ⎜ +400 ⎟⎜ −2000 ⎟ = −8.00 x10 V − s ⎠⎝ cm ⎠ s ⎝ ⎛ 1 ⎞⎛ cm ⎞ −10 A jn = − qnvn = −1.60 x10−19 C ⎜103 3 ⎟⎜ +2.00 x106 ⎟ = −3.20 x10 s ⎠ cm2 ⎝ cm ⎠⎝ ⎛ 1 ⎞⎛ cm ⎞ 4 A j p = qnv p = 1.60 x10−19 C ⎜1017 3 ⎟⎜ −8.00 x105 ⎟ = −1.28 x10 s ⎠ cm ⎠⎝ cm2 ⎝ ( ) ( ) 2.12 (a ) 2.13 E= V 5V = 5000 −4 cm 10 x10 cm (b ) V ⎞ ⎛ −4 V = ⎜105 ⎟ 10 x10 cm = 100 V cm ⎝ ⎠ ( ) ⎛ 1019 ⎞⎛ cm ⎞ 7 A j p = qpv p = 1.60 x10−19 C ⎜ 3 ⎟⎜10 7 ⎟ = 1.60 x10 s ⎠ cm2 ⎝ cm ⎠⎝ ⎛ A ⎞ i p = j p A = ⎜1.60 x10 7 2 ⎟ 1x10−4 cm 25 x10−4 cm = 4.00 A cm ⎠ ⎝ ( ) ( )( ) 2.14 For intrinsic silicon, σ = q (μn ni + μ p ni )= qni (μn + μ p ) σ ≥ 1000(Ω − cm) for a conductor −1 ni ≥ cm 2 1.602 x10−19 C ( 100 + 50) v − sec 39 ⎛ ⎞ 1.73 x10 E n2 = BT 3 exp⎜ − G ⎟ with i = 6 cm ⎝ kT ⎠ q (μn + μ p ) σ = 1000(Ω − cm) −1 = 4.16 x1019 cm3 B = 1.08 x1031 K −3cm−6 , k = 8.62x10-5 eV/K and EG = 1.12eV This is a transcendental equation and must be solved numerically by iteration. Using the HP solver routine or a spread sheet yields T = 2701 K. Note that this temperature is far above the melting temperature of silicon. 22 2.15 For intrinsic silicon, σ = q (μn ni + μ p ni )= qni (μn + μ p ) σ ≤ 10−5 (Ω − cm) for an insulator −1 ni ≥ ⎛ cm 2 ⎞ 1.602 x10 C (2000 + 750) ⎜ ⎟ ⎝ v − sec ⎠ ⎛ E ⎞ 5.152 x1020 n2 = = BT 3 exp⎜ − G ⎟ with i 6 cm ⎝ kT ⎠ q (μn + μ p ) σ = 10−5 (Ω − cm) −1 ( −19 ) = 2.270 x1010 cm 3 B = 1.08 x1031 K −3cm−6 , k = 8.62x10-5 eV/K and EG = 1.12eV Using MATLAB as in Problem 2.5 yields T = 316.6 K. 2.16 Si Si Si Donor electron fills acceptor vacancy P B Si Si Si Si No free electrons or holes (except those corresponding to ni). 2.17 (a) Gallium is from column 3 and silicon is from column 4. Thus silicon has an extra electron and will act as a donor impurity. (b) Arsenic is from column 5 and silicon is from column 4. Thus silicon is deficient in one electron and will act as an acceptor impurity. 2.18 Since Ge is from column IV, acceptors come from column III and donors come from column V. (a) Acceptors: B, Al, Ga, In, Tl (b) Donors: N, P, As, Sb, Bi 23 2.19 (a) Germanium is from column IV and indium is from column III. Thus germanium has one extra electron and will act as a donor impurity. (b) Germanium is from column IV and phosphorus is from column V. Thus germanium has one less electron and will act as an acceptor impurity. 2.20 E= A ⎞ V ⎛ = jρ = ⎜10000 2 ⎟(0.02Ω − cm ) = 200 , a small electric field. cm ⎠ σ cm ⎝ j 2.21 ⎛ C ⎞⎛ cm ⎞ A jndrift = qnμn E = qnv n = 1.602 x10−19 1016 ⎜ 3 ⎟⎜10 7 ⎟ = 16000 2 s ⎠ cm ⎝ cm ⎠⎝ ( )( ) 2.22 ⎛ 1015 atoms ⎞ ⎛ 10−4 cm ⎞3 N =⎜ 10μm)(0.5μm ) 1μm)( ⎟( ⎜ ⎟ = 5,000 atoms 3 ⎝ cm ⎠ ⎝ μm ⎠ 2.23 N A > N D : N A − N D = 1015 − 1014 = 9 x1014 /cm3 If we assume N A − N D >> 2 ni = 1014 / cm 3 : p = N A − N D = 9 x1014 /cm3 | n = If we use Eq. 2.12 : p = 12 3 ni2 251026 = = 2.78 x1012 /cm3 p 9 x1014 9 x1014 ± 2 and n = 2.77 x10 /cm . The answers are essentially the same. (9 x10 ) + 4(5x10 ) = 9.03x10 14 2 13 2 14 2.24 N A > N D: N A − N D = 5 x1016 − 1016 = 4 x1016 /cm 3 >> 2ni = 2 x1011 /cm 3 p = N A − N D = 4 x1014 /cm 3 | n = ni2 10 22 = = 2.50 x10 5 /cm 3 p 4 x1016 2.25 N D > N A: N D − N A = 3 x1017 − 2 x1017 = 1x1017 /cm3 2ni = 2 x1017 /cm3 ; Need to use Eq. (2.11) n= p= 1017 ± 2 i ( ) ( ) = 1.62 x10 1017 + 4 1017 2 34 2 2 17 /cm3 n 10 = = 6.18 x1016 /cm3 n 1.62 x1017 24 2.26 N D − N A = −2.5 x1018 / cm 3 Using Eq. 2.11: n = −2.5 x1018 ± (−2.5x10 ) + 4(10 ) 18 2 10 2 2 Evaluating this with a calculator yields n = 0, and n = ni2 = ∞. p No, the result is incorrect because of loss of significant digits within the calculator. It does not have enough digits. 2.27 (a) Since boron is an acceptor, NA = 6 x 1018/cm3. Assume ND = 0, since it is not specified. The material is p-type. At room temperature, ni = 1010 /cm3 and N A − N D = 6 x1018 / cm3 >> 2n i ni2 10 20 /cm 6 So p = 6 x10 /cm and n = = = 16.7 /cm3 18 3 p 6 x10 /cm (b) ⎛ ⎞ 3 1.12 9 6 ⎟ At 200K, ni2 = 1.08 x1031 (200) exp⎜ − ⎜ 8.62 x10−5 (200)⎟ = 5.28 x10 /cm ⎝ ⎠ 18 3 ni = 7.27 x10 4 /cm 3 N A − N D >> 2 ni , so p = 6 x1018 /cm3 and n = 5.28 x109 = 8.80 x10−10 /cm 3 18 6 x10 2.28 (a) Since arsenic is a donor, ND = 3 x 1017/cm3. Assume NA = 0, since it is not specified. The material is n-type. At room temperature, n i = 1010 / cm 3 and N D − N A = 3 x1017 / cm 3 >> 2n i So n = 3 x1017 /cm 3 and p = 10 20 /cm 6 ni2 = = 333 /cm 3 17 3 n 3x10 /cm ⎛ ⎞ 3 1.12 15 6 ⎟ − (b) At 250K, ni2 = 1.08 x1031 (250) exp⎜ ⎜ 8.62 x10−5 (250)⎟ = 4.53 x10 /cm ⎝ ⎠ ni = 6.73 x10 7 /cm 3 N D − N A >> 2 ni , so n = 3 x1017 / cm 3 and n = 4.53x1015 = 0.0151/ cm 3 17 3x10 2.29 (a) Arsenic is a donor, and boron is an acceptor. ND = 2 x 1018/cm3, and NA = 8 x 1018/cm3. Since NA > ND, the material is p-type. 25 (b) At room temperature, n i = 1010 / cm3 and N A − N D = 6 x1018 / cm3 >> 2n i ni2 10 20 /cm 6 So p = 6 x10 /cm and n = = = 16.7 /cm3 18 3 p 6 x10 /cm 18 3 2.30 (a) Phosphorus is a donor, and boron is an acceptor. ND = 2 x 1017/cm3, and NA = 5 x 1017/cm3. Since NA > ND, the material is p-type. (b) At room temperature, ni = 1010 /cm3 and N A − N D = 3x1017 / cm3 >> 2n i ni2 10 20 /cm 6 So p = 3x10 /cm and n = = = 333 /cm3 17 3 p 3x10 /cm 17 3 2.31 ND = 4 x 1016/cm3. Assume NA = 0, since it is not specified. N D > N A : material is n - type | N D − N A = 4 x1016 / cm3 >> 2 ni = 2 x1010 / cm 3 n = 4 x1016 / cm3 | p = n2 1020 i = = 2.5 x103 / cm 3 16 n 4 x10 cm2 cm 2 and μp = 310 V −s V −s N D + N A = 4 x1016 / cm3 | Using Fig. 2.13, μn = 1030 ρ= 1 qμn n = ( ⎛ cm 2 ⎞⎛ 4 x1016 ⎞ 1.602 x10 C ⎜1030 ⎟⎜ ⎟ V − s ⎠⎝ cm 3 ⎠ ⎝ −19 1 ) = 0.152 Ω − cm 26 2.32 NA = 1018/cm3. Assume ND = 0, since it is not specified. N A > N D : material is p - type | N A − N D = 1018 / cm3 >> 2 ni = 2 x1010 / cm 3 p = 1018 / cm3 18 | n= 3 n2 1020 i = 18 = 100 / cm 3 p 10 cm2 cm 2 N D + N A = 10 / cm | Using Fig. 2.13, μn = 375 and μp = 100 V −s V −s 1 1 ρ= = = 0.0624 Ω − cm ⎛ qμ p p cm 2 ⎞⎛ 1018 ⎞ −19 1.602 x10 C⎜100 ⎟⎜ ⎟ V − s ⎠⎝ cm 3 ⎠ ⎝ 2.33 Indium is from column 3 and is an acceptor. NA = 7 x 1019/cm3. Assume ND = 0, since it is not specified. N A > N D : material is p - type | N A − N D = 7 x1019 /cm3 >> 2 ni = 2 x1010 /cm3 p = 7 x1019 /cm3 | 19 n= 3 ni2 1020 = = 1.43 /cm3 19 p 7 x10 cm2 cm 2 N D + N A = 7 x10 / cm | Using Fig. 2.13, μn = 120 and μp = 60 V −s V −s 1 1 ρ= = = 1.49 mΩ − cm ⎛ qμ p p cm 2 ⎞⎛ 7 x1019 ⎞ −19 1.602 x10 C⎜ 60 ⎟⎜ 3 ⎟ ⎝ V − s ⎠⎝ cm ⎠ 2.34 Phosphorus is a donor : N D = 5.5 x1016 / cm 3 | Boron is an acceptor : N A = 4.5 x1016 / cm 3 N D > N A : material is n - type 16 3 | N D − N A = 1016 / cm3 >> 2 ni = 2 x1010 / cm 3 ni2 1020 n = 10 /cm | p = = 16 = 10 4 /cm3 p 10 cm2 cm 2 N D + N A = 10 / cm | Using Fig. 2.13, μn = 800 and μp = 230 V −s V −s 1 1 ρ= = = 0.781 Ω − cm ⎛ qμ n n cm 2 ⎞⎛ 1016 ⎞ −19 1.602 x10 C⎜ 800 ⎟⎜ ⎟ V − s ⎠⎝ cm 3 ⎠ ⎝ 17 3 27 2.35 ρ= 1 qμ p p | μp p = (1.602 x10 C)(0.054Ω − cm) −19 1 = 1.16 x1020 V − cm − s An iterative solution is required. Using the equations in Fig. 2.8: NA 1018 1.1 x1018 1.2 x 1017 1.3 x 1019 μp 96.7 93.7 91.0 88.7 μp p 9.67 x 1020 1.03 x 1020 1.09 x 1020 1.15 x 1020 2.36 8.32 x1018 ρ= | μp p = = qμ p p 1.602 x10−19 C (0.75Ω − cm) V − cm − s 1 ( 1 ) An iterative solution is required. Using the equations in Fig. 2.8: NA 1016 2 x 1016 3 x 1016 2.4 x 1016 μp 406 μp p 4.06 x 1018 363 7.26 x 1018 333 1.00 x 1019 350 8.40 x 1018 2.37 Based upon the value of its resistivity, the material is an insulator. However, it is not intrinsic because it contains impurities. Addition of the impurities has increased the resistivity. 2.38 ρ= 1 qμn n | μ n n ≈ μn N D = (1.602 x10 C)(2Ω − cm) −19 1 = 3.12 x1018 V − cm − s An iterative solution is required. Using the equations in Fig. 2.8: ND 1015 2 x 1015 2.5 x 1015 μn 1350 1330 1330 μnn 1.35 x 1018 2.67 x 1018 3.32 x 1018 28 2.3 x 1015 1330 3.06 x 1018 29 2.39 (a) 1 1 6.24 x1021 ρ= | μn n ≈ μn N D = = qμn n 1.602 x10−19 C (0.001Ω − cm) V − cm − s ( ) An iterative solution is required. Using the equations in Fig. 2.8: ND 1019 7 x 1019 6.5 x 1019 μn 116 96.1 96.4 μnn 1.16 x 1021 6.73 x 1021 6.3 x 1021 (b) ρ= 1 qμ p p | μp p ≈ μp N A = 1 6.24 x10 21 = (1.602 x10−19 C)(0.001Ω − cm) V − cm − s An iterative solution is required using the equations in Fig. 2.8: NA 1.3 x 1020 μp 49.3 μp p 6.4 x 1021 2.40 Yes, by adding equal amounts of donor and acceptor impurities the mobilities are reduced, but the hole and electron concentrations remain unchanged. See Problem 2.37 for example. However, it is physically impossible to add exactly equal amounts of the two impurities. 2.41 (a) For the 1 ohm-cm starting material: 1 1 6.25x1018 ρ= | μp p ≈ μpN A = = qμ p p 1.602 x10−19 C ( 1Ω − cm) V − cm − s ( ) An iterative solution is required. Using the equations in Fig. 2.8: NA 1016 1.5 x 1016 1.7 x 1016 μp 406 383 374 μp p 4.1 x 1018 5.7 x 1018 6.4 x 1019 30 To change the resistivity to 0.25 ohm-cm: 1 1 2.5 x1019 ρ= | μp p ≈ μpN A = = qμ p p 1.602 x10−19 C (0.25Ω − cm) V − cm − s ( ) NA 6 x 1016 8 x 1016 1.1 x 1017 μp 276 233 225 μp p 1.7 x 1019 2.3 x 1019 2.5 x 1019 17 16 16 3 Additional acceptor concentration = 1.1x10 - 1.7x10 = 9.3 x 10 /cm (b) If donors are added: ND 2 x 1016 1 x 1017 8 x 1016 4.1 x 1016 ND + NA 3.7 x 1016 1.2 x 1017 9.7 x 1016 5.8 x 1016 16 3 μn 1060 757 811 950 ND - NA 3 x 1015 8.3 x 1016 6.3 x 1016 2.4 x 1016 μnn 3.2 x 1018 6.3 x 1019 5.1 x 1019 2.3 x 1019 So ND = 4.1 x 10 /cm must be added to change achieve a resistivity of 0.25 ohm-cm. The silicon is converted to n-type material. 2.42 Phosphorus is a donor: ND = 1016/cm3 and μn = 1250 cm2/V-s from Fig. 2.8. 2.00 σ = qμn n ≈ qμn N D = 1.602 x10−19 C ( 1250) 1016 = Ω − cm -1 Now we add acceptors until σ = 5.0 (Ω-cm) : ( ) ( ) σ = qμ p p NA 1 x 1017 2 x 1017 1.8 x 1017 | 3.12 x1019 μ p p ≈ μ p (N A − N D )= = 1.602 x10−19 C V − cm − s ND + NA 1.1 x 1017 2.1 x 1017 1.9 x 1017 μp 250 176 183 NA - ND 9 x 1016 1.9 x 1017 1.7 x 1016 μp p 2.3 x 1019 3.3 x 1019 3.1 x 1019 5(Ω − cm) −1 31 2.43 Boron is an acceptor: NA = 1016/cm3 and μp = 405 cm2/V-s from Fig. 2.8. 0.649 σ = qμ p p ≈ qμ p N A = 1.602 x10−19 C (405) 1016 = Ω − cm -1 Now we add donors until σ = 5.5 (Ω-cm) : ( ) ( ) σ = qμ n n ND 8 x 1016 6 x 1016 4.5 x 1016 | μn n ≈ μn (N D − N A )= ND + NA 9 x 1016 7 x 1016 5.5 x 1016 μn 5.5(Ω − cm) −1 1.602 x10−19 C = 3.43 x1019 V − cm − s μp p 5.8 x 1019 4.5 x 1019 3.4 x 1019 ND - NA 7 x 1016 5 x 1016 3.5 x 1016 832 901 964 2.44 VT = kT 1.38 x10−23 T = = 8.62 x10−5 T q 1.602 x10−19 50 4.31 75 6.46 100 8.61 150 12.9 200 17.2 250 21.5 300 25.8 350 30.1 400 34.5 T (K) VT (mV) 2.45 ⎛ dn ⎞ dn j = −qDn ⎜− ⎟ = qVT μn dx ⎝ dx ⎠ ⎛ cm2 ⎞⎛ 1018 − 0 ⎞ 1 kA j = 1.602 x10−19 C (0.025V ) = −14.0 2 ⎜ 350 ⎟⎜ −4 ⎟ 4 V − s ⎠⎝ 0 − 10 ⎠ cm cm ⎝ ( ) 2.46 ⎛ cm2 ⎞⎛ 1019 / cm 3 ⎞ ⎛ ⎞ dp x = −1.602 x10−19 C ⎜15 ⎟⎜ − ⎟ exp⎜− ⎟ −4 −4 s ⎠⎝ 2 x10 cm ⎠ ⎝ 2 x10 cm ⎠ dx ⎝ ⎛ x ⎞ A j = 1.20 x105 exp⎜−5000 ⎟ 2 cm ⎠ cm ⎝ ⎛ 10−8 cm2 ⎞ ⎛ A ⎞ I (0)= j (0)A = ⎜1.20 x105 2 ⎟ 10μm2 ⎜ ⎟ = 12.0 mA 2 cm ⎠ ⎝ ⎝ μm ⎠ j = −qD p ( ) ( ) 32 2.47 j p = qμ p pE − qD p ⎛ dp 1 dp ⎞ 1 dp = qμ p p⎜ E − VT ⎟ = 0 → E = VT dx p dx ⎠ p dx ⎝ −1022 exp −10 4 x 1 dN A E ≈ VT = 0.025 14 N A dx 10 + 1018 exp −10 4 x ( ( ) ) E (0)= −0.025 10 V = −250 18 cm 10 + 10 22 10 exp(−5) V E 5x10−4 cm = −0.025 14 = −246 18 cm 10 + 10 exp(−5) 14 22 ( ) 2.48 ⎛ V ⎞ cm2 ⎞⎛ 1016 ⎞⎛ A jndrift = qμn nE = 1.60 x10−19 C ⎜350 ⎟⎜ 3 ⎟⎜ −20 ⎟ = −11.2 2 cm ⎠ V − s ⎠⎝ cm ⎠⎝ cm ⎝ ⎛ V ⎞ cm2 ⎞⎛1.01x1018 ⎞⎛ A drift −19 − 20 j p = qμ p pE = 1.60 x10 C ⎜150 ⎟⎜ ⎟ ⎜ ⎟ = −484 2 3 cm ⎠ V − s ⎠⎝ cm cm ⎝ ⎠⎝ ⎛ A cm2 ⎞⎛ 10 4 − 1016 ⎞ dn = −70.0 2 jndiff = qDn = 1.60 x10−19 C ⎜ 350 ⋅ 0.025 ⎟⎜ −4 4⎟ s ⎠⎝ 2 x10 cm ⎠ dx cm ⎝ ⎛ A cm2 ⎞⎛ 1018 − 1.01x1018 ⎞ dp diff jp = −qD p = −1.60 x10−19 C ⎜150 ⋅ 0.025 ⎟⎜ ⎟ = 30.0 2 −4 4 s ⎠⎝ 2 x10 cm ⎠ dx cm ⎝ A jT = −11.2 − 484 − 70.0 + 30.0 = −535 2 cm ( ) ( ) ( ) ( ) 2.49 EC ED NA = 2ND N D ND ND EA E V NA NA NA NA Holes 2.50 −34 8 hc 6.626 x10 J − s 3x10 m / s λ= = = 1.108 μm E (1.12eV ) 1.602 x10−19 J / eV ( ( )( ) ) 33 2.51 Al - Anode Al - Cathode Si02 n-type silicon p-type silicon 2.52 An n-type ion implantation step could be used to form the n+ region following step (f) in Fig. 2.17. A mask would be used to cover up the opening over the p-type region and leave the opening over the n-type silicon. The masking layer for the implantation could just be photoresist. Mask Photoresist Si02 p-type silicon n+ p-type silicon Ion implantation n-type silicon n-type silicon Structure after exposure and development of photoresist layer Structure following ion implantation of n-type impurity Mask for ion implantation Side view Top View 2.53 ⎛ 1⎞ ⎛ 1⎞ 1 = 8 atoms ( a ) N = 8⎜ ⎟ + 6⎜ ⎟ + 4() ⎝ 8⎠ ⎝ 2⎠ ( b ) V = l 3 = 0.543x10−9 m = 0.543x10−7 cm = 1.60 x10−22 cm3 8 atoms atoms = 5.00 x1022 1.60 x1022 cm3 cm3 ⎛ g ⎞ ( d ) m = ⎜ 2.33 3 ⎟1.60 x1022 cm3 = 3.73 x10−22 g cm ⎠ ⎝ (c ) D = ( e ) From Table 2.2, silicon has a mass of 28.086 protons. mp = g 3.73x10−22 g = 1.66 x10−24 proton 28.082(8)protons ( ) ( 3 ) 3 Yes, near the actual proton rest mass. 34 CHAPTER 3 3.1 1019 ⋅ cm−3 )( 1018 ⋅ cm −3 ) ( NA ND φ j = VT ln 2 = (0.025V )ln = 0.979V ni 10 20 ⋅ cm −6 2( 11.7 ⋅ 8.854 x10−14 F ⋅ cm−1 )⎛ ⎞ 2εs ⎛ 1 1 ⎞ 1 1 w do = + ⎜ 19 −3 + 18 −3 ⎟ (0.979V) ⎜ ⎟ φj = −19 ⎝ 10 cm q ⎝ NA ND ⎠ 1.602 x10 C 10 cm ⎠ w do = 3.73 x 10−6 cm = 0.0373μm w do 0.0373μm w do 0.0373μm -3 xn = = = μm 18 −3 = 0.0339 μm | x p = 19 −3 = 3.39 x 10 ND N 10 10 cm cm A 1+ 1+ 1 + 19 −3 1 + 18 −3 NA ND 10 cm 10 cm E MAX = qN A x p εs 1018 cm 3 (1.60 x10 = | n po = −19 C) (1019 cm−3 )(3.39 x10−7 cm) −14 11.7 ⋅ 8.854 x10 F / cm = 5.24 x 10 5 V cm 3.2 p po = N A = n i2 10 20 10 2 = = p po 1018 cm 3 1015 cm −3 ) 1018 cm−3 )( ( NA ND φ j = VT ln 2 = (0.025V )ln = 0.748 V ni 10 20 cm−6 w do = 2( 11.7 ⋅ 8.854 x10−14 F ⋅ cm−1 )⎛ ⎞ 2εs ⎛ 1 1 ⎞ 1 1 + φ = ⎜ 18 −3 + 15 −3 ⎟ (0.748V) ⎜ ⎟ j −19 ⎝ 10 cm 10 cm ⎠ q ⎝ NA ND ⎠ 1.602 x10 C 1015 n no = N D = cm 3 n i2 10 20 10 5 | p no = = = n no 1015 cm 3 w do = 98.4 x 10−6 cm = 0.984 μm 3.3 p po = N A = 1018 ni2 1020 102 | n = = = po p po 1018 cm3 cm3 1018 ni2 1020 102 nno = N D = 3 | p no = = = nno 1018 cm3 cm 1018 ⋅ cm −3 ) 1018 ⋅ cm−3 )( ( NA ND φ j = VT ln 2 = (0.025V )ln = 0.921V ni 10 20 ⋅ cm −6 w do = 2( 11.7 ⋅ 8.854 x10−14 F ⋅ cm−1 )⎛ ⎞ 2εs ⎛ 1 1 ⎞ 1 1 + φ = ⎜ 18 −3 + 18 −3 ⎟ (0.921V) ⎜ ⎟ j −19 ⎝ 10 cm 10 cm ⎠ q ⎝ NA ND ⎠ 1.602 x10 C w do = 4.881x10−6 cm = 0.0488 μm 34 3.4 p po = N A = nno = N D = 1018 ni2 1020 102 | n = = = po p po 1018 cm3 cm3 1018 ni2 1020 102 | p = = = no nno 1018 cm3 cm3 (1018 ⋅ cm−3 )(1020 ⋅ cm−3 ) = 1.04V N N φ j = VT ln A 2 D = (0.025V )ln ni 10 20 ⋅ cm −6 2( 11.7 ⋅ 8.854 x10−14 F ⋅ cm−1 )⎛ ⎞ 2εs ⎛ 1 1 ⎞ 1 1 w do = + ⎜ 18 −3 + 20 −3 ⎟ (1.04V) ⎜ ⎟ φj = −19 ⎝ 10 cm 10 cm ⎠ q ⎝ NA ND ⎠ 1.602 x10 C w do = 0.0369 μm 3.5 p po = N A = nno = N D = 1016 ni2 1020 10 4 | n = = = po p po 1016 cm3 cm3 1019 ni2 1020 10 | p = = = no nno 1019 cm3 cm3 1019 ⋅ cm−3 1016 ⋅ cm −3 N AND φ j = VT ln = (0.025V )ln = 0.864V ni2 1020 ⋅ cm −6 ( )( ) 2 11.7 ⋅ 8.854 x10−14 F ⋅ cm−1 ⎛ ⎞ 2εs ⎛ 1 1 ⎞ 1 1 wdo = + ⎜ ⎟ φj = ⎜ 19 −3 + 16 −3 ⎟ (0.864V) −19 q ⎝ N A ND ⎠ 10 cm ⎠ 1.602 x10 C ⎝ 10 cm wdo = 0.334 μm 3.6 wd = wdo 1 + VR ( ) φj | (a) wd = 2 wdo requires VR = 3φ j = 2.55 V | wd = 0.4μm 1 + 5 = 1.05 μm 0.85 3.7 wd = wdo 1 + VR φj | (a) wd = 3wdo requires VR = 8φ j = 4.80 V | wd = 1μm 1 + 10 = 4.20 μm 0.6 3.8 jn = σE , σ = 1 ρ = 1 2 j 1000 A ⋅ cm −2 V = | E= n = = 500 −1 0.5 Ω ⋅ cm Ω ⋅ cm cm σ 2(Ω ⋅ cm) 35 3.9 j p = σE 3.10 | E= σ jn = jn ρ = 5000 A ⋅ cm −2 (2Ω ⋅ cm)= 10.0 ( ) kV cm ⎛ 4 x1015 ⎞⎛ 10 7 cm ⎞ A j ≅ jn = qnv = 1.60 x10−19 C ⎜ ⎟ = 6400 2 3 ⎟⎜ cm ⎝ cm ⎠⎝ s ⎠ ( ) 3.11 ⎛ 5 x1017 ⎞⎛ 10 7 cm ⎞ kA j ≅ j p = qpv = 1.60 x10−19 C ⎜ ⎟ = 800 2 3 ⎟⎜ cm ⎝ cm ⎠⎝ s ⎠ ( ) 3.12 j p = qμ p pE − qD p ⎛ D ⎞ 1 dp ⎛ kT ⎞ 1 dp dp p = 0 → E = −⎜ = −⎜ ⎟ ⎟ ⎜ ⎟ dx ⎝ q ⎠ p dx ⎝ μ p ⎠ p dx 1 dp 1 V V 0.025V | E = − T = − −4 = = −250 L p dx L cm 10 cm ⎛ x⎞ p ( x ) = N o exp⎜ − ⎟ | ⎝ L⎠ The exponential doping results in a constant electric field. 3.13 j p = qDn dn dn dn 2000 A / cm 2 1.00 x 1021 = qμnVT | = = dx dx dx 1.60 x10−19 C 500cm2 / V − s (0.025V ) cm 4 ( )( ) 3.14 10 = 10 4 ⋅ 10−16 exp(40VD )− 1 + VD [ ] and the solver yields VD = 0.7464 V 3.15 f = 10 − 10 4 I D − 0.025ln ID + IS IS | f ' = −10 4 − 0.025 ID + IS | I'D = I D − -13 f f' Starting the iteration process with ID = 100 μA and IS = 10 A: ID 1.000E-04 9.275E-04 9.426E-04 9.426E-04 f 8.482E+0 0 1.512E01 3.268E06 9.992E16 f' -1.025E+04 -1.003E+04 -1.003E+04 -1.003E+04 36 3.16 (a) Create the following m-file: function fd=current(id) fd=10-1e4*id-0.025*log(1+id/1e-13); Then: fzero('current',1) yields ans = 9.4258e-04 -15 (b) Changing IS to 10 A: function fd=current(id) fd=10-1e4*id-0.025*log(1+id/1e-15); Then: fzero('current',1) yields ans = 9.3110e-04 3.17 −19 qVT 1.60 x10 C (0.025V ) T= = = 290 K k 1.38 x10−23 J / K 3.18 −23 kT 1.38 x10 J / K T VT = = = 8.63 x10−5 T −19 q 1.60 x10 C For T = 218 K, 273 K and 358 K, VT = 18.8 mV, 23.6 mV and 30.9 mV ( ) 3.19 ⎡ ⎛ 40V ⎞ ⎤ D Graphing I D = I S ⎢exp⎜ ⎟ − 1⎥ yields : n ⎝ ⎠ ⎦ ⎣ 6 5 (b) 4 3 (a) 2 1 (c) 0 0.0 0.2 0.4 0.6 0.8 1.0 1.2 37 3.20 1.38 x10−23 J / K (300) kT = 1.04 = 26.88 mV nVT = n q 1.60 x10−19 C ( ) T = 26.88mV 1.602x10-19 = 312 K 1.38x10-23 3.21 ⎡ ⎛v ⎞ ⎤ ⎛ i ⎞ vD = ln⎜1 + D ⎟ iD = I S ⎢exp⎜ D ⎟ − 1⎥ or nVT ⎝ IS ⎠ ⎣ ⎝ nVT ⎠ ⎦ ⎛i ⎞ ⎛ 1 ⎞ v For i D >> I S , D ≅ ln⎜ D ⎟ or ln (I D )= ⎜ ⎟v D + ln (IS ) nVT ⎝ IS ⎠ ⎝ nVT ⎠ -4 which is the equation of a straight line with slope 1/nVT and x-axis intercept at -ln (IS). The values of n and IS can be found from any two points on the line in the figure: e. g. iD = 10 A for vD = 0.60 V and iD = 10 A for vD = 0.20 V. Then there are two equations in two unknowns: ⎛ 40 ⎞ ⎛8⎞ ln 10-9 = ⎜ ⎟.20 + ln (IS ) or 9.21 = ⎜ ⎟ + ln (IS ) ⎝n⎠ ⎝n⎠ ⎛ 40 ⎞ ⎛ 24 ⎞ ln 10-4 = ⎜ ⎟.60 + ln (IS ) or 20.72 = ⎜ ⎟ + ln (IS ) ⎝n⎠ ⎝n⎠ -12 Solving for n and IS yields n = 1.39 and IS = 3.17 x 10 A = 3.17 pA. -9 ( ) ( ) 3.22 ⎡ ⎛V ⎞ ⎤ ⎛ I ⎞ VD = nVT ln⎜1 + D ⎟ | I D = I S ⎢exp⎜ D ⎟ − 1⎥ ⎝ IS ⎠ ⎣ ⎝ nVT ⎠ ⎦ ⎛ 7 x10−5 A ⎞ ⎛ 5 x10−6 A ⎞ = 1.05 0.025 V ln (a) VD = 1.05(0.025V )ln⎜1 + = 0.837 V | (b) V ⎟ ⎜1 + ⎟ = 0.768V ( ) D 10−18 A ⎠ 10−18 A ⎠ ⎝ ⎝ ⎡ ⎛ ⎞ ⎤ 0 (c) I D = 10−18 A⎢exp⎜ ⎟ − 1⎥ = 0 A ⎣ ⎝1.05 ⋅ 0.025V ⎠ ⎦ ⎡ ⎛ −0.075V ⎞ ⎤ −19 (d) I D = 10−18 A⎢exp⎜ ⎟ − 1⎥ = −0.943x10 A ⎣ ⎝ 1.05 ⋅ 0.025V ⎠ ⎦ ⎡ ⎛ ⎞ ⎤ −5V −18 (e) I D = 10−18 A⎢exp⎜ ⎟ − 1⎥ = −1.00 x10 A ⎣ ⎝1.05 ⋅ 0.025V ⎠ ⎦ ⎡ ⎛V ⎞ ⎤ ⎛ I ⎞ VD = nVT ln⎜1 + D ⎟ | I D = I S ⎢exp⎜ D ⎟ − 1⎥ ⎝ IS ⎠ ⎣ ⎝ nVT ⎠ ⎦ ⎛ 10−4 A ⎞ ⎛ 10−5 A ⎞ (a) VD = 0.025V ln⎜1 + −17 ⎟ = 0.748V | (b) VD = 0.025V ln⎜1 + −17 ⎟ = 0.691V ⎝ 10 A ⎠ ⎝ 10 A ⎠ 3.23 38 ⎡ ⎛ 0 ⎞ ⎤ (c) ID = 10−17 A⎢exp⎜ ⎟ − 1⎥ = 0 A ⎣ ⎝ 0.025V ⎠ ⎦ ⎡ ⎛ −0.06V ⎞ ⎤ −17 (d) ID = 10−17 A⎢exp⎜ ⎟ − 1⎥ = −0.909 x10 A ⎣ ⎝ 0.025V ⎠ ⎦ ⎡ ⎛ −4V ⎞ ⎤ −17 (e) ID = 10−17 A⎢exp⎜ ⎟ − 1⎥ = −1.00 x10 A ⎣ ⎝ 0.025V ⎠ ⎦ 3.24 ⎡ ⎛V ⎞ ⎤ ⎡ ⎛ 0.675 ⎞ ⎤ −6 I D = I S ⎢exp⎜ D ⎟ − 1⎥ = 10−17 A ⎢exp⎜ ⎟ − 1⎥ = 5.32 x10 A = 5.32 μA ⎣ ⎝ 0.025 ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎦ ⎛ 15.9 x10−6 A ⎞ ⎞ ⎛I VD = VT ln⎜ D + 1⎟ = (0.025V )ln⎜ + 1⎟ = 0.703 V −17 ⎠ ⎝ IS ⎝ 10 A ⎠ 3.25 ⎛ I ⎞ ⎛ 40 A ⎞ VD = nVT ln⎜1 + D ⎟ = 2(0.025V )ln⎜1 + −10 ⎟ = 1.34 V ⎝ 10 A ⎠ ⎝ IS ⎠ ⎛ 100 A ⎞ VD = 2(0.025V )ln⎜1 + −10 ⎟ = 1.38 V ⎝ 10 A ⎠ 3.26 ID 2mA = = 1.14 x10−17 A ⎡ ⎛ V ⎞ ⎤ ⎡ ⎛ 0.82 ⎞ ⎤ ⎢exp⎜ D ⎟ − 1⎥ ⎢exp⎜ ⎟ − 1⎥ ⎣ ⎝ nVT ⎠ ⎦ ⎣ ⎝ 0.025 ⎠ ⎦ ⎡ ⎛ −5 ⎞ ⎤ −17 ( b ) I D = 1.14 x10−17 A ⎢exp⎜ ⎟ − 1⎥ = −1.14 x10 A 0.025 ⎠ ⎦ ⎣ ⎝ (a ) IS = 3.27 ID 300μA = = 2.81x10−17 A ⎡ ⎛ V ⎞ ⎤ ⎡ ⎛ 0.75 ⎞ ⎤ ⎢exp⎜ D ⎟ − 1⎥ ⎢exp⎜ ⎟ − 1⎥ ⎣ ⎝ nVT ⎠ ⎦ ⎣ ⎝ 0.025 ⎠ ⎦ ⎡ ⎛ −3 ⎞ ⎤ −17 ( b ) I D = 2.81x10−17 A ⎢exp⎜ ⎟ − 1⎥ = −2.81x10 A ⎣ ⎝ 0.025 ⎠ ⎦ (a ) IS = 39 3.28 ⎛ I ⎞ VD = nVT ln⎜1 + D ⎟ | 10-14 ≤ I S ≤ 10−12 ⎝ IS ⎠ ⎛ 10−3 A ⎞ VD = (0.025V )ln⎜1 + −14 ⎟ = 0.633 V ⎝ 10 A ⎠ 1.38 x10−23 (307) | | ⎛ 10−3 A ⎞ VD = (0.025V )ln⎜1 + −12 ⎟ = 0.518 V ⎝ 10 A ⎠ So, 0.518 V ≤ VD ≤ 0.633 V 3.29 ⎡ ⎛ V ⎞ ⎤ D = 0.0264 V | I = I exp ⎢ ⎟ − 1⎥ ⎜ D S 1.60 x10−19 ⎣ ⎝ 0.0264 n ⎠ ⎦ Varying n and IS by trial-and-error with a spreadsheet: VT = n 1.039 VD 0.500 0.550 0.600 0.650 0.675 0.700 0.725 0.750 0.775 IS 7.606E-15 ID-Measured 6.591E-07 3.647E-06 2.158E-05 1.780E-04 3.601E-04 8.963E-04 2.335E-03 6.035E-03 1.316E-02 ID-Calculated 6.276E-07 3.885E-06 2.404E-05 1.488E-04 3.702E-04 9.211E-04 2.292E-03 5.701E-03 1.418E-02 Error Squared 9.9198E-16 5.6422E-14 6.0672E-12 8.518E-10 1.0261E-10 6.1409E-10 1.8902E-09 1.1156E-07 1.0471E-06 Total Squared Error 1.1622E-06 3.30 −23 kT 1.38 x10 J / K T = = 8.63 x10−5 T VT = q 1.60 x10−19 C For T = 233 K, 273 K and 323 K, VT = 20.1 mV, 23.6 mV and 27.9 mV ( ) 3.31 −23 ⎛ 10−3 ⎞ kT 1.38 x10 (303) = = 26.1 mV | V = 0.0261 V ln 1 + = 0.757 V ⎜ ( ) D −16 ⎟ q 1.60 x10−19 ⎝ 2.5 x10 ⎠ ΔV = (−1.8 mV / K )(20 K ) = −36.0 mV | VD = 0.757 − 0.036 = 0.721 V 40 3.32 ΔV = (−2.0 mV / K )(25 K ) = −50.0 mV ( b ) VD = 0.650 − 0.050 = 0.600 V −23 ⎛ 10−4 ⎞ kT 1.38 x10 (298) = = 25.67 mV | ( a ) VD = (0.02567V )ln⎜1 + −15 ⎟ = 0.650 V q 1.602 x10−19 ⎝ 10 ⎠ 3.33 (b) ΔV = (−1.8mV / K )(60 K )= −50.0 mV ( c ) ΔV = (−1.8 mV / K )(−80 K )= +144 mV 3.34 −23 ⎛ 2.5 x10−4 ⎞ kT 1.38 x10 (298) = = 25.67 mV | ( a ) V = 0.02567 V ln ( ) ⎜1 + 10−14 ⎟ = 0.615 V D q 1.602 x10−19 ⎝ ⎠ VD = 0.615 − 0.108 = 0.507 V VD = 0.615 + 0.144 = 0.758 V mV dvD vD − VG − 3VT 0.7 − 1.21 − 3(0.0259) = = = −1.96 dT T 300 K 41 3.35 3 ⎡ ⎛ E ⎞⎛ 1 1 ⎞⎤ ⎛ T ⎞3 ⎡⎛ E ⎞⎛ T ⎞⎤ I S 2 ⎛ T2 ⎞ = ⎜ ⎟ exp⎢−⎜ G ⎟⎜ − ⎟⎥ = ⎜ 2 ⎟ exp⎢⎜ G ⎟⎜1 − 1 ⎟⎥ I S1 ⎝ T1 ⎠ ⎣ ⎝ k ⎠⎝ T2 T1 ⎠⎦ ⎝ T1 ⎠ ⎣⎝ kT1 ⎠⎝ T2 ⎠⎦ ⎡⎛ E ⎞⎛ 1 ⎞⎤ 3 T x= 2 f (x ) = (x ) exp⎢⎜ G ⎟⎜1 − ⎟⎥ T1 ⎣⎝ kT1 ⎠⎝ x ⎠⎦ Using trial and error with a spreadsheet yields T = 4.27 K, 14.6 K, and 30.7 K to increase the saturation current by 2X, 10X, and 100X respectively. x 1.00000 1.00500 1.01000 1.01500 1.01400 1.01422 1.01922 1.02422 1.02922 1.03422 1.03922 1.04422 1.04922 1.04880 1.10000 1.10239 f(x) 1.00000 1.27888 1.63167 2.07694 1.97945 2.00051 2.54151 3.22151 4.07433 5.14160 6.47438 8.13522 10.20058 10.00936 90.67434 100.0012 Delta T 0.00000 1.50000 3.00000 4.50000 4.20000 4.26600 5.76600 7.26600 8.76600 10.26600 11.76600 13.26600 14.76600 14.64000 30.00000 30.71610 3.36 wd = wdo 1 + VR φj | (a) wd = 1μm 1 + 5 10 = 2.69 μm (b) wd = 1μm 1 + = 3.67 μm 0.8 0.8 3.37 1016 cm −3 1015 cm −3 N AND φ j = VT ln = (0.025V )ln = 0.633 V ni2 1020 cm−6 2 11.7 ⋅ 8.854 x10−14 F ⋅ cm−1 ⎛ ⎞ 2εs ⎛ 1 1 ⎞ 1 1 + wdo = ⎜ ⎟ φj = ⎜ 16 −3 + 15 −3 ⎟ (0.633V) −19 q ⎝ N A ND ⎠ 10 cm ⎠ 1.602 x10 C ⎝10 cm wdo = 0.949 μm wd = 0.949μm 1 + | wd = wdo 1 + VR ( )( ) ( ) φj 10V = 3.89 μm 0.633V | wd = 0.949μm 1 + 100V = 12.0 μm 0.633V 42 3.38 1018 cm −3 1020 cm −3 N AND φ j = VT ln = (0.025V )ln = 1.04 V ni2 1020 cm−6 2 11.7 ⋅ 8.854 x10−14 F ⋅ cm−1 2εs ⎛ 1 1 ⎞ wdo = + ⎟ φj = ⎜ q ⎝ N A ND ⎠ 1.602 x10−19 C wdo = 0.0368 μm wd = 0.0368μm 1 + 3.39 ( )( ) ( )⎛ ⎜ ⎞ 1 1 1.04V) + ⎟( 18 −3 1020 cm−3 ⎠ ⎝ 10 cm | wd = wdo 1 + VR φj 5 = 0.0887 μm 1.04 2(φ j + VR ) wdo 1 + VR | wd = 0.0368μm 1 + 25 = 0.184 μm 1.04 Emax = 2(φ j + VR ) wd = = 2φ j V 1+ R wdo φj φj 3 x105 3.40 2(0.6V ) V V = −4 1 + R → VR = 374 V cm 10 cm 0.6 2(0.748V ) 2φ kV = 15.2 | E= j = −4 wdo 0.984 x10 cm cm VR = 291.3 − 0.748 = 291 V 3.41 E w φ j + VR = max do = 2 φj 3 x105 V 0.984 x10−4 cm cm 2 0.748V ( ) VZ = 4 V; RZ = 0 Ω since the reverse breakdown slope is infinite. 3.42 Since NA >> ND, the depletion layer is all on the lightly-doped side of the junction. Also, VR >> φj, so φj can be neglected. Emax = qN A x p εS = qN A wd εS = qN A εS 2εS VR q NA 2 3 x105 ( 11.7) 8.854 x10−14 Emax εS NA = = = 2.91 x 1014 / cm3 −19 2qVR 2 1.602 x10 1000 2 ( ) ( ( ) ) 43 3.43 φ j = VT ln N AND 10151020 = 0.025ln = 0.864V ni2 1020 2( 11.7) 8.854 x10−14 ⎛ 1 2εS ⎛ 1 1 ⎞ 1 ⎞ −4 wdo = + + ⎟φ j = ⎜ ⎟0.864 = 1.057 x10 cm ⎜ q ⎝ N A ND ⎠ 1.602 x10−19 ⎝ 1015 1020 ⎠ C = " jo ( ) εS wdo = 11.7 8.854 x10−14 1.057 x10−4 ( ) = 9.80x10 -9 F / cm 2 | Cj = C" jo A 1+ VR = 9.80x10-9 (0.05) 5 1+ 0.864 = 188 pF φj 3.44 φ j = VT ln N AND 10181015 = 0.025ln = 0.748V ni2 1020 2( 11.7) 8.854 x10−14 2εS ⎛ 1 1 ⎞ wdo = + ⎟φ j = ⎜ q ⎝ N A ND ⎠ 1.602 x10−19 C = " jo ( )⎛ ⎜ 1 ⎞ 1 0.748 = 0.984 x10−4 cm + 18 15 ⎟ 10 10 ⎠ ⎝ | Cj = C" jo A 1+ VR = 10.5x10-9 (0.02) 10 1+ 0.748 = 55.4 pF εS wdo = 11.7 8.854 x10−14 0.984 x10−4 ( ) = 10.5x10 -9 F / cm 2 φj 3.45 −4 −10 I D τ T 10 A 10 s (a) CD = = = 400 fF VT 0.025V ( ) (b) Q = I D τ T = 10−4 A 10−10 s = 10 fC ( ) (c) CD = 3.46 25x10−3 A 10−10 s 0.025V ( ) = 100 pF | Q = I D τ T = 5 x10−3 A 10−10 s = 0.50 pC ( ) −8 I D τ T 1A 10 s (a) CD = = = 0.400 μF VT 0.025V ( ) (b) Q = I D τ T = 1A 10−8 s = 10.0 nC ( ) (c) CD = 100mA 10−8 s 0.025V ( ) = 0.04 μF | Q = I D τ T = 100 mA 10−8 s = 1.00 nC ( ) 44 3.47 φ j = VT ln N AND 10191017 = 0.025ln = 0.921V ni2 1020 2( 11.7) 8.854 x10−14 ⎛ 1 2εS ⎛ 1 1 ⎞ 1 ⎞ wdo = + + ⎟φ j = ⎜ ⎟0.921 = 0.110 μm ⎜ q ⎝ N A ND ⎠ 1.602 x10−19 ⎝ 1019 1017 ⎠ C jo = ( ) εS A wdo = 11.7 8.854 x10−14 10−4 0.110 x10 −4 ( )( ) = 9.42 pF / cm 2 | Cj = C jo 1+ VR = 9.42 pF 5 1+ 0.921 = 3.72 pF φj 3.48 φ j = VT ln N AND 10191016 = 0.025ln = 0.864V ni2 1020 2( 11.7) 8.854 x10−14 2εS ⎛ 1 1 ⎞ wdo = + ⎟φ j = ⎜ q ⎝ N A ND ⎠ 1.602 x10−19 C jo = ( )⎛ ⎜ 1 ⎞ 1 + ⎟0.864 = 0.334μm ⎝ 1019 1016 ⎠ | Cj = C jo 1+ VR = 7750 pF 3 1+ 0.864 = 3670 pF εS A wdo = 11.7 8.854 x10−14 0.25cm2 0.334 x10 −4 ( )( ) = 7750 pF φj 3.49 L= RFC C VDC C 10 μH 10 μH C= C jo 1+ VR (a) C = 39 pF 1V 1+ 0.75V = 25.5 pF | f o = 1 2π LC 1 = φj 2π ( 1 10−5 H 25.5 pF = 15.7 MHz ) = 9.97 MHz (b) C = 39 pF 10V 1+ 0.75V = 10.3 pF | f o = 1 2π LC = 2π ( 10−5 H 10.3 pF ) 3.50 ⎛ ⎛ 50 A ⎞ 50 A ⎞ (a) VD = (0.025V )ln⎜1 + −7 ⎟ = 0.501 V | (b) VD = (0.025V )ln⎜1 + −15 ⎟ = 0.961 V ⎝ 10 A ⎠ ⎝ 10 A ⎠ 45 3.51 ⎛ 4 x10−3 A ⎞ ⎛ 4 x10−3 A ⎞ (a) VD = (0.025V )ln⎜1 + = 0.025 V ln = 0.495 V | (b) V ⎟ ⎜ ( ) 1 + 10−14 A ⎟ = 0.668 V D 10−11 A ⎠ ⎝ ⎝ ⎠ 3.52 RS = R p + Rn Rn = ρ n 3.53 Rp = ρ p Lp 0.025cm =( 1Ω − cm) = 2.5Ω Ap 0.01cm 2 RS = 2.53 Ω Ln 0.025cm = (0.01Ω − cm) = 0.025Ω An 0.01cm 2 ' VD = VD + I D RS = 0.708V + 10−3 A( 10Ω)= 0.718 V ⎛ ⎛ I ⎞ 10−3 ⎞ ' (a) VD = VT ln⎜1 + D ⎟ = (0.025V )ln⎜1 + = 0.708V −16 ⎟ ⎝ IS ⎠ ⎝ 5x10 ⎠ ' (b) VD = VD + I D RS = 0.708V + 10−3 A( 100Ω)= 0.808 V 3.54 10Ω − μm2 ρ c = 10Ω − μm Ac = 1μm RC = = = 10Ω / contact Ac 1μm2 5 anode contacts and 14 cathode contacts 2 2 ρc 10Ω = 2Ω 5 10Ω = 0.71Ω Resistance of cathode contacts = 14 Resistance of anode contacts = 3.55 (a) From Fig. 3.21a, the diode is approximately 10.5 μm long x 8 μm wide. Area = 84 μm2. (b) Area = (10.5x0.13 μm) x (8x0.13μm) = 1.42 μm2. 46 3.56 (a) 5 = 10 4 I D + VD | VD = 0 I D = 0.500 mA | I D = 0 VD = 5V Forward biased - VD = 0.5 V I D = (b) − 6 = 3000 I D + VD 4.5V = 0.450 mA 10 4 Ω | VD = 0 I D = −2.00 mA | I D = 0 VD = −6V −2V = −0.667 mA 3kΩ | VD = 0 I D = −1.00 mA | I D = 0 VD = −3V iD 2 mA In reverse breakdown - VD = −4 V I D = (c) − 3 = −3000 I D + VD Reverse biased - VD = −3 V I D = 0 1 mA (c) Q-point -6 -5 -4 -3 -2 -1 1 (b) Q-point -1 mA 2 3 4 5 6 (a) Q-point vD -2 mA 3.57 (a) 10 = 5000 I D + VD | VD = 0 I D = 2.00 mA | VD = 5 V I D = 1.00 mA 9.5V = 1.90 mA 5kΩ | VD = 0 I D = −2.00 mA | VD = −5 V I D = −1.00 mA Forward biased - VD = 0.5V I D = (b) -10 = 5000 I D + VD In reverse breakdown - VD = −4V I D = (c) − 2 = 2000 I D + VD Reverse biased - VD = −2 V I D = 0 iD 2 mA (a) Q-point −6V = −1.20 mA 5kΩ | VD = 0 I D = −1.00 mA | I D = 0 VD = −2 V 1 mA (c) Q-point -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 v D -1 mA (b) Q-point -2 mA 47 3.58 *Problem 3.58 - Diode Circuit V 1 0 DC 5 R 1 2 10K D1 2 0 DIODE1 .OP .MODEL DIODE1 D IS=1E-15 .END SPICE Results VD = 0.693 V ID = 0.431 μA 3.59 (a) − 10 = 10 4 I D + VD | VD = 0 I D = −1.00 mA | VD = −5 V I D = −0.500 mA In reverse breakdown - VD = −4 V I D = (b) 10 = 10 4 I D + VD −10 − ( −4)V = −0.600 mA 10 kΩ | VD = 0 I D = 1.00 mA | VD = 5 V I D = 0.500 mA 10 − 0.5V = 0.950 mA 10 kΩ | VD = 0 I D = −2.00 mA | I D = 0 iD 2 mA Forward biased - VD = 0.5 V I D = (c) − 4 = 2000 I D + VD VD = − 4 V Reverse biased - VD = −4 V I D = 0 1 mA (c) Q-point -6 -5 -4 -3 -2 -1 (b) Q-point vD 1 2 3 4 5 6 (a) Q-point -1 mA -2 mA 48 iD (A) 0.002 0.001 -7 -6 -5 -4 -3 -2 -1 v D (V) 1 2 3 4 5 6 7 -0.001 -0.002 49 3.60 R i V + D - + v D - The load line equation: V = iD R + vD We need two points to plot the load line. (a) V = 6 V and R = 4kΩ: For vD = 0, iD = 6V/4 kΩ = 1.5 mA and for iD = 0, vD = 6V. Plotting this line on the graph yields the Q-pt: (0.5 V, 1.4 mA). (b) V = -6 V and R = 3kΩ: For vD = 0, iD = -6V/3 kΩ = -2 mA and for iD = 0, vD = -6V. Plotting this line on the graph yields the Q-pt: (-4 V, -0.67 mA). (c) V = -3 V and R = 3kΩ: Two points: (0V, -1mA), (-3V, 0mA); Q-pt: (-3 V, 0 mA) (d) V = +12 V and R = 8kΩ: Two points: (0V, 1.5mA), (4V, 1mA); Q-pt: (0.5 V, 1.4 mA) (e) V = -25 V and R = 10kΩ: Two points: (0V, -2.5mA), (-5V, -2mA); Q-pt: (-4 V, -2.1 mA) i (A) D .002 Q-Point (0.5V,1.45 mA) (d) Q-Point (-3V,0 mA) -7 -6 -5 -4 -3 -2 .001 Q-Point (0.5V,1.4 mA) -1 1 2 3 Load line for (a) 4 5 6 7 v (V) D Q-Point (-4V,-0.67 mA) (c) -.001 Load line for (b) -.002 Q-Point (-4V,-2.1 mA) (e) 50 3.61 -9 Using the equations from Table 3.1, (f = 10-10 exp ..., etc.) VD = 0.7 V requires 12 iterations, VD = 0.5 V requires 22 iterations, VD = 0.2 V requires 384 iterations - very poor convergence because the second iteration (VD = 9.9988 V) is very bad. 3.62 Using Eqn. (3.28), ⎛i ⎞ V = iD R + VT ln⎜ D ⎟ ⎝ IS ⎠ or 10 = 10 4 iD + 0.025ln 1013 iD . ( ) 4 13 We want to find the zero of the function f = 10 − 10 iD − 0.025ln 10 iD ( ) iD .001 .0001 .0009 .00094 f -0.576 8.48 0.427 0.0259 - converged 3.63 ⎛ I ⎞ 0.025 f = 10 − 10 4 ID − 0.025ln⎜1 + D ⎟ | f ' = −10 4 − ID + IS ⎝ IS ⎠ x 1.0000E+00 9.2766E-04 9.4258E-04 9.4258E-04 9.4258E-04 f(x) -9.991E+03 1.496E-01 3.199E-06 9.992E-16 9.992E-16 f'(x) -1.000E+04 -1.003E+04 -1.003E+04 -1.003E+04 -1.003E+04 3.64 Create the following m-file: function fd=current(id) fd=10-1e4*id-0.025*log(1+id/1e-13); Then: fzero('current',1) yields ans = 9.4258e-04 + 1.0216e-21i 51 3.65 The one-volt source will forward bias the diode. Load line: 1 = 10 4 I D + VD | I D = 0 VD = 1V | VD = 0 I D = 0.1mA → (50 μA, 0.5 V ) −9 Mathematical model: f = 1 − 10 exp(40VD )− 1 + VD → (49.9 μA, 0.501 V ) [ ] Ideal diode model: ID = 1V/10kΩ = 100μA; (100μA, 0 V) Constant voltage drop model: ID = (1-0.6)V/10kΩ = 40.0μA; (40.0μA, 0.6 V) 3.66 Using Thévenin equivalent circuits yields and then combining the sources 1.2 k Ω 1.2 V + I V + 1k Ω + 1.5 V I V + 2.2 k Ω + 0.3 V - - - (a) Ideal diode model: The 0.3 V source appears to be forward biasing the diode, so we will assume it is "on". Substituting the ideal diode model for the forward region yields 0.3V I= = 0.136 mA . This current is greater than zero, which is consistent with the diode 2.2 kΩ being "on". Thus the Q-pt is (0 V, +0.136 mA). I V + I 0.6 V 2.2 k Ω + 0.3 V V on + 2.2 k Ω + 0.3 V - - Ideal Diode: CVD: (b) CVD model: The 0.3 V source appears to be forward biasing the diode so we will assume it 0.3V − 0.6V = −136 μA . is "on". Substituting the CVD model with Von = 0.6 V yields I = 2.2 kΩ This current is negative which is not consistent with the assumption that the diode is "on". Thus the diode must be off. The resulting Q-pt is: (0 mA, -0.3 V). V + I=0 2.2 k Ω + 0.3 V 52 (c) The second estimate is more realistic. 0.3 V is not sufficient to forward bias the diode into -15 significant conduction. For example, let us assume that IS = 10 A, and assume that the full 0.3 V appears across the diode. Then ⎡ ⎛ 0.3V ⎞ ⎤ iD = 10−15 A⎢exp⎜ ⎟ − 1⎥ = 163 pA , a very small current. ⎣ ⎝ 0.025V ⎠ ⎦ 3.67 The nominal values are: ⎛ R2 ⎞ ⎛ 2 kΩ ⎞ VA = 3V ⎜ ⎟ = 3V ⎜ ⎟ = 1.20V ⎝ 2 kΩ + 3kΩ ⎠ ⎝ R1 + R2 ⎠ and RTHA = ⎛ R4 ⎞ ⎛ 2 kΩ ⎞ 2 kΩ(2 kΩ) R3 R4 VC = 3V ⎜ = = 1.00 kΩ ⎟ = 3V ⎜ ⎟ = 1.50V and RTHC = R3 + R4 2 kΩ + 2 kΩ ⎝ 2 kΩ + 2 kΩ ⎠ ⎝ R3 + R4 ⎠ ⎛ 1.50 − 1.20 ⎞ V nom ID =⎜ = 136 μA ⎟ ⎝1.20 + 1.00 ⎠ kΩ For maximum current, we make the Thévenin equivalent voltage at the diode anode as large as possible and that at the cathode as small as possible. VA = 3V 3V = = 1.65V and R1 2 kΩ(0.9) 1+ R2 1 + 2kΩ( 1.1) 3V 3V = = 1.06V and R3 3kΩ( 1.1) 1+ R4 1 + 2kΩ(0.9) RTHA = 2 kΩ(0.9)2 kΩ( 1.1) R1 R2 = = 0.990 kΩ R1 + R2 2 kΩ(0.9)+ 2 kΩ( 1.1) 3kΩ( 1.1)2 kΩ(0.9) R3 R4 = = 1.17 kΩ R3 + R4 3kΩ( 1.1)+ 2 kΩ(0.9) 2 kΩ(3kΩ) R1 R2 = = 1.20 kΩ R1 + R2 2 kΩ + 3kΩ VC = RTHC = ⎛ 1.65 − 1.06 ⎞ V max ID =⎜ = 274 μA ⎟ ⎝ 0.990 + 1.17 ⎠ kΩ For minimum current, we make the Thévenin equivalent voltage at the diode anode as small as possible and that at the cathode as large as possible. VA = 3V 3V = = 1.350V and R1 2kΩ( 1.1) 1+ R2 1 + 2 kΩ(0.9) 3V 3V = = 1.347V and R3 3kΩ(0.9) 1+ R4 1 + 2 kΩ( 1.1) RTHA = 2 kΩ( 1.1)2 kΩ(0.9) R1 R2 = = 0.990kΩ R1 + R2 2 kΩ( 1.1)+ 2 kΩ(0.9) 3kΩ(0.9)2 kΩ( 1.1) R3 R4 = = 1.21kΩ R3 + R4 3kΩ(0.9)+ 2 kΩ( 1.1) VC = RTHC = ⎛1.350 − 1.347 ⎞ V min =⎜ = 1.39 μA ≅ 0 ID ⎟ ⎝ 0.990 + 1.21 ⎠ kΩ 53 3.68 SPICE Input *Problem 3.68 V1 1 0 DC 4 R1 1 2 2K R2 2 0 2K R3 1 3 3K R4 3 0 2K D1 2 3 DIODE .MODEL DIODE D IS=1E-15 RS=0 .OP .END Results NAME D1 MODEL ID VD DIODE 1.09E-10 3.00E-01 The diode is essentially off - VD = 0.3 V and ID = 0.109 nA. This result agrees with the CVD model results. 3.69 (a) ( a ) Diode is forward biased : V = 3 − 0 = 3 V | I = ( d ) Diode is reverse biased : I = 0 | V = 7 − 16kΩ(I ) = 7 V | VD = −10 V (b) ( a ) Diode is forward biased : V = 3 − 0.7 = 2.3 V | I = 2.3 − (−7) = 0.625 mA 16kΩ 5 − (−5) = 0.625 mA ( b ) Diode is forward biased : V = −5 + 0 = −5 V | I = 16kΩ ( c ) Diode is reverse biased : I = 0 | V = −5 + 16 kΩ(I )= −5 V | VD = −10 V 3 − (−7) ( d ) Diode is reverse biased : I = 0 | V = 7 − 16kΩ(I ) = 7 V | VD = −10 V 3.70 (a) = 0.581 mA 16kΩ 5 − (−4.3) ( b ) Diode is forward biased : V = −5 + 0.7 = −4.3 V | I = = 0.581 mA 16kΩ ( c ) Diode is reverse biased : I = 0 | V = −5 + 16 kΩ(I )= −5 V | VD = −10 V ( a ) Diode is forward biased : V = 3 − 0 = 3 V | I = ( d ) Diode is reverse biased : I = 0 A | V = 7 − 100 kΩ(I )= 7 V | VD = −10 V (b) = 100 μA 100kΩ 5 − (−5) ( b ) Diode is forward biased : V = −5 + 0 = −5 V | I = = 100 μA 100 kΩ ( c ) Diode is reverse biased : I = 0 A | V = −5 + 100kΩ(I )= −5 V | VD = −10 V 3 − (−7) 54 = 94.0 μA 100 kΩ 5 − (−4.4) ( b ) Diode is forward biased : V = −5 + 0.6 = −4.4 V | I = = 94.0 μA 100kΩ ( c ) Diode is reverse biased : I = 0 | V = −5 + 100 kΩ(I ) = −5 V | VD = −10 V ( a ) Diode is forward biased : V = 3 − 0.6 = 2.4 V | I = ( d ) Diode is reverse biased : I = 0 | V = 7 − 100kΩ(I ) = 7 V | VD = −10 V 0 − (−9) 2.4 − (−7) 3.71 (a) ( a ) D1 on , D2 on : I D2 = D1 : (409 μA, 0 V ) 22 kΩ D2 : (270 μA, 0 V ) = 409μA | I D1 = 409μA − 6−0 = 270μA 43kΩ ( b ) D1 on , D2 off : I D 2 = 0 | I D1 = 140 μA, 0 V ) D1 : ( 6−0 = 140 μA | VD2 = −9 − 0 = −9V 43kΩ D2 : (0 A, − 9 V ) ( c ) D1 off, D2 on : I D1 = 0 | I D2 = D1 : (0 A, −3.92 V ) 0 − (−6) 6 − (−9) 65kΩ D2 : (230 μA,0 V ) = 230 μA | VD1 = 6 − 43 x103 I D2 = −3.92 V ( d ) D1 on, D2 on : I D 2 = 140 μA,0 V ) D1 : ( (b) (a) D1 on, D2 on : I D2 = 43kΩ D2 : (270 μA,0 V ) = 140 μA | I D1 = 9−0 − 140 μA = 270 μA 22kΩ -0.75 − 0.75 − (−9) 22 kΩ 184 μA, 0.75 V) D2 : (341 μA, 0.75 V) D1 : ( (b) D1 on, D2 off : = 341μA | I D1 = 341μA − 6 − (−0.75) 43kΩ = 184μA 6 − 0.75 = 122μA | VD 2 = −9 − 0.75 = −9.75V 43kΩ 122 μA, 0.75 V) D2 : (0 A, − 9.75 V) D1 : ( I D2 = 0 | I D1 = 55 ( c ) D1 off, D2 on : I D1 = 0 | I D2 = 6 − 0.75 − (−9) 65kΩ D1 : (0 A, − 3.43 V ) D2 : (219 μA, 0.75 V ) (d) D1 on, D2 on : I D2 = = 219μA | VD1 = 6 − 43 x103 I D2 = −3.43V 0.75 − 0.75 − (−6) 43kΩ D1 : (235 μA, 0.75 V) 9 − 0.75 − 400μA = 235μA 22 kΩ D2 : ( 140 μA, 0.75 V) = 140μA | I D1 = 3.72 (a) (a) D1 and D2 forward biased 15 kΩ D1 : (0 V, 200 μA) I D2 = 0 − (−9) V = 600μA I D1 = I D2 − D2 : (0 V, 600 μA) 6 − (0) V = 200μA 15 kΩ (b) D1 forward biased, D2 reverse biased 6−0 V = 400μA VD2 = −9 − 0 = −9 V 15 kΩ D1 : (0 V, 400 μA) D2 : (-9 V, 0 A ) I D1 = (c) D1 reverse biased, D2 forward biased = 500μA VD1 = 6 − 15000 I D2 = −1.50V 30 kΩ D1 : (−1.50 V, 0 A) D2 : (0 V, 500 μA) I D2 = 6V − (−9V ) (d) D1 and D2 forward biased 15 kΩ D1 : (0 V, 200 μA) I D2 = 0 − (−6) V = 400μA I D1 = 9 − (0) V 15 kΩ D2 : (0 V, 400 μA) − I D 2 = 200μA (b) (a) D1 on, D2 on : I D2 = -0.75 − 0.75 − (−9) 6 − (−0.75) 15kΩ 15kΩ D1 : (50.0 μA, 0.75 V) D2 : (500 μA, 0.75 V) = 500μA | I D1 = 500μA − = 50.0μA 56 (b) D1 on, D2 off : 6 − 0.75 = 350μA | VD 2 = −9 − 0.75 = −9.75V 15kΩ D1 : (350 μA, 0.75 V) D2 : (0 A, − 9.75 V) I D2 = 0 | I D1 = ( c ) D1 off, D2 on : I D1 = 0 | I D2 = 6 − 0.75 − (−9) = 475μA | VD1 = 6 − 15 x103 I D2 = −1.13V 30 kΩ D1 : (0 A, − 1.13 V ) D2 : (475 μA, 0.75 V ) (d) D1 on, D2 on : I D2 = 0.75 − 0.75 − (−6) 15kΩ D1 : ( 150 μA, 0.75 V) 9 − 0.75 − 400μA = 150μA 15kΩ D2 : (400 μA, 0.75 V) = 400μA | I D1 = 3.73 Diodes are labeled from left to right (a) D1 on, D2 off, D3 on : I D 2 = 0 | I D1 = I D3 + 0.990 mA = 0 − (−5) 10 − 0 = 0.990mA 3.3kΩ + 6.8 kΩ → I D 3 = 1.09 mA | VD 2 = 5 − ( 10 − 3300 I D1 ) = −1.73V 2.4 kΩ 1.09 mA, 0 V) D1 : (0.990 mA, 0 V) D2 : (0 mA, − 1.73 V) D3 : ( ( b ) D1 on, D2 off, D3 on : I D2 = 0 | I D3 = 0 = 0.495mA | VD 2 = 5 − ( 10 − 8200 I D1 )= −0.941V 8.2 kΩ + 12 kΩ 0 − (−5V ) − I D1 = 0.005mA I D3 = 10kΩ D1 : (0.495 mA, 0 V ) D2 : (0 A, − 0.941 V ) D3 : (0.005 mA, 0 V ) I D1 = (10 − 0)V 57 (c) D1 on, D2 on, D3 on I D1 = 0 − (−10) 0 − (2) V = 1.22mA > 0 | I12 K = V = −0.167 mA | I D 2 = I D1 + I12 K = 1.05mA > 0 8.2 kΩ 12 kΩ 2 − (−5) V = 0.700mA | I D3 = I10 K − I12 K = 0.533mA > 0 I10 K = 10kΩ D1 : ( 1.22 mA, 0 V ) D2 : ( 1.05 mA, 0 V ) D3 : (0.533 mA, 0 V ) (d) D1 off, D2 off, D3 on : I D1 = 0, I D 2 = 0 V = 1.21mA > 0 | VD1 = 0 − (−5 + 4700 I D3 ) = −0.667V < 0 4.7 + 4.7 + 4.7 kΩ 12 − 4700 I D 3 ) = −1.33V < 0 VD2 = 5 − ( I D3 = 1.21 mA, 0 V ) D1 : (0 A, − 0.667 V ) D2 : (0 A, − 1.33 V ) D3 : ( 12 − (−5) 3.74 Diodes are labeled from left to right (a) D1 on, D2 off, D3 on : I D 2 = 0 | I D1 = I D3 + 0.990 mA = −0.6 − (−5) 10 − 0.6 − (−0.6) 3.3kΩ + 6.8 kΩ = 0.990 mA → I D3 = 0.843mA | VD2 = 5 − ( 10 − 0.6 − 3300 I D1 )= −1.13V 2.4 kΩ D1 : (0.990 mA, 0.600 V) D2 : (0 A, − 1.13 V) D3 : (0.843 mA, 0.600V) (b) D1 on, D2 off, D3 off : I D 2 = 0 | I D 3 = 0 V = 0.477 mA | VD 2 = 5 − ( 10 − 0.6 − 8200 I D1 )= −0.490V 8.2 kΩ + 12 kΩ + 10 kΩ VD3 = 0 − (−5 + 10000 I D1 )= +0.230V < 0.6V so the diode is off I D1 = D1 : (0.477 mA, 0.600 V) D2 : (0 A, − 0.490 V) D3 : (0 A, 0.230 V) −0.6 − (−9.4) V 8.2 −0.6 − ( 1.4) V 12 kΩ 10 − 0.6 − (−5) ( c ) D1 on, D2 on, D3 on I D1 = kΩ = 1.07mA > 0 | I12 K = = −0.167 mA I D2 = I D1 + I12 K = 0.906 mA > 0 | I10 K = D1 : ( 1.07 mA, 0.600 V) 1.4 − (−5) V = 0.640 mA | I D 3 = I10 K − I12 K = 0.807 mA > 0 kΩ 10 D2 : (0.906 mA, 0.600 V) D3 : (0.807 mA, 0.600 V) 58 11.4 − (−5) V = 1.16mA > 0 | VD1 = 0 − (−5 + 4700 I D3 ) = −0.452V < 0 4.7 + 4.7 + 4.7 kΩ VD2 = 5 − ( 11.4 − 4700 I D 3 ) = −0.948V < 0 I D3 = D1 : (0 A, − 0.452 V ) D2 : (0 A, − 0.948 V ) D3 : ( 1.16 mA, 0.600 V ) (d) D1 off, D2 off, D3 on : I D1 = 0, I D 2 = 0 3.75 *Problem 3.75(a) (Similar circuits are used for the other three cases.) V1 1 0 DC 10 V2 4 0 DC 5 V3 6 0 DC -5 R1 2 3 3.3K R2 3 5 6.8K R3 5 6 2.4K D1 1 2 DIODE D2 4 3 DIODE D3 0 5 DIODE .MODEL DIODE D IS=1E-15 RS=0 .OP .END NAME D1 D2 D3 MODEL DIODE DIODE DIODE ID 9.90E-04 -1.92E-12 7.98E-04 VD 7.14E-01 -1.02E+00 7.09E-01 NAME D1 D2 MODEL DIODE DIODE ID 4.74E-04 -4.22E-13 VD 6.95E-01 -4.21E-01 NAME D1 MODEL DIODE ID 8.79E-03 VD 7.11E-01 D3 DIODE 2.67E-11 2.63E-01 D2 D3 DIODE DIODE 1.05E-03 7.96E-04 7.16E-01 7.09E-01 NAME D1 D2 D3 MODEL DIODE DIODE DIODE ID -4.28E-13 -8.55E-13 1.15E-03 VD -4.27E-01 -8.54E-01 7.18E-01 For all cases, the results are very similar to the hand analysis. 3.76 59 = 1.50 mA | I D 2 = 0 10 kΩ + 10 kΩ 0 − (−10) I D3 = = 1.00mA | VD2 = 10 − 10 4 I D1 − 0 = −5.00V 10kΩ D1 : ( 1.50 mA, 0 V) D2 : (0 A, −5.00 V) D3 : ( 1.00 mA, 0 V) 3.77 *Problem 3.77 V1 1 0 DC -20 V2 4 0 DC 10 V3 6 0 DC -10 R1 1 2 10K R2 4 3 10K R3 5 6 10K D1 3 2 DIODE D2 3 5 DIODE D3 0 5 DIODE .MODEL DIODE D IS=1E-14 RS=0 .OP .END NAME D1 D2 D3 MODEL DIODE DIODE DIODE ID 1.47E-03 -4.02E-12 9.35E-04 VD 6.65E-01 -4.01E+00 6.53E-01 I D1 = 10 − (−20) The simulation results are very close to those given in Ex. 3.8. 3.78 3.9 kΩ = 6.28V | RTH = 11kΩ 3.9 kΩ = 2.88 kΩ 3.9 kΩ + 11kΩ 6.28 − 4 IZ = = 0.792 mA > 0 | (I Z , VZ )= (0.792 mA,4 V ) 2.88 kΩ VTH = 24V 60 3.79 −6.28 = 2880 I D + VD | I D = 0, VD = −6.28V | VD = 0, I D = -6 -5 -4 -3 -2 -1 v D −6.28 = −2.18 mA 2880 Q-point -1 mA -2 mA i D Q-Point: (-0.8 mA, -4 V) 3.80 IS = 3.81 27 − 9 9V = 1.20mA → I L < 1.20 mA | RL > = 7.50 kΩ 15kΩ 1.2mA 27 − 9 = 1.20mA | P = (9V )( 1.20 mA) = 10.8 mW 15kΩ IS = 3.82 PZmax = 9V ( .95)(0.796 mA)= 6.81 mW I min Z ⎛ 1 VS − VZ VZ VS 1⎞ − = − VZ ⎜ + ⎟ | PZ = VZ I Z RS RL RS ⎝ RS RL ⎠ ⎛ 1 30V 1 ⎞ nom nom IZ = − 9V ⎜ + ⎟ = 0.500 mA | PZ = 9V (0.500 mA)= 4.5 mW 15kΩ 15 k Ω 10 k Ω ⎝ ⎠ ⎞ ⎛ 30V ( 1.05) 1 1 max ⎟ ⎜ IZ = − 9V (0.95) + ⎜15kΩ 0.95 10kΩ(1.05) ⎟ = 0.796 mA 15kΩ(0.95) ( ) ⎠ ⎝ IZ = ⎞ ⎛ 1 1 ⎟ ⎜ = − 9V ( 1.05) ⎜15kΩ 1.05 + 10 kΩ(0.95) ⎟ = 0.215 mA 15kΩ( 1.05) ( ) ⎠ ⎝ 30V (0.95) PZmin = 9V ( 1.05)(0.215mA)= 2.03 mW 3.83 ( a ) VTH = 60V 100Ω 24 − 15 = 24.0V | RTH = 150Ω 100Ω = 60Ω | I Z = = 150 mA 150Ω + 100Ω 60 60 − 15 = 300 mA | P = 15 I Z = 4.50 W P = 15 I Z = 2.25 W | ( b ) I Z = 150 61 3.84 IZ = ⎛ 1 VS − VZ VZ VS 1⎞ − = − VZ ⎜ + ⎟ RS RL RS ⎝ RS RL ⎠ | PZ = VZ I Z nom IZ = max IZ PZmax = 15V (0.90)(266mA)= 3.59 W min = IZ = 150 mA | PZnom = 15V ( 150 mA) = 2.25 W 150Ω 100Ω ⎛ ⎞ 60V ( 1.1) 1 1 ⎜ ⎟ = − 15V (0.90) + ⎜ 150Ω 0.90 100Ω(1.1) ⎟ = 266 mA 150Ω(0.90) ( ) ⎝ ⎠ 60V (0.90) ⎛ ⎞ 1 1 ⎜ ⎟ − 15V ( 1.1) + ⎜ 150Ω 1.1 100Ω(0.9) ⎟ = 43.9 mA 150Ω( 1.1) ( ) ⎝ ⎠ (60 − 15)V − 15V 1.1)(43.9mA)= 0.724 W PZmin = 15V ( 3.85 Using MATLAB, create the following m-file with f = 60 Hz: function f=ctime(t) f=5*exp(-10*t)-6*cos(2*pi*60*t)+1; Then: fzero('ctime',1/60) yields ans = 0.01536129698461 and T = (1/60)-0.0153613 = 1.305 ms. ΔT = ΔT = 3.86 1 120π 1 120π IT 5 2Vr | Vr = = = 0.8333V VP C 0.1(60) 2(0.8333) 6 = 1.40 ms ⎛ I ⎞ ⎛ 48.6 A ⎞ VD = nVT ln⎜1 + D ⎟ = 2(0.025V )ln⎜1 + −9 ⎟ = 1.230 V ⎝ 10 A ⎠ ⎝ IS ⎠ 62 3.87 ⎛ I ⎞ Von = nVT ln⎜1 + D ⎟ | VD = Von + I D RS ⎝ IS ⎠ ⎛ 100 A ⎞ VD = 1.6(0.025V )ln⎜1 + −8 ⎟ + 100 A(0.01Ω) = 1.92 V ⎝ 10 A ⎠ ⎛100 A ⎞⎛ 1ms ⎞ I ΔT = 0.92V ⎜ Pjunction ≅ Von I DC = Von P ⎟⎜ ⎟ = 2.75 W 2T ⎝ 2 ⎠⎝16.7 ms ⎠ 2 4⎛ T ⎞ 2 4 ⎛16.7 ms ⎞ PR ≅ ⎜ ⎟ I DC RS = ⎜ ⎟(3 A) 0.01Ω = 2.00 W 3 ⎝ ΔT ⎠ 3 ⎝ 1ms ⎠ Ptotal = 4.76 W 3.88 VDC = 1 T T ∫ v (t )dt = T ⎢ (V ⎣ 0 1⎡ P − Von )T − VDC = 0.975( 18V )= 17.6 V 3.89 0.05( VP − Von )⎤ TVr ⎤ ⎡ ⎥ = 0.975( VP − Von )− VP − Von ) ⎥ = ⎢( 2 ⎦ ⎢ 2 ⎥ ⎣ ⎦ 1 PD = T T 2 1 ΔT 2 ⎛ t ⎞ 1− ⎟ RS dt ∫ i (t )RS dt = T ∫ I P ⎜ ⎝ ΔT ⎠ 0 0 2 D ΔT ΔT 2 2 ⎛ 2t t2 ⎞ IP t2 t3 ⎞ RS ⎛ 1 − + dt = t − + ⎟ ⎟ ⎜ ⎜ ∫ ΔT ΔT 2 T ⎝ ΔT 3ΔT 2 ⎠ ⎠ 0 ⎝ 0 2 I R ⎛ ΔT ⎞ 1 2 ⎛ ΔT ⎞ PD = P S ⎜ΔT − ΔT + ⎟ = I P RS ⎜ ⎟ T ⎝ 3 ⎠ 3 ⎝ T ⎠ I2R PD = P S T 3.90 Using SPICE with VP = 10 V. 15V Voltage 10V 5V 0V -5V -10V t -15V 0s 10ms 20ms 30ms 40ms 50ms 63 3.91 VP − Von ) = − 6.3 2 − 1 = −7.91V (a) Vdc = −( (c) PIV ≥ 2VP = 2 ⋅ 6.3 2 = 17.8V (e) ΔT = 1 ( ) (b) C = I T 7.91 1 1 = = 1.05F Vr 0.55 0.5 60 (d) I surge = ωCVP = 2π (60)( 1.05) 6.3 2 = 3530 A ( ) 2( .25) 2Vr 2T 7.91 2 1 1 = = 0.628 ms | I P = I dc = = 841 A ΔT .5 60 .628 ms ω VP 2π (60) 6.3 2 3.92 VOnom = −( VP − Von )= − 6.3 2 − 1 = −7.91V 1.1) 2 − 1 = −8.80V VOmax = − VPmax − Von = − 6.3( VOmin min P ( = −( V − Von ) [ ] )= −[6.3(0.9) 2 − 1]= −7.02V ( ) 3.93 *Problem 3.93 VS 1 0 DC 0 AC 0 SIN(0 10 60) D1 2 1 DIODE R 2 0 0.25 C 2 0 0.5 .MODEL DIODE D IS=1E-10 RS=0 .OPTIONS RELTOL=1E-6 .TRAN 1US 80MS .PRINT TRAN V(1) V(2) I(VS) .PROBE V(1) V(2) I(VS) .END Circuit3_93b-Transient-8 +0.000e+000 +10.000m +20.000m +30.000m +40.000m +50.000m +60.000m +70.000m Time (s) +10.000 +5.000 +0.000e+000 -5.000 -10.000 V(2) *REAL(Rectifier)* SPICE Graph Results: VDC = 9.29 V, Vr = 1.05 V, IP = 811 A, ISC = 1860 A I T 9.00V 1 1 Vdc = −( VP − Von ) = −( 10 − 1)= −9.00V | Vr = = = 1.20V C 0.25Ω 60 s 0.5F 10)= 1890 A | ΔT = I SC = ωCVP = 2π (60)(0.5)( I P = I dc 9 2 1 2T = = 923 A ΔT 0.25 60 1.3ms 2( 1.2) 1 2Vr = = 1.30ms ω VP 2π (60) 10 1 64 Circuit3_93b-Transient-11 (Amp) +0.000e+000 +20.000m +40.000m +60.000m +80.000m +100.000m +120.000m Time (s) +140.000m +10.000 +5.000 +0.000e+000 -5.000 -10.000 V(1) V(2) *REAL(Rectifier)* SPICE Graph Results: VDC = -6.55 V, Vr = 0.58 V, IP = 150 A, ISC = 370 A Note that a significant difference is caused by the diode series resistance. 3.94 VP − Von )= − 6.3 2 − 1 = −7.91V ( a ) Vdc = −( ( c ) PIV ≥ 2VP = 2 ⋅ 6.3 2 = 17.8V ( e ) ΔT = 1 ( ) ( b) C = I T 7.91 1 1 = = 0.158 F Vr 0.25 0.5 400 ( d ) I surge = ωCVP = 2π (400)(0.158) 6.3 2 = 3540 A ( ) 2( .25) 2Vr 2T 7.91 2 1 1 = = 94.3μs | I P = I dc = = 839 A ΔT ω VP 2π (400) 6.3 2 .5 400 94.3μs 3.95 ( a ) Vdc = −( VP − Von )= − 6.3 2 − 1 = −7.91V ( c ) PIV ≥ 2VP = 2 ⋅ 6.3 2 = 17.8V ( e ) ΔT = 3.96 ( ) ( b) C = I T 7.91 1 1 = = 633μF Vr 0.25 0.5 105 ( d ) I surge = ωCVP = 2π 105 (633μF ) 6.3 2 = 3540 A = 0.377μs | I P = I dc 2T 7.91 2 1 = = 839 A 5 .5 10 0.377μs ΔT ( ) ( ) 1 2Vr = ω VP 2π 105 1 ( ) 2( .25) 6.3 2 65 (a) C = 1 IT 1 = = 556 μF Vr 3000(0.01) 60 3000 = 2120 V ( d ) ΔT = 1 ( b) PIV ≥ 2VP = 2 ⋅ 3000 = 6000V ( c ) Vrms = 2 ⎞ 2T ⎛ 2 ⎞⎛ 1 = 1⎜ ⎟⎜ I P = I dc ⎟ = 88.9 A ΔT ⎝ 60 ⎠⎝ 0.375ms ⎠ 1 2Vr = 2(0.01) = 0.375ms ω VP 2π (60) ( e ) I surge = ωCVP = 2π (60)(556μF )(3000)= 629 A 3.97 Assuming Von = 1 V: V − Von 1 1 ⎛ 1 ⎞⎛ 30 ⎞ 3.3 + 1 C= P T = V = 8.6 V | Vrms = = 3.04 V ⎜ ⎟⎜ ⎟ = 6.06 F | PIV = 2VP = 2(3.3 + 1) Vr R 0.025 ⎝ 60 ⎠⎝ 3.3⎠ 2 ΔT = 1 ω ⎛ 1 ⎞⎛ 3.3V ⎞ 1 2T VP − Von 2 = ⎜ s⎟⎜ ⎟ = 0.520 ms RC VP 2π (60) 0.110Ω(6.06 F )⎝ 60 ⎠⎝ 4.3V ⎠ ⎛ 2 ⎞⎛ ⎞ 1 2T = 30⎜ s ⎟⎜ ⎟ = 1920 A | I surge = ωCVP = 2π (60 / s)(6.06 F )(4.3V ) = 9820 A ΔT ⎝ 60 ⎠⎝ 0.520 ms ⎠ I P = I dc 3.98 40V vO 20V v1 0V vS Time -20V 0s 5ms 10ms 15ms 20ms 25ms 30ms VDC = 2(VP - Von) = 2(17 - 1) = 32 V. 3.99 *Problem 3.99 VS 2 1 DC 0 AC 0 SIN(0 1500 60) D1 2 3 DIODE D2 0 2 DIODE C1 1 0 500U C2 3 1 500U RL 3 0 3K .MODEL DIODE D IS=1E-15 RS=0 .OPTIONS RELTOL=1E-6 .TRAN 0.1MS 100MS .PRINT TRAN V(2,1) V(3) I(VS) 66 .PROBE V(3) V(2,1) I(VS) .END 4.0kV 3.0kV vO 2.0kV 1.0kV vS 0V -1.0kV Time -2.0kV 0s 20ms 40ms 60ms 80ms 100ms Simulation Results: VDC = 2981 V, Vr = 63 V The doubler circuit is effectively two half-wave rectifiers connected in series. Each capacitor is discharged by I = 3000V/3000 = 1 A for 1/60 second. The ripple voltage on each capacitor is 33.3 V. With two capacitors in series, the output ripple should be 66.6 V, which is close to the simulation result. 3.100 ( a ) Vdc = −( VP − Von ) = − 15 2 − 1 = −20.2 V ( b) C = ( c ) PIV ≥ 2VP = 2 ⋅ 15 2 = 42.4 V ( e ) ΔT = 3.101 ( ) I ⎛ T ⎞ 20.2V ⎛ 1 ⎞⎛ 1 ⎞ ⎜ ⎟= ⎜ ⎟⎜ ⎟ = 1.35 F Vr ⎝ 2 ⎠ 0.5Ω ⎝ 0.25V ⎠⎝120 s ⎠ ( d ) I surge = ωCVP = 2π (60)( 1.35) 15 2 = 10800 A ( ) 2( .25) 1 20.2V ⎛ 1 ⎞ 1 2Vr T = = 0.407 ms | I P = I dc = = 1650 A ⎜ s⎟ ω VP 2π (60) 15 2 ΔT 0.5Ω ⎝ 60 ⎠ 0.407 ms 1 ( a ) Vdc = −( VP − Von )= − 9 2 − 1 = −11.7 V ( b ) C = ( c ) PIV ≥ 2VP = 2 ⋅ 9 2 = 25.5 V ( e ) ΔT = 1 ( ) I ⎛ T ⎞ 11.7V ⎛ 1 ⎞⎛ 1 ⎞ ⎜ ⎟= ⎜ ⎟⎜ ⎟ = 0.780 F Vr ⎝ 2 ⎠ 0.5Ω ⎝ 0.25V ⎠⎝120 s ⎠ ( d ) I surge = ωCVP = 2π (60)(0.780) 9 2 = 3740 A ( ) ⎞ 2( .25) 1 1 2Vr T 11.7V ⎛ 1 ⎞⎛ = = 0.526 ms | I P = I dc = ⎜ s⎟⎜ ⎟ = 958 A ω VP 2π (60) 9 2 ΔT 0.5Ω ⎝ 60 ⎠⎝ 0.407 ms ⎠ 67 3.102 *Problem 3.102 VS1 1 0 DC 0 AC 0 SIN(0 14.14 400) VS2 0 2 DC 0 AC 0 SIN(0 14.14 400) D1 3 1 DIODE D2 3 2 DIODE C 3 0 22000U R303 .MODEL DIODE D IS=1E-10 RS=0 .OPTIONS RELTOL=1E-6 .TRAN 1US 5MS .PRINT TRAN V(1) V(2) V(3) I(VS1) .PROBE V(1) V(2) V(3) I(VS1) .END 20V 10V vS 0V -10V vO Time -20V 0s 1.0ms 2.0ms 3.0ms 4.0ms 5.0ms Simulation Results: VDC = -13.4 V, Vr = 0.23 V, IP = 108 A VDC = VP − Von = 10 2 − 0.7 = 13.4 V | Vr = ΔT = 1 120π 2Vr 1 = VP 120π 2(0.254) 14.1 1 13.4 1 = 0.254 V 3 800 22000μF = 0.504 ms I P = I dc 1 T 13.4V 1 = s = 150 A 3Ω 60 0.504 ms ΔT Circuit3_102-Transient-15 Time (s) +10.000m +12.000m +14.000m Simulation with RS = 0.02 Ω. +0.000e+000 +2.000m +4.000m +6.000m +8.000m +15.000 +10.000 +5.000 +0.000e+000 -5.000 -10.000 -15.000 V(1) V(2) *REAL(Rectifier)* Simulation Results: VDC = -12.9 V, Vr = 0.20 V, IP = 33.3 A, ISC = 362 A. RS results in a significant reduction in the values of IP and ISC. 68 3.103 (a) C = 1 ⎛ 1s ⎞⎛ 30 A ⎞ VP − Von 1 T = V = 8.6 V ⎜ ⎟⎜ ⎟ = 3.03 F (b) PIV = 2VP = 2(3.3 + 1) R 0.025 ⎝120 ⎠⎝ 3.3V ⎠ Vr 3.3 + 1 2 = 3.04 V (d) ΔT = 2(0.025)(3.3) 2Vr 1 = = 0.520 ms ω VP 2π (60) 4.3 1 ( c ) Vrms = ( e ) I P = I dc ⎞ ⎛ 1 ⎞⎛ T 1 = 30 A⎜ s ⎟⎜ ⎟ = 962 A | I surge = ωCVP = 2π (60 / s)(3.03F )(4.3V )= 4910 A ΔT ⎝ 60 ⎠⎝ 0.520 ms ⎠ 3.104 (a) C = ( c ) VS = I P = I dc I T 1 1 = = 139 μF Vr 2 3000(0.01) 2 ⋅ 120 3000 2 = 2120 V ( d ) ΔT = 1 ( b ) PIV ≥ 2VP = 6000 V ω 2 Vr 1 = 2(0.01) = 0.375 ms VP 2π (60) 139μF )(3000V )= 157 A ( e ) I surge = ωCVP = 2π (60 / s)( ⎛ 1 ⎞⎛ ⎞ 1 T = 1⎜ s ⎟⎜ ⎟ = 44.4 A ΔT ⎝ 60 ⎠⎝ 0.375ms ⎠ 3.105 The circuit is behaving like a half-wave rectifier. The capacitor should charge during the first 1/2 cycle, but it is not. Therefore, diode D1 is not functioning properly. It behaves as an open circuit. 3.106 ( a ) Vdc = −( VP − 2Von )= − 15 2 − 2 = −19.2 V ( b ) C = ( c ) PIV ≥ VP = 15 2 = 21.2 V ( e ) ΔT = 3.107 ( ) I ⎛ T ⎞ 19.2V ⎛ 1 ⎞⎛ 1 ⎞ s⎟ = 1.28 F ⎜ ⎟= ⎜ ⎟⎜ Vr ⎝ 2 ⎠ 0.5Ω ⎝ 0.25V ⎠⎝ 120 ⎠ ( d ) I surge = ωCVP = 2π (60 / s)( 1.28 F ) 15 2 = 10200 A ( ) ⎞ 2( .25) 1 2Vr T 19.2V ⎛ 1s ⎞⎛ 1 = = 0.407 ms | I P = I dc = ⎜ ⎟⎜ ⎟ = 1570 A ω VP 2π (60) 15 2 ΔT 0.5Ω ⎝ 60 ⎠⎝ 0.407 ms ⎠ 1 ⎛ 1 ⎞ I ⎛T ⎞ 1A s⎟ = 278 μF ⎜ ⎟= ⎜ Vr ⎝ 2 ⎠ 3000V (0.01)⎝ 120 ⎠ 3000 2 = 2120 V ( d ) ΔT = 1 (a) C = ( c ) VS = I P = I dc ( b ) PIV ≥ VP = 3000 V 1 2Vr = 2(0.01) = 0.375 ms ω VP 2π (60) ( e ) I surge = ωCVP = 2π (60)(278μF )(3000) = 314 A ⎛1 ⎞ 1 T = 1 A⎜ s⎟ = 44.4 A ΔT ⎝ 60 ⎠ 0.375ms 69 3.108 (a) C = ⎛ 1 ⎞ I ⎛T ⎞ 30 A s⎟ = 3.03 F ⎜ ⎟= ⎜ Vr ⎝ 2 ⎠ (0.025)(3.3V )⎝ 120 ⎠ 5.3 2 = 3.75 V ( d ) ΔT = 1 ( b ) PIV ≥ Vdc + 2Von = (3.3 + 2) = 5.3 V ( c ) Vrms = I P = I dc ⎡ 0.025(3.3)⎤ 1 2Vr ⎥ = 0.468 ms = 2⎢ ω VP 2π (60) ⎢ 5.3 ⎥ ⎣ ⎦ ( e ) I surge = ωCVP = 2π (60 / s)(3.03F )(3.3V )= 3770 A ⎛1 ⎞ T 1 = 30 A⎜ s⎟ = 1070 A ΔT ⎝ 60 ⎠ 0.468 ms 3.109 V1 = VP - Von = 49.3 V and V2 = -(VP -Von) = -49.3V. 40V 3.110 *Problem 3.110 VS1 1 0 DC 0 AC 0 SIN(0 35 60) VS2 0 2 DC 0 AC 0 SIN(0 35 60) D1 1 3 DIODE D4 2 3 DIODE D2 4 1 DIODE D3 4 2 DIODE C1 3 0 0.1 C2 4 0 0.1 R1 3 0 500 R2 4 0 500 .MODEL DIODE D IS=1E-10 RS=0 .OPTIONS RELTOL=1E-6 .TRAN 10US 50MS .PRINT TRAN V(3) V(4) .PROBE V(3) V(4) .END 3.111 v1 20V 0V -20V v2 Time -40V 0s 10ms 20ms 30ms 40ms 50ms VP − 2Von )= − 15 2 − 2 = −19.2 V ( a ) Vdc = −( ( c ) PIV ≥ VP = 15 2 = 21.2 V ( e ) ΔT = 1 ( ) ( b) C = I ⎛ T ⎞ 19.2V ⎛ 1 ⎞⎛ 1 ⎞ ⎜ ⎟= ⎜ ⎟⎜ ⎟ = 1.28 F Vr ⎝ 2 ⎠ 0.5Ω ⎝ 0.25⎠⎝120 ⎠ ( d ) I surge = ωCVP = 2π (60 / s)( 1.28 F ) 15V 2 = 10200 A ( ) ⎞ 2( .25) 1 2Vr T 19.2V ⎛ 1 ⎞⎛ 1 = = 0.407 ms | I P = I dc = ⎜ s ⎟⎜ ⎟ = 1570 A ΔT 0.5Ω ⎝ 60 ⎠⎝ 0.407 ms ⎠ ω VP 2π (60) 15 2 70 3.112 3.3-V, 15-A power supply with Vr ≤ 10 mV. Assume Von = 1 V. Rectifier Type Peak Current PIV Filter Capacitor Half Wave 533 A 8.6 V 25 F Full Wave 266 A 8. 6 V 12.5 F Full Wave Bridge 266 A 5.3 V 12.5 F (i) The large value of C suggests we avoid the half-wave rectifier. This will reduce the cost and size of the circuit. (ii) The PIV ratings are all low and do not indicate a preference for one circuit over another. (iii) The peak current values are lower for the full-wave and full-wave bridge rectifiers and also indicate an advantage for these circuits. (iv) We must choose between use of a center-tapped transformer (full-wave) or two extra diodes (bridge). At a current of 15 A, the diodes are not expensive and a four-diode bridge should be easily found. The final choice would be made based upon cost of available components. 3.113 200-V, 3-A power supply with Vr ≤ 4 V. Assume Von = 1 V. Rectifier Type Peak Current PIV Filter Capacitor Half Wave 189 A 402 V 12,500 μF Full Wave 94.3 A 402 V 6250 μF Full Wave Bridge 94.3 A 202 V 6250 μF (i) The the half-wave rectifier requires a larger value of C which may lead to more cost. (ii) The PIV ratings are all low enough that they do not indicate a preference for one circuit over another. (iii) The peak current values are lower for the full-wave and full-wave bridge rectifiers and also indicate an advantage for these circuits. (iv) We must choose between use of a center-tapped transformer (full-wave) or two extra diodes (bridge). At a current of 3 A, the diodes are not expensive and a four-diode bridge should be easily found. The final choice would be made based upon cost of available components. 71 3.114 3000-V, 1-A power supply with Vr ≤ 120 V. Assume Von = 1 V. Rectifier Type Peak Current PIV Filter Capacitor Half Wave 133 A 6000 V 41.7 μF Full Wave 66.6 A 6000 V 20.8 μF Full Wave Bridge 66.6 A 3000 V 20.8 μF (i) A series string of multiple capacitors will normally be required to achieve the voltage rating. (ii) The PIV ratings are high, and the bridge circuit offers an advantage here. (iii) The peak current values are lower for the full-wave and full-wave bridge rectifiers but neither is prohibitively large. (iv) We must choose between use of a center-tapped transformer (full wave) or extra diodes (bridge). With a PIV of 3000 or 6000 volts, multiple diodes may be required to achieve the require PIV rating. 3.115 iD 0 + = 5V 5 − VD 5 − 0.6 = 5 mA | I F = = = 4.4 mA 1kΩ 1kΩ 1kΩ ⎛ −3 − 0.6 4.4 mA ⎞ Ir = = −3.6 mA | τ S = (7 ns) ln⎜1 − ⎟ = 5.59 ns 1kΩ ⎝ −3.6 mA ⎠ ( ) 3.116 *Problem 3.143 - Diode Switching Delay V1 1 0 PWL(0 0 0.01N 5 10N 5 10.02N -3 20N -3) R1 1 2 1K D1 2 0 DIODE .TRAN .01NS 20NS .MODEL DIODE D TT=7NS IS=1E-15 .PROBE V(1) V(2) I(V1) .OPTIONS RELTOL=1E-6 .OP .END 10 5 v1 vD 0 -5 Time -10 0s 5ns 10ns 15ns 20ns Simulation results give S = 4.4 ns. 72 3.117 iD 0 + = 5V 5 − Von 5 − 0.6 = 1 A | IF = = = 0.880 A 5Ω 1Ω 5Ω ⎛ −3 − 0.6 0.880 A ⎞ = −0.720 A | τ S = (250 ns) ln⎜1 − IR = ⎟ = 200 ns 5Ω ⎝ −0.720 A ⎠ ( ) 3.118 *Problem 3.145(a) - Diode Switching Delay V1 1 0 DC 1.5 PWL(0 0 .01N 1.5 7.5N 1.5 7.52N -1.5 15N -1.5) R1 1 2 0.75K D1 2 0 DIODE .TRAN .02NS 100NS .MODEL DIODE D TT=50NS IS=1E-15 CJO=0.5PF .PROBE V(1) V(2) I(V1) .OPTIONS RELTOL=1E-6 .OP .END 2.0 v1 1.0 vD 0 -1.0 Time -2.0 0s 5ns 10ns 15ns 20ns 25ns For this case, simulation yields S = 3 ns. *Problem 3.145(b) - Diode Switching Delay V1 1 0 DC 1.5 PWL(0 1.5 7.5N 1.5 7.52N -1.5 15N -1.5) R1 1 2 0.75K D1 2 0 DIODE .TRAN .02NS 100NS .MODEL DIODE D TT=50NS IS=1E-15 CJO=0.5PF .PROBE V(1) V(2) I(V1) .OPTIONS RELTOL=1E-6 .OP .END 2.0 v1 1.0 vD 0 -1.0 Time -2.0 0s 10ns 20ns 30ns 40ns For this case, simulation yields S = 15.5 ns. 73 In case (a), the charge in the diode does not have time to reach the steady-state value given by Q = (1mA)(50ns) = 50 pC. At most, only 1mA(7.5ns) = 7.5 pC can be stored in the diode. Thus is turns off more rapidly than predicted by the storage time formula. It should turn off in approximately t = 7.5pC/3mA = 2.5 ns which agrees with the simulation results. In (b), the diode charge has had time to reach its steady-state value. Eq. (3.103) gives: (50 ns) ln (1-1mA/(3mA)) = 14.4 ns which is close to the simulation result. 3.119 IC = 1 − 10−15 exp(40VC )− 1 A | For VC = 0, I SC = 1 A VOC = 1 ⎛ 1 ⎞ ln⎜1 + −15 ⎟ = 0.864 V 40 ⎝ 10 ⎠ [ ] P = VC IC = VC 1 − 10−15 exp(40VC )− 1 [ [ ]] dP = 1 − 10−15 exp(40VC )− 1 − 40 x10−15 VC exp(40VC ) = 0 dVC Using the computer to find VC yields VC = 0.7768 V, IC = 0.9688 A, and Pmax = 7.53 Watts 3.120 ( a) For VOC, each of the three diode teminal currents must be zero, and 1 VOC = VC1 + VC 2 + VC 3 = V ln( 1.05 x1015 )+ ln( 1.00 x1015 )+ ln(0.95 x1015 ) = 2.59 V 40 (b) For ISC, the external currents cannot exceed the smallest of the short circuit current [ ] [ ] of the individual diodes. Thus, ISC = min[1.05 A,1.00 A,0.95 A] = 0.95 A Note that diode three will be reversed biased in part (b). Using the computer to find VC yields VC = 0.7768 V, IC = 0.9688 A, and Pmax = 7.53 Watts 3.121 hc λ= E ) = 1.11 μm - far infrared 1.12eV ( 1.602 x10 j / eV ) 3 x10 m / s) 6.625 x10 J − s( = 0.875 μm - near infrared ( b) λ = 1.42eV ( 1.602 x10 j / eV ) (a) λ = 6.625 x10−34 J − s 3 x108 m / s −19 −34 8 −19 ( 74 CHAPTER 4 4.1 (a) VG > VTN corresponds to the inversion region (b) VG 0 → enhancement - mode transistor 395 (4 − VTN ) μA = 2 → VTN = 1.5 V → K n = 125 140 (3 − VTN ) V2 2 78 4.19 Using the parameter values from problem 4.22: 800uA VGS = 5 V 600uA VGS = 4.5 V Drain Current (A) 400uA VGS = 4 V VGS = 3.5 V 200uA VGS = 3 V VGS = 2.5 V VGS = 2 V 0 A 0V 1.0V 2.0V 3.0V 4.0V 5.0V 6.0V Drain-Source Voltage (V) 4.20 (a) For VGS = 0, VGS ≤ VTN and ID = 0 (b) For VGS = 1 V , VGS = VTN and ID = 0 (c ) VGS − VTN ID = = 2 -1 =1V and VDS = 3.3 | VDS > (VGS − VTN ) so the saturation region is correct mA 375 μA ⎛ 5μm ⎞ μA ⎛ 5μm ⎞ 2 2 ' W = 375 2 ⎜ ⎟ = 3.75 2 ⎟(2 − 1) V = 1.88 mA | K n = K n 2 ⎜ V 2 V ⎝ 0.5μum ⎠ L V ⎝ 0.5um ⎠ = 3 -1 = 2V and VDS = 3.3 | VDS > (VGS − VTN ) so the saturation region is correct 375 μA ⎛ 5μm ⎞ 2 2 ⎟(3 − 1) V = 7.50 mA 2 ⎜ 2 V ⎝ 0.5μm ⎠ (d) VGS − VTN ID = 4.21 (a) For VGS = 0, VGS < VTN and ID = 0 (b) For VGS = 1 V , VGS < VTN and ID = 0 (c ) VGS − VTN ID = = 2 -1.5 = 0.5V and VDS = 4 | VDS > (VGS − VTN ) so the saturation region is correct 200 μA ⎛10μm ⎞ μA ⎛ 10μm ⎞ mA 2 2 ' W V = 250 μ A | K = K = 200 2 − 1.5 ⎜ ⎟ ⎟ = 2.00 2 ( ) n n 2 2 ⎜ 2 V ⎝ 1um ⎠ L V ⎝ 1um ⎠ V = 3 -1.5 =1.5V and VDS = 4 | VDS > (VGS − VTN ) so the saturation region is correct 200 μA ⎛10μm ⎞ 2 2 ⎟(3 − 1.5) V = 2.25 mA 2 ⎜ 2 V ⎝ 1μm ⎠ (d) VGS − VTN ID = 79 4.22 (a) VGS - VTN = 2 - 0.75 = 1.25 V and VDS = 0.2 V. VDS < VGS - VTN so the transistor is operating in the triode region. ⎛ V ⎞ 0.2 ⎞ W⎛ μA ⎞⎛10 ⎞⎛ ⎜VGS − VTN − DS ⎟VDS = ⎜200 2 ⎟⎜ ⎟⎜ 2 − 0.75 − ⎟0.2 = 460 μA ⎝ 2 ⎠ 2 ⎠ L⎝ V ⎠⎝ 1 ⎠⎝ (b) VGS - VTN = 2 - 0.75 = 1.25 V and VDS = 2.5 V. VDS > VGS - VTN so the transistor is operating in the saturation region. ⎛ 200 μA ⎞⎛10 ⎞ K' W 2 2 ID = n ⎟(2 − 0.75) = 1.56 mA (VGS − VTN ) = ⎜ 2 ⎟⎜ ⎝ 2 V ⎠⎝ 1 ⎠ 2 L (c) VGS < VTN so the transistor is cutoff with ID = 0. ⎛ 300 ⎞ ' ID ∝ K n so (a) ID = ⎜ (d) ⎟460μA = 690μA (b) ID = 2.34 mA (c ) ID = 0 ⎝ 200 ⎠ ' ID = K n 4.23 (a) VGS - VTN = 4 V, VDS = 6 V. VDS > VGS - VTN --> Saturation region (b) VGS < VTN --> Cutoff region (c) VGS - VTN = 1 V, VDS = 2 V. VDS > VGS - VTN --> Saturation region (d) VGS - VTN = 0.5 V, VDS = 0.5 V. VDS = VGS - VTN --> Boundary between triode and saturation regions (e) The source and drain of the transistor are now reversed because of the sign change in VDS. Assuming the voltages are defined relative to the original S and D terminals as in Fig. P4.11(b), VGS = 2 - (-0.5) = 2.5 V, VGS - VTN = 2.5 - 1 = 1.5 V, and VDS = 0.5 V --> triode region (f) The source and drain of the transistor are again reversed because of the sign change in VDS. Assuming the voltages are defined relative to the original S and D terminals as in Fig. P4.11(b), VGS = 3 - (-6) = 9 V, VGS - VTN = 9 - 1 = 8.0 V, and VDS = 6 V --> triode region D --> 'S' G + 2V (e) + S --> 'D' VG'S' = +2.5 V VD'S' = +0.5 V G 0.5 V + 3V (f) + S --> 'D' V G'S' = +9.0 V V D'S' = +6.0 V D --> 'S' - 6.0 V 80 4.24 (a) VGS - VTN = 2.6 V, VDS = 3.3 V. VDS > VGS - VTN --> Saturation region (b) VGS < VTN --> Cutoff region (c) VGS - VTN = 1.3 V, VDS = 2 V. VDS > VGS - VTN --> Saturation region (d) VGS - VTN = 0.8 V, VDS = 0.5 V. VDS < VGS - VTN --> triode region (e) The source and drain of the transistor are now reversed because of the sign change in VDS. Assuming the voltages are defined relative to the original S and D terminals as in Fig. 4.54(b), VGS = 2 - (-0.5) = 2.5 V, VGS - VTN = 2.5 – 0.7 = 1.8 V, and VDS = 0.5 V --> triode region (f) The source and drain of the transistor are again reversed because of the sign change in VDS. Assuming the voltages are defined relative to the original S and D terminals as in Fig. 4.54(b), VGS = 3 - (-3) = 6 V, VGS - VTN = 6 – 0.7 = 5.3 V, and VDS = 3 V --> triode region 4.25 R R 2 4 D G B S R R V DD + - 1 3 4.26 +V DD D G I B S +V DD D D G S D B G S B I B G S D B G S (a) 4.27 (b) 81 VDS = 3.3V, VGS – VTN = 1.3 V; VDS > VGS - VTN so the transistor is saturated. (a) gm = K n (VGS − VTN ) = 250 μA ⎛ 20μm ⎞ ⎜ ⎟(2 − 0.7) = 6.50 mS V 2 ⎝ 1μm ⎠ μA ⎛ 20μm ⎞ ⎜ ⎟(3.3 − 0.7) = 13.0 mS V 2 ⎝ 1μm ⎠ (b) gm = K n (VGS − VTN ) = 250 4.28 (a) gm = ΔiD 760 − 140 μA = = 310 μS | As a check, we can use the results from Problem 4.22. ΔvGS 5−3 V gm = K n (VGS − VTN ) = 125 (b) gm = ΔiD ΔvGS (4 − 1.5)V = 313 μS V2 μA 390 − 15 μA = = 188 μS | Checking : gm = 125 2 (3 − 1.5)V = 188 μS V 4−2 V μA 4.29 VDS > VGS - VTN so the transistor is saturated. Kn 250 μA 2 2 5 − 0.75) ( 1 + 0.025(6)) = 2.60 mA (VGS − VTN ) (1 + λVDS ) = 2 ( 2 2 V K 250 μA 2 2 (b) ID = n (VGS − VTN ) = 5 − 0.75) = 2.26 mA 2 ( 2 2 V ( a) ID = 4.30 VDS > VGS - VTN so the transistor is saturated. Kn 500 μA 2 2 4 − 1) ( 1 + 0.02(5)) = 2.48 mA (VGS − VTN ) (1 + λVDS ) = 2 ( 2 2 V K 500 μA 2 2 (b) ID = n (VGS − VTN ) = 4 − 1) = 2.25 mA 2 ( 2 2 V ( a) ID = 4.31 (a) The transistor is saturated by connection. ID = 12V − VGS 100 x10−6 ⎛10 ⎞ A 2 = ⎜ ⎟ 2 (VGS − 0.75V ) 5 ⎝ 1 ⎠V 10 Ω 2 2 12.5VGS − 17.8VGS − 4.97 = 0 VGS = 0.266V , 1.214V ⇒ VGS = 1.214 V since it must exceed 0.75V ID = 12 − 1.214 = 108 μA 10 5 Checking : 100 x10−6 ⎛10 ⎞ A 2 ⎜ ⎟ 2 (1.214 − 0.75V ) = 108 μA ⎝ 1 ⎠V 2 (b) ID = 12V − VGS 1000 x10−6 A 2 = V − 0.75V ) (1 + 0.025VGS ) 5 2 ( GS 10 Ω 2 V 82 Starting with the solution from part (a) and solving iteratively yields VGS = 1.20772 V and ID = 108 μA, essentially no change. (c) 12V − VGS 100 x10−6 ⎛ 25 ⎞ A 2 ID = = ⎜ ⎟ 2 (VGS − 0.75V ) 5 ⎝ 1 ⎠V 10 Ω 2 2 62.5VGS − 91.75VGS + 11.16 = 0 VGS = 0.446V , 1.046V ⇒ VGS = 1.046 V since VGS must exceed the threshold voltage. 12 − 1.046 ID = = 110 μA 10 5 100 x10−6 ⎛ 25 ⎞ A 2 Checking : ID = ⎜ ⎟ 2 (1.046 − 0.75V ) = 110 μA ⎝ 1 ⎠V 2 4.32 (a) The transistor is saturated by connection. 12V − VGS 100 x10−6 ⎛10 ⎞ A 2 ID = = ⎜ ⎟ 2 (VGS − 0.75V ) 4 ⎝ 1 ⎠V 5 x10 Ω 2 2 31.25VGS − 45.88VGS + 5.58 = 0 VGS = 0.0588V , 1.401V ⇒ VGS = 1.401 V since VGS must exceed the threshold voltage. 12 − 1.401 100 x10−6 ⎛10 ⎞ A 2 = 212 μ A Checking : I = ⎜ ⎟ 2 (1.401 − 0.75V ) = 212 μA D 4 ⎝ 1 ⎠V 5 x10 2 −6 12V − VGS 1000 x10 A 2 = V − 0.75V ) (1 + 0.02VGS ) (b) ID = 4 2 ( GS 5 x10 Ω 2 V ID = Starting with the solution from part (a) and solving iteratively yields VGS = 1.3925 V and IDS = 212 μA, essentially no change 4.33 (a) Since VDS = VGS and VTN > 0 for both transistors, both devices are saturated. ' ' W W Kn Kn 2 2 I = V − V and I = ( GS1 TN ) (VGS 2 − VTN ) . Therefore D1 D2 2 L 2 L From the circuit, however, ID2 must equal ID1 since IG = 0 for the MOSFET: ' ' Kn Kn W W 2 2 I = ID1 = ID 2 or (VGS1 − VTN ) = (VGS 2 − VTN ) 2 L 2 L which requires VGS1 = VGS2. Using KVL: VDD = VDS1 + VDS 2 = VGS1 + VGS 2 = 2VGS 2 V VGS1 = VGS 2 = DD = 5V 2 ' K W 100 μA 10 2 2 I= n (VGS1 − VTN ) = (5 − 0.75) V 2 = 9.03 mA 2 2 V 1 2 L (b) The current simply scales by a factor of two (see last equation above), and ID = 18.1 mA. (c) For this case, 83 ' ' Kn Kn W W 2 2 V − V 1 + λ V and I = ( GS1 TN ) ( (VGS 2 − VTN ) (1 + λVDS2 ). DS1 ) D2 2 L 2 L Since VGS = VDS for both transistors ' ' Kn Kn W W 2 2 ID1 = (VGS1 − VTN ) (1 + λVGS1 ) and ID 2 = (VGS 2 − VTN ) (1 + λVGS2 ) 2 L 2 L and ID1 = ID2 = I ' K' W Kn W 2 2 (VGS1 − VTN ) (1 + λVGS1 ) = n (VGS 2 − VTN ) (1 + λVGS2 ) 2 L 2 L which again requires VGS1 = VGS2 = VDD/2 = 5V. K' W 100 μA 10 2 2 I= n (VGS1 − VTN ) (1 + λVDS ) = (5 − 0.75) V 2 (1 + (.04 )5) = 10.8 mA 2 2 L 2 V 1 ID1 = 4.34 (a) Since VDS = VGS and VTN > 0 for both transistors, both devices are saturated (“by ' ' ⎛W ⎞ ⎛W ⎞ Kn Kn 2 2 I = V − V and I = ⎜ ⎟ ⎜ ⎟ (VGS 2 − VTN ) . ( ) D 1 GS 1 TN D 2 Therefore 2 ⎝ L ⎠1 2 ⎝ L ⎠2 From the circuit, however, ID2 must equal ID1 since IG = 0 for the MOSFET: ' ' ⎛ 10 ⎞ ⎛ 40 ⎞ Kn Kn 2 2 V − V = I = ID1 = ID 2 or ⎜ ⎟( GS1 TN ) ⎜ ⎟(VGS 2 − VTN ) 2 ⎝1⎠ 2 ⎝1⎠ which requires VGS1 = 2VGS2 - VTN. Using KVL: connection”). VDD = VDS1 + VDS 2 = VGS 2 + VGS1 = 3VGS 2 − VTN V + VTN 10 + 0.75 VGS 2 = DD = = 3.583V VGS1 = 6.417 3 3 100 μA 10 K' W 2 2 I= n (VGS1 − VTN ) = (6.417 − 0.75) V 2 = 16.1 mA 2 2 V 1 2 L 100 μA 40 2 Checking : I = (3.583 − 0.75) V 2 = 16.1 mA which agrees. 2 2 V 1 (b) For this case with VGS = VDS for both transistors and ID1 = ID2, K' ⎛W ⎞ K' ⎛W ⎞ 2 2 ID1 = n ⎜ ⎟ (VGS1 − VTN ) (1 + λVGS1 ) and ID 2 = n ⎜ ⎟ (VGS 2 − VTN ) (1 + λVGS2 ) 2 ⎝ L ⎠1 2 ⎝ L ⎠2 where VGS2 = VDD – VGS1. Therefore, ' ⎛10 ⎞ K ' ⎛ 40 ⎞ Kn 2 2 1 + λ(10 − VGS1 )) ⎜ ⎟(VGS1 − VTN ) (1 + λVGS1 ) = n ⎜ ⎟(10 − VGS1 − VTN ) ( 2 ⎝1⎠ 2 ⎝1⎠ VGS1 = 6.3163, VGS2 = 3.6837, ID1 = 20.4 mA, Checking: ID2 = 20.4 mA which agrees. 84 4.35 (a) Since VDS = VGS and VTN > 0 for both transistors, both devices are saturated (“by ' ' ⎛W ⎞ ⎛W ⎞ Kn Kn 2 2 V − V and I = ⎜ ⎟ ⎜ ⎟ (VGS 2 − VTN ) . ( ) GS1 TN D2 Therefore 2 ⎝ L ⎠1 2 ⎝ L ⎠2 From the circuit, however, ID2 must equal ID1 since IG = 0 for the MOSFET: ' ' ⎛ 25 ⎞ ⎛12.5 ⎞ Kn Kn 2 2 I = ID1 = ID 2 or ⎜ ⎟(VGS1 − VTN ) = ⎜ ⎟(VGS 2 − VTN ) 2 ⎝1⎠ 2 ⎝ 1 ⎠ Solving for VGS2 yields: VGS 2 = 2VGS1 − 2 − 1 VTN connection”). ID1 = ( ) Also, VDD = VDS1 + VDS 2 or VGS1 = 10 − VGS 2 VGS1 = I= 10 + ( 2 − 1)V 1+ 2 TN = 4.271V VGS 2 = 5.729V ' 100 μA ⎛ 25 ⎞ Kn W 2 2 2 ⎟(4.271 − 0.75) V = 15.5 mA (VGS1 − VTN ) = 2 ⎜ 2 V ⎝1⎠ 2 L 100 μA ⎛ 12.5 ⎞ 2 2 Checking : I = ⎟(5.729 − 0.75) V = 15.5 mA - agrees. 2 ⎜ 2 V ⎝ 1 ⎠ (b) For this case with VGS = VDS for both transistors and ID1 = ID2, K' ⎛W ⎞ K' ⎛W ⎞ 2 2 ID1 = n ⎜ ⎟ (VGS1 − VTN ) (1 + λVGS1 ) and ID 2 = n ⎜ ⎟ (VGS 2 − VTN ) (1 + λVGS2 ) 2 ⎝ L ⎠1 2 ⎝ L ⎠2 where VGS2 = VDD – VGS1. Therefore, ' ' ⎛10 ⎞ ⎛ 40 ⎞ Kn Kn 2 2 1 + λ(10 − VGS1 )) ⎜ ⎟(VGS1 − VTN ) (1 + λVGS1 ) = ⎜ ⎟(10 − VGS1 − VTN ) ( 2 ⎝1⎠ 2 ⎝1⎠ VGS1 = 4.3265 V, VGS2 = 5.6735 V, ID1 = 19.4 mA, Checking: ID2 = 19.4 mA – both agree 4.36 VGS - VTN = 5 - (-2) = 7 V > VDS = 6 V so the transistor is operating in the triode region. 6⎞ −6 ⎛ (a) ID = 250 x10 ⎜5 − (−2) − ⎟6 = 6.00 mA ⎝ 2⎠ (b) Our triode region model is independent of λ, so ID = 6.00 mA. 4.37 Since VDS = VGS, and VTN < 0 for an NMOS depletion mode device, VGS - VTN will be greater than VDS and the transistor will be operating in the triode region. 85 4.38 ( a) VDS = 6V | VGS − VTN = 0 − (−3) = 3V so the transistor is saturated 2 Kn 250 μA 2 0 − (−3V )] = 1.13 mA (VGS1 − VTN ) = 2 [ 2 V 2 2 250 μA 0 − (−3V )] ( 1 + 0.025(6)) = 1.29 mA (b) ID = 2 [ 2 V ID = 4.39 +10 V 100 k Ω D G S (a) (b) I DS W = 10 L 1 G D -10 V 100 k Ω S I DS W = 10 L 1 (a) If the transistor were saturated, then 100 x10−6 ⎛ 10 ⎞ 2 ID = ⎜ ⎟(−2) = 2.00 mA ⎝1⎠ 2 but this would require a power supply of greater than 200 V (2 mA x 100 kΩ). Thus the transistor must be operating in the triode region. 10V − VDS VDS ⎞ −3 ⎛ = 10 0 − − 2 − ⎜ ⎟VDS ( ) ⎝ 10 5 Ω 2 ⎠ 10 − VDS = 50VDS (4 − VDS ) and VDS = 0.0504V using the quadratic equation. ⎛ 0.0504 ⎞ 10V − VDS ID = 10−3 ⎜2 − = 99.5 μA ⎟0.0504 = 99.5 μA Checking : ⎝ 2 ⎠ 10 5 Ω (b) For R = 50 kΩ and W/L = 20/1, ⎛ 10V − VDS V ⎞ = 2 x10−3⎜ 0 − (−2) − DS ⎟VDS 4 ⎝ 5 x10 Ω 2 ⎠ 10 − VDS = 50VDS (4 − VDS ), the same as part (a). ⎛ 0.0504 ⎞ 10V − VDS ID = 2 x10−3 ⎜2 − = 199 μA ⎟0.0504 = 199 μA Checking : ⎝ 5 x10 4 Ω 2 ⎠ (c) In this circuit, the drain and source terminals of the transistor are reversed because of the power supply voltage, and the current direction is also reversed. However, now VDS = VGS and since the transistor is a depletion-mode device, it is still operating in the triode region. 86 ⎛ 10V − VDS V ⎞ = 1000 x10−6 ⎜VDS − (−2) − DS ⎟VDS 5 ⎝ 10 Ω 2 ⎠ 10 − VDS = 50VDS (4 + VDS ) and VDS = 0.04915V using the quadratic equation. ⎛ 0.04915 ⎞ 10V − VDS ID = 10−3 ⎜0.04915 − (−2) − = 99.5 μA ⎟0.04915 = 99.5 μA Checking : ⎝ 10 5 Ω 2 ⎠ (d) In this circuit, the drain and source terminals of the transistor are reversed because of the power supply voltage, and the current direction is also reversed. However, now VDS = VGS and since the transistor is a depletion-mode device, it is still operating in the triode region. ⎛ 10V − VDS V ⎞ = 2000 x10−6⎜VDS − (−2) − DS ⎟VDS 4 ⎝ 5 x10 Ω 2 ⎠ 10 − VDS = 50VDS (4 + VDS ) Same as part (c). VDS = 0.04915V using the quadratic equation. ⎛ 0.04915 ⎞ 10V − VDS ID = 10−3 ⎜0.04915 − (−2) − = 99.5 μA ⎟0.04915 = 99.5 μA Checking : ⎝ 10 5 Ω 2 ⎠ 4.40 See figures in previous problem but use W/L = 20/1. 25 x10−6 ⎛ 20 ⎞ 2 I = ⎜ ⎟(−1) = 250 μA but this would require (a) If the transistor were saturated, then D 2 ⎝1⎠ a power supply of greater than 25 V. Thus the transistor must be operating in the triode region. ⎛ 20 ⎞⎛ 10V − VDS V ⎞ = 100 x10−6 ⎜ ⎟⎜0 − (−1) − DS ⎟VDS 5 ⎝ 1 ⎠⎝ 10 Ω 2 ⎠ 10 − VDS = 100VDS (2 − VDS ) and VDS = 0.05105V using the quadratic equation. ⎛ 0.05105 ⎞ 10 − 0.0510 ID = 2.00 x10−3 ⎜1 − V = 99.5 μA ⎟0.05105 = 99.5 μA Checking : ⎝ 2 ⎠ 10 5 Ω (b) In this circuit, the drain and source terminals of the transistor are reversed because of the power supply voltage, and the current direction is also reversed. However, now VDS = VGS and since the transistor is a depletion-mode device, it is still operating in the triode region. ⎛ 20 ⎞⎛ V ⎞ VDS = 10 − ( 10 5 )( 100 x10−6 ) ⎜ ⎟⎜VDS − (−1) − DS ⎟VDS ⎝ 1 ⎠⎝ 2 ⎠ ⎛ V ⎞ VDS = 10 − 200VDS ⎜1 + DS ⎟ and VDS = 0.04858V using the quadratic equation. ⎝ 2 ⎠ ⎛ 0.04858 ⎞ 10 - 0.04858 V = 99.5 μA ID = 2000 x10−6 ⎜1 + ⎟0.04858 = 99.5 μA Checking : ⎝ Ω 2 ⎠ 10 5 87 4.41 ( a) VTN = 0.75 + 0.75 1.5 + 0.6 − 0.6 = 1.26V ( ) VGS − VTN = 2 − 1.26 = 0.74V > VDS = 0.2V ⇒ Triode region ⎛10 ⎞⎛ 0.2 ⎞ ID = 200 x10−6 ⎜ ⎟⎜ 2 − 1.26 − ⎟0.2 = 256 μA (compared to 460 μA) ⎝ 1 ⎠⎝ 2 ⎠ (b) VGS − VTN = 2 − 1.26 = 0.74V < VDS = 2.5V ⇒ Saturation region 200 x10−6 ⎛ 10 ⎞ 2 ID = ⎜ ⎟(2 − 1.26) = 548 μA (compared to 1.56 mA) ⎝1⎠ 2 (c) VGS < VTN so the transistor is cut off, and ID = 0. ⎛ 300 ⎞ ' so ( a) ID = ⎜ ( d ) ID ∝ K n ⎟256μA = 384 μA (b) ID = 822 μA (c ) ID = 0 ⎝ 200 ⎠ 4.42 ( a) VTN = 1.5 + 0.5 ( 4 + 0.75 − 0.75 = 2.16V | VGS < VTN ⇒ Cutoff & ID = 0 ) (b) ID = 0. The result is independent of VDS . 4.43 ( a) VTN = 1 + 0.7 ( 3 + 0.6 − 0.6 = 1.79V ) VGS − VTN = 2.5 − 1.79 = 0.71V < VDS = 5V ⇒ Saturation region 100 x10−6 ⎛ 8 ⎞ 2 ⎜ ⎟(0.71) = 202 μA ⎝ 1⎠ 2 (b) 0.5 < 0.71 ⇒ Triode region ⎛ 8 ⎞⎛ 0.5 ⎞ ID = 100 x10−6 ⎜ ⎟⎜ 0.71 − ⎟0.5 = 184 μA ⎝ 1 ⎠⎝ 2 ⎠ ID = 4.44 0.85 = −1.5 + 1.5 VSB + 0.75 − 0.75 ( Checking : VTN = −1.5 + 1.5 ( 5.17 + 0.75 − ) | Solving for VSB yields VSB = 5.17 V 0.75 = 0.85 V ) 4.45 Using trial and error with a spreadsheet yielded: VTO = 0.74V γ = 0.84 V 2φ F = 0.87V RMS Error = 51.9 mV 88 4.46 −14 3.9εo ⎛ cm 2 ⎞ 3.9(8.854 x10 F / cm) ( a) K = μ p C = μ p = μp = ⎜200 ⎟ Tox Tox V − sec ⎠ 50 x10−9 m(100cm / m) ⎝ F μA K 'p = 13.8 x10−6 = 13.8 2 V − sec V μA 50nm μA ' (b) Scaling the result from Part (a) yields: K n = 13.8 2 = 34.5 2 V 20nm V μA 50 nm μA ' = 13.8 2 = 69.0 2 (c) K n V 10 nm V μA 50 nm μA ' = 13.8 2 = 138 2 (d) K n V 5 nm V ' p " ox εox 4.47 The pinchoff points and threshold voltage can be estimated directly from the graph: e. g. VGS = -3 V curve gives VTP = 2.5 - 3 = - 0.5 V or from the VGS = -5 V curve gives VTP = 4.5 - 5 = 0.5 V. Alternately, choosing two points in saturation, say ID = 1.25 mA for VGS = -3 V and ID = 4.05 mA for VGS = -5 V: ID1 (VGS1 − VTP ) = or ID 2 (VGS 2 − VTP ) 1.25 (−3 − VTP ) = 4.05 (−5 − VTP ) Solving for VTP yields : 0.8VTP = −0.4V and VTP = −0.500V . Solving for K p : K p = 2 ID = 2(1.25mA) = 0.400 mA W V 2 = 10 | = = μA V2 L K 'p 1 40 2 V Kp 400 μA (VGS − VTP ) 2 (−3 + 0.5) 2 4.48 Using the values from the previous problem PMOS Output Characteristics 0.0045 0.004 0.0035 0.003 0.0025 0.002 0.0015 0.001 0.0005 0 -6 -5 -4 -3 VDS -2 -1 0 -2 V -3 V -3.5 V -4 V -4.5 V -5 V (IDSAT, VDSAT): (0.45 mA,-1.5 V) (1.25 mA,-2.5 V) (1.8 mA,-3 V) (2.45 mA,-3.5 V) (3.7 mA,-4V) (4.05 mA,-4 V) 89 4.49 ( a ) VGS − VTP = −1.1 + 0.75 = −0.35V | VDS = −0.2V → Triode region (−0.2)⎤ 40μA ⎛ 20 ⎞⎡ ⎢ ⎥(−0.2) = 40.0 μA ID = − 1.1 − − 0.75 − ⎜ ⎟ ( ) 2 ⎥ V 2 ⎝ 1 ⎠⎢ ⎣ ⎦ ( b ) VGS − VTP = −1.3 + 0.75 = −0.55V | VDS = −0.2V → Triode region (−0.2)⎤ 40μA ⎛ 20 ⎞⎡ ⎢ ID = − 1.3 − − 0.75 − ⎜ ⎟ ( ) 2 ⎥ (−0.2)= 72.0 μA V 2 ⎝ 1 ⎠⎢ ⎥ ⎣ ⎦ (c ) VTP = − 0.75 + .5 1 + .6 − −6 = −0.995V VGS − VTP = −1.1 − (−0.995) = −0.105V | VDS = −0.2V → saturation region 1 ⎛ 40μA ⎞⎛ 20 ⎞ 2 ID = ⎜ 2 ⎟⎜ ⎟(−1.1 + 0.995) = 4.41 μA 2 ⎝ V ⎠⎝ 1 ⎠ ( d ) VGS − VTP = −1.3 + 0.995 = −0.305V | VDS = −0.2V → triode region 10μA ⎛ 10 ⎞⎡ (−0.2)⎤ −0.2 = 32.8 μA ID = ⎟⎢−1.3 − (−0.995) − ) ⎥( 2 ⎜ 2 ⎦ V ⎝ 1 ⎠⎣ 4.50 [ ( )] W VGS − VTP L W 1 5810 W 1 2330 ( a) = = | (b) = = −6 −6 L 40 x10 −5 + 0.70 (1) 1 L 100 x10 (5 − 0.70)(1) 1 K 'p 4.51 For PMOS : Ron = 1 or W 1 = ' L K p VGS − VTN Ron For PMOS : Ron = 1 K 'p W VGS − VTP L W 1 2.91 W 1 1.16 ( a) = = | (b) = = −6 −6 L 40 x10 −5 + 0.70 (2000) 1 L 100 x10 (5 − 0.70)(2000) 1 or W 1 = ' L K p VGS − VTN Ron 90 4.52 For PMOS : Ron = K 'p (b) Ron = 1 W VGS − VTP L ( a) Ron = 1 = 29.4 Ω −6 ⎛ 200 ⎞ 40 x10 ⎜ ⎟ −5 − (−0.75) ⎝ 1 ⎠ W 1 499 (c ) = = −6 L 40 x10 −5 − (−0.75)(11.8) 1 1 = 11.8 Ω ⎛ −6 200 ⎞ 100 x10 ⎜ ⎟(5 − 0.75) ⎝ 1 ⎠ ⎛W ⎞ K' ⎛W ⎞ 500 Checking : ⎜ ⎟ = n ⎟ = 2.5(200) = ' ⎜ ⎝ L ⎠ p K p ⎝ L ⎠n 1 4.53 +18 V R 2 R4 R S VDD G D G B + - S B D R 1 R 3 (a) (b) 4.54 (a) For VIN = 0, the NMOS device is on with VGS = 5 and VSB = 0, and the PMOS transistor is off with VGS = 0, VO = 0, and VSB = 0. 1 Ron = = 235 Ω −6 (100 x10 )(10)(5 − 0.75) (b) For VIN = 5V, the NMOS device is off with VGS = 0, and the PMOS transistor is on with VGS = -5V, VO = 5V, and VSB = 0. 1 Ron = = 235 Ω −6 (40 x10 )(25)(−5 + 0.75) 4.55 Ron ≤ 0.1V 0.5 A ID A = 0.2Ω K p = = = 0.629 2 0.5 A V (VGS − VTP − 0.5VDS )VDS [−10V − (−2V ) − 0.5(−0.1V )](−0.1V ) 4.56 VTP = −0.75 − 0.5 ( 4 + 0.6 − 0.6 = −1.44V ) VGS - VTP = −1.5 − (−1.44 ) = −0.065 | VDS = −4V ⇒ Saturation region ID = 2 40 x10−6 A ⎛ 25 ⎞ ⎟[−1.5 − (−1.44 )] = 1.80 μA 2⎜ V ⎝1⎠ 2 91 4.57 VGS − VTP = −1.5 − (−0.75) = −0.75V | VDS = −0.5V VGS - VTN < VDS ⇒ Triode region | ID = 40 x10−6 A ⎛ 40 ⎞⎡ (−0.5)⎤ −0.5 = 400 μA ⎟⎢−1.5 − (−0.75) − ) ⎥( 2⎜ V ⎝ 1 ⎠⎣ 2 ⎦ 4.58 The PMOS transistor could be either an enhancement-mode or a depletion-mode device depending upon the specific values of R1, R2 and R4. Thus an enhancement device with VTP < 0 is correct and the symbol is correct. 4.59 If this PMOS transistor is conducting, then its threshold voltage must be greater than zero and it is a depletion-mode device. The symbol is that of an enhancement-mode device and is incorrect. 4.60 +18 V R2 S D R4 R G VDD + G S R1 R 3 D (a) 4.61 R2 1.5 M Ω G S R1 R 1 MΩ S (b) RD 75 k Ω D VDD + -3 V 10 V 39 k Ω 92 4.62 R D 75 k Ω R EQ G D -5 V VDD + 10 V 600 k Ω V EQ 4V RS S 39 k Ω 93 4.63 n+ 22 λ Metal Polysilicon 12 λ 4.64 [(2x20)/(12x22)]= 0.152 or 15.2% n+ 14 λ Metal Polysilicon 18 λ [2x10/(18x14)]= 0.079 or 7.9% 94 4.65 Metal 12 λ n+ Polysilicon 12 λ (2x10/122)= 0.139 or 13.9% 4.66 Metal 12 λ n+ Polysilicon 14 λ [2x10/(14x12)] = 0.119 or 11.9% 95 4.67 (a) " Cox = εox Tox = (3.9)⎜8.854 x10−14 5 x10 cm −6 ⎛ ⎝ F⎞ ⎟ cm ⎠ = 6.906 x10−8 F cm 2 ⎛ F ⎞ " CGC = Cox WL = ⎜6.906 x10−8 2 ⎟(20 x10−4 cm)(2 x10−4 cm)= 27.6 fF ⎝ cm ⎠ F " = 1.73 x 10−7 2 | CGC = 69.1 fF (b) Cox cm F " = 3.45 x 10−7 2 | CGC = 138 fF (c ) Cox cm F " = 7.90 x 10−7 2 | CGC = 276 fF ( d ) Cox cm 4.68 " Cox = εox Tox = (3.9)⎜8.854 x10−14 1x10 cm −6 ⎛ ⎝ F⎞ ⎟ cm ⎠ = 3.46 x10−7 F cm 2 ⎛ F ⎞ " CGC = Cox WL = ⎜ 3.46 x10−7 2 ⎟(5 x10−4 cm)(5 x10−5 cm)= 8.64 fF ⎝ cm ⎠ 4.69 ' COL F⎞ ⎟ pF ε cm ⎠ = ox L = 0.5 x10−4 cm)= 17.3 ( ⎛ cm ⎞ cm Tox 10 x10−9 m⎜10 2 ⎟ ⎝ m⎠ (3.9)⎜8.854 x10−14 ⎛ ⎝ 4.70 ⎛ ⎞ −15 F 10μm)(1μm) ⎜1.4 x10 2 ⎟( ⎛ μm ⎠ C WL F ⎞ ⎝ ' + COLW = + ⎜ 4 x10−15 (a) CGS = CGD = ⎟(10μm) = 47 fF 2 2 μm ⎠ ⎝ 2 " 2 ' WL + COL W = 14 fF + 40 fF = 49 fF (b) CGS = COX 3 3 ⎛ F ⎞ ' CGD = COL W = ⎜ 4 x10−15 ⎟(10μm) = 40 fF μm ⎠ ⎝ ⎛ F ⎞ ' W = ⎜ 4 x10−15 (c ) CGS = CGD = COL ⎟(10μm ) = 40 fF μm ⎠ ⎝ " OX 96 4.71 F⎞ ⎟ εox F cm ⎠ " Cox = = = 3.453 x10−8 2 ⎛ ⎞ cm Tox cm 100 x10−9 m⎜10 2 ⎟ ⎝ m⎠ (3.9)⎜8.854 x10−14 ⎛ ⎝ CGC 4.72 ⎛ −4 cm ⎞ 2 ⎛ −8 F ⎞ 6 2 = C WL = ⎜ 3.453 x10 ⎟(50 x10 μm ) ⎟ = 17.3 nF ⎜10 ⎝ μm ⎠ cm 2 ⎠ ⎝ " ox " L = 2Λ = 1μm | W =10 L = 5μm | Cox = εox Tox = 3.9(8.854 x10−14 F / cm) 150 x10 cm −7 = 0.23 fF / μm 2 Triode region : " 0.23 fF / μm 2 )(5μm 2 ) ( Cox WL + CGSOW = + (0.02 fF / μm)(5μm) = 0.675 fF CGS = CGD = 2 2 2 " WL + CGSOW = 0.867 fF | CGS = CGSOW = 0.10 fF Saturation region : CGS = Cox 3 Cutoff : CGS = CGD = CGSOW = 0.10 fF 4.73 F⎞ ⎟ cm ⎠ εox F " = 3.453x10−8 2 (a) Cox = T = ⎛ cm ⎞ cm ox 100 x10−9 m⎜102 ⎟ m⎠ ⎝ ⎛ F ⎞ " CGC = Cox WL = ⎜3.453 x10−8 2 ⎟ 10 x10−4 cm 1x10−4 cm = 3.45 fF cm ⎠ ⎝ ⎛ F ⎞ " WL = ⎜ 3.453x10−8 2 ⎟ 100 x10−4 cm 1x10−4 cm = 34.5 fF (b) CGC = Cox cm ⎠ ⎝ 8.854 x10 (3.9)⎜ ⎝ −14 ⎛ ( )( ) ( )( ) 97 4.74 CSB = C j AS + C jsw PS | CDB = C j AD + C jsw PD | AS = 50 Λ2 = 12.5μm 2 | PS = 30 Λ = 15μm Cj = εs w do | w do = ⎛ NA ND ⎞ ⎛10 201016 ⎞ 2εs ⎛ 1 1 ⎞ = 0.025ln + φ | φ = V ln ⎜ ⎟ j ⎜ 2 ⎟ ⎜ ⎟ = 0.921V j T 20 q ⎝ NA NA ⎠ ⎝ 10 ⎠ ⎝ ni ⎠ 2(11.7)(8.854 x10−14 )⎛ 1 1 ⎞ -5 w do = ⎜ 20 + 16 ⎟0.921 = 3.45x10 cm −19 ⎝ 10 10 ⎠ 1.602 x10 Cj = CSB (11.7)(8.854 x10−14 F / cm) −5 3.45 x10 cm = (3.00 x10−8 F / cm 2 )( 12.5 x10−8 cm 2 )+ 5 x10−4 cm(3.00 x10−8 F / cm 2 )( 15 x10−4 cm)= 26.3 fF = 3.00 x10−8 F / cm 2 CDB = CSB = 26.3 fF 4.75 KP = K 'n = K n L μA ⎛ 0.25μm ⎞ = 175 2 ⎜ ⎟ = 8.75U | VTO = VTN = 0.7 W V ⎝ 5μm ⎠ PHI = 2φ F = 0.8V | L = 0.25U | W = 5U | LAMBDA = 0.02 4.76 (a) VTO = 0.7 | PHI = 2φ F = 0.6 | GAMMA = 0.75 (b) VTO = 0.74 | PHI = 0.87 | GAMMA = 0.84 4.77 KP = K 'n = 50U VTO = VTN = 1 V L = 0.5U W = 2.5U LAMBDA = 0 4.78 KP = K 'n = 10U VTO = VTN = 1 V L = 0.6U W = 1.5U LAMBDA = 0 4.79 KP = K 'p = 10U VTO = VTP = −1 V L = 0.5U Using the - 3 - V curve, K P = 2 50μA [-3 - (-1)] 2 = 25 μA V2 W = 1.25U LAMBDA = 0 4.80 KP = K 'n = 25U VTO = VTN = 1 V L = 0.6U W = 0.6U LAMBDA = 0 98 NMOS i-v Characteristics for Load-Line Problems 800 5V 600 Drain Current (uA) 400 4V 200 3V 2V 0 0 1 2 3 4 5 6 Drain Voltage (V) 99 4.81 For VDS = 0, ID = 4V = 0.588mA. For ID = 0,VDS = 4V . 6.8 kΩ Also, VGS = 4V. From the graph, the transistor is operating below pinchoff in the triode region and the Q-point is Q-point: (350 μA, 1.7V) 800 5V 600 Drain Current (uA) Q-point (4.82) 400 4V 200 Q-point (4.81) 3V 2V 0 0 1 2 3 4 5 6 Drain Voltage (V) 4.82 For VDS = 0, ID = 5V = 0.602mA. For ID = 0,VDS = 5V . 8.3kΩ For VGS = 5V, the Q-point is (450 μA, 1.25 V). From the graph in Prob. 4.81, the transistor is operating below pinchoff in the triode region. 4.83 800 5V 600 Drain Current (uA) Q-point (4.84) 400 4V 200 Q-point (4.83) 3V 2V 0 0 1 2 3 4 5 6 Drain Voltage (V) 100 VDD = 3V | 6 =10 4 ID +V DS | VDS = 0, ID = 0.6 mA | ID = 0, VDS = 6V 2 From the graph, Q-pt: (140 μA, 4.6V) in the saturation region. VGS = 4.84 VDD = 4V | 8 =10 4 ID +V DS | VDS = 6, ID = 0.2 mA | VDS = 0, ID = 0.8 mA 2 See graph for Problem 4.83: Q-pt: (390 μA, 4.1 V) in saturation region. VGS = 4.85 (a) 100 kΩ 12V = 3.75V | Assume saturation 100 kΩ + 220 kΩ −6 ⎞ ⎛ ⎛ 5⎞ 2 3 3 100 x10 3.75 = VGS + 24 x10 ID = VGS + 24 x10 ⎜ ⎟⎜ ⎟(VGS − 1) 2 ⎝ ⎠⎝ 1 ⎠ VGG = 2 6VGS − 11VGS + 2.25 = 0 → VGS = 1.599V and ID = 89.7μA VDS = 12 − 36 x10 3 ID = 8.77V | VDS > VGS − VTN Saturation is correct. Checking : VGG = 24 x10 3 ID + VGS = 3.75V which is correct. Q − po int : (89.7 μA, 8.77 V ) (b) Assume saturation 3 3 ⎛ 100 x10−6 ⎞⎛ 10 ⎞ 2 3.75 = VGS + 24 x10 ID = VGS + 24 x10 ⎜ ⎟⎜ ⎟(VGS − 1) 2 ⎠⎝ 1 ⎠ ⎝ 2 12VGS − 23VGS + 8.25 = 0 → VGS = 1.439V and ID = 96.4 μA VDS = 12 − 36 x10 3 ID = 8.53V | VDS > VGS − VTN Saturation is correct. Checking : VGG = 24 x10 3 ID + VGS = 3.75V which is correct. Q − po int : (96.4 μA, 8.53 V ) 4.86 VGG = 10kΩ 12V = 3.75V | Assume saturation 10kΩ + 22kΩ ⎛ 100 x10−6 ⎞⎛ 20 ⎞ 2 3.75 = VGS + 2.4 x10 3 ID = VGS + 24 x10 3⎜ ⎟⎜ ⎟(VGS − 1) 2 ⎠⎝ 1 ⎠ ⎝ 2 2.4VGS − 3.8VGS + 1.35 = 0 → VGS = 1.882V and ID = 778μA VDS = 12 − 3.6 x10 3 ID = 9.20V | VDS > VGS − VTN Saturation is correct. Checking : VGG = 2.4 x10 3 ID + VGS = 3.75V which is correct. Q − po int : (778 μA, 9.20 V ) 101 4.87 VGG = 1MΩ 12V = 3.75V | Assume saturation 1MΩ + 2.2 MΩ ⎛ 100 x10−6 ⎞⎛ 5 ⎞ 2 3.75 = VGS + 2.4 x10 3 ID = VGS + 2.4 x10 5⎜ ⎟⎜ ⎟(VGS − 1) 2 ⎠⎝ 1 ⎠ ⎝ 2 60VGS − 119VGS + 56.25 = 0 → VGS = 1.206V and ID = 10.6μA VDS = 12 − 3.6 x10 5 ID = 8.18V | VDS > VGS − VTN Saturation is correct. Checking : VGG = 2.4 x10 5 ID + VGS = 3.75V which is correct. Q − po int : (10.6 μA, 8.18 V ) 4.88 (a) 100 kΩ 15V = 4.69V | Assume Saturation 100 kΩ + 220 kΩ ⎛ 100 x10−6 ⎞⎛ 5 ⎞ 2 4.69 = VGS + 24 x10 3 ID = VGS + 24 x10 3⎜ ⎟⎜ ⎟(VGS − 1) 2 ⎠⎝ 1 ⎠ ⎝ VGG = 2 6VGS − 11VGS + 1.31 = 0 → VGS = 1.705V and ID = 124 μA VDS = 15 − 36 x10 3 ID = 10.5 V | VDS > VGS − VTN Saturation is correct. Checking : VGG = 24 x10 3 ID + VGS = 4.68V which is correct. Q − po int : (124 μA, 10.5 V ) (b) VGG = 4.69V | Assume Saturation ⎛ 100 x10−6 ⎞⎛ 10 ⎞ 2 4.69 = VGS + 24 x10 3 ID = VGS + 24 x10 3⎜ ⎟⎜ ⎟(VGS − 1) 2 ⎝ ⎠⎝ 1 ⎠ 2 12VGS − 23VGS + 7.31 = 0 → VGS = 1.514V and ID = 132 μA VDS = 15 − 36 x10 3 ID = 10.3 V | VDS > VGS − VTN Saturation is correct. Checking : VGG = 24 x10 3 ID + VGS = 4.68V which is correct. Q − po int : (132 μA, 10.3 V ) 4.89 (a) VGG = 200 kΩ 12V = 3.81V | Assume Saturation 200 kΩ + 430 kΩ ⎛ 100 x10−6 ⎞⎛ 5 ⎞ 2 3.81 = VGS + 47 x10 3 ID = VGS + 47 x10 3 ⎜ ⎟⎜ ⎟(VGS − 1) 2 ⎝ ⎠⎝ 1 ⎠ 2 23.5VGS − 45VGS + 15.88 = 0 → VGS = 1.448V and ID = 50.3 μA VDS = 12 − 71x10 3 ID = 8.43 V | VDS > VGS − VTN Saturation is correct. Checking : VGG = 47 x10 3 ID + VGS = 3.81V which is correct. Q − po int : (50.3 μA, 8.43 V ) 102 (b) VGG = 3.81V | Assume Saturation 3 3 ⎛ 100 x10−6 ⎞⎛ 15 ⎞ 2 3.81 = VGS + 47 x10 ID = VGS + 47 x10 ⎜ ⎟⎜ ⎟(VGS − 1) 2 ⎝ ⎠⎝ 1 ⎠ 2 70.5VGS − 139VGS + 62.88 = 0 → VGS = 1.269V and ID = 54.3 μA VDS = 12 − 71x10 3 ID = 8.15 V | VDS > VGS − VTN Saturation is correct. Checking : VGG = 47 x10 3 ID + VGS = 3.82V which is correct. Q − po int : (54.3 μA, 8.15 V ) 4.90 (a) Setting KP = 500U, VTO = 1, and GAMMA = 0 yields ID = 89.6 μA, VGS = 1.60 V and VDS = 8.77 V 4 (a) Setting KP = 1000U, VTO = 1, and GAMMA = 0 yields ID = 96.3 μA, VGS = 1.44 V and VDS = 8.53 V 4 4.91 (a) Setting KP = 500U, VTO = 1, and GAMMA = 0 yields ID = 124 μA, VGS = 1.71 V and VDS = 10.5 V 4 (a) Setting KP = 1000U, VTO = 1, and GAMMA = 0 yields ID = 132 μA, VGS = 1.51 V and VDS = 10.2 V 4 4.92 (a) Setting KP = 500U, VTO = 1, and GAMMA = 0 yields ID = 50.2 μA, VGS = 1.45 V and VDS = 8.43 V 4 (a) Setting KP = 1000U, VTO = 1, and GAMMA = 0 yields ID = 54.1 μA, VGS = 1.27 V and VDS = 8.16 V 4 4.93 (300 kΩ, 700 kΩ) or (1.2 MΩ, 2.8 MΩ). We normally desire the current in the gate bias network to be much less than ID. We also usually like the parallel combination of R1 and R2 to be as large as possible. 4.94 35 x10−6 2 (a) ID = (4 − 1 − 1700 ID ) and using the quadratic equation, 2 ID = 134μA. VDS =10 -134 x10−6 (1700 + 38300) = 4.64V (b) 25 x10−6 2 (4 − 0.75 − 1700ID ) and using the quadratic equation, 2 ID = 116μA. VDS =10 -116 x10−6 (1700 + 38300) = 5.36V ID = 103 4.95 (a) Example 4.3 Setting KP = 25U and VTO = 1 yields ID = 34.4 μA, VGS = 2.66 V and VDS = 6.08 V 4 Results agree with hand calculations (b) Example 4.4 Setting KP = 25U and VTO = 1 yields ID = 99.2 μA, VGS = 3.82 V and VDS = 6.03 V 4 Results are almost identical to hand calculations 4.96 ' Kn = 100μA /V 2 | VTN = 0.75V | Choose VDS = VR D = VR S = 4V and VGS − VTN = 1V RS = 4 4.1 = 40 kΩ ⇒ 39 kΩ and VR S = 3.9V | RD = = 41kΩ ⇒ 43kΩ 100μA 100μA 2 ID 2I W 2 = 1V and K n = D = 200μA /V 2 ⇒ = | 2 Kn 1V L 1 VGS − VTN = VG = VS + VGS = 3.9 + 1 + 0.75 = 5.65V 5.65V = 5.65V = ⎛ 12V ⎞ R1 R R ⎛ 12V ⎞ 12V | 5.65V = 1 2 ⎜ ⎟ = 530 kΩ ⇒ 560 kΩ ⎟ | R2 = 250 kΩ⎜ ⎝ 5.65V ⎠ R1 + R2 R1 + R2 ⎝ R2 ⎠ R1 12V ⇒ R1 = 500 kΩ ⇒ 510 kΩ R1 + 560 kΩ W 2 = L 1 R1 = 510 kΩ, R2 = 560 kΩ, RS = 39 kΩ, RD = 43kΩ, 4.97 ' Kn = 100μA /V 2 | VTN = 0.75V | Choose VDS = VR D = VR S = 3V and VGS − VTN = 1V 3 3 = 12 kΩ | RD = = 12 kΩ 0.25 mA 0.25 mA 2 ID 2I W 5 VGS − VTN = = 1V and K n = D = 500μA /V 2 ⇒ = 2 Kn 1V L 1 RS = VG = VS + VGS = 3 + 1 + 0.75 = 4.75V 4.75V = 4.75V = ⎛ 9V ⎞ R1 R R ⎛ 9V ⎞ 9V | 4.75V = 1 2 ⎜ ⎟ | R2 = 250 kΩ⎜ ⎟ = 473kΩ ⇒ 470 kΩ ⎝ 4.75V ⎠ R1 + R2 R1 + R2 ⎝ R2 ⎠ R1 9V ⇒ R1 = 525 kΩ ⇒ 510 kΩ R1 + 470 kΩ W 5 = L 1 R1 = 510 kΩ, R2 = 470 kΩ, RS = 12 kΩ, RD = 12 kΩ, 104 4.98 ' Kn = 100μA /V 2 | VTN = 0.75V | Choose VDS = VR D = VR S = 5V and VGS − VTN = 1V 5 5 = 10 kΩ | RD = = 10 kΩ 0.5 mA 0.5 mA 2 ID 2I W 10 VGS − VTN = = 1V and K n = D = 1mA /V 2 ⇒ = 2 Kn 1V L 1 RS = VG = VS + VGS = 5 + 1 + 0.75 = 6.75V 6.75V = 6.75V = ⎛ 15V ⎞ R1 R R ⎛ 15V ⎞ 15V | 6.75V = 1 2 ⎜ ⎟ = 1.33 MΩ ⇒ 1.5 MΩ ⎟ | R2 = 600 kΩ⎜ ⎝ 6.75V ⎠ R1 + R2 R1 + R2 ⎝ R2 ⎠ R1 15V ⇒ R1 = 1.23 MΩ ⇒ 1.2 MΩ R1 + 1.5 MΩ W 10 R1 = 1.2 MΩ, R2 = 1.5 MΩ, RS = 10 kΩ, RD = 10 kΩ, = L 1 4.99 ⎛10−3 ⎞ 2 Assume Saturation. For IG = 0, VGS = −10 4 ID = −10 4 ⎜ ⎟(VGS + 5) ⎝ 2 ⎠ 2 5VGS + 51VGS + 125 = 0 ⇒ VGS = −4.10 V and ID = 410 μA VDS = 15 − 15000ID = 8.85 V | VDS > VGS − VTN so saturation is ok. Q - Point : (410μA, 8.85V) 4.100 ⎛ 6 x10−4 ⎞ 2 Assume Saturation. For IG = 0, VGS = −27 x10 3 ID = −27 x10 4 ⎜ ⎟(VGS + 4 ) ⎝ 2 ⎠ 2 8.1VGS + 65.8VGS + 129.6 = 0 ⇒ VGS = −3.36 V and ID = 124 μA VDS = 12 − 78000ID = 2.36 V | VDS > VGS − VTN so saturation is ok. Q - Point : (124 μA, 2.36V) 4.101 Kn = 1 mA/V 2 V TN = -5 V ID RD VDD + IG = 0 + V RG 5V R + 15 V GS S 105 ⎛ 1mA/V 2 ⎞ 2 Assume Saturation. IG = 0. 250μA = ⎜ ⎟(VGS + 5) 2 ⎠ ⎝ 0.25 mA 4.29V = −4.29V | RS = = 17.2 kΩ ⇒ 18 kΩ 0.5 mA 0.25mA 15 − 5 − 4.29 V VDS = 15 − ID (RD + RS ) ⇒ RD = = 22.88kΩ ⇒ 24 kΩ 0.25 mA RG is arbitrary but normally fairly large. Choose RG = 510 kΩ. VGS = −5 + 4.102 + R 2 R D 5V + 5V + 15 V VDD + R1 R S 5V - Assume Saturation. IG = 0. Assume power supply is split in thirds: VDS = VR D = VRS = 5V Note that although, this is a depletion - mode device, I D exceeds Kn 2 VTN and this will require VGS > 0. 2 ⎛ 0.25mA/V 2 ⎞ 5V 2 RS = = 2.5 kΩ ⇒ 2.4 kΩ and VRS will be 4.8 V | 2mA = ⎜ ⎟(VGS + 2) 2 2 mA ⎝ ⎠ 2 mA = +2.00V | VG = VS + VGS = 4.8 + 2 = 6.8V 0.125 mA VGS = −2 + 6.8 = 15 R1 ⇒ R1 = 680 kΩ and R2 = 820 kΩ is one convenient possibility. R1 + R2 Another is R1 = 68 kΩ and R2 = 82 kΩ. Both choices have IR 2 0, so the transistor is saturated by connection. +12 V W = L 1 10 330 k Ω IDS I + G ID = 10 M Ω ' ⎛100 μA ⎞⎛10 ⎞ W Kn 2 2 ⎟(VGS − 0.75) (VGS − VTN ) = ⎜ 2 ⎟⎜ ⎝ 2 V ⎠⎝ 1 ⎠ 2 L VGS = 12 − 330 kΩ(ID + IG ) − 10 MΩ(IG ) but I G = 0 VGS = 12 − 330 kΩ(ID ) ⎛ 1.00 x10−3 A ⎞ 2 VGS = 12 − (3.30 x10 5 ) V − 0.75) ⎜ 2 ⎟( GS 2 V ⎠ ⎝ 2 165VGS − 246.5VGS + 80.81 = 0 yields VGS = 1.008V , 0.486V + V GS - VDS - VGS must be 1.008 V since 0.486 V is below threshold. ⎛100 μA ⎞ 10 2 ID = ⎜ (1.008 − 0.75) = 33.3 μA and VDS = VGS 2⎟ ⎝ 2 V ⎠1 Q-Point: (33.3 μA, 1.01 V) 4.104 Checking: ID =(12-1.01)V/330kΩ = 33.3 μA. 4 (a) The transistor is saturated by connection. and ID = 100 x10−6 ⎛ 20 ⎞⎛ A ⎞ 2 ⎜ ⎟⎜ 2 ⎟(VGS − 0.75V ) ⎝ 1 ⎠⎝ V ⎠ 2 100 x10−6 ⎛ 20 ⎞⎛ A ⎞ 2 ⎜ ⎟⎜ 2 ⎟(1.08 − 0.75V ) = 109μA ⎝ 1 ⎠⎝ V ⎠ 2 VGS = 12 − 10 5 ID 2 100VGS − 149VGS + 44.25 = 0 ⇒ VGS = 1.08V , 0.410V ⇒ VGS = 1.08 V since VGS must exceed the threshold voltage. ID = Checking : ID = 12 − 1.08 = 109 μA | Q - Point : (109 μA, 1.08 V) 10 5 7 (b) Using KVL, VDS = 10 IG +VGS. But, since IG = 0, VGS = VDS. Also VTN = 0.75 V > 0, so the transistor is saturated by connection. 107 +12 V W = L 1 10 330 k Ω IDS I + G ID = 10 M Ω ' ⎛100 μA ⎞⎛ 20 ⎞ W Kn 2 2 ⎟(VGS − 0.75) (VGS − VTN ) = ⎜ 2 ⎟⎜ ⎝ 2 V ⎠⎝ 1 ⎠ 2 L VGS = 12 − 330 kΩ(ID + IG ) − 10 MΩ(IG ) but VGS = 12 − 330 kΩ(ID ) ⎛ −3 A ⎞ 2 VGS = 12 − (3.30 x10 5 ) V − 0.75) ⎜10 2 ⎟( GS ⎝ V ⎠ IG = 0 + V GS - VDS - 2 3.3VGS − 4.94VGS + 1.736 = 0 yields VGS = 0.933V , 0.564V VGS must be 0.933 V since 0.564 V is below threshold. ⎛100 μA ⎞ 20 12 − 0.933 2 ID = ⎜ V = 33.5μA (0.933 − 0.75) = 33.5 μA Checking : 2⎟ ⎝ 2 V ⎠ 1 330kΩ and VDS = VGS : 4.105 Q-Point: (33.5 μA, 0.933 V) ' ⎛100 μA ⎞⎛ 10 ⎞ Kn W 2 2 ⎟(VGS − 0.75) (VGS − VTN ) = ⎜ 2 ⎟⎜ ⎝ 2 V ⎠⎝ 1 ⎠ 2 L (a) Assume saturation : ID = VGS = 15 − 330kΩ(ID + IG ) − 10 MΩ(IG ) but I G = 0 VGS = 15 − 330kΩ(ID ) and VGS = VDS so saturation regioin operation is correct ⎛ 10−3 A ⎞ 2 VGS = 15 − (3.30 x10 5 ) V − 0.75) ⎜ 2 ⎟( GS ⎝ 2 V ⎠ 2 3.30VGS − 4.93VGS + 1.556 = 0 yields VGS = 1.041V , 0.453V ⎛100 μA ⎞⎛ 10 ⎞ 15 − 1.041 2 ID = ⎜ V = 42.3μA ⎟(1.041 − 0.75) = 42.3μA | Checking : ID = 2 ⎟⎜ ⎝ 2 V ⎠⎝ 1 ⎠ 330kΩ Q − Point : (42.3 μA, 1.04 V) (b) Assume saturation ID = ' ⎛100 μA ⎞⎛ 25 ⎞ W Kn 2 2 ⎟(VGS − 0.75) (VGS − VTN ) = ⎜ 2 ⎟⎜ ⎝ 2 V ⎠⎝ 1 ⎠ 2 L VGS = 15 − 330 kΩ(ID + IG ) − 10 MΩ(IG ) but I G = 0 VGS = 15 − 330 kΩ(ID ) and VGS = VDS so saturation region operation is correct. ⎛ 2.50 x10−3 A ⎞ 2 V − 0.75) VGS = 15 − (3.30 x10 5 ) ⎜ 2 ⎟( GS V ⎠ 2 ⎝ 2 8.25VGS − 12.355VGS + 4.341 = 0 yields VGS = 0.9345V , 0.563V ⎛100 μA ⎞⎛ 25 ⎞ 15 − 0.9345 2 V = 42.6μA ID = ⎜ ⎟(0.9345 − 0.75) = 42.6μA | Checking : ID = 2 ⎟⎜ ⎝ 2 V ⎠⎝ 1 ⎠ 330 kΩ Q - Point : (42.6 μA, 0.935 V) 108 4.106 (a) Asssume saturation ID = ' ⎛100 μA ⎞⎛10 ⎞ W Kn 2 2 ⎟(VGS − 0.75) (VGS − VTN ) = ⎜ 2 ⎟⎜ ⎝ 2 V ⎠⎝ 1 ⎠ 2 L VGS = 12 − 470kΩ(ID + IG ) − 10 MΩ(IG ) but IG = 0 VGS = 12 − 470kΩ(ID ) and VGS = VDS so saturation region operation is correct. ⎛ 10−3 A ⎞ 2 VGS = 12 − (4.70 x10 5 ) V − 0.75) ⎜ 2 ⎟( GS ⎝ 2 V ⎠ 2 4.70VGS − 7.03VGS + 2.404 = 0 yields VGS = 0.9666V , 0.529V ⎛100 μA ⎞⎛ 10 ⎞ 12 − 0.967 2 ID = ⎜ = 23.5μA ⎟(0.9666 − 0.75) = 23.5μA | Checking : ID = 2 ⎟⎜ ⎝ 2 V ⎠⎝ 1 ⎠ 470kΩ Q - Point : (23.5 μA, 0.967 V) (b) Since the current in RG is zero, the drain current is independent of RG. 4.107 (a) Create an M-file: function f=bias(id) vtn=1+0.5*(sqrt(22e3*id)-sqrt(0.6)); f=id-(25e-6/2)*(6-22e3*id-vtn)^2; fzero('bias',1e-4) yields ans = 8.8043e-05 (b) Modify the M-file: function f=bias(id) vtn=1+0.75*(sqrt(22e3*id)-sqrt(0.6)); f=id-(25e-6/2)*(6-22e3*id-vtn)^2; fzero('bias',1e-4) yields ans = 8.3233e-05 4.108 Using a spreadsheet similar to Table 4.2 yields: (a) 88.04 μA, (b) 83.23 μA. 4.109 100kΩ 12V = 3.75V | Assume saturation 100 kΩ + 220kΩ 2 ID VTN = 1 + 0.6 24 x10 3 ID + 0.6 − 0.6 | VGS = VTN + 5 x10−4 3.75 − VGS ID = | Solving iteratively yields ID = 73.1μA with VTN = 1.460V 24 kΩ V = 12V − ID (24 kΩ + 12 kΩ) = 9.37 V Transistor is saturated. Q - Point : (73.1 μA, 9.37 V) VGG = ( ) 109 4.110 VGG = 100 kΩ 12V = 3.75V | Assume saturation 100 kΩ + 220 kΩ 0.6 (a) VTN = 1 + 0.75( 24 x10 3 ID + 0.6 − ID = VDS )| VGS = VTN + 2 ID 5 x10−4 3.75 − VGS | Solving iteratively yields ID = 69.7μA with VTN = 1.550V 24 kΩ = 12V − ID (24 kΩ + 12 kΩ) = 9.49 V The transistor is saturated. Q - Point : (69.7 μA, 9.49 V) 0.6 (b) VTN = 1 + 0.6( 24 x10 3 ID + 0.6 − ID = )| VGS = VTN + 2 ID 5 x10−4 3.75 − VGS | Solving iteratively yields ID = 73.1μA with VTN = 1.460V 24 kΩ V = 12V − ID (24 kΩ + 24 kΩ) = 8.49 V The transistor is saturated. Q - Point : (73.1 μA, 8.49 V) 4.111 (a) γ = 0.6 VTN = 1.46 V VTN = 1.55 V VTN = 1.46 V ID = 73.1 μA ID = 69.7 μA ID = 73.1 μA VDS = 9.37 V VDS = 9.49 V VDS = 8.49 V (b) γ = 0.75 (c ) γ = 0.6 These results all agree with the hand calculations. (They should - they are all solving the same sets of equations.) 4.112 (a) γ = 0 VTN = 1.00 V ID = 50.2 μA ID = 42.2 μA ID = 54.1 μA ID = 45.2 μA VDS = 8.43 V VDS = 9.01 V VDS = 8.16 V VDS = 8.79 V γ = 0.5 VTN = 1.42 V (b) γ = 0 VTN = 1.00 V γ = 0.5 VTN = 1.44 V The γ = 0 values agree with the hand calculations in the original problem. Including body effect in the simulations reduces the Q-point current by approximately 15%. Although this may sound large, it is within the error that will be introduced by the use of 5% resistors and typical device tolerances. So, we normally omit γ in our hand calculations, and then refine the results using SPICE. 4.113 (a) γ = 0 VTN = 1.00 V ID = 89.6 μA ID = 75.5 μA ID = 96.3 μA ID = 80.8 μA VDS = 8.77 V VDS = 9.28 V VDS = 8.53 V VDS = 9.09 V γ = 0.5 VTN = 1.39 V (b) γ = 0 VTN = 1.00 V γ = 0.5 VTN = 1.41 V The γ = 0 values agree with the hand calculations in the original problem. Including body effect in the simulations reduces the Q-point current by approximately 15%. Although this may sound 110 large, it is within the error that will be introduced by the use of 5% resistors and typical device tolerances. So, we normally omit γ in our hand calculations, and then refine the results using SPICE. 4.114 γ =0 γ = 0.5 VTN = 1.00 V VTN = 1.35 V ID = 778 μA ID = 661 μA VDS = 9.20 V VDS = 9.62 V The γ = 0 values agree with the hand calculations in the original problem. Including body effect in the simulations reduces the Q-point current by approximately 15%. Although this may sound large, it is within the error that will be introduced by the use of 5% resistors and typical device tolerances. So, we normally omit γ in our hand calculations, and then refine the results using SPICE. 4.115 γ =0 γ = 0.5 VTN = 1.00 V VTN = 1.45 V ID = 10.5 μA ID = 9.18 μA VDS = 8.03 V VDS = 8.69 V The γ = 0 values agree with the hand calculations in the original problem. Including body effect in the simulations reduces the Q-point current by approximately 15%. Although this may sound large, it is within the error that will be introduced by the use of 5% resistors and typical device tolerances. So, we normally omit γ in our hand calculations, and then refine the results using SPICE. 4.116 (a) Both transistors are saturated by connection and the two drain currents must be equal. K K 2 2 ID1 = n1 (VGS1 − VTN1 ) and ID 2 = n 2 (VGS 2 − VTN 2 ) 2 2 But since the transistors are identical, ID1 = ID2 requires VGS1 = VGS2 = VDD/2 = 2.5V. 100 x10−6 ⎛ 20 ⎞ 2 ID1 = ID 2 = ⎜ ⎟(2.5 − 1) = 2.25 mA ⎝1⎠ 2 (b) For this case, the same arguments hold, and VGS1 = VGS2 = VDD/2 = 5V. 100 x10−6 ⎛ 20 ⎞ 2 ID1 = ID 2 = ⎜ ⎟(5 − 1) =16.0 mA ⎝1⎠ 2 . (c) For this case, the threshold voltages will be different due to the body-effect in the upper transistor. The drain currents must be the same, but the gate-source voltages will be different: VGS1 = VTN1 + VTN1 =1V 2 ID 2 ID ; VGS 2 = VTN 2 + ; VGS1 + VGS 2 = 5V . Kn Kn VTN 2 = 1 + 0.5 VGS1 + 0.6 − 0.6 ( ) 111 Combining these equations yields 5 - 2VGS1 − 0.5 VGS1 + 0.6 − 0.6 = 0 ⇒ VGS1 = 2.27V ; VGS 2 = 5 − VGS1 = 2.73V ID 2 = ID1 = 100 x10−6 ⎛ 20 ⎞ 2 ⎜ ⎟(2.27 − 1) = 1.61 mA. ⎝1⎠ 2 ( ) Checking : VTN 2 = 1 + 0.5 2.27 + 0.6 − 0.6 = 1.46V ID 2 = 100 x10−6 ⎛ 20 ⎞ 2 ⎜ ⎟(2.73 − 1.46) = 1.61 mA. ⎝1⎠ 2 ( ) 4.117 If we assume saturation, we find ID = 234 μA and VDS = 0.65 V, and the transistor is not saturated. Assuming triode region operation, VGS = 10 − 2 x10 4 ID | VDS = 10 − 4 x10 4 ID μA ⎛ 2 ⎞⎛ 10 − 4 x10 4 ID ⎞ ID = 100 2 ⎜ ⎟⎜10 − 2 x10 4 ID − 1 − 10 − 4 x10 4 ID ) ⎟( ⎝ ⎠ 2 V 1 ⎝ ⎠ VDS = 10 − 4 x10 4 (2.42 x10−4 )= 0.320V | Q - Pt : (242 μA, 0.320V ) Checking the operating region : VGS − VTN = 4.16V > VDS and the triode region assumption is correct. Checking : ID = 10 − 0.32 V = 242μA 40 kΩ Collecting terms : 16.5 x10 4 ID = 40 → ID = 242 μA 4.118 If we assume saturation, we find an inconsistent answer. Assuming triode region operation, VGS = 10 − 2 x10 4 ID | VDS = 10 − 3 x10 4 ID μA ⎛ 4 ⎞⎛ 10 − 3 x10 4 ID ⎞ ID = 100 2 ⎜ ⎟⎜10 − 2 x10 4 ID − 1 − 10 − 3 x10 4 ID ) ⎟( 2 V ⎝ 1 ⎠⎝ ⎠ VDS = 10 − 3 x10 4 (3.22 x10−4 )= 0.340V | Q - Pt : (322 μA, 3.18 V ) Checking the operating region : VGS − VTN = 2.56V > VDS and the triode region assumption is correct. Checking : ID = 10 − 0.34 V = 322μA 30 kΩ 2 Collecting terms : 1.5x10 8 ID − 1.725 x10 5 ID + 40 = 0 → ID = 322μA 112 4.119 For (a) and (b), γ = 0. The transistor parameters are identical so 3VGS = 15V or VGS = 5V. ⎛ 20 ⎞ 1 2 ( a) ID = ( 100 x10−6 ) ⎜ ⎟(5 − 0.75) = 18.1 mA ⎝1⎠ 2 ⎛ 50 ⎞ 1 2 100 x10−6 ) (b) ID = ( ⎜ ⎟(5 − 0.75) = 45.2 mA ⎝1⎠ 2 (c) Now we have three different threshold voltages and need an iterative solution. Using MATLAB: function f=Prob112(id) gamma=0.5; vgs1=.75+sqrt(2*id/2e-3); vtn2=0.75+gamma*(sqrt(vgs1+0.6)-sqrt(0.6)); vgs2=vtn2+sqrt(2*id/2e-3); vtn3=0.75+gamma*(sqrt(vgs1+vgs2+0.6)-sqrt(0.6)); vgs3=vtn3+sqrt(2*id/2e-3); f=15-vgs1-vgs2-vgs3; fzero('Prob112',1e-4) --> ans = 0.0130 ID = 13.0 mA 4.120 W 20 = VTN = 0.75 V ID = 18.1 mA VDS = 5.00 V 1 L W 50 = VTN = 0.75 V ID = 45.2 mA VDS = 5.00 V (b) γ = 0 L 1 W 20 = VTN 3 = 1.95 V ID 3 = 13.0 mA VDS 3 = 5.56 V (b) γ = 0.5 L 1 VTN 2 = 1.48 V ID 2 = 13.0 mA VDS 2 = 5.09 V VTN1 = 0.75 V ID1 = 13.0 mA (a) γ = 0 VDS1 = 4.36 V Results are identical to calculations in Prob. 4.119 4.121 For VGS = 5 V and VDS = 0.5 V, the transistor will be in the triode region. (5 − 0.5)V = 54.88μA | 54.88 x10−6 = 100 x10−6⎛ W ⎞⎛5 − 0.75 − 0.5 ⎞0.5 | W = 0.274 = 1 ID = ⎜ ⎟⎜ ⎟ ⎝ L ⎠⎝ 2 ⎠ L 1 3.64 82 kΩ 4.122 For VGS = 3.3 V and VDS = 0.25 V, the transistor will be in the triode region. (3.3 − 0.25)V = 16.94 μA | 16.94 x10−6 = 100 x10−6⎛ W ⎞⎛ 3.3 − 0.75 − 0.25 ⎞0.25 | W = 0.280 = 1 ID = ⎜ ⎟⎜ ⎟ ⎝ L ⎠⎝ 2 ⎠ L 1 3.57 180kΩ 113 4.123 (a) The transistor is saturated by connection. For this circuit, VGS = VDD + ID R = −15 + 75000 ID 4 x10−5 ⎛ 1⎞ 2 ID = ⎜ ⎟(−15 + 75000 ID + 0.75) ⇒ 153 μA 2 ⎝ 1⎠ VGS = −15 + 75000 ID = −3.525V VDS = VGS = −3.525V | Q - point : (153 μA, −3.53 V ) (b) Here the transistor has VGS = -15 V, a large value, so the transistor is most likely operating in the triode region. ⎛ VDS − (−15) V ⎞ = 4 x10−5 ⎜ −15 − (−0.75) − DS ⎟VDS ⇒ VDS = −0.347 V and ID = 195 μA. ⎝ 75000 2 ⎠ 15 − 0.347 V = 195μA Q - point : (195 μA,-0.347 V ) Checking : ID = 785 kΩ Checking the region of operation: VDS = −0.347V > VGS − VTP = −15 + 0.75 = −14.25V ID = Triode region is correct 4.124 Set W=1U L=1U KP=40U VTO=-0.75 GAMMA=0 Results are almost identical to hand calculations for both parts. 4.125 ( a) IDP = IDN , and both transistors are saturated by connection. 10 = -VGSP + VGSN 1 ⎛ 100μA ⎞⎛ 20 ⎞ 1 ⎛ 40μA ⎞⎛ 20 ⎞ 2 2 ⎜ 2 ⎟⎜ ⎟(−10 + VGSN + 0.75) = ⎜ ⎟⎜ ⎟(VGSN − 0.75) 2 2 ⎝ V ⎠⎝ 1 ⎠ 2 ⎝ V ⎠⎝ 1 ⎠ (9.25 − VGSN ) = 2.5 (VGSN − 0.75) → VGSN = 4.04V | VGSP = −5.96V IDP = IDN = 10.8 mA | VO = VGSN = 4.04V (b) Everything is the same except the currents scale by 80/20: IDP = IDN = 43.2 mA 4.126 For (a) and (b), γ = 0. The transistor parameters are identical so -3VGS = 15V or VGS = -5V. 1 40 2 40 x10−6 ) (−5 + 0.75) = 14.4 mA ( 2 1 1 75 2 (b) ID = (40 x10−6 ) (−5 + 0.75) = 27.1 mA 2 1 ( a) ID = (c) Now we have three different threshold voltages and need an iterative solution. Using MATLAB: function f=PMOSStack(id) gamma=0.5; 114 vsg1=.75+sqrt(2*id/1.6e-3); vtp2=-0.75-gamma*(sqrt(vsg1+0.6)-sqrt(0.6)); vsg2=-vtp2+sqrt(2*id/1.6e-3); vtp3=-0.75-gamma*(sqrt(vsg1+vsg2+0.6)-sqrt(0.6)); vsg3=-vtp3+sqrt(2*id/1.6e-3); f=15-vsg1-vsg2-vsg3; fzero('PMOSStack',1e-1) --> ans = 0.0104 ID = 10.4 mA. 4.127 (a) W=40U L=1U KP=40U VTO=-0.75 GAMMA=0 (b) W=75U L=1U KP=40U VTO=-0.75 GAMMA=0 (c) W=75U L=1U KP=40U VTO=-0.75 GAMMA=0.5 Results agree with hand calculations. 4.128 4V = 2 mA. For ID = 0,VDS = −4V . (VSD = +4V ) 2 kΩ 300kΩ VGS = VEQ = −4V = −3V (VSG = +3V ) 300kΩ + 100 kΩ From the graph, the transistor is operating below pinchoff in the linear region. For VDS = 0, ID = 5000 VSG = 5 V 4000 μ A) 3000 V Drain Current ( SG =4V 2000 Q-point 1000 VSG = 3 V V =2V SG 0 -1000 -1 0 1 2 3 4 5 6 Source-Drain Voltage (V) Q-point: (1.15 mA, 1.7V) 115 PMOS Transistor Output Characteristics 5000 V GS = -5 V 4000 Drain Current ( μA) 3000 VGS = -4 V 2000 V GS = -3 V 1000 V GS = -2 V 0 -1000 -1 0 1 2 3 4 5 6 Source-Drain Voltage (V) 116 4.129 ( a) VGG = ⎛ 4 x10−5 ⎞ 20 15V 2 = 7.5V | 7.5 = 10 5 ID - VGS | 7.5 = 10 5 ⎜ ⎟ (VGS + 0.75) - VGS 2 ⎝ 2 ⎠ 1 15 − (100 kΩ + 50 kΩ)ID ) = −5.47V | Q - point : (59.78 μA, −5.47 V ) VDS = −( (b) For saturation, VDS ≤ VGS − VTP or VSD ≥ VSG + VTP 15 − (100 kΩ + R)ID ≥ 7.5 − 100 kΩID − 0.75 → R ≤ 130 kΩ 4.130 Setting W=20U, L=1U, LEVEL=1, KP=40U, VTO=-0.75 yields results identical to the previous problem. 4.131 (a) Using MATLAB: function f=bias4(id) gamma=0.5; vbs=1e5*id; vgs=-7.5+vbs; vtp=-0.75-gamma*(sqrt(vbs+0.6)-sqrt(0.6)); f=id-(8e-4/2)*(vgs-vtp)^2; fzero('bias4',4e-5) --> ans = 5.5278e-05 --> ID = 55.3 μA VDS = -15 + (100kΩ+R) ID (b) VDS ≤ VGS - VTP | -15 + (100kΩ+R)ID ≤ -1.972 + 1.600 | 2 4VGS + 5.9VGS − 1.5 = 0 → VGS = −1.148V and ID = 63.5μA R ≤ 164 kΩ 4.132 Setting W=20U, L=1U, LEVEL=1, KP=40U, VTO=-0.75 GAMMA=0.5 yields results identical to the previous problem. 4.133 (a) V DD R V + - GS S VDS G D I SD + The arrow identifies the transistor as a PMOS device. Since γ = 0, we do not need to worry about body effect: VTP = VTO. Since VDS = VGS, and VTP < 0, the transistor is saturated. 117 K 'p W 2 ID = (VGS − VTP ) and - VGS = 12 − 105 ID 2 L ⎛ 4 x10−5 ⎞⎛ 10 ⎞ 2 VGS − (−0.75)) -VGS = 12 − 10 5 ⎜ ⎟⎜ ⎟( ⎝ 2 ⎠⎝ 1 ⎠ 2 20VGS + 29VGS − 0.75 = 0 yields VGS = −1.475V , +0.0255V We require VGS < VTP = -0.75 V for the transistor to be conducting so 2 4 x10−5 ⎛ 10 ⎞ A VGS = −1.475V and ID = ⎜ ⎟ 2 (−1.475 − (−0.75)) = 105 μA 2 ⎝ 1 ⎠V Since VDS = VGS, the Q-point is given by: Q-Point = (105 μA, -1.475 V). (b) Using MATLAB for the second part (Set gamma = 0 for part (a)): function f=bias2(id) gamma=1.0; vgs=-12+1e5*id; vsb=-vgs; vtp=-0.75-gamma*(sqrt(vsb+0.6)-sqrt(0.6)); f=id-5e-5*(vgs-vtp)^2; fzero('bias2',1e-4) --> ans = 9.5996e-05 and Q-Point = (96.0 μA, 2.40 V). 4.134 −5 ⎞ ⎛ 2 15V 40 4 4 4 x10 = = 7.5V | 7.5 = 5x10 I D - VGS | 7.5 = 5x10 ⎜ VGS + 0.75) - VGS ⎟ ( 2 ⎝ 2 ⎠ 1 VGG VDS = − 15 − (R + 50 kΩ)I D = −5V → R = 22.3 kΩ 4.135 2 890VGS + 119VGS − 30 = 0 → VGS = −1.166V and I D = 138 μA ( ) I D = 138 μA VGG = ⎛ 4 x10−5 ⎞ 40 2 15V = 7.5V | 7.5 = 15 - 5x10 4 I D + VGS | 7.5 = 5x10 4 ⎜ VGS − VTP ) - VGS ⎟ ( 2 ⎝ 2 ⎠ 1 VTP = −0.75 − 0.5 5x10 4 I D + 0.6 − 0.6 ( ) VDS = − 15 − (R + 50 kΩ)I D = −5V → R = 40.1 kΩ Solving iteratively yields I D = 111 μA with VTP = −1.60V ( ) 118 4.136 ( a ) VGG = 15V 510 kΩ = 9.81V | 9.81 = 15 -105 I D + VGS 510 kΩ + 270 kΩ ⎛ 4 x10−5 ⎞ 20 2 5.19 = 105⎜ VGS + 0.75) − VGS ⎟ ( ⎝ 2 ⎠ 1 2 40VGS + 59VGS + 17.31 = 0 → VGS = −1.071 V and I D = 41.2 μA −15 + ( 100kΩ + R)I D ≤ −1.071 + 0.75 → R ≤ 256 kΩ 4.137 ( b ) For saturation, VDS ≤ VGS − VTP 510kΩ = 9.81V | 9.81 = 15 -105 I D + VGS 510 kΩ + 270kΩ ⎛ 4 x10−5 ⎞ 20 2 5.19 = 105 ⎜ VGS − VTP ) − VGS | VTP = −0.75 − 0.5 105 I D + 0.6 − 0.6 ⎟ ( ⎝ 2 ⎠ 1 ( a ) VGG = 15V ( ) Solving iteratively yields I D = 35.2 μA with VTP = −1.38 V and VGS = −1.67 V −15 + ( 100 kΩ + R)I D ≤ −1.67 + 1.38 → R ≤ 318 kΩ 4.138 (a) Assume an equal voltage (5V) split between R D , RS and VDS. We need VDS ≤ VGS − VTP or - 5 ≤ VGS − VTP . Choose VGS − VTP = −2V . Kn = VGS = −2 − 0.75 = −2.75V . VEQ = 5 − VGS = 7.75V . 7.75 = 15 7.75 = 15 RS = R1 15 R1 R2 15 = = 100 kΩ. R2 = 193.5kΩ → 200 kΩ. R1 + R2 R2 R1 + R2 R2 220 kΩ R1 → R1 = 214 kΩ → 220 kΩ. VEQ = 15 = 7.86V 220 kΩ + 200 kΩ R1 + R2 ( b ) For saturation, VDS ≤ VGS − VTP (V 2ID GS − VTP ) 2 = 2 mA W 12.5 → = . 4 L 1 7.86V − 2.75V 15 − 5 − 5.1 = 5.11kΩ → 5.1kΩ | RD = = 4.9 kΩ → 4.7 kΩ 1mA 1mA Note that R1 is connected between the gate and + 15 V, and R 2 is connected between the gate and ground. (b) For the NMOS case, choose W/L = 5/1. The resistors now have the same values except R2 is now connected between the gate and +15 V, R1 is connected between the gate and ground, and RD = 15 − 6 − 5.1 = 3.9 kΩ 1mA 119 4.139 (a) Assume an equal voltage (3V) split between R D , RS and VDS. We need VDS ≤ VGS − VTP or - 3 ≤ VGS − VTP . Choose VGS − VTP = −1V . Kn = VGS = −1 − 0.75 = −1.75V . VEQ = 3 − VGS = 4.75V . 4.75 = 9 4.75 = 9 RS = R1 9 R1 R2 9 = = 1MΩ. R2 = 1.7 MΩ → 1.8 MΩ. R1 + R2 R2 R1 + R2 R2 2 MΩ R1 → R1 = 2.01 MΩ → 2 MΩ | VEQ = 9 = 4.74V 1.8 MΩ + 2 MΩ R1 + R2 (V 2ID GS − VTP ) 2 = 1mA W 25 → = . 1 L 1 4.74V − 1.75V 9 −3−3 = 5.97 kΩ → 6.2 kΩ | RD = = 6.0 kΩ → 6.2 kΩ 0.5mA 0.5mA Note that R1 is connected between the gate and + 9 V, and R 2 is connected between the gate and ground. R1 = 2 MΩ, R2 = 1.8 MΩ, RS = RD = 6.2 kΩ, W / L = 25/1 (b) For the NMOS case, choose W/L the gate and ground. 4.140 = 40/1. The resistors now have the same values except R2 is now connected between the gate and + 9 V, and R1 is connected between ⎛ 40 μA ⎞⎛ 10 ⎞ 4 VGS = 10 4 I D | Assume saturation : I D = ⎜ ⎟ 10 I D − 4 2 ⎟⎜ ⎝ 2 V ⎠⎝ 1 ⎠ ( ) 2 VDS = − 15 − 10 4 I D = −12.2V | Q - Pt : (281 μA, −12.2 V ) 2 Collecting terms : 108 I D − 8.5 x10 4 I D + 16 = 0 → I D = 281 μA ( ) Checking : VGS − VTP = 2 − 4 = −1.19 V | VDS = −12.2 | Saturation is correct. 4.141 VGS = 10 4 I D | VTP = 4 − 0.25 VGS + 0.6 − 0.6 ( ) | ID = 2 4 x10−4 VGS − VTP ) ( 2 VDS = − 15 − 10 4 I D = −12.4V | Q - Pt : (260 μA, −12.4 V ) Solving these equations iteratively yields I D = 260 μA ( ) 120 4.142 Note: The answers are very sensitive to round-off error and are best solved iteratively using MATLAB, a spreadsheet, HP solver, etc. Hand calculations using the quadratic equation will generally yield poor results. Saturated by connection with V TP = −1 −5 ⎛ ⎞ 2 4 x10 10 5 6 11 2 ID = ⎜ ⎟ 3.3 x10 I D − 12 − (−1) → 121 − 7.265x10 I D + 1.089 x10 I D = 0 2 ⎝1⎠ [ ] I D = 34.6μA, 32.1μA | VDS = 3.3 x105 I D − 12 = −0.582V , −1.407V | Q - point : (32.1 μA, −1.41 V ) since the transistor would not be conducting for V GS = −0.582V . 4.143 Note: The answers are very sensitive to round-off error and are best solved iteratively using MATLAB, a spreadsheet, HP solver, etc. Hand calculations using the quadratic equation will generally yield poor results. Saturated by connection with V TP = −3 ⎛ 4 x10−5 ⎞⎛ 30 ⎞ 2 5 6 11 2 ID = ⎜ ⎟⎜ ⎟ 3.3 x10 I D − 12 − (−3) → 81 − 5.941x10 I D + 1.089 x10 I D = 0 ⎝ 2 ⎠⎝ 1 ⎠ [ ] I D = 27.07μA | VDS = 3.3 x105 I D − 12 = −3.067V | Q - point : (27.1 μA, −3.07 V ) 4.144 Note: The answers are very sensitive to round-off error and are best solved iteratively using MATLAB, a spreadsheet, HP solver, etc. Hand calculations using the quadratic equation will generally yield poor results. (a) Large VGS − Assume triode region. VDS = 12 − 3.3x105 I D Q − po int : (36.1 μA, 80.6 mV) | VDS < VGS − VTN so triode region is correct. 2 10 x10−6 ⎛ 25 ⎞ 5 Saturated by connection : I = b ⎜ ⎟ 3.3x10 I D − 12 + 0.75 () D 2 ⎝1⎠ ⎛ 10 ⎞⎛ V ⎞ | I D = 40 x10−6 ⎜ ⎟⎜12 − 0.75 − DS ⎟VDS 2 ⎠ ⎝ 1 ⎠⎝ ( ) Q − point : (32.4μA,-1.32V) (c) V ID = TP = −0.75 − 0.5 3.3 x105 I D + 0.6 − 0.6 ( ) ( ) Q − point : (28.8 μA,-2.49 V) 40 x10−6 ⎛ 25 ⎞ 5 ⎜ ⎟ 3.3 x10 I D − 12 − VTP 2 ⎝1⎠ ( ) 2 | VDS = − 12 − 3.3 x105 I D 121 4.145 (a) Kn = μn ID = εox W Tox −14 ⎡ cm2 ⎢ 3.9 8.854 x10 F / cm = 500 V −s⎢ L 40 x10−7 cm ⎣ ( ⎤⎛ )⎥ 20μm ⎞ μA ⎜ ⎟ = 432 ⎥⎝ 2μm ⎠ ⎦ V2 2 432μA 4 − 1) = 1.94 mA ( 2 (b) K 4.146 ' n = 2 Kn V | V = 2 ' 2 864μA ⎛ 4 1 ⎞ | I = ⎜ − ⎟ = 0.972 mA 2 ⎝ 2 2⎠ ' D (a) C GC ⎡ 3.9 8.854 x10−14 F / cm ⎢ = C WL = ⎢ 20 x10−7 cm ⎣ " OX ' | CGC = ( ⎤ )⎥ (20 x10 ⎥ ⎦ −4 cm 10−4 cm = 34.5 fF )( ) (b) α = 2 4.147 CGC α = 17.3 fF fT = fT = 1 gm 2π CGC | gm = ∂iD " W " = KP ( VGS − VTP ) = μ pCOX | CGC = COX WL ∂vGS L 1 μp (VGS − VTP ), but (VGS − VTP )< 0 for PMOS transistor. 2π L2 1 μp Since f T should be positive, f T = (VGS − VTP ) 2π L2 4.148 1 ⎛μ⎞ VGS − VTN ) ⎜ ⎟( 2π ⎝ L2 ⎠ ⎡ ⎤ 1 ⎢400cm2 / V − s ⎥ fTN = (1V )= 6.37 GHz | fTP = 0.4 fTN = 2.55 GHz 2π ⎢ 10−4 cm 2 ⎥ ⎢ ⎥ ⎣ ⎦ ⎡ ⎤ 1 ⎢ 400cm 2 / V − s ⎥ (b) fTN = 2π ⎢ −5 2 ⎥(1V )= 637 GHz | fTP = 0.4 fTN = 255 GHz ⎢ ⎥ ⎣ 10 cm ⎦ (a) f T = ( ) ( ) 122 4.149 (a) Kn = μn ID = εox W Tox −14 ⎡ cm2 ⎢ 3.9 8.854 x10 F / cm = 400 V − s⎢ L 80 x10−7 cm ⎣ ( ⎤⎛ )⎥ 2μm ⎞ μA ⎜ ⎟ = 345 ⎥⎝ 0.1μm ⎠ ⎦ V2 345μA 2 (2) = 690μA 2 " ox 3.9 8.854 x10−14 F / cm W (b) I D = C 2 (VGS − VTN )vsat = 80 x10−7 cm 4.150 ( )⎛ 2 x10 ⎜ ⎝ cm ⎞ 7 ⎟(2V ) 10 cm / s = 86.3μA 2 ⎠ −4 ( ) For VGS = 0, ID = 10-22 A. For VTN = 0.5 V and VGS = 0, ID = 10-15 A. 123 CHAPTER 5 5.1 Base Contact = B n-type Emitter = D 5.2 v BC iB + B + C iC Collector Contact = A n-type Collector = F Emitter Contact = C Active Region = E For VBE > 0 and VBC = 0, IC = β F I B or β F = IC 275μA = = 68.8 4μA IB βR = 0.5 αR = =1 1 − α R 1 − 0.5 IC 275μA = = 2.10 fA ⎛ VBE ⎞ ⎛ 0.64 ⎞ exp⎜ ⎟ exp⎜ ⎟ ⎝ 0.025 ⎠ ⎝ VT ⎠ ⎛ −275μA ⎞ IE = −⎜ ⎟ = 2.20 IB ⎝ 125μA ⎠ V + - vBE - E iE ⎛V ⎞ IC = I S exp⎜ BE ⎟ or I S = ⎝ VT ⎠ 5.3 i vBE i B - E For VBC > 0 and VBE = 0, I E = −β R I B or β R = − E + B βF = αF 0.975 = = 39 1 − α F 1 − 0.975 IC 275μA = = 3.13 fA ⎛ VBE ⎞ ⎛ 0.63 ⎞ exp⎜ ⎟ exp⎜ ⎟ ⎝ 0.025 ⎠ ⎝ VT ⎠ V + - + v BC C iC ⎛V ⎞ I E = − I S exp⎜ BC ⎟ or I S = ⎝ VT ⎠ 123 5.4 Using β = α β and α = : 1− α β +1 Table 5.P1 0.167 0.400 0.750 0.200 0.667 3.00 10.0 0.909 0.980 49.0 200 1000 0.995 0.999 0.9998 5000 5.5 (a) For this circuit, VBE = 0 V, VBC = -5 V and I = IC. Substituting these values into the collector current expression in Eq. (5.13): ⎡ ⎛ −5 ⎞⎤ I S ⎡ ⎛ −5 ⎞ ⎤ IC = I S ⎢exp(0)− exp⎜ ⎟⎥ − ⎢exp⎜ ⎟ − 1⎥ ⎝ .025 ⎠⎦ β R ⎣ ⎝ .025 ⎠ ⎦ ⎣ ⎛ ⎛ 1⎞ 1⎞ I = IC = I S ⎜1 + ⎟ = 10−15 A⎜1 + ⎟ = 2 fA. ⎝ 1⎠ ⎝ βR ⎠ (b) For this circuit, the constraints are VBC = -5 V and IE = 0. Substituting these values into the emitter current expression in Eq. (5.13): ⎡ ⎛V ⎞ ⎛ V ⎞⎤ I ⎡ ⎛ V ⎞ ⎤ I E = I S ⎢exp⎜ BE ⎟ − exp⎜ BC ⎟⎥ + S ⎢exp⎜ BE ⎟ − 1⎥ = 0 which gives ⎝ VT ⎠⎦ β F ⎣ ⎝ VT ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎛V ⎞ ⎛V ⎞ βF 1 + exp⎜ BC ⎟. Substituting this result into IC : exp⎜ BE ⎟ = ⎝ VT ⎠ 1 + β F 1 + β F ⎝ VT ⎠ ⎛ V ⎞⎤ I ⎡ ⎛ V ⎞ ⎤ I ⎡ ICBO = S ⎢1 − exp⎜ BC ⎟⎥ − S ⎢exp⎜ BC ⎟ − 1⎥. 1 + βF ⎣ ⎝ VT ⎠⎦ β R ⎣ ⎝ VT ⎠ ⎦ ⎡ 1 ⎡ 1 1⎤ 1⎤ + ⎥ = 10−15 A⎢ + ⎥ = 1.01 fA, and For VBC = -5V , ICBO = I S ⎢ ⎣101 1⎦ ⎣1 + β F β R ⎦ ⎛ 1 ⎞ ⎛ 1 ⎞ VBE = VT ln⎜ ⎟ = 0.025V ln⎜ ⎟ = −0.115 V ≠ 0! ⎝101⎠ ⎝1 + β F ⎠ 5.6 124 + 150 μA VBC IC C IB B E + V BE IE (a) - (c) (b) npn transistor (d) VBE = VBC (e) I I S ⎡ ⎛ VBE ⎞ ⎤ ⎢exp⎜ ⎟ − 1⎥ β R ⎣ ⎝ VT ⎠ ⎦ I ⎡ ⎛V ⎞ ⎤ I E = + S ⎢exp⎜ BE ⎟ − 1⎥ β F ⎣ ⎝ VT ⎠ ⎦ ⎛ 1 1 ⎞ ⎡ ⎛V ⎞ ⎤ IB = IS ⎜ + ⎟ ⎢exp⎜ BE ⎟ − 1⎥ ⎝ β F β R ⎠ ⎣ ⎝ VT ⎠ ⎦ C =− IE = IB 1 1+ βF I = − 400 I E and I B = I E − IC = 401 I E βR E For the circuit I B = 150μA 150μA Therefore I E = = 0.374 μA, and IC = −149.6 μA. βF βR and IE β =− R IC βF (f ) Using IC = − VBC = VBE 401 ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ IB 150μA ⎜ ⎟ ⎜ ⎟ = 0.591 V = (0.025V )ln = VT ln ⎞ ⎛ ⎞ ⎜ ⎛ 1 ⎟ ⎜ 1 1 ⎟ 1 + + ⎟⎟ ⎟⎟ ⎜ IS ⎜ ⎜ 2 fA⎜ ⎝ 100 0.25 ⎠ ⎠ ⎝ ⎝ ⎝ βF βR ⎠ ⎠ 5.7 VBC IB + B + V BE 175 μA E IE - C IC npn transistor ⎛ 1 ⎞⎡ ⎛ VBE ⎞ ⎤ IE | IC = β F I B For VBC = 0, I E = I S ⎜1 + ⎟⎢exp⎜ ⎟ − 1⎥ | I B = βF + 1 ⎝ β F ⎠⎣ ⎝ VT ⎠ ⎦ I E = 175 μA | I B = VBE 175μA 100 = 1.73 μA | IC = 175μA = 173 μA 101 101 ⎛ β ⎛100 175μA ⎞ IE ⎞ = VT ln⎜ F + 1⎟ = 0.025V ln⎜ + 1⎟ = 0.630 V ⎠ ⎝ 101 2 fA ⎠ ⎝ βF + 1 IS 125 5.8 VBE IB + B + V BC 175 μA C IC - E IE npn transistor ⎛ 1 ⎞⎡ ⎛ V ⎞ ⎤ I For VBE = 0, IC = − I S ⎜1 + ⎟⎢exp⎜ BC ⎟ − 1⎥ | I B = − C | I E = −β R I B βR + 1 ⎝ β R ⎠⎣ ⎝ VT ⎠ ⎦ ⎛ −175μA ⎞ 0.25 IC = −175 μA | I B = −⎜ 175μA = −35 μA ⎟ = 140 μA | I E = − 1.25 ⎝ 1.25 ⎠ ⎛ β ⎞ ⎛ 0.25 175μA ⎞ I + 1⎟ = 0.590 V VBC = VT ln⎜− R C + 1⎟ = 0.025V ln⎜ ⎝ 1.25 2 fA ⎠ ⎝ βR + 1 IS ⎠ 5.9 Using vBC = 0 in Eq. 5.13 and recognizing that i = iC + iB = iE : ⎛ 1 ⎞⎡ ⎛ vBE ⎞ ⎤ i = iE = I S ⎜1 + ⎟⎢exp⎜ ⎟ − 1⎥ , and the reverse saturation current ⎝ β F ⎠⎣ ⎝ VT ⎠ ⎦ ⎛ ⎛ 1 ⎞ 1 ⎞ ' = I S ⎜1 + of the diode connected transistor is I S ⎟ = (2 fA) ⎜1 + ⎟ = 2.02 fA ⎝ 100 ⎠ ⎝ βF ⎠ 5.10 Using vBE = 0 in Eq. 5.13 and recognizing that i = −iC : ⎛ 1 ⎞⎡ ⎛ v ⎞ ⎤ i = −iC = − I S ⎜1 + ⎟⎢exp⎜ BC ⎟ − 1⎥ , and the reverse saturation current ⎝ β R ⎠⎣ ⎝ VT ⎠ ⎦ ⎛ ⎛ 1⎞ 1⎞ ' = I S ⎜1 + ⎟ = (5 fA) of the diode connected transistor is I S ⎜1 + ⎟ = 6.67 fA ⎝ 3⎠ ⎝ βR ⎠ 5.11 ⎡ ⎛v ⎞ ⎡ ⎛ 0.75 ⎞ ⎛ v ⎞⎤ ⎛ −3 ⎞⎤ = I S ⎢exp⎜ BE ⎟ − exp⎜ BC ⎟⎥ = 5 x10−16 A⎢exp⎜ ⎟ − exp⎜ ⎟⎥ = 5.34 mA ⎝ 0.025 ⎠⎦ ⎝ VT ⎠⎦ ⎣ ⎝ 0.025 ⎠ ⎣ ⎝ VT ⎠ (b) The current is symmetric: For VBC = 0.75 V and VBE = -3 V, iT = -5.34 mA. (a) i T ⎡ ⎛v ⎞ ⎡ ⎛ 0.70 ⎞ ⎛ vBC ⎞⎤ ⎛ −3 ⎞⎤ −15 BE a i = I exp A exp = 10 − exp − exp ⎢ ⎥ ⎢ ⎜ ⎟ ⎜ ⎟ ⎟ ⎜ ⎟⎥ = 1.45 mA ()T S ⎜ ⎝ 0.025 ⎠⎦ ⎝ VT ⎠⎦ ⎣ ⎝ 0.025 ⎠ ⎣ ⎝ VT ⎠ (b) The current is symmetric: For VBC = 0.70 V and VBE = -3 V, iT = -1.45 mA. 5.12 5.13 Base Contact = F p-type Emitter = D 5.14 Collector Contact = G p-type Collector = A Emitter Contact = E Active Region = C 126 (a) pnp transistor + v iB B v 100 μA CB EB i E E C + iC (b)-(c) (d) Using Eq. (5.17) with vEB = 0 and dropping the "-1" terms: ⎛ ⎛v ⎞ ⎛v ⎞ 1 ⎞ ⎛v ⎞ I iC = − I S ⎜1 + ⎟ exp⎜ CB ⎟ iE = − I S exp⎜ CB ⎟ iB = S exp⎜ CB ⎟ βR ⎝ β R ⎠ ⎝ VT ⎠ ⎝ VT ⎠ ⎝ VT ⎠ IE βR 1 IE = = = αR = −β R 1 IC βR + 1 IB 1+ ⎛V ⎞ ⎛ I ⎞ VEB = 0 and I E = − I S exp⎜ CB ⎟ VCB = VT ln⎜ − E ⎟ ⎝ VT ⎠ ⎝ Is ⎠ ⎛ −25 x10−6 A ⎞ VCB = 0.025V ln⎜ − ⎟ = 0.599 V 10−15 A ⎠ ⎝ βR IC = −100μA, I E = α R IC = 0.25 IC = −25.0μA αR I 0.25 1 IB = − E β R = = = I B = +75μA βR 1 − α R 1 − 0.25 3 α 0.985 βF = F = = 65.7 1 − α F 1 − 0.985 127 5.15 (a) pnp VCB IB + B V V EB E + IE + IC C (b)-(c) (d) Using Eq. (5.17) with vCB = 0 and droping the "-1" terms: ⎛ ⎛v ⎞ 1 ⎞ ⎛ vEB ⎞ iE = I S ⎜1 + iC = − I S exp⎜ EB ⎟ ⎟ exp⎜ ⎟ ⎝ β F ⎠ ⎝ VT ⎠ ⎝ VT ⎠ IC 300μA = = 2.29 fA IS = ⎛ VEB ⎞ ⎛ 0.640 ⎞ exp⎜ ⎟ exp⎜ ⎟ ⎝ 0.025V ⎠ ⎝ VT ⎠ iB = ⎛v ⎞ exp⎜ EB ⎟ βF ⎝ VT ⎠ IS βF = αR IC 300μA 0.2 = = 75 | β R = = = 0.25 4μA IB 1 − α R 1 − 0.2 5.16 Using VCB = 0 in Eq. 5.17 and recognizing that i = iE : ⎛ 1 ⎞⎡ ⎛ vEB ⎞ ⎤ i = iE = I S ⎜1 + ⎟⎢exp⎜ ⎟ − 1⎥ , and the reverse saturation current ⎝ β F ⎠⎣ ⎝ VT ⎠ ⎦ ⎛ ⎛ 1 ⎞ 1 ⎞ ' = I S ⎜1 + of the diode connected transistor is I S ⎟ = (2 fA) ⎜1 + ⎟ = 2.02 fA ⎝ 100 ⎠ ⎝ βF ⎠ 5.17 v 35 μ A i CB i B B - v EB + C E + iE (a)-(c) (b) pnp transistor(d) ⎛ 1 I ⎡ ⎛v ⎞ ⎤ I ⎡ ⎛v ⎞ ⎤ 1 ⎞⎡ ⎛ v ⎞ ⎤ + ⎟⎢exp⎜ EB ⎟ − 1⎥ vEB = vCB iC = − S ⎢exp⎜ EB ⎟ − 1⎥ iE = + S ⎢exp⎜ EB ⎟ − 1⎥ iB = + I S ⎜ β R ⎣ ⎝ VT ⎠ ⎦ β F ⎣ ⎝ VT ⎠ ⎦ ⎝ β F β R ⎠⎣ ⎝ VT ⎠ ⎦ 1 IE βR 4 IE β 4 βF = = = = 0.0506 = − R = − = −0.0533 1 1 IB β F + β R 79 IB βF 75 + C βF βR 128 I B = 35 μA VEB 4 75 I B = 1.77 μA IC = − I E = −33.2 μA 79 4 ⎛ 4 −33.2 x10−6 A ⎛ β R IC ⎞ VCB = VEB = 0.025V ln⎜1 − = VT ln⎜1 − ⎟ ⎜ Is ⎠ 2 x10−15 A ⎝ ⎝ IE = ( ⎞ )⎟ = 0.623 V ⎟ ⎠ 5.18 C IC VCB + IB B V EB I + E IE pnp transistor ⎛ 1 ⎞ ⎡ ⎛ VEB ⎞ ⎤ For VCB = 0, I E = I S ⎜1 + ⎟ ⎢exp⎜ ⎟ − 1⎥ ⎝ β F ⎠ ⎣ ⎝ VT ⎠ ⎦ I E = 300 μA | I B = VEB | IB = IE | IC = β F I B βF + 1 300μA = 2.97 μA | IC = 100(2.97μA)= 297 μA 101 ⎤ ⎡⎛ β ⎞ I ⎡⎛ 100 ⎞ 300μA ⎤ = VT ln⎢⎜ F ⎟ E + 1⎥ = 0.025V ln⎢⎜ + 1⎥( )= 0.626 V ⎟ ⎦ ⎣⎝ 101⎠ 4 fA ⎣⎝ β F + 1⎠ I S ⎦ 5.19 VEB IB B V CB I C + IC + E IE pnp transistor ⎛ 1 ⎞⎡ ⎛ V ⎞ ⎤ I | I E = −β R I B For VEC = 0, IC = − I S ⎜1 + ⎟⎢exp⎜ CB ⎟ − 1⎥ | I B = − C βR + 1 ⎝ β R ⎠⎣ ⎝ VT ⎠ ⎦ ⎛ −300μA ⎞ IC = −300 μA | I B = −⎜ 150μA)= −150 μA ⎟ = 150 μA | I E = −1( ⎝ 2 ⎠ ⎡ ⎛ β ⎞I ⎤ ⎡ 1 ⎛ −300μA ⎞ ⎤ VCB = VT ln⎢−⎜ R ⎟ C + 1⎥ = 0.025V ln⎢− ⎜ ⎟ + 1⎥ = 0.603 V ⎣ 2 ⎝ 5 fA ⎠ ⎦ ⎣ ⎝ β R + 1⎠ I S ⎦ 5.20 (a) i ⎡ ⎛v ⎞ ⎡ ⎛ 0.70 ⎞ ⎛ v ⎞⎤ ⎛ −3 ⎞⎤ = I S ⎢exp⎜ EB ⎟ − exp⎜ CB ⎟⎥ = 5 x10−16 A⎢exp⎜ ⎟ − exp⎜ ⎟⎥ = 723 μA ⎝ 0.025 ⎠⎦ ⎝ VT ⎠⎦ ⎣ ⎝ 0.025 ⎠ ⎣ ⎝ VT ⎠ (b) The current is symmetric: For VCB = 0.75 V and VEB = -3 V, iT = -723 μA. T 129 5.21 ⎡ ⎛v ⎞ ⎤ ⎡ ⎛ 0.73V ⎞ ⎤ −15 BE i = I a ⎢ ⎟ − 1⎥ = 4 x10 A⎢exp⎜ ⎟ − 1⎥ = 19.2 mA ( ) F S exp⎜ ⎣ ⎝ 0.025V ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎦ ⎡ ⎛v ⎞ ⎤ ⎡ ⎛ −3V ⎞ ⎤ iR = I S ⎢exp⎜ BC ⎟ − 1⎥ = 4 x10−15 A⎢exp⎜ ⎟ − 1⎥ = −4.00 fA ⎣ ⎝ 0.025V ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎦ 19.2 mA i −4.00 fA = 240 μA | R = = −2.00 μA βF βR 80 2 ⎡ ⎛v ⎞ ⎤ ⎡ ⎛ −3V ⎞ ⎤ −15 BE ⎟ − 1⎥ = 4 x10 A⎢exp⎜ ⎟ − 1⎥ = −4.00 fA (b) iF = I S ⎢exp⎜ ⎣ ⎝ 0.025V ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎦ ⎡ ⎛v ⎞ ⎤ ⎡ ⎛ 0.73V ⎞ ⎤ iR = I S ⎢exp⎜ BC ⎟ − 1⎥ = 4 x10−15 A⎢exp⎜ ⎟ − 1⎥ = 19.2 mA ⎣ ⎝ 0.025V ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎦ iT = iF − iR = 19.2 mA | iF = iT = iF − iR = −19,2 mA | 5.22 βF iF = −4.00 fA i 19.2mA = −0.05 μA | R = = 9.60 mA βR 80 2 (a) i ⎡ ⎛v ⎞ ⎤ ⎡ ⎛ 0.68V ⎞ ⎤ = I S ⎢exp⎜ EB ⎟ − 1⎥ = 6 x10−15 A⎢exp⎜ ⎟ − 1⎥ = 3.90 mA ⎣ ⎝ 0.025V ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎦ ⎡ ⎛v ⎞ ⎤ ⎡ ⎛ −3V ⎞ ⎤ iR = I S ⎢exp⎜ CB ⎟ − 1⎥ = 6 x10−15 A⎢exp⎜ ⎟ − 1⎥ = −6.00 fA ⎣ ⎝ 0.025V ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎦ F 3.90 mA i −6.00 fA = 65.0 μA | R = = −2.00 μA βF βR 60 3 ⎡ ⎛v ⎞ ⎤ ⎡ ⎛ −3V ⎞ ⎤ −15 EB ⎟ − 1⎥ = 6 x10 A⎢exp⎜ ⎟ − 1⎥ = −6.00 fA (b) iF = I S ⎢exp⎜ ⎣ ⎝ 0.025V ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎦ ⎡ ⎛v ⎞ ⎤ ⎡ ⎛ 0.68V ⎞ ⎤ iR = I S ⎢exp⎜ CB ⎟ − 1⎥ = 6 x10−15 A⎢exp⎜ ⎟ − 1⎥ = 3.90 mA ⎣ ⎝ 0.025V ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎦ iT = iF − iR = 3.90 mA | iF = iT = iF − iR = −3.90 mA | βF iF = −6.00 fA i 3.90 mA = −0.100 fA | R = = 1.30 mA 60 3 βR 130 5.23 ⎡ ⎛v ⎞ ⎡ ⎛v ⎞ ⎛ v ⎞⎤ I ⎡ ⎛ v ⎞ ⎤ ⎛v ⎞ ⎤ I ⎡ ⎛v ⎞ ⎤ iE = I S ⎢exp⎜ BE ⎟ − exp⎜ BC ⎟⎥ + S ⎢exp⎜ BE ⎟ − 1⎥ = I S ⎢exp⎜ BE ⎟ − 1 − exp⎜ BC ⎟ + 1⎥ + S ⎢exp⎜ BE ⎟ − 1⎥ ⎝ VT ⎠⎦ β F ⎣ ⎝ VT ⎠ ⎦ ⎝ VT ⎠ ⎦ β F ⎣ ⎝ VT ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎣ ⎝ VT ⎠ ⎡ ⎛v ⎞ ⎤ I ⎡ ⎛v ⎞ ⎤ ⎡ ⎛v ⎞ ⎤ ⎛ ⎛ v BC ⎞ ⎤ 1 ⎞⎡ ⎛ vBE ⎞ BC S BE BC iE = I S ⎜1 + exp exp exp − 1 − exp + 1 − I − 1 = − 1 − I ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎟ − 1⎥ S S ⎢exp⎜ ⎝ β F ⎠⎣ ⎝ VT ⎠ ⎝ VT ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎦ α F ⎣ ⎝ VT ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎦ ⎡ ⎛v ⎞ ⎡ ⎛v ⎞ ⎛ v ⎞⎤ I ⎡ ⎛ v ⎞ ⎤ ⎛v ⎞ ⎤ I ⎡ ⎛v ⎞ ⎤ iC = I S ⎢exp⎜ BE ⎟ − exp⎜ BC ⎟⎥ − S ⎢exp⎜ BC ⎟ − 1⎥ = I S ⎢exp⎜ BE ⎟ − 1 − exp⎜ BC ⎟ + 1⎥ − S ⎢exp⎜ BC ⎟ − 1⎥ ⎝ VT ⎠⎦ β R ⎣ ⎝ VT ⎠ ⎦ ⎝ VT ⎠ ⎦ β R ⎣ ⎝ VT ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎣ ⎝ VT ⎠ ⎡ ⎛v ⎞ ⎤ ⎡ ⎛v ⎞ ⎤ I ⎡ ⎛v ⎞ ⎤ ⎛ 1 ⎞⎡ ⎛ v ⎞ ⎤ iC = I S ⎢exp⎜ BE ⎟ − 1⎥ − I S ⎜1 + ⎟⎢exp⎜ BC ⎟ − 1⎥ = I S ⎢exp⎜ BE ⎟ − 1⎥ − S ⎢exp⎜ BC ⎟ − 1⎥ ⎝ β R ⎠⎣ ⎝ VT ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎦ α R ⎣ ⎝ VT ⎠ ⎦ Defining I ES = αF IS and ICS = αR IS , then we see I S = α F I ES = α R ICS and ⎡ ⎛v ⎞ ⎤ ⎡ ⎛v ⎞ ⎤ iE = I ES ⎢exp⎜ BE ⎟ − 1⎥ − α R I S ⎢exp⎜ BC ⎟ − 1⎥ ⎣ ⎝ VT ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎦ ⎡ ⎛v ⎞ ⎤ ⎡ ⎛v ⎞ ⎤ iC = α F I ES ⎢exp⎜ BE ⎟ − 1⎥ − ICS ⎢exp⎜ BC ⎟ − 1⎥ ⎣ ⎝ VT ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎦ ⎡ ⎛v ⎞ ⎤ ⎡ ⎛v ⎞ ⎤ iB = iE − iC = ( 1 − α R )I S ⎢exp⎜ BC ⎟ − 1⎥ 1 − α F )I ES ⎢exp⎜ BE ⎟ − 1⎥ + ( ⎣ ⎝ VT ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎦ 5.24 αF = ICS = βR 100 0.5 I 2 fA = 0.990 | α R = = = 0.333 | I ES = S = = 2.02 fA β F + 1 101 β R + 1 1.5 α F 0.990 = βF αR IS = 2 fA = 6.00 fA | α F I ES = α R ICS = I S 0.333 131 5.25 ⎡ ⎛v ⎞ ⎡ ⎛v ⎞ ⎤ ⎛v ⎞ ⎤ I ⎡ ⎛v ⎞ ⎤ ⎛ 1 ⎞⎡ ⎛ v EB ⎞ ⎤ CB iE = I S ⎢exp⎜ EB ⎟ − 1 − exp⎜ CB ⎟ + 1⎥ + S ⎢exp⎜ EB ⎟ − 1⎥ = I S ⎜1 + ⎟⎢exp⎜ ⎟ − 1⎥ − I S ⎢exp⎜ ⎟ − 1⎥ ⎝ VT ⎠ ⎦ β F ⎣ ⎝ VT ⎠ ⎦ ⎝ β F ⎠⎣ ⎝ VT ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎣ ⎝ VT ⎠ ⎦ ⎡ ⎛v ⎞ ⎡ ⎛v ⎞ ⎤ ⎛v ⎞ ⎤ I ⎡ ⎛v ⎞ ⎤ ⎛ 1 ⎞⎡ ⎛ v ⎞ ⎤ iC = I S ⎢exp⎜ EB ⎟ − 1 − exp⎜ CB ⎟ + 1⎥ − S ⎢exp⎜ CB ⎟ − 1⎥ = I S ⎢exp⎜ EB ⎟ − 1⎥ − I S ⎜1 + ⎟⎢exp⎜ CB ⎟ − 1⎥ ⎝ VT ⎠ ⎦ β R ⎣ ⎝ VT ⎠ ⎦ ⎝ β R ⎠⎣ ⎝ VT ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎣ ⎝ VT ⎠ ⎦ ⎡ ⎛v ⎞ ⎤ ⎡ ⎛v ⎞ ⎤ iE = I ES ⎢exp⎜ EB ⎟ − 1⎥ − α R ICS ⎢exp⎜ CB ⎟ − 1⎥ ⎣ ⎝ VT ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎦ ⎡ ⎛v ⎞ ⎤ ⎡ ⎛v ⎞ ⎤ iC = α F I ES ⎢exp⎜ EB ⎟ − 1⎥ − ICS ⎢exp⎜ CB ⎟ − 1⎥ ⎣ ⎝ VT ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎦ ⎡ ⎛v ⎞ ⎤ ⎡ ⎛v ⎞ ⎤ iB = iE − iC = ( 1 − α F )I ES ⎢exp⎜ EB ⎟ − 1⎥ + ( 1 − α R )ICS ⎢exp⎜ CB ⎟ − 1⎥ ⎣ ⎝ VT ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎦ 5.26 At IC = 5 mA and VCE = 5 V , I B = 60μA : β F = IC 5mA = = 83.3 I B 60μA IC 7 mA = = 87.5 I B 80μA IC 10 mA = = 100 I B 100μA At IC = 7 mA and VCE = 7.5 V , I B = 80μA : β F = At IC = 10 mA and VCE = 14 V , I B = 100μA : β F = 5.27 See Problem 5.28 5.28 20mA 10mA 0A -4mA 0V -I(VCC) V_VCC 2V 4V 6V 8V 10V 132 5.29 3.0mA 2.0mA 1.0mA 0A -1.0mA -2V -I(VCB) 0V 2V 4V V_VCB 6V 8V 10V 5.30 See Problem 5.31 5.31 20mA 10mA 0A -4mA 0V -I(VEC) 2V 4V V_VEC 6V 8V 10V 133 5.32 3.0mA 2.0mA 0A -1.0mA -2V -I(VBC) 0V 2V 4V V_VBC 6V 8V 10V 5.33 The change in vBE for a decade change in iC is ΔVBE = VT ln ( 10) = 2.30VT . ⎛1.38 x10-23 ⎞ kT V/dec) The reciprocal of the slope is 2.30VT = 2.30 = 2.30⎜ ⎟T ( -19 q ⎝1.60 x10 ⎠ (a) 39.6 mV/dec (b) 49.5 mV/dec (c) 59.4 mV/dec (d) 69.3 mV/dec 5.34 (a) The break down voltage is equal to that of the emitter-base junction: VZ = 6 V. (b) The break down voltage is determined by the base-collector junction: VZ = 50 V. (c) The break down voltage is set by the emitter-base junction: VZ = 6 V. 5.35 (a) The base-emitter junction breaks down with VEB = 6.3 V. 5 − 6.3 − (−5) V = 2.31 mA IR = Ω 1600 (b) The base-emitter junction is forward biased; VBE = 0.7 V 5 − 0.7 − (−5) V IR = = 388 μA 24000 Ω (c) VBE = 0, and the collector-base junction is reversed biased with VBC ≈ -10V which is less than the breakdown voltage of 75 V. The transistor is operating in cutoff. ⎛ I 1⎞ Using Eq. (5.13), I R = IC = I S ( 1 − 0)− S (0 − 1)= I S ⎜1 + ⎟ ≈ 0 βR ⎝ βR ⎠ 5.36 VCE = VCB + VBE = VCB + 0.7 ≤ 65.7 V 134 5.37 (a) IB is forced to be negative by the current source, and the largest negative base current according to the Transport model is ⎛ 1 ⎛1 1⎞ 1 ⎞ IB = −IS ⎜ + ⎟ = −10−15 A⎜ + ⎟ = −2.02 fA ⎝ 50 0.5 ⎠ ⎝ βF βR ⎠ . (b) IB is forced to be -1 mA by the current source. One or both of the junctions must enter the breakdown region in order to supply this current. For the case of a normal BJT, the base-emitter junction will break down and supply the current since it has the lower reverse breakdown voltage. 5.38 Base-Emitter Voltage Base-Collector Voltage 0.7 V -5.0 V Cutoff Forward Active -5.0 V 0.7 V Reverse Active Saturation 5.39 (a) vBE > 0, vBC = 0, forward-active region; vBE = 0, vBC > 0, reverse-active region; vBE > 0, vBC = 0, forward-active region (b) vEB < 0, vCB < 0, cutoff region (c) vEB > 0, vCB < 0, forward-active region (d) vBE > 0, vBC < 0, forward-active region; vBE > 0, vBC > 0, saturation region 5.40 (a) vBE = 0, vBC < 0 cutoff region (b) vBC < 0, IE = 0, cutoff region 5.41 (a) vBE > 0, vBC > 0 saturation region (b) vBE > 0, vBC = 0, forward-active region (c) vBE = 0, vBC > 0, reverse-active region 135 5.42 Emitter-Base Voltage Collector-Base Voltage 0.7 V -0.65 V Forward Active Cutoff 0.7 V -0.65 V Saturation Reverse Active 5.43 (a) vBE > 0, vBC = 0, forward-active region (b) vBE = 0, vBC > 0, reverse-active region 5.44 (a) vEB = 0, vCB > 0, reverse-active region (b) vEB > 0, vCB = 0, forward-active region 5.45 (a) vEB > 0, vCB > 0, saturation region (b) vEB > 0, vCB = 0, forward-active region (c) vEB = 0, vCB > 0, reverse-active region 5.46 (a ) pnp transistor with VEB = −3V and VCB = −3V → Cutoff | Using Eq. (5.17) : IC = + 10−15 A I 10−15 A = 0.5x10-15 = 0.5 fA | I E = − S = = 13.3x10-18 = 13.3 aA 2 75 βR βF ⎛ 1 ⎛ 1 1⎞ 1⎞ + ⎟ = 10−15 A⎜ + ⎟ = 0.263 x10−15 = 0.263 fA IB = −IS ⎜ ⎝ 75 4 ⎠ ⎝ βF βR ⎠ IS = BE (b) npn transistor with V in part (a). = −5V and VBC = −5V → Cutoff | The currents are the same as 5.47 ⎡ ⎛ 0.3 ⎞ ⎛ −5 ⎞⎤ 10−16 ⎡ ⎛ −5 ⎞ ⎤ iC = 10−16 ⎢exp⎜ − exp ⎢exp⎜ ⎟ ⎜ ⎟⎥ − ⎟ − 1⎥ = 16.3 pA 1 ⎣ ⎝ 0.025⎠ ⎦ ⎝ 0.025 ⎠⎦ ⎣ ⎝ 0.025 ⎠ ⎡ ⎛ 0.3 ⎞ ⎛ −5 ⎞⎤ 10−16 ⎡ ⎛ 0.3 ⎞ ⎤ iE = 10−16 ⎢exp⎜ ⎢exp⎜ ⎟ − exp⎜ ⎟⎥ + ⎟ − 1⎥ = 17.1 pA ⎝ 0.025 ⎠⎦ 19 ⎣ ⎝ 0.025⎠ ⎦ ⎣ ⎝ 0.025 ⎠ 10−16 ⎡ ⎛ 0.3 ⎞ ⎤ 10−16 ⎡ ⎛ −5 ⎞ ⎤ iB = ⎢exp⎜ ⎢exp⎜ ⎟ − 1⎥ + ⎟ − 1⎥ = 0.857 pA 19 ⎣ ⎝ 0.025 ⎠ ⎦ 1 ⎣ ⎝ 0.025 ⎠ ⎦ 136 These currents are all very small - for most practical purposes it still appears to be cutoff. Since VBE > 0 and VBC < 0, the transistor is actually operating in the forward-active region. Note that I C = β FI B . 5.48 An npn transistor with VBE = 0.7V and VBC = −0.7V → Forward - active region Using Eq. (5.45) : I E = (β F + 1)I B | β F = IE 10mA −1 = − 1 = 65.7 0.15mA IB ⎛ 1 ⎞ ⎛V ⎞ 0.01 A = 6.81x10−15 A = 6.81 fA I E = I S ⎜1 + ⎟ exp⎜ BE ⎟ | I S = ⎛ ⎞ ⎛ ⎞ 1 0.7 ⎝ β F ⎠ ⎝ VT ⎠ ⎜1 + ⎟ exp⎜ ⎟ ⎝ 65.7 ⎠ ⎝ 0.025⎠ 5.49 A pnp transistor with VEB = 0.7V and VCB = −0.7V ⇒ Forward - active region ⎛V ⎞ I 2.5mA 2.5mA Using Eq. (5.44) : β F = C = = 62.5 | IC = I S exp⎜ EB ⎟ | I S = = 1.73 fA ⎛ 0.7V ⎞ I B 0.04 mA ⎝ VT ⎠ exp⎜ ⎟ ⎝ 0.025V ⎠ 5.50 IE = −0.7V − (−3.3V ) 47 kΩ = 55.3μA | I B = IC = β F I B = 80(0.683μA)= 54.6μA | Check : I B + IC = I E is ok 5.51 IE 55.3μA = = 0.683μA 81 βF + 1 (a) fβ = βF fT = 500 MHz = 6.67 MHz 759 ( b ) The graph represents the Bode magnitude plot. Thus β (s) = βF 1+ s = ωβ βFωβ ωT = s + ωβ s + ωβ ωT βF β (s) s + ωβ βFωβ ωT αF βF + 1 α (s)= = = = = ≈ ωT s s s + ωT + ω β s + (β F + 1) β (s)+ 1 ωβ +1 1+ 1+ s + ωβ ωT (β F + 1)ω β α ( jω ) = αF ⎛ ω ⎞2 1+ ⎜ ⎟ ⎝ ωT ⎠ 137 5.52 vEB > 0 vCB < −4VT ⎛V ⎞ I ⎛V ⎞ iC = I S exp⎜ EB ⎟ + S ≈ I S exp⎜ EB ⎟ ⎝ VT ⎠ β R ⎝ VT ⎠ ⎛V ⎞ I ⎛V ⎞ I ⎛V ⎞ iE = I S exp⎜ EB ⎟ + S exp⎜ EB ⎟ = S exp⎜ EB ⎟ ⎝ VT ⎠ β F ⎝ VT ⎠ α F ⎝ VT ⎠ ⎛V ⎞ I ⎛V ⎞ I I iB = S exp⎜ EB ⎟ − S ≈ S exp⎜ EB ⎟ βF ⎝ VT ⎠ β R β F ⎝ VT ⎠ iC = β F iB | iC = α F iE iB B iC C + 0.7 V vEB i = β i C F B i E E 5.53 An npn transistor with VBE = −0.7V and VBC = +0.7V → Reverse - active region Using Eq. (5.51) : IC = −(β R + 1)I B | β R = − ⎛V ⎞ I E = − I S exp⎜ BC ⎟ | I E = −35μA | I S = − ⎝ VT ⎠ IC −75μA −1 = − − 1 = 0.875 40μA IB −35μA = 2.42 x10−17 A = 0.0242 fA = 24.2 aA ⎛ 0.7 ⎞ exp⎜ ⎟ ⎝ 0.025⎠ 5.54 A pnp transistor with VEB = −0.7 V and VCB = +0.7 V → Reverse − active region ⎛V ⎞ I ⎛V ⎞ ⎛V ⎞ I iC = − I S exp⎜ CB ⎟ − S exp⎜ CB ⎟ = − S exp⎜ CB ⎟ αR ⎝ VT ⎠ β R ⎝ VT ⎠ ⎝ VT ⎠ ⎛V ⎞ I ⎛V ⎞ iE = − I S exp⎜ CB ⎟ − S ≅ − I S exp⎜ CB ⎟ ⎝ VT ⎠ β F ⎝ VT ⎠ ⎛V ⎞ I ⎛V ⎞ I I iB = − S + S exp⎜ CB ⎟ ≅ S exp⎜ CB ⎟ βF βR ⎝ VT ⎠ β R ⎝ VT ⎠ βR = − (−0.1mA) = 0.667 | I = − iE =− S iB 0.15mA iE −10−4 A =− = 6.91x10−17 A ⎛ VCB ⎞ ⎛ 0.7 ⎞ exp⎜ exp⎜ ⎟ ⎟ ⎝ 0.025⎠ ⎝ VT ⎠ IC −46.4μA =− = 26.5 μA 1.75 βR + 1 5.55 IC = − −0.7V − (−3.3V ) 56 kΩ = −46.4 μA | I B = − I E = IC + I B = −46.4μA + 26.5μA = −19.9 μA 138 5.56 ⎡ ⎤ β 1 + FOR ⎥ ⎢ ⎛ 1 ⎞ (β R + 1)⎥ βR I 1mA 2 | αR = β FOR = C = = 1 | VCESAT = VT ln⎢⎜ ⎟ = ⎢⎝ α R ⎠ ⎛β ⎞ ⎥ I B 1mA βR + 1 3 1 − ⎜ FOR ⎟ ⎥ ⎢ ⎝ βF ⎠ ⎦ ⎣ ⎡ 1 ⎤ ⎢ 1+ ⎥ ⎛ 3 ⎞ (2 + 1)⎥ ⎢ = 17.8 mV VCESAT = 0.025ln ⎜ ⎟ ⎢⎝ 2 ⎠ ⎛1⎞⎥ 1− ⎜ ⎟ ⎥ ⎢ ⎝ 50 ⎠ ⎦ ⎣ ⎡ ⎤ ⎢ ⎥ ⎡ 1mA + ( 1 − α R )IC ⎥ 1mA ⎤ IB + ( 1 − 0.667) ⎢ ⎢ ⎥ = 0.724 V = 0.025V )ln −15 VBE = VT ln ⎞⎥ ( ⎢ ⎛ 1 10 A 0.02 + 1 − .0.667 ⎢ ( )⎥ ⎣ ⎦ + 1 − α R ⎟⎥ ⎢ IS ⎜ β ⎠⎦ ⎣ ⎝ F 5.57 ⎛v ⎞ I ⎛v ⎞ ⎛v ⎞ I ⎛v ⎞ I iC = I S exp⎜ EB ⎟ − S exp⎜ CB ⎟ | iB = S exp⎜ EB ⎟ + S exp⎜ CB ⎟ | Simultaneous βF ⎝ VT ⎠ α R ⎝ VT ⎠ ⎝ VT ⎠ β R ⎝ VT ⎠ i iB − C iB + ( 1 − α R )iC βF solution yields : vEB = VT ln | vCB = VT ln ⎤ ⎤ ⎡1 ⎡ 1 ⎤⎡ 1 IS ⎢ + ( 1− αR ) I S ⎢ ⎥⎢ + ( 1− αR ) ⎥ ⎥ ⎦ ⎦ ⎣β F ⎣α R ⎦⎣β F ⎡ ⎤ iC 1+ ⎢ ⎥ ⎛ 1 ⎞ (β R + 1) iB ⎥ i ⎢ vECSAT = vEB − vCB = VT ln ⎜ ⎟ for i B > C ⎢⎝ α R ⎠ ⎥ i βF 1− C ⎢ ⎥ β F iB ⎦ ⎣ 5.58 (a) Substituting iC = 0 in Eq. 5.30 gives ⎛ 1 ⎞ ⎛ 1 ⎞ VCESAT = VT ln⎜ ⎟ = (0.025V )ln⎜ ⎟ = 0.0173 V = 17.3 mV ⎝ 0.5⎠ ⎝αR ⎠ (b) By symmetry ⎛ 1 ⎞ VECSAT = VT ln⎜ ⎟ ⎝αF ⎠ or by using iE = 0 and iC = -iB, 139 VCESAT βR 1 1− ⎛ 1 ⎞ β +1 ⎛ 1 ⎞ β +1 ⎛ 1 ⎞α R = VT ln⎜ ⎟ = VT ln⎜ ⎟ R = VT ln⎜ ⎟ R ⎝αR ⎠ 1 + 1 ⎝ α R ⎠ βF + 1 ⎝αR ⎠ 1 βF βF αF ⎛ 1 ⎞ VCESAT = VT ln ( α F ) and VECSAT = VT ln⎜ ⎟ ⎝αF ⎠ ⎛ 1 ⎞ ⎛ 1 ⎞ VECSAT = VT ln⎜ ⎟ = (0.025V )ln⎜ ⎟ = 0.000251 V = 0.251 mV ⎝ 0.99 ⎠ ⎝αF ⎠ 5.59 (a) Substituting iC = 0 in Eq. 5.30 gives ⎛ 1 ⎞ ⎛ 1 ⎞ VCESAT = VT ln⎜ ⎟ = (0.025V )ln⎜ ⎟ = 27.7 mV ⎝ 0.33⎠ ⎝αR ⎠ (b) By symmetry ⎛ 1 ⎞ ⎛ 1 ⎞ VECSAT = VT ln⎜ ⎟ = (0.025V )ln⎜ ⎟ = 1.28 mV ⎝ 0.95⎠ ⎝αF ⎠ 5.60 ( a ) VCESAT ⎡ ⎤ β 1 + FOR ⎥ ⎢ ⎛ 1 ⎞ (β R + 1)⎥ 0.9 βR = VT ln⎢⎜ ⎟ = = 0.4737 | αR = ⎢⎝ α R ⎠ ⎛ β FOR ⎞ ⎥ β R + 1 0.9 + 1 1− ⎜ ⎢ ⎟⎥ ⎝ βF ⎠ ⎦ ⎣ ⎡ β FOR ⎤ 1+ ⎢ ⎥ ⎛ 1 ⎞ (0.9 + 1)⎥ ⎢ 0.1 = 0.025ln ⎜ → β FOR = 11.05 ⎢⎝ 0.4737 ⎟ ⎛ β FOR ⎞ ⎥ ⎠ 1− ⎜ ⎢ ⎟⎥ ⎝ 15 ⎠ ⎦ ⎣ 1+ 0.4737exp(4) = ⎛β ⎞ 1 − ⎜ FOR ⎟ ⎝ 15 ⎠ (0.9 + 1) → β β FOR FOR = 11.05 | I B = β FOR IC = 20 A = 1.81A 11.05 ⎡ β FOR ⎤ 1+ ⎢⎛ ⎥ 1 ⎞ (0.9 + 1) ⎥ I 20 A (b) 0.04 = 0.025ln⎢⎜ = 10.1A → β FOR = 1.97 | I B = C = ⎟ ⎞ ⎛ β FOR 1.97 ⎢⎝ 0.4737 ⎠ 1 − β FOR ⎥ ⎟ ⎜ ⎢ ⎝ 15 ⎠ ⎥ ⎣ ⎦ 140 5.61 With VBE = 0.7 and VBC = 0.5, the transistor is technically in the saturation region, but calculating the currents using the transport model in Eq. (5.13) yields ⎡ ⎛ 0.7 ⎞ ⎛ 0.5 ⎞⎤ 10−16 ⎡ ⎛ 0.5 ⎞ ⎤ iC = 10−16 ⎢exp⎜ ⎢exp⎜ ⎟ − exp⎜ ⎟⎥ − ⎟ − 1⎥ = 144.5 μA 1 ⎣ ⎝ 0.025 ⎠ ⎦ ⎝ 0.025 ⎠⎦ ⎣ ⎝ 0.025 ⎠ ⎡ ⎛ 0.7 ⎞ ⎛ 0.5 ⎞⎤ 10−16 ⎡ ⎛ 0.7 ⎞ ⎤ −16 iE = 10 ⎢exp⎜ ⎢exp⎜ ⎟ − exp⎜ ⎟⎥ + ⎟ − 1⎥ = 148.3 μA ⎝ 0.025 ⎠⎦ 39 ⎣ ⎝ 0.025⎠ ⎦ ⎣ ⎝ 0.025 ⎠ iB = 10−16 ⎡ ⎛ 0.7 ⎞ ⎤ 10−16 ⎡ ⎛ 0.5 ⎞ ⎤ ⎢exp⎜ ⎢exp⎜ ⎟ − 1⎥ + ⎟ − 1⎥ = 3.757 μA 39 ⎣ ⎝ 0.025⎠ ⎦ 1 ⎣ ⎝ 0.025 ⎠ ⎦ At 0.5 V, the collector-base junction is not heavily forward biased compared to the base-emitter junction, and IC = 38.5 IB ≅ β F IB . The transistor still acts as if it is operating in the forwardactive region. 5.62 (a) The current source will forward bias the base - emitter junction (VBE ≅ 0.7V ) and the collector - base junction will then be reverse biased (VBC ≅ −2.3V ). Therefore, the npn transistor is in the forward - active region. ⎛ 50 175 x10−6 A ⎞ ⎛ VBE ⎞ ⎟ = 0.803 V IC = β F I B = I S exp⎜ ⎟ | VBE = 0.025ln⎜ −16 ⎜ ⎟ V 10 A ⎝ T ⎠ ⎝ ⎠ ( b ) Since I B = 175μA and IC = 0, IC < β F I B , and the transistor is saturated. ( ) 175x10−6 + 0 Using Eq. (5.53) : VBE = 0.025ln = 0.714 V | Using Eq. (5.54) with iC = 0, ⎡ ⎛ .5 ⎞⎤ −16 1 10 ⎢ + ⎜1 − ⎟⎥ ⎣ 50 ⎝ 1.5 ⎠⎦ ⎛ 1 ⎞ ⎛ β + 1⎞ ⎛ 1.5 ⎞ VCESAT = 0.025ln⎜ ⎟ = 0.025ln⎜ R ⎟ = 0.025ln⎜ ⎟ = 27.5 mV ⎝ 0.5 ⎠ ⎝αR ⎠ ⎝ βR ⎠ 141 5.63 ⎡ ⎛v ⎞ ⎛ v ⎞⎤ I ⎡ ⎛ v ⎞ ⎤ iC = I S ⎢exp⎜ BE ⎟ − exp⎜ BC ⎟⎥ + S ⎢exp⎜ BC ⎟ − 1⎥ ⎝ VT ⎠⎦ β R ⎣ ⎝ VT ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎡ ⎛v ⎞ ⎛ v ⎞⎤ I ⎡ ⎛ v ⎞ ⎤ iE = I S ⎢exp⎜ BE ⎟ − exp⎜ BC ⎟⎥ + S ⎢exp⎜ BE ⎟ − 1⎥ ⎝ VT ⎠⎦ β F ⎣ ⎝ VT ⎠ ⎦ ⎣ ⎝ VT ⎠ vBE > 4VT and vBC < −4VT ⎡ ⎛ v ⎞⎤ ⎛ 1 ⎞⎡ ⎛ vBE ⎞⎤ I S ⎡ ⎛ v BE ⎞⎤ iC ≅ I S ⎢exp⎜ BE ⎟⎥ and iE = I S ⎜1 + ⎢exp⎜ ⎟⎢exp⎜ ⎟⎥ = ⎟⎥ → iC ≅ α F iE ⎝ β F ⎠⎣ ⎝ VT ⎠⎦ α F ⎣ ⎝ VT ⎠⎦ ⎣ ⎝ VT ⎠⎦ ⎛i ⎞ ⎛α i ⎞ vBE ≅ VT ln⎜ C ⎟ = VT ln⎜ F E ⎟ ⎝ IS ⎠ ⎝ IS ⎠ 5.64 I SD = αF IS = 1 fA = 1.02 fA 0.98 5.65 Both transistors are in the forward - active region. For simplicity, assume VA = ∞. IC 2 = IC 1 and I B1 = I B 2 | I = IC1 + 2 I B1 = (β F + 2)I B1 | IC 2 = β F I B 2 = β F I B1 IC 2 = I = IC1 + I B1 + I B 2 | Since the transistors are identical and have the same VBE , βF + 2 βF I= 25 25μA | IC 2 = 23.2 μA | See the Current Mirror in Chapter 15. 25 + 2 5.66 CD = 5.67 IC 50 x10−12 τF = IC = 2 x10−9 IC (F) (a) 4 fF (b) 0.4 pF (c) 40 pF VT 0.025 Using Fig. 2.8 with N = 1018 cm2 cm2 , μ = 260 and μ = 100 n p v-s v-s cm3 ⎛ cm2 ⎞ 2(0.025V ) ⎟ ⎜260 v - s⎠ ⎝ ⎛ cm2 ⎞ 2(0.025V ) ⎟ ⎜100 v - s⎠ ⎝ W2 WB2 ( a ) npn : τ F = B = = 2 Dn 2VT μn (1x10 −4 cm ) 2 = 0.769 ns W2 WB2 = ( b ) pnp : τ F = B = 2 D p 2VT μ p (1x10 −4 cm ) 2 = 2.00 ns 142 5.68 For f >> f β , f T = β ⋅ f = 10(75 MHz)= 750 MHz | f β = 5.69 βF fT = 750 MHz = 3.75 MHz 200 βF = 5.70 f T 900 MHz f 900 MHz = = 180 | For f >> 5 MHz, β (f ) = T = = 18 5 MHz 50 MHz fβ f 6 x1018 cm2 cm2 cm2 → μ = 130 using Fig. 2.8. D = μ V = 130 0.025 V = 3.25 ( ) n n n T v−s v−s s cm3 2 ⎞⎛ 20 ⎞ ⎛ cm 10 1.60 x10−19 C 25 x10−8 cm 2 ⎜3.25 ⎟⎜ ⎟ 2 s ⎠⎝ cm6 ⎠ qADn ni ⎝ IS = = = 5.42 x10−20 A 18 N AWB 6 x10 0.4 x10−4 cm cm3 NA = ( ) ( ) 5.71 WB = 2 Dnτ F | τ F ≤ NA = 18 1 1 = = 31.8 ps 2πf 2π 5x109 ( ) 5 x10 cm → μn = 135 using Fig. 2.8. 3 v−s cm cm 2 cm2 0.025 V = 3.38 Dn = μnVT = 135 ( ) v−s s ⎛ cm2 ⎞ −12 WB ≤ 2⎜ 3.38 ⎟31.8 x10 s = 0.147 μm s ⎠ ⎝ 5.72 2 ⎛ V ⎞ ⎛ ⎛ 10 ⎞ 265μA 5 ⎞ 240μA and β FO ⎜1 + ⎟ = IC = β F I B = β FO ⎜1 + CE ⎟ I B | β FO ⎜1 + ⎟ = 3μA 3μA ⎝ VA ⎠ ⎝ VA ⎠ ⎝ VA ⎠ ⎛ 10 ⎞ ⎜1 + ⎟ 80 ⎝ VA ⎠ 265μA = ⇒ VA = 43.1 V | β FO = = 71.7 ⎛ ⎛ 5 ⎞ 5 ⎞ 240μA ⎜1 + ⎟ ⎜1 + ⎟ ⎝ 43.1⎠ ⎝ VA ⎠ 5.73 143 ⎡ ⎛ V ⎞ ⎤⎛ V ⎞ ⎡ ⎛ 0.72V ⎞ ⎤⎛ 10V ⎞ ( a ) IC = I S ⎢exp⎜ BE ⎟ − 1⎥⎜1 + CE ⎟ = 10−16 A⎢exp⎜ ⎟ − 1⎥⎜1 + ⎟ = 371 μA ⎣ ⎝ 0.025V ⎠ ⎦⎝ 65V ⎠ ⎣ ⎝ VT ⎠ ⎦⎝ VA ⎠ ⎡ ⎛V ⎞ ⎤ ⎡ ⎛ 0.72V ⎞ ⎤ ( b ) IC = I S ⎢exp⎜ BE ⎟ − 1⎥ = 10−16 A⎢exp⎜ ⎟ − 1⎥ = 322 μA ⎣ ⎝ 0.025V ⎠ ⎦ ⎣ ⎝ VT ⎠ ⎦ ( c ) 1.15 :1 (a) is 15% larger than (b) due to the Early effect. 5.74 For example : ( 10 mA, 14 V) and ⎛ V ⎞ IC = β F I B = β FO ⎜1 + CE ⎟ I B | We need two Q - points from the output characteristics. ⎝ VA ⎠ (5 mA, 5 V) ⎛ 14 ⎞ ⎛ 5⎞ 10mA = β FO ⎜1 + ⎟0.1mA and 5mA = β FO ⎜1 + ⎟0.06 mA yields ⎝ VA ⎠ ⎝ VA ⎠ ⎛ 14 ⎞ ⎛ 5⎞ 100 = β FO ⎜1 + ⎟ and 83.3 = β FO ⎜1 + ⎟. Solving these two equations yields ⎝ VA ⎠ ⎝ VA ⎠ β FO = 72.9 and VA = 37.6 V . 5.75 ⎡ ⎛ V ⎞ ⎤ ⎡ ⎛ V +V ⎞ ⎤ I ⎛V ⎞ BE Fig. 5.16(a) : I E = IC + I B = ⎢β FO ⎜1 + CE ⎟ + 1⎥ I B ≅ ⎢β FO ⎜1 + CB ⎟ + 1⎥ s exp⎜ BE ⎟ VA ⎠ ⎦ β FO ⎝ VT ⎠ ⎣ ⎝ VA ⎠ ⎦ ⎣ ⎝ ⎛ V ⎞ ⎡ 5 + VBE 1 ⎤ + ⎥ 5 x10−15 exp⎜ BE ⎟ = 100μA → VBE = 0.589V by iteration ⎢1 + 19 ⎦ 50 ⎣ ⎝ 0.025 ⎠ ⎛ 5.589 ⎞ 100μA = 4.52 μA | IC = 19⎜1 + IB = ⎟ I B = 95.48 μA ⎡ ⎛ 5.589 ⎞ ⎤ 50 ⎠ ⎝ ⎢19⎜1 + ⎟ + 1⎥ 50 ⎠ ⎦ ⎣ ⎝ ⎛V ⎞ 100μA For VA = ∞, I E = I s exp⎜ BE ⎟ | VBE = 0.025ln = 0.593 V 5 fA ⎝ VT ⎠ ( ) Fig. 5.16(b) : I B = ⎛V ⎞ 19( 100μA) exp⎜ BE ⎟ → VBE = 0.025ln = 0.667 V β FO 5 fA ⎝ VT ⎠ ⎛ V ⎞ ⎛ 5⎞ IC = β FO ⎜1 + CE ⎟ I B = 19⎜1 + ⎟100μA = 2.09 mA | I E = IC + I B = 2.19 mA ⎝ 50 ⎠ ⎝ VA ⎠ Is VBE is independent of VA in the equation above. 5.76 144 ⎛ V ⎞ ⎛ 9 + 0.7 ⎞ IC = β F I B | I E = (β F + 1)I B | β F = β FO ⎜1 + CE ⎟ = 50⎜1 + ⎟ = 59.7 50 ⎠ ⎝ ⎝ VA ⎠ IE = (9 − 0.7)V = 1.01 mA | 8200Ω IB = IE 1.01mA = = 16.7 μA | IC = 59.7 I B = 0.996 mA βF + 1 60.7 5.77 I gm = C VT −23 kT 1.38 x10 J / K (300 K ) | VT = = = 25.9 mV q 1.60 x10−19 C ( ) (a) g (c) g 5.78 m = 10−5 A 10−4 A = 0.387 mS (b) g m = = 3.87 mS VT VT 10−3 A 10−2 A = 38.7 mS (d ) g m = = 0.387 S VT VT m m = (e) The values of g are the same for the pnp. IC 10 x10−12 CD = τ F = IC = 3.88 x10−10 IC (F) (a) 0.388 fF (b) 0.388 pF (c) 3.88 pF VT 0.0258 5.79 The following are from the Cadence website and the file psrefman.pdf: IS = 10fA, BF = 100, BR = 1, VAF = ∞, VAR = ∞, TF = 0, TR = 0, NF = 1, NE = 1.5, RB = 0, RC = 0, RE = 0, ISE = 0, ISC = 0, ISS = 0, IKF = ∞, IKR = ∞, CJE = 0, CJC = 0. These default values apply to both npn and pnp transistors. 5.80 ⎡ ⎛ i ⎞⎤ ⎛ 1mA ⎞ 1 + ⎢1 + 4⎜ F ⎟⎥ 1 + 1 + 4⎜ ⎟ ⎝ IKF ⎠⎦ ⎝10 mA ⎠ ⎣ = = 1.09 → 8.3% reduction (a) KBQ = 2 2 ⎛ 10 mA ⎞ 1 + 1 + 4⎜ ⎟ ⎝ 10 mA ⎠ i = 1.62 | iC = F = 0.62iF → 38% reduction (a) KBQ = 2 1.62 ⎛ 50mA ⎞ 1 + 1 + 4⎜ ⎟ ⎝ 10 mA ⎠ i = 2.79 | iC = F = 0.36iF → 64% reduction (a) KBQ = 2 2.79 NK 5.81 145 12.000 10.000 8.000 6.000 4.000 2.000 0.000 0.1 1 10 100 1000 Collector Current Series1 5.82 ( a ) VEQ = VCE = 10 − 43000 IC − 33000 I E = 3.797V | Q - point : (80.9 μA,3.80 V) ( b ) VEQ = 36 kΩ 10V = 3.462V | REQ = 36kΩ 68kΩ = 23.54 kΩ 36 kΩ + 68 kΩ 3.462 − 0.7 V IB = = 1.618μA | IC = 50 I B = 80.9 μA | I E = 51I B = 82.5 μA 23.54 + (50 + 1)33 kΩ VCE = 10 − 8600 IC − 6600 I E = 3.7976V | Q - point : (405 μA,3.80 V) 7.2 kΩ 10V = 3.462V | REQ = 7.2 kΩ 13.6 kΩ = 4.708kΩ 7.2 kΩ + 13.6 kΩ 3.462 − 0.7 V IB = = 8.092μA | IC = 50 I B = 404.6μA | I E = 51I B = 412.7 μA 4.708 + (50 + 1)6.6 kΩ ( c ) VEQ = VEC = 10 − 33000 IC − 43000 I E = 3.797V | Q - point : (80.9 μA,3.80 V) ( b ) VEQ = 68 kΩ 10V = 6.538V | REQ = 36kΩ 68kΩ = 23.54 kΩ 36 kΩ + 68 kΩ 10 − 0.7 − 6.538 V = 1.618μA | IC = 50 I B = 80.9 μA | I E = 51I B = 82.5 μA IB = 23.54 + (50 + 1)33 kΩ 13.6 kΩ 10V = 6.538V | REQ = 7.2kΩ 13.6 kΩ = 4.708kΩ 7.2 kΩ + 13.6 kΩ 10 − 0.7 − 6.538 V = 8.092μA | IC = 50 I B = 404.6μA | I E = 51I B = 412.7 μA IB = 4.708 + (50 + 1)6.6 kΩ VEC = 10 − 6600 IC − 8600 I E = 3.7976V | Q - point : (405 μA,3.80 V) 5.83 146 ( a ) VEQ = 122μA,2.02V) VCE = 10 − 43000 IC − 22000 I E = 2.022V | Q - point : ( ( b ) VEQ = 36 kΩ 10V = 3.462V | REQ = 36kΩ 68kΩ = 23.54kΩ 36 kΩ + 68 kΩ V 3.462 − 0.7 = 1.629μA | IC = 75 I B = 122.2μA | I E = 76 I B = 123.8μA IB = 23.54 + (75 + 1)22 kΩ 68 kΩ 10V = 6.538V | REQ = 36kΩ 68kΩ = 23.54kΩ 36 kΩ + 68 kΩ 10 − 0.7 − 6.538 V = 1.629μA | IC = 75 I B = 122.2μA | I E = 76 I B = 123.8μA IB = 23.54 + (75 + 1)22 kΩ 122μA,2.02V) VEC = 10 − 22000 IC − 43000 I E = 2.022V | Q - point : ( 5.84 *Problem 5.83(a) VCC 1 0 10 R1 3 0 36K R2 1 3 68K RC 1 2 43K RE 4 0 33K Q1 2 3 4 NPN .MODEL NPN NPN IS=1E-16 BF=50 BR=0.25 .OP .END *Problem 5.83(b) VCC 1 0 10 R1 3 0 36K R2 1 3 68K RC 1 2 43K RE 4 0 33K Q1 2 3 4 NPN .MODEL NPN NPN IS=1E-16 BF=50 BR=0.25 VAF=60 .OP .END *Problem 5.83(c) VCC 1 0 10 R1 1 3 36K R2 3 0 68K RC 4 0 43K RE 1 2 33K Q1 4 3 2 PNP .MODEL PNP PNP IS=1E-16 BF=50 BR=0.25 .OP .END *Problem 5.83(d) VCC 1 0 10 147 R1 1 3 36K R2 3 0 68K RC 4 0 43K RE 1 2 33K Q1 4 3 2 PNP .MODEL PNP PNP IS=1E-16 BF=50 BR=0.25 VAF=60 .OP .END 5.85 6.2 kΩ = 3.41V and REQ = 6.2 kΩ 12 kΩ = 4.09 kΩ 6.2kΩ + 12 kΩ 3.41 − 0.7 IC = β F I B = 100 = 0.356 mA. 4090 + 101(7500) VEQ = 10 VCE = 10 − 0.356 mA(5.1kΩ)− 101 0.356 mA(7.5kΩ)= 5.49V 100 Q − po int : (0.356 mA, 5.49 V ) 120 kΩ = 5.00V and REQ = 120 kΩ 240 kΩ = 80 kΩ 120kΩ + 240 kΩ 5.00 − 0.700 IC = β F I B = 100 = 42.2μA. 80000 + 101( 100000) 5.86 VEQ = 15 VCE = 15 − 42.2 x10−6 A 105 Ω − Q − po int : (42.2 μA, 4.39 V ) ( ) 101 42.2 x10−6 A 1.5 x105 Ω = 4.39V 100 ( ) 5.87 (a) I RC = E = | VB = 2 + 0.7 = 2.7V 1.00mA V 2.7V = 27 kΩ → 27 kΩ Set R1 = B = 10 I B 10(0.01mA) R2 = (12 − 5 − 2)V = 5kΩ → 5.1 kΩ ⎛ 101⎞ 2V =⎜ = 1.98 kΩ → 2.0 kΩ ⎟1mA = 1.01mA | RE = 1.01mA α F ⎝ 100 ⎠ IC (12 − 2.7)V = 11I B 9.3V = 84.55kΩ → 82 kΩ 11(0.01mA) 148 ( b ) VEQ = 1.02 mA,4.72V ) VCE = 12 − 5100 IC − 2000 I E = 4.723V | Q - po int : ( 5.88 27 kΩ 12V = 2.972V | REQ = 27kΩ 82 kΩ = 20.31kΩ 27 kΩ + 82 kΩ V 2.972 − 0.7 = 10.22μA | IC = 100 I B = 1.022 mA | I E = 101I B = 1.033mA IB = 20.31 + ( 100 + 1)2 kΩ (a) I ⎛ 76 ⎞ = ⎜ ⎟10μA = 10.13μA | Let VRC = VR E = VCE = 6V α F ⎝ 75 ⎠ 6V = 592 kΩ → 620 kΩ RE = 10.13μA 6V = 600 kΩ → 620 kΩ | VB = 6 + 0.7 = 6.7V RC = 10μA V 6.7V = 5.03 MΩ → 5.1 MΩ Set R1 = B = ⎛ 10 I B 10μA ⎞ 10⎜ ⎟ ⎝ 75 ⎠ E = IC VCE = 18 − 620000 IC − 620000 I E = 5.707V | Q - point : (9.85 μA, 5.71 V) 5.89 11.3V = 7.71 MΩ → 7.5 MΩ ⎛10μA ⎞ 11I B 11⎜ ⎟ ⎝ 75 ⎠ 5.1 MΩ 18V = 7.286V | REQ = 5.1 MΩ 7.5 MΩ = 3.036 MΩ ( b ) VEQ = 5.1 MΩ + 7.5 MΩ 7.286 − 0.7 V = 0.1313μA | IC = 75 I B = 9.848μA | I E = 76 I B = 9.980μA IB = 3036 + (75 + 1)620 kΩ R2 = (18 − 6.7)V = (a) I RC = E = (5 − 2 − 1)V = 2.35kΩ → 2.4 kΩ 850μA VR1 10 I B = = ⎛ 61 ⎞ 1V = ⎜ ⎟850μA = 864.2μA | RE = = 1.16 kΩ → 1.2 kΩ 864.2μA α F ⎝ 60 ⎠ IC | VB = 5 − 1 − 0.7 = 3.3V Set R1 = 5 − 3.3V = 12.0 kΩ → 12 kΩ ⎛ 850μA ⎞ 10⎜ ⎟ ⎝ 60 ⎠ R2 = VR 2 11I B 3.3V = 21.2 kΩ → 22 kΩ ⎛ 850μA ⎞ 11⎜ ⎟ ⎝ 60 ⎠ 149 ( b ) VEQ = VCE = 5 − 1200 I E − 2400 IC = 2.14V | Q - point : (786 μA, 2.14 V) 5.90 22 kΩ 5V = 3.24V | REQ = 12 kΩ 22 kΩ = 7.77 kΩ 12 kΩ + 22 kΩ 5 − 0.7 − 3.24 V = 13.1μA | IC = 60 I B = 786μA | I E = 61I B = 799μA IB = 1.2 kΩ 7.77 + (60 + 1) (a ) V 11mA = 0.220 mA 50 ⎛ 51 ⎞ I 1V I E = C = ⎜ ⎟11mA = 11.22 mA | RE = = 89.1Ω → 91 Ω α F ⎝ 50 ⎠ 11.22 mA RE = 1 V , VRC = 9 V | I B = 9V = 818Ω → 820 Ω | VB = −15 + 1 + 0.7 = −13.3V 11.0 mA −13.3V − (−15V ) V Set R1 = R1 = = 772Ω → 750 Ω 10 I B 10(0.220 mA) RC = R2 = 0−( 13.3V ) 11I B = 13.3V = 5.50 kΩ → 5.6 kΩ 11(0.220 mA) ( b ) VEQ = 10.2 mA, 5.59 V) VEC = 0 − 820 IC − 91I E − (−15V ) = 5.59V | Q - point : ( 5.91 Problem numbers on graph 3.3kΩ VEQ = 10V = 3.056V | REQ = 7.5kΩ 3.3kΩ = 2.292 kΩ 3.3kΩ + 7.5kΩ 5.6 kΩ (−15V )= −13.2V | REQ = 0.75kΩ 5.6kΩ = 0.661kΩ 0.75kΩ + 5.6 kΩ −13.2 − 0.7 − (−15) V = 0.207 mA | IC = 50 I B = 10.3mA | I E = 51I B = 10.6 mA IB = 661 + (50 + 1)(91) Ω VCE = 10 − 820 IC − 1200 I E | From characteristics at VCE = 5V : β F ≅ VCE = 10 − 820 IC − 1200 84 IC = 10 - 2034IC 83 Load line points : IC = 0, VCE = 10V and VCE = 0, IC = 4.9 mA IB = 5mA = 83 60μA 3.056 − 0.7 = 23μA | From Graph : Q - point : ( 1.9 mA, 6.0 V) 2292 + (83 + 1) 1200 150 10mA IB = 100 μA C o l l e c t o r 5mA C u r r e n t IB = 92 μA Q-Point Prob. 5.92 I = 80 μA B IB = 60 μA I = 40 μA B IB = 23 μA Q-Point Prob. 5.91 0A 0V IB = 20 μ A 5V VCE 10V 15V 5.92 VEQ = 6.8 kΩ (10)= 6.538V | REQ = 6.8kΩ 3.6kΩ = 2.354kΩ 6.8kΩ + 3.6kΩ 5mA = 83 60μA VEC = 10 − 420 IC − 330 I E | From characteristics at VEC = 5V : β F ≅ VEC = 10 − 420 IC − 3300 84 IC = 10 − 754 IC 83 10 − 0.7 − 6.538 = 92μA 2354 + (83 + 1)330 Load line points : IC = 0, VEC = 10V and VEC = 0, IC = 13.3mA − off the graph VEC = 5V , IC = 6.63mA | I B = From Graph : Q - point : (7.5 mA, 4.3 V) 151 5.93 Writing a loop equation starting at the 9 V supply gives: 9 = 1500(IC + IB ) + 10000 IB + VBE Assuming forward-active region operation, VBE = 0.7 V and IC = βFIB. 9 = 1500(β F I B + I B )+ 10000 I B + 0.7 IB = β F (9 − 0.7) 9 − 0.7 and IC = β F I B = 1500(β F + 1)+ 1000 1500(β F + 1)+ 1000 = = = 1.5kW(30 + 1)+ 10kW 1.5kΩ( 100 + 1)+ 10 kΩ 1.5kΩ(250 + 1)+ 10 kΩ V 250(9 − 0.7) V 100(9 − 0.7) 30(9 − 0.7)V = 4.41 mA | VCE = 9 − 1500I E = 2.17V | Q - pt : (4.41mA,2.17V) = 5.14 mA | VCE = 9 − 1500 I E = 1.21V | Q - pt : (5.14mA,1.21V) = 5.37 mA | VCE = 9 − 1500 I E = 0.913V | Q - pt : (5.37mA,0.913V) | VCE = 9 − 1500 I E = 0.705V | Q - pt : (5.53mA,0.705V) (a ) I C (b) I (c) I C C (d ) I 5.94 C = (9 − 0.7)V = 5.53 mA 1500Ω ⎛ I ⎞ V − 0.7 VCE = 9 − (IC + I B ) 1500 | VCE = 9 − ⎜ IC + C ⎟1500 | I B = CE 4 βF ⎠ 10 ⎝ 5mA From Fig. P5.26 at 5V : β F = = 83.3 | VCE = 9 − 1518 IC 60μA IC = 0, VCE = 9V | VCE = 0, IC = 5.93mA VCE = 0.9V , I B = 20μA | VCE = 1.3V , I B = 60μA | VCE = 1.7V , I B = 100μA From graph : Q - point = (5.0 mA, 1.3 V) 10mA IB = 100 I = 100 B μA μA VCE = 1.7 V IB = 80 μA VCE = 1.5 V I = 80 μA B Collector Current Q-Point 5mA IB = 60 μA VCE = 1.3 V IB = 40 μA V CE = 1.1 V I = 20 μA B V CE = 0.9 V 0A 0V IB = 60 μA IB = 40 μA IB = 20 μA 5V V 10V 15V CE 152 5.95 ( a ) VEC = 10 − (IC + I B )RC = 10 − I E RC | I E = RC = 10.17 mA (10 − 3)V = 689Ω → 680 Ω | RB = ( b ) 5 − 0.7 − 14000 I B − 680(IC + I B )− (−5)= 0 IB = V VEC − VEB (3 − 0.7) = = 13.8 kΩ → 14 kΩ 0.1667 mA IB αF IC = 61 βF + 1 IC = 10 mA = 10.17 mA 60 βF VEC = 10V − (8.88mA)680Ω = 3.96 V | Q - po int : (8.88 mA, 3.96 V ) 5.96 10 − 0.7 V = 222.1μA | IC = β F I B = 8.88 mA 14000 + 41(680) Ω VCE = 1.5 − (IC + I B )RC → RC = 1.5 − 0.9 = 29.4kΩ → 30 kΩ 20μA 20μA + 50 RB = VCE − VBE 0.9 − 0.65 = = 625kΩ → 620 kΩ 20μA IB 50 126)I B | I B = For RC = 30 kΩ : VCE = 1.5 − 30 kΩ(IC + I B )RC = 1.5 − 30 kΩ( VCE − 0.65 620 kΩ V − 0.65 VCE = 1.5 − 30 kΩ( 126) CE → VCE = 0.770V 620 kΩ 0.770 − 0.65 IC = 125 I B = 125 = 24.2μA | Q - po int : (24.2 μA, 0.770 V ) 620 kΩ 5.97 12 = RC (IC + I B )+ VZ + VBE = 500(I E )+ 7.7 | I E = IB = 12 − 7.7 = 8.60 mA 500 Q - point = (8.52 mA, 7.70 V) 5.95 IE 8.60 mA = = 85.2μA | IC = β F I B = 8.52 mA | VCE = 7.70V 101 βF + 1 VEQ = 6 + 100 15 − 6 = 6.114V | REQ = 100Ω 7800Ω = 98.73Ω 7800 + 100 20mA VO 20mA 6.14 − 98.7I B − VBE 101.1− VBE + = + → I C = 50I B = 50 IB = 51 51 51(4700Ω) 51(4700Ω) 2.398x105 IC 10−16 153 VBE = 0.025ln Using MATLAB: fzero('IC107',.02) ---> ans =0.0207 function f=IC107(ic) vbe=0.025*log(ic/1e-16); f=ic-50*(101.1-vbe)/2.398e5; VO = 6.14 − 98.7 20.7mA 20.7mA − .025ln = 5.276 V 51 10−16 5.99 *Problem 5.98 VCC 1 0 DC 15 R1 1 2 7.8K RZ 2 4 100 VZ 4 0 DC 6 Q1 1 2 3 NPN RE 3 0 4.7K IL 3 0 20MA .MODEL NPN NPN IS=1E-16 BF=50 BR=0.25 .OP .END Output voltages will differ slightly due to different value of VT. 5.100 vO = 7 − 100iB − vBE = 7 − 100iB − VT ln vO = 7 − 100iB − VT ln iL − VT ln iC α i = 7 − 100iB − VT ln F L IS IS αF IS ⎛ dv di V ⎞ 100Ω 0.025V Ro = − O = −⎜ −100Ω B − T ⎟ = + = 3.21Ω 51 0.02 A diL diL iL ⎠ ⎝ 5.101 Since the voltage across the op - amp input must be zero, vO = VZ = 10 V . Since the input current to the op amp is zero, I E = I +15 = I Z + IC = I Z + α F I E = VO = 100 mA 100 15V − 10V 60 + 100 mA = 98.5 mA 47 kΩ 61 154 5.102 47Ω = VZ 47Ω + 47Ω and vO = 10 V . Since the input current to the op amp is zero, I 10V ⎛ 41⎞ 15V − 5V + 109mA = 109 mA IE = C = ⎜ ⎟ = 109 mA I +15 = I Z + I E = α F 94Ω ⎝ 40 ⎠ 82 kΩ Since the voltage across the op - amp input must be zero, vO VEQ − VBE 5.103 IC = β F I B = β F R1 = 0.95(82 kΩ)= 77.9 kΩ | R2 = 1.05( 120 kΩ)= 126 kΩ | RE = 1.05(6.8 kΩ)= 7.14 kΩ VEQ = min IC REQ + (β F + 1)RE min | For IC : VCC = 0.95( 15) = 14.25 V 77.9 14.25V = 5.44V | REQ = 77.9 kΩ 126 kΩ = 48.1kΩ 77.9 + 126 5.44V − 0.7V = 100 = 616 μA 48.1kΩ + ( 101)7.14 kΩ max min min VCE = 14.25 − IC 0.95(6.8 kΩ) − I E 7.14 kΩ max = 14.25 − 3.98 − 4.44 = 5.83V | Q - po int : (616 μA, 5.83 V ) VCE max : VCC = 1.05( 15) = 15.75 V For IC [ ] 120 kΩ)= 114 kΩ | RE = 0.95(6.8 kΩ)= 6.46 kΩ R1 = 1.05(82 kΩ) = 86.1kΩ | R2 = 0.95( 86.1 15.75V = 6.78V | REQ = 86.1kΩ 114 kΩ = 49.0 kΩ 86.1 + 114 6.78V − 0.7V = 100 = 867μA 49.0 kΩ + ( 101)6.46 kΩ VEQ = max IC min max max VCE = 15.75 − IC 1.05(6.8 kΩ) − I E 6.46 kΩ min = 15.75 − 6.19 − 5.66V = 3.90V | Q - po int : (867 μA, 3.90 V ) VCE [ ] 155 5.104 Using the Spreadsheet approach in Fig. 5.40, Eq. set (5.66) becomes: 1. 3. 5. VCC = 15 1 + 0.1 (RAND() − 0.5) ( R2 = 120000 1 + 0.1 (RAND() − 0.5) | 4. RC = 6800 1 + 0.1 (RAND() − 0.5) IC (A) 7.69E-04 4.02E-05 6.62E-04 8.93E-04 VEQ − VBE ( ( ) ) | 2. ) R1 = 82000 1 + 0.1 (RAND() − 0.5) RE ( ) = 6800 ( 1 + 0.1 (RAND() − 0.5)) | 6. β F = 100 500 Cases Average Std. Dev. Min Max 5.105 IC = β F I B = β F VCE (V) 4.51 0.40 3.31 5.55 R1 = 0.95( 18 kΩ) = 17.1kΩ | R2 = 1.05(36 kΩ) = 37.8 kΩ | RE = 1.05( 16 kΩ)= 16.8 kΩ VEQ = min IC REQ + (β F + 1)RE min | For IC : VCC = 0.95( 12) = 11.4 V 17.1 11.4V = 3.55V | REQ = 17.1kΩ 37.8 kΩ = 11.8 kΩ 17.1 + 37.8 3.55V − 0.7V = 50 = 164 μA 11.8 kΩ + (51) 16.8 kΩ max min min VCE = 11.4 − IC 0.95(22 kΩ) − I E 16.8 kΩ max = 11.4 − 3.43 − 2.81 = 5.16V | Q - po int : ( 164 μA, 5.16 V ) VCE max : VCC = 1.05( 12) = 12.6 V For IC [ ] 18 kΩ) = 18.9 kΩ | R2 = 0.95(36 kΩ)= 34.2 kΩ | RE = 0.95( 16 kΩ)= 15.2 kΩ R1 = 1.05( 18.9 12.6V = 4.49V | REQ = 18.9 kΩ 34.2 kΩ = 12.2 kΩ 18.9 + 34.2 4.49V − 0.7V = 150 = 246μA 12.2 kΩ + ( 151) 15.2 kΩ VEQ = max IC min max max VCE = 12.6 − IC 1.05(22 kΩ) − I E 15.2 kΩ min = 12.6 − 5.68 − 3.77V = 3.15V | Q - po int : (246 μA, 3.15 V ) VCE 500 Cases IC (A) VCE (V) [ ] Average Std. Dev. Min Max 156 2.02E-04 1.14E-05 1.71E-04 2.35E-04 4.26 0.32 3.43 5.21 The averages are close to the hand calculations that go with Fig. 5.35. The minimum and maximum values fall within the worst-case analysis as we expect. 5.106 Using the Spreadsheet approach with zero tolerance on the current gain, Eq. set (5.66) becomes: 1. 3. 5. VCC = 12 1 + 0.0 (RAND() − 0.5) R2 = 36000 1 + 0.1 (RAND() − 0.5) RC ( ) = 22000 ( 1 + 0.1 (RAND() − 0.5)) IC (A) 2.03E-04 1.10E-05 1.74E-04 2.36E-04 VCE (V) 4.29 0.32 3.46 5.27 ( ) | 2. | 4. | 6. ( ) R = 16000 ( 1 + 0.1 (RAND() − 0.5)) β = 100 ( 1 + 0.0(RAND() − 0.5)) R1 = 18000 1 + 0.1 (RAND() − 0.5) E F 500 Cases Average Std. Dev. Min Max Note that the current gain tolerance has little effect on the results. 5.107 ( a ) Approximately 22 cases fall outside the interval [ 170μA,250μA]: 100% ( b ) Approximately 125 cases fall inside the interval [3.2V,4.8V]: 100% 22 = 4.4% fail 500 125 = 25% fail 500 5.108 Using the Spreadsheet approach with 50% tolerance on the current gain, a tolerance TP on VCC, and a tolerance TR on resistor values, Eq. set (5.66) becomes: 1. 3. VCC = 12 * ( 1 + 2 * TP * (RAND() − 0.5)) | 2. 4. R1 = 18000 * ( 1 + 2 * TR * (RAND() − 0.5)) RE = 16000 * ( 1 + 2 * TR * (RAND() − 0.5)) R2 = 36000 * ( 1 + 2 * TR * (RAND() − 0.5)) | 5. RC = 22000 * ( 1 + 2 TR (RAND() − 0.5)) | 6. β F = 100 * ( 1 + 1* (RAND() − 0.5)) 10,000 case Monte Carlo runs indicate that the specifications cannot be achieved even with ideal resistors. For TP = 5% and TR = 0%, 18 % of the circuits fail. With TP = 2% and TR = 0%, 1.5% percent fail. The specifications can be met with TP = 1% and TR = 1%. 157 5.109 IC = β F I B = β F R1 = 0.8( 18 kΩ) = 14.4 kΩ | R2 = 1.2(36 kΩ)= 43.2 kΩ | RE = 1.2( 16 kΩ)= 19.2 kΩ VEQ = min IC REQ + (β F + 1)RE VEQ − VBE min | For IC : VCC = 0.95( 12) = 11.4 V 14.4 11.4V = 2.85V | REQ = 14.4 kΩ 43.2 kΩ = 10.8 kΩ 14.4 + 43.2 2.85V − 0.7V = 50 = 109 μA 10.8 kΩ + (51) 19.2 kΩ max min min VCE = 11.4 − IC 0.8(22 kΩ) − I E 19.2 kΩ max = 11.4 − 1.91 − 2.13 = 7.36V | Q - point : ( 109 mA, 7.36 V) VCE max : VCC = 1.05( 12) = 12.6 V For IC [ ] 18kΩ)= 21.6 kΩ | R2 = 0.80(36kΩ)= 28.8 kΩ | RE = 0.80( 16kΩ)= 12.8 kΩ R1 = 1.2( 21.6 12.6V = 5.40V | REQ = 21.6 kΩ 28.8 kΩ = 12.3kΩ 21.6 + 28.8 5.4V − 0.7V = 150 = 362μA 12.3kΩ + ( 151) 12.8 kΩ VEQ = max IC min max max = 12.6 − IC 1.2(22kΩ) − I E 12.8kΩ VCE min = 12.6 − 9.57 − 4.67V = −1.64V ! Saturated! VCE [ ] The forward - active region assumption is violated. See the next problem. Based upon a Monte Carlo analysis, only about 1% of the circuits actually have this problem, although VCE will be relatively small in many circuits. 158 5.110 Using the Spreadsheet approach: 1 + .1* (RAND() − 0.5)) 1. VCC = 12 * ( | 2. R1 = 18000 * ( 1 + 0.4 * (RAND() − 0.5)) RE = 16000 * ( 1 + 0.4 * (RAND() − 0.5)) 3. 5. 7. R2 = 36000 * ( 1 + 0.4 * (RAND() − 0.5)) | 4. RC = 22000 * ( 1 + 0.4 * (RAND() − 0.5)) VA = 75 * ( 1 + 0.66 * (RAND() − 0.5)) | 6. β F = 100 * ( 1 + 1* (RAND() − 0.5)) In order to avoid an iterative solution at each step, assume that VCE does not influence the base current. Then, IB = VEQ − 0.7 and VCE REQ + (β FO + 1)RE 500 Cases Average Std. Dev. Min** Max ⎛ R ⎞ VCC − β FO IB ⎜ RC + E ⎟ ⎛ V ⎞ αF ⎠ ⎝ = | IC = β FO IB ⎜1 + CE ⎟ ⎛ β R ⎞ ⎝ VA ⎠ 1 + FO IB ⎜ RC + E ⎟ VA ⎝ αF ⎠ VCE (V) 3.81E+00 1.26E+00 -2.07E-01 6.94E+00 IC (A) 2.049E-04 3.785E-05 1.264E-04 3.229E-04 **Note: In this particular simulation, there were 4 cases in which the transistor was saturated. 159 CHAPTER 6 6.1 (a ) Pavg = 1W 10-5W / gate = 10 μW / gate (b) I = = 4 μA / gate 105 gates 2.5V 6.2 ( a ) Pavg = 100 5 x10-6W / gate = 5 μ W / gate ( b ) I = = 2 μA / gate 2.5V 2 x10 7 gates ( c ) I total = 2 (2 x10 gates)= 40 A gate 7 μA 6.3 ⎛ 2.5 − 0 ⎞ 5 (a ) VH = 2.5 V | VL = 0 V | P ⎟ 10 = 62.5 μW V H = I R = 0 mW | P VL = ⎜ 5 ⎝ 10 ⎠ 2 2 ⎛ 3.3 − 0 ⎞ 5 (b) VH = 3.3 V | VL = 0 V | P ⎟ 10 = 109 μW V H = I R = 0 mW | P VL = ⎜ 5 ⎝ 10 ⎠ 2 2 6.4 vO VH (3.3 V) vI V (0V) L 1.1 V (V REF) 3.3V V+ 6.5 v V H (3.3 V) O V L (0V) 1.1 V (V REF) 3.3V V+ vI Z= A =A () 6-1 6.6 V REF vI AV 6.7 V H = 3 V | VL = 0 V | VIH = 2 V | VIL = 1 V | AV = dvO −3V = = −3 dv I 1V 6.8 V (3 V) H v O Slope = +9 1.5 V v V (0V) L I 1.33 V 1.67 V 1.5 V 3V V + 6.9 VOH = 5 V VOL = 0 V VIH = VREF = 2 V VIL = VREF = 2 V NM H = 5 − 2 = 3 V NM L = 2 − 0 = 2 V 6.10 We would like to achieve the highest possible noise margins for both states and have them be symmetrical. Therefore VREF = 3.3/2=1.65 V. 6.11 V H = 3.3 V | VL = 0 V | VIH = 1.8 V | VOL ≅ 0.25 V | VIL = 1.5 V | VIH ≅ 3.0 V NM H = 3.0 − 1.8 = 1.2 V | NM L = 1.5 − 0.25 = 1.25 V 6.12 VH = 2.5 V | VL = 0.20 V 6-2 6.13 V H = −0.80 V | VL = −1.35 V 6.14 VIH = VOH − NM H = −0.8 − 0.5 = −1.3 V | VIL = NM L + VOL = 0.5 + (−2) = −1.5 V 6.15 -13 -4 -9 τP = PDP/P = 10 J/10 W = 10 s = 1 ns 6.16 4 x10-6W / gate 1W = μ = 1.60 μA / gate 4 W / gate ( b ) I = 2.5V 2.5 x105 gates (c) PDP = 2ns (4 μW ) = 8 fJ (a ) Pavg = 6.17 1μW / gate 100W = 1 μW / gate (b) I = = 0.4 μA / gate 8 2.5V 10 gates (c) PDP = 1ns (1μW ) = 1 fJ (a) Pavg = 6.18 PDP 250 100 Slope = 2 10 Slope = 1 1 P 1 10 50 100 6-3 6.19 dv (t ) dvc (t ) | v(t ) = RC c + vC (t ) | v(t ) = 1 for t ≥ 0 dt dt ⎛ t ⎞ ⎛ t ⎞ v(t ) = 1 − exp⎜ − ⎟ | 0.9 = 1 − exp⎜ − 90% ⎟ → t90% = − RC ln (0.1) ⎝ RC ⎠ ⎝ RC ⎠ ⎛ t ⎞ 0.1 = 1 − exp⎜ − 10% ⎟ → t10% = − RC ln (0.9 ) | tr = t90% − t10% = RC ln (9) = 2.20 RC ⎝ RC ⎠ ⎛ t ⎞ ⎛ t ⎞ (b) v(t ) = 0 vC (0 ) = 1 v(t ) = exp⎜ − ⎟ | 0.9 = exp⎜ − 90% ⎟ → t90% = − RC ln (0.9) ⎝ RC ⎠ ⎝ RC ⎠ ⎛ t ⎞ 0.1 = exp⎜ − 10% ⎟ → t10% = − RC ln (0.1) | t f = t10% − t90% = RC ln (9 ) = 2.20 RC ⎝ RC ⎠ (a ) v(t ) = i (t )R + vC (t ) | i (t ) = C 6.20 ( a ) VH = 2.5V | VL = 0.20V ( b ) V10% = VL + 0.1ΔV = 0.20 + 0.23 = 0.43V → t10% ≅ 23 ns for vO V90% = VL + 0.9ΔV = 0.20 + 2.07 = 2.27V → t90% ≅ 33 ns for vO → t r = 33 − 23 = 10 ns For fall time : t10% ≅ 2.5 ns for vO t90% ≅ 0.8 ns for vO → t f = 1.7 ns For v I , t10% ≅ 0 ns t90% ≅ 1 ns t r = 1 ns | t f ≅ 1 ns (c) τ PHL ≅ 1.5ns − 0.5ns = 1 ns | τ PLH ≅ 26 ns − 21ns = 5 ns ( d ) τ P = 1+ 5 ns = 3 ns 2 6.21 (a) VH = −0.78V | VL = −1.36V V90% = VL + 0.9ΔV = −1.36 + 0.9(0.58) = −0.84V → t90% ≅ 42 ns for vO tr = 42 − 32.5 = 9.5 ns For fall time : t10% ≅ 11.5 ns for vO t90% ≅ 2 ns for vO → t f = 9.5 ns For vI , t10% ≅ 0 ns t90% ≅ 1 ns tr = 1 ns | t f ≅ 1 ns (b) V10% = VL + 0.1ΔV = −1.36 + 0.1(0.58) = −1.30V → t10% ≅ 32.5 ns for vO (c ) V50% = − 0.78 − 1.36 = −1.07V | τ PHL ≅ 4 ns | τ PLH 2 ≅ 4 ns (d ) τ P = 4+4 ns = 4 ns 2 6-4 6.22 (A + B)(A + C) AA + AC + BA + BC A + AC + BA + BC A(1 + C) + AB + BC A + AB + BC A(1 + B) + BC A + BC 6.23 Z = ABC + ABC + ABC Z = ABC + + ABC + ABC + ABC Z = AB C + C + A + A BC Z = AB(1) + (1)BC Z = AB + BC ( ) ( ) 6.24 A B C Z 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 0 0 1 0 0 1 1 Z = AB+BC 6-5 6.25 Z = ABC + ABC + ABC + ABC Z = C AB + AB + AB + AB ( ) Z = C (AB + AB + AB + AB) Z = C (A(B + B)+ A(B + B) ) Z = C (A(1) + A(1)) Z = C (A + A) Z = C (1) Z =C 6.26 A B C 0 0 0 0 0 0 1 1 1 0 1 1 0 0 1 1 0 1 0 1 0 Z 0 1 0 1 0 1 0 1 1 1 1 Z =C 6-6 6.27 A B C D Z 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 1 1 0 0 0 1 1 0 1 0 1 1 1 0 0 1 1 1 1 1 6.28 A B C Z1 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 0 1 1 0 1 1 1 1 1 6.29 (a) Fanout = 2 Z = AB + CD ⎛ ⎞⎛ ⎞ Z = ⎜ AB⎟⎜CD⎟ ⎝ ⎠⎝ ⎠ Z = ABCD Z2 1 1 0 1 0 1 1 1 Z1 = AB = AB Z 2 = AB + C (b) Fanout = 1 6-7 6.30 i(t) 0.132 A For each line : i = C i = 40 x10−12 F 1 ns 0 t dv dt 3.3V = 132mA = 0.132 A 10− 9 s For all 64 lines, I = 64(0.132 A) = 8.45 A! 6.31 i(t) 1.32 A For each line : i = C i = 40 x10−12 F t 0 dv dt 0.1 ns 3.3V = 1.32 A 10−10 s For all 64 lines, I = 64(1.32 A) = 84.5 A! 6.32 F ⎞⎛ 7.5mm 0.1cm ⎞ ⎛ 3.9⎜ 8.854 x10 −14 ⎟(1.5μm ) ⎟⎜ ⎛ ε ox A ⎞ 3.9ε o LW cm ⎠⎝ 2 mm ⎠ ⎝ = 0.583 pF C = 3⎜ =3 ⎜ t ⎟ ⎟=3 t 1μm ox ⎝ ox ⎠ 6.33 " CΔV KCox WLΔV | Let W* = αW and L* = αL ΔT = = 1 I 2 " ⎛W ⎞ μ nCox ⎜ ⎟(VGS − VTN ) 2 ⎝L⎠ C *ΔV * K (αW )(αL )(αΔV ) ΔT = = = α ΔT * 1 ⎛ αW ⎞ I 2 μn ⎜ ⎟(αVGS − αVTN ) 2 ⎝ αL ⎠ * V ε ⎛W ⎞ 2 ⎟(VGS − VTN ) = μ n ox ⎜ 2 Tox ⎝ L ⎠ αV ε ⎛ αW ⎞ 2 2 P* = μn ox ⎜ ⎟(αVGS − αVTN ) = α P 2 αTox ⎝ αL ⎠ P = VI = V " ⎛W μ nCox ⎜ 2 ⎝L P P P* α 2P P = | *= = A WL A αW (αL ) A ⎞ 2 ⎟(VGS − VTN ) ⎠ PDP* = P*ΔT * = (αΔT )α 2 P = α 3 PΔT = α 3 PDP Power density = 6-8 6.34 1 1 ε ⎛W ⎞ 2 2 " ⎛W ⎞ μnCox ⎜ ⎟(VGS − VTN ) = μ n ox ⎜ ⎟(VGS − VTN ) 2 2 Tox ⎝ L ⎠ ⎝L⎠ ⎛W ⎞ ⎜ ⎟ ε 1 2 * ID = μn ox ⎜ 2 ⎟(VGS − VTN ) = 2 I D T L 2 ox ⎜ ⎜ ⎟ ⎟ 2 ⎝ 2 ⎠ (b) P* = V (2 I ) = 2VI = 2 P - The power has increased by a factor of two. ε ε W L CG " * WL = ox WL | CG (c) CG = Cox = = ox Tox 2 2 Tox 2 2 The capacitance has decreased by a factor of two. (a) I D = ⎛ W ⎞⎛ L ⎞ K ⎜ ⎟⎜ ⎟(ΔV ) ΔT C ΔV ⎝ 2 ⎠⎝ 2 ⎠ = (d ) ΔT * = = * 4 I ⎛W ⎞ 1 ⎜ 2 ⎟ 2 μ n ⎜ ⎟(VGS − VTN ) 2 ⎜ L ⎟ ⎜ ⎟ ⎝ 2 ⎠ * * 6.35 ( a ) Pavg = 0.5 2 x10 gates ( 1W 6 ) = 1 μW / gate ( b ) I = 1μW / gate = 0.556 μA / gate 1.8V 6.36 ( a ) Pavg = 20W 1.5μW / gate = 1.5 μW / gate ( b ) I = = 0.833 μA / gate ⎛2⎞ 1.8V 6 ⎜ ⎟20 x10 gates ⎝ 3⎠ 6.37 V 2.5 - 0.2 50 μW = 20 μA | Let VL = TN = 0.2V | R = = 115 kΩ 2x10-5 2.5V 3 0.926 1 0.2 ⎞ ⎛W ⎞ ⎛W ⎞ ⎛ = M S is in the triode region : 20x10-6 = 60x10-6 ⎜ ⎟ ⎜ 2.5 − 0.6 − ⎟0.2 → ⎜ ⎟ = 1 1.08 2 ⎠ ⎝ L ⎠S ⎝ L ⎠S ⎝ I DD = 6-9 6.38 (a ) For MS off, I D = 0 and VH = 2.5V . ⎛ ⎛ 3⎞⎛ μA ⎞ 2.5 − VL V ⎞ μA = Kn ⎜VH − VTN − L ⎟VL | Kn = ⎜ ⎟⎜60 2 ⎟ = 180 2 200 kΩ 2⎠ V ⎝ ⎝ 1 ⎠⎝ V ⎠ ⎛ V ⎞ μA ⎞⎛ 2.5 − VL = 2 x105 ⎜180 2 ⎟⎜2.5 − 0.6 − L ⎟VL → 36VL2 − 138.8VL + 5 = 0 2⎠ V ⎠⎝ ⎝ 2.5 − 0.0364 = 12.3 μA | P = 2.5V ( 12.3 μA) = 30.8 μW VL = 0.0364 V | I D = 200kΩ 0.0364 ⎞ μA ⎛ Checking : I D = 180 2 ⎜ 2.5 − 0.6 − ⎟0.0364 = 12.3 μA 2 ⎠ V ⎝ (b) For MS off, I D = 0 and VH = 2.5V . For VL , I D = ( ) ⎛ ⎛ 6 ⎞⎛ μA ⎞ 2.5 − VL V ⎞ μA = Kn ⎜VH − VTN − L ⎟VL | Kn = ⎜ ⎟⎜60 2 ⎟ = 360 2 400 kΩ 2⎠ V ⎝ ⎝ 1 ⎠⎝ V ⎠ ⎛ V ⎞ μA ⎞⎛ 2.5 − VL = 4 x105 ⎜ 360 2 ⎟⎜2.5 − 0.6 − L ⎟VL → 144VL2 − 549.2VL + 5 = 0 2⎠ V ⎠⎝ ⎝ 2.5 − 0.00913 = 6.23 μA | P = 2.5V (6.23 μA)= 15.6 μW VL = 9.13 mV | I D = 400 kΩ 0.00913⎞ μA ⎛ Checking : I D = 360 2 ⎜ 2.5 − 0.6 − ⎟0.00913 = 6.21 μA 2 ⎠ V ⎝ For VL , I D = ( ) 6.39 (a ) For MS off, I D = 0 and VH = 2.5V . ⎛ ⎛ 3⎞⎛ μA ⎞ 2.5 − VL V ⎞ μA = Kn ⎜VH − VTN − L ⎟VL | Kn = ⎜ ⎟⎜60 2 ⎟ = 180 2 200 kΩ 2⎠ V ⎝ ⎝ 1 ⎠⎝ V ⎠ ⎛ V ⎞ μA ⎞⎛ 2.5 − VL = 2 x105 ⎜180 2 ⎟⎜2.5 − 0.8 − L ⎟VL → 36VL2 − 124.4VL + 5 = 0 2⎠ V ⎠⎝ ⎝ 2.5 − 0.0407 = 12.3 μA | P = 2.5V ( 12.3 μA) = 30.7 μW VL = 0.0407 V | I D = 200kΩ 0.0407 ⎞ μA ⎛ Checking : I D = 180 2 ⎜ 2.5 − 0.8 − ⎟0.0407 = 12.3 μA 2 ⎠ V ⎝ For VL , I D = ( ) 180 (b) 2.5 − V = (2 x10 )⎜ V ⎝ 5 L ⎛ μA ⎞⎛ VL ⎞ 2 2.5 − 0.4 − ⎟ ⎜ ⎟VL → 36VL − 153.2VL + 5 = 0 2 2⎠ ⎠⎝ VL = 0.0329 V | I D = 2.5 − 0.0329 = 12.3 μA | P = 2.5V ( 12.3 μA) = 30.8 μW 200 kΩ 0.0329 ⎞ μA ⎛ Checking : I D = 180 2 ⎜ 2.5 − 0.4 − ⎟0.0329 = 12.3 μA 2 ⎠ V ⎝ 6-10 6.40 (a) V IL = VTN + 1 1 1 = 0.6V + = 0.6 + = 0.627 V Kn R 36 3 ⎛ μA ⎞ ⎜ 60 ⎟(200 kΩ) 1⎝ V 2 ⎠ VOH = VDD − VIH = VTN − 1 1 2VDD 5 = 2.5 − = 2.49V | VOL = = = 0.215V 3Kn R 2 Kn R 72 108 1 V 1 2.5 + 1.63 DD = 0.6 − + 1.63 = 1.00V Kn R 36 36 Kn R NM L = 0.627 − 0.215 = 0.412 V | NM H = 2.49 − 1.00 = 1.49 V 1 1 1 = 0.6 + = 0.607 V (b) VIL = VTN + K R = 0.6V + 6 ⎛ μA⎞ 144 n ⎜ 60 ⎟(400 kΩ) 1⎝ V2 ⎠ VOH = VDD − VIH = VTN − 1 1 2VDD 5 = 2.5 − = 2.50V | VOL = = = 0.108V 3 Kn R 2 Kn R 288 432 1 V 1 2.5 + 1.63 DD = 0.6 − + 1.63 = 0.807V Kn R 144 144 Kn R NM L = 0.607 − 0.108 = 0.499 V | NM H = 2.50 − 0.807 = 1.69 V 6.41 (a ) For MS off, I D = 0 and VH = 2.5V . ⎛ ⎛ 6 ⎞⎛ μA ⎞ 2.5 − VL V ⎞ μA = Kn ⎜VH − VTN − L ⎟VL | Kn = ⎜ ⎟⎜60 2 ⎟ = 360 2 400 kΩ 2⎠ V ⎝ ⎝ 1 ⎠⎝ V ⎠ ⎛ V ⎞ μA ⎞⎛ 2.5 − VL = 4 x105 ⎜ 360 2 ⎟⎜2.5 − 0.6 − L ⎟VL → 144VL2 − 549.2VL + 5 = 0 2⎠ V ⎠⎝ ⎝ 2.5 − 0.00913 = 6.23 μA | P = 2.5V (6.23 μA)= 15.6 μW VL = 9.13 mV | I D = 400 kΩ 0.00913⎞ μA ⎛ Checking : I D = 360 2 ⎜ 2.5 − 0.6 − ⎟0.00913 = 6.23 μA 2 ⎠ V ⎝ ⎛ V ⎞ 2.5 − 0.5 − L ⎟VL → 144VL2 − 578VL + 5 = 0 (b) 2.5 − VL = 144⎜ 2⎠ ⎝ 2.5 − 0.00867 = 6.33 μA | P = 2.5V (6.33 μA)= 15.8 μW VL = 8.67 mV | I D = 400 kΩ μA ⎛ 0.00867 ⎞ Checking : I D = 360 2 ⎜ 2.5 − 0.5 − ⎟0.00867 = 6.23 μA 2 ⎠ V ⎝ For VL , I D = ( ) 6-11 (c) 2.5 − V L ⎛ V ⎞ = 144⎜2.5 − 0.7 − L ⎟VL → 144VL2 − 520.4VL + 5 = 0 2⎠ ⎝ VL = 9.63 mV | I D = 2.5 − 0.00963 = 6.23 μA | P = 2.5V (6.23 μA)= 15.6 μW 400 kΩ μA ⎛ 0.00963 ⎞ Checking : I D = 360 2 ⎜ 2.5 − 0.7 − ⎟0.00963 = 6.22 μA 2 ⎠ V ⎝ In this design, we see that VL is not sensitive to VTN. 6.42 (a) V IL = VTN + 1 1 1 = 0.6V + = 0.6 + = 0.607V Kn R 144 6 ⎛ μA ⎞ ⎜ 60 ⎟(400 kΩ) 1⎝ V2 ⎠ VOH = VDD − VIH = VTN − 1 1 2VDD 5 = 2.5 − = 2.50V | VOL = = = 0.108V 2 Kn R 288 432 3 Kn R 1 V 1 2.5 + 1.63 DD = 0.6 − + 1.63 = 0.807V Kn R 144 144 Kn R NM L = 0.607 − 0.108 = 0.499 V | NM H = 2.50 − 0.807 = 1.69 V (b) V 6.43 IL = 0.6 + 1 5 | VOL = | NM L = VIL-VOL = 0 Kn R 3 Kn R Solving for Kn R yields no solution. Zero noise margin will not occur. (a) I D = P 0.25mW V − VL 2.5 − 0.5 = = 100μA | R = DD = = 20.0 kΩ ID VDD 2.5V 1x10-4 Using the values corresponding to Fig. 6.12, K 'p = 100μA / V 2 ⎛W ⎞ ⎛ ⎛ W ⎞ 1.21 0.5 ⎞ 100μA = 100 x10-6 ⎜ ⎟ ⎜2.5 − 0.6 − ⎟0.5 → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠S ⎝ ⎝ L ⎠S ( ) (b) V IL = VTN + 1 1 1 = 0.6V + = 0.6 + = 1.01 V ⎛ Kn R 2.42 μA ⎞ ⎜1.21x100 2 ⎟(20 kΩ) V ⎠ ⎝ VOH = VDD − VIH = VTN − 1 1 2VDD 5 = 2.5 − = 2.29V | VOL = = = 0.830V 2 Kn R 4.84 7.26 3 Kn R 1 V 1 2.5 + 1.63 DD = 0.6 − + 1.63 = 1.84V Kn R 2.42 2.42 Kn R NM L = 1.01 − 0.83 = 0.18 V | NM H = 2.29 − 1.84 = 0.450 V 6-12 6.44 P 0.25mW V − VL 3.3 − 0.2 = = 75.76μA | R = DD = = 40.9 kΩ ID VDD 3.3V 75.76 x10-6 ⎛W ⎞ ⎛ ⎛ W ⎞ 1.52 0.2 ⎞ 75.76 x10-6 = 100 x10-6 ⎜ ⎟ ⎜3.3 − 0.7 − ⎟0.2 → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠S ⎝ ⎝ L ⎠S (a) I D = ( ) (b) V IL = VTN + 1 = 0.7V + Kn R 1 1 = 0.7 + = 0.861 V ⎛ 6.22 μA ⎞ 100 2 ⎟(40.9 kΩ) (1.52)⎜ V ⎠ ⎝ VOH = VDD − VIH = VTN − 1 1 2VDD 6.6 = 3.3 − = 3.22 | VOL = = = 0.594V 3 Kn R 2 Kn R 12.4 18.7 1 V 1 3.3 + 1.63 DD = 0.7 − + 1.63 = 1.73V Kn R Kn R 6.22 6.22 NM L = 0.861 − 0.594 = 0.267 V | NM H = 3.22 − 1.73 = 1.49 V 6.45 VDD − VL 3 − 0.25 = = 83.3 kΩ ID 33x10-6 ⎛W ⎞ ⎛ ⎛ W ⎞ 1.04 0.25 ⎞ 33x10−6 = 60 x10-6 ⎜ ⎟ ⎜3 − 0.75 − ⎟0.25 → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠S ⎝ ⎝ L ⎠S (a) R = ( ) (b) SPICE yields VL = 0.249 V with ID = 33.0 μA. 6.46 VDD − VL 2 − 0.15 = = 185 kΩ ID 10 x10-6 ⎛W ⎞ ⎛ ⎛W ⎞ 0.15 ⎞ 1 10−5 = 75x10-6 ⎜ ⎟ ⎜2 − 0.6 − ⎟0.15 → ⎜ ⎟ = 2 ⎠ ⎝ L ⎠S ⎝ ⎝ L ⎠ S 1.49 (a) R = ( ) (b) SPICE yields VL = 0.15 V with ID = 10 μA. 6-13 6.47 1 = 417 Ω ⎛ ⎞ ' W −6 10 VGS − VTN ) Kn ( 60 x10 ⎜ ⎟(5 − 1) L ⎝1⎠ 1 1 = = 1000 Ω ( b ) Ron = ⎛ ⎞ ' W −6 10 VSG + VTP ) Kp ( 25 x10 ⎜ ⎟(5 − 1) L ⎝1⎠ ( c ) A resistive connection exists between the source and drain. 1 1 20 W = ' = = (d ) −6 L Kn ( VGS − VTN )Ron 60 x10 (3 − 1)(417) 1 ( a ) Ron = = W 1 1 20 = ' = = −6 L K p( VSG + VTP )Ron 25x10 (3 − 1)( 1000) 1 6.48 1 VH = VDD − VTO + γ ( (V SB + 2φ F − 2φ F ))→ V H = 3.3 − 0.75 + 0.75 VH + 0.7 − 0.7 ( ( )) (V H 2 − 4.88) = 0.5625( VH + 0.7) → VH − 6.918VH + 9.706 = 0 2 VH = 4.962V , 1.956V → VH = 1.96 V Checking : VTN = 0.75 + 0.75 1.956 + 0.75 − 0.75 = 1.345V | VH = 3.3 − 1.345 = 1.96V ( ) 6.49 VH = VDD − VTO + γ ( 2 (V SB + 2φ F − 2φ F ))→ V H = 3.3 − 0.6 + 0.6 VH + 0.6 − 0.6 ( ( )) (V H 2 − 3.165) = 0.36( VH + 0.6)→ VH − 6.69VH + 9.80 = 0 VH = 2.166V , 4.524V → VH = 2.17 V Checking : VTN = 0.5 + 0.6 ( 2.166 + 0.6 − 0.6 = 1.133V | VH = 3.3 − 1.133 = 2.167V ) 6.50 VH = VDD − VTO + γ ( (V SB + 2φ F − 2φ F ))→ V H = 2.5 − 0.5 + 0.85 VH + 0.6 − 0.6 ( ( )) (V H 2 − 2.659) = 0.7225( VH + 0.6)→ VH − 6.04VH + 6.634 = 0 2 VH = 1.444V , 4.596V → VH = 1.44 V Checking : VTN = 0.5 + 0.85 1.444 + 0.6 − 0.6 = 1.057V | VH = 2.5 − 1.057 = 1.443V ( ) 6-14 6.51 For γ = 0, VH = VDD − VTN = 3.3 − 0.6 = 2.7V | For VL : I DL = I DS ' ⎛ ⎞ ⎛ ⎞⎛ 2 Kn 1 VL ⎞ ' 4 2 ⎜ ⎟(3.3 − VL − 0.6) = Kn ⎜ ⎟⎜ 2.7 − 0.6 + − ⎟VL → 9VL − 39VL + 7.29 = 0 2 ⎝ 2⎠ 2⎠ ⎝ 1 ⎠⎝ 2 60 x10−6 ⎛ 1 ⎞ VL = 0.1958V | I DD = ⎜ ⎟(3.3 − 0.1958 − 0.6) = 94.1μA 2 ⎝ 2⎠ P = (3.3V )(94.08μA)= 0.311 mW ⎛ 4 ⎞⎛ 0.1958 ⎞ Checking : I DD = 60 x10−6⎜ ⎟⎜2.7 − 0.6 − ⎟0.1958 = 94.1μA 2 ⎠ ⎝ 1 ⎠⎝ 6.52 (a ) For γ = 0, VH = VDD − VTN = 3.3 − 0.8 = 2.5V | For VL : I DL = I DS ' ⎛ ⎞⎛ 2 Kn 1 VL ⎞ ' 4 2 3.3 − VL − 0.8) = Kn ( ⎜ ⎟⎜2.5 − 0.8 − ⎟VL → 9VL − 32.2VL + 6.25 = 0 | VL = 0.206V 2 2 2⎠ ⎝ 1 ⎠⎝ 2 60 x10−6 1 I DD = 3.3 − 0.206 − .8) = 78.9μA | P = 3.3V(78.9μA)= 0.260 mW ( 2 2 ⎛ 4 ⎞⎛ 0.206 ⎞ Checking : I DD = 60 x10−6⎜ ⎟⎜2.5 − 0.8 − ⎟0.206 = 79.0μA 2 ⎠ ⎝ 1 ⎠⎝ (b) For γ = 0, VH = VDD − VTN = 3.3 − 0.4 = 2.9V | For VL : I DL = I DS ' ⎛ ⎞⎛ 2 Kn 1 VL ⎞ ' 4 2 3.3 − VL − 0.4) = Kn ( ⎜ ⎟⎜2.9 − 0.4 − ⎟VL → 9VL − 45.8VL + 8.41 = 0 | VL = 0.191 V 1 2 2 2 ⎝ ⎠⎝ ⎠ 2 60 x10−6 1 3.3 − 0.191 − 0.4) = 110μA | P = 3.3V(6.55μA)= 0.363 mW ( 2 2 ⎛ 4 ⎞⎛ 0.191⎞ Checking : I DD = 60 x10−6⎜ ⎟⎜2.9 − 0.4 − ⎟0.191 = 110μA 2 ⎠ ⎝ 1 ⎠⎝ I DD = 6.53 VIL = VTNS = 0.6 V | VOH = VH = VDD − VTNL = 3.3 − 0.6 = 2.7V At VIH (See Eq. 6.29 Second Edition) VOL = 2 VDD − VTNL 3.3 − 0.6 1 + 3 KR 4 1+ 3 0.5 = 0.540V VIH = VTNS (VDD − VOL − VTNL ) = 0.6 + 0.54 + 0.5 1 3.3 − 0.54 − 0.6 2 = 1.41V V + OL + ( ) 2 2 2 KRVOL 2(4) 0.54 NM H = 2.7 − 1.41 = 1.29 V | NM L = 0.60 − 0.54 = 0.06 V These values are readily confirmed with SPICE. 6-15 4.0 2.0 0 -2.0 6.54 (a ) For γ = 0, VH = VDD − VTN = 3.3 − 0.6 = 2.7V | For VL : I DL = I DS ' ⎛ ⎞⎛ 2 Kn 1 VL ⎞ ' 8 2 3.3 − VL − 0.6) = Kn ⎜ ⎟⎜2.7 − 0.6 − ⎟VL → 9VL − 39VL + 7.29 = 0 ( 2 1 2⎠ ⎝ 1 ⎠⎝ −6 ⎛ ⎞ 2 60 x10 1 VL = 0.1958V | I DD = ⎜ ⎟(3.3 − 0.1958 − 0.6) = 188μA 2 ⎝ 1⎠ 188μA)= 0.621 mW P = (3.3V )( ⎛ 8 ⎞⎛ 0.1958 ⎞ Checking : I DD = 60 x10−6⎜ ⎟⎜2.7 − 0.6 − ⎟0.1958 = 188μA - check is ok 2 ⎠ ⎝ 1 ⎠⎝ (b) VIL = VTNS = 0.6 V | VOH = VH = VDD − VTNL = 3.3 − 0.6 = 2.7V At VIH (See Eq. 6.29 Second Edition) VOL = 2 VDD − VTNL 3.3 − 0.6 1 + 3 KR 4 1+ 3 0.5 = 0.540V VIH = VTNS (VDD − VOL − VTNL ) = 0.6 + 0.54 + 0.5 1 3.3 − 0.54 − 0.6 2 = 1.41V V + OL + ( ) 2 2 KRVOL 2 2(4) 0.54 NM H = 2.7 − 1.41 = 1.29 V | NM L = 0.60 − 0.54 = 0.06 V These values are easily checked with SPICE. See Prob. 6.53, (c) For γ = 0, VH = VDD − VTN = 3.3 − 0.7 = 2.6V | For VL : I DL = I DS ' ⎛ ⎞⎛ 2 Kn 1 VL ⎞ ' 8 2 3.3 − VL − 0.7) = Kn ⎜ ⎟⎜2.6 − 0.7 − ⎟VL → 9VL − 32.2VL + 6.25 = 0 ( 2 1 2⎠ ⎝ 1 ⎠⎝ VL = 0.200V | I DD = 2 60 x10−6 1 3.3 − 0.200 − 0.7) = 173μA ( 1 2 173μA)= 570 μW P = (3.3V )( ⎛ 8 ⎞⎛ 0.200 ⎞ Checking : I DD = 60 x10−6⎜ ⎟⎜2.6 − 0.7 − ⎟0.200 = 173μA - check is ok 2 ⎠ ⎝ 1 ⎠⎝ 6-16 6.55 ( a ) VH = VDD − VTO + γ ( (V SB + 2φ F − 2φ F ))→ V H = 3.3 − 0.7 + 0.5 VH + 0.6 − 0.6 ( ( )) (V H 2 − 2.987) = 0.25( VH + 0.6)→ VH − 6.225VH + 8.772 = 0 → VH = 2.156 V 2 VL = 0.20V | I D = ⎛W ⎞ ⎛ 0.25mW 0.20 ⎞ = 75.76μA | 75.76 = 100⎜ ⎟ ⎜ 2.156 − 0.7 − ⎟0.20 3.3V 2 ⎠ ⎝ L ⎠S ⎝ ⎛W ⎞ 2.79 | VTNL = 0.7 + 0.5 0.2 + 0.6 − 0.6 = 0.760V ⎜ ⎟ = 1 ⎝ L ⎠S ⎛W ⎞ 2 1 100 ⎛W ⎞ 75.76 = ⎜ ⎟ (3.3 − 0.20 − 0.760) → ⎜ ⎟ = 2 ⎝ L ⎠L ⎝ L ⎠ L 3.61 ( b ) VIL = VTNS = 0.70V | VOH = VH = 2.16V ( ) Finding VIH (See Eq 6.29 in 2nd Ed.) : VOL = VDD − VTNL 1+ 3 (W / L) (W / L) = S L 1 + 3(2.79)(3.61) 3.3 − VTNL = 3.3 − VTNL 5.587 VTNL = 0.7 + 0.5 VOL + 0.6 − 0.6 ( )| 5.587VOL = 3.3 − 0.7 + 0.5 VOL + 0.6 − 0.6 ( ( )) Using the quadratic equation : VOL = 0.4432V → VTNL = 0.8234V VIH = VTNS + VIH = 0.7 + 4.0 W / L) 1 2 VOL ( L + VDD − VOL − VTNL ) ( 2 (W / L) 2VOL S 2 0.443 1 ⎛ 1⎞ 1 + 3.3 − 0.443 − 0.823) | VIH = 1.39V ⎜ ⎟ ( 2 10.07 ⎝ 2 ⎠ 0.443 NM H = 2.16 − 1.39 = 0.77 V | NM L = 0.7 − 0.443 = 0.26 V 2.0 0 -2.0 6-17 6.56 0.4 mW = 160μA | VTNL = 0.6 + 0.5 0.3 + 0.6 − 0.6 = 0.687V 2.5V ⎛ W ⎞ 1.40 2 100 x10−6 ⎛W ⎞ 160 x10−6 = ⎜ ⎟ (2.5 − 0.3 − 0.687) → ⎜ ⎟ = 2 1 ⎝ L ⎠L ⎝ L ⎠L ⎛W ⎞ ⎛ ⎛W ⎞ 0.3 ⎞ 6.67 160 x10−6 = 100 x10−6 ⎜ ⎟ ⎜1.55 − 0.6 − ⎟0.3 → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠S ⎝ ⎝ L ⎠S I DD = 6.57 (a) VDD = 3.3 V VTN =1 V ID = 75 μA VL = 0.2 V VH = VDD - VTN = 3.3 V - 0.6V = 2.7 V +3.3 V ( ) + M L V + V = 3.1 V GS DSL = 3.1 V ID VL = 0.2V + M S V H = 2.7 V V DSS = 0.2 V - I DS = I DL = 75μA ⎛ ⎞ ⎛ VDSS ⎞ ' W I DS = Kn ⎜ ⎟ ⎜VGSS − VTHS − ⎟VDSS 2 ⎠ ⎝ L ⎠S ⎝ 2 K ' ⎛W ⎞ I DL = n ⎜ ⎟ ( VGSL − VTNL ) 2 ⎝ L ⎠L 75μA = 100 100 ⎛W ⎞ 1.88 0.2 ⎞ ⎜ ⎟ ⎜ 2.7 − 0.6 − ⎟0.2 → ⎜ ⎟ = 2 ⎠ 1 V ⎝ L ⎠S ⎝ ⎝ L ⎠S 2 μA ⎛W ⎞ ⎛ μA 75μA = V2 2 ⎛W ⎞ ⎛W ⎞ 2 1 ⎜ ⎟ (3.3 − 0.2 − 0.6) → ⎜ ⎟ = ⎝ L ⎠L ⎝ L ⎠ L 4.17 H ( b ) VH = VDD − VTO + γ ( (V SB + 2φ F − 2φ F ))→ V = 3.3 − 0.6 + 0.5 VH + 0.6 − 0.6 ( ( )) (V H 2 − 3.087) = 0.25( VH + 0.6)→ VH − 6.424VH + 9.381 = 0 → VH = 2.245 2 75μA = 100 ⎛W ⎞ 0.2 ⎞ 2.43 2.245 − 0.6 − 0.2 → ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = 2 2 ⎠ 1 V ⎝ L ⎠S ⎝ ⎝ L ⎠S 2 K ' ⎛W ⎞ I DL = n ⎜ ⎟ ( VGSL − VTNL ) | VTNL = 0.6 + 0.5 0.2 + 0.6 − 0.6 = 0.660V 2 ⎝ L ⎠L μA ⎛ W ⎞ ⎛ ( ) 75μA = 100 μA V2 2 ⎛W ⎞ ⎛W ⎞ 2 1 ⎜ ⎟ (3.3 − 0.2 − 0.66) → ⎜ ⎟ = ⎝ L ⎠L ⎝ L ⎠ L 3.97 6-18 6.58 (a ) For γ = 0, V H = VDD − VTO = 2 − 0.6 = 1.4V (b) For γ = 0.6, V ⎛ ⎞ ⎛ ⎛ ⎞ ⎛ ⎛W ⎞ VL ⎞ 0.15⎞ 2.30 ' W −6 −6 W I DS = Kn ⎜ ⎟ ⎜VH − VTNS − ⎟VL | 25 x10 = 100 x10 ⎜ ⎟ ⎜1.4 − 0.6 − ⎟0.15 → ⎜ ⎟ = 2 ⎠ 1 2⎠ ⎝ L ⎠S ⎝ ⎝ L ⎠S ⎝ ⎝ L ⎠S ⎛W ⎞ 2 2 K ' ⎛W ⎞ 100 x10−6 ⎛W ⎞ 1 I DL = n ⎜ ⎟ ( VGSL − VTNL ) | 25 x10−6 = ⎜ ⎟ (2 − 0.15 − 0.6) → ⎜ ⎟ = 2 ⎝ L ⎠L 2 ⎝ L ⎠L ⎝ L ⎠ L 3.13 H = VDD − VTNL = 2 − 0.6 + 0.6 VH + 0.6 − 0.6 → VH = 1.09V [ ( )] ⎛ ⎞ ⎛ ⎛ ⎞ ⎛ ⎛W ⎞ VL ⎞ 0.15 ⎞ 4.02 ' W −6 −6 W I DS = Kn ⎜ ⎟ ⎜VH − VTNS − ⎟VL | 25 x10 = 100 x10 ⎜ ⎟ ⎜1.09 − 0.6 − ⎟0.15 → ⎜ ⎟ = 2 ⎠ 1 2⎠ ⎝ L ⎠S ⎝ ⎝ L ⎠S ⎝ ⎝ L ⎠S For vO = VL = 0.15V , VTN = 0.6 + 0.6 I DL = ( 0.15 + 0.6 − 0.6 = 0.655V ) ' ⎛ ⎛W ⎞ 2 2 Kn W⎞ 100 x10−6 ⎛W ⎞ 1 −6 V − V | 25x10 = ⎜ ⎟ ( GSL TNL ) ⎜ ⎟ (2 − 0.15 − 0.655) → ⎜ ⎟ = 2 ⎝ L ⎠L 2 ⎝ L ⎠L ⎝ L ⎠ L 2.86 (c) Using LEVEL=1 KP=100U VTO=0.6 GAMMA=0, the values of ID and VL agree with our hand calculations. The results also agree for GAMMA=0.6. 6.59 ' ⎛ ⎛W ⎞ ⎛ 2 VDSL ⎞ Kn W⎞ I DS = I DL | K ⎜ ⎟ ⎜VGSS − VTNS − VGSL − VTNL ) ⎟VDSL = ⎜ ⎟ ( 2 ⎠ 2 ⎝ L ⎠L ⎝ L ⎠S ⎝ ' ⎛ ⎛ ⎞⎛ 2 Kn 1 ⎞ VO ⎞ ' 4.71 Kn ⎜ ⎜ ⎟⎜2.5 − 0.6 − ⎟VO = ⎟(2.5 − VO − VTNL ) 2⎠ 2 ⎝ 1.68 ⎠ ⎝ 1 ⎠⎝ ' n VTNL = 0.6 + 0.5 VO + 0.6 − 0.6 6.60 ( ) An iterative solution yields VO = 0.1061 V ' ⎛ ⎛ ⎞ ⎛ 2 VL ⎞ Kn W⎞ ' W I DS = I DL | Kn V − V − = V ⎜ ⎟ ⎜ H ⎟ L ⎜ ⎟ (2.5 − VL − VTNL ) TNS 2⎠ 2 ⎝ L ⎠L ⎝ L ⎠S ⎝ which is independent of K'n . Ratioed logic maintains VL and VH independent ' : of K'n . So VH = 1.55V and VL = 0.20V . However, I DS = I DL ∝ Kn So, I D = 80μA 80 μA V 2 = 64.0 μA P = 2.5V (64μA)= 0.160 mW 100 μA V 2 μA ⎛ 4.71⎞⎛ 0.2 ⎞ Checking : I DS = 80 2 ⎜ ⎟⎜1.55 − 0.6 − ⎟0.2 = 64.1μA 2 ⎠ V ⎝ 1 ⎠⎝ 6-19 6.61 ' ⎛ ⎛ ⎞ ⎛ 2 VL ⎞ Kn W⎞ ' W I DS = I DL | Kn ⎜ ⎟ ⎜VH − VTNS − ⎟VL = ⎜ ⎟ (2.5 − VL − VTNL ) 2⎠ 2 ⎝ L ⎠L ⎝ L ⎠S ⎝ which is independent of K'n . Ratioed logic maintains VL and VH independent ' of K'n . So VH = 1.55V and VL = 0.20V . However, I DS = I DL ∝ Kn : So, I D = 80μA 120 μA V 2 = 96.0 μA P = 2.5V (96μA)= 0.240 mW 100 μA V 2 0.2 ⎞ μA ⎛ 4.71⎞⎛ Checking : I DS = 120 2 ⎜ ⎟⎜1.55 − 0.6 − ⎟0.2 = 96.1μA 2 ⎠ V ⎝ 1 ⎠⎝ 6.62 Noise Margins vs. KR 2.5 2 1.5 NMH NML 1 0.5 0 0 2 4 6 8 10 12 -0.5 KR 6.63 (a) VH = VDD – VTNL does not depend upon λ. However, VL is dependent upon λ. (b) SPICE yields VL = 0.20 V, 0.207 V, 0.217 V, and 0.232 V for λ = 0, 0.02/V, 0.05/V, and 0.1/V respectively. The current also increases: IDD = 80.1, 82.8, 86.9 and 93.3 μA, respectively. 6.64 VTNL = 0.6 + 0.5 0.20 + 0.6 − 0.6 = 0.660V ( ) VGSL − VTNL = 4 − 0.2 − 0.66 = 3.14V | VDSL = 2.5 − 0.2 = 2.30V → Triode region ⎛W ⎞ 2.3⎞ 1 μA ⎛ W ⎞ ⎛ 80μA = 100 2 ⎜ ⎟ ⎜ 4 − 0.2 − 0.66 − ⎟2.3 → ⎜ ⎟ = 2 ⎠ V ⎝ L ⎠L⎝ ⎝ L ⎠ L 5.72 ⎛W ⎞ 2.22 0.2 ⎞ μA ⎛ W ⎞ ⎛ 80μA = 100 2 ⎜ ⎟ ⎜ 2.5 − 0.6 − ⎟0.2 → ⎜ ⎟ = 2 ⎠ 1 V ⎝ L ⎠S ⎝ ⎝ L ⎠L 6-20 6.65 For linear operation at vo = VL : VTNL = 0.8 + 0.5 0.2 + 0.6 − 0.6 = 0.860V VGSL − VTNL ≥ VDSL : VGG − 0.20 − 0.860 ≥ 2.5 − 0.2 → VGG ≥ 3.36V We also require : VGG ≥ 2.5 + VTNL = 2.5 + 0.8 + 0.5 2.5 + 0.6 − 0.6 = 3.79V so VGG ≥ 3.79V 6.66 ( ) ( ) If VH = 3.3 V , VTNL = 0.6 + 0.5 3.3 + 0.6 − 0.6 = 1.2V 5 -1.2 = 3.8V > 3.3V so VH = 3.3 V is correct. ' ⎛ ⎞ ⎛ ⎞⎛ 2 VL ⎞ Kn 1 ' 5 I DS = I DL | Kn ⎜ ⎟⎜ 3.3 − 0.6 − ⎟VL = ⎜ ⎟(3.3 − VL − VTNL ) 2⎠ 2 ⎝ 2⎠ ⎝ 1 ⎠⎝ VTNL = 0.6 + 0.5 VL + 0.6 − 0.6 I DS = ( ) ( ) An interative solution gives V = 0.1222 V , V L TNL = 0.6376 V 2 100μA ⎛ 1 ⎞ 161μA) = 0.532 mW ⎜ ⎟(3.3 − 0.1222 − 0.6376) = 161 μA | P = 3.3V ( 2 ⎝ 2⎠ 6.67 We require VGG ≥ VDD + VTNL so VH = VDD VTNL = VTO + γ (V SB + 0.6 − 0.6 = 0.6 + 0.6 ) ( 3.3 + 0.6 − 0.6 = 1.32V ) VGG ≥ 3.3 + 1.32 = 4.62V 6.68 We require VGG ≥ VDD + VTNL so VH = VDD VTNL = VTO + γ (V SB + 0.6 − 0.6 = 0.6 + 0.6 3.3 + 0.6 − 0.6 = 1.32V ) ( ) VGG ≥ 3.3 + 1.32 = 4.62V | Design decision - Choose VGG = 5 V I DD = 300μW = 90.9μA 3.3V ⎛W ⎞ ⎛ ⎛ W ⎞ 1.75 0.2 ⎞ For MS : 90.9μA = 100μA⎜ ⎟ ⎜3.3 − 0.6 − ⎟0.2 → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠S ⎝ ⎝ L ⎠S For ML : VTNL = 0.6 + 0.6 0.2 + 0.6 − 0.6 = 0.672V ⎛W ⎞ ⎛ ⎛W ⎞ 3.3 − 0.2 ⎞ 1 90.9μA = 100μA⎜ ⎟ ⎜ 5 − .2 − 0.672 − ⎟(3.3 − 0.2)→ ⎜ ⎟ = 2 ⎠ ⎝ L ⎠L⎝ ⎝ L ⎠ L 8.79 6.69 We require VTNL ≤ 0 : − 1 + γ ( ) ( 2.5 + 0.6 − 0.6 ≤ 0 → γ ≤ 1.014 ) 6-21 6.70 (a) V H ' ⎛ ⎛ ⎞ ⎛ 2 W⎞ VL ⎞ Kn ' W = VDD | I DS = I DL | Kn VTNL ) ⎜ ⎟ ⎜VDD − VTNS − ⎟VL = ⎜ ⎟ ( 2⎠ 2 ⎝ L ⎠L ⎝ L ⎠S ⎝ For ratioed logic, both VH and VLare independent of K'n . VH = 2.5 V | VL = 0.2 V ⎛ 80 ⎞ ' | I DS = 80μA⎜ However, I D ∝ Kn ⎟ = 64μA | P = 2.5V (64μA) = 0.160 mW ⎝ 100 ⎠ ⎛120 ⎞ ⎟ = 96μA | P = 2.5V (96μA)= 0.240 mW (b) VH = 2.5 V VL = 0.2 V I DS = 80μA⎜ ⎝100 ⎠ 6.71 0.20mW = 60.1μA VTNL = −1 + 0.5 3.3 + 0.6 − 0.6 = −0.400V → VH = 3.3V 3.3V ⎛W ⎞ ⎛ ⎛W ⎞ 1.16 0.20 ⎞ | For VO = VL = 0.2V , 60.1μA = 100μA⎜ ⎟ ⎜ 3.3 − 0.6 − ⎟0.20 → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠S ⎝ ⎝ L ⎠S ⎛ W ⎞ 1.36 2 100μA ⎛ W ⎞ VTNL = −1 + 0.5 0.20 + 0.6 − 0.6 = −0.940V | 60.1μA = ⎜ ⎟ (−0.940) | ⎜ ⎟ = 1 2 ⎝ L ⎠L ⎝ L ⎠L I DD = ( ) ( ) 6.72 Assume VH = VDD = 3.3V | Checking : VTNL = −2 + 0.5 3.3 + 0.6 − 0.6 = −1.40 VTNL < 0, so our assumption is correct. | I DD = P 250μW = = 75.8μA VDD 3.3V ⎛W ⎞ ⎛ ⎛W ⎞ 1.46 0.2 ⎞ For MS in the triode region, 75.8μA = 100μA⎜ ⎟ ⎜ 3.3 − 0.6 − ⎟0.2 → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠S ⎝ ⎝ L ⎠S For ML in the saturation region, VTNL = −2 + 0.5 0.2 + 0.6 − 0.6 = −1.94V and 75.8μA = 2 100μA ⎛ W ⎞ ⎜ ⎟ (0 − VTNL ) → 2 ⎝ L ⎠L ( ) ( ) ⎛W ⎞ 1 ⎜ ⎟ = ⎝ L ⎠ L 2.48 6.73 (a) No, VH does not depend upon λ. (b) As λ increases, IDD increases in ML, and VL increases. λ 0 0.02/V 0.05/V 0.1/v IDD 78.2 μA 81.4 μA 86.0 μA 93.6 μA VL 195 mV 203 mV 214 mV 231 mV 6-22 6.74 ( a ) The PMOS load is still saturated, so I DD remains the same : I DD = 80μA. ⎛W ⎞ ⎛ ⎛ W ⎞ 1.80 0.25 ⎞ Also, VH = 2.5 V . 80 x10-6 = 100 x10-6⎜ ⎟ ⎜2.5 − 0.6 − ⎟0.25 → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠n ⎝ ⎝ L ⎠n 1.80( 100) V +V K 2.5 − 0.6 = 1.02 V (b) VIL = VTNS + DD2 TP KR = KS = 1.11 40 = 4.05 VIL = 0.6 + ( ) 4.052 + 4.05 K R + KR L ⎛ ⎛ KR ⎞ 4.05 ⎞ ⎜ ⎜ ⎟ ⎟ = 2.30V VOH = VDD − ( VDD + VTP ) 1 − 1 − = 2.5 − 2.5 − 0.6 ( ) ⎜ ⎜ 5.05 ⎟ KR + 1 ⎟ ⎝ ⎝ ⎠ ⎠ V + VTP 2.5 − 0.6 = = 0.545V VIH = VTNS + 2VOL = 0.6 + 2(0.545) = 1.69V VOL = DD 3 KR 3(4.05) NM H = VOH − VIH = 2.30 − 1.69 = 0.610 V NM L = VIL − VOL = 1.02 − 0.545 = 0.475 V 6.75 ( a ) The PMOS load is still saturated, so I DD remains the same : I DD = 80μA. ⎛ 2.22 ⎞ ⎛ VL ⎞ Also, VH = 2.5 V . 80 x10-6 = 100 x10-6⎜ ⎟ ⎜2.5 − 0.5 − ⎟VL → VL = 0.189 V 2⎠ ⎝ 1 ⎠n ⎝ V +V K 2.22 ⎛100 ⎞ 2.5 − 0.6 = 0.847 V (b) VIL = VTNS + DD2 TP KR = KS = 1.11 ⎜ ⎟ = 5.00 VIL = 0.5 + ⎝ 40 ⎠ 52 + 5 KR + KR L ⎛ ⎛ KR ⎞ 5 ⎞ ⎜ ⎜ ⎟ ⎟ = 2.33V VOH = VDD − ( VDD + VTP ) 1 − 1 − = 2.5 − 2.5 − 0.6 ( ) ⎜ ⎜ 5 + 1⎟ KR + 1 ⎟ ⎝ ⎝ ⎠ ⎠ V + VTP 2.5 − 0.6 = = 0.491V VIH = VTNS + 2VOL = 0.5 + 2(0.491) = 1.48V VOL = DD 3 KR 3(5) NM H = VOH − VIH = 2.33 − 1.48 = 0.849 V NM L = VIL − VOL = 0.847 − 0.491 = 0.356 V ( c ) The PMOS load is still saturated, so I DD remains the same : I DD = 80μA. ⎛ 2.22 ⎞ ⎛ VL ⎞ Also, VH = 2.5 V . 80 x10-6 = 100 x10-6⎜ ⎟ ⎜2.5 − 0.7 − ⎟VL → VL = 0.213 V 2⎠ ⎝ 1 ⎠n ⎝ V +V K 2.22 ⎛100 ⎞ 2.5 − 0.6 = 1.05 V (d) VIL = VTNS + DD2 TP KR = KS = 1.11 ⎜ ⎟ = 5.00 VIL = 0.7 + ⎝ 40 ⎠ 52 + 5 KR + KR L ⎛ ⎛ KR ⎞ 5 ⎞ ⎜ ⎜ ⎟ ⎟ = 2.33V VOH = VDD − ( VDD + VTP ) 1 − 1 − = 2.5 − 2.5 − 0.6 ( ) ⎜ ⎜ 5 + 1⎟ KR + 1 ⎟ ⎝ ⎝ ⎠ ⎠ V + VTP 2.5 − 0.6 = = 0.490V VIH = VTNS + 2VOL = 0.7 + 2(0.490)= 1.68V VOL = DD 3 KR 3(5) NM H = VOH − VIH = 2.33 − 1.68 = 0.650 V NM L = VIL − VOL = 1.05 − 0.490 = 0.560 V 6-23 6.76 ( a ) The PMOS load is still saturated, so I DD remains the same : I DD = 80μA. ⎛ 2.22 ⎞ ⎛ VL ⎞ Also, VH = 2.5 V . 80 x10-6 = 120 x10-6⎜ ⎟ ⎜2.5 − 0.6 − ⎟VL → VL = 0.165 V 2⎠ ⎝ 1 ⎠n ⎝ V +V K 2.22 ⎛120 ⎞ 2.5 − 0.6 = 0.893V (b) VIL = VTN + DD2 TP KR = KS = 1.11 ⎜ ⎟ = 6 VIL = 0.6 + ⎝ 40 ⎠ 62 + 6 KR + KR L VOH ⎛ ⎜ = VDD − ( VDD + VTP ) ⎜1 − ⎝ VDD + VTP 3 KR = 2.5 − 0.6 3(6) ⎛ KR ⎞ 6 ⎞ ⎜ ⎟ = 2.5 − (2.5 − 0.6) ⎜1 − 6 + 1 ⎟ ⎟ = 2.36V KR + 1 ⎟ ⎝ ⎠ ⎠ = 0.448V VIH = VTN + 2VOL = 0.6 + 2(0.448)= 1.50V VOL = NM H = VOH − VIH = 2.36 − 1.50 = 0.860 V NM L = VIL − VOL = 0.893 − 0.448 = 0.445 V ( c ) The PMOS load is still saturated, so I DD remains the same : I DD = 80μA. ⎛ 2.22 ⎞ ⎛ VL ⎞ Also, VH = 2.5 V . 80 x10-6 = 80 x10-6 ⎜ ⎟ ⎜2.5 − 0.6 − ⎟VL → VL = 0.254 V 2⎠ ⎝ 1 ⎠n ⎝ V +V K 2.22 ⎛ 80 ⎞ 2.5 − 0.6 = 1.03V (d) VIL = VTN + DD2 TP KR = KS = 1.11 ⎜ ⎟ = 4 VIL = 0.6 + ⎝ 40 ⎠ 42 + 4 KR + KR L ⎛ ⎜ VOH = VDD − ( VDD + VTP ) ⎜1 − ⎝ VOL = VDD + VTP 3 KR = 2.5 − 0.6 3(4) ⎛ KR ⎞ 4 ⎞ ⎜ ⎟ 1 − = 2.5 − 2.5 − 0.6 ( )⎜ 4 + 1 ⎟ ⎟ = 2.30V KR + 1 ⎟ ⎝ ⎠ ⎠ = 0.549V VIH = VTN + 2VOL = 0.6 + 2(0.549)= 1.70V NM L = VIL − VOL = 1.03 − 0.549 = 0.481 V NM H = VOH − VIH = 2.30 − 1.70 = 0.600 V 6-24 6.77 ( a ) The PMOS load is still saturated, and VH = 2.5 V . 2 40 x10-6 ⎛ 1.11⎞ I DD = ⎜ ⎟ (2.5 − 0.5) → I DD = 88.8 μA 2 ⎝ 1 ⎠n ⎛ 2.22 ⎞ ⎛ VL ⎞ For MS : 88.8 x10-6 = 100 x10-6 ⎜ ⎟ ⎜ 2.5 − 0.6 − ⎟VL → VL = 0.224 V 2⎠ ⎝ 1 ⎠n⎝ V +V K 2.22 ⎛100 ⎞ 2.5 − 0.5 = 0.965V (b) VIL = VTN + DD2 TP KR = KS = 1.11 ⎜ ⎟ = 5 VIL = 0.6 + ⎝ 40 ⎠ 52 + 5 KR + KR L ⎛ ⎜ VOH = VDD − ( VDD + VTP ) ⎜1 − ⎝ VOL = VDD + VTP 3 KR = 2.5 − 0.5 3(5) ⎛ KR ⎞ 5 ⎞ ⎜ ⎟ 1 − = 2.5 − 2.5 − 0.5 ( )⎜ 5 + 1 ⎟ ⎟ = 2.33V KR + 1 ⎟ ⎝ ⎠ ⎠ = 0.516V VIH = VTN + 2VOL = 0.6 + 2(0.516) = 1.63V NM H = VOH − VIH = 2.33 − 1.63 = 0.700 V NM L = VIL − VOL = 0.965 − 0.516 = 0.449 V ( c ) The PMOS load is still saturated, and VH = 2.5 V . 2 40 x10-6 ⎛ 1.11⎞ I DD = ⎜ ⎟ (2.5 − 0.7) → I DD = 71.9 μA 2 ⎝ 1 ⎠n ⎛ 2.22 ⎞ ⎛ VL ⎞ For the NMOS device : 71.9 x10-6 = 100 x10-6 ⎜ ⎟ ⎜2.5 − 0.6 − ⎟VL → VL = 0.179 V 2⎠ ⎝ 1 ⎠n⎝ V +V K 2.22 ⎛100 ⎞ 2.5 − 0.7 = 0.929V (d) VIL = VTN + DD2 TP KR = KS = 1.11 ⎜ ⎟ = 5 VIL = 0.6 + 2 40 ⎝ ⎠ 5 +5 KR + KR L ⎛ ⎜ VOH = VDD − ( VDD + VTP ) ⎜1 − ⎝ VOL = VDD + VTP 3 KR = 2.5 − 0.7 3(5) ⎛ KR ⎞ 5 ⎞ ⎜ ⎟ 1 − = 2.5 − 2.5 − 0.7 ( )⎜ 5 + 1 ⎟ ⎟ = 2.34V KR + 1 ⎟ ⎝ ⎠ ⎠ = 0.465V VIH = VTN + 2VOL = 0.6 + 2(0.465) = 1.53V NM L = VIL − VOL = 0.929 − 0.465 = 0.464 V NM H = VOH − VIH = 2.34 − 1.53 = 0.810 V 6-25 6.78 ( a ) The PMOS load is still saturated, and VH = 2.5 V . 2 50 x10-6 ⎛ 1.11⎞ I DD = ⎜ ⎟ (2.5 − 0.6) → I DD = 100 μA 2 ⎝ 1 ⎠n ⎛ 2.22 ⎞ ⎛ VL ⎞ For MS : 100 x10-6 = 100 x10-6 ⎜ ⎟ ⎜ 2.5 − 0.6 − ⎟VL → VL = 0.254 V 2⎠ ⎝ 1 ⎠n⎝ V +V K 2.22 ⎛100 ⎞ 2.5 − 0.6 = 1.03V (b) VIL = VTN + DD2 TP KR = KS = 1.11 ⎜ ⎟ = 4 VIL = 0.6 + ⎝ 50 ⎠ 42 + 4 KR + KR L ⎛ ⎜ VOH = VDD − ( VDD + VTP ) ⎜1 − ⎝ VOL = VDD + VTP 3 KR = 2.5 − 0.6 3(4) ⎛ KR ⎞ 4 ⎞ ⎜ ⎟ 1 − = 2.5 − 2.5 − 0.6 ( )⎜ 4 + 1 ⎟ ⎟ = 2.30V KR + 1 ⎟ ⎝ ⎠ ⎠ = 0.549V VIH = VTN + 2VOL = 0.6 + 2(0.549)= 1.70V NM H = VOH − VIH = 2.30 − 1.70 = 0.600 V NM L = VIL − VOL = 1.03 − 0.549 = 0.481 V ( c ) The PMOS load is still saturated, and VH = 2.5 V . 2 30 x10-6 ⎛ 1.11⎞ I DD = ⎜ ⎟ (2.5 − 0.6) → I DD = 60.0 μA 2 ⎝ 1 ⎠n ⎛ 2.22 ⎞ ⎛ VL ⎞ For the NMOS device : 60 x10-6 = 100 x10-6 ⎜ ⎟ ⎜2.5 − 0.6 − ⎟VL → VL = 0.148 V 2⎠ ⎝ 1 ⎠n⎝ V +V K 2.22 ⎛100 ⎞ 2.5 − 0.6 = 0.866V (d) VIL = VTN + DD2 TP KR = KS = 1.11 ⎜ ⎟ = 6.67 VIL = 0.6 + 2 30 ⎝ ⎠ 6.67 + 6.67 KR + KR L ⎛ ⎜ VOH = VDD − ( VDD + VTP ) ⎜1 − ⎝ VOL = VDD + VTP 3 KR = 2.5 − 0.6 3(6.67) ⎛ KR ⎞ 6.67 ⎞ ⎜ ⎟ 1 − = 2.5 − 2.5 − 0.6 ( )⎜ 6.67 + 1 ⎟ ⎟ = 2.37V KR + 1 ⎟ ⎝ ⎠ ⎠ = 0.425V VIH = VTN + 2VOL = 0.6 + 2(0.425)= 1.45V NM L = VIL − VOL = 0.866 − 0.425 = 0.441 V NM H = VOH − VIH = 2.37 − 1.45 = 0.920 V 6.79 ( a ) I DD = P 100μW = = 55.6μA VDD 1.8V ⎛W ⎞ 2 25x10-6 ⎛W ⎞ 2.63 1.8 − 0.5) → ⎜ ⎟ = ⎜ ⎟ ( 2 ⎝ L ⎠p 1 ⎝ L ⎠p For the saturated PMOS load : 55.6x10-6 = ⎛W ⎞ ⎛ ⎛W ⎞ 3.86 0.2 ⎞ For the linear NMOS switch : 55.6x10-6 = 60x10-6 ⎜ ⎟ ⎜1.8 − 0.5 − ⎟0.2 → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠ n⎝ ⎝ L ⎠n 6-26 (b) V VOH K + KR ⎛ ⎜ = VDD − ( VDD + VTP ) ⎜1 − ⎝ 2 R IL = VTN + VDD + VTP KR = KS 3.86 ⎛ 60 ⎞ = ⎜ ⎟ = 3.52 KL 2.63 ⎝ 25 ⎠ VIL = 0.5 + 1.8 − 0.5 3.522 + 3.52 = 0.826V ⎛ KR ⎞ 3.52 ⎞ ⎟ ⎜ 1 − = 1.8 − 1.8 − 0.5 ( )⎜ 3.52 + 1 ⎟ ⎟ = 1.65V KR + 1 ⎟ ⎠ ⎝ ⎠ = 0.400V VIH = VTN + 2VOL = 0.5 + 2(0.400) = 1.30V NM L = VIL − VOL = 0.826 − 0.400 = 0.426 V VOL = VDD + VTP 3 KR = 1.8 − 0.5 3(3.52) NM H = VOH − VIH = 1.60 − 1.30 = 0.300 V 6.80 (a ) ID = P 200μW = = 66.7μA VDD 3V ⎛W ⎞ 2 25x10-6 ⎛W ⎞ 0.926 1 = ⎜ ⎟ (3 − 0.6) → ⎜ ⎟ = 2 ⎝ L ⎠p 1 1.08 ⎝ L ⎠p For the saturated PMOS load : 66.7x10-6 = ⎛W ⎞ ⎛ ⎛W ⎞ 1.65 0.3 ⎞ For the linear NMOS switch : 66.7x10-6 = 60x10-6 ⎜ ⎟ ⎜ 3 − 0.6 − ⎟0.3 → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠ n⎝ ⎝ L ⎠n ⎛ 60 ⎞ V +V K 3 − 0.6 = 1.11V ⎟ = 4.28 VIL = 0.6 + (b) VIL = VTN + DD2 TP KR = KS = 1.65(1.08)⎜ ⎝ 25 ⎠ 4.282 + 4.28 KR + KR L VOH ⎛ ⎜ = VDD − ( VDD + VTP ) ⎜1 − ⎝ VDD + VTP 3 KR = 3 − 0.6 3(4.28) ⎛ KR ⎞ 4.28 ⎞ ⎟ ⎜ = 3 − (3 − 0.6) ⎜1 − 4.28 + 1 ⎟ ⎟ = 2.76V KR + 1 ⎟ ⎝ ⎠ ⎠ = 0.670V VIH = VTN + 2VOL = 0.6 + 2(0.670)= 1.94V NM L = VIL − VOL = 1.11 − 0.670 = 0.440 V VOL = NM H = VOH − VIH = 2.76 − 1.94 = 0.821 V 6.81 With A = 1 = B, the circuit is equivalent to a single 4.44/1 switching device. ⎛ 4.44 ⎞⎛ 2 VL ⎞ 100μA ⎛1.81⎞ VTNL ) | VTNL = −1 + 0.5 VL + 0.6 − 0.6 100μA⎜ ⎟⎜ 2.5 − 0.6 − ⎟VL = ⎜ ⎟( 2⎠ 2 ⎝ 1 ⎠ ⎝ 1 ⎠⎝ 2 100μA ⎛ 1.81⎞ Solving iteratively → VL = 0.1033V | VTNL = −0.968V (b) I DD = ⎜ ⎟(0.968) = 84.8 μA 2 ⎝ 1 ⎠ ( ) 6-27 6.82 ⎛W ⎞ ⎛ ⎛W ⎞ 0.1⎞ 4.32 80μA = 100μA⎜ ⎟ ⎜ 2.5 − 0.6 − ⎟0.1 → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠ A⎝ ⎝ L ⎠A VTNB = 0.6 + 0.5 0.1 + 0.6 − 0.6 = 0.631 ⎛W ⎞ ⎛ ⎛W ⎞ 0.1⎞ 4.65 80μA = 100μA⎜ ⎟ ⎜ 2.5 − 0.1 − 0.631 − ⎟0.1 → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠B ⎝ ⎝ L ⎠A 6.83 ( ) We require Ron R R + on = on and the total area AT ∝ (WL)A + (WL)B ⎛W ⎞ ⎛W ⎞ K ⎜ ⎟ ⎜ ⎟ ⎝ L ⎠A ⎝ L ⎠B Setting L = 1, 1 1 1 KW B KW B W B2 + = → WA = → AT ∝ + WB = WB − K WB − K WB − K WA WB K d ⎛ W B2 ⎞ W B2 − 2KW B Finding the minimum : = 0 → W B = 2K & WA = 2K . ⎜ ⎟= dWB ⎝ W B − K ⎠ (W B − K )2 6.84 +2.5 V ML 1.81 1 Y M B MC C 2.22 1 D MD 2.22 1 MA A 2.22 1 B 2.22 1 6-28 6.85 +2.5 V ML 1.81 1 Y D MD 8.88 1 8.88 1 8.88 1 8.88 1 C MC B MB A MA 6.86 +2.5 V ML 1.11 1 Y MB B 2.22 1 C MC 2.22 1 MA A 2.22 1 (a ) With A = B = C = 1, the circuit is equivalent to a single 6.66/1 switching device. ⎛ 6.66 ⎞⎛ 2 40μA ⎛1.81⎞ VL ⎞ 100μA⎜ ⎟⎜ 2.5 − 0.6 − ⎟VL = ⎜ ⎟(0.6) → VL = 0.1033V 2⎠ 2 ⎝ 1 ⎠ ⎝ 1 ⎠⎝ 2 40μA ⎛ 1.81⎞ ⎟(0.6) = 13.0 μA (b) I DD = 2 ⎜ ⎝ 1 ⎠ 6-29 6.87 +2.5 V (a) ML 1.11 1 Y C MC 6.66 1 6.66 1 6.66 1 B MB A MA (b) The PMOS device remains saturated with I DD = 80μA. ⎛ 6.66 ⎞⎛ VDSA ⎞ For MA : 80μA = 100μA⎜ ⎟⎜ 2.5 − 0.6 − ⎟VDSA → VDSA = 0.0643V 2 ⎠ ⎝ 1 ⎠⎝ For MB : VTNB = 0.6 + 0.5 0.0643 + 0.6 − 0.6 = 0.620 ⎛ 6.66 ⎞⎛ VDSB ⎞ 80μA = 100μA⎜ ⎟⎜ 2.5 − .0643 − 0.620 − ⎟VDSB → VDSB = 0.0674V 2 ⎠ ⎝ 1 ⎠⎝ For MC : VTNC = 0.6 + 0.5 0.0643 + 0.674 + 0.6 − 0.6 = 0.640 ⎛ 6.66 ⎞⎛ VDSC ⎞ 80μA = 100μA⎜ ⎟⎜ 2.5 − .0674 − .0643 − 0.640 − ⎟VDSC → VDSC = 0.0709V 2 ⎠ ⎝ 1 ⎠⎝ VL = VDSA + VDSB + VDSC = 0.203 V ( ) ( ) (c) Assume equal values of 0.2 = 0.0667V 3 ⎛W ⎞ ⎛ ⎛W ⎞ 0.0667 ⎞ 6.43 For MA : 80μA = 100μA⎜ ⎟ ⎜ 2.5 − 0.6 − ⎟0.0667 → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠ A⎝ ⎝ L ⎠A VDS = For MB : VTNB = 0.6 + 0.5 0.0667 + 0.6 − 0.6 = 0.621 ⎛W ⎞ ⎛ ⎛W ⎞ 0.0667 ⎞ 6.74 80μA = 100μA⎜ ⎟ ⎜ 2.5 − 0.0667 − 0.621 − ⎟0.0667 → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠B ⎝ ⎝ L ⎠B For MC : VTNC = 0.6 + 0.5 0.1334 + 0.6 − 0.6 = 0.641 ⎛W ⎞ ⎛ ⎛W ⎞ 0.0667 ⎞ 7.09 80μA = 100μA⎜ ⎟ ⎜ 2.5 − 0.1334 − 0.641 − ⎟0.0667 → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠B ⎝ ⎝ L ⎠B ( ) ( ) 6-30 6.88 VDD 1/1.7 vO A 2/14.7/1 4.7/1 2/1 B Ground 6.89 VDD 1/1.7 A 2/14.7/1 4.7/1 2/1 4.7/1 2/1 vO C Ground 6.90 B ⎛ W⎞ ⎛ 2.22 ⎞ 6.66 | ⎜ ⎟ =3⎜ ⎟= 1 ⎝ L ⎠ A− F ⎝ 1 ⎠ ⎛ W ⎞ 1.81 Y = (A + B)(C + D)(E + F ) | ⎜ ⎟ = 1 ⎝ L ⎠L 6.91 (a) The only change to the schematic is to connect the gate of load transistor ML to its drain instead of its source. (b) There is no change to the logic function Y = (A + B)(C + D)(E + F ) ⎛ W⎞ ⎛ W⎞ ⎛ 4.71⎞ 14.1 1 | ⎜ ⎟ =3⎜ ⎟= (c) ⎜ ⎟ = 1 ⎝ L ⎠ L 1.68 ⎝ L ⎠ ABCDEF ⎝ 1 ⎠ 6-31 6.92 ⎛ W ⎞ 1.11 Y = (A + B)(C + D)E | ⎜ ⎟ = 1 ⎝ L ⎠L ⎛ W⎞ ⎛ 2.22 ⎞ 6.66 | ⎜ ⎟ =3⎜ ⎟= 1 ⎝ L ⎠ A− E ⎝ 1 ⎠ 6.93 (a) In the new circuit schematic, the PMOS transistor is replaced with a saturated NMOS load device as in Fig. 6.29(b). (b) The logic function is unchanged: Y = (A + B)(C + D)E ⎛ W⎞ ⎛ W⎞ ⎛ 4.71⎞ 14.1 1 (c ) ⎜ ⎟ = | ⎜ ⎟ =3⎜ ⎟= 1 ⎝ L ⎠ L 1.68 ⎝ L ⎠ ABCDE ⎝ 1 ⎠ 6.94 ⎛ W⎞ 1.11 3.33 = | ACDF path contains 4 devices ⎜ ⎟ =3 1 1 ⎝ L ⎠L (a) Y = ACE + ACDF + BF + BDE (b) ⎡ ⎛ 2.22 ⎞⎤ 26.6 ⎛ W⎞ ⎛ W⎞ 1 1 1 1 17.8 = 3 | + + = →⎜ ⎟ = ⎢4 ⎜ ⎜ ⎟ ⎟⎥ = ⎛ W⎞ ⎛ 26.6 ⎞ ⎛ W ⎞ 2.22 1 1 ⎝ L ⎠ A, C , D , F ⎝ L ⎠B , E ⎣ ⎝ 1 ⎠⎦ 3 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 1 ⎝ L ⎠B ⎝ 1 ⎠ ⎝ L ⎠E 6.95 (a) In the new circuit schematic, the PMOS transistor is replaced with a saturated NMOS load device as in Fig. 6.29(b). (b) There is no change to the logic function Y = ACDF + ACE + BDE + BF ⎛ W⎞ ⎛ W⎞ ⎛ 4.71⎞ 18.8 1 | ⎜ ⎟ =4 ⎜ ⎟ = ⎟= (c) ⎜ 1 ⎝ L ⎠ L 1.68 ⎝ L ⎠ ACDF ⎝ 1 ⎠ RoB + RonD + RonE setting RoB = RonE : ⎛ W ⎞ 12.6 1 1 1 2 1 1 + + = + = →⎜ ⎟ = ⎛ W⎞ ⎛ W⎞ 18.8 ⎛ W ⎞ 18.8 4.71 ⎝ L ⎠ B 1 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ L ⎠B ⎝ L ⎠E ⎝ L ⎠B Checking : RoB + RonF = 1 1 1 1 + = ≤ so path BF is ok. 18.8 12.6 7.42 4.71 6-32 6.96 +2.5 V Y D E B C A ⎛W ⎞ 1.81 ⎜ ⎟ = 1 ⎝ L ⎠L DCA and ECA paths contain three devices ⎛W ⎞ ⎛ 2.22 ⎞ 6.66 = 3⎜ ⎜ ⎟ ⎟= 1 ⎝ L ⎠ A, C , D , E ⎝ 1 ⎠ 1 1 1 + = ⎛W ⎞ ⎛W ⎞ ⎛ 2.22 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ L ⎠ A ⎝ L ⎠B ⎝ 1 ⎠ ⎛W ⎞ 1 1 1 3.33 + = →⎜ ⎟ = ⎛ 6.66 ⎞ ⎛W ⎞ ⎛ 2.22 ⎞ ⎝ L ⎠ 1 B ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 1 ⎠ A ⎝ 1 ⎠B ⎝ 1 ⎠ 6-33 6.97 +2.5 V C E B D A ⎛W ⎞ 1 ⎛1.11⎞ 1 ⎜ ⎟ = ⎜ ⎟= ⎝ L ⎠ L 2 ⎝ 1 ⎠ 1.80 CBA and EDA paths contain three devices ⎛W ⎞ 1 ⎛ 2.22 ⎞ 3.33 = (3) ⎜ ⎟ ⎜ ⎟= 1 ⎝ L ⎠ A− E 2 ⎝ 1 ⎠ 6-34 6.98 +2.5 V Y C E B D A ⎛W ⎞ 1 ⎜ ⎟ = ⎝ L ⎠ L 1.68 CBA and EDA paths contain three devices ⎛W ⎞ ⎛ 4.71⎞ 14.1 = 3⎜ ⎜ ⎟ ⎟= 1 ⎝ L ⎠ A− E ⎝ 1 ⎠ 6-35 6.99 +2.5 V ML Y D B C E A ⎛W ⎞ 1.11 ⎜ ⎟ = 1 ⎝ L ⎠L ⎛W ⎞ 2.22 ⎜ ⎟ = 1 ⎝ L ⎠E DCA path contains three devices ⎛W ⎞ ⎛ 2.22 ⎞ 6.66 = 3⎜ ⎜ ⎟ ⎟= 1 ⎝ L ⎠ A, C , D ⎝ 1 ⎠ ⎛W ⎞ 1 1 1 3.33 + = →⎜ ⎟ = ⎛ 6.66 ⎞ ⎛W ⎞ ⎛ 2.22 ⎞ ⎝ L ⎠ 1 B ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 1 ⎠ A ⎝ 1 ⎠B ⎝ 1 ⎠ 6.100 Y = A(B + D)(C + E )+ (C + E )G + F = (C + E ) A(B + D)+ G + F ⎡ ⎛ 2.22 ⎞⎤ 13.3 ⎛ W⎞ = 2 ⎢3⎜ ⎜ ⎟ ⎟⎥ = 1 ⎝ L ⎠ A− E ⎣ ⎝ 1 ⎠⎦ [ ] ⎛ W⎞ 1.81 3.62 = | ⎜ ⎟ =2 1 1 ⎝ L ⎠L ⎛ W⎞ ⎛ 2.22 ⎞ 4.44 ⎜ ⎟ = 2⎜ ⎟= 1 ⎝ L ⎠F ⎝ 1 ⎠ | ⎛ W⎞ 1 1 1 6.67 + = →⎜ ⎟ = ⎛ W⎞ ⎛13.3 ⎞ 2.22 ⎝ L ⎠G 1 ⎜ ⎟ ⎜ ⎟ 2 1 ⎝ L ⎠G ⎝ 1 ⎠ 6-36 6.101 (a) +2.5 V ML Y D B C E A ⎛W ⎞ 1.81 ⎜ ⎟ = 1 ⎝ L ⎠L ⎛W ⎞ 2.22 ⎜ ⎟ = 1 ⎝ L ⎠E DCA path contains three devices ⎛W ⎞ ⎛ 2.22 ⎞ 6.66 = 3⎜ ⎜ ⎟ ⎟= 1 ⎝ L ⎠ A, C , D ⎝ 1 ⎠ ⎛W ⎞ 1 1 1 3.33 + = →⎜ ⎟ = ⎛ 6.66 ⎞ ⎛W ⎞ ⎛ 2.22 ⎞ ⎝ L ⎠ 1 B ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 1 ⎠ A ⎝ 1 ⎠B ⎝ 1 ⎠ (b) Device E remains the same. 0.20 VL 0.20 V = = 0.0667V | B : VDS = 2 L = 2 = 0.133V 3 3 3 3 ⎛W ⎞ ⎛ ⎛W ⎞ 0.0667 ⎞ 6.43 100μA⎜ ⎟ ⎜ 2.5 − .6 − ⎟0.0667 = 80μA → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠ A⎝ ⎝ L ⎠A A, C , D : VDS = VTNB = VTNC = 0.6 + 0.5 0.0667 + 0.6 − 0.6 = 0.621V ⎛W ⎞ ⎛ ⎛W ⎞ 0.133⎞ 3.45 100μA⎜ ⎟ ⎜ 2.5 − 0.0667 − 0.621 − ⎟0.133 = 80μA → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠B ⎝ ⎝ L ⎠B ⎛W ⎞ ⎛ ⎛W ⎞ 0.0667 ⎞ 6.74 100μA⎜ ⎟ ⎜ 2.5 − 0.0667 − 0.621 − ⎟0.0667 = 80μA → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠C ⎝ ⎝ L ⎠C VTND = 0.6 + 0.5 0.133 + 0.6 − 0.6 = 0.641V ⎛W ⎞ ⎛ ⎛W ⎞ 0.0667 ⎞ 7.09 100μA⎜ ⎟ ⎜ 2.5 − 0.133 − 0.641 − ⎟0.0667 = 80μA → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠D ⎝ ⎝ L ⎠D ( ) ( ) 6-37 6.102 ⎛W ⎞ 2.22 Device A remains the same. ⎜ ⎟ = 1 ⎝ L ⎠A B, C , D : VDS = VL 0.20 = = 0.100V 2 2 ⎛W ⎞ ⎛ ⎛W ⎞ 0.100 ⎞ 4.32 100μA⎜ ⎟ ⎜2.5 − 0.6 − ⎟0.100 = 80μA → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠C , D ⎝ ⎝ L ⎠C , D VTNB = 0.6 + 0.5 0.100 + 0.6 − 0.6 = 0.631V ⎛W ⎞ ⎛ ⎛W ⎞ 0.100 ⎞ 4.65 100μA⎜ ⎟ ⎜ 2.5 − 0.100 − 0.631 − ⎟0.100 = 80μA → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠B ⎝ ⎝ L ⎠B ( ) 6.103 The load device remains the same. VL 0.20 V 0.20 = = 0.0667V | A : VDS = 2 L = 2 = 0.133V 3 3 3 3 ⎛W ⎞ ⎛ ⎛W ⎞ 0.0667 ⎞ 6.43 100μA⎜ ⎟ ⎜ 2.5 − .6 − ⎟0.0667 = 80μA → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠B ⎝ ⎝ L ⎠B B, C , D : VDS = VTNA = VTND = 0.6 + 0.5 0.0667 + 0.6 − 0.6 = 0.621V ⎛W ⎞ ⎛ ⎛W ⎞ 0.133⎞ 3.45 100μA⎜ ⎟ ⎜ 2.5 − 0.0667 − 0.621 − ⎟0.133 = 80μA → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠ A⎝ ⎝ L ⎠A ⎛W ⎞ ⎛ ⎛W ⎞ 0.0667 ⎞ 6.74 100μA⎜ ⎟ ⎜ 2.5 − 0.0667 − 0.621 − ⎟0.0667 = 80μA → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠D ⎝ ⎝ L ⎠D VTNC = 0.6 + 0.5 0.133 + 0.6 − 0.6 = 0.641V ⎛W ⎞ ⎛ ⎛W ⎞ 0.0667 ⎞ 7.09 100μA⎜ ⎟ ⎜ 2.5 − 0.133 − 0.641 − ⎟0.0667 = 80μA → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠C ⎝ ⎝ L ⎠C ( ) ( ) 6-38 6.104 The load device remains the same. VL 0.20 V 0.20 = = 0.10V | C , D : VDS = L = = 0.050V 2 4 2 4 ⎛W ⎞ ⎛ ⎛W ⎞ 0.10 ⎞ 4.32 100μA⎜ ⎟ ⎜ 2.5 − .6 − ⎟0.10 = 80μA → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠B ⎝ ⎝ L ⎠B A, B : VDS = VTNA = VTND = 0.6 + 0.5 0.1 + 0.6 − 0.6 = 0.631V ⎛W ⎞ ⎛ ⎛W ⎞ 0.10 ⎞ 4.65 100μA⎜ ⎟ ⎜ 2.5 − 0.10 − 0.631 − ⎟0.10 = 80μA → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠ A⎝ ⎝ L ⎠A ⎛W ⎞ ⎛ ⎛W ⎞ 0.05⎞ 9.17 100μA⎜ ⎟ ⎜ 2.5 − 0.10 − 0.631 − ⎟0.05 = 80μA → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠D ⎝ ⎝ L ⎠D VTNC = 0.6 + 0.5 0.15 + 0.6 − 0.6 = 0.646V ⎛W ⎞ ⎛ ⎛W ⎞ 0.05⎞ 9.53 100μA⎜ ⎟ ⎜ 2.5 − 0.15 − 0.646 − ⎟0.05 = 80μA → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠C ⎝ ⎝ L ⎠C 6.105 Device E and the load device remain the same. In the worst case, for paths BCD or ADE ( ) ( ) VL 0.20 = = 0.0667V 3 3 ⎛W ⎞ ⎛ ⎛W ⎞ 0.0667 ⎞ 6.43 100μA⎜ ⎟ ⎜2.5 − .6 − ⎟0.0667 = 80μA → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠B , E ⎝ ⎝ L ⎠B , E A, B, C , D, E : VDS = VTND = 0.6 + 0.5 0.0667 + 0.6 − 0.6 = 0.621V ⎛W ⎞ ⎛ ⎛W ⎞ 0.0667 ⎞ 6.74 100μA⎜ ⎟ ⎜ 2.5 − 0.0667 − 0.621 − ⎟0.0667 = 80μA → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠D ⎝ ⎝ L ⎠D VTNA = VTNC = 0.6 + 0.5 0.133 + 0.6 − 0.6 = 0.641V ⎛W ⎞ ⎛ ⎛W ⎞ 0.0667 ⎞ 7.09 100μA⎜ ⎟ ⎜2.5 − 0.133 − 0.641 − ⎟0.0667 = 80μA → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠ A, C ⎝ ⎝ L ⎠ A,C ( ) ( ) 6-39 6.106 A 0 (a) 0 1 1 B Y 0 1 1 0 0 1 0 1 (b) Y = A B + AB = A ⊕ B (c) Assuming equal voltage drops (0.10V) across MP and MS : MP must carry one unit of load current with one - half the drain - source ⎛ W⎞ 4.44 voltage (VDS = 0.10V ) of the switching transistor in Fig.6.29(d). → ⎜ ⎟ = 1 ⎝ L ⎠P MS must carry two units of load current with one - half the drain - source ⎛ W⎞ 8.88 voltage (VDS = 0.10V ) of the switching transistor in Fig.6.29(d). → ⎜ ⎟ = 1 ⎝ L ⎠S (d) MS will not change. MP will need to be somewhat larger. (e) Coincidence gate (Exclusive NOR) 6.107 Original design 0.20 mW - 1 mW requires 5 times larger current. ⎛ W⎞ 28.8kΩ 2.22 11.1 (a) R = = 5.76 kΩ ⎜ ⎟ = 5 = 5 1 1 ⎝ L ⎠S ⎛ W⎞ 1 2.98 (b) ⎜ ⎟ = 5 = 1.68 1 ⎝ L ⎠L ⎛ W⎞ 1 1 (c) ⎜ ⎟ = 5 = 5.72 1.14 ⎝ L ⎠L ⎛ W⎞ 1.81 9.05 (d) ⎜ ⎟ = 5 = 1 1 ⎝ L ⎠L ⎛ W⎞ 4.71 23.6 = ⎜ ⎟ =5 1 1 ⎝ L ⎠S ⎛ W⎞ 2.22 11.1 = ⎜ ⎟ =5 1 1 ⎝ L ⎠S ⎛ W⎞ 2.22 11.1 = ⎜ ⎟ =5 1 1 ⎝ L ⎠S ⎛ W⎞ 1.11 5.55 ⎛ W ⎞ 2.22 11.1 (e) ⎜ ⎟ = 5 = = ⎜ ⎟ =5 1 1 1 1 ⎝ L ⎠L ⎝ L ⎠S 6.108 ⎛ W⎞ 1.81 7.24 = | ⎜ ⎟ =4 1 1 ⎝ L ⎠L ⎛ W⎞ ⎛ 2.22 ⎞ 8.88 ⎜ ⎟ = 4⎜ ⎟= 1 ⎝ L ⎠F ⎝ 1 ⎠ ⎡ ⎛ 2.22 ⎞⎤ 26.6 ⎛ W⎞ = 4 ⎢3 ⎜ ⎜ ⎟ ⎟⎥ = 1 ⎝ L ⎠ A− E ⎣ ⎝ 1 ⎠⎦ ⎛ W ⎞ 13.3 1 1 1 | + = →⎜ ⎟ = ⎛ W⎞ ⎛ 26.6 ⎞ 2.22 1 ⎝ L ⎠G ⎜ ⎟ ⎜ ⎟ 4 1 ⎝ L ⎠G ⎝ 1 ⎠ 6-40 6.109 ⎛ W ⎞ 1 ⎛1.81⎞ 1 ⎜ ⎟ = ⎜ ⎟= ⎝ L ⎠ L 4 ⎝ 1 ⎠ 2.21 ⎛ W⎞ 1 ⎡ ⎛ 2.22 ⎞⎤ 1.67 | ⎜ ⎟ = ⎢3 ⎜ ⎟⎥ = 1 ⎝ L ⎠ A− F 4 ⎣ ⎝ 1 ⎠⎦ 6.110 ⎛ W⎞ ⎛ 1.81⎞ 5.43 ⎛ W⎞ ⎛ 6.66 ⎞ 20.0 ⎛ W⎞ ⎛ 3.33 ⎞ 9.99 | ⎜ ⎟ = 3⎜ | ⎜ ⎟ = 3⎜ ⎟ = 3⎜ ⎟= ⎟= ⎟= (a) ⎜ 1 1 1 ⎝ L ⎠L ⎝ 1 ⎠ ⎝ L ⎠ BCD ⎝ 1 ⎠ ⎝ L ⎠A ⎝ 1 ⎠ ⎛ W ⎞ 1 ⎛1.81⎞ ⎛ W⎞ ⎛ W⎞ 1 1 1 ⎛ 6.66 ⎞ 1.33 1 ⎛ 3.33 ⎞ | ⎜ ⎟ = ⎜ | ⎜ ⎟ = ⎜ ⎟ = ⎜ ⎟= ⎟= ⎟= (b) ⎜ 1 ⎝ L ⎠ L 5 ⎝ 1 ⎠ 2.76 ⎝ L ⎠ BCD 5 ⎝ 1 ⎠ ⎝ L ⎠ A 5 ⎝ 1 ⎠ 1.50 6.111 ⎛ W⎞ ⎛ W⎞ ⎛ W⎞ ⎛ 1 ⎞ 1 1 1 1 ⎛1.81⎞ 1 ⎛ 4.44 ⎞ | ⎜ ⎟ = ⎜ | ⎜ ⎟ = 2⎜ ⎟ = ⎜ ⎟= ⎟= ⎟= (a) ⎜ ⎝ L ⎠ L 10 ⎝ 1 ⎠ 5.53 ⎝ L ⎠ AB 10 ⎝ 1 ⎠ 2.25 ⎝ L ⎠CD ⎝ 2.25 ⎠ 1.13 ⎛ W ⎞ 2.5 ⎛1.81⎞ 4.53 ⎛ W⎞ ⎛ W⎞ ⎛ 11.1⎞ 22.2 2.5 ⎛ 4.44 ⎞ 11.1 | ⎜ ⎟ = | ⎜ ⎟ = 2⎜ ⎟ = ⎟= ⎟= ⎟= ⎜ ⎜ (b) ⎜ 1 1 1 1 ⎝ 1 ⎠ ⎝ L ⎠L 1 ⎝ 1 ⎠ ⎝ L ⎠ AB ⎝ L ⎠CD ⎝ 1 ⎠ 6.112 ⎛ W⎞ ⎛ 1.11⎞ 4.44 ⎛ W⎞ ⎛ 6.66 ⎞ 26.6 | ⎜ ⎟ = 4⎜ ⎟ = 4⎜ ⎟= ⎟= (a) ⎜ 1 1 ⎝ L ⎠L ⎝ 1 ⎠ ⎝ L ⎠ ABCDE ⎝ 1 ⎠ ⎛ W ⎞ 1 ⎛1.11⎞ ⎛ W⎞ 1 1 ⎛ 6.66 ⎞ 2.22 | ⎜ ⎟ = ⎜ ⎟ = ⎜ ⎟= ⎟= (b) ⎜ 1 ⎝ L ⎠ L 3 ⎝ 1 ⎠ 2.70 ⎝ L ⎠ ABCDE 3 ⎝ 1 ⎠ 6.113 ⎛ ⎞ 2 2 1 1 ε ⎛W ⎞ " W VGS − VTN ) = μn ox ⎜ ⎟( VGS − VTN ) ( a ) I D = μnCox ⎜ ⎟( 2 2 Tox ⎝ L ⎠ ⎝ L⎠ ⎛W ⎞ 2 1 ε ⎜ ⎟ I* * ID = μn ox ⎜ 2 ⎟( VGS − VTN ) = 2 I D | D = 2 ID 2 Tox ⎜ L ⎟ 2 ⎝2⎠ * = V (2 I )= 2VI = 2 PD - Power dissipation has increased by a factor of two. ( b ) PD 6-41 6.114 For each line : i = C dv | Assume the transition occurs in ΔT seconds generating dt 2.5V a current pulse with constant amplitude I = 10 x10−12 F . ΔT 2.5 x10−11 ΔT Then I avg = = 500μA and P = 64(2.5V )I avg = 64(2.5)(0.50mA)= 80 mW 50ns ΔT ⎛ 3.3 ⎞2 2 ( b ) P ∝ V so P = 80mW ⎜ ⎟ = 139 mW ⎝ 2.5 ⎠ 6.115 C C τ PHL ∝ and τ PLH ∝ KS KL C C" L2 oxWL | For either case, τ PHL ∝ = = μn KS " W μnCox L 6.116 τP = 6.117 PDP 100 fJ 10−13 J = = −4 = 1 ns PD 100μW 10 W 2.5 + 0.20 = 1.35V 2 = 0.25 + 0.23 = 0.48V VH = 2.5V | VL = 0.20V | V50% = V90% = 2.5 − 0.23 = 2.27V | V10% vI : t f = 62 − 55 = 7 ns ( a ) vI : t r = 22.5 − 1.5 = 21 ns | vO : t r = 81 − 58 = 23 ns | vO : t r = 12.5 − 6 = 6.5 ns 2.5 + 7 = 4.8 ns 2 ( b ) τ PHL = 2.5 ns | τ PLH = 7 ns ( c ) τ P = 6.118 ( a) T = 301(τ PHL + τ PLH ) = 602 P = 602(0.1ns) = 60.2 ns 2 (b) An even number of inverters has a potential steady state and may not oscillate. (τ PHL + τ PLH ) = 602τ 6-42 6.119 t r = 2.2 RC = 2.2(28.8 kΩ)(0.5 pF )= 31.7 ns t f ≅ 3.7 RonS C = 3.7(0.5 pF ) 3.7C = = 4.39 ns VGS − VTN ) 2.22 10−4 (2.5 − 0.6) Kn ( τ PLH = 0.69 RC = 0.69(28.8 kΩ)(0.5 pF )= 9.94 ns τ PHL ≅ 1.2 RonS C = ( ) 1.2(0.5 pF ) 1.2C = = 1.78 ns VGS − VTN ) 2.22 10−4 (2.5 − 0.6) Kn ( ( ) τP = 9.94 + 1.78 = 5.86 ns 2 6.120 t r = 2.2 RC = 2.2(28.8 kΩ)(0.5 pF )= 31.7 ns t f ≅ 3.7 RonS C = 3.7(0.5 pF ) 3.7C = = 3.09 ns VGS − VTN ) 2.22 10−4 (3.3 − 0.6) Kn ( τ PLH = 0.69 RC = 0.69(28.8 kΩ)(0.5 pF )= 9.94 ns τ PHL ≅ 1.2 RonS C = 6.121 ( ) 1.2(0.5 pF ) 1.2C = = 1.00 ns VGS − VTN ) 2.22 10−4 (3.3 − 0.6) Kn ( ( ) τP = 9.94 + 1.00 = 5.47 ns 2 Resistive Load : τ P = τ PLH + τ PHL 2 VH = 2.5V VL = 0.20V VTNS = 0.6V τ PLH = 0.69 RC and τ PHL = 1.2 RonS C for RonS = C C 0.526C = = KS KS ( VH − VTNS ) KS (2.5 − 0.6) ⎛ V ⎞ 2.5 − 0.20 2.5 − VL = KS ⎜VGS − VTN − L ⎟VL → KS R = = 6.39 ⎛ R 2⎠ 0.20 ⎞ ⎝ ⎜2.5 − 0.6 − ⎟0.20 2 ⎠ ⎝ ⎡ ⎛ 6.39 ⎞ 0.526⎤ ⎛W ⎞ 1 987 16.5 μA 2.5ns = ( 1 pF ) = and R = 6.47 kΩ ⎢0.69⎜ ⎥ → KS = 987 2 | ⎜ ⎟ = ⎟+ KS ⎦ 2 1 V ⎝ L ⎠ S 60 ⎝ KS ⎠ ⎣ I DDL = (2.5 − 0.20)V = 356μA 6.47 kΩ | P= 2.5(356μA) 2 = 0.444 mW 6.122 Resistive load inverter – λ has very little effect on the results: λ = 0 : t r = 3.8ns t f = 1.3ns τ PLH = 10.0 ns τ PHL = 1.6 ns λ = 0.04/V : t r = 31.6 ns t f = 3.6ns τ PLH = 9.9ns τ PHL = 1.5ns 6.123 6-43 Ignore body effect for simplicity. Equate drain currents to find VL : ' ⎛ 2 VL ⎞ 1 Kn ' 2.5 − VL − 0.6) = 4 Kn ⎜2.5 − 0.6 − 0.6 − ⎟VL → VL = 0.156V ( 2 2 2⎠ ⎝ 1 1 RonL = = = 19.1kΩ KL ( VGS − VTN ) 0.5 60 x10−6 (2.5 − 0.156 − 0.6) ( ) RonS = KS ( VGS − VTN ) 4 60 x10−6 (2.5 − 0.6 − 0.6) 1 = τ PLH ≅ 3.0 RonLC = 3.0(0.5 pF )( 19.1kΩ)= 28.7 ns τ PHL ≅ 1.2 RonS C = 1.2(0.5 pF )(3.21kΩ)= 1.93 ns τP = 28.7 + 1.93 = 15.3 ns 2 t f ≅ 3.7 RonS C = 3.7(0.5 pF )(3.21kΩ) = 5.94 ns t r ≅ 11.9 RonLC = 11.9(0.5 pF )( 19.1kΩ) = 114 ns ( ) 1 = 3.21kΩ 6.124 Ignore body effect for simplicity. Equate drain currents to find VL : VTN = 0.6 → VH = 3.3 − 0.6 = 2.7V −6 ⎛ 2 4 VL ⎞ 1 60 x10 −6 60 x10 ⎜2.7 − 0.6 − ⎟VL = 3.3 − VL − 0.6) ( 2⎠ 2 1 2 ⎝ ( ) ( ) 9VL2 − 39.0VL + 7.29 = 0 → VL = 0.196V RonL = RonS = 0.5 60 x10 4 60 x10 ( −6 ) (3.3 − 0.196 − 0.6) = 1.98 kΩ 1 = 13.3kΩ t r = 11.9 RonLC = 11.9( 13.3kΩ)(0.3 pF )= 47.5 ns 1.98 kΩ)(0.3 pF )= 2.20 ns t f = 3.7 RonS C = 3.7( 12.0 + 0.713 = 6.36 ns 2 ( 1 −6 )(2.7 − 0.6) τ PLH = 3.0 RonLC = 3.0( 13.3kΩ)(0.3 pF ) = 12.0 ns τ PHL = 1.2 RonS C = 1.2( 1.98 kΩ)(0.3 pF )= 0.713 ns τP = 6-44 6.125 2 2 x10−9 s 3.0 RonL + 1.2 RonS τP = = C → 3.0 RonL + 1.2 RonS = = 4000Ω 2 2 10−12 F ⎛ 2 V ⎞ K KS ⎜VGSS − VTNS − L ⎟VL = L ( VGSL − VTNL ) Ignore body effect for simplicity. 2⎠ 2 ⎝ ⎛ 2 0.25 ⎞ KL KS ⎜ 2.5 − 0.6 − 0.6 − 2.5 − 0.25 − 0.6) → KS = 4.63 KL ⎟0.25 = ( 2 2 ⎠ ⎝ τ PLH + τ PHL ( ) 3.0 1.2 + = 4000Ω → KL = 5.04 x10−4 A / V 2 KL (2.5 − .25 − 0.6) 4.63 KL (2.5 − 0.6 − 0.6) ⎛W ⎞ 5.04 x10−4 8.41 = ⎜ ⎟ = −5 1 ⎝ L ⎠ L 6.0 x10 ⎛W ⎞ 8.41 = 38.9 ⎜ ⎟ = 4.63 1 ⎝ L ⎠S 6.126 VH = 2.5V VL = 0.2V RonL = RonS = VGS − VTN ) KL ( 1 1 = = VTNL = 0.6 + 0.5 0.2 + 0.6 − 0.6 = 0.66V ( ) ( 6 x10 −5 ) (4 − 0.2 − 0.66) 1 −5 5.72 = 30.4 kΩ = 3.95kΩ VGS − VTN ) 2.22 6 x10 KS ( t r ≅ 3.7 RonLC = 3.7(0.7 pF )(30.4 kΩ) = 78.7 ns ( ) (2.5 − 0.6) τ PLH ≅ 0.69 RonLC = 0.69(0.7 pF )(30.4 kΩ)= 14.7 ns τ PHL ≅ 1.2 RonS C = 1.2(0.7 pF )(3.95kΩ)= 3.32 ns τP = 14.7 + 3.32 = 9.00 ns 2 t f ≅ 3.7 RonS C = 3.7(0.7 pF )(3.95kΩ) = 10.2 ns 6-45 6.127 3.0V 2.0V 1.0V 0V 3.0V 2.0V 1.0V 0V Results: tf = 1.0 ns, tr = 22.3 ns, τPHL = 0.47 ns, τPLH = 4.0 ns, τP = 4.2 ns 6.128 2 3 x10−9 s 3.6 RonL + 1.2 RonS τP = = C → 3.6 RonL + 1.2 RonS = = 6000Ω 2 2 10−12 F ⎛ 2 V ⎞ K KS ⎜VGSS − VTNS − L ⎟VL = L (−VTNL ) Ignore body effect for simplicity. 2⎠ 2 ⎝ ⎛ 0.25 ⎞ KL 2 KS ⎜ 3 − 0.6 − ⎟0.25 = (3) → KS = 7.91KL 2 2 ⎠ ⎝ 3.6 1.2 + = 6000Ω → KL = 2.11x10−4 A / V 2 KL (3) 7.91KL (3 − 0.6) τ PLH + τ PHL ( ) ⎛W ⎞ 2.11x10−4 3.52 = ⎜ ⎟ = −5 1 ⎝ L ⎠ L 6.0 x10 t r = 8.1RonL ( ) = 12.8 ns C= 3.52( 6 x10 ) (3) 8.1 10−12 −5 ⎛W ⎞ 3.52 = 27.8 ⎜ ⎟ = 7.91 1 ⎝ L ⎠S t f = 3.7 RonS C = ( ) = 0.924 ns 27.8( 6 x10 ) (3 − 0.6) 3.7 10−12 −5 6-46 6.129 2 10−9 s 3.6 RonL + 1.2 RonS τP = = C → 3.6 RonL + 1.2 RonS = = 10.0kΩ 2 2 0.2 x10−12 F ⎛ 2 V ⎞ K KS ⎜VGSS − VTNS − L ⎟VL = L (−VTNL ) Ignore body effect for simplicity. 2⎠ 2 ⎝ ⎛ KL 2 0.20 ⎞ KS ⎜ 3.3 − 0.75 − ⎟0.20 = (2) → KS = 4.08 KL 2 2 ⎠ ⎝ τ PLH + τ PHL ( ) 1.2 3.6 + = 10 kΩ → KL = 1.92 x10−4 A / V 2 KL (2) 4.08 KL (3.3 − 0.75) ⎛W ⎞ 1.92 x10−4 3.19 = ⎜ ⎟ = −5 1 ⎝ L ⎠ L 6.0 x10 I DD = ⎛W ⎞ 3.19 = 13.0 ⎜ ⎟ = 4.08 1 ⎝ L ⎠S 3.3(384μA) 2 KL 1.92 x10−4 2 − V = 2) = 384μA P = = 0.634 mW ( ( TNL ) 2 2 2 6.130 (a) VTN = 0.6 + 0.5 0.20 + 0.6 − 0.6 = 0.660V ( ) 80 x10−6 = ⎛W ⎞ 2 100 x10−6 ⎛W ⎞ 1 ⎜ ⎟ (2.5 − 0.20 − 0.66) → ⎜ ⎟ = 2 ⎝ L ⎠L ⎝ L ⎠ L 1.68 −6 (b) 80 x10 (c) 80 x10 = ⎛W ⎞ 2 100 x10−6 ⎛W ⎞ 1 ⎜ ⎟ (2.5 − 0.20 − 0.6) → ⎜ ⎟ = 2 ⎝ L ⎠L ⎝ L ⎠ L 1.81 ⎛W ⎞ ⎛ ⎛W ⎞ 2.3 ⎞ 1 = 100 x10−6⎜ ⎟ ⎜4 − 0.20 − 0.66 − ⎟2.3 → ⎜ ⎟ = 2 ⎠ ⎝ L ⎠L ⎝ ⎝ L ⎠ L 5.72 ⎛W ⎞ ⎛ ⎛W ⎞ 2.3 ⎞ 1 ⎟ ⎜4 − 0.20 − 0.6 − ⎟2.3 → ⎜ ⎟ = (d ) 80 x10−6 = 100 x10−6⎜ 2 ⎠ ⎝ L ⎠L ⎝ ⎝ L ⎠ L 5.89 −6 (e) V TN = −1 + 0.5 0.20 + 0.6 − 0.6 = −0.940V −6 ( ) ⎛W ⎞ 1.81 2 100 x10 ⎛W ⎞ ⎜ ⎟ (−0.940) → ⎜ ⎟ = 2 1 ⎝ L ⎠L ⎝ L ⎠L ⎛W ⎞ 1.60 2 100 x10−6 ⎛W ⎞ ⎟ (−1) → ⎜ ⎟ = (f ) 80 x10−6 = 2 ⎜ 1 ⎝ L ⎠L ⎝ L ⎠L 80 x10−6 = 6-47 6.131 For VDD = -2.5 V, we have VH = -0.20 V with a power dissipation of 0.20 mW. Since these gates are all ratioed logic design, the ratio of the W/L ratios of the load and switching transistors does not change. We only need to scale both equally to achieve the power level. ⎛ W ⎞ 100 2.22 5.55 = ( a ) RL = 28.8 kΩ | ⎜ ⎟ = 1 ⎝ L ⎠ S 40 1 ⎛W ⎞ 100 1 ⎛ W ⎞ 100 4.71 11.8 1.49 ( b) ⎜ ⎟ = = | ⎜ ⎟ = = 1 1 ⎝ L ⎠ L 60 1.68 ⎝ L ⎠ S 60 1 ⎛W ⎞ 100 1 ⎛ W ⎞ 100 1 5.55 (c ) ⎜ ⎟ = = | ⎜ ⎟ = 2.22 = 1 ⎝ L ⎠ L 60 5.72 2.29 ⎝ L ⎠ S 60 ⎛W ⎞ 100 1.81 4.53 ⎛ W ⎞ 100 2.22 5.55 (d ) ⎜ ⎟ = = | ⎜ ⎟ = = 1 1 ⎝ L ⎠ L 60 1 ⎝ L ⎠ S 60 1 ⎛W ⎞ 100 1.11 2.78 ⎛ W ⎞ 100 2.22 5.55 (e ) ⎜ ⎟ = = | ⎜ ⎟ = = 1 1 ⎝ L ⎠ L 60 1 ⎝ L ⎠ S 60 1 6.132 ⎛2⎞ ⎛ 2 −V ⎞ 1 ⎛ 25 x10−6 ⎞ VL = −2.5 + 0.6 = −1.9 V | ⎜ ⎟ 25x10-6 ⎜ −1.9 − (−0.6)− H ⎟(−VH )= ⎜ ⎟ −2.5 − VH − (−0.6) 2 ⎠ 4⎝ 2 ⎠ ⎝1⎠ ⎝ ( ) ( ) 2 9VH − 24.6VH + 3.61 = 0 → VH = −0.156 V 6.133 Pretend this is an NMOS gate with VDD = 3.3V and VL = 0.33V VH = 3.3 − 0.6 + 0.75 VH + 0.7 − 0.7 → VH = 2.08V VTNL = 0.6 + 0.75 0.33 + 0.7 − 0.7 = 0.734 | I DD = 0.1mW = 30.3μA 3.3V ⎛W ⎞ 0.303 2 40μA ⎛ W ⎞ 1 30.3μA = = ⎜ ⎟ (3.3 − 0.33 − 0.734) → ⎜ ⎟ = 1 3.30 2 ⎝ L ⎠L ⎝ L ⎠L ⎛W ⎞ ⎛ ⎛W ⎞ 1.75 0.33⎞ 30.3μA = 40μA⎜ ⎟ ⎜ 2.08 − 0.60 − ⎟0.33 → ⎜ ⎟ = 2 ⎠ 1 ⎝ L ⎠S ⎝ ⎝ L ⎠S [ ( )] ( ) 6-48 6.134 VL = −VTPL | VL = − −0.6 − 0.75 2.5 − VL + 0.7 − 0.7 → VL = 1.07V ⎛W ⎞ ⎛ K 'p ⎛W ⎞ 2 VDS ⎞ K ⎜ ⎟ ⎜VGS − VTPS − VGSL − VTPL ) ⎟VDS = ⎜ ⎟ ( 2 ⎠ 2 ⎝ L ⎠L ⎝ L ⎠S ⎝ 2 VH − 2.5⎞ 1 ⎛ 1⎞ 3⎛ VH − 2.5) = ⎜ ⎟( VH + VTPL ) ⎜1.07 − 2.5 − (−0.6)− ⎟( 2 ⎝ 3⎠ 1⎝ 2 ⎠ ' p [ ( )] VTPL = −0.6 − 0.75 VH + 0.7 − 0.7 Solving the last two equations iteratively : VH = 2.30 V 6.135 Y is low only when both A and B are high : Y = AB or Y = AB. ( ) Alternatively, Y is high when either A or B is low : Y = A + B = AB 6.136 Y is high only when both A and B are low : Y = AB or Y = A + B 6.137 0.0V -0.5V -1.0V -1.5V VL = -1.90 and VH = -0.156 agree with the hand calculations in Prob. 6.132 6.138 -0.0V 2 0V -1.0V -2.0V 3 0V VH = -0.33 and VL -2.08 agree with the design values in Prob. 6.133. 6-49 6.139 0.0V -0.5V -1.0V -1.5V tr = 16.8 ns, tf = 560 μs, τPLH = 11.7 ns, τPHL = 60 ns, τP = 35.9 ns 6.140 -0.0V 2 0V -1.0V -2.0V -3.0V 0s 0 2us 0 4us 0 6us 0 8us 1 0us 1 2us 1 4us 1 6us 1 8us 2 0us tr = 46 ns, tf = 1.1 s, τPLH = 21 ns, τPHL = 122 ns, τP = 72 ns 6-50 CHAPTER 7 7.1 ' n −14 3.9εo ⎛ cm 2 ⎞ (3.9) 8.854 x10 F / cm K = µ nC = µ n = µn = ⎜ 500 ⎟ Tox Tox V − sec ⎠ 10 x10−9 m( 100cm / m) ⎝ " ox εox ( ) F A µA = 173 x 10−6 2 = 173 2 V − sec V V ⎛ ⎞ µ 200 µA µA " ' K 'p = µ pCox = p Kn =⎜ ⎟173 2 = 69.1 2 µn V V ⎝ 500 ⎠ ' Kn = 173x10−6 7.2 V DD (5 V) B n+ S vI D V SS (0 V) B vo D S p+ PMOS transistor p+ n+ p-well NMOS transistor n+ p+ Ohmic contact n-type substrate Ohmic contact 7.3 ⎛ pA ⎞ A = ⎜ 500 2 ⎟( 1cm x 0.5cm)= 250 pA cm ⎠ ⎝ ⎛ pA ⎞ 100 2 ⎟ 2 x10−4 cm 5 x10−4 cm = 250 + 200 = 450 pA (b) I = I S A + 20 x106 ⎜ cm ⎠ ⎝ (a) I = I S ( ) ( )( ) (c) Same as (b) 7.4 F ⎞⎛ 10mm 0.1cm ⎞ ⎛ 3.9⎜ 8.854 x10 −14 ⎟(1µm ) ⎟⎜ ⎛ ε ox A ⎞ 3.9ε o LW cm ⎠⎝ 2 mm ⎠ ⎝ C = 3⎜ = 0.518 pF =3 ⎟=3 ⎜ t ⎟ t ox 1µm ⎝ ox ⎠ 7.5 (a) VH = 2.5 V, VL = 0 V (b) VH = 1.8 V, VL = 0 V 7-1 7.6 (a) VH = 2.5 V, VL = 0 V (b) Same as (a). VH and VL don't depend upon W/L in a CMOS gate. 7.7 (a) VH = 3.3 V, VL = 0 V (b) Same as (a). VH and VL don't depend upon W/L in a CMOS gate. 7.8 (a) VH = 2.5V | VL = 0V | For MN , VGS = 0, so MN is cut off. For MP , VGS = −2.5, VDS = 0V and VTP = -0.60V. For VDS < VGS - VTP , M P is in the triode region. (b) For M (c) For M N , VGS = 2.5, VDS = 0V and VTN = 0.60V. For VDS < VGS - VTN , M N is in the triode region. , VGS = 1.25, VDS = 1.25 V and VTN = 0.60V. For VDS > VGS - VTN , MN is saturated. For MP , VGS = 0, so M P is cut off. N For MP , VGS = −1.25, VDS = -1.25V and VTP = -0.75V. For VDS > VGS - VTP , MP is saturated. 7.9 (a) VH = 3.3V | VL = 0V | For MN , VGS = 0, so MN is cut off. For MP , VGS = −3.3, VDS = 0V and VTP = -0.75V. For VDS < VGS - VTP , M P is in the triode region. (b) For M (c) For M N , VGS = 3.3, VDS = 0V and VTN = 0.75V. For VDS < VGS - VTN , MN is in the triode region. , VGS = 1.65, VDS = 1.65 V and VTN = 0.75V. For VDS > VGS - VTN , MN is saturated. For MP , VGS = 0, so MP is cut off. N For MP , VGS = −1.65, VDS = -1.65V and VTP = -0.75V. For VDS > VGS - VTP , MP is saturated. 7.10 (a) VH = 0 V, VL = -5.2 V (b) Same as (a). VH and VL don't depend upon W/L in a CMOS gate. 7-2 7.11 For vI = vO, both transistors will be saturated since vGS = vDS for each device. Equating the drain currents with Kn = Kp yields: (a) Both transistors are saturated with VDS = VGS K 2 2 Kn vI − VTN ) = p (v I − VDD − VTP ) so vI − VTN = VDD − vI + VTP ( 2 2 V + VTN + VTP 2.5 + 0.6 − .6 vO = vI = DD = = 1.25V 2 2 2 2 K 100µA ⎛ 2 ⎞ 1.25 − 0.6) = 42.3µA (b) I DN = 2n (vI − VTN ) = 2 ⎜ ⎟( ⎝ 1⎠ K 2 2 40µA ⎛ 5 ⎞ Checking I DP = p (vI − VDD − VTp ) = 1.25 − 2.5 + 0.6) = 42.3µA ⎜ ⎟( 2 ⎝1⎠ 2 (c) For K n = 2.5 K p , K 2.5 K p 2 2 vI − VTN ) = p (vI − VDD − VTP ) or 1.58(vI − VTN )= VDD − vI + VTP ( 2 2 V + 1.58VTN + VTP 2.5 + 1.58(0.6)+ (−0.6) vO = vI = DD = = 1.104V 2.58 2.58 2 100µA ⎛ 2 ⎞ 1.104 − 0.6) = 25.4µA | Check by finding IDP : (d ) I DN = 2 ⎜ ⎟( ⎝ 1⎠ 2 40µA ⎛ 2 ⎞ I DP = 1.104 − 2.5 + 0.6) = 25.3µA ⎜ ⎟( 2 ⎝ 1⎠ 7.12 (a) For vI = vO, both transistors will be saturated since vGS = vDS for each device. Equating the drain currents with Kn = Kp yields: K 2 2 Kn vI − VTN ) = p (vI − VDD − VTP ) and vI − VTN = VDD − v I + VTP i () ( 2 2 V + VTN + VTP 3.3 + 0.75 − 0.75 vO = vI = DD = = 1.65V 2 2 2 2 K 100µA ⎛ 2 ⎞ 1.65 − 0.75) = 81µA ⎟( (ii) I DN = 2n (vI − VTN ) = 2 ⎜ ⎝1⎠ 2 2 K 40µA ⎛ 5 ⎞ Checking : I DP = P (vI − VDD + VTP ) = 1.65 − 3.3 + 0.75) = 81µA ⎜ ⎟( 2 ⎝ 1⎠ 2 For Kn = 2.5 Kp, 7-3 2.5 K p K 2 2 vI − VTN ) = p (v I − VDD − VTP ) so 1.58(vI − VTN )= VDD − vI + VTP ( 2 2 V + 1.58VTN + VTP 3.3 + 1.58(0.75)+ (−0.75) vO = vI = DD = = 1.448V 2.58 2.58 2 100µA ⎛ 2 ⎞ 1.448 − 0.75) = 48.7µA | Check by finding I DP : ⎟( (iv ) I DN = 2 ⎜ ⎝1⎠ 2 40µA ⎛ 2 ⎞ I DP = 1.448 − 3.3 + 0.75) = 48.6µA ⎜ ⎟( 2 ⎝1⎠ (iii ) (b) For vI = vO, both transistors will be saturated since vGS = vDS for each device. Equating the drain currents with Kn = Kp yields: K 2 2 Kn vI − VTN ) = p (vI − VDD − VTP ) and vI − VTN = VDD − v I + VTP i () ( 2 2 V + VTN + VTP 2.5 + 0.6 − 0.6 vO = vI = DD = = 1.25V 2 2 2 2 K 100µA ⎛ 2 ⎞ 1.25 − 0.6) = 42.3µA ⎟( (ii ) I DN = 2n (vI − VTN ) = 2 ⎜ ⎝1⎠ K 2 2 100µA ⎛ 2 ⎞ Checking : I DP = p (vI − VDD + VTP ) = 1.25 − 2.5 + 0.6) = 42.3µA ⎜ ⎟( 2 ⎝1⎠ 2 For Kn = 2.5 Kp, 2.5 K p K 2 2 vI − VTN ) = p (v I − VDD − VTP ) and 1.58(v I − VTN ) = VDD − v I + VTP ( 2 2 V + 1.58VTN + VTP 2.5 + 1.58(0.60)+ (−0.60) vO = vI = DD = = 1.104V 2.58 2.58 2 100µA ⎛ 2 ⎞ 1.104 − 0.60) = 25.4µA | Check by finding I DP : ⎟( (iv ) I DN = 2 ⎜ ⎝1⎠ 2 40µA ⎛ 2 ⎞ I DP = 1.104 − 2.5 + 0.60) = 25.3µA ⎜ ⎟( 2 ⎝1⎠ (iii ) 7-4 7.13 (a) For vI = vO, both transistors will be saturated since vGS = vDS for each device. Equating the drain currents with Kn = Kp yields: K 2 2 Kn vI − VTN ) = p (vI − VDD − VTP ) and vI − VTN = VDD − v I + VTP i () ( 2 2 V + VTN + VTP 1.8 + 0.5 − 0.5 vO = vI = DD = = 0.90V 2 2 2 2 K 100µA ⎛ 2 ⎞ ⎟(0.9 − 0.5) = 16.0µA (ii ) I DN = 2n (vI − VTN ) = 2 ⎜ ⎝1⎠ 2 2 K 40µA ⎛ 5 ⎞ Checking : I DP = P (vI − VDD − VTP ) = ⎜ ⎟(0.9 − 1.8 + 0.5) = 16.0µA 2 ⎝ 1⎠ 2 For Kn = 2.5 Kp, 2.5 K p K 2 2 vI − VTN ) = p (v I − VDD − VTP ) so 1.58(vI − VTN )= VDD − vI + VTP ( 2 2 V + 1.58VTN + VTP 1.8 + 1.58(0.5)+ (−0.5) vO = vI = DD = = 0.810V 2.58 2.58 2 100µA ⎛ 2 ⎞ ⎟(0.810 − 0.5) = 96.2µA | Check by finding I DP : (iv ) I DN = 2 ⎜ ⎝1⎠ 2 40µA ⎛ 2 ⎞ I DP = ⎜ ⎟(0.8101 − 1.8 + 0.5) = 96.0µA 2 ⎝1⎠ (iii ) (b) For vI = vO, both transistors will be saturated since vGS = vDS for each device. Equating the drain currents with Kn = Kp yields: K 2 2 Kn vI − VTN ) = p (vI − VDD − VTP ) and vI − VTN = VDD − v I + VTP i () ( 2 2 V + VTN + VTP 2.5 + 0.75 − 0.65 vO = vI = DD = = 1.30V 2 2 2 2 K 100µA ⎛ 2 ⎞ 1.30 − 0.75) = 30.3µA ⎟( (ii ) I DN = 2n (vI − VTN ) = 2 ⎜ ⎝1⎠ K 2 2 40µA ⎛ 5 ⎞ Checking : I DP = p (vI − VDD + VTP ) = 1.30 − 2.5 + 0.65) = 30.3µA ⎜ ⎟( 2 ⎝ 1⎠ 2 7-5 For Kn = 2.5 Kp, 2.5 K K 2 2 (iii ) 2 p (vI − VTN ) = 2p (vI − VDD − VTP ) and 1.58(vI − VTN )= VDD − vI + VTP V + 1.58VTN + VTP 2.5 + 1.58(0.75)+ (−0.65) vO = vI = DD = = 1.176V 2.58 2.58 2 100µA ⎛ 2 ⎞ 1.176 − 0.75) = 18.2µA | Check by finding I DP : ⎟( (iv ) I DN = 2 ⎜ ⎝1⎠ 2 40µA ⎛ 2 ⎞ I DP = 1.176 − 2.5 + 0.65) = 18.2µA ⎜ ⎟( 2 ⎝1⎠ (c) For vI = vO, both transistors will be saturated since vGS = vDS for each device. Equating the drain currents with Kn = Kp yields: K 2 2 Kn vI − VTN ) = p (vI − VDD − VTP ) and vI − VTN = VDD − v I + VTP i () ( 2 2 V + VTN + VTP 2.5 + 0.65 − 0.75 vO = vI = DD = = 1.20V 2 2 2 2 K 100µA ⎛ 2 ⎞ 1.20 − 0.65) = 30.3µA ⎟( (ii ) I DN = 2n (vI − VTN ) = 2 ⎜ ⎝1⎠ K 2 2 40µA ⎛ 5 ⎞ Checking : I DP = p (vI − VDD + VTP ) = 1.20 − 2.5 + 0.75) = 30.3µA ⎜ ⎟( 2 ⎝ 1⎠ 2 For Kn = 2.5 Kp, 2.5 K p K 2 2 vI − VTN ) = p (v I − VDD − VTP ) and 1.58(v I − VTN ) = VDD − v I + VTP ( 2 2 V + 1.58VTN + VTP 2.5 + 1.58(0.65)+ (−0.75) vO = vI = DD = = 1.076V 2.58 2.58 2 100µA ⎛ 2 ⎞ 1.076 − 0.65) = 18.2µA | Check by finding I DP : ⎟( (iv ) I DN = 2 ⎜ ⎝1⎠ 2 40µA ⎛ 2 ⎞ I DP = 1.076 − 2.5 + 0.75) = 18.2µA ⎜ ⎟( 2 ⎝1⎠ (iii ) 7.14 *PROBLEM 7.14 - CMOS INVERTER TRANSFER CHARACTERISTICS VIN 1 0 DC 0 VDD 3 0 DC 2.5 M1 2 1 0 0 MOSN W=2U L=1U M2 2 1 3 3 MOSP W=2U L=1U .DC VIN 0 2.5 .01 *.DC VIN 2.16 2.17 .0001 .MODEL MOSN NMOS KP=10E-5 VTO=0.6 GAMMA=0 .MODEL MOSP PMOS KP=4E-5 VTO=-0.6 GAMMA=0 7-6 .PRINT DC V(2) .END Result: vI = 1.104 V K 2.5 K p 2 2 vI − VTN ) = p ( VDD − vI + VTP ) and 1.58(vI − VTN )= VDD − vI + VTP ( 2 2 V + 1.58VTN + VTP 2.5 + 1.58(0.6)+ (−0.6) vO = vI = DD = = 1.10 V 2.58 2.58 7.15 (a) VH = 3.3 V. For vO = VL , assume MP is saturated and MN is in the triode region. ⎛ ⎞⎛ K 'p ⎛1⎞ 2 VL ⎞ ' 4 ⎜ ⎟(−3.3 + 0.6) = Kn⎜ ⎟⎜3.3 − 0.6 + ⎟VL 2 ⎝1⎠ 2⎠ ⎝ 1 ⎠⎝ −5 ⎛ 7.29 ⎞ ⎛ 4 x10 VL ⎞ 2 = 2.7 + ⎜ ⎟ ⎜ ⎟VL and rearranging : VL + 5.4VL − 0.729 = 0 −5 2⎠ 2 10 x10 ⎝ 4 ⎠ ⎝ ( ) VL = 0.132V . Checking the assumptions - For MN , 3.3 - 0.6 > 0.132. Triode region is correct. For MP , VGS - VTP = -3.3 + 0.6 = -2.7V and VDS = 0.132 - 3.3 = -3.17V. Saturation region operation is correct. H (b) V = 2.5 V. For vO = VL , assume MP is saturated and MN is in the triode region. ⎛ ⎞⎛ K 'p ⎛1⎞ 2 VL ⎞ ' 4 ⎜ ⎟(−2.5 + 0.6) = Kn⎜ ⎟⎜2.5 − 0.6 + ⎟VL 2 ⎝1⎠ 2⎠ ⎝ 1 ⎠⎝ −5 ⎛ 3.61⎞ ⎛ 4 x10 VL ⎞ 2 = 1.9 + ⎜ ⎟ ⎜ ⎟VL and rearranging : VL + 3.8VL − 0.361 = 0 −5 2⎠ 2 10 x10 ⎝ 4 ⎠ ⎝ ( ) VL = 0.0928V . Checking the assumptions - For MN , 2.5 - 0.6 > 0.0928. Triode region is correct. For MP , VGS - VTP = -2.5 + 0.6 = -1.9V and VDS = 0.0928 - 2.5 = -2.41V. Saturation region operation is correct. 7-7 7.16 For the NMOS device 1.5 mA +5 V + 0.6 V - W 0.6 ⎟ ⎜ 5 − 0.6 − ⎟0.6 = 1.5x10 (100 x10 )⎜ 2 ⎠ ⎝ L⎠ ⎝ −6 n ⎛ ⎞⎛ ⎞ −3 ⎛W ⎞ 61.0 ⎜ ⎟ = 1 ⎝ L ⎠n For the PMOS device +5 V + 2.6 V 60 µ A ( ⎛W ⎞ ⎛ 2.6 ⎞ −5 40 x10−6 ⎜ ⎟ ⎜ 5 − 0.6 − ⎟2.6 = 6 x10 L 2 ⎝ ⎠ p⎝ ⎠ ) ⎛W ⎞ 1 ⎜ ⎟ = ⎝ L ⎠ p 5.37 7-8 7.17 +2.5 V vO +2.5 V . V V2 Therefore the output will be forced below VDD /2. 2 Kn = 2000 µA and K p = 1600 µA For both transistors, VGS − VTN = 1.9V . Assume that both devices are in the linear region. ⎛ 40 ⎞ ⎛ ⎛ 20 ⎞ VO − 2.5 ⎞ -5 VO − 2.5)= ⎜ ⎟ 10x10-5 ⎜ ⎟ 4x10 ⎜ −2.5 + 0.6 − ⎟( 2 ⎠ ⎝1⎠ ⎝ ⎝1⎠ ( ) ( V 2.5 − 0.6 − ⎟V )⎜ 2⎠ ⎝ O ⎛ ⎞ O Rearranging : VO2 − 14.2VO + 13 = 0 ⇒ VO = 0.9836V | VDSN = 0.984V | VDSP = −1.52V and the assumed operating regions are correct. ⎛ 40 ⎞ ⎛ 0.9836 − 2.5 ⎞ I DP = ⎜ ⎟ 4x10-5 ⎜ −2.5 + 0.6 − ⎟(0.9836 − 2.5)= 2.77 mA, and checking 2 ⎝1⎠ ⎝ ⎠ ⎛ 20 ⎞ ⎛ 0.9836 ⎞ I DN = ⎜ ⎟ 10x10-5 ⎜ 2.5 − 0.6 − ⎟0.9836 = 2.77 mA. 2 ⎠ ⎝1⎠ ⎝ ( ) ( ) 7.18 KR = 2(2.5)(2.5 − 0.6 − 0.6) 2.5 − 2.5(0.6)− 0.6 Kn = 2.5 | VIH = − = 1.22V Kp 2.5 − 1 2.5 − 1 1 + 3 2.5 ( ) ( ) VOL = VIL VOH (2.5 + 1)(1.22)− 2.5 − 2.5(0.6)+ 0.6 = 0.174V 2(2.5) 2( 2.5 ) (2.5 − 0.6 − 0.6) 2.5 − 2.5(0.6)− 0.6 = − = 0.902V 2.5 − 1 (2.5 − 1)( 2.5 + 3) (2.5 + 1)(0.902)+ 2.5 − 2.5(0.6)+ 0.6 = 2.38V = ( ) NMH = VOH 2 − VIH = 2.38 − 1.22 = 1.16 V NML = VIL − VOL = 0.902 − 0.174 = 0.728 V 7-9 7.19 VDD − VTN − 3VTP 3.3-0.75-3(-0.75) = = 1.20 V 4 4 V + 3VTN + VTP 3.3+3(0.75)-0.75 and NM L = DD = = 1.20 V 4 4 2(2.5)(3.3 − 0.75 − 0.75) 3.3 − 2.5(0.75)− 0.75 K − = 1.61V (b) KR = K n = 2.5 | VIH = 2 . 5 − 1 p (2.5 − 1) 1 + 3(2.5) (a ) For K R = 1: NM H = VOL = VIL VOH (2.5 + 1)(1.61)− 3.3 − 2.5(0.75)+ 0.75 = 0.242V 2(2.5) 2( 2.5 ) (3.3 − 0.75 − 0.75) 3.3 − 2.5(0.75)− 0.75 = − = 1.17V 2 . 5 − 1 2 . 5 − 1 2 . 5 + 3 ( )( ) (2.5 + 1)(1.17)+ 3.3 − 2.5(0.75)+ 0.75 = 3.14V = ( ) NM H = VOH 7.20 4.0V 2 − VIH = 3.14 − 1.61 = 1.53 V NM L = VIL − VOL = 1.17 − 0.242 = 0.928V 3.0V 2.0V 1.0V 0V 0V V(M2:d) 0.5V V(M4:d) V(M6:d) 1.0V 1.5V V_VI 2.0V 2.5V 3.0V 3.5V 7-10 7.21 (a) CMOS Noise Margins (VDD = 3.3 V, VTN = 0.75 V, VTP = -.75 V) 2.5 2 1.5 NMH NML 1 0.5 0 0 2 4 6 KR 8 10 12 (b) CMOS Noise Margins (VDD = 2.0 V, VTN = 0.5 V, VTP = -0.5 V) 1.4 1.2 1 0.8 NML NMH 0.6 0.4 0.2 0 0 2 4 6 KR 8 10 12 7-11 7.22 (a) t r ≅ 3.6 RonP C = 3.6(0.25 pF ) 3.6C = = 2.36 ns K p VGS − VTP 5 4 x10−5 −2.5 + 0.6 t f ≅ 3.6 RonN C = 3.6(0.25 pF ) 3.6C = = 2.36 ns Kn ( VGS − VTN ) 2 10−4 (2.5 − 0.6) ( ) ( ) tr = 0.788 ns 3 t 0.788 + 0.788 τ PHL ≅ 1.2 RonN C = f = 0.788 ns τ P = = 0.788 ns 2 3 3.6(0.25 pF ) 3.6C (b) tr ≅ 3.6 RonP C = K V − V = 5 4 x10−5 −2.0 + 0.6 = 3.21 ns p GS TP τ PLH = 1.2 RonP C = t f ≅ 3.6 RonN C = 3.6(0.25 pF ) 3.6C = = 3.21 ns VGS − VTN ) 2 10−4 (2 − 0.6) Kn ( ( ) ( ) tr = 1.07 ns 3 t 1.07 + 1.07 τ PHL ≅ 1.2 RonN C = f = 1.07 ns τ P = = 1.07 ns 2 3 3.6(0.25 pF ) 3.6C (c) tr ≅ 3.6 RonP C = K V − V = 5 4 x10−5 −1.8 + 0.6 = 3.75 ns p GS TP τ PLH = 1.2 RonP C = t f ≅ 3.6 RonN C = 3.6(0.25 pF ) 3.6C = = 3.75 ns Kn ( VGS − VTN ) 2 10−4 ( 1.8 − 0.6) ( ) ( ) tr = 1.25 ns 3 t 1.25 + 1.25 τ PHL ≅ 1.2 RonN C = f = 1.25 ns τ P = = 1.25 ns 2 3 τ PLH = 1.2 RonP C = 7.23 t r ≅ 3.6 RonP C = t f ≅ 3.6 RonN C = 3.6(0.5 pF ) 3.6C = = 11.9 ns K p VGS − VTP 2 4 x10−5 −2.5 + 0.6 3.6(0.5 pF ) 3.6C = = 4.74 ns VGS − VTN ) 2 10−4 (2.5 − 0.6) Kn ( ( ) ( ) tr = 3.96 ns 3 t 3.96 + 1.58 τ PHL ≅ 1.2 RonN C = f = 1.58 ns τ P = = 2.77 ns 2 3 τ PLH = 1.2 RonP C = 7-12 7.24 t r ≅ 3.6 RonP C = t f ≅ 3.6 RonN C = 3.6(0.15 pF ) 3.6C = = 1.42 ns K p VGS − VTP 5 4 x10−5 −2.5 + 0.6 3.6(0.15 pF ) 3.6C = = 1.42 ns VGS − VTN ) 2 10−4 (2.5 − 0.6) Kn ( ( ) ( ) tr = 0.474 ns 3 t 0.474 + 0.474 τ PHL ≅ 1.2 RonN C = f = 0.474 ns τ P = = 0.474 ns 2 3 τ PLH = 1.2 RonP C = 7.25 t r ≅ 3.6 RonP C = t f ≅ 3.6 RonN C = 3.6(0.2 pF ) 3.6C = = 1.41 ns K p VGS − VTP 5 4 x10−5 −3.3 + 0.75 3.6(0.2 pF ) 3.6C = = 1.41 ns VGS − VTN ) 2 10−4 (3.3 − 0.75) Kn ( ( ) ( ) tr = 0.470 ns 3 t 0.470 + 0.470 τ PHL ≅ 1.2 RonN C = f = 0.470 ns τ P = = 0.470 ns 2 3 τ PLH = 1.2 RonP C = 7.26 For the symmetrical design, τ PLH = τ PHL and τ P = τ PHL 3ns = ⎛W ⎞ −6 ⎜ ⎟ 100 x10 (2.5 − 0.6) ⎝ L ⎠n 1.2( 1pF) ( ) ⎛ W ⎞ 2.11 ⎛W ⎞ ⎛ W ⎞ 5.26 →⎜ ⎟ = | ⎜ ⎟ = 2.5⎜ ⎟ = 1 1 ⎝ L ⎠n ⎝ L ⎠p ⎝ L ⎠n t r = t f = 3τ PHL = 9.00 ns 7-13 7.27 (a ) For the symmetrical design, τ PLH = τ PHL and τ P = τ PHL 1ns = ⎛W ⎞ −6 ⎜ ⎟ 100 x10 (2.5 − 0.6) ⎝ L ⎠n 1.2( 10pF) ( ) ⎛ W ⎞ 63.2 ⎛W ⎞ ⎛ W ⎞ 158 →⎜ ⎟ = | ⎜ ⎟ = 2.5⎜ ⎟ = 1 1 ⎝ L ⎠n ⎝ L ⎠p ⎝ L ⎠n t r = t f = 3τ PHL = 3.00 ns (b) 1ns = ⎛W ⎞ 1.2( 10pF) −6 ⎜ ⎟ 100 x10 (3.3 − 0.7) ⎝ L ⎠n ( ) ⎛W ⎞ 46.2 ⎛W ⎞ ⎛W ⎞ 115 →⎜ ⎟ = | ⎜ ⎟ = 2.5⎜ ⎟ = 1 1 ⎝ L ⎠n ⎝ L ⎠p ⎝ L ⎠n t r = t f = 3τ PHL = 3.00 ns 7.28 For the symmetrical design, τ PLH = τ PHL and τ P = τ PHL 0.2ns = ⎛W ⎞ −6 1.5 − 0.5) ⎜ ⎟ 100 x10 ( ⎝ L ⎠n 1.2(0.1pF ) ( ) ⎛ W ⎞ 6.00 ⎛W ⎞ ⎛ W ⎞ 15.0 →⎜ ⎟ = | ⎜ ⎟ = 2.5⎜ ⎟ = 1 1 ⎝ L ⎠n ⎝ L ⎠p ⎝ L ⎠n t r = t f = 3τ PHL = 0.600 ns 7.29 For the symmetrical design, τ PLH = τ PHL and τ P = τ PHL 0.4ns = ⎛W ⎞ −6 ⎜ ⎟ 100 x10 (2.5 − 0.6) ⎝ L ⎠n 1.2(0.1pF) ( ) ⎛ W ⎞ 1.58 ⎛W ⎞ ⎛ W ⎞ 3.95 →⎜ ⎟ = | ⎜ ⎟ = 2.5⎜ ⎟ = 1 1 ⎝ L ⎠n ⎝ L ⎠p ⎝ L ⎠n t r = t f = 3τ PHL = 1.20 ns 7.30 *PROBLEM 7.30 - CMOS INVERTER DELAY *SIMULATION USES THE MODELS IN APPENDIX B VIN 1 0 PULSE (0 2.5 0 0.1N 0.1N 10N 20N) VDD 3 0 DC 2.5 M1 2 1 0 0 MOSN W=4U L=2U AS=16P AD=16P M2 2 1 3 3 MOSP W=10U L=2U AS=40P AD=40P CL 2 0 100FF .OP .TRAN 0.1N 20N .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99 +LAMBDA=.02 TOX=41.5N +CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5 +LAMBDA=.05 TOX=41.5N 7-14 +CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P .PRINT TRAN V(2) .PROBE V(1) V(2) .END Results: tr = 1.7 ns, tf = 2.25 ns, τPHL = 1.1 ns, τPLH = 0.9 ns τ PHL = 1.2 RonnC τ PHL τ PLH = 1.2 RonpC 1.1x10−9 ⎛ 2 ⎞ −6 C1 = = ⎜ ⎟ 50 x10 (2.5 − 0.91)= 146 fF | Inverter is symmetrical, so 1.29 Ronn 1.2 ⎝ 1 ⎠ τ 9 x10−10 ⎛ 5 ⎞ 146 + 130 −6 fF = 138 fF C2 = PLH = ⎜ ⎟ 20 x10 (2.5 − 0.77) = 130 fF | C = 1.2 Ronp 1.2 ⎝ 1 ⎠ 2 ( ) ( ) 7.31 *PROBLEM 7.31 - FIVE CASCADED INVERTERS *SIMULATION USES THE MODELS IN APPENDIX B VDD 1 0 DC 2.5 VIN 2 0 PULSE (0 2.5 0 0.1N 0.1N 10N 20N) * MN1 3 2 0 0 MOSN W=16U L=2U AS=64P AD=64P MP1 3 2 1 1 MOSP W=40U L=2U AS=160P AD=160P *AS=4UM*W - AD=4UM*W CL1 3 0 100FF * MN2 4 3 0 0 MOSN W=16U L=2U AS=64P AD=64P MP2 4 3 1 1 MOSP W=40U L=2U AS=160P AD=160P CL2 4 0 100FF * MN3 5 4 0 0 MOSN W=16U L=2U AS=64P AD=64P MP3 5 4 1 1 MOSP W=40U L=2U AS=160P AD=160P CL3 5 0 100FF * MN4 6 5 0 0 MOSN W=16U L=2U AS=64P AD=64P MP4 6 5 1 1 MOSP W=40U L=2U AS=160P AD=160P CL4 6 0 100FF * MN5 7 6 0 0 MOSN W=16U L=2U AS=64P AD=64P MP5 7 6 1 1 MOSP W=40U L=2U AS=160P AD=160P CL5 7 0 100FF .OP .TRAN 0.025N 20N .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99 +LAMBDA=.02 TOX=41.5N +CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5 +LAMBDA=.05 TOX=41.5N +CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P 7-15 .PROBE V(2) V(3) V(5) V(6) .END First inverter : t r = 0.83 ns, t f = 1.4 ns, τ PLH = 0.35 ns, τ PHL = 0.42 ns Fourth inverter : t r = 0.96 ns, t f = 1.0 ns, τ PLH = 0.64 ns, τ PHL = 1.1 ns τ PHL = 1.2 RonnC C1 = C2 = τ PLH = 1.2 RonpC τ PHL 1.2 Ronn = τ PLH 1.2 Ronp 0.42 x10−9 ⎛ 8 ⎞ −6 ⎜ ⎟ 50 x10 (2.5 − 0.91)= 223 fF | The inverter is symmetrical, so 1 1.2 ⎝ ⎠ −9 ⎛ 0.35 x10 20 ⎞ 223 + 202 −6 = fF = 212 fF ⎜ ⎟ 20 x10 (2.5 − 0.77)= 202 fF | C = 2 1.2 ⎝ 1 ⎠ ( ) ( ) The average capacitance of 212 fF that is required to fit the results is consistent with the device capacitances calculated by SPICE. The approximate 3:1 relationship holds between rise/fall times and the propagation delay times in the first inverter. The first inverter response is faster than that of the fourth inverter because of the rapid rise and fall times on the input signal. The first inverter response is closest to our model used for hand calculations. However, the response of inverter four would be more representative of the actual logic situation. 7.32 (a) 100W = 1µW / gate 100 x10 6 gates (b) I = 100W = 55.6 A 1.8V 7.33 (a) 5W = 2.5µW / gate 2 x106 gates 2.5 x10-6 2.52 5 x106 2 P = CVDD f ;C = 2.5 x10-6 3.32 5 x106 ( ) = 45.9 fF ( b) C = ( ) = 80.0 fF 7.34 ⎛ pA ⎞ A = ⎜ 400 2 ⎟(0.5cm x 1cm)= 200 pA cm ⎠ ⎝ ⎛ pA ⎞ 150 2 ⎟ 2.5 x10−4 cm 1x10−4 cm = 200 + 281 = 481 pA (b) I = I S A + 75x106 ⎜ cm ⎠ ⎝ (a ) I = I S ( ) ( )( ) (c) Same as (b) 7.35 2 f = 64 25x10-12 2.52 ( a ) P = 64CVDD ( 1 )( )10 −8 = 1.00 W ( b ) P = 64 25 x10-12 3.32 ( 1 )( )10 −8 = 1.74 W 7.36 7-16 Peak current occurs for vO = v I . Assume both transistors are saturated since vO = vI . 2 2 20 ⎛100 x10−6 ⎞ 20 ⎛ 40 x10−6 ⎞ v − V = ⎜ ⎟( I ⎜ ⎟(v I − VDD − VTP ) → 1.58(v I − VTN ) = VDD − v I + VTP TN ) 2 1⎝ 1⎝ 2 ⎠ ⎠ (a) v 3.3 + 1.58(0.6)+ (−0.6) VDD + 1.58VTN + VTP = = 1.414 V 2.58 2.58 2 20 ⎛ 100 x10−6 ⎞ iD = ⎜ 1.414 − 0.6) = 663 µA ⎟( 2 1⎝ ⎠ O = vI = Checking the current : iD = (b) v O = vI = 2.5 + 1.58(0.6)+ (−0.6) 2.58 2 20 ⎛ 40 x10−6 ⎞ 1.414 − 3.3 + 0.6) = 662 µA | Within roundoff error. ⎜ ⎟( 1⎝ 2 ⎠ = 1.104 V | iD = 2 20 ⎛ 100 x10−6 ⎞ 1.104 − 0.6) = 254 µA ⎜ ⎟( 2 1⎝ ⎠ 2 20 ⎛ 40 x10−6 ⎞ 1.104 − 2.5 + 0.6) = 253 µA | Within roundoff error. Checking the current : iD = ⎜ ⎟( 1⎝ 2 ⎠ 7.37 For a symmetrical inverter, the peak current occurs for vO = v I = VDD . 2 Assume both transistors are saturated since vO = vI . 2 2 2 ⎛100 x10−6 ⎞ 5 ⎛ 40 x10−6 ⎞ ⎜ ⎟(vI − VTN ) = ⎜ ⎟(v I − VDD − VTP ) → (vI − VTN )= VDD − vI + VTP 1⎝ 1⎝ 2 ⎠ 2 ⎠ (a) v iDN VDD + VTN + VTP 3.3 + (0.7)+ (−0.7) = = 1.65 V 2 2 2 2 2 ⎛100 x10−6 ⎞ 5 ⎛ 40 x10−6 ⎞ = ⎜ 1.65 − 0.7) = 90.3 µA Checking : iDP = ⎜ 1.65 − 3.3 + 0.7) = 90.3 µA ⎟( ⎟( 1⎝ 1⎝ 2 ⎠ 2 ⎠ O = vI = (b) v iDN VDD + VTN + VTP 2 + (0.5)+ (−0.5) = = 1.00 V 2 2 2 2 2 ⎛100 x10−6 ⎞ 5 ⎛ 40 x10−6 ⎞ = ⎜ 1.00 − 0.5) = 25.0 µA Checking : iDP = ⎜ 1.0 − 2 + 0.5) = 25.0 µA ⎟( ⎟( 1⎝ 1⎝ 2 ⎠ 2 ⎠ O = vI = 7-17 7.38 For a symmetrical inverter, the peak current occurs for vO = v I = VDD . 2 Assume both transistors are saturated since vO = vI . 2 2 2 ⎛100 x10−6 ⎞ 5 ⎛ 40 x10−6 ⎞ ⎜ ⎟(vI − VTN ) = ⎜ ⎟(v I − VDD − VTP ) → (vI − VTN )= VDD − vI + VTP 2 1⎝ 1⎝ 2 ⎠ ⎠ (a) v iDN VDD + VTN + VTP 2.5 + (0.7)+ (−0.7) = = 1.25 V 2 2 2 2 2 ⎛100 x10−6 ⎞ 5 ⎛ 40 x10−6 ⎞ = ⎜ 1.25 − 0.7) = 30.3 µA Checking : iDP = ⎜ 1.25 − 2.5 + 0.7) = 30.3 µA ⎟( ⎟( 2 1⎝ 1⎝ 2 ⎠ ⎠ O = vI = (b) v iDN VDD + VTN + VTP 2.5 + (0.65)+ (−0.55) = = 1.30 V 2 2 2 2 2 ⎛100 x10−6 ⎞ 5 ⎛ 40 x10−6 ⎞ 1.30 − 0.65 1.3 − 2.5 + 0.55) = 42.3 µA = ⎜ = 42.3 µ A Checking : i = ⎟( ⎜ ⎟( ) DP 2 1⎝ 1⎝ 2 ⎠ ⎠ O = vI = 7.39 For the inverter, the peak current occurs for vO = vI . Assume both transistors are saturated since vO = vI . 2 2 2 ⎛100 x10−6 ⎞ 5 ⎛ 40 x10−6 ⎞ ⎜ ⎟(vI − VTN ) = ⎜ ⎟(v I − VDD − VTP ) → (vI − VTN )= VDD − vI + VTP 2 1⎝ 1⎝ 2 ⎠ ⎠ (a) v iDN VDD + VTN + VTP 2 + (0.55)+ (−0.45) = = 1.05 V 2 2 2 2 2 ⎛100 x10−6 ⎞ 5 ⎛ 40 x10−6 ⎞ 1.05 − 0.55 1.05 − 2 + 0.45) = 25.0 µA = ⎜ = 25.0 µ A Checking : i = ⎟( ⎜ ⎟( ) DP 1⎝ 1⎝ 2 ⎠ 2 ⎠ O = vI = (b) v iDN VDD + VTN + VTP 2 + (0.45)+ (−0.55) = = 0.950 V 2 2 2 2 2 ⎛100 x10−6 ⎞ 5 ⎛ 40 x10−6 ⎞ = ⎜ ⎟(0.95 − 0.45) = 25.0 µA Checking : iDP = ⎜ ⎟(0.95 − 2 + 0.55) = 25.0 µA 2 1⎝ 1⎝ 2 ⎠ ⎠ O = vI = 7.40 2 (0.25 pF ) 2.5 = 0.313 pJ P = CV 2 f = 0.25 pF 2.52 108 = 156 µW CVDD = ( ) DD 5 5 2 0.25 pF ) 22 ( CVDD 2 = 0.200 pJ P = CVDD f = (0.25 pF ) 22 108 = 100 µW (b) PDP ≅ 5 = 5 2 0.25 pF ) 1.82 ( CVDD 2 = 0.163 pJ P = CVDD f = (0.25 pF ) 1.82 108 = 81.0 µW (c) PDP ≅ 5 = 5 (a) PDP ≅ ( ) 2 ( )( ) () ( )( ) ( ) ( )( ) 7-18 7.41 2 0.2 pF ) 3.32 ( CVDD = 0.290 pJ (a) PDP ≅ 7.5 = 7.5 ( ) (b) f max ≅ τ PLH = 1.2 RonP C = τ PHL ≅ 1.2 RonN C = τP = 1.2(0.2 pF ) 1.2C = = 0.471 ns K p VGS − VTP 5 4 x10−5 −3.3 + 0.75 1.2(0.2 pF ) 1.2C = = 0.471 ns Kn ( VGS − VTN ) 2 10−4 (3.3 − 0.75) 1 7.5τ P ( ) ( ) 0.471 + 0.471 = 0.471 ns 2 f max ≅ 2 ( c ) P = CVDD f = (0.2 pF ) 3.32 2.83 x108 = 616 µW ( )( 1 1 = = 283 MHz 7.5τ P 7.5(0.471ns) ) 7.42 2 (0.15 pF ) 2.5 = 0.125 pJ CVDD PDP ≅ = a () 7.5 7.5 2 ( ) (b) f max ≅ τ PLH = 1.2 RonP C = τ PHL ≅ 1.2 RonN C = τP = 1.2(0.15 pF ) 1.2C = = 0.474 ns K p VGS − VTP 5 4 x10−5 −2.5 + 0.6 1.2(0.15 pF ) 1.2C = = 0.474 ns Kn ( VGS − VTN ) 2 10−4 (2.5 − 0.6) 1 7.5τ P ( ) ( ) 0.474 + 0.474 = 0.474 ns 2 f max ≅ 2 ( c ) P = CVDD f = (0.15 pF ) 2.52 2.82 x108 = 264 µW ( )( 1 1 = = 282 MHz 7.5τ P 7.5(0.474 ns) ) 7.43 *PROBLEM 7.41 - INVERTER PDP VDD 1 0 DC 2.5 VIN 2 0 PULSE (0 2.5 0 0.1N 0.1N 15N 30N) * MN1 3 2 0 0 MOSN W=4U L=1U AS=8P AD=8P MP1 3 2 1 1 MOSP W=4U L=1U AS=8P AD=8P C1 3 0 200fF *AS=2UM*W - AD=2UM*W * MN2 4 2 0 0 MOSN W=8U L=1U AS=16P AD=16P MP2 4 2 1 1 MOSP W=8U L=2U AS=16P AD=16P C2 4 0 200fF * MN3 5 2 0 0 MOSN W=16U L=1U AS=32P AD=32P MP3 5 2 1 1 MOSP W=16U L=1U AS=32P AD=32P 7-19 C3 5 0 200fF * MN4 6 2 0 0 MOSN W=32U L=1U AS=64P AD=64P MP4 6 2 1 1 MOSP W=32U L=1U AS=64P AD=64P C4 6 0 200fF * MN5 7 2 0 0 MOSN W=64U L=1U AS=128P AD=128P MP5 7 2 1 1 MOSP W=64U L=1U AS=128P AD=128P C5 7 0 200fF * MN6 8 2 0 0 MOSN W=100U L=1U AS=200P AD=200P MP6 8 2 1 1 MOSP W=100U L=1U AS=200P AD=200P C6 8 0 200fF * MN7 9 2 0 0 MOSN W=320U L=1U AS=640P AD=640P MP7 9 2 1 1 MOSP W=2320U L=1U AS=640P AD=640P C7 9 0 200fF * MN9 11 2 0 0 MOSN W=1000U L=1U AS=2000P AD=2000P MP9 11 2 1 1 MOSP W=1000U L=1U AS=2000P AD=2000P C9 11 0 200fF * .OP .TRAN 0.025N 50N .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99 +LAMBDA=.02 TOX=41.5N +CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5 +LAMBDA=.05 TOX=41.5N +CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P .PROBE V(2) V(3) V(4) V(5) V(6) V(7) V(8) V(9) V(10) V(11) .END At small device sizes, the power-delay product will be CVDD2 = 1.25 pJ. 7-20 Delay versus Device Size 1.00E+01 1.00E+00 Series1 1.00E-01 1 10 100 1000 W/L of NMOS & PMOS Devices 7.44 ∆T = ∆T ' = PDP ' = P ' ∆T ' = ( α∆T ) α 2 P = α 3 P∆T = α 3 PDP αW )( αL)( α∆V ) K( C '∆V ' = = α∆T ' 2 I 1 ⎛ αW ⎞ µn ⎜ αVGS − αVTN ) ⎟( 2 ⎝ αL ⎠ ⎛ ⎞ 2 2 V V ε ⎛W ⎞ " W P = VI = µnCox VGS − VTN ) = µn ox ⎜ ⎟( VGS − VTN ) ⎜ ⎟( 2 2 Tox ⎝ L ⎠ ⎝ L⎠ 2 αV ε ⎛ αW ⎞ P' = µn ox ⎜ αVGS − αVTN ) = α 2 P ⎟( αTox ⎝ αL ⎠ 2 " WL∆V C∆V KCox = ⎛ ⎞ 2 I 1 " W VGS − VTN ) µnCox ⎜ ⎟( 2 ⎝ L⎠ ' | Let W' = αW , L' = αL, Tox = αTox , V' = αV 7-21 7.45 ∆T = ∆T ' = αW )( αL)(∆V ) K( C '∆V ' = = α 2∆T ' ⎛ ⎞ 2 I 1 αW µn ⎜ VGS − VTN ) ⎟( 2 ⎝ αL ⎠ ⎛ ⎞ 2 2 V V ε ⎛W ⎞ " W P = VI = µnCox VGS − VTN ) = µn ox ⎜ ⎟( VGS − VTN ) ⎜ ⎟( 2 2 Tox ⎝ L ⎠ ⎝ L⎠ 2 ε ⎛ αW ⎞ V P P ' = µn ox ⎜ VGS − VTN ) = ⎟( 2 αTox ⎝ αL ⎠ α PDP ' = P ' ∆T ' = α 2∆T " WL∆V KCox C∆V ' = | Let W' = αW , L' = αL, Tox = αTox ⎛ ⎞ 2 I 1 W " µnCox ⎜ ⎟( VGS − VTN ) 2 ⎝ L⎠ ( P = αP∆T = α PDP )α 7.46 (Note: Simulation time needs to be extended.) *PROBLEM 7.46 - FIVE CASCADED INVERTERS VDD 1 0 DC 2.5 VIN 2 0 PULSE (0 2.5 0 0.1N 0.1N 25N 50N) * MN1 3 2 0 0 MOSN W=2U L=1U AS=16P AD=16P MP1 3 2 1 1 MOSP W=5U L=1U AS=40P AD=40P C1 3 0 0.25P *AS=8UM*W - AD=8UM*W * MN2 4 3 0 0 MOSN W=2U L=1U AS=16P AD=16P MP2 4 3 1 1 MOSP W=5U L=1U AS=40P AD=40P C2 4 0 0.25P * MN3 5 4 0 0 MOSN W=2U L=1U AS=16P AD=16P MP3 5 4 1 1 MOSP W=5U L=1U AS=40P AD=40P C3 5 0 0.25P * MN4 6 5 0 0 MOSN W=2U L=1U AS=16P AD=16P MP4 6 5 1 1 MOSP W=5U L=1U AS=40P AD=40P C4 6 0 0.25P * MN5 7 6 0 0 MOSN W=2U L=1U AS=16P AD=16P MP5 7 6 1 1 MOSP W=5U L8U AS=40P AD=40P C5 7 0 0.25P .OP .TRAN 0.025N 50N .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99 +LAMBDA=.02 TOX=41.5N 7-22 +CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5 +LAMBDA=.05 TOX=41.5N +CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P .PROBE V(2) V(3) V(5) V(6) .END 3.0V 2.0V 1.0V 0V First inverter : t r = 4.6 ns, t f = 5.4 ns, τ PLH = 2.6 ns, τ PHL = 2.1 ns Fourth inverter : t r = 5.8 ns, t f = 6.3 ns, τ PLH = 4.2 ns, τ PHL = 4.7 ns τ PHL = 1.2 RonnC = ⎛ 2⎞ −6 ⎜ ⎟ 50 x10 (2.5 − 0.91) 1 ⎝ ⎠ 0.25 x10−12 ( ) = 1.58 ns τ PLH = 1.2 RonpC = ⎛ 5⎞ −6 ⎜ ⎟ 20 x10 (2.5 − 0.77) ⎝ 1⎠ 0.25 x10−12 ( ) = 1.45ns The inverters are slower than the equations predict because of the additional capacitances in the transistor models. The effective capacitance appears to be approximately 0.4 pF. The delay of the interior inverter is substantially slower than predicted by the formula because of the slow rise and fall times of the driving signals. 7-23 7.47 (Note: Simulation time needs to be extended.) *PROBLEM 7.47(a) - FIVE CASCADED SYMMETRICAL INVERTERS VDD 1 0 DC 5 VIN 2 0 PULSE (0 5 0 0.1N 0.1N 75N 150N) * MN1 3 2 0 0 MOSN W=2U L=1U AS=16P AD=16P MP1 3 2 1 1 MOSP W=5U L=1U AS=40P AD=40P C1 3 0 1P *AS=8UM*W - AD=8UM*W * MN2 4 3 0 0 MOSN W=2U L=1U AS=16P AD=16P MP2 4 3 1 1 MOSP W=5U L=1U AS=40P AD=40P C2 4 0 1P * MN3 5 4 0 0 MOSN W=2U L=1U AS=16P AD=16P MP3 5 4 1 1 MOSP W=5U L=1U AS=40P AD=40P C3 5 0 1P * MN4 6 5 0 0 MOSN W=2U L=1U AS=16P AD=16P MP4 6 5 1 1 MOSP W=5U L=1U AS=40P AD=40P C4 6 0 1P * MN5 7 6 0 0 MOSN W=2U L=1U AS=16P AD=16P MP5 7 6 1 1 MOSP W=5U L=1U AS=40P AD=40P C5 7 0 1P .OP .TRAN 0.025N 150N .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99 +LAMBDA=.02 TOX=41.5N CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5 +LAMBDA=.05 TOX=41.5N CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P .PROBE V(2) V(3) V(5) V(6) .END 7-24 First inverter : t r = 17.7 ns, t f = 20.7 ns, τ PLH = 8.1 ns, τ PHL = 9.9 ns Fourth inverter : t r = 22.3 ns, t f = 24.2 ns, τ PLH = 16.3 ns, τ PHL = 18.3 ns τ PHL = 1.2 RonnC = ⎛ 2⎞ −6 ⎜ ⎟ 50 x10 (2.5 − 0.91) ⎝ 1⎠ 1.2 x10−12 ( ) = 7.6 ns τ PLH = 1.2 RonpC = ⎛ 5⎞ −6 ⎜ ⎟ 20 x10 (2.5 − 0.77) ⎝ 1⎠ 1.2 x10−12 ( ) = 6.9 ns 3.0V 2.0V 1.0V 0V -1.0V 0s V(CL1:2) 20ns V(CL2:2) V(CL3:2) 40ns V(CL4:2) 60ns V(CL5:2) V(VI:+) 80ns Time 100ns 120ns 140ns 160ns The inverters are slower than the equations predict because of the additional capacitances in the transistor models. The delay of the interior inverter is substantially slower than predicted by the formula because of the slow rise and fall times of the driving signals. *PROBLEM 7.47(b) - FIVE CASCADED MINIMUM SIZE INVERTERS VDD 1 0 DC 5 VIN 2 0 PULSE (0 5 0 0.1N 0.1N 125N 250N) * MN1 3 2 0 0 MOSN W=4U L=2U AS=16P AD=16P MP1 3 2 1 1 MOSP W=4U L=2U AS=16P AD=16P C1 3 0 1P *AS=4UM*W - AD=4UM*W * MN2 4 3 0 0 MOSN W=4U L=2U AS=16P AD=16P MP2 4 3 1 1 MOSP W=4U L=2U AS=16P AD=16P C2 4 0 1P * MN3 5 4 0 0 MOSN W=4U L=2U AS=16P AD=16P MP3 5 4 1 1 MOSP W=4U L=2U AS=16P AD=16P C3 5 0 1P * MN4 6 5 0 0 MOSN W=4U L=2U AS=16P AD=16P MP4 6 5 1 1 MOSP W=4U L=2U AS=16P AD=16P C4 6 0 1P * MN5 7 6 0 0 MOSN W=4U L=2U AS=16P AD=16P 7-25 MP5 7 6 1 1 MOSP W=4U L=2U AS=16P AD=16P C5 7 0 1P .OP .TRAN 0.025N 250N .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99 +LAMBDA=.02 TOX=41.5N +CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5 +LAMBDA=.05 TOX=41.5N +CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P .PROBE V(2) V(3) V(5) V(6) .END 3.0V 2.0V 1.0V 0V -1.0V 0s V(CL1:2) V(CL2:2) 50ns V(CL3:2) V(CL4:2) V(CL5:2) 100ns V(VI:+) Time 150ns 200ns 250ns First inverter : t r = 43.7 ns, t f = 20.6 ns, τ PLH = 19.7 ns, τ PHL = 9.7 ns Fourth inverter : t r = 47.5 ns, t f = 31.4 ns, τ PLH = 30.4 ns, τ PHL = 25.7 ns τ PHL = 1.2 RonnC = ⎛ 2⎞ −6 ⎜ ⎟ 50 x10 (2.5 − 0.91) ⎝ 1⎠ 1.2 x10−12 ( ) = 7.5 ns τ PLH = 1.2 RonpC = ⎛ 2⎞ −6 ⎜ ⎟ 20 x10 (2.5 − 0.77) ⎝ 1⎠ 1.2 x10−12 ( ) = 17.3 ns The inverters are slower than the equations predict because of the additional capacitances in the transistor models. The delay of the interior inverter is substantially slower than predicted by the formula because of the slow rise and fall times of the driving signals. 7-26 7.48 (a) V DD 20 1 D 20 1 C 20 1 B 20 1 vo 2 1 B 2 1 C 2 1 D 2 1 2⎞ 6 (b ) NMOS : 3 ⎛ ⎜ ⎟= ⎝1⎠ ⎛ 20 ⎞ 60 | PMOS : 3⎜ ⎟ = 1 ⎝ 1 ⎠ 1 7-27 7.49 (a) V DD D C B A vO D C B A W ⎛2⎞ 8 = 4⎜ ⎟ = L ⎝1⎠ 1 W 5 PMOS : = L 1 2 ⎞ 16 (b ) NMOS : W = 2(4)⎛ ⎜ ⎟= L ⎝1⎠ 1 (a ) NMOS : PMOS : W ⎛ 5 ⎞ 10 = 2⎜ ⎟ = L ⎝1⎠ 1 7-28 7.50 V DD D 1 5 2 1 NMOS: C PMOS: B Z = A+B+C+D A B C D PMOS: 7.51 0.4 ⎛ 2 ⎞ 1 2 | NMOS: ⎜ ⎟= 4 ⎝1⎠ 5 1 VDD C 1 3.75 2 1 NMOS: B PMOS: Z = A+B+C A B C 7-29 7.52 VDD A B C Z = ABC A B C (W/L)N = 2/1, (W/L)P = 2.5(2/1)/3, = 1.67/1 7.53 VDD D C B A Z = ABCD D C B A (W/L)N = 2/1, (W/L)P = 2.5(2/1)/4, = 1.25/1 7-30 7.54 Output Z is A multiplied by B. From the truth table, Z = AB, a two input AND gate. A B 0 0 0 1 1 0 1 1 Z 0 0 0 1 Assuming complemented variables are available, VDD B M = ( A + B ) = AB A NMOS: B PMOS: 4 1 5 1 7.55 (The dc input should be 0 V.) *PROBLEM 7.55 - TWO-INPUT CMOS NOR GATE VDD 1 0 DC 2.5 VA 2 0 DC 0 PULSE (0 2.5 0 0.1N 0.1N 25N 50N) VB 5 0 DC 0 * MNA 4 2 0 0 MOSN W=2U L=1U AS=16P AD=16P MPA 4 2 3 1 MOSP W=10U L=1U AS=80P AD=80P MNB 4 5 0 0 MOSN W=2U L=1U AS=16P AD=16P MPB 3 5 1 1 MOSP W=10U L=1U AS=80P AD=80P CL 4 0 1PF * MNC 6 5 0 0 MOSN W=2U L=1U AS=16P AD=16P MPC 6 5 7 1 MOSP W=10U L=1U AS=80P AD=80P MND 6 2 0 0 MOSN W=2U L=1U AS=16P AD=16P MPD 7 2 1 1 MOSP W=10U L=1U AS=80P AD=80P CL 6 0 1PF * 7-31 .OP .DC VDD 0 2.5 0.01 .TRAN 0.1N 50N .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0 +LAMBDA=.02 TOX=41.5N +CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0 +LAMBDA=.05 TOX=41.5N +CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P .PROBE V(2) V(4) V(6) .END 3.0V 2.0V 1.0V 0V 0V V2(C2) 3.0V V2(C3) 0.5V 1.0V V_VA 1.5V 2.0V 2.5V 2.0V 1.0V 0V 0s V(VA:+) 10ns V2(C3) 20ns V2(C3) 30ns 40ns 50ns Time 60ns 70ns 80ns 90ns 100ns The transitions of the two VTCs are separated by approximately 50 mV. The dynamic characteristics for switching one input with the other constant are essentially identical. The two transitions are virtually identical because of the ideal step inputs: τPHL = 3.6 ns, τPLH = 3.6 ns, tf = 8.1 ns, tr = 7.9 ns. With the inputs switched together, τPHL and tf are reduced by 50% because the two NMOS devices are working in parallel. 7.56 The simulation results show only slight changes from those of Problem 7.55. 7.57 *PROBLEM 7.54 - TWO-INPUT CMOS NAND GATE VDD 1 0 DC 2.5 VA 2 0 DC 0 PULSE (0 2.5 0 0.1N 0.1N 25N 50N) VB 4 0 DC 2.5 * MNA 3 4 0 0 MOSN W=4U L=1U AS=16P AD=16P MPA 5 4 1 1 MOSP W=5U L=1U AS=80P AD=80P 7-32 MNB 5 2 3 0 MOSN W=4U L=1U AS=16P AD=16P MPB 5 2 1 1 MOSP W=5U L=1U AS=80P AD=80P CL1 5 0 1PF * MNC 6 2 0 0 MOSN W=4U L=1U AS=16P AD=16P MPC 7 2 1 1 MOSP W=5U L=1U AS=80P AD=80P MND 7 4 6 0 MOSN W=4U L=1U AS=16P AD=16P MPD 7 4 1 1 MOSP W=5U L=1U AS=80P AD=80P CL2 7 0 1PF * .OP .DC VA 0 2.5 0.01 .TRAN 0.05N 50N .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0 +LAMBDA=.02 TOX=41.5N CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0 LAMBDA=.05 TOX=41.5N +CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P .PROBE V(2) V(5) V(7) .END 3.0V 2.0V 1.0V 0V 0V 3.0V 0 5V 1 0V 1 5V 2 0V 2 5V 2.0V 1.0V 0V 0s 10ns 20ns 30ns 40ns 50ns 60ns 70ns 80ns 90ns 100ns The transitions of the two VTCs are separated by approximately 50 mV. The dynamic characteristics for switching one input with the other constant are essentially identical. The two transitions are virtually identical because of the ideal step inputs: τPHL = 3.6 ns, τPLH = 3.7 ns, tf = 8.0 ns, tr = 8.1 ns. With the inputs switched together, τPLH and tr are reduced by 50% because the two PMOS devices are working in parallel 7.58 The simulation results show only slight changes. 7.59 7-33 Worst-case paths are the same as the symmetrical reference inverter: ⎛ 1 ⎞⎛ 15⎞ ⎛ 5⎞ ⎛ 1 ⎞⎛ 4 ⎞ ⎛ 2 ⎞ PMOS tree : ⎜ ⎟⎜ ⎟ = ⎜ ⎟ | NMOS tree : ⎜ ⎟⎜ ⎟ = ⎜ ⎟ | For VDD = 2.5V , ⎝ 3 ⎠⎝ 1 ⎠ ⎝ 1 ⎠ ⎝ 2 ⎠⎝ 1 ⎠ ⎝ 1 ⎠ τ PLH = 1.2 RonP C = τ PHL ≅ 1.2 RonN C = 1.2( 1.25 pF ) 1.2C = = 3.95ns K p VGS − VTP 5 4 x10−5 −2.5 + 0.6 1.2( 1.25 pF ) 1.2C = = 3.95 ns Kn ( VGS − VTN ) 2 10−4 (2.5 − 0.6) ( ) ( ) Since there is a 2.5 :1 ratio in transistor sizes, τ PLH = τ PHL , τ P = τ PHL = 3.95 ns and t r = t f = 3τ PHL = 3τ P = 11.8 ns 7-34 7.60 (a) A depletion-mode design requires the same number of NMOS transistors in the switching network, but only one load transistor. The depletion-mode design requires 5 transistors total. The CMOS design requires 8 transistors. (b) For the CMOS design, first find the worst - case delay of the circuit in Fig. 7.30 and then scale the result to achieve the desired delay. For VDD = 2.5 V, 1.2 10−12 1.2C = = = 6.32 ns Kn ( VDD − VTN ) 1 ⎛ 2 ⎞ −6 ⎜ ⎟ 100 x10 (2.5 − 0.6) 2 ⎝1⎠ −12 τ PHL 1.2C τ PLH = K p( VDD + VTP ( ) ( ) 1.2( 10 ) = = 23.7 ns 2⎞ ) 1⎛ 40 x10 )−2.5 + 0.6 ⎜ ⎟( 3⎝ 1⎠ −6 τP = Relative Area = 8(3.00)() 1 = 24.0 ⎛ W⎞ 6.32 + 23.7 15.0 ns ⎛ 2 ⎞ ⎛ 3.00 ⎞ ns = 15.0 ns → ⎜ ⎟ = ⎜ ⎟=⎜ ⎟ 2 ⎝ L ⎠ all 10ns ⎝ 1 ⎠ ⎝ 1 ⎠ −−−−− For the depletion - mode design, first assume that τ P is dominated by τ PLH. τ PLH ≅ 2τ P | τ PLH = 3.6 RonLC 3.6 10 ⎛W ⎞ 4.50 W = = | Since NMOS is ratioed logic, the ratios ⎜ ⎟ −8 −6 1 L ⎝ L ⎠ L 2 10 40 x10 () 1 ⎛W ⎞ (2.22 /1) 4.50 = 5.52 must maintain the ratio in Fig. 7.29(d) : ⎜ ⎟ = 1 1.81/1) 1 ⎝ L ⎠S ( Using this value, τ PHL = ( ) ( )( ) −12 ( ) = 1.14ns and 5.52( 100x10 ) (2.5 - 0.6) 1.2 10-12 -6 20 + 1.14 ns = 10.6ns 2 Rescaling to achieve τ P = 10 ns : ⎛W ⎞ 10.6 ⎛ 4.50 ⎞ 4.77 ⎛ W ⎞ 10.6 ⎛ 5.52 ⎞ 5.85 ⎛W ⎞ ⎛ W ⎞ 11.7 | ⎜ ⎟ = | ⎜ ⎟ = 2⎜ ⎟ = ⎜ ⎟ = ⎜ ⎟= ⎜ ⎟= 1 1 1 ⎝ L ⎠ L 10 ⎝ 1 ⎠ ⎝ L ⎠ A 10 ⎝ 1 ⎠ ⎝ L ⎠B − D ⎝ L ⎠A τP = Relative area = (4.50)() 1 + 3( 11.7)() 1 + (5.85)() 1 = 45.5 The CMOS design uses 47% less area and consumes no static power. 7-35 7.61 (a) Y = (A + B)(C + D)(E + F ) (b) NMOS: W ⎛ 2⎞ 6 = 3⎜ ⎟ = L ⎝ 1⎠ 1 PMOS: ⎛ 5 ⎞ 10 W = 2⎜ ⎟ = L ⎝1⎠ 1 A C E B D F 7.62 (a) Y = (A + B)(C + D)E (b) NMOS: ⎛ 2 ⎞ 18 W = 3(3) ⎜ ⎟= L ⎝ 1⎠ 1 ⎛ W⎞ ⎛ 5⎞ 30 PMOS: ⎜ ⎟ = 2(3) ⎜ ⎟= ⎝ L ⎠ A− D ⎝ 1⎠ 1 ⎛ W⎞ ⎛ 5⎞ 15 ⎜ ⎟ = (3) ⎜ ⎟= ⎝ L ⎠E ⎝ 1⎠ 1 A E C B D 7-36 7.63 F G C E E D B D F G A C A B (a) Y = F + G(C + E )+ A(B + D)(C + E )= F + (C + E )(G + A(B + D)) ⎛ W⎞ ⎛ 2 ⎞ 12 ⎛ W ⎞ ⎛ 2⎞ 4 ⎛ W⎞ 1 6 = 3(2) | ⎜ ⎟ = (2) = ⎜ ⎟ ⎜ ⎟= ⎜ ⎟= | ⎜ ⎟ = ⎝ L ⎠ A− E ⎝ 1⎠ 1 ⎝ L ⎠F ⎝ 1 ⎠ 1 ⎝ L ⎠G 1 − 1 1 4 12 ⎛ W⎞ ⎛ 5 ⎞ 40 ⎛ W ⎞ ⎛ ⎞ 1 20 W 1 26.7 PMOS : ⎜ ⎟ = 4(2) | ⎜ ⎟ = = | ⎜ ⎟ =2 = ⎜ ⎟= 1 1 1 1 ⎝ L ⎠ F ,G , B , D ⎝ 1⎠ 1 ⎝ L ⎠A 1 − 2 ⎝ L ⎠C , E − 10 40 10 40 (b) NMOS : 7.64 (a) Y = A + B C + D E + F = AB + CD + EF ( )( )( ) (b) NMOS : ⎛ W⎞ ⎛2⎞ 4 = 2⎜ ⎟ = ⎜ ⎟ ⎝ L ⎠ A− F ⎝1⎠ 1 ⎛ W⎞ ⎛ 5 ⎞ 15 PMOS : ⎜ ⎟ = 3⎜ ⎟ = ⎝ L ⎠ A− F ⎝ 1⎠ 1 A C E B D F 7-37 7.65 (a) Y = ACE + ACDF + BDE + BF = (A + C)(B + DF )+ E(F + DB) ( ) ⎛ W⎞ ⎛ 2 ⎞ 18 = 3(3) ⎜ ⎟ ⎜ ⎟= ⎝ L ⎠ A− F ⎝ 1⎠ 1 ⎛ W⎞ ⎛ 5 ⎞ 60 PMOS : ⎜ ⎟ = 4(3) ⎜ ⎟= ⎝ L ⎠ A, C , D , F ⎝ 1⎠ 1 (b) NMOS : ⎛ W⎞ 1 40 = ⎜ ⎟ =2 1 1 1 ⎝ L ⎠B , E − 15 60 C Y E A F D B 7-38 7.66 VDD B 15 1 A 15 1 D 15 1 C 15 1 15 1 E F 15 1 Y A 4 1 C 6 1 F 2 1 B 4 1 D 6 1 E 6 1 7-39 7.67 V DD A 20 1 B 20 1 C 20 1 D 20 1 E 20 1 F 20 1 G 20 1 H 20 1 I 20 1 Y A 6 1 D 4 1 F 6 1 I 2 1 B 6 1 E 4 1 G 6 1 C 6 1 H 6 1 7-40 7.68 Y A C Y E V DD D B Gnd (a) An Euler path does not appear to exist. Y A C E Y V DD D F B Gnd (b) An Euler path does not appear to exist. . 7-41 7.69 10011 VDD D A B C E Y D A C B E 10001 VDD D A B C E Y D A C B E 7-42 11101 V DD D A B C E Y D A C B E 00010 V D A B DD C E Y D A C B E 7-43 7.70 V B D DD F A C E A B C D E F (a) V DD B D F A C E A B C D E F (b) 7-44 V B D DD F A C E A B C D E F (c) VDD B D F A C E A B C D E F (d) 7-45 7.71 + 10 1 10 1 10 1 A B C D 10 1 E 10 1 Y A 6 1 D B 6 1 E 4 1 4 1 C 6 1 7.72 V DD B 20 1 40 1 C 40 1 A D 40 1 E 40 1 Y A 24 1 B 24 1 D 24 1 C 24 1 E 24 1 7-46 7.73 VDD B 5 1 15 1 A D C 7.5 1 E 15 1 15 1 Y A 6 1 B 3 1 C 6 1 D 6 1 E 6 1 7.74 V DD E 60 1 A 30 1 C 60 1 D 60 1 B 60 1 Y A 24 1 8 1 E 12 1 B C 24 1 24 1 D 7-47 7.75 S = X ⊕ Y = XY + XY and C = XY V DD Y 10 1 Y 10 1 V DD X 10 1 X 10 1 10 1 Y S = (X + Y)(X + Y) 4 1 4 1 10 1 C=X+Y X Y X 4 1 Y 4 1 X 2 1 2 1 Y 7-48 7.76 Xi 0 0 0 0 1 1 1 1 Yi 0 0 1 1 0 0 1 1 Ci-1 0 1 0 1 0 1 0 1 Si 0 1 1 0 1 0 0 1 Ci 0 0 0 1 0 1 1 1 Let X = X i ( ) ( S =( C + (X + Y ) (X + Y ))(C + (X + Y )(X + Y )) i Y = Yi C = Ci -1 Si = XYC + XYC + XYC + XYC = C XY + XY + C XY + XY ) Ci = XY + XC + YC = XY + C (X + Y ) Ci = X + Y C + XY 15 ( )( ) VDD C 15 1 C 1 Y 15 1 Y 15 1 X 15 1 X 15 1 V Y 10 1 C 10 1 DD X 15 1 X 15 1 Y 15 1 Y 15 1 Si X 8 1 X 10 1 X 10 1 Y 10 1 Y 8 1 C X 8 1 4 1 Ci X 6 1 C Y 6 1 3 1 Y 8 1 X 8 1 Y 8 1 C 8 1 8 1 4 1 X Y X 6 1 Y 6 1 7-49 7.77 N A B 00 00 00 00 01 01 01 01 M C D 00 01 10 11 00 01 10 11 Output O3 O2 O1 O0 0000 0000 0000 0000 0000 0001 0010 0011 N A B 10 10 10 10 11 11 11 11 M C D 00 01 10 11 00 01 10 11 Output O3 O2 O1 O0 0000 0010 0100 0110 0000 0011 0110 1001 O3 = ABCD = A + B + C + D O2 = AC = A + C ( ) ( )( )( )( ) ( ) O1 = ABC + ABD + BCD + ACD = A + B + C A + B + D B + C + D A + C + D O0 = BD = B + D VDD D ( ) V DD V DD C C D B O2 O3 O4 A C B D A B C D ⎛W ⎞ 2 ⎛W ⎞ 20 ⎜ ⎟ = ⎜ ⎟ = ⎝ L ⎠N 1 ⎝ L ⎠P 1 | ⎛W ⎞ 2 ⎛W ⎞ 10 ⎜ ⎟ = ⎜ ⎟ = ⎝ L ⎠N 1 ⎝ L ⎠P 1 | ⎛W ⎞ 2 ⎛W ⎞ 10 ⎜ ⎟ = ⎜ ⎟ = ⎝ L ⎠N 1 ⎝ L ⎠P 1 7-50 V D DD D D C C C B B A B A A O1 A C D NMOS: 8 1 B C D PMOS: 15 1 A B D A B C 7.78 (a) τ PHL = 1.2 RonnC = 1.2 0.4 x10−12 −6 100 x10 ) (2 1)( (2.5 − 0.6) 1.2 0.4 x10−12 −6 ( ) = 1.26 ns τ PLH = 1.2 RonpC = (2 / 3)(40 x10 )−2.5 + 0.6 1.2 0.4 x10−12 ( ) = 9.47 ns | τ P = = 1.26 ns τ PLH + τ PLH 2 = 5.37 ns (b) τ PHL = 1.2 RonnC = τ PLH = 1.2 Ronp ( ) 1.2( 0.4 x10 ) C= = 3.16 ns (2 1)(40 x10 )−2.5 + 0.6 (2 1) 100 x10−6 (2.5 − 0.6) −12 −6 ( ) | τP = τ PLH + τ PLH 2 = 2.21 ns 7-51 7.79 (a) τ PHL = 1.2 RonnC = 1.2 0.18 x10−12 −6 100 x10 ) (2 5)( (2.5 − 0.6) 1.2 0.18 x10−12 ( ) = 2.84 ns τ PLH = 1.2 RonpC = (2 /1) 40 x10−6 −2.5 + 0.6 ( ( ) ) = 1.42 ns | τ P = τ PLH + τ PLH 2 = 2.13 ns (b) τ PHL = 1.2 RonnC = 1.2 0.18 x10−12 −6 100 x10 ) (2 1)( (2.5 − 0.6) 1.2 0.4 x10−12 ( ) = 0.568 ns τ PLH = 1.2 RonpC = (2 1) 40 x10 ( ( −6 ) ) −2.5 + 0.6 = 3.16 ns | τ P = τ PLH + τ PLH 2 = 1.86 ns 7.80 The worst-case NMOS path contains 2 transistors. The worst-case PMOS path contains 3 transistors. 1.2 250 x10−15 τ PHL = 1.2 RonnC = = 1.58 ns (2 2) 100 x10−6 (2.5 − 0.6) ( ( τ PLH = 1.2 RonpC = 1.2 250 x10−15 −6 (2 / 3)(40 x10 )−2.5 + 0.6 ( ) ) ) = 5.92 ns 7.81 The worst-case NMOS path contains 3 transistors. The worst-case PMOS path contains 3 transistors. 1.2 400 x10−15 τ PHL = 1.2 RonnC = = 3.79 ns (2 3) 100 x10−6 (2.5 − 0.6) ( ( τ PLH = 1.2 RonpC = 1.2 400 x10−15 −6 (2 / 3)(40 x10 )−2.5 + 0.6 ( ) ) ) = 9.47 ns 7.82 The worst-case NMOS path contains 3 transistors (ABE or CBD). The worst-case PMOS path also contains 3 transistors. τ PHL = 1.2 RonnC = τ PLH = 1.2 Ronp ( ) = 9.47 ns 100 x10 ) (2 3)( (2.5 − 0.6) 1.2( 10 ) C= = 23.7 ns (2 / 3)(40 x10 )−2.5 + 0.6 1.2 10−12 −6 −12 −6 | τP = τ PLH + τ PLH 2 = 16.6 ns 7-52 7.83 The worst-case NMOS path contains 3 transistors. 1.2 10−12 τ PHL = 1.2 RonnC = = 9.47 ns −6 2 3 100 x 10 2.5 − 0.6 ( ) ( ) ( ( ) ) 7.84 Student PSPICE will only accept 5 inverters. *PROBLEM 7.84(a) - FIVE CASCADED INVERTERS VDD 1 0 DC 2.5 VIN 2 0 PULSE (0 2.55 0 0.1N 0.1N 20N 40N) * MN1 3 2 0 0 MOSN W=2U L=1U AS=16P AD=16P MP1 3 2 1 1 MOSP W=2U L=1U AS=16P AD=16P C1 3 0 200fF *AS=8UM*W - AD=8UM*W * MN2 4 3 0 0 MOSN W=2U L=1U AS=16P AD=16P MP2 4 3 1 1 MOSP W=2U L=1U AS=16P AD=16P C2 4 0 200fF * MN3 5 4 0 0 MOSN W=2U L=1U AS=16P AD=16P MP3 5 4 1 1 MOSP W=2U L=1U AS=16P AD=16P C3 5 0 200fF * MN4 6 5 0 0 MOSN W=2U L=1U AS=16P AD=16P MP4 6 5 1 1 MOSP W=2U L=1U AS=16P AD=16P C4 6 0 200fF * MN5 7 6 0 0 MOSN W=2U L=1U AS=16P AD=16P MP5 7 6 1 1 MOSP W=2U L=1U AS=16P AD=16P C5 7 0 200fF * .OP .TRAN 0.025N 40N .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99 +LAMBDA=.02 TOX=41.5N +CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5 +LAMBDA=.05 TOX=41.5N +CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P .PROBE V(2) V(3) V(4) V(10) V(11) .END (a) 7-53 3.0V 2.0V 1.0V 0V -1.0V 0s V(CL1:2) 5ns V(CL2:2) 10ns V(CL3:2) 15ns 20ns V(CL4:2) V(CL5:2) V(VI:+) 25ns Time 30ns 35ns 40ns 45ns 50ns (b) 3.0V 2.0V 1.0V 0V -1.0V 0s V(VI:+) 5ns V(CL1:2) 10ns V(CL2:2) 15ns V(CL3:2) 20ns V(CL4:2) 25ns V(CL5:2) 30ns Time 35ns 40ns 45ns 50ns 55ns 60ns (a) The minimum size inverters yield P = 5.8 ns. (b) The symmetrical inverters yield τP = 3.7 ns. Note that these results are larger than the delay equation estimate because of the slope of the waveforms. 7-54 7.85 (a) (b) V DD CLK V DD CLK Z = AB B A Z=A+B B A CLK CLK 7-55 7.86 (a) V DD CLK (b) V DD CLK A B A Z = A + B = AB B Z = A + B = AB CLK CLK 7.87 (a) (b) VDD CLK A V DD CLK Z Z = ABC B A B C Z = A+B+C C CLK CLK 7-56 7.88 (a) V DD (b) CLK A Z = A B C = A+B+C CLK VDD B Z A B C Z = A + B + C = ABC C CLK CLK 7.89 VDD CLK A B Z = (AB+CD+EF) C D E F CLK 7-57 7.90 VDD CLK A B C Z = (ABC+DE+F) D E F CLK 7.91 Charge sharing occurs. Assuming C2 and C3 are discharged (the worst case) (a ) V 2C2VDD 2 2 = VDD | Node B drops to VDD . C1 + C2 2C2 + C2 3 3 ⎛2 ⎞ 2 3 C V ⎜ ⎟ 2 DD + C V + C 0 C ( 1 2 )3 DD 3 ( ) V 1 ⎝3 ⎠ = = DD | Node B drops to VDD . (b) VB = 2C2 + C2 + C2 C1 + C2 + C3 2 2 B = C1VDD + C2 (0) = (c) V R≥ B = C1VDD RC2VDD R = = VDD ≥ VIH → R( VDD − VIH )≥ 2VIH C1 + C2 + C3 RC2 + C2 + C2 R + 2 2VIH 2VIH = VDD − VIH NM H Using VDD = 2.5V , VTN = 0.6V , VTP = −0.6V in Eq. (8.9) : VIH = R≥ 5(2.5)+ 3(0.6)+ 5(−0.6) 8 2( 1.41) = 1.41V | NM H = 2.5 − 0.6 − 3(−0.6) 4 = 0.925V 2VIH = = 3.05 | C1 ≥ 83.05C2 NM H 0.925 7-58 7.92 Z = A0 + A1 + A2 V DD Clock A 0 A1 A 2 Z C1 7.93 V DD CLK A B Z=(A+B)(C+D) C D CLK 7-59 7.94 V DD CLK A C Z = AB + CD B D CLK 7-60 7.95 V DD CLK Z A B C D E F CLK 7.96 V DD CLK Z A C E B D F CLK 7.97 N opt = ln 1 CL = ln (4000)= 8.29 → N = 8 | β = (4000)8 = 2.82. Co Each inverter has a delay of 2.82τ o. The total delay is 8(2.82τ o )= 22.6τ o The relative sizes of the 8 inverters are : 1, 2.82, 7.95, 22.4, 63.2, 178, 503, 1420. 7-61 7.98 N opt = ln 1 ⎛ 10 pF ⎞ CL 4 = 3.16. = 4.61 → N = 4 | = ln⎜ β = 100 ( ) ⎟ 100 fF Co ⎝ ⎠ Each inverter has a delay of 3.16τ o. The total delay is 4(3.16τ o )= 12.6τ o Note, N = 5 yields β = 2.51 and the total delay is 5(2.51τ o ) = 12.6τ o . However, the area will be significantly larger. See Prob. 7.100. 7.99 1 ⎛ 40 pF ⎞ CL = ln = ln⎜ ⎟ = 6.69 → N = 6 | β = (800)6 = 2.82. Co ⎝ 50 fF ⎠ The relative sizes of the 4 inverters are : 1, 3.16, 10.0, 31.6 N opt Each inverter has a delay of 2.82τ o. The total delay is 8(2.82τ o )= 22.6τ o . Note, N = 7 yields β = 2.60 and the total delay is 7(2.60τ o ) = 18.2τ o. However, the area will be significantly larger. See Prob. 7.100. 7.100 AT = Ao ( 1 + β + β 2 + ... + β N −1 )= Ao The relative sizes of the 8 inverters are: 1, 2.82, 7.95, 22.4, 63.2, 178, 503, 1420. β N −1 β −1 1000 − 1 = 462 Ao 3.1623 − 1 1000 − 1 For N = 7, β = 10001/7 = 2.6827 | A = Ao = 594 Ao 2.6827 − 1 Since the two values of N give similar delays, N = 6 would be used For N = 6, β = 10001/6 = 3.1623 | A = Ao because of it requires significantly less area. 7.101 ⎡⎛ 20 ⎞ ⎤−1 −6 ⎥ = 263 Ω ⎟ 100 x10 (2.5 − 0.6) (a) Ronn = K V − V = ⎢⎜ ⎦ n ( GS TN ) ⎣⎝ 1 ⎠ 1 ( ) (b) R onp = 1 K p VGS − VTP ⎡⎛ 20 ⎞ ⎤ = ⎢⎜ ⎟ 40 x10−6 −2.5 + 0.6 ⎥ = 658 Ω ⎣⎝ 1 ⎠ ⎦ ( ) −1 (c) A resistive channel exists connecting the source and drain. 7-62 7.102 Gon = Gonn + Gonp = Kn ( VGSN − VTN )* ( ( VGSN − VTN )> 0) + K p ( VTP − VGSP )* ( ( VTP − VGSP )> 0) VGSN = 2.5 − VI VTN = 0.6 + 0.5 VI + 0.6 − 0.6 (a) R Gon = [ ( VSBN = VI VGSP = −VI )] VTP = −0.6 − 0.5 2.5 − VI + 0.6 − 0.6 [ VBSN = 2.5 − VI ( )] on will be the largest for VI = 1 V VTP = −0.937V VTN = 0.845V 10 −4 10 10 (2.5 − 1 − 0.845)+ 4 x10−5 (−0.937 + 1) → Ron = 1470Ω 1 1 (b) Ron will be the largest for the VI at which the NMOS transistor just cuts off : 2.5 − VI = 0.6 + 0.5 VI + 0.6 − 0.6 → VI = 1.554V Gon = 10 4 x10−5 (−0.834 + 1.554)→ Ron = 3470Ω 1 ( ) ( ) [ ( )] VTN = 0.947V VTP = −0.834V ( ) 7.103 Gon = Gonn + Gonp = Kn ( VGSN − VTN )* ( ( VGSN − VTN )> 0) + K p ( VTP − VGSP )* ( ( VTP − VGSP ) > 0) VGSN = 2.5 − VI VTN = 0.75 + 0.5 VI + 0.6 − 0.6 [ ( VSBN = VI VGSP = −VI )] VBSN = 2.5 − VI VTP = −0.75 − 0.5 2.5 − VI + 0.6 − 0.6 [ ( )] The worst cases occur approximately at the point where the PMOS or NMOS transistors just cut off. The NMOS transistor cuts off for 2.5 − VI = 0.75 + 0.5 VI + 0.6 − 0.6 → VI = 1.426V VTP = −1.01V ⎛W ⎞ ⎛W ⎞ 1 240 = ⎜ ⎟ 4 x10−5 (−1.01 + 1.426)→ ⎜ ⎟ = 250 ⎝ L ⎠ P 1 ⎝ L ⎠P [ ( )] ( ) The PMOS transistor cuts off for VI = 0.75 + 0.5 2.5 − VI + 0.6 − 0.6 → VI = 1.074V VTN = −1.01V ⎛W ⎞ ⎛W ⎞ 1 96.2 = ⎜ ⎟ 10−4 (2.5 − 1.074 − 1.01)→ ⎜ ⎟ = 250 ⎝ L ⎠ N 1 ⎝ L ⎠P [ ( )] ( ) 7.104 (a ) The output of the first NMOS transistor will be VI = 2.5 − VTN = 2.5 − 0.75 + 0.55 VI + 0.6 − 0.6 → VI = 1.399V | VTN = 1.10V The output of the other gates reaches this same value. All three nodes = 1.40 V. For the PMOS transistors, the node voltages will all be 2.5 V. + 2.5 V. [ ( )] (b) The node voltages would all be 7-63 7.105 *Figure 7.38(b) - CMOS Latchup VDD 1 0 RC 1 2 25 RL 3 4 2000 RN 2 3 2000 RP 4 0 500 Q1 3 4 0 NBJT Q2 4 3 2 PBJT .DC VDD 0 5 .01 .MODEL NBJT NPN BF=60 BR=.25 IS=1E-15 .MODEL PBJT PNP BF=60 BR=.25 IS=1E-15 .PROBE I(VDD) V(1) V(2) V(3) V(4) .OPTIONS ABSTOL=1E-12 RELTOL=1E-6 VNTOL=1E-6 .END Simulation results from B2SPICE Circuit 7_91-DC Transfer-1 (V) +0.000e+000 +1.000 +2.000 +3.000 +4.000 VDD +2.000 +1.500 +1.000 +500.000m +0.000e+000 V(1) V(2) V(3) 7.106 *Figure 7.39(b) - CMOS Latchup VDD 1 0 RC 1 2 25 RL 3 4 2000 RN 2 3 200 RP 4 0 50 Q1 3 4 0 NBJT Q2 4 3 2 PBJT .DC VDD 0 5 .01 .MODEL NBJT NPN BF=60 BR=.25 IS=1E-15 .MODEL PBJT PNP BF=60 BR=.25 IS=1E-15 .PROBE I(VDD) V(1) V(2) V(3) V(4) 7-64 .OPTIONS ABSTOL=1E-12 RELTOL=1E-6 VNTOL=1E-6 .END Simulation results from B2SPICE – Latchup does not occur! Circuit 7_92-DC Transfer-2 (V) +0.000e+000 +1.000 +2.000 +3.000 +4.000 VDD +5.000 +4.000 +3.000 +2.000 +1.000 +0.000e+000 V(1) V(2) V(3) 7.107 V DD B n+ Ohmic contact S p+ v I V SS D p+ PMOS transistor vo D n+ p-well NMOS transistor n+ S p+ B Ohmic contact n-type substrate 7-65 7.108 (a) VIH = 2K R (VDD − VTN + VTP ) (VDD − K RVTN + VTP ) − (K R − 1) (K R − 1) 1 + 3K R VIH = 2K R (VDD − VTN + VTP ) − (VDD − K RVTN + VTP ) 1 + 3K R 0 = 0 (K R − 1) 1 + 3K R 2(VDD − VTN + VTP ) − (VDD − K RVTN + VTP ) 3 + VTN 1 + 3K R 2 1 + 3K R 3 (K R − 1) 1 + 3K R + 2 1 + 3K R K R →1 lim VIH = lim K R →1 3 2(VDD − VTN + VTP ) − (VDD − VTN + VTP ) + 2VTN 5V + 3VTN + 5VTP 4 lim VIH = = DD K R →1 2 8 2V − VDD − VTN − VTP VDD − VTN + VTP VOL = IH = 2 8 (b) VIL = 2 K R (VDD − VTN + VTP ) 2 K R →1 lim VIL = (VDD − K RVTN + VTP ) (K R − 1) (K R − 1) K R + 3 K R (VDD − VTN + VTP ) − (VDD − K RVTN + VTP ) (K R − 1) K R + 3 − KR + 3 = 0 0 2 1 + VTN K R + 3 (VDD − VTN + VTP ) − (VDD − K RVTN + VTP ) 2 KR 2 KR + 3 lim VIL = lim K R →1 K R →1 1 (K R − 1) KR + 3 + 2 KR + 3 lim VIL = (VDD − VTN + VTP ) − (VDD − VTN + VTP ) 2 2V + VDD − VTN − VTP 7VDD + VTN − VTP VOH = IH = 2 8 3V + 5VTN + 3VTP VDD − VTN + VTP VDD + 3VTN + VTP NM L = VIL − VOL = DD − = 8 8 4 7V + VTN − VTP 5VDD + 3VTN + 5VTP VDD − VTN − 3VTP NM H = VOH − VIH = DD − = 8 8 4 K R →1 1 + 2VTN 3V + 5V + 3V TN TP 4 = DD 8 7-66 7.109 (a) ∆τ P For VDD = 5V , VTN = 1V , VTP = −1V τP = τ 0.322C dτ P 0.322C | =− =− P 2 Kn Kn dK n Kn P ≈ Sτ Kn P | Sτ Kn = K n dτ P = −1 τ P dK n ∆K n ∆K = − n = −(−0.25) = +0.25 | A 25% decrease in K n will cause τP Kn Kn a 25% increase in propagation delay. (b) Assuming a symmetrical inverter with VDD = 5V and VTN = 0.75V, ⎡ ⎛ V −V ⎞ 2VTN ⎤ TN − 1⎟ + ⎢ln⎜ 4 DD ⎥ K n (VDD − VTN ) ⎣ ⎝ VDD ⎠ VDD − VTN ⎦ C ⎡ ⎛ 5 − 0.75 ⎞ 2(0.75) ⎤ 0.289C C − 1⎟ + ⎢ln⎜ 4 ⎥= ⎠ 5 − 0.75 ⎦ Kn K n (5 − 0.75) ⎣ ⎝ 5 ⎡ ⎤ −4 ⎢ ⎥ τP C 2VDD VDD ⎢ ⎥ = + + 2 ⎛ ⎞ ⎢ (VDD − VTN ) K n (VDD − VTN ) 4 VDD − VTN − 1 (VDD − VTN ) ⎥ ⎟ ⎢⎜ ⎥ VDD ⎠ ⎣⎝ ⎦ ⎡ ⎤ ⎛ −4 ⎞ ⎜ ⎟ ⎢ 2(5) ⎥ 0.120C 0.289C C ⎝ 5⎠ ⎢ ⎥= = + + 5 − 0.75 ⎞ (5 − 0.75)2 ⎥ Kn (5 − 0.75)K n K n (5 − 0.75)⎢⎛ − 1⎟ ⎜4 ⎢ ⎥ ⎝ ⎠ 5 ⎣ ⎦ VTN dτ P 0.75K n ⎛ 0.120C ⎞ = ⎜ ⎟ = 0.311 τ P dVTN 0.289C ⎝ K n ⎠ τP = τP = dτ P dVTN dτ P dVTN P Sτ VTN = ⎛ 0.1 ⎞ ∆VTN = 0.311⎜ ⎟ = 0.0415 = 4.15%. ⎝ 0.75 ⎠ τP VTN A 13% increase in VTN causes an 4.2% increase in τ P . ∆τ P P ≅ Sτ Kn 7.110 *PROBLEM 7.110 - INVERTER DELAY VS RISETIME VDD 1 0 DC 2.5 V1 2 0 PULSE (0 2.5 0 0.1N 0.1N 50N 100N) MN1 3 2 0 0 MOSN W=1U L=1U AS=4P AD=4P MP1 3 2 1 1 MOSP W=1U L=1U AS=4P AD=4P C1 3 0 1PF * V2 4 0 PULSE (0 2.5 0 0.2N 0.2N 50N 100N) MN3 5 4 0 0 MOSN W=1U L=1U AS=4P AD=4P MP3 5 4 1 1 MOSP W=1U L=1U AS=4P AD=4P C3 5 0 1PF * 7-67 V3 6 0 PULSE (0 2.5 0 0.5N 0.5N 50N 100N) MN5 7 6 0 0 MOSN W=1U L=1U AS=4P AD=4P MP5 7 6 1 1 MOSP W=1U L=1U AS=4P AD=4P C5 7 0 1PF * V4 8 0 PULSE (0 2.5 0 1N 1N 50N 100N) MN7 9 8 0 0 MOSN W=1U L=1U AS=4P AD=4P MP7 9 8 1 1 MOSP W=1U L=1U AS=4P AD=4P C7 9 0 1PF * V5 10 0 PULSE (0 2.5 0 2N 2N 50N 100N) MN9 11 10 0 0 MOSN W=1U L=1U AS=4P AD=4P MP9 11 10 1 1 MOSP W=1U L=1U AS=4P AD=4P C9 11 0 1PF * .OP .TRAN 0.025N 50N .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99 +LAMBDA=.02 TOX=41.5N +CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5 +LAMBDA=.05 TOX=41.5N +CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P .PROBE V(2) V(3) V(4) V(5) V(6) V(7) V(8) V(9) V(10) V(11) .END tr 0.1 ns 0.2 ns 0.5 ns 1 ns 5 ns τP 14.6 ns 14.7 ns 14.7 ns 14.8 ns 15.8 ns 7-68 CHAPTER 8 8.1 ( a ) 256Mb = 28 210 210 = 268,435,456 bits (b) 1Gb = 210 = 1,073,741,824 bits 8 10 10 28 ( )( ) (c) 256Mb = 2 (2 )(2 )= 2 I≤ pA 1mA = 3.73 28 bit 2 bits ( ) 3 | 128kb = 2 7 210 = 217 | ( ) 228 = 211 = 2048 blocks 17 2 8.2 8.3 (a) P = CV (b) P = CV 2 DD f = 64( 10pF)(3.3) ( 1GHz)= 6.97 W 2 2 DD f = 64( 10 pF )(2.5) (3GHz)= 12 W 2 8.4 ⎛ 230 ⎞ 2⎛ 1 ⎞ 2 P = CVDD f = ⎜ ⎟( 100fF)(2.5V ) ⎜ ⎟ = 28.0 mW ⎝ 0.012 s ⎠ ⎝ 2 ⎠ 8.5 "1" = VDD = 3 V | "0": ⎛ 3 − VO 2 VO ⎞ −6 = 100 x 10 3 − 0.75 − ⎜ ⎟VO → VO = 0.667µV | "0"= 0.667 µV 2⎠ 1 1010 ⎝ ( ) 8-1 8.6 *PROBLEM 8.6 - 6-T Cell VDD 1 0 DC 3 MN1 3 2 0 0 MOSN W=4U L=2U AS=16P AD=16P MP1 3 2 1 1 MOSP W=10U L=2U AS=40P AD=40P MN2 2 3 0 0 MOSN W=4U L=2U AS=16P AD=16P MP2 2 3 1 1 MOSP W=10U L=2U AS=40P AD=40P MN3 3 0 0 0 MOSN W=4U L=2U AS=16P AD=16P MN4 2 0 0 0 MOSN W=4U L=2U AS=16P AD=16P .IC V(3)=1.55V V(2)=1.45V V(1)=3 .OP .TRAN 0.025N 10N UIC .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99 +LAMBDA=.02 TOX=41.5N +CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5 +LAMBDA=.05 TOX=41.5N +CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P .PRINT TRAN V(2) V(3) .PROBE V(2) V(3) .END 4.0V 3.0V 2.0V 1.0V 0V Time -1.0V 0s 2ns 4ns 6ns 8ns 10ns Result: t = 1.5 ns 8-2 8.7 (a) 3V (b) 0.7 V +3 V 3V 0.7 V 1.5 V 2.3 V 1.5 V 2.3 V First Case : Both transistors are in the linear region ⎛1⎞⎛ 0.7 ⎞ I DS = 25 x10−6 ⎜ ⎟⎜ 2.3 − 0.7 − ⎟0.7 = 21.88µA 2 ⎠ ⎝1⎠⎝ ⎛ W ⎞⎛ ⎛W ⎞ 0.8 ⎞ 1 21.88µA = 25 x10−6 ⎜ ⎟⎜3 − 0.7 − 0.7 − ⎟0.8 → ⎜ ⎟ ≤ 0.911 = 2 ⎠ 1.10 ⎝ L ⎠⎝ ⎝ L⎠ Second Case : Both transistors are in the linear region ⎛ ⎞⎛ ⎛ W ⎞⎛ ⎛ W ⎞ 1.09 −0.7)⎞ ( 0.8 ⎞ −6 1 ⎜ ⎟(−0.7)= 25 x10−6 ⎜ ⎟⎜3 − 1.5 − 0.7 − 0.8 → 10 x10 ⎜ ⎟⎜ 0.7 − 3 − (−0.7)− ⎟ ⎜ ⎟≤ 2 ⎟ 2 ⎠ 1 ⎝1⎠⎝ ⎝ L ⎠⎝ ⎝ L⎠ ⎠ ⎛W ⎞ 1 So ⎜ ⎟ ≤ ⎝ L ⎠ 1.10 8-3 8.8 *Problem 8.8 - WRITING THE CMOS SRAM VWL 6 0 DC 0 PULSE(0 3 1NS 1NS 1NS 100NS) VDD 3 0 DC 3 VBL1 4 0 DC 0 VBL2 5 0 DC 3 CBL1 4 0 500FF CBL2 5 0 500FF *Storage Cell MCN1 2 1 0 0 MOSN W=1U L=1U AS=4P AD=4P MCP1 2 1 3 3 MOSP W=1U L=1U AS=4P AD=4P MCN2 1 2 0 0 MOSN W=1U L=1U AS=4P AD=4P MCP2 1 2 3 3 MOSP W=1U L=1U AS=4P AD=4P MA1 4 6 2 0 MOSN W=1U L=1U AS=4P AD=4P MA2 5 6 1 0 MOSN W=1U L=1U AS=4P AD=4P * .OP .TRAN 0.01NS 20NS .NODESET V(1)=3 V(2)=0 .MODEL MOSN NMOS KP=2.5E-5 VTO=.70 GAMMA=0.5 +LAMBDA=.05 TOX=20N +CGSO=4E-9 CGDO=4E-9 CJ=2.0E-4 CJSW=5.0E-10 .MODEL MOSP PMOS KP=1.0E-5 VTO=-.70 GAMMA=0.75 +LAMBDA=.05 TOX=20N +CGSO=4E-9 CGDO=4E-9 CJ=2.0E-4 CJSW=5.0E-10 .PROBE V(1) V(2) V(3) V(4) V(5) V(6) .END 4.0V 3.0V 2.0V 1.0V 0V Time -1.0V 0s 2ns 4ns 6ns 8ns 10ns Small voltage transients occur on both cell storage nodes which die out in 5 - 7 ns. 8-4 8.9 (a ) The transistor will fully discharge CC : VC 0 = 0 V VC1 = 2.5 − VTN = 2.5 − 0.6 − 0.5 VC1 + 0.6 − 0.6 → VC1 = 1.55V | "1"= 1.55 V For VC1 = 2.5V , VTN W/L ( = 0.6 + 0.5( 2.5 + 0.6 − ) 0.6 )= 1.09V | V ≥ 2.5 + 1.09 = 3.59 V 8.10 For "0" = 0V, the bias across the source-substrate junction is 0 V, so the leakage current would be 0 and the "0" state is undisturbed. For a "1" corresponding to a positive voltage, a reverse bias across the source-substrate junction, and the diode leakage current will tend to destroy the "1" state. iL C n + "OFF" n + 8.11 CDG VG C GS VC = 2.5 − VTN = 2.5 − 0.7 − 0.5 VC + 0.6 − 0.6 → VC = 1.47V 1 sCC 1 1 + sCC sCGS ∆VW / L −2.5 = = −1.43 V CC 75 fF 1+ 1+ CGS 100 fF ( ) ∆VC = ∆VW / L = CC 8.12 QI = 60 fF (0V )+ 7.5 pF (2.5V ) | QF = 7.56 pF ( VF ) | QF = QI → VF = VF = 2.48 V | ∆V = 7.5 pF 0.06 2.5V − 2.5V = −2.5 = −19.8 mV 7.56 pF 1.56 7.5 pF 2.5V 7.56 pF 8.13 (a) "1"= +3.3 V (b) "1"= +2.5 V | VC = −VTP = 0.7 + 0.5 3.3 − VC + 0.6 − 0.6 → VC = 1.14V | "0"= 1.14 V | VC = −VTP C ( = 0.7 + 0.5( 2.5 − V + 0.6 − ) 0.6 )→ V C = 1.03V | "0"= 1.03 V 8-5 8.14 Note that the simulation results in Fig. 9.28 assume that the word line is also driven higher than 3 V. For this case: (a) V (b) V C = 5 − VTN = 5 − 0.7 − 0.5 VC + 0.6 − 0.6 → VC = 3.66 V = 0.7 − 0.5 1.3 + 0.6 − 0.6 = 1.00V TN ( ( ) ) VGS − VTN = 5 − 1.3 − 1.00 = 2.7V | VDS = 3.7 − 1.3 = 2.4V → linear region ⎛1⎞⎛ 2.4 ⎞ iDS = 60 x10−6 ⎜ ⎟⎜5 − 1.3 − 1.00 − ⎟2.4 = 216 µA which agrees with the text. 2 ⎠ ⎝1⎠⎝ (a) "0"= +0 V | V = 3 − V = 3 − 0.7 − 0.5( V + 0.6 − 0.6 )→ V = 1.90V | "1"= 1.90 V (b) A "0" will have 0 V across the drain - substrate junction, so no leakage occurs. C TN C C 8.15 A "1" will have a reverse bias of 1.9 V across the junction, so the junction leakage will tend to destroy the "1" level. (Note that this discussion ignores subthreshold leakage through the FET which has not been discussed in the text.) 8.16 (a) "1"= +5 V | V = −V = 0.8 + 0.65( 5 − V (b) V = −0.8 − 0.65( 5 + 0.6 − 0.6 )= −1.83V C TP TP C + 0.6 − 0.6 → VC = 1.60V | "0"= +1.60 V | VW/L ≤ −1.83V ) 8.17 ⎛1⎞⎛ 0.6 ⎞ Original : iD = 60 x10−6 ⎜ ⎟⎜ 3 − 1.3 − 1 − ⎟0.6 = 14.4µA 2 ⎠ ⎝1⎠⎝ ⎛ 1⎞⎛ 2.4 ⎞ New : iD = 60 x10−6 ⎜ ⎟⎜5 − 1.3 − 1 − ⎟2.4 = 216µA. 2 ⎠ ⎝ 1⎠⎝ 2.4 5 − 1.3 − 1 − 1.2 1.5 = = 3.75 | VDS ratio : =4 Gate drive terms: 0.6 3 − 1.3 − 1 − 0.3 0.4 Improved gate drive yields a 3.75 times improvement, although it is reduced by the larger VDS term. Improved drain - source voltage yields a 4 times improvement. 4 x 3.75 = 15. 2⎛ 1 ⎞ 2 f = (0.5) 230 ( 100fF)(2.5V ) ⎜ P = CVDD ⎟ = 33.6 mW ⎝ 0.01s ⎠ 8.18 ( ) 8-6 8.19 *Problem 8.19 - 4-T Refresh SRAM VWL 3 0 DC 0 PULSE(0 3 1NS 1NS 1NS 6NS) VBL 4 0 DC 3 VBLB 5 0 DC 3 CC1 1 0 50FF CC2 2 0 50FF *Storage Cell MCN1 2 1 0 0 MOSN W=4U L=2U AS=16P AD=16P MCN2 1 2 0 0 MOSN W=4U L=2U AS=16P AD=16P MA1 4 3 2 0 MOSN W=4U L=2U AS=16P AD=16P MA2 5 3 1 0 MOSN W=4U L=2U AS=16P AD=16P .IC V(1)=0 V(2)=1 V(3)=0 V(4)=3 V(5)=3 .OP .TRAN 0.01NS 20NS UIC .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99 +LAMBDA=.02 TOX=41.5N +CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .PROBE V(1) V(2) V(3) V(4) V(5) .END 4.0V Note the very slow recovery due to relatively high threshold and gamma values relative to the power supply voltage. 3.0V vWL 2.0V D 1.0V D 0V Time -1.0V 0s 5ns 10ns 15ns 20ns 8.20 *Problem 8.20 4-T READ ACCESS VPC 7 0 DC 3 PULSE(3 0 1NS .5NS .5NS 100NS) VWL 6 0 DC 0 PULSE(0 3 2NS .5NS .5NS 100NS) VDD 3 0 DC 3 CBL1 4 0 1PF CBL2 5 0 1PF *Storage Cell MCN1 2 1 0 0 MOSN W=4U L=2U AS=16P AD=16P MCN2 1 2 0 0 MOSN W=4U L=2U AS=16P AD=16P 8-7 MA1 4 6 2 0 MOSN W=4U L=2U AS=16P AD=16P MA2 5 6 1 0 MOSN W=4U L=2U AS=16P AD=16P CC1 1 0 50FF CC2 2 0 50FF * *Sense Amplifier MSN1 4 5 0 0 MOSN W=4U L=2U AS=16P AD=16P MSP1 4 5 3 3 MOSP W=4U L=2U AS=16P AD=16P MSN2 5 4 0 0 MOSN W=4U L=2U AS=16P AD=16P MSP2 5 4 3 3 MOSP W=4U L=2U AS=16P AD=16P MRS 5 7 4 0 MOSN W=4U L=2U AS=16P AD=16P * .OP .TRAN 0.01NS 40NS UIC .IC V(1)=1.5 V(2)=0 V(3)=3 V(4)=1.7 V(5)=1.7 V(6)=0 V(7)=3 .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99 +LAMBDA=.02 TOX=41.5N +CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5 +LAMBDA=.05 TOX=41.5N +CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P .PROBE V(1) V(2) V(3) V(4) V(5) V(6) V(7) .END vPC 3.0V vWL BL 2.0V D 1.0V D BL 0V Time 0s 10ns 20ns 30ns 40ns 8-8 8.21 +3.3 V 10 1 Each inverter will have vI = v O . vI 5 1 vO Equating inverter drain currents : ⎛100x10-6 ⎞⎛ 5 ⎞ ⎛ 40x10-6 ⎞⎛ 10 ⎞ 2 2 − 0.7 = v ⎜ ⎟⎜ ⎟( I ⎜ ⎟⎜ ⎟(3.3 − v I − 0.7) ) 2 ⎝ ⎠⎝ 1 ⎠ ⎝ 2 ⎠⎝ 1 ⎠ → vO = vI = 1.597 V Sense amp current = 2iDS = 402 µA P = 1024(3.3V)(402 µA)= 1.36 W 8.22 V C G DG C GS C BL C 500 fF 500 fF BL The precharge transistor is operating in the linear region with VDS = 0 1 " WL + CGSOW CGS = Cox 2 −14 1 3.9 8.854 x10 F / cm CGS = 10−3 cm 10−4 cm + 4 x10−11 F / cm 10−3 cm = 48.6 fF 2 2 x10−6 cm 1 ∆VG -3 sC BL ∆V(s) = ∆VG (s) → ∆V = = = 0.266 V 500 1 C BL 1 + +1 +1 48.6 sCGS sC BL CGS ( ) ( )( )( )( ) This value provides a good estimate of the drop observed in Fig. 8.25. 8-9 8.23 The precharge transistor is operating in the linear region with VDS = 0 1 " WL + CGSOW CGS = Cox 2 −14 1 3.9 8.854 x10 F / cm 10−3 cm 10−4 cm + 4 x10−11 F / cm 10−3 cm = 48.6 fF CGS = −6 2 2 x10 cm Using CBL = 500 fF as in the previous problem, 1 ∆VG 3 sC BL ∆VG (s)→ ∆V = = = 0.266 V ∆V(s) = 500 1 1 C BL + +1 +1 48.6 CGS sCGS sC BL ( ) ( )( )( )( ) This value provides a good estimate of the observed change in Fig. 8.29. The source - substrate diode will clamp the voltage to ∆V ≤ 0.7 V. 8.24 The bitline will charge to an initial voltage of VBL = 3 - VTN VTN = 0.7 + 0.5 3 − VTN + 0.6 − 0.6 → VTN = 1.10V | VBL = 1.90V 1.9V )= 1.9 pC The initial charge QI on C BL : QI = 10−12 F ( After charge sharing : VBL = 1.9 pC = 1.81V . The voltage will be restored to 1.90V 1.05 pF ( ) by the transistor. The total charge delivered through the transistor is 9.45 x10−14 C ∆Q = 0.09V ( 1.05 pF )= 0.0945 pC | ∆vO = = 0.945 V 10−13 F 0.945 This sense amplifier provides a voltage gain of AV = = 10.5 0.09 8.25 *PROBLEM 8.24 Charge Transfer Sense Amplifier VSW 5 0 DC 0 PULSE(0 3 2NS .5NS .5NS 100NS) VGG 3 0 DC 3 CL 2 0 100FF CBL 4 0 1PF CC 6 0 50FF M1 2 3 4 0 MOSN W=100U L=2U M2 4 5 6 0 MOSN W=8U L=2U .IC V(2)=3 V(4)=1.9 .TRAN 0.02NS 100NS UIC .MODEL MOSN NMOS KP=25U VTO=0.7 GAMMA=0.5 PHI=0.6 .PROBE V(2) V(3) V(4) V(5) V(6) .END 8-10 3.0V vSW vO 2.0V vBL 1.0V 0V Time 0s 20ns 40ns 60ns 80ns 100ns 8-11 8.26 *PROBLEM 8.26 - Cross-Coupled Latch VDD 1 0 DC 3.3 MN1 3 2 0 0 MOSN W=2U L=1U AS=16P AD=16P MP1 3 2 1 1 MOSP W=2U L=1U AS=16P AD=16P MN2 2 3 0 0 MOSN W=2U L=1U AS=16P AD=16P MP2 2 3 1 1 MOSP W=2U L=1U AS=16P AD=16P CBL1 3 0 1PF CBL2 2 0 1PF .IC V(3)=1V V(2)=1.25V V(1)=5 .OP .TRAN 0.05N 50N UIC .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99 +LAMBDA=.02 TOX=41.5N +CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5 +LAMBDA=.05 TOX=41.5N +CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P .PRINT TRAN V(2) V(3) .PROBE V(2) V(3) .END 4.0 3.0 2.0 1.0 0 0s Time 5ns V2(CBL1) V2(CBL2) 10ns 15ns 20ns 25ns 30ns 35ns 40ns Time 8-12 8.27 *PROBLEM 8.27 - Cross-Coupled Latch VDD 1 0 DC 3 VSW 4 0 DC 0 PULSE(3 0 5NS 1NS 1NS 100NS) MPC 3 4 2 0 MOSN W=20U L=2U AS=80P AD=80P MN1 3 2 0 0 MOSN W=4U L=2U AS=16P AD=16P MP1 3 2 1 1 MOSP W=8U L=2U AS=32P AD=32P MN2 2 3 0 0 MOSN W=4U L=2U AS=16P AD=16P *MN2 2 3 0 0 MOSN W=4.4U L=2U AS=17.6P AD=17.6P MP2 2 3 1 1 MOSP W=8U L=2U AS=32P AD=32P CBL1 3 0 400FF CBL2 2 0 400FF .OP .TRAN 0.05N 50N .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99 +LAMBDA=.02 TOX=41.5N +CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5 +LAMBDA=.05 TOX=41.5N +CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P .PROBE V(2) V(3) V(4) .END 3.0V vPC 2.0V D1 and D 2 1.0V 0V Time 0s 10ns 20ns 30ns 40ns 50ns The latch is perfectly balanced in Part (a) and the voltage levels remain symmetrical even after the PC transistor turns off. This would not happen in the real case because of small asymmetries and noise in the latch. Even a small capacitive imbalance will cause the latch to assume a preferred state. Try setting CBL2 = 425 fF in Part (a) for example. The asymmetry in the latch in Part (b) causes it to switch to a preferred state. 8-13 8.28 *PROBLEM 8.28 - Clocked NMOS Sense Amplifier VPC 2 0 DC 0 PULSE(3 0 1NS .5NS .5NS 250NS) VWL 6 0 DC 0 PULSE(0 3 2NS .5NS .5NS 250NS) VLC 9 0 DC 0 PULSE(0 3 3NS .5NS .5NS 250NS) VDD 3 0 DC 3 CBL1 5 0 2PF CBL2 4 0 2PF *Storage Cell MA1 5 6 1 0 MOSN W=2U L=2U AS=8P AD=8P CC 1 0 100FF *Dummy Cell MA2 4 6 7 0 MOSN W=2U L=2U AS=8P AD=8P CD 7 0 50FF *Sense Amplifier MPC 5 2 4 0 MOSN W=10U L=2U AS=40P AD=40P ML1 3 2 4 0 MOSN W=10U L=2U AS=40P AD=40P ML2 3 2 5 0 MOSN W=10U L=2U AS=40P AD=40P MS1 5 4 8 0 MOSN W=50U L=2U AS=200P AD=200P MS2 4 5 8 0 MOSN W=50U L=2U AS=200P AD=200P MLC 8 9 0 0 MOSN W=50U L=2U AS=200P AD=200P * .OP .TRAN 0.01NS 250NS .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99 +LAMBDA=.02 TOX=41.5N +CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .PROBE V(1) V(2) V(3) V(4) V(5) V(6) V(7) V(8) V(9) .END 3.0V 2.0V 1.0V 0V Time 0s 50ns 100ns 125ns 8-14 With only a 3 V power supply, the maximum bit-line differential is only 1.14 V which is achieved in 120 ns. (This is relatively slow due to the discharge of large bitline capacitances and the relatively large threshold voltage of the NMOS transistors.) 8-15 8.29 *PROBLEM 8.29 - Clocked NMOS Sense Amplifier VPC 2 0 DC 0 PULSE(5 0 1NS .5NS .5NS 250NS) VWL 6 0 DC 0 PULSE(0 5 2NS .5NS .5NS 250NS) VLC 9 0 DC 0 PULSE(0 5 3NS .5NS .5NS 250NS) VDD 3 0 DC 5 CBL1 5 0 2PF CBL2 4 0 2PF *Storage Cell MA1 5 6 1 0 MOSN W=2U L=2U AS=8P AD=8P CC 1 0 100FF *Dummy Cell MA2 4 6 7 0 MOSN W=2U L=2U AS=8P AD=8P CD 7 0 50FF *Sense Amplifier MPC 5 2 4 0 MOSN W=10U L=2U AS=40P AD=40P ML1 3 2 4 0 MOSN W=10U L=2U AS=40P AD=40P ML2 3 2 5 0 MOSN W=10U L=2U AS=40P AD=40P MS1 5 4 8 0 MOSN W=50U L=2U AS=200P AD=200P MS2 4 5 8 0 MOSN W=50U L=2U AS=200P AD=200P MLC 8 9 0 0 MOSN W=50U L=2U AS=200P AD=200P * .OP .TRAN 0.01NS 250NS .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99 +LAMBDA=.02 TOX=41.5N +CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .PROBE V(1) V(2) V(3) V(4) V(5) V(6) V(7) V(8) V(9) .END 6.0V 4.0V 2.0V 0V Time 0s 5ns 10ns 15ns 20ns 25ns 30ns With the 5 V power supply, the maximum bit-line differential is 1.75 V. A 1.5 V differential is achieved in approximately 15 ns, which is much faster than the 3 V case. 8.30 8-16 *PROBLEM 8.30 - Cascaded Inverter Pair VDD 1 0 DC 3 VI 2 0 DC 0 MN1 3 2 0 0 MOSN W=2U L=1U AS=16P AD=16P MP1 3 2 1 1 MOSP W=2U L=1U AS=16P AD=16P MN2 4 3 0 0 MOSN W=2U L=1U AS=16P AD=16P MP2 4 3 1 1 MOSP W=2U L=1U AS=16P AD=16P .OP .DC VI 0 3 0.001 .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99 +LAMBDA=.02 TOX=41.5N +CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5 +LAMBDA=.05 TOX=41.5N +CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P .PROBE V(2) V(3) V(4) .END 3.0V 2.0V 1.0V vO 0V 0V 0.5V 1.0V 1.5V 2.0V 2.5V 3.0V Results: 0 V, 1.429 V, 3 V 8.31 (a) The array requires: ( 12 transistors/row) 212 rows + ( 1 load transistor/row) 212 rows = 13 212 = 53,248 transistors. The 24 inverters require an additional 48 transistors. N = 53,296 transistors | (b) The number is the same. ( ) ( ) ( ) 8-17 8.32 (a) NMOS Pass Transistor Tree : 2 x 20 + 21 + 22 + 23 + 2 4 + 25 + 26 = 254 Transistors ( ) 2 logic inverters per level = 14 inverters = 28 transistors. Total = 282 Transistors (b) An estimate : 128 data bits requires 128 7 - input gates for data selectors; 1 - 128 input NOR gate; 14 address bit inverters Total = 128(8)+ 1( 129)+ 14(2) = 1181 transistors without looking closely at the logic detail. A number of additional inverters may be needed, and the 128 input gate can likely be replaced with a smaller NOR tree. 8.33 A0 A1 A2 Clock V Clock DD 0 V DD 1 V DD 2 V DD 3 V DD 4 V DD 5 V DD 6 V DD 7 NMOS Transistor 8-18 8.34 (a) The output of the first NMOS transistor will be V1 = 5 − VTN = 5 − 0.75 + 0.55 V1 + 0.6 − 0.6 → V1 = 3.55V | VTN = 1.45V [ ( )] (b) The node voltages will all be C1VDD + C2 (0) The output of the other gates reaches this same value. All three nodes = 3.55V. + 5 V. 8.35 Charge sharing occurs. Assuming C2 and C3 are discharged (the worst case) (a) V 2C2VDD 2 2 = VDD | Node B drops to VDD . C1 + C2 2C2 + C2 3 3 ⎛2 ⎞ 2 3 C V ⎜ ⎟ 2 DD + C V + C 0 C ( 1 2 )3 DD 3 ( ) V 1 ⎝3 ⎠ b V = = = DD | Node B drops to VDD . () B 2C2 + C2 + C2 C1 + C2 + C3 2 2 B = = (c) V B = C1VDD RC2VDD R C = = VDD ≥ VIH where R = 1 C1 + C2 + C3 RC2 + C2 + C2 R + 2 C2 or R ≥ 2VIH VDD − VIH R( VDD − VIH ) ≥ 2VIH Using VDD = 5V , VTN = 0.7V , VTP = −0.7V in Eq. (7.9) : VIH = 5(5)+ 3(0.7)+ 5(−0.7) 8 = 2.95V | R ≥ 2(2.95) 2VIH = = 2.88 | C1 ≥ 2.88C2 VDD − VIH 2.05 8.36 Z = A0 + A1 + A2 VDD Clock A0 A1 A 2 Z C1 8-19 8.37 B7 W0 W1 W2 W3 W4 W5 1 0 1 0 0 0 B6 0 1 1 0 0 1 B5 1 0 0 1 0 0 B4 1 0 0 0 0 0 VDD Clock W L B3 0 0 0 1 1 0 B2 0 1 1 0 1 0 B1 0 1 0 1 1 0 B0 0 0 0 1 0 0 8.38 W 5 W L 4 W L W L *PROBLEM 8.38 - Simplified ROM Cross-Section VCLK 1 0 DC 0 PULSE(0 5 2.5NS 1NS 1NS 25NS) VW5 3 0 DC 0 PULSE(0 5 4.5NS 1NS 1NS 25NS) VDD 5 0 DC 5 MPC 4 1 5 5 MOSP W=4U L=2U AS=16P AD=16P MNC 2 1 0 0 MOSN W=4U L=2U AS=16P AD=16P MW5 4 3 2 0 MOSN W=4U L=2U AS=16P AD=16P MWW 4 0 2 0 MOSN W=16U L=2U AS=64P AD=64P .OP .TRAN 0.01NS 15NS .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99 +LAMBDA=.02 TOX=41.5N +CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5 +LAMBDA=.05 TOX=41.5N +CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P .PROBE V(1) V(2) V(3) V(4) V(5) .END 8-20 6.0V vO 4.0V vCLK 2.0V vW5 0V Time 0s 2ns 4ns 6ns 8ns 10ns 12ns 8.39 W1 W2 W3 B5 0 1 0 B4 0 0 1 B3 1 0 1 B2 0 1 1 B1 1 0 0 B0 0 1 1 8.40 B2 W0 W1 W2 W3 1 1 1 0 B1 0 1 0 1 B0 1 0 1 0 Note that the input lines are active low. 8-21 8.41 +5 V 4 1 (2.64V) Q 2 1 OFF 2 1 Q 0.7V MR 4 1 +5 V Regenerative switching of the cell will take place when the voltage at Q is pulled low enough by transistor MR that the voltage at Q rises above the NMOS transistor threshold voltage. Equating drain currents for this condition yields the value of V Q . It appears that the NMOS transistor will be in the linear region, and the PMOS transistor will be saturated. ⎛ ⎞⎛ 2 4 x10−5 ⎛ 4 ⎞ 0.7 ⎞ −4 2 For VDD = 5V, ⎜ ⎟(5 − VQ − 0.7) = 10 ⎜ ⎟⎜VQ − 0.7 − ⎟0.7 → VQ = 2.64V which agrees with 2 ⎝1⎠ 2 ⎠ ⎝ 1 ⎠⎝ the assumptions. Now, MR must be large enough to force VQ = 2.64V . MR and the PMOS load transistor are both in the linear region. ⎛ 4 ⎞⎛ ⎛ ⎞ ⎛ ⎛W ⎞ 1.16 2.36 ⎞ 2.64 ⎞ −4 W 4 x10−5 ⎜ ⎟⎜ 5 − 0.7 − 0.7 − ⎟2.36 ≤ 10 ⎜ ⎟ ⎜ 5 − 0.7 − ⎟2.64 → ⎜ ⎟ ≥ 2 ⎠ 2 ⎠ 1 ⎝ 1 ⎠⎝ ⎝ L ⎠R ⎝ ⎝ L ⎠R 8.42 The inputs are active in the low voltage state. V1 low sets the latch and V2 low resets the latch. V1 = S V2 = R . 8-22 8.43 *PROBLEM 8.43 - D-Latch VDD 7 0 DC 2.5 VI 1 0 DC 2.5 VCLK 2 0 DC 0 PULSE(0 2.5 3NS 1NS 1NS 5NS) VNCLK 3 0 DC 0 PULSE(2.5 0 3NS 1NS 1NS 5NS) MTN1 1 2 4 0 MOSN W=2U L=1U AS=16P AD=16P MTP1 1 3 4 7 MOSP W=2U L=1U AS=16P AD=16P MIN1 5 4 0 0 MOSN W=2U L=1U AS=16P AD=16P MIP1 5 4 7 7 MOSP W=2U L=1U AS=16P AD=16P MIN2 6 5 0 0 MOSN W=2U L=1U AS=16P AD=16P MIP2 6 5 7 7 MOSP W=2U L=1U AS=16P AD=16P MTN2 6 3 4 0 MOSN W=2U L=1U AS=16P AD=16P MTP2 6 2 4 7 MOSP W=2U L=1U AS=16P AD=16P .OP .TRAN 0.01N 12N .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99 +LAMBDA=.02 TOX=41.5N +CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5 +LAMBDA=.05 TOX=41.5N +CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P .PROBE V(1) V(2) V(3) V(4) V(5) V(6) .END 3.0V 2.0V 1.0V 0V -1.0V 0s 1ns 2ns 3ns 4ns 5ns 6ns 7ns 8ns 9ns 10ns 11ns 12ns 8-23 8.44 *PROBLEM 8.44 - Master-Slave Flip-Flop VDD 10 0 DC 5 VI 1 0 DC 5 PWL(0 5 17.4NS 5 17.6NS 0 30NS 0) VCLK 2 0 DC 0 PULSE(0 5 0NS 2.5NS 2.5NS 7.5NS) VNCLK 3 0 DC 0 PULSE(5 0 0NS 2.5NS 2.5NS 7.5NS) MTN1 1 2 4 0 MOSN W=4U L=2U AS=16P AD=16P MTP1 1 3 4 10 MOSP W=4U L=2U AS=16P AD=16P MIN1 5 4 0 0 MOSN W=4U L=2U AS=16P AD=16P MIP1 5 4 10 10 MOSP W=4U L=2U AS=16P AD=16P MIN2 6 5 0 0 MOSN W=4U L=2U AS=16P AD=16P MIP2 6 5 10 10 MOSP W=4U L=2U AS=16P AD=16P MTN2 6 3 4 0 MOSN W=4U L=2U AS=16P AD=16P MTP2 6 2 4 10 MOSP W=4U L=2U AS=16P AD=16P * MTN3 6 3 7 0 MOSN W=4U L=2U AS=16P AD=16P MTP3 6 2 7 10 MOSP W=4U L=2U AS=16P AD=16P MIN3 8 7 0 0 MOSN W=4U L=2U AS=16P AD=16P MIP3 8 7 10 10 MOSP W=4U L=2U AS=16P AD=16P MIN4 9 8 0 0 MOSN W=4U L=2U AS=16P AD=16P MIP4 9 8 10 10 MOSP W=4U L=2U AS=16P AD=16P MTN4 9 2 7 0 MOSN W=4U L=2U AS=16P AD=16P MTP4 9 3 7 10 MOSP W=4U L=2U AS=16P AD=16P .IC V(1)=5 V(2)=0 V(3)=5 V(4)=0 V(5)=5 V(6)=0 V(7)=0 + V(8)=5 V(9)=0 V(10)=5 .TRAN 0.05N 20N UIC .MODEL MOSN NMOS KP=5E-5 VTO=0.91 GAMMA=0.99 +LAMBDA=.02 TOX=41.5N +CGSO=330P CGDO=330P CJ=3.9E-4 CJSW=510P .MODEL MOSP PMOS KP=2E-5 VTO=-0.77 GAMMA=0.5 +LAMBDA=.05 TOX=41.5N +CGSO=315P CGDO=315P CJ=2.0E-4 CJSW=180P .PROBE V(1) V(2) V(3) V(4) V(6) V(7) V(9) .END The flip-flop operates normally. Data is transferred to the master following the first clock transition and to the slave after the second clock transition. The maximum rise and fall times are highly dependent upon the position of the data transition edge. It is interesting to experiment with the data delay to see the effect. At some point the flip-flop will fail. 8-24 6.0V Data 4.0V CLK CLK Data 2.0V Master Slave 0V Time 0s 5ns 10ns 15ns 20ns 8-25 CHAPTER 9 9.1 Since VREF = −1.25V , and v I = −1.6V , Q1 is off and Q2 is conducting. vC1 = 0 V and vC 2 = −α F I EE RC ≅ − I EE RC = −(2 mA)(350Ω) = −0.700 V 9.2 ⎛ ∆V ⎞ IC 2 0.995α F I EE = exp⎜ BE ⎟ ⇒ ∆VBE = 0.025ln = 0.132V I C1 0.005α F I EE ⎝ VT ⎠ (a) v I = VREF + ∆VBE = −1.25 + 0.132 = −1.12 V v I = VREF + ∆VBE = −1.25 − 0.132 = −1.38 V (b) v I = VREF + ∆VBE = −2.00 + 0.132 = −1.87 V v I = VREF + ∆VBE = −2.00 − 0.132 = −2.13 V 9.3 Since VREF = −2V , and v I = −1.6V , Q2 is off and Q1 is conducting. vC 2 = 0 V and vC1 = −α F I EE RC ≅ − I EE RC = −(2.5mA)(700Ω)= −1.75 V Note that Q1 is beginning to enter the saturation region of operation, but VBC = +0.15 V is not really enough to turn on the collector-base diode. (See Problems 9.5 or 5.61.) 9.4 vI = VREF + 0.3V ⇒ Q1 on; Q2 off. IC1 = α F I EE ≅ I EE = 0.3mA | IC 2 = 0 vC1 = 0 − IC1 (R1 + RC )= −0.3mA(3.33kΩ + 2 kΩ) = −1.60 V vC 2 = 0 − IC 1 R1 = −0.3mA(3.33kΩ) = −0.999 V 9-1 9.5 With VBE = 0.7 and VBC = 0.3, the transistor is technically in the saturation region, but calculating the currents using the transport model in Eq. (5.13) yields βF = 0.98 0.2 αF αR = = 49 | β R = = = 0.25 1 − α F 1 − 0.98 1 − α R 1 − 0.2 ⎡ ⎛ 0.7 ⎞ ⎛ 0.3 ⎞⎤ 10−15 ⎡ ⎛ 0.3 ⎞ ⎤ iC = 10−15 ⎢exp⎜ ⎟ − exp⎜ ⎟⎥ − ⎟ − 1⎥ = 1.446 mA ⎢exp⎜ ⎝ 0.025 ⎠⎦ 0.25 ⎣ ⎝ 0.025 ⎠ ⎦ ⎣ ⎝ 0.025 ⎠ ⎡ ⎛ 0.7 ⎞ ⎛ 0.3 ⎞⎤ 10−15 ⎡ ⎛ 0.7 ⎞ ⎤ iE = 10−15 ⎢exp⎜ − exp ⎟ ⎟⎥ + ⎟ − 1⎥ = 1.476 mA ⎜ ⎢exp⎜ ⎝ 0.025 ⎠⎦ 49 ⎣ ⎝ 0.025 ⎠ ⎦ ⎣ ⎝ 0.025 ⎠ 10−15 ⎡ ⎛ 0.7 ⎞ ⎤ 10−15 ⎡ ⎛ 0.3 ⎞ ⎤ iB = ⎟ − 1⎥ + ⎟ − 1⎥ = 29.52 µA ⎢exp⎜ ⎢exp⎜ 49 ⎣ ⎝ 0.025 ⎠ ⎦ 0.25 ⎣ ⎝ 0.025 ⎠ ⎦ At 0.3 V, the collector-base junction is not heavily forward-biased compared to the baseemitter junction, and IC = 48.99 IB ≅ β F IB . The transistor still acts as if it is operating in the forward-active region. 9.6 (a) For Q2 off, VH = 0 V. For Q2 on, IC ≈ IE and = 100 µA VL ≅ −4000 I E = −0.400 V 1.1x10 4 (b) Yes, these voltages are symmetrically positioned above and below VREF, i. e. VREF ± 0.2 V, and the current will be fully switched. See Parts (d) and (e). 0 − 0.7 − (−2) 0.4V = 118 µA R = = 3.39 kΩ (c) For v I = 0, IC ≅ I E = 4 118µA 1.1x10 IE = (d) Q2 is cutoff. Q1 is saturated with VBC = +0.4 V. (e) Q1 is cutoff. Q2 is saturated with VBC = +0.2 V. (f) 0.2 V and 0.4 V are not large enough to heavily saturate Q1 or Q2. Although the transistors are technically operating in the saturation region, the transistors still behave as if they are in the forward-active region. (See problem 10.5). −0.2 − 0.7 − (−2) 9-2 9.7 (a) For v = 100 µA | VL ≅ −4000 I E = −0.400 V 1.1x10 4 0 − 0.7 − (−2) 0.4V = 118 µA | R = = 3.39 kΩ For vI = VH = 0V , IC ≅ I E = 4 118µA 1.1x10 ⎛100µA + 118µA ⎞ P = 2V ⎜ ⎟ = 218 µW 2 ⎝ ⎠ R 11kΩ RC1 4 kΩ RC 2 3.39 kΩ ' ' ' = EE = = 2.20 kΩ | RC = = 800 Ω | RC = = 678 Ω (b) REE 1 = 2 = 5 5 5 5 5 5 I = VL , I E = −0.2 − 0.7 − (−2) 9.8 VH = 0 − VBE = −0.7 V | VL = −(5mA)(200Ω)− 0.7 = −1.70 V VREF = VH + VL = −1.2 V | ∆V = (5mA)(200Ω)= 1.00 V 2 9.9 VH = 0 − VBE = −0.7 V | VL = −( 1mA)(600Ω)− 0.7 = −1.30 V VREF = 9.10 VH + VL = −1.0 V | ∆V = ( 1mA)(600Ω) = 0.600 V 2 2 kΩ = 500Ω 4 I EE = 4(0.3mA) = 1.2 mA | I3 = I 4 = 4(0.1mA)= 0.4 mA | RC = 9.11 (a) R C = ∆V 0.8V = = 2.67 kΩ | VH = 0 − VBE = −0.7 V | VL = −0.8 − VBE = −1.5 V I EE 0.3mA VH + VL = −1.10 V 2 ⎡ ⎡ ⎛ ∆V ⎞⎤ 0.8 ⎞⎤ ⎛ 0.8 ∆V − 1⎟⎥ = − 0.025V ⎢1 + ln⎜ − 1⎟⎥ = 0.289 V (b) NM H = NM L = 2 − VT ⎢1 + ln⎜ ⎝ 0.025 ⎠⎦ ⎠⎦ 2 ⎝ VT ⎣ ⎣ (c) For Q1: VCB = -0.8 - (-0.7) = -0.1 V which represents a slight forward bias, but it is not enough to turn on the diode. For Q2: VCB = -0.8 - (-1.10) = +0.3 V which represents a reverse bias. Both values are satisfactory for operation of the logic gate. VREF = 9-3 9.12 (b) For Q1 on and Q2 off, IC1 = α F I EE ≅ I EE = 0.3mA | IC2 = 0 VH = 0 − IC1 R1 − 0.7V = −0.3mA(3.33kΩ)− 0.7 = −1.70 V ∆V = VH − VL = 0.600 V VL = 0 − IC1 (R1 + RC )− 0.7V = −0.3mA(3.33kΩ + 2 kΩ)− 0.7V = −2.30 V VH + VL = −2.0V = VREF | Yes, the input and output voltage levels are 2 compatible with each other and are symmetrically placed around VREF. (c) R1 RC RC Q 3 Q 4 v O2 v O1 vI 0.1 mA Q 1 Q 2 V REF 0.1 mA IEE - 5.2 V 9.13 (a) See Prob. 9.12 (b) ∆V = α F EE I RC ≅ I EE RC | RC = VH = 0 − α F I EE R1 − VBE VL = 0 − α F I EE (R1 + RC )− VBE ≅ − I EE (R1 + RC )− VBE = −1.5mA( 1067)− 0.7V = −2.30 V VREF = VH + VL −1.90 − 2.30 = = −2.10 V 2 2 0.4V = 267 Ω 1.5mA ≅ − I EE R1 − VBE = −1.5mA(800)− 0.7V = −1.90 V 9-4 9.14 ∆V = ∆VBE + ∆iB 4 RC | Let the Fanout = N; β F = 30. Then there will be N base currents that must be supplied from emitter - follower transistor Q4 : ∆i E4 = N ∆VBE = VT ln ∆iB 4 = I EE βF + 1 ⎡ ⎤ ⎛ ∆I ⎞ I EE I C 4 + ∆I C 4 I + ∆I E 4 ⎥ = VT ln E 4 = VT ln⎜1 + E 4 ⎟ = 0.025ln⎢1 + N IC 4 IE 4 IE 4 ⎠ β + 1 I ⎝ ⎢ ⎥ ( ) F E 4 ⎣ ⎦ ∆i E4 I EE =N 2 βF + 1 (β F + 1) ⎛ 0.3mA 0.3mA ⎞ ⎟+ N ∆V = ∆VBE + ∆iB 4 RC = 0.025ln⎜ 1 + N 2 kΩ | ∆V ≤ 0.025 2 ⎜ ⎟ 31 0.1 mA ( ) ⎝ ⎠ (31) ⎛ 3 N ⎞ 0.6 N 0.025 = 0.025ln⎜1 + ⎟+ 31 ⎠ (31)2 ⎝ | Using MATLAB or HP - Solver : N ≤ 11.01 → N = 11 9.15 ' ' 1.85kΩ) = 14.8 kΩ | RC 2 kΩ)= 16.0 kΩ RC 1 = 8 RC 1 = 8( 2 = 8 RC 2 = 8( ' REE = 8 REE = 8( 11.7 kΩ)= 93.6 kΩ | R' = 8 R = 8(42 kΩ)= 336 kΩ 9.16 ' ( a ) RC 1 = RC1 1850Ω RC 2 2000Ω ' = = 231 Ω | RC = = 250 Ω 2 = 8 8 8 8 R 11.7kΩ R 42 kΩ ' REE = EE = = 1.46 kΩ | R' = = = 5.25 kΩ 8 8 8 8 ' ' ( b ) RC 1.85kΩ) = 9.25 kΩ | RC 2 kΩ)= 10.0 kΩ 1 = 5 RC 1 = 5( 2 = 5 RC 2 = 5( ' REE = 5 REE = 5( 11.7 kΩ) = 58.5 kΩ | R' = 5R = 5(42 kΩ)= 210 kΩ 9-5 9.17 ∆V = α F I EE RC ≅ I EE RC = 0.2 mA(2kΩ) = 0.400 V VL = 0 − α F I EE (R1 + RC )− VBE ≅ − I EE (R1 + RC )− VBE = −0.2 mA(4 kΩ)− 0.7V = −1.50 V VREF = VH + VL −1.10 − 1.50 = = −1.30 V 2 2 ⎡ ⎡ ⎛ ∆V ⎞⎤ 0.4 ⎞⎤ ⎛ 0.4 ∆V − VT ⎢1 + ln⎜ − 1⎟⎥ = − 0.025V ⎢1 + ln⎜ − 1⎟⎥ = 0.107 V NM L = NM H = 2 ⎝ 0.025 ⎠⎦ ⎠⎦ 2 ⎝ VT ⎣ ⎣ IE3 + IE 4 = VH = 0 − α F I EE R1 − VBE ≅ − I EE R1 − VBE = −0.2 mA(2 kΩ)− 0.7V = −1.10 V R P = 28µA(2V )+ 0.2 mA(5.2V )= 1.10 mW 9.18 NM H = ⎡ ⎛ ∆V ⎞⎤ ∆V − VT ⎢1 + ln⎜ − 1⎟⎥ | 2 ⎝ VT ⎠⎦ ⎣ [V − (−2)]+ [V − (−2)] = (4 − 1.10 − 1.50)V = 28.0 µA H L 50kΩ 0.1V = ∆V − 0.025V 1 + ln (40∆V − 1) 2 [ ] Solving by trial - and - error, HP - Solver, or MATLAB : ∆V = 0.383 V function f=dv15(v) f=4-20*v+1+log(40*v-1); fzero('dv15',0.5) yields ans = 0.3831 9.19 (a) The change in vBE will be neglected : ∆v BE = VT ln 0.8IC = −5.6 mV IC V H = 0 − VBE = 0 − 0.7 = −0.7 V - no change VL = 0 − α F IEE RC − VBE ≅ −IEE RC − VBE = −0.3mA(1.2)(2kΩ) − 0.7V = −1.42 V VL has dropped by 0.12V. | ∆V = 0.3mA(1.2)(2kΩ) = 0.72 V ⎡ ⎛ ∆V ⎞⎤ 0.72 ⎡ ⎛ 0.72 ⎞⎤ ∆V − VT ⎢1 + ln⎜ − 1⎟⎥ = − 0.025V ⎢1 + ln⎜ − 1⎟⎥ = 0.252 V ⎝ 0.025 ⎠⎦ 2 2 ⎣ ⎝ VT ⎠⎦ ⎣ (b) At node A : VH = 0 − VBE = 0 − 0.7 = −0.7 V - no change NM H = NM L = VL = 0 − α F IEE RC − VBE ≅ −IEE RC − VBE = − −1.0 − 0.7 − (−5.2) V (1.2)(2 kΩ) − 0.7V = −1.30 V 1.2(11.7kΩ) VL also has not changed! | Similar results hold at node B because the voltages are set by resistor ratios. NM H = NM L = ⎡ ⎛ ∆V ⎞⎤ 0.6 ⎡ ⎛ 0.6 ⎞⎤ ∆V − VT ⎢1 + ln⎜ − 1⎟⎥ = − 0.025V ⎢1 + ln⎜ − 1⎟⎥ = 0.197 V , unchanged ⎝ 0.025 ⎠⎦ 2 ⎣ ⎝ VT ⎠⎦ 2 ⎣ 9-6 9.20 (a) V H = −0.7V | ∆V = 0.8V | VL = −0.8 − 0.7 = −1.5V | VREF = −1.1 − 0.7 − (−5.2) V 0.8V = 11.3 kΩ | RC 2 = = 2.67 kΩ mA 0.3mA 0.3 −0.7 − 0.7 − (−5.2) V 0.8V I E1 = = 0.336 mA | RC1 = = 2.38 kΩ kΩ 0.336 mA 11.3 ⎡ ⎛ 0.8 ⎞⎤ 0.8 b NM = NM = − 0.025 V 1 + ln − 1 ⎢ () H ⎜ ⎟⎥ = 0.289 V L 2 ⎝ 0.025 ⎠⎦ ⎣ REE = VH + VL = −1.1V 2 (c) V CB1 = −0.8 − (−0.7)= −0.1V | VCB 2 = −0.8 − (−1.1)= +0.3V The collector - base junction of Q2 is reverse - biased by 0.3 V. Although the collector - base junction of Q1 is forward - biased by 0.1 V, this is not large enough to cause a problem. Therefore the voltages are acceptable. 9.21 NM H = ⎡ ⎛ ∆V ⎞⎤ ∆V − VT ⎢1 + ln⎜ − 1⎟⎥ 2 ⎝ VT ⎠⎦ ⎣ ⎡ ⎛ ∆V ⎞⎤ ∆V − 0.025V ⎢1 + ln⎜ − 1⎟⎥ → ∆V = 0.383V 2 ⎝ 0.025 ⎠⎦ ⎣ For room temperature, VT = 0.025V : 0.1V = For - 55C, VT = 0.0188V : 0.1V = ⎡ ⎛ ∆V ⎞⎤ ∆V − 0.0188V ⎢1 + ln⎜ − 1⎟⎥ → ∆V = 0.346V 2 ⎝ 0.0188 ⎠⎦ ⎣ ⎡ ⎛ ∆V ⎞⎤ ∆V − 0.0300V ⎢1 + ln⎜ − 1⎟⎥ → ∆V = 0.413V 2 ⎝ 0.0300 ⎠⎦ ⎣ For + 75C, VT = 0.0300V : 0.1V = ∆V = 0.413 V 9-7 9.22 In the original circuit : VH = −2 mA(2 kΩ)− 0.7V = −1.1V | ∆V = 2 mA(2 kΩ)= 0.4V −1.3 − 0.7 − (−5.2) V = 16.0 kΩ. R1 and R C2 remain unchanged. mA 0.2 −1.1 − 0.7 − (−5.2) V For Q1 on and and Q2 off : I EE = = 0.2125mA kΩ 16.0 VL1 = −(0.2125mA)(2 kΩ + RC 1)− 0.7V | VL1 = −1.5V → RC1 = 1.77 kΩ. REE = Thus we cannot force them all to the desired level. For this design, VL = −1.1V − ∆V = −1.5V . VH and VL are symmetrically placed about VREF . Note that there are only 3 variables (R1 , RC1 and R C2 ) and four voltage levels. VH 2 = −(0.2125mA)(2 kΩ)− 0.7V = −1.125V rather than the desired -1.10V 9.23 VEQ = I BS 60 kΩ (−5.2V )= −3.0V | REQ = 60kΩ 44kΩ = 25.38kΩ 60kΩ + 44kΩ −3.0 − 0.7 − (−5.2) V −3.0 − 0.7 − (−5.2) V = | I EE = β F I BS = β F 25.38 + (β F + 1)30 kΩ 25.38 + (β F + 1)30 kΩ −3.0 − 0.7 − (−5.2) V = 50.0 µA | Active region operation kΩ 30 = VREF − VBE 2 − VBS = VREF − 0.7V − (−3V ) For large β F , I EE = requires VCBS ≥ 0V | VCBS VREF − 0.7V − (−3V )≥ 0 → VREF ≥ −2.30 V 9.24 The base of QS must not be higher than VL − 0.7 = −0.2 mA(4kΩ)− 0.7 − 0.7 = −2.2V Design choice - Choose VB = −3 V . Assume β F = 50. | I B = RE = VB − VEE −3 − (−5.2) = = 10.8 kΩ → RE = 11 kΩ IE 0.204 mA R2 = 0 − (−3) 20µA = 150 kΩ → R2 = 150 kΩ = 138 kΩ → R1 = 136 kΩ 200µA = 4µA 50 Choose I R 2 = 20µA. I R1 = I R 2 − I B = 16µA. R1 = −3 − (−5.2) 16µA 9-8 9.25 *PROBLEM 9.25 - ECL INVERTER VTC VIN 2 0 DC -1.3 VREF 4 0 -1.0 VEE 8 0 -5.2 Q1 1 2 3 NBJT Q2 5 4 3 NBJT Q3 0 1 6 NBJT Q4 0 5 7 NBJT REE 3 8 11.7K RC1 0 1 1.85K RC2 0 5 2K R3 6 8 42K R4 7 8 42K .DC VIN -1.3 -0.7 .01 .TEMP -55 25 85 .MODEL NBJT NPN BF=40 BR=0.25 VA=50 .PROBE V(2) V(1) V(5) V(6) V(7) .PRINT DC V(2) V(6) V(7) .END T VT VH VL ∆V VREF VIH VOH VIL VOL NMH NML -55C 0.0188 V -0.846 V -1.40 V 0.554 V -1.00 V -0.918 V -0.865 V -1.08 V -1.38 V 0.053 V 0.300 V +25C 0.0257 V -0.724 V -1.30 V 0.576 V -1.00 V -0.895 V -0.750 V -1.10 V -1.27 V 0.145 V 0.170 V +85C 0.0309 V -0.629 V -1.22 V 0.591 V -1.00 V -0.880 V -0.660 V -1.12 V -1.19 V 0.220 V 0.070 V VIH , VOH , VOL , and VIL were calculated from Eqns. 9.27 - 9.30. With a fixed reference voltage, the noise margins change with temperature and can become zero for a large enough temperature change. 9-9 9.26 RC Q 4 Q A B C D E 2 V REF Y =A+B+C+D+E I EE R -V EE 9.27 R Q C 3 Q A B C D 2 V REF Y=A+B+C+D I EE R -V EE 9.28 840 VL = 1.0V − (2.14mA)(390Ω)− 0.7V = −0.540 V (a ) For Q 4 on, IC 4 = α F I E 4 ≅ I E 4 = −0.7 − (−2.5) = 2.14 mA For Q4 off, and neglecting the base current in Q5 , VH = 1.0 − 0.7 = +0.300 V (b) For v = 2.50 mA 840 ∆V 0.84V ∆V = 0.30 − (−0.54)= 0.84V | R = = = 336 Ω IC 2 2.50 mA A = 0.3V, IC 2 = α F I E 2 ≅ I E 2 = 0.3 − 0.7 − (−2.5) 9-10 9.29 840 VL = 1.3V − (2.98 mA)(390Ω)− 0.7V = −0.56 V (a) For Q 4 on , IC 4 = α F I E 4 ≅ I E 4 = −0.7 − (−3.2) = 2.98 mA For Q4 off, and neglecting the base current in Q5 , VH ≅ 1.3 − 0.7 = +0.60 V (b) For v = 3.69 mA 840 ∆V 1.16V ∆V = 0.60 − (−0.56)= 1.16V | R = = = 314 Ω IC 2 3.69 mA A = 0.6V , IC 2 = α F I E 2 ≅ I E 2 = 0.6 − 0.7 − (−3.2) 9.30 V CC 390 Ω A Q 2 B Q 3 Q 4 Q 5 Y=A+B 840 Ω 600 Ω (a) -V EE (b) The NOR output is taken from the collectors of Q2/Q3, and the 390Ω resistor, Q5, and the 600-Ω resistor are removed. 9.31 min = − I EE RL = −(2.5mA)( 1.2 kΩ) = −3.00 V | I E = I EE + vO vO 4 − 0.7 = 2.5mA + = 5.25 mA 1.2 kΩ RL VBC = 4 − 5 = −1 V , so the transistor is in the forward - active region. IB = 9.32 IE 5.25mA = = 0.103 mA and IC = β F I B = 5.15 mA. β F + 1 50 + 1 (a-b) See Problem 9.33 (c) IEE ≥ − (VI − 0.7)V 1kΩ = 3.7V = 3.7 mA 1kΩ 9-11 9.33 Simulation Results from B2SPICE Circuit 9.32-Transient-2 Circuit 9.32-Transient-1 (V) +0.000e+000 +500.000u +1.000m +1.500m Time (s) +2.000m Time (s) +1.500m +2.000m (V) +0.000e+000 +500.000u +1.000m +3.000 +3.000 +2.000 +2.000 +1.000 +1.000 +0.000e+000 +0.000e+000 -1.000 -1.000 -2.000 -2.000 -3.000 -3.000 -4.000 IEE = 4 mA V(1) V(3) IEE = 2 mA V(1) V(3) 9.34 (a) vO = v I − 0.7V = (−1.7 + sin2000πt ) V min = −2.7V (b) vO 2.7V = 0.13 mA with no safety margin. 20kΩ The transistor will cut off at the bottom of the input waveform for IEE = 0.13 mA. | - IEE RL ≤ −2.7V → IEE ≥ 9.35 Simulation results from B2SPICE Circuit 9.35-Transient-1 (V) +0.000e+000 +500.000u +1.000m +1.500m Time (s) +2.000m +0.000e+000 -500.000m -1.000 -1.500 -2.000 -2.500 -3.000 V(1) V(3) 9.36 min = - I EE RL = -(0.5mA)( 1kΩ) = -0.5V. So v I ≥ -0.5 + 0.7 = +0.2V. (a ) The transistor cuts off for vO For v O > 1.5 V, the transistor enters the saturation region of operation. Therefore : 0.2 V ≤ v I ≤ 1.5 V . (b) v 9.37 min O = −1.5 − 0.7 = −2.2 V . We need - I EE RL ≤ −2.2V → I EE ≥ 2.2V = 2.2 mA 1kΩ 10.7V = 10.7 mA 1kΩ min vO = −10 − 0.7 = −10.7 V . We need - IEE RL ≤ −10.7V → IEE ≥ 9-12 9.38 Assuming Q1 off and using voltage division, − 12 = −15 IE = 12 − (−15) 12 + = 60 mA ! 2000 500 2000 ⇒ RE = 500 Ω 2000 + RE 9.39 (a) v RE ≤ min O (15 − 10.7)(4.7kΩ) = 1.89 kΩ 10.7 = ⎛ 4.7 kΩ ⎞ = −10 − 0.7 = −10.7 V . We need -15V ⎜ ⎟ ≤ −10.7V ⎝ 4.7 kΩ + RE ⎠ | IE = vI − 0.7 vI − 0.7 − VEE + RE RL (b) I E −0.7 −0.7 − (−15) + = 7.43 mA | 4700 1890 (c) I E = −10 − 0.7 −10 − 0.7 − (−15) + = 0 mA 4700 1890 9.40 (a) See the solution to Problem 9.41. (b) v (c) v IE = O = v I − 0.7V = (−2.2 + 1.5sin 2000πt ) V max = −1.5 + 1.5 = 0V | vO = −0.7V | max | IE = max I −3.7 −3.7 − (−6) + = 0.982 mA 1300 4700 ⎛ 4.7 kΩ ⎞ 4.7 kΩ(6 − 3.7) = 2920 Ω ⎟ ≤ −3.7V → RE ≤ (e) We need - 6V ⎜ 3.7 ⎝ 4.7 kΩ + RE ⎠ vO vO − VEE + RL RE min O −0.7 −0.7 − (−6) + = 3.93 mA 1300 4700 (d ) v min = −2.2 − 1.5 = −3.7V | I E = 9.41 Simulation results from B2SPICE Circuit 9.37-Transient-2 +0.000e+000 +500.000u +1.000m +1.500m +2.000m +2.500m Time (s) +3.000m +0.000e+000 -1.000 -2.000 -3.000 -4.000 V(1) V(3) V(5) 9-13 9.42 The outputs act as a "wired - or" connection. For v I = −0.7V , vO1 = vO 2 = −0.7 V | IE 3 = 0 | IE 4 = 0.1mA + 0.1mA = 0.200 mA For v I = −1.3V , vO1 = vO 2 = −0.7 V | IE 3 = 0.1mA + 0.1mA = 0.200 mA | IE 4 = 0 9.43 Y = A+ B | Z = A+ B 9.44 A OR B Y C NOR D 9.45 For Fig. 9.21, P ≅ 0.5mA(5.2V)= 2.6 mW = 2600µW . For 20µW, the power must be reduced by 130X. The currents must be reduced by 130X and the resistors must increase by this factor to keep the logic swing the same : R C = 130(2kΩ)= 260 kΩ. Using Eq. (9.54), τ P = 0.69(260kΩ)(2 pF ) = 359 ns - rather slow! 9.46 RC 2kΩ = = 1kΩ | ∆V = 0.3mA( 1kΩ) = 0.3V | VH = 0 − 0.7 = −0.7V 2 2 −0.7 − 1.0 V = −0.850 V | P ≅ 0.5mA(5.2V )= 2.6 mW VL = VH − 0.3V = −1.0V | VREF = 2 τ P = 0.69( 1kΩ)(2 pF ) = 1.38 ns | PDP = 2.6 mW ( 1.38 ns)= 3.59 pJ ' RC = 9.47 ∆V = 0.15mA(2 kΩ) = 0.3V | VH = 0 − 0.7 = −0.7V VL = VH − 0.3V = −1.0V | VREF = −0.7 − 1.0 V = −0.850 V | P ≅ 0.25mA(5.2V )= 1.30 mW 2 τ P = 0.69(2kΩ)(2 pF ) = 2.76 ns | PDP = 1.30mW (2.76 ns)= 3.59 pJ 9-14 9.48 ∆V → ∆V = 2 0 − 0.7 − (−1) = 0.6V 2 VL = VH − ∆V = 0 − .6 = −0.600 V . Ignoring the base currents, the average power is ⎡ −1.7 − −3.3 V −1.0 − (−3.3) V ⎤ ( ) ⎢ ⎥ 3.3V = 5.67 mW + P≈ ⎢ ⎥ 1.6 kΩ 3.2kΩ ⎣ ⎦ (a) At the outputs : V H = 0 V | VREF = VH − 0.7 − ( ) ( ) ( ) RC 2 = ∆V 0.6 ∆V 0.6 = = 600 Ω | RC1 = = = 505 Ω I EE 2 −1 − 0.7 − (−3.3) I EE1 −0.7 − 0.7 − (−3.3) 1600 (b) Y = A + B + C Y = A + B + C (c) 5 versus 6 transistors 1600 9.49 At the outputs : VH = 0 V | VL = VH − ∆V = 0 − .4 = −0.400 V . At the base of QD : VH → VBD = 0 − 0.7 = −0.7V | VL → VBD = −0.4 − 0.7 = −1.10V VREF = −0.7 − 1.1 = −0.90V | VEE ≤ VREF − 0.7 − 0.6 = −0.9 − 0.7 − 0.6 = −2.20 V 2 −0.9 − (−2.2) V For VEE = −2.20V : RB = = 1.30 kΩ 1 mA −0.9 − 0.7 − (−2.2) + −0.7 − 0.7 − (−2.2) V RE = = 700 Ω 2 1 mA 0.4V 0.4V = 350 Ω | RC 2 = = 467 Ω RC1 = −0.7 − 0.7 − (−2.2) −0.9 − 0.7 − (−2.2) A A 700 700 [ ][ ] 9-15 9.50 *PROBLEM 9.50 - ECL DELAY VIN 1 0 PULSE(-0.6 0 0 .01NS .01NS 15NS) VB 8 0 -0.6 VREF 6 0 -1.0 VEE 7 0 -3.3 QA 0 1 2 NBJT QB 0 8 2 NBJT QC 0 8 2 NBJT QD 4 2 3 NBJT QE 5 6 3 NBJT RB 2 7 3.2K RE 3 7 1.6K RC1 0 4 505 RC2 0 5 600 .OP .TRAN 0.1N 30N .MODEL NBJT NPN BF=40 BR=0.25 +IS=5E-16 TF =0.15NS TR=15NS +CJC=0.5PF CJE=.25PF CJS=1.0PF +RB=100 RC=5 RE=1 .PROBE V(2) V(1) V(4) V(5) V(6) .END 200mV 0V -200mV -400mV -600mV vI Time -800mV 0s 5ns 10ns 15ns 20ns 25ns 30ns Result: τP = 0.95 ns 9-16 9.51 One approach is to scale all the resistor values. To reduce the power from 2.7 mW to 1.0 mW, the resistor values should all be increased a factor of 2.7. REE = 2.7( 11.7kΩ) = 31.6 kΩ | R = 2.7(42kΩ) = 113 kΩ 1.85kΩ)= 5.00 kΩ | RC 2 = 2.7(2 kΩ)= 5.40 kΩ RC1 = 2.7( 9.52 Voltage levels remain unchanged : VREF = −1 V , VH = −0.7 V , VL = −1.3 V , I EE = 0.3 mA −1 − 0.7 − (−2) V 0.6V 0.6V = 1 kΩ | RC1 = = = 1 kΩ mA 0.6mA 0.3 −0.7 − 0.7 − (−2) A 1 kΩ −1 − (−2) V 0.3 + 0.6 I ≅2 + mA = 0.650 mA | P = 0.65mA(2V )= 1.30 mW (-28%) kΩ 2 10 Note that this gate will now have quite asymmetrical delays at the two outputs since the two collector resistors differ by a factor of two in value. REE = 9.53 The circuit is the pnp version of the ECL gate in Fig. P9.48. | Y = ABC 9.54 VL = 0 | VH = VL + ∆V = +0.6V | VREF = 0.7 + 1.3 1mW = +1.0V | I = = 333µA 3V 2 (1 + 0.7)+ (0.7 + 0.7) = 1.55V The average voltage at the emitter of QD is 2 (3 − 1.55)V = 4.84 kΩ | R = (3 − 1)V = 60.1 kΩ | R = 0.6V = 2.23 kΩ RE = B C 0.9(333µA) 0.1(333µA) (3 − 1.7)V 4.84 kΩ 9.55 *Problem 9.55(a) - PNP ECL GATE DELAY VI 4 0 PULSE(0.6 0 0 .01NS .01NS 25NS) VB 7 0 DC 0.6 VREF 6 0 DC 1.0 VEE 1 0 DC 3 QA 0 4 3 PBJT QB 0 7 3 PBJT QC 0 7 3 PBJT QD 0 3 2 PBJT QE 5 6 2 PBJT RB 1 3 60.1K RE 1 2 4.84K 9-17 RC 5 0 2.23K .OP .TRAN 0.1N 50N .MODEL PBJT PNP BF=40 BR=0.25 IS=5E-16 +TF =0.15NS TR=15NS +CJC=0.5PF CJE=.25PF CJS=1.0PF +RB=100 RC=5 RE=1 .PROBE V(4) V(3) V(5) .END 800mV 600mV vI 400mV 200mV vO 0V Time -200mV 0s 10ns 20ns 30ns 40ns 50ns Result: τP = 6.0ns. This delay is dominated by a slow charge up at the base of QD. *Problem 9.55(b) - Prob. 9.4 VIN 1 0 PULSE( -2.3 -1.7 0 .01NS .01NS 15NS) VREF 6 0 -2.0 IEE 2 0 0.0003 Q1 3 1 2 NBJT Q2 4 6 2 NBJT R1 0 5 3.33K RC1 5 3 2K RC2 5 4 2K .OP .TRAN 0.1N 30N .MODEL NBJT NPN BF=40 BR=0.25 IS=5E-16 TF =0.15NS TR=15NS +CJC=0.5PF CJE=.25PF CJS=1.0PF RB=100 RC=5 RE=1 .PROBE V(1) V(3) V(4) .END Result: τP = 2.4 ns. *Problem 9.55(c) - Fig. P9.16 VIN 1 0 PULSE( -1.5 -1.1 0 .01NS .01NS 15NS) VREF 6 0 DC -1.30 IEE 2 0 DC 0.0002 9-18 Q1 3 1 2 NBJT Q2 4 6 2 NBJT Q3 0 3 7 NBJT Q4 0 4 8 NBJT R1 0 5 2K RC1 5 3 2K RC2 5 4 2K RE1 7 9 50K RE2 8 9 50K VEE 9 0 DC -2 .OP .TRAN 0.1N 30N .MODEL NBJT NPN BF=40 BR=0.25 IS=1E-17 TF =0.15NS TR=15NS +CJC=0.5PF CJE=.25PF CJS=1.0PF RB=100 RC=5 RE=1 .PROBE V(1) V(3) V(4) V(7) V(8) .END Result: τP = 3.0 ns. 9.56 Applying the transport model, ⎡ ⎛V ⎞ ⎛ V ⎞⎤ I ⎡ ⎛ V ⎞ ⎤ IC = IS ⎢exp⎜ BE ⎟ − exp⎜ BC ⎟⎥ − S ⎢exp⎜ BC ⎟ − 1⎥ ⎝ VT ⎠⎦ β R ⎣ ⎝ VT ⎠ ⎦ ⎣ ⎝ VT ⎠ I ⎡ ⎛V ⎞ ⎤ I ⎡ ⎛V ⎞ ⎤ IB = S ⎢exp⎜ BE ⎟ − 1⎥ + S ⎢exp⎜ BC ⎟ − 1⎥ β F ⎣ ⎝ VT ⎠ ⎦ β R ⎣ ⎝ VT ⎠ ⎦ ⎡ ⎛ 0.2 ⎞ ⎛ −4.8 ⎞⎤ 10−15 ⎡ ⎛ −4.8 ⎞ ⎤ −15 IC = 10 ⎢exp⎜ ⎟ − exp⎜ ⎟⎥ − ⎟ − 1⎥ = 2.98 pA ⎢exp⎜ ⎝ 0.025 ⎠⎦ 0.25 ⎣ ⎝ 0.025 ⎠ ⎦ ⎣ ⎝ 0.025 ⎠ IB = 10−15 ⎡ ⎛ 0.2 ⎞ ⎤ 10−15 ⎡ ⎛ −4.8 ⎞ ⎤ ⎟ − 1⎥ + ⎟ − 1⎥ = 74.5 fA ⎢exp⎜ ⎢exp⎜ 40 ⎣ ⎝ 0.025 ⎠ ⎦ 0.25 ⎣ ⎝ 0.025 ⎠ ⎦ Although the transistor is technically in the forward-active region, (and operating with IC = βF IB), it is esentially off - its terminal currents are zero for most practical purposes. 9.57 ⎛ 1 ⎞ For IC = 0, VCESAT = VT ln⎜ ⎟ ⎝αR ⎠ 1+ IC (β R + 1)I B VCESAT IC βF I B ⎛ β + 1⎞ ⎛ 1.25 ⎞ = VT ln⎜ R ⎟ = 0.025ln⎜ ⎟ = 0.402 V ⎝ 0.25 ⎠ ⎝ βR ⎠ 1− ⎛ 1 ⎞ = VT ln⎜ ⎟ ⎝αR ⎠ 9-19 9.58 (a) For the Transport model with VBE = VBC, the transport current iT = 0: β 40 I ⎡ ⎛V ⎞ ⎤ I ⎡ ⎛V ⎞ ⎤ I IC = S ⎢exp⎜ BC ⎟ − 1⎥ and IE = S ⎢exp⎜ BE ⎟ − 1⎥ ⇒ C = F = = 160 β R ⎣ ⎝ VT ⎠ ⎦ β F ⎣ ⎝ VT ⎠ ⎦ IE β R 0.25 (b) v BE = VB − 0.6 | v BC = VB − 0.8 = v BE − 0.2 ⎤ ⎡ v v ⎤ I ⎡ v iE = IS ⎢exp BE − exp BC ⎥ + S ⎢exp BE − 1⎥ VT VT ⎦ β F ⎣ VT ⎦ ⎣ ⎡ v v −0.2 ⎤ IS ⎡ v BE ⎤ − 1⎥ iE = IS ⎢exp BE − exp BE exp ⎢exp ⎥+ VT VT VT VT ⎦ β F ⎣ ⎦ ⎣ ⎛ ⎡ v ⎤ I ⎡ v ⎤ 1 ⎞⎡ v ⎤ v ⎤ I ⎡ iE ≅ IS ⎢exp BE ⎥ + S ⎢exp BE ⎥ = IS ⎜1 + ⎟⎢exp BE ⎥ = S ⎢exp BE ⎥ VT ⎦ β F ⎣ VT ⎦ VT ⎦ α F ⎣ VT ⎦ ⎣ ⎝ β F ⎠⎣ −−−−− ⎡ v v −0.2 ⎤ IS ⎡ v BE − 0.2 ⎤ iC = IS ⎢exp BE − exp BE exp − 1⎥ ⎥ − ⎢exp VT ⎦ β R ⎣ VT VT VT ⎦ ⎣ ⎡ v ⎤ 40 i iC ≅ IS ⎢exp BE ⎥ | C = α F = = 0.976 41 VT ⎦ iE ⎣ i (c ) C = −1 → iB = iE − iC = 2iE | Both junctions will be forward - biased. Neglect iE ⎛ v I v v v ⎞ v I I the -1 terms : S exp BE + S exp BC = 2 IS ⎜ exp BE − exp BC ⎟ + 2 S exp BE βF VT β R VT VT VT ⎠ βF VT ⎝ v BE − v BC 1 0.25 = 27.2 mV = VT ln = 0.025V ln 1 1 2+ 2+ βF 40 2+ 1 βR 2+ (v B − v I ) − (v B − 0.8) = 27.7mV | v I = 0.773 V 9-20 9.59 (a) For the default value of β F = 40, IC = β F I B ⇒ Forward − active region | VBE ≅ VT ln (b) I VBE IC 10−3 A = 0.025V ln −15 = 0.691 V IS 10 A C < β F I B ⇒ saturation region; VBE is given by Eqn. 5.45 ⎛ 1 ⎞ IB + ⎜ ⎟ IC IB + ( 1 − α R )IC ⎝ β R + 1⎠ = VT ln = V ln ⎡ 1 ⎛ 1 ⎞⎤ ⎡1 ⎤ T IS ⎢ + ( 1− αR ) IS ⎢ + ⎜ ⎥ ⎟⎥ ⎣β F ⎦ ⎣ β F ⎝ β R + 1⎠⎦ ⎛ 1 ⎞ −3 25 x10−6 + ⎜ ⎟10 ⎝ 0.25 + 1⎠ = 0.025V ln = 0.691 V ⎡ ⎛ 1 ⎞⎤ −15 1 10 ⎢ + ⎜ ⎟⎥ ⎣80 ⎝ 0.25 + 1⎠⎦ < β F I B ⇒ saturation region | VBE ⎛ 1 ⎞ −3 10−3 + ⎜ ⎟10 ⎝ 0.25 + 1⎠ = 0.025V ln = 0.710 V ⎡ ⎛ 1 ⎞⎤ −15 1 10 ⎢ + ⎜ ⎟⎥ ⎣40 ⎝ 0.25 + 1⎠⎦ VBE (c) I 9.60 C αR = βR + 1 βR = 1 | 2 IC 1mA = = 40 I B 25µA 1+ IC (β R + 1)I B IC βF I B ⎡ 40 ⎤ ⎢⎛ 3 ⎞ 1 + ⎥ 2 ⎥ = 114 mV = (0.025V )ln ⎢⎜ ⎟ 40 2 ⎝ ⎠ ⎢ 1− ⎥ ⎣ 60 ⎦ (a) V (b) I CESAT ⎛ 1 ⎞ = VT ln⎜ ⎟ ⎝αR ⎠ 1− IC B = 1mA = 25 | VCESAT 40µA ⎡ 25 ⎤ ⎢⎛ 3 ⎞ 1 + ⎥ 2 ⎥ = 88.7 mV = (0.025V )ln ⎢⎜ ⎟ 25 2 ⎢⎝ ⎠ 1 − ⎥ ⎣ 60 ⎦ 9-21 9.61 αR = 2 1mA I | C = = 40 βR + 1 3 IB 25µA IC ⎡ 40 ⎤ 1+ ⎛ 1 ⎞ (β R + 1)IB ⎢⎛ 3 ⎞ 1 + 3 ⎥ = (0.025V )ln⎢⎜ ⎟ = 117 mV (a) VCESAT = VT ln⎜ ⎟ 40 ⎥ ⎠ ⎝ 2 ⎝ α R ⎠ 1 − IC ⎢ 1− ⎥ ⎣ 50 ⎦ β F IB = βR (b) VCESAT 9.62 ⎡ 40 ⎤ ⎢⎛ 3⎞ 1 + 3 ⎥ = (0.025V )ln⎢⎜ ⎟ = 89.5 mV 40 ⎥ ⎠ ⎝ 2 ⎢ 1− ⎥ ⎣ 100 ⎦ αR = 0.25 = 0.2 β R + 1 1.25 ⎛V ⎞ ⎛V ⎞ ⎛ 0.2V ⎞ ⎛ 0.1V ⎞ ⎟ = 2980 ⎟ = 54.6 (a) Γ = exp⎜ CESAT ⎟ = exp⎜ (b) Γ = exp⎜ CESAT ⎟ = exp⎜ ⎝ 0.025V ⎠ ⎝ 0.025V ⎠ ⎝ VT ⎠ ⎝ VT ⎠ ⎤ ⎤ ⎡ ⎡ ⎡ ⎡ 40 40 β ⎤ β ⎤ 1+ F ⎥ 1+ F ⎥ ⎢1 + 0.25 2980 ⎥ ⎢1 + 0.25 54.6 ⎥ ⎢ ⎢ I I I I βRΓ ( )⎥ = IC βRΓ ( )⎥ = IC ⎥= C ⎢ ⎥= C ⎢ IB ≥ C ⎢ IB ≥ C ⎢ 1 1 1 1 β F ⎢1 − β F ⎢1 − ⎥ 37.9 ⎥ 9.25 ⎥ 40 ⎢ 1 − ⎥ 40 ⎢ 1 − ⎥ ⎢ ⎢ ⎥ ⎥ ⎢ ⎢ 0.2(2980) ⎦ 0.2(54.6) ⎥ ⎣ αRΓ ⎦ ⎣ α RΓ ⎦ ⎦ ⎣ ⎣ I 1 ⎛ 5 − 0.2 ⎞ I 1 ⎛ 5 − 0.1⎞ IB ≥ C = IB ≥ C = ⎜ ⎜ ⎟ = 63.3 µA ⎟ = 265 µA β FOR 37.9 ⎝ 2 kΩ ⎠ β FOR 9.25 ⎝ 2 kΩ ⎠ = βR 9.63 αR = ⎛V ⎞ ⎛ 0.1V ⎞ 0.25 = 0.2 Γ = exp⎜ CESAT ⎟ = exp⎜ ⎟ = 54.6 ⎝ 0.025V ⎠ β R + 1 1.25 ⎝ VT ⎠ βR = ⎡ ⎤ 40 ⎡ β ⎤ 1+ F ⎥ ⎢1 + 0.25 54.6 ⎥ ⎢ I I βRΓ ( )⎥ = IC ⎥= C ⎢ IB ≥ C ⎢ 1 1 β F ⎢1 − ⎥ 9.25 ⎥ 40 ⎢ 1 − ⎢ ⎢ ⎥ 0.2(54.6) ⎥ ⎣ αRΓ ⎦ ⎣ ⎦ I 1 ⎛ 5 − 0.1⎞ IB ≥ C = ⎟ = 147 µA ⎜ β FOR 9.25 ⎝ 3.6 kΩ ⎠ 9-22 9.64 ⎛ 1 ⎞ ⎛ β + 1⎞ ⎛ 1.25 ⎞ For IC = 0, VCESAT = VT ln⎜ ⎟ = VT ln⎜ R ⎟ = 0.025ln⎜ ⎟ = 40.2 mV ⎝ 0.25 ⎠ ⎝αR ⎠ ⎝ βR ⎠ ⎛ 1 ⎞ ⎛ β + 1⎞ ⎛ 41 ⎞ For IE = 0, VECSAT = VT ln⎜ ⎟ = VT ln⎜ F ⎟ = 0.025ln⎜ ⎟ = 0.617 mV ⎝ 40 ⎠ ⎝αF ⎠ ⎝ βF ⎠ 9.65 αF = τS = 0.976 0.4 ns + 0.2( 12 ns) 1 − 0.976(0.2) βF + 1 βF = ( βR 40 0.25 = 0.976 α R = = = 0.200 β R + 1 1.25 41 ) = 3.40ns iCMAX ≅ 5V = 2.5 mA 2 kΩ t S = (3.40 ns)ln 2mA − (−0.5mA) 2.5mA − (−0.5mA) 40 = 5.07 ns 9.66 V H = VCC = 3.0 V VL = VCESAT = 0.15 VIL = 0.7 − VCESAT = 0.7 − 0.04V = 0.66V VIH ≅ VBESAT 2 = 0.8 V 3 − 0.7 − 0.8 V v I = 3V : IB1 = = 375µA | IB 2 = 1.25 IB1 = 469µA kΩ 4 3 − 0.8 − 0.15 V 3 − 0.15 = −513µA | IC 2 SAT = A = 1.43mA v I = 0.15V : IIL = − kΩ 4 2000 1.43mA + N (513µA) ≤ 40(469µA) → N ≤ 33.8 → N ≤ 33. 9.67 vI = VH : I = vI = VL : I = 5 − 0.7 − 0.8 5 − 0.15 + = 4.13mA | P = 5(4.13mA) = 20.6 mW (0.8)4kΩ (0.8)2kΩ 5 − 0.8 − 0.15 = 0.844mA | P = 5(0.844 mA)= 4.22 mW (1.2)4kΩ Pmax = 20.6 mW | Pmin = 4.22 mW 9-23 9.68 ⎛ 1 ⎞ Using Eqs. 9.44 and 9.47 : VCC − iC RC = VT ln⎜ ⎟ ⎝αR ⎠ 1+ iC 1.25( 1.09 mA) iC 40( 1.09mA) 1+ iC (β R + 1)iB iC β F iB 1− ⎛1⎞ 5 − 2000iC = 0.025ln⎜ ⎟ ⎝ .2 ⎠ 1 − → iC = 2.4659 mA vCESAT = 5 − 2000iC = 0.0682V 9.69 V H = 2.5 V | VL = VCESAT = 0.15 V VIL = 0.7 − VCESAT = 0.55V | VOL ≅ VL = 0.15V VIH ≅ VBESAT 2 = 0.8 V | VOH ≅ V H = 2.5 V NM L = 0.55 − 0.15 = 0.40 V NM H = 2.5 − 0.8 = 1.7 V 9.70 For vI = VH , we require VCC = VBE 2SAT + VBC1 + I B1 RB = 0.8 + 0.7 + ∆V = 1.5V + ∆V where ∆V is the voltage across the base resistor. ∆V must be large enough to absorb VBE process variations and to establish the base current. 0.5 V should be sufficient. Thus VCC = 2.0 V or more is acceptable. 9.71 The VTC transitions are set by the values of vBE and vBESAT and are not changed by the power supply voltage. (b) VIL = 0.66 V and VIH = 0.80 V. But VOH ≅ VH = 3 V and VOL ≅ VL = 0.15 V. (c) NMH = 3 - 0.8 = 2.2 V | NML = 0.66-0.15 = 0.51 V. 9.72 We need to reduce the currents by a factor of 11.2. Thus, RB = 11.2 (4kΩ) = 44.8 kΩ and RC = 11.2 (2kΩ) = 22.4 kΩ 9.73 (a) *Problem 9.73 - Prototype TTL Inverter +Delay VI 1 0 DC 0 PWL(0 0 0.2N 5 25N 5 25.2N 0 +50N 0) VCC 5 0 DC 5 Q1 3 2 1 NBJT Q2 4 3 0 NBJT RB 5 2 4K RC 5 4 2K *RB 5 2 45.2K 9-24 *RC 5 4 22.6K .OP .TRAN .1N 80N .MODEL NBJT NPN BF=40 BR=0.25 +IS=5E-16 TF =0.15NS TR=15NS +CJC=0.5PF CJE=.25PF CJS=1.0PF +RB=100 RC=5 RE=1 .PROBE V(1) V(2) V(3) V(4) .END 6.0V vI 4.0V 2.0V vO 0V Time (a) -2.0V 0s 20ns 40ns 60ns 80ns Results: (a) τP = 2.9 ns (b) τP = 15.8 ns. 9.74 (a) VH = 5V | VL = VCE 2SAT = 0.15V vI = VL = 0.15V , I IN = − vI = VH = 5V , I IN B 5 − 0.15 − 0.6 = −1.06 mA 4000 = − I S ≅ 0 where IS is the diode saturation current. 5 - 0.15 (b) I = 4000 = 0.90mA; 2000 + N (1.06mA)≤ 40(0.9mA); N ≤ 31. (c) -1.06 mA compared to -1.01 mA and 0 mA compared to 0.22 mA. 9.75 If we assume that the diode on-voltage is 0.7 V to match the base-emitter voltage of the BJT, then the VTC will be the same as that in Fig. 9.35. Both VTCs will be the same. 5 − 0.8 − 0.6 9-25 9.76 *Figure 9.76 - Prototype TTL Inverter VTC's VI 1 0 DC 0 VCC 5 0 DC 5 *DTL D1A 6 1 D1 D2A 6 7 D1 RBA 5 6 4K RCA 5 8 2K Q2A 8 7 0 NBJT *TTL Q1B 3 2 1 NBJT Q2B 4 3 0 NBJT RBB 5 2 4K RCB 5 4 2K .DC VI 0 5 .01 .MODEL NBJT NPN BF=40 BR=0.25 IS=5E-16 TF =0.15NS TR=15NS +CJC=0.5PF CJE=.25PF CJS=1.0PF RB=100 RC=5 RE=1 .MODEL D1 D IS=5E-16 TT=0.15NS CJO=1PF .PROBE V(1) V(2) V(3) V(4) V(6) V(7) V(8) .END 5.0V 4.0V vO 3.0V 2.0V 1.0V 0V 0V 1.0V 2.0V vI 3.0V 4.0V 5.0V The TTL transition is sharper (more abrupt) and is shifted by approximately 50 mV. 9.77 *Figure 9.77 - Prototype Inverter Delays VI 1 0 DC 0 PWL(0 0 0.2N 5 25N 5 25.2N 0 5 50N 0) VCC 5 0 DC 5 *DTL D1A 6 1 D1 D2A 6 7 D1 RBA 5 6 4K RCA 5 8 2K Q2A 8 7 0 NBJT *TTL Q1B 3 2 1 NBJT Q2B 4 3 0 NBJT RBB 5 2 4K RCB 5 4 2K .OP .TRAN 0.1N 100N .MODEL NBJT NPN BF=40 BR=0.25 IS=5E-16 TF =0.15NS TR=15NS +CJC=0.5PF CJE=.25PF CJS=1.0PF RB=100 RC=5 RE=1 .MODEL D1 D IS=5E-16 TT=0.15NS CJO=1PF 9-26 .PROBE V(1) V(2) V(3) V(4) V(6) V(7) V(8) .END 6.0V vI 4.0V vO TTL 2.0V vO DTL 0V Time -2.0V 0s 20ns 40ns 60ns 80ns 100ns The fall time of the output of the TTL gate is somewhat slower than the DTL gate since transistor Q1 must come out of saturation. However, the rise time of the DTL gate is extremely slow because there is no reverse base current to remove the charge from the transistor base. 9.78 *Figure 9.78 - DTL Inverter Delays VI 1 0 DC 0 PWL(0 0 0.2N 5 25N 5 25.2N 0 50N 0) VCC 5 0 DC 5 *DTLA D1A 6 1 D1 D2A 6 7 D1 RBA 5 6 4K RCA 5 8 2K Q2A 8 7 0 NBJT *DTL-B D1B 2 1 D1 D2B 2 3 D1 Q2B 4 3 0 NBJT RBB 5 2 4K RCB 5 4 2K RB1 3 0 1K .OP .TRAN 0.1N 100N .MODEL NBJT NPN BF=40 BR=0.25 IS=5E-16 TF =0.15NS TR=15NS +CJC=0.5PF CJE=.25PF CJS=1.0PF RB=100 RC=5 RE=1 .MODEL D1 D IS=5E-16 TT=0.15NS CJO=1PF .PROBE V(1) V(2) V(3) V(4) V(6) V(7) V(8) .END 9-27 ©R. C. Jaeger - February 7, 2007 6.0V vO (b) 4.0V vI 2.0V vO (a) 0V Time -2.0V 0s 20ns 40ns 60ns 80ns 100ns Without the 1-kΩ resistor, the rise time of the DTL gate is extremely slow because there is no reverse base current to remove the charge from the transistor base. The resistor provides an initial reverse base current of -0.7 mA to turn off the transistor and significantly reduces the rise time and propagation delay. 9.79 See problem 9.80. 9.80 *Figure 9.79 - Inverter VTC VI 1 0 DC 0 VCC 6 0 DC 3.3 Q1 3 2 1 NBJT Q2 5 3 4 NBJT Q3 5 4 0 NBJT R1 6 2 4K R2 6 5 2K R3 4 0 3K .OP .DC VI 0 3.3 0.01 .MODEL NBJT NPN BF=40 BR=0.25 IS=1E-16 TF =0.15NS TR=15NS +CJC=0.5PF CJE=.25PF CJS=1.0PF RB=100 RC=5 RE=1 .PROBE V(1) V(2) V(3) V(4) V(5) .END Circuit 9.108-DC Transfer-1 VO (V) +3.50 +0.00e+000 +500.00m +1.00 +1.50 +2.00 +2.50 +3.00 VI +3.00 +2.50 +2.00 +1.50 +1.00 +500.00m +0.00e+000 V(2) V(7) The first break point occurs when the input reaches a voltage large enough to just start turning on Q2, at approximately VCESAT1 + VBE2 = 0.04V + 0.6 V = 0.64V. The second breakpoint begins when the input reaches VCESAT1 + VBE2 +VBE3 = 0.04V + 0.7 + 0.6 V = 1.34V. Note 9-28 that the shallow slope is set by the ratio of R2/R3 = 2/3, and also note that Q3 cannot saturate. From the B2SPICE simulation, VH = 3.3 V, VL ≅ VBE3 + VCESAT2 = 0.82V, VIH = 1.38 V, VOL = 0.84 V, VIL = 1.38 V, VOH = 2.82 V. NMH = 2.82 – 1.38 = 1.54 V. NML = 1.38 - 0.84 = 0.54 V. 9.81 VCC = 5 V 0.875 mA 4kΩ iR 2k Ω N(0.875 mA) iC = 0 vH < 5 V N VL 0.15 V Q1 Q2 + 0.19 V "Off" 0.875 mA From the analysis in the text, we see that the fanout is limited by the VH condition. 5 − 0.7 − 0.8 V iB1 = = 0.875 mA | iE1 = −β R iB1 = −0.875 mA 4 kΩ 5 − 2000(N )( 0.875 x10−3 )≥ 1.5 → N ≤ 2 → Fanout = 2 9.82 V CC = 5 V 1.2(4 k Ω ) 1.2(2k Ω ) 1.5 V iB1 + 0.7 V VH iIH = β RiB1 Q1 N ( βR + 1)i B1 Q2 + 0.8 V 1.2(2k Ω) + 0.15 V - 9-29 From the analysis in the text, we see that the fanout is limited by the VH condition. 5 − 0.7 − 0.8 V iB1 = = 0.729 mA | iE1 = −β R iB1 = −0.25(0.729 mA) = 0.182 mA 4 (1.2) kΩ 5 − 2000(1.2)(N )(0.182 x10−3 )≥ 1.5 → N ≤ 8.01 → Fanout = 8 iB1 = 5 − 0.7 − 0.8 V = 1.09 mA | iE1 = −β R iB1 = −0.25(1.09 mA) = 0.273mA 4 (0.8) kΩ 5 − 2000(0.8)(N )(0.273 x10−3 )≥ 1.5 → N ≤ 8.01 → Fanout = 8 The result is independent of the tolerance if the resistors track each other. Note that Eq. 9.83 also yields N = 8 if more digits are used in the calculation. 9.83 From the analysis in the text, we see that the fanout is limited by the VH condition. iB1 = 5 − 0.7 − 0.8 RB iE1 = −β R iB1 = −0.25iB1 5 − 0.7 − 0.8 5 − 2000(N )(0.25) ≥ 1.5 → RB ≥ 5 kΩ RB 9.84 (a) Q4 is in the forward - active region with I E = (β F + 1)I B 5 − 0.7 − 0.6 = 234 mA 1600 5 − 0.8 − 0.6 5 − 0.6 − 0.15 (b) Q4 saturates; I E = I B + IC = 1600 + 130 = 34.9 mA I E = 101 9.85 ( a) PD = 5V (234 mA) = 1.17 W (b) PD = 5V (34.9 mA) = 0.175 W 9-30 9.86 *Figure 9.86 - TTL Output Current VCC 5 0 DC 5 RB 5 3 1.6K RS 5 4 130 Q1 4 3 2 NBJT D1 2 1 D1 IL 1 0 DC 0 .DC IL 0 30MA 0.01MA .MODEL NBJT NPN BF=40 BR=0.25 IS=5E-16 TF =0.15NS TR=15NS +CJC=0.5PF CJE=.25PF CJS=1.0PF RB=100 RC=5 RE=1 .MODEL D1 D IS=5E-16 .PROBE V(1) V(2) V(3) V(4) .END 5.0V vO 4.0V 3.0V 2.0V iL 1.0V 0A 5mA 10mA 15mA 20mA 25mA 30mA 9-31 9.87 *Problem 9.87 - Modified TTL Inverter VTC VI 1 0 DC 0 VCC 9 0 DC 5 Q1 2 8 1 NBJT Q2 4 3 0 NBJT Q3 6 2 3 NBJT Q4 7 6 5 NBJT D1 5 4 DN RB 9 8 4K RC 9 6 1.6K RS 9 7 130 RL 4 0 100K Q5 10 11 0 NBJT RB5 3 11 3K RC5 3 10 1K .DC VI 0 5 .01 .MODEL NBJT NPN BF=40 BR=0.25 IS=1E-17 TF =0.25NS TR=25NS +CJC=0.6PF CJE=.6PF CJS=1.25PF RB=100 RC=5 RE=1 .MODEL DN D .PROBE V(1) V(2) V(3) V(4) V(5) V(6) .END 4.0V vO 3.0V 2.0V 1.0V 0V 0V 1.0V 2.0V vI 3.0V 4.0V 5.0V In the modified TTL circuit, Q3 cannot start conducting until its base reaches at least VBE5 + VBE6 = 1.2 V. 9-32 9.88 + 5V 20 k Ω 8k Ω + Q 1 0.7 V + 0.8 V Q 3 Q + 5k Ω 0.8 V 2 (a) v I = VH : Q4 off − IB 4 = 0 = IC 4 | Q2 saturated with IC 4 = 0 (5 − 0.7 − 0.8 − 0.8)V = 135 µA | I = −β I = −0.25 135 µA = −33.8 µA I B1 = ( ) E1 R B1 20 kΩ IC1 = −169 µA | IC 3 = (5 − 0.15 − 0.8)V 8 kΩ = 506 µA 0.8V = 515 µA 5 kΩ IE 3 = 506µA + 169µA = 675 µA | IB 2 = 675µA − (b) v I I B1 = = VL : Q2 , Q3 off ; Q4 on (5 − 0.8 − 0.15)V 20 kΩ = 203 µA = IE1 | IC1 = 0 9-33 9.89 See Problem 9.90. 9.90 4.0V *Problem 9.90 - Low Power TTL Inverter VTC versus Temperature VI 1 0 DC 0 VCC 9 0 DC 5 Q1 2 8 1 NBJT Q2 4 3 0 NBJT Q3 6 2 3 NBJT Q4 7 6 5 NBJT D1 5 4 DN RB 9 8 20K RC 9 6 8K RS 9 7 650 RL 4 0 100K RE 3 0 5K .DC VI 0 5 .01 .TEMP -55 25 85 .MODEL NBJT NPN BF=40 BR=0.25 IS=1E17 TF =0.25NS TR=25NS +CJC=0.6PF CJE=.6PF CJS=1.25PF RB=100 RC=5 RE=1 .MODEL DN D .PROBE V(1) V(2) V(3) V(4) V(5) V(6) .END 9.91 +5V RC RS vO 3.0V 2.0V -55 o C 1.0V +25 o C +85 o C 0V 0V 1.0V 2.0V vI 3.0V 4.0V 5.0V 1.6 k Ω 130 Ω Q4 V H = 5 − 0.7 − 0.7 − N x 0.17 mA N (IIH ) R βF + 1 C N (0.17mA) (1600) 40 + 1 V H ≥ 2.4V → N ≤ 180 V H = 3.6 − 9-34 9.92 For small β R , fanout is limited by the vO = VL case (v I = V H ). = 567µA 5 kΩ (5 − 0.15 − 0.8)V + 567µA − 0.8V = 1.95 mA iB 2 = iE 3 − iR E = 1.25 kΩ 2 kΩ (5 − 0.8 − 0.15)V = 0.810 mA | α = β R = 0.05 = 0.0476 iIL = −iE1 = −iB1 = R 1 + β R 1.05 5 kΩ 1 1− ⎛ 0.15V ⎞ 0.0476(403.4) = 9.52 Using Eq. 9.61, Γ = exp⎜ ⎟ = 403.4 → β FOR = 20 20 1 ⎝ 0.025V ⎠ 1+ 0.05 403.4 N (0.810mA) ≤ 9.52(1.95mA) → N = 22 9.93 For the vO = VL case, the equations are given in Problem 9.92. For vO = VH, V H = 5 − IB 4 RC − 0.7 − 0.7 ≥ 2.4V ⎛ 5 − 0.7 − 0.8 − 0.8 ⎞ ⎛ 2.7 ⎞ IIH = β R ⎜ ⎟ ⎟ = βR ⎜ ⎝ ⎠ ⎝ 4000 ⎠ 4000 NIIH Nβ R ⎛ 2.7 ⎞ IB 4 = = ⎜ ⎟ β F + 1 β F + 1 ⎝ 4000 ⎠ iB 3 = (β R + 1)iB1 = 1.05 (5 − 0.7 − 0.8 − 0.8)V N≤ 1.2(β F + 1)(4000) 2.7β R (1600) 100 80 Fanout 60 40 20 0 0 1 2 3 4 Inverse Current Gain 5 9-35 function [N,X]=P993 br=0; bf=40; g=exp(.15/.025); for i=1:50 br=br+.1; ar=br/(1+br); ib3=(1+br)*675; ib2=1730+ib3; bfor=40*(1-1/(ar*g))/(1+bf/(br*g)); N1=fix(bfor*ib2/1013); N2=1.2*(bf+1)*4000/(2.7*br*1600); N(i)=min(N1,N2); X(i)=0.1*i; end »[Y,X]=p993; »plot(X,Y) 9.94 +2 V 2 kΩ +2 V + 2kΩ 0.8 V 2kΩ + Q1 Q2 + VH iIN Q1 0.7 V + 0.8 V - "Saturated" Q3 0.8 V 0.15 V (a) VH = 2 − VECSAT 2 = 2 − 0.15 = 1.85 V | VL = VCESAT 3 = 0.15 V (2 − 0.7 − 0.8)V = 62.5 µA (b) iIH : iB 2 ≅ 0 | iIH = 0.25 2 kΩ (2 − 0.8 − 0.15)V − (2 − 0.8 − 0.8 − 0.15)V = −650 µA iIL = −iB1 = − 2 kΩ 2 kΩ (c ) Assume β FOR ≤ 28.3 For the pnp transistor : N (62.5µA) ≤ 28.3 → N = 56 2 kΩ (2 − 0.7 − 0.8)V → N = 13 For the npn transistor : N (650µA) ≤ 28.3(1.25) 2 kΩ (2 − 0.8 − 0.8 − 0.15)V 9-36 9.95 (a) VL = VCESAT 3 = 0.15 V | VH = 2 − VBE 2 = 2 − 0.7 = 1.3 V ⎛ 2 − 0.8 − 0.15 2 − 0.15 − 0.15 ⎞ = 0.15V : iIL = −(iB1 + iC1 ) = −⎜ + ⎟ = −247 µA ⎝ 10000 ⎠ 12000 ⎛ 2 − 0.7 − 0.8 ⎞ v I = 1.3V : iIL = β R iB1 = 0.25⎜ ⎟ = 12.5 µA ⎝ ⎠ 10 4 2 − 0.8 − 0.15 2 − 0.15 − 0.15 + = 1.875mA (c ) Using β FOR = 28.3 : iL = iB 2 + iC 2 = 6000 1000 ⎛ 2 − 0.7 − 0.8 ⎞ 2 − 0.8 iB 3 = + 1.25⎜ ⎟ = 162.5µA ⎝ 10000 ⎠ 12000 (b) v I 28.3(0.1625 mA) ≥ N (0.247 mA) + 1.875mA → N = 11 9.96 (a) Y = ABC (b) VL = VCESAT 3 = 0.15 V | V H = 3.3 − VBE1 − VD = 3.3 − 1.4 = 1.9 V 3.3 − 0.7 − 0.15 = −408 µA 6000 (c ) v I 2.0V = 1.9V , input diode is off and iIH = 0. v I = 0.15V , iIL = − vO 1.5V 1.0V 0.5V vI 0V 0V 1.0V 2.0V 3.0V 4.0V The VTC starts to decrease immediately because Q2 is ready to conduct due to the 0.7-V drop across the input diode. When the input has increased to approximately 0.7 V, Q3 begins to conduct and the output drops rapidly. The VTC is much sloppier than that of the corresponding TTL gate. For this particular circuit VIL = 0 and VIH = 0.8 V. Based upon our definitions, NML = 0. However, the initial slope can be reduced by changing the ratio RC/R2 so that VIL = 0.7 V. 9-37 9.97 (a) If either input A is low or input B is low, VB2 will be low. Q2 will be off, Q3 will be on and Y will be low. Therefore Y = A + B → Y = AB . +5 V +5 V 4kΩ 4 kΩ 0.8 V +0.3 V 0.15 V + 10 k Ω -5 V 10 k Ω 5 kΩ + +0.45 V +4.3 V Q1 VB Q2 -5 V (b) VH = 5 − IB 5 R2 − VBE 5 ≈ 5 − VBE 5 = 5 − 0.7 = 4.30 V VL = 5 − α F IE 3 R2 − IB 5 R2 − VBE 5 ≈ 5 − IE 3 R2 − VBE 5 +0.7 − 0.7 − (−5) = 1.00 mA | VL = 5 − 0.001(4000) − 0.7 = +0.300 V 5000 ⎛ 5 − 0.7 − VB ⎞ VB − (−5) VB − 0.7 − (−5) 1 + → VB = 1.97V ⎟= (c ) v I = 4.3V : 1.25⎜ ⎝ 4000 ⎠ 10000 5000 41 5 − 0.7 − 1.97 I B1 = = 582 µA | IE1 = −0.25 IB1 = −146 µA | IIH = 146 µA 4000 0.3 + 0.15 − (−5) 5 − 0.8 − 0.3 v I = 0.3V : IB1 = = 975 µA | IC1 = − = −545 µA 10000 4000 IE1 = IB1 + IC1 = 430 µA | IIL = −430 µA IE 3 = 9.98 1.5 V 1.5 V 800 Ω 1kΩ 800 Ω 1kΩ 0.70 V 0.25 V - 0.45 V + Off Q 1 VH + 0.45 V + 0.7 V Q 1 Off - 9-38 (a) VH = VCC = 1.5 V | VL = "VCESAT 1" = 0.7 − 0.45 = 0.25 V (b) For v I = 1.5V , the input diode is off, and IIH = 0. 1.5 − 0.45 − 0.25 = 1.00 mA 800 (c ) Note that Q1 operates as if it were in the forward - active region : For v I = 0.25V , IIL = β F IB1 ≥ NIIL + IR 9.99 vBE + vD 2 − vD1 = vCE 2 ⎛ 1.5 − 0.45 − 0.70 ⎞ 1.5 − 0.25 | 40⎜ → N = 16 ⎟ ≥ N (0.001) + ⎝ ⎠ 800 1000 | For vD 2 ≅ vD1 , vCE = v BE = 0.7 V iC = iCC + iD1 | iB = iBB − iD1 | iC = β F iB iCC + iD1 = β F (iBB − iD1) → iD1 = β F iBB − iCC 20(0.25)− 1 = mA = 0.191 mA 21 βF + 1 iD 2 = iBB − iD1 = 0.25 − 0.191 = 0.059 mA | iC = 20iB = 20iD 2 = 1.18 mA 9.100 In this circuit as drawn, the collector - base junction of Q1 is bypassed by a Schottky diode. Q1 will be "off" with VBC = +0.45 V . 5 − 0.45 − 0.7 iB1 = = 963 µA | IIN = 0 | iB 2 = iB1 = 963 µA 4000 9.101 4.0V 4.0V vO vI 3.0V 3.0V 2.0V 2.0V vO 1.0V 1.0V Time 0V 0V 1.0V 2.0V vI 3.0V 4.0V 5.0V 0V 0s 5ns 10ns 15ns 20ns 25ns 30ns Result: τP = 3.0 ns *Problem 9.101 - Schottky TTL Inverter VTC VI 1 0 DC 3.5 PWL(0 3.5 0.2N 0.25 15N 0.25 15.2N 3.5 30N 3.5) 9-39 VCC 9 0 DC 5 Q1 2 8 1 NBJT D1 2 8 DS Q2 4 3 0 NBJT D2 3 4 DS Q3 6 2 3 NBJT D3 2 6 DS Q4 7 5 4 NBJT Q5 7 6 5 NBJT D5 6 7 DS RB 9 8 2.8K RC 9 6 900 RS 9 7 50 R5 5 0 3.5K RL 4 0 100K Q6 10 11 0 NBJT D6 11 10 DS R2 3 11 500 R6 3 10 250 .OP .DC VI 0 5 .01 .TRAN .025N 30N .MODEL NBJT NPN BF=40 BR=0.25 IS=1E-17 TF =0.15NS TR=15NS +CJC=1PF CJE=.5PF CJS=1PF RB=100 RC=10 RE=1 .MODEL DS D IS=1E-12 .PROBE V(1) V(2) V(3) V(4) V(5) V(6) .END 9-40 9.102 RC 0.54 k Ω Y C VREF + 0.7 V 0.75 k Ω RE -3 V Y = A + B + C | VH = 0 V | VL = −540 IC = −540 VREF − 0.7 − (−3) = −0.72(VREF + 2.3) 750 V H + VL = VREF − 0.7 + 0.4 + 0.7 = VREF + 0.4 2 0 − 0.72(VREF + 2.3) = VREF + 0.4 → VREF = −0.903V | VL = −0.72(−.903 + 2.3) = −1.01 V 2 9.103 RC V REF- 0.4V 3.3 k Ω Y V REF V REF- 0.7V 2.4 k Ω RE -3 V Y = A + B + C | V H = 0 V | VL = VREF − 0.4 0 + (VREF − 0.4 ) V H + VL = VREF | = VREF → VREF = −0.40 V | VL = −0.80V 2 2 9-41 9.104 The circuit can be modeled by a normal BJT with a Schottky diode in parallel with the collector base junction. If iC and iB are defined to be the collector- and base-currents of the BJT, iC + iB = +5 V i=0 4 kΩ iC 5 − 0.7 1.075 mA = 1.075 mA | iC ≅ β F iB → iB = = 26.9 µA | iC = 1.05 mA 4000 40 iB Q1 9.105 If we assume 50% of the gates are switching (an over estimate), 50W = 2µW / gate | τ P = 1ns | PDP = (2µW )(1ns) = 2 fJ (a) P = 0.5(50 x10 6 ) (b) PDP = (0.1mW )(0.1ns) = 10 fJ | The result in Part (a) is off the graph! 9.106 If we assume 50% of the gates are switching (an over estimate), 100W = 1µW / gate | τ P = 0.25 ns | PDP = (1µW )(0.25ns) = 0.25 fJ (a) P = 0.5(200 x10 6 ) (b) PDP = (0.1mW )(0.1ns) = 10 fJ 9.107 | The result in Part (a) is off the graph! (a) τ P = 9.109 PDP 0.5 pJ = = 1.67 ns P 0.3mW (b) P = PDP τP = 0.5 pJ = 0.5 mW 1ns (a) PDP = (0.7 ns)(40 mW ) = 28 pJ (b) P = PDP | τP = PDP 28 pJ = = 2.8 ns P 10 mW τP = 28 pJ = 140 mW 0.2 ns 9-42 9.110 Results from B2SPICE: VH = 4.48 V, VL = 0.54 V, τPHL = 3.3 ns, τPLH = 4.4 ns. Circuit 9_127-DC Transfer-2 (V) +5.000 +0.000e+000 +500.000m +1.000 +1.500 +2.000 +2.500 +3.000 +3.500 +4.000 +4.500 VI +4.000 +3.000 +2.000 +1.000 +0.000e+000 V(3) Circuit 9_127-Transient-2 (V) +5.000 +0.000e+000 +10.000n +20.000n +30.000n +40.000n +50.000n +60.000n +70.000n +80.000n +90.000n Time (s) +100.000n +4.000 +3.000 +2.000 +1.000 +0.000e+000 V(3) 9.111 V M4 A DD B M3 Q4 vO R v I B M2 Q 3 A M1 9-43 9.112 VDD M A 4 B M 3 Q4 vO B M2 B M 6 A M 1 A M7 Q3 M 5 9.113 Results from B2SPICE with R = 4 kΩ: VH = 5 V, VL = 0 V, τPHL = 1.9 ns, τPLH = 4.1 ns. Circuit 9_130-DC Transfer-1 (V) +0.000e+000 +500.000m +1.000 +1.500 +2.000 +2.500 +3.000 +3.500 +4.000 +4.500 VI +5.000 +4.000 +3.000 +2.000 +1.000 +0.000e+000 V(3) Circuit 9_130-Transient-1 (V) +0.000e+000 +10.000n +20.000n +30.000n +40.000n +50.000n +60.000n +70.000n +80.000n +90.000n Time (s) +100.000n +5.000 +4.000 +3.000 +2.000 +1.000 +0.000e+000 V(3) 9-44 9.114 V V A DD DD M 3 A M3 M B 4 Q B M4 4 B Q M2 4 M 2 R v O A M B 1 R vO A M1 B M 6 A M 5 M6 B A M5 (a) (b) 9.115 Results from B2SPICE: VH = 4.7 V, VL = 0.34 V, τPHL = 7.0 ns, τPLH = 14 ns. Circuit 9_133-DC Transfer-4 (V) +5.000 +0.000e+000 +500.000m +1.000 +1.500 +2.000 +2.500 +3.000 +3.500 +4.000 +4.500 VI +4.000 +3.000 +2.000 +1.000 +0.000e+000 V(5) Circuit 9_133-Transient-4 (V) +5.000 +0.000e+000 +10.000n +20.000n +30.000n +40.000n +50.000n +60.000n +70.000n +80.000n +90.000n Time (s) +100.000n +4.000 +3.000 +2.000 +1.000 V(5) 9-45 CHAPTER 10 10.1 A/C temperature Automobile coolant temperature gasoline level oil pressure sound intensity inside temperature Battery charge level Battery voltage Fluid level Computer display hue contrast brightness Electrical variables voltage amplitude voltage phase current amplitude current phase power power factor spectrum Fan speed Humidity Lawn mower speed Light intensity Oven temperature Refrigerator temperature Sewing machine speed Stereo volume Stove temperature Time TV picture brightness TV sound level Wind velocity 10.2 ( a ) 20 log (120) = 41.6 dB | 20 log (60) = 35.6 dB | 20 log (50000) = 94.0 dB 20 log(100000) = 100 dB | 20 log(0.90) = −0.915 dB ( b ) 20 log (600) = 55.6 dB | 20 log (3000) = 69.5 dB | 20 log (106 ) = 120 dB 20 log(200000) = 106 dB | 20 log(0.95) = −0.446 dB ( c ) 10 log (2x109 ) = 93.0 dB | 10 log (4x105 ) = 56.0 dB 10 log (6x108 ) = 87.8 dB | 10 log(1010 ) = 100 dB 10.3 (a) 4 2 vO 0 -2 vS -4 0 0.5 1 1.5 2 2.5 3 3.5 4 x10 -3 10-1 (b) 500 Hz : (c) 500 Hz : (d) 500 Hz : (e) Yes 1∠0 o | 1500 Hz : 0.333∠0 o | 2500 Hz : 0.200∠0 o 2∠30 o | 1500 Hz : 1∠30 o | 2500 Hz : 1∠30 o 2∠30 o | 1500 Hz : 3∠30 o | 2500 Hz : 5∠30 o 10.4 Vs = 0.0025V | PO = 40W | Vo = 2PO RL = 2(40)(8) = 25.3V 25.3 = 10100 | 20 log ( 10100)= 80.1 dB .0025 0.0025V V 25.3V = 45.45nA | I o = o = = 3.162 A Is = 5kΩ + 50kΩ 8Ω 8Ω 3.162 A = 6.96 x 10 7 | 20 log 3.48 x 10 7 = 157 dB Ai = 45.45nA 40W = 7.04 x 1011 | 10 log 7.04 x 1011 = 118 dB Ap = .0025V (45.45nA) Av = ( ) ( ) 2 10.5 Vs = 0.01V | PO = 20 mW | Vo = 2 PO RL = 2(.02)(8) = 0.566V 0.566 = 56.6 | 20 log (56.6) = 35.0 dB .01 0.01V V 0.566V Is = = 192 nA | Io = o = = 70.8 mA 8Ω 2 kΩ + 50 kΩ 8Ω 70.8 mA Ai = = 3.68 x 10 5 | 20 log (3.68 x 10 5 )= 111 dB 192 nA 0.02W Ap = = 2.08 x 10 7 | 10 log (2.08 x 10 7 )= 73.2 dB .01V (192 nA) 2 Av = 10.6 Rth v th (a) v + - th = voc = 0.768 2 = 1.09 V vo = ⎛ 0.768 − 0.721⎞ v −v RL v th → Rth = RL th o = 430⎜ ⎟ = 28.0 Ω 0.721 Rth + RL vo ⎝ ⎠ th (b) v vo = = voc = 0.760 2 = 1.08 V (c) 1.09 V and 1.08 V → 9% error and 8% error 10-2 ⎛ 0.760 − 0.740 ⎞ v −v RL v th → Rth = RL th o = 1040⎜ ⎟ = 28.1 Ω 0.740 Rth + RL vo ⎝ ⎠ 10.7 G4 laptop – 1 V, 28 Ω. 10.8 (a) Vo = 2 RL PO = 2(8)(20) = 17.9V Pi = Av = Vi2 12 1V 17.9V = = 25.0µW | I i = = 49.9µA | I o = = 2.24 A 2 Ri 40066 20000Ω + 32Ω 8Ω Vo 17.9V 20W 2.24 A = = 17.9 | AP = = 8.00 x105 | Ai = = 4.49 x10 4 Vi 1V 25µW 49.9µA o (b) V = 17.9 V ; recommend ± 20 - V supplies 10.9 V = 2 PR | I= 2P R The 24-Ω case represents a good trade off between voltage and current. R ( Ω) 8 24 1000 V (V) 1.27 2.19 14.1 I (mA) 158 91.3 14.1 10.10 In the dc steady state, the internal circuit voltages cannot exceed the power supply limits. (a) +15 V (b) -9 V 10.11 (a) For VB = 0.6V , VO = +8V | Av = Av = 32 dB ∠AV = 180 o dvO dvI = v I = 0.6V 12 − 4 = −40 0.5 − 0.7 | VM ≤ 0.100 V for linear operation vO (t )= (8 − 4sin1000t ) V (b) vI (t )= (0.6 + 0.1sin1000t ) V 10-3 10.12 (a) For VB = 0.5V , VO = +12V | dvO is different for positive and negative values of dvI VM sin1000t . Thus, the gain is different for positive and negative signal excursions and the output will always be a distorted sine wave. This is not a useful choice of bias point for the amplifier. (b) For VB = 1.1V , VO = +2V and dvO = 0. The gain is zero for this bias point. dvI Thus this is also not a useful choice of bias point for the amplifier. 10.13 (a) For V (b) For V 10.14 14 B = 0.8V , VO = +3V | Av = dvO dv I = v I = 0.8V 4−2 = −10 0.7 − 0.9 Av = 20dB ∠AV = 180 o | VM ≤ 0.100 V for linear operation B = 0.2V , VO = +14V | Av = dvO dv I =0 v I = 0.8V The output signal will be distorted regardless of the value of V M. vO 12 10 8 6 4 Time 2 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 The amplifier is operating in a linear region. vO = 8 - 4 sin 1000t volts There are only two spectral components: 8 V at dc and -4 V at 159 Hz 10-4 10.15 For sin 1000t ≥ 0, vO = 12 − 4 sin 1000t For sin 1000t < 0, vO = 12 − 1 sin 1000t 14 vO 12 10 8 Time 6 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 Using the MATLAB FFT capability with a fundamental frequency of 1000/2 t=linspace(0,2*pi/1000,1000); y=12-4*sin(1000*t).*(sin(1000*t)>=0)-sin(1000*t).*(sin(1000*t)=0)sin(1000*t).*(sin(1000*t)=0)+2*sin(w).*sin(w)> 50π , Av (s) ≈ 2π x 10 5 fH ≈ = 100 kHz | However, we are not that lucky at low frequencies. 2π 2 10 3 s2 10 3 ω L 10 3 For s =0.95*anom & a> -400d SPICE yields: fo = 38.9 kHz, Q = 8.1, Center frequency gain = 38.8 dB. These values are off due to the finite bandwidth of the op-amp and its excess phase shift at the center frequency of the filter. The center frequency is substantially shifted. 11-25 11.85 (a) BW = ωo Q = 1 1 rad rad | ω L = 0.833 | ω H = 1.167 3 s s rad rad 1 | ω 'o = ω o = 1 | Q' = = 4.65 s s 0.215 1 1 R2 1 | Q= | = 2Qω o C1 = C2 = C : ω o = 2 Rth Rth C C Rth R2 BW ' = BW 2 2 − 1 = 0.215 ⎛ ⎞2 ⎛ ⎞2 ⎛ ⎞2 2 ⎛ Rth ⎞ ⎜ −2Qsω o ⎟ ⎛ Rth ⎞ ⎜ −6 s ⎟ ⎜ −6 s ⎟ b A s = | For R3 = ∞, ABP (s) = ⎜ ⎟ =⎜ ⎟ ⎜ ( ) BP ( ) ⎜ ⎟ ⎜ ⎟ ωo s s ⎟ 2 R ⎝ R1 ⎠ ⎜ ⎜ s 2 + + 1⎟ s 2 + + 1⎟ ⎜ s + s + 1⎟ ⎟ ⎝ 1⎠ ⎜ ⎝ ⎝ 3 ⎠ 3 ⎠ Q ⎝ ⎠ 2 11.86 n=conv([-6 0],[-6,0]); d=conv([1 1/3 1],[1 1/3 1]); bode(n,d) 60 50 40 Gain dB 30 20 10 0 -10 10 -1 10 0 Frequency (rad/sec) 10 1 200 150 100 Phase deg 50 0 -50 -100 -150 -200 10 -1 10 0 Frequency (rad/sec) 10 1 11-26 11.87 Using normalized frequency and R3 = ∞ : 5 kHz → ω o = 1 and 6 kHz → ω o = 1.1 ⎛ −10 s ⎞⎛ ⎞ −12s 120 s 2 ABP (s) = ⎜ 2 ⎟⎜ ⎟= ⎝ s + 0.2s + 1⎠⎝ s 2 + 0.24s + 1.44 ⎠ s 4 + 0.44s3 + 2.484 s2 + 0.528s + 1.44 and A(jω o ) = 1429 = 63.1 dB At the new center frequency s = jω o , − 0.44ω 3 o + 0.528ω o = 0 → ω o = 1.095 The bandwidth points can be found using MATLAB: w=linspace(.9,1.5,250); [m,p,w]=bode([120 0 0],[1 .44 2.484 .528 1.44],w); 20*log10(max(m)) ans = 63.098 ((20*log10(a))>60.098).*(w.'); From this last vector one can easily find: ωo = 1.095 or fo = 5.48 kHz, ωL = 0.970 or fo = 4.85 kHz, ωH = 1.237 or fo = 6.19 kHz, BW = 1.34 kHz, Q = 4.09 11-27 11.88 w1=2*pi*5000; q1=5; w2=2*pi*6000; q2=5; n1=[-2*q1*w1 0]; d1=[1 w1/q1 w1*w1]; n2=[-2*q2*w2 0]; d2=[1 w2/q2 w2*w2]; n=conv(n1,n2); d=conv(d1,d2); w=logspace(4,5,100); bode(n,d,w) 70 60 50 Gain dB 40 30 20 10 0 10 4 200 150 100 Phase deg Frequency (rad/sec) 10 5 50 0 -50 -100 -150 -200 10 4 Frequency (rad/sec) 10 5 11.89 Using ABP = 20 dB at the center frequency : Rin = R1 = 10 kΩ | 10 = KQ = R2 = 100 kΩ | K = R2 R1 10 1 1 = 2 | R = KR1 = 20 kΩ | C = = = 0.0133 µF . ω o R 2π (600 Hz)20 kΩ Q SQ C = 0. 11.90 Q is independent of C in the Tow-Thomas biquad. 11-28 11.91 vO 0 t T/2 T 3T/2 2T -5 V 11.92 The waveform going into the low - pass filter is the same as that in Prob. 11.91 ⎛ 8.2 kΩ ⎞ except the amplitude will be VM = 1V ⎜− ⎟ = −3.037 V . ⎝ 2.7 kΩ ⎠ 1T ⎛ 10 kΩ ⎞ 2 2 (−3.037) The average value of the waveform is V = ⎜ − = +0.759 V . ⎟ T ⎝ 10 kΩ ⎠ 11.93 The Fourier series converges very rapidly since only the even terms exist for n ≥ 2 and the terms decrease as 1/n2. Thus the RMS value will be dominated by the first term (n = 1). ⎛ ω ⎞2 π 1 ω 120π 2 Require : ≤ 0.01 | 50 π ≤ 1 + | ω ≥ = = 2.40 Hz ( ) ⎟ ⎜ o 2 157 157 ⎝ωo ⎠ ⎛ ω ⎞2 1+ ⎜ ⎟ ⎝ωo ⎠ 11.94 vO 0.5 V t 0 0 T/2 T 3T/2 2T 11.95 vO 1.0 V t 0 0 T/2 T 3T/2 2T 11-29 11.96 *Figure P11.94 - RECTIFIER VS 1 0 PWL(0 0 1M 1 3M -1 5M 1 7M -1 8M 0) R1 1 2 10K R2 4 5 10K R3 5 6 10K R4 2 4 10K R5 1 5 20K D1 3 2 DIODE D2 4 3 DIODE EOP1 3 0 0 2 1E5 EOP2 6 0 0 5 1E5 .MODEL DIODE D IS=1E-12A .TRAN .01M 8M .PRINT TRAN V(6) .PROBE V(1) V(2) V(3) V(4) V(5) V(6) .END 11.97 *Figure P11.95 - RECTIFIER VS 1 0 PWL(0 0 1M 1 3M -1 5M 1 7M -1 8M 0) R1 0 2 10K R2 4 5 10K R3 5 6 20K R4 2 4 10K D1 3 2 DIODE D2 4 3 DIODE EOP1 3 0 1 2 1E5 EOP2 6 0 1 5 1E5 .MODEL DIODE D IS=1E-12A .TRAN .01M 8M .PRINT TRAN V(6) .PROBE V(1) V(2) V(3) V(4) V(5) V(6) .END 11.98 −V V V V1 = I S exp o1 | VO1 = −VT ln 41 | VO 2 = −VT ln 42 10kΩ VT 10 I S 10 I S ⎛ V V ⎞ VV VO1 + VO2 )= VT ⎜ ln 41 + ln 42 ⎟ = VT ln 18 2 2 VO3 = −( 10 I S ⎠ 10 I S ⎝ 10 I S ⎛ VV ⎞ VV V VO = −10 4 I D = −10 4 I S exp D = −10 4 I S exp⎜ ln 18 2 2 ⎟ = − 14 2 VT 10 I S ⎝ 10 I S ⎠ 11-30 11.99 Simplify the circuit by taking a Thevenin equivalent of the 5V source and two 10 kΩ 10 kΩ resistors : VTH = 5V = 2.5V | RTH = 10 kΩ 10 kΩ = 5kΩ 10 kΩ + 10 kΩ 100 kΩ 5kΩ VO = 5V − Using superposition : V+ = 2.5 +5 = 2.62V 100 kΩ + 5kΩ 100 kΩ + 5kΩ 100 kΩ VO = 0V : V+ = 2.5 = 2.38V VN = 2.62 − 2.38 = 0.24V 100 kΩ + 5kΩ 11.100 4.3kΩ = −0.993 V 4.3kΩ + 39 kΩ 4.3kΩ = 0.993 V VO = 10V : V+ = 10 4.3kΩ + 39 kΩ VN = 0.993 − (−0.993)= 1.99 V VO = −10V : V+ = −10 11.101 4.3kΩ = 0.487 V 4.3kΩ + 39 kΩ 4.3kΩ = −0.487 V For VO = −4.3 − 0.6 = −4.9V : V+ = −4.9 4.3kΩ + 39 kΩ VN = 0.487 − (−0.487) = 0.974 V For VO = 4.3 + 0.6 = 4.9V : V+ = 4.9 11.102 R4 R2 R2 +5 V +5 V v + +5 V R3 + R TH VTH O v O R C R C 11-31 For VO = 0 : V+ = VTH For VO = 5 : V+ = VTH Subtracting : 5 VTH 0.05 R2 R4 = 1− = 0.975V | RTH = R3 R4 | VTH = 5 2 RTH + R2 R3 + R4 RTH 0.05 R2 +5 = 1+ = 1.025V RTH + R2 RTH + R2 2 RTH RTH R = 0.05V → = 0.01 → 2 = 99 RTH + R2 RTH + R2 RTH R2 97.5 R RTH + R2 0.975 = → VTH 2 = 97.5 → VTH = = 0.985V RTH RTH 0.01 99 RTH + R2 R4 R → 3 = 4.077 | Choosing R4 = 2 kΩ → R3 = 8.154 kΩ R3 + R4 R4 0.985 = 5 RTH = 8.154 kΩ 2 kΩ = 1.606 kΩ | R2 = 99(1.606 kΩ) = 159 kΩ Choosing standard values : R2 = 160 kΩ | R3 = 8.2 kΩ | R4 = 2 kΩ 11.103 24 kΩ 3.4 kΩ + 12 = 6.74 V 3.4 kΩ + 24 kΩ 3.4 kΩ + 24 kΩ 24 kΩ = 5.26 V For vO = 0V : V+ = 6 3.4 kΩ + 24 kΩ ⎛ t ⎞ VF − VI )exp⎜ − v (t )= VF − ( ⎟ ⎝ RC ⎠ ⎛ T ⎞ 6.74 = 50.7 µs 6.74 = 12 − ( 12 − 5.26)exp⎜− 1 ⎟ → T1 = 6200 3.3 x10−8 ln 5.26 ⎝ RC ⎠ ⎛ T ⎞ 6.74 = 50.7 µs 5.26 = 0 − (0 − 6.74)exp⎜− 2 ⎟ → T2 = 6200 3.3 x10−8 ln 5.26 ⎝ RC ⎠ For vO = +12V : V+ = 6 ( ) ( ) f = 1 = 9.86 kHz 50.7µs + 50.7µs 11.104 f = 0. The circuit does not oscillate. VO = 0 is a stable state. 11-32 11.105 1 1+β R1 = | T = 2RC ln = 2 RC ln 3 = 2.197 RC R1 + R2 2 1− β During steady - state oscillation, the maximum output current from the op- amp is (a) Let R1 = R2 | β= 5 − (−2.5) 5 7.5 5 + | Let R = R1 = R2 | + ≤ 1mA → R ≥ 10kΩ 2R R R R1 + R2 0.001s RC = = 4.55 x10−4 s | Selecting C = 0.015µF , R = 30.3kΩ → 30 kΩ (5% values) 2.197 1 Final values : R = R1 = R2 = 30 kΩ | C = 0.015 µF | f = = 1.01 kHz 2.197(30 kΩ)(0.015µF ) 1 1 1 (b) β = R | β max = 30 kΩ 0.95 = 0.525 | β min = 30kΩ 1.05 = 0.475 ( ) ( ) 1+ 2 1+ 1+ R1 30 kΩ(1.05) 30 kΩ(0.95) I= Tmax = 2(30 kΩ)(1.05)(0.015µF )(1.1)ln 1 + 0.525 = 1.213 x10−3 s → f min = 825 Hz 1 − 0.525 1 + 0.475 Tmin = 2(30kΩ)(0.95)(0.015µF )(0.90)ln = 7.949 x10−4 s → f min = 1.26 kHz 1 − 0.475 4.75 = 2.375V (c ) For vO = +4.75V : V+ = 4.75β = 2 −5.25 For vO = -5.25V : V+ = −5.25β = = −2.625V 2 ⎛ t ⎞ v (t ) = VF − (VF − VI )exp⎜ − ⎟ ⎝ RC ⎠ ⎛ T ⎞ 7.375 2.375 = 4.75 − (4.75 − (−2.625))exp⎜ − 1 ⎟ → T1 = RC ln = 1.133RC ⎝ RC ⎠ 2.375 ⎛ T ⎞ 7.625 = 1.066RC −2.625 = −5.25 − (−5.25 − 2.375)exp⎜ − 2 ⎟ → T2 = RC ln ⎝ RC ⎠ 2.625 T = 2.199RC = 2.199(30kΩ)(0.015µF ) = 9.896 x10 -4 s → f = 1.01 kHz − Very little change 11-33 11.106 For a triangular waverform with peak amplitude VS and ω o = 2000π : v (t )= ∑ 8VS nπ sin sin nω ot 2 2 2 n=1 n π 1 1+ s 3000π | Av ( jf ) = 1 ⎛ f ⎞2 1+ ⎜ ⎟ ⎝ 1500 ⎠ ∞ For the low - pass filter : Av (s)= − Av (j1000) = 0.832 | Av (j 3000) = 0.447 | Av (j 5000) = 0.287 For a 5 - V fundamental : 1 5 2 π = 7.41 V 0.832 8 π This series contains only odd harmonics : For n = 2, V2000 = 0. ⎛ 5 ⎞1 For f = 3 kHz, n = 3 : V3000 = 0.447⎜ ⎟ = 0.298 V ⎝ .832 ⎠ 32 ⎛ 5 ⎞1 For f = 5 kHz, n = 5 : V5000 = 0.287⎜ ⎟ = 69.0 mV ⎝ .832 ⎠ 52 8VS 2 0.832 = 5 → VS = 11.107 Von VCC T = RC ln 1− β 1+ VCC VEE Tr = RC ln V 1 − on VEE 1+ β 15kΩ | β= = 0.357 | 15kΩ + 27kΩ 0.6 10 = 841 µs (51kΩ)(0.033µF )ln 1 − 0.357 ⎛ 10 ⎞ 1 + 0.357⎜ ⎟ ⎝ 10 ⎠ = (51kΩ)(0.033µF )ln = 416 µs 0.6 1− 10 1+ 11-34 11.108 Von VCC T = RC ln 1− β 1+ V VCC V 1 + on 1 + β CC VEE VCC T VEE | Tr = RC ln | ln = ln Von V Tr 1− β 1− 1 − on VEE VEE ⎛5⎞ 0.6 1 + β⎜ ⎟ 1+ ⎛ 1.12 ⎞ ⎛ 1.12 ⎞ ⎛1 + β ⎞ 2 1+ β 10µs ⎝5⎠ 5 ln = ln | ln⎜ | ⎜ ⎟ ⎟ = 2ln ⎟=⎜ 0.6 ⎝ ⎠ 1 − 1 − 1− β β 0.88 β 0.88 5µs ⎝ ⎠ ⎝ ⎠ 1− 5 R MATLAB gives β = 0.6998 → 1 = 2.33 | R2 = 13 kΩ | R2 = 30.3kΩ → 30 kΩ R2 1+ β 10−5 RC = = 7.595µs | C =150 pF | R = 50.6kΩ → 51 kΩ 0.6 1+ 5 ln 1 − 0.6998 11-35 CHAPTER 12 12.1 (a) A = 10 20 = 2.00 x104 | Av- ideal = 1 + A Av = = 1 + Aβ FGE = 86 150 kΩ = 13.5 12 kΩ 2.00 x10 4 = 13.49 ⎛ ⎞ 4 12 kΩ 1 + 2.00 x10 ⎜ ⎟ ⎝162 kΩ ⎠ 1 13.5 − 13.49 = 6.75 x10−4 or 0.0675% | Note : FGE ≅ = 6.75 x10−4 Aβ 13.5 2.00 x10 4 = 125 ⎛ 1.2 kΩ ⎞ 4 1 + 2.00 x10 ⎜ ⎟ ⎝ 151.2 kΩ ⎠ 150 kΩ (b) Av- ideal = 1 + 1.2kΩ = 126 | Av = ⎛ 1.2 kΩ ⎞ 1 = 6.30 x10−3 Aβ = 2.00 x10 4 ⎜ ⎟ = 159 >> 1 | FGE ≅ 151.2 k Ω β A ⎝ ⎠ 12.2 (a) A = 10 Av = 100 20 = 105 | Av - ideal = 1 + 47kΩ = 9.393 5.6kΩ 105 = 9.392 ⎛ ⎞ 5 5.6 kΩ 1 + 10 ⎜ ⎟ ⎝ 52.6 kΩ ⎠ 1⎛ 1 ⎞ 1 −4 GE = ⎜ FGE = = 9.39 x10−5 ⎟ = 8.82 x10 β ⎝ 1 + Aβ ⎠ 1 + Aβ A = 1 + Aβ (b) A = 10 20 = 5.01x104 | Av- ideal = 1 + A = Av = 1 + Aβ 94 47 kΩ = 9.393 5.6 kΩ 5.01x10 4 = 9.391 ⎛ ⎞ 4 5.6 kΩ 1 + 5.01x10 ⎜ ⎟ ⎝ 52.6 kΩ ⎠ 1⎛ 1 ⎞ 1 −3 GE = ⎜ FGE = = 1.87 x10−4 ⎟ = 1.76 x10 β ⎝ 1 + Aβ ⎠ 1 + Aβ 12-1 12.3 (a) A = 10 20 = 3.98 x104 | Av- ideal = − 220kΩ = −10 22kΩ ⎡ ⎛ 22 kΩ ⎞ ⎤ 3.98 x10 4 ⎜ ⎢ ⎟ ⎥ R2 ⎛ Aβ ⎞ ⎝ 242 kΩ ⎠ ⎥ ⎢ = −9.997 Av = − ⎜ ⎟ = −10 ⎛ ⎞⎥ ⎢ R1 ⎝1 + Aβ ⎠ 4 22 kΩ ⎟⎥ ⎢1 + 3.98 x10 ⎜ ⎝ 242 kΩ ⎠⎦ ⎣ 92 1 10 − 9.997 = 2.76 x10−4 or 0.0276% | Note : FGE ≅ = 2.76 x10−4 Aβ 10 ⎡ ⎛ 1.1kΩ ⎞ ⎤ 3.98 x10 4 ⎜ ⎢ ⎟ ⎥ 220 kΩ R2 ⎛ Aβ ⎞ ⎝ 221.1kΩ ⎠ ⎥ ⎢ = −199.0 ⎟ = −200 (b) Av- ideal = − 1.1kΩ = −200 | Av = − R ⎜ ⎛ 1.1kΩ ⎞ ⎥ ⎢ 4 1 ⎝ 1 + Aβ ⎠ ⎢1 + 3.98 x10 ⎜ 221.1kΩ ⎟ ⎥ ⎝ ⎠⎦ ⎣ ⎛ 1.1kΩ ⎞ 1 = 5.05 x10−3 Aβ = 3.98 x10 4 ⎜ ⎟ = 198 >> 1 | FGE ≅ β 221.1 k Ω A ⎝ ⎠ FGE = 12.4 (a) A = 10 47 kΩ = −10 4.7 kΩ ⎡ ⎛ 4.7 kΩ ⎞ ⎤ 5.01x10 4 ⎜ ⎢ ⎟ ⎥ R2 ⎛ Aβ ⎞ ⎝ 51.7 kΩ ⎠ ⎥ ⎢ Av = − ⎜ = −9.998 ⎟ = −10 ⎛ ⎞⎥ ⎢ R1 ⎝1 + Aβ ⎠ 4 4.7 kΩ ⎟⎥ ⎢1 + 5.01x10 ⎜ ⎝ 51.7 kΩ ⎠⎦ ⎣ −10 1 R ⎛ 1 ⎞ = -2.20x10-3 | FGE = = 2.20 x10−4 GE = − 2 ⎜ ⎟= ⎛ ⎞ R1 ⎝1 + Aβ ⎠ 1 + Aβ 4.7 kΩ 1 + 5.01x10 4 ⎜ ⎟ ⎝ 51.7 kΩ ⎠ 94 20 = 5.01x10 4 | Av - ideal = − 47kΩ = −10 4.7kΩ ⎡ ⎛ 4.7 kΩ ⎞ ⎤ 105 ⎜ ⎢ ⎟ ⎥ R2 ⎛ Aβ ⎞ ⎝ 51.7 kΩ ⎠ ⎥ ⎢ Av = − ⎜ = −9.999 ⎟ = −10 ⎛ ⎞⎥ ⎢ R1 ⎝1 + Aβ ⎠ 5 4.7 kΩ ⎢1 + 10 ⎜ 51.7 kΩ ⎟⎥ ⎝ ⎠⎦ ⎣ −10 1 R ⎛ 1 ⎞ = -1.10x10-3 | FGE = = 1.10 x10−4 GE = − 2 ⎜ ⎟= ⎛ 4.7 kΩ ⎞ R1 ⎝1 + Aβ ⎠ 1 + Aβ 1 + 105 ⎜ ⎟ ⎝ 51.7 kΩ ⎠ (b) A = 10 20 = 105 | Av- ideal = − 100 12-2 12.5 46 ACL = 10 20 = 200 → R2 = 200 | β ≅ R1 1 1 = R 201 1+ 2 R1 FGE ≅ 1 201 < 0.001 | A > = 2.01x105 or 106 dB Aβ 0.001 12.6 FGE = 1 − 1 A = ≤ 10−4 requires A ≥ 10,000 (80 dB) 1+ A 1+ A 12.7 32 1 1 = Av = 10 20 = 39.8 | FGE ≅ < 0.002 Aβ β 12.8 Av = 1 + R2 R1 | Avnom = 1 + | A> 39.8 = 1.99 x10 4 or 86 dB 0.002 Avmin = 1 + 1 − 0.0001) 99 R1 ( R1 ( 1 + 0.0001) 1 + 0.0001) 99 R1 ( 99 R1 = 100 | Avmax = 1 + = 100.02 R1 R1( 1 − 0.0001) = 98.98 | 1 < .0001 | A > 106 or 120 dB Aβ 12.9 Driving the output of the circuit in Fig. 12.3 with a current source of value iX: vX v − Av id iX = iO + i2 | i2 = ; iO = X ; v id = −i 2 R1 R1 + R2 RO 1 + Aβ v + Ai 2 R1 R1 iO = X = vX where β = RO RO R1 + R2 1 + Aβ vX v RO i X = vX + and R out = X = (R1 + R2 ) RO R1 + R2 i X 1 + Aβ 12.10 Assuming i− 1 R1 + R2 200 200 1 + Aβ )≅ 500kΩ(250)= 125 MΩ does not meet the requirements. R in = Rid ( So the specifications cannot be met using a single-stage amplifier built using the op-amp that was specified in the problem. 12.17 The non-inverting amplifier is the only one that can hope to achieve both the required gain and with such a high value of input resistance: Ro 35 ≅ = 0.14Ω meets the specification 1 + Aβ 250 Av = 1 β = 200 and Aβ = 10 4 = 50 200 Rin = Rid (1 + Aβ ) = 1MΩ(51) = 51 MΩ - too small R out = Ro 100Ω = = 1.96Ω - too large (1 + Aβ ) 51 If the gain specification is met, the input and output resistance specifications will not be met. ⎛ R2 ⎞⎛ Aβ ⎞ The open-circuit voltage is v th = v s ⎜− ⎟⎜ ⎟ . Checking the loop-gain: ⎝ R1 ⎠⎝ 1 + Aβ ⎠ ⎛ R ⎞ ⎛ ⎛110 kΩ ⎞ ⎞ 6.8 kΩ Aβ = (5 x10 4 ) ⎜ ⎟ = 2910 >> 1 so v th = v S ⎜− 2 ⎟ = −v S ⎜ ⎟ = −16.2v S ⎝ 6.8 kΩ + 110 kΩ ⎠ ⎝ 6.8 kΩ ⎠ ⎝ R1 ⎠ 12.18 Rth = Rout = Ro R 250Ω ≅ o = = 85.9 mΩ 1 + Aβ Aβ 2910 12-6 12.19 The open circuit voltage is vth = A v S . Checking the loop gain : 1 + Aβ ⎛ R2 ⎞ ⎛ ⎛ ⎞ 56 kΩ ⎞ 0.39 kΩ Aβ = 10 4 ⎜ ⎟ = 69.2 >> 1 so vth ≅ vS ⎜1 + ⎟ = v S ⎜1 + ⎟ = 145 vS ⎝ 0.39 kΩ + 56 kΩ ⎠ ⎝ 0.39 kΩ ⎠ ⎝ R1 ⎠ Ro 200Ω A 10 4 = vS = 143 vS | Rth = Rout = = = 2.85 Ω 1 + 69.2 1 + Aβ 70.2 1 + Aβ or more exactly : vth = v S 12.20 Applying the definintion of fractional gain error: 1 ± ε)⎤ Aβ R ⎡ R2 ( ⎥ − 2 − ⎢− ⎡( R1 ⎢ 1 m ε)⎥ 1 ± ε)⎤ Aβ Aβ ⎣ R1( ⎦ 1 + Aβ ⎥ FGE = = 1− ⎢ ≅ 1− ( 1 ± 2ε) R 1 + Aβ 1 m ε)⎥ ⎢ ⎣( ⎦ 1 + Aβ − 2 R1 FGE ≅ 1 − Aβ 1 Aβ Aβ m 2ε = m 2ε 1 + Aβ 1 + Aβ 1 + Aβ 1 + Aβ 1 m 2ε which must be ≤ 0.01. A = 10 20 = 2.00 x105 Aβ m 2ε ≤ 0.01 | Taking the positive sign, 2ε ≤ 0.005 and ε ≤ 0.25% 106 For Aβ >> 1, FGE ≅ 1 m 2ε = Aβ 1 ( 1 2 x10 1000 5 ) 12.21 Using the results from Prob. 12.20, Av = 10 20 = 501 | For Aβ = 54 4 x10 4 1 = 79.8 >> 1, so FGE ≅ m 2ε which must be ≤ 0.02 501 Aβ 1 1 m 2ε = m 2ε ≤ 0.02 | Taking the positive sign, 2ε ≤ 0.00747 and ε ≤ 0.374% 79.8 Aβ 12.22 V+ = 3V 99 kΩ 3 - 2.722 2.722 − VO = 2.722 V | = 10.1kΩ + 99 kΩ 9.9kΩ 101kΩ | VO = −0.111 V | VOideal = 0 V 12-7 12.23 V+ = 4.05V 99 kΩ 3.95 - 3.675 3.675 − VO = 3.675 V | = | VO = 0.869 V 101kΩ 10.1kΩ + 99 kΩ 9.9 kΩ V1 − V2 ) = −10(3.95 − 4.05)= +1.00 V For matched resistors, VOideal = −10( ⎛ 1 − 0.869 ⎞ ⎟ = 13.1% 1 ⎝ ⎠ ε = 100%⎜ 12.24 v1 + v 2 = 10sin120πt V and v id = v1 − v 2 = 0.50sin5000πt V 2 v − v− v ic − v + 0.09258 99 kΩ = 0.90742v ic | i = ic = = v ic (b) v + = v ic 9.9kΩ 9.9 kΩ 9.9 kΩ 10.1kΩ + 99kΩ 0.09258 v O = v− − i(101kΩ) = v + − i(101kΩ) = 0.90742v ic − v ic (101kΩ) 9.9kΩ v v O = −0.037 v ic and Acm = O = −0.037 | The value of Adm = −10 is not v ic (a) v ic = affected by the small tolerances. (c) CMRR = Adm = 270 - a paltry 48.6 dB Acm (d ) vO = Admv id + Acmv ic = −0.370sin(120πt ) − 5.00sin(5000πt ) V 12.25 VIC = 5 + 5.01 = 5.005V . The maximum equivalent input error is 2 5.005 VIC = = 0.500 mV , but the sign is unknown. Therefore the meter reading CMRR 10 4 may be anywhere in the range 9.50 mV ≤ Vmeter ≤ 10.5 mV . 12-8 12.26 10 kΩ = 3.75 V 10 kΩ + 30 kΩ 20 kΩ V + V 11.25 + 3.75 V1 - V2 = 15V = 7.50 V | VCM = 1 2 = = 7.50V . 2 10 kΩ + 30 kΩ 2 VCM VCM The maximum equivalent input error is . We need < 10−4 ( V1 - V2 ) CMRR CMRR 10 4 (7.5V ) CMRR > = 10 4 or 80 dB. 7.5V 10.2 kΩ 10 kΩ (b) V1 = 15V 20.2kΩ = 7.574 V | V2 = 15V 20.2kΩ = 7.426 V 200Ω V +V V1 - V2 = 15V = 0.1485 V | VCM = 1 2 = 7.50V . 2 20.2 kΩ 10 4 (7.5V ) VCM −4 We need < 10 ( V1 - V2 ) or CMRR > = 5.05 x105 or 114 dB. CMRR 0.1485V (a) V = 15V 10kΩ + 30kΩ = 11.25 V 1 30 kΩ | V2 = 15V 12.27 One worst-case tolerance assignment is given below. The second is found by reversing the pairs of resistor values. 9.995 k Ω i 10.005 k Ω v v i c O i + 10.005 k Ω 9.995 k Ω 9.995 v −v v − v+ 0.50025 = 0.49975v ic | i = ic − = ic = v ic 9.995kΩ 9.995kΩ 9.995kΩ 10.005 + 9.995 0.50025 vO = v− − i( 10.005kΩ)= v + − i( 10.005kΩ)= 0.49975v ic − v ic ( 10.005kΩ) 9.995kΩ v vO = −0.001vic and Acm = O = −0.001 | The value of Adm = 1 is not affected by the vic v+ = v ic small tolerances. CMRR = Adm = 1000 | CMRRdB = 60 dB Acm 12-9 12.28 ⎛ ⎞ 2 kΩ Setting v2 = 0, Rin1 = Rid (1 + Aβ ) 2Ric = (1MΩ)⎜1 + 7.5 x10 4 ⎟ 2(500 MΩ) = 852 MΩ ⎝ 2 kΩ + 24 kΩ ⎠ Ro 100 By symmetry, Rin 2 = Rin1 = 852 MΩ | Rout = Rout 3 = = = 2.67 mΩ 75000 ⎞ (1 + Aβ ) ⎛ ⎟ ⎜1 + ⎝ 2 ⎠ 12.29 V2 = Va = 4.99V | V3 = Vb = 5.01V | V1 = V2 + V2 − V3 (4.9kΩ)= 4.500 200Ω 9.99 kΩ V −V V4 = V3 − 2 3 (4.9 kΩ)= 5.500 | V6 = V4 = 2.7473V 200Ω 10.01kΩ + 9.99 kΩ V1 - V5 V −V = 5 O | V5 = V6 | VO = 0.991 V 9.99kΩ 10.01kΩ V −V V - V V −V V4 i1 = - 2 3 - 1 5 = -75.4 µA | i2 = 2 3 = -375 µA 200Ω 9.99kΩ 200Ω 10.01kΩ + 9.99kΩ V −V i3 = 5 O = +175 µA 10.01kΩ The common - mode and differential - mode inputs to the differential subtractor are V +V Vcm = 1 4 = 5V & Vdm = V1 − V4 = −1V with V1 = −0.5V and V4 = +0.5V 2 The subtractor outputs for the common - mode and differential - mode inputs are : 9.99kΩ V1 - V5 V5 − VOcm = 2.4975V | = | VOcm = −0.0100 V CM : V5 = V6 = 5 9.99kΩ 10.01kΩ 10.01kΩ + 9.99kΩ 9.99 kΩ V1 - V5 V − V dm DM : V5 = V6 = 0.5 = 0.24975V | = 5 O | VOdm = 1.00 V 9.99 kΩ 10.01kΩ 10.01kΩ + 9.99 kΩ 1.00 −0.01 A Adm = = −50.0 | Acm = = −0.002 | CMRR = dm = 25000 or 88.0 dB Acm −0.02 5 12-10 12.30 V2 − V3 (4.9kΩ)= 3.00V 200Ω V −V 9.99 kΩ = 1.4985V V4 = V3 − 2 3 (4.9 kΩ)= 3.00V | V6 = V4 200Ω 10.01kΩ + 9.99 kΩ V −V V1 - V5 = 5 O | V5 = V6 | VO = −6.01 mV 9.99 kΩ 10.01kΩ V −V V - V V −V V4 = -150 µA i1 = - 2 3 - 1 5 = -150 µA | i2 = 2 3 200Ω 9.99kΩ 200Ω 10.01kΩ + 9.99kΩ V −V i3 = 5 O = +149 µA 10.01kΩ 3+ 3 | Acm = −2.00 x10−3 VO = Admvid + Acmvic | − 6.01x10−3 = -50(3 − 3)+ Acm 2 A 50 CMRR = dm = = 2.50 x10 4 or 88.0 dB | It has been assumed that −3 Acm 2.00 x10 V2 = Va = 3.00V | V3 = Vb = 3.00V | V1 = V2 + the value of Adm is not affected by the small tolerances. See solution to Prob. 12.29. 12.31 ⎛ R⎞ ⎛ R⎞ ⎛ R⎞ VOS − I B1 R1 ) VO = ( VOS − I B1 R1) ⎜1 + 2 ⎟ − I B 2 R3 ⎜− 2 ⎟ = ( ⎜1 + 2 ⎟ + I B 2 R2 ⎝ R3 ⎠ ⎝ R3 ⎠ ⎝ R3 ⎠ ⎛ 106 ⎞ VO = ±0.001 − 10−7105 ⎜1 + 5 ⎟ + 0.95 x10−7106 = ±.011 − .015V . Worst case VO = −0.026 V . ⎝ 10 ⎠ ( ) Ideal output = 0 V. Error = -26 mV.Yes, R1 should be R2 R3 = 90.9 kΩ. I B2 1M Ω R 1M Ω R 100 k Ω R 3 2 -I B2 R 3 100 k Ω R 2 V + 3 O V O + V I B1 OS + + R 1 + I B1 V R 1 100 k Ω OS R 1 100 k Ω 12-11 12.32 ⎛ R ⎞ ⎛ R ⎞ ⎛ R ⎞ VO = (VOS − IB1R1 )⎜1 + 2 ⎟ − IB 2 R3 ⎜− 2 ⎟ = (VOS − IB1R1 )⎜1 + 2 ⎟ + IB 2 R2 ⎝ R3 ⎠ ⎝ R3 ⎠ ⎝ R3 ⎠ ⎛ 510 kΩ ⎞ −7 VO = ±0.01 − 2 x10−7 ( 10 5 ) ⎜1 + ⎟ + 2.5 x10 (510 kΩ) = ±0.061 + 0.0055V ⎝ 100 kΩ ⎠ [ ] Worst case VO = +0.066.5 mV. Ideal output = 0 V. Error = 66.5 mV. Yes, R1 should be R2||R3 = 90.9 kΩ. 12.33 vO = Av (vid + VOS ) | Av = dvO 10 − (−5) V = = +7500 dvid 2 − 0 mV When vO = 0, vid = −VOS and so VOS = − 0.667 mV . 12.34 A 7,500 v -6 -4 -2 2 4 6 vid (mV) 12.35 For I B 2 = 0 : Since v + must = v- = VOS , the current through C is iC (t ) = 1 t 1 tV V iC (t ) dt = VOS + ∫0 OS dt = VOS + OS t ∫ 0 R RC C C For VOS = 0, iC (t ) = I B 2 since v- = v+ = 0. vO (t )= IB2 t | Summing these two results yields 0 C V I Eq. (11.103) : vO (t ) = VOS + OS t + B 2 t | Note that vC (0) = 0 for both cases. RC C t t 0 C VOS R vO (t )= VOS + 1 C ∫ i (t )dt = C ∫ 1 I B 2 dt = 12-12 12.36 R2 R or 2 = 99 | For bias current compensation, R1 R2 = 10 kΩ R1 R1 R1R2 R2 1MΩ = = 10kΩ | R2 = 10 kΩ(1 + 99) = 1.00 MΩ and R1 = = 10.1kΩ R1 + R2 1 + R2 99 R1 The nearest 5% values would be 1 MΩ and 10 kΩ. 40dB = 100 = 1 + 12.37 (a ) Ideal O ⎛ 100 kΩ ⎞ VO = −0.005V ⎜1 + ⎟ = −0.460V ⎝ 1.1kΩ ⎠ A 10 4 = −0.546V 4 1.1kΩ 1 + 10 101.1kΩ (b) V = (−0.005V − 0.001V )1 + Aβ = (−0.005V − 0.001V ) (c) Error = -0.460 - (-0.546) = −0.187 or − 18.7% -0.460 12.38 Inverting Amplifier : vO = Av v S = −6.2vS as long as v O ≤ 10V as constrained ( a ) VO = −6.2() 1 = -6.2V, feedback loop is working and V- = 0 ( b ) VO = −6.2(−3) = +18V; VO saturates at VO = +10V The feedback loop is "broken" since the open - loop gain is now 0. (The output voltage does not change when the input changes so A = 0) 6.2 kΩ 1kΩ + 10 = −1.19V By superposition, V− = −3 7.2 kΩ 7.2 kΩ by the op - amp power supply voltages 12.39 vO 10 V Gain = -6.2 5V vS -2 V -1 V 1V 2V -5 V -10 V 12-13 12.40 Inverting Amplifier : vO = Av v S = −10vS as long as v O ≤ 10V as constrained ( a ) VO = −10(0.5) = -5.00 V, the feedback loop is working, and V- = 0 ( b ) VO = −10( 1.2) = -12.0V; VO saturates at VO = +10V The feedback loop is "broken" since the open - loop gain is now 0. (The output voltage does not change when the input changes so A = 0) 10kΩ 1kΩ − 10 = 0.182 V By superposition, V− = 1.2 11kΩ 11kΩ by the op - amp power supply voltages 12.41 vO 10 V 5V Gain = -10.0 vS -2 V -1 V 1V 2V -5 V -10 V 12.42 Noninverting Amplifier : vO = Av v S = +40vS as long as v O ≤ 15V ( b ) VO = 40(0.5V ) = 20V; VO saturates at VO = +15V ( a ) VO = 40(0.25V ) = +10V, the feedback loop is working, and VID = 0 The feedback loop is "broken" since the open - loop gain is now 0. (The output voltage does not change when the input changes so A = 0) 1kΩ = 0.125V . VID = V+ − V− = 0.5V − 15 1kΩ + 39kΩ 12-14 12.43 vO 15 V 7.5 V Gain = +40.0 vS -1 V -0.5 V 0.5 V 1V -7.5 V -15 V 12.44 Noninverting Amplifier : vO = Av v S = +43.9vS as long as v O ≤ 15V ( a ) VO = 43.9(0.25V ) = +11.0V, feedback loop is working, and VID = 0 ( b ) VO = 43.9(0.5V ) = 22.0V; VO saturates at VO = +15V The feedback loop is "broken" since the open - loop gain is now 0. (The output voltage does not change when the input changes so A = 0) 0.91kΩ = 0.158 V . VID = V+ − V− = 0.5V − 15 0.91kΩ + 39kΩ 12-15 12.45 v S + i O i L v O i2 10 k Ω R R 1 2 iO = iL + i2 and iO ≤ 1.5mA. The output voltage requirement gives i L ≤ which leaves 0.500mA as the maximum value of i2 . i2 = The closed - loop gain of 40 db (A v = 100) requires 10V = 1.00 mA 10 kΩ 10V gives (R1 + R2 )≥ 20 kΩ. R1 + R2 R2 = 99. R1 R2 = 100 which is within 1% R1 The closest ratio from the resistor tables appears to be of the desired ratio. (This is close enough since we are using 5% resistors.) There are many many choices that meet both R2 = 100 and (R1 + R2 )≥ 20 kΩ. R1 However, the choice, R1 = 200Ω and R 2 = 20 kΩ is not acceptable because its minimum value does not meet the requirements : 20.2kΩ( 1 − 0.05) = 19.2 kΩ. The smallest acceptable pair is R1 = 220Ω and R 2 = 22 kΩ. 12-16 12.46 R R v S 1 2 i 2 v i O 5k Ω R + i O L L 15V = 3 mA so i 2 ≤ 1 mA 5 kΩ v 15 i 2 = O ≤ 1 mA requires R 2 ≥ = 15 kΩ R2 .001 i O = i L + i 2 ≤ 4 mA and i L = To account for the resistor tolerance, 0.95 R2 ≥ 15 kΩ requires R 2 ≥ 15.8 kΩ . For AV = 46 dB = 200, R2 = 200 R1, and one acceptable resistor pair would be R1 = 1 kΩ and R2 = 20 kΩ. Many acceptable choices exist. An input resistance constraint might set a lower limit on R1. 12.47 The maximum base current is limited to 5 mA, so the maximum emitter current is limited to I E = (β F + 1)I B = 51(5mA)= 255 mA. Since I E = 12.48 R R v S 1 2 10V 10V , R≥ = 39.2 Ω. R 255mA i 2 v i O 5k Ω R + i O L L iO = iL + i2 ≤ 10V 10V = 2.5 mA and iL = = 2 mA so i2 ≤ 0.5 mA 4 kΩ 5kΩ 10V R R2 ≥ = 20 kΩ Av = 46dB ⇒ 2 = 200 R1 0.5mA One possible choice would be R2 = 20 kΩ and R1 = 100 Ω. However, the op-amp would not be able to supply enough output current if tolerances are take into account. Better choices would be R2 = 22 kΩ and R1 = 110 Ω or R2 = 200 kΩ and R1 = 1 kΩ which would give the amplifier a much higher input resistance. (b) V = vo max 200 = 10V = 50 mV (c) Rin = R1 = 110 Ω and 1 kΩ for the two designs given above. 200 12.49 12-17 Using the expressions in Table 12.1: ⎛ 105 ⎞ ⎜ 24kΩ 1 R2 ⎛ Aβ ⎞ 240kΩ 11 ⎟ ⎜ ⎟ = −10.0 = | A v1 = − ⎜ First stage : β = ⎟=− 24kΩ + 240kΩ 11 24kΩ ⎜ 105 ⎟ R1 ⎝ 1 + Aβ ⎠ ⎜1 + ⎟ ⎝ 11 ⎠ ⎛ Ro 100 R2 ⎞ 240 kΩ = 24.0 kΩ | Rout1 = = = 11.0 mΩ Rin1 = R1 + ⎜ Rid ⎟ = 24 kΩ + 500 kΩ 5 1 + Aβ ⎠ 1 + Aβ 105 1 + 10 ⎝ 1+ 11 ⎛ 105 ⎞ 10 kΩ 1 50 kΩ ⎜ 6 ⎟ ⎜ ⎟ = −5.00 Second stage : β = = | Av 2 = − 10 kΩ + 50 kΩ 6 10 kΩ ⎜ 105 ⎟ ⎜1 + ⎟ ⎝ 6 ⎠ Rin2 = 10 kΩ + 500 kΩ Overall amplifier : 10 kΩ (−5.00)= +50.0 | Rin = 24.0 kΩ | Rout = 6.00 mΩ 10.0 kΩ + 11.0 mΩ For all practical purposes, the numbers the same. R out = 6.00 mΩ is a good Av = −10.0 approximation of 0 Ω, and Av = −10.0(−5.00)= +50.0 100 240 kΩ = 10.0 kΩ | Rout 2 = = 6.00 mΩ 5 105 1 + 10 1+ 6 12.50 Use the expressions in Table 12.1. β1 = 47 kΩ A 105 105 = 0.010755 | Av1 = = = = +9.30 47 kΩ + 390 kΩ 1 + Aβ1 1 + 105 (0.010755) 10756 24 kΩ A 105 105 β2 = = 0.19355 | Av1 = = = = +5.17 24 kΩ + 100 kΩ 1 + Aβ2 1 + 105 (0.19355) 19356 Av = (9.30)(5.17)= 48.1 1 + Aβ )= 250 kΩ( 10756)= 2.69 GΩ | Rout = Rin = Rid1 ( Ro2 200 = = 10.3 mΩ 1 + Aβ2 19356 12-18 12.51 Use the expressions in Table 12.1. β1 = β2 = Av2 = − 20 kΩ 1 A 105 105 = | Av1 = = = = +7.00 20 kΩ + 120 kΩ 7 1 + Aβ1 1 + 105 7 14287 ( ) ⎛ 14286 ⎞ R2 ⎛ Aβ2 ⎞ ⎜ ⎟ = −6⎜ ⎟ = −6.00 | R1 ⎝ 1 + Aβ2 ⎠ ⎝ 14287 ⎠ Av = (7.00)(−6.00) = −42.0 Ro2 200 = = 14.0 mΩ 1 + Aβ2 14287 Rin = Rid1 ( 1 + Aβ1 )= 250 kΩ( 14287)= 3.57 GΩ | Rout = 12.52 Use the expressions in Table 12.1. The three individual amplifier stages are the same. 5.01x10 4 92 2 kΩ 1 R ⎛ Aβ ⎞ 40kΩ 21 β= = | A = 10 20 = 5.01x10 4 | Av = − 2 ⎜ = −20.0 ⎟=− R1 ⎝ 1 + Aβ ⎠ 2 kΩ + 40kΩ 21 2 kΩ 5.01x10 4 1+ 21 ⎛ ⎞ R2 40 kΩ Ro 300Ω = 2.00 kΩ | Rout = = = 126 mΩ Rin = R1 + ⎜ Rid ⎟ = 2 kΩ + 500 kΩ 4 1 + A⎠ 1 + Aβ 1 + 5.01x10 5.01x10 4 ⎝ 1+ 21 ⎛ ⎞⎛ ⎞ 2 kΩ 2 kΩ For the overall amplifier : Av = ⎜−20.0 ⎟⎜−20.0 ⎟(−20.0) = −8000 2 kΩ + 126 mΩ ⎠⎝ 2kΩ + 126mΩ ⎠ ⎝ Rin = 2.00 kΩ | Rout = 126 mΩ 12.53 50 2 < 5000 < 50 3 | Three stages will be required to keep the gain of each stage ≤ 50. However, the input and output resistance requirements may further constrain the gains and must be checked as well. A =10 = 1.778 x 10 4 Ro 100Ω 1 For Rout = : ≤ 0.1Ω → Aβ ≥ 999 → β ≥ 0.0562 → ≤ 17.8. 1 + Aβ 1 + Aβ β 1 For Rin = Rid (1 + Aβ ) 2 Ric : 1MΩ(1 + Aβ ) 2GΩ ≥ 10 MΩ → Aβ ≥ 9 → ≤ 1976 85 20 β 17.8(50)(50) > 5000 so three stages is still sufficient. 12-19 12.54 VS = VO Z1 = VO Z1 + Z2 R1 + R1 R2 SC R2 + = VO (SCR + 1)R (SCR + 1)R + R 2 1 2 1 2 1 SC V ⎛ R ⎞ SC R1 R2 + 1 Av (s)= O = ⎜1 + 2 ⎟ VS ⎝ R1 ⎠ SCR2 + 1 ( ) 12.55 Av (s) = − Avnom = − f Hnom = f Hmin = 330kΩ( 1.1) 330kΩ(0.9) 330kΩ = −33 | Avmax = − = −40.3 | Avmax = − = −27.0 10kΩ 10kΩ(0.9) 10kΩ( 1.1) 1 R2 R 1 | Av (0)= − 2 | f H = R1 sCR2 + 1 R1 2πCR2 2π 10 2π 10 ( )3.3x10 −10 1 5 = 4.83kHz | f Hmax = = 3.65kHz 2π 10 ( )(0.5)3.3x10 (0.9) −10 5 1 = 10.7 kHz ( )(1.2)3.3x10 (1.1) −10 5 1 12.56 −60db/decade requires 3 poles - 3 x (-20db/decade). Using three identical fH3 amplifiers : Av = 3 1000 = 10 and f H1 = = 1.96(20 kHz) = 39.2 kHz. 1 R2 = 10 R1 | f H = 1 | R2C = = 4.06 x10−6 s 3 2πR2C 2π 39.2 x10 ( 23 − 1 1 ) Try C = 270 pF . R2 = 1 = 15.0 kΩ, R1 = 1.5kΩ 2πf H C 12-20 12.57 A(s) = Z out s + ωB Ro Ro = = Ro ω 1 + A(s)β 1 + s + ω B + ωT β T β s + ωB s s 1+ 1+ s + ωB Ro Ro ωB ωB = Ro = ≅ s s + ω B (1 + Aoβ ) (1 + Aoβ ) 1 + (1 + Aoβ ) 1 + s ω B (1 + Aoβ ) βωT | ωT = Aoω B | Z out = R OUT ωT s + ωB RO RO 1+A o β ωB Inductive region ω βω T 12.58 Using MATLAB: b=1/11; ro=100; wt=2*pi*1e6; wb=wt/1e5; n=ro*[1 wb]; d=[1 b*wt];w=logspace(0,7); r=freqs(n,d,w); mag=abs(r); phase=angle(r)*180/pi; subplot(212);semilogx(w,phase) subplot(211);loglog(w,mag) 10 2 10 Magnitud e 1 10 0 10 -1 10 -2 10 0 100 80 60 10 1 10 2 10 3 10 4 10 5 10 6 10 7 Phase 40 20 0 10 0 10 1 10 2 10 3 10 4 Frequency (rad/s) 10 5 10 6 10 7 12.59 12-21 ⎛ ⎜ ⎛ R2 ⎞ R2 ⎟ = R1 + ⎜ Rid Z in = R1 + ⎜ R id ⎜ ⎟ ωT 1 + A(s)⎠ ⎜ ⎝ 1+ ⎜ s + ωB ⎝ Z in = R1 + Rid R2 ⎞ ⎟ ⎛ s + ωB ) ⎞ ⎟ = R1 + ⎜ Rid R2 ( ⎟ s + ω B + ωT ⎠ ⎟ ⎝ ⎟ ⎠ Rid R2 (s + ω B ) s + ω B + ωT = R1 + (s + ω B ) Rid (s + ω B + ωT ) + R2 (s + ω B ) Rid + R2 s + ω B + ωT ⎛ ⎛ R2 s ⎞ s ⎞ Rid R2ω B ⎜1 + Rid ⎟ ⎜1 + ⎟ 1 + Ao ) ( ⎝ ωB ⎠ ⎝ ωB ⎠ Z in = R1 + = R1 + Rid + R2 s R2 Rid ω B (1 + Ao ) + R2ω B + s(Rid + R2 ) 1+ Rid + (1 + Ao ) ω B (1 + Ao ) Rid + R2 (1 + Ao ) ⎛ s ⎞ 1 + ⎜ ⎟ ⎛ R2 ⎞ ⎝ ωB ⎠ Z in = R1 + ⎜ ⎜ Rid (1 + A )⎟ ⎟ Rid + R2 s o ⎠1+ ⎝ ω B (1 + Ao ) R + R2 id (1 + Ao ) (s + ω B ) 12-22 12.60 Using MATLAB: n1=1e6; d1=[1 2000]; n2=1e6; d2=[1 4000]; n3=1e12; d3=[1 6000 8e6]; w=logspace(2,5); [m1,p1,w]=bode(n1,d1,w); [m2,p2,w]=bode(n2,d2,w); [m3,p3,w]=bode(n3,d3,w); subplot(211) loglog(w,m1,w,m2,w,m3) subplot(212) semilogx(w,p1,w,p2,w,p3) . 10 6 10 5 AV 10 Magnitude 10 10 10 10 4 3 AV1 AV2 2 1 0 10 2 10 3 Frequency (rad/s) 10 4 10 5 0 AV2 -50 Phase -100 A V1 -150 AV -200 10 2 10 3 Frequency (rad/s) 10 4 10 5 12-23 12.61 Av = − Z 2 Aβ Z1 1 + Aβ | β= Z1 Z1 + Z 2 | A= ωT R2 | Z1 = R1 | Z 2 = s +ωo sCR2 + 1 Av (s) = − ⎞ ⎛ R ⎛ R ⎞ s2 R2C + s⎜1 + 2 + R2C (ω o + ωT )⎟ + ω o ⎜1 + 2 ⎟ + ωT ⎠ ⎝ R1 ⎝ R1 ⎠ 3.653 x1013 s2 + 3.142 x10 7 s + 1.916 x1012 R2 ωT R1 Av (s) = − Using MATLAB: bode(-3.653e13,[1 3.142e7 1.916e12]) 30 20 10 0 Gain dB -10 -20 -30 -40 -50 10 200 150 100 3 10 4 5 10 6 10 Frequency (rad/sec) 10 7 10 8 Phase deg 50 0 -50 -100 -150 -200 3 10 10 4 10 5 10 6 10 7 10 8 Frequency (rad/sec) 12-24 12.62 V (s) V (s) 1 which is the transfer function of an integrator (a) S = −sCVO (s) | Av (s) = O = − R VS (s) sRC (b) Generalizing Eq. (11.122) : Av (s) = − Z 2 A(s)β 1 Z1 | Z2 = | Z1 = R1 | β = Z1 1 + A(s)β Z1 + Z 2 sC ωT sRC R1 sCR 1 ωT sRC s + ω B sRC + 1 β= = | A(s)β = | Av (s) = − 1 sRC s + ω B sRC + 1 sCR + 1 sRC 1 + ωT R1 + s + ω B sRC + 1 sC 1 sRCωT ωT Av (s) = − =− sRC (s + ω B )(sRC + 1) + sRCωT (s + ω B )(sRC + 1) + sRCωT ωT ωT ωT Av (s) = − RC RC RC ≅− =− ⎛ ⎛ ⎞ ⎛ 1 ⎞ ωB 1 ⎞ ω B s + s⎜ω B + ωT + ⎟+ s + ωT )⎜ s + s + ωT )⎜ s + ( ( ⎟ ⎟ ⎝ RC ⎠ RC Ao RC ⎠ ⎝ ωT RC ⎠ ⎝ 2 using dominant root factorization where it is assumed ωT >> ω B and ωT >> |A ( ω)| v 1 . RC A o Integrator region 1 A RC o ω T ω 12-25 12.63 wrc=1/(1e4*470e-12); wt=2*pi*5e6; wb=2*pi*50; n=wt*wrc; d=[1 wt+wb+wrc wb*wrc]; bode(n,d) 100 50 Gain dB 0 -50 -100 10 -1 0 -50 Phase deg -100 -150 -200 10 -1 10 0 1 10 0 10 1 10 2 10 3 10 4 10 Frequency (rad/sec) 5 10 6 10 7 10 8 10 10 2 10 3 10 4 10 Frequency (rad/sec) 5 10 6 10 7 10 8 12.64 β= 2 kΩ 1 105 = | Aβ = = 4760 >> 1 2kΩ + 40kΩ 21 21 40 kΩ 3x106 Hz R ( a ) Av = − 2 = − = −20 | f H = βf T = = 143kHz R1 2 kΩ 21 3 ( b ) Av = (−20) = −8000 (78dB) | f H 3 = 0.51 f H = 72.9 kHz 12-26 12.65 The table below follows the approach used in Table 12.6. The only change is the required gain is Av = 10 20 = 1.778 x 10 4 . Cascade of Identical Non-Inverting Amplifiers # of Stages AV(0) Gain per Stage 1/β 2.00E+04 1.41E+02 2.71E+01 1.19E+01 7.25E+00 5.21E+00 4.12E+00 3.45E+00 3.01E+00 2.69E+00 2.46E+00 2.28E+00 fH Single Stage β x fT 5.00E+01 7.07E+03 3.68E+04 8.41E+04 1.38E+05 1.92E+05 2.43E+05 2.90E+05 3.33E+05 3.71E+05 4.06E+05 4.38E+05 fH N Stages RIN ROUT 85 1 2 3 4 5 6 7 8 9 10 11 12 5.000E+01 4.551E+03 1.878E+04 3.658E+04 5.320E+04 6.717E+04 7.839E+04 8.724E+04 9.415E+04 9.951E+04 1.037E+05 1.068E+05 6.00E+09 7.08E+11 3.69E+12 8.41E+12 1.38E+13 1.92E+13 2.43E+13 2.90E+13 3.33E+13 3.71E+13 4.06E+13 4.38E+13 8.33E+00 7.06E-02 1.36E-02 5.95E-03 3.62E-03 2.60E-03 2.06E-03 1.72E-03 1.50E-03 1.35E-03 1.23E-03 1.14E-03 We see from the spreadsheet that a cascade of seven identical stages is required to achieve the bandwidth specification. Fortuitously, it also meets the input and output resistance specs. For the non-inverting amplifier cascade: R R AV = 1 + 2 = 4.12 → 2 = 3.12 R1 R1 A similar spreadsheet for the cascade of identical inverting amplifiers indicates that it is impossible to meet the bandwidth requirement. 12-27 12.66 (a) From Problem 11.99, Av = 1 + R2 R = 4.12 → 2 = 3.12 | Exploring the 5% resistor R1 R1 R tables, we find R2 = 62kΩ and R1 = 20kΩ yields 2 = 3.10 as a reasonable pair. R1 The nominal gain of the cascade is then Av = (4.10) =1.948 x 10 4 . 7 Av = 86db ± 1dB ⇒ 1.778 x 10 4 ≤ AV ≤ 2.239 x 10 4 and the gain is well within this range. Many amplifiers will probably fail due to tolerances with 5% resistors. A Monte Carlo analysis would tell us. If we resort to 1% resistors to limit the tolerance spread, R2 = 30.9 kΩ and R1 = 10.0 kΩ is one of many possible pairs. 1 5 x10 6 | f H 1 = βf T = = 1.22 MHz (b) For R2 = 62kΩ and R1 = 20kΩ, β = 4.1 4.1 f H = 1.22 MHz 2 7 − 1 = 394 kHz 1 12-28 12.67/12.68 One possibility: Use a cascade of two non-inverting amplifiers, and shunt the input of the first amplifier to define the input resistance. 60db → Av = 1000 | A single - stage amplifier with a gain of 1000 would have a bandwidth of only 5 kHz using this op - amp. Two stages should be sufficient if Rin and R out can also be met. A design with f H 2 >> f H1 will be tried. First stage : Non - inverting with bandwidth of 20 kHz f 20kHz 1 β1 = H 1 = = 0.004 | Av1 = = 250 → Av 2 = 4 → β 2 = 0.25 → f H 2 = 1.25 MHz fT 5 MHz β1 Since f H 2 >> f H1, f H = f H 1 = 20kHz. Ao = 85 dB = 17800 100 Ro 2 = = 0.0225Ω which is ok. Checking Rout = 1 + Aoβ 2 1 + 17800(0.25) Choosing resistors from the Appendix, a possible set is Amplifier 1: R1 = 1.2kΩ, R 2 = 300 kΩ and shunt the input with R 3 = 27kΩ Amplifier 2 : R1 = 3.3kΩ, R 2 = 10 kΩ ⎛ ⎞ ⎜ 17800 17800 ⎟ Checking gain : Av = = 997.5 = 60.0 dB ⎜ 17800 17800 ⎟ ⎜1 + ⎟ 1+ 251 ⎝ 4.03 ⎠ vS + + 27 k Ω vO 300 k Ω 10 k Ω 1.2 k Ω 3.3 k Ω 12-29 12.69 function sg=Prob103(tol); sg=0; for j=1:10 ao=100000; ft=1e6*sqrt(2^(1/6)-1); for i=1:500, r1=22000*(1+2*tol*(rand-0.5));r2=130000*(1+2*tol*(rand-0.5)); beta=r1/(r1+r2);g1=ao/(1+ao*beta); b1=beta*ft; r1=22000*(1+2*tol*(rand-0.5));r2=130000*(1+2*tol*(rand-0.5)); beta=r1/(r1+r2);g2=ao/(1+ao*beta); b2=beta*ft; r1=22000*(1+2*tol*(rand-0.5));r2=130000*(1+2*tol*(rand-0.5)); beta=r1/(r1+r2);g3=ao/(1+ao*beta); b3=beta*ft; r1=22000*(1+2*tol*(rand-0.5));r2=130000*(1+2*tol*(rand-0.5)); beta=r1/(r1+r2);g4=ao/(1+ao*beta); b4=beta*ft; r1=22000*(1+2*tol*(rand-0.5));r2=130000*(1+2*tol*(rand-0.5)); beta=r1/(r1+r2);g5=ao/(1+ao*beta); b5=beta*ft; r1=22000*(1+2*tol*(rand-0.5));r2=130000*(1+2*tol*(rand-0.5)); beta=r1/(r1+r2);g6=ao/(1+ao*beta); b6=beta*ft; gain(i)=g1*g2*g3*g4*g5*g6; bw(i)=(b1+b2+b3+b4+b5+b6)/6; end; sg=sg+sum(gain 1 22 kΩ + 130 kΩ 6.91 6.91 R 130 kΩ = 6.91 ( a ) Avnom = 1 + 2 = 1 + 22 kΩ R1 Avmax = 1 + 130kΩ( 1.05) 130kΩ(0.95) R2 R = 1+ = 7.53 | Avmin = 1 + 2 = 1 + = 6.35 R1 R1 22kΩ(0.95) 22 kΩ( 1.05) 106 Hz 106 Hz 106 Hz = 145 kHz | f Hmax = = 157 kHz | f Hmin = = 133 kHz 6.91 6.36 7.53 f Hnom = β nom f T = 12-31 12.72 function [gain,bw]=Prob1272a ao=50000; ft=1e6; for i=1:500, r1=22000*(1+0.1*(rand-0.5)); r2=130000*(1+0.1*(rand-0.5)); beta=r1/(r1+r2); gain(i)=ao/(1+ao*beta); bw(i)=beta*ft; end; end [gain,bw]=prob1272a; mean(gain) ans = 6.9140 std(gain) ans = 0.2339 mean(bw) ans = 1.4478e+05 std(bw) ans = 4.8969e+03 Three sigma limits: 6.21 ≤ Av ≤ 7.62 130 kHz ≤ BW ≤ 159 kHz function [gain,bw]=Prob1272b for i=1:500, ao=100000*(1+1.0*(rand-0.5)); ft=2e6*(1+1.0*(rand-0.5)); r1=22000*(1+0.1*(rand-0.5)); r2=130000*(1+0.1*(rand-0.5)); beta=r1/(r1+r2); gain(i)=ao/(1+ao*beta); bw(i)=beta*ft; end; end [gain,bw]=prob1272b; mean(gain) ans = 6.9201 std(gain) ans = 0.2414 mean(bw) ans = 2.8925e+05 std(bw) ans = 8.5536e+04 Note that the bandwidth is essentially a uniform distribution. 3σ: 6.20 ≤ Av ≤ 7.64 98.9% of the values fall between: 146 kHz ≤ BW ≤ 439 kHz 12.73 SR ≥ VOω = ( 15V )(2π ) 2 x10 4 Hz = 1.89 x106 ( ) V V or 1.89 µs s 12.74 f = SR 10V 1 = −6 = 159 kHz 2πVo 10 s 20πV 12.75 12-32 The negative transition requires the largest slew rate : SR = V ∆V 20V = = 10 µs 2µs ∆t 12.76 (a) For the circuit in Fig. 12.26 : Rid = 250 kΩ | R = 1 kΩ − an arbitrary choice ωB = 2π 5 x106 8 x10 4 ( ) = 125π rad/s | C= 1 ωB R = (b) Add a resistor from each input terminal to ground of value 2 R See Prob. 11.111 for schematics. (125π )1000 1 = 2.55 µF | Ro = 50Ω | Ao = 80,000 IC = 1 GΩ. 12.77 Two possibilities: 105 nA 1 mV - + 200 MΩ 1k Ω 50 Ω + v 1 + 250 k Ω + 80,000v 2.55 µ F 2 v 1 v 2 v o 200 M Ω 95 nA 1 mV - + + 125 k Ω 105 nA 100 M Ω 1k Ω 50 Ω + v 1 + 80,000v 2.55 µ F 2 v v 1 2 vo 125 k Ω 95 nA 12-33 12.78 + v1 400 k Ω 100 Ω + v 2 1 Ω v2 µF + v3 µF 75 Ω 80,000v 3 + vo v 1 1.592 1.592 ω1 : C = 1 1 = = 1.592 µF − setting R1 arbitrarily to 100 Ω. ω1R1 2π ( 10 3 ) (100) 1 1 = = 1 Ω − Using the same value of C. 5 ω 2C 2π ( 10 )(1.592 µF ) ω 2 : R2 = 12.79 *PROBLEM 12.79 - Six-Stage Amplifier VS 1 0 AC 1 XA1 1 2 0 AMP XA2 2 3 0 AMP XA3 3 4 0 AMP XA4 4 5 0 AMP XA5 5 6 0 AMP XA6 6 7 0 AMP .SUBCKT AMP 1 2 7 RID 1 3 1E9 RO 6 2 50 E2 6 7 5 7 1E5 E1 4 7 1 3 1 R 4 5 1K C 5 7 15.915UF R2 2 3 130K R1 3 7 22K .ENDS .TF V(7) VS .AC DEC 40 1 1MEG .PRINT AC V(1) V(2) V(3) V(4) V(5) V(6) V(7) .PROBE V(1) V(2) V(3) V(4) V(5) V(6) V(7) .END 12-34 12.80 Student PSPICE cannot handle 6 copies of the uA741 op amp, but since all the stages are the same, we can square the output from a 3-stage version or cube the output from a two-stage model. 200 100 0 -100 100Hz 300Hz 1.0KHz DB(V(I1:+))+ DB(V(I1:+)) 3.0KHz 10KHz 30KHz Frequency 100KHz 300KHz 1.0MHz 3.0MHz 10MHz From the SPICE graph, BW = 54.3 kHz. 12-35 12.81 *PROBLEM 12.81 - Six Stage Amplifier VS 1 0 AC 1 XA1 1 2 0 AMP XA2 2 3 0 AMP XA3 3 4 0 AMP XA4 4 5 0 AMP XA5 5 6 0 AMP XA6 6 7 0 AMP .SUBCKT AMP 1 2 8 RID 1 3 1E9 RO 7 2 50 E2 7 8 6 8 1E5 *Two dummy loops provide separate control of Gain & BW tolerances G1 8 4 1 3 .001 R11 4 8 RG 1000 E1 5 8 4 8 1 RC 5 6 1000 C 6 8 CC 15.915UF * R2 2 3 RR 130K R1 3 8 RR 22K .ENDS .MODEL RR RES (R=1 DEV=5%) .MODEL RG RES (R=1 DEV=50%) .MODEL CC CAP (C=1 DEV=50%) .AC DEC 20 1E3 1E6 .PROBE V(7) .PRINT AC V(7) .MC 1000 AC V(7) MAX OUTPUT(EVERY 20) *.MC 1000 AC V(7) MAX OUTPUT(RUNS 77 573 597 777) .END Maximum gain = 103 dB; Minimum gain = 98.5 dB Maximum Bandwidth = 65 kHz; Minimum bandwidth = 38 kHz (These are approximate.) 12.82 Rid = 1010 Ω 12.83 A ≥ 118 dB I B ≤ 8.8 nA C= 1 = 7.96 pF 2π ( 1000Ω)(20 MHz) A = 4 x106 Ro not specified CMRR ≥ 80 dB IOS ≤ 2.2 nA PSRR ≥ 100 dB VOS ≤ 1.5 mV Power Supplies ± 18 V maximum SR = 12.5 V/µs nominal f T = 20 MHz Nominal values only : Rid = 1010 Ω 12.84 12-36 (a ) Use the expressions in Tables 12.1 and 12.2. A = 10 20 = 105 | β A = 47 kΩ 1 15kΩ 1 18 kΩ 1 = | βB = = | βC = = 47kΩ + 470kΩ 11 15kΩ + 150kΩ 11 18 kΩ + 270 kΩ 16 5 10 105 R ⎛ Aβ ⎞ R ⎛ Aβ ⎞ 470kΩ 11 150 kΩ 11 AvA = − 2 ⎜ = −10.0 AvB = − 2 ⎜ = −10.0 ⎟=− ⎟=− 5 R1 ⎝ 1 + Aβ ⎠ 47kΩ R1 ⎝1 + Aβ ⎠ 15kΩ 10 105 1+ 1+ 11 11 5 10 R2 ⎛ Aβ ⎞ 270 kΩ 16 AvC = − ⎜ = −15.0 Av = −10.02 (−15)= −1500 ⎟=− 5 R1 ⎝ 1 + Aβ ⎠ 18kΩ 10 1+ 16 ⎛ ⎞ R2 470 kΩ Rin = RinA = R1 + ⎜ Rid = 47.0 kΩ ⎟ = 47 kΩ + 1MΩ 1 + A⎠ 1 + 105 ⎝ 100 Ro 250Ω = = 40.0 mΩ 1 + Aβ 105 1+ 16 2 MHz 2 MHz = 182kHz f HC = βC fT = = 125kHz f HA = f HB = β A fT = 11 16 For the overall amplifier : Av = −1500 | Rin = 47.0 kΩ | Rout = 40.0 mΩ Rout = RoutC = Using the definition of bandwidth : 10 10 15 ⎛ f H ⎞2 1+ ⎜ ⎟ ⎝182 kHz ⎠ ⎛ f H ⎞2 1+ ⎜ ⎟ ⎝182 kHz ⎠ ⎛ f H ⎞2 1+ ⎜ ⎟ ⎝ 125kHz ⎠ = 1500 2 → f H = 79.9 kHz (b) v S = 50.0 mV | vOA = −10vS = −500 mV | vOB = −10vOA = +5.00 V | vO = −15vOB = −75.0V > 18V → vO = −18 V | v -A = vOA v = +5.00 µV | v -B = OB5 = −50.0 µV 5 -10 -10 The output of the third amplifier is saturated at -18 V, and the inverting input is no longer near 270kΩ 18 kΩ + (−18V ) = +3.56 V ground potential. Using superposition, v -C = 5V 18kΩ + 270kΩ 18kΩ + 270kΩ The remaining three nodes are V+ = +18 V V− = −18 V and Vgnd = 0 V . 12-37 12.85 47 kΩ 1 15kΩ 1 = | βB = = 47kΩ + 470kΩ 11 15kΩ + 150kΩ 11 ⎛ 470 kΩ ⎞⎛ 150kΩ ⎞⎛ 270 kΩ ⎞ 18 kΩ 1 βC = = Aβ >> 1 Avnom ≅ ⎜− ⎟⎜ − ⎟⎜− ⎟ = −1500 18 kΩ + 270 kΩ 16 ⎝ 47 kΩ ⎠⎝ 15kΩ ⎠⎝ 18 kΩ ⎠ A = 10 20 = 105 | β A = 100 A max v max max Rin ≅ RinA = 47.0 kΩ( 1.05)= 49.4 kΩ nom nom Rout ≅ RoutC = ⎛ 1.05 ⎞3 ≅ −1500⎜ ⎟ = −2025 ⎝ 0.95 ⎠ A min v ⎛ 0.95 ⎞3 ≅ −1500⎜ ⎟ = −1111 ⎝ 1.05 ⎠ nom nom Rin ≅ RinA = 47.0 kΩ min min Rin ≅ RinA = 47.0 kΩ(0.95)= 44.7 kΩ Ro 250Ω = = 40.0 mΩ 1 + Aβ 105 1+ 16 R 250Ω Ro 250Ω max max min min o Rout ≅ RoutC = = = 43.9 mΩ Rout ≅ RoutC = = = 36.4 mΩ min 5 max 10 105 1 + Aβ 1 + Aβ 1+ 1+ 17.6 14.6 2 MHz 2 MHz f HA = f HB = β A fT = = 182kHz f HC = βC fT = = 125kHz 11 16 Using the definition of bandwidth : 10 10 15 1500 = → f H = 79.8 kHz 2 2 2 2 ⎛ fH ⎞ ⎛ fH ⎞ ⎛ fH ⎞ 1+ ⎜ ⎟ 1+ ⎜ ⎟ 1+ ⎜ ⎟ ⎝182 kHz ⎠ ⎝182 kHz ⎠ ⎝ 125kHz ⎠ max max f HA = f HB = β A fT = 9.048 ⎛ f max ⎞2 1+ ⎜ H ⎟ ⎝ 200 kHz ⎠ 2 MHz 2 MHz = 200kHz f HC = βC fT = = 137 kHz 10.0 14.6 9.048 13.57 1111 = → f Hmax = 87.5 kHz 2 2 2 ⎛ f max ⎞ ⎛ f max ⎞ 1+ ⎜ H ⎟ 1+ ⎜ H ⎟ ⎝ 200 kHz ⎠ ⎝ 137 kHz ⎠ max max f HA = f HB = β A fT = 11.05 ⎛ f max ⎞ 1+ ⎜ H ⎟ ⎝167 kHz ⎠ 2 2 MHz 2 MHz = 167kHz f HC = βC fT = = 114kHz 12.0 17.6 11.05 16.58 2025 = → f Hmax = 73.0 kHz 2 2 2 ⎛ f max ⎞ ⎛ f max ⎞ H H 1+ ⎜ ⎟ 1+ ⎜ ⎟ ⎝167 kHz ⎠ ⎝ 114kHz ⎠ 12-38 12.86 (a ) Use the expressions in Tables 12.1 and 12.2. Three identical gain blocks. A = 10 106 20 = 2.00 x105 | β = 3kΩ 1 A = | Aβ = 14286 | Av1 = = 14.0 3kΩ + 39 kΩ 14 1 + Aβ 1 + Aβ )= 1 MΩ 500 kΩ( 1 + 14286)= 1.00 MΩ Rin = RinA = 1 MΩ Rid ( Rout = RoutC = Ro 300Ω = = 21.0 mΩ 1 + Aβ 1 + 14286 1 Av = 14.03 = 2744 (b) v 5 MHz = 357 kHz f H = f H1 2 3 − 1 = 182 kHz f H1 = β A fT = 14 For the overall amplifier : Av = +2740 | Rin = 1.00 MΩ | Rout = 21.0 mΩ | f H = 182 kHz I = 5.00 mV | vOA = 14vI = 70.0 mV | vOB = 14vOA = 980 mV | vO = 14vOB = 13.7V > 12V → vO = 12.0 V | v- A = vOA v = +5.00 mV | v- B = OB = 70.0 mV 14 14 The output of the third amplifier is saturated at 12 V, and the inverting and non - inverting inputs 3kΩ = 0.857 V . V+ = +12 V 3kΩ + 39 kΩ V− = −12 V Vgnd = 0 V are no longer equal. v-C = 12V 12.87 (a ) Use the expressions in Tables 12.1 and 12.2. Three identical gain blocks. A = 10 106 20 = 2.00 x105 | β = 1.5kΩ 1 A = | Aβ = 7407 | Av1 = = 27.0 1.5kΩ + 39 kΩ 27 1 + Aβ 1 + Aβ ) = 1.5 MΩ 500 kΩ( 1 + 14286) = 1.50 MΩ Rin = RinA = 1.5 MΩ Rid ( Rout = RoutC = Ro 300Ω = = 38.8 mΩ 1 + Aβ 1 + 7407 1 Av = 27.03 = 19700 5 MHz = 185kHz f H = f H1 2 3 − 1 = 94.3kHz f H1 = β A fT = 27 For the overall amplifier : Av = +19700 | Rin = 1.50 MΩ | Rout = 38.8 mΩ | f H = 94.3 kHz (b) v I = 5.00 mV | vOA = 27vI = 135 mV | vOB = 27vOA = 3.65 V | vO = 27vOB = 98.4V > 12V → vO = 12.0 V | v- A = vOA v = +5.00 mV | v- B = OB = 135 mV 27 27 The output of the third amplifier is saturated at 12 V, and the inverting and non - inverting inputs 1.5kΩ = 0.444 V . V+ = +12 V 1.5kΩ + 39 kΩ V− = −12 V Vgnd = 0 V are no longer equal. v-C = 12V 12-39 12.88 Three identical stages : β nom = 3kΩ 1 = 3kΩ + 39 kΩ 14 Aβ = 2 x105 = 14286 >> 1 14 A nom v ⎛ R2 ⎞3 ⎛ 39 kΩ ⎞3 = ⎜1 + ⎟ = ⎜1 + ⎟ = 2740 3kΩ ⎠ ⎝ R1 ⎠ ⎝ ⎡ 39 kΩ( ⎡ 39 kΩ(0.98)⎤ 1.02)⎤ ⎥ = 3070 | Avmin = ⎢1 + ⎥ = 2460 = ⎢1 + 3kΩ( 1.02) ⎥ ⎢ ⎢ ⎣ 3kΩ(0.98) ⎥ ⎦ ⎣ ⎦ 3 3 Avmax nom max min Rin = 1 MΩ | Rin = 1 MΩ( 1.02) = 1.02 MΩ | Rin = 1 MΩ(0.98) = 980 kΩ nom Rout = 300 300 min = 21.8 MΩ | Rout = = 20.2 MΩ 5 2 x10 2 x105 1+ 1+ 14.53 13.49 5 MHz 5 MHz 5 MHz = 357 kHz | f Hmax = = 371 kHz | f Hmin = = 344 kHz f Hnom = β nom f T = 14 13.49 14.53 300 max = 21.0 MΩ | Rout = 1 + 14286 12-40 12.89 (a ) Use the expressions in Tables 12.1 and 12.2. A = 10 20 = 10 4 | β A = 10kΩ 1 2 kΩ 1 10 kΩ 1 = | βB = = | βC = = 10 kΩ + 39 kΩ 4.9 2 kΩ + 200 kΩ 101 10 kΩ + 39 kΩ 4.9 4 10 A 10000 R2 ⎛ Aβ ⎞ 200 kΩ 101 AvC = AvA = = = +4.90 AvB = − ⎜ = −99.0 ⎟=− 10000 R1 ⎝ 1 + Aβ ⎠ 1 + Aβ 2 kΩ 10 4 1+ 1+ 4.9 101 ⎛ ⎞ 10000 Av = 4.902 (−99.0)= −2380 Rin = RinA = Rid ( 1 + Aβ )= 300 kΩ⎜1 + ⎟ = 613 MΩ 4.9 ⎠ ⎝ 80 Ro 250Ω = = 98.0 mΩ 1 + Aβ 10 4 1+ 4.9 3 MHz 3 MHz = 612 kHz f HC = βC fT = = 29.7 kHz f HC = f HA = β A fT = 4.9 101 For the overall amplifier : Av = −2380 | Rin = 613 MΩ | Rout = 98.0 mΩ Rout = RoutC = Using the definition of bandwidth : 4.9 ⎛ fH ⎞ 1+ ⎜ ⎟ ⎝ 612 kHz ⎠ 2 99 ⎛ fH ⎞ 1+ ⎜ ⎟ ⎝ 29.7 kHz ⎠ 2 4.9 ⎛ fH ⎞ 1+ ⎜ ⎟ ⎝ 612 kHz ⎠ 2 = 2380 2 → f H = 29.6 kHz Note that the bandwidth is controlled by amplifier B because it's bandwidth is much smaller (b) v than the others. I = 00.0 mV | vOA = 4.9vOS = +49.0 mV | vOB = −100vOA + 101(10 mV ) = −3.89V | vOA v = 10.0 mV | v -B = OB4 = +389 µV 4.9 -10 V− = −15 V and Vgnd = 0 V . vO = 4.9( vOB + .010) = −19.0V < -15V → vO = −15 V | v -A = v -C = − 15V = −3.06 V The remaining three nodes are V+ = +15 V 4.9 12-41 12.90 10kΩ 1 2 kΩ 1 A = 10 = 10 4 | βC = β A = = | βB = = 10kΩ + 39 kΩ 4.9 2 kΩ + 200 kΩ 101 2 ⎛ 39 kΩ ⎞ ⎛ 200kΩ ⎞ Aβ >> 1 Avnom ≅ ⎜1 + ⎟ ⎜− ⎟ = −2401 ⎝ 10 kΩ ⎠ ⎝ 2 kΩ ⎠ 80 20 A max v ⎛ 39 kΩ ⎛ 1.10 ⎞⎞ ⎛ 200kΩ ⎛ 1.10 ⎞⎞ ≅ ⎜1 + ⎜ ⎟⎟ ⎜ − ⎜ ⎟⎟ = −4064 ⎝ 10 kΩ ⎝ 0.90 ⎠⎠ ⎝ 2 kΩ ⎝ 0.90 ⎠⎠ 2 ⎛ 39 kΩ ⎛ 0.90 ⎞⎞2⎛ 200 kΩ ⎛ 0.90 ⎞⎞ A ≅ ⎜1 + ⎜ ⎟⎟ ⎜ − ⎜ ⎟⎟ = −1437 ⎝ 10 kΩ ⎝ 1.10 ⎠⎠ ⎝ 2 kΩ ⎝ 1.10 ⎠⎠ ⎡ ⎛ 1 ⎞⎤ nom Rin = Rid ( 1 + Aβ ) = 300 kΩ ⎢1 + 10 4 ⎜ ⎟⎥ = 612 MΩ ⎝ 4.9 ⎠⎦ ⎣ ⎡ ⎡ ⎛ 1 ⎞⎤ ⎛ 1 ⎞⎤ 4 max min Rin = 300kΩ ⎢1 + 10 4 ⎜ ⎟⎥ = 716 MΩ Rin = 300 kΩ ⎢1 + 10 ⎜ ⎟⎥ = 521 MΩ ⎝ 4.19 ⎠⎦ ⎝ 5.77 ⎠⎦ ⎣ ⎣ min v Ro 200Ω = = 98.0 mΩ 1 + Aβ 10 4 1+ 4.9 Ro 200Ω Ro 200Ω max max min min Rout ≅ RoutC = = = 115 mΩ Rout ≅ RoutC = = = 83.4 mΩ min 4 max 10 10 4 1 + Aβ 1 + Aβ 1+ 1+ 5.77 4.19 3 MHz 3 MHz f HA = f HB = β A fT = = 612 kHz f HC = β B fT = = 29.7 kHz 4.9 101 The bandwidth is controlled by the narrow bandwidth stage. nom nom Rout ≅ RoutC = f Hnom = 29.7 kHz min f Hmin = β B fT = 3 MHz = 24.3 kHz 123 max f Hmax = β B fT = 3 MHz = 36.2 kHz 82.8 12-42 CHAPTER 13 13.1 Assuming linear operation : vBE = 0.700 + 0.005sin 2000πt V ⎡⎛ 5mV ⎞ ⎤ vce = ⎢⎜ ⎥ sin 2000πt = −1.03sin 2000πt V ⎟(−1.65V ) ⎣⎝ 8 mV ⎠ ⎦ vCE = 5.00 − 1.03sin 2000πt V ; 10 - 3300 IC ≥ 0.700 → IC ≤ 2.82 mA 13.2 Assuming linear region operation : vGS = 3.50 + 0.25sin 2000πt V ⎡ 0.25V ⎤ −2V ) vds = ⎢ ( ⎥ sin 2000πt = −1.00sin 2000πt V ⎣0.50V ⎦ vDS = 4.80 − 1.00sin 2000πt V vDS ≥ vGS − VTN → 10 − 3300 I D − 1.00sin 2000πt ≥ 3.50 + 0.25sin 2000πt − 1 For sin 2000πt = 1, I D ≤ 10 − 1 − 3.5 − 0.25 + 1 V = 1.89 mA 3300 Ω 13.3 (a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a coupling capacitor that couples the ac component of the signal at the collector to the output vO. C3 is a bypass capacitor. (b) The signal voltage at the top of resistor R4 will be zero. 13.4 (a) C1 is a bypass capacitor. C2 is a coupling capacitor that couples the ac component of vI into the amplifier. C3 is a coupling capacitor that couples the ac component of the signal at the collector to output vO. (b) The signal voltage at the base will be vb = 0. 13.5 (a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a bypass capacitor. C3 is a coupling capacitor that couples the ac component of the signal at the collector to output vO. (b) The signal voltage at the emitter will be ve = 0. 13.6 (a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a bypass capacitor. C3 is a coupling capacitor that couples the ac component of the signal at the drain to output vO. (b) The signal voltage at the source will be vs = 0. 13.7 13-1 (a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a coupling capacitor that couples the ac component of the signal at the drain to output vO. 13.8 (a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a bypass capacitor. C3 is a coupling capacitor that couples the ac component of the signal at the drain to output vO. (b) The signal voltage at the source will be vs = 0. 13.9 (a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a bypass capacitor. C3 is a coupling capacitor that couples the ac component of the signal at the collector to output vO. (b) The signal voltage at the emitter will be ve = 0. 13.10 (a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a coupling capacitor that couples the signal from the emitter of Q1 back to the node joining R1 and R2. C3 is a coupling capacitor that couples the ac component of the signal at the emitter to the output vO. (b) The signal voltage at the collector will be zero. 13.11 (a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a bypass capacitor. C3 is a coupling capacitor that couples the ac component of the signal at the drain to the output vO. (b) The signal voltage at the top of R4 will be zero. 13.12 (a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a coupling capacitor that couples the ac component of the signal at the drain to output vO. 13.13 (a) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a coupling capacitor that couples the ac component of the signal at the drain to the output vO. 13.14 dc voltage sources produce constant values of output voltage. Hence no signal voltage can appear at the terminals of the source. The signal component of the voltage is forced to be zero, and a direct path to ground is provided for signal currents. 13-2 13.15 NPN Common-Emitter Amplifier +18 V 270 k Ω VEQ = 18V REQ 360 kΩ = 5.84 V 360 kΩ + 750 kΩ = R1 R2 = 360 kΩ 750 kΩ = 243 kΩ REQ Q1 243 k Ω VEQ 5.84 V 228.2 k Ω 5.84 = 243 x103 I B + 0.7 + 91 228.2 x103 I B I B = 0.245 µA | IC = 90 I B = 22.0 µA VCE = 18 − 2.7 x105 IC − 2.28 x105 I E = 6.99 V Q − point : (22.0 µA, 6.99 V ) ( ) 13.16 SPICE results: (a) (22.5 µA, 6.71 V) (b) (22.6 µA, 6.69 V) The discrepancies between the results in Probs. 13.15 and 13.16 arise because VBE = 0.575 V with IS = 5 fA. Very little changes occurs with the addition of VA. 13.17 NPN Common-Emitter Amplifier +12 V VEQ = −12V + 24V 6k Ω REQ R EQ Q1 3.33 k Ω VEQ -4 V 4k Ω 5kΩ = −4 V 5kΩ + 10 kΩ = R1 R2 = 5kΩ 10kΩ = 3.33 kΩ −4 = 3300 I B + 0.7 + 76(4000)I B − 12 VCE = 12 − 6000 IC − 4000 I E − (−12) = 6.08 V Q − point : ( 1.78 mA, 6.08 V ) I B = 23.8 µA | IC = 75I B = 1.78 mA -12 V 13-3 13.18 *Problem 13.17 - NPN Common-Emitter Amplifier VCC 7 0 DC 12 VEE 8 0 DC -12 R1 3 8 5K R2 7 3 10K RE 4 8 4K RC 7 5 6K Q1 5 3 4 NBJT .OP .MODEL NBJT NPN IS=1E-15 BF=75 VA=75 .END Results: IC 1.78E-03 VCE 6.14E+00 VBE 7.28E-01 13.19 PNP Common-Base Amplifier -7.5 V 33 k Ω 7.5 = 3000IB + 66(68000)IB + 0.7 Q1 3k Ω 68 k Ω IB = 1.51 µA | IC = 98.4 µA VEC = 15 − 33000IC − 68000 IE = 4.96 V Q − point : (98.4 µA,4.96 V ) +7.5 V 13-4 13.20 *Problem 13.19 - PNP Common-Base Amplifier VEE 1 0 DC 7.5 VCC 5 0 DC -7.5 RC 5 4 33K RB 3 0 3K RE 1 2 68K Q1 4 3 2 PBJT .OP .MODEL PBJT PNP IS=1E-16 BF=65 VA=75 .END Results: IC -9.83E-05 VCE -4.97E+00 VBE 13.21 PNP Common-Emitter Amplifier +9 V -7.13E-01 VEQ = 9V REQ 62 kΩ = 6.80V 62 kΩ + 20 kΩ = 62kΩ 20kΩ = 15.1kΩ 3.9 k Ω R EQ 9 = 3900(136)IB + 0.7 + 15100IB + 6.80 Q1 VEQ 15.1 k Ω IB = 2.75 µA | IC = 371 µA | IE = 374 µA 13 k Ω +6.80 V VEC = 9 − 3900IE − 13000 IC = 2.72 V Q − point : (371 µA, 2.72 V ) 13.22 *Problem 13.21 - PNP Common-Emitter Amplifier VCC 4 0 DC 9 RC 1 0 13K R2 2 0 62K R1 4 2 20K RE 4 3 3.9K Q1 1 2 3 PBJT .OP .MODEL PBJT PNP IS=1E-15 BF=135 VA=75 .END Results: IC -3.73E-04 VCE -2.68E+00 VBE -6.88E-01 13-5 13.23 NMOS Common-Source Amplifier +15 V VEQ = 15V REQ 82 k Ω R EQ 1MΩ = 4.05V 1MΩ + 2.7 MΩ = 1MΩ 2.7 MΩ = 730 kΩ 0.25mA 2 (VGS − 1) 2 M1 VEQ 730 k Ω 27 k Ω 4.05 V ID = VGS = 4.05 − 27000 ID 2 ID = 0.125mA(3.05 − 27000 IDS ) → ID = 82.2 µA VDS = 15 − 82000ID − 27000 ID = 6.04V Q − point : (82.2 µA, 6.04 V ) 13.24 *Problem 13.23 - NMOS Common-Source Amplifier VDD 4 0 DC 15 RD 4 3 82K R2 4 2 2.7MEG R1 2 0 1MEG R4 1 0 27K M1 3 2 1 1 NFET .OP .MODEL NFET NMOS KP=250U VTO=1 .END Results: ID 8.29E-05 VDS 5.96E+00 VGS 13.25 Depletion-mode NMOS Common-Gate Amplifier +15 V 1.81E+00 4.3 k Ω M1 3.9 k Ω 5 x10−4 2 (VGS + 2) | VGS = −3900ID 2 5 x10−4 2 VGS = −3900 (VGS + 2) → VGS = −0.990V 2 V ID = − GS = 254 µA 3900 VDS = 15 − 4300 ID − 3900 ID = 12.9V ID = Q − point : (254 µA, 12.9 V ) 13-6 13.26 *Problem 13.25 - Depletion-mode NMOS Common-Gate Amplifier VDD 3 0 DC 15 RD 3 2 4.3K R1 1 0 3.9K M1 2 0 1 1 NDMOS .OP .MODEL NDMOS NMOS KP=500U VTO=-2 .END Results: ID 2.54E-04 VDS 1.29E+01 VGS 13.27 PMOS Common-Source Amplifier +18 V -9.92E-01 VEQ = 18V REQ 22 k Ω R EQ M1 VEQ 1.65 M Ω 24 k Ω 9.00 V 3.3MΩ = 9.00V 3.3 MΩ + 3.3 MΩ = 3.3MΩ 3.3MΩ = 1.65 MΩ 18 = 22000ID − VGS + 9 ⎛ 4 x10−4 ⎞ 2 9 = 22000⎜ ⎟(VGS + 1) − VGS ⎝ 2 ⎠ VGS = −2.24V | ID = 307 µA VDS = 18 − 22000ID − 24000 ID = −3.88 V Q − point : (307 µA, 3.88 V ) 13.28 *Problem 13.27 - PMOS Common-Source Amplifier VDD 4 0 DC 18 RD 1 0 24K R2 4 2 3.3MEG R1 2 0 3.3MEG R4 4 3 22K M1 1 2 3 3 PFET .OP .MODEL PFET PMOS KP=400U VTO=-1 .END Results: ID -3.07E-04 VDS -3.86E+00 VGS -2.24E+00 13-7 13.29 NPN Common-Collector Amplifier +9 V 9 = 86000 IB + 101(82000)IB + 0.7 IB = 0.992 µA | IC = 99.2 µA VCE = 18 − 82000 IE = 18 − 82000(100µA) = 9.80 V Q − point : (99.2 µA,9.80 V ) Q1 86 k Ω 82 k Ω -9 V 13.30 *Problem 13.29 - NPN Common-Collector Amplifier VCC 5 0 DC 9 VEE 8 0 DC -9 R1 3 6 43K R2 6 0 43K RE 4 8 82K Q1 5 3 4 NBJT .OP .MODEL NBJT NPN IS=1E-16 BF=100 VA=75 .END Results: IC 9.93E-05 VCE 9.79E+00 VBE 13.31 PMOS Common-Gate Amplifier 7.12E-01 12 + VGS 200 x10−6 2 = (VGS − 1) 33000 2 VGS = −0.84V | ID = 338 µA ID = VDS = −(12 − 33000ID − 22000 ID + 12) VDS = −5.41 V Q − point : (338 µA, - 5.41 V ) M1 +12 V 33 k Ω 22 k Ω -12 V 13-8 13.32 *Problem 13.31 - PMOS Common-Gate Amplifier VDD 4 0 DC 12 VSS 1 0 DC -12 RD 2 1 22K R1 4 3 33K M1 2 0 3 3 PFET .OP .MODEL PFET PMOS KP=200U VTO=+1 .END Results: ID -3.38E-04 VDS -5.40E+00 VGS -8.39E-01 13.33 RS = 1 kΩ, R4 = 1 kΩ +18 V 3.9 k Ω IG = 0, so VG = 0. Assume active region operation. 2 4 x10−4 VGS + 5) | VGS = −2000 I D ( 2 M 2 4 x10−4 VGS = −2000 VGS + 5) → VGS = −2.50V ( 2 V I D = − GS = 1.25 mA 2000 2 k Ω VDS = 18 − 3900 I D − 2000 I D = 10.6V ID = RG 10 M Ω Q − point : ( 1.25 mA, 10.6 V ) 13.34 SPICE results: The Q-point is the same (1.25 mA, 10.6 V) . 13-9 13.35 +15 V 7.5 k Ω M RG 2.2 M Ω 2 225 x10−6 0 − (−3) = 1.01 mA IG = 0, so VG = 0 and VGS = 0. I D = 2 Q − point : ( 1.01 mA, 7.41 V ) VDS = 15 − 7500 I D = 7.41V [ ] 13.36 SPICE results: The Q-point is the same (1.01 mA, 7.41 V). 13.37 (a) + Q RI vi R C R 3 vo - R1 R2 RE (b) + vi RI + RB v rπ gmv ro RC R 3 vo - RE 13-10 (c) C1 is a coupling capacitor that couples the ac component of vI into the amplifier. C2 is a coupling capacitor that couples the ac component of the signal at the collector to the output vO. C3 is a bypass capacitor. 13.38(a) Figure P13.4 Q1 RI v (a) i + R C RE R 3 vo - (c ) ro RI v (b) i C1 − Bypass + R3 vo - R v E g v m RC rπ C2 − Coupling C3 − Coupling + 13.38(b) Figure P13.5 RI Q1 vi (a) R1 R2 R R + C 3 v o - (c ) C1 − Coupling RI vi (b) R1 R2 rπ + v g v m ro R C + vo 3 C2 − Bypass C3 − Coupling R - 13-11 13.39(a) Figure P13.9 + Q RI vi (a) R 1 R R 2 C R 3 v o - (c ) C1 − Coupling RI vi (b) R 1 + v R 2 + v g v m o C2 − Bypass C3 − Coupling rπ - r o R C R 3 - 13.39(b) Figure P13.10 Q1 RI R vi 1 + R R v 3 o (a) 2 RE - (c ) C1 − Coupling RI R1 vi rπ + v g v m ro + C2 − Coupling o (b) R2 RE R v 3 - C3 − Coupling 13-12 13.40(a) Figure P13.6 R I M vi (a) + R D 1 R 3 v o R1 R 2 - (c ) C1 − Coupling R vi (b) + v R1 R 2 I + vo 3 C2 − Bypass C3 − Coupling - gm v r o R D R - 13.40(b) Figure P13.7 M RI vi (a) R1 R R D 3 1 + v o - ro RI vi (b) R 1 (c ) + R vo - v + gmv RD 3 C1 − Coupling C2 − Coupling 13-13 13.41(a) Figure P13.8 M RI vi + RD R 3 1 v o R1 R2 - (a) (c ) C1 − Coupling RI vi (b) R1 R2 + v gmv ro RD + vo 3 C2 − Bypass C3 − Coupling R - 13.42(a) Figure P13.12 M1 RI vi (a) R 1 + RD R 3 vo - r o (c ) + R v o RI vi (b) v R 1 gmv RD 3 C1 − Coupling C2 − Coupling + 13.43 RI: Thévinen equivalent source resistance; R1: base bias voltage divider; R2: base bias voltage divider; RE and R4: emitter bias resistors - determine the emitter current; RC: collector bias resistor - sets the collector-emitter voltage; R3: load resistor 13.44 RI: Thévinen equivalent source resistance; R1: gate bias voltage divider; R2: gate bias voltage divider; R4: source bias resistor - sets source current; RD: drain bias resistor - sets drain-source voltage; R3: load resistor 13-14 13.45 RI: Thévinen equivalent source resistance; R1: base bias voltage divider; R2: base bias voltage divider; RE: emitter bias resistor - determines the emitter current; RC: collector bias resistor - sets the collector-emitter voltage; R3: load resistor 13.46 (a) r (b) I d = VT ID + IS ⎛ ⎛ 0.6 ⎞ ⎞ 0.025 | I D = 10−14 ⎜ exp⎜ = 94.4 Ω ⎟ − 1⎟ = 264.9µA | rd = 264.9µA + 10 fA ⎝ ⎝ 0.025 ⎠ ⎠ 0.025 0.025 = 2.50 TΩ (c) > 1015 → I D + I S < 2.5 x10−17 A 10 fA ID + IS D = 0 rd = VD < VT ln 2.5x10−17 ID + IS = 0.025ln = −0.150 V IS 10−14 13.47 kT 1.38 x10−23 V = T = 8.63x10−5 T | rd ≅ T = 1000VT VT = −19 ID q 1.60 x10 T VT rd 13.48 ⎛ 0.005 ⎞ 75K 6.47 mV 6.47 Ω 100K 8.63 mV 8.63 Ω 200K 17.3 mV 17.3 Ω 300K 25.9 mV 25.9 Ω 400K 34.5 mV 34.5 Ω (a) exp⎜ ⎟ − 1 = 0.221 ⎝ 0.025 ⎠ | 0.005 = 0.200 → +10.7% error 0.025 | − 0.005 = −0.200 → −9.37% error 0.025 ⎛ 0.005 ⎞ exp⎜− ⎟ − 1 = −0.181 ⎝ 0.025 ⎠ ⎛ 0.010 ⎞ (b) exp⎜ ⎟ − 1 = 0.492 | ⎝ 0.025 ⎠ ⎛ 0.010 ⎞ exp⎜− ⎟ − 1 = −0.330 ⎝ 0.025 ⎠ 0.010 = 0.400 → +23.0% error 0.025 | − 0.010 = −0.400 → −17.5% error 0.025 13.49 (a) IC = g m 0.03 = = 0.750 mA = 750 µA 40 40 g m 50 x10−6 c I = ( ) C 40 = 40 = 1.25 µA (b) IC = g m 250 x10−6 = = 6.25 µA 40 40 13-15 13.50 IC = β oVT rπ = 75(0.025V ) 10 4 Ω = 187.5 µA | Q - point : ( 188 µA, VCE ≥ 0.7 V ) g m = 40 IC = 40 1.875 x10−4 = 7.50 mS | ro = 13.51 ( ) VA + VCE VA 100V ≅ = = 533 kΩ IC IC 187.5 µA IC = β oVT rπ = 125(0.025V ) 2 x106 Ω = 1.56 µA | Q - point : ( 1.56 µA, VCE ≥ 0.7 V ) g m = 40 IC = 40 1.56 x10−6 = 62.4 µS | ro = 13.52 ( ) VA + VCE VA 75V ≅ = = 48.1 MΩ IC IC 1.56 µA IC = β oVT rπ = 100(0.025V ) 250 kΩ = 10 µA | Q - point : ( 10 µA, VCE ≥ 0.7 V ) VA + VCE VA 100V ≅ = = 10 MΩ IC IC 10µA g m = 40 IC = 40 10−5 = 0.400 mS | ro = 13.53 ( ) IC = β oVT rπ = 75(0.025V ) 106 Ω = 1.875 µA | Q - point : ( 1.88 µA, VCE ≥ 0.7 V ) g m = 40 IC = 40 1.875 x10−6 = 75.0 µS | ro = ( ) VA + VCE VA 100V ≅ = = 53.3 MΩ IC IC 1.875 µA 13-16 13.54 V + VCE ; solving for VA : VA = IC ro − VCE ro = A IC Using the values from row 1: VA = 0.002(40000)− 10 = 70 V Using the values from the second row : β o = g mrπ = 0.12(500)= 60 and β F = β o = 60. Row 1: g m = 40 IC = 40(0.002) = 0.08 S | rπ = βo gm = 60 = 750 Ω 0.08 µ f = g mro = 0.08(40000) = 3200 Row 2 : IC = µ f = g mro = 0.12(26700) = 3200 Row 3 : g m = ro = g m 0.12 V + VCE 80 = = 3 mA | ro = A = = 26.7 kΩ 40 0.003 40 IC βo rπ = 60 g m 1.25 x 10-4 -4 = 1.25 x 10 S | I = = = 3.13 µA C 40 40 4.8 x 105 VA + VCE 80 = = 25.6 MΩ | µ f = g mro = 1.25 x 10-4 25.6 x 106 = 3200 IC 3.13 x 10-6 ( ) 13.55 ⎟ − 1 = 0.221 (a) exp⎜ ⎝ 0.025 ⎠ ⎛ 0.005 ⎞ | 0.005 = 0.200 → +10.7% error 0.025 ⎛ 0.005 ⎞ 0.005 = −0.200 → −9.37% error exp⎜− ⎟ − 1 = −0.181 | − 0.025 ⎝ 0.025 ⎠ ⎛ 0.0075⎞ 0.0075 = 0.300 → +16.7% error ⎟ − 1 = 0.350 | (b) exp⎜ 0.025 ⎝ 0.025 ⎠ ⎛ 0.0075 ⎞ 0.0075 = −0.300 → −13.6% error exp⎜− ⎟ − 1 = −0.259 | − 0.025 ⎝ 0.025 ⎠ ⎛ 0.0025 ⎞ 0.0025 = 0.100 → +5.17% error ⎟ − 1 = 0.105 | (c) exp⎜ 0.025 ⎝ 0.025 ⎠ ⎛ 0.0025 ⎞ 0.0025 = −0.100 → −4.84% error exp⎜− ⎟ − 1 = −0.0952 | − 0.025 ⎝ 0.025 ⎠ 13.56 ( a ) βF = ( b) β F ≅ IC 350µA ∆I 600µA − 125µA ≅ ≅ 90 | β o = C ≅ ≅ 120 4µA 6µA − 2µA IB ∆I B 750µA 900µA − 600µA ≅ 95 | β o ≅ ≅ 75 8µA 4µA 13.57(a) 13-17 t=linspace(0,.004,1024); ic=.001*exp(40*.005*sin(2000*pi*t)); IC=fft(ic); z=abs(IC(1:26)/1024); z(1) ans = 0.001 plot(t,ic) x10-3 1.2 1 0.8 0 1 2 3 4 x10-3 z([5 9 13]) ans = 0.0001 0.0000 0.0000 13.57(b) t=linspace(0,.004,1024); ic=.001*exp(40*.005*sin(2000*pi*t)); IC=fft(ic); z=abs(IC(1:26)/1024); z(1) ans = 0.0023 plot(t,ic) -3 8 x10 6 4 2 0 0 1 2 3 4 x10-3 z([5 9 13]) ans = 0.0016 0.0007 0.0002 13-18 13.58 (a) NAME Q1 MODEL NBJT IB 2.21E-05 IC 1.78E-03 VBE 7.28E-01 VBC -5.41E+00 VCE 6.14E+00 BETADC 8.04E+01 GM 6.87E-02 RPI 1.17E+03 RX 0.00E+00 RO 4.52E+04 BETAAC 8.04E+01 T = 27 o C | VT = 8.625 x10−5 (300) = 25.9 mV IC 1.78 mA = = 68.7 mS VT 25.9mV ⎛ V ⎞ ⎛ 6.14 ⎞ β o = β FO ⎜1 + CE ⎟ = 75⎜1 + ⎟ = 81.1 ⎝ 75 ⎠ ⎝ VA ⎠ β 81.1 rπ = o = = 1180 Ω gm 0.0687 gm = ro = VA + VCE 75 + 6.14 = = 45.6 kΩ 1.78 mA IC 13 - 19 13.58(b) MODEL PBJT IB -2.69E-06 IC -3.73E-04 VBE -6.88E-01 VBC 1.99E+00 VCE -2.68E+00 BETADC 1.39E+02 GM 1.44E-02 RPI 9.61E+03 RX 0.00E+00 RO 2.06E+05 BETAAC 1.39E+02 T = 27 o C | VT = 8.625 x10−5 (300) = 25.9 mV IC 0.373mA = = 14.4 mS VT 25.9mV ⎛ V ⎞ ⎛ 2.68 ⎞ β o = β FO ⎜1 + CE ⎟ = 135⎜1 + ⎟ = 140 ⎝ 75 ⎠ ⎝ VA ⎠ gm = rπ = ro = βo gm = 140 = 9.72 kΩ 0.0144 VA + VCE 75 + 2.68 = = 208 kΩ 0.373mA IC ⎛ ⎝ 5.41⎞ ⎟ = 80.4 75 ⎠ ⎛ ⎝ 1.99 ⎞ ⎟ = 139 75 ⎠ Note: The SPICE model actually is using VCB instead of VCE in the current gain calculations. (a) β o = 75⎜1 + (b) β o = 135⎜1 + 13-20 13.59 + y 11 ix vx re ie αo ie rπ v gm v ro 1 rπ v 1− αo v For the T - model : ix = x − α o x = vx re re re For the hybrid pi model: y11 = y11 = re = ix 1 − α o = = vx re 1− βo + 1 re βo = 1 → r = (β o + 1)re (β o + 1)re π βo α αV V rπ = = o= o T = T IE (β o + 1) gm (β o + 1) gm IC 100 75V + 10V = 50 kΩ | ro = = 1.70 MΩ 2.00 mS 50µA 13.60 g m = 40(50µA) = 2.00 mS | rπ = RBB = RB rπ = 100 kΩ 50 kΩ = 33.3kΩ ⎛ ⎞ 33.3kΩ Av = −⎜ ⎟(2 mS ) 1.70 MΩ 100 kΩ 100 kΩ = −95.0 ⎝ 33.3kΩ + 0.75kΩ ⎠ ( ) 13.61 For β o = 100, see Prob. 13.60. 60 = 30 kΩ | RBB = RB rπ = 100 kΩ 30 kΩ = 23.1kΩ 2.00 mS ⎛ ⎞ 23.1kΩ Av = −⎜ ⎟(2 mS ) 1.70 MΩ 100 kΩ 100 kΩ = −94.1 ⎝ 23.1kΩ + 0.75kΩ ⎠ −95.0 ≤ Av ≤ −94.1 − only a small variation rπ = ( ) 13.62 g m = 40(2.5mA)= 0.100 S | rπ = RBB 75 50 + 7.5 V = 750Ω | ro = = 23.0 kΩ 2.5 mA 0.1S = RB rπ = 4.7 kΩ 750Ω = 647Ω ⎛ 647Ω ⎞ Av = ⎜ ⎟(−0.100 S ) 23.0 kΩ 4.3kΩ 10 kΩ = −247 ⎝ 50Ω + 647Ω ⎠ ( ) 13-21 13.63 g m = 40( 1µA)= 40µS | rπ = 40 50 + 1.5 V = 1 MΩ | ro = = 51.5 MΩ µA 40µS 1 RBB = RB rπ = 5 MΩ 1 MΩ = 833kΩ ⎛ ⎞ 833kΩ Av = ⎜ ⎟(−40µS ) 51.5 MΩ 1.5 MΩ 3.3 MΩ = −40.0 ⎝10 kΩ + 833kΩ ⎠ ( ) 13.64 SPICE Results: IC = 248 µA, VCE = 3.30 V, AV = -15.1 dB -- IC differs by 1.2% - AV is off by 0.5% 13.65 [10(V )] = [10(9)] N CC N ≥ 20000 → N ≥ log(20000) log(90) = 2.20 → N = 3 (or quite possibly 2 by droping a larger fraction of the supply voltage across RC ) 13.66 *Problem 13.21 - PNP Common-Emitter Amplifier - Figure P13.5 VCC 7 0 DC 9 VI 1 0 AC 1 RI 1 2 1K C1 2 3 100U R1 7 3 20K R2 3 0 62K RE 7 4 3.9K C 7 4 100U RC 5 0 13K C3 5 6 100U R3 6 0 100K Q1 5 3 4 PBJT .OP .MODEL PBJT PNP IS=1E-15 BF=135 VA=75 .AC DEC 10 100Hz 10000Hz .PRINT AC VM(6) VDB(6) VP(6) .END Results: IC = 373 µA, VEC = 2.68V, AV = -134 Hand calculations in Prob. 13.21 yielded (371 µA, 2.72V) gm = 40(371µA) = 14.8 mS | rπ = 135 = 9.12 kΩ | ro = ∞ 14.8 mS RB = R1 R2 = 20 kΩ 62 kΩ = 15.1kΩ | RBB = 15.1kΩ 9.12 kΩ = 5.69 kΩ ⎛ 5.69 kΩ ⎞ Av = −⎜ ⎟(14.8 mS ) ∞ 13kΩ 100 kΩ = −145 ⎝ 1kΩ + 5.69 kΩ ⎠ ( ) 13-22 SPICE AV result is somewhat lower because ro is included. 13.67 Av ≅ −10VCC = −10( 12) = −120 13.68 Av ≅ −10(VCC + VEE ) = −10(15 + 15) = −300 13.69 VCC + VEE )= −10( 1.5 + 1.5)= −30; This estimate says no. Av = −10( However, if we look a bit deeper, Av = −40(IC RC ) = −40VRC ,and we let VRC = 1.5)= −60. then we can achieve A v = −40( (V CC + VEE ) 2 = 1.5V , So, with careful design, we can probably achieve a gain of 50. 13.70 Using our rule - of - thumb estimate, 1.5)= −15 | Av = −10() 1 = −10 Av ≅ −10VCC = −10( Note that this result assumes that IC varies with VCC. 13.71 (a) i = 10kΩ = 0.5 mA, but i ≤ 0.2 I for small - signal operation. (b) V ≥ V + i R + I R = 0.7 + 5 + 25 = 30.7 V 5V c c C CC BE c L C L So IC ≥ 5ic = 2.5 mA. 13.72 Av = 40 dB = 100 | v o = 100v be = 100(0.005V ) = 0.500V . 13-23 13.73 For common - emitter stage : Av = 50 dB → Av = −316 15V = 47.5mV which is far too big for small- signal operation. 316 The will be significant distortion of the sine wave. vbe = 13.74 20 kΩ 18 = −4.61V | REQ = 20 kΩ 62 kΩ = 15.1kΩ 62 kΩ + 20 kΩ −4.61 − 0.7 − (−9) IB = = 6.76µA | IC = 135 I B = 913µA 15.1kΩ + 136(3.9 kΩ) EQ (a) V = −9 + VCE = 9 − 13000 IC − 3900 I E − (−9) = 2.54V g m = 40 IC = 0.0365S | rπ = 135 = 3.70 kΩ | ro = ∞ | R L = 13kΩ 100 kΩ = 11.5kΩ gm (b) For V 1 ⎛ 2.97 kΩ ⎞ Av = −⎜ 11.5kΩ)= −314 ⎟(0.0365)( ⎝ 1kΩ + 2.97 kΩ ⎠ CC = 18V , the answers are the same : IC = 913µA | VEC = 2.54V | Av = −314 13.75 Using the information from Row 1: λ = I D ro − VDS = 8 x10−4 4 x10 4 − 6 = 26V ; λ = 0.0385V -1 ( )( ) 2 gm From Row 2 : K n = = 2 ID (1 + λVDS ) (2 x10 ) = 3.25 x10 ⎛ 6⎞ 2(5 x10 ) ⎜1 + ⎟ ⎝ 26 ⎠ −4 2 −5 −4 A V2 ⎛ 6⎞ −4 Row 1: gm = 2K n ID (1 + λVDS ) = 2(3.25 x10−4 )(8 x10−4 ) ⎜1 + ⎟ = 8 x10 S ⎝ 26 ⎠ µ f = gm ro = 8 x10−4 (4 x10 4 )= 32 2 ID 0.2(VGS − VTN ) = 0.2 = 0.2 K n (1 + λVDS ) 2(8 x10−4 ) = 0.40V 6⎞ −4 ⎛ 3.25 x10 ⎜1 + ⎟ ⎝ 26 ⎠ 13-24 1 Row 2 : ro = λ + VDS ID = 26 + 6 = 640 kΩ | µ f = gm ro = 2 x10−4 (6.4 x10 5 )= 128 −5 5 x10 2(5 x10−5 ) = 0.10V 6⎞ −4 ⎛ 3.25 x10 ⎜1 + ⎟ ⎝ 26 ⎠ ⎛ 6⎞ −3 Row 3 : gm = 2K n ID (1 + λVDS ) = 2(3.25 x10−4 )( 10−2 ) ⎜1 + ⎟ = 2.83 x10 S ⎝ 26 ⎠ 2 ID = 0.2 0.2(VGS − VTN ) = 0.2 K n (1 + λVDS ) 1 ro = λ + VDS ID = 26 + 6 = 3.2kΩ | µ f = gm ro = 2.83 x10−3 (3.2 x10 3 )= 9.06 0.01 2 ID = 0.2 K n (1 + λVDS ) 2( 10−2 ) = 1.41V 6⎞ −4 ⎛ 3.25 x10 ⎜1 + ⎟ ⎝ 26 ⎠ 0.2(VGS − VTN ) = 0.2 MOSFET Small-Signal Parameters ID gm (S) ro (Ω) µf Small-Signal Limit vgs (V) 0.8 mA 50 µA 0.0008 40,000 32 0.40 0.0002 640,000 128 0.10 10 mA 0.00283 3200 9.06 1.41 13.76 ' ⎛ ⎛1 ⎞ 2 Kn ( 1 + λVDS ) ⎛ 1 ⎞ 2 Kn ⎛ 1 ⎞ 2 Kn W⎞ µ f = ⎜ + VDS ⎟ ≅⎜ ⎟ =⎜ ⎟ ⎜ ⎟ ID ⎝λ ⎠ ⎝ λ ⎠ ID ⎝ λ ⎠ ID ⎝ L ⎠ 2 2 x10-4 2 I W 50 = (µ f λ ) D ' = 250(0.02) = L 1 2 Kn 2 5 x10-5 [ ] ( ) VGS − VTN 2 2 x10-4 2ID ≅ = = 0.160 V Kn 50 5 x10-5 ( ( ) ) 13-25 13.77 2.5 x10−4 ) ⎛1 ⎞ 2K n (1 + λVDS ) ⎛ 1 ⎞ 2K n ⎛ 1 ⎞ 2( µ f = ⎜ + VDS ⎟ ≅⎜ ⎟ | ⎜ ⎟ ≤ 1 → ID ≥ 1.25 A ⎝λ ⎠ ⎝ λ ⎠ ID ⎝ .02 ⎠ ID ID 13.78 gm = 2ID VGS − VTN | ID = (.005)(0.5) = 1.25 mA 2 | 2( 1.25mA) 250 W 2ID = = = 2 2 −5 L K' ( 1 4 x10 (0.5) n VGS − VTN ) 13.79 2 (1 + 0.2) − 1 = 0.44 | 2(0.2) = 0.40 → 10% error (1 + 0.4 ) 13.80 2 − 1 = 0.96 | 2(0.4) = 0.80 → 20% error From the results of Problem 13.24: ID = 8.29E-05, VGS = 1.81E+00, VDS=5.96E+00, GM = 2.04E-04, GDS = 0.00E+00 gm = 2(82.9µA) 2ID 1 = = 205 µS | λ = 0 → ro = ∞ | g ds = = 0 VGS − VTN 1.81 − 1 ro 13.81 From the results of Problem 13.28: ID = 3.07E-04, VGS = -2.24E+00, VDS=-3.86E+00, GM = 4.96E-04, GDS = 0.00E+00 gm = 13.82 2(307µA) 2 ID 1 = = 495 µS | λ = 0 → ro = ∞ | g ds = = 0 ro VGS − VTP −2.24 + 1 RD ro V 9V 50V + 9V 7.8V | Using VDS = DD | RD = | ro = → Rout = RD + ro 2 ID ID ID Rout = ID = 7.8V = 156µA → RD = 57.6 kΩ | ro = 378 kΩ | Q - point : ( 156 µA, 9 V ) 50kΩ 13.83 Virtually any Q-point is possible. RIN is set by RG which can be any value desired since there is no gate current. (Note this is not the case with a BJT for which base current must be considered.) 13-26 13.84 Note that iG ≅ 0 for this device. Load line : 400 = 133000iP + v PK and vGK = −1.5V Two points (i P , v PK ) : (3mA, 0V) and (0mA, 400V) → Q - pt : (1.4 mA, 215 V) ro = 250V − 200V 2.3mA − 0.7 mA = 55.6 kΩ | g m = = 1.6mS | µ f = 89.0 2.15mA − 1.25mA −1V − ( −2V ) Av = −g m RP ro = −1.6mS 133kΩ 55.6 kΩ = −62.7 13.85 2 gm (0.5 S) = 5 A! BJT : IC = gmVT = 0.5S (0.025V ) = 12.5 mA | MOSFET : ID = = 2K n 2(25 mA /V 2 ) 2 ( ) ( ) The BJT can achieve the required transconductance at a 400 times lower current than the MOSFET. For a given power supply voltage, the BJT will therefore use 400 times less power. 60 = 120Ω (versus ∞ for the FET). Note, however, that rπ is small for the BJT: rπ = 0.5 13.86 Since a relatively high input resistance is required at a relatively high current, a FET should be used. If a BJT were selected, it would be very difficult to achieve the required input resistance because its value of rπ is low: β V 100(.025V ) rπ = o T = = 250 Ω IC 10 mA 13.87 ⎛1 ⎞ 2 Kn ( 1 + λVDS ) 40( VA + VCE )= ⎜ + VDS ⎟ ID ⎝λ ⎠ 1.2) 2(0.025)( ID → I D = 111 µA | µ f = 40(35) = 1400 40(35) = 60 13.88 V I µ f ≅ A = 40VA = 40(50)= 2000 | g m = C = 40 IC = 40 2 x10−4 = 8.00 mS VT VT ( ) 2 2 x10−4 2 2 2ID µf ≅ = = 200 | g m = = = 0.800 mS VGS − VTN 0.5 λ( VGS − VTN ) 0.02(0.5) ( ) 13-27 13.89 Either transistor could be used. For a BJT operating in the common-emitter configuration or an FET operating in the common-source configuration: 100(0.025V ) = 33.3 mA - A fairly high current rπ 75Ω For the FET : Rin = RG and setting RG = 75 Ω is satisfactory, particularly if a depletion mode FET is available. The input of a CE circuit operating at a much lower current could also be swamped by the addition of a 75-Ω resistor in parallel with its input. (Note that common-base and common-gate amplifiers from Chapter 14 could also be used.) For the BJT : Rin ≅ rπ | BJT : IC = β oVT = 13.90 26 (a) Av = 10 20 = 20.0. Either a BJT or MOSFET can achieve the required gain. However, based upon the material in Chapter 13, an FET should be chosen since the input voltage of 0.25 V is 50 time larger than the permissible value for vbe (0.005 V) for the BJT. For the FET, a value of VGS – VTN = 1.25 V will satisfy the small-signal limit with vgs = 0.25 V. (The generalized commonemitter stage with emitter degeneration can also satisfy the requirements.) (b) The FET is also best for this case, since the amplifier will see input signals much greater than the 5-mV limit of the BJT. 13.91 Av ≅ − (12) = −12 or 21.6 dB VDD =− VGS − VTN 1 13.92 15 7.5V vd = = 7.5V peak | 15dB → Av = −5.62 | vgs = = 1.34V | VGS − VTN ≥ 5( 1.34)= 6.70V 2 5.62 Yes, it is possible although the required value of (VGS − VTN ) is getting rather large. 13.93 Av ≅ 13.94 VDD VGS − VTN | 9 ≥ 30 → VGS − VTN ≤ 0.300 V VGS − VTN For VDS = ID = VDD VDD , Av = − 2 VGS − VTN | 30 = 15 VGS − VTN | VGS − VTN = 0.5 V 2 1mA VGS − VTN ) = 125 µA | Q - point : ( 125 µA,7.5 V ) ( 2 13-28 13.95 vgs ≤ 0.2( VGS − VTN ) requires ( VGS − VTN ) ≥ VDD VGS − VTN 0.1 = 0.5V 0.2 Av = 35dB → Av = −56.2. Using the rule - of - thumb estimate to select VDD : Av = − and VDD = 56.2(0.5V )= 28 V 13.96 vgs ≤ 0.2( VGS − VTN ) requires ( VGS − VTN ) ≥ VDD VGS − VTN 0.5 = 2.5V 0.2 Av = 20 dB → Av = −10. Using the rule - of - thumb estimate to select VDD : Av = − and VDD = 10(2.5) = 25 V 13.97 ⎛ VDD ⎞ N ⎛ 10 ⎞ N We desire ⎜ ⎟ ≥ 1000 | ⎜ ⎟ ≥ 1000 ⎝ VGS − VTN ⎠ ⎝ VGS − VTN ⎠ For VGS − VTN = 1 V , N = 3 meets the requirements, but with no safety margin. For VGS − VTN = 0.75 V , N = 3 easily meets the requirements. 13.98 For the bias network : VEQ = 10V ID = VDS 430 kΩ = 4.343V | REQ = 430 kΩ 560 kΩ = 243kΩ 430 kΩ + 560 kΩ 2 5 x10−4 VGS − 1) | VGS = 4.343 − 2 x10 4 I D → VGS = 1.72 V | I D = 131 µA ( 2 = 10 − 63kΩ( 131µA)= 1.75V ≥ VGS − VTN so active region assumption is ok. ⎛ 1 ⎞ + 1.75⎟ ⎜ ⎝ 0.0133 ⎠ 131µA) = 362µS | ro = = 586 kΩ g m = 2 5 x10−4 ( 131µA 243kΩ Av = − (362µS ) 586kΩ 43kΩ 100 kΩ = −10.3 243kΩ + 1kΩ ( ) ( ) 13.99 ⎛ 50 + 5V µA ⎞ 100µA)( 1 + 0.02(5)) = 332µS | ro = = 550 kΩ g m = 2⎜ 500 2 ⎟( 100µA V ⎠ ⎝ ⎞ ⎛ 6.8 MΩ Av = −⎜ ⎟(332µS ) 550kΩ 50kΩ 120kΩ = −10.9 ⎝ 6.8 MΩ + 0.1MΩ ⎠ ( ) 13-29 13.100 max min = 2 700µA / V 2 ( 100µA) = 374µS | g m = 2 300µA / V 2 ( 100µA) = 245µS gm ( ) ( ) ⎞ ⎛ 6.8 MΩ Av = −⎜ ⎟(g m ) 550 kΩ 50 kΩ 120 kΩ = (−32.7kΩ)(g m ) ⎝ 6.8 MΩ + 0.1 MΩ ⎠ ( ) Avmax = −12.2 | Avmin = −8.01 13.101 10µA)( 1 + 0.02(5)) = 46.9µS | ro = g m = 2 100µA / V 2 ( ( ) 50 + 5V = 5.50 MΩ 10µA ⎛ ⎞ 10 MΩ Av = −⎜ ⎟(46.9µS ) 5.50 MΩ 560 kΩ 2.2 MΩ = −19.2 ⎝ 10 MΩ + 0.1MΩ ⎠ ( ) 13.102 1 + λVDS ) = 2(0.001)(0.002) 1 + 0.015(7.5) = 2.11x10−3 S g m = 2 Kn I DS ( 1 + VDS I DS 1 + 7.5 0.015 = = 37.1kΩ 0.002 ( ) ro = λ Avt = − g m ro RD R3 = −2.11x10−3 37.1kΩ 3.9kΩ 270kΩ = −7.35 Av = 10 MΩ Avt = −7.34 10kΩ + 10 MΩ ( ) ( ) 13-30 13.103 *Problem 13.103 - NMOS Common-Source Amplifier - Figure P13.6 VDD 7 0 DC 15 *FOR OUTPUT RESISTANCE *VO 6 0 AC 1 *VI 1 0 AC 0 VI 1 0 AC 1 RI 1 2 1K C1 2 3 100UF R1 3 0 1MEG R2 7 3 2.7MEG R4 4 0 27K C2 4 0 100UF RD 7 5 82K C3 5 6 100UF R3 6 0 470K M1 5 3 4 4 NFET .OP .MODEL NFET NMOS KP=250U VTO=1 .AC LIN 1 1000 1000 .PRINT AC VM(6) VDB(6) VP(6) IM(VI) IP(VI) *.PRINT AC IM(C3) IP(C3) .END Results: ID = 8.29E-05 VGS = 1.81E+00 VDS = 5.96E+00 VM(6) = 1.420E+01 VP(6) = -1.800E+02 IM(VI) = 1.369E-06 IP(VI) = -1.800E+02 VM(3) = 9.986E-01 VP(3) = 1.248E-04 IM(C3) = 1.220E-05 IP(C3) = -1.800E+02 Av = −14.6 | Rin = VM (3) 0.9986 1 1 = = 729 kΩ | Rout = = = 82.0 kΩ IM (C 3) 12.20µA IM (VI ) 1.369µA 13-31 13.104 *Problem 13.104 - PMOS Common-Source Amplifier - Figure P13.8 VDD 7 0 DC 18 *FOR OUTPUT RESISTANCE *VO 6 0 AC 1 *VI 1 0 AC 0 * VI 1 0 AC 1 RI 1 2 1K C1 2 3 100U R2 7 3 3.3MEG R1 3 0 3.3MEG R4 7 4 22K C2 7 4 100U RD 5 0 24K C3 5 6 100U R3 6 0 470K M1 5 3 4 4 PFET .OP .MODEL PFET PMOS KP=400U VTO=-1 .AC LIN 1 1000 1000 .PRINT AC VM(6) VDB(6) VP(6) IM(VI) IP(VI) VM(3) VP(3) *.PRINT AC IM(C3) IP(C3) .END Results: ID = 3.07E-04 VGS = -2.24E+00 VDS = -3.86E+00 IP(C3) = -1.800E+02 VM(6) = 12.76E+01 VP(6) = -1.799E+02 IM(VI) = 6.057E-07 IP(VI) = -1.800E+02 VM(3) = 9.994E-01 VP(3) = 5.523E-05 IM(C3) = 4.167E-05 Av = −12.8 | Rin = VM (3) 0.9994V 1 1 = = 1.65 MΩ | Rout = = = 24.0 kΩ IM (C 3) 41.67µA IM (VI ) 0.6057µA 13-32 13.105 *Problem 13.105 - Depletion-mode NMOS Common-Source Amplifier - Figure P13.11 VDD 7 0 DC 18 *FOR OUTPUT RESISTANCE *VO 6 0 AC 1 *VI 1 0 AC 0 * VI 1 0 AC 1 RI 1 2 10K C1 2 3 100UF RG 3 0 10MEG R1 4 0 2K C3 4 0 100UF RD 7 5 3.9K C2 5 6 100UF R3 6 0 36K J1 5 3 4 NFET .OP .MODEL NFET NMOS KP=400U VTO=-5 .AC LIN 1 1000 1000 .PRINT AC VM(6) VDB(6) VP(6) IM(VI) IP(VI) VM(3) VP(3) *.PRINT AC IM(C2) IP(C2) .END Results: ID = 1.25E-03 VGS = -2.50E+00 VDS = -1.06E+01 VM(6) = 3.515E+00 VP(6) = -1.799E+02 IM(VI) = 9.991E-08 IP(VI) = -1.800E+02 VM(3) = 9.990E-01 VP(3) = 9.106E-06 IM(C3) = 2.564E-04 IP(C3) = -1.800E+02 Av = −3.52 | Rin = VM (3) 0.9990V 1 1 = = 10.0 MΩ | Rout = = = 3.90 kΩ IM (C 3) 256.4 µA IM (VI ) 99.91nA 13.106 SPICE Results: Q-point: (1.01 mA, 7.41 V), Av = 15.1 dB, Rin = 2.20 MΩ, Rout = 7.50 kΩ 13.107 g m = 40(50µA) = 2.00 mS | rπ = 100 75V + 10V = 50kΩ | ro = = 1.70 MΩ 2.00 mS 50µA Rin = RB rπ = 100 kΩ 50 kΩ = 33.3kΩ | Rout = 1.7 MΩ 100 kΩ = 94.4 kΩ 13-33 13.108 Rin = RB rπ | rπmin = 60 100 = 30 kΩ | rπmax = = 50 kΩ 40(50µA) 40(50µA) 75 + 10 = 100 kΩ 1.7 MΩ = 94.4 kΩ independent of βo 50µA min max = RB rπ = 100 kΩ 30 kΩ = 23.1kΩ | Rin = RB rπ = 100 kΩ 50 kΩ = 33.3kΩ Rin Rout = RC ro = 100 kΩ 13.109 40(0.025V ) 50 + 1.5 V = 1.00 MΩ | ro = = 51.5 MΩ rπ = 1 1µA µA Rin = RB rπ = 5 MΩ 1MΩ = 833 kΩ | Rout = RC ro = 1.5 MΩ 51.5 MΩ = 1.46 MΩ 13.110 75(0.025V ) 50 + 7.5 V rπ = = 750Ω | ro = = 23.0 kΩ 2.5mA 2.5 mA Rin = RB rπ = 4.7 kΩ 0.75kΩ = 647 Ω | Rout = RC ro = 4.3kΩ 23kΩ = 3.62 kΩ 13.111 From Prob. 13.98 : Q - Point = ( 131µA, 1.75V) Rin = R1 R2 = 430 kΩ 560 kΩ = 243 kΩ | Rout = 43kΩ ro ⎛ 1 ⎞ + 1.75⎟V ⎜ ⎝ 0.0133 ⎠ ro = = 587 kΩ | Rout = 43kΩ 587 kΩ = 40.1kΩ 0.131mA 13.112 Rin = RG = 6.8 MΩ | Rout = 50 kΩ ro ro = (50 + 5)V = 550 kΩ 0.1mA | Rout = 50 kΩ 550 kΩ = 45.8 kΩ 13.113 Rin = RG = 6.8 MΩ which is independent of Kn | Rout = RD ro ⎛ 1 ⎞ + 5⎟V ⎜ ⎝ 0.02 ⎠ ro = = 550 kΩ | Rout = 50 kΩ 550kΩ = 45.8 kΩ, also independent of Kn 0.1mA 13-34 13.114 Rin = RG = 10 MΩ | Rout = RD ro Rout = 560kΩ 5.50 MΩ = 508 kΩ 13.115 ⎛ 1 ⎞ + 5⎟V ⎜ ⎝ 0.02 ⎠ | ro = = 5.50 MΩ 10µA Rin = RG = 1 MΩ | Rout = RD ro Rout = 3.9kΩ 37.1kΩ = 3.53 kΩ 13.116 ⎛ 1 ⎞ + 7.5⎟V ⎜ ⎝ 0.015 ⎠ | ro = = 37.1 kΩ 2mA g m = 40(50µA) = 2.00 mS | rπ = 100 75V + 10V = 50kΩ | ro = = 1.70 MΩ 2.00 mS 50µA RBB = RB rπ = 100kΩ 50 kΩ = 33.3kΩ ⎛ ⎞ 33.3kΩ vth = −vi ⎜ ⎟(2 mS ) 1.70 MΩ 100 kΩ = −185vi ⎝ 33.3kΩ + 0.75kΩ ⎠ ( ) Rth = 1.70 MΩ 100 kΩ = 94.4 kΩ 13.117 g m = 40(2.5mA) = 100 mS | rπ = RBB 75 50V + 7.5V = 750Ω | ro = = 23.0 kΩ 100 mS 2.5mA = RB rπ = 4.7kΩ 0.75kΩ = 647Ω ⎛ 647Ω ⎞ vth = −vi ⎜ 100mS ) 23.0kΩ 4.3kΩ = −336v i ⎟( ⎝ 647Ω + 50Ω ⎠ ( ) Rth = 23.0 kΩ 4.3kΩ = 3.62 kΩ 13.118 100µA)( 1 + 0.02(5)) = 332µS | ro = g m = 2 500µA / V 2 ( ( ) (50 + 5)V = 550kΩ 100µA ⎛ ⎞ 6.8 MΩ vth = −vi ⎜ ⎟(332µS ) 550 kΩ 50 kΩ = −15.0vi ⎝ 6.8 MΩ + 0.1 MΩ ⎠ ( ) Rth = 550 kΩ 50 kΩ = 45.8 kΩ 13-35 13.119 10µA)( 1 + 0.02(5)) = 46.9µS | ro = g m = 2 100µA / V 2 ( ( ) (50 + 5)V = 5.50 MΩ 10µA ⎛ ⎞ 10 MΩ vth = −vi ⎜ ⎟(46.9µS ) 5.50 MΩ 560kΩ = −23.6vi ⎝10 MΩ + 0.1 MΩ ⎠ ( ) Rth = 5.50 MΩ 560 kΩ = 508 kΩ 13.120 SPICE Results: Q-point: (242 µA, 3.61 V), Av = 31.1 dB, Rin = 14.8 kΩ, Rout = 9.81 kΩ 13.121 0 = 10000 I B + 0.7 + 66( 1615)I B − 5 I B = 36.9 µA | IC = 65I B = 2.40 mA Q − point : (2.40 mA, 3.66 V ) VCE = 5 − 1000 IC − 1615 I E − (−5)= 3.66 V g m = 40(2.40mA)= 96.0 mS | rπ = 65 50V + 3.66V = 677 Ω | ro = = 22.4 kΩ 96.0 mS 2.40 mA Rin = RB rπ ( 1 + g m RE )= 10 kΩ 677Ω 1 + 0.096( 15) = 1.42 kΩ [ ] ⎛ ⎞ 96.0 mS 1.42kΩ 1kΩ 220 kΩ = −31.8 Av = −⎜ ⎟ 15) ⎝ 0.33kΩ + 1.42kΩ ⎠ 1 + 0.096( ( ) Rout = 1kΩ 22.4kΩ 1 + 0.096( 15) = 982 Ω 13.122 SPICE Results: Q-point: (2.39 µA, 3.69 V), Av = 29.7 dB, Rin = 1.49 kΩ, Rout = 977 Ω The results agree closely with the hand calculations in Prob. 13.121. The small disagreements arise from  m atu us in  13.123 161.5kΩ)I B − 5 0 = 106 I B + 0.7 + 66( I B = 0.369 µA | IC = 65 I B = 24.0 µA Q − point : (24.0 µA, 3.66 V ) 65 50V + 3.66V = 67.8 kΩ | ro = = 2.24 MΩ 0.959 mS 24.0µA [ ] VCE = 5 − 1000 IC − 1615 I E − (−5)= 3.66 V g m = 40(24.0µA) = 0.959 mS | rπ = Rin = RB rπ ( 1 + g m RE )= 1 MΩ 67.8 kΩ 1 + 0.959 mS ( 1.5kΩ) = 142 kΩ ⎛ ⎞ 0.959 mS 142 kΩ Av = −⎜ 100 kΩ 220 kΩ = −27.0 ⎟ 1.5kΩ) ⎝ 0.33kΩ + 142 kΩ ⎠ 1 + 0.959 mS ( [ ] ( ) Rout = 100 kΩ 2.24 MΩ 1 + 0.959 mS ( 1.5kΩ) = 98.2 kΩ [ ] 13-36 13.124 SPICE Results: Q-point: (24.6 µA, 3.52 V), Av = 28.4 dB, Rin = 148 kΩ, Rout = 98.1 kΩ The results agree closely with the hand calculations in Prob. 13.123. The small disagreements arise from   m atu us in  13.125 SPICE Results: Q-point: (251 µA, 4.45 V), Av = 16.3 dB, Rin = 1.00 MΩ, Rout = 28.7 kΩ The results agree closely with the hand calculations in Ex. 13.10. 13.126 *Problem 13.126 - NMOS Common-Source Amplifier - Figure P13.98 VDD 7 0 DC 10 *FOR OUTPUT RESISTANCE *VO 6 0 AC 1 *VI 1 0 AC 0 * VI 1 0 AC 1 RI 1 2 1K C1 2 3 100U R1 3 0 430K R2 7 3 560K R4 4 0 20K C3 4 0 100U RD 7 5 43K C2 5 6 100U R3 6 0 100K M1 5 3 4 4 NFET .OP .MODEL NFET NMOS KP=500U VTO=1 LAMBDA=0.0133 .AC LIN 1 1000 1000 .PRINT AC VM(6) VDB(6) VP(6) VM(3) VP(3) IM(VI) IP(VI) *.PRINT AC IM(C2) IP(C2) .END Results: ID = 1.31E-04 VDS = 1.73E+00 VM(6) = 1.044E+01 VP(6) = -1.800E+02 IM(VI) = 4.094E-06 IP(VI) = -1.800E+02 VM(3) = 9.959E-01 VP(3) = 3.734E-04 IM(C3) = 2.496E-05 IP(C3) = -1.800E+02 Av = −10.4 | Rin = VM (3) 0.9959V 1 1 = = 243 kΩ | Rout = = = 40.1 kΩ IM (C 3) 24.96µA IM (VI ) 4.094 µA 13-37 13.127 I B = 3.71µA IC = 241µA I E = 245µA VCE = 3.67V 2 2 PR B = I B RB = (3.71µA) ( 100 kΩ)= 1.38 µW | PRC = IC RC = (241µA) ( 10 kΩ)= 0.581 mW 2 2 2 RE = (245µA) ( 16 kΩ) = 0.960 mW PR E = I E 2 PS = 5V (241µA)+ (−5V )(−245µA) = 2.43 mW | PS = PR B + PRC + PR E + PQ 13.128 I D = 250µA VDS = 4.75V 2 PJFET = I DVDS = (250µA)(4.75V )= 1.19 mW | PR D = I D RD = (250µA) (27 kΩ)= 1.69 mW 2 2 R4 = (250µA) (2 kΩ) = 0.125 mW | PRG = 0 PR 4 = I D 2 PBJT = ICVCE + I BVBE = (241µA)(3.67V )+ (3.71µA)(0.7V )= 0.887 mW PS = 12V (250µA) = 3.00 mW | PJFET + PR D + PR 4 + PRG = 3.00 mW 13.129 Using the values from Prob. 13.17 : I B = 23.8µA IC = 1.78mA I E = 1.81mA VCE = 6.08V VB = −4.09V 2 12 − ( −4.09)) V 2 V12 (−4.09 − ( −12)) V V22 ( PR1 = = = 12.5 mW | PR 2 = = = 25.9 mW R1 5000Ω R2 10000Ω 2 2 2 2 PRC = IC RC = ( 1.78 mA) (6 kΩ) = 19.0 mW | PR E = I E RE = ( 1.81mA) (4 kΩ)= 13.0 mW 2 2 PBJT = ICVCE + I BVBE = ( 1.78 mA)(6.08V )+ (23.8µA)(0.7V ) = 10.8 mW ⎡ ⎡ −4.09 − ( −12)V ⎤ 12 − ( −4.09)V ⎤ PS = 12V ⎢1.78 mA + ⎥ + 12V ⎢1.81mA + ⎥ = 81.3 mW 10000Ω ⎦ 5000Ω ⎦ ⎣ ⎣ PS = PR1 + PR 2 + PRC + PR E + PBJT = 81.2 mW 13.130 Using the values from Prob. 13.19 : I B = 1.51µA IC = 98.4µA I E = 99.9µA VCE = 4.96V 2 2 PR B = I B RB = ( 1.51µA) 3kΩ = 6.84 nW | PRC = IC RC = (98.4µA) (33kΩ)= 0.320 mW 2 2 2 RE = (99.9µA) (68 kΩ)= 0.679 mW PR E = I E 2 PBJT = ICVCE + I BVBE = (98.4µA)(4.96V )+ ( 1.51µA)(0.7V ) = 0.489 mW PS = 7.5(98.4µA)+ 7.5V (99.9µA)= 1.49 mW | PS = PR B + PRC + PR E + PBJT = 1.49 mW 13-38 13.131 Using the values from Prob. 13.23: I D = 82.2µA VDS = 6.04V 2 RD = (82.2µA) (82kΩ) = 0.554 mW PFET = I DVDS = (82.2µA)(6.04V ) = 0.497 mW | PR D = I D 2 2 R4 = (82.2µA) (27 kΩ)= 0.182 mW | I2 = PR 4 = I D 2 2 15V = 4.05µA 3.7 MΩ 2 1 MΩ)= 16.4 µW | PR 2 = I22 R2 = (4.05µA) (2.7 MΩ) = 44.3 µW PR1 = I22 R1 = (4.05µA) ( PS = 15V (82.2µA + 4.05µA)= 1.29 mW | PFET + PR D + PR 4 + PR1 + PR 2 = 1.29 mW 13.132 Using the values from Prob. 13.27 : I D = 307µA VSD = 3.88V 2 RD = (307µA) (24kΩ)= 2.26 mW PFET = I DVSD = (307µA)(3.88V ) = 1.19 mW | PR D = I D 2 2 R4 = (307µA) (22 kΩ) = 2.07 mW | I2 = PR 4 = I D 2 2 18V = 2.73µA 6.6 MΩ 2 PS = 18V (307µA + 2.73µA)= 5.58 mW | PFET + PR D + PR 4 + PR1 + PR 2 = 5.57 mW 13.133 Using the values from Prob. 13.33: I D = 1.25mA VDS = 10.6V PR1 = I22 R1 = (2.73µA) (3.3 MΩ) = 24.6 µW | PR 2 = I22 R2 = (2.73µA) (3.3 MΩ)= 24.6 µW 2 1.25mA)( 10.6V ) = 13.3 mW | PR D = I D RD = ( 1.25mA) (3.9kΩ) = 6.09 mW PFET = I DVDS = ( 2 2 RS = ( 1.25mA) ( 1kΩ)= 1.56 mW PR S = I D 2 2 PRG = IG RG = 0 2 | PR 4 = I D R4 = ( 1.25mA) ( 1kΩ) = 1.56 mW 2 PS = 18V ( 1.25mA) = 22.5 mW | PFET + PR D + PR S + PR1 = 22.5 mW 13.134 V IC = CC 3RC 13.135 | ic ≤ 0.2 IC | v c = ic RC ≤ 0.2 VCC V RC = CC 3 RC 15 id ≤ 0.4 I D | I D = 2 500 x10−6 0 − (−1.5) = 563 µA 2 VGS − VTN )= 0.2 0 − (−1.5) = 0.3 V vgs ≤ 0.2( ( ) vds = id RD ≤ 0.4(563µA)( 15kΩ)= 3.38 V VDD ≥ 1.8 + 3.38 + (563µA)( 15kΩ)= 13.6 V To insure saturation : vDS ≥ vGS − VTN = vgs − VTN = 0.3 − ( −1.5) = 1.8 V ( ) 13.136 13-39 Assuming VCC >> VCESAT : 2 2 VCC 8R | ε = 100% 2 L = 25 % VCC 2 RL 2 2 2 ⎛ VCC ⎞ VCC VCC 1 ⎛ VCC ⎞ 1 VCC | Pdc = VCC ⎜ | Pac = ⎜ = vo ≤ ⎟= ⎟ 2 ⎝ 2 ⎠ RL 8 RL 2 ⎝ 2 RL ⎠ 2 RL 13.137 The Q - point from problem 13.21 is (371µA, 2.72V). vo ≤ g m vbe ro RC R3 = 40(371µA)(5mV ) ∞ 13kΩ 100 kΩ = 0.854 V 13.138 The Q - point from Problem 13.23 is (82.2µA,6.04V ). ( ) ( ) id ≤ 0.4 I D = 32.9µA | vds ≤ 0.4 I D ro RD R3 = 32.9µA ∞ 82kΩ 470kΩ = 2.30V 13.139 The Q - point from Problem 13.27 is (307µA,3.88V ). id ≤ 0.4 I D = 307µA | vo ≤ 0.4 I D ro RL R3 = 307µA ∞ 24 kΩ 470 kΩ = 2.80V Checking the bias point : VR D = 307µA(24 kΩ) = 7.37V | 2.80 < 7.37 & 2.80 < 3.88 -1 ( ) ( ) ( ) ( ) 13.140 The Q - point from problem 13.17 is (1.78 mA, 6.08V ). ic ≤ 0.2 IC = 0.356 mA | vc ≤ 0.2 IC ro RC R3 = 0.356 mA ∞ 6kΩ 100kΩ = 2.02V 13.141 The Q - point from problem 13.33 is (1.25 mA, 10.6 V ). ( ) ( ) id ≤ 0.4 I D = 0.500mA | vd ≤ 0.4 I D RD R3 = 0.500mA 3.9 kΩ 36kΩ = 1.76 V 13.142 The Q - point from problem 13.35 is (1.01 mA, 7.41 V ). ( ) ( Neglecting ro , ) id ≤ 0.4 I D = 0.404 mA | vd ≤ 0.4 I D RD R3 = 0.404 mA 7.5kΩ 220 kΩ = 2.93 V ( ) ( ) 13-40 13.143 VCE = 20 − 20000IC : Two points on the load line (0 mA, 20V ) , (1mA, 0V ) At IB = 2µA, the maximum swing is approximately 2.5 V limited by VRC . For IB = 5µA,the maximum swing is approximately - 8.5 V limited by VCE . IC 1000 µA IB = 10 µA I = 8 µA B 750 µA I = 6 µA B 500 µA I B = 5 µA I = 4 µA B 250 µA 8.5 V 9V I = 2 µA B 0 0 10 V 2.5 V 20 V V CE 13-41 CHAPTER 14 14.1 (a) Common-collector Amplifier (npn) (emitter-follower) RI Q1 vi R 1 R 2 + R E R 3 vo - (b) Not a useful circuit because the signal is injected into the drain of the transistor. RI vi RD R1 M 1 + R3 vo - (c) Common-emitter Amplifier (pnp) R RI Q 1 1 R 2 1kΩ vi RC R + vo 3 100 k Ω - © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-1 14.1 (d) Common-source Amplifier (NMOS) RI + R M 1 RD 3 vo 1k Ω vi R 1 R 470 k Ω 2 - (e) Common-gate Amplifier (PMOS) RI vi M1 + RD R3 vo R 1 - (f) Common-collector Amplifier (emitter-follower) (npn) RI Q 1 vi R 1 + R 2 RE R3 v o - 14-2 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.1(h) Common-base Amplifier (pnp) + RC Q1 R3 v o - R RE I v i (i) Not a useful circuit since the signal is being taken out of the base terminal. Q1 + vo RI RE vi R3 RB (j) Common-source Amplifier (PMOS) RI 1k Ω vi R 1 M1 R 2 + RD R3 470 k Ω v o - © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-3 14.1(k) Common-gate Amplifier (Depletion-mode NMOS) RI vi RS RD M1 R 3 + vo - (l) Not a useful circuit because the signal is injected into the drain of the transistor. M RI vi RD RS 1 + R3 vo - (n) Common-emitter Amplifier (npn) + RI Q1 R R3 vo vi B (o) Common-drain Amplifier (Source-follower) (NMOS) RI M R vi G + R3 vo - 14-4 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.2 V DD RI 1k Ω vi CB RG RD -V SS CC2 + RL vo 100 k Ω CC1 RS 14.3 VDD RS RI 1k Ω RG RD -V SS 14.4 V DD RS RI 1k Ω vi RG RD RL -V SS CC1 CB CC1 RL CC2 + vo 100 k Ω vi CB CC2 + vo 100 k Ω © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-5 14.5 +15 V RC 15 kΩ CC2 RL Q1 RI + v O CC1 3.3 k Ω vI 150 k Ω RB - 220 k Ω RE 22 k Ω CB -15 V 14.6 +15 V RC 15 kΩ +15 V RI CC1 Q1 3.3 k Ω vI CB RI CC1 Q1 3.3 k Ω RB 220 k Ω CC2 RB 220 k Ω CC2 vI RE 22 k Ω RL 150 k Ω -15 V + vO RE 22 k Ω RL 150 k Ω -15 V + vO - - (a) 14.7 (b) +15 V RC CC2 +15 V RC 15 k Ω 15 kΩ CC2 RL CB Q1 + vO RL Q1 150 k Ω CC1 RI 3.3 kΩ vI -15 V + vO 150 kΩ CC1 RI - - RB 220 kΩ RE 22 kΩ 3.3 k Ω vI RE 22 kΩ (a) -15 V (b) 14-6 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.8 (a) Neglecting R Av = Avt out : Avt = − (0.005S )(2000Ω) = −3.77 g m RL =− 1 + g m RS 1 + (0.005S )(330Ω) | Rin = RG = 2 MΩ ⎛ ⎞ RG 2 MΩ = -3.77⎜ ⎟ = -3.64 RI + RG ⎝ 75kΩ + 2 MΩ ⎠ Rout = ro ( 1 + g m RS ) = 10 kΩ 1 + (0.005S )(330Ω) = 26.5 kΩ >> 2 kΩ Ai = − RG gm 0.005S = − 2 MΩ = −3770 1 + g m RS 1 + (0.005S )(330Ω) [ ] (b) A v Rin = RG = 2 MΩ | Rout = ro = 10.0 kΩ | Ai = − RG g m = −2 MΩ(0.005S ) = −10000 ⎛ RG ⎞ ⎛ ⎞ 2 MΩ = − g m RL ro ⎜ ⎟ = −(0.005S ) 2 kΩ 10 kΩ ⎜ ⎟ = −8.03 ⎝ 75kΩ + 2 MΩ ⎠ ⎝ RI + RG ⎠ ( ) ( ) © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-7 14.9 (a) g m = 0.02 S | rπ = Rin = RB [ 75 = 3750Ω .02 rπ + (β o + 1)RE = 15kΩ 3750Ω + 76(300Ω) = 9.58 kΩ ] [ ] ⎛ Rin ⎞ ⎛ Rin ⎞ β o RL Neglecting Rout : Av = Avt ⎜ ⎟=− ⎜ ⎟ rπ + (β o + 1)RE ⎝ RI + Rin ⎠ ⎝ RI + Rin ⎠ Av = − ⎛ ⎞ 9.58 kΩ ⎜ ⎟ = −32.2 3750Ω + 76(300Ω)⎝ 500Ω + 9.58 kΩ ⎠ 75( 12 kΩ) Rout ⎡ ⎤ ⎛ ⎞ 75(300Ω) β o RE ⎢ ⎥ = 596 kΩ >> 12 kΩ ≅ ro ⎜1 + ⎟ = 100 kΩ 1 + ⎢ ⎥ 15 k Ω 500 Ω + 3750 Ω + 300 Ω ⎝ Rth + rπ + RE ⎠ ⎣ ⎦ ( ) Ai = −β o RB 15kΩ = −75 = −27.1 15kΩ + 3750Ω + 76(300Ω) RB + rπ + (β o + 1)RE 75 = 3750Ω .02 rπ + (β o + 1)RE = 15kΩ 3750Ω + 76(620Ω) = 11.6 kΩ (b) g m = 0.02 S | rπ = Rin = RB [ ] [ ] ⎛ Rin ⎞ ⎛ Rin ⎞ β o RL Neglecting Rout : Av = Avt ⎜ ⎟=− ⎜ ⎟ rπ + (β o + 1)RE ⎝ RI + Rin ⎠ ⎝ RI + Rin ⎠ Av = − ⎛ ⎞ 11.6 kΩ ⎜ ⎟ = −17.0 3750Ω + 76(620Ω)⎝ 500Ω + 11.6 kΩ ⎠ 75( 12 kΩ) ⎡ ⎤ ⎛ ⎞ 75(620Ω) β o RE ⎢ ⎥ = 1060 kΩ >> 12 kΩ Rout ≅ ro ⎜1 + = 100 k Ω 1 + ⎟ ⎢ ⎥ 15 k Ω 500 Ω + 3750 Ω + 620 Ω ⎝ Rth + rπ + RE ⎠ ⎣ ⎦ ( ) Ai = −β o 14.10 RB 15kΩ = −75 = −17.1 15kΩ + 3750Ω + 76(620Ω) RB + rπ + (β o + 1)RE 8.2 kΩ 47 kΩ RL =− = −6.91 | RE 330Ω + 680Ω (a ) For large β o : Av ≅ − (b) Place a bypass capacitor in (c) Place a parallel with the 330Ω resistor. Then Av ≅ − 8.2 kΩ 47 kΩ RL =− = −10.3 | RE 680Ω = −21.2 330Ω (d) Place a bypass capacitor from the emitter to ground. (e) Av ≅ −10(VCC + VEE )= −240. bypass capacitor in parallel with the 680Ω resistor. Then Av ≅ − 8.2 kΩ 47 kΩ 14-8 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.11 Rin = rπ + (β o + 1)RE = 250 kΩ Avt = − β o RL 75 RL =− = −10 → RL = 33.3 kΩ → 33 kΩ 250 kΩ rπ + (β o + 1)RE 250 kΩ 250 kΩ = = 3.29 kΩ → 3.3 kΩ βo + 1 76 Assuming (β o + 1)RE >> rπ , RE ≅ As indicated above, the nearest 5% values would be 33 kΩ and 3.3 kΩ. 14.12 Rin = rπ = 250 kΩ | rπ = βo gm = β oVT IC | IC = β oVT rπ = 75(0.025V ) 250 kΩ = 7.50 µA ⎛ r ⎞ β R 75 RL = −10 | RL = 33.3 kΩ → 33 kΩ Av = −g m RL ⎜ π ⎟ = − o L = − Rth + rπ 100Ω + 250 kΩ ⎝ Rth + rπ ⎠ The closest 5% value is RL = 33 kΩ. 14.13 vd gm vgs + vgs - r ⎡ix − gm v gs⎤ ⎡ go − go ⎤⎡v d ⎤ ⎢ ⎥=⎢ ⎥⎢ ⎥ | v gs = −v s ⎣ + gm v gs ⎦ ⎣−go go + GS ⎦⎣ v s ⎦ ⎡ix ⎤ ⎡ go −(gm + go ) ⎤⎡v d ⎤ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎣0 ⎦ ⎣−go gm + go + GS ⎦⎣ v s ⎦ o ∆ = goGS | v d = (gm + go + GS ) ix R R th 5 Rout ix (gm + go + GS ) = ix ∆ go G S ⎛ g ⎛ v G ⎞ r ⎞ = d = RS ⎜1 + m + S ⎟ = RS ⎜1 + µ f + o ⎟ ix RS ⎠ ⎝ go go ⎠ ⎝ Rout = ro + ( 1 + µ f )RS ≅ ro + µ f RS = ro (1 + gm RS ) © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-9 14.14 IB = (15 − 0.7 )V = 1.29 µA | I C = 129 µA | VCE = 30 − 39000 I C − 100000 I E 1MΩ + (100 + 1)100kΩ 100(0.025V ) = 19.4kΩ | r - no V specified - neglect Active region is correct. | r = π = 11.9 V 129 µA o A RL = 500kΩ 39kΩ = 36.2kΩ | Rin = RB rπ = 1MΩ 19.4kΩ = 19.0kΩ ⎛ Rin ⎞ 19.0kΩ ⎞ ⎛ Av = − g m RL ⎜ ⎟ = −182 ⎜R +R ⎟ ⎟ = 40(0.129mA)(36.2kΩ )⎜ ⎝ 500Ω + 19.0kΩ ⎠ in ⎠ ⎝ I Rout = RC ro = 39 kΩ ⎛ Rin ⎞ 0.005V 19.0kΩ ⎞ ⎛ vi ≤ = 5.13 mV vbe = vi ⎜ ⎟ = 0.974vi ⎜R +R ⎟ ⎟ = vi ⎜ k 0 . 974 500 Ω + 19 . 0 Ω ⎠ ⎝ I in ⎝ ⎠ Av ≅ −10(VCC + VEE ) = −10(30 ) = −300. | A closer estimate is - 40VR C = -40(5.03) = −201 14.15 VEQ = 9 62kΩ = 6.80V | REQ = 20kΩ 62kΩ = 15.1kΩ 20 kΩ + 62kΩ (9 − 0.7 − 6.80)V = 4.82µA | I = 361 µA | V = 9 − 3900 I − 8200 I = 4.61 V IB = C EC E C 15.1kΩ + (75 + 1)3.9 kΩ Active region is correct. | rπ = 75(0.025V ) 361µA = 5.19 kΩ | VA not specified, choose ro = ∞ Rin = 15.1kΩ 5.19 kΩ = 3.86 kΩ | Rout = ro 8.2kΩ = 8.2 kΩ | g m = 40 IC = 12.6 mS RL = ro 8.2 kΩ 100 kΩ = 8.2kΩ 100kΩ = 7.58kΩ ⎛ Rin ⎞ ⎛ 3.86 kΩ ⎞ 12.6 mS )(7.58kΩ) Av = −g m RL ⎜ ⎟ = −( ⎜ ⎟ = −75.9 ⎝1kΩ + 3.86 kΩ ⎠ ⎝ RI + Rin ⎠ Rout 8.2kΩ RB 15.1kΩ −β o ) = −75) = −4.23 Ai = ( ( RB + rπ Rout + R3 15.1kΩ + 5.19 kΩ 8.2 kΩ + 100kΩ vbe = vi Rin 3.86kΩ 5.00 mV = vi = 0.794v i | vi = = 6.30 mV 1kΩ + 3.86kΩ 0.794 RI + Rin Av ≅ −10VCC = −10(9)= −90. | The voltage gain is slightly below the rule - of - thumb estimate. 14-10 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.16 VEQ = 15 500 kΩ = 3.95V | REQ = 500 kΩ 1.4 MΩ = 368 kΩ 1.4 MΩ + 500 kΩ 3.95 = VGS + 27000 I D = 1 + VDS 2 ID + 27000 I D → I D = 79.7µA 250 x10−6 = 15 − I D (75kΩ + 27 kΩ) = 6.87 V | Active region operation is correct. g m = 2 250 x10−6 79.7 x10−6 = 0.200 mS | Assume λ = 0, ro = ∞. RL = ro 75kΩ 470 kΩ ≅75kΩ 470 kΩ = 64.7 kΩ Rin = RG = R1 R2 = 368 kΩ | Rout = ro 75kΩ ≅ 75kΩ Av = −g m RL ⎛ 368 kΩ ⎞ Rin = −(0.200 mS )(64.7 kΩ) ⎟ = −12.9 ⎜ RI + Rin ⎝ 1kΩ + 368 kΩ ⎠ RD 75kΩ = 368 kΩ(−0.200 mS ) = −10.1 75kΩ + 470 kΩ RD + R3 2(79.7µA) 250µA / V 2 = 0.798V ( )( ) Ai = RG (−g m ) vgs = vi Rin 368 kΩ = vi = 0.997v i | VGS − VTN = 1kΩ + 368 kΩ RI + Rin VGS − VTN )→ vi ≤ 0.2 vgs ≤ 0.2( 0.798V VDD 15 = 0.160 V | Av ≅ − =− = −18.8 0.997 0.798 VGS − VTN VDD . We have VR L = 79.7µA(75kΩ)= 5.98V = 0.399VDD 2 The estimate also doesn't account for the presence of R3 . The rule - of - thumb estimate assumes VR L = © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-11 14.17 VEQ = 22 2.2 MΩ = 11.0V | RG = REQ = 2.2 MΩ 2.2 MΩ = 1.10 MΩ 2.2 MΩ + 2.2 MΩ Assume active region operation. 2ID → I D = 391 µA 400 x10−6 22 = 22000 I D − VGS + 11 | 11 = 22000 I D + 1 + VDS = − 22 − I D (22kΩ + 18kΩ) = −6.36 V | Active region operation is correct. gm = −6 −6 [ ] 2( 400 x10 ) (391x10 ) = 0.559mS | Assume λ = 0, ro = ∞ RL = ro 18kΩ 470 kΩ ≅18 kΩ 470 kΩ = 17.3kΩ | Rin = 1.10 MΩ | Rout = ro 18 kΩ = 18 kΩ ⎛ Rin ⎞ 1.1 MΩ Av = −g m RL ⎜ 17.3kΩ) = −9.48 ⎟ = −(0.559 mS )( 22 kΩ + 1.1 MΩ ⎝ RI + Rin ⎠ RD 18kΩ = −0.559 mS ( 1.1 MΩ) = −22.7 Ai = −g m Rin 18 kΩ + 470kΩ RD + R3 Rin 1.1MΩ = vi = 0.980vi | VGS − VTN = RI + Rin 22kΩ + 1.1MΩ ⎛1.40V ⎞ VGS − VTN ) | vi ≤ 0.2⎜ vgs ≤ 0.2( ⎟ = 0.286 V ⎝ 0.980 ⎠ vgs = vi 14.18 2(391µA) 400µA / V 2 = 1.40V VGS = 0 → I D = VDS 2 2 Kn 4 x10−4 V = −5) = 5.00 mA ( ( TN ) 2 2 = 16 − 1800 I D = 7.00V | Active region operation is correct. g m = 2 4 x10 4 5 x10−3 = 2.00 mS | Assume λ = 0, ro = ∞. ⎛ RG ⎞ ⎛ 10 MΩ ⎞ Av = - g m RL ⎜ ⎟ = −(2.00mS ) 1.8 kΩ 36 kΩ ⎜ ⎟ = −3.43 ⎝ 10 MΩ + 5kΩ ⎠ ⎝ RI + RG ⎠ ( )( ) ( ) Rin = 10.0 MΩ | Rout = RD ro = 1.80 kΩ ⎛ RD ⎞ ⎛ ⎞ 1.8kΩ 10 MΩ) Ai = −g m RG ⎜ ⎟ = −2.00 mS ( ⎜ ⎟ = −952 ⎝ 1.8kΩ + 36 kΩ ⎠ ⎝ RD + R3 ⎠ ⎛ 10 MΩ ⎞ vgs = vi ⎜ ⎟ ≤ 0.2 VGS − VTN → vi ≤ 1 V ⎝10 MΩ + 5kΩ ⎠ 14-12 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.19 (12 − 0.7)V = 14.9µA | I = 1.19 mA | V = 24 − 9100 I = 13.2 V 20kΩ + (80 + 1)9.1kΩ 80(0.025V ) (100 + 13.2)V = 95.1kΩ = 1.68 kΩ | r = Active region is correct. | r = IB = C CE E 1.19mA 1.19mA RL = ro 1MΩ = 95.1kΩ 1 MΩ = 86.9 kΩ | Rin = RB rπ = 20 kΩ 1.68 kΩ = 1.55 kΩ | Rout = ro = 95.1 kΩ π o ⎛ Rin ⎞ 1.55kΩ = −3560 Av = - g m RL ⎜ 1.19mA)(86.9 kΩ) ⎟ = −40( 250Ω + 1.55kΩ ⎝ RI + Rin ⎠ ⎛ RB ⎞⎛ ro ⎞ ⎛ ⎞⎛ ⎞ 95.1kΩ 20kΩ Ai = −β o ⎜ ⎟⎜ ⎟ = −80⎜ ⎟⎜ ⎟ = −6.41 ⎝ 20 kΩ + 1.68kΩ ⎠⎝ 95.1kΩ + 1 MΩ ⎠ ⎝ RB + rπ ⎠⎝ ro + R3 ⎠ ⎛ Rin ⎞ 1.55kΩ 5.00 mV vbe = vi ⎜ = 0.861v i | vi ≤ = 5.81 mV ⎟ = vi 250Ω + 1.55kΩ 0.861 ⎝ RI + Rin ⎠ 14.20 rπ = 80 = 200Ω | Assume VA = ∞, ro = ∞. 0.4 S Rin = RB rπ + (β o + 1)RL = 47kΩ 200Ω + 81( 1kΩ) = 29.8 kΩ [ Rout = Av = + 47kΩ 10kΩ + 200Ω Rth + rπ = = 104 Ω βo + 1 81 rπ ( ] ) [ ] (β + 1)R + (β + 1)R o L o ⎛ Rin ⎞ ⎛ ⎞ 81( 1kΩ) 29.8 kΩ ⎜ ⎟= ⎜ ⎟ = 0.747 200Ω + 81( 1kΩ)⎝10 kΩ + 29.8 kΩ ⎠ L ⎝ RI + Rin ⎠ ⎡ ⎤⎛ r ⎞ RB 47 kΩ ⎢ ⎥⎜ o ⎟ = 81 = 29.7 Ai = +(β o + 1) 47kΩ + 200Ω + 81( 1kΩ) ⎢ ⎣ RB + rπ + (β o + 1)RL ⎥ ⎦⎝ ro + RL ⎠ 14.21 Assume λ = 0, ro = ∞. | Rin = RG = 2 MΩ | Rout = Av = + ⎞ 0.01(2 kΩ) ⎛ g m RL ⎛ Rin ⎞ 2 MΩ ⎜ ⎟= ⎜ ⎟ = 0.907 1 + g m RL ⎝ RI + Rin ⎠ 1 + 0.01(2 kΩ)⎝100 kΩ + 2 MΩ ⎠ 1 = 100 Ω gm ⎛ r ⎞ Ai = + g m RG ⎜ o ⎟ = 0.01(2 MΩ)= 2 x10 4 ⎝ ro + RL ⎠ © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-13 14.22 Defining v1 as the source node: (a) 2kΩ 100kΩ = 1.96kΩ (v − v ) + 3.54 x10 v − v = v ( ) 1960 10 i 1 −3 1 6 i 1 + 1 MΩ + RG v v 1 gs g v m gs 3.541x10 vi = 4.051x10 v1 v1 = 0.874v i | Av = 0.874 Rin = vi v = −6 i = 7.94 MΩ ii 10 (vi − v1 ) v1 v + 1 + 3.54 x10−3 v1 6 10 2000 −3 −3 vi 2kΩ 100 k Ω Driving the output with current source ix : Rout : ix = Rout = v1 = 247 Ω | ix (b) R in =∞ 14.23 VEQ = −12 + 12 100 kΩ = −6.00V | REQ = 100 kΩ 100 kΩ = 50.0 kΩ 100 kΩ + 100 kΩ (−0.7 + 6)V IB = = 8.25µA | IC = 1.03 mA | VCE = 24 − 2000 IC − 4700 I E = 17.1 V 50.0 kΩ + ( 126)(4.7 kΩ) 125(0.025V ) 1.03mA 1.03mA RB = R1 R2 = 100 kΩ 100 kΩ = 50.0 kΩ | RL = R3 RE ro = 24 kΩ 4.7 kΩ 65.1kΩ = 3.71kΩ Rin = RB rπ + (β o + 1)RL = 50.0 kΩ 3.03kΩ + ( 126)3.71kΩ = 45.2 kΩ Av = + rπ = 3.03kΩ | ro = Active region is correct. | rπ = (50 + 17.1)V = 65.1kΩ [ (β + 1)R + (β + 1)R o L o ⎛ R ⎞ ⎛ ⎞ 126(3.71kΩ) 50.0 kΩ in ⎜ ⎟= ⎜ ⎟ = 0.984 R + R 500 Ω + 50.0 k Ω 3.03 k Ω + 126 3.71 k Ω ⎝ ⎠ ⎝ ⎠ ( ) I in L ] [ ] ⎞ ⎛ ⎤ ⎛ R ⎞⎛ ⎞⎡ rπ 3.03kΩ 50.0 kΩ in ⎟=⎜ ⎢ ⎥ = 6.34 x103 vi vbe = vi ⎜ ⎟⎜ ⎟ ⎜ ⎟ R + R 500 Ω + 50.0 k Ω r + β + 1 R 3.03 k Ω + 126 3.71 k Ω ⎠⎢ ⎝ I ( o ) L⎠ ⎝ ( )⎥ in ⎠⎝ π ⎣ ⎦ vi ≤ 0.005V = 0.784 V | R out = RE 6.34 x10−3 (R B RI + rπ ) βo + 1 = 4.7 kΩ (50.0kΩ 500Ω)+ 3.03kΩ = 27.8 Ω 126 14-14 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.24 VGS = 5V | I D = 2 5x10−4 5 − 1.5) = 3.06 mA | VDS = 5 − (−5)= 10V - Pinchoff region ( 2 operation is correct. | g m = 2 5x10−4 (3.06mA) 1 + 0.02( 10) = 1.92 mS 1 + 10 V = 19.6 kΩ − Cannot neglect! | RL = 19.6 kΩ 100 kΩ = 16.4 kΩ ro = 0.02 3.06 mA 1 Rin = RG = 1 MΩ | Rout = ro = 507 Ω gm ⎛ 1.92 mS ( 16.4 kΩ) ⎞ 1 MΩ Rin ⎛ g m RL ⎞ ⎟ = 0.960 ⎜ Av = + = + ⎜ ⎟ ⎟ 10 kΩ + 1 MΩ ⎜ RI + Rin ⎝ 1 + g m RL ⎠ 1 + 1.92 mS 16.4 k Ω ( ) ⎠ ⎝ ⎤ ⎛ ⎞⎡ ⎛ Rin ⎞⎛ 106 Ω 1 ⎞ 1 ⎥ = 0.0305v i ⎢ v gs = v i ⎜ ⎟ ⎟⎜ ⎟ = vi⎜ 4 6 16.4kΩ)⎥ ⎝ RI + Rin ⎠⎝ 1 + g m RL ⎠ ⎝ 10 Ω + 10 Ω ⎠⎢ ⎦ ⎣1 + 1.92 mS ( vi ≤ VDS = 23.0 V But, vDS must exceed vGS − VTN ≅ VGS − VTN = 4V for pinchoff. 0.0305 = 10 − vo = 10 − 0.970vi ≥ 4 → vi ≤ 6.19 V − Limited by the Q - point voltages 0.2(5 − 1.5) ( ) [ ] 14.25 IB = (9 − 0.7)V = 187 nA | I = 18.7 µA | V 1MΩ + ( 100 + 1)430 kΩ 100(0.025V ) Active region is correct. | r = = 134 kΩ | C CE = 18 − 430000 I E = 9.89 V π 18.7µA ro = (60 + 9.89)V = 3.74 MΩ - neglected 18.7µA In the ac model, R1 appears in parallel with rπ . The circuit appears to be using a transistor with rπ' = 500kΩ rπ = 106 kΩ ' and β o = g mrπ' = 40( 18.7µA) 106 kΩ = 79.0 ' + 1 RL = 106 kΩ + 79.0( 158 kΩ)= 12.6 MΩ RL = 500kΩ 430 kΩ 500 kΩ = 158kΩ | Rin = rπ' + β o (β + 1)R ⎛ R A = ⎜ r +( β + 1)R ⎝ R + R ' o L in v ( ) π ' ' o L I ⎞ ⎛ ⎞ 80.0( 158kΩ) 12.6 MΩ ⎟= ⎜ ⎟ = +0.992 106kΩ + 80.0( 158kΩ)⎝ 500Ω + 12.6 MΩ ⎠ in ⎠ Rout RI + rπ' 500Ω + 106 kΩ = RE R2 ' = 430 kΩ 500 kΩ = 1.34 kΩ 79.0 βo + 1 rπ' ' vbe = vi vi ≤ RI + rπ + β + 1 RL ' o ( ) = vi 106kΩ = 8.32 x10−3 vi 500Ω + 106 kΩ + 80.0( 158 kΩ) 0.005V = 0.601 V 8.32 x10−3 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-15 14.26 vi ≤ 0.005( 1 + g m RL ) | RL = RE R3 ≅ RE ⎛ I R ⎞ vi ≤ 0.005( 1 + g m RL ) = 0.005( 1 + g m RE ) = 0.005⎜1 + C E ⎟ VT ⎠ ⎝ ⎛ ⎛ I R ⎞ I R ⎞ vi ≤ 0.005⎜1 + α F E E ⎟ ≅ 0.005⎜1 + E E ⎟ VT ⎠ VT ⎠ ⎝ ⎝ ⎛ VR ⎞ ⎛ VR E ⎞ vi ≤ 0.005⎜1 + E ⎟ = 0.005⎜1 + ⎟ = 0.005 + 0.2VR E VT ⎠ ⎝ ⎝ 0.025 ⎠ 14.27 β o = g mrπ = 3.54 mS ( 1 MΩ) = 3540 | RL = 2 kΩ 100 kΩ = 1.96 kΩ Av = Rin (β + 1)R r + (β + 1)R = r + (β + 1)R o L π o L L π o (3540 + 1)(1.96kΩ) = 0.874 1 MΩ + (3540 + 1)( 1.96 kΩ) = 1 MΩ + (3540 + 1)( 1.96 kΩ)= 7.94 MΩ = Rout = 2 kΩ rπ 106 = 2 kΩ = 247 Ω (β o + 1) (3541) 14.28 (a) v be = v i − vo | 0.005 ≤ 5 − v o → Av = = (b) A v rπ (β + 1)R + (β + 1)R o E o vo 4.995 ≥ = 0.999 vi 5 = E 1 1 1 1 = = = βo rπ αo rπ V 1+ 1+ 1+ 1+ T g m RE I E RE (β o + 1)RE (β o + 1) βo RE 1 V 0.025V ≥ 0.999 → T ≤ 0.001 → I E RE ≥ = 25.0 V VT I E RE 0.001 1+ I E RE 14-16 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.29 1 − Av )vi | 0.005 ≤ ( 1 − Av )7.5 → Av = vbe = vi − vo = ( From Prob. 14.28, Av = vo 7.5 − 0.005 ≥ = 0.999333 vi 7.5 1 500 RE 1 | RL = RE 500Ω = | Av = ⎛ V 500 + RE 500 + RE ⎞ V 1+ T 1+ T ⎜ ⎟ I E RL I E RE ⎝ 500 ⎠ 1 V ⎛ 500 + RE ⎞ −4 ≥ 0.999333 → T ⎜ ⎟ ≤ 6.67 x10 ⎛ ⎞ I R 500 V 500 + RE ⎠ E E ⎝ 1+ T ⎜ ⎟ I E RE ⎝ 500 ⎠ 500 I E RE 0.025V ≥ = 37.5V | VCC ≥ I E RE + 0.7 + 7.5 500 + RE 6.67 x10−4 Some design possibilities are listed in the table below. RE 100 Ω 250 Ω 360 Ω 500 Ω 750 Ω 1000 Ω 2000 Ω IE 450 mA 225 mA 179mA 150 mA 125 mA 113mA 93.8 mA VCC 53 V 64 V 73V 83 V 102 V 120 V 196 V VCC IE 24 W 16 W 13 W 12 W 13 W 14 W 18 W Using a result near the minimum-power case in the table: RE = 510 Ω  E= 149 mA and VCC = 85 V. 149 mA For β F = 50 : I B ≅ = 2.92 mA | Set I R1 = 5 I B = 14.6 mA ≅ 15mA 51 V + VBE 149 mA(510Ω)+ 0.7 R1 = E = = 5.07 kΩ → 5.1 kΩ | I R 2 = I R1 + I B ≅ 18 mA I R1 15mA 85 − VBE − VBE 8.3V = = 462Ω → 470 Ω I R2 18 mA It is obviously very difficult to achieve the required level of linearity! R2 = © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-17 14.30 (a) g m = 40(12.5µA)= 0.5mS Rin = R4 Av = rπ α 0.99 = R4 o = 100 kΩ = 1.94 kΩ βo + 1 gm 0.5mS g m RL ⎛ R4 ⎞ ⎛ 100 kΩ ⎞ 0.5mS ( 100 kΩ) ⎜ ⎟= ⎜ ⎟ = 48.7 1 + g m RI R4 ⎝ RI + R4 ⎠ 1 + 0.5mS 50Ω 100kΩ ⎝ 50Ω + 100 kΩ ⎠ ( ) ( ) Rout = ro 1 + g m RI R4 = 60V 1 + 0.5mS (50Ω) = 4.92 MΩ 12.5µA ⎛ R4 ⎞ ⎛ 100 kΩ ⎞ Ai = α o ⎜ ⎟ = 0.990⎜ ⎟ = 0.990 ⎝ 50Ω + 100kΩ ⎠ ⎝ RI + R4 ⎠ v [ ( )] [ ] (b) A = ⎛ ⎞ 100kΩ ⎜ ⎟ = 23.6 | Rin = 1.94 kΩ - no change 1 + 0.5mS 2.2 kΩ 100kΩ ⎝ 2.2kΩ + 100kΩ ⎠ 0.5mS ( 100 kΩ) Rout = ro 1 + g m [ ) 60V (R R )]= 12.5 [1 + 0.5mS (2.2Ω)]= 10.1 MΩ µA I 4 ( 14.31 (a) R Av = in = R4 ⎛ R ⎞ 0.5mS (60 kΩ) ⎛ 3kΩ ⎞ 4 ⎜ ⎟= ⎜ ⎟ = +28.8 1 + g m RI R4 ⎝ RI + R4 ⎠ 1 + 0.5mS 50Ω 3kΩ ⎝ 50Ω + 3kΩ ⎠ g m RL 1 1 = 3kΩ = 1.20 kΩ | Rout = ∞ (assume λ = 0) gm 0.5mS ( ) ( ) Ai = 1 R4 R4 + = 1 gm = 3kΩ = 0.600 3kΩ + 2 kΩ (b) A v ⎛ 3kΩ ⎞ 1 = 1.43 kΩ ⎜ ⎟ = +5.81 | Rin = 5kΩ 0.5mS 1 + 0.5mS 5kΩ 3kΩ ⎝ 5kΩ + 3kΩ ⎠ 0.5mS (60 kΩ) ( ) Rout = ∞ | Ai = 5kΩ = 0.714 5kΩ + 2 kΩ 14.32 The voltage gain is approximately 0. The signal is injected into the collector and taken out of the emitter. This is not a useful amplifier circuit. 14-18 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.33 IB = (12 − 0.7)V = 2.64µA | 100kΩ + (50 + 1)82 kΩ βo gm IC = 132 µA VCE = 24 − 82000 I E − 39000 IC = 7.81 V | Active region operation is correct. g m = 40 IC = 5.28 mS | rπ = = 9.47kΩ | ro = 132µA (50 + 7.81)V = 438kΩ - neglected RI RE = 0.5kΩ 82kΩ = 497Ω | RL = RC R3 = 39kΩ 100kΩ = 28.1kΩ Av = ⎛ RE ⎞ 5.28 mS (28.1kΩ) ⎛ ⎞ 82 kΩ ⎜ ⎟= ⎜ ⎟ = 40.7 1 + g m RI RE ⎝ RI + RE ⎠ 1 + 5.28 mS (497Ω)⎝ 500Ω + 82 kΩ ⎠ g m RL ( ) Rin = 82 kΩ R + Rin rπ 500Ω + 185Ω = 185 Ω | Ai = Av I = 40.7 = 0.279 βo + 1 R3 100kΩ Rin ≤ 5.00 mV | 0.270vi ≤ 5.00 mV | vi ≤ 18.5 mV RI + Rin Rout = RC = 39.0 kΩ | veb = vi 14.34 IB = (9 − 0.7)V = 194 nA | 1000kΩ + (50 + 1)820 kΩ βo gm IC = 9.69 µA VCE = 18 − 820000 I E − 390000 IC = 6.12 V | Active region is correct. g m = 40 IC = 0.388 mS | rπ = = 129kΩ | ro = 9.69µA (50 + 6.12)V = 5.79 MΩ - neglected RI RE = 5kΩ 820kΩ = 4.97kΩ | RL = RC R3 = 390 kΩ 1 MΩ = 281kΩ Av = ⎛ RE ⎞ 0.388 mS (281kΩ) ⎛ 820 kΩ ⎞ ⎜ ⎟= ⎜ ⎟ = 37.0 1 + g m RI RE ⎝ RI + RE ⎠ 1 + 0.388 mS (4.97 kΩ)⎝ 5kΩ + 820 kΩ ⎠ g m RL ( ) Rin = 820 kΩ R + Rin rπ 5kΩ + 2.52kΩ = 2.52 kΩ | Ai = Av I = 37.0 = 0.278 βo + 1 R3 1 MΩ Rin ≤ 5.00mV | 0.335vi ≤ 5.00 mV | vi ≤ 14.9 mV RI + Rin Rout ≅ RC = 390 kΩ | veb = vi © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-19 14.35 VGS = −3900 I D −4 (5x10 )(V = −3900 −4 ID = gm = 2 2(254µA) 2 − 0.992 (5x10 )(V 2 GS + 2) | VGS = −0.975( VGS + 2) → VGS = −0.9915V 2 2 GS + 2) = 254µA | VDS = 16 − 23.9 kΩI D = 9.92V - Pinched off. 2 = 0.504 mS | Rin = 3.9 kΩ 1 = 1.32 kΩ | Rout ≅ RD = 20 kΩ gm RL = 20 kΩ 51kΩ = 14.4 kΩ Av = ⎛ RS ⎞ 0.504 mS ( 14.4 kΩ) ⎛ 3.9 kΩ ⎞ ⎜ ⎟= ⎜ ⎟ = 4.12 1 + g m RI RS ⎝ RI + RS ⎠ 1 + 0.504 mS (0.796 kΩ)⎝1kΩ + 3.9 kΩ ⎠ g m RL ( ) Ai = Av vgs = vi 1kΩ + 1.32 kΩ RI + Rin = 4.12 = 0.187 R3 51kΩ 1.32 kΩ 1.32 kΩ ≤ 0.2( VGS + 2) | v i ≤ 0.2(−0.992 + 2)→ vi ≤ 0.354 V 1kΩ + 1.32 kΩ 1kΩ + 1.32 kΩ −4 14.36 ID 2 15 + VGS ID = = 186µA | VDS = − 30 − (68 kΩ + 43kΩ)I D = −9.35V | 68 kΩ 2( 186µA) 1 Pinchoff region is correct. | g m = = 0.274 mS | Rin = 68 kΩ = 3.46 kΩ 2.36 − 1 gm (2 x10 )(V = GS + 1) | 2 2 15 + VGS = 10−4 ( VGS + 1) → VGS = −2.363V 68 kΩ [ ] Rout ≅ RD = 43kΩ | RL = 43kΩ 200 kΩ = 35.4 kΩ Rin 3.46 kΩ g m RL = (0.274mS )(35.4kΩ)= 9.05 RI + Rin 0.250 kΩ + 3.46 kΩ ⎛ 3.71kΩ ⎞ R + Rin 3.46 kΩ Ai = Av I = 9.05⎜ ≤ 0.2( VSG − 1) ⎟ = 0.168 | vgs = vi R3 0.250 kΩ + 3.46 kΩ ⎝ 200 kΩ ⎠ 3.46 kΩ vi ≤ 0.2(2.36 − 1)→ v i ≤ 0.292 V 0.250 kΩ + 3.46 kΩ Av = 14-20 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.37 VGS = −10 + (33kΩ)I D | VGS VGS = −2.507 V & I D = VDS (3.3x10 )(2 x10 )(V = −10 + 4 −4 2 = − 20 − I D (33kΩ + 24kΩ) = −7.06 V - Active region operation is correct. (2 x10 )(V −4 −4 2 GS + 1) 2 GS + 1) = 227µA 2 gm = [ ] 2( 2 x10 ) (2.27 x10 ) = 3.01x10 −4 −4 S | RI RS = 0.5kΩ 33kΩ = 493Ω Assume λ = 0, ro = ∞ | RL = RD R3 = 24 kΩ 100 kΩ = 19.4 kΩ Av = ⎛ RS ⎞ 0.301mS ( ⎞ 19.4kΩ) ⎛ 33kΩ ⎜ ⎟= ⎜ ⎟ = 5.01 1 + g m RI RS ⎝ RI + RS ⎠ 1 + 0.301mS (493Ω)⎝ 500Ω + 33kΩ ⎠ g m RL ( ) Ai = 1 ⎛ RD ⎞ ⎛ ⎞ 24 kΩ 33kΩ ⎜ ⎟= ⎜ ⎟ = 0.176 1 ⎝ RD + R3 ⎠ 33kΩ + 3.32kΩ ⎝ 24 kΩ + 100 kΩ ⎠ RS + gm RS 1 = 3.02 kΩ | Rout = RD = 24 kΩ gm RIN 3.02 kΩ ≤ 0.2 VGS + 1 | vi ≤ 0.2( 1.51)→ vi ≤ 0.352 V RI + RIN 0.5kΩ + 3.02 kΩ 1 , Av ≅ g m RL All of the input voltage appears across the gate - source gm Rin = RS vgs = vi 14.38 For Rth > , Av ≅ L For large Rth , all of the Thevenin equivalent source Rth gm current, vth , goes into the transistor source terminal. Rth 14.39 Rin = 14.40 75(0.025V ) rπ + 1.5kΩ 1.88kΩ + 1.5kΩ | rπ = = 1.88kΩ | Rin = = 44.5 Ω βo + 1 1mA 76 g m = 2( 1.25mA)( 1mA) = 1.58 mS | Rin = 1 = 633 Ω gm © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-21 14.41 ⎡ β o RE ⎤ 15V − 0.7V a | IE = R = r 1 + = 100µA | For β F = 100, IC = 99.0µA ⎢ ( ) out o⎣ r + R ⎥ 143kΩ π E⎦ 100(0.025V ) 99.0µA ⎡ 100( 143kΩ) ⎤ ⎢ ⎥ = 43.4 MΩ = 505kΩ 1 + 25.3 k Ω + 143 k Ω ⎢ ⎥ ⎣ ⎦ E rπ = Rout = 25.3kΩ | ro ≅ 50V = 505kΩ 99.0µA (b) 0 V 100(0.025V ) 15V − 0.7V = 953µA | For β F = 100, IC = 944µA | rπ = = 2.65kΩ 944µA 15kΩ ⎡ 100( 15kΩ) ⎤ 50V ⎥ = 4.56 MΩ | VCB ≥ 0 V = 53.0 kΩ | Rout = 53.0 kΩ ⎢1 + ro ≅ 944µA ⎢ ⎣ 15kΩ + 2.65kΩ⎥ ⎦ (c) I = 14.42 ⎛V + V ⎞ ⎛ 50 + 10.7 ⎞ Rout = (β o + 1)ro = (β o + 1) ⎜ A CE ⎟ = 126 ⎜ ⎟ = 154 MΩ ⎝ 49.6µA ⎠ ⎝ IC ⎠ 14.43 43 Rin = 350Ω | Av = 10 20 = 141 | Low Rin , large gain A common - base amplifier can achieve these speocfications. Rin ≅ 1 1 → IC ≅ = 71.4 µA gm 40(350) A common emitter amplifier operating at a higher current is an alternate choice. 100 Rin ≅ rπ → IC ≅ = 7.14 mA 40(350) For both cases, Av ≅ 10VCC → VCC = 14 V 14.44 Rin = 0.3 MΩ | Av = 10 46 20 = 200 | Fairly large Rin , large gain A common - emitter amplifier operating at a low current can achieve both a large gain and input resistance. Av ≅ 20VCC → VCC = 10V Achieving this gain with an FET is much more difficult : Av ≅ VDD V = DD → VDD ≅ 50V which is unreasonably large. VGS −V TN 0.25V 14-22 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.45 26 Rin = 10 MΩ | Av = 10 20 = 20 | Large Rin , moderate gain These requirements are readily met by a common - source amplifier. VDD 15V = = 30. For example, Av ≅ VGS −V TN 0.5V A common - emitter stage operating at a low collector current with ⎛ ⎞ 10 MΩ an unbypassed emitter resistor ⎜ R E ≅ = 100 kΩ⎟is a second possibility, 100 ⎝ ⎠ but the circuit will require careful design. 14.46 Rin = 50 kΩ | Av = 10 = 792 | A bipolar transistor would be required for such a large gain. However, this is a large fraction of the BJT amplification factor [i. e. (40/V)(75V) = 3000] and will be very difficult to achieve with the information thus far (the active load discussed later is a possibility). Using our rule - of - thumb for the common - emitter amplifier, Av ≅ 10VCC → VCC = 80 V which is too large. Thus, it is not possible is the best answer. 14.47 An inverting amplifier with a gain of 40 dB is most easily achieved with a common - emitter stage : Av ≅ 10VCC → VCC = 10 V . The input resistance can be achieved by shunting the = 0.5 A and would 5Ω waste a large amount of power to achieve the required input resistance. It would be better to operate the transistor at a much lower current and "swamp" the input resistance by shunting the input with a 5 - Ω resistor input with a 5 - Ω resistor. Setting rπ = 5 Ω would require IC ≅ 100(0.025V ) 58 20 14.48 A non - inverting amplifier with a gain of 20 and an input resistance of 5 kΩ should be readily achievable with either a common - base or common - gate amplifier with proper choice of operating point. The gain of 10 is easily achieved with either the VDD 1 or Av ≅ 10VCC . Rin ≅ = 5 kΩ is within FET or BJT design estimate : Av ≅ gm VGS − VTN easy reach of either device. The gain and input resistance can also be easily met with either a common - emitter, or common - source stage with a resistor shunt at the input. 14.49 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-23 0 - dB gain corresponds to a follower ( Av = 1). For an emitter - follower, R in ≅ (β o + 1)RL ≅ 101(20 kΩ) = 2.02 MΩ. So an BJT cannot meet the input resistance requirement. A source follower provides a gain of approximately 1 and can easily achieve the required input resistance. 14.50 A gain of 0.97 and an input resistance of 400kΩ should be achievable with g m RL = 0.97 either a source - follower or an emitter - follower. For the FET, Av ≅ 1 + g m RL requires g m RL = 33.3 : 2 I D RL = 33.3 → I D RL = 8.3V for a design with VGS − VTN = 0.5V . VGS − VTN can still meet the Rin requirement : Rin ≅ β o RL ≅ 100(5kΩ) = 500 kΩ. Av ≅ The BJT can achieve the required gain with a much lower power supply and g m RL = 0.97 | g m RL = 33.3 → IC RE = 33.3(0.025)= 0.833 V . 1 + g m RL The requirements can be met with careful bias circuit design and specification of a BJT with minimum current gain of at least 100. 14.51 66 of the BJTs : Av ≤ µ f = 40VA = 40(75)= 3000. Such a large gain Av = 10 20 = 2,000. This value of voltage gain approaches the amplification factor requirement cannot be met with single - transistor BJT amplifiers using the resistive loaded amplifiers in this chapter (Remember the 10VCC limit). FETs typically have much lower values of µf and are at an even worse disadvantage. None of the single - transistor amplifier configurations can meet the gain requirements. 14-24 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.52 Such a large output resistance will require either a CE or CB stage or a CS or CG stage. For a BJT, Rout ≤ β oro = 7.5µA using typical 109 Ω values for β o and VA . We need to also see how much voltage is required. IC = 109 Ω or IC = β oVA 100(75V ) 75 We also need ro ( 1 + g m RE )> 109 Ω or 40(IC RE ) > 109 Ω → IC RE > 2.5 V whihc is reasonable. 7.5µA For a MOSFET, Rout = ro ( 1 + g m RS ) ≅ g mro RS = RS = 109 (0.01/ V )(0.25V ) 2 = 1.25 MΩ 2 RS = 109 Ω λ( VGS − VTN ) For typical values, Using this value to estimate the required voltage, 2 2 2 Kn ⎛ RS ⎞ 2(0.001)⎛1.25 x106 ⎞ RS 9 g mro RS = 2 Kn I D = 10 Ω → I D = 18 ⎜ ⎟ = ⎜ ⎟ = 31.3µA and λI D 10 ⎝ λ ⎠ 1018 ⎝ 0.01 ⎠ VR S = 39 V which is getting large. So the BJT appears to be the best choice. 14.53 Rout = Rth + rπ βo + 1 | Assuming Rth ≅ RI and rπ = 0, Rout ≥ RI 250 = = 1.66 Ω β o + 1 151 14.54 Rin = rπ + (β o + 1)RE ≅ rπ + β o RE = rπ ( 1 + g m RE ) | rπ' = rπ ( 1 + g m RE ) ' gm = βo βo βo ic gm = ≅ = = vi rπ + (β o + 1)RE rπ + β o RE rπ ( 1 + g m RE ) 1 + g m RE ⎛ βR ⎞ ⎛ β R ⎞ = ro⎜1 + o E ⎟ ≅ ro⎜1 + o E ⎟ = ro ( 1 + g m RE ) for rπ >> RE r + R r ⎝ ⎠ ⎝ ⎠ π E π vi = 0 ⎛ ⎛ gm ⎞ gm ⎞ ' ' 1 + g m RE ) = β o | µ'f = g m ro = ⎜ 1 + g m RE ) = µ f ⎟rπ ( ⎟ro ( ⎝1 + g m RE ⎠ ⎝1 + g m RE ⎠ ro' = ic vc ' ' ' βo = gm rπ = ⎜ 14.55 *Problem 14.55 - Common-Emitter Amplifier 5mV VCC 6 0 DC 9 VEE 4 0 DC -9 VS 1 0 SIN(0 0.005 1K) C1 1 2 1U RB 2 0 10K RC 6 5 3.6K RE 3 4 2K C2 3 0 50U C3 5 7 1U © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-25 R3 7 0 10K Q1 5 2 3 NBJT .OP .TRAN 1U 5M .FOUR 1KHZ V(7) .MODEL NBJT NPN IS=1E-16 BF=100 VA=70 .PROBE V(7) .END *Problem 14.55 - Common-Emitter Amplifier 10mV VCC 6 0 DC 9 VEE 4 0 DC -9 VS 1 0 SIN(0 0.01 1K) C1 1 2 1U RB 2 0 10K RC 6 5 3.6K RE 3 4 2K C2 3 0 50U C3 5 7 1U R3 7 0 10K Q1 5 2 3 NBJT .OP .TRAN 1U 5M .FOUR 1KHZ V(7) .MODEL NBJT NPN IS=1E-16 BF=100 VA=70 .PROBE V(7) .END *Problem 14.55 - Common-Emitter Amplifier 15mV VCC 6 0 DC 9 VEE 4 0 DC -9 VS 1 0 SIN(0 0.015 1K) C1 1 2 1U RB 2 0 10K RC 6 5 3.6K RE 3 4 2K C2 3 0 50U C3 5 7 1U R3 7 0 10K Q1 5 2 3 NBJT .OP .TRAN 1U 5M .FOUR 1KHZ V(7) .MODEL NBJT NPN IS=1E-16 BF=100 VA=70 .PROBE V(7) .END Results: 1 kHz 2 kHz 3 kHz vi THD 14-26 © R. C. Jaeger & T. N. Blalock - February 20, 2007 5 mV 10 mV 15 mV 5.8 mV 12.4 mV 20.6 mV 0.335 mV (5.7%) 0.043 mV (0.74% 1.54 mV (12.5%) 0.258 mV (2.1%) 4.32 mV (21%) 1.18 mV (5.4%) 5.9% 12.8% 22% 14.56 *Problem 14.56 - Output Resistance VCC 2 0 DC 10 IB1 0 1 DC 10U Q1 2 1 0 NBJT IB2 0 3 DC 10U RE 4 0 10K Q2 2 3 4 NBJT .OP .DC VCC 10 20 .025 .MODEL NBJT NPN IS=1E-16 BF=60 VA=20 .PRINT DC IC(Q1) IC(Q2) .PROBE IC(Q1) IC(Q2) .END Results: A small value of Early voltage has been used deliberately to accentuate the results. Note that the transistors have significantly different values of F because of the collectoremitter voltage differences and low value of VA. NAME Q1 Q2 MODEL NBJT NBJT IB 1.00E-05 1.00E-05 IC 8.77E-04 6.72E-04 VBE 7.61E-01 7.61E-01 VBC -9.24E+00 -2.41E+00 VCE 1.00E+01 3.18E+00 BETADC 8.77E+01 6.72E+01 GM 3.39E-02 2.60E-02 RPI 2.59E+03 2.59E+03 RO 3.33E+04 3.33E+04 From SPICE : Rout1 = V (20 − 10) V = 43.5 kΩ = 34.1 kΩ | Rout 2 = (1.17 − 0.877) mA (903 − 673) µA (20 − 10) For circuit 1: Rout1 = ro1 = 33.3 kΩ ⎛ ⎞ β o RE For circuit 2 : Rout 2 = ro 2⎜1 + ⎟ + (Rth + rπ ) RE (See Eq. 14.28) ⎝ Rth + rπ + RE ⎠ But Rth = ∞ → Rout 2 = ro 2 + RE = 33.3kΩ + 10kΩ = 43.3 kΩ © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-27 14.57 ⎛ ⎛ 100(270Ω) ⎞ β o RI ⎞ ⎜ ⎟ = 384 kΩ Rth ≅ ro ⎜1 + ⎟ = 250 kΩ⎜1 + ⎟ 270 Ω + 50 k Ω ⎝ RI + rπ ⎠ ⎝ ⎠ vi RI + vth = isc Rth = αo gm α o Rth = vi 0.990 270Ω + 0.002 0.990(384 kΩ) → vth = 479v i 14.58 vth = Gm Rth = − gm ro ( 1 + g m RS ) vi = −µ f v i = −(0.5mS )(250 kΩ)vi = −125vi 1 + g m RS [ ] Rth = ro ( 1 + g m RS ) = ro + µ f RS = 250 kΩ + 125( 18 kΩ)= 2.50 MΩ 14.59 vth = v i RI + rπ + (β o + 1)ro (β o + 1)ro = vi 1 rπ RI + +1 (β o + 1)ro (βo + 1)ro = vi 1 βo g m RI + +1 (β o + 1)µ f (β o + 1)µ f ≅ vi Rth ≅ α RI + rπ R +r RI ro ≅ I π = + o βo + 1 βo + 1 βo + 1 gm 14.60 (a) g g12 = 21 = v2 v1 = i2 = 0 g m RE ≅1 1 + g m RE i1 i2 21 v1 = 0 (b) g ⎛ 1 ⎞ ie ⎛ 1 ⎞ R E 1 ⎛ g m RE ⎞ 1 =⎜ = ⎟ ≅⎜ ⎟ ⎜ ⎟≅ βo + 1 ⎝1 + g m RE ⎠ βo ⎝ β o + 1⎠ i 2 ⎝ β o + 1⎠ R + 1 E gm | g12 = 9.51x10−3 | g21 >> g12 = 0.960 14.61 (a) g g12 = 21 = v2 v1 | v2 = i2 = 0 g m RD ro ( 1 + g m RD ro ( ) ) v1 | g21 = g m RD ro ( 1 + g m RD ro ( ) ) ≅1 i1 i2 | i1 = v1 = 0 (b) g21 = 0.2 mS 50 kΩ 450 kΩ ( 1 + 0.2 mS 50 kΩ 450 kΩ ( ) ) = 0.947 | g12 = 0 | g21 >> g12 14-28 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.62 (a) g g12 = 21 = v2 v1 | v2 = g m RC ro v1 | g21 = g m RC ro i2 = 0 ( ) ( ) i1 i2 21 =− v1 = 0 RC ro + RC | R r r + RC g21 = gm C o o = g mro = µ f >> 1 g12 ro + RC RC (b) g = 3mS 18 kΩ 800 kΩ = 52.8 ( ) g12 = − 18 kΩ = 0.0220 18 kΩ + 800 kΩ 14.63 (a) g (b) g 14.64 21 = v2 v1 = + g m RD ro i2 = 0 ( )| g12 = i1 i2 =− v1 = 0 RD | g21 = −g mro g12 ≅ −µ f g12 RD + ro 100kΩ = −0.167 | g21 >> g12 100kΩ + 500kΩ i1 i2 21 = +0.5mS 100kΩ 500kΩ = 41.7 | g12 = − ( ) (a) g i1 ≅ − 21 = v2 v1 | v2 ≅ − i2 = 0 g m RC g R v1 | g21 = − m C 1 + g m RE 1 + g m RE | g12 = v1 = 0 ⎛ RE ⎞ ⎛ RE ⎞ ⎛ RE ⎞ v2 RC i2 RC ⎜ ⎟≅− ⎜ ⎟ | g12 = − ⎜ ⎟ ro ( 1 + g m RE )⎝ RE + rπ ⎠ ro ( 1 + g m RE )⎝ RE + rπ ⎠ ro ( 1 + g m RE )⎝ RE + rπ ⎠ =− 1 + 2 mS ( 12 kΩ) 2 mS ( 130 kΩ) = -10.4 | g12 = − ⎛ ⎞ 12 kΩ -3 ⎜ ⎟ = -1.01x10 1 MΩ 1 + 2 mS ( 12 kΩ) ⎝12 kΩ + 50 kΩ ⎠ (b) g 21 [ 130 kΩ ] g21 >> g12 14.65 (a) g g12 = 21 = v2 v1 | v2 ≅ − i2 = 0 g m RD g R v1 | g21 ≅ − m D 1 + g m RS 1 + g m RS g12 = 0 | = −9.75 g12 =0 g21 | g21 >> g12 i1 i2 | i1 = 0 | v1 = 0 (b) g 21 = − 1 + 0.75mS ( 12 kΩ) 0.75mS ( 130 kΩ) | g12 = 0 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-29 14.66 At the output node v o : g mvx = (g o + GL )vo − g ovx | vo = ix = g mvx + g o (vx − v o ) gm + go vx g o + GL ⎛ g + go ⎞ ⎛ GL − g m ⎞ i | x = g m + g o ⎜1 − m ⎟ = g m + g o⎜ ⎟ vx ⎝ g o + GL ⎠ ⎝ g o + GL ⎠ RL v 1 1 ⎛ RL ⎞ ro | Rin = x = ≅ ⎜1 + ⎟ 1 gm ⎝ ix g m ro ⎠ 1+ 1+ ⎛G − g ⎞ gm + go ix m = g m + g o⎜ L ⎟ = GL vx g o + GL ⎝ g o + GL ⎠ µf 14.67 IC = 100 5 − 0.7 4 10 + 101 10 ( ) 3 = 3.87 mA | g m = 40 IC = 0.155S | rπ = 100 = 645Ω gm RL = 1kΩ 20 kΩ = 952Ω | RE = 1kΩ 20 kΩ = 952Ω Av1 = − Av2 = 100(952Ω) β o RL =− = −0.984 rπ + (β o + 1)RE 645Ω + 101(952Ω) rπ (β + 1)R + (β + 1)R o E o = E 645Ω + 101(952Ω) 101(952Ω) = 0.993 The small - signal requirement limits the output signal to : 1 − 0.993) = 0.007vi | v i ≤ vbe = vi − vo2 = vi ( vo1 ≤ 0.984(0.714V ) = 0.703V 0.005 = 0.714V 0.007 1kΩ)= 1.13V and VB = −10 4 I B = −0.387V . We also need to check VCB : VC = 5 − 3.87 mA( The total collector - base voltage of the transistor is therefore : VCB = 1.52V − 0.984v i − vi . We require VCB ≥ 0 for forward - active region operation. Therefore : vi ≤ 0.766 V . The small - signal limit is the most restrictive. 14-30 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.68 ( a ) VEQ = −15 + 30 100 kΩ = 0V | REQ = 100 kΩ 100 kΩ = 50 kΩ 100 kΩ + 100 kΩ 0 − 0.7 − (−15) IB = = 22.3µA | IC = 2.78 mA | I E = 2.81 mA 50 kΩ + 126(4.7 kΩ) 125(0.025V ) VCE = 30 − 2000 IC − 4700 I E = 11.4 V rπ = = 1.12 kΩ | ro = 2.78 mA 2.78 mA RB = R1 R2 = 100 kΩ 100 kΩ = 50 kΩ | RL = R3 RE ro = 24 kΩ 4.7 kΩ 22.1kΩ = 3.34 kΩ (50 + 11.4)V = 22.1kΩ 126)3.34 kΩ = 44.7 kΩ Rin = RB rπ + (β o + 1)RL = 50 kΩ 1.12 kΩ + ( Av = + rπ [ (β + 1)R + (β + 1)R o L o βo + 1 126 (b) SPICE Results: Q-point: (2.81 mA, 11.1 V), Av = 0.984, Rin = 45.5 kΩ Rout = 13.0 Ω 14.69 Rout = RE ro (R B RI + rπ ) ⎛ R ⎞ ⎛ ⎞ 126(3.34 kΩ) 44.7 kΩ in ⎜ ⎟= ⎜ ⎟ = 0.984 1.12 kΩ + 126(3.34 kΩ)⎝ 0.600 kΩ + 44.7 kΩ ⎠ L ⎝ RI + Rin ⎠ = 4.7 kΩ 22.1kΩ ] [ ] (44.7kΩ 600Ω)+ 1.12kΩ = 13.5 Ω (10 − 0.7)V = 1.43 µA | I = 114 µA | V 1MΩ + (80 + 1)68 kΩ 81(0.025V ) Active region is correct. | r = = 17.8 kΩ IB = C CE = 20 − 39000 IC − 68000 I E = 7.71 V | ro = 75 + 7.71 = 726 kΩ 114µA π 114µA RL = 500kΩ 39kΩ 726 kΩ = 34.5kΩ | Rin = RB rπ = 1MΩ 17.8 kΩ = 17.5 kΩ ⎛ Rin ⎞ ⎛ ⎞ 17.5kΩ Av = −g m RL ⎜ ⎟ = 40(0.114 mA)(34.5kΩ) ⎜ ⎟ = −153 ⎝ 500Ω + 17.5kΩ ⎠ ⎝ RI + Rin ⎠ Rout = RC ro = 39 kΩ 726 kΩ = 37.0 kΩ ⎛ Rin ⎞ ⎛ ⎞ 17.5kΩ vbe = vi ⎜ ⎟ = vi ⎜ ⎟ = 0.972vi ⎝ 500Ω + 17.5kΩ ⎠ ⎝ RI + Rin ⎠ vi ≤ 0.005V = 5.14 mV 0.972 Av ≅ −10( VCC + VEE ) = −10(20)= −200. | A closer estimate is - 40VR C = -40(4.47) = −178 (b) SPICE Results: Q-point: (116 µA, 7.53 V), Av = -150, Rin = 19.6 kΩ, Rout = 37.0 kΩ © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-31 14.70 ( a ) VEQ = 12 62 kΩ = 9.07V | REQ = 62 kΩ 20 kΩ = 15.1kΩ 62 kΩ + 20 kΩ 12 − 0.7 − 9.07 IB = = 4.19µA | IC = 314 µA | I E = 319 µA 15.1kΩ + 76(6.8kΩ) VEC = 12 − 16000 IC − 6800 I E = 4.81 V ro = 75(0.025) 60 + 4.81 = 206 kΩ | rπ = = 5.97kΩ −6 314 x10−6 314 x10 RL = ro 16kΩ 100kΩ = 206kΩ 16 kΩ 100kΩ = 12.9kΩ Rin = 15.1kΩ 5.97 kΩ = 4.28 kΩ | Rout = ro 16 kΩ = 14.8 kΩ | g m = 40 IC = 12.6 mS ⎛ Rin ⎞ ⎛ 4.28 kΩ ⎞ Av = −g m RL ⎜ 12.9kΩ) 12.6 mS )( ⎟ = −( ⎜ ⎟ = −132 ⎝1kΩ + 4.28 kΩ ⎠ ⎝ RI + Rin ⎠ (b) SPICE Results: Q-point: (309 µA, 4.93 V), Av = -127, Rin = 4.65 kΩ Rout = 14.9 kΩ 14.71 (a) V 500 kΩ = 4.73V | REQ = R1 R2 = 500 kΩ 1.4 MΩ = 368kΩ 500 kΩ + 1.4 MΩ Assume Active Region Operation EQ = 18 4.73 = VGS = 27000 I D → 4.73 = 1 + VDS ro = 2ID + 27000 I D → I D = 113 µA 5 x10−4 = 18 − (27000 + 75000)I D = 6.47 V > 3.73 V - Active region is correct. 50 + 6.47 = 500 kΩ | g m = 2 500 x10−6 113x10−6 1 + 0.02(6.47) = 357µS −6 113 x10 RL = ro RD 470kΩ = 500kΩ 75kΩ 470kΩ = 57.3kΩ ( )( )[ ] Rin = R1 R2 = 500 kΩ 1.4 MΩ = 368 kΩ | Rout = ro 75kΩ = 65.2 kΩ ⎛ Rin ⎞ ⎛ 368kΩ ⎞ Av = −g m RL ⎜ ⎟ = −(0.357 mS )(57.3kΩ) ⎜ ⎟ = −20.4 ⎝ 1kΩ + 368 kΩ ⎠ ⎝ RI + Rin ⎠ (b) SPICE Results: Q-point: (115 µA, 6.30 V), Av = -20.5, Rin = 368 kΩ Rout = 65.1 kΩ 14-32 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.72 ID 2 10 + VGS = 232µA | VDS = −10 + 0.232 mA(24 kΩ)− 2.36 = −6.79V | Pinchoff region is correct. ID = 33kΩ (2.5x10 )(V = −4 GS + 1) | 2 2 10 + VGS = 1.25 x10−4 ( VGS + 1) → VGS = −2.358V 33kΩ g m = 2 2.5 x10−4 2.32 x10−4 1 + 0.02(6.79) = 0.363mS | RI RS = 0.5kΩ 33kΩ = 493Ω ro= 50 + 6.79 1 =245kΩ RL = ro RD R3 = 245kΩ 24 kΩ 100 kΩ = 17.9 kΩ | Rin = 33kΩ = 2.69 kΩ −4 gm 2.32 x10 g m RL ( )( )[ ] Rout = ro RD = 21.9 kΩ Av = ⎛ RS ⎞ ⎞ 0.363mS ( 17.9 kΩ) ⎛ 33kΩ ⎜ ⎟= ⎜ ⎟ = 5.42 1 + g m RI RS ⎝ RI + RS ⎠ 1 + 0.363mS (0.493kΩ)⎝ 500Ω + 33kΩ ⎠ ( ) (b) SPICE Results: Q-point: (234 µA, -6.67V), Av = +5.56, Rin = 2.69 kΩ, Rout= 18.1 kΩ 14.73 IC = 100 rπ = 5 − 0.7 = 12.5µA | VCE = 5 − IC (330 kΩ)− (−5)= 5.87V 500 kΩ + 500 kΩ + ( 101)330 kΩ 100 60 + 5.87 = 200 kΩ | ro = = 5.27 MΩ | RL = 500 kΩ 330 kΩ 500 kΩ ro = 139 kΩ 40( 12.5µA) 12.5 x10−6 ' = g mrπ' = 71.4 Absorb R1 into the transistor : rπ' = rπ R1 = 143kΩ | β o ' + 1 RL = 143kΩ + 71.4( 139 kΩ)= 10.1 MΩ Rin = rπ' + β o ( ) ' o Rout = 330 kΩ 500 kΩ ro L (R I RB + rπ' ' βo +1 ) = 1.97 kΩ 72.4( 139 kΩ) = 0.986 (β + 1)R A =+ (R R )+ r + (β + 1)R v I B π ' ' o = L 1kΩ + 143kΩ + 72.4( 139 kΩ) (b) SPICE Results: Q-point: (12.7 µA, 5.78 V), Av = +0.986, Rin = 10.7 MΩ, Rout= 2.00 kΩ © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-33 14.74 IB = (12 − 0.7)V = 2.69 µA | 100 kΩ + (51)82 kΩ IC = 135 µA | VCE = 24 − 39000 IC − 82000 I E = 7.58 V Active region is correct. 135µA) = 5.40 mS | rπ = g m = 40( 50(0.025V ) 135µA = 9.26 kΩ | ro = 50 + 7.58 = 427 kΩ 135µA 1 + g m Rth ) RC R3 = 1.57 MΩ 39 k Ω 100 kΩ = 27.6 kΩ Rth = RI RE = 0.5kΩ 82 kΩ = 497Ω | RL = ro ( Av = ⎛ RE ⎞ 5.40 mS (27.6 kΩ) ⎛ ⎞ 82 kΩ ⎜ ⎟= ⎜ ⎟ = 40.2 1 + g m RI RE ⎝ RI + RE ⎠ 1 + 5.40 mS (497Ω)⎝ 500Ω + 82 kΩ ⎠ g m RL ( ) Rin = 82 kΩ rπ = 181 Ω | Rout = ro ( 1 + g m Rth ) RC = 38.1 kΩ βo + 1 (b) SPICE Results: Q-point: (132 µA, 7.79V), Av = 39.0, Rin = 204 Ω Rout = 38.0 kΩ 14.75 VEQ = 18 2.2 MΩ = 9.00V | REQ = R1 R2 = 2.2 mΩ 2.2 MΩ = 1.10 MΩ 2.2 MΩ + 2.2 MΩ Assume Active Region Operation 2 ID + 110000 I D → I D = 67.5 µA 4 x10−4 = 18 − ( 110000 + 90000)I D = 4.50 V > 0.575 V - Active region is correct. 18 - 9 = 110000 I D − VGS → 9 = 1 + VDS ro = 50 + 4.50 = 807 kΩ | g m = 2 400 x10−6 67.5 x10−6 1 + 0.02(4.50) = 243µS −6 67.5x10 RL = ro RD 470kΩ = 807kΩ 90kΩ 470 kΩ = 69.7kΩ ( )( )[ ] Rin = R1 R2 = 2.2 MΩ 2.2 MΩ = 1.1 MΩ | Rout = ro 90kΩ = 81.0 kΩ ⎛ Rin ⎞ ⎛ 1.1MΩ ⎞ Av = −g m RL ⎜ ⎟ = −(0.243mS )(69.7kΩ) ⎜ ⎟ = −16.9 ⎝ 1kΩ + 1.1MΩ ⎠ ⎝ RI + Rin ⎠ (b) SPICE Results: Q-point: (66.7 µA, 4.47V), Av = -16.8, Rin = 1.10 MΩ Rout = 81.0 kΩ 14-34 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.76 (a ) Assume Active Region operation. V = −51000 I I = 2 (V + 2) → I V = 15 − (20000 + 51000)I = 12.7 V > 0.36 V - Active Region is correct. 2 GS D D GS DS D 5 x10−4 D = 32.2 µA 12.70) = 0.201mS | Rin = 51kΩ g m = 2 5 x10−4 32.2 x10−6 1 + 0.02( ro = Rout 50 + 12.7 = 1.95 MΩ | Rth = RI RS = 981Ω 32.2 x10−6 = ro ( 1 + g m Rth ) RD = 19.8 kΩ | RL = Rout 10 kΩ = 6.65kΩ g m RL ( )( )[ ] 1 = 4.53kΩ gm Av = ⎛ R ⎞ 0.201mS (6.65kΩ) ⎛ 51kΩ ⎞ S ⎜ ⎟= ⎜ ⎟ = 1.10 1 + g m RI RS ⎝ RI + RS ⎠ 1 + 0.201mS (0.981kΩ)⎝1kΩ + 51kΩ ⎠ ( ) (b) SPICE Results: Q-point: (32.9 µA, 12.7 V), Av = +1.10, Rin = 4.50 kΩ, Rout= 19.8 kΩ 14.77 The power supply should be +16 V. (a ) Assume Active Region operation. Since there is no negative feedback (RS = 0), we should include the effect of channel - length modulation. VGS = 0 ID = VDS 2 4 x10−4 −5) ( 1 + 0.02VDS ) and VDS = 16 − 1800 I D → I D = 5.59 mA ( 2 = 16 − 1800 I D = 5.93 V > 5 V - Active region is correct. g m = 2 4 x10−4 (5.59mA) 1 + 0.02(5.93) = 2.24 mS | ro = 50 + 5.93 = 10.0kΩ 5.59 x10−3 ⎛ RG ⎞ ⎛ 10 MΩ ⎞ Av = - g m RL ⎜ ⎟ = −(2.24 mS ) 10.0 kΩ 1.8 kΩ 36 kΩ ⎜ ⎟ = −3.27 ⎝ 10 MΩ + 5kΩ ⎠ ⎝ RI + RG ⎠ ( ) [ ] ( ) Rin = 10.0 MΩ | Rout = RD ro = 1.52 kΩ (b) SPICE Results: Q-point: (5.59 mA, -5.93V), Av = -3.27, Rin = 10.0 MΩ, Rout= 1.52 kΩ 14.78 IB = (10 − 0.7)V = 14.0µA | I = 1.12 mA | V = 20 − 7800 I = 11.3 V 33kΩ + (80 + 1)7.8 kΩ 80(0.025V ) (100 + 11.3)V = 99.4kΩ = 1.79 kΩ | r = Active region is correct. | r = C CE E 1.12 mA 1.12 mA RL = ro 1 MΩ = 95.1kΩ 1 MΩ = 86.8 kΩ | Rin = RB rπ = 33kΩ 1.79 kΩ = 1.70 kΩ | Rout = ro = 99.4 kΩ π o ⎛ Rin ⎞ 1.70 kΩ = −3390 1.12 mA)(86.8 kΩ) Av = - g m RL ⎜ ⎟ = −40( 250Ω + 1.70 kΩ ⎝ RI + Rin ⎠ (b) Results: Q-point: (1.12 mA, 11.2 V), Av = -3440, Rin = 1.93 kΩ Rout= 98.9 kΩ © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-35 14.79 Bias forces Active Region operation : VDS = 6 − (−6)= 12 V VGS = 6 V 2 2 Kn 4 x10−4 VGS − VTN ) ( 1 + λVDS ) = 6 − 1) 1 + 0.02( 12) = 6.20 mA ID = ( ( 2 2 [ ] 12) = 2.48 mS g m = 2 4 x10−4 6.2 x10−3 1 + 0.02( 1 + 12 V 0.02 = 10.0 kΩ − Cannot neglect! | RL = 10.0 kΩ 100 kΩ = 9.09 kΩ ro = 6.20 mA 1 Rin = RG = 2 MΩ | Rout = ro = 388 Ω gm ⎛ 2.48 mS (9.09 kΩ) ⎞ Rin ⎛ g m RL ⎞ 2 MΩ ⎜ ⎟ = 0.953 Av = + ⎜ ⎟=+ 10 kΩ + 2 MΩ ⎜ RI + Rin ⎝ 1 + g m RL ⎠ 1 + 2.48 mS 9.09 k Ω ( )⎟ ⎝ ⎠ (b) SPICE Results: Q-point: (6.20 mA, 12.0V), Av = 0.953, Rin = 2.00 MΩ, Rout= 388 Ω ( )( )[ ] 14.80 (a) V EQ = 15 500kΩ = 3.95 V | REQ = 500 kΩ 1.4 MΩ = 368 kΩ 1.4 MΩ + 500 kΩ Assume active region | 3.95 = VGS + 27000 I D = 1 + VDS 2ID + 27000 I D → I D = 85.1 µA 400 x10−6 = 15 − I D (75kΩ + 27 kΩ) = 6.32V | Active region operation is correct. g m = 2 400 x10−6 85.1x10−6 = 0.261 mS | ro = C1 ≥ 10 C2 ≥ 10 ( )( ) 50 + 6.32 = 662 kΩ | RG = R1 R2 = 368 kΩ 85.1µA 1 10 = | C1 ≥ 0.0108 µF → 0.01 µF 2πf (RI + RG ) 2π (400 Hz)( 1kΩ + 368 kΩ) 1 10 = | C2 ≥ 1.19 µF → 1.2 µF ⎛ ⎞ ⎛ ⎞ 1 1 2πf ⎜ RS ⎟ 2π (400 Hz) ⎜27 kΩ ⎟ 0.261mS ⎠ gm ⎠ ⎝ ⎝ 1 10 C3 ≥ 10 = | C3 ≥ 7.40 nF → 8200 pF 2π (400 Hz)(67.4 kΩ + 470 kΩ) 2πf RD ro + R3 [( ) ] 1 = (b) C 2 = ⎛ 1⎞ 2πf ⎜ RS ⎟ gm ⎠ ⎝ ⎛ ⎞ 1 2π (4000 Hz) ⎜27 kΩ ⎟ 0.261mS ⎠ ⎝ 1 = 0.0119 µF → 0.012 µF 14-36 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.81 (a) V 62 kΩ = 9.07V | REQ = 20 kΩ 62 kΩ = 15.1kΩ 20 kΩ + 62 kΩ (12 − 0.7 − 9.07)V = 7.16µA | I = 537 µA | V = 12 − 3900 I − 8200 I = 5.47 V IB = C EC E C 15.1kΩ + (75 + 1)3.9 kΩ EQ = 12 Active region is correct. rπ = 75(0.025V ) 537µA = 3.49 kΩ | ro = 60 + 5.47 = 122kΩ | g m = 40 IC = 21.5 mS 537µA Rin = R1 R2 rπ = 15.1kΩ 3.49 kΩ = 2.83 kΩ | Rout = ro RC = 122 kΩ 8.2 kΩ = 7.68 kΩ C1 ≥ 10 C2 ≥ 10 1 10 = | C1 ≥ 4.16 µF → 0.01 µF 2πf (RI + Rin ) 2π ( 100 Hz)( 1kΩ + 2.83kΩ) 1 = 10 | C2 ≥ 273 µF → 270 µF ⎛R R +r ⎞ ⎛1kΩ 15.1kΩ + 3.49 kΩ ⎞ I EQ π ⎟ ⎟ ⎜ 2πf ⎜ 100 Hz) ⎟ ⎜ β + 1 ⎟ 2π ( ⎜ 76 o ⎠ ⎠ ⎝ ⎝ 1 10 = | C3 ≥ 0.148 µF → 0.15 µF C3 ≥ 10 2πf [Rout + R3 ] 2π ( 100 Hz)(7.68kΩ + 100kΩ) (b) C 2 = ⎛R R +r ⎞ I EQ π ⎟ 2πf ⎜ ⎜ β +1 ⎟ o ⎝ ⎠ 5 − 0.7 1 = ⎛ 1kΩ 15.1kΩ + 3.49 kΩ ⎞ ⎜ ⎟ 2π ( 1000 Hz) ⎜ ⎟ 76 ⎝ ⎠ 1 = 2.73 µF → 2.7 µF 14.82 IC = 100 10 4 + 101 103 ( ) = 3.87 mA | g m = 40 IC = 0.155S | rπ = 100 = 645Ω | ro = ∞ gm The capacitors will be negligible at a frequency 10 times the individual break frequencies : Req1 = 10 kΩ rπ + (β o + 1) 1kΩ 20 kΩ = 9.06 kΩ | f1 = [ ( )] 10 = 87.8 Hz 2π (9.06 kΩ)2µF ⎡ 10 rπ ⎤ ⎥ = 20.0 kΩ | f2 = = 7.96 Hz Req 2 = 20 kΩ + ⎢1kΩ β + 1 π 20.0 k Ω 10 µ F 2 ⎢ ⎥ ( ) ( ) o ⎣ ⎦ Req 3 = Rout + 20 kΩ = 21.0 kΩ | f 3 = 10 = 7.56 Hz 10µF 2π (21.0 kΩ) © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-37 14.83 (a) I 80 ⎛ 5 − 0.7 ⎞ 80 = 6.26 kΩ | ro = ∞ = 319µA | g m = 40 IC = 12.8 mS | rπ = ⎜ 3⎟ 81 ⎝ 13.3 x10 ⎠ gm ⎡ rπ ⎤ 10 ⎥ = 152Ω | C1 = = 0.209 µF → 0.20 µF Req1 = 75Ω + ⎢13.3kΩ β + 1 π 50 kHz 152 Ω 2 ⎢ ⎥ ( ) ( ) o ⎣ ⎦ C = αF IE = Req 2 = Rout + 100 kΩ = 8.25kΩ + 100kΩ = 108kΩ | C2 = ⎟ = 105 µF → 100 µF (b) C = 0.209µF⎜ ⎝ 100 Hz ⎠ 1 10 = 295 pF → 270 pF 108 kΩ 2π (50kHz) ⎛ 50kHz ⎞ ⎛ 50 kHz ⎞ | C2 = 295 pF ⎜ ⎟ = 0.148 µF → 0.15 µF ⎝ 100 Hz ⎠ 14.84 VEQ = −15 + 15 1.85mA)= 7.40mS VCE = 30 − 4700 I E = 21.2 V | g m = 40( Active region is correct. | rπ = 100(0.025V ) 51kΩ = −4.87V | REQ = 51kΩ 100kΩ = 33.8kΩ 51kΩ + 100 kΩ −4.87 − 0.7 − (−15) V = 18.5µA | IC = 1.85 mA Assume active region operation | I B = 33.8 kΩ + ( 101)(4.7 kΩ) [ ] 1.85mA 1.85mA RB = R1 R2 = 51kΩ 100kΩ = 33.8kΩ | RL = R3 RE ro = 24 kΩ 4.7 kΩ 38.5kΩ = 3.57 kΩ = 1.35kΩ | ro = (50 + 21.2)V = 38.5kΩ Rin = RB rπ + (β o + 1)RL = 33.8kΩ 1.35kΩ + ( 101)3.57kΩ = 30.9 kΩ Rout C1 = C2 = B [ (R =R E RI + rπ ) ] βo + 1 = 4.7 kΩ (33.8kΩ 500Ω)+ 1.35kΩ = 18.2 Ω 101 [ ] 10 10 = = 1.01µF → 1.0 µF 2πf (RI + Rin ) 2π (50 Hz)(500Ω + 30.9 kΩ) 10 10 = = 1.33µF → 1.5 µF 2πf (R3 + Rout ) 2π (50 Hz)(24kΩ + 18.2Ω) 14-38 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.85 Note : RS ≡ R1 = 3.9 kΩ | Assume active region operation. VGS = −3900 I D −4 (5x10 )(V = −3900 −4 ID = gm = 2 2(254µA) 2 − 0.992 (5x10 )(V 2 GS + 2) | VGS = −0.975( VGS + 2) → VGS = −0.9915V 2 2 GS + 2) = 254µA | VDS = 15 − 23.9 kΩI D = 8.92V - Active region is correct. 2 = 0.504 mS | ro = Rout = RD ro 1 + g m RS RI C1 = C2 = [ ( 1 + 0.504 mS ( 3.9 kΩ 1kΩ) )]= 20kΩ 232kΩ[ ]= 18.8kΩ 50 + 8.92 1 = 232 kΩ | Rin = 3.9 kΩ = 1.32 kΩ −6 gm 254 x10 10 10 = = 1.72µF → 1.8 µF 1kΩ + 1.32 kΩ) 2πf (RI + Rin ) 2π (400 Hz)( 10 10 = = 0.0335µF → 0.033 µF 2πf (R3 + Rout ) 2π (400 Hz)( 100 kΩ + 18.8kΩ) 14.86 (a) Use C2 to set the lower cutoff frequency to 1 kHz. C1 and C3 remain negligible at 1 kHz. C2 = 0.056 µF , C1 = 1800 pF , C3 = 0.015 µF (b) SPICE Results: fL = 925 Hz 14.87 (a ) Use C3 to set the lower cutoff frequency to 2 kHz. C1 remains negligible at 2 kHz. C1 = 8200 pF , C3 = 820 pF 1 (b) Use C to set the lower cutoff frequency to 1 kHz. C 2 and C3 remain negligible at 1 kHz. C1 = 0.042 µF , C2 = 1800 pF , C3 = 0.015 µF (c) SPICE Results: For (a) fL = 1.96 kHz. For (b) fL = 1.02 kHz 14.88 Av = g m RL 2ID ≥ 0.95 → g m RL ≥ 19 | g m = 2 Kn I D | VGS − VTN = = 0.5V 1 + g m RL Kn 2 ID (0.5) (0.03) = 3.75mA = 2 | RL ≥ 19 2(0.03)(0.00375) = 1.27 kΩ RL = RS 3kΩ → RS ≥ 2.19 kΩ | VSS = VGS + (3.75mA)RS | Possible designs : 2.4 kΩ, 11.5V ; 2.7 kΩ, 12.6V ; 3.0 kΩ, 13.75V - Making a choice which uses a nearly minimum value of supply voltage gives : VSS = 12 V , RS = 2.4 kΩ. © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-39 14.89 For a common - emitter amplifier with RE = 0, Rin ≅ rπ = 14.90 Using Eqn. 14.73 : 50 = 50 = g m RL β oVT IC | IC = 100(0.025V ) 75Ω = 33.3 mA 2(50) g R Rin | Assuming RI = 50Ω, 50 = m L → RL = = 5.03kΩ RI + Rin 2 40(497µA) 1 RE 4.3V → RE = 8.65kΩ | IC ≅ = 497µA RE 1 + 40(4.3) RL = RC 100 kΩ → RC = 5.30 kΩ | VEC = 5 + 0.7 − IC RC = 3.07V - Active region is ok. C1 >> C2 >> 2π (500 kHz)(50Ω + 50Ω) = 3.18 nF → C1 = 0.033 µF 1 = 3.02 pF → C2 = 33 pF 2π (500 kHz)( 105kΩ) 14.91 The base voltage should remain half way between the positive and negative power supply voltages. If VEE = +10V and VCC = 0V, then VB should = 5 V which can be obtained using a resistive voltage divider from the +10V supply. We now have the standard four-resistor bias circuit. The base current is 327 µA/80 = 4.08 µA. C1 75 Ω v 13 k Ω S C2 + RE 120 k Ω 8.2 k Ω 110 k Ω R2 RC 100 k Ω v O - + 10 V R1 CB Setting the current in R1 to 10I B = 40µA, R1 = The current in R2 = 11I B = 44µA, and R 2 = 5V = 125kΩ → 120kΩ. 40µA 5V = 114kΩ → 110kΩ. 44µA Note that the base terminal must now be bypassed with a capacitor. 14-40 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.92 Using Eqn. 14.73 : 75 = 50 = g m RL RE 8.3V → RE = 25.0 kΩ | IC ≅ = 332µA RE 1 + 40(8.3) 2(50) g R Rin | 50 = m L → RL = = 7.52 kΩ | RL = RC 100 kΩ → RC = 8.13kΩ RI + Rin 2 40(332µA) 1 = 2.12 nF → C2 = 0.022 µF VEC = 9 + 0.7 − IC RC = 6.98V - Active region is ok. C1 >> C2 >> 2π (500 kHz)(75Ω + 75Ω) 1 = 2.95 pF → C2 = 30 pF 2π (500 kHz)( 108 kΩ) 14.93 Rin ≅ 1 0.01 → g m = 2 Kn I D = 0.1S | I D = | gm 2 Kn = (a) I D = 0.01 =1 A 2(0.005) (b) I D 0.01 = 10.0 mA | The second FET achieves the desired input resistance 2(0.5) at much lower current and hence much lower power for a given supply voltage. 14.94 1 VT kT = = ∝T g m IC qIC At − 40 o C = 233 K , ⎛ 233k ⎞ ⎛ 323 K ⎞ 1 1 o = 50Ω⎜ = 50Ω⎜ ⎟ = 38.8 Ω. | At + 50 C = 323 K , ⎟ = 53.8 Ω. gm gm ⎝ 300 K ⎠ ⎝ 300 K ⎠ 300 k 6 k = . 50 q q Another approach : At 27 o C = 300 K , IC = At − 40 o C = 233 K , 1 233 1 323 = = 38.8 Ω. | At + 50 o C = 323 K , = = 53.8 Ω. gm 6 gm 6 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-41 14.95 This analysis assumes that the source and load resistors are fixed, and that only ⎛ Rin ⎞ the amplifier parameters are changing. Av = g m RL ⎜ ⎟ ⎝ RI + Rin ⎠ 1 g R 1 ≅ | Av ≅ m L Since RE >> 75Ω, RE RI ≅ 75Ω and Rin ≅ RE 1 + g m RI gm gm max max To achieve Avmax , RL → RL , gm → gm which requires max IC → IC = 0.988 (5.25 − 0.7)V = 364µA 13kΩ(0.95) max | RL = 8.2kΩ( 1.05)100 kΩ = 7.93kΩ g m = 40(364µA) = 14.6 mS | Avmax = 14.6 mS (7.93kΩ) 1 + 0.0146(75) = 55.3 min min which requires To achieve A min v , R L → RL , g m → g m min IC → IC = 0.988 (4.75 − 0.7)V = 293µA 13kΩ( 1.05) min | RL = 8.2 kΩ(0.95)100kΩ = 7.23kΩ g m = 40(293µA) = 11.7 mS | Avmin = 11.7 mS (7.23kΩ) 1 + 0.0117(75) = 45.1 | 45.1 ≤ Av ≤ 55.3 The range is only slightly larger than that observed in the Monte Carlo analysis in Table 14.15. 14.96 *Problem 14.96 - Common-Base Amplifier - Monte Carlo Analysis *Generate Voltage Sources with 5% Tolerances IEE 0 8 DC 5 REE 8 0 RTOL 1 EEE 6 0 8 0 1 * ICC 0 9 DC 5 RCC 9 0 RTOL 1 ECC 7 0 9 0 -1 * VS 1 0 AC 1 RS 1 2 75 C1 2 3 47U RE 3 6 RTOL 13K Q1 4 0 3 PBJT RC 4 7 RTOL 8.2K C2 4 5 4.7U R3 5 0 100K .OP .AC LIN 1 10KHZ 10KHZ .PRINT AC VM(5) VP(5) .MODEL PBJT PNP (BF=80 DEV 25%) (VA = 60 DEV 33.33%) 14-42 © R. C. Jaeger & T. N. Blalock - February 20, 2007 .MODEL RTOL RES (R=1 DEV 5%) .MC 1000 AC VM(5) YMAX .END Results: Mean value Av = 47.5; 3σ limits: 42.5 ≤ Av ≤ 52.5. However, the worst-case values observed in the analysis are Avmin = 43.2 and Avmax = 51.9. The mean is 5% lower than the design value. The width of the distribution is approximately the same as that in Table 14.15. 14.97 (a ) This analysis assumes that the source and load resistors are fixed, and that only the amplifier parameters are changing. Av = Rth = RE RI and RE = RI + RE g m RL ⎛ RE ⎞ ⎜ ⎟ 1 + g m Rth ⎝ RI + RE ⎠ 1 where RE = 13.3kΩ and RI = 75Ω RI 1+ RE Since R E >> RI , Rth and RE max max are essentially constant. To achieve Avmax , RL → RL , gm → gm RI + RE max which requires IC → IC = 0.988 g m = 40(330µA) = 13.2 mS | Avmax (5.10 − 0.7)V = 330µA | R = 8.25kΩ 1.01 100kΩ = 7.69kΩ ( ) 13.3kΩ(0.99) 13.2 mS (7.69kΩ)⎡ 13.3(0.99) ⎤ ⎢ ⎥ = 50.7 = 1 + 0.0132(75) ⎢ ⎣ 75 + 13.3(0.99)⎥ ⎦ max L min | RL = 8.25kΩ(0.99)100kΩ = 7.55kΩ min min , gm → gm which requires To achieve Avmin , RL → RL min IC → IC = 0.988 (4.90 − 0.7)V = 309µA 13.3kΩ( 1.01) 1.01) ⎤ 12.4 mS (7.55kΩ)⎡ 13.3( ⎢ ⎥ = 48.2 1 + 0.0124(75) ⎢ 1.01)⎥ ⎣ 75 + 13.3( ⎦ (b) Using a Spreadsheet similar to Table 14.16: Mean value Av = 49.6; 3σ limits: 48.2 ≤ Av ≤ max 50.9. The worst-case values observed in the analysis are Avmin = 48.4 and Av = 50.8. g m = 40(309µA) = 12.4 mS | Avmax = 14.98 *Problem 14.98 - Common-Base Amplifier - Monte Carlo Analysis *Generate Voltage Sources with 2% Tolerances IEE 0 8 DC 5 REE 8 0 RTOL 1 EEE 6 0 8 0 1 * ICC 0 9 DC 5 RCC 9 0 RTOL 1 ECC 7 0 9 0 -1 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-43 * VS 1 0 AC 1 RS 1 2 75 C1 2 3 100U RE 3 6 RR 13.3K Q1 4 0 3 PBJT RC 4 7 RR 8.25K C2 4 5 1U R3 5 0 100K .OP .AC LIN 1 10KHZ 10KHZ .PRINT AC VM(5) VP(5) IM(VS) IP(VS) .MODEL PBJT PNP (BF=80 DEV 25%) (VA = 60 DEV 33.33%) .MODEL RTOL RES (R=1 DEV 2%) .MODEL RR RES (R=1 DEV 1%) .MC 1000 AC VM(5) YMAX *.MC 1000 AC IM(VS) YMAX .END Results: Mean value: Av = 47.2; 3σ limits: 45.7 ≤ Av ≤ 48.5 Mean value: Rin = 83.4 Ω; 3σ limits: 79.5 Ω ≤ Av ≤ 87.6 Ω 14.99 Rin = RE RE 1 RE = = = g m 1 + g m RE 1 + 40 IC RE RE RE = 80 1 + 39.5 I E RE 1 + 40 I E RE 81 I E RE = 2.5 − 0.7 = 1.8V | 75 = IC = RE → RE = 5.41kΩ 1 + 39.5( 1.8) 80 1.8V = 329µA | Rth = 75Ω 5.41kΩ = 74.0Ω | g m = 40(329µA)= 13.2 mS 81 5.41kΩ g m RL ⎛ RE ⎞ Av = ⎜ ⎟ → RL = 7.59 kΩ 1 + g m Rth ⎝ RI + RE ⎠ 7.59 kΩ = RC 100 kΩ → RC = 8.21kΩ | VC = −2.5V + IC RC = +0.201V | Oops! We are violating our definition of the forward - active region. If we use the nearest 5% values, RE = 5.6 kΩ and RC = 8.2 kΩ, IC = 318µA and VC = +0.108V . The transistor is just entering saturation. 14-44 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.100 Using a Spreadsheet similar to Table 14.15: Mean value: Av = 0.960; 3σ limits: 0.942 ≤ Av ≤ 0.979 Mean value: ID = 4.91 mA; 3σ limits: 4.27 mA ≤ ID ≤ 5.55 mA Mean value: VDS = 7.03 V; 3σ limits: 4.52 V ≤ VDS ≤ 9.54 V *Problem 14.102 - Common-Drain Amplifier - Fig. 14.34 *Generate Voltage Sources with 5% Tolerances IDD 0 7 DC 5 RDD 7 0 RTOL 1 EDD 5 0 7 0 1 * ISS 0 8 DC 20 RSS 8 0 RTOL 1 ESS 6 0 8 0 -1 * VGG 1 0 DC 0 AC 1 C1 1 2 4.7U RG 2 0 RTOL 22MEG RS 3 6 RTOL 3.6K C2 3 4 68U R3 4 0 3K M1 5 2 3 3 NMOSFET .OP .AC LIN 1 10KHZ 10KHZ .DC VGG 0 0 1 .MODEL NMOSFET NMOS (VTO=1.5 DEV 33.33%) (KP=20M DEV 50%) LAMBDA=0.02 .MODEL RTOL RES (R=1 DEV 5%) .PRINT AC VM(4) VP(4) IM(VGG) IP(VGG) .MC 1000 DC ID(M1) YMAX *.MC 1000 DC VDS(M1) YMAX *.MC 1000 AC IM(VGG) YMAX *.MC 1000 AC VM(4) YMAX .END Results: Mean value: ID = 4.97 mA; 3σ limits: 4.32 mA ≤ ID ≤ 5.62 mA Mean value: VDS = 7.19 V; 3σ limits: 6.18 V ≤ VDS ≤ 8.20 V Mean value: Rin = 22.0 MΩ; 3σ limits: 20.3 Ω ≤ Rin ≤ 24.0 Ω Mean value: Av = 0.956; 3σ limits: 0.936 ≤ Av ≤ 0.976 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-45 14.101 ac equivalent circuit Rin3 Rin2 RIN RI Q Q 3 R 2 OUT M1 10 k Ω R vi 1 MΩ G R C2 R R B3 D1 R B2 4.7 k Ω RE2 4.7 k Ω 51.8 k Ω R R E3 L RS1 200 Ω 620 Ω 17.2 k Ω 3.3 k Ω 250 Ω The Q - points and small - signal parameter values have already been found in the text. ⎛ ⎞ Rin The bypass capacitors do not affect Rin : Rin = RG = 1 MΩ | Av = ⎜ ⎟ Avt1 Avt 2 Avt 3 ⎝ 10kΩ + Rin ⎠ RL1 = 620Ω 17.2 kΩ rπ 2 + (β o2 + 1) 1.6kΩ = 598Ω 2.39 kΩ + ( 151) 1.6 kΩ = 597Ω Avt1 = − 0.01(597) g m1 RL1 =− = −1.99 | RL2 = 3.54 kΩ (Eq. 15.7) 1 + g m1 RS 1 1 + (0.01)200 150(3.54 kΩ) β o2 RL2 =− = −2.18 rπ 2 + (β o2 + 1) 1.6 kΩ 2.39 kΩ + ( 151) 1.6kΩ ( )( ) ( ) Avt 2 = − ⎛ ⎞ 1MΩ Avt 3 = +0.950 | Av = −⎜ 1.99)(−2.18)(0.950)= +4.08 ⎟( ⎝ 10 kΩ + 1MΩ ⎠ ⎛R +r ⎞ ⎛ ⎞ β o2 RE 2 Rout = (3300Ω) ⎜ th3 π 3 ⎟ | Rth3 = RI 3 Ro2 = RI 3 ro2 ⎜1 + ⎟ ≅ RI 3 = 4.31kΩ ⎝ β o3 + 1 ⎠ ⎝ Rth2 + rπ 2 + RE 2 ⎠ ⎛ 4.31 + 1.00 ⎞ Rout = (3.30 kΩ) ⎜ kΩ⎟ = 64.3Ω 81 ⎝ ⎠ 14-46 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.102 M1 : ID = ID = 0.01 2 (VGS + 2) | VGS = −9000ID → VGS = −1.80V 2 0.01 2 (−1.8 + 2) = 200 µA | VDS = 15 − (15kΩ + 9 kΩ)ID = 10.2 V 2 (50 + 10.2)V = 301kΩ gm = 2( 10mA /V 2 )(0.2mA) = 2 mS | ro = 0.2mA Q2 : VEQ 2 = 15 43kΩ = 3.18V | REQ 2 = 160 kΩ 43kΩ = 33.9 kΩ 43kΩ + 160 kΩ ⎛ ⎞ 3.18 - 0.7 151 IC 2 = 150 =1.35 mA | VCE 2 = 15 − ⎜ 4.7 kΩ + 1.6 kΩ⎟IC = 6.49 V ⎝ ⎠ 33.9 kΩ +151(1.6 kΩ) 150 gm 2 = 40(1.35 mA) = 54.0 mS | rπ 2 = 150 (80 + 6.49)V = 64.1kΩ = 2.78 kΩ | ro2 = 54.0mS 1.35 mA Q3 : VEQ 3 = 15 120 kΩ = 8.53V | REQ 3 = 120 kΩ 91kΩ = 51.8 kΩ 120 kΩ + 91kΩ ⎞ ⎛ 81 8.53 - 0.7 IC 3 = 80 = 2.72 mA | VCE 3 = 15 − ⎜ 2.2 kΩ⎟IC = 8.93 V ⎝ 80 ⎠ 51.8 kΩ + 81(2.2 kΩ) gm 3 = 40(2.72 mA) = 109 mS | ro 3 = ⎛ ⎞ RG Av = ⎜ ⎟ Avt1 Avt 2 Avt 3 ⎝10 kΩ + RG ⎠ (60 + 8.93)V 2.72mA = 25.3kΩ | rπ 3 = 80 = 734Ω 109 mS Avt1 = −(2mS ) 301kΩ 15 kΩ 33.9 kΩ 2.78 kΩ = −4.36 ⎤ ⎡ Avt 2 = (−54.0mS )⎢64.1kΩ 4.7 kΩ 51.8 kΩ 734 + 81(2.2kΩ 250Ω) ⎥ = −180 ⎦ ⎣ ( ) ( ) Avt 3 = ⎛ ⎞ 1MΩ = 0.961 | Av = −4.36(−180)(0.961)⎜ ⎟ = 747 ⎝10 kΩ + 1MΩ ⎠ 734 + 81(2.2kΩ 250Ω) 0.005 = 165µV 0.99(4.36)(180)(1 − 0.961) 81(2.2kΩ 250Ω) v be 3 = v b 3 (1 − Avt 3 ) = 0.99 Avt1 Avt 2v s (1 − Avt 3 ) ≤ 5 mV | v s ≤ The gain is actually reduced rather than improved. The signal range increased since the gain was reduced. © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-47 14.103 *Problem 14.102/14.103 - Multistage Amplifier – Figure P14.102 VCC 12 0 DC 15 VI 1 0 AC 1 *For output resistance *VI 1 0 AC 0 *VO 11 0 AC 1 RS 1 2 10K C1 2 3 22U RG 3 0 1MEG M1 5 3 4 4 NMOSFET RS1 4 0 9K C2 4 0 22U RD 12 5 15K C3 5 6 22U R1 12 6 160K R2 6 0 43K Q2 8 6 7 NBJT1 RC 12 8 4.7K RE2 7 0 1.6K C4 7 0 22U C5 8 9 22U R3 12 9 91K R4 9 0 120K Q3 12 9 10 NBJT2 RE3 10 0 2.2K C6 10 11 22U RL 11 0 250 .MODEL NMOSFET NMOS VTO=-2 KP=.01 LAMBDA=0.02 .MODEL NBJT1 NPN IS=1E-16 BF=150 VA=80 .MODEL NBJT2 NPN IS=1E-16 BF=80 VA=60 .OPTIONS TNOM=17.2 .OP .AC LIN 1 2KHZ 2KHZ .PRINT AC VM(3) VP(3) IM(VI) IP(VI) VM(11) VP(11) IM(C6) IP(C6) .END VM (3) 1 Results : Av = VM (11) = +879 | Rin = = 1.00 MΩ | Rout = = 51.8 Ω IM (VI) IM (C 6) 14-48 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.104 ⎛ ⎞ Rin The bypass capacitors do not affect Rin : Rin = RG = 1 MΩ | Av = ⎜ ⎟ Avt1 Avt 2 Avt 3 ⎝ 10kΩ + Rin ⎠ RL1 = ( 15kΩ 160kΩ 43kΩ) (rπ 2 + (β o 2 + 1)1.6kΩ) = 10.4 kΩ [2.78 kΩ + 151(1.6kΩ)] = 9.98kΩ Avt1 = − 2(9.98 kΩ) gm1RL1 =− = −1.05 1 + gm1 RS1 1 + 2(9 kΩ) RL 2 = (4.7kΩ 91kΩ 120kΩ) rπ 3 + (β o 3 + 1)(2.2 kΩ 250Ω) = 4.31kΩ [734 Ω + 81(225Ω)] = 3.51kΩ Avt 2 = − Avt 3 = 150(3.51kΩ) β o2 RL 2 =− = −2.17 rπ 2 + (β o 2 + 1)1.6 kΩ 2.39 kΩ + (151)1.6kΩ 81(2.2kΩ 250Ω) [ ] ⎛ ⎞ 1MΩ = 0.961 | Av = −⎜ ⎟(−1.05)(−2.17)(0.961) = +2.17 ⎝10 kΩ + 1MΩ ⎠ 734 + 81(2.2kΩ 250Ω) ⎛R + r ⎞ ⎛ ⎞ β o 2 RE 2 Rout = (3300Ω) ⎜ th 3 π 3 ⎟ | Rth 3 = RI 3 Ro 2 = RI 3 ro 2 ⎜1 + ⎟ ≅ RI 3 = 4.31kΩ ⎝ βo3 + 1 ⎠ ⎝ Rth 2 + rπ 2 + RE 2 ⎠ ⎛ 4.31 + 1.00 ⎞ Rout = (3.30 kΩ) ⎜ kΩ⎟ = 64.3Ω ⎝ ⎠ 81 14.105 *Problem 14.105 - Use the listing from Problem 14.103, but remove C2 and C4. Result: Av = VM (11) = +2.20 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-49 14.106 Q1 : VEQ1 = 15 100 kΩ = 1.63V | REQ1 = 100 kΩ 820 kΩ = 89.1kΩ 100 kΩ + 820 kΩ ⎛ ⎞ 1.63 - 0.7 101 IC1 = 100 = 319 µA | VCE1 = 15 − ⎜18 kΩ + 2 kΩ⎟IC = 8.61 V ⎝ ⎠ 89.1kΩ +101(2 kΩ) 100 gm1 = 40(319 µA) = 12.8 mS | rπ 1 = 100 (70 + 8.61)V = 246kΩ = 7.81kΩ | ro1 = 12.8 mS 319 µA Q2 : VEQ 2 = 15 43kΩ = 3.18V | REQ 2 = 160 kΩ 43kΩ = 33.9 kΩ 43kΩ + 160 kΩ ⎛ ⎞ 3.18 - 0.7 101 IC 2 = 100 =1.27 mA | VCE 2 = 15 − ⎜ 4.7 kΩ + 1.6 kΩ⎟IC = 6.98 V ⎝ ⎠ 33.9 kΩ +101(1.6 kΩ) 100 gm 2 = 40(1.27 mA) = 50.8 mS | rπ 2 = M 3 : VEQ 3 = 15 100 (70 + 6.98)V = 60.6kΩ = 1.97 kΩ | ro 2 = 50.8mS 1.27 mA 1.2 MΩ = 8.53V | REQ 3 = 1.2 MΩ 910 kΩ = 518 kΩ 1.2 MΩ + 910 kΩ 2 ID 3 + 3000 ID 3 → ID 3 = 1.87mA | VGS 3 − VTN 3 = 1.93V 8.53 = VGS 3 + 3000 ID 3 =1 + 0.001 VDS 3 = 15 − 3000 ID 3 = 9.39V | gm 3 = 2(0.001)(0.00187) = 1.93mS ⎛ Rin ⎞ Av = ⎜ ⎟ Avt1 Avt 2 Avt 3 | Rin = 820kΩ 100kΩ rπ 1 = 820 kΩ 100 kΩ 7.81kΩ = 7.18 kΩ ⎝ RI + Rin ⎠ 18 kΩ 33.9 kΩ 1.97 kΩ) = −21.6 Avt1 = −gm1 (RC1 RB 2 rπ 2 ) = −12.8 mS ( Avt 2 = −gm 2 (RC 2 RG 3 ) = −50.8mS (4.7kΩ 518kΩ) = −237 Avt 3 = 1 + gm 3 (RE 3 RL ) 1 + 1.93mS (3.0 kΩ 250Ω) gm 3 (RE 3 RL ) = 1.93mS (3.0 kΩ 250Ω) = 0.308 ⎛ ⎞ 7.18 kΩ Av = ⎜ ⎟(−21.6)(−237)(0.308) = 659 ⎝10 kΩ + 7.18 kΩ ⎠ ⎛ Rin ⎞ v gs3 = v g 3 (1 − Avt 3 ) = ⎜ ⎟ Avt1 Avt 2v i (1 − Av 3 ) ≤ 0.2(VGS 3 − VTN 3 ) ⎝ RI + Rin ⎠ vi ≤ 0.2(1.93) = 261µV 0.418(21.6)(237)(1 − 0.308) The gain is reduced rather than improved. The signal range increased since the gain was reduced. 14-50 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.107 *Problem 14.107 - Multistage Amplifier – Figure P14.106 VCC 12 0 DC 15 VI 1 0 AC 1 RI 1 2 10K *For output resistance *VI 1 0 AC 0 *VO 11 0 AC 1 C1 2 3 22U R1 12 3 820K R2 3 0 100K Q1 5 3 4 NBJT RE1 4 0 2K C2 4 0 22U RC1 12 5 18K C3 5 6 22U R3 12 6 160K R4 6 0 43K Q2 8 6 7 NBJT RC 12 8 4.7K RE2 7 0 1.6K C4 7 0 22U C5 8 9 22U R5 12 9 910K R6 9 0 1.2MEG M3 12 9 10 10 NMOSFET RE3 10 0 3K C6 10 11 22U RL 11 0 250 .OP .AC LIN 1 3KHZ 3KHZ .MODEL NMOSFET NMOS VTO=1 KP=.001 LAMBDA=0.02 .MODEL NBJT NPN IS=1E-16 BF=100 VA=70 .PRINT AC VM(3) VP(3) IM(VI) IP(VI) VM(11) VP(11) IM(C6) IP(C6) .END VM (3) 1 Results : Av = VM (11) = +711 | Rin = = 8.29 kΩ | Rout = = 401 Ω IM (VI ) IM (C6) © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-51 14.108 Rin = R1 R2 [rπ 1 + (β o1 + 1)RE1 ] = 100kΩ 820kΩ [7.81kΩ + 101(2 kΩ)] = 62.66 kΩ Avt1 = − β o1 (RC1 RB 2 RI 2 ) | RI 2 = rπ 2 + (β o 2 + 1)RE 2 = 1.97 kΩ + 101(1.6 kΩ) = 164 kΩ rπ 1 + (β o1 + 1)RE1 100( 18 kΩ 33.9 kΩ 164 kΩ) 7.81kΩ + 101(2 kΩ) = −6.69 R3 R4 = 43kΩ 160kΩ = 33.9kΩ | Avt1 = − RG 3 = R5 R6 = 1.2 MΩ 910 kΩ = 518 kΩ | Avt 2 = − Avt 3 = 1 + gm 3 (RE 3 RL ) 1 + 1.93mS (3.0 kΩ 250Ω) gm 3 (RE 3 RL ) = 1.93mS (3.0 kΩ 250Ω) β o 2 (RC 2 RG 3 ) 100(4.7 kΩ 518 kΩ) =− = −2.85 rπ 2 + (β o 2 + 1)RE1 1.97 kΩ + 101(1.6kΩ) = 0.308 ⎛ ⎞ ⎛ ⎞ Rin 62.6 kΩ Av = ⎜ ⎟(−6.69)(−2.85)(0.308) = 5.05 ⎟ Avt1 Avt 2 Avt 3 = ⎜ ⎝ 10 kΩ + 62.6kΩ ⎠ ⎝10 kΩ + Rin ⎠ 14-52 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.109 Note that the equivalent circuits are the same for Q1 and Q2 . VEQ = 180kΩ 15V = 5.63V | REQ = 180 kΩ 300 kΩ = 113 kΩ 180kΩ + 300kΩ 5.63 − 0.7 V IB = = 2.31µA | IC = 100 I B1 = 232µA | I E = 101I B1 = 234µA 113 + 101(20) kΩ 100(0.025V ) 232µA RI VCE = 15 − 2 x10 4 I E − 2 x10 4 IC = 5.71V rπ = = 10.8 kΩ | ro = v1 (70 + 5.71)V = 326kΩ 232µA Q 20 k Ω 1 v2 20 k Ω Q 2 2kΩ 300 k Ω 180 k Ω 100 k Ω 16.7 k Ω R L vi 2kΩ 113 k Ω 180 k Ω R I1 300 k Ω =17.0 k Ω ⎛ Rin ⎞ Av = ⎜ ⎟ Avt1 Avt 2 ⎝ 2 kΩ + Rin ⎠ Avt1 = Avt 2 = β o1 (RI 1 rπ 2 ) 100( 17 kΩ 10.8kΩ) v2 =− =− = −3.10 v1 rπ 1 + (β o1 + 1)R5 10.8 kΩ + (101)2kΩ vo = −gm 2 RL | RL = 100kΩ 20 kΩ = 16.7kΩ | Avt 2 = −40(232µA)(16.7 kΩ) = −155 v2 10.8 kΩ + (101)2kΩ] = 73.6 kΩ Rin = RB1 (rπ 1 + (β o1 + 1)R5 ) = 300 kΩ 180kΩ [ ⎛ 73.6 kΩ ⎞ Av = ⎜ ⎟(−3.10)(−155) = +468 | Rout = 20kΩ ro2 = 20 kΩ 326 kΩ = 18.8 kΩ ⎝ 2 kΩ + 73.6 kΩ ⎠ © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-53 14.110 The ac equivalent circuit from Problem 14.109 becomes: v1 Q 2k Ω 300 k Ω 180 k Ω 20 k Ω 180 k Ω 300 k Ω 20 k Ω 1 20 k Ω v2 20 k Ω Q 2 vo 100 k Ω 16.7 k Ω R L vi Avt1 = Avt 2 = Av = 100 17 kΩ ( 10.8kΩ + (101)20 kΩ) β o RI 1 [rπ 2 + (β o 2 + 1)R6 ] v2 =− =− = −0.830 v b1 rπ 1 + (β o1 + 1)R5 10.8kΩ + (101)20 kΩ 100(16.7 kΩ) vo β o2 RL =− =− = −0.822 v2 rπ 2 + (β o 2 + 1)R6 10.8 kΩ + (101)20kΩ ( 113 k Ω ) RI1 =17.0 k Ω [ ] Rin Avt1 Avt 2 | Rin = 113kΩ (rπ 1 + (β o1 + 1)R5 ) = 113kΩ ( 10.8kΩ + (101)20 kΩ) = 107kΩ RI + Rin 107kΩ Av = (−0.830)(−0.822) = +0.670 2kΩ + 107 kΩ The voltage gain is completely lost. | Rout ≅ 20 kΩ 14-54 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.111 *Problem 14.111 - Multistage Amplifier – Figure P14.109 VCC 11 0 DC 15 VI 1 0 AC 1 RI 1 2 2K *For output resistance *VI 1 0 AC 0 *VO 10 0 AC 1 C1 2 3 10U R1 3 0 180K R2 11 3 300K Q1 6 3 4 NBJT RE1 4 5 2K RE2 5 0 18K C2 5 0 10U RC1 11 6 20K C3 6 7 10U R3 7 0 180K R4 11 7 300K Q2 9 7 8 NBJT RC2 11 9 20K RE3 8 0 20K C4 8 0 10U C5 9 10 10U RL 10 0 100K .OP .AC LIN 1 5KHZ 5KHZ .MODEL NMOSFET NMOS VTO=1 KP=.001 LAMBDA=0.02 .MODEL NBJT NPN IS=1E-16 BF=100 VA=70 .PRINT AC VM(3) VP(3) IM(VI) IP(VI) VM(10) VP(10) IM(C5) IP(C5) .END VM (3) 1 Results : Av = VM (10) = +454 | Rin = = 74.7 kΩ | Rout = = 18.8 kΩ IM (VI) IM (C5) © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-55 14.112 ⎛ Rin ⎞ Av = ⎜ ⎟ Avt1 Avt 2 ⎝ 2 kΩ + Rin ⎠ v Avt1 = 2 = −gm1 (RI 1 rπ 2 ) = −40(232µA)( 17 kΩ 10.8 kΩ) = −61.3 v1 v Avt 2 = o = −gm 2 RL | RL = 100kΩ 20 kΩ = 16.7kΩ | Avt 2 = −40(232µA)(16.7 kΩ) = −155 v2 Rin = RB1 rπ 1 = 300 kΩ 180 kΩ 10.8 kΩ = 10.0 kΩ ⎛ 10.0 kΩ ⎞ Av = ⎜ ⎟(−61.3)(−155) = +7920 | Rout = 20kΩ ro 2 = 20 kΩ 326 kΩ = 18.8 kΩ ⎝ 2 kΩ + 10.0 kΩ ⎠ 14.113 0.05 2 (VGS + 2) and VGS = −1800 ID M1: Assume saturation: ID = 2 0.05 2 2 + 181VGS + 180 = 0 VGS = −1800 (VGS + 2) or 45VGS 2 VGS = −2.22V , −1.80V | VGS = −1.80 V and ID =1 mA VDS = 20 − 15000(0.001) − 1800(0.001) = 3.2 V > VGS − VTN 0.05 2 (VGS + 2) and VGS = −2500 ID M2: Assume saturation: ID = 2 0.05 2 2 VGS = −2500 + 251VGS + 250 = 0 (VGS + 2) or 62.5VGS 2 VGS = −1.83 V and ID = 0.723 mA VDS = 20 − 2500(0.723mA) =18.2 V > VGS − VTN gm1 = 2(0.05)(0.001) = 10.0 mS | gm 2 = 2(0.05)(7.23x10−4 ) = 8.50 mS Rin = 1800 1 1 = 1800 100 = 94.7Ω | Rout = 2500 = 2500 118 = 113Ω gm1 gm 2 8.5 x10−3 (2.5 kΩ 10 kΩ) 15kΩ 1MΩ) = 0.01S (14.8 kΩ) = 148 Avt1 = + gm1 ( Avt 2 = + 1 + gm 2 (2.5 kΩ 10 kΩ) gm 2 (2.5 kΩ 10 kΩ) =+ 1 + 8.5 x10−3 (2.5 kΩ 10 kΩ) = 0.944 Av = Avt1 Avt 2 = +140 14-56 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.114 VEQ = −15 + 30 100 kΩ = 0V | REQ = 100 kΩ 100 kΩ = 50 kΩ 100 kΩ + 100 kΩ 0 − 0.7 − (−15) IB = = 22.3µA | IC = 2.78 mA | I E = 2.81 mA 50kΩ + 126(4.7 kΩ) 125(0.025V ) 2.78mA = 1.12 kΩ VCE = 30 − 2000 IC − 4700 I E = 11.4 V | rπ = ro = | RB = R1 R2 = 100 kΩ 100kΩ = 50 kΩ 2.78mA 126)3.34kΩ = 44.7 kΩ Rin = RB rπ + (β o + 1)RL = 50kΩ 1.12 kΩ + ( (50 + 11.4)V = 22.1kΩ [ Rout = RE ro (R B RI + rπ ) ] βo + 1 ] (44.7kΩ 600Ω)+ 1.12kΩ = 13.5 Ω = 4.7 kΩ 22.1kΩ 126 [ R1S = RI + Rin = 45.3kΩ R2 S = R3 + Rout = 24.0kΩ R3 S ≅ RC = 2 kΩ ⎤ 1 ⎡ 1 1 ⎢ −5 ⎥ = 0.492 Hz SPICE result : f L = 0.39 Hz + fL ≅ −6 2π ⎢ ⎣10 (45.3kΩ) 47 x10 (24.0kΩ)⎥ ⎦ Note that C3 is not in the signal path and doesn't contribute to f L . 14.115 Use C3 = 2.2 µF (10 − 0.7)V = 1.43 µA | I = 114 µA | V = 20 − 39000 I − 68000 I = 7.71 V IB = C CE C E 1 MΩ + (80 + 1)68 kΩ Active region is correct. | rπ = 81(0.025V ) 114µA = 17.8 kΩ | ro = 75 + 7.71 = 726 kΩ 114µA RL = 500 kΩ 39 kΩ 726 kΩ = 34.5kΩ | Rin = RB rπ = 1 MΩ 17.8 kΩ = 17.5 kΩ ⎛ Rin ⎞ ⎛ ⎞ 17.5kΩ Av = −g m RL ⎜ ⎟ = 40(0.114 mA)(34.5kΩ) ⎜ ⎟ = −153 ⎝ 500Ω + 17.5kΩ ⎠ ⎝ RI + Rin ⎠ Rout = RC ro = 39 kΩ 726 kΩ = 37.0 kΩ R1S = RI + Rin = 18.0 kΩ R3 S = Rout + R3 = 537 kΩ R2S = RE (R B RI + rπ ) βo + 1 = 68 kΩ (1MΩ 500Ω)+ 17.8kΩ = 225Ω 81 SPICE result : f L = 18 Hz ⎤ 1 ⎡ 1 1 1 ⎥ = 19.2 Hz ⎢ fL ≅ + + 2π ⎢ 47 2.2 µ F 18.0 k Ω µ F 225 Ω µ F 537 k Ω 2.2 ( ) ( ) ( )⎥ ⎦ ⎣ © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-57 14.116 ( a ) VEQ = 12 62 kΩ = 9.07V | REQ = 62 kΩ 20 kΩ = 15.1kΩ 62 kΩ + 20 kΩ 12 − 0.7 − 9.07 = 4.19µA | IC = 314 µA | I E = 319 µA IB = 15.1kΩ + 76(6.8 kΩ) VEC = 12 − 16000 IC − 6800 I E = 4.81 V ro = 75(0.025) 60 + 4.81 = 206 k Ω | r = = 5.97 kΩ π 314 x10−6 314 x10−6 Rin = 15.1kΩ 5.97 kΩ = 4.28 kΩ | Rout = ro 16 kΩ = 14.8 kΩ | g m = 40 IC = 12.6 mS R1S = RI + Rin = 5.28 kΩ R3 S = Rout + R3 = 115kΩ R2S = RE fL ≅ (R B RI + rπ ) βo + 1 = 6.8 kΩ (15.1kΩ 1kΩ)+ 5.97kΩ = 90.0Ω 76 SPICE result : f L = 43.8 Hz ⎤ 1 ⎡ 1 1 1 ⎥ = 51.5 Hz ⎢ + + 2π ⎢ 47 10 µ F 5.28 k Ω µ F 90.0 Ω µ F 115 k Ω 2.2 ⎥ ( ) ( ) ( ) ⎦ ⎣ 14.117 VEQ = 18 500 kΩ = 4.73V | REQ = R1 R2 = 500kΩ 1.4 MΩ = 368 kΩ 500kΩ + 1.4 MΩ Assume Active Region Operation 2ID + 27000 I D → I D = 113 µA 5 x10−4 = 18 − (27000 + 75000)I D = 6.47 V > 3.73 V - Active region is correct. 4.73 = VGS = 27000 I D → 4.73 = 1 + VDS ro = 50 + 6.47 = 500 kΩ | g m = 2 500 x10−6 113x10−6 1 + 0.02(6.47) = 357µS −6 113 x10 RL = ro RD 470kΩ = 500kΩ 75kΩ 470kΩ = 57.3kΩ ( )( )[ ] Rin = R1 R2 = 500 kΩ 1.4 MΩ = 368 kΩ | Rout = ro 75kΩ = 65.2 kΩ R1S = RI + Rin = 369kΩ R3 S = Rout + R3 = 535kΩ R2 S = RS 1 1 = 27 kΩ = 2.54kΩ gm 0.357mS SPICE result : f L = 1.22 Hz ⎤ 1 1 ⎡ 1 1 ⎢ ⎥ = 1.56 Hz fL ≅ + + 2π ⎢ 2.2 47 10 µ F 369 k Ω µ F 2.54 k Ω µ F 535 k Ω ( ) ( ) ( )⎥ ⎣ ⎦ 14-58 © R. C. Jaeger & T. N. Blalock - February 20, 2007 14.118 2.5 x10−4 2 2 10 + VGS VGS + 1) | = 1.25 x10−4 ( VGS + 1) → VGS = −2.358V ID = ( 2 33kΩ 10 + VGS ID = = 232µA | VDS = −10 + 0.232mA(24 kΩ)− 2.36 = −6.79V | Pinchoff region is correct. 33kΩ 50 + 6.79 = 245kΩ g m = 2 2.5 x10−4 2.32 x10−4 1 + 0.02(6.79) = 0.363mS ro = 2.32 x10−4 1 Rin = 33kΩ = 2.69 kΩ | R1S = RI + Rin = 3.19 kΩ | R2 S ≅ RD + R3 = 124 kΩ gm ⎤ 1 1 ⎡ 1 ⎢ ⎥ = 5.94 Hz SPICE result : f L = 5.00 Hz + fL ≅ 2π ⎢ 124 kΩ)⎥ ⎣10µF (2.69kΩ) 47µF ( ⎦ ( ) ( )( )[ ] 14.119 (12 − 0.7)V = 2.69 µA | I = 135 µA | V 100kΩ + (51)82 kΩ 50(0.025V ) = 9.26 kΩ Active region is correct. r = IB = C CE = 24 − 39000 IC − 82000 I E = 7.58 V | ro = 50 + 7.58 = 427 kΩ 135µA π 135µA R1S = RB rπ + (β o + 1) RE RI R2S = RI + RE [ ( )]= 100kΩ [9.26kΩ + (51)(82kΩ 500Ω)]= 25.7kΩ rπ 9.26 kΩ = 500 + 82 kΩ = 681Ω | R3 S = RC + R3 = 139 kΩ 51 βo + 1 ⎤ 1 1 ⎡ 1 1 ⎢ ⎥ = 6.40 Hz SPICE result : f L = 5.72 Hz fL ≅ + + 2π ⎢ 4.7 47 10 µ F 25.7 k Ω µ F 681 Ω µ F 139 k Ω ⎥ ( ) ( ) ( ) ⎣ ⎦ 2 5 x10−4 VGS + 2) → I D = 32.2 µA ( 2 = 15 − (20000 + 51000)I D = 12.7 V > 0.36 V - Active Region is correct. 14.120 Assume Active Region operation. VGS = −51000 I D I D = VDS g m = 2 5 x10−4 32.2 x10−6 1 + 0.02( 12.70) = 0.201mS | Rin = 51kΩ ro = ( )( )[ ] 1 = 4.53kΩ gm 50 + 12.7 = 1.95 MΩ R1S = RI + Rin = 5.53kΩ R2 S ≅ RD + R3 = 30 kΩ 32.2 x10−6 ⎤ 1 ⎡ 1 1 ⎥ = 13.2 Hz SPICE result : f L = 12.8 Hz ⎢ fL ≅ + 2π ⎢ ⎦ ⎣2.2µF (5.53kΩ) 47µF (30 kΩ)⎥ © R. C. Jaeger & T. N. Blalock - February 20, 2007 14-59 14.121 The power supply should be +16 V. Assume Active Region operation. Since there is no negative feedback (RS = 0), we should include the effect of channel - length modulation. VGS = 0 ID = VDS ro = 2 4 x10−4 −5) ( 1 + 0.02VDS ) and VDS = 16 − 1800 I D → I D = 5.59 mA ( 2 = 16 − 1800 I D = 5.93 V > 5 V - Active region is correct. 50 + 5.93 = 10.0 kΩ Rin = 10.0 MΩ | Rout = RD ro = 1.52 kΩ 5.59 x10−3 R1S = RI + Rin = 10.0 MΩ R2 S = Rout + R3 = 37.5kΩ ⎤ 1 1 ⎡ 1 ⎢ ⎥ = 0.497 Hz SPICE result : f L = 0.427 Hz + fL ≅ 2π ⎢ 2.2 10 µ F 10.0 M Ω µ F 37.5 k Ω ⎥ ( ) ( ) ⎣ ⎦ VDS = 20 − 15000(0.001)− 1800(0.001) = 3.2 V > VGS − VTN 0.05 2 (VGS + 2) and VGS = −2500 ID M2: Assume saturation: ID = 2 2 0.05 2 VGS + 2) or 62.5VGS + 251VGS + 250 = 0 VGS = −2500 ( 2 VGS = −1.83 V and I D = 0.723 mA VDS = 20 − 2500(0.723mA) = 18.2 V > VGS − VTN g m1 = 2(0.05)(0.001) = 10.0 mS | g m2 = 2(0.05) 7.23 x10−4 = 8.50 mS R1S = Rin = 1800 Rout = 2500 1 = 1800 100 = 94.7Ω | R2 S = 15kΩ + 1MΩ = 1.02 MΩ g m1 14.122 Use C1 = C2 = C3 = 1 µF 0.05 2 (VGS + 2) and VGS = −1800 ID M1: Assume saturation: ID = 2 2 0.05 2 VGS + 2) or 45VGS + 181VGS + 180 = 0 VGS = −1800 ( 2 VGS = −2.22V , −1.80V | VGS = −1.80 V and I D = 1 mA ( ) 1 = 2500 118 = 113Ω | R3 S = 10kΩ + Rout = 10.1kΩ g m2 ⎤ 1 ⎡ 1 1 1 ⎢ ⎥ = 1.42 kHz SPICE result : f L = 1.68 kHz fL ≅ + + 2π ⎢ 113Ω) 1µF ( 1.02 MΩ) 1µF ( 10.1kΩ)⎥ ⎣1µF ( ⎦ 14-60 © R. C. Jaeger & T. N. Blalock - February 20, 2007 CHAPTER 15 15.1 (a) I C = αF IE = VCE = VC − (−0.7V ) = 5.87V | Q − Point = (20.7µA, 5.87V ) 1 β F 12 − VBE 1 ⎛100 ⎞⎛ 12 − 0.7 ⎞ = 20.7 µA | VC = 12 − 3.3x105 IC = 5.17V = ⎜ ⎟⎜ 5⎟ 2 β F + 1 REE 2 ⎝ 101 ⎠⎝ 2.7 x10 ⎠ (b) A dd = − g m RC = −40(20.7µA)(330kΩ)= −273 Rid = 2rπ = 2 β oVT IC =2 100(0.025V ) 20.7µA = 243 kΩ | Rod = 2 RC = 660 kΩ (c) A cc =− 100(330kΩ) β o RC =− = −0.604 rπ + (β o + 1)2 REE 122kΩ + 2( 101)270kΩ Add = − Ric = 15.2 g m RC −137 = −137 | Acd = Acc | CMRR = = 227 or 47.1 dB (very low) 2 −0.604 2 = 122 kΩ + 2( 101)270 kΩ 2 = 27.3 MΩ rπ + (β o + 1)2 REE (a) I E VCE = 1.5 − 105 IC − (−0.7)= 1.68V | Q - Pt : (5.25µA, 1.68V ) 1 ⎛ 1.5 − 0.7 ⎞ V 60 = ⎜ = 5.33µA | IC = α F I E = I E = 5.25µA 3 ⎟ 2 ⎝ 75 x10 ⎠ Ω 61 60 = 286 kΩ | Add = −g m RC = −0.210 mS ( 100 kΩ)= −21.0 gm (b) g m = 40 IC = 0.210 mS | rπ = Acc = − 60( 100 kΩ) β o RC =− = −0.636 rπ + (β o + 1)2 REE 286 kΩ + 61( 150 kΩ) −21.0 =∞ 0 For differential output : CMRR = −21.0 2 For single - ended output : CMRR = = 16.5, a paltry 24.4 dB! −0.636 Rid = 2rπ = 572 kΩ | Ric = rπ + (β o + 1)2 REE 2 RC = 50 kΩ 2 = 286 + 61( 150) 2 kΩ = 4.72 MΩ Rod = 2 RC = 200 kΩ | Roc = 15.3 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-1 *Problem 15.3 VCC 2 0 DC 12 VEE 1 0 DC -12 VIC 8 0 DC 0 VID1 4 8 AC 0.5 VID2 6 8 AC -0.5 RC1 2 3 330K RC2 2 7 330K Q1 3 4 5 NBJT Q2 7 6 5 NBJT REE 5 1 270K .MODEL NBJT NPN BF=100 VA=60 IS=1FA .OP .AC LIN 1 1KHZ 1KHZ .PRINT AC IM(VID1) IP(VID1) VM(3,7) VP(3,7) .TF V(7) VIC .END 1 Results : Add = VM (3,7) = −241 | Rid = = 269 kΩ | Acc = -0.602 | Ric = 23.2 MΩ IM (VID1) Problem15.45(b)-Transient-7 +0.000e+000 +1.000m +2.000m +3.000m +4.000m Time (s) +6.000 +4.000 +2.000 +0.000e+000 -2.000 -4.000 -6.000 V(IVOUT) Simulation results from B2SPICE. Problem15.3(b)-Fourier-Table FREQ mag phase norm_mag norm_phase +0.000 +49.786n +0.000 +0.00 +0.000 +1.000k +5.766 +180.000 +1.000 +0.000 +2.000k +99.572n +93.600 +17.268n -86.400 +3.000k +80.305m -180.000 +13.927m -360.000 +4.000k +99.572n +97.200 +17.268n -82.800 +5.000k +1.161m +179.993 +201.326u -7.528m +6.000k +99.572n +100.800 +17.268n -79.200 +7.000k +13.351u -179.005 +2.315u -359.005 +8.000k +99.572n +104.400 +17.268n -75.600 Using the Fourier analysis capability of SPICE, THD = 1.39% 15-2 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.4 (a) I E = 18V − 0.7V 2 4.7 x10 Ω 4 VCE = 18 − 105 IC − (−0.7)= 0.92 V | Q - point : ( 182 µA, 0.92 V ) ( ) = 184 µA | IC = α F I E = 100 I E = 182 µA 101 Note that RC is quite large and the common - mode input range is poor. More realistic choices might be 47 kΩ or 51 kΩ 100 (b) g m = 40 IC = 7.28 mS | rπ = g = 13.7 kΩ | Add = −g m RC = −7.28mS (100kΩ)= −728 m Acc = − 100( 100 kΩ) β o RC =− = −1.05 rπ + (β o + 1)2 REE 13.7 kΩ + 101(94 kΩ) −33.7 =∞ 0 For differential output : CMRR = −728 For single - ended output : CMRR = 2 = 346, a paltry 50.8 dB! −1.05 Rid = 2rπ = 27.4 kΩ | Ric = rπ + (β o + 1)2 REE = 13.7 + 101(94) 2 2 RC = 50 kΩ 2 kΩ = 4.75 MΩ Rod = 2 RC = 200 kΩ | Roc = ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-3 15.5 (a) I C VC1 = VC 2 = 12 − 2.4 x105 IC = 7.03V | VCE = VC − (−0.7V )= 7.73V 1 ⎛ β ⎞⎛ 12 − VBE ⎞ 1 ⎛ 100 ⎞⎛ 12 − 0.7 ⎞ = α F I E = ⎜ F ⎟⎜ ⎟= ⎜ ⎟⎜ ⎟ = 20.7 µA 2 ⎝ β F + 1⎠⎝ REE ⎠ 2 ⎝ 101 ⎠⎝ 2.7 x105 ⎠ 100(0.025V ) 20.7 µA Q − Point = (20.7µA, 7.73V ) | rπ = Acc = − = 121kΩ vic = 5 V , vC1 = vC 2 = 7.03 + Accvic = 7.03 − 0.439(5) = 4.84 V 100(240 kΩ) β o RC 5.000 + 5.000 = = −0.439 | vic = = 5.00V 2 rπ + (β o + 1)2 REE 121kΩ + ( 101)540kΩ Note that the BJT's are just beyond the edge of saturation! 1 β F ⎛ 5V − VBE − (−12V )⎞ 1 ⎛ 100 ⎞⎛17V − 0.7V ⎞ ⎟= ⎜ = 29.9 µA b I = α I = ⎜ ( ) C F E 2 β + 1⎜ 5 ⎟ ⎜ ⎟ 2 ⎝ 101⎟ R 2.7 x 10 ⎠ ⎝ ⎠ F EE ⎝ ⎠ VC1 = VC 2 = 12 − 2.4 x105 IC = 4.82 V | Part (a) has a small error of 0.02 mV rπ 121kΩ =5 = 11.1 mV > 5mV . rπ + (β o + 1)2 REE 121kΩ + ( 101)540 kΩ (c) The common - mode signal voltage applied to the base - emitter junction is vbe = vic A common - mode input voltage of 5 volts exceeds the small - signal limit. 15.6 We should first check the feasibility of the design using the Rule- of - Thumb estimates similar to those developed in Chapter 13 (Eq. (13.55)). The required Add = 794 (58 db) . (This sounds fairly large - a significant fraction of the BJT amplification factor µf .) (which provides no common - mode input range) : Add = g m RC = 40 IC RL ≤ 40VCC = 40(9)= 360. Thus, a gain of 794 is not feasible with this topology! Even assuming we choose to drop all of the positive power supply voltage across RC 15-4 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.7 We should first check the feasibility of the design using the Rule- of - Thumb estimates VCC + VEE ) = 240. Thus, a gain of 200 appears feasible. For symmetric supplies, Add ≅ 10( Rid = 2rπ = 1MΩ → rπ = 500 kΩ | IC = IE = IC = similar to those developed in Chapter 13 (Eq. (13.55)). The required Add = 200 (46 db) . β oVT rπ = 100(0.025V ) 500kΩ = 5.00 µA αF V 12 − 0.7) 101 V − VBE ( IC = 5.05µA | REE = EE = = 1.12 MΩ 100 2 IE 2(5.05µA) 200 200 = = 1.00 MΩ g m 40 5x10−6 Add = −g m RC = −200 (46 dB) | RC = Checking the collector voltage : VC = 12 − (990kΩ)(5µA)= 7V | Picking the closest 5% values. These values give IC = 5.09µA and Add = −204 (46.2dB) . valuesfrom the table in the Appendix : REE = 1.1 MΩ and RC = 1 MΩ are the final design ( ) ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-5 15.8 (a) I (b) C = αF IE = αF Q - point : ( 198µA,4.98V ) I EE 100 ⎛ 400µA ⎞ 4 = ⎜ ⎟ = 198µA | VCE = 12 − 3.9 x10 IC − (−0.7)= 4.98V 2 101 ⎝ 2 ⎠ 100 = 12.6kΩ | Add = − g m RC = −7.92mS (39 kΩ)= −309 gm g m = 40 IC = 7.92 mS | rπ = Acc = − 100(39 kΩ) β o RC =− = −0.0965 rπ + (β o + 1)2 REE 12.6 kΩ + 101(400 kΩ) −309 =∞ 0 For a differential output : CMRR = −309 2 For a single - ended output : CMRR = = 1600 or 64.1 dB −0.0965 kΩ = 20.2 MΩ 2 2 ( Note that this value is approaching the βoro limit and hence is not really correct.) Rod = 2 RC = 78.0 kΩ | Roc = RC = 19.5 kΩ 2 Rid = 2rπ = 25.2 kΩ | Ric = rπ + (β o + 1)2 REE = 12.6kΩ + 101(400kΩ) (c) r 50 + 4.98 = 278 kΩ 198 x10−6 Add = −g m RC ro = −7.92mS 39 kΩ 278 kΩ = −271 o = Acc ≅ − 100(39 kΩ) β o RC =− = −0.0965 rπ + (β o + 1)2 REE 12.6 kΩ + 101(400 kΩ) −271 =∞ 0 ( ) ( ) For differential output : CMRR = −271 2 For single - ended output : CMRR = = 1400 or 62.9 dB −0.0965 rπ + (β o + 1) 2 REE ro Rid = 2rπ = 25.2 kΩ | Ric = ( Rod = 2 RC ro = 68.4 kΩ | Roc = ( ) ( 2 CB RC Rout 2 ) = 12.6kΩ + 101(164kΩ)kΩ = 8.29 MΩ 2 CB = 19.5 kΩ since Rout ≅ µ f 2 REE rπ = 24.4 MΩ )≅ R C 2 ( ) 15-6 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.9 IC = α F I E = α F VCE I EE 75 400µA = = 197µA | VC1 = VC 2 = 12 − 3.9 x10 4 IC = 4.32V 76 2 2 = 4.32 − (−0.7)= 5.02V | Q - point : ( 197µA,5.02V ) 75 = 9.52kΩ | Add = − g m RC = −7.88 mS (39 kΩ) = −307 gm g m = 40 IC = 7.88 mS | rπ = Acc = − 75(39 kΩ) β o RC =− = −0.0962 rπ + (β o + 1)2 REE 9.52 kΩ + 76(400 kΩ) 2.005 + 1.995 = 2.00V 2 0.01V v − 0.0962(2V ) = 2.593 V vC1 = VC1 + Add id + Accv ic = 4.32V − 307 2 2 0.01V v − 0.0962(2V )= 5.663 V vC 2 = VC 2 − Add id + Accv ic = 4.32V + 307 2 2 vOD = 2.593 − 5.663 = −3.07 V vid = 2.005 − 1.995 = 0.01V | v ic = VCB = VC1 + AccVIC − VIC ≥ 0 | VIC ≤ 15.10 4.32 = 3.94 V 1 + 0.0962 Rid = 2rπ = 2( 100)(0.025V ) 2β oVT I 101 → IC = 1.00µA | I EE = 2 C = 2 (1µA)= 2.02 µA 100 IC 5 MΩ αF 105 = 2.5 GΩ ! 40( 1.00µA) CMRR = g m REE = 105 → REE = 15.11 (a ) I C = αF IE = αF g m = 40 IC = 0.396 mS | Add = − g m RC = −0.396 mS (910 kΩ) = −360 | REE = ∞ → Acc ≅ 0 vC 2 = VC 2 − Add vid + Accv ic | For vs = 0 : vC 2 = VC 2 = 0.991 2 For v s = 2mV : vC 2 = 0.991V + 360(0.001V )− 0(0.001V )= +1.35 V BC I EE 100 ⎛ 20µA ⎞ 5 = ⎜ ⎟ = 9.90µA | VC 2 = 10 − 9.1x10 IC = 0.991V 101 ⎝ 2 ⎠ 2 (b) For v ≥ 0, vs − (0.991 − 180vs )≥ 0 → vs ≤ 0.991V = 5.48 mV 181 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-7 15.12 (a ) I C = αF IE = αF g m = 40 IC = 3.97 mS | Add = − g m RC = −3.97 mS ( 110 kΩ)= −437 | REE = ∞ → Acc ≅ 0 For vs = 0 : VO = 1.09 V and v o = − Add vid =0 2 I EE 120 ⎛ 200µA ⎞ 5 = ⎜ ⎟ = 99.2µA | VO = 12 − 1.10 x10 IC = 1.09 V 2 121 ⎝ 2 ⎠ 0.001V For vs = 1 mV : VO = 1.09 V and vo = −(−437) = 0.219 V 2 ⎛ ⎞ A VO − dd v s ⎟ ≥ 0 → v s ≤ 4.96 mV (b) For vCB ≥ 0, v s − ⎜ 2 ⎝ ⎠ 15.13 *Problem 15.13 - Figure P15.11 VCC 2 0 DC 12 VEE 1 0 DC -12 V1 3 7 AC 1 V2 5 7 AC 0 VIC 7 0 DC 0 RC 2 6 110K Q1 2 3 4 NBJT Q2 6 5 4 NBJT IEE 4 1 DC 200U .MODEL NBJT NPN VA=60V BF=120 .OP .AC LIN 1 1KHz 1KHZ .PRINT AC IM(V1) IP(V1) VM(6) VP(6) .TF V(6) VIC .END 1 Results : Add = VM (6) = −193 | Rid = = 82.0 kΩ | Acc = +0.0123 | Ric = 45.8 MΩ IM (V 1) Circuit 15.55-Transient-3 +0.000e+000 +1.000m +2.000m +3.000m +4.000m +5.000m +6.000m +7.000m +8.000m +9.000m Time (s) +13.000 +12.000 +11.000 +10.000 +9.000 +8.000 +7.000 +6.000 +5.000 *REAL(Vout)* Simulation results from B2SPICE. The amplifier is over driven causing the output to be distorted. Using the Fourier analysis capability of SPICE, THD = 16.9% 15.14 15-8 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 150 ⎡ 15 − 0.7 ⎤ ⎢ ⎥ = 47.4µA | VEC = 0.7 − −15 + 2 x105 IC = 6.22V IC = α F I E = 151 ⎢ 150 kΩ)⎥ ⎣2( ⎦ ( ) Q - points : (47.4µA, 6.22V ) | g m = 40 IC = 1.90mS | rπ = Add = −g m RC = −1.90mS (200 kΩ) = −380 Acc = − 150(200 kΩ) β o RC =− = −0.661 rπ + (β o + 1)2 REE 79.0 kΩ + 151(300 kΩ) rπ + (β o + 1)2 REE 2 = 150 = 79.0kΩ gm = 22.7 MΩ 2 For a differential output : Adm = Add = −380 | Acm = 0 | CMRR = ∞ For a single - ended output : Adm = CMRR = 15.15 Rid = 2rπ = 158kΩ | Ric = 79.0 kΩ + 151(300kΩ) Add = −190 | Acm = Acc = −0.661 2 −190 = 287 or 49.2dB −0.661 100 ⎡ 10 − 0.7 ⎤ ⎢ ⎥ = 10.7µA | VC1 = VC 2 = −10 + 5.6 x105 IC = −4.01V 101 ⎢ ⎣2(430 kΩ)⎥ ⎦ 100 = 234kΩ gm IC = α F I E = VEC = 0.7 − (−4.01) = 4.71V | g m = 40 IC = 0.428mS | rπ = Add = −g m RC = −0.428mS (560 kΩ)= −240 Acc = − 100(560 kΩ) β o RC =− = −0.643 rπ + (β o + 1)2 REE 234 kΩ + 101(860 kΩ) 1 + 0.99 = 0.995V 2 0.01V v − 0.643(0.995V )= −5.850 V vC1 = VC1 + Add id + Accv ic = −4.01V − 240 2 2 0.01V v − 0.643(0.995V ) = −3.450 V vC 2 = VC 2 − Add id + Accv ic = −4.01V + 240 2 2 −5.850 − 3.450 = −4.65 vOD = −5.850 − (−3.450)= −2.40 V | Note : Add vid = −2.40V and vOC = 2 Also note : vOC = VC + Accv ic = −4.01 − 0.643(0.995V ) = −4.65V vid = 1 − 0.99 = 0.01V | v ic = ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-9 15.16 *Problem 15.16 – Figure P15.14 VCC 2 0 DC 10 VEE 1 0 DC -10 V1 4 8 AC 1 V2 6 8 AC 0 VIC 8 0 DC 0 RC1 5 1 560K RC2 7 1 560K Q1 5 4 3 PBJT Q2 7 6 3 PBJT REE 2 3 430K .MODEL PBJT PNP VA=60V BF=100 .OP .AC LIN 1 5KHz 5KHZ .PRINT AC IM(V1) IP(V1) VM(5,7) VP(5,7) .TF V(7) VIC .END 1 Add = VM (5,7) = −213 | Rid = = 511 kΩ IM (V 1) 213 Acc = −0.642 | Ric = 37.5 MΩ | CMRR = = 332 → 50.4 dB 0.642 15-10 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.17 IC = α F I EE 80 ⎛10µA ⎞ 5 = ⎜ ⎟ = 4.94µA | VEC = 0.7 − −3 + 3.9 x10 IC = 1.77V 81 ⎝ 2 ⎠ 2 ( ) Q - points : (4.94µA, 1.77V ) | g m = 40 IC = 0.198 mS | rπ = Add = −g m RC = −0.198 mS (390 kΩ)= −77.2 Acc = − 80(390 kΩ) β o RC =− = −0.0385 rπ + (β o + 1)2 REE 404 kΩ + 81( 10 MΩ) rπ + (β o + 1)2 REE 2 2 = 2 80 = 404 kΩ gm Rid = 2rπ = 808 kΩ | Ric = Note that Ric is similar to 808 kΩ + 81( 10 MΩ) = 405 MΩ βoro so that Ric = 405 MΩ will not be fully acheived. For example, if VA were 80V , β oro 2 ≅ 80 ⎛ 80 ⎞ ⎟ = 648 MΩ ⎜ 2 ⎝ 4.94µA ⎠ For a differential output : Adm = Add = −77.2 | Acm = 0 | CMRR = ∞ For a single - ended output : Adm = CMRR = Add = −38.6 | Acm = Acc = −0.661 2 −38.6 = 1000 or 60.0 dB | VBC ≥ 0 requires VIC ≥ VC = −1.07V and −0.0385 Without detailed knowledge of the circuit for IEE , we can only estimate that VIC should not exceed VIC + 0.7 ≤ VCC − 0.7V which allows 0.7V for biasing I EE → −1.07V ≤ VIC ≤ +1.6V . ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-11 15.18 IC = α F I EE 120 ⎛1mA ⎞ 4 = ⎜ ⎟ = 496µA | VC1 = VC 2 = −22 + 1.5 x10 IC = −14.6V 2 121 ⎝ 2 ⎠ 120 = 6.06 kΩ gm Checking VEC = 0.7 − (−14.6)= 15.3V | g m = 40 IC = 19.8 mS | rπ = 15kΩ)= −297 Add = −g m RC = −19.8 mS ( Acc = − 120( 15kΩ) β o RC =− = −0.0149 rπ + (β o + 1)2 REE 6.06 kΩ + 121( 1 MΩ) 0.01 + 0 = 0.005V 2 0.01V v − 0.0149(0.005V )= −16.09 V vC1 = VC1 + Add id + Accv ic = −14.6V − 297 2 2 v 0.01V − 0.0149(0.005V )= −13.12 V vC 2 = VC 2 − Add id + Accv ic = −14.6V + 297 2 2 vOD = −16.09 − (−13.12)= −2.97 V | Note : Add v id = −2.97V vid = 0.01 − 0 = 0.01V | v ic = Also note : vOC = vOC −16.09 − 13.12 = −14.6 and 2 = VC + Accvic = −14.60 − 0.0149(0.005V )= −14.6V 100 ⎡15V − 0.7V ⎤ 100 ⎢ ⎥ = 70.8µA | g m = 40 IC = 2.83mS | rπ = = 35.3kΩ 101 ⎢ g 2 100 k Ω ⎥ ( ) m ⎣ ⎦ 15.19 IC = α F I E = Add = vod vid ⎛ ⎛ ∆R ⎞ ∆R ⎞ | vod = vc1 − vc2 = ic1⎜ R + ⎟ − ic2 ⎜ R − ⎟ 2 ⎠ 2 ⎠ ⎝ ⎝ vid ⎞⎛ ∆R ⎞ v ⎛ ∆R ⎞ ⎛ vod = −gm id ⎜ R + ⎟ − ⎜− gm ⎟⎜ R − ⎟ = −g m R vid | Add = −g m R = −283 2 ⎠ 2 ⎠ ⎝ 2⎝ 2 ⎠⎝ ⎛ ⎛ v ∆R ⎞ ∆R ⎞ Acd = od | vod = vc1 − vc2 = ic1⎜ R + ⎟ − ic2 ⎜ R − ⎟ 2 ⎠ 2 ⎠ vic ⎝ ⎝ For a common - mode input, ic1 = ic2 = vod = − Acd = − rπ + (β o + 1)2 REE βo vic ⎡⎛ ∆R ⎞⎤ ∆R ⎞ ⎛ β o∆R vic⎢⎜ R + v ic ⎟ −⎜R − ⎟⎥ = vod = − 2 ⎠⎦ 2 ⎠ ⎝ rπ + (β o + 1)2 REE ⎣⎝ rπ + (β o + 1)2 REE βo 100( 100 kΩ) ∆R βo R = −0.01 = −.00494 R rπ + (β o + 1)2 REE 35.3kΩ + ( 101)200 kΩ −283 = 57300 or 95.2 dB −0.00494 CMRR = 15-12 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.20 *Problem 15.20 – Figure P15.19 VCC 2 0 DC 15 VEE 1 0 DC -15 V1 4 8 AC 0.5 V2 6 8 AC -0.5 VIC 8 0 DC 0 RC1 2 5 100.5K RC2 2 7 99.5K Q1 5 4 3 NBJT Q2 7 6 3 NBJT REE 3 1 100K .MODEL NBJT NPN BF=100 .OP .AC LIN 1 100 100 .PRINT AC IM(V1) IP(V1) VM(5,7) VP(5,7) .TF V(5,7) VIC .END Results: Add = VM (5,7) = −274 | A cd = −0.00494 | CMRR = 55500 or 94.9 dB ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-13 15.21 For a differential - mode input : ⎞⎛ ⎞⎛ ⎛v ⎛ v ∆g ⎞ ∆g ⎞ vod = −⎜ id − ve ⎟⎜ g m + m ⎟ R + ⎜− id − ve ⎟⎜ g m − m ⎟ R = − g m R vid + ∆g m Rve 2 ⎠ 2 ⎠ ⎠⎝ ⎠⎝ ⎝2 ⎝ 2 ⎛ ⎞ ∆g vod = −g m R⎜ vid + m ve ⎟ | At the emitter node : gm ⎠ ⎝ ⎞⎛ ⎞ ⎛ v ⎞⎛ ⎞ ⎛ vid ∆g m ∆g + g π ⎟ + ⎜− id − ve ⎟⎜ g m − m + g π ⎟ − GEE ve = 0 ⎜ − ve ⎟⎜ g m + 2 2 ⎠⎝ ⎠ ⎝ 2 ⎠⎝ ⎠ ⎝2 ve = 1 ∆g m 1 ∆g m β o REE vid ≅ vid VGS − VTN ) 2 24.2 x10−6 2ID gm = = = 1.39 mS | Add = −g m RD = −1.39 ms(330 kΩ)= −45.9 VGS − VTN 0.348 Acc = − 1.39 ms(330 kΩ) g m RD = = −0.738 1 + 2 g m RSS 1 + 2( 1.39 ms)(220 kΩ) 1 ⎛12 − 1.35 ⎞ 5 ID = IS = ⎜ ⎟ = 24.2µA. VD = 12 − 3.3 x10 (I D ) = 4.01V 2 ⎝ 220 kΩ ⎠ ( ) For a differential output : Adm = Add = −45.9 | Acm = 0 | CMRR = ∞ For a single - ended output : Adm = CMRR = Add = −23.0 | Acm = Acc = −0.738 2 23.0 = 31.2 | CMRRdb = 29.8 dB | Rid = ∞ | Ric = ∞ 0.738 15.23 VSS − VGS = 2 I D RSS | VGS = VTN + 15 = 2 I SS 62 x103 + 1 + VDS 2ID 2ID | VSS = 2 I D RSS + VTN + Kn Kn 2ID → I D = 107µA | VGS − VTN = 0.731V 4 x10−4 = 15 − (62 kΩ)I D − (−VGS )= 10.1V > 0.731V - Active | Q - pt : ( 107µA, 10.1V ) ( ) gm = 2( 107µA) 2ID = = 0.293mS | Add = − g m RD = −(0.293mS )(62 kΩ)= −18.2 VGS − VTN 0.731V Acc = − (0.293mS )(62kΩ) = −0.487 g m RD =− 1 + 2 g m RSS 1 + 2(0.293mS )(62 kΩ) For a differential output : Adm = Add = −18.2 | Acm = 0 | CMRR = ∞ For a single - ended output : Adm = CMRR = Add = −9.10 | Acm = Acc = −0.487 2 9.10 = 18.7 | CMRRdb = 25.4 dB | Rid = ∞ | Ric = ∞ 0.487 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-15 15.24 *Problem 15.24 - Figure P15.22 VCC 2 0 DC 12 VEE 1 0 DC -12 VIC 8 0 DC 0 VID1 4 8 AC 0.5 VID2 6 8 AC -0.5 RD1 2 3 330K RD2 2 7 330K M1 3 4 5 5 NFET M2 7 6 5 5 NFET REE 5 1 220K .MODEL NFET NMOS KP=400U VTO=1 .OP .AC LIN 1 1KHZ 1KHZ .PRINT AC IM(VID1) IP(VID1) VM(3,7) VP(3,7) .TF V(7) VIC .END Results : Add = VM (3,7) = −45.9 | Acc = −0.738 | CMRR = 31.2 | Rid = ∞ | Ric = ∞ (b) Results from B2SPICE Problem15.67(b)-Transient-0 (V) +0.000e+000 +1.000m +2.000m +3.000m +4.000m Time (s) +1.000 +500.000m +0.000e+000 -500.000m -1.000 V(IVOUT) Problem 15.24(b)-Fourier-Table FREQ +0.000 +1.000k +2.000k +3.000k +4.000k +5.000k mag phase norm_mag norm_phase -4.011 +586.074m +82.747u +90.858u +14.251n +9.695n +0.000 +0.000 +0.000 +180.000 +1.000 +0.000e+000 -90.000 +141.188u -270.000 +179.996 +155.029u -3.577m +93.958 +24.316n -86.042 +53.059 +16.541n -126.941 THD = 0.021% is very low due to the Level-1 square law model used in the simulation 15.25 15-16 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 First, we should check our rule - of - thumb. Since we have symmetric power supplies, Add ≅ VDD + VSS = 10 or 20 dB. We should be ok. Rod = 2 RD = 5kΩ → RD = 2.5kΩ | Selecting closest 5% value : RD = 2.4 kΩ 20 Add = −g m RD = 10 20 = 10 | g m = ID (4.17 x10 ) = 348µA = 2( 25x10 ) −3 2 −3 10 = 4.17 mS = 2 Kn I D 2400 2 I DS = 1.16V Kn | VGS = VTN + RSS = VSS − VGS 5 − 1.16 = = 5.52 kΩ | Selecting closest 5% value : RSS = 5.6 kΩ 2ID 2(348µA) 15.26 (a ) This solution made use of the m - file listed above Prob. 15.22. VSS − VGS = 2 I S RSS = 2 I D RSS | VGS = VTN + VTN = VTO + γ 2 ID 2 ID | VSS = 2 I D RSS + VTN + Kn Kn D (V SB + 0.6 − 0.6 = VTO + γ ) ( 2I RSS + 0.6 − 0.6 ) Solving iteratively with RSS = 62 kΩ | Kn = 400 I D = 91.3µA | VGS − VTN = 0.676V | VTN VDS = 15 − (62 kΩ)I D − (−VGS )= 12.9V > 0.676V - Saturated | Q - pt : (91.3µA, 12.9V ) | VTO = 1V | γ = 0.75 V yields V2 = 3.01V | VGS = 3.69V µA (b) g m = Acc = − (0.270mS )(62kΩ) = −0.486 assuming η = 0 g m RD =− 1 + 2gm ( 1 + η)RSS 1 + 2(0.270 mS )(62 kΩ) Add = −8.35 | Acm = Acc = −0.486 2 2(91.3µA) 2 ID = = 0.270 mS | Add = −g m RD = −(0.270 mS )(62 kΩ)= −16.7 VGS − VTN 0.676V For a differential output : Adm = Add = −16.7 | Acm = 0 | CMRR = ∞ For a single - ended output : Adm = CMRR = 8.35 = 17.2 | CMRRdb = 24.7 dB | Rid = ∞ | Ric = ∞ 0.486 TN (c) For γ = 0, V I D = 107 µA = VTO | VSS = 2 I D RSS + VTO + VDS = 30 − 62000 I D − 128000 I S = 10.1 V Saturated | Q - pt : ( 107 µA, 10.1 V ) 2 ID 2ID | 15 = 124000 I D + 1 + Kn 4 x10−4 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-17 15.27 *Problem 15.27 - Figure P15.26 VCC 2 0 DC 15 VEE 1 0 DC -15 VIC 8 0 DC 0 VID1 4 8 AC 0.5 VID2 6 8 AC -0.5 RD1 2 3 62K RD2 2 7 62K M1 3 4 5 1 NFET M2 7 6 5 1 NFET REE 5 1 62K .MODEL NFET NMOS KP=400U VTO=1 PHI=0.6 GAMMA=0.75 .OP .AC LIN 1 1KHZ 1KHZ .PRINT AC IM(VID1) IP(VID1) VM(3,7) VP(3,7) .TF V(7) VIC .END Results: Add = VM (3,7) = −16.8 | Acc = −0.439 | CMRRdB = 25.6 dB | Rid = ∞ | Ric = ∞ (b) Problem15.70(b)-Transient-1 (V) +0.000e+000 +1.000m +2.000m +3.000m +4.000m Time (s) +400.000m +200.000m +0.000e+000 -200.000m -400.000m V(IVOUT) Problem15.27(b)-Fourier-Table FREQ +0.000 +1.000k +2.000k +3.000k +4.000k +5.000k mag phase +0.000 +180.0 -30.192 -179.929 -50.644 -57.083 -345.019n +444.350m +35.476n +15.056u +23.789n +22.342n THD = 0.0034% norm_mag norm_phase +0.000 +0.000 +1.000 +0.000 +79.838n -210.192 +33.884u -359.929 +53.536n -230.643 +50.280n -237.083 15-18 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.28 This solution made use of the m - file listed above Prob. 15.22. (a) V SS − VGS = 2 I S RSS = 2 I D RSS | VGS = VTN + VTN = VTO + γ (V SB + 0.6 − 0.6 = VTO + γ ) ( 2I 2ID 2ID | VSS = 2 I D RSS + VTN + Kn Kn RSS + 0.6 − 0.6 D ) Solving iteratively with RSS = 220 kΩ | Kn = 400 I D = 20.3µA | VGS − VTN = 0.319V | VTN VDS = 12 − (330 kΩ)I D − (−VGS )= 8.35V > 0.319V - Active region | Q - pt : (20.3µA, 8.35V ) | VTO = 1V | γ = 0.75 V yields V2 = 2.74V | VGS = 3.05V µA (b) g m = Acc = − (0.127 mS )(330kΩ) = −0.737 assuming η = 0 g m RD =− 1 + 2g m ( 1 + η)RSS 1 + 2(0.127 mS )(220 kΩ) Add = −21.0 | Acm = Acc = −0.737 2 2(20.3µA) 2 ID = = 0.127 mS | Add = −g m RD = −(0.127 mS )(330 kΩ)= −41.9 VGS − VTN 0.319V For a differential output : Adm = Add = −41.9 | Acm = 0 | CMRR = ∞ For a single - ended output : Adm = CMRR = 21.0 = 28.4 | CMRRdb = 29.1 dB | Rid = ∞ | Ric = ∞ 0.737 TN (c) For γ = 0, V I D = 24.2 µA = VTO | VSS = 2 I D RSS + VTO + VDS = 24 − 330000 I D − 440000 I S = 5.37 V Saturated | Q - pt : (25.2 µA, 5.37 V ) 2ID 2ID | 12 = 440000 I D + 1 + Kn 4 x10−4 15.29 VDS = 9 − (300 kΩ)I D − (−VGS )= 4.32V > 0.316V - Active region | Q - pt : (20µA, 4.32V ) 2 2 x10−5 I SS 2ID (a) I D = 2 = 20µA | VGS = VTN + K = 1 + 4 x10−4 n ( ) = 1.316V | V GS − VTN = 0.316V (b) g m = Acc = − (0.127mS )(300kΩ) = −0.120 g m RD =− 1 + 2 g m RSS 1 + 2(0.127 mS )( 1.25 MΩ) 2(20µA) 2 ID = = 0.127 mS | Add = −g m RD = −(0.127 mS )(300 kΩ) = −38.0 VGS − VTN 0.316V For a differential output : Adm = Add = −38.0 | Acm = 0 | CMRR = ∞ For a single - ended output : Adm = CMRR = Add = −19.0 | Acm = Acc = −0.120 2 19.0 = 158 | CMRRdb = 44.0 dB | Rid = ∞ | Ric = ∞ 0.120 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-19 15.30 VDS = 15 − (75kΩ)I D − (−VGS ) = 5.62V > 0.866V - Active region | Q - pt : ( 150µA, 5.62V ) 2 1.5 x10−4 I SS 2ID (a) I D = 2 = 150µA | VGS = VTN + K = 1 + 4 x10−4 n ( ) = 1.866V | VGS − VTN = 0.866V (b) g m = Acc = − (0.346mS )(75kΩ) = −0.232 g m RD =− 1 + 2 g m RSS 1 + 2(0.346 mS )( 160 kΩ) 2( 150µA) 2 ID = = 0.346 mS | Add = −g m RD = −(0.346 mS )(75kΩ)= −26.0 VGS − VTN 0.866V For a differential output : Adm = Add = −26.0 | Acm = 0 | CMRR = ∞ For a single - ended output : Adm = CMRR = Add = −13.0 | Acm = Acc = −0.232 2 13.0 = 56.0 | CMRRdb = 35.0 dB | Rid = ∞ | Ric = ∞ 0.232 15.31 2 2 x10−5 I SS 2ID (a) I D = 2 = 20µA | VGS = VTN + K = VTN + 4 x10−4 = VTN + 0.316V n VTN = VTO + γ ( ) (V SB + 0.6 − 0.6 = 1 + 0.75 9 − VGS + 0.6 − 0.6 VGS − 0.316 = 1 + 0.75 9 − VGS + 0.6 − VDS = 9 − (300 kΩ)I DS − (−VGS ) = 5.71V > 0.316V - Active region | Q - pt : (20µA, 5.71V ) ( ) ( 0.6 )→ V ) GS = 2.71V | VTN = 2.39V (b) g m = Acc = − (0.127mS )(300kΩ) = −0.120 assuming η = 0 g m RD =− 1 + 2gm ( 1 + η)RSS 1 + 2(0.127 mS )( 1.25 MΩ) Add = −19.0 | Acm = Acc = −0.120 2 2(20µA) 2 ID = = 0.127 mS | Add = −g m RD = −(0.127 mS )(300 kΩ)= −38.1 VGS − VTN 0.316V For a differential output : Adm = Add = −38.1 | Acm = 0 | CMRR = ∞ For a single - ended output : Adm = CMRR = 19.0 = 158 | CMRRdb = 44.0 dB | Rid = ∞ | Ric = ∞ 0.120 15-20 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.32 2 1.5 x10−4 I SS 2ID (a) I D = 2 = 150µA | VGS = VTN + K = VTN + 4 x10−4 = VTN + 0.866V n VTN = VTO + γ ( ) (V SB + 0.6 − 0.6 = VTO + γ VGS − 0.866 = 1 + 0.75 15 − VGS + 0.6 − 150µA, 7.61V ) VDS = 15 − (75kΩ)I DS − (−VGS )= 7.61V > 0.866V - Active region | Q - pt : ( ( ) ( 15 − V 0.6 )→ V GS + 0.6 − 0.6 ) GS = 3.86V | VTN = 2.99V (b) g m = Acc = − (0.346 mS )(75kΩ) = −0.233 assuming η = 0 g m RD =− 1 + 2gm ( 1 + η)RSS 1 + 2(0.346 mS )( 160 kΩ) Add = −13.0 | Acm = Acc = −0.233 2 2( 150µA) 2 ID = = 0.346 mS | Add = −g m RD = −(0.346 mS )(75kΩ)= −26.0 VGS − VTN 0.866V For a differential output : Adm = Add = −26.0 | Acm = 0 | CMRR = ∞ For a single - ended output : Adm = CMRR = 13.0 = 55.8 | CMRRdb = 34.9 dB | Rid = ∞ | Ric = ∞ 0.233 30 15.33 Since we have symmetric power supplies, Add ≅ ( VDD + VSS ) = 15, within a factor of about 2. We should be ok if we reduce the value of VGS − VTN . (Remember, our rule - of - thumb used VGS − VTN = 1V .) | g m RD = 31.6 = 2 I D RD VGS − VTN Add = −g m RD = 10 20 = 31.6 | First, we should check our rule - of - thumb. Maximum common - mode range requires minimum I D RD ⇒ minimum VGS − VTN Choosing VGS − VTN = 0.25V to insure strong inversion operation, I D RD = 0.25(31.6) 2 = 3.95V | 0.25V = (0.25) (0.005) = 156µA 2ID → ID = Kn 2 2 I SS = 2 I D = 312 µA | RD = 3.95V = 25.3kΩ → 27 kΩ, the nearest 5% value. 156µA ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-21 15.34 (a) I D Kp 2 µA 1 ⎛ 18 − VGS ⎞ 18 + VGS = ⎜ VGS − VTP ) = and for K p = 200 2 and VTP = −1V ⎟⇒ ( 2 112 kΩ 2 ⎝ 56kΩ ⎠ V 2 18 + VGS = 11.2( VGS + 1) → VGS = −2.19V | VGS − VTP = −1.19V | I D = 142µA VDS = − −VGS − (91kΩ)I D − 18 = −7.27V ≤ −1.19V − Active region | Q - Point = ( 142µA, 7.27V ) { [ (b) g m = 2 2 x10−4 1.42 x10−4 ( )( ]} ) = 0.238mS | Add = − g m RD = −0.238 mS (91kΩ)= −21.7 Acc = − 0.238 mS (91kΩ) g m RD = = −0.785 1 + 2 g m RSS 1 + 2(0.238 mS )(56 kΩ) Add = −10.9 | Acm = Acc = −0.785 2 For a differential output : Adm = Add = −21.7 | Acm = 0 | CMRR = ∞ For a single - ended output : Adm = CMRR = 10.9 = 13.9 | CMRRdb = 22.9 dB | Rid = ∞ | Ric = ∞ 0.785 15.35 *Problem 15.35 – Figure P15.34 VCC 2 0 DC 18 VEE 1 0 DC -18 VIC 8 0 DC 0 V1 4 8 AC 0.5 V2 6 8 AC -0.5 RD1 5 1 91K RD2 7 1 91K M1 5 4 3 3 PFET M2 7 6 3 3 PFET REE 2 3 56K .MODEL PFET PMOS KP=200U VTO=-1 .OP .AC LIN 1 3KHZ 3KHZ .PRINT AC IM(V1) IP(V1) VM(5,7) VP(5,7) .TF V(7) VIC .END Results : Add = VM (5,7) = −21.6 | Acc = −0.783 | CMRR = 13.8 | Rid = ∞ | Rid = ∞ 15-22 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.36 2 2 x10−5 I SS 2ID (a) I D = 2 = 20µA | VGS = VTP − K = VTP − 2 x10−4 = VTP − 0.447V p VTP = − VTO − γ VGS ( ) ( V + 0.6 − 0.6 )= −1 − 0.6( 10 + V + 0.6 − 0.6 )} = −0.447 − { 1 + 0.6( 10.6 + V − 0.6 ) }→ V = −2.67V | V = −2.23V BS GS GS GS TP m { VDS = VGS + −10 + (300 kΩ)I D = −6.67V ≤ −0.447V - Active region | Q - pt : (20µA,-6.67V ) (b) g = [ ] 2( 2 x10 ) (2 x10 )= 89.4µS −5 −4 | Add = − g m RD = −(89.4µS )(300 kΩ)= −26.8 Acc = − (89.4µS )(300kΩ) = −0.119 assuming η = 0 g m RD =− 1 + 2g m ( 1 + η)RSS 1 + 2(89.4µS )( 1.25 MΩ) Add = −13.4 | Acm = Acc = −0.119 2 For a differential output : Adm = Add = −26.8 | Acm = 0 | CMRR = ∞ For a single - ended output : Adm = CMRR = 13.4 = 113 | CMRRdb = 41.0 dB | Rid = ∞ | Ric = ∞ 0.119 15.37 (a) I D = I SS = 10µA | VO = −12 + (820 kΩ)I D = −3.80V | For vI = 0, vO = VO = −3.80 V 2 2 10−5 2ID = 1− = 0.86 V | VGS − VP = −0.14V VGS = VTP + KP 10−3 VDS ≤ −0.14V for pinchoff. So VD ≤ −1 V for pinchoff. 1mA)( 10µA) = 141µS | Add = − g m RD = −( 141µS )(820 kΩ)= −116 g m = 2( Acc = 0 for RSS and ro = ∞ | vO = VO − Add 116.0 v1 = −3.80 + (0.02)= −2.64V 2 2 ( ) (b) 2 ≤ 0.2 V v1 GS − VP = 0.2(0.14) = 28.0mV | v1 ≤ 56 mV based upon the small - signal limit 116.0 v1 → v1 = 48.3 mV | So, v1 ≤ 48.3 mV 2 Also vO ≤ −1 V for pinchoff. -1 ≤ −3.80 + ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-23 15.38 +22 V RC 200 k Ω RC 200 k Ω RC 200 k Ω Q 1 Q R1 2 k Ω + 1 Q 1 vic R 1 +2R EE + vid 2 2R EE 400 k Ω -22 V dc half-circuit 402 k Ω R1 2kΩ Common-mode half circuit Differential-mode half circuit (a) (b) IC = α F IE = 150 22 − 0.7 = 52.6 µA | VCE = 22 − (200 kΩ)IC − (−0.7) = 12.2 V 151 402 kΩ 150(0.025V ) Q − Po int = (52.6µA, 12.2V ) for both transistors | rπ = = 71.3kΩ 52.6µA Acc = − Add = − rπ + (β o + 1)(R1 + 2 REE ) β o RC =− 150(200 kΩ) = −0.494 71.3kΩ + (151)402 kΩ 150(200 kΩ) β o RC =− = −80.4 rπ + (β o + 1)R1 71.3kΩ + (151)2 kΩ Rid = 2[rπ + (β o + 1)R1 ] = 2[71.3kΩ + (151)2 kΩ] = 747 kΩ Note also : Ric = 0.5[rπ + (β o + 1)(R1 + 2 REE )] = 0.5[71.3kΩ + (151)(402 kΩ)] = 30 MΩ and single - ended CMRR = 40.2 = 81.4, a paltry 38.2 dB 0.494 15-24 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.39 *Problem 15.39 – Figure P15.38 VCC 2 0 DC 22 VEE 1 0 DC -22 VIC 10 0 DC 0 V1 4 10 AC 0.5 V2 8 10 AC -0.5 RC1 2 5 200K RC2 2 9 200K Q1 5 4 3 NBJT Q2 9 8 7 NBJT RE1 3 6 2K RE2 7 6 2K REE 6 1 200K .MODEL NBJT NPN BF=150 .OP .AC LIN 1 1KHZ 1KHZ .PRINT AC IM(V1) IP(V1) VM(5,9) VP(5,9) .TF V(9) VIC .END Results : Add = VM (5,9) = −79.9 | Acc = −0.494 | Rid = 1 = 751 kΩ IM (V 1) ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-25 15.40 +V CC 100 k Ω 100 k Ω 100 k Ω 500 k Ω Q 1 Q + 100 µA 1 Q + - 1 2.5 k Ω v ic v id I 2 EE REE 600 k Ω REE 600 k Ω -V EE dc half-circuit (a) Common-mode half circuit Differential-mode half circuit (b) I 100 (100µA)= 99.0 µA VCE = 20 − 105 IC − (−0.7)= 10.8 V 101 100(0.025V ) Q − Po int = (99.0µA, 10.8V ) for both transistors | rπ = = 25.3kΩ 99.0 µA C = αF IE = Acc = − Add = − Add = − ' 100 kΩ β o RL =− = −0.165 rπ + (β o + 1)REE 25.3kΩ + 101(600 kΩ) β o RL | RL = 100 kΩ 500 kΩ =83.3kΩ | R5 = 600 kΩ 2.5kΩ =2.49 kΩ rπ + (β o + 1)R5 25.3kΩ + 101(2.49 kΩ) 100(83.3kΩ) = −30.1 | Rid = 2 rπ + (β o + 1)R5 = 2 25.3kΩ + 101(2.49 kΩ) = 554 kΩ [ ] [ ] ⎛ 30.1 ⎞ Note : Single - ended CMRR = 0.5⎜ ⎟ = 91.2 and ⎝ 0.165 ⎠ Ric = 0.5 rπ + (β o + 1)R5 = 0.5 25.3kΩ + 101(600 kΩ) = 30.3 MΩ [ ] [ ] 15-26 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.41 *Problem 15.41 – Figure P15.40 VCC 2 0 DC 20 VEE 1 0 DC -20 VIC 9 0 DC 0 V1 4 9 AC 0.5 V2 7 9 AC -0.5 RC1 2 5 100K RC2 2 8 100K RL 5 8 1MEG Q1 5 4 3 NBJT Q2 8 7 6 NBJT REE 3 6 5K IEE1 3 1 67.8U RE1 3 1 600K IEE2 6 1 67.8U RE2 6 1 600K .MODEL NBJT NPN BF=100 .OP .AC LIN 1 1KHZ 1KHZ .PRINT AC IM(V1) IP(V1) VM(5,8) VP(5,8) .TF V(8) VIC .END Results : Add = VM (5,8) = −30.0 | Acc = −0.165 | Rid = 1 = 555 kΩ IM (V 1) ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-27 15.42 +V R CC D R D R D M 1 M 1 M 1 Q 1 Q + v 1 Q + v id 2 1 ic I EE 2 dc half-circuit -V EE 2R EE Differential-mode half circuit (b) I C = αF VCE = −VGS = 3.29V | VDS = 15 − 7.5x10 4 IC − VCE − (−0.7)= 8.70 V 100(0.025V ) 49.5 µA 2 2 x10−4 I EE 100 ⎛ 100µA ⎞ = VGS − (−4) → VGS = −3.29V ⎜ ⎟ = 49.5 µA | 49.5µA = 2 101 ⎝ 2 ⎠ 2 Common-mode half circuit [ ] BJT Q − Points = (49.5µA, 3.29V ) | JFET Q − Points = (49.5µA, 8.70V ) = 50.5kΩ | Acc = − − rπ + (β o + 1)(2 REE ) rπ = β o RL =− Add = −g m1 RD = −40(49.5µA)(75kΩ)= −149 | Rid = 2rπ = 101kΩ 50.5kΩ + 101( 1.2 MΩ) 100(75kΩ) = −0.0619 ⎛ 149 ⎞ Note : Single - ended CMRR = 0.5⎜ ⎟ = 1200 or 61.6 dB and ⎝ 0.0619 ⎠ Ric = 0.5 rπ + (β o + 1)2 REE = 0.5 50.5kΩ + 101( 1.2 MΩ) = 60.6 MΩ (c) From (a) V [ ] [ ] CE = −VGS = 3.29V | VBE = 0.7V | VCE ≥ VBE → Active − region operation 15-28 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.43 +V DD = 6 V 200 µA I1 M1 M3 v1 I2 100 µA RD 30 k Ω - V SS = -6 V v ic M1 M3 v id RD 30 k Ω 2 M1 M3 RD 30 k Ω (a) dc half-circuit Common-mode half circuit Differential-mode half circuit (b) I D1 = I2 = 100µA | VGS1 = VTN + 2 I D1 10−4 = 1 + 2 −3 = 1.447V | VGS 1 − VTN = 0.447V Kn 10 10−4 = 1.632 V 5 x10−4 I D3 = I1 − I D1 = 200µA − 100µA = 100µA | VDS 1 = −VGS 3 = 1 + 2 VGS 3 − VTP = −0.632V | VDS 3 = − VS 1 − VGS 3 − −6 + (30 kΩ)I D 3 { [ ]}= −(−1.45 + 1.63 + 6 − 3)= −3.18V Both M1 and M3 are saturated. Q - points : M1 : ( 100µA, 1.63V ) M3 : ( 100µA, − 3.18V ) Add = −g m RD = − 2 10−4 10−3 (30 kΩ) = −13.4 | For ro = ∞, Acc = 0 | Rid = ∞ ( )( ) ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-29 15.44 (a) +V DD = 1.5 V 200 µA I1 +0.18 V M1 M3 -0.5 V I2 100 µA RD 10 k Ω -V SS V S3 M1 -1.45 V M1 M3 M3 vid R 2 D V S1 v ic RD 10 k Ω 10 k Ω = 1.5 V dc half-circuit Common-mode half circuit Differential-mode half circuit (b) I D1 = I 2 = 100µA | VGS 1 = VTN + 2 I D1 10−4 = 1 + 2 −3 = 1.447V | VGS 1 − VTN = 0.447V Kn 10 10−4 = 1.632 V 5 x10−4 I D3 = I1 − I D1 = 200µA − 100µA = 100µA | VDS 1 = −VGS 3 = 1 + 2 VGS 3 − VTP = −0.632V | VDS 3 = − VS 1 − VGS 3 − −1.5 + (30 kΩ)I D 3 { [ ]}= −(−1.45 + 1.63 + 1.5 − 3)= −1.32V These voltages can barely be supported by the 1.5- V negative power supply. VS 1 = −VGS1 = −1.45V which is more negative than the -1.5 - V supply. Also, VS 2 = VS 1 − VGS 3 = −1.45V + 1.63 = +0.18V , but for I D3 = 100µA, VD3 = −1.5 + 10−4 ( 10 kΩ)= −0.5V . M3 is just beyond pinchoff, and current source I2 has a very small voltage across it. 15-30 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.45 (a) I C1 = αF I1 100 ⎛ 50µA ⎞ = ⎜ ⎟ = 24.8 µA | VCE 2 = 12 − VEB 3 − (−VBE 2 )= 12 V 2 101 ⎝ 2 ⎠ 12V = 500 µA 24 kΩ Q - points : (24.8 µA, 12V ) (24.8 µA, 12V ) (500 µA, 12V ) For VO = 0 → VEC 3 = 12 V | IC 3 = (b) R rπ 2 = rπ 3 = C1 = 100(0.025V ) 100(0.025V ) 500µA 24.8µA 0.7V VEB 3 0.7V = 28.2 kΩ | RC 2 = = = 35.4 kΩ IC 2 − I B 3 24.8µA − 5µA 24.8µA = 101kΩ | ro2 = = 5kΩ | ro3 = 60 + 12 = 2.90 MΩ 24.8µA 60 + 12 = 144 kΩ 500µA g m2 ro2 RC 2 rπ 3 g m3 ro3 R 2 40(24.8µA) Adm = 2.90 MΩ 35.4 kΩ 5kΩ (40)(500µA) 144 kΩ 24 kΩ = 893 2 Rid = 2rπ 2 = 202 kΩ | (c) Rout = ro3 R = 20.6 kΩ Adm = ( ) ( ) ( ) ( ) (d ) R ic ≅ (β o + 1)ro2 2 = (101)2.90 MΩ = 147 MΩ 2 | (e) v 2 is the non - inverting input 15.46 vic ≥ −VEE + 0.75V + VBE1 = −12 + 0.7 + 0.75 = −10.6V vic ≤ VCC − VEB 3 = 12 − 0.7 = 11.3V | − 10.6 V ≤ vic ≤ 11.3 V ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-31 15.47 Note that the parameters of the transistors and values of RC have been carefully adjusted to permit open-loop operation and achieve VO = 0. *Problem 15.47 - Two Stage Amplifier – Figure P15.45 VCC 1 0 DC 12 VEE 2 0 DC -12 RC1 1 5 28.2K RC2 1 7 33.9K Q1 5 4 3 NBJT Q2 7 6 3 NBJT I1 3 2 DC 50U Q3 8 7 1 PBJT R 8 2 24K V1 4 10 AC 0.5 V2 6 10 AC -0.5 VIC 10 0 DC 0 .MODEL NBJT NPN BF=100 VA=60 .MODEL PBJT PNP BF=100 VA=60 IS=0.288F .OPTIONS TNOM=17.2 .OP .AC LIN 1 1KHZ 1KHZ .TF V(8) VIC .PRINT AC VM(8) VP(8) IM(V1) IP(V1) .END Adm = VM (8) = 1030 | Acm = −6.07 x 10−3 | CMRRdB = 105 dB 1 Results: = 239 kΩ | Rout = 20.6 kΩ Rid = IM (V 1) 15-32 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.48 IC1 = β F 1 I B1 = β F 1 I B IC 2 = β F 2 I B 2 = β F 2 (β F 1 + 1)I B ≅ β F 1β F 2 I B F1 B F1 F 2 B I C = I C 1 + I C 2 = β F 1 I B1 + β F 2 g m1 = 40 IC1 = β F 1 I B rπ 1 = ro1 ≅ [ [(β + 1)I ]≅ β = ] β I Assume β o2 = β o1 = β F 2 = β F 1 g m2 = 40 IC 2 ≅ β F 2β F 1 I B g m2 = β o2 g m1 = β o1g m1 = β o1 g m1 rπ 2 = β o2 g m2 β o1g m1 β o1 1 rπ 1 = β o1rπ 2 g m1 ro1 ≅ β o2ro2 = β o1ro2 VA V = A IC 1 β F 1 I B rπ 1 ro2 ≅ VA VA ≅ IC 2 β F 1β F 2 I B ≅ vbe vbe1 = v be rπ 1 + (β o1 + 1)rπ 2 β o1rπ 2 v v ≅ be → vbe2 ≅ be 2 β o1rπ 2 + (β o1 + 1)rπ 2 2 Gmvbe = g m1vbe1 + g m2vbe2 ≅ g m2 g vbe → Gm ≅ m2 2 2 RiB = rπ 1 + (β o1 + 1)rπ 2 = β o1rπ 2 + (β o1 + 1)rπ 2 ≅ 2β o1rπ 2 ⎡ (β o2 + 1) 1 ic = v ce2⎢ + ⎢ ro2 r 1 + g r r o1 m1 π 2 π 1 ⎣ RiC ≅ ro2 ro1 1 + g m1 rπ 2 rπ 1 [ ( ( ( β o2 )]≅ r )) o2 ⎤ ⎡ β o2 ⎥≅v ⎢ 1 + ce2 ⎥ ⎢ ro2 r 1 + g r r o1 m1 π 2 π 1 ⎦ ⎣ [ ( )] ⎤ ⎥ ⎥ ⎦ ro1 1 + g m1 (rπ 2 ) [ β o2 ]= r o2 2 β o2 ro1 2 = ro2 2ro2 = ro2 3 2 Note that βo = Gm RiB = β o1β o2 ≅ β o µf = Gm RiC = µf 2 3 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-33 15.49 +V CC RC RC Q Q3 4 Q vO 1 Q 2 v1 v2 I2 I1 -V EE (a) I C1 = αF I1 100 ⎛ 50µA ⎞ = ⎜ ⎟ = 24.8 µA | VCE 2 = VCC − VEB 3 − VEB 4 − (−VBE 2 )= 12 − 0.7 = 11.3 V 2 101 ⎝ 2 ⎠ Q - points : (24.8 µA, 11.3V ) (24.8 µA, 11.3V ) (4.95 µA, − 11.3V ) (495 µA, 12V ) 145 MΩ | For VO = 0 → VEC 4 = 12 V and VEC 3 = 12 − 0.7 = 11.3 V | For balance, VCE1 = VCE 2 = 11.3 V ⎛ I 1 ⎞ 12V IC 4 + IC 3 = IC 4 + α F 3 I B 4 = IC 4 + α F 3 C 4 = IC 4 ⎜1 + | IC 4 = 495 µA | IC 3 = 4.95 µA ⎟= βF ⎝ β F + 1⎠ 24 kΩ (e) v 2 is the non - inverting input (b) R rπ 2 = ro3 = C2 = 100(0.025V ) 24.8µA VEB 3 + VEB 4 1.4V 1.4V = = 56.6 kΩ | RC1 = = 56.5kΩ IC 2 − I B 3 24.8µA − 0.0495µA 24.8µA = 101kΩ | ro2 = 100(0.025V ) 60 + 11.3 60 + 11.3 = 14.4 MΩ | rπ 4 = = 5.05kΩ | ro 4 = = 144 kΩ 4.95µA 495µA 495µA ⎛ g ⎞⎛ 2r ⎞ g g Darlington Darlington Darlington gm Rout R = m2 ro2 RC 2 2β orπ 4 ⎜ m 4 ⎟⎜ o 4 R⎟ Adm = m2 ro2 RC 2 Rin 2 2 ⎝ 2 ⎠⎝ 3 ⎠ 100(0.025V ) 60 + 11.3 = 2.88 MΩ | rπ 3 = = 505kΩ 4.95µA 24.8µA ( ) ( ) ( ) Adm = 40(24.8µA) 2 40 495µA (2.88 MΩ 56.6kΩ 101kΩ)( )(2 )(96kΩ 24kΩ)= 9180 2ro4 R = 19.2 kΩ 3 (β + 1)r (101)2.88 MΩ = 145 MΩ | e v is the non - inverting input (d ) Ric ≅ o 2 o2 = () 2 2 Rid = 2rπ 2 = 202 kΩ | (c) Rout = 15-34 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.50 (a) For V VCE1 = VCE 2 = 13.6 − (−0.7)= 14.3V | VEC 3 = 15 − 2400 I E 3 − VO = 14.3V Q - points : (98.8µA, 14.3V ) (98.8µA, 14.3V ) (300µA, 14.3V ) RC = 80(0.025V ) 15 − 13.6 V = 15.1kΩ | rπ 3 = = 6.67 kΩ 0.3mA (98.8 − 3.75) µA 15V = 300µA | VC 2 = 15 − 2400 I E 3 − VEB 3 = 15 − 0.729 − 0.7 = 13.6V 50 kΩ ⎛ 200µA ⎞ 80 ⎛ 200µA ⎞ IC 3 300µA I C1 = I C 2 = α F ⎜ = = 3.75µA ⎟= ⎜ ⎟ = 98.8µA | I B 3 = βF 3 80 ⎝ 2 ⎠ 81 ⎝ 2 ⎠ O = 0, IC 3 = (b) A ⎛g ⎞ vc2 = −⎜ m1 ⎟ RC rπ 3 + (β o3 + 1)RE vid ⎝ 2 ⎠ ⎛ 40(98.8µA)⎞ ⎟ 15.1kΩ 6.67 kΩ + 81(2.4 kΩ) = −27.7 Av1 = −⎜ ⎜ ⎟ 2 ⎝ ⎠ v1 = [ [ ]] [ ( )] Av2 = Av = 80(50 kΩ) β o3 RL vo =− =− = −19.9 vc2 rπ 3 + (β o3 + 1)RE 6.67 kΩ + 81(2.4 kΩ) vc2 vo = −27.7(−19.9)= 551 vid v c2 Rid = 2rπ 1 = 2 β o1VT IC1 =2 80(0.025V ) 98.8µA = 40.5kΩ | ro3 = 70 + 14.3 = 281kΩ 0.3mA (c) R out ⎡ ⎤ ⎛ ⎞ 80(2.4 kΩ) β o RE ⎥ =49.0 kΩ = 50 kΩ ro3 ⎜1 + ⎟ =50kΩ 281kΩ⎢1 + ⎝ RC + rπ 3 + RE ⎠ ⎢ ⎣ 15.1kΩ + 6.67kΩ + 2.4 kΩ⎥ ⎦ (d ) R ic = (β o1 + 1)ro1 2 = 81 ⎛ 70 + 14.3⎞ ⎜ ⎟ = 34.6 MΩ | 2 ⎝ 98.8µA ⎠ (e) v 2 is the non - inverting (+) input. 15.51 vic ≥ −VEE + 0.75V + VBE1 = −15 + 0.7 + 0.75 = −13.6V | vic ≤ VCC − I E 3 RE − VEB 3 ⎛ 81 ⎞⎛ 15V ⎞ From Prob. 15.92, I E 3 RE = ⎜ ⎟⎜ ⎟(2.4 kΩ) = 0.729V ⎝ 80 ⎠⎝ 50kΩ ⎠ vic ≤ 15 − 0.729 − 0.7 = 13.6V | − 13.6 V ≤ vic ≤ 13.6 V ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-35 15.52 (a) For V VCE1 = VCE 2 = 15 − VEB 3 − (−VBE1 ) = 15.0V 15V = 300µA VEC 3 = 15 − VO = 15 − 0 = 15V 50 kΩ ⎛ 200µA ⎞ 80 ⎛ 200µA ⎞ IC 3 300µA I C1 = I C 2 = α F ⎜ = = 3.75µA ⎟= ⎜ ⎟ = 98.8µA I B 3 = βF 3 80 ⎝ 2 ⎠ 81 ⎝ 2 ⎠ O = 0, IC 3 = Q - points : (98.8µA, 15.0V ) (98.8µA, 15.0V ) (300µA, 15.0V ) RC 2 = rπ 2 = rπ 3 = V 0.7 0.7 V = 7.37 kΩ | For balance, RC1 = = 7.09 kΩ 98.8 µA (98.8 − 3.75) µA 98.8µA 0.3mA = 20.2 kΩ | ro2 = = 6.67 kΩ | ro3 = 70V + 15V = 860kΩ 98.8µA 70V + 15V = 283kΩ 0.3mA 80(0.025V ) 80(0.025V ) 15.53 For VO = 0, IC 3 = ⎛ 200µA ⎞ 80 ⎛ 200µA ⎞ 0.7V + I E 3 RE 0.7V + (304µA)RE I C1 = I C 2 = α F ⎜ = ⎟= ⎜ ⎟ = 98.8µA | RC = IC1 − I B 3 98.8µA − 3.75µA ⎝ 2 ⎠ 81 ⎝ 2 ⎠ VC 2 = 15 − 2400 I E 3 − VEB 3 = 15 − 0.729 − 0.7 = 13.6V Avt1 = − 15V I 300µA 81 = 300µA | I B 3 = C 3 = = 3.75µA | I E 3 = IC 3 = 304µA 80 50kΩ 80 βF 3 g m1 RC Rin3 = −20(98.8µA) RC Rin3 = -1.976 x10-3 RC Rin3 | Rin3 = rπ 3 + (β o + 1)RE 2 ⎛ ⎞ β o RL β o RE | RL = R ro3 ⎜1 + Avt 2 = − ⎟ | Av = Avt1 Avt 2 rπ 3 + (β o + 1)RE ⎝ Rth + rπ 3 + RE ⎠ 1000 ( ) ( ) ( ) Voltage Gain 100 10 1 0 10000 20000 30000 Emitter Resistance 15-36 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.54 VCC + VEE ) = 32,400 First, use our gain estimates to check feasibility of the design : Av ≅ 10( 2 [ ] There is plenty of margin available. Rout = R ro3 | 10−3 = 1 1 1 1 IC 3 1 1⎛ I R ⎞ = + = + = + ⎜ C3 ⎟ | VO = 0V Rout R ro3 R VA + VEC 3 R R ⎝ VA + VEC 3 ⎠ 100(0.025V ) 1⎛ 9 ⎞ 9 = 308Ω ⎜1 + ⎟ → R = 1.11 kΩ | IC 3 = = 8.11mA | rπ 3 = R ⎝ 70 + 9 ⎠ R 8.11mA Av 2000 = = −6.165 Avt 2 −324 Avt 2 = −g m3 Rout = −40(8.11mA)( 1kΩ)= −324 | Avt1 = Avt1 = − g m2 I Rr I R 0.7 RC rπ 3 = −20 C 2 C π 3 = −20 C 2 C ≅ −20 neglecting IB3 RC RC 2 RC + rπ 3 +1 +1 rπ 3 rπ 3 ( ) 0.7 0.7V 0.7V 8.11mA = −6.165 → RC = 391Ω | IC1 = + IB3 = + = 1.87 mA RC 391Ω 391Ω 100 +1 308 Selecting the closest 5% values : R = 1.1 kΩ , RC = 390 Ω , I1 = 3.74 mA −20 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-37 15.55 (a) For VO = 0, IC 3 = I2 = 300µA VC 2 = 15 − 2400 I E 3 − VEB 3 = 15 − 0.729 − 0.7 = 13.6V ⎛ 200µA ⎞ 80 ⎛ 200µA ⎞ IC 3 300µA IC1 = IC 2 = α F ⎜ = = 3.75µA ⎟= ⎜ ⎟ = 98.8µA | I B 3 = βF 3 80 ⎝ 2 ⎠ 81 ⎝ 2 ⎠ VCE1 = VCE 2 = 13.6 − (−0.7)= 14.3V | VEC 3 = 15 − 2400 I E 3 − VO = 14.3V Q - points : (98.8µA, 14.3V ) (98.8µA, 14.3V ) (300µA, 14.3V ) RC 2 = rπ 3 = 80(0.025V ) 0.3mA vt1 15 − 13.6 V 15 − 13.6 V = 14.7 kΩ | For balance, RC1 = = 14.2 kΩ 98.8 µA (98.8 − 3.75) µA = 6.67 kΩ | ro2 = 80V = 810 kΩ 98.8µA (b) A ⎛g ⎞ v c2 = −⎜ m1 ⎟ RC 2ro2 rπ 3 + (β o3 + 1)RE v id ⎝ 2 ⎠ ⎛ 40(98.8µA)⎞ ⎟ 15.1kΩ 1.62 MΩ 6.67 kΩ + 81(2.4 kΩ) = −27.5 Avt1 = −⎜ ⎜ ⎟ 2 ⎝ ⎠ ⎛ ⎞ β o3 RL β o RE vo ⎜ ⎟ | ro3 = 70 + 14.3 = 281kΩ =− | RL = ro3⎜1 + Avt 2 = ⎟ vc2 300µA rπ 3 + (β o3 + 1)RE ⎝ RC 2ro2 + rπ 3 + RE ⎠ ⎛ ⎞ 80(2.4 kΩ) 80(2.53 MΩ) ⎟ = 2.53 MΩ | Avt 2 = − 1 + = −1010 RL = 281kΩ⎜ ⎜ 14.8 kΩ + 6.67 kΩ + 2.4 kΩ ⎟ 6.67 kΩ + 81(2.4 kΩ) ⎝ ⎠ = { [ ]} [ ( )] Av = Avt1 Avt 2 = −27.5(−1010)= 27800 | Rid = 2rπ 1 = 2 Rout = RL = 2.51 MΩ β o1VT I C1 =2 80(0.025V ) 98.8µA = 40.5kΩ 15.56 The amplifier has an offset voltage of approximately 3.92 mV. Use this value to force the output to nearly zero. A transfer function analysis then yields Av = +28,627 Rout = 2.868 MΩ Rin = +50.051 kΩ These values are similar to the hand calculations in Prob. 15.55. Rin and Rout are larger because the hand calculations did not adjust the value of current gain based upon the Early voltage. 15.57 ⎛ 200µA ⎞ 100 ⎛ 200µA ⎞ I C1 = I C 2 = α F ⎜ ⎟= ⎜ ⎟ = 99.0µA | VCE1 = VCE 2 = 15 − VEB 3 − (−VBE1 ) = 15V ⎝ 2 ⎠ 101 ⎝ 2 ⎠ For VO = 0, IC 3 = I2 = 300µA | VEC 3 = 15 − VO = 15V Q - points : (99.0µA, 15.0V ) (99.0µA, 15.0V ) (300µA, 15.0V ) 15-38 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.58 g m1 RC 2ro2 rπ 3 + (β o3 + 1)RE 2 ⎡ ⎤ β o Rout βo β R o E ⎥ Avt 2 = − =− ro3⎢1 + ⎥ rπ 3 + (β o + 1)RE rπ 3 + (β o + 1)RE ⎢ R 2 r + r + R C o2 π3 E⎦ ⎣ Av = Avt1 Avt 2 | Avt1 = − { [ ]} ( ) I C1 = I C 2 = α F IB3 = RC 2 = rπ 3 = βF 3 IC 3 I1 80 200µA = = 98.8µA | For VO = 0, IC 3 = I2 = 300µA 2 81 2 300µA 81 = = 3.75µA | I E 3 = IC 3 = 303.8µA | VEC 3 = 15 − I E 3 RE 80 80 80(0.025V ) 300µA 0.7V + I E 3 RE 0.7V + (303.8µA)RE = IC1 − I B 3 98.8µA − 3.75µA = 6.67 kΩ | ro3 = 70 + 15 − (303.8µA)RE 70 + 14.3 − (303.8µA)RE 98.8µA 300µA g m1 = 40(98.8µA)= 3.95mS | ro2 = 30000 Voltage Gain 20000 10000 0 0 10000 20000 30000 40000 Emitter Resistance ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-39 15.59 2 2.5 x10-4 500µA 2 I D2 (a) I D2 = 2 = 250µA | VS = VSG = −VTP + K = 1 + 5x10-3 = 1.32V p VD = −15 + 0.7 = −14.3V | VDS = VD − VS = −14.3 − 1.32 = −15.6V | Q - pt : (250µA, − 15.6V ) IC 3 = 500µA | VCE 3 = VC 3 - VE 3 = 0 - (-15) = 15V | Q - pt : (500µA, 15V ) VBE = I D2 − I B 3 0.7V = 2.87 kΩ 500µA 250µA − 80 ( ) RD = (b) g = 4 kΩ 0.5mA v v g 1.58 mS Av = d 2 o = Avt1 Avt 2 | Avt1 = − m2 RD rπ 3 = − 2.87kΩ 4kΩ = −1.30 vid v d2 2 2 ⎛ 75V + 15V ⎞ Avt 2 = −g m3 ro3 R2 = −40(0.5mA) 2 MΩ⎟ = −0.02 180kΩ 2 MΩ = −3300 ⎜ ⎝ 0.5mA ⎠ m2 = 2(0.005)(0.00025) = 1.58 x10−3 S | rπ 3 = 80(0.025V ) ( ) ( ) ( ) ( ) (c) v Av = −1.30(−3300) = 4300 | Rin = ∞ | Rout = ro3 R2 = 180 kΩ 2 MΩ = 165kΩ 2 is the non - inverting (+) input (d ) v 1 is the inverting (-) input 15-40 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.60 Note that the parameters of the transistors and values of RD have been carefully adjusted to permit open-loop operation and achieve VO = 0. *Problem 15.60 – Figure P15.59 VCC 7 0 DC 15 VEE 8 0 DC -15 V1 1 9 AC 0.5 V2 3 9 AC -0.5 VIC 9 0 DC 0 I1 7 2 DC 493.2U R1 7 2 2MEG M1 4 1 2 2 PFET M2 5 3 2 2 PFET RD1 4 8 2.863K RD2 5 8 2.863K Q3 6 5 8 NBJT I2 7 6 DC 492.5U R2 7 6 2MEG .MODEL PFET PMOS KP=5M VTO=-1 .MODEL NBJT NPN BF=80 VA=75 IS=0.2881FA .OP .AC LIN 1 1000 1000 .TF V(6) VIC .PRINT AC IM(V1) IP(V1) VM(6) VP(6) .OPTIONS TNOM=17.2 .END Adm = VM (6) = 4630 | Acm = −1.46 | CMRRdB = 70.0 dB Results: R = id 1 = ∞ | Rout = 164 kΩ IM (V 1) 80(0.025V ) 100µA 15.61 v v g Av = d 2 o = Avt1 Avt 2 | Avt1 = − m2 RD rπ 3 vid v d2 2 ( )| IC 3 = 100µA | rπ 3 = = 20 kΩ I D2 = 500µA = 250µA | g m2 = 2(0.005)(0.00025) = 1.58 x10−3 S 2 VBE 0.7V 1.58 mS = = 2.81kΩ | Avt1 = − 2.81kΩ 20 kΩ = −1.95 RD = 100µA I D2 − I B 3 2 250µA − 80 ⎛ 75V + 5V ⎞ Avt 2 = −g m3 ro3 R2 = −40( 100µA) 10 MΩ⎟ = −0.02 800 kΩ 10 MΩ = −2960 ⎜ ⎝ 100µA ⎠ ( ) ( ) ( ) Av = −1.95(−2960)= 5770 15.62 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-41 VGS 2 = VTP − 2ID 5 x10 4 = −1 − = −1.316V KP 5 x103 For PMOS active region : VDS 2 ≤ VGS 2 − VTP = −0.316V vic ≥ −VEE + VBE 3 − VDS 2 + VGS 2 = −15 + 0.7 + 0.316 − 1.316 = −15.3V vic ≤ VCC − 0.75 + VGS 2 = 15 − 0.75 − 1.316 = 12.9V | − 15.3 V ≤ vic ≤ 12.9 V 15.63 2 2.5 x10-4 500µA 2 I D2 (a) I D2 = 2 = 250µA | VS = −VGS = −VTP + K = −1 + 5x10-3 = 1.32V p VD = −5 + 0.7 + 0.7 = −3.6V | VDS = VD − VS = −3.6 − 1.32 = −4.92V | Q - pt : (250µA, - 4.92V ) IC 3 + IC 4 = 500µA | IC 3 + β F I E 3 = (β F + 2)IC 3 = 500µA → IC 3 = 6.10µA | IC 4 = 494µA For VO = 0, VCE 4 = 5V and VCE 3 = 5 - 0.7 = 4.30V VBE 3 + VBE 4 = I D2 − I B 3 ( ) Q - pts : (250µA, - 4.92V ) (250µA, - 4.92V ) (6.10µA, 4.30V ) (494µA, 5.00V ) 1.4V = 5.60 kΩ | Based upon results for the Darlington 6.10µA 250µA − 80 80(0.025V ) g = 328kΩ circuit in Prob. 15.48 : Avt1 = m1 RD 2rπ 3 | rπ 3 = 2 6.10µA 1.58 mS 5.60 kΩ 656 kΩ = −4.39 g m1 = 2(0.005)(0.00025) = 1.58 mS | Avt1 = − 2 ⎤ ⎞ 40(494µA)⎡2 ⎛ 75V + 5V ⎞ g ⎛2 Avt 2 = − m 4 ⎜ ro 4 R2 ⎟ = − ⎢ ⎜ ⎟ 1 MΩ⎥ = −9.88 mS 108 kΩ 1 MΩ = −963 2 ⎝3 2 ⎠ ⎣ 3 ⎝ 494µA ⎠ ⎦ ⎛2 ⎞ Av = −4.39(−963)= 4230 | Rid = ∞ | Rout = ⎜ ro 4 ⎟ R2 = 108 kΩ 1 MΩ = 97.5 kΩ ⎝3 ⎠ RD = ( ) ( ) ( ) 15.64 *Problem 15.64 – Figure P15.63 *Vos (the dc value of V2) has been carefully adjusted to set Vo ≈ 0 VCC 8 0 DC 5 VEE 9 0 DC -5 V1 1 10 AC 0.5 V2 3 10 DC 1.21M AC -0.5 VIC 10 0 DC 0 I1 8 2 DC 496.3U R1 8 2 1MEG M1 4 1 2 2 PFET M2 5 3 2 2 PFET RD1 4 9 5.6K RD2 5 9 5.6K 15-42 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 Q3 7 5 6 NBJT Q4 7 6 9 NBJT I2 8 7 DC 495U R2 8 7 1MEG .OP .MODEL PFET PMOS KP=5M VTO=-1 .MODEL NBJT NPN BF=80 VA=75 .AC LIN 1 1000 1000 .TF V(7) VIC .PRINT AC IM(V1) IP(V1) VM(7) VP(7) .END Adm = VM (7) = 4080 | Acm = −2.58 | CMRRdB = 64.0 dB 1 Results: R = = ∞ | Rout = 96.2 kΩ id IM (V 1) ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-43 15.65 +V CC R C R C Q 3 v 1 v Q 2 Q4 Q5 vO = 0 1 Q 2 R I1 I 2 I L 3 2kΩ -V EE The new output stage can be treated as an improved single transistor using the results from Prob. 15.48. Using the results from Ex. 15.5: 2 RL = 2( 100)(505Ω)+ 1002 (2 kΩ)= 20.1 MΩ | Av 2 becomes Rin 4-5 = 2β orπ 4 + β o Av2 = − g m3 ro3 Rin 4−5 = −22 mS 161kΩ 20.1MΩ = −3510 | The gain of the g m4 RL 40 4.95 x10−3 (2kΩ) g m 4 RL 2 = = 0 = 0.995 emitter follower becomes Av3 ≅ g m4 RL 2 + g m 4 RL 2 + 40 4.95x10−3 (2 kΩ) 1+ 2 Av = −3.50(−3510)(0.995)= 12200. ( ) ( ) ( ( ) ) CMRR and Rid do not change : CMRR = 63.5 dB and Rid = 101 kΩ Rout = 2 r 2 161kΩ + o3 = + = 26.2 Ω 2 −3 g m4 β o 40 4.95 x10 10 4 ( ) 15-44 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.66 +V CC R C R C Q 3 v 1 v Q 2 Q4 Q5 vO = 0 1 Q 2 R I1 I 2 I L 3 2kΩ -V EE The new output stage can be treated as an improved single transistor using the equation set from prob. 15.48 and the results from Ex. 15.4 2 RL = 2( 100)(505Ω)+ 1002 (2 kΩ) = 20.1MΩ | Av 2 becomes Rin 4-5 = 2β orπ 4 + β o Av2 = − g m3 ro3 Rin 4−5 = −22 mS 161kΩ 20.1MΩ = −3510 | The gain of the g m4 RL 40 4.95x10−3 (2kΩ) g m 4 RL 2 = = 0 = 0.995 emitter follower becomes Av3 ≅ g m4 RL 2 + g m 4 RL 2 + 40 4.95 x10−3 (2kΩ) 1+ 2 Av = −3.50(−3510)(0.995)= 12200. ( ) ( ) ( ( ) ) CMRR and Rid do not change : CMRR = 63.5 dB and Rid = 101 kΩ Rout = 2 r 2 161kΩ + o3 = + = 26.2 Ω 2 −3 g m4 β o 40 4.95 x10 10 4 ( ) ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-45 15.67 I1 100 ⎛ 100µA ⎞ = ⎜ ⎟ = 49.5µA | VEC 2 = +0.7V − (−15V + 0.7V )= 15.0V 2 101 ⎝ 2 ⎠ ⎛100 ⎞ For VO = 0, IC 4 = α F I3 = ⎜ ⎟1.00 mA = 990µA | VEC 4 = 0 − (−15V )= 15.0V ⎝ 101 ⎠ 1mA = 360µA | VCE 3 = VO − 0.7V − (−15) = 14.3 IC 3 = I2 + I B 4 = 350µA + 101 Q − pts : (49.5µA, 15.0V ) (49.5µA, 15.0V ) (360µA, 14.3V ) (990µA, 15.0V ) (a) I C1 = IC 2 = α F (b) R ro2 = C = 100(0.025V ) 50V = 1.01 MΩ | rπ 3 = = 6.94kΩ 49.5µA 360µA 50 + 14.3 g g = 179 kΩ | Av = Avt1 Avt 2 Avt 3 = m1 RC 2ro2 rπ 3 (g m3ro3 )() 1 = m1 RC 2ro2 rπ 3 µ f 3 ro3 = 360µA 2 2 0.7V 0.7V = = 15.3kΩ IC 2 − I B 3 (49.5 − 3.60)µA ( ) ( ) Av = 40(49.5µA) 2 15.3kΩ 2.02 MΩ 6.94kΩ) (40)(64.3)= 12100 ( = 101 kΩ 100(0.025V ) 990µA = 2.53kΩ | Rout = 179 kΩ + 2.53kΩ = 1.80 kΩ 101 Rid = 2rπ 1 = 2 100(0.025V ) 49.5µA (c) R out = = (d ) R (β ro3 + rπ 4 β o4 + 1 o1 | rπ 4 = + 1)ro1 2 ic | ro1 = 101( 1.31 MΩ) 50V + 15V 1.31MΩ | Ric = = 66.3 MΩ (e) v2 49.5µA 2 15-46 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.68 *Problem 15.68 – Figure P15.67 *RC and Vos (see V2) have been carefully adjusted to set Vo ≈ 0 VCC 7 0 DC 15 VEE 8 0 DC -15 V1 1 9 DC 0.117M AC 0.5 V2 3 9 AC -0.5 VIC 9 0 DC 0 I1 7 2 DC 100U Q1 4 1 2 PBJT Q2 5 3 2 PBJT RC1 4 8 15.8K RC2 5 8 15.8K Q3 6 5 8 NBJT I2 7 6 DC 350U Q4 8 6 10 PBJT I3 7 10 DC 1M .MODEL PBJT PNP BF=100 VA=50 .MODEL NBJT NPN BF=100 VA=50 .NODESET V(10)=0 .OP .AC LIN 1 1000 1000 .TF V(10) VIC .PRINT AC IM(V1) IP(V1) VM(10) VP(10) .END Adm = VM (10) = 13800 | Acm = −0.0804 | CMRRdB =105 dB Results: R = id 1 = 133 kΩ | Rout = 1.37 kΩ IM (V 1) ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-47 15.69 (a) Working backwards from the output : V = V − V = 12 − 0 = 12.0V | I = I = 5.00mA 2(0.005) 2I V =V + = 0.75 + = 2.16V | V = −( V − V )= −( 12 − 2.16)= −9.84V K 0.005 DS 4 DD O D4 3 D4 n GS 4 TN DS 3 DD GS 4 I D3 = I2 = 2.00 mA | VGS 3 = VTP − 2(0.002) 2 ID4 = −0.75V − = −2.16V Kn 0.002 I1 = 250µA 2 VD2 = VDD + VGS 3 = 12V − 2.16V = 9.84V | I D1 = I D2 = VGS 2 = 0.75V + 2 2.5 x10−4 −3 | VDS 1 = VDS 2 = 9.84V − (−1.07V )= 10.9V 5 x10 Q − pts : (250µA,10.9V) (250µA,10.9V) (2.00 mA,-9.84V) (5.00mA,12.0V) dm ( ) = 1.07V (b) A = µf 4 g m1 g g r RD (g m3ro3 ) m 4 o4 = m1 RD µ f 3 2 1 + g m 4ro4 2 1+ µf 4 | RD = 2.16V = 8.64 kΩ 0.25mA g m1 = 2 5 x10−3 ( )( g m3 = 2 2 x10−3 g m4 = Adm = −3 ( )( )[ ] 2( 5 x10 ) (5x10 )[1 + 0.02(12)] = 7.87mS −3 1 + 9.84 −4 0.015 2.5 x10 1 + 0.02( 10.9) = 1.75mS | ro3 = = 38.3kΩ 2 mA 1 + 12 2 x10−3 1 + 0.015(9.84) = 3.03mS | ro 4 = 0.02 = 12.4 kΩ 5mA )[ ] | µ f 3 = g m3ro3 = 116 | µ f 4 = 97.6 1.75ms 1 97.6 8.64 kΩ)( 116) = 868 | Rid = ∞ | Rout = = 127Ω ( 2 gm4 1 + 97.6 15-48 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.70 *Problem 15.70 – Figure P15.69 *The values of RD have been adjusted to bring the offset voltage to ≈ 0 VCC 8 0 DC 12 VEE 9 0 DC -12 V1 1 10 AC 1 V2 3 10 AC 1 VIC 10 0 DC 0 I1 2 9 DC 500U M1 4 1 2 2 NFET M2 5 3 2 2 NFET RD1 8 4 8.28K RD2 8 5 8.28K M3 6 5 8 8 PFET M4 8 6 7 7 NFET I2 6 9 DC 2M I3 7 9 DC 5M .MODEL PFET PMOS KP=2M VTO=-0.75 LAMBDA=0.015 .MODEL NFET NMOS KP=5M VTO=0.75 LAMBDA=0.02 .OP .AC LIN 1 1000 1000 .TF V(7) VIC .PRINT AC VM(7) VP(7) IM(V1) IP(V1) .END Adm = VM (7) = 802 | Acm = −4.74 x10 -7 ≅ 0 | CMRRdB = ∞ 1 Results: Rid = = 10 30 ≅ ∞ | Rout = 126 Ω IM (V 1) ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-49 15.71 (a) Working backwards from the output : VDS 4 = −(VO − VSS )= 0 + (−5)= −5.00V I D 4 = I3 = 2.00 mA | VGS 4 = VTP − 2(0.002) 2ID4 = −0.7 − = −2.11V Kp 0.002 VDS 3 = VO + VGS 4 − (−VSS ) = 0 − 2.11 + 5 = 2.89V 2 5 x10−4 2 ID4 = 0.75V + = 1.15V I D3 = I2 = 500µA | VGS 3 = VTN + Kn 5 x10−3 VD2 = −VSS + VGS 3 = −5 + 1.15V = −3.85V | I D1 = I D 2 = VGS1 = VGS 2 = −0.7V − 2 3 x10−4 −3 ( ) | VDS 2 = VDS 2 = − 1.25 − (3.85) = −5.10V 2 x10 Q − pts : (300µA, −5.10V ) (300µA, −5.10V ) (500µA,2.89V ) (2.00 mA,5.00V ) dm ( ) = −1.25V I1 = 300µA 2 [ ] (b) A = µf 4 g m1 g 1.15V g r RD (g m3ro3 ) m 4 o4 = m1 RD µ f 3 | RD = = 3.83kΩ 2 1 + g m 4ro4 2 1+ µf 4 0.3mA g m1 = 2 2 x10−3 3 x10−4 g m3 = 2 5 x10−3 5 x10−4 g m4 = Adm = −3 ( )( )[ ( )( )[ ] 2( 2 x10 ) (2 x10 )[1 + 0.015(15)] = 2.93mS −3 1 + 2.89 0.02 1 + 0.015(5.10) = 1.14 mS | ro3 = = 106kΩ 0.5mA 1 + 5.00 1 + 0.02(2.89) = 2.30 mS | ro 4 = 0.015 = 35.8 kΩ 2mA ] | µ f 3 = g m3ro3 = 244 | µ f 4 = 105 1.14ms 1 105 3.83kΩ)(244) = 528 | Rid = ∞ | Rout = = 341Ω ( 2 g m4 1 + 105 15.72 The amplifier has an offset voltage of approximately –6.69 mV. This value is used to force the output to nearly zero. A transfer function analysis then yields Av = +517 Rout = 339 Ω Rin = +1.00 x 1020 Ω These values are similar to the hand calculations in Prob. 15.71. 15-50 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.73 (a) Working backwards from the output : VDS 4 = −(VO − VSS )= 0 + (−5)= −5.00V I D 4 = I3 = 2.00 mA | VGS 4 = VTP − 2(0.002) 2 ID4 = −0.7 − = −2.11V Kp 0.002 VCE 3 = VO + VGS 4 − (−VSS )= 0 − 2.11 + 5 = 2.89V IC 3 = I2 = 500µA | VD 2 = −VSS + VBE 3 = −5 + 0.7V = −4.30V | I D1 = I D2 = VGS1 = VGS 2 = −0.7V − 2 3 x10−4 −3 | VDS1 = VDS 2 = − 1.25 − (−4.30) = −5.55V 2 x10 Q − pts : (300µA, −5.55V ) (300µA, −5.55V ) (500µA,2.89V ) (2.00 mA,5.00V ) dm = ( ) = −1.25V I1 = 300µA 2 [ ] (b) A RD = 150(0.025V ) 0.7V = 2.33kΩ | rπ 3 = = 7.53kΩ 500µA 0.3mA ⎛ g m4ro 4 ⎞ g m1 µf 4 g m1 RD rπ 3 (g m3ro3 ) RD rπ 3 µ f 3 ⎜ ⎟= 2 1+ µf 4 ⎝ 1 + g m4ro 4 ⎠ 2 ( ) ( ) 1 + 2.89 g m1 = 2 2 x10−3 3 x10−4 1 + 0.015(5.10) = 1.14 mS | ro3 = 0.02 = 106 kΩ 0.5mA 1 + 5.00 g m4 = 2 2 x10−3 2 x10−3 1 + 0.015( 15) = 2.93mS | ro 4 = 0.015 = 35.8 kΩ 2 mA µ f 3 = g m3ro3 = 40(70) = 2800 | µ f 4 = 105 ( )( )[ ] ( )( )[ ] Adm = 1.14 ms 1 105 2.33kΩ 7.53kΩ (2800) = 2810 | Rid = ∞ | Rout = = 341Ω 2 g m4 1 + 105 ( ) 15.74 The amplifier has an offset voltage of approximately 48.69 mV. This value is used to force the output to nearly zero. A transfer function analysis then yields Av = +2810 Rout = 339 Ω Rin = +1.00 x 1020 Ω These values are similar to the hand calculations in Prob. 15.73 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-51 15.75 (a) Working backwards from the output with VO = 0 : VDS 4 = VCC − VO = 5 − 0 = 5V ID 4 = I3 = 2 mA | VGS 4 = VTN + 2(0.002) 2 ID 4 = 0.7 + = 1.59V | IC 3 = I2 = 500µA Kn 0.005 VEC 3 = 5 − VGS 4 = 5 − 1.59 = 3.41V | VCE 2 = 5 − VEB 3 − (−VBE 2 ) = 5 − 0.7 + 0.7 = 5.00V IC1 = IC 2 = α F VEB 3 0.7V I1 100 200µA = = 99.0µA | RC = = = 7.45 kΩ 2 101 2 IC 2 − IB 3 (99.0 − 5.00)µA VCE1 = 5 − IC 1RC − (−VBE 2 ) = 5 − 99.0µA(7.45 kΩ) + 0.7 = 4.96V Q = pts : (99.0µA, 4.96V ) (99.0µA, 5.00V ) (500µA, 3.41V ) (2.00 mA, 5.00V ) (b) Using current division at the collector of Q2 : Adm = rπ 3 = ⎛ gm 4 RL ⎞ gm1 gm1 ⎛ RC ⎞ g R RC rπ 3 )µ f 3 m 4 L ( ⎜ ⎟β o3 ro 3 ⎜ ⎟= 2 ⎝ RC + rπ 3 ⎠ 1 + gm 4 RL ⎝ 1 + gm 4 RL ⎠ 2 100(0.025V ) = 5.00 kΩ | gm 4 = 2(0.005)(0.002) = 4.47 mS 500µA 40(99.0µA) 4.47 mS (2 kΩ) 7.45 kΩ 5.00 kΩ)(40)(50 + 3.41) = 11400 ( 2 1 + 4.47 mS (2 kΩ) 100(0.025V ) 1 = 50.5 kΩ | Rout = = 224 Ω 99.0µA gm 4 Adm = Rid = 2 rπ 1 = 2 (c) *Problem 15.75 – Figure P15.75 *The values of RC have been adjusted to set Vo ≈ 0. VCC 8 0 DC 5 VEE 9 0 DC -5 VIC 10 0 DC 0 V1 1 10 AC 0.5 V2 3 10 AC -0.5 I1 2 9 DC 200U Q1 4 1 2 NBJT Q2 5 3 2 NBJT RC1 8 4 8.00K RC2 8 5 8.00K Q3 6 5 8 PBJT I2 6 9 DC 500U M4 8 6 7 7 NFET I3 7 9 DC 2M RL 7 0 2K .MODEL NBJT NPN BF=100 VA=50 .MODEL PBJT PNP BF=100 VA=50 .MODEl NFET NMOS KP=5M VTO=0.70 .OP .AC LIN 1 2KHZ 2KHZ 15-52 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 .PRINT AC VM(7) VP(7) IM(V1) IP(V1) .TF V(7) VIC .END Adm = VM (7) = 11200 | Acm = −0.0957 | CMRRdB = 101 dB 1 Results: R = = 56.4 kΩ ≅ ∞ | Rout = 201 Ω id IM (V 1) 15.76 (a) I C1 = IC 2 = α F VCE 2 = VCC 50 50µA I1 100 10µA I = = 4.95 µA | IC 3 = IC 4 = α F 2 = = 24.5 µA 2 101 2 2 51 2 ⎛ 24.5 µA ⎞ − (IC 2 − I B 3 )RC − (−VBE 2 )= 3V − ⎜4.95µA − ⎟300kΩ − (−0.7)= 2.36V 50 ⎠ ⎝ 50 (250µA)= 245µA | VEC 5 = 3.00 V | VC 4 = −0.7V 51 ⎛ 24.5 µA ⎞ VC1 = 3 − ⎜ 4.95µA − ⎟300 kΩ = 1.66V | VEC 3 = VEC 4 = 1.66 + 0.7 − (−0.7)= 3.06V 50 ⎠ ⎝ For VO = 0 : IC 5 = α F I3 Q - pts : (4.95µA,2.36V ) (4.95µA,2.36V ) (24.5µA,3.06V ) (24.5µA,3.06V ) (245µA,3.00V ) dm (b) A rπ 3 = ro1 = = g m1 RC1 rπ 3 ro1 ( ) ( g m3 RC 2 2ro 4 rπ 5 + (β o5 + 1)RL 2 [ ]) 50(0.025V ) 24.5µA = 51.0 kΩ | ro 4 = (β o5 + 1)RL = 51(5kΩ) = 0.980 50 = 10.1 MΩ | 4.95µA rπ 5 + (β o5 + 1)RL 5.10 kΩ + 51(5kΩ) 50(0.025V ) 70 = 2.86 MΩ | rπ 5 = = 5.10 kΩ 245µA 24.5µA ⎛ (β + 1)R ⎞ o5 L ⎟ ⎜ ⎜r + β +1 R ⎟ ⎝ π 5 ( o5 ) L ⎠ Adm = 40(4.95µA) 300kΩ 51.0kΩ 10.1MΩ • ( (40)(24.5µA) 78kΩ 5.72 MΩ 2 ) ( 0.980 = 235 [5.10kΩ + 51(5kΩ)]) Rid = 2rπ 1 = 2 A v 100(0.025V ) 4.95µA = 1.01 MΩ | Rout = B (c) v is the non - inverting input - v is the inverting input (d) A = (10V )(10V )= 30 = 900 | r > ro 4 = ⎟ 2 3 ⎝ 20.1 MΩ ⎠ 490µA ( ) µf 4 3 = 40(70 + 11.6) 3 = 1155 | Adm = 26500 or (88.5dB) | Rid = 2rπ 1 = 2 100(0.025V ) 24.8µA = 202 kΩ Rout 100(0.025V ) 2 2 118 kΩ + 2 ro 4 + 2rπ 5 R + 2rπ 5 3 3 49.0µA = th5 = = = 18.1 Ω β o5β o6 β o5β o6 100( 100) 15-56 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.82 (a) For V IC 3 O = 0, IC 6 = α F I E 6 = α F I3 = 100 (5mA)= 4.95mA 101 IC 5 = α F I E 5 = α F βF 6 IC 6 = 4.95mA 49.0µA = 49.0µA | IC 4 + IC 3 = I2 + I B 5 = 500µA + = 500µA 101 100 βF 4 αF3 + IC 3 = (β F 3 + 2)IC 3 = 500µA → IC 3 = 9.62µA | IC 4 = 500µA − IC 3 = 490µA I1 100 ⎛ 50µA ⎞ = ⎜ ⎟ = 24.8µA | VCE 6 = 22 − 0 = 22V | VCE 5 = VCE 6 − VBE 6 = 21.3V 2 101 ⎝ 2 ⎠ I C1 = I C 2 = α F VCE1 = VCE 2 = 22 − VEB 4 − VEB 3 − (−VEB 2 ) = 21.3V VEC 4 = 22 − VBE 5 − VBE 6 = 20.6V | VEC 3 = VEC 4 − VEB 4 = 19.9V Q − pts : (24.8µA,21.3V ) (24.8µA,21.3V ) (9.62µA,19.9V ) (490µA,20.6V ) (49.0µA,21.3V ) (4.95mA,22.0V ) | RC = (b) Using the properties of the Darlington configuration from Prob. 15.48 : 50(0.025V ) = 255kΩ Rin3 ≅ 2β o3rπ 4 = 2(50) 490µA ⎞⎤⎡ β o5β o6 RL ⎤ ⎡g ⎤⎡ g ⎛ 2 Adm = AV 1 AV 2 AV 3 = ⎢ m2 RC Rin3 ⎥⎢ m 4 ⎜ ro 4 Rin5 ⎟⎥⎢ ⎥ ⎣ 2 ⎦⎣ 2 ⎝ 3 ⎠⎦⎣ 2rπ 5 + β o5β o6 RL ⎦ 1.4V 1.4V = = 56.9 kΩ IC 2 − I B 3 ⎛ 9.62 ⎞ ⎜24.8 + ⎟µA 50 ⎠ ⎝ ( ) 100(0.025V ) 100) + 100( 100)(2 kΩ)= 20.1 MΩ Rin5 ≅ 2β o5rπ 6 + β o5β o6 RL = 2( 4.95mA 40(24.8µA) µ ⎛ 20 MΩ ⎞ 70V + 20.6V = 185kΩ | Adm = 56.9 kΩ 255kΩ f 4 ⎜ Rin5 >> ro 4 = ⎟ 2 3 ⎝ 20.1 MΩ ⎠ 490µA ( ) µf 4 3 = 40(70 + 20.6) 3 = 1208 | Adm = 27700 (88.9dB) | Rid = 2rπ 1 = 2 100(0.025V ) 24.8µA = 202 kΩ Rout 100(0.025V ) 2 2 185kΩ + 2 ro 4 + 2rπ 5 R + 2rπ 5 3 3 49.0µA = th5 = = = 22.5 Ω β o5β o6 β o5β o6 100( 100) 15.83 Since the transistor parameters are the same, VGS1 = −VGS 2 = I D2 = I D1 = 2 6 x10−4 1.1 − 0.75) = 36.8 µA ( 2 2.2V = 1.1V 2 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-57 15.84 2 I D1 ⎛ 2 I D2 ⎞ ⎜ ⎟ where I D 2 = I D1 − VTP − 2.2V = VGS 1 − VGS 2 = VTN + ⎟ Kn ⎜ K p ⎠ ⎝ ⎛ 2 ⎞ 0.7 2 ⎟ | 2.2 = 0.7 + 0.8 + I D1 ⎜ + → I D 2 = I D1 = 29.7µA I D1 = −4 ⎟ ⎜ 6 x10−4 128.5 4 x10 ⎠ ⎝ 15.85 Since the values of IS and IE are the same, VBE1 = VEB 2 I 1.30 1.30 = VBE1 + VEB 2 = 2VT ln C | IC = 10−15 exp = 196 µA IS 2(0.025V ) 15.86 2 IC IC IC 1.30 = VBE1 + VEB 2 = VT ln + VT ln = VT ln I S1 IS 2 I S1I S 2 IC = 1.30 = 391 µA (4 x10 )(10 )exp 0.025 −15 −15 15-58 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.87 vO Slope = 1 +0.7 V vS -0.7 V 15.88 *Problem 15.88 – Figure P15.87 VCC 3 0 DC 10 VEE 5 0 DC -10 VBB 2 1 DC 1.3 VS 1 0 DC 0 Q1 3 2 4 NBJT Q2 5 1 4 PBJT RL 4 0 1K .MODEL NBJT NPN IS=5FA BF=60 .MODEL PBJT PNP IS=1FA BF=50 .OP .DC VS -10 +8.7 0.01 .PROBE .END 10V vO 0V vS -10V -10V -5V 0V 5V 10V 15.89 Since the base currents are zero (β F = ∞), VBE1 + VEB 2 = (250µA)(5kΩ) = 1.25V 1.25V = VT ln IC I I2 + VT ln C = VT ln C IS1 IS 2 IS1IS 2 | IC = 1.25 = 22.8 µA (10 )(10 )exp 0.025 −15 −16 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-59 15.90 VGS1 + VGS 2 = (0.5mA)(4kΩ)= 2.00V | 2.00V = VTN + ⎛ 2 ⎞ 2 ⎟ | 2.00 = 0.75 + 0.75 + I D1 ⎜ + ⎜ 5 x10−4 2 x10−4 ⎟ ⎠ ⎝ 15.91 I SS ≥ 5V 5V = = 5.00 mA | iS = I SS + iL RL 1kΩ 2 I D1 ⎛ 2 I D2 ⎞ ⎟ | I D2 = I D1 −⎜ V − TP ⎟ Kn ⎜ K p ⎝ ⎠ I D1 = 0.5 → I D 2 = I D1 = 9.38 µA 163.3 iSmax = I SS + For I SS Power delivered from the supplies : P (t )= 10V (iD )+ 10V (I SS )= 0.05(2 + sin 2000πt ) W Pav = 1 T T 5V 5V min = I SS + 5.00 mA | iS = I SS − = I S − 5.00 mA 1kΩ 1kΩ = 5.00 mA, iSmax = 10.0 mA | iSmin = 0 | iD = 0.005( 1 + sin 2000πt ) A ∫ 0.05(2 + sin 2000πt )dt = 100 mW 0 ⎛ 5 ⎞2 1 12.5mW Signal power developed in RL : Pac = ⎜ ⎟ = 12.5mW | η = 100% = 12.5% 100 mW ⎝ 2 ⎠ 1kΩ 15.92 +5V T 0 -5V t 2 2 1 ⎡(+5V ) T (−5V ) T ⎤ ⎥ = 10.0 mW Pac = ⎢ + 5 kΩ 2 ⎥ T⎢ ⎣ 5 kΩ 2 ⎦ 1⎡ 5V T −5V T ⎤ Pav = ⎢5V − 5V = 10.0 mW ⎦ T ⎣ 5 kΩ 2 5 kΩ 2 ⎥ | η = 100% 15-60 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.93 +10 V t 0 T 4 T 2 2 T 4 T -10 V 1 Pac = T Pav = 1 T T ∫ 0 v 2 (t ) 4 dt = R T ∫ 0 ⎛ 40 t ⎞ ⎟ ⎜ 6400 ⎝ T ⎠ dt = 3 R T R T 4 ∫ t dt = 2 0 100 3R T 4 T ∫ 10i(t )dt = 0 20 T T 2 ∫ i(t )dt = 0 40 T T 4 ∫ 0 1600 40 t dt = 2 T R TR ∫ 0 100 50 tdt = | η = 100% 3R = 66.7% 50 R R 15.94 *Problem 15.94(a) VBB = 0 V VCC 3 0 DC 10 VEE 5 0 DC -10 VBB 2 1 DC 0 VS 1 0 DC 0 SIN(0 4 2000) Q1 3 2 4 NBJT Q2 5 1 4 PBJT RL 4 0 2K .MODEL NBJT NPN IS=5FA BF=60 .MODEL PBJT PNP IS=1FA BF=50 .OP .TRAN 1U 2M .FOUR 2000 V(4) .PROBE .END *Problem 15.94(b) VBB = 1.3 V VCC 3 0 DC 10 VEE 5 0 DC -10 VBB 2 1 DC 1.3 VS 1 0 DC 0 SIN(0 4 2000) Q1 3 2 4 NBJT Q2 5 1 4 PBJT RL 4 0 2K .MODEL NBJT NPN IS=5FA BF=60 .MODEL PBJT PNP IS=1FA BF=50 .OP .TRAN 1U 2M .FOUR 2000 V(4) .PROBE .END ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-61 (a) HARMONIC FREQUENCY FOURIER NORMALIZED NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) 1 2 3 4 5 6 7 8 9 2.000E+03 4.000E+03 6.000E+03 8.000E+03 1.000E+04 1.200E+04 1.400E+04 1.600E+04 1.800E+04 3.056E+00 2.693E-02 2.112E-01 3.473E-02 7.718E-02 4.064E-02 3.179E-02 4.109E-02 2.127E-02 1.000E+00 8.811E-03 6.910E-02 1.136E-02 2.525E-02 1.330E-02 1.040E-02 1.345E-02 6.960E-03 -4.347E-01 -1.300E+02 -1.744E+02 -1.550E+02 -1.678E+02 -1.679E+02 -1.580E+02 -1.736E+02 -1.568E+02 0.000E+00 -1.296E+02 -1.740E+02 -1.545E+02 -1.674E+02 -1.675E+02 -1.576E+02 -1.731E+02 -1.564E+02 PHASE PHASE (DEG) TOTAL HARMONIC DISTORTION = 7.831458E+00 PERCENT with VBB = 0 (b) HARMONIC FREQUENCY FOURIER NORMALIZED NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) 1 2 3 4 5 6 7 8 9 2.000E+03 4.000E+03 6.000E+03 8.000E+03 1.000E+04 1.200E+04 1.400E+04 1.600E+04 1.800E+04 3.853E+00 1.221E-02 1.537E-02 1.504E-02 1.501E-02 1.531E-02 1.435E-02 1.467E-02 1.382E-02 1.000E+00 3.169E-03 3.990E-03 3.903E-03 3.897E-03 3.973E-03 3.726E-03 3.807E-03 3.587E-03 2.544E-01 6.765E+01 9.046E+01 5.520E+01 5.500E+01 4.231E+01 3.680E+01 2.823E+01 2.087E+01 0.000E+00 6.740E+01 9.020E+01 5.495E+01 5.475E+01 4.206E+01 3.654E+01 2.798E+01 2.062E+01 PHASE PHASE (DEG) TOTAL HARMONIC DISTORTION = 1.064939E+00 PERCENT with VBB = 1.3 V 15-62 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.95 VBE 2 0.7V = = 70.0 mA. R 10Ω 0.07 v S = 1000iB + VBE1 + VBE 2 + 250iE = 1000 + 0.7 + 0.7 + 250(0.07) = 19.6 V 101 The current begins to limit at iE = 15.96 *Problem 15.96 – Figure 15.38 VCC 3 0 DC 50 VS 1 0 DC 1 R1 1 2 1K Q1 3 2 4 NBJT Q2 2 4 5 NBJT R 4 5 10 RL 5 0 250 .MODEL NBJT NPN IS=1FA BF=100 .OP .DC VS 1 50 .05 .PROBE .END 120mA 80mA Slope = 1k Ω iL 40mA Slope = 250 0A 0V 10V Ω 20V vS 30V 40V 50V The results agree well with hand calculations. 15.97 I2 RG = VGS 4 − VGS 5 | (0.25mA)(7kΩ)= VTN 4 + 2ID 4 ⎛ 2 I D5 ⎞ ⎟ | I D5 = I D 4 −⎜ V − TP 5 0.005 ⎜ 0.002 ⎟ ⎝ ⎠ ⎛ 2 ⎞ 2 ⎟ → I D5 = I D 4 = 23.5 µA 1.75 − 0.75 − 0.75 = I D 4 ⎜ + ⎜ 0.005 0.002 ⎟ ⎠ ⎝ ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-63 15.98 +15 V 0.2 V 50 Ω Q 4 + 2.4 k Ω 1.2 V + 0.7 V 0.5 V + 2k Ω Q 5 For VBE 4 = 0.7V , VEB 5 = I2 RB - VBE 4 VEB 5 = 1.2 − 0.7 = 0.5V and Q 5 is off. VEQ = 15 − 0.2 − 500µA(50Ω) = 14.8V REQ = 50Ω IC 4 = 100 -15 V - - 500 µ A (14.8 − 0.7)V = 6.98 mA 50Ω + 101(2 kΩ) 15.99 Rout = 1 ⎛ rπ ⎞ 1 ⎛ β o ⎞⎛ VT ⎞ 1 ⎛ VT ⎞ 1 ⎛ 0.025V ⎞ ⎜ ⎟ = 25.0 mΩ ⎜ ⎟= ⎜ ⎟⎜ ⎟ = ⎜ ⎟= n 2 ⎝ β o + 1⎠ 100 ⎝ β o + 1⎠⎝ IC ⎠ 100 ⎝ IE ⎠ 100 ⎝ 10 mA ⎠ V 9 − 0.7 = 97.9 µA | Looking back into the 200 kΩ + 101(82 kΩ) Ω rπ 1 ⎛ rπ ⎞ 1 100(0.025V ) = = 253Ω ⎟ | 2⎜ β o + 1 101 97.9µA n ⎝ β o + 1⎠ 15.100 IC = 100 I B = 100 transformer : Rth = Desire to match the Thevenin equivalent resistance to RL : vth = rπ (β + 1)n R + (β + 1)n R 2 o L 2 o vs = L (101)253 v 25.6 kΩ + ( 101)253 1 253Ω = 10Ω → n = 5.03 n2 s = 0.500v s | Using the ideal transformer | v th = 1 vo Rth + nvo n RL relationships : vth = i1 Rth + nvo | i1 = vo = vth R n + th nRL | vo = 1 1 vo i2 = n n RL 0.500v s = 0.0497vs | vo = 0.0497sin 2000πt 253Ω 5.03 + 5.03( 10Ω) ⎛ 0.0497 ⎞2 1 Po = ⎜ = 0.124 mW ⎟ ⎝ 2 ⎠ 10 15-64 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.101 (a) V 2 MΩ = −6V | REQ = 2 MΩ 2 MΩ = 1MΩ 2 MΩ + 2 MΩ −6 − 0.7 − (−12) V 100(0.025V ) = 22.8 µA | rπ = = 110 kΩ IO = 100 I B = 100 22.8µA 1 MΩ + 101(220 kΩ) Ω EQ = −12V VCE = 12 − I E (220 kΩ)= 12 − Rout (b) Using the CVD model for the diode, (50 + 6.98)V = 2.50 MΩ 101 22.8µA)(220 kΩ)= 6.98V | ro = ( 22.8µA 100 ⎞ ⎛ ⎛ ⎞ 100(220kΩ) β o RE ⎟ = 43.9 MΩ ⎜ = ro ⎜1 + ⎟ = 2.50 MΩ⎜1 + ⎟ 1 M Ω + 110 k Ω + 220 k Ω ⎝ Rth + rπ + RE ⎠ ⎠ ⎝ 2 MΩ VEQ = −12 + ( 12V − 0.7V ) + 0.7 = −5.65V | REQ = 2 MΩ 2 MΩ = 1MΩ 2 MΩ + 2 MΩ −5.65 − 0.7 − (−12) V 100(0.025V ) = 24.3 µA | rπ = = 103kΩ IO = 100 I B = 100 24.3µA 1 MΩ + 101(220 kΩ) Ω VCE = 12 − I E (220 kΩ)= 12 − Rout (50 + 6.60)V = 2.33 MΩ 101 24.3µA)(220 kΩ)= 6.60V | ro = ( 24.3µA 100 ⎞ ⎛ ⎛ ⎞ 100(220kΩ) β o RE ⎟ 1 + = ro ⎜1 + ⎟ = 2.33 MΩ⎜ ⎜ 1 MΩ + 103kΩ + 220 kΩ ⎟ = 41.1 MΩ ⎝ Rth + rπ + RE ⎠ ⎠ ⎝ 15.102 The dc analysis is the same as Problem 15.101. However, the bypass capacitor provides as ac ground at the base of the transistor so that Rth = 0. ⎛ ⎛ 100(220kΩ) ⎞ β o RE ⎞ ⎟ = 169 MΩ ⎜ Rout = ro ⎜1 + ⎟ = 2.50 MΩ⎜1 + ⎟ 110 k Ω + 220 k Ω ⎝ rπ + RE ⎠ ⎠ ⎝ ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-65 15.103 (a) V 430 kΩ = −5.53V | REQ = 270 kΩ 430 kΩ = 166 kΩ 270 kΩ + 430 kΩ −5.53 − 0.7 − (−9) 150(0.025V ) V = 144 µA | rπ = = 26.0 kΩ IO = 100 I B = 150 144µA 166 kΩ + 151( 18 kΩ) EQ = −9V 18kΩ) = 9 − VCE = 9 − I E ( Rout (b) Using the CVD model for the diode, −5.10 − 0.7 − (−9) (75 + 6.39)V = 565kΩ 151 144µA)( 18 kΩ)= 6.39V | ro = ( 150 144µA ⎛ ⎞ ⎛ ⎞ 150( 18 kΩ) β o RE ⎟ = ro⎜1 + 1 + ⎟ = 565kΩ⎜ ⎜ 166 kΩ + 26.0 kΩ + 18 kΩ ⎟ = 7.83 MΩ ⎝ Rth + rπ + RE ⎠ ⎝ ⎠ ⎛ ⎞ 270 kΩ VEQ = −9 + (9V − 0.7V ) ⎜ ⎟ + 0.7 = −5.10V | REQ = 270 kΩ 430 kΩ = 166 kΩ ⎝ 270 kΩ + 430 kΩ ⎠ IO = 100 I B = 150 166 kΩ + 151( 18 kΩ) V = 166 µA | rπ = 150(0.025V ) 166µA = 22.6 kΩ 18kΩ) = 9 − VCE = 9 − I E ( (75 + 5.99)V = 488kΩ 151 166µA)( 18 kΩ)= 5.99V | ro = ( 150 166µA ⎛ ⎞ ⎛ ⎞ 150( 18 kΩ) β o RE ⎟ Rout = ro⎜1 + 1 + ⎟ = 488 kΩ⎜ ⎜ 166 kΩ + 22.6 kΩ + 18 kΩ ⎟ = 6.87 MΩ ⎝ Rth + rπ + RE ⎠ ⎝ ⎠ 200 kΩ (c) VEQ = −5V 100kΩ + 200kΩ = −3.33V | REQ = 100kΩ 200kΩ = 66.7kΩ −3.33 − 0.7 − (−5) 100(0.025V ) IO = 100 I B = 100 V = 61.3 µA | rπ = = 40.8 kΩ 61.3µA 66.7 kΩ + 101( 15kΩ) 15kΩ)= 5 − VCE = 5 − I E ( Rout (d ) Using the CVD model for the diode, −2.87 − 0.7 − (−5) (75 + 4.07)V = 1.29 MΩ 101 61.3µA)( 15kΩ)= 4.07V | ro = ( 61.3µA 100 ⎛ ⎞ ⎛ ⎞ 100( 15kΩ) β o RE ⎜ ⎟ = ro⎜1 + = 1.29 M Ω 1 + ⎟ ⎜ 66.7 kΩ + 40.8 kΩ + 15kΩ ⎟ = 17.1 MΩ ⎝ Rth + rπ + RE ⎠ ⎝ ⎠ ⎛ ⎞ 100 kΩ VEQ = −5 + (5V − 0.7V ) ⎜ ⎟ + 0.7 = −2.87V | REQ = 100 kΩ 200 kΩ = 66.7 kΩ ⎝100 kΩ + 200 kΩ ⎠ IO = 100 I B = 100 66.7 kΩ + 101( 15kΩ) V = 90.4 µA | rπ = 100(0.025V ) 90.4µA = 27.7 kΩ 15kΩ)= 5 − VCE = 5 − I E ( Rout (75 + 3.63)V = 870 kΩ 101 90.4µA)( 15kΩ)= 3.63V | ro = ( 90.4µA 100 ⎛ ⎞ ⎛ ⎞ 100( 15kΩ) β o RE ⎟ = ro⎜1 + 1 + ⎟ = 870 kΩ⎜ ⎜ 66.7 kΩ + 27.7 kΩ + 15kΩ ⎟ = 12.8 MΩ ⎝ Rth + rπ + RE ⎠ ⎝ ⎠ 15-66 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.104 (a) VEQ = −12V 2 MΩ = −6V | REQ = 2 MΩ 2 MΩ = 1MΩ 2 MΩ + 2 MΩ −6 − 0.7 − (−12) V 100(0.025V ) = 22.8 µA | rπ = = 110 kΩ IO = 100 IB = 100 1MΩ + 101(220 kΩ) Ω 22.8µA VCE = 12 − IE (220 kΩ) = 12 − 101 (50 + 6.98)V = 2.50 MΩ (22.8µA)(220 kΩ) = 6.98V | ro = 100 22.8µA ⎛ ⎞ ⎛ ⎞ 100(220 kΩ) β o RE Rout = ro⎜1 + ⎟ = 2.50 MΩ⎜1 + ⎟ = 43.9 MΩ ⎝ Rth + rπ + RE ⎠ ⎝ 1MΩ + 110 kΩ + 220 kΩ ⎠ (b) Using the CVD model for the diode (diode - connected transistor), 2 MΩ VEQ = −12 + ( 12V − 0.7V ) + 0.7 = −5.65V | REQ = 2 MΩ 2 MΩ = 1MΩ 2 MΩ + 2 MΩ −5.65 − 0.7 − (−12) V 100(0.025V ) = 24.3 µA | rπ = = 103kΩ IO = 100 I B = 100 24.3µA 1 MΩ + 101(220 kΩ) Ω VCE = 12 − I E (220 kΩ) = 12 − Rout (50 + 6.60)V = 2.33 MΩ 101 24.3µA)(220 kΩ)= 6.60V | ro = ( 24.3µA 100 ⎞ ⎛ ⎛ ⎞ 100(220 kΩ) β o RE ⎟ ⎜ = 2.33 M Ω 1 + = ro ⎜1 + ⎟ ⎜ 1 MΩ + 103kΩ + 220kΩ ⎟ = 41.1 MΩ ⎝ Rth + rπ + RE ⎠ ⎠ ⎝ 15.105 IC R 1 IC RB Q Q R V 2 B RE -12 V RE -12 V ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-67 A spread sheet will be used to assist in this design using β F = β o = 100 & VA = 70V The maximum current in the two bias resistors is 0.2mA. To allow some room for tolerances, choose I1 ≅ 0.15mA. Neglecting the transistor base current, 12V V V = 80 kΩ | R2 = B (R1 + R2 ) = B 80kΩ | RB = R1 R2 0.15mA 12 12 ⎞ 100( VB − 0.7) (VB − 0.7) or R = 1 ⎛ ⎟ ⎜ IC = 100 − R E B ⎟ | VCE = 12 − I E RE 101 ⎜ RB + 101RE I C ⎠ ⎝ ⎛ ⎞ β o RE 70 + VCE ro = | Rout = ro ⎜1 + ⎟ IC ⎝ RB + rπ + RE ⎠ Now, a spreadsheet MATLAB, MATHCAD, etc. can be used to explore the design space with VB as the primary design variable. R1 + R2 = VB 0.500 0.600 0.700 0.800 0.900 1.000 1.100 1.200 1.300 1.400 1.500 1.600 1.700 1.800 1.900 2.000 R2 3.33E+03 4.00E+03 4.67E+03 5.33E+03 6.00E+03 6.67E+03 7.33E+03 8.00E+03 8.67E+03 9.33E+03 1.00E+04 1.07E+04 1.13E+04 1.20E+04 1.27E+04 1.33E+04 Two possible solutions 0.916 1.800 6.20E+03 1.20E+04 7.50E+04 6.80E+04 5.73E+03 1.02E+04 1.50E+02 1.00E+03 2.10E+05 1.07E+06 5.85E+05 8.86E+06 R1 7.67E+04 7.60E+04 7.53E+04 7.47E+04 7.40E+04 7.33E+04 7.27E+04 7.20E+04 7.13E+04 7.07E+04 7.00E+04 6.93E+04 6.87E+04 6.80E+04 6.73E+04 6.67E+04 RB 3.19E+03 3.80E+03 4.39E+03 4.98E+03 5.55E+03 6.11E+03 6.66E+03 7.20E+03 7.73E+03 8.24E+03 8.75E+03 9.24E+03 9.73E+03 1.02E+04 1.07E+04 1.11E+04 RE ro Rout 5.57E+05 9.73E+04 5.13E+03 1.80E+05 5.56E+05 1.09E+06 1.75E+06 2.52E+06 3.38E+06 4.31E+06 5.32E+06 6.38E+06 7.50E+06 8.68E+06 9.90E+06 1.12E+07 -2.30E+02 -1.74E+05 -1.37E+02 -8.00E+04 -4.35E+01 1.41E+04 4.97E+01 1.08E+05 1.43E+02 2.03E+05 2.37E+02 2.97E+05 3.30E+02 3.91E+05 4.24E+02 4.86E+05 5.18E+02 5.81E+05 6.11E+02 6.76E+05 7.05E+02 7.71E+05 8.00E+02 8.66E+05 8.94E+02 9.61E+05 9.88E+02 1.06E+06 1.08E+03 1.15E+06 1.18E+03 1.25E+06 IO 1.04E-03 9.89E-04 The first solution is the lowest value of VB that was found to meet the output specification using the nearest 5% values. The second is one in which the values were found to be very close to existing standard 5% resistor values, but it uses twice the value of VB and has a smaller output voltage compliance range. 15-68 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.106 330 kΩ 10V = 3.27V | REQ = 330 kΩ 680 kΩ = 222 kΩ 330kΩ + 680kΩ ⎛ 5x10−4 ⎞ 2 Assume active region operation : I D = ⎜ VGS − 1) ⎟( ⎝ 2 ⎠ ⎛ 5 x10−4 ⎞ 2 VGS − 1) ⇒ VGS = 1.467V | IO = I D = 54.5 µA VGS = 3.27 − (33kΩ)I D = 3.27 − 33 x103 ⎜ ⎟( ⎝ 2 ⎠ VEQ = VDS = 10 − (33kΩ)I D = 8.20V | Active region operation is correct. ro = 100 + 8.20 V = 1.99 MΩ | g m = 2 5 x10−4 54.5 x10−6 1 + (0.01)8.20 = 0.243 mS 54.5 µA ( )( )[ ] Rout = ro 1 + g m 33 x103 = 1.99 MΩ 1 + 0.243mS (33kΩ) = 17.9 MΩ 15.107 [ ( )] [ ] VEQ = VEQ 68kΩ 3V = 0.760V | REQ = 68 kΩ 200 kΩ = 50.8 kΩ 68kΩ + 200kΩ < VTN , (0.76V < 1V) so the transistor is off, and IO = ID = 0. The circuit designer made an error and failed to check the final design. 15.108 VEQ = 100kΩ 6V = 2V | REQ = 100 kΩ 200 kΩ = 66.7 kΩ 100kΩ + 200 kΩ ⎛ 5 x10−4 ⎞ 2 Assume active region operation : 2 = VGS + ( VGS − 1) 16 kΩ)I D | I D = ⎜ ⎟( ⎝ 2 ⎠ VDS = VO − ( 16 kΩ)I D = 6 − ( 16 kΩ)(38.1µA)= 5.39V | Active region is correct. ro = 100 + 5.39 V = 2.77 MΩ | g m ≅ 2 5 x10−4 38.1x10−6 1 + 0.01(5.39) = 0.200 mS 38.1 µA 2 4VGS − 7VGS + 2 = 0 ⇒ VGS = 1.390V | IO = I D = 38.1 µA ( )( )[ ] Rout = ro 1 + g m 16 x103 = 2.77 MΩ 1 + 0.200 mS ( 16 kΩ) = 11.6 MΩ [ ( )] [ ] ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-69 15.109 VEQ = 15V 200 kΩ = 10V | REQ = 200 kΩ 100 kΩ = 66.7 kΩ 200 kΩ + 100 kΩ (15 − 0.7 − 10)V = 88.6 µA | r = 75(0.025V ) = 21.2kΩ IO = 75I B = 75 π 88.6µA 66.7 kΩ + 76(47 kΩ) VEC = 15 − I E RE = 15 − Rout (50 + 10.7)V = 685kΩ 76 88.6 µA)(47 kΩ) = 10.7V | ro = ( 75 88.6µA ⎛ ⎞ ⎛ ⎞ 75(47kΩ) β o RE ⎟ = ro ⎜1 + 1 + ⎟ = 685kΩ⎜ ⎜ 66.7 kΩ + 21.2 kΩ + 47 kΩ ⎟ = 18.6 MΩ ⎝ Rth + rπ + RE ⎠ ⎝ ⎠ 15.110 VEQ = 5V 33kΩ = 3.84V | REQ = 33kΩ 10 kΩ = 7.67 kΩ 33kΩ + 10 kΩ 75(0.025V ) 5 − 0.7 − 3.84 V = 284 µA | rπ = = 6.60 kΩ IO = 75 IB = 75 7.67 kΩ + 76(1.5 kΩ) 284 µA 76 (60 + 4.57)V = 227 kΩ (284 µA)(1.5 kΩ) = 4.57V | ro = 75 284 µA ⎛ ⎞ ⎛ ⎞ 75(1.5 kΩ) β o RE = ro⎜1 + ⎟ = 227 kΩ⎜1 + ⎟ = 1.85 MΩ ⎝ Rth + rπ + RE ⎠ ⎝ 7.67 kΩ + 6.60 kΩ + 1.5 kΩ ⎠ VEC = 5 − IE RE = 5 − Rout 15.111 VEQ = 10V 300 kΩ = 7.50V | REQ = 300 kΩ 100 kΩ = 75.0 kΩ 300 kΩ + 100 kΩ 90(0.025V ) 10 − 0.7 − 7.50 IO = 90 I B = 90 V = 94.6 µA | rπ = = 23.4 kΩ 94.6µA 75.0 kΩ + 91( 18 kΩ) VEC = 10 − I E RE = 10 − Rout (75 + 8.28)V = 880kΩ 91 94.6µA)( 18 kΩ)= 8.28V | ro = ( 90 94.6µA ⎛ ⎞ ⎛ ⎞ 90( 18kΩ) β o RE ⎟ = ro⎜1 + 1 + ⎟ = 880 kΩ⎜ ⎜ 75.0 kΩ + 23.4 kΩ + 18 kΩ ⎟ = 13.1 MΩ ⎝ Rth + rπ + RE ⎠ ⎝ ⎠ 15-70 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.112 VEQ = 200kΩ 6V = 4V | REQ = 200 kΩ 100 kΩ = 66.7 kΩ | 6 − 162 − 16e3* x103 I D + VGS = 4 200kΩ + 100 kΩ 2 7.5 x10−4 VGS + 0.75) Assume active region operation : I D = ( 2 3 2 − 16 x10 I D + VGS = 0 ⇒ VGS = −1.13V | IO = I D = 54.3 µA VDS = − 6 − 16 x103 I D = −5.13V | Active region is correct. ro = 100 + 5.13 V = 1.94 MΩ | g m = 2 7.5 x10−4 54.3 x10−6 1 + 0.01(5.13) = 0.292 mS 54.3 µA ( ) ( )( )[ ] Rout = ro ( 1 + g m RS )= 1.94 MΩ 1 + 0.292mS ( 16 kΩ) = 11.0 MΩ 15.113 [ ] VEQ = 9V 2 MΩ = 6V | REQ = 2 MΩ 1MΩ = 667 kΩ | 9 − 105 I D + VGS = 6 2 MΩ + 1 MΩ 2ID Kp Assume active region operation : VGS = VTP − ⎛ 2ID ⎞ ⎟ ⇒ IO = I D = 17.0 µA 1.2 x105 I D = 3 + ⎜ − 0.75 − ⎜ 7.5 x10−4 ⎟ ⎝ ⎠ VDS = − 9 − 1.2 x105 I D = −6.96V | Active region is correct. ro = 100 + 6.96 V = 6.29 MΩ | g m = 2 7.5 x10−4 17.0 x10−6 1 + 0.01(6.96) = 0.165mS 17.0 µA ( ) Rout = ro ( 1 + gm 15.114 ( )( )[ R )= 6.29 MΩ[ 1 + 0.165mS ( 1.2 x10 ) ]= 131 MΩ 5 S ] VEQ = 200 kΩ 4V = 3.05V | REQ = 200 kΩ 62 kΩ = 47.3kΩ 200kΩ + 62kΩ 2 7.5 x10−4 VGS + 0.75) ( 2 Assume active region operation : 4 − 43 x103 I D + VGS = 3.05 | I D = 0.95 − 43x103 I D + VGS = 0 ⇒ VGS = −0.6034V | IO = I D = 8.06 µA VDS = − 4 − 43x103 I D = −3.65V | Active region is correct. ro = ( ) 100 + 3.65 V = 12.9 MΩ | g m = 2 7.5 x10−4 8.06 x10−6 1 + 0.01(3.65) = 112 µS 8.06 µA ( )( )[ ] 1 + g m RS )= 12.9 MΩ 1 + 112µS (43kΩ) = 75.0 MΩ Rout = ro ( [ ] ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-71 15.115 1 + g m RS )≅ Estimating Rout ≅ ro ( 50V ⎛ ⎞ ⎜1 + 2 2 x10−4 1.75 x10−4 RS ⎟ −4 ⎝ ⎠ 1.75 x10 Note that including λ in the g m expression will increase Rout above this estimate. ( )( ) Hence neglecting λ represents a conservative simplification. 286 kΩ 1 + 2.65 x10−4 RS ≥ 2.5 MΩ ⇒ RS ≥ 29.2 kΩ | Choose RS = 33kΩ ⎛ ⎞ 2 1.75 x10−4 ⎟ ⎜ VG = VDD − I D RS + VGS = 12 − 1.75 x10 3.3x10 + −1 − = 3.90V ⎜ 2 x10−4 ⎟ ⎝ ⎠ 12V R4 12 = 3.90 | I2 ≤ 25µA | Assign I2 = 20µA | R3 + R4 = = 600kΩ 20µA R3 + R4 −4 ( ) ( 4 ) ( ) 3.90 (R3 + R4 ) = 195kΩ ⇒ R4 = 200kΩ | R3 = 430kΩ 12 15.116 68 kΩ (a) VEQ = −12V 68kΩ + 33kΩ = −8.08V | REQ = 68kΩ 33kΩ = 22.2kΩ | VB = −8.08 − (I B1 + I B 2 )RTH ⎛ V − 0.7 − (−12) V − 0.7 − (−12)⎞ B B ⎟22.2kΩ → VB = −8.11V VB = −8.08 − ⎜ + ⎜ 126 20 kΩ 126( 100kΩ) ⎟ ( ) ⎠ ⎝ ⎛ V − 0.7 − (−12)⎞ V − 0.7 − (−12)⎞ 125 ⎛ ⎟ = 158 µA | IC 2 = α F I E 2 = 125 ⎜ B ⎟ = 31.7 µA ⎜ B I C1 = α F I E1 = ⎟ ⎟ 126 ⎜ 126 ⎜ 20kΩ 100kΩ ⎠ ⎠ ⎝ ⎝ R4 = VCE = 0 − (−8.11 − 0.7)= 8.87V | ro1 = rπ 1 = 125(0.025V ) 158 µA = 19.8 kΩ | rπ 2 = (50 + 8.11)V = 368kΩ | 158µA 31.7µA = 98.6 kΩ Rth1 = REQ rπ 2 + (β o + 1)( 100 kΩ) [ ] 125(0.025V ) 126)( 100 kΩ) = 22.2kΩ R th1 = 22.2 kΩ 98.6kΩ + ( ⎞ ⎛ ⎛ ⎞ 125(20 kΩ) β o RE ⎟ ⎜ = 368 k Ω 1 + Rout1 = ro1⎜1 + ⎟ ⎜ 22.2kΩ + 19.8kΩ + 20kΩ ⎟ = 15.2 MΩ ⎝ Rth + rπ 1 + RE ⎠ ⎠ ⎝ ro2 = [ ] (50 + 8.11)V = 1.83 MΩ 31.7µA | Rth2 = RTH rπ 1 + (β o + 1)(20 kΩ) [ ] Rth2 = 22.2 kΩ 19.8 kΩ + ( 126)(20 kΩ) = 22.0 kΩ ⎞ ⎛ ⎛ ⎞ 125( 100kΩ) β o RE ⎟ ⎜ = 1.83 M Ω 1 + Rout 2 = ro2⎜1 + ⎟ ⎜ 22.0kΩ + 98.6kΩ + 100kΩ ⎟ = 106 MΩ ⎝ Rth + rπ 2 + RE ⎠ ⎠ ⎝ [ ] 15-72 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.116 cont. (b) Using the CVD model for diode Q for dc calculations, (12V − 0.7) 68kΩ = −7.61V | R = 68kΩ 33kΩ = 22.2kΩ | V = −8.08 − I + I R V = ( ) ( ) 68 kΩ + 33kΩ ⎛ V − 0.7 − (−12) V − 0.7 − (−12)⎞ ⎟22.2 kΩ → V = −7.65V V = −7.61 − ⎜ + ⎜ 126 20 kΩ ⎟ 126 100 k Ω ( ) ( ) ⎝ ⎠ 125 ⎛ V − 0.7 − (−12)⎞ 125 ⎛ V − 0.7 − (−12)⎞ ⎜ ⎜ ⎟ = 181 µA | I = α I = ⎟ = 36.2 µA I =α I = 3 EQ EQ B B1 B2 B B B B C1 F E1 EQ 126 ⎜ ⎝ B 20 kΩ ⎟ ⎠ C2 F E2 VCE = 0 − (−7.65 − 0.7)= 8.35V | ro1 = rπ 1 = 125(0.025V ) 181 µA = 17.3kΩ | rπ 2 = (50 + 8.35)V = 322kΩ | 181µA 36.2µA = 86.3kΩ 126 ⎜ ⎝ B 100 kΩ ⎟ ⎠ Rth1 = REQ rπ 2 + (β o + 1)( 100 kΩ) [ ] 125(0.025V ) 126)( 100 kΩ) = 22.2 kΩ Rth1 = 22.2 kΩ 86.3kΩ + ( ⎛ ⎞ ⎛ ⎞ 125(20 kΩ) β o RE ⎜ ⎟ Rout1 = ro1⎜1 + = 322 k Ω 1 + ⎟ ⎜ 22.2 kΩ + 17.3kΩ + 20 kΩ ⎟ = 13.9 MΩ ⎝ Rth + rπ 1 + RE ⎠ ⎝ ⎠ ro2 = [ ] (50 + 8.35)V = 1.61MΩ 36.2µA | Rth2 = RTH rπ 1 + (β o + 1)(20 kΩ) [ ] Rth2 = 22.2 kΩ 17.3kΩ + ( 126)(20 kΩ) = 22.0 kΩ ⎛ ⎞ ⎛ ⎞ 125( 100 kΩ) β o RE ⎜ ⎟ = 98.2 MΩ Rout 2 = ro2 ⎜1 + ⎟ = 1.61 MΩ⎜1 + ⎟ 22.0 k Ω + 86.3 k Ω + 100 k Ω ⎝ Rth + rπ 2 + RE ⎠ ⎝ ⎠ [ ] ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-73 15.117 VEQ = −12V 20kΩ = −4.07V | REQ = 20 kΩ 39 kΩ = 13.2 kΩ 20 kΩ + 39kΩ ⎛ 1 VB + 0.7 1 VB + 0.7 1 VB + 0.7 ⎞ + + I B = −⎜ ⎟ | VB = −4.07V + 13200 I B → VB = −3.95V 76 16kΩ 76 8.2kΩ ⎠ ⎝ 76 33kΩ ⎛ β o (33kΩ) ⎞ 75 ⎛ 0 − 0.7V − (−3.95V )⎞ ⎟ = 97.2µA | Rout1 = ro1⎜1 + ⎟ I C1 = ⎜ ⎟ ⎜ R + r + 33kΩ ⎟ 76 ⎜ 33 k Ω th 1 π 1 ⎠ ⎠ ⎝ ⎝ Rth1 = 13.2kΩ rπ 2 + (β o2 + 1) 16kΩ rπ 3 + (β o3 + 1)8.2 kΩ ≅ 13.2kΩ Rout1 = IC 2 = ⎞ 75(33kΩ) 60 + 8.75 ⎛ ⎟ = 27.4 MΩ ⎜1 + ⎟ 97.2µA ⎜ ⎝ 13.2kΩ + 19.3kΩ + 33kΩ ⎠ [ ][ ] ⎛ βo( 16kΩ) ⎞ 75 ⎛ 0 − 0.7V − (−3.95V )⎞ ⎟ = 201µA | Rout 2 = ro2 ⎜1 + ⎟ ⎜ ⎟ ⎜ R + r + 16kΩ ⎟ 76 ⎜ 16kΩ th2 π2 ⎠ ⎠ ⎝ ⎝ Rth2 = 13.2kΩ rπ 1 + (β o1 + 1)33kΩ rπ 3 + (β o3 + 1)8.2 kΩ ≅ 13.2kΩ ⎞ 75( 16 kΩ) 60 + 8.75 ⎛ ⎟ = 11.0 MΩ ⎜ Rout 2 = 1+ ⎟ 201µA ⎜ 13.2 k Ω + 9.33 k Ω + 16 k Ω ⎠ ⎝ ⎛ β o (8.2kΩ) ⎞ 75 ⎛ 0 − 0.7V − (−3.95V )⎞ ⎟ ⎟ ⎜ ⎜ IC 3 = ⎜ ⎟ = 391µA | Rout 3 = ro3 ⎜1 + R + r + 8.2kΩ ⎟ 76 ⎝ 8.2kΩ th2 π2 ⎠ ⎠ ⎝ Rth3 = 13.2kΩ rπ 1 + (β o1 + 1)33kΩ rπ 2 + (β o2 + 1) 16 kΩ ≅ 13.2kΩ ⎞ 75(8.2 kΩ) 60 + 8.75 ⎛ ⎟ = 4.30 MΩ ⎜ Rout 3 = 1+ ⎟ 391µA ⎜ 13.2 k Ω + 4.80 k Ω + 8.2 k Ω ⎠ ⎝ [ ][ ] [ ][ ] 15-74 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.118 VEQ = 12V 2 MΩ = 6.00V | REQ = 2 MΩ 2 MΩ = 1.00 MΩ 2 MΩ + 2 MΩ 2 I D1 Assume saturation : 12 − 105 I D1 + VGS 1 = 6 | VGS1 = −1 − 2.5x10−4 I D1 = 44.1 µA, VGS 1 = −1.59V , VDS1 = −(6 − VGS 1 )= −7.59V 1 + g m1 R1 )= Rout1 = ro1( 50V + 7.59V ⎛ ⎞ ⎜1 + 100kΩ 2(250µA)(44.1µA) 1 + 0.02(7.69) ⎟ = 22.2 MΩ ⎝ ⎠ 44.1µA [ ] 12 − 4.7 x105 I D 2 + VGS 2 = 6 | VGS 2 = −1 − 2 I D2 2.5 x10−4 I D2 = 10.0 µA, VGS 2 = −1.28 V , VDS 2 = −(6 − VGS 2 ) = −7.28V 50V + 7.28V ⎛ ⎞ ⎜1 + 470kΩ 2(250µA)( 10.0µA) 1 + 0.02(7.28) ⎟ = 210 MΩ ⎝ ⎠ 10.0µA 1 + g m2 R2 )= Rout 2 = ro2 ( [ ] 15.119 *Problem 15.119 – Figure P15.118 VCC 1 0 DC 12 R1 1 2 100K R4 1 3 2MEG R3 3 0 2MEG R2 1 4 470K M1 5 3 2 2 PFET M2 6 3 4 4 PFET VD1 5 0 DC 0 VD2 6 0 DC 0 .MODEL PFET PMOS VTO=-1 KP=250U LAMBDA=0.02 .OP *.TF I(VD1) VD1 .TF I(VD2) VD2 .END Results: IO1 = 44.4 µA, ROUT1 = 22.1 MΩ, IO1 = 10.1 µA, ROUT1 = 209 MΩ ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-75 15.120 For large A, IO ≅ α F + -A v e v rπ gmv ro VREF 120 5V = = 99.2 µA R 121 50 kΩ vx ix ve R For the small - signal model above, vx = ve + (ix − g mv )ro | v = (− Av e )− ve = −ve ( 1 + A) | ix = Gve + g π ( 1 + A)ve | Combining : Rout = ro = Rout 1 + A) 1 vx 1 + µ f ( = + ≅ ro ( 1 + β o ) for g π ( 1 + A)>> G and µ f ( 1 + A)>> 1 ix G + g π ( 1 + A) g o 50V + 10V = 605kΩ | Rout = 605kΩ( 121)= 73.2 MΩ 99.2µA cannot exceed β oro because of the loss of base current through rπ . 15.121 ROUT is limited to βoro of the BJT. We need to increase the effective current gain of the transistor which can be done by replacing Q1 with a Darlington configuration of two transistors. +V CC I o + V REF A Q 2 Q1 R -V EE 2 2 Now ROUT can approach the βoro product of the Darlington which is Rout ≅ β o ro 2 . See Prob. 3 15.48 15-76 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.122 For large A, IO ≅ VREF 5V = = 100µA R 50 kΩ + -A v s v vx ix vs gm v ro R For the small - signal model above, v x = v s + (ix − gm v )ro | v = (− Av s ) − v s = −v s (1 + A) | v s = ix R | Combining : Rout = vx 50V + 10V = R + ro [ 1 + gm R(1 + A)] | ro = = 600 kΩ 100µA ix 10−4 )[ 1 + 0.02(10)] = 0.438 mS gm = 2(8 x10−4 )( Rout = 50 kΩ + 600 kΩ 1 + 0.438mS (50 kΩ)( 1 + 5 x10 4 ) = 6.57 x1011 Ω !! 15.123 [ ] (a) V 91kΩ = 9.03V | REQ = 91kΩ 30 kΩ = 22.6kΩ 91kΩ + 30 kΩ 12 − 0.7 − 9.03 V = 9.34 µA IC 3 = 85 I B 3 = 85 22.6 kΩ + 86(240 kΩ) Ω EQ = 12V VEC 3 = 12 − I E RE − 0.7 = 12 − Q − po int s : (4.62µA,7.62V ) (4.62µA,7.62V ) (9.34µA,9.03V ) 86 (9.34µA)(240kΩ)− 0.7 = 9.03V 85 85 ⎛ 9.34µA ⎞ I IC1 = I C 2 = α F C 3 = ⎜ ⎟ = 4.62µA | VEC1 = VEC 2 = 0.7 − −12 + 1.2 MΩ(4.62µA) = 7.16V 86 ⎝ 2 ⎠ 2 [ ] (b) r π3 = 85(0.025V ) 9.34µA = 228 kΩ | ro3 = (70 + 9.03)V = 8.46 MΩ 9.34µA ⎞ ⎛ ⎛ ⎞ 85(240kΩ) β o RE ⎟ 1 + Rout 3 = ro3⎜1 + ⎟ = 8.46 MΩ⎜ ⎜ 22.6kΩ + 228kΩ + 240kΩ ⎟ = 360 MΩ ⎝ Rth + rπ 3 + RE ⎠ ⎠ ⎝ For a single - ended output, Av = g m RC = 20(4.62µA)( 1.2 MΩ) = +111 (40.9 dB) 2 CMRR = g m1 Rout 3 = 40(4.62µA)(360 MΩ)= 6.65 x10 4 (96.5dB) (c) The answers are the same as parts (a) and (b). ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-77 15.124 (a) V EQ = −15V 100 kΩ = −9.93V | REQ = 100 kΩ 51kΩ = 33.8 kΩ | Assume saturation : 100 kΩ + 51kΩ 2 I D3 | I D3 = 363 µA, VGS 3 = 2.35V 4 x10−4 −9.93 = −15 + 7500 I D 3 + VGS 3 | VGS 3 = 1 + I D1 = I D 2 = 2(363µA) I D3 = 182µA | VGS 1 = 1 + = 1.95V 2 2(400µA) VDS 3 = −VGS 1 − 7500 I D 3 − (−15)= 10.3V VDS1 = VDS 2 = VD1 − VS 1 = 15 − 36000 I D1 − (−VGS 1)= 10.4V (182µA, 10.4V ) (182µA, 10.4V ) (363µA, 10.3V ) (b) r o3 50V + 10.3V = 166 kΩ 363µA ⎛ ⎞ Rout 3 = ro3 ( 1 + g m3 RS )= 166 kΩ⎜1 + 2 4 x10−4 3.63 x10−4 1 + 0.02( 10.3) (7.5kΩ)⎟ = 903 kΩ ⎝ ⎠ = ( Add = −g m RD ro2 = − 2 4 x10−4 1.82 x10−4 For a single - ended output, Acd ≅ − CMRR = ( ) ( )( )( )[ ] )[1 + 0.02(10.4)](36kΩ 332kΩ)= 0.419mS (325kΩ)= −13.6 RD 36 kΩ =− = −0.199 2 Rout 3 2(903kΩ) 13.6 / 2 = 342 or 50.7 dB 0.0199 The approximate CMRR estimate is CMRR ≅ g m1 Rout 3 = 0.419 mS (903kΩ) = 378 (51.6 dB) 15.125 ro1 since the collector current of the 2 current source is twice that of the input transistors. For a single- ended output, Assuming all devices are identical, Rout = β o1 Add = − Using our default paramters: CMRR ≅ 20β o1VA1 = 20( 100)(70)= 140,000 ( 103dB) (Note that this analysis neglects the contribution of the output resistance ro of the input pair. If this resistance is included, a theoretical cancellation occurs and Acc = 0! Of course the output β o ro resistance expression Rout = is not precise, but an improvement over the CMRR expression 2 above is possible.) β µ g m1 RC RC R g β r | Acc = − = − C | CMRR = m1 o1 o1 = o1 f 1 ⎛ r ⎞ 2 β o1ro1 2 2 2⎜β o1 o1 ⎟ 2⎠ ⎝ 15-78 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.126 Rout = ro ( 1 + g m RS ) ≅ µF RS = g mro RS ≅ 2 Kn I D VR S = I D RS = 1.5 λ ID Rout 1 R λI D S ( ) = 0.02 10−4 2 Kn ( ) (5x10 ) = 3.16 V 2( 5 x10 ) 1.5 6 −4 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-79 15.127 *Problem 15.127 - Fig. 15.49(a) - BJT Current Source Monte Carlo Analysis *Generate a Voltage Source with 5% Tolerances IEE 0 5 DC 1 REE 5 0 RTOL 15 EEE 1 0 5 0 -1 * VO 4 0 AC 1 RE 1 2 RTOL 18.4K R1 1 3 RTOL 113K R2 3 0 RTOL 263K Q1 4 3 2 NBJT .OP .DC VO 0 0 .01 .AC LIN 1 1000 1000 .PRINT AC IM(VO) IP(VO) .MODEL NBJT NPN BF=150 VA=75 .MODEL RTOL RES (R=1 DEV 5%) .MC 1000 DC I(VO) YMAX *.MC 1000 AC IM(VO) YMAX .END Results - 3σ limits: IO = 199 µA ± 32.5 µA, ROUT = 11.8 MΩ ± 2.6 MΩ *Problem 15.127 - Fig. 15.49(b) - MOSFET Current Source *Generate a Voltage Source with 5% Tolerance IEE 0 5 DC 1 REE 5 0 RTOL 15 EEE 1 0 5 0 -1 * VO 4 0 AC 1 RS 1 2 RTOL 18K R3 1 3 RTOL 240K R4 3 0 RTOL 510K M1 4 3 2 2 NFET .OP .DC VO 0 0 .01 .AC LIN 1 1000 1000 .PRINT AC IM(VO) IP(VO) .MODEL NFET NMOS KP=9.95M VTO=1 LAMBDA=0.01 .MODEL RTOL RES (R=1 DEV 5%) .MC 1000 DC I(VO) YMAX *.MC 1000 AC IM(VO) YMAX .END Results - 3σ limits: IO = 201 µA ± 34.7 µA, ROUT = 21.7 MΩ ± 3.6 MΩ 15-80 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.128 4.02 kΩ( 1 + 0.15)( 1 − 0.03) ≤ R ≤ 4.02 kΩ( 1 + 0.15)( 1 + 0.03) | 4.48kΩ ≤ R ≤ 4.76kΩ 15.129 ⎛V ⎞ ⎛V ⎞ I ⎛V − V ⎞ I IC1 = I S 1 exp⎜ BE1 ⎟ | IC 2 = I S 2 exp⎜ BE 2 ⎟ | C 2 = S 2 exp⎜ BE 2 BE1 ⎟ | ∆VBE = VBE 2 − VBE1 VT ⎝ VT ⎠ ⎝ VT ⎠ IC1 I S 1 ⎝ ⎠ ⎛ ∆I ⎞ ⎛ ∆I ⎞ I +I ∆I S = I S 1 − I S 2 | I S = S 1 S 2 | I S 1 = I S ⎜1 + S ⎟ | I S 2 = I S ⎜1 − S ⎟ 2 ⎝ 2 IS ⎠ ⎝ 2IS ⎠ ⎡ ⎛ ∆I ⎞ ⎤ ⎢ I S ⎜1 + S ⎟ ⎥ ⎡ 1.05)⎤ ⎛ IC 2 I S 1 ⎞ ⎢ 1 ⎝ 2 I S ⎠ ⎥ = 0.025ln⎢ ( ⎥ = 2.50 mV = 0.025ln I = I : ∆ V = V ln a ⎜ ⎟ ( ) C 2 C1 BE T ⎢() ⎛ ∆I S ⎞ ⎥ 0.95 ⎝ IC1 I S 2 ⎠ ⎢ ⎥ ( ) ⎦ ⎣ ⎟⎥ ⎢ I S ⎜1 − ⎝ 2IS ⎠⎦ ⎣ ⎡( 1.10)⎤ ⎥ = 5.02 mV (b) ∆VBE = 0.025ln⎢ ⎢ ⎦ ⎣(0.90)⎥ ⎛ ∆I S ⎞ ⎜1 + ⎟ ⎛ VBE 2 − VBE1 ⎞ ⎛ 0.001⎞ IS ⎠ ∆I S I S1 ⎝ = 0.02 or 2% ⎟ = exp⎜ ⎟ = 1.04 → (c) I = ⎛ ∆I ⎞ = exp⎜ 0.025 V I ⎝ ⎠ ⎝ ⎠ S2 T S S ⎜1 − ⎟ IS ⎠ ⎝ ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-81 15.130 (a) For v1 = 0 = v2 , VBE1 = VBE 2 and the collector currents are the same. So, VOS = 0. Only the base currents will be mismatched. ⎛ VBE ⎞⎛ VCE ⎞ ⎛ VBE ⎞⎛ VCE ⎞ b 1 + | I I = I 1 − 0.025 exp = I 1 + 0.025 exp ⎜ ⎟ ⎜ ⎟ ⎟⎜1 + ⎟ ( ) C1 S ( ) ⎝ V ⎠⎝ V ⎠ C 2 S ( ) ⎜ VA ⎠ ⎝ VT ⎠⎝ T A ⎛ V ⎞⎛ V ⎞ ⎛ V ⎞⎛ V ⎞ I +I ∆IC = IC 2 − IC1 = 0.05 I S exp⎜ BE ⎟⎜1 + CE ⎟ | IC = C1 C 2 = I S exp⎜ BE ⎟⎜1 + CE ⎟ VA ⎠ 2 VA ⎠ ⎝ VT ⎠⎝ ⎝ VT ⎠⎝ ∆I ∆I VOS = C = VT C = 0.025V (0.05)= 1.25 mV gm IC ⎞ ⎞ ⎛ VBE ⎞⎛ ⎛ VBE ⎞⎛ VCE VCE ⎜ ⎟ ⎜ ⎟ 1 + 1 + c I = I exp = I exp | I ⎟ ⎜ ⎟⎜ ( ) C1 S ⎜ S ⎟ C2 ⎟ V V 1 + 0.025 V 1 − 0.025 ⎝ VT ⎠⎜ ⎝ ⎠ ( ) ( ) T A A ⎝ ⎠ ⎝ ⎠ ⎛ V ⎞⎛ ⎛ V ⎞⎛ V V ⎞ V ⎞ ∆IC = IC 2 − IC1 ≅ I S exp⎜ BE ⎟⎜1-1.025 CE − 1 + 0.975 CE ⎟ = I S exp⎜ BE ⎟⎜0.05 CE ⎟ VA VA ⎠ VA ⎠ ⎝ VT ⎠⎝ ⎝ VT ⎠⎝ ⎛ VCE ⎞ ⎜ ⎟ ⎛ VBE ⎞⎛ VCE ⎞ IC1 + IC 2 ∆I C ∆IC ⎝ VA ⎠ = I S exp⎜ = VT = 0.025V (0.05) IC = ⎟⎜1 + ⎟ | VOS = ⎛ VCE ⎞ 2 VA ⎠ gm IC ⎝ VT ⎠⎝ ⎜1 + ⎟ VA ⎠ ⎝ V For CE = 0.1, VOS = 114 µV VA (d ) V OD = IC (RC + 0.025RC )− (RC − 0.025 RC ) = 0.05IC RC [ ] VOS = VOD V = VT OD = 0.025V (0.05) = 1.25 mV g m RC IC RC 15.131 ⎛V ⎞ ⎛ V + 0.002 ⎞ IS1 ⎛ 0.002 ⎞ IS1 exp⎜ BE1 ⎟ = IS 2 exp⎜ BE1 = exp⎜ ⎟ = 1.08 | IS1 = 1.08IS 2 ⎟ | ⎝ 0.025 ⎠ VT ⎝ VT ⎠ ⎝ ⎠ IS 2 I +I ∆IS 0.08 ∆IS = IS1 − IS 2 = 0.08 IS 2 | IS = S1 S 2 = 1.04 IS 2 | = = 7.7% 2 IS 1.04 ⎛ 10V ⎞ ⎛ 10V ⎞ β F1 = 100(1 + 0.025)⎜1 + ⎟ = 123 | β F 2 = 100(1 − 0.025)⎜1 + ⎟ = 117 ⎝ 50V ⎠ ⎝ 50V ⎠ 100µA 100µA I B1 = = 0.813 µA | IB 2 = = 0.855 µA 123 117 Note: IOS = -42.0 nA. 15-82 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.132 (a) ID = (250)(1 ± 0.05) µA 2 2 2 V V (250)(1 − 0.05) µA 1 − 0.025 2 = 113µA min ID = [ ] 2 V2 138µA +113µA ∆ID ID = = 125.5µA | ∆ID = 138µA − 113µA = 25µA | = 19.8% 2 ID [2 − (1 ± 0.025)] 2 max | ID = (250)(1 + 0.05) µA 2 [1 + 0.025] 2 =138µA max = (b) ID =1.20 mA V 2 (250)(1 − 0.05) µA 3 − 0.025 2 = 1.05 mA min ID = [ ] V2 2 1.20mA +1.05 mA ∆ID = 1.125 mA | ∆ID = 1.20 mA − 1.05mA = 0.150 mA | = 13.3% ID = ID 2 2 (250)(1 + 0.05) µA [3 + 0.025] 2 15.133 VGS1 = VTN + 1 2 IDS1 2 IDS1 2 IDS1 ⎛ ∆ (W / L)⎞ = VTN + ≅ VTN + ⎜1 − ⎟ ∆ (W / L) 4 (W / L) ⎠ ' ⎛W ⎞ ' ⎛W ⎞ ' ⎛W ⎞ ⎝ Kn K K ⎜ ⎟ ⎟ 1+ ⎟ n⎜ n⎜ ⎝ L ⎠1 ⎝L⎠ ⎝L⎠ 2(W / L) 1 2 IDS 2 2 IDS 2 2 IDS 2 ⎛ ∆ (W / L)⎞ = VTN + ≅ VTN + ⎜1 + ⎟ ∆ (W / L) 4 (W / L) ⎠ ' ⎛W ⎞ ' ⎛W ⎞ ' ⎛W ⎞ ⎝ Kn ⎜ ⎟ Kn⎜ ⎟ 1 − Kn ⎜ ⎟ ⎝ L ⎠2 ⎝L⎠ ⎝L⎠ 2(W / L) ⎛ ∆ (W / L)⎞ 2 IDS 2 ⎛ ∆ (W / L)⎞ ⎜ ⎟ = (VGS − VTN )⎜ ⎟ W / L) ⎠ 2(W / L) ⎠ ' ⎛ W ⎞ ⎝ 2( ⎝ Kn⎜ ⎟ ⎝L⎠ VGS 2 = VTN + IDS 2 = IDS1 : VGS 2 − VGS1 = (a) ∆VGS = (VGS − VTN )⎜ (b) ⎛ ∆ (W / L)⎞ ⎛ 0.10 ⎞ ⎟ = (0.5)⎜ ⎟ = 25 mV ⎝ 2 ⎠ ⎝ 2(W / L ) ⎠ ∆ (W / L) ∆VGS 0.003 =2 =2 = 1.2 % | 0.5 (W / L) (VGS − VTN ) (c ) ∆ (W / L ) ∆VGS 0.001 =2 =2 = 0.4 % 0.5 (W / L) (VGS − VTN ) ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-83 15.134 ⎡⎛ W ⎞ ⎛ W ⎞ ⎤ K ' 2 ∆ID = ID 2 − ID1 = ⎢⎜ ⎟ − ⎜ ⎟ ⎥ n (VGS − VTN ) (1 + λVDS ) ⎣⎝ L ⎠ 2 ⎝ L ⎠1⎦ 2 ⎛ 0.05 ⎞ ∆I ∆I ∆ID = 0.05 ID | VOS = D = (VGS − VTN ) D = 0.75V ⎜ ⎟ = 18.8 mV ⎝ 2 ⎠ gm 2 ID K 2 2 (b) ∆ID = ID 2 − ID1 = n (1 + λVDS ) (VGS − VTN + 0.025VTN ) − (VGS − VTN − 0.025VTN ) 2 K ID ∆ID = n (1 + λVDS )(0.1VTN )(VGS − VTN ) = (0.1VTN ) 2 (VGS − VTN ) (a) Assume active region operation : [ ] VOS = ∆ID ∆I = (VGS − VTN ) D = 0.05 VTN | If VTN = 1V , VOS = 50 mV gm 2 ID K 2 (c ) ∆ID = ID 2 − ID1 = n (VGS − VTN ) [(1 + λVDS + 0.025 λVDS ) − (1 + λVDS − 0.025 λVDS )] 2 ⎛ 0.05 ⎞⎛ λVDS ⎞ ∆I 0.05 λVDS ∆I ∆ID = ID 2 − ID1 = ID | VOS = D = (VGS − VTN ) D = 0.75V ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 1 + λVDS ⎠ 1 + λVDS gm 2 ID (d) VOD = ID [(RD + 0.025RD ) − (RD − 0.025 RD )] = 0.05ID RD VOS = VOD V 0.05 ID = OD = (VGS − VTN ) = 0.025(0.75V ) = 18.8 mV Avt gm RD 2 ID ⎛W ⎞ ⎜ ⎟ 1 + λVDSX ⎝ L ⎠X = I ⎛ W ⎞ REF 1 + λVDS 1 ⎜ ⎟ ⎝ L ⎠1 If λVDS = 0.1, VOS = 1.71 mV 15.135 1 | RoutX = λ (a) I + VDSX IOX OX 2 30 x10−6 2 I D1 VDS1 = VGS 1 = VTN + = 0.75 + = 1.52V Kn 4 25 x10−6 1 + 10 10 0.015 IO2 = (30µA) = 84.3µA | Rout 2 = = 909 kΩ 4 84.3µA 1 + 0.015( 1.52) 1 + 0.015( 10) 1 + 0.015(8) 1 +8 20 0.015 = 164µA | Rout 3 = = 455kΩ IO3 = (30µA) 4 164µA 1 + 0.015( 1.52) 1 + 12 40 0.015 = 346µA | Rout 4 = = 227 kΩ IO 4 = (30µA) 4 346µA 1 + 0.015( 1.52) 1 + 0.015( 12) ( ( ) ) 15-84 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.135 cont. ⎛W ⎞ ⎜ ⎟ L 1 + λV (b) IOX = ⎝⎛W⎠⎞X I REF 1 + λVDSX DS 1 ⎜ ⎟ ⎝ L ⎠1 1 | RoutX = λ + VDSX IOX 2 50 x10−6 2 I D1 VDS1 = VGS 1 = VTN + = 0.75 + = 1.75V Kn 4 25 x10−6 1 + 10 10 IO2 = (50µA) = 140µA | Rout 2 = 0.015 = 548 kΩ 4 140µA 1 + 0.015( 1.75) 1 + 0.015( 10) 1 + 0.015(8) 1 +8 20 IO3 = (50µA) = 273µA | Rout 3 = 0.015 = 274 kΩ 4 273µA 1 + 0.015( 1.75) 1 + 12 1 + 0.015( 12) 40 = 575µA | Rout 4 = 0.015 = 137 kΩ IO 4 = (50µA) 4 575µA 1 + 0.015( 1.75) ( ( ) ) (c) I O2 = 10 20 30µA) = 75 µA | Rout 2 = ∞ | IO 3 = (30µA)= 150 µA | Rout 3 = ∞ ( 4 4 IO 4 = 40 (30µA)= 300 µA | Rout 4 = ∞ 4 15.136 (a) I OX IO2 = IO 4 = 1 + 0.015( 10) 1 + 0.015(8) 10 20 30µA) = 135 µA | IO3 = 30µA) = 262 µA ( ( 2.5 2.5 1 + 0.015( 1.73) 1 + 0.015( 1.73) 1 + 0.015( 12) 40 30µA) = 552 µA ( 2.5 1 + 0.015( 1.73) = 0.75 + 2 20 x10−6 −6 ⎛W ⎞ ⎜ ⎟ 1 + λVDSX ⎝ L ⎠X = I REF ⎛W ⎞ 1 + λVDS 1 ⎜ ⎟ ⎝ L ⎠1 2 30 x10−6 2 I D1 | VDS 1 = VGS 1 = VTN + = 0.75 + = 1.73V Kn 2.5 25 x10−6 ( ( ) ) (b) V IO3 = DS 1 ( ) = 1.27V 6( 25x10 ) | IO2 = 1 + 0.015( 10) 10 20µA) = 37.6 µA ( 6 1 + 0.015( 1.27) 1 + 0.015(8) 1 + 0.015( 12) 20 40 20µA) = 73.3 µA | IO 4 = (20µA) = 154 µA ( 6 6 1 + 0.015( 1.27) 1 + 0.015( 1.27) ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-85 15.137 (a) For λ = 0, I O2 (b) From Prob. 16.8, I ⎛ 10 ⎞ ⎛ 20 ⎞ ⎛ 40 ⎞ = 30µA⎜ ⎟ = 75µA, IO 3 = 30µA⎜ ⎟ = 150µA, IO 2 = 30µA⎜ ⎟ = 300µA ⎝4⎠ ⎝4⎠ ⎝4⎠ O2 = 84.3µA, IO 3 = 164µA, IO 4 = 346µA ∆IO 2 84.3 − 75 ∆IO 3 164 − 150 ∆IO 4 346 − 300 = = 0.124 LSB, = = 0.187 LSB, = = 0.613 LSB IO 2 IO 2 IO 2 75 75 75 15.138 *Problem 15.135(a) - NMOS Current Source Array IREF 0 1 DC 30U VD2 2 0 DC 10 AC 1 VD3 3 0 DC 8 AC 1 VD4 4 0 DC 12 AC 1 M1 1 1 0 0 NFET W=4U L=1U M2 2 1 0 0 NFET W=10U L=1U M3 3 1 0 0 NFET W=20U L=1U M4 4 1 0 0 NFET W=40U L=1U .MODEL NFET NMOS KP=25U VTO=0.75 LAMBDA=0.015 .OP .AC LIN 1 1000 1000 .PRINT AC IM(VD2) IM(VD3) IM(VD4) IP(VD2) IP(VD3) IP(VD4) .END The results are identical to the hand calculations. 15.139 2 5 + VGS 1 15x10-6 ⎛ 2 ⎞ 5 + VGS 1 VGS 1 + 0.9) ( | 1 − 0.01VGS 1 )= → VGS 1 = −2.985V I D1 = ⎜ ⎟( R 2 ⎝ 1⎠ 3x10 4 I REF = 5 − 2.985 = 67.2µA 3 x10 4 1 + VDS 2 2 15 x10-6 ⎛ 8 ⎞ 100 + 5 λ = = 383kΩ IO2 = ⎜ ⎟(−2.985 + 0.9) 1 − 0.01(−5) = 274µA | Rout 2 = 274µA 2 ⎝ 1⎠ IO 2 2 15 x10-6 ⎛16 ⎞ 100 + 10 = 192kΩ IO3 = ⎜ ⎟(−2.985 + 0.9) 1 − 0.01(−10) = 574µA | Rout 3 = 574µA 2 ⎝1⎠ [ ] [ ] 15-86 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.140 5 + VGS1 15 x10-6 ⎛ 3.3 ⎞ 5 + VGS 1 VGS 1 + 0.9) | 12 ( 1 − 0.01VGS 1 )= → VGS 1 = −2.655V ⎜ ⎟( R 2 ⎝ 1 ⎠ 3x10 4 2 5 − 2.655 15 x10-6 ⎛ 8 ⎞ = 78.2 µ A | I = I REF = ⎜ ⎟(−2.655 + 0.9) 1 − 0.01(−5) = 194µA O2 4 2 ⎝1⎠ 3 x10 -6 ⎛ 2 15 x10 16 ⎞ IO3 = ⎜ ⎟(−2.655 + 0.9) 1 − 0.01(−10) = 407µA 2 ⎝1⎠ 5+V 15 x10-6 ⎛ 4 ⎞ 5 + VGS 1 VGS 1 + 0.9) 12 ( 1 − 0.01VGS 1 )= → VGS 1 = −2.241V (b) I D1 = R GS1 | 2 ⎜ ⎟( 5x10 4 ⎝1⎠ 2 5 − 2.241 15 x10-6 ⎛ 8 ⎞ = 55.2 µ A | I = I REF = ⎜ ⎟(−2.241 + 0.9) 1 − 0.01(−5) = 113µA O2 4 2 ⎝1⎠ 3 x10 2 15 x10-6 ⎛16 ⎞ IO3 = ⎜ ⎟(−2.241 + 0.9) 1 − 0.01(−10) = 237µA 2 ⎝1⎠ (a) I D1 = [ ] [ ] [ ] [ ] 15.141 *Problem 15.141 - PMOS Current Source Array RREF 0 1 30K VSS 4 0 DC 5 VD2 2 0 DC 0 AC 1 VD3 3 0 DC -5 AC 1 M1 1 1 4 4 PFET W=2U L=1U M2 2 1 4 4 PFET W=8U L=1U M3 3 1 4 4 PFET W=16U L=1U .MODEL PFET PMOS KP=15U VTO=-0.9 LAMBDA=0.01 .OP .AC LIN 1 1000 1000 .PRINT AC IM(VD2) IM(VD3) IP(VD2) IP(VD3) .END The results are identical to the hand calculations. 15.142 K ' ⎛W ⎞ 2 2 15x10-6 ⎛ 8 ⎞ VGS 2 − VTP ) 1 + λ VDS 2 | 55x10-6 = VGS1 + 0.9) 1 + 0.01 −5 I D2 = p ⎜ ⎟( ⎜ ⎟( 2 ⎝ L⎠ 2 ⎝ 1⎠ ⎛W ⎞ ⎛ 8⎞ ⎜ ⎟ 1 + λ VDS 2 ⎜ ⎟ 1 + 0.01(5) I D 2 ⎝ L ⎠2 55µA ⎝ 1 ⎠2 VGS1 = −1.834V | = | = → I REF = 13.3µA ⎛ 2⎞ I REF ⎛W ⎞ I REF 1.834) ⎜ ⎟ 1 + λ VDS 1 ⎜ ⎟ 1 + 0.01( ⎝ L ⎠1 ⎝ 1 ⎠1 5 + VGS1 5 − 1.834 I REF = | R= = 238 kΩ 13.3µA R [ ] [ ] [ [ ] [ ] ] [ ] 15.143 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-87 (a) I REF = ⎛ V ⎞⎛ 14.3 VBE ⎞ 12 − 0.7 = 151 µA | I REF = IC1 + ( 1 + 5 + 8.3)I B | I REF = I S exp⎜ BE ⎟⎜1 + + ⎟ 4 7.5 x10 ⎝ VT ⎠⎝ β FO VA ⎠ V 5 1 + CE 2 1+ ⎛ VBE ⎞⎛ VCE 2 ⎞ VA 60 IO2 = 5 I S exp⎜ | IO 2 = 5( 151µA) = 631 µA ⎟⎜1 + ⎟ = 5I REF 14.3 VBE 14.3 0.7 VA ⎠ ⎝ VT ⎠⎝ 1+ + 1+ + 50 60 β FO VA VA + VCE 2 60 + 5 = = 103 kΩ IC 2 6.31x10−4 V 3 1 + CE 3 1+ VA 60 IO3 = 8.3 I REF = 8.3( 151µA) = 1.02 mA 14.3 VBE 14.3 0.7 1+ + 1+ + 50 60 β FO VA Rout 2 = ro2 = Rout 3 = ro3 = VA + VCE 3 60 + 3 = = 61.8 kΩ IC 3 1.02 x10−3 (b) Since all areas are scaled equally, the current ratios stay the same, and there is no change from part (a). This ignores the slight change in VBE of Q1 due to its area change. (c) I 12 − 0.7 − 0.7 I = 141 µA | I REF = IC1 + ( 1 + 5 + 8.3) B 4 β FO + 1 7.5 x10 ⎛ V ⎞⎛ 2VBE ⎞ 14.3 ⎟ I REF = I S exp⎜ BE ⎟⎜ + 1 + ⎟ ⎝ VT ⎠⎜ ⎝ β FO (β FO + 1) VA ⎠ V 1 + CE 2 ⎛ VBE ⎞⎛ VCE 2 ⎞ VA IO2 = 5 I S exp⎜ ⎟⎜1 + ⎟ = 5I REF 14.3 2V VA ⎠ ⎝ VT ⎠⎝ 1+ + BE β FO (β FO + 1) VA REF = 5 V + VCE 2 60V + 5V 60 IO2 = 5( 141µA) = 745 µA | Rout 2 = ro2 = A = = 87.2 kΩ 14.3 1.4 745µA IC 2 1+ + 50(51) 60 1+ VCE 3 3 1+ VA 60 IO3 = 8.3 I REF = 8.3( 141µA) = 1.20 mA 14.3 2VBE 14.3 1.4 1+ + 1+ + 50(51) 60 β FO (β FO + 1) VA 1+ Rout 3 = ro3 = VA + VCE 3 60V + 3V = = 52.5 kΩ IC 3 1.20 mA 15-88 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.144 *Problem 15.144 – Figure P15.143(a) - NPN Current Source Array RREF 2 1 75K VCC 2 0 DC 12 VC2 3 0 DC 5 AC 1 VC3 4 0 DC 3 AC 1 Q1 1 1 0 NBJT 1 Q2 3 1 0 NBJT 5 Q3 4 1 0 NBJT 8.3 .MODEL NBJT NPN BF=50 VA=60 .OP .AC LIN 1 1000 1000 .PRINT AC IM(VC2) IM(VC3) IP(VC2) IP(VC3) .END *Problem 15.144 – Figure 15.143(b) - Buffered NPN Current Source Array RREF 2 5 75K VCC 2 0 DC 12 VC2 3 0 DC 5 AC 1 VC3 4 0 DC 3 AC 1 Q1 5 1 0 NBJT 1 Q2 3 1 0 NBJT 5 Q3 4 1 0 NBJT 8.3 Q4 2 5 1 NBJT 1 .MODEL NBJT NPN BF=50 VA=60 .OP .AC LIN 1 1000 1000 .PRINT AC IM(VC2) IM(VC3) IP(VC2) IP(VC3) .END The results are almost identical to the hand calculations. ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-89 15.145 (a) I I REF ⎛ VBE ⎞⎛ 14.3 VBE ⎞ = I + 1 + 5 + 8.3 I = I + 14.3 I = I exp + ⎜ ⎟⎜1 + ⎟ ( ) REF C1 B C1 B S ⎝ VT ⎠⎝ β FO VA ⎠ ⎛ V ⎞⎛ 14.3 0.7 ⎞ ⎛ VBE ⎞ ≅ I S exp⎜ BE ⎟⎜1 + + ⎟ ⎟ = 1.298 I S exp⎜ 50 60 ⎠ ⎝ VT ⎠⎝ ⎝ VT ⎠ | I REF = 150µA( 1.298) = 22.3µA ⎛ 3⎞ 8.3⎜1 + ⎟ ⎝ 60 ⎠ V 1 + CE 2 ⎛ VBE ⎞⎛ VCE 3 ⎞ VA IO3 = 8.3 I S exp⎜ ⎟⎜1 + ⎟ = 8.3I REF VA ⎠ 1.298 ⎝ VT ⎠⎝ VCE 2 5 1+ 12 − 0.7 12 − 0.7 VA I REF = | R= = 507 kΩ | IO 2 = 5I REF = 5(22.3µA) 60 = 93.1 µA 1.298 1.298 R 22.3µA ⎛ VBE ⎞⎛ IB 14.3 I B 2VBE ⎞ 14.3 ⎜ ⎟ b I = I + 1 + 5 + 8.3 = I + = I exp + 1 + ⎟ ( ) REF C1 ( )β + 1 C1 β + 1 S ⎜ ⎟ V β β + 1 ⎝ VT ⎠⎜ ( ) FO FO A FO FO ⎝ ⎠ ⎛ V ⎞⎛ ⎛V ⎞ 14.3 1.4 ⎞ ⎟ = 1.029 I S exp⎜ BE ⎟ + 1 + I REF ≅ I S exp⎜ BE ⎟⎜ ⎟ ⎝ VT ⎠⎜ ⎝ VT ⎠ ⎝ 50(51) 60 ⎠ 1+ V 1 + CE 2 ⎛ VBE ⎞⎛ VCE 3 ⎞ VA IO3 = 8.3 I S exp⎜ ⎟⎜1 + ⎟ = 8.3 I REF VA ⎠ 1.029 ⎝ VT ⎠⎝ | I REF = 1.029) 150µA( = 17.7µA ⎛ 3⎞ 8.3⎜1 + ⎟ ⎝ 60 ⎠ 1+ VCE 2 5 1+ VA = 5( 17.7µA) 60 = 93.2 µA 1.029 1.029 I REF = 12 − 0.7 − 0.7 12 − 1.4 | R= = 599 kΩ | IO 2 = 5 I REF R 17.7µA 15-90 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.146 (a) I REF = ⎛ VBE ⎞⎛ 7.65 VBE ⎞ ⎛ 5 8.3 ⎞ 12 − 0.7 = 151 µA | I REF = IC1 + ⎜1 + + + ⎟ ⎟⎜1 + ⎟ I B | I REF = I S exp⎜ 3 75x10 ⎝ 2 2 ⎠ ⎝ VT ⎠⎝ β FO VA ⎠ V 5 1 + CE 2 1+ ⎛ ⎞ ⎛ ⎞ 5 VBE VCE 2 VA 75 IO2 = I S exp⎜ | IO2 = 2.5( 151µA) = 376 µA ⎟⎜1 + ⎟ = 2.5 I REF 7.65 VBE 7.65 0.7 2 VA ⎠ ⎝ VT ⎠⎝ 1+ + 1+ + β FO VA 125 75 VCE 3 3 1+ 8.3 VA 75 IO3 = I REF = 4.15( 151µA) = 609 µA 7.65 VBE 7.65 0.7 2 1+ + 1+ + β FO VA 125 75 ⎛ 5 8.3 ⎞ I B 12 − 0.7 − 0.7 1+ + (b) I REF = 75x104 = 141 µA | I REF = IC1 + ⎜ ⎟ ⎝ 2 2 ⎠ β FO + 1 ⎛ V ⎞⎛ 2VBE ⎞ 7.65 ⎟ I REF = I S exp⎜ BE ⎟⎜ + 1 + V β β + 1 ⎝ VT ⎠⎜ ) A⎟ FO ( FO ⎝ ⎠ V 1 + CE 2 ⎛ ⎞ ⎛ ⎞ 5 V V VA IO2 = I S exp⎜ BE ⎟⎜1 + CE 2 ⎟ = 2.5 I REF 7.65 2V 2 VA ⎠ ⎝ VT ⎠⎝ 1+ + BE β FO (β FO + 1) VA 1+ 5 75 IO2 = 2.5( 141µA) = 370 µA 7.65 1.4 1+ + 125( 126) 75 1+ VCE 3 3 1+ 8.3 VA 75 I REF = 4.15( 141µA) = 596 µA IO3 = 7.65 2VBE 6.65 1.4 2 1+ + 1+ + β FO (β FO + 1) VA 125( 126) 75 1+ ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-91 15.147 ⎛ 5 8.3⎞ I B 12 − 0.7 − 0.7 = 106 µ A | I = I + 1+ + ⎜ ⎟ REF C1 100 x103 ⎝ 3 3 ⎠ β FO + 1 ⎛ V ⎞⎛ 5.43 2VBE ⎞ ⎟ 1 + I REF = I S exp⎜ BE ⎟⎜ + ⎟ ⎝ VT ⎠⎜ ⎝ β FO (β FO + 1) VA ⎠ V 1 + CE 2 ⎛ ⎞ ⎛ ⎞ V 5 5 V VA IO2 = I S exp⎜ BE ⎟⎜1 + CE 2 ⎟ = I REF 5.43 2V VA ⎠ 3 3 ⎝ VT ⎠⎝ 1+ + BE β FO (β FO + 1) VA I REF = 5 IO2 = ( 106µA) 3 1+ 5 75 = 185 µA 5.43 1.4 + 100( 101) 75 1+ 1+ VCE 3 3 1 + 8.3 8.3 VA I REF = IO3 = (106µA) 5.4375 1.4 = 299 µA 5.43 2VBE 3 3 1+ + 1+ + 100( 101) 75 β FO (β FO + 1) VA 15-92 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.148 (a) I REF = ⎛ VBE ⎞⎛ 14.3 VBE ⎞ 12 − 0.7 = 80.7 µ A | I = I + 1 + 5 + 8.3 I | I = I exp + ⎟ ⎜ ⎟⎜1 + ( ) REF C 1 B REF S 1.4 x105 ⎝ VT ⎠⎝ β FO VA ⎠ 1+ VCE 2 5 1+ ⎛ V ⎞⎛ V ⎞ V A 75 IO 2 = 5 I S exp⎜ BE ⎟⎜1 + CE 2 ⎟ = 5 I REF | IO 2 = 5(80.7µA) = 383 µA 14.3 VBE 14.3 0.7 VA ⎠ ⎝ VT ⎠⎝ 1+ + 1+ + β FO VA 125 75 VCE 3 3 1+ VA 75 IO 3 = 8.3 I REF = 8.3(80.7µA) = 620 µA 14.3 VBE 14.3 0.7 1+ + 1+ + β FO VA 125 75 ⎛ VBE ⎞⎛ IB 14.3 I B 2VBE ⎞ 14.3 ⎜ ⎟ + ⎟ 1+ (b) I REF = IC1 + (1 + 5 + 8.3)β + 1 = IC1 + β + 1 = I S exp⎜ ⎟ V β β + 1 ⎝ VT ⎠⎜ ( ) FO FO A FO FO ⎝ ⎠ ⎛ V ⎞⎛ ⎛V ⎞ 1.4 ⎞ 14.3 ⎟ = 1.029 I S exp⎜ BE ⎟ + 1 + I REF = I S exp⎜ BE ⎟⎜ 126) 75 ⎟ ⎝ VT ⎠⎜ ⎝ VT ⎠ ⎝ 125( ⎠ V 1 + CE 2 ⎛ V ⎞⎛ V ⎞ 620µA( 1.02) VA IO 3 = 8.3 I S exp⎜ BE ⎟⎜1 + CE 3 ⎟ = 8.3I REF | I REF = = 73.3µA ⎛ VA ⎠ 1.02 3⎞ ⎝ VT ⎠⎝ 8.3⎜1 + ⎟ ⎝ 75 ⎠ 1+ 12 − 0.7 − 0.7 12 − 1.4 | R= = 145 kΩ R 73.3µA V 5 1 + CE 2 1+ VA = 5(73.3µA) 75 = 383 µA - Correct. Checking : IO 2 = 5 I REF 1.02 1.02 I REF = ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-93 15.149 Use βFO = 50 and VA = 60 V. (a) I REF = ⎛ V ⎞⎛ 14.3 VBE ⎞ 12 − 0.7 = 113 µA | I REF = IC1 + ( 1 + 5 + 8.3)I B | I REF = I S exp⎜ BE ⎟⎜1 + + ⎟ 5 10 ⎝ VT ⎠⎝ β FO VA ⎠ V 5 1 + CE 2 1+ ⎛ VBE ⎞⎛ VCE 2 ⎞ VA 60 IO2 = 5 I S exp⎜ | IO 2 = 5( 113µA) = 472 µA ⎟⎜1 + ⎟ = 5I REF 14.3 VBE 14.3 0.7 VA ⎠ ⎝ VT ⎠⎝ 1+ + 1+ + 50 60 β FO VA VCE 3 3 1+ VA 60 IO3 = 8.3 I REF = 8.3( 113µA) = 759 µA 14.3 VBE 14.3 0.7 1+ + 1+ + 50 60 β FO VA 1+ 6 (b) IO2 = 5(113µA) 14.360 0.7 = 479 µA | No change in IO3. 1+ + 50 60 11 − 0.7 (c) I REF = 105 = 103 µA | IO2 and IO3 are proportional to I REF 103µA 103µA IO2 = 472µA = 430 µA | IO3 = 472µA = 430 µA 113µA 113µA 1+ IO3 ∝ I REF | IO 2 = 759µA 103µA = 692 µA 113µA 60 + 5 ∆V 1V = = 138 kΩ | ∆IO 2 = = = 7.25µA 472µA ro2 138 kΩ (d ) R out 2 = ro2 = VA + VCE 2 IC 2 IO2−6V − IO2−5V = 479µA − 472µA = 7µA - Agrees within the calculation precision. 15.150 I REF = IB = ⎛ V ⎞ 15 − 0.7 = 238µA | I REF = 2 IC 1 + (2 + 1 + 6 + 9)I B = 2β FO ⎜1 + EC1 ⎟ I B + 18 I B 4 VA ⎠ 6 x10 ⎝ 238µA = 2.00µA ⎛ 0.7 ⎞ 18 + 2(50) ⎜1 + ⎟ ⎝ 60 ⎠ ⎛ V ⎞ ⎛ 15 ⎞ 60 + 15 IO2 = β FO ⎜1 + EC 2 ⎟ I B = 50⎜1 + ⎟(2.00µA) = 125µA | Rout 2 = ro2 = = 600 kΩ VA ⎠ 1.25x10−4 ⎝ 60 ⎠ ⎝ ⎛ V ⎞ ⎛ 9⎞ 60 + 9 = 100 kΩ IO3 = 6β FO ⎜1 + EC 3 ⎟ I B = 300⎜1 + ⎟(2.00µA) = 690µA | Rout 3 = ro3 = VA ⎠ 6.90 x10−4 ⎝ 60 ⎠ ⎝ ⎛ V ⎞ ⎛ 27 ⎞ 60 + 27 = 66.4 kΩ IO 4 = 9β FO ⎜1 + EC 4 ⎟ I B = 450⎜1 + ⎟(2.00µA)= 1.31mA | Rout 4 = ro 4 = VA ⎠ 1.31x10−3 ⎝ 60 ⎠ ⎝ 15-94 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.151 I REF = ⎛ V ⎞ 15 − 0.7 = 238µA | I REF = 2 IC 1 + (2 + 1 + 6 + 9)I B = 2β FO ⎜1 + EC1 ⎟ I B + 18 I B VA ⎠ R ⎝ 65µA = 0.189µA ⎛ 9⎞ 300⎜1 + ⎟ ⎝ 60 ⎠ ⎡ ⎛ VEC 3 ⎞ ⎛ 0.7 ⎞⎤ I REF = I B ⎢18 + 2(50) ⎟IB | IB = ⎜1 + ⎟⎥ = 119 I B | IO3 = 6β FO ⎜1 + VA ⎠ ⎝ 60 ⎠⎦ ⎝ ⎣ 15 − 0.7 = 63.8 kΩ 22.4µA ⎛ V ⎞ ⎛ 15 ⎞ IO2 = β FO ⎜1 + EC 2 ⎟ I B = 50⎜1 + ⎟(0.189µA)= 11.8 µA VA ⎠ ⎝ 60 ⎠ ⎝ ⎛ V ⎞ ⎛ 27 ⎞ IO 4 = 9β FO ⎜1 + EC 4 ⎟ I B = 450⎜1 + ⎟(0.189µA)= 123 µA VA ⎠ ⎝ 60 ⎠ ⎝ I REF = 119 I B = 22.4µA | R = 15.152 15 V 2A A 6A 9A Q 1 Q A 2 Q 3 Q 4 Q 5 R I O2 I O3 +6 V I O4 -12 V (2 + 1 + 6 + 9)I B 15V − 0.7V − 0.7V = 544 kΩ | I REF = IC1 + β FO + 1 25µA ⎛ V ⎞ 18 I B 25µA I REF = 2β FO ⎜1 + EC1 ⎟ I B + | IB = = 0.2435µA ⎛ VA ⎠ β FO + 1 1.4 ⎞ 18 ⎝ 2(50) ⎜1 + ⎟+ ⎝ 60 ⎠ 51 ⎛ V ⎞ ⎛ 15 ⎞ IO2 = β FO ⎜1 + EC 2 ⎟ I B = 50⎜1 + ⎟ I B = 15.2 µA VA ⎠ ⎝ 60 ⎠ ⎝ ⎛ V ⎞ ⎛ 9⎞ IO3 = 6β FO ⎜1 + EC 3 ⎟ I B = 300⎜1 + ⎟ I B = 84.0 µA VA ⎠ ⎝ 60 ⎠ ⎝ ⎛ V ⎞ ⎛ 27 ⎞ 50 IO 4 = 9β FO ⎜1 + EC 4 ⎟ I B = 450⎜1 + ⎟ I B = 159 µA | IC 5 = α F I E 5 = 18 I B = 4.30 µA 51 VA ⎠ ⎝ 60 ⎠ ⎝ R= ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-95 15.153 VBE 2 + I E 2 R2 = VBE 3 + I E 3 R3 → I E 3 = R2 V −V I E 2 + BE 2 BE 3 R3 R3 I E 2 I E 3 5I E 2 = = → n = 10 2 A nA nA In order to have equal base - emitter voltages, the two transistors must operate at the equal collector − current densities : 15.154 ⎞ I I REF I I 76 ⎛ I = REF | I E1 = C1 = ⎜ REF ⎟ = REF 7 1.093 α F 75 ⎝1.093 ⎠ 1.079 1+ 75 I 12V - 0.7V 12 = I REF ( 10kΩ)+ 0.7V + REF ( 10 kΩ) | I REF = = 586µA | I E1 = 543µA 1.079 10 kΩ( 1.927) I REF = IC1 + 7 I B | IC1 ≅ I E 2 = 2 I E1 | IO 2 = α F (2 I E1 )= 2 I E 3 = 4 I E1 75 (543µA)= 1.07mA | VE1 = 543µA(10kΩ)= 5.43V 76 75 1 1 | IO 3 = α F (4 I E1 )= 4 (543µA) = 2.14mA | = = 46.6Ω 76 g m1 40(536µA) ⎛ 1 ⎞ Rth = R ⎜ + R1 ⎟ = 10 kΩ (46.6Ω + 10kΩ) = 5.01kΩ | Rth2 = Rth rπ 3 + (β o3 + 1)(2.5kΩ) ⎝ g m1 ⎠ [ ] rπ 2 = 75(0.025V ) 1.07 mA = 1.75kΩ | rπ 3 = 75(0.025V ) 2.14 mA = 0.876 kΩ | ro2 = 60 + ( 10 − 5.43) 1.07 mA = 60.4 kΩ Rth2 = 5.01kΩ 0.876 kΩ + (76)(2.5kΩ) = 4.88 kΩ [ ] ⎛ ⎞ ⎛ ⎞ 75(5kΩ) β o R2 ⎟ 1 + Rout 2 = ro2⎜1 + ⎟ = 60.4 kΩ⎜ ⎜ 4.88 kΩ + 1.75kΩ + 5kΩ ⎟ = 2.01 MΩ ⎝ Rth2 + rπ 2 + R2 ⎠ ⎝ ⎠ ro3 = (60 + 10 − 5.43)V = 30.2kΩ 2.14 mA | Rth2 = Rth rπ 2 + (β o + 1)(5kΩ) [ ] Rth3 = 5.01kΩ 1.75kΩ + (76)(5kΩ) = 4.95kΩ ⎛ ⎞ ⎛ ⎞ 75(2.5kΩ) β o R3 ⎜ ⎟ = 30.2 k Ω 1 + Rout 3 = ro3⎜1 + ⎟ ⎜ 4.95kΩ + 0.876kΩ + 2.5kΩ ⎟ = 710 kΩ ⎝ Rth + rπ 3 + R3 ⎠ ⎝ ⎠ 15.155 [ ] For VBE 2 = VBE 3, IO 3 R3 = IO2 R2 | R3 = IO 2 I I 2 R2 = 3(5kΩ) = 15 kΩ | O2 = O 3 | n = 3 IO 3 2 A nA 15-96 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.156 ⎞ I I REF I I 76 ⎛ I = REF | I E1 = C1 = ⎜ REF ⎟ = REF 13 1.173 α F 75 ⎝1.173 ⎠ 1.158 1+ 75 I 12V - 0.7V 12 = I REF ( 10kΩ)+ 0.7V + REF (20 kΩ) | I REF = = 414µA | I E1 = 357µA 1.079 10 kΩ(2.73) I REF = IC1 + 13 I B | IC1 ≅ I E 2 = 4 I E1 | IO 2 = α F (4 I E1 )= 4 I E 3 = 8 I E1 75 (357µA)= 1.41 mA | VE1 = 357µA(20kΩ)= 7.14V 76 75 1 1 | IO 3 = α F (8 I E1 )= 8 (357µA)= 2.82 mA | = = 71.0Ω 76 g m1 40(352µA) ⎛ 1 ⎞ Rth = R ⎜ + R1 ⎟ = 10 kΩ (71.0Ω + 20 kΩ)= 6.68 kΩ | Rth2 = Rth rπ 3 + (β o3 + 1)(2.5kΩ) ⎝ g m1 ⎠ [ ] rπ 2 = 75(0.025V ) 1.41mA = 1.33kΩ | rπ 3 = 75(0.025V ) 2.82 mA = 0.665kΩ | ro2 = 60 + ( 10 − 7.14) 1.41mA = 44.6 kΩ Rth2 = 6.68 kΩ 0.665kΩ + (76)(2.5kΩ) = 6.45kΩ [ ] ⎛ ⎞ ⎛ ⎞ 75(5kΩ) β o R2 ⎜ ⎟ = 1.35 MΩ Rout 2 = ro2⎜1 + ⎟ = 44.2 kΩ⎜1 + ⎟ 6.45 k Ω + 1.33 k Ω + 5 k Ω ⎝ Rth2 + rπ 2 + R2 ⎠ ⎝ ⎠ ro3 = (60 + 10 − 7.14)V = 22.3kΩ 2.82 mA | Rth2 = Rth rπ 2 + (β o + 1)(5kΩ) [ ] Rth3 = 6.68 kΩ 1.33kΩ + (76)(5kΩ) = 6.57 kΩ ⎛ ⎞ ⎛ ⎞ 75(2.5kΩ) β o R3 ⎜ ⎟ = 452 kΩ Rout 3 = ro3⎜1 + ⎟ = 22.3kΩ⎜1 + ⎟ 6.57 k Ω + 0.665 k Ω + 2.5 k Ω ⎝ Rth + rπ 3 + R3 ⎠ ⎝ ⎠ 15.157 IC1 = IREF = 15µA | IB = IC1 2 ID | VCE1 = VBE1 + VGS 3 = 0.7 + VTN + ⎛ V ⎞ Kn β FO ⎜1 + CE1 ⎟ VA ⎠ ⎝ [ ] 15µA 4 IB | VCE1 = 1.45 + | Solving iteratively yields IB = 0.147µA ⎛ VCE1 ⎞ 50 x10−6 100⎜1 + ⎟ VA ⎠ ⎝ ⎛ V ⎞ ⎛ 5⎞ (75 + 5)V = 5.10 MΩ IO 2 = β FO ⎜1 + CE 2 ⎟ IB = 100⎜1 + ⎟(0.147µA) = 15.7 µA | Rout = ro 2 = ⎝ 75 ⎠ 15.7µA VA ⎠ ⎝ ⎛ I 5⎞ Note : If one assumes IB ≅ C1 = 0.15µA, then IO 2 = 100⎜1 + ⎟(0.15µA) = 16.0 µA, a very ⎝ 75 ⎠ β FO IB = similar result and a valid approximation, since VCE1 > n ⎜ IREF − O ⎟ → IO = n 1 n REF βF + 1 ⎝ βF ⎠ 1+ + 1+ β F β F (β F + 1) βF 1 1+ 50µA = 49.6 µA | Rout ≅ (a) IO ≅ β o ro 1 2 125 β r 125 44.3 V 3 50µA = 146 µA | Rout ≅ o o = = 19.0 MΩ (b) IO ≅ 3 146 2 2 µ A 1+ 125 (c ) VCS = IO Rout = 146µA(19.0 MΩ) = 2770V (d) VCB 3 = VEE − 0.7V − 0.7V ≥ 0 → VEE ≥ 1.40V = 125 40 − 0.7 − (−5) V = 55.8 MΩ 2 49.6 µA 15-100 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.166 i1 vb r π3 βo i 1 r o3 ix ve + v1 ix 1 gm1 + vx - r o2 i i ⎡ix ⎤ ⎡gm1 + gπ 3 − gπ 3 ⎤⎡v e ⎤ | ⎢ ⎥=⎢ ⎥⎢ ⎥ ⎣0 ⎦ ⎣gm1 − gπ 3 gπ 3 + go 2 ⎦⎣v b ⎦ IC1 ≅ IC 2 ≅ IC 3 so the small - signal parameters are matched ⎛ βo 1 ⎞ 2 + + ⎟ ≅ 2 gm1gπ 3 for µ f >> β o >> 1 ∆ = 2 gm1gπ 3 + gm1go 2 + gπ 3 go2 = gm1gπ 3 ⎜ ⎜ µf µf ⎟ ⎝ ⎠ g + go 2 g −g i ⎛ βo ⎞ i = x ⎜ 1+ ⎟ | v b = −ix m1 π 3 v e = ix π 3 ≅ x ⎜ ⎟ ∆ 2 gm1 ⎝ µ f ⎠ 2 gm1 ∆ g + go 2 i ⎛ 1 ⎞ ix i v b − v e = −ix m1 =− x ⎜ 1 + ≅ rπ 3 | i1 = gπ 3 (v b − v e ) = x ⎟ ⎜ ⎟ ∆ 2 gπ 3 ⎝ µ f ⎠ 2 2 i βr βr i v 1 v x = v e + (ix − β oi1)ro3 = x + ix ro 3 + β o ro 3 x | Rout = x = + ro 3 + o o 3 ≅ o o3 2 gm1 2 ix 2 gm1 2 2 Rout = vx ix 15.l67 Rout ≅ β oro 2 ≅ β oVA 2 IO = β oVA 2 nI REF | For Prob. 15.165, Rout ≅ 2n (50µA) 125(40) = 50 MΩ n 15.168 VCB 3 = VC 3 − VBE 3 − VBE 2 − (−VEE )≥ 0 → VC 3 ≥ −VEE + VBE 3 + VBE 2 VBE 3 + VBE1 = VT ln 15.165 : IC 3 ≅ n 1+ β +1 IC 3 I I I I + VT ln C1 | IC1 + C1 + C1 = C 3 ≅ F I | From Prob. IS3 I S1 β F nβ F α F βF C3 n I REF = 5 1+ 5 125 15µA = 72.1µA | IC1 = βF βF + 1 I = 72.0µA 1 1 βF C3 + 1+ β F nβ F 1 ⎛ 72.1µA 72.0µA ⎞ VBE 3 + VBE 2 = 0.025V ⎜ ln + ln ⎟ = 1.16V | VC 3 ≥ −VEE + 1.16 V 3 fA 15 fA ⎠ ⎝ ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-101 15.169 (a) Assuming balanced drain voltages, I VGS1 = VTN + 2 I D1 2 I REF = 0.75 + ' Kn1 4 5 Kn 2 I REF 4 5K D3 = I D1 = I REF ⎛W ⎞ ⎜ ⎟ ⎝ L ⎠1 I REF = ⎛W ⎞ 4 ⎜ ⎟ ⎝ L ⎠2 ' 4 20 Kn | I REF = 5 − VGS1 − VGS 3 30 kΩ ( ) | VGS 3 = 0.75 + 2 I REF ( 2 I REF ) I REF ' 10 Kn 5 − 0.75 − I REF = ( ) ' n − 0.75 − ' 4 20 Kn ( ) 30 kΩ | (30kΩ)I REF = 3.5 − 1.5 ' 2 Using Kn = 25 x10-6 and rearranging : 9 x108 I REF − 2.19 x105 I REF + 12.25 = 0 I REF = 87.2 µA and IO = requires VGS 4 = VGS 3 (b) This part requires an iterative solution or the use of a computer solver. Assuming VDS balance between M1 and M2 , λ ≠ 0 will not affect the current mirror ratio, but it will change VGS and hence I REF slightly. One iterative approach : Guess VGS 1 | Then I D1 = ' 2 5 Kn VGS 1 − VTN ) ( 1 + λVGS 1 ) ( 2 I REF = 21.8 µA. Drain voltage balance on M1 and M2 4 ⎛ W ⎞ 80 2 I REF 2 I REF | VTN + = VTN + | ⎜ ⎟ = ' ⎛W ⎞ ' ⎝ L ⎠4 1 4 20 Kn ⎜ ⎟ Kn ⎝ L ⎠4 ( ) Since I D 3 = I D1 , VGS 3 = VTN + I REF = 20 K 1 + λ (5 − VGS 1 ) ' n [ 2 I DS1 ] 5 − VGS1 − VGS 3 I and I D1 = REF . 30 kΩ 4 If the second value of I D1 does not agree with the first, then try a new VGS 1. A spreadsheet yields : VGS 1 = 1.336V , VGS 3 = 1.381V , I REF = 87.5µA, IO = 21.7µA. Note: There is essentially no change from the first answer! 15-102 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.169 cont. (c) *Problem 15.169 - NMOS Wilson Source RREF 1 0 30K VSS 4 0 DC -5 M1 3 3 4 4 NFET W=5U L=1U M2 2 3 4 4 NFET W=20U L=1U M3 0 1 3 3 NFET W=20U L=1U M4 1 1 2 2 NFET W=80U L=1U .MODEL NFET NMOS KP=25U VTO=0.75 LAMBDA=0.015 .OP .END 15.170 IO = I REF ⎛W ⎞ ⎜ ⎟ ⎝ L ⎠1 ⎛W ⎞ ⎜ ⎟ ⎝ L ⎠2 | Rout = µ f 2ro3 = µ f 2 1 1 = λ3 IO λ2 2 Kn 1 I D2 λ3 IO 3 Rout = 1 λ2 ⎛W ⎞ ' 2⎜ ⎟ Kn 1 ⎝ L ⎠2 I REF λ3 I REF ⎡⎛ W ⎞ ⎤2 ⎛W ⎞ ⎢⎜ ⎟ ⎥ ⎜ ⎟ 1 ⎢⎝ L ⎠2 ⎥ ⎝ L ⎠2 = ⎛ W ⎞ λ2 λ3 ⎢ I REF ⎥ ⎜ ⎟ ⎥ ⎢ ⎝ L ⎠1 ⎦ ⎣ ' 2 Kn ⎛W ⎞ ⎜ ⎟ ⎝ L ⎠1 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-103 15.171 ix M3 + vx + ro2 gm2 v v 1 g m1 ix = go 2v x + gm 2v1 ⎛ 1 ⎞ gm 3 ⎜ ⎟ v ⎝ gm1 ⎠ | v1 = v x = x ⎛ 1 ⎞ 2 1 + gm 3 ⎜ ⎟ ⎝ gm1 ⎠ | Rin = vx 2 2 = ro ≅ ix gm 2 gm 2 | gm1 = gm 3 | v1 = vx 2 ix = go 2v x + gm 2 15.172 vx 2 ⎛W ⎞ ⎜ ⎟ ⎝ L ⎠1 150µA Since λ = 0, ID 3 = ID1 = IREF = = 37.5µA ⎛W ⎞ 4 ⎜ ⎟ ⎝ L ⎠2 2 ID1 2 ID 3 + VTN 3 + − VTN 3 K n1 Kn 3 VDS 3 ≥ VGS 3 − VTN 3 | VD 3 − (−10 + VGS1 ) ≥ VGS 3 − VTN 3 | VD 3 ≥ −10 + VGS1 + VGS 3 − VTN 3 VD 3 ≥ −10 + VTN1 + 2(37.5µA)V 2 2(37.5µA)V 2 VD 3 ≥ −10V + 0.75V + + = −8.09 V 5(25µA) 20(25µA) 15.173 ⎛W ⎞ ⎛W ⎞ 1 1 Rout ≅ µ f 2 ro 3 | ⎜ ⎟ = ⎜ ⎟ ⇒ IO = IREF = 50µA | ro 3 ≅ = = 1.60 MΩ ⎝ L ⎠ 2 ⎝ L ⎠1 λID 3 0.0125(50µA) Rout 250 MΩ 1 2K n = = 156 | Using Eq. 13.71, µ f 2 = ro 3 1.60 MΩ λ ID 2 µf 2 = ⎛W ⎞ 2 I 2 3.80 5 x10−5 = ⎜ ⎟ = (λµ f 2 ) D 2' = [0.0125(156)] −5 ⎝ L ⎠2 2K n 1 2(2.5 x10 ) 15.174 15-104 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 The circuit is the same as Fig. 16.20 with the addition of RREF in parallel with ro2. We require RREF >> ro2 in order not to reduce the gain of the feedback loop. A current source with a source resistor which achieves ROUT = ro(1+gmRs) should be sufficient. A cascode or Wilson source will also work. 15.175 M1 and M2 are voltage balanced. ( a ) Rout = µ f 4ro2 | All Kn are the same : IO = I REF = 17.5µA VGS1 = 0.75 + 2 17.5 x10−6 75 x10 −6 ( ) = 1.43V | ∆V3 = VGS 3 − VTN 3 = 1.43V − 0.75V = 0.680 1 1 + 1.43 + (5 − 1.43) ro2 = 0.0125 = 4.65 MΩ | ro 4 = 0.0125 = 4.78 MΩ 17.5µA 17.5µA g m4 = 2 75 x10−6 17.5 x10−6 1 + 0.0125(5 − 1.43) = 5.24 x10−5 S ( )( )( ) µ f 4 = 5.24 x10−5 S (4.78 MΩ) = 250 | Rout = 1.16 GΩ min = VGS1 + ∆V4 = 1.43 + 0.680 = 2.11V ( b ) VCS = IO Rout = 20.3 kV ! ( c ) VDD 15.176 *Problem 15.176 - NMOS Cascode Source IREF 0 1 DC 17.5U VDD 2 0 DC 5 M1 3 3 0 0 NFET W=3U L=1U M2 4 3 0 0 NFET W=3U L=1U M3 1 1 3 3 NFET W=3U L=1U M4 2 1 4 4 NFET W=3U L=1U .MODEL NFET NMOS KP=25U VTO=0.75 LAMBDA=0.0125 .OP .TF I(VDD) VDD .END Results: IO = 17.5 µA ROUT = 1.17 GΩThe same as the hand analysis. ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-105 15.177 ⎛W ⎞ ⎜ ⎟ ⎝L⎠ (a) M1 and M 2 are voltage balanced, so IO = ⎛ W ⎞2 IREF = 1.05 IREF = 18.4µA, a 5% error. ⎜ ⎟ ⎝ L ⎠1 (b) To first order, IO does not depend upon W/L of M 3 and M 4 . The mismatch will create a small VDS mismatch between M1 and M 2 , but this error will be negligible. An estimate of this effect is ∆Io = go 2∆VDS where ⎛ 2 ID 3 ⎞ ⎛ 2 ID 4 ⎞ 2 ID 2 ID 2 ID − = 0.026 ∆VDS = VGS 3 − VGS 4 = ⎜VTN + ⎟ − ⎜VTN + ⎟= Kn 3 ⎠ ⎝ Kn 4 ⎠ Kn3 0.95K n 3 Kn3 ⎝ ∆VDS = 0.026 15.178 2(17.5) = 0.0218V | ∆Io = 0.0125( 17.5 x10−6 ) (0.0218) = 3.89 nA 50 ⎛W ⎞ ⎛W ⎞ 1 1 Rout = µ f 4 ro 2 | ⎜ ⎟ = ⎜ ⎟ ⇒ IO = IREF = 50µA | ro 2 ≅ = = 1.60 MΩ ⎝ L ⎠ 2 ⎝ L ⎠1 λID 2 0.0125(50µA) Rout 250 MΩ 1 2K n 4 = = 156 | Using Eq. 13.71, µ f 4 = ro 2 1.60 MΩ λ ID 4 µf 4 = ⎛W ⎞ 2 I 2 3.80 5 x10−5 = 0.0125 156 = ⎜ ⎟ = (λµ f 4 ) DS 4 ( ) [ ] ' −5 ⎝ L ⎠4 2K n 1 2(2.5 x10 ) 15-106 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.179 ( a − a) Rout = µ f 4 ro2 | All K n are the same : IO = IREF = 25µA VGS1 = 0.75 + 2(25 x10−6 ) 75 x10−6 = 1.57V | ∆V3 = VGS 3 − VTN 3 = 1.57V − 0.75V = 0.817V 1 1 + 1.57 + (5 − 1.57) ro2 = 0.0125 = 3.26 MΩ | ro 4 = 0.0125 = 3.38 MΩ 25µA 25µA 25 x10−6 )( 1 + 0.0125(5 − 1.57)) = 6.25 x10−5 S gm 4 = 2(75 x10−6 )( µ f 4 = 6.25 x10−5 S (3.38 MΩ) = 211 | Rout = 689 MΩ min = VGS1 + ∆V4 = 1.57 + 0.817 = 2.39 V ( a − b) VCS = IO Rout = 17.2 kV ! ( a - c ) VDD (b − a) Rout = µ f 4 ro 2 | All K n are the same : IO = IREF = 25µA VGS1 = 0.75 + 2( 50 x10−6 ) 75 x10−6 = 1.91V | ∆V3 = VGS 3 − VTN 3 = 1.91V − 0.75V = 1.16V 1 1 + 1.91 + (5 − 1.91) ro2 = 0.0125 = 1.64 MΩ | ro 4 = 0.0125 = 1.66 MΩ 50µA 50µA 50 x10−6 )( 1 + 0.0125(5 − 1.91)) = 8.83 x10−5 S gm 4 = 2(75 x10−6 )( µ f 4 = 8.83 x10−5 S (1.66 MΩ) = 147 | Rout = 240 MΩ min = VGS1 + ∆V4 = 1.91 + 1.16 = 3.07 V (b − b) VCS = IO Rout = 12.0 kV ! (b - c ) VDD 15.180 1 1 2 2 R= + = = = 39.0 kΩ − 6 gm 3 gm1 gm1 17.5 x10−6 ) 2(75 x10 )( ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-107 15.181 (a ) I REF = I C 3 + I B 3 + I B 4 = I E 3 + I B 4 = I C1 + I REF 2 I C1 βF + IC 2 2I I = I C1 + C1 + C 1 βF + 1 βF βF + 1 I C1 = 1+ 2 βF + 1 βF + 1 | IO = IC 4 = α F IC 2 = α F IC1 = βF + 1 βF I REF 1+ 2 βF + 1 βF + 1 (b) V ⎛ ⎞ ⎜ 110(50) 110 17.5µA ⎟ βr IO = = 163 MΩ ⎜ ⎟ = 16.9 µA | Rout = o o ≅ 2 1 ⎟ 2 111 ⎜ 2( 16.9µA) 1+ + ⎝ 110 111 ⎠ CS = IO Rout = 16.9µA( 163 MΩ) = 2750 V (c) V CC ≥ 2VBE = 1.40 V 15.182 *Figure 15.182 - NPN Cascode Current Source IREF 0 1 17.5U VCC 2 0 DC 5 Q1 3 3 0 NBJT 1 Q2 4 3 0 NBJT 1 Q3 1 1 3 NBJT 1 Q4 2 1 4 NBJT 1 .MODEL NBJT NPN BF=110 VA=50 .OP .TF I(VCC) VCC .END Results: IO = 16.9 µA Rout = 164 MΩ  the same as the hand analysis 15.183 ⎛ 1 ⎞ 1 R =⎜ + rπ 2 ⎟ ⎝ g m3 g m1 ⎠ [r + (β π4 o4 + 1)ro2 ≅ ] 1 1 2 2 + ≅ = = 2.86 kΩ g m3 g m1 g m1 40( 17.5µA) 15-108 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.184 IO = ro2 = (a) Assuming β o = 80 and VA = 60V : 100(0.025) 60V + 10 − 0.0635 = 551kΩ | rπ 2 = = 19.7 kΩ | g m1 = 40(80µA) = 3.2 mS 0.127 mA 0.127 mA ⎛ ⎞ ⎜ ⎟ ⎛ ⎞ 100(0.5kΩ) β o R2 ⎟ = 551kΩ⎜1 + ⎟ Rout = ro2 ⎜1 + ⎜ 0.313kΩ + 19.7 kΩ + 0.5kΩ ⎟ = 1.89 MΩ 1 ⎜ ⎟ ⎝ ⎠ + rπ 2 + R2 ⎟ ⎜ g ⎝ ⎠ m1 ⎛ ⎞ 0.025V 80µA ln 20( 1.05)⎟ → IO = 129 µA (b) IO = 500Ω ⎜ IO ⎝ ⎠ V ⎛ I REF AE 2 ⎞ 0.025V ⎛ 80µA ⎞ ln 14⎟ → IO = 114 µA ⎜ ln ⎟= (c) IO = RT ⎜ IO AE1 ⎠ 500Ω ⎝ IO ⎠ 2⎝ ro2 = Rout 100(0.025) 60V + 10 − 0.057 = 614kΩ | rπ 2 = = 21.9kΩ | g m1 = 40(80µA)= 3.2 mS 0.114 mA 0.114mA ⎛ ⎞ 100(0.5kΩ) ⎟ = 614kΩ⎜ 1 + ⎜ 0.313kΩ + 21.9kΩ + 0.5kΩ ⎟ = 1.97 MΩ ⎝ ⎠ VT ⎛ I REF AE 2 ⎞ 0.025V ⎛ 80µA ⎞ 20⎟ → IO = 127 µA ⎜ ln ⎜ ln ⎟= R2 ⎝ IO AE1 ⎠ 500Ω ⎝ IO ⎠ 15.185 Assuming βo = 80 and VA = 60V : V ⎛ I REF AE 2 ⎞ 0.025V ⎛ 35µA ⎞ ln 20⎟ → IO = 64 µA ⎜ ln ⎟= (a) IO = RT ⎜ IO AE1 ⎠ 935Ω ⎝ IO ⎠ 2⎝ 100(0.025) 60V + 10 − 0.0598 = 1.09 MΩ | rπ 2 = = 39.1kΩ | g m1 = 40(35µA) = 1.40 mS 0.064 mA 0.064 mA ⎛ ⎞ ⎜ ⎟ ⎛ ⎞ 100(0.935kΩ) β o R2 ⎟ ⎜ ⎜ ⎟ = 3.59 MΩ Rout = ro2 1 + = 1.09 MΩ⎜1 + ⎟ 1 0.714 k Ω + 39.1 k Ω + 0.935 k Ω ⎜ ⎟ ⎝ ⎠ + rπ 2 + R2 ⎟ ⎜ ⎝ g m1 ⎠ V ⎛ I REF AE 2 ⎞ 0.025V ⎛ 35µA ⎞ ln 10⎟ → IO = 51.3 µA ⎜ ln ⎟= (b) IO = RT ⎜ IO AE1 ⎠ 935Ω ⎝ IO ⎠ 2⎝ ro2 = ro2 = Rout 100(0.025) 60V + 10 − 0.0480 = 1.36 MΩ | rπ 2 = = 48.7 kΩ | g m1 = 40(35µA)= 1.40 mS 51.3µA 51.3µA ⎛ ⎞ ⎜ ⎟ ⎛ ⎞ 100(0.935kΩ) β R o 2 ⎟ = 1.36 MΩ⎜1 + ⎟ = ro2 ⎜1 + ⎜ 0.714 kΩ + 48.7 kΩ + 0.935kΩ ⎟ = 3.89 MΩ 1 ⎜ ⎟ ⎝ ⎠ + rπ 2 + R2 ⎟ ⎜ ⎝ g m1 ⎠ ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-109 15.186 VT ⎛ IREF ⎜ ln IO ⎝ IO V ⎛ I (b) R2 = T ⎜ln REF IO ⎝ IO (a) R2 = AE 2 ⎞ 0.025V ⎛ 73µA ⎞ 20⎟ = 4.77 kΩ ⎟= ⎜ln AE1 ⎠ 22µA ⎝ 22µA ⎠ AE 2 ⎞ 0.025V ⎛ 73µA ⎞ 20⎟ = 24.3 kΩ ⎟= ⎜ln AE1 ⎠ 5.7µA ⎝ 5.7µA ⎠ (c ) R2 = 15.187 VT IO ⎛ IREF AE 2 ⎞ 0.025V ⎛ 73µA ⎞ 10⎟ = 21.3 kΩ ⎜ ln ⎟= ⎜ln ⎝ IO AE1 ⎠ 5.7µA ⎝ 5.7µA ⎠ AE 2 ⎞ 0.025V ⎛ 62µA ⎞ 10⎟ = 8.22 kΩ ⎟= ⎜ln AE1 ⎠ 12µA ⎝ 12µA ⎠ AE 2 ⎞ 0.025V ⎛ 62µA ⎞ 10⎟ = 8.22 kΩ ⎟= ⎜ln AE1 ⎠ 512µA ⎝ 512µA ⎠ VT ⎛ IREF ⎜ ln IO ⎝ IO V ⎛ I (b) R2 = T ⎜ln REF IO ⎝ IO (a) R2 = 15.188 *Problem 15.188 - NPN Widlar Current Source IREF 2 1 50U VCC 2 0 DC 10 Q1 1 1 0 NBJT 1 Q2 2 1 3 NBJT 20 R2 3 0 4K .MODEL NBJT NPN BF=110 .OP .DC IREF 50U 5M 50U .PROBE IC(Q2) .END 50uA I O2 40uA 30uA I REF 20uA 0A 1.0mA 2.0mA 3.0mA 4.0mA 5.0mA 15-110 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.189 ⎛V ⎞ V V I IO = α F IE 2 = α F ⎜ BE1 + IB1 ⎟ ≅ BE1 = T ln C1 R2 IS1 ⎝ R2 ⎠ R2 V − VBE 2 − VBE1 V − VBE 2 − VBE1 IC1 = EE − IB 2 ≅ EE R1 R1 0.025V 15 − 1.4 V 0.7V ln 4 −15 = 318 µA | Note : BE1 ≅ = 318 µA (a) IC 2 ≅ R2 2.2 kΩ 10 ( 2.2 kΩ 10 ) (b) IC 2 ≅ (c ) IC 2 ≅ 15.190 0.025V 3.3 − 1.4 V 0.7V ln 4 −15 = 295 µA | Note : BE1 ≅ = 318 µA R2 2.2 kΩ 10 ( 2.2 kΩ 10 ) VT VCC − VEB1 − VEB 2 0.025V 5 − 1.4 0.7V ln = ln 4 −15 = 66.5 µA | = 70 µA R2 IS1R1 10 kΩ 10 ( 10kΩ 10 ) (a) Choose IC1 = 0.2 IO VEE − VBE 2 − VBE1 3.3 − 1.4 − IB 2 ≅ = 317 kΩ IC1 6µA ⎛ 6µA ⎞ I VT ln C1 0.0258V ⎜ln ⎟ ⎛V ⎞ IS1 ⎝ 0.1 fA ⎠ IO = α F IE 2 = α F ⎜ BE1 + IB1 ⎟ | R2 = = = 21.2 kΩ 131 6µA IO ⎝ R2 ⎠ −I (30µA) − 130 α F B1 130 | R1 = VCC − VEB 2 − VEB1 3.3 − 1.4 − IB 2 ≅ = 317 kΩ 6µA IC1 ⎛ 6µA ⎞ I VT ln C1 0.0258V ⎜ln ⎟ ⎛V ⎞ IS1 ⎝ 0.1 fA ⎠ EB 1 IO = α F IE 2 = α F ⎜ + IB1 ⎟ | R2 = = = 21.2 kΩ 6µA 131 IO ⎝ R2 ⎠ −I (30µA) − 130 α F B1 130 (b) This is the same circuit implemented with pnp transistors. Choose IC1 = 0.2 IO | R1 = ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-111 15.191 i r π4 −β o i r o4 ix + v1 ix + vx R th i r o2 + ve - - v x = v1 + v e = (ix − β oi1)ro 4 + (ix − 2i)ro2 | i1 = −i | v x = (ix + β oi)ro4 + (ix − 2i)ro2 i= (ix − 2i)ro 2 rπ 4 + Rth where Rth = ≅ 1 1 (i − 2i)ro2 + > β o | v x = ix ⎜ ro2 + ro 4 + o o4 − ro 2 ⎟ ⎝ ⎠ 2 2 for Rth > β o ⎛β ⎞ βr Rout ≅ ro4 ⎜ o + 1⎟ ≅ o o 4 ⎝2 ⎠ 2 15.192 An iterative solution is required : 1. Choose VGS 2 . Then I D 2 = 2 Kn2 VGS 2 − VTN 2 ) 1 + λ (VDD − I D2 R1 ) ( 2 [ ] 2 Kn2 VGS 2 − VTN 2 ) ( 1 + λVDD ) ( or I D 2 = 2 and 2 K n2 1+ (VGS 2 − VTN 2 ) (λR1) 2 2 I D1 2. VGS 1 = VTN 1 + Kn1 1 + λ( VDD − I D1 R2 ) I D1 = VGS 2 R2 [ ] 3. I D 2 = VDD − VGS 1 − VGS 2 Compare to I D 2 in step 1 and choose new VGS 2 R1 −−−−− I D2 2 2.5 x10-4 VGS 2 − 0.75) ( 1.17) ( V 2 = | I D1 = GS 2 -4 2 15kΩ 2.5 x10 1+ VGS 2 − 0.75) (0.017 *10000) ( 2 VGS1 = 0.75 + 2.5 x10 1 + 0.017( 10 − 15000 I D1) -4 [ 2 I D1 ] | I D2 = 10 − VGS 1 − VGS 2 10 kΩ Iteration yields VGS 2 = 2.744V , IO = I D 2 = 536 µA, and I D1 = 183 µA. 15-112 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.193 Choose ID 2 = 0.2IO = 15µA | The effect of λ1 can be ignored because of the presence of 2(75 x10−6 ) 2 ID1 resistor R2 : VGS1 = VTN1 + = 0.75 + = 1.53V | Then, K n1 250 x10−6 ID 2 = Kn 2 250µA 2 2 (VGS 2 − VTN 2 ) [1 + λ(VGS 2 + VGS1 )] | 15µA = (VGS 2 − 0.75) [1 + λ(VGS 2 + 1.53)] 2 2 V 1.09V An iterative solution gives VGS 2 = 1.089V | R2 = GS 2 = = 14.5 kΩ IO 75µA V − VGS1 − VGS 2 6 − 1.53 − 1.09 R1 = DD = = 225 kΩ ID 2 15µA 15.194 An iterative solution is required : K 2 1. Choose VGS 1. Then I D1 = p1 ( VGS1 − VTP1) 1 + λ (VDD − I D1 R1 ) 2 K p1 2 VGS 1 − VTP1 ) ( 1 + λVDD ) ( V 2 or I D 2 = and I D 2 = SG1 K 2 R2 1 + p1 ( VGS 1 − VTP1 ) (λR1 ) 2 2 I D2 2. VGS 2 = VTP 2 − KP 2 1 + λ( VDD − I D 2 R2 ) [ ] [ ] 3. I D1 = VDD − VSG1 − VSG 2 Compare to I D2 in step 1 and choose new VGS 1 R1 2 10-4 VGS 1 + 0.75) ( 1.10) ( 2 −−−−− I D1 = | I D2 = VGS1 18 kΩ 5 − VSG1 − VSG 2 10 kΩ 2 10 VGS 1 + 0.75) (0.02 *10000) ( 2 2 I D2 VGS 2 = −0.75 − -4 10 1 + 0.02(5 − 18000 I D2 ) 1+ -4 [ ] | I D1 = Iteration yields VGS 1 = −1.984V , IO = I D2 = 110 µA, and I D1 = 82.4 µA. ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-113 15.195 Choose ID1 = 0.2 IO = 25µA | The effect of λ2 can be ignored because of the presence of resistor R2 : VGS 2 = VTP 2 − ID1 = 2( 125 x10−6 ) 2 ID 2 = −0.75 − = −2.33V | Then, KP 2 100 x10−6 K p1 100µA 2 2 (VGS1 − VTP1 ) 1 + λ(VGS1 + VGS 2 ) | 25µA = (VGS1 + 0.75) 1 + λ(VGS1 + 2.33) 2 2 VGS 1.43V An iterative solution gives VGS1 = −1.432V | R2 = − = = 11.4 kΩ IO 125µA V + VGS1 + VGS 2 9 − 1.43 − 2.33 R1 = DD = = 210 kΩ 25µA ID1 [ ] [ ] 15.196 IC1 = IC 3 = 3IC 4 | IC 4 = IC 2 = IC 2 = VBE1 − VBE 2 VT ⎛ IC1 I ⎞ V ⎛ I I ⎞ = ⎜ln − ln C 2 ⎟ = T ⎜ln C1 S 2 ⎟ R R ⎝ IS1 IS 2 ⎠ R ⎝ IC 2 IS1 ⎠ 0.025V ⎛ 3IC 2 20 A ⎞ ⎜ln ⎟ = 46.5 µA | IC1 = 3IC 2 = 140 µA 2.2 kΩ ⎝ IC 2 A ⎠ 15.197 *Problem 15.197 - BJT reference current cell P15.196 VCC 1 0 DC 1.5 AC 1 VEE 5 0 DC -1.5 Q4 2 2 1 PBJT 1 Q3 3 2 1 PBJT 3 Q1 3 3 5 NBJT 1 Q2 2 3 4 NBJT 20 R 4 5 2.2K .MODEL NBJT NPN BF=100 VA=50 .MODEL PBJT PNP BF=100 VA=50 .OP .AC LIN 1 1000 1000 .PRINT AC IC(Q1) IC(Q2) .END Results: IC1 = 140 µA 15.198 IC1 = IC 3 = 3IC 4 | IC 4 = IC 2 = IC 4 = IC 2 = VBE1 − VBE 2 VT = R R IC2 = 47.8 µA C1 S IV = 2. 92 x10 CC −2 −3 C2 SI V CC = 9 . 92 x 10 ⎛ IC1 I ⎞ V − ln C 2 ⎟ = T ⎜ln IS 2 ⎠ R ⎝ IS1 ⎛ IC1 IS 2 ⎞ ⎜ln ⎟ ⎝ IC 2 IS1 ⎠ 0.025V ⎛ 3IC 2 8 A ⎞ ⎜ ln ⎟ = 19.9 µA | IC 1 = 3IC 2 = 59.6 µA | IC 3 = IC1 = 59.6 µA 4 kΩ ⎝ IC 2 A ⎠ 15.199 15-114 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 VBE1 − VBE 2 must be greater than 0. | VBE1 − VBE 2 = VT ( ) > 0 ln nA 1 IC1IS 2 > 0 or 3 >1 | n > A 3 IC 2 IS1 15.200 VBE1 − VBE 2 VT ⎛ IC1 I ⎞ V ⎛I I ⎞ = ⎜ ln − ln C 2 ⎟ = T ln⎜ C1 S 2 ⎟ R R ⎝ IS1 IS 2 ⎠ R ⎝ IC 2 IS1 ⎠ −23 kT 1.38 x10 (323) V ⎛ I I ⎞ 27.9mV VT = = = 0.0279V | R = T ln⎜ C1 S 2 ⎟ = ln(3 ⋅ 5) = 2.16 kΩ −19 35µA 1.6 x10 IC 2 ⎝ IC 2 IS1 ⎠ q −23 kT 1.38 x10 (273) VT ⎛ IC1 IS 2 ⎞ 23.6 mV = 0.0236 V | R = ln⎜ ln(3 ⋅ 10) = 2.29 kΩ (b) VT = = ⎟= 35µA 1.6 x10−19 IC 2 ⎝ IC 2 IS1 ⎠ q (a) IC1 = IC 3 = 3IC 4 | IC 4 = IC 2 = 15.201 The M 3 - M 4 current mirror forces ID1 = ID 2 . VGS1 − VGS 2 = ID 2 R | ID 2 R = VTN + ID 2 = 2 ID1 2 ID 2 − VTN − ' ' 10K n 20K n | ID 2 R = 2 ID 2 ⎛ 1 ⎞ 1− ⎟ ' ⎜ 10K n ⎝ 2⎠ 1 0.293 1 0.293 = ⇒ ID 2 = 26.4 µA ' 5K n 5100 5(25 x10−6 ) R 15.202 (a) The M 3 - M 4 current mirror forces ID1 = ID 2 . VGS1 − VGS 2 = ID 2 R | ID 2 R = VTN1 + ID 2 = 2 ID1 2 ID 2 2 ID 2 ⎛ 1 ⎞ − VTN 2 − | ID 2 R = 1− ⎟ ' ' ' ⎜ 10K n 20K n 10K n ⎝ 2⎠ 1 0.293 1 0.293 = ⇒ ID 2 = 6.86 µA 4 ' 5K n R 10 5(25 x10−6 ) | VTN 2 = VTO + γ (b) VTN1 = VTO ID 2 R = VTO + ID 2 = ( 2φ ( F + VSB − 2φ F = VTO + 0.5 0.6 + ID 2 R − 0.6 ) ( ) 2 ID1 2ID 2 − VTO − 0.5 0.6 + ID 2 R − 0.6 − ' ' 10K n 20K n ) 0.293 2 IDS 2 0.5 − 4 ' 10 4 10K n 10 ( 0.6 + 10 I 4 D2 − 0.6 → ID 2 = 3.96 µA ) ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-115 15.203 This problem should refer to Prob. 15.202 (a) and (b) *Problem 15.203(a) - MOS reference current cell VDD 1 0 DC 5 AC 1 VSS 5 0 DC -5 M1 3 3 5 5 NFET W=10U L=1U M2 2 3 4 5 NFET W=20U L=1U M3 3 2 1 1 PFET W=10U L=1U M4 2 2 1 1 PFET W=10U L=1U R 4 5 10K .MODEL NFET NMOS KP=25U VTO=0.75 PHI=0.6 GAMMA=0 LAMBDA=0.017 .MODEL PFET PMOS KP=10U VTO=-0.75 PHI=0.6 GAMMA=0 LAMBDA=0.017 *.MODEL NFET NMOS KP=25U VTO=0.75 PHI=0.6 GAMMA=0 LAMBDA=0 *.MODEL PFET PMOS KP=10U VTO=-0.75 PHI=0.6 GAMMA=0 LAMBDA=0 *Problem 15.203(b) - MOS reference current cell *.MODEL NFET NMOS KP=25U VTO=0.75 PHI=0.6 GAMMA=0.5 LAMBDA=0.017 *.MODEL PFET PMOS KP=10U VTO=-0.75 PHI=0.6 GAMMA=0.75 LAMBDA=0.017 .OP .AC LIN 1 1000 1000 .PRINT AC ID(M1) ID(M2) .END Results: (a) ID2 = 13.9 µA ID2 = 12.3 µA I D1 ID2 SV = 7.64 x10−2 SV = 6.23 x10−2 DD DD The currents differ considerably from the hand calculations. I D1 ID2 Results: (b) ID1 = 8.19 µA ID2 = 7.24 µA SV = 7.75 x10−2 SV = 6.31x10−2 DD DD The currents differ considerably from the hand calculations. The currents are quite sensitive to the value of λ. The hand calculations used λ = 0. If the simulations are run with λ = 0, then the results are identical to the hand calculations. 15.204 V ⎛ I 5A⎞ 0.025V ⎛ 2 IC 2 5 A ⎞ IC 2 = T ln⎜ C1 ln⎜ ⎟ | IC1 = IC 3 = 2IC 4 = 2IC 2 | IC 2 = ⎟ = 5.23 µA R ⎝ A IC 2 ⎠ 11kΩ ⎝ A IC 2 ⎠ V ⎛ I 3 A ⎞ 0.025V ⎛15.7µA ⎞ ln⎜ IC 7 = 5 IC 4 = 5(5.23µA) = 26.2 µA | IC 8 = T ln⎜ C 4 ⎟= ⎟ → IC 8 = 6.00 µA 4 kΩ ⎝ IC 8 ⎠ R8 ⎝ A IC 8 ⎠ V ⎛ I A ⎞ 0.025V ⎛ 10.4µA ⎞ ln⎜ IC 5 = 2.5 IC1 = 5 IC 2 = 26.2 µA | IC 6 = T ln⎜ C1 ⎟= ⎟ → IC 6 = 5.42 µA 3kΩ ⎝ IC 6 ⎠ R6 ⎝ A IC 6 ⎠ 15-116 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.205 V ⎛ I 10 A ⎞ 0.025V ⎛ IC 2 10 A ⎞ IC 2 = T ln⎜ C1 ln⎜ ⎟ | IC1 = IC 3 = IC 4 = IC 2 | IC 2 = ⎟ = 5.23 µA R ⎝ A IC 2 ⎠ 11kΩ ⎝ A IC 2 ⎠ V ⎛ I 3 A ⎞ 0.025V ⎛15.7µA ⎞ ln⎜ IC 7 = 5 IC 4 = 5(5.23µA) = 26.2 µA | IC 8 = T ln⎜ C 4 ⎟= ⎟ → IC 8 = 6.00 µA 4 kΩ ⎝ IC 8 ⎠ R8 ⎝ A IC 8 ⎠ V ⎛ I A ⎞ 0.025V ⎛ 5.23µA ⎞ ln⎜ IC 5 = 2.5 IC1 = 13.1 µA | IC 6 = T ln⎜ C 1 ⎟= ⎟ → IC 6 = 3.45 µA 3kΩ ⎝ IC 6 ⎠ R6 ⎝ A IC 6 ⎠ 15.206 (a) IC1 = 2IC 2 set by Q 4 and Q 3 IC1 = 2 IC 2 = 30.6 µA | IC 4 VT ⎛ IC1 IS 2 ⎞ 0.025V ⎛ 2 IC 2 7 IS1 ⎞ ⎜ln ⎟= ⎜ln ⎟ = 15.3 µA R ⎝ IC 2 IS1 ⎠ 4.3kΩ ⎝ IC 2 IS1 ⎠ = IC 5 = IC 6 = IC1 = 30.6 µA | IC 3 = IC 7 = IC 2 = 15.3 µA | IC 2 = (b) No change. The currents are independent of the areas of Q5 , Q6 , and Q7 . 15.207 *Problem 15.207 - NPN Cascode Current Source VCC 1 0 DC 5 AC 1 Q4 2 2 1 PBJT 2 Q3 3 2 1 PBJT 1 Q5 4 3 2 PBJT 1 Q1 6 6 0 NBJT 1 Q2 5 6 7 NBJT 7 Q6 4 4 6 NBJT 1 Q7 3 4 5 NBJT 1 R 7 0 4.3K .MODEL NBJT NPN BF=100 VA=50 .MODEL PBJT PNP BF=50 VA=50 .OP .AC LIN 1 1000 1000 .PRINT AC IC(Q7) IC(Q5) .END Results: IC2 = 15.2 µA IC1 = 28.5 µA - Similar to hand calculations. IC1 IC 2 SV = 1.81x10−3 SV = 7.07 x10−4 CC CC 15.208 (a) IC1 = IC 2 set by Q 4 and Q 3 IC1 = IC 2 = 11.3 µA | IC 4 VT ⎛ IC1 IS 2 ⎞ 0.025V ⎛ IC 2 7IS1 ⎞ ⎜ ln ⎟= ⎜ ln ⎟ = 11.3 µA R ⎝ IC 2 IS1 ⎠ 4.3kΩ ⎝ IC 2 IS1 ⎠ = IC 5 = IC 6 = IC1 = 11.3 µA | IC 3 = IC 7 = IC 2 = 11.3 µA | IC 2 = (b) No change. The currents are independent of the areas of Q5 , Q6, and Q7 . ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-117 15.209 (a) The M 3 - M 4 current mirror forces ID1 = 1.5ID 2 . VGS1 − VGS 2 = ID 2 R | ID 2 R = VTN + 2 ID1 2 ID 2 − VTN − ' ' 10K n 30K n ⎞ ⎛ 1 ⎜ 1 ⎛ 3ID 2 2 ID 2 ⎞ 2 ID 2 3ID 2 ⎟ = ID 2 = ⎜ − − ' ' ⎟ −6 −6 ⎟ ⎜ 3300 R ⎝ 10K n 30K n 10 25 x 10 30 25 x 10 ⎠ ( ) ( )⎠ ⎝ 57.9 ID 2 = ⇒ ID 2 = 308 µA ID1 = 462 µA 3300 ID 4 = ID 5 = ID 6 = ID1 = 462 µA | ID 3 = ID 7 = ID 2 = 308 µA (b) With λ = 0, the currents do not depend upon the W/L ratios of M5 , M 6 or M 7 as long as all transistors remain in the active region. For λ ≠ 0, there will be a weak dependence, since the drain - source voltages of M2 and M 3 will change slightly. 15.210 *Problem 15.210 - MOS reference current cell VDD 1 0 DC 15 AC 1 M3 3 2 1 1 PFET W=10U L=1U M4 2 2 1 1 PFET W=15U L=1U M5 4 3 2 2 PFET W=10U L=1U M6 4 4 6 6 NFET W=10U L=1U M7 3 4 5 5 NFET W=10U L=1U M1 6 6 0 0 NFET W=10U L=1U M2 5 6 7 7 NFET W=30U L=1U R 7 0 3.3K *.MODEL NFET NMOS KP=25U VTO=0.75 LAMBDA=0 *.MODEL PFET PMOS KP=10U VTO=-0.75 LAMBDA=0 .MODEL NFET NMOS KP=25U VTO=0.75 LAMBDA=0.017 .MODEL PFET PMOS KP=10U VTO=-0.75 LAMBDA=0.017 .OP .AC LIN 1 1000 1000 .PRINT AC ID(M1) ID(M2) .END Results: ID2 = 265 µA ID1 = 377 µA These differ from the hand calculations due to the nonzero value of λ. Simulation with λ = 0 gives results very close to the hand calculations. ID2 ID2 = 9.82 x 10−4 SV = 6.99 x 10−4 SV DD DD 15-118 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.211 The M 3 - M 4 current mirror forces I D1 = I D 2. ⎛ 2 I D1 ⎞ ⎛ 2 I D2 ⎞ ⎟ ⎜ ⎟ − V V + + VGS 1 − VGS 2 = I D 2 R | I D 2 R = ⎜ ' ⎟ ⎜ TN ' ⎟ ⎜ TN 10 Kn 30 Kn ⎝ ⎠ ⎝ ⎠ ⎛ 1 ⎜ 1 ⎛ 2 I D2 2 I D2 ⎞ I D2 I D2 ⎜ ⎟ = I D2 = ⎜ − − ' ' ⎟ − 6 R ⎝ 10 Kn 30 Kn ⎠ 3300 ⎜ 5 25 x10 15 25 x10−6 ⎝ 34.1 I D2 = ⇒ I D 2 = 107 µA I D1 = 107 µA 3300 I D 4 = I D 5 = I D 6 = I D1 = 107 µA | I D3 = I D 7 = I D 2 = 107 µA ( ) ( ) ⎞ ⎟ ⎟ ⎠ 15.212 1 g m3 v3 r ο3 r o4 gm4 v3 + icc i cc Current mirror model: 1 Assuming VDS ro2 g m2ro2 µ f 2 Acd = vo 1 ∆g m2 ≅ v ic 2 g m2 The collector current imbalance can be found as follows: Assume that VEC 4 = VEC 3 + ∆V and equal Early voltages: IC 4 0.7 + ∆V ⎛ ∆V ⎞ V 75V VA | ∆V ≅ A = = IC 2 ⇒ IC1 = IC1⎜1 − = 0.60V ⎟ 2 0.7 β F 125 ⎝ VA ⎠ 1+ + β F VA 1+ ⎛ V ⎞ ⎛ V − ∆V ⎞ V ∆V ∆IC = IC1 − IC 2 = IC 0 ⎜1 + C1 ⎟ − IC 0 ⎜1 + C1 | IC 0 = IS exp BE ≅ IC ⎟ = IC 0 VT VA ⎠ VA ⎝ VA ⎠ ⎝ ∆IC ∆V 1 ∆gm 2 ∆IC 1 1 ∆V ∆V ∆IC = IC 0 ≅ IC | ≅ = | = = | Acd = VA β F 2β F IC gm 2 IC βF VA VA 1 1500 = 4 x10 -3 | CMRR = = 3.75 x10 5 (112 dB) -3 2(125) 4 x10 15.218 continued on next page Acd = ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-125 15.218 cont. Note that VOS ≅ ∆V Add | ∆V ≅ 75 0.60 = 0.60V | VOS ≅ = 0.400 mV 125 1500 For VIC : VCB1 = VCC − VEB 3 − VIC ≥ 0 | VCC ≥ VIC + VEB 3 = 1.5 + 0.7 = 2.2 V Assume that the current source needs VCS = 0.7 V across it to operate properly. VIC − VBE − (−VEE ) ≥ VCS | VEE ≥ VCS + VBE − VIC = 0.7 + 0.7 − (−1.5) = 2.9V We need ± 2.9 - V supplies or approximately ± 3V. (b) Add = g m2 ro2 ro4 ≅ g m2 ( ) ⎛ 100 ro 4 100 ⎞ ≅ 40 5 x10−5 ⎜ = 2.00 mS ( 1.00 MΩ) = 2000 −5 −5 ⎟ 2 ⎝ 5 x10 5 x10 ⎠ ( ) 0.7 + ∆V ⎛ ∆V ⎞ v 1 ∆g m2 VA 100V VA | IC 4 = IC 2 ⇒ IC 1 = IC1⎜1 − = = 0.80V Acd = o ≅ ⎟ | ∆V ≅ 2 0.7 v ic 2 g m2 βF VA ⎠ 125 ⎝ 1+ + β F VA ⎛ V ⎞ ⎛ V − ∆V ⎞ V ∆V | IC 0 = I S exp BE ≅ IC ∆IC = IC 1 − IC 2 = IC 0 ⎜1 + C1 ⎟ − IC 0 ⎜1 + C1 ⎟ = IC 0 VT VA ⎠ VA ⎝ VA ⎠ ⎝ ∆IC ∆V 1 ∆g m2 ∆IC 1 1 ∆V ∆V ∆IC = IC 0 ≅ IC | ≅ = | = = | Acd = IC g m2 IC βF VA β F 2β F VA VA 1+ Acd = 1 2000 = 4 x10-3 | CMRR = = 5.00 x105 ( 114 dB) -3 2( 125) 4 x10 For VIC : VCB1 = VCC − VEB 3 − VIC ≥ 0 | VCC ≥ VIC + VEB 3 = 1.5 + 0.7 = 2.2 V VIC − VBE − (−VEE )≥ VCS | VEE ≥ VCS + VBE − VIC = 0.7 + 0.7 − (−1.5) = 2.9V We need ± 2.9 - V supplies or approximately ± 3V. Assume that the current source needs VCS = 0.7 V across it to operate properly. 15.219 Results: For VA = 75 V, Adm = 1470 and Acd = 6.92x10-3. CMRR = 106 dB. The results are similar to hand calculations. Note that a very high CMRR is achieved when the circuit is brought back to balance (with VOS = 0.728mV), as is the case in operational amplifier input stages with feedback applied. For the case with the offset voltage applied, Acd = 2.71x10-7. 15-126 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.220 (a) I D5 = I D 4 = I D 3 = I D2 = I D1 = I SS = 100 µA 2 −4 2 10 2 I D1 VGS 2 = VGS 1 = VTN + = 0.75 + = 1.20V Kn1 40 2.5 x10−5 ( ( ) ) 2 10−4 2ID 4 VGS 5 = VGS 4 = VGS 3 = VTP − = −0.75 − = −1.25V | VDS 3 = VDS 4 + VGS 5 = −2.50V K p4 80 10−5 VDS1 = VDS 2 = 10 + VGS 4 + VGS 5 − (−VGS1 )= 8.70V | VDS 5 = VDS 4 = VGS 4 = −1.25V 100µA,8.70V ) ( 100µA,-2.50V ) ( 100µA,-1.25V ) ( 100µA,-1.25V ) Q − Pts : ( 100µA,8.70V ) ( ( ) ( ) (b) A dd ⎛ 2 ⎞ = g m2 ⎜ ro2 µ f 3ro5 ⎟ ≅ g m2ro2 ⎝ 3 ⎠ ⎛ 1 ⎞ + 8.7 ⎟ ⎜ Add = 2(40) 2.5 x10−5 10−4 1 + 0.017(8.7) ⎜ 0.017 −4 ⎟ = 0.479mS (675kΩ) = 323 10 ⎜ ⎟ ⎝ ⎠ (Note that the loop - gain of the Wilson source is reduced by the presence of Rout1 ≅ 2ro1.) ⎞ ⎛ 1 + 1.25⎟ ⎜ ⎟ = 601kΩ (c) Add = g m1 ro1 ro3 | ro3 = ⎜ 0.017 −4 10 ⎟ ⎜ ⎠ ⎝ ( )( )[ ] ( ) Add = 0.479mS 675kΩ 601kΩ = 152 - The Wilson source yields a 2X improvement. 15.221 M 1 2 3 4 5 ID(µA) 101 99.0 101 99.0 99.0 VDS 8.69 7.33 -2.48 -1.24 -2.60 (V) ( ) Results: Adm = 313. The gain and Q-points are similar to hand calculations. ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-127 15.222 ID 2 = ID1 = I1 I = 125 µA | ID 3 = ID 4 = ID 5 = ID 6 = ID 7 = I2 − 1 = 125 µA 2 2 2( 1.25 x10−4 ) 2 ID For the NMOS transistors VGS = VTN + = 0.75 + = 1.25V Kn 40(2.5 x10−5 ) 2( 1.25 x10−4 ) 2 ID For the PMOS transistors VGS = VTP − = −0.75 − = −1.54V Kp 40( 10−5 ) VDS1 = VDS 2 = −VGS 3 = 1.54V | VDS 7 = VDS 6 = VGS 6 = 1.25V | VDS 5 = VGS 6 + VGS 7 = 2.50V VDS 4 = VDS 3 = (−5 + VDS 5 ) − (−VGS1 − VGS 3 ) = −5 + 2.50 − (−1.25 + 1.54) = −2.79V Q − Pts : (125µA,1.54V) (125µA,1.54V) (125µA,-2.79V) (125µA,-2.79V) (125µA,2.50V) (125µA,1.25V) (125µA,1.25V) M3 v id 2 M1 RL (b) Add = gm 2 (µ f 4 ro 2 µ f 5 ro 7 )≅ µ f 2µ f 4 2 ⎛ 1 ⎞ + 1.54 ⎟ ⎜ gm 2 = 2(40)(2.5 x10−5 )( 1.25 x10−4 )[ 1 + 0.017(1.54 )] = 0.507 mS | ro 3 = ⎜ 0.017 −4 ⎟ = 483kΩ ⎜ 1.25 x10 ⎟ ⎝ ⎠ ⎛ 1 ⎞ + 2.79 ⎟ ⎜ 10−5 )( 1.25 x10−4 )[ 1 + 0.017(2.79)] = 0.324 mS | ro 4 = ⎜ 0.017 −4 ⎟ = 493kΩ gm 4 = 2(40)( ⎜ 1.25 x10 ⎟ ⎝ ⎠ Add ≅ µ f 2µ f 4 2 = (0.507 mS)(483kΩ)(0.324 mS )(493kΩ) = 19600 2 15-128 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.223 *Figure P15.222 - CMOS Folded Cascode Amplifier with Active Load VDD 1 0 DC 5 VSS 10 0 DC -5 *An offset voltage must be applied to bring output to -2.5V V1 4 11 DC -5.085M AC 0.5 V2 5 11 DC 0 AC -0.5 VIC 11 0 DC 0 M1 2 4 6 6 NFET W=40U L=1U M2 3 5 6 6 NFET W=40U L=1U M3 8 6 2 2 PFET W=40U L=1U M4 7 6 3 3 PFET W=40U L=1U M5 8 9 10 10 NFET W=40U L=1U M6 9 9 10 10 NFET W=40U L=1U M7 7 8 9 9 NFET W=40U L=1U I2A 1 2 DC 250U I2B 1 3 DC 250U I1 6 10 DC 250U .MODEL NFET NMOS KP=25U VTO=0.75 LAMBDA=0.017 .MODEL PFET PMOS KP=10U VTO=-0.75 LAMBDA=0.017 .OP .AC LIN 1 1000 1000 .PRINT AC VM(7) VP(7) VM(6) VP(6) VM(8) VP(8) .TF V(7) VIC .END 8 Results: Add = 23700, Acd = 1.81x10-4. Rout = 47.7 MΩ, CMRR = 1.31 x 10 . The values of Add and Rout are similar to hand calculations. Acd and the CMRR are limited by the small residual mismatches in device parameters. From Problem 15.222, g m2 = 0.507 mS | ro3 = 483kΩ | g m4 = 0.324 mS | ro 4 = 493kΩ Rout = µ f 4ro2 µ f 5ro 7 ≅ µ f 4ro2 2 = 0.324 mS (493kΩ)(493kΩ) 2 = 39.4 MΩ ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-129 15.224 M 25 1 25 1 R 25 1 I2 I2 1 M 2 M +5 V 3 I1 M 1 M 10 1 2 10 1 -5V Using a reference current of 250µA and 1:1 current mirrors with − VGSP = VGSN : −VGSP = VGSN = VTN + 15.225 2( 2.5 x10−4 ) 2 ID 10 − 2.16 − 2.16 V = 0.75 + = 2.16V | R = = 22.7 kΩ −6 2.5 x10−4 A Kn 10(25 x10 ) ⎛ R ⎞ VBE 2 R2 = VBE 2 ⎜1 + 2 ⎟ R1 ⎝ R1 ⎠ ⎛ VBE 2 ⎞ ⎜ 200µA − R ⎟ 1 ⎟ | VBE 2 = 0.025ln⎜ 10 fA ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ For β F = ∞, IB = 0. | VBE 3 + VEB 4 = VBE 2 + ⎛ 200µA − I1 ⎞ VBE 2 VBE 2 = 0.025ln⎜ ⎟ | I1 = R1 ⎝ 10 fA ⎠ ⎛ VBE 2 ⎞ ⎜ 200µA − 20 kΩ ⎟ 0.589 VBE 2 = 0.025ln⎜ = 171µA ⎟ → VBE 2 = 0.589V | IC 2 = 200µA − 10 fA 20 k Ω ⎜ ⎟ ⎝ ⎠ ⎛ 20 kΩ ⎞ 1 Since IS 4 = IS 3 , VBE 3 = VEB 4 = (0.589)⎜1 + ⎟ = 0.589V and IC 4 = IC 3 = 171 µA. ⎝ 20 kΩ ⎠ 2 15-130 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.226 (a) β F = ∞ : VBE 3 + VEB 4 = VBE1 + VEB 2 VT ln IO IO I2 I2 + VT ln - V ln - V ln =0 AE 3 AE 4 T AE1 T A ISON ISOP ISON ISOP E 2 AEO AEO AEO AEO 2 2 2 ⎡ IO IO AEO ISON ISOP AE1 AE 2 ⎤ AE1 AE 2 AE 3 AE 4 = 0 → = 1 | IO = I2 VT ln⎢ ⎥ 2 2 2 I2 AEO I2 AE 3 AE 4 AE1 AE 2 ⎣ ISON ISOP AE 3 AE 4 ⎦ (b) IO = 300µA AE 3 AE 4 = 100 µA 3 AE 3 3 AE 4 15.227 (a) I D8 = I REF = 250µA | I D10 = I D9 = 2 I D8 = 500µA I D1 = I D 2 = I D3 = I D 4 = 2(250µA) I D9 = 250µA | VDS 8 = VGS 8 = 0.75 + = 2.16V 2 10 25x10−6 ( ) I D5 = I D11 = I D10 = 500µA | VGS 11 = 0.75 + V VGS 6 = −VGS 7 = GS11 = 1.789V | I D 7 = I D6 2 −VDS 4 = −VDS 3 = −VGS 3 = VGS 2 = VTN + ( ) 10( 25 x10 ) = (1.789 − 0.75) = 135µA 5 25 x10−6 −6 2 2(500µA) = 3.58V 2(250µA) 2 I D2 = 0.75 + = 1.75V Kn2 20 25 x10−6 2 ( ) VDS1 = VDS 2 = 5 − VSD 4 − (−VGS 2 )= 5.00V | VDS 10 = −VDS 5 = 5 − VDS 6 = −VDS 7 = 5.00V | VDS 9 = 5 − VGS 2 = 5 − 1.75 = 3.25V 1 ID (µA) VDS (V) SPICE ID (µA) VDS (V) 250 5.00 255 4.97 2 250 5.00 255 4.99 3 250 -1.75 255 -1.74 4 250 -1.75 255 -1.73 5 500 -3.21 509 -3.24 6 135 5.00 139 5.01 7 135 -5.00 139 -5.00 VGS 11 = 3.21V 2 8 250 2.16 250 2.14 9 500 3.25 509 3.28 10 500 3.21 509 3.23 11 500 3.58 509 3.52 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-131 (b) Adm = Avt1 Avt 2 Avt 3 = [gm 2 (ro2 1 = ro 4 ) gm 5 (ro 5 ro12 ) [1] ][ ] gm 2 = 2(20)(25 x10−6 )(2.50 x10−4 )[ 1 + 0.017(5.00)] = 0.521mS 1 58.8 + 5.00 V 58.8 + 1.75 V = 58.8V | ro 2 = = 255 kΩ | ro4 = = 242 kΩ | AV 1 = 64.7 −4 2.5 x10 2.5 x10−4 A A λ 0.017 58.8 + 3.21 V 10−5 )(5.00 x10−4 )[ 1 + 0.017(3.21)] = 1.03mS | ro12 = ro5 = = 124 kΩ gm 5 = 2(100)( 5.00 x10−4 A Avt 2 = 63.9 | Adm = Avt1 Avt 2 Avt 3 = [64.7] [63.9] [1] = 4130 (c ) The amplification factor is inversely proportional to the square root of current. ⎡ 64.7⎤ ⎡ 63.9 ⎤ Adm = Avt1 Avt 2 Avt 3 ≅ ⎢ [1] = 2065 ⎣ 2⎥ ⎣ 2⎥ ⎦⎢ ⎦ SPICE Results: Adm = 4000, Acm = 0.509, Rout = 1.81 kΩ. 15.228 Thus, Adm ≅ µ f 2µ f 5 4 | For the MOSFET, µ f 2 ∝ 1 IREF 1 ID ∴ Adm ∝ 1 ID 2 1 ID 5 But, ID 2 ∝ IREF and ID 5 ∝ IREF → Adm ∝ (a) Adm = 16000 100µA = 6400 250µA (b) Adm = 16000 100µA = 80000 20µA 15.229 Adm = Av1 Av 2 Av 3 = gm 2 (ro 2 ro 4 ) gm 5 (ro 5 ro12 ) [1] [ ][ ] ID10 = IREF =100µA | ID12 = ID11 = 2 ID10 = 200µA | ID1 = ID 2 = ID 3 = ID 4 = ID 5 = ID12 = 200µA | VDS 4 = VGS 3 = −VGS 2 = −VTN − ID11 = 100µA 2 2(100µA) 2 IDS 2 = −0.75 − = −1.38V Kn2 20(25 x10−6 ) VDS 2 = 10 + VDS 4 − (−VGS 2 ) = 10.0V | gm 2 = 2(20)(25 x10−6 )(100µA)[ 1 + 0.017(10)] = 0.342 mS 1 λ = 1 58.8 + 10 V 58.8 + 1.38 V = 58.8V | ro 2 = = 688 kΩ | ro 4 = = 602 kΩ | Av1 = 110 −4 0.017 10 10−4 A A VGSGG 2 | VGSGG = 0.75 + 2(200µA) = 2.54V | VDS12 = −VDS 5 = 8.73V 5(25 x10−6 ) 58.8 + 8.73 V = 338 kΩ 2 x10−4 A VDS12 = −VDS 5 = 10 − 10−5 )(2 x10−4 )[ 1 + 0.017(8.73)] = 0.678mS | ro12 = ro5 = gm 5 = 2(100)( Av 2 = 115 | Adm = Av1 Av 2 Av 3 = [110] [115] [1] = 12600 Adm = 10900 if (1 + λVDS ) is neglected in the gm calculation. 15-132 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.230 *Figure P15.229 - CMOS Amplifier with Active Load VDD 8 0 DC 10 VSS 14 0 DC -10 *An offset voltage is used to set Vo to approximately zero volts. V2 1 15 DC 0.3506M AC 0.5 V1 2 15 DC 0 AC -0.5 VIC 15 0 DC 0 M1 3 1 5 14 NFET W=20U L=1U M2 4 2 5 14 NFET W=20U L=1U M3 3 3 8 8 PFET W=50U L=1U M4 4 3 8 8 PFET W=50U L=1U M5 6 4 8 8 PFET W=100U L=1U *The offset can be adjusted to zero by correcting the value of W/L *M5 6 4 8 8 PFET W=89.5U L=1U M6 8 6 13 14 NFET W=10U L=1U M7 14 7 13 8 PFET W=25U L=1U MGG 6 6 7 14 NFET W=5U L=1U M10 9 9 14 14 NFET W=10U L=1U M11 5 9 14 14 NFET W=20U L=1U M12 7 9 14 14 NFET W=20U L=1U IREF 0 9 DC 100U .MODEL NFET NMOS KP=25U VTO=0.75 LAMBDA=0.017 .MODEL PFET PMOS KP=10U VTO=-0.75 LAMBDA=0.017 *.MODEL NFET NMOS KP=25U VTO=0.75 GAMMA=0.6 LAMBDA=0.017 *.MODEL PFET PMOS KP=10U VTO=-0.75 GAMMA=0.75 LAMBDA=0.017 .OP .AC LIN 1 1000 1000 .PRINT AC VM(13) VP(13) VM(4) VP(4) .TF V(13) VIC .END Results: Adm = 11200, Acm = 0.604, Rout = 3.10 kΩ. (a) 1 2 3 4 5 6 7 GG 10 11 12 IDS (µA) 112 112 112 112 223 44.2 44.2 223 100 223 223 VDS (V) 9.96 9.99 -1.41 -1.37 -8.70 10.0 -10.0 -2.60 1.63 8.63 8.70 (b) IDS (µA) 110 110 110 110 219 0 0 219 100 219 219 VDS (V) 11.2 11.2 -1.40 -1.37 -8.85 10.0 -9.97 -3.79 1.63 7.41 7.35 Note that the body effect has increased the threshold voltages of M6 and M7 to the point that they are no longer conducting. VTN6 = 2.24 V and VTP7 = -2.61V. The W/L ratio of MGG needs to be redesigned to solve this problem. ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-133 15.231 ID10 = IREF = 250µA | ID11 = 2 ID10 = 500µA | ID12 = 4 ID10 = 1000µA ID1 = ID 2 = ID 3 = ID 4 = 2(250µA) ID11 = 250µA | VDS10 = VGS10 = 0.75 + = 2.16V 2 10(25 x10−6 ) 2(1000µA) = 4.75V 5(25 x10−6 ) −6 ID 5 = IDGG = ID12 = 1000µA | VGSGG = 0.75 + 10( 25 x10 V VGS 6 = −VGS 7 = GSGG = 2.375V | ID 7 = ID 6 = 2 2 VDS 4 = VDS 3 = VGS 3 = −VGS 2 = −VTN − )(2.375 − 0.75) 2 = 330µA 2(250µA) 2 ID 2 = −0.75 − = −1.75V Kn 2 20( 25 x10−6 ) VGSGG = 5.13V 2 VDS1 = VDS 2 = 7.5 + VDS 4 − (−VGS 2 ) = 7.50V | VDS12 = −VDS 5 = 7.5 + VDS 6 = −VDS 7 = 7.5V | VDS11 = 7.5 − VGS 2 = 7.5 − 1.75 = 5.75V M IDS (µA) VDS (V) SPICE IDS (µA) VDS (V) 1 2 3 4 5 6 330 7.50 359 7.54 7 GG 10 11 12 250 250 250 250 1000 7.50 7.50 -1.75 -1.75 -5.13 264 266 -264 -266 1050 7.46 7.09 -1.76 -2.14 -5.20 330 1000 250 500 1000 -7.50 4.75 2.16 5.75 5.13 -359 1050 250 530 1050 -7.46 4.69 2.14 5.78 5.11 Adm = Av1 Av 2 Av 3 = gm 2 (ro 2 ro 4 ) gm 5 (ro 5 ro12 ) [1] gm 2 = 2(20)(25 x10−6 ) (250µA)[1 + 0.017(7.50)] = 0.531mS 1 58.8 + 7.50 V 58.8 + 1.75 V = 58.8V | ro 2 = = 265 kΩ | ro 4 = = 242 kΩ | Av1 = 67.2 −4 2.5 x10 2.5 x10−4 A A λ 0.017 58.8 + 5.13 V 10−5 )( 10−3 ) gm 5 = 2(100)( [1 + 0.017(5.13)] = 1.48mS | ro12 = ro5 = 10−3 A = 63.9kΩ Av 2 = 47.3 | Adm = Av1 Av 2 Av 3 = [67.2] [47.3] [1] = 3180 1 = SPICE Results: Adm = 2950, Acm = 0.03, Rout = 1.10 kΩ. [ ][ ] 15-134 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.232 (a) For saturation of M11 : VDS11 = 0 − VGS 2 − (−VSS ) = VSS − VGS 2 ≥ VGS 2 = VTN + 2(200µA) 2 IDS11 = = 0.894 K n11 20(25 x10−6 ) 2(100µA) 2 IDS 2 = 0.75 + = 1.38V | VSS − 1.38 ≥ 0.894 → VSS ≥ 2.27V Kn 2 20(25 x10−6 ) For saturation of M12 : VDS12 = 0 − 2(200µA) VGSGG V 2ID12 − (−VSS ) = VSS − GSGG ≥ = = 0.894 2 2 K n12 20(25 x10−6 ) 2(200µA) 2.54 = 2.54V | VSS − ≥ 0.894 → VSS ≥ 2.16V −6 2 5( 25 x10 ) 2(100µA) 2 ID1 = = 0.633 K n1 20( 25 x10−6 ) VGSGG = 0.75 + For saturation of M1 and M 2 : VDS1 = VDD + VGS 3 − (−VGS1 ) = VDD + VGS 3 + VGS1 ≥ VGS 3 = −VGS1 : VDD ≥ 0.633V For saturation of M 5 : −VDS 5 = VDD − 2(200µA) 2.54 2 ID 5 ≥ = = 0.633V → VDD ≥ 1.90V K p5 2 100( 10 x10−6 ) M 6 and M 7 are always saturated : e.g. VDS 6 ≥ VGS 6 The minimum supply voltages are : VDD ≥ 1.90V VSS ≥ 2.27V For the symmetrical supply case, VDD = VSS ≥ 2.27V (b) The values of VDD and VSS in part (a) do not provide any significant common - mode input voltage range. For saturation of M11 with VIC = −5V , VDS11 = VIC − VGS 2 − (−VSS ) = VSS − 1.38 − 5 ≥ 0.894 → VSS ≥ 7.27V For saturation of M1 and M 2 : VDS1 = VDD − VSG 3 − (VIC − VGS1) = VDD − 5 − 1.38 +1.38 ≥ 0.633 → VDD ≥ 5.63V For an output range of 5V, saturation of M12 requires V 2.54 ≥ 0.894 → VSS ≥ 7.25V VDS12 = −5 − GSGG − (−VSS ) = VSS − 5 − 2 2 For Saturation of M 5 : 2.54 − 5 ≥ 0.633V → VDD ≥ 6.90V VSD 5 = VDD − 2 The minimum supply voltages are : VDD ≥ 6.90V VSS ≥ 7.25V For the symmetrical supply case, VDD = VSS ≥ 7.25V ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-135 15.233 ID 9 = ID10 = ID12 = ID11 = IREF = 250µA | ID 6 = ID 7 = ID 8 = 3ID12 = 750µA ID1 = ID 2 = ID13 = ID 5 = ID 3 = ID 4 = VDS10 = VDS12 = VDS11 = VGS12 = VTN + VGS1 = 0.75 + ID 9 = 125µA 2 2(250µA) 2 ID12 = 0.75 + = 2.75V K n12 5(25 x10−6 ) 2(125µA) = 1.25V | VDS 9 = −VGS1 − (−10 + VGS12 ) = 6V 40( 25 x10−6 ) 2(750µA) = 7.25V 15(25 x10−6 ) 2(125µA) = −1.31V 80( 10 x10−6 ) 11 12 13 VDS 7 = VO − (−10 + VGS12 + VGS11 − VGS 7 ) = 0 + 10 − 2.75 − 2.75 + 0.75 + VDS 8 = 10 − VDS 7 = 2.75V VDS 6 = −(10 − VO ) = −10V + 0 = −10V | VDS 5 = VDS13 = VDS 3 = VDS 4 = −0.75 − M 1 2 3 4 5 6 7 8 9 10 ID (µA) VDS (V) 125 8.63 125 8.63 125 -1.31 125 -1.31 125 -1.31 750 -10 750 7.25 750 2.75 250 6.00 250 2.75 250 2.75 250 2.75 125 -1.31 ' ⎛W ⎞ Kn 10−5 ⎛ W ⎞ 2 2 ⎜ ⎟ (−VDS 4 − VDS 5 + 0.75) = 750µA (b) ⎜ ⎟ (VGS 6 − VTP ) = 2 ⎝ L ⎠6 2 ⎝ L ⎠6 ⎛W ⎞ 10−5 ⎛ W ⎞ 2 ⎜ ⎟ (−2.62 + 0.75) = 750µA → ⎜ ⎟ = 42.9 ⎝ L ⎠6 2 ⎝ L ⎠6 Add = Av1 Av 2 = (gm 2 ro 2 )(gm 6 ro 6 ) = µ f 2µ f 6 | µ f 2 ≅ 1 λn 2(40)(25 x10−6 ) 2K n 2 1 = = 235 0.017 ID 2 125 x10−6 µf 6 ≅ 1 λp 2(42.9)( 10 x10−6 ) 2K p 6 1 = = 62.9 | Add = 235(62.9) = 14800 0.017 ID 6 750 x10−6 15-136 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.234 *Figure P15.233 - CMOS Amplifier with Active Load VDD 8 0 DC 10 VSS 14 0 DC -10 *Connect feedback to determine Vos *V1 1 13 DC 0 *The offset voltage must be used to set Vo to approximately zero voltages. V1 1 15 DC 0.4423M AC 0.5 V2 2 15 DC 0 AC -0.5 VIC 15 0 DC 0 M1 3 1 5 5 NFET W=40U L=1U M2 4 2 5 5 NFET W=40U L=1U M3 6 7 8 8 PFET W=80U L=1U M4 7 7 8 8 PFET W=80U L=1U M5 4 3 7 7 PFET W=80U L=1U M6 13 4 8 8 PFET W=42.9U L=1U *The offset can be adjusted to zero by correcting the value of W/L *M6 13 4 8 8 PFET W=37.25U L=1U M7 13 9 12 12 NFET W=15U L=1U M8 12 10 14 14 NFET W=15U L=1U M9 5 9 11 11 NFET W=5U L=1U M10 11 10 14 14 NFET W=5U L=1U M11 9 9 10 10 NFET W=5U L=1U M12 10 10 14 14 NFET W=5U L=1U M13 3 3 6 6 PFET W=80U L=1U IREF 0 9 DC 250U .MODEL NFET NMOS KP=25U VTO=0.75 LAMBDA=0.017 .MODEL PFET PMOS KP=10U VTO=-0.75 LAMBDA=0.017 .OP .AC LIN 1 1000 1000 .PRINT AC VM(13) VP(13) VM(4) VP(4) .TF V(13) VIC .END Results: Vos = 0.4423 mV, Adm = 22500, Acm = 0.2305, CMRR = 99.9 dB, ROUT = 90.3 MΩ. The values of Add and Rout are similar to hand calculations. Acd and the CMRR are limited by the offset induced mismatches in the devices. With the W/L of M6 corrected, Vos ≈ 0, Adm = 20800, A = 9.28 x 10-3. R = 90.3 MΩ, CMRR = 127 dB. cm out ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-137 15.235 +V 25 1 M8 50 1 M9 I I REF I 10 1 M6 12.5 M1 50 1 2 M 10 50 1 DD 2 1 M2 50 1 1 M 11 vO v v 1 M7 25 1 40 1 M5 M3 20 1 M4 20 1 -V SS The W/L ratios have been scaled to keep the Q-points and gain the same. Note that the output stage should remain a source follower pair and is not mirrored. 15-138 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.236 A A 5A VCC Q 10 Q Q 9 8 I I I1 REF 2 Q A 5A Q 11 7 v O 5A Q 1 Q 12 A R Q 7 L Q 2 v v 2 1 Q 5 AE5 Q 3 Q 4 A A -V EE Note that the output stage should remain complementary emitter − followers. I The gain of the first stage is approximately Av1 = gm1rπ 5 = C 1 β o5 , the mirror IC 5 image amplifier with an npn transistor for Q5 will have the highest gain. The voltage gain of the rest of the amplifier is the same. ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-139 15.237 VEB 7 + VEB 8 = VEB 6 + VEB 4 = 2VT ln VEB 7 + VEB 8 = VT ln VEB 7 + VEB 8 = VT ln ⎡ 250µA ⎤ IC 4 I = 2VT ln C14 = 0.05ln⎢ ⎥ = 1.142V IS 4 2 IS 4 ⎣ 2(15 fA)⎦ IC 7 I 60 IC 8 IC 8 I + VT ln C 8 IC 7 = α F IB 8 = α F C 8 = = IS 7 IS 8 β F 61 60 61 2 IC 8 IC 8 IC 8 + VT ln → 0.025ln = 1.142 61(15 fA) 4 (15 fA) 61(15 fA)(4 )(15 fA) IC 8 = 1.946 mA | IC 16 ≅ IC 8 | AE16 = IC16 1946µA AE12 = (1) = 7.78 250µA IC12 ⎛ 75µA ⎞ 2(0.5583V ) = 574 Ω VBE 6 = VEB10 = 0.025ln⎜ ⎟ = 0.558V | RBB = 1.946 mA ⎝ 15 fA ⎠ ⎡ (β + 1)r ⎤ o π8 Adm = Av1 Av 2 Av 3 ≅ gm 2 ro 2 [rπ 7 + (β o + 1)rπ 8 ] ⎢ gm 8 (ro 8 ro16 ) ⎥ [1] ⎣ rπ 7 + (β o + 1)rπ 8 ⎦ [ { [ ( ]4 µf 8 } ] ⎡ Adm = Av1 Av 2 Av 3 ≅ gm 2 ro 2 rπ 7 + (β o + 1)rπ 8 Adm ≅ gm 2 (ro2 2 rπ 7 ) Rid = 2 rπ 1 = 2 [ ≅ µ IC 2 β o7 f 8 = 31.8µA IC 7 2 (β o + 1)rπ 8 ⎛ µ f 8 ⎞⎤ 1 ⎜ ⎟⎥ [ ] π 7 + (β o + 1)rπ 8 ⎝ 2 ⎠⎦ 125µA (40)(60 + 4.3) = 3.03 x 10 5 60 )]⎢ ⎣r 2 150(0.025V ) = 60 kΩ 125µA 15-140 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.238 *Figure P15.237 - Bipolar Op-Amp VCC 1 0 DC 5 VEE 14 0 DC -5 V1 6 15 DC -74.17U AC 0.5 V2 7 15 DC 0 AC -0.5 VIC 15 0 DC 0 Q1 4 6 8 NBJT 1 Q2 5 7 8 NBJT 1 Q3 4 4 2 PBJT 1 Q4 5 4 3 PBJT 1 Q5 2 3 1 PBJT 1 Q6 3 3 1 PBJT 1 Q7 14 5 10 PBJT 1 Q8 11 10 1 PBJT 4 Q9 1 11 12 NBJT 1 Q10 14 13 12 PBJT 1 Q12 9 9 14 NBJT 1 Q14 8 9 14 NBJT 1 Q16 13 9 14 NBJT 7.78 IB 0 9 250U RBB 11 13 574 .MODEL NBJT NPN BF=150 VA=60 IS=15F .MODEL PBJT PNP BF=60 VA=60 IS=15F .OP .AC LIN 1 1000 1000 .PRINT AC VM(12) VP(12) .TF V(12) VIC .END SPICE Results: Vos = -74.17µV, Adm = 2.83 x 105, Acm = 0.507, CMRR = 115 dB, Rid = 81.6 kΩ, Rout = 523 Ω. ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-141 15.239 (a) We require forward - active region operation of all transistors. For Q14 : VCB14 = 0 − VBE1 − (−VEE + VBE14 ) ≥ 0 → VEE ≥ 1.4V For Q1 : VCB1 = VCC − VEB 6 − VEB 4 ≥ 0 → VCC ≥ 1.4V For Q16 : VCB16 = 0 − VEB10 − (−VEE + VBE14 ) ≥ 0 → VEE ≥ 1.4V For Q 8 : VBC 8 = VCC − VEB 8 − VBE 6 ≥ 0 → VCC ≥ 1.4V So VCC ≥ 1.4V and VEE ≥ 1.4V (b) We require forward - active region operation of all transistors with VIC present. For Q14 : VCB14 = VIC − VBE1 − (−VEE + VBE14 ) = −1 − 0.7 − (−VEE + 0.7) =≥ 0 → VEE ≥ 2.4V For Q1 : VCB1 = VCC − VEB 6 − VEB 4 − VIC = VCC − 0.7 − 0.7 − 1 ≥ 0 → VCC ≥ 2.4V For an output range of ± 1V, For Q16 : VCB16 = VO − VEB10 − (−VEE + VBE14 ) = −1 ≥ 0 → VEE ≥ 1.4V For Q 8 : VBC 8 = VCC − VEB 8 − (VO + VBE 6 ) = VCC − 0.7 − (1 + 0.7) ≥ 0 → VCC ≥ 2.4V So VCC ≥ 2.4V and VEE ≥ 2.4V 15.240 VCC − VEB 22 − VBE 20 − (−VEE ) 3V − 0.7 − 0.7 − (−3V ) = = 46.0µA R1 100 kΩ (a) I C 22 = IC 20 = IC 23 = 3 IC 22 = 138 µA | IC 24 = IC 22 = 46.0 µA | ⎛ 46.0µA ⎞ V ⎛ I ⎞ 0.025V ⎛ 46.0µA ⎞ I1 = T ln⎜ C 20 ⎟ = ln⎜ ⎟ = 6.25µA ln⎜ ⎟ =→ I1 = 9.72µA 4 kΩ R ⎝ I1 ⎠ ⎝ I1 ⎠ ⎝ I1 ⎠ (b) I (c) The input bias current and input resistance of the amplifier are directly dependent upon I1 , whereas the gain of the interior amplifier stages is approximately independent of bias current. = 426µA 100 kΩ IC 23 = 3 IC 22 = 128 µA | IC 24 = IC 22 = 426 µA ⎛ 426µA ⎞ V ⎛I ⎞ I1 = T ln⎜ C 20 ⎟ = 6.25µA ln⎜ ⎟ → I1 = 19.3µA R ⎝ I1 ⎠ ⎝ I1 ⎠ C 22 = IC 20 = 22V − 0.7 − 0.7 − (−22V ) 15-142 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.241 250µA = 83.3 µA | I3 = IREF = 83.3 µA 3 V − VEB 22 − VBE 20 − (−VEE ) 12 − 0.7V − 0.7 − (−12) V | R1 = = 271 kΩ IREF = CC R1 83.3 µA V ⎛ I ⎞ 0.025V ⎛ 83.3µA ⎞ ln⎜ R2 = T ln⎜ REF ⎟ = ⎟ = 255 Ω I1 ⎝ I1 ⎠ 50µA ⎝ 50µA ⎠ I2 = 3IREF → IREF = 15.242 IREF = I3 = 300 µA | I2 = 3IREF = 900 µA IREF = VCC − VEB 22 − VBE 20 − (−VEE ) 15 − 0.7V − 0.7 − (−15) V | R1 = = 95.3 kΩ R1 300 µA V ⎛ I ⎞ 0.025V ⎛ 300µA ⎞ ln⎜ R2 = T ln⎜ REF ⎟ = ⎟ = 462 Ω I1 ⎝ I1 ⎠ 75µA ⎝ 75µA ⎠ 15.243 For forward - active region operation of Q3 , VBC 3 ≥ 0 VEE ≥ VIC + VBE1 + VBE 3 + VBE 7 + VBE 5 + VR1 For forward - active region operation of Q1, VCB1 ≥ 0 VCC − VEB 9 ≥ VIC For the output stage, VCC ≥ VBE15 + VI 3 = 0.7 + 0.7 = 1.4 V −VEB16 − VEB12 ≥ VBE11 + VR 5 − (−VEE ) → VEE ≥ 3VBE + VR1 ≅ 2.1V (a) VIC = 0, VEE ≥ 4VBE + VR 1 ≅ 2.8V | VCC − VEB 9 ≥ 0 → VCC ≥ 0.7V ≅ 3.8V | VCC − VEB 9 ≥ 1 → VCC ≥ 1.7V Combining these results yields : VCC ≥ 1.4V and VEE ≥ 2.8V (b) VIC = ±1, VEE ≥ 1 + 4VBE + VR 1 If also account for the output stage, VCC ≥ VO + VBE15 + VI 3 = 1 + 0.7 + 0.7 = 2.4 V Combining these results yields : VCC ≥ 2.4V and VEE ≥ 3.8V ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-143 15.244 50µA (7.32µA) = 20.3µA 18µA 50µA Using Eq. 16.139 : io = −20(20.3µA)v id = (−0.406 mS )v id | IC 4 = (7.25µA) = 20.1µA 18µA ⎛ 20.3µA(1kΩ)⎞ 60V = 2.84 MΩ Rout 6 ≅ ro 6⎜1 + ⎟ = 1.81ro 6 | Rth = 1.81ro 6 2ro 4 = 0.952ro 4 = 0.95 0.025V ⎠ 20.1µA ⎝ The input stage current is proportional to I1 : IC 2 = io = (−4.06 x 10 -4 ) v id | Rth = 2.84 MΩ As a check, we know that g m ∝ IC and ro ∝ 1 . Using the results from Fig. 16.60, IC ⎛ 18µA ⎞ and Rth = 6.54 MΩ⎜ ⎟ = 2.35 MΩ ⎝ 50µA ⎠ ⎛ 50µA ⎞ -4 io = 1.46 x10−4 v id ⎜ ⎟ = −4.06 x 10 v id 18 µ A ⎝ ⎠ which underestimates Rth . 15.245 (a) R2 = β o2 ro 2 2 −1 = 50 60 + 15.7 = 2.84 MΩ | The cascode source uses up an extra VEB . 2 0.666 mA | The other y - parameters are unchanged. (b) [y 22 ] = Rout11 R2 = 407 kΩ 2.84 MΩ = 356kΩ (c ) Adm = 256(6.70 mS)(356 kΩ) = 6.11 x 10 5 15.246 +VCC A Q 23A A Q 22A 3A Q 24 I A Q 22B 3A Q 23B 3 I 2 R 1 I 1 Q 20 A A Q 21 R 2 -V EE 15-144 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.247 gm10 = 40(19.8µA) = 0.792 mS | rπ 10 = gm11 = 40(0.666 mA) = 26.6 mS | rπ 11 = 150(0.025V ) 60V = 189 kΩ | ro10 = = 3.03 MΩ 19.8µA 19.8µA 150(0.025V ) 60V = 5.63kΩ | ro11 = = 90.1kΩ 0.666mA 0.666 mA *Problem 15.247 - Small Signal Parameters. V1 1 0 DC 0 V2 4 0 AC 1 RPI10 1 2 189K RO10 2 0 3.03MEG GM10 0 2 1 2 0.792M RE10 2 0 50K RPI11 2 3 5.63K RO11 4 3 90.1K GM11 4 3 2 3 26.6M RE11 3 0 100 R2 4 0 115K .TF I(V2) V1 .AC LIN 1 1000 1000 .PRINT AC IM(V2) IP(V2) IM(V1) IP(V1) .END -1 -1 Results : y11 = 2.38 MΩ | y12 = 3.27 x 10−10 S ≅ 0 | y 21 = 6.66 mS | y 22 = 81.9 kΩ ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-145 15.248 (a) Assume large β F : IC11 = 50µA | IC 3 = IC 6 = IC 4 = 50µA 2 I IC1 = IC 3 = IC 7 = 50µA | IC 2 = IC 6 = IC 8 = 50µA | IC 9 = 2 C 8 IC11 = IREF = 100µA | IC 4 = IC 5 = βF VCE1 = VCE 2 = 15 − (−0.7) = 15.7V | VEC 4 = VEC 5 = 0.7V VCE 7 = VCE 8 = 0.7 + 0.7 = 1.4V | VCE 9 = 15 − (−15 + 0.7) = 29.3V | VCE10 = 0.7V VEC 3 = VEC 2 = (0 − 0.7) − (−15 + 1.4) = 12.9 | VCE11 = 0 − 0.7 − 0.7 − (−15) = 13.6V Q 1 2 3 4 -50 -0.7 5 -50 -0.7 6 -50 -12.9 7 50 1.4 8 50 1.4 9 --29.3 10 100 0.7 11 100 13.6 IC (µA) 100 100 -50 VCE (V) 15.7 15.7 -12.9 (b) Transistor Q11 replicates the reference current. This current divides in two and controls two matched current mirrors formed of Q4-Q3 and Q5-Q6. The currents of Q1 and Q7, and Q2 and Q8 are equal to the output current of Q3 and Q4. (c) v1 is the inverting input; v2 is the non-inverting input. g m5 = g m6 | g m2 = 2 g m6 | ro8 = ro6 | io = g m6ve6 ve6 = v id Differential-mode Half Circuit io 2 g m2 1 1 = v | io = g m6 vid ⎛ 1 1 ⎞ 2 id 2 1 + g m2 ⎜ ⎟ ⎝ g m5 g m6 ⎠ Q2 v id 2 + 1 g m5 Q6 1 1 1 Gm = g m6 = g m2 = (40)( 100µA)= 1.00 mS 2 4 4 ⎡ ⎞⎤ ⎛ 1 6 = ro8 ro6 ⎢1 + g m6⎜ Rout = ro8 Rout ⎟⎥ ⎝ g m5 + g m2 ⎠⎦ ⎣ ⎛ 61.4V ⎞ ⎛ 72.9V ⎞ Rout = ro8 1.33r06 = ⎜ ⎟ 1.33⎜ ⎟ = 752 kΩ ⎝ 50µA ⎠ ⎝ 50µA ⎠ 15-146 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15.249 (a) Assume large β F : IC 3 = IC 4 = β F IC 8 = IREF = 100µA | IC10 = IC 9 = IB 8 = IC 8 βF IC 10 IC 8 I = = 50µAIC1 = IC 3 = IC 5 = 50µA | IC 2 = IC 4 = IC 6 = 50µA | IC 7 = 2 C 5 2 2 βF VCE1 = VCE 2 = 15 − (−0.7) = 15.7V | VCE 5 = VCE 6 = 0.7 + 0.7 = 1.4V VEC 3 = VEC 4 = 0 − 0.7 − (−15 + 1.4 ) = 12.9V | VCE 7 = 15 − (−15 + 0.7) = 29.3V VEC 8 = 0.7 + 0.7 = 1.4V | VCE 9 = 0.7V | VCE10 = 0 − 0.7 − 0.7 − (−15) = 13.6V Q 1 2 3 4 5 50 1.4 6 50 1.4 7 --29.3 8 -100 1.4 9 --0.7 10 --13.6 IC (µA) 50 50 -50 -50 VCE (V) 15.7 15.7 -12.9 -12.9 (b) Transistors Q9 and Q10 form a current mirror that replicates the base current of transistor Q8. The output current divides in two and forms the base currents of Q3 and Q4. Since Q3 and Q4 match Q8, the collector currents of Q1-Q6 will all be equal to IREF/2. (c) v1 is the inverting input; v2 is the non-inverting input. io = g m 4ve 4 | io = 2g m 4ve 4 2 g m2 v v 1 ve4 = id = vid | io = g m 4 id ⎛ 1 ⎞ 4 2 2 1 + g m2 ⎜ ⎟ ⎝ g m4 ⎠ g m2 = g m4 | ro6 = ro 4 | Q2 Differential-mode Half Circuit io 2 v id 2 + - Q4 1 1 1 Gm = g m 4 = g m2 = (40)(50µA)= 1.00 mS 2 2 2 ⎡ ⎛ 1 ⎞⎤ 4 = ro6 ro4 ⎢1 + g m4 ⎜ Rout = ro6 ROUT ⎟⎥ ⎝ g m2 ⎠⎦ ⎣ ⎛ 61.4V ⎞ ⎛ 72.9V ⎞ Rout = ro6 2ro 4 = ⎜ ⎟ 2⎜ ⎟ = 864 kΩ ⎝ 50µA ⎠ ⎝ 50µA ⎠ ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 15-147 CHAPTER 16 16.1 Av (s) = 50 s2 s2 | Amid = 50 | FL (s)= | Poles : - 2,-30 | Zeros : 0,0 (s + 2)(s + 30) (s + 2)(s + 30) s rad | ω L ≅ 30 s (s + 30) | fL = Yes, s = −30 | Av (s) ≈ 50 fL = ω L 30 ≅ = 4.77 Hz 2π 2π 2 2 1 302 + 22 − 2(0) − 2(0) = 4.79 Hz 2π 50ω 2 | MATLAB : f L = −4.80 Hz Av (jω ) = ω 2 + 22 ω 2 + 302 16.2 Av (s) = 400 s2 s2 s2 = 200 | A = 200 | F s = ( ) mid L 2 s2 + 1400 s + 100,000 (s + 619)(s + 80.8) (s + 619)(s + 80.8) rad s | Zeros : 0, 0 | Yes, a 5 :1 split is sufficient | s = −619 | fL ≅ 619 = 98.5 Hz 2π Poles : - 619,-80.8 Av (s) ≈ 200 fL ≅ s rad | ω L ≅ 619 s (s + 619) 2 2 1 80.82 + 6192 − 2(0) − 2(0) = 99.4 Hz 2π 200ω 2 | MATLAB : 100 Hz Av (jω ) = ω 2 + 80.82 ω 2 + 6192 16.3 Av (s) = −150 (s + 12)(s + 20) s (s + 15) | Amid = −150 | FL (s)= (s + 12)(s + 20) s(s + 15) Poles : -12, - 20 fL ≅ rad rad | Zeros : 0, -15 | No, the poles and zeros are closely spaced. s s 2 2 1 122 + 202 − 2(0) − 2( 15) = 1.54 Hz 2π Av (jω ) = 150ω ω 2 + 152 ω 2 + 122 ω 2 + 202 | MATLAB : f L = 2.72 Hz | ω L = 17.1 rad s Note that ωL =16.1 rad/s does not satisfy the assumption used to obtain Eq. (16.15), and the estimate using Eq. (16.15) is rather poor. 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-1 16.4 3 x1011 10−4 10−5 9 x1011 300 Av (s) = 2 = = 5 9 ⎛ s ⎞⎛ s ⎞ ⎛ s ⎞⎛ s ⎞ 3s + 3.3 x10 s + 3 x10 ⎜ 4 + 1⎟⎜ 5 + 1⎟ ⎜ 4 + 1⎟⎜ 5 + 1⎟ ⎝ 10 ⎠⎝10 ⎠ ⎝ 10 ⎠⎝10 ⎠ 1 Amid = 300 | FH (s)= ⎛ s ⎞⎛ s ⎞ ⎜ 4 + 1⎟⎜ 5 + 1⎟ ⎝ 10 ⎠⎝10 ⎠ Poles : -10 4 ,-105 ⎛ 1 ⎜ fH ≅ 2π ⎜ ⎝ Av (jω ) = 300 rad | ω H ≅ 10 4 s s +1 4 10 ⎞−1 ⎛ 1 ⎞2 ⎛ 1 ⎞2 ⎛ 1 ⎞2 ⎛ 1 ⎞2 ⎟ ⎜ 4 ⎟ + ⎜ 5 ⎟ − 2⎜ ⎟ − 2⎜ ⎟ ⎟ = 1.58 kHz ⎝ 10 ⎠ ⎝ 10 ⎠ ⎝ ∞⎠ ⎝ ∞⎠ ⎠ 3x1011 rad s | Yes : Av (s) ≈ | fH ≅ 10 4 = 1.59 kHz 2π ( )( )( ) ω + 10 2 ( ) 4 2 ω + 10 2 ( ) 5 2 | MATLAB : f H = 1.58 kHz 16.5 ⎛ s ⎞ ⎜1 + 9⎟ ⎝ 2 x10 ⎠ Av (s) = = 300 ⎛ ⎛ s ⎞⎛ s ⎞ s ⎞ s ⎞⎛ 10 7 ⎜1 + 7 ⎟⎜1 + 9 ⎟ ⎜1 + 7 ⎟⎜1 + 9 ⎟ ⎝ 10 ⎠⎝ 10 ⎠ ⎝ 10 ⎠⎝ 10 ⎠ ⎛ s ⎞ ⎜1 + 9⎟ ⎝ 3 x10 ⎠ Amid = 300 | FH (s)= | Poles : -10 7 , -109 Zeros : - 3 x109 , ∞ ⎛ ⎞ ⎛ ⎞ s s ⎜1 + 7 ⎟⎜1 + 9 ⎟ ⎝ 10 ⎠⎝ 10 ⎠ s 1+ ⎟ (3x10 )⎜ ⎝ 3 x10 ⎠ 9 9 ⎛ ⎞ 300 rad | ω H ≅ 10 7 Yes : Av (s ) ≈ ⎛ ⎞ s s ⎜1 + 7 ⎟ ⎝ 10 ⎠ 10 4 | fH ≅ = 1.59 MHz 2π −1 ⎛ 2 2 2 2⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 1 ⎜ 1 1 1 1 − 2⎜ ⎟ ⎟ = 1.59 MHz fH = ⎜ 7 ⎟ + ⎜ 9 ⎟ − 2⎜ 9⎟ ⎜ 2π ⎝ 10 ⎠ ⎝ 10 ⎠ ⎝ 3 x10 ⎠ ⎝ ∞⎠ ⎟ ⎝ ⎠ Av (jω ) = 2 x109 ω 2 + 3 x109 ( ) 2 ω + 10 2 ( ) 7 2 ω + 10 2 ( ) 9 2 | MATLAB : f H = 1.59 MHz 17-2 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.6 ⎛ ⎛ s ⎞ s ⎞ 2 x109 5 x105 ⎜1 + 1 + ⎟ ⎜ 5 5⎟ ⎝ 5 x10 ⎠ ⎝ 5x10 ⎠ Av (s) = = 3333 ⎛ ⎛ s ⎞⎛ s ⎞ s ⎞ s ⎞⎛ 5 6 1 + 1 + 1.5 x10 2 x10 ⎜1 + 1 + ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 5 6 5 6⎟ ⎝ 1.3 x10 ⎠⎝ 2 x10 ⎠ ⎝ 1.5 x10 ⎠⎝ 2 x10 ⎠ ⎛ s ⎞ ⎜1 + 5⎟ rad ⎝ 5x10 ⎠ Amid = 3333 | FH (s)= | Poles : -1.5x105 , - 2x106 ⎛ s s ⎞ s ⎞⎛ 1 + ⎜1 + ⎟ ⎜ ⎟ 5 6 ⎝ 1.5x10 ⎠⎝ 2 x10 ⎠ rad Zeros : - 5 x105 , ∞ | No, the poles and zeros are closely spaced and will interact. s −1 ⎛ 2 2 2 2⎞ ⎛ 1 ⎞ ⎛1⎞ 1 ⎜ ⎛ 1 ⎞ ⎛ 1 ⎞ +⎜ − 2⎜ − 2⎜ ⎟ ⎟ = 26.3 kHz fH ≅ ⎜ ⎟ ⎟ ⎟ 5 6 5 2π ⎜ ⎝ 1.5 x10 ⎠ ⎝ 2 x10 ⎠ ⎝ 5 x10 ⎠ ⎝ ∞⎠ ⎟ ⎝ ⎠ ( )( ) ( )( ) Av (jω ) = 2 x109 ω 2 + 5 x105 ( ω + 1.3 x10 2 ( 5 ) 2 ω 2 ) +( 2 x10 ) 2 6 2 | MATLAB : 26.3 kHz 16.7 6 x108 s2 1 1000(2000) (s + 1)(s + 2) ⎛ s ⎞ s ⎞⎛ ⎜1 + ⎟⎜1 + ⎟ ⎝ 1000 ⎠⎝ 2000 ⎠ ⎡ ⎤ ⎢ ⎥ ⎡ ⎤ s2 1 ⎥ = 200 F (s)F (s) ⎢ ⎥ Av (s) = 300 ⎢ L H ⎛ ⎞ ⎛ ⎞ ⎢ ⎥ s + 1 s + 2 s s ⎢ )( )⎥ ⎣( ⎦ ⎜1 + 1 + ⎟⎜ ⎟⎥ ⎢ ⎣⎝ 500 ⎠⎝ 1000 ⎠⎦ Av (s) = Poles; -1, - 2, -1000, - 2000 Av (jω ) = fL = 1 2π rad | Zeros : 0, 0, ∞, ∞ | No. | No. s 6 x108 ω 2 2 12 + ω 2 22 + ω 2 10002 + ω 2 20002 + ω 2 1 + (2) − 2(0) − 2(0) () 2 2 2 = 0.356 Hz | MATLAB : 0.380 Hz 1 fH = 2π ⎛ 1 ⎞2 ⎛ 1 ⎞2 ⎛ 1 ⎞2 ⎛ 1 ⎞2 ⎜ ⎟ +⎜ ⎟ − 2⎜ ⎟ − 2⎜ ⎟ = 142 Hz | MATLAB : 133 Hz ⎝1000 ⎠ ⎝ 2000 ⎠ ⎝ ∞⎠ ⎝ ∞⎠ 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-3 16.8 ⎛ s ⎞ 1+ ⎜ ⎟ 10 (200) s (s + 1) ⎝ 200 ⎠ Av (s) = 2 2 s ⎞ s ⎞⎛ (100) (300) (s + 3)(s + 5)(s + 7)⎛ ⎜1 + ⎟ ⎜1 + ⎟ ⎝ 100 ⎠ ⎝ 300 ⎠ ⎡ ⎤ ⎛ s ⎞ ⎢ ⎥ ⎜1 + ⎟ ⎡ ⎤ s2 (s + 1) 200 ⎠ ⎢ ⎥ ⎝ 5 ⎥ Av (s) = 6.67 x10 ⎢ = 6.67 x105 FL (s )FH (s) 2 ⎢ ⎥ ⎢ s ⎞⎥ s ⎞⎛ ⎣(s + 3)(s + 5)(s + 7)⎥ ⎦⎢⎛ ⎜1 + ⎟ ⎜1 + ⎟ ⎢ ⎣⎝ 100 ⎠ ⎝ 300 ⎠ ⎥ ⎦ 10 2 No dominant pole at either low or high frequencies. Av (jω ) = fL = 1 2π 1010 ω 2 ω 2 + 12 ω 2 + 2002 ω 2 + 32 ω 2 + 52 ω 2 + 72 ω 2 + 1002 ω 2 + 3002 1 − 2(0) (3) + (5) + (7) − 2() 2 2 2 2 2 ( ) = 1.43 Hz | MATLAB : 1.62 Hz −1 ⎛ 2 2 2 2 2⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 1 ⎜ 1 1 1 1 1 +⎜ +⎜ − 2⎜ − 2⎜ ⎟ ⎟ = 12.5 Hz | MATLAB : 10.6 Hz fH = ⎜ ⎟ ⎟ ⎟ ⎟ 2π ⎜ ⎝ 100 ⎠ ⎝ 100 ⎠ ⎝ 300 ⎠ ⎝ 200 ⎠ ⎝ ∞⎠ ⎟ ⎝ ⎠ 17-4 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.9 Low frequency: 2kΩ 0.1 µF 0.1 µF Rout Rin 2.43 M Ω 10 µF 43 k Ω v i 1 MΩ 13 k Ω Mid-band: 2kΩ Rout Rin i 43 k Ω 2.43 M Ω 1 MΩ v ( b ) Avt = vd = − g m RL = −g m Rout R3 vg ( )| ( Amid = 2(0.2 mA) Rin 2ID Avt | g m = = = 0.400 mS RI + Rin VGS − VTN 1V Rin = 2.43 MΩ | Rout = RD ro ≅ RD = 43kΩ assuming λ = 0 since it is not specified. ⎛ 2.43 MΩ ⎞ Amid = −⎜ ⎟(0.400 mS ) 43kΩ 1 MΩ = −16.5 ⎝ 2 kΩ + 2.43 MΩ ⎠ 1 rad 1 rad ω1 = −7 = 4.11 | ω2 = = 9.59 −7 s s 10 F (2.43 MΩ + 2kΩ) 10 F (43kΩ + 1MΩ) ) ( ) ( ) ω3 = ( 1 1 rad = = 47.7 −5 ⎛ ⎞ s 10 F 13kΩ 2.5kΩ 1 10−5 F ⎜13kΩ ⎟ gm ⎠ ⎝ ) ( )( ) | ωZ = (10 F )(13kΩ) −5 1 = 7.69 rad s ω3 is dominant : f L ≅ ω3 = 7.59 Hz 2π 2 2 2 2 1 4.11) + (9.59) + (47.7) − 2(7.69) = 7.58 Hz ( 2π = 0.2 mA(56kΩ)+ 5V = 16.2 V Using Eq. (16.15) yields : f L ≅ ( c ) VDD = I D (RD + RS )+ VDS 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-5 16.10 5kΩ 0.1 µF 0.1 µF Rout Rin i v 43 k Ω 13 k Ω 220 k Ω 430 k Ω 560 k Ω 10 µF Low frequency: 5k Ω Rout Rin vi 243 k Ω 43 k Ω 220 k Ω Mid-band: Avt = vd = −g m RL = − g m Rout R3 vg ( )| ( Amid = 2(0.2mA) Rin 2 ID Avt | g m = = = 0.400mS RI + Rin VGS − VTN 1V Rin = 243kΩ | Rout = RD ro ≅ RD = 43kΩ assuming λ = 0 since it is not specified. ⎛ 243kΩ ⎞ Amid = −⎜ ⎟(0.400 mS ) 43kΩ 220 kΩ = −14.1 ⎝ 5kΩ + 243kΩ ⎠ ) ω1 = ω3 = (10 F )(243kΩ + 5kΩ) −7 1 = 40.3 rad s | ω2 = (10 F )(43kΩ + 220kΩ) −7 1 = 38.0 1 rad s = 7.69 rad s ( 1 1 rad = = 47.7 −5 ⎛ ⎞ s 10 F 13kΩ 2.5kΩ 1 10−5 F ⎜13kΩ ⎟ gm ⎠ ⎝ ) ( )( ) | ωZ = (10 F )(13kΩ) −5 Using Eq. (17.16) : f L ≅ 1 2π (40.3) + (38.0) + (47.7) − 2(7.69) 2 2 2 2 = 11.5 Hz 17-6 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.11 (a ) Assume that ω C3 = 3 is dominant : f L ≅ ω3 = 2π (50) = 314 rad s 1 1 2 ID 0.4 mA = = 1.52 µF where g m = = = 0.4 mS ⎛ ⎞ 1V VGS − VTN 314 13kΩ 2.5kΩ 1 ω3⎜ RS ⎟ gm ⎠ ⎝ ( ) (b) Choose C ω1 = −7 3 = 1.5 µF | ω3 = 1 1.5 µF 13kΩ 2.5kΩ rad s | ω2 = 2 ( 1 ) = 318 rad 1 rad | ωZ = = 51.3 s s (1.5µF )(13kΩ) 1 = 9.59 rad s (10 F )(2.43 MΩ + 2kΩ) = 4.11 1 2π (10 F )(43kΩ + 1MΩ) −7 2 2 2 Using Eq. (17.16) yields : f L ≅ (4.11) + (9.59) + (318) − 2(51.3) rad s = 1.52 µF where g m = = 49.3 Hz (c) Assume that ω C3 = 3 is dominant : f L ≅ ω3 = 2π (50)= 314 ω3⎜ RS ⎝ ⎛ 1 1⎞ ⎟ gm ⎠ = 314 13kΩ 2.5kΩ ( 1 ) 2 ID 0.4 mA = = 0.4 mS 1V VGS − VTN Choose C3 = 1.5 µF | ω3 = 1.5 µF 13kΩ 2.5kΩ = 40.3 1 2π rad s ( 1 ) = 318 rad 1 rad | ωZ = = 51.3 s s (1.5µF )(13kΩ) 1 = 38.0 rad s ω1 = (10 F )(243kΩ + 5kΩ) −7 1 | ω2 = 2 (10 F )(43kΩ + 220kΩ) −7 2 2 2 Using Eq. (16.15) yields : f L ≅ (40.3) + (38.0) + (318) − 2(51.3) = 50.1 Hz 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-7 16.12 RI 100 Ω Rin vi C1 4.7 µF C2 1 µF Rout 1.3 k Ω RD 4.3 k Ω 100 k Ω R3 R S Low Frequency: (b) Av (s) = Amid (c )Amid | 2 zeros at ω = 0 ⎛ C2 (RD + R3 ) 1⎞ C1⎜ RI + RS ⎟ gm ⎠ ⎝ ⎛ Rin ⎞ ⎛ Rin ⎞ ⎛ Rin ⎞ =⎜ ⎟ Avt = ⎜ ⎟ gm RL = ⎜ ⎟ gm (Rout R3 ) | gm = 5 mS ⎝ RI + Rin ⎠ ⎝ RI + Rin ⎠ ⎝ RI + Rin ⎠ 1 = 173Ω | RL = Rout R3 | Rout = RD ro = RD = 4.3kΩ (assuming ro = ∞) gm s2 | ω1 = (s + ω1)(s + ω 2 ) 1 | ω2 = 1 Rin = RS ⎛ ⎞ 173Ω Amid = ⎜ ⎟(0.005)(4.3kΩ 100 kΩ) = +13.1 → 22.3 dB ⎝ 100Ω + 173Ω ⎠ 1 rad 1 rad ω1 = = 779 | ω 2 = −6 = 9.59 −6 4.7 x10 (100 + 173) s 10 (4.3kΩ + 100 kΩ) s ω1 is dominant : f L ≅ 16.13 ω1 = 124 Hz 2π rad 1 | C1 = ω1 (RI + Rin ) s (a) Assume ω1 Rin = RS is dominant : ω L ≅ ω1 = 2π (1000 Hz) = 6280 1 1 = 1300Ω 200Ω = 173Ω | C1 = = 0.583 µF 3 6.28 x10 (100 + 173) gm C1 = 0.56 µF | ω1 = 1 rad = 6540 0.56 x10 (100 + 173) s −6 (b) Choose ω2 = −6 1 rad ω = 10.3 | ω1 is dominant : f L ≅ 1 = 1040 Hz 2π 10 (22 kΩ + 75 kΩ) s 17-8 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.14 200 Ω v i 4.7 µF 1 µF Rin 4.3 k Ω 2.2 k Ω Rout 51 k Ω Low frequency: 200 Ω vi Rin Rout 4300 Ω 2.2 k Ω 51 k Ω Mid-band: (b) A (s)= A v (c) A mid | 2 zeros at ω = 0 ⎛ C2 (RC + R3 ) 1⎞ C1⎜ RS + RE ⎟ gm ⎠ ⎝ ⎛ Rin ⎞ ⎛ Rin ⎞ ⎛ Rin ⎞ =⎜ 1mA)= 0.04S ⎟ Avt = ⎜ ⎟ g m RL = ⎜ ⎟ g m Rout R3 | g m = 40( ⎝ RI + Rin ⎠ ⎝ RI + Rin ⎠ ⎝ RI + Rin ⎠ s2 | ω = mid (s + ω1)(s + ω2 ) 1 1 | ω2 = 1 ( ) 1 = 24.9Ω | RL = Rout R3 | Rout = RC ro = RC = 2.2 kΩ gm ⎛ ⎞ 24.9Ω Amid = ⎜ ⎟(0.04) 2.2 kΩ 51kΩ = +9.34 → 19.4 dB ⎝ 200Ω + 24.9Ω ⎠ 1 rad 1 rad ω1 = = 946 | ω2 = −6 = 18.8 −6 s s 4.7 x10 (200 + 24.9) 10 (2.2 kΩ + 51kΩ) Rin = RE ( ) ω1 is dominant : f L ≅ (d) g m = 40( 10µA)= 0.0004 S | Rin = 2.49kΩ | Rout = RC ro = RC = 220 kΩ ω1 = 151 Hz 2π ⎛ ⎞ 2.49 kΩ Amid = ⎜ ⎟(0.0004) 220 kΩ 510 kΩ = +56.9 → 35.1 dB ⎝ 200Ω + 2.49 kΩ ⎠ 1 rad 1 rad ω1 = = 79.1 | ω2 = −6 = 1.37 −6 s s 4.7 x10 (200 + 2.49 kΩ) 10 (220 kΩ + 510 kΩ) ( ) ω1 is dominant : f L ≅ ω1 = 12.6 Hz 2π 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-9 16.15 (a ) Assume ω C1 = C1 = 1 is dominant : ω L ≅ ω1 = 2π (500 Hz)= 3140 rad s 1 1 | Rin = RE | g m = 40( 1mA) = 0.04 S | Rin = 4300Ω 25Ω = 24.9Ω gm ω1(RI + Rin ) 1 = 1.42 µF 3.14 x10 (200 + 24.9) 3 (b) Choose ω2 = −6 C1 = 1.5 µF | ω1 = 1 rad = 2960 s 1.5 x10 (200 + 24.9) −6 1 rad = 18.8 s 10 (2.2 kΩ + 51kΩ) 1 | ω1 is dominant : f L ≅ ω1 = 472 Hz 2π (c) Assume ω C1 = C1 = is dominant : ω L ≅ ω1 = 2π (500 Hz)= 3140 rad s 1 1 | Rin = RE | g m = 40( 10µA) = 0.0004 S | Rin = 430 kΩ 2500Ω = 2.49 kΩ gm ω1(RI + Rin ) 1 = 0.118 µF | Choose C1 = 0.12 µF 3.14 x10 (200 + 2490) 3 ω1 = 1 rad = 2100 s 0.12 x10 (200 + 2490) −6 | ω2 = 1 rad = 18.8 s 10 (2.2 kΩ + 51kΩ) −6 ω1 is dominant : f L ≅ ω1 = 493 Hz 2π 17-10 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.16 (a) gm = 40 IC = 40(0.175mA) = 7.00mS | rπ = βo gm = 100 = 14.3kΩ | ro = ∞ (VA not given) 7.00mS Rin = R1 R2 rπ = 100 kΩ 300 kΩ 14.3kΩ = 12.0 kΩ | RL = RC R3 = 43kΩ 100 kΩ = 30.1kΩ Amid = Rin 12.0 kΩ gm RL = (7.00 mS)(30.1kΩ) = −194 1kΩ + 12.0kΩ RI + Rin 987Ω + 14.3kΩ Rth + rπ = 15 kΩ = 150Ω | R3 S = RC + R3 = 43kΩ + 100kΩ = 143kΩ 101 βo + 1 SCTC : R1S = RI + Rin = 1kΩ + 12.0 kΩ = 13.0 kΩ | Rth = R1 R2 RI = 100 kΩ 300 kΩ 1kΩ = 987Ω R2 S = RE fL ≅ ⎤ (38.5 + 667 + 69.9) 1 ⎡ 1 1 1 + + = 123 Hz ⎥= ⎢ 2π ⎣2 x10−6 (13.0kΩ) 10 x10−6 (150Ω) 1x10−7 (143kΩ)⎦ 2π (b) Note that the Q-point assumed in part (a) is not quite correct. SPICE yields: (144 µA, 3.67 V), Amid = 43.9 dB, fL = 91 Hz R1 100 kΩ = 12 = 3V | REQ = R1 R2 = 100 kΩ 300 kΩ = 75.0 kΩ 100 kΩ + 300 kΩ R1 + R2 = 100 3 − 0.7 = 145µA 75kΩ + (101)15 kΩ 101 (0.145mA)(15 kΩ) = 3.57 V 100 (c ) VEQ = VCC IC = β F REQ + (β F + 1)RE VEQ − VBE VCE = VCC − IC RC − IE RE = 12 − (0.145 mA)(43kΩ) − These values agree with the SPICE results listed above in part (b). 16.17 (a ) Use the values from Section 16.3.1, and assume ω3 is dominant. ω L = 2π (2500 Hz)= 15700 C3 = rad s | ω3 = ω L − ω1 − ω2 = 15700 − 225 + 96.1 = 15390 rad s 1 1 = = 2.86 µF ω3 R3S 15390(22.7Ω) C3 = 2.7 µF | ω3 = 1 2.7 x10 −6 (b) Choose ω2 = 96.1 (22.7Ω) = 16320 rad s rad rad 225 + 96.1 + 16320 | ω1 = 225 | fL ≅ = 2650 Hz s s 2π or if ω L must be no more than 2500 Hz, choose C3 = 3.3 µF 1 3.3x10 −6 ω3 = (22.7Ω) = 13350 rad s | fL ≅ 225 + 96.1 + 13350 = 2180 Hz 2π 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-11 16.18 1k Ω 43 k Ω Rin v i 43 k Ω 100k Ω 300k Ω 75 k Ω (a) Mid-band: 1k Ω 5 µF 21.5 k Ω 1 µF 43 k Ω 43 k Ω vi 75 k Ω 13 k Ω 22 µF Low frequency: (b) g m = 40 IC = 40(0.164mA)= 6.56 mS | rπ = βo gm = 100 = 15.2kΩ | ro = ∞ (VA not given) 6.56 mS Rin = R1 R2 rπ = 100 kΩ 300 kΩ 15.2 kΩ = 12.6 kΩ | RL = RC R3 = 43kΩ 43kΩ = 21.5kΩ Amid = Rin 12.6 kΩ g m RL = (6.56mS )(21.5kΩ)= −131 1kΩ + 12.6 kΩ RI + Rin SCTC : R1S = RI + Rin = 1kΩ + 12.6 kΩ = 13.6 kΩ | Rth = R1 R2 RI = 100 kΩ 300 kΩ 1kΩ = 987Ω 987Ω + 15.2 kΩ Rth + rπ = 13kΩ = 158Ω | R3 S = RC + R3 = 43kΩ + 43kΩ = 86kΩ 101 βo + 1 ⎤ ( 14.7 + 288 + 11.6) 1 ⎡ 1 1 1 ⎢ ⎥= + + = 50.0 Hz fL ≅ −6 −6 −6 2π ⎢ 2 π 22 x 10 1 x 10 5 x 10 13.6 k Ω 158 Ω 86 k Ω ⎥ ( ) ( ) ( ) ⎣ ⎦ ⎛ 101 ⎞ 13kΩ)+ 2.79V + 0.164mA(43kΩ)= 12.0 V ⎟( (c) VCC = I E RE + VEC + IC RC = 0.164mA⎜ ⎝100 ⎠ R2S = RE 17-12 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.19 SCTC : R1S = RI + RG = 1kΩ + 1MΩ = 1.00 MΩ | ω1 = R2S = RS 1 rad = 10.0 s 1.00 MΩ(0.1µF ) 1 1 rad 1 = 6.8 kΩ = 607Ω | ω2 = = 165 1.5mS s gm 607Ω( 10µF ) 1 rad = 111 s 90kΩ(0.1µF ) R3S = RD + R3 = 22 kΩ + 68 kΩ = 90 kΩ | ω3 = fL ≅ (10.0 + 165 + 111) = 45.5 Hz 2π 16.20 SCTC : R1S = RI + RG = 1kΩ + 500kΩ = 501kΩ | ω1 = R2S = RS 1 rad = 20.0 s 501kΩ(0.1µF ) 1 1 rad 1 = 10 kΩ = 1.18 kΩ | ω2 = = 84.8 0.75mS s gm 1.18 kΩ( 10µF ) 1 rad = 189 s 53kΩ(0.1µF ) R3S = RD + R3 = 43kΩ + 10 kΩ = 53kΩ | ω3 = fL ≅ (20.0 + 84.8 + 189) = 46.8 Hz 2π 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-13 16.21 C1 R v I 2k Ω C 3 4.7 µF 12 k Ω C R4 892 k Ω R 2 1 µF R7 22 k Ω 100 k Ω 3 i 0.1 µF (a) Low Frequency: 2kΩ Rin vi 12 k Ω 22 k Ω 100 k Ω Mid-band: 2(0.1mA) 1 = 0.200mS | = 5000Ω gm 1V 1 = 12kΩ 5kΩ = 3.53kΩ | RL = 22 kΩ 100 kΩ = 18.0 kΩ Rin = RS gm Rin 3.53kΩ gm RL = Amid = (0.200 mS)(18kΩ) = 2.30 (7.24dB) 2kΩ + 3.53kΩ RI + Rin 1 1 rad ω1 = = = 38.5 | ω 2 = doesn't matter since ig = 0! −6 C1(RI + Rin ) 4.7 x10 (2 kΩ + 3.53kΩ) s gm = ω3 = 1 1 rad 1 = −7 = 82.0 | fL ≅ (38.5 + 82.0) = 19.2Hz C3 (R3 + R7 ) 10 (100kΩ + 22 kΩ) s 2π 17-14 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.22 2kΩ 2kΩ 4.7 µF vi 10 µF 75 k Ω 13 k Ω 100 k Ω Rin v 75 k Ω i 11.5 k Ω (a) Low frequency Mid-band 100 = 10.0 kΩ 40(0.25mA) (b) R in = R1 R2 rπ + (β o + 1)RL | RL = 13kΩ 100 kΩ = 11.5kΩ | rπ = 101) 11.5kΩ = 70.5kΩ Rin = R1 R2 rπ + (β o + 1)RL = 100kΩ 300kΩ 10.0 kΩ + ( Amid = 101( 11.5kΩ) Rin (β o + 1)RL = 0.972 = 0.963 | RB = R1 R2 = 75kΩ RI + Rin Rin 2 + 10.0 + 101( 11.5) kΩ [ ] [ ] 101) 11.5kΩ = 72.5kΩ R1S = RI + RB rπ + (β o + 1)RL = 2 kΩ + 75kΩ 10.0 kΩ + ( [ ] [ [ ] ] ω1 = 1 rad = 2.94 −6 s (72.5kΩ)4.7 x10 RB RI + rπ R3S = R7 + RE (β o + 1) = 100 kΩ + 13kΩ fL ≅ 1.95kΩ + 10.0 kΩ = 100 kΩ 101 ω3 = 1 10−5 105 ( ) =1 rad s (2.94 + 1) = 0.627 Hz 2π 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-15 16.23 2k Ω 4.7 µF v i 892 k Ω 0.1 µF 12 k Ω 100 k Ω (a) Low Frequency: 2k Ω Rin v i 892 k Ω 10.7 k Ω Mid-band: (b) g m = 2(0.1mA) 0.75V = 0.267 mS | Rin = R1 R2 = 892 kΩ | RL = 12 kΩ 100 kΩ = 10.7 kΩ Amid = (0.267mS )(10.7kΩ) = +0.739 Rin g m RL = 0.998 RI + Rin 1 + g m RL 1 + (0.267 mS )( 10.7 kΩ) (-2.62 dB) ω1 = ω3 = 1 1 rad = = 0.238 −6 s C1(RI + Rin ) 4.7 x10 (2 kΩ + 892 kΩ) 1 1 rad = = 97.2 ⎡ ⎤ ⎡ ⎤ ⎛ ⎛ ⎞ s 1 1⎞ −7 C3⎢ R7 + ⎜ RS ⎟⎥ 10 ⎢100 kΩ + ⎜12 kΩ ⎟⎥ 0.267 mS ⎠⎥ g m ⎠⎥ ⎢ ⎢ ⎝ ⎝ ⎣ ⎦ ⎣ ⎦ 1 (0.238 + 97.2)= 15.5 Hz 2π (c) VDD = VDS + I S RS = 8.8V + 0.1mA(12kΩ)= 12.0 V fL ≅ 17-16 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.24 SCTC requires : ω L ≅ ∑ 1 rad = 2π (500) = 3140 s i =1 RisC i | ω2 = 1 rad = 9.59 s (10 F )(43kΩ + 1MΩ) −7 3 ω1 = 1 rad = 4.11 s (10 F )(2.43MΩ + 1kΩ) −7 ω1 + ω 2 > ω1 → ω 3 is dominant −6 C1(RI + RG ) 4.7 x10 (2kΩ + 892kΩ) s ωL ≅ ω3 = 1 1 0.75V | = = 3.75 kΩ ⎡ ⎛ gm 2(0.1mA) 1 ⎞⎤ C3⎢R7 + ⎜ RS ⎟⎥ ⎢ ⎝ gm ⎠⎥ ⎣ ⎦ 1 rad C3 = = 0.155 → 0.15 µF using Appendix C s 62.8 100 kΩ + ( 12 kΩ 3.75 kΩ) [ ] 17-18 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.29 SCTC requires : ω L ≅ ∑ i =1 3 1 rad = 2π (5) = 31.4 RisCi s 100 = 10.0kΩ 40(0.25mA) RL = 13kΩ 100kΩ = 11.5kΩ | rπ = 101) 11.5kΩ = 72.5kΩ R1S = RI + RB rπ + (β o + 1)RL = 2kΩ + 75kΩ 10.0 kΩ + ( [ ] [ ] ω1 = 1 rad = 2.94 −6 s (72.5kΩ)4.7 x10 | ω3 = 31.4 − 2.94 = 28.5 rad s R3S = R7 + RE C3 = (R (β B RI + rπ o + 1) ) = 100 kΩ + 13kΩ 1.95kΩ + 10.0 kΩ = 100 kΩ 101 1 = 0.351 µF → 0.39 µF using the values from Appendix C. 28.5( 100 kΩ) 1 ⎛ gm ⎞ g | Cπ = m − Cµ | gm = 40 IC ⎜ ⎟ ⎜ ⎟ 2πf T 2π ⎝ C π + C µ ⎠ IC 10 µA fT 50 MHz 300 MHz 1 GHz 6.06 GHz 3.18 MHz 5 GHz Cπ 0.773 pF 0.75 pF 2.93 pF 10 pF 1 pF 1 pF Cµ 0.5 pF 1.37 pF 0.25 pF 0.500 pF 1 pF 0.5 pF 1/2πrxCµ 1.59 GHz 580 MHz 3.19 GHz 1.59 GHz 795 MHz 1.59 GHz 16.30 fT = 100 µA 50 µA 10 mA 1 µA 1.18 mA 16.31 Cπ = gm τ F | Cπ = 40(2 x10−3 ) ωT gm − Cµ | VCB = 5 − 0.7 = 4.3V | Cµ = Cµo 1+ VCB = φ jc 2 pF = 0.832 pF 4.3V 1+ 0.9V Cπ 24.6 x10−12 Cπ = − 0.832 pF = 24.6 pF | τ F = = = 0.308 ns = 308 ps gm 40( 2π (5 x10 8 ) 2 x10−3 ) 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-19 16.32 fT = ⎞ 1 ⎛ gm ⎜ ⎟ | gm = 2K n ID 2π ⎝ CGS + CGD ⎠ ID fT 15.9 MHz 79.6 MHz 250 MHz CGS 1.5 pF 1.5 pF 1.5 pF CGD 0.5 pF 0.5 pF 0.5 pF 10 µA 250 µA 2.47 mA 16.33 (a) fT = (b) fT = VGS − VTN ) 3 600(0.25V ) cm 2 3 µn ( = = 22.5 GHz 2 2 10−4 2 V − s L2 VGS − VTN ) 3 250(0.25V ) cm 2 3 µn ( = = 9.38 GHz 2 2 10−4 2 V − s L2 T ( ) (c) NMOS : f = VGS − VTN ) 3 600(0.25V ) cm2 3 µn ( = = 2.25 THz 2 2 10−5 2 V − s L2 VGS − VTN ) 3 250(0.25V ) cm2 3 µn ( = = 938 GHz 2 2 10−5 2 V − s L2 VGS − VTN ) 3 600(0.25V ) cm2 3 µn ( = = 36.0 THz 2 2 2.5 x10−6 2 V − s L2 ( ) ( ) PMOS : f T = ( ) (d ) NMOS : fT = PMOS : f T = VGS − VTN ) 3 250(0.25V ) cm2 3 µn ( = = 15.0 GHz 2 2 2.5x10−6 2 V − s L2 ( ) ( ) 16.34 (a) r π = 125(0.025V ) 1mA = 3.13kΩ | Rin = 7.5kΩ (rx + rπ ) = 2.44 kΩ | RL = 4.3kΩ 100 kΩ = 4.12 kΩ ⎛ 2.44 kΩ ⎞ Rin g m RL = −⎜ ⎟(40mS )(4.12 kΩ)= −117 RI + Rin ⎝1kΩ + 2.44 kΩ ⎠ ⎛ 2.21kΩ ⎞ ⎟(40 mS )(4.12 kΩ)= −113 (b) Rin = 7.5kΩ rπ = 2.21kΩ | Amid = −⎜ ⎝1kΩ + 2.21kΩ ⎠ g m = 40 10−3 = 40mS | Amid = − ( ) 17-20 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.35 125(0.025V ) rπ = = 3.13kΩ | g m = 40( 1mA) = 40mS | RL = 3kΩ 47 kΩ = 2.82kΩ 1mA 126)2.82 kΩ = 78.2 kΩ Rin = RB rx + rπ + (β o + 1)RL = 100 kΩ 0.35kΩ + 3.13kΩ + ( [ ] [ ] (a) A (b) R Amid mid = Amid in = RB rπ + (β o + 1)RL = 100 kΩ 3.13kΩ + ( 126)2.82 kΩ = 78.2kΩ [ ⎤ ⎤ β o + 1)RL 126(2820) ( Rin ⎡ 78.2 kΩ ⎡ ⎢ ⎥ ⎢ ⎥ = 0.978 = = RI + Rin ⎢ 1 k Ω + 78.2 k Ω r + r + β + 1 R 350 + 3130 + 126 2820 ⎥ ⎢ ( )⎥ ⎣ x π ( o ) L⎦ ⎣ ⎦ ] [ ] ⎤ 126(2820) 78.2 kΩ ⎡ ⎢ ⎥ = 0.978 = 1kΩ + 78.2 kΩ ⎢ 350 + 3130 + 126 2820 ⎥ ( ) ⎣ ⎦ 16.36 125(0.025V ) rπ = = 31.25kΩ | g m = 40(0.1mA)= 4mS 0.1mA r +r 200Ω + 31.25kΩ = 248Ω | RL = 22 kΩ 75kΩ = 17.0 kΩ Rin = RE x π = 43kΩ βo + 1 126 (a) A (b) R mid = Rin 248Ω g m RL = (4mS )(17.0kΩ)= 48.5 RI + Rin 100Ω + 248Ω rπ 31.25kΩ 247Ω = 43kΩ = 247Ω | Amid = (4mS )(17.0kΩ)= 48.4 βo + 1 126 100Ω + 247Ω in = RE 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-21 16.37 (a) s s= 2 + 5100 s + 500000 | s1 ≅ − −5100 ± 51002 − 4 5 x105 ( 2 2 2 2 (b) 2s + 700s + 30000 = 2 s + 350s + 15000 ) = −5100 ± 4900 → −100, ) 5100 5 x105 = −5100 | s2 ≅ − = −98.0 5100 1 − 5000 | 2% error ( s1 ≅ − s= −350 ± 250 → −50, − 300 | 14% error 2 2 3300 3 x105 2 3s + 3300 s + 300000 | s ≅ − = − 1100 | s ≅ − = −90.9 c () 1 2 3300 3 = s= −3300 ± 33002 − 4(3) 3x105 15000) −350 ± 3502 − 4( 350 = −350 1 | s2 ≅ − 15000 = −42.9 350 ( 6 6 2 2 (d ) 0.5s + 300s + 40000 = 0.5 s + 600s + 80000 ) = −3300 ± 2700 → −100, ) − 1000 | 11% error ( s1 ≅ − s= −600 ± 6002 − 4(80000) 2 600 = −600 1 | s2 ≅ − 80000 = −133 600 = −600 ± 200 → −200, − 400 | 34%, 50% error 2 16.38 s3 + 1110s2 + 111000 s + 1000000 1110 111000 1000000 = −1110 | s2 ≅ − = −100 | s3 ≅ − = −9.01 s1 ≅ − 1 1110 111000 Factoring the polynomial : s3 + 1110s2 + 111000 s + 1000000 = (s + 10)(s + 100)(s + 1000) s = −1000, − 100, − 10 | 11% error in s1 , 10% error in s3 In MATLAB: roots([1 1110 111000 1000000]) 16.39 f (s) = s6 + 142 s 5 + 4757 s 4 + 58230 s3 + 256950 s2 + 398000 s + 300000 f ' (s)= 6 s5 + 710 s 4 + 19028 s 3 + 174690 s2 + 513900 s + 398000 i +1 s =s − i () f (s ) f si ' i | Using a spreadsheet, four real roots are found : -100, - 20, -15, - 5 Using MATLAB: roots([1 142 4757 58230 256950 398000 300000]) ans = -100, -20, -15, -5, -1+i, -1-i 17-22 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.40 RI 1kΩ r x Cµ Rin RB 300 Ω r 2500 π + v 1 + Cπ v RC 2 R3 v i 7.5 k Ω Ω - gm v1 - 4.3 k Ω 100 k Ω (a) r π = 100(0.025) 0.001 = 2500Ω | Cµ = 0.75 pF | Cπ = Rin = 7.5kΩ (rx + rπ ) = 2.03kΩ | RL ( ) − 0.75 pF = 12.0 pF 2π ( 5 x10 ) = 4.3kΩ 100 kΩ = 4.12 kΩ | g = 40( 10 )= 40mS 40 10−3 8 −3 m ⎛ Rin ⎞⎛ rπ ⎞ ⎛ 2.03kΩ ⎞⎛ ⎞ 2500Ω Amid = −⎜ ⎟⎜ ⎟ g m RL = −⎜ ⎟⎜ ⎟(40 mS )(4.12 kΩ)= −98.6 ⎝1kΩ + 2.03kΩ ⎠⎝ 300Ω + 2500Ω ⎠ ⎝ RI + Rin ⎠⎝ rx + rπ ⎠ 1 ωH = | rπo = rπ rx + RB RI = 2500 300 + 7500 1000 = 803 Ω rπoCT ⎡ 4120⎤ 1 = 1.42 MHz CT = 12.0 + 0.75 ⎢1 + 40 10−3 (4120)+ ⎥ = 140 pF | f H = 803 ⎦ ⎣ 2π (803) 1.4 x10−10 [ ( )] [ ( )] ( ) (b) GBW = 98.6(1.42 MHz)= 140 MHz ( ) 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-23 16.41 +12 V R2 15 k Ω RC 2.15 k Ω R1 RE 5k Ω 650 Ω (a) VEQ = 12V 5 kΩ = 3V | REQ = 5 kΩ 15 kΩ = 3.75 kΩ 5 kΩ + 15 kΩ VEQ − VBE (3 − 0.7)V IC = β F IB = β F = 100 = 3.31 mA REQ + (β F + 1)RE 3.75 kΩ + (101)(0.65 kΩ) ⎛ ⎞ 101 0.65 kΩ⎟ = 2.71 V | Q - Point : (3.31 mA, 2.71 V ) VCE = VCC − IC RC − IE RE = 12 − (3.31mA)⎜ 2.15 kΩ + ⎝ ⎠ 100 (b) RI 1kΩ r x Cµ Rin 300 Ω RB + v 1 + Cπ v RC R 3 v rπ 2 i 3.75 k Ω - gm v 1 - 2.15 k Ω 100 k Ω Note: As designers, we are free to change the amplifier design, but we typically cannot change the characteristics of the source and load resistances. 40(3.31x10−3 ) 100(0.025) rπ = = 755Ω | Cµ = 0.75 pF | Cπ = − 0.75 pF = 41.4 pF 3.31mA 2π ( 5 x10 8 ) gm = 40(3.31x10−3 )= 132 mS Amid = − Rin = 3.75 kΩ (rx + rπ ) = 823Ω | RL = 2.15 kΩ 100 kΩ = 2.11kΩ ⎛ ⎞⎛ ⎞ Rin ⎛ rπ ⎞ 823Ω 755Ω ⎟⎜ ⎟(132 mS )(2.11kΩ) = −90.0 ⎜ ⎟gm RL = −⎜ ⎝1000Ω + 823Ω ⎠⎝ 300Ω + 755Ω ⎠ RI + Rin ⎝ rx + rπ ⎠ | rπo = rπ rx + (RB RI ) = 755Ω 300 + (3.75 kΩ 1kΩ) = 260Ω ωH = 1 rπoCT [ ] [ ] ⎡ 2.11kΩ ⎤ 1 = 1.96 MHz CT = 41.4 + 0.75 ⎢1 + (132 mS )(2.11kΩ) + = 312 pF | f H = ⎥ ⎣ 0.260 kΩ⎦ 2π (260Ω)(3.12 x10−10 F ) (c ) GBW = 90.0(1.96 MHz) = 176 MHz | GBW ≤ 1 1 ⎛ 1 ⎞ = = 707 MHz ⎟ ⎜ ⎟ ⎜ 2π ⎝ rx Cµ ⎠ 2π (300Ω)(0.75 pF ) 17-24 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.42 +12 V R2 1.5 M Ω RC 215 k Ω R1 RE 500 k Ω 65 k Ω (a) RI 50 k Ω r x Cµ Rin 300 Ω RB rπ + v 1 + Cπ v RC 2 R3 vi 375 k Ω - gm v1 - 215 k Ω 5 M Ω (b) VEQ = 12V 500 kΩ = 3V | REQ = 500 kΩ 1.5 MΩ = 3.75kΩ 500 kΩ + 1.5 MΩ VEQ − VBE (3 − 0.7)V = 100 = 33.1 µA IC = β F I B = β F 375kΩ + ( 101)(65kΩ) REQ + (β F + 1)RE ⎛ ⎞ 101 65kΩ⎟ = 2.71 V | Q - Point : (33.1 µA, 2.71 V ) VCE = VCC − IC RC − I E RE = 12 − (33.1µA) ⎜ 215kΩ + 100 ⎝ ⎠ Note: As designers, we are free to change the amplifier design, but we typically cannot change the characteristics of the source and load resistances. However, the problem statement indicated changing all resistors. 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-25 rπ = 100(0.025) 33.1µA = 75.5kΩ | Cµ = 0.75 pF | Cπ = 40 33.1x10−6 2π 5 x10 ( ( 8 ) )− 0.75 pF = −0.329 pF ( ) - not possible. The constant fT model is breaking down at low currents. Set Cπ = 0 Rin = 375kΩ (rx + rπ ) = 63.1kΩ | RL = 215kΩ 5 MΩ = 211kΩ Amid = − | g m = 40 33.1x10−6 = 1.32 mS ⎛ ⎞⎛ ⎞ Rin ⎛ rπ ⎞ 63.1kΩ 75.5kΩ 1.32 mS )(211kΩ) = −155.0 ⎜ ⎟ g m RL = −⎜ ⎟⎜ ⎟( RI + Rin ⎝ rx + rπ ⎠ ⎝ 50 kΩ + 63.1kΩ ⎠⎝ 300Ω + 75.5kΩ ⎠ | rπo = rπ rx + RB RI ωH = 1 rπoCT [ ( )]= 75.5kΩ [300 + (375kΩ 50kΩ)]= 28.0kΩ ( ) 1 ⎛ 1 ⎞ 1 = 707 MHz = ⎜ ⎟ ⎜ ⎟ 2π ⎝ rx Cµ ⎠ 2π (300Ω)(0.75 pF ) ⎡ 211kΩ ⎤ 1 1.32 mS )(211kΩ)+ = 24.9 kHz CT = 0 + 0.75 ⎢1 + ( ⎥ = 228 pF | f H = 8.86 kΩ⎦ ⎣ 2π (28.0 kΩ) 2.28 x10−10 F (c) GBW = 155(24.9kHz)= 3.86 MHz | GBW ≤ 17-26 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.43 Rin = R1 R2 = 4.3 MΩ 5.6 MΩ = 2.43 MΩ | RL = 43kΩ 470 kΩ = 39.4 kΩ gm = 2(0.2 mA) 2ID = = 0.400 mS | VGS − VTN 1 Rin 2.43 MΩ g m RL = − 0.400 mS (39.4 kΩ)= −15.7 RI + Rin 2 kΩ + 2.43 MΩ | rπo = R1 R2 RI = 2.00 kΩ Amid = − fH = 1 2πrπoCT ⎡ 39.4 kΩ⎤ CT = 2.5 pF + 2.5 pF ⎢1 + (0.400 mS )(39.4 kΩ)+ ⎥ = 93.7 pF 2 kΩ ⎦ ⎣ fH = 2π (2 kΩ) 93.7 x10−12 F ( 1 ) = 849 kHz 16.44 *Problem 16.44 - Common-Source Amplifier VDD 7 0 DC 0 VS 1 0 AC 1 RS 1 2 2K C1 2 3 0.1UF R1 3 0 4.3MEG R2 3 7 5.6MEG RD 7 5 43K R4 4 0 13K C3 4 0 10UF C2 5 6 0.1UF R3 6 0 1MEG *Small-Signal FET Model GM 5 4 3 4 0.4MS CGS 3 4 2.5PF CGD 3 5 2.5PF * .AC DEC 20 1 10MEG .PRINT AC VM(6) .PROBE .END Results: Amid = -15.7, fL = 8.52 Hz, fH = 866 MHz 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-27 16.45 V DD = 12 V R2 560 k Ω RD 10 k Ω R1 RS 430 k Ω 3.9 k Ω (a) VEQ = 12V 430 kΩ = 5.21V | REQ = 430 kΩ 560 kΩ = 243kΩ 430 kΩ + 560 kΩ K 2 Assume active region operation : ID = n (VGS − VTN ) | VEQ = VGS + ID RS 2 ⎛ 0.5 mA ⎞ 2 5.21 = VGS + 3.9kΩ⎜ ⎟(VGS − 1) → VGS = 2.629V and ID = 663µA ⎝ 2 ⎠ The transistor is not in pinch off! Reduce RD to 10 kΩ. VDS = VDD − ID RD − IS RS = 12 − (663µA)(13kΩ + 3.9kΩ) = 0.795 V VDS = VDD − ID RD − IS RS = 12 − (663µA)(10 kΩ + 3.9kΩ) = 2.78 V - Active region is correct. RI 1kΩ CGD Rin RG + v 1 + v C GS RD 2 R3 v i 243 k Ω - g mv 1 - 10 k Ω 100 k Ω (b) Rin = R1 R2 = 430 kΩ 560 kΩ = 243kΩ | RL = 10 kΩ 100 kΩ = 9.09 kΩ gm = 2(0.663mA) 2 ID = = 1.33mS VGS − VTN 1 Rin 243kΩ gm RL = − (1.33mS )(9.09kΩ) = −12.0 RI + Rin 1kΩ + 243kΩ | rπo = R1 R2 RI = 243kΩ 1kΩ = 0.996 kΩ Amid = − fH = 1 2πrπoCT ⎡ 9.09 kΩ ⎤ CT = 2.5 pF + 2.5 pF ⎢1 + (1.33mS)(9.09 kΩ) + = 58.1 pF ⎣ ⎦ 0.996 kΩ⎥ 1 = 2.75 MHz fH = 2π (0.996 kΩ)(58.1x10−11 F ) (c ) GBW 17-28 = 12.0(2.75 MHz) = 33 MHz ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.46 RI 1kΩ r x Cµ Rin RB 75 k Ω 300 Ω rπ + v 1 + Cπ v RC R3 2 vi - gm v 1 - 43 k Ω 43 k Ω g m = 40 IC = 40(0.164 mA) = 6.56mS | rπ = βo gm = 100 = 15.2 kΩ | ro = ∞ ( VA not given ) 6.56 mS Rin = R1 R2 rπ = 100 kΩ 300 kΩ 15.2 kΩ = 12.6 kΩ | RL = RC R3 = 43kΩ 43kΩ = 21.5kΩ Amid = − Cπ = Rin 12.6 kΩ g m RL = − (6.56mS )(21.5kΩ)= −131 RI + Rin 1kΩ + 12.6 kΩ − Cµ = 6.56 mS 2π 5 x108 Hz ωT gm ( ) − 0.75 = 1.34 pF ωH = 1 rπoCT | rπo = rπ rx + R1 R2 RI = 15.2 kΩ (300 + 987)= 1.19 kΩ ( ) ⎛ ⎡ R ⎞ 21.5kΩ⎤ CT = Cπ + Cµ ⎜1 + g m RL + L ⎟ = 1.34 pF + 0.75 pF ⎢1 + 6.56 mS (21.5kΩ)+ ⎥ = 121 pF rπo ⎠ 1.19kΩ ⎦ ⎣ ⎝ 1 = 1.10 MHz fH ≅ 1.19 kΩ) 1.21x10−10 F 2π ( ( ) 16.47 *Problem 16.47 - Common-Emitter Amplifier VCC 7 0 DC 0 VS 1 0 AC 1 RS 1 2 1K C1 2 3 5UF R1 3 0 300K R2 3 7 100K RC 5 0 43K R4 7 4 13K C2 7 4 22UF C3 5 6 1UF R3 6 0 43K *Small-signal Model for the BJT GM 5 4 8 4 6.56MS RX 3 8 0.3K RPI 8 4 15.24K CPI 8 4 1.34PF 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-29 CU 8 5 0.75PF * .AC DEC 100 1 10MEG .PRINT AC VM(6) .PROBE .END Results: Amid = -128, fL = 47 Hz, fH = 1.10 MHz 16.48 (a ) See Eqs. (16.88 -16.96). ⎤ ⎡V (s)⎤ ⎡I (s) ⎤ ⎡s(C + C )+ g − sC ⎥⎢ ⎥ ⎢ ⎥=⎢ − sC − g s C + C + g V s ⎥ ⎥ ⎢ ( ) ( ) ( ) ⎣ 0 ⎦ ⎢ ⎦⎣ ⎦ ⎣ ∆=s [ C (C + C )+ C C ]+ s[ C g + C (g + g + g )+ C g ]+ g g S π µ πo µ 1 2 µ m µ L L 2 π µ L µ L π L µ m πo L L πo L πo (b) ω P1 ≅ g L g πo = Cπ g L + Cµ (g m + g πo + g L )+ CL g πo = 1 ⎡ R⎤ rπo ⎢Cπ + Cµ ( 1 + g m RL )+ (Cµ + CL ) L ⎥ rπo ⎦ ⎣ gm ω P2 ≅ Cπ g L + Cµ (g m + g πo + g L )+ CL g πo Cπ (Cµ + CL )+ Cµ CL (c) The three capacitors form a loop, and there are only two independent voltages among the three capacitors. 16.49 1 + g m RL )= 20 pF + 1 pF 1 + 40( 1mA)( 1kΩ) = 61 pF CT = Cπ + Cµ ( ⎛ C ⎞ L Cπ ⎜ ⎜1 + C ⎟ ⎟ + CL µ⎠ ⎝ [ ] fT = 1mA) ⎤ 1 ⎛ g m ⎞ 1 ⎡ 40( ⎢ ⎥ = 303 MHz = ⎜ ⎟ ⎟ 2π ⎜ ⎝ Cπ + Cµ ⎠ 2π ⎢ ⎣ 20 pF + 1 pF ⎥ ⎦ 17-30 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.50 1+ A 1+ A = = sC( 1 + A) | Cin = C ( 1 + A)= 10−10 F 1 + 105 = 10 µF 1 Z (s) sC Z (s) s + 10 s + 10 1 105 (b) Zin = Y = 1 + A s = 106 = 105 s + 10 + 106 ≅ 105 s + 106 ( ) 1+ in s + 10 Using MATLAB : Zin (j 2000π )= (4.95 + j 6.28) Ω (a) Y in = ( ) Zin j105 π = (8.98 + j 28.6) kΩ | Zin j 2πx106 = (97.5 + j15.5) kΩ ( ) ( ) 16.51 10 Ao s + 10 ⎛ 1 ⎞ A(s) 10 Ao ⎜ ⎟ 1+ ⎛ 1 ⎞ ⎝ RC ⎠ 1 + A(s) 10 Ao s + 10 | A(s) = | Av (s)= ⎜ ⎟ (a) Av (s)= 1 1 RC ⎠ s + 10 ⎝ s+ s+ ⎛ 10 Ao ⎞ RC 1 + A(s) RC⎜1 + ⎟ ⎝ s + 10 ⎠ ⎛ 10 Ao ⎞ ⎜ ⎟ ⎛ 1 ⎞ 10 Ao ⎝ RC ⎠ Av (s) = ⎜ = ⎟ ⎡ 1 ⎤ 10 ⎝ RC ⎠ s2 + s 1 + A 10 + s + 10 + 10( 1 + Ao ) ( o ) RC s2 + s⎢ RC ⎥+ ⎣ ⎦ RC ⎛ 106 ⎞ ⎛ 106 ⎞ ⎜ ⎟ ⎜ ⎟ 1 ⎝ RC ⎠ ⎝ RC ⎠ Av (s) = ≅ ; ωL = 5 ⎛ ⎡ 1 ⎤ 10 10 RC 1 ⎞ s2 + s⎢ s + 106 ⎜ s + 5 + 106⎥ + ⎟ ⎣ RC ⎦ RC ⎝ 10 RC ⎠ 7 ⎛ 10 ⎞ ⎛ 106 ⎞ ⎜ ⎟ ⎜ ⎟ 1 ⎝ RC ⎠ ⎝ RC ⎠ ≅ ; ωL = 6 (b) Av (s)= ⎛ ⎡ 1 ⎤ 10 10 RC 1 ⎞ s2 + s⎢ s + 10 7 ⎜ s + 6 + 10 7 ⎥ + ⎟ ⎣ RC ⎦ RC ⎝ 10 RC ⎠ ⎛10 Ao ⎞ ⎜ ⎟ 1 ⎝ RC ⎠ (c) lim Av (s)= 10 A s = sRC Ao →∞ o [ ] ( ) ( ) 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-31 16.52 r vi x 250 Ω rπ 2500 Ω C T ⎡ 2500⎤ rπo = 2500Ω 250Ω = 227Ω | CT = 15 + 1⎢1 + 0.04(2500)+ ⎥ = 127 pF 227 ⎦ ⎣ (a) At 1 kHz, Z C = 127 pF ) j (2π ) 10 ( 3 ( ) ( 1 = − j 1.25 x106 ( ) Using MATLAB : Z = 250 + 2500ZC = (2750 − j 4.99) Ω | SPICE : (2750 − j 4.56) Ω 2500 + ZC 4 (b) At 50 kHz, Z C = 127 pF ) j (2π ) 5 x10 ( 1 ) = − j 2.51x10 4 Ω Using MATLAB : Z = 250 + 2500ZC = (2730 − j 247) Ω | SPICE : (2730 − j 226) Ω 2500 + ZC = − j( 12.53) (c) At 1 MHz, Z C = 127 pF ) j (2π ) 106 ( ( ) 1 2500ZC = (752 − j1000) Ω | SPICE : (836 − j1040) Ω 2500 + ZC (d) *Problem 16.52 - Common-Emitter Amplifier IS 0 1 AC 1 RX 1 2 0.25K RPI 2 0 2.5K CPI 2 0 15PF CU 2 3 1PF GM 3 0 2 0 40MS RL 3 0 2.5K .AC LIN 1 1KHZ 1KHZ *.AC LIN 1 50KHZ 50KHZ *.AC LIN 1 1MEG 1MEG .PRINT AC VR(1) VI(1) VM(1) VP(1) .END Using MATLAB : Z = 250 + Note that the CT approximation does not provide as good an estimate of Zin at high frequencies (note the discrepancy at 1 MHz). 16.53 Amid = 39.2 dB, fL = 0 Hz, fH = 5.53 MHz 16.54 17-32 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 (a) g Cπ = m = 40 IC = 40( 1mA) = 40.0 mS | rπ = β oVT IC = 125(0.025) 1mA = 3.13kΩ | ro = ∞ ( VA not given ) | RL = RC R3 = 4.3kΩ 100 kΩ = 4.12 kΩ rπo = rπ rx + RB RI [ ( )]= 3.13kΩ (500 + (7.5kΩ 1kΩ))= 959Ω 40.0 mS ωT gm − Cµ = 2π 5 x10 Hz 8 ( ) − 0.75 pF = 12.0 pF | f H ≅ 1 2π rπoCT ⎛ ⎡ R ⎞ 4.12 kΩ ⎤ CT = Cπ + Cµ ⎜1 + g m RL + L ⎟ = 12.0 pF + 0.75 pF ⎢1 + 40.0 mS (4.12 kΩ)+ ⎥ = 140 pF rπo ⎠ 0.959 kΩ⎦ ⎣ ⎝ fH ≅ 2π (959Ω) 1.40 x10−10 F = rπ x B (b) r πo ( [r + (R 1 1 RI )]= 3.13kΩ (0 + (7.5kΩ 1kΩ))= 688Ω = 1.64 MHz ) = 1.19 MHz ⎡ 4.12 kΩ ⎤ CT = 12.0 pF + 0.75 pF ⎢1 + 40.0 mS (4.12 kΩ)+ ⎥ = 141 pF 0.688 kΩ⎦ ⎣ fH ≅ 2π (688Ω) 1.41x10−10 F ( ) 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-33 16.55 RI 1kΩ r x Cµ Rin RB 75 k Ω 400 Ω rπ + v 1 + Cπ v RC R3 vi 2 - gm v1 - 43 k Ω 100 k Ω (a) g m = 40 IC = 40(0.1mA)= 4.00 mS | rπ = β oVT IC = 100(0.025) 0.1mA = 25kΩ | ro = ∞ ( VA not given ) rπo = rπ rx + RB RI [ ( )]= 25kΩ (400 + (75kΩ 1kΩ))= 1.31kΩ Rin = R1 R2 (rx + rπ ) = 100kΩ 300 kΩ 25.4 kΩ = 19.0 kΩ | RL = RC R3 = 43kΩ 100 kΩ = 30.1kΩ Amid = − Cπ = Rin 19.0 kΩ g m RL = − (4.00mS )(30.1kΩ)= −114 RI + Rin 1kΩ + 19.0 kΩ − Cµ = 4.00 mS 2π 5 x10 Hz 8 ωT gm ( ) − 0.75 pF = 0.523 pF | f H ≅ 1 2π rπoCT ⎛ ⎡ R ⎞ 30.1kΩ⎤ CT = Cπ + Cµ ⎜1 + g m RL + L ⎟ = 0.523 pF + 0.75 pF ⎢1 + 4.00 mS (30.1kΩ)+ ⎥ = 109 pF rπo ⎠ 1.31kΩ ⎦ ⎣ ⎝ fH ≅ 1.31kΩ) 1.09 x10−10 F 2π ( ( 1 ) = 1.12 MHz | 1 1 = = 531 MHz 2π rx Cµ 2π (400Ω)(0.75 pF ) (b) GBW = 114(1.12 MHz)= 128 MHz 16.56 Rin = R1 R2 = 4.3 MΩ 5.6 MΩ = 2.43 MΩ | RL = 43kΩ 470 kΩ = 39.4 kΩ gm = 2(0.2 mA) 2ID = = 0.400 mS | VGS − VTN 1 Rin 2.43 MΩ g m RL = − 0.400 mS (39.4 kΩ) = −15.7 2 kΩ + 2.43 MΩ RI + Rin | rπo = R1 R2 RI = 2.00 kΩ Amid = − fH = 1 2πrπoCT ⎡ 39.7 kΩ⎤ CT = 5 pF + 2 pF ⎢1 + (0.400 mS )(39.7 kΩ)+ ⎥ = 78.5 pF 2 kΩ ⎦ ⎣ fH = 2π (2 kΩ) 78.5 x10−12 F ( 1 ) = 1.01 MHz | GBW = 15.7( 1.01 MHz)= 15.9 MHz 17-34 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.57 Problem 16.57 should refer to Fig. 16.34, ot Fig. 1631. 1 1 1 = | CT = = 48.5 pF fH = 2πrπoCT 2π (656Ω)CT 2π (656Ω)(5 MHz) ⎛ ⎡ R⎤ 1 ⎞ C − Cπ 48.5 pF − 19.9 pF CT = Cπ + Cµ ⎢1 + g m RL + L ⎥ | RL ⎜ g m + ⎟ = T −1 = − 1 = 56.2 rπo ⎦ Cµ rπo ⎠ 0.5 pF ⎣ ⎝ RL = ⎛ 1 ⎞ ⎜.064 S + ⎟ 656Ω ⎠ ⎝ 56.2 = 858Ω | RL = RC 100kΩ → RC = 865Ω = −31.9 | GBW = 31.9(5 MHz)= 160 MHz 882Ω + 250Ω + 1560Ω The nearest 5% value is RC = 820 Ω | RL = 820Ω 100 kΩ = 813Ω Amid = − Amid = − fH = ⎡ 813⎤ = −30.2 | CT = 19.9 + 0.5⎢1 + 0.064(813)+ ⎥ = 47.0 pF 882Ω + 250Ω + 1560Ω 656 ⎦ ⎣ 100(813Ω) 100(858Ω) 1 1 = = 5.16 MHz | GBW = 156 MHz 2πrπoCT 2π (656Ω)(47.0 pF ) 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-35 16.58 +12 V 43 k Ω 100 Ω 43 k Ω 100 k Ω Rin 75 k Ω 30.1 k Ω vi 3V 13 k Ω 75 k Ω 3kΩ IC = 100 3 − 0.7 = 0.166 mA | VCE = 12 − 43kΩIC − 13kΩI E = 2.70 V 75kΩ + 101( 13kΩ) rπ 0 = 100 = 15.1kΩ 6.64 mS Cπ = g m = 40(0.166mA)= 6.64 mS Short-Circuit Time Constants ωT gm − Cµ = 3.02 pF R1S = 100Ω + 75kΩ 300Ω + 15.1kΩ + 101(3kΩ) = 60.8kΩ ⎛ 15.1kΩ + 99.9Ω ⎞ R2S = 10 kΩ ⎜ 3kΩ + ⎟ = 2.40kΩ 101 ⎝ ⎠ R3S = 43kΩ + 100 kΩ = 143kΩ ⎤ 1 ⎡ 1 1 1 ⎢ ⎥ = 43.9 Hz fL ≈ + + 2π ⎢ 143kΩ)(0.1µF )⎥ 1µF ) (2.40kΩ)(2.2µF ) ( ⎣(60.8kΩ)( ⎦ Open-Circuit Time Constants Using the results from Table 16.2 on page 1037 : rπ 0 = 15.1kΩ 300 + 100 75kΩ = 390 Ω ⎡ (6.64mS )(30.1kΩ) 30.1kΩ⎤ 3.02 pF ⎥ CTB = + 0.5 pF ⎢1 + + 390 Ω 1 + (6.64 mS )(3kΩ) 1 + 6.64 mS 3 k Ω ⎢ ⎥ ( )( ) ⎣ ⎦ CTB = 44.0 pF | f H = 1 = 9.27 MHz 2π (390Ω)(44.0 pF ) [ ] [ ( )] (b) A mid ⎛ 60.7 kΩ ⎞ (6.64 mS )(30.1kΩ) = −⎜ = −9.54 GBW = 9.54(9.27 MHz − 43.9 Hz)= 88.0 MHz ⎟ ⎝ 60.8 kΩ ⎠ 1 + (6.64 mS )(3kΩ) 17-36 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.59 Using the results from Table 16.2 on page 1037 and Prob. 16.58 1 = 54.4 rπ 0 = 15.1kΩ 300 + 100 75kΩ = 390 Ω CTB = 2π (7.5 MHz)(390Ω) [ ( )] CTB = ⎡ (6.64mS )(30.1kΩ) 30.1kΩ⎤ 3.02 pF ⎥ = 54.4 pF + 0.5 pF ⎢1 + + 390 Ω 1 + (6.64 mS )RE 1 + 6.64 mS R ⎢ ⎥ ( ) E ⎣ ⎦ −100(30.1kΩ) Using MATLAB : RE = 862 Ω Amid = 0.999 99.9Ω + 300Ω + 15.1kΩ + 101(862Ω) = −29.3 | GBW = 220 MHz The closest 5% resistor values are RE = 820 Ω and R6 = 12 kΩ ⎡ (6.64mS )(30.1kΩ) + 30.1kΩ⎤ 3.02 pF ⎥ = 55.1 pF CTB = + 0.5 pF ⎢1 + 1 + (6.64 mS )0.82kΩ ⎢ ⎣ 1 + (6.64mS )0.82 kΩ 390Ω ⎥ ⎦ fH = 1 = 7.41MHz 2π (390Ω)(55.1 pF ) = −30.6 | GBW = 227 MHz 99.9Ω + 300Ω + 15.1kΩ + 101(0.82 kΩ) Note: The Q-point will actually change somewhat and this has been neglected. Amid = 0.999 −100(30.1kΩ) 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-37 16.60 ⎡ ⎤ ⎛ IC ⎞ 3 − 0.7 ⎢ ⎥ ⎟ = 2.69V (a) IC = 100⎢7.5kΩ + 101 1.3kΩ ⎥ = 1.66mA | VCE = 12 − 4.3kΩ(IC )− 1.3kΩ⎜ ⎝αF ⎠ ( )⎦ ⎣ 2.69V ≥ 0.7V Active region operation is correct. | rπ = g m = 40( 1.66 mA)= 66.4 mS | Cπ = 66.4 mS 2π 2 x108 +12 V 100(0.025) 1.66 mA = 1.51 kΩ ( ) − 1 = 51.8 pF | rx = 300Ω | Cµ = 1.0 pF 4.3 k Ω 250 Ω 7.5 k Ω Rin vi 3V 1.3 k Ω 7.5 k Ω 200 Ω 4.3 k Ω 47 k Ω 3.94 k Ω 101)200Ω = 5.60 kΩ Rin = R1 R2 rx + rπ + (β o + 1)RE1 = 10 kΩ 30 kΩ 0.350kΩ + 1.51kΩ + ( Rth = 7.5kΩ 250Ω = 242Ω | RL = 4.3kΩ 47kΩ = 3.94kΩ Amid = − ⎤ ⎡100(3.94 kΩ)⎤ Rin ⎡ β o RL 5.60kΩ ⎢ ⎥=− ⎢ ⎥ = −16.9 RI + Rin ⎢ 350Ω + 5.60kΩ ⎢ ⎣rx + rπ + (β o + 1)RE1 ⎥ ⎦ ⎣ 22.0 kΩ ⎥ ⎦ [ ] [ ] (b) Using the Short-Circuit Time Constants: R1S = 250Ω + 7.5kΩ 350Ω + 1.51kΩ + 101(200Ω) = 5.85kΩ [ ] R2S = 4.3kΩ + 43kΩ = 47.3kΩ ⎛ 1.51kΩ + 350 + 242Ω ⎞ R3S = 1.1kΩ ⎜ 200Ω + ⎟ = 184Ω 101 ⎝ ⎠ ⎤ 1 ⎡ 1 1 1 ⎢ ⎥ = 193 Hz + + fL ≅ 2π ⎢ 1µF ) ( 184Ω)(4.7µF )⎥ ⎣(5.85kΩ)(5µF ) (47.3kΩ)( ⎦ (c) Using the Open-Circuit Time Constants: Using the results from Table 16.2 on page 1037 : rπ 0 ≅ Rth + rx = 242Ω + 350Ω = 592Ω ⎡ (66.4mS )(3.94kΩ) + 3.94kΩ⎤ 51.8 pF ⎥ = 29.6 pF + 1 pF ⎢1 + CTB = 1 + (66.4 mS )(200Ω) ⎢ ⎣ 1 + (66.4 mS )(200Ω) 592Ω ⎥ ⎦ CTB = 29.6 pF fL = 1 = 9.08 MHz 2π (592Ω)(29.6 pF ) 17-38 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.61 Using the results in Table 16.2 on page 1037 and the values from Prob. 16.60 : rπ 0 ≅ Rth + rx = 242Ω + 350Ω = 592Ω | CTB = CTB = 1 = 22.4 pF 12 MHz) 2π (592Ω)( ⎡ (66.4 mS )(3.94 kΩ) 3.94 kΩ⎤ 51.8 ⎥ = 22.4 pF + 1 pF ⎢1 + + 592 Ω 1 + 66.4 mS R 1 + (66.4 mS )RE ⎢ ⎥ ( ) E ⎣ ⎦ Using MATLAB : RE = 305 Ω The closest 5% resistor values are RE = 300 Ω and R6 = 1 kΩ ⎡ (66.4 mS )(3.94 kΩ) 3.94 kΩ⎤ 51.8 pF ⎥ = 22.6 pF CTB = + 1 pF ⎢1 + + 592 Ω 1 + (66.4 mS )300Ω 1 + 66.4 mS 300 Ω ⎢ ⎥ ( ) ⎣ ⎦ fH = 1 = 11.9 MHz 2π (592Ω)(22.6 pF ) Rin = R1 R2 rx + rπ + (β o + 1)RE1 = 10 kΩ 30 kΩ 0.300 kΩ + 1.51kΩ + ( 101)300Ω = 6.08 kΩ Rth = 7.5kΩ 250Ω = 242Ω | RL = 4.3kΩ 47kΩ = 3.94 kΩ Amid = − ⎤ ⎡100(3.94 kΩ)⎤ 6.08kΩ β o RL Rin ⎡ ⎢ ⎥=− ⎢ ⎥ = −11.8 250 Ω + 6.08 k Ω RI + Rin ⎢ 32.1 k Ω r + r + β + 1 R ⎥ ⎢ ⎥ ( ) o E1 ⎦ ⎣x π ⎣ ⎦ [ ] [ ] 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-39 16.62 1k Ω R 1S 10 k Ω 1k Ω R 2S 1k Ω 1k Ω R 1O 10 k Ω 1k Ω 10 k Ω 1k Ω R 2O 1k Ω (a) SCTC: R1S = 10 kΩ 1kΩ = 909Ω | R2 S = 1kΩ 1kΩ = 500Ω | ω L = 1 1 rad + = 1300 −6 −5 s 909( 10 ) 500( 10 ) 1 rad = 92.3 −5 s 10 ) 1670( 10 )+ 917( −6 (b) OCTC: R1O = 10 kΩ 2 kΩ = 1.67 kΩ | R2O = 1kΩ 11kΩ = 917Ω | ω H = (c) There are two poles. The SCTC technique assumes both are at low frequency and yields the largest pole. The OCTC assumes both are at high frequency and yields the smallest pole. (d ) ⎡(sC1 + G1 + G2 ) ⎢ −G2 ⎣ ⎤⎡V1 ⎤ −G2 ⎥⎢ ⎥ = 0 (sC2 + G2 + G3 )⎦⎣V2 ⎦ ∆ = s 2C1C2 + s[C2 (G1 + G2 ) + C1 (G2 + G3 )]+ G1G2 + G2G3 + G1G3 ∆ = s 210−11 + s( 1.30 x10−8 )+ 1.20 x10−6 ∆ = s 2 + 1300 s + 1.20 x10 5 → s = −1200, −100 rad s 17-40 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.63 100 Ω R in vi 1.3 k Ω 4.3 k Ω 100 k Ω RL = 4.3kΩ 100 kΩ = 4.12 kΩ | CGS = 3.0 pF | CGD = 0.6 pF Rin = RS Amid = 1 1 = 1.3kΩ = 173Ω gm 5mS Rin 173Ω g m RL = (5ms)(4.12kΩ)= +13.1 RI + Rin 100Ω + 173Ω ⎞ ⎛ 1 1 ⎛ 1 ⎞ 1 ⎟ = 64.4 MHz ⎜ = fH ≅ ⎜ ⎟ ⎟ 2π ⎝ CGD RL ⎠ 2π ⎜ 0.6 pF 4.12 k Ω ( ) ⎠ ⎝ 16.64 *Problem 16.12 - Common-Gate Amplifier - ac small-signal model VI 1 0 AC 1 RI 1 2 100 C1 2 3 4.7UF RS 3 0 1.3K RD 4 0 4.3K C2 4 5 1UF R3 5 0 100K G1 4 3 0 3 5mS .OP .AC DEC 100 0.01 100MEG .PRINT AC VM(5) VP(5) .END Results: Amid = +13.3, fL = 123 Hz, fH = 64.4 MHz 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-41 16.65 200 Ω vi Rin 4.3 k Ω 2.2k Ω 51k Ω 2.11 k Ω gm = 40(1 mA) = 0.04 S | rx = 300Ω | rπ = Cπ = 2π (5 x10 8 ) 40( 10−3 ) 100(0.025) = 2.50 kΩ | Cµ = 0.6 pF 1 mA − 0.6 = 12.1 pF | Rth = 4.3kΩ 200Ω = 191Ω | RL = 2.2 kΩ 51kΩ = 2.11kΩ Rin = RE Amid 101 100(2.11kΩ) 27.6Ω Rin ⎛ β o RL ⎞ = = +9.14 ⎜ ⎟= RI + Rin ⎝ rx + rπ ⎠ 200Ω + 27.6Ω 2.80 kΩ 1 ⎡ 0.04 (2110) ⎤ 12.1 pF ⎛ 300 ⎞ 191 ⎥ + 0.6 pF (2110Ω) ⎜1 + ⎟ + 0.6 pF (300Ω)⎢1 + 1 + 0.04 (191) ⎝ 191 ⎠ ⎣ 1 + 0.04 (191)⎦ ⎞ 1 1 ⎛ = 40.9 MHz ⎜ −10 −9 −9 ⎟ 2π ⎝ 6.876 x10 + 1.938 x10 + 1.266 x10 ⎠ (rx + rπ ) = 4.3kΩ (0.3kΩ + 2.50kΩ) = 27.6Ω βo + 1 ωH = fH = 17-42 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.66 First, estimate the required SPICE parameters: 0.333 ⎛ CJC 2.8 ⎞ | CJC = CJC = 0.6 pF ⎜1 + ≅ 1.01pF Cµ = ⎟ ME ⎝ 0.75 ⎠ ⎛ VCB ⎞ ⎜1 + ⎟ ⎝ PHIE ⎠ C 1 Cµ 1 0.6 pF τF = π = − = 9 − = 303 ps gm ωT gm 10 π 40(1mA) *Figure P16.14 - Common-Base Amplifier VCC 6 0 DC 5 VEE 7 0 DC -5 VI 1 0 AC 1 RI 1 2 200 C1 2 3 4.7UF RE 3 7 4.3K Q1 4 0 3 NBJT RC 4 6 2.2K C2 4 5 1UF R3 5 0 51K .MODEL NBJT NPN BF=100 RB=300 CJC=1.01PF TF=303PS .OP .AC DEC 50 1 50MEG .PRINT AC VM(5) .PROBE .END Results: Amid = 19.1 dB, fL = 149 Hz, fH = 43.8 MHz 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-43 16.67 200 Ω vi Rin 4.3 k Ω 2.2k Ω 51k Ω 2.11 k Ω IC = α F IE = 100 ⎡−0.7 − (−10)⎤ ⎢ ⎥ = 2.14 mA | VCE = 10 − (2.14 mA)(2.2 kΩ) − (−0.7) = 5.99 V 101 ⎣ 4300 ⎦ 100(0.025) = 1.17 kΩ | Cµ = 0.6 pF 2.14 mA gm = 40(2.14 mA) = 85.6 mS | rx = 300Ω | rπ = Cπ = 85.6 mS − 0.6 = 26.7 pF | Rth = 4.3kΩ 200Ω = 191Ω | RL = 2.2 kΩ 51kΩ = 2.11kΩ 2π (5 x10 8 ) Rin = RE Amid = (rx + rπ ) = 4.3kΩ (0.3kΩ + 1.17kΩ) = 14.5Ω βo + 1 101 100(2.11kΩ) Rin ⎛ β o RL ⎞ 14.5Ω = +9.70 ⎜ ⎟= RI + Rin ⎝ rx + rπ ⎠ 200Ω + 14.5Ω 1.47 kΩ 1 ⎡ ⎛ 300 ⎞ 0.0856(2110) ⎤ 26.7 pF ⎥ + 0.6 pF (2110Ω) ⎜1 + ⎟ + 0.6 pF (300Ω)⎢1 + 1 + 0.0856(191) ⎝ 191 ⎠ ⎣ 1 + 0.0856(191)⎦ ωH = 191 fH = ⎞ 1 1 ⎛ = 39.1 MHz ⎜ −10 −9 −9 ⎟ 2π ⎝ 7.556 x10 + 2.054 x10 + 1.266 x10 ⎠ 17-44 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.68 2kΩ Rin v i 12 k Ω 22 k Ω 100 k Ω Rth = 12 kΩ 2 kΩ = 1.71kΩ | RL = 22 kΩ 100 kΩ = 18.0 kΩ | CGS = 3.0 pF | CGD = 0.6 pF gm = 2(0.1mA) 1V = 0.200 mS | Rin = 12 kΩ 1 1 = 12 kΩ = 3.53kΩ gm 0.200 mS Amid = Rin 3.53kΩ g m RL = (0.200ms)(18.0kΩ)= +2.30 2 kΩ + 3.53kΩ RI + Rin ⎛ ⎞ ⎞ ⎛ ⎜ ⎟ ⎟ ⎜ 1 1 1 1 ⎜ ⎜ ⎟ = 10.9 MHz ⎟= fH = ⎜ ⎟ C 3.0 pF 2π ⎜ ⎟ 2π GS + CGD RL ⎟ + 0.6 pF ( 18.0 kΩ)⎟ ⎜ ⎜ ⎠ ⎝ Gth + g m ⎝ (0.5848 + 0.200)mS ⎠ 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-45 16.69 (a) First find VTN and K n based upon Problem 16.21 1.5 MΩ 12V = 4.87V | VGG − VGS = 0.1mA( 12 kΩ)→ VGS = 3.67V 1.5 MΩ + 2.2 MΩ 2(0.1mA) 2ID VGS − VTN = 1V → VTN = 2.67V | Kn = = = 0.2mS 2 2 1 V − V ( ) VGG = GS TN Now, find the Q - point for VDD = 18V : VGG = RGG 1.5 MΩ 18V = 7.30V 1.5 MΩ + 2.2 MΩ = 1.5 MΩ 2.2 MΩ = 892 kΩ | VGG − VGS = I D RS ⎛ 0.2mS ⎞ 2 VGS − 2.67) → VGS = 4.73V | I D = 0.254 mA | VDS = 9.37V ok 7.30 - VGS = ( 12 kΩ) ⎜ ⎟( ⎝ 2 ⎠ 2kΩ Rin v i 12 k Ω 22 k Ω 100 k Ω Rth = 12 kΩ 2 kΩ = 1.71kΩ | RL = 22 kΩ 100 kΩ = 18.0 kΩ | CGS = 3.0 pF | CGD = 0.6 pF gm = (4.26 − 2.67)V 2(0.254 mA) = 0.320 mS | Rin = 12 kΩ 1 1 = 12 kΩ = 2.48 kΩ gm 0.320 mS Rin 2.48 kΩ g m RL = (0.320ms)(18.0kΩ)= +3.19 2 kΩ + 2.48 kΩ RI + Rin ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ 1 1 1 ⎜ 1 ⎜ ⎟ = 11.3 MHz ⎟= fH = ⎜ ⎟ 3.0 pF 2π ⎜ CGS π 2 ⎟ + CGD RL ⎟ + 0.6 pF ( 18.0 kΩ)⎟ ⎜ ⎜ ⎝ Gth + g m ⎠ ⎝ (0.5848 + 0.32)mS ⎠ Note that the contribution of the input pole cannot be neglected because of the low fT of the MOSFET. Amid = R1S = RI + Rin = 4.48 kΩ | R3 S = R7 + Rout ≅ 100 kΩ + 22 kΩ = 122 kΩ 1 ⎛ 1 1 ⎞ fL = + ⎜ ⎟ = 20.6 Hz 2π ⎝ R1S C1 R3S C3 ⎠ Note that the is no signal current in C2 , so it does not contribute to f L . 17-46 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.70 g m = 40(0.25 mA) = 10.0 mS | rx = 300Ω | rπ = Cµ = 0.6 pF | Cπ = 2π 5 x108 100(0.025) 0.25 mA = 10.0 kΩ ( 0.01 ) − 0.6 = 2.58 pF | RB = 100 kΩ 300 kΩ = 75.0 kΩ RL = 13kΩ 100 kΩ = 11.5kΩ | Rth = 75kΩ 2 kΩ = 1.95kΩ 101) 11.5kΩ = 70.5kΩ Rin = RB rx + rπ + (β o + 1)RL = 75.0 kΩ 300Ω + 10.0 kΩ + ( ⎛ Rin ⎞ 101( 11.5kΩ) (β o + 1)RL = 0.972 = 0.964 Amid = ⎜ ⎟ 0.300 + 10.0 + 101( 11.5) kΩ ⎝ RI + Rin ⎠ rx + rπ + (β o + 1)RL [ ] [ ] [ ] fH ≅ 1 2π ⎡ ⎤ 2.58 pF ⎢ ⎥ + 0.6 pF 1950 + 300 ( )⎢1 + 10mS 11.5kΩ ⎥ ( ) ⎣ ⎦ 1 = 1 1 = 114 MHz 2π (2250)(0.622 pF ) (b) Calculating the required SPICE parameters: ⎛ 11.8 ⎞ 0.333 CJC Cµ = | CJC = 0.6 pF ≅ 1.54 pF ⎜1 + ⎟ ME ⎝ 0.75 ⎠ ⎛ VCB ⎞ ⎜1 + ⎟ ⎝ PHIE ⎠ C 1 Cµ 1 0.6 pF τF = π = − = 9 − = 260 ps | TF = 260 ps gm ωT gm 10 π 40(0.25 mA) *Problem 16.70 - Common-Collector Amplifier VCC 6 0 DC 15 VS 1 0 AC 1 RS 1 2 2K C1 2 3 4.7UF R1 3 0 100K R2 6 3 300K Q1 6 3 4 NBJT R4 4 0 13K C3 4 5 10UF R7 5 0 100K .MODEL NBJT NPN BF=100 TF=260PS CJC=1.54PF RB=300 .OP .AC DEC 100 0.1 200MEG .PRINT AC VM(5) VP(5) .END Results: Amid = 0.962, fL = 0.52 Hz, fH = 110 MHz 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-47 16.71 VBB = 9V 100 kΩ = 2.25V | RB = 100 kΩ 300 kΩ = 75.0 kΩ 100 kΩ + 300 kΩ (2.25 − 0.7)V = 0.251mA IC = 100 75.0kΩ + 101(13kΩ) 100(0.025) = 9.96 kΩ 0.251mA gm = 40(0.251mA) = 10.0 mS | rx = 300Ω | rπ = Cµ = 0.6 pF | Cπ = 0.01 − 0.6 = 2.58 pF 5 x10 8 ) 2π ( RL = 13kΩ 100kΩ = 11.5kΩ | Rth = 75 kΩ 2 kΩ = 1.95 kΩ Rin = RB [rx + rπ + (β o + 1)RL ] = 75.0 kΩ [300Ω + 9.96kΩ + (101)11.5 kΩ] = 70.5kΩ ⎛ Rin ⎞ 101(11.5kΩ) (β o + 1)RL Amid = ⎜ = 0.972 = 0.964 ⎟ ⎝ RI + Rin ⎠ rx + rπ + (β o + 1)RL [0.300 + 9.96 + 101(11.5)]kΩ fH ≅ 1 2π 1 (1950 + 300)⎢ ⎤ 2.58 pF + 0.6 pF ⎥ ⎣1 + 10mS (11.5 kΩ) ⎦ ⎡ = 1 1 = 114 MHz 2π (2250)(0.622 pF ) 17-48 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.72 First, find the value of Kn required for I D = 0.1 mA VGG = 1.5 MΩ 10V = 4.05V | VGG − VGS = 0.1mA( 12 kΩ) → VGS = 2.85V 1.5 MΩ + 2.2 MΩ 2(0.1mA) 2ID mA VGS − VTN = 0.75V → VTN = 2.10V | Kn = = = 0.356 2 2 2 V (VGS − VTN ) (0.75) gm = 2(0.1mA) 0.75V = 0.267 mS | Rin = R1 R2 = 892 kΩ | RL = 12 kΩ 100 kΩ = 10.7kΩ Amid = (0.267mS )(10.7kΩ) = +0.739 Rin g m RL = 0.998 RI + Rin 1 + g m RL 1 + (0.267mS )( 10.7 kΩ) 1 (-2.62 dB) = 57.9 MHz From Table 16.2 on page 1037 : f H = ⎡ ⎤ 3 pF 2π 2 kΩ 892 kΩ ⎢ + 0.6 pF ⎥ 10.7 kΩ) ⎢ ⎥ ⎣1 + (0.267 mS )( ⎦ ( ) f P1 = 1 = 0.379 Hz 2π (894 kΩ)(4.7µF ) fP2 = ⎡ ⎛ ⎞⎤ 1 2π ⎢100 kΩ + ⎜12 kΩ ⎟⎥(0.1µF ) 0.267 mS ⎠⎥ ⎢ ⎝ ⎣ ⎦ 1 = 15.5 Hz f L ≅ 15.5 Hz Note that a low frequency RHP zero makes the calculation of fH a very poor estimate for the FET case. See the analysis in Prob. 17.73 which shows ωz = gm/CGS. *Problem 16.72 - Common-Drain Amplifier VDD 6 0 DC 10 VS 1 0 AC 1 RS 1 2 2K C1 2 3 4.7UF R1 3 0 1.5MEG R2 6 3 2.2MEG M1 6 3 4 4 NFET R4 4 0 12K C3 4 5 0.1UF R7 5 0 100K .MODEL NFET NMOS VTO=2.10 KP=0.356MA CGSO=30NF CGDO=6NF .OP .AC DEC 100 1 500MEG .PRINT AC VM(5) VP(5) .END Results: Amid = 0.740, fL = 15.5 Hz, fH = 195MHz - Note that there is peaking in the response. 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-49 16.73 First find VDD , VTN and Kn from Prob. 17.22 : VDD = VDS + I D RS = 10 V 1.5 MΩ 10V = 4.05V | VGG − VGS = 0.1mA( 12 kΩ)→ VGS = 2.85V VGG = 1.5 MΩ + 2.2 MΩ 2(0.1mA) 2ID mA = = 0.356 2 VGS − VTN = 0.75V → VTN = 2.10V | Kn = 2 2 V (VGS − VTN ) (0.75) Now find the new Q - point with VDD = 20 V . VGG = 1.5 MΩ 20V = 8.11V | RGG = 1.5 MΩ 2.2 MΩ = 892kΩ | VGG − VGS = I D RS 1.5 MΩ + 2.2 MΩ ⎛ 0.356mA ⎞ 2 VGS − 2.10) → VGS = 3.56V | I D = 0.379 mA | VDS = 15.5 V ok 8.11- VGS = ( 12 kΩ) ⎜ ⎟( 2 ⎝ 2V ⎠ gm = (3.56 − 2.10)V 2(0.379mA) = 0.519 mS | Rin = R1 R2 = 892kΩ | RL = 12 kΩ 100 kΩ = 10.7 kΩ Amid = (0.519mS )(10.7kΩ) = +0.846 g m RL Rin = 0.998 RI + Rin 1 + g m RL 1 + (0.519 mS )( 10.7kΩ) 1 (-1.46 dB) = 75.4 MHz From Table 16.2 on page 1037 : f H = ⎡ ⎤ 3 pF 2π 2 kΩ 892 kΩ ⎢ + 0.6 pF ⎥ 10.7 kΩ) ⎢ ⎥ ⎣1 + (0.519 mS )( ⎦ ( ) 17-50 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 16.74 ix i1 rπ + v vx gm v Cπ Cµ - RL Ix = sCµVx + I1 | Vx = Z1 = ⎛ ⎞ I1 I1 + ⎜ I1 + gm ⎟ RL (sCπ + gπ ) ⎝ (sCπ + gπ )⎠ Vx sCπ rπ RL + RL + rπ + β o RL sCπ rπ RL + rπ + (β o + 1)RL = = I1 sCπ rπ + 1 sCπ rπ + 1 1 1 sCπ rπ sCπ + + 1 r + (β o + 1)RL rπ + (β o + 1)RL (1 + gm RL ) rπ + (β o + 1)RL Y1 = = π ≅ for β o >> 1 Cπ rπ RL Cπ RL Z1 s +1 s +1 rπ + (β o + 1)RL (1 + gm RL ) ⎞ ⎞ 1 ⎛ 1 Cπ RL 1 ⎛ 1 rx | rπo = rπ rx ≅ rx ωH = 1 ⎡ ⎛ R ⎞⎤ rx ⎢Cπ + Cµ ⎜1 + g m RL + L ⎟⎥ rx ⎠⎦ ⎝ ⎣ 1 ≅ ⎛ R ⎞ rx Cµ ⎜1 + g m RL + L ⎟ rx ⎠ ⎝ = rx Cµ ( 1 + g m RL )+ RLCµ = Cµ = 0.884 pF 1 + 1.01x10−3 rx ωH | rx Cµ ( 1 + 100)+ 105 Cµ = 2π 1.8 x10 ( ) = 8.84 x10−8 | Cµ cannot exceed 0.884 pF for an ideal transistor with rx = 0. Other more realistic possibilities (Cu , rx ): 16.79 (0.75pF, 177Ω) (0.5pF,760Ω) ⎛ 1 ⎞ ⎛ 1 ⎞ | RL = Amid ⎜ + RI ⎟ = 20⎜ + 100⎟ ⎝ gm ⎠ ⎝ gm ⎠ ωH ≅ 1 RLCGD | Amid = Rin g R g m RL ≅ m L RI + Rin 1 + g m RI 2π 25 x106 = ( ) 1 → g m = 164mS ⎛ 1 ⎞ −12 20⎜ + 100Ω⎟3 x10 F ⎝ gm ⎠ 2 ⎛ 1 ⎞ (164mS ) = 672 mA g2 RL = 20⎜ + 100⎟ = 2.12 kΩ | I D = m = ⎛ mS ⎞ 2 Kn ⎝ 0.164 ⎠ 2⎜20 ⎟ ⎝ V ⎠ Note that we cannot supply I D through RL since I D RL = 1420V >> VDD . RI vi C 1 C2 L R S RL One Possibility: +V DD This is really not a realistic design. The current and power are far too high. We need to find an FET with a much higher Kn. 03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-53 16.80 ωH = 1 1 = ⎡ ⎡ ⎤ ⎡ ⎡ R ⎤⎤ R ⎤ Rth ⎢CGS + CGD ⎢1 + g m RL + L ⎥⎥ 100⎢12 pF + 5 pF ⎢1 + g m RL + L ⎥⎥ 100 ⎦⎦ Rth ⎦⎦ ⎣ ⎣ ⎣ ⎣ ⎡ ⎤ 1 RL ⎢ 1 6 ⎥1 2π 25 x10 = | g m RL + = − 17 ⎡ ⎡ ⎥5 100 ⎢ 2π 25 x106 ( 100) 10−12 R ⎤⎤ ⎣ ⎦ 100⎢17 pF + 5 pF ⎢g m RL + L ⎥⎥ 100 ⎣ ⎦ ⎣ ⎦ ( ) ( ) gm = 9.33 g2 g2 2 − 0.01 → RL ≤ 933Ω | I D = m = m = 20 g m RL 2 Kn 0.05 2 0.025 0.25) = 781 µA. ( 2 For strong inversion (for the square - law model to be valid), we desire (V GS − VTN ) ≥ 0.25V → I D ≥ g m must exceed 0.01S. Choose g m = 0.015S . I D RL = 9.33 = 1.87 kΩ | g m RL = 28.0 g m − .01 (0.015) = .05 2 = 4.5 mA. 16.81 fH ≤ 1 1 = → f H ≤ 8.33 MHz 2πRLCµ 2π 12 kΩ 47 kΩ (2 pF ) ( ) 16.82 i r x1 rπ1 + v1 b c u1 c π1 g m1 Q v RL 1 1 i 1 g m1 b Q Q2 Cπ & C µ 1 1 r +rπ 2 R L x2 2 Cπ2 & C µ2 Use the open-circuit time-constant approach: 17-54 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 (a) ⎡ (β o + 1)(rx 2 + rπ 2 ) ⎤⎛ ⎞ βo Rµ1O : v x ≅ ix rx1 − ix rx1⎢ RL ⎟ − (−ix RL ) ⎥⎜− ⎣ rπ 1 + (β o + 1)(rx 2 + rπ 2 )⎦⎝ rx 2 + rπ 2 ⎠ ⎡ ⎞⎤ ⎛ β o rπ 2 Rµ1O ≅ ix ⎢RL + rx1⎜1 + gm 2 RL ⎟⎥ assuming rx 2


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