Rizzoni5eSM_CH10
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G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 Chapter 10: Bipolar Junction Transistors: Operation, Circuit Models, and Applications – Instructor Notes Chapter 10 introduces bipolar junction transistors. The material on transistors is divided into two independent chapters, one on bipolar devices, and one on field-effect devices. The two chapters are functionally independent, except for the fact that Section 10.1, introducing the concept of transistors as amplifiers and switches, can be covered prior to starting Chapter 11 if the instructor decides to only teach field-effect devices, or to cover them before bipolar devices. Section 10.2 introduces the fundamental ideas behind the operation of bipolar transistors, and illustrates the calculation of the state and operating point of basic transistor circuits. The discussion of the properties of the BJT in Section 10.2 is centered around a description of the base and collector characteristics, and purposely avoids a detailed description of the physics of the device, with the intent of providing an intuitive understanding of the transistor as an amplifier and electronic switch. The second part of the chapter has been reorganized for clarity. Section 10.3 introduces large-signal models of the BJT, and also includes the box Focus on Methodology: Using device data sheets (pp. 559-561). Example 10.4 (LED Driver) and the box Focus on Measurements: Large Signal Amplifier for Diode Thermometer (pp. 566-568) provide two application examples. New to the 5th Edition are examples 10.5 and 10.6, that present simple but practically useful battery charger and DC motor drive BJT circuits. These examples are accompanied by related homework problems (10.25-10.27). Section 10.4 defines the concept of operating point and illustrates the selection of a bias point, introducing the idea of a small-signal amplifier in the most basic way. Finally, Section 10.5 introduces the analysis of BJT switches and presents TTL gates. The end-of-chapter problems are straightforward applications of the concepts illustrated in the chapter. The 5th Edition of this book includes 17 new problems; some of the 4th Edition problems were removed, increasing the end-of-chapter problem count from 40 to 51. Learning Objectives 1. 2. 3. 4. 5. Understand the basic principles of amplification and switching. Section 10.1. Understand the physical operation of bipolar transistors, and identify their state. Section 10.2 Understand the large-signal model of the bipolar transistor, and apply it to simple amplifier circuits. Section 10.3. Determine and select the operating point of a bipolar transistor circuit; understand the principle of small signal amplifiers. Section 10.4. Understand the operation of bipolar transistor as a switch and analyze basic analog and digital gate circuits. Section 10.5. 10.1 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 Section 10.2: Operation of the Bipolar Junction Transistor Problem 10.1 Solution: Known quantities: Transistor diagrams, as shown in Figure P10.1: (a) pnp, VEB = 0.6 V and VEC = 4.0 V (b) npn, VCB = 0.7 V and VCE = 0.2 V (c) npn, VBE = 0.7 V and VCE = 0.3 V (d) pnp, VBC = 0.6 V and VEC = 5.4 V Find: For each transistor shown in Figure P10.1, determine whether the BE and BC junctions are forward or reverse biased, and determine the operating region. Analysis: (a) VBE = - 0.6 V for a pnp transistor implies that the BE junction is forward-biased. VBC = VEC - VEB = 3.4 V. The CB junction is reverse-biased. Therefore, the transistor is in the active region. (b) VBC = - 0.7 V for a npn transistor implies that the CB junction is reverse-biased. VBE = VBC - VEC = -0.5 V. The BE junction is reverse-biased. Therefore, the transistor is in the cutoff region. (c) VBE = 0.7 V for a npn transistor implies that the BE junction is forward-biased. VBC = VEC - VEB = 0.4 V. The CB junction is forward-biased. Therefore, the transistor is in the saturation region. (d) VBC = 0.6 V for a pnp transistor implies that the CB junction is reverse-biased. VBE = VBC – VEC = - 4.8 V. The BE junction is forward-biased. Therefore, the transistor is in the active region. Problem 10.2 Solution: Known quantities: Transistor type and operating characteristics: a) npn, VBE = 0.8 V and VCE = 0.4 V b) npn, VCB = 1.4 V and VCE = 2.1 V c) pnp, VCB = 0.9 V and VCE = 0.4 V d) npn, VBE = - 1.2 V and VCB = 0.6 V Find: The region of operation for each transistor. Analysis: a) Since VBE = 0.8 V, the BE junction is forward-biased. VCB = VCE + VEB = - 0.4 V. Thus, the CB junction is forward-biased. Therefore, the transistor is in the saturation region. 10.2 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 b) VBE = VBC + VCE = 0.7 V. The BE junction is forward-biased. VCB = 1.4 V. The CB junction is reverse-biased. Therefore, the transistor is in the active region. c) VCB = 0.9 V for a pnp transistor implies that the CB junction is forward-biased. VBE = VBC – VCE = - 1.3 V. The BE junction is forward-biased. Therefore, the transistor is in the saturation region. d) With VBE = - 1.2 V, the BE junction is reverse-biased. VCB = - 0.6 V. The CB junction is reverse-biased. Therefore, the transistor is in the cutoff region. Problem 10.3 Solution: Known quantities: I The circuit of Figure P10.3: β = C = 100 . IB Find: The operating point and the state of the transistor. Analysis: VBE = 0.6 V and the BE junction is forward biased. IB = VCC − VBE 12 − 0.6 = = 13.9 µA R1 820 I C = β ⋅ I B = 1.39mA Writing KVL around the right-hand side of the circuit: −VCC + I C RC + VCE + I E RE = 0 VCE = VCC − I C RC − (I C + I B )RE = 12 − (1.39)(2.2) − (1.39 + 0.0139)(0.910) = 7.664 V VBC = VBE + VCE = 0.6 _ 7.664 = 8.264 V VCE > VBE ⇒ The transistor is in the active region. 10.3 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 Problem 10.4 Solution: Known quantities: The magnitude of a pnp transistor's emitter and base current, and the magnitudes of the voltages across the emitterbase and collector-base junctions: IE = 6 mA, IB = 0.1 mA and VEB = 0.65 V, VCB = 7.3 V. Find: a) VCE. b) IC. c) The total power dissipated in the transistor, defined as P = VCE I C + VBE I B . Analysis: a) VCE = VCB - VEB = 7.3 - 0.65 = 6.65 V. b) IC = IE - IB = 6 - 0.1 = 5.9 mA. c) The total power dissipated in the transistor can be found to be: P ≈ VCE I C = 6.65 × 5.9 × 10 −3 = 39 mW Problem 10.5 Solution: Known quantities: The circuit of Figure P10.5, assuming the BJT has Vγ = 0.6 V. Find: The emitter current and the collector-base voltage. Analysis: V + 15 0.6 + 15 Applying KVL to the right-hand side of the circuit, I E = − BE = − = −520 µA 30000 30000 Then, on the left-hand side, assuming β >> 1: −10 + I C RC + VCB = 0⇒ VCB = 10 − I C RC = 10 − − 520 × 10 − 6 × 10 × 103 = 17.8 V ( ) 10.4 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 Problem 10.6 Solution: Known quantities: The circuit of Figure P10.6, assuming the BJT has VBE = 0.6 V and β =150. Find: The operating point and the region in which the transistor operates. Analysis: Define RC = 3.3 kΩ, RE = 1.2 kΩ, R1 = 62 kΩ, R2 = 15 kΩ, VCC = 18 V By applying Thevenin’s theorem from base and mass, we have RB = R1 || R2 = 12.078 k VBB = IB = R2 VCC ≅ 3.5 V R1 + R2 VBB − VBE ≅ 15 µA RB + RE (1 + β ) I C = β I B = 2.25 mA VCE = VCC − RC I C − RE I E = 18 − 3300 ⋅ 2.25 ⋅ 10 − 3 − 1200 ⋅ 151 ⋅ 15 ⋅ 10 − 6 = 7.857 V From the value of VCE it is clear that the BJT is in the active region. Problem 10.7 Solution: Known quantities: The circuit of Figure P10.7, assuming the BJT has Vγ = 0.6 V . Find: The emitter current and the collector-base voltage. Analysis: Applying KVL to the right-hand side of the circuit, −VCC + I E RE + VEB = 0 IE = VCC − VEB 20 − 0.6 = = 497.4 µA . Since β >> 1 , I C ≈ I E = 497.4 µA RE 39 ⋅ 103 VCB + I C RC − VDD = 0 Applying KVL to the left-hand side: VCB = VDD − I C RC = 20 − 497.4 ⋅ 20 ⋅10−3 = 10.05V 10.5 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 Problem 10.8 Solution: Known quantities: The circuit of Figure P10.7, assuming the emitter resistor is changed to 22 kΩ and the BJT has Vγ = 0.6 V . Find: The operating point of the transistor. Analysis: IE = VCB VCC − VEB 20 − 0.6 = = 881.8 µA , I C ≈ I E = 881.8 µA RE 22 ⋅ 103 =V − I R = 20 − 881.8 ⋅ 20 ⋅ 10 −3 = 2.364 V DD C C Problem 10.9 Solution: Known quantities: The collector characteristics for a certain transistor, as shown in Figure P10.9. Find: The ratio IC/IB for VCE = 10 V and I B = 100 µA, 200 µA, and 600 µA b) VCE, assuming the maximum allowable collector power dissipation is 0.5 W for I B = 500 µA . a) Analysis: a) For IB = 100 A and VCE = 10 V, from the characteristics, we have IC = 17 mA. The ratio IC / IB is 170. For IB = 200 A and VCE = 10 V, from the characteristics, we have IC = 33 mA. The ratio IC / IB is 165. For IB = 600 A and VCE = 10 V, from the characteristics, we have IC = 86 mA. The ratio IC / IB is 143. b) For IB = 500 µA, and if we consider an average β from a., we have IC = 159·500 10-3= 79.5 mA. The power P 0.5 dissipated by the transistor is P = VCE I C + VBE I B ≈ VCE I C , therefore: VCE ≈ = = 6.29 V . I C 79.5 ⋅ 10−3 10.6 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 Problem 10.10 Solution: Known quantities: Figure P10.10, assuming both transistors are silicon-based with β = 100 . Find: a) IC1, VC1, VCE1. b) IC2, VC2, VCE2. Analysis: a) From KVL: −30 + I B1 RB1 + VBE1 = 0 30 − 0.7 I B1 = = 39.07 µA 750 ⋅ 103 ⇒ I C1 = β ⋅ I B1 = 3.907 mA VCE1 = VC1 = 5.779 V . ⇒ VC1 = 30 − RC1I C1 = 30 − 3.907 ⋅ 6.2 = 5.779 V 5.779 − 0.7 4.7 ⋅ 103 b) Again, from KVL: −5.779 + VBE 2 + I E 2 RE 2 = 0 ⇒ I E2 = = 1.081 mA β 100 and I C 2 = I E 2 β + 1 = 1.081 ⋅ 101 = 1.07 mA . Also, −30 + I C 2 ( RC 2 + RE 2 ) + VCE 2 = 0 ⇒ VCE 2 = 30 − (1.07) ⋅ (20 + 4.7) = 3.574 V . 30 − VC 2 Finally, I C 2 = ⇒ VC 2 = 30 − (1.07) ⋅ (20) = 8.603 V . RC 2 Problem 10.11 Solution: Known quantities: Collector characteristics of the 2N3904 npn transistor, see data sheet pg. 560. Find: The operating point of the transistor in Figure P10.11, and the value of β at this point. Analysis: Construct a load line. Writing KVL, we have: −50 + 5000 ⋅ I C + VCE = 0 . Then, if I C = 0 , VCE = 50 V ; and if VCE = 0 , I C = 10 mA . The load line is shown superimposed on the collector characteristic below: 10.7 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 The operating point is at the intersection of the load line and the I B = 20 µA line of the characteristic. Therefore, I CQ ≈ 5 mA and VCEQ ≈ 20 V . Under these conditions, an 5 µA increase in I B yields an increase in I C of approximately 6 − 5 = 1 mA . Therefore, ∆I 1 ⋅ 10 −3 β≈ C = = 200 ∆I B 5 ⋅ 10 − 6 The same result can be obtained by checking the hFE gain from the data-sheets corresponding to 5 mA. Load line Problem 10.12 Solution: Known quantities: The circuit shown in Figure P10.12. With reference to Figure 10.20, assume Vγ = 0.6 V , Vsat = 0.2 V . Find: The operating point of the transistor, by computing the ratio of collector current to base current. Analysis: VCE = Vsat = 0.2 V , therefore I C = IC 9.8 ⋅ 10 −3 = = 96.08 RC = VCC − Vγ LED − VCEsat ≤ 5 − 1 .4 − 0 .2 = 340 Ω 0.01 VCC − Vγ LED − VCEsat = 47 Ω I LED max Therefore, RC ∈[47, 340] Ω Problem 10.18 Solution: Known quantities: For the circuit shown in Figure 10.18 in the text: VD = 1.1 V, RB = 33 k , VCC = 12 V, VBE = 0.75 V, VCEQ = 6 V, β = 188.5, RS = 500 Ω Find: The resistance RC. Analysis: The current through the resistance RB is given by VD − VBEQ 1.1 − 0.75 IB = = = 10.6 µA RB 33000 The current through RS is: I S = 6 − 1.1 = 9.8 mA RS 500 It follows that the current through the resistance RC is I CQ = β I B + I S = 11.8 mA = Finally, RC = VCEQ − VD VCC − VCEQ 12 − 6 = = 508.5 Ω I CQ 0.0118 10.12 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 Problem 10.19 Solution: Known quantities: For the circuit shown in Figure 10.14 in the text: Voff = 0 V, Von = 5 V, I B max = 5 mA, RC = 340 , VCC = 5 V, Vγ = 0.7V, VCEsat = 0.2 V, β = 95, Vγ LED = 1.4 V, I LED ≥ 10 mA, Pmax = 100 mW Find: Range of RB. Analysis: If the BJT is in saturation VCC − Vγ LED − VCEsat = 10 mA IC = RC In order to guarantee that the BJT is in saturation Von − Vγ 5 − 0.7 RB ≤ = = 40.85 kΩ 0.01 IC / β 95 Von − Vγ RB ≥ = 860 Ω I B max Problem 10.20 Solution: Known quantities: For the circuit shown in Figure 10.14 in the text: Voff = 0 V, Von = 5 V, I B max = 5 mA, RB = 10 k , RC = 340 , VCC = 5 V, Vγ = 0.7V, VCEsat = 0.2 V, Vγ LED = 1.4 V, I LED ≥ 10 mA, Pmax = 100mW Find: Minimum value of β that will ensure the correct operation of the LED. Analysis: 4 .3 = 0.43 mA RB 10000 I 0.01 β min = LED min = = 23.25 IB 0.43 ⋅ 10 − 3 IB = = Von − Vγ 10.13 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 Problem 10.21 Solution: Known quantities: For the circuit shown in Figure 10.14 in the text: Voff = 0 V, Von = 3.3V, I B max = 5mA, RB = 10k , RC = 340 , VCC = 5 V, Vγ = 0.7V,VCEsat = 0.2 V, Vγ LED = 1.4 V, I LED ≥ 10mA, Pmax = 100mW Find: Minimum value of β that will ensure the correct operation of the LED. Analysis: IB = Von − Vγ 3.3 − 0.7 = = 0.26 mA RB 10000 I LED min 0.01 = = 38.5 IB 0.26 ⋅ 10 −3 β min = Problem 10.22 Solution: Known quantities: For the circuit shown in Figure 10.14 in the text: Voff = 0 V, Von = 5 V, I B max = 1 mA, RB = 1 k , R = 12 , VCC = 13 V, Vγ = 0.7V, VCEsat = 1 V, IC ≥ 1A Find: Minimum value of β that will ensure the correct operation of the fuel injector. Analysis: IC = VCC − VCEsat 13 − 1 = = 1A R 12 IC I B max = 1 1 ⋅ 10 − 3 = 1000 β min = 10.14 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 Problem 10.23 Solution: Known quantities: For the circuit shown in Figure 10.14 in the text: Voff = 0 V, Von = 5 V, I B max = 1 mA, β = 2000, R = 12 , VCC = 13 V, Vγ = 0.7V, VCEsat = 1 V, IC ≥ 1A Find: The range of RB that will ensure the correct operation of the fuel injector. Analysis: If the BJT is in saturation V − VCEsat I C = CC = 1A R Because this is the minimum value allowed for the current to drive the fuel injector, it is necessary to guarantee that the BJT is in saturation. In order to guarantee that the BJT is in saturation Von − Vγ 5 − 0.7 RB ≤ = = 8.6 kΩ 1 IC / β 2000 Von − Vγ RB ≥ = 4 .3 k Ω I B max Problem 10.24 Solution: Known quantities: For the circuit shown in Figure 10.14 in the text: Voff = 0 V, Von = 3.3 V, I B max = 1 mA, β = 2000, R = 12 , VCC = 13 V, Vγ = 0.7V, VCEsat = 1 V, IC ≥ 1A Find: The range of RB that will ensure the correct operation of the fuel injector. Analysis: If the BJT is in saturation V − VCEsat I C = CC = 1A R Because this is the minimum value allowed for the current to drive the fuel injector, it is necessary to guarantee that the BJT is in saturation. In order to guarantee that the BJT is in saturation Von − Vγ 3.3 − 0.7 RB ≤ = = 5.2 kΩ 1 IC / β 2000 Von − Vγ RB ≥ = 2.6 kΩ I B max 10.15 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 Problem 10.25 Solution: Known quantities: The circuit of Figure P10.25: IC = 40 mA; Transistor large signal parameters. Find: Design a constant-current battery charging circuit, that is, find the values of VCC, R1, R2 that will cause the transistor Q1 to act as a 40mA constant current source. Assumptions: Assume that the transistor is forward biased. Use the large-signal model with β = 100. Analysis: The battery charging current is 40 mA, IC = 40 mA. β +1 I E = 40.4mA . Thus, the emitter current must be I E = β Since the base-emitter junction voltage is assumed to be 0.6 V, then resistor R2 has a voltage: V2 = Vz − Vγ = 5.6 − 0.6 = 5 V , so the required value of R2 to be: 5 V = = 123.8Ω I E 0.0404 Since the only purpose of R1 is to bias the Zener diode, we can select a value that will supply enough current fro the Zener to operate, for example R1 > 100 Ω, so that there will be as little current flow through this resistance as possible. Finally, we need to select an appropriate supply voltage. VCC must be greater than or equal to the sum of the battery voltage, the CE junction voltage and the voltage across R2. That is, VCC ≥ 9 + VCE + 5 . A collector supply of 24 V will be more than adequate for this task. R2 = 10.16 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 Problem 10.26 Solution: Known quantities: The circuit of Figure of P10.26. Find: Analyze the operation of the circuit and explain how I E is decreasing until the battery is full. Find the values of VCC, R1 that will result in a practical design. Assumptions: Assume that the transistor is forward biased. Analysis: When the Zener Diode works in its reverse breakdown area, it provides a constant voltage: Vz = 11 V . That means: VB = VZ = 11 V . When the transistor is forward biased, according to KVL, VZ = I BE ⋅ RBE + Vγ + Vbattery , where RBE is the base resistance. As the battery gets charged, the actual battery charging voltage Vbattery will increase from 9.6 V to 10.4 V. As Vbattery increases gradually, VZ and Vγ stay unchanged, then we can see that I BE will decrease gradually. So I E = (β + 1)I BE will also decrease at the same time. Since the only purpose of R1 is to bias the Zener diode, we can select a value that will supply enough current fro the Zener to operate, for example R1 > 100 Ω, so that there will be as little current flow through this resistance as possible. Finally, we need to select an appropriate supply voltage. VCC must be greater than or equal to the sum of the battery voltage, the CE junction voltage. That is, VCC ≥ 11 + VCE . A collector supply of 12 V should be adequate for this task. 10.17 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 Problem 10.27 Solution: Known quantities: The circuit of Figure P10.27: Vin= 5 V Find: Values of Rb. Assumptions: Assume that the transistors are in the active region. Use the large-signal model with β = 40 for each transistor. Analysis: The emitter current from Q1, iE1 = (β+1) iB1 becomes the base current for Q2, and therefore, iC2 = β iE1 = β (β+1) iB1. The Q1 base current is given by the expression Vin − Vγ − Vγ iB1 = Rb Therefore the motor current will reach maximum when Vin= 5 V: Vin − Vγ − Vγ = 0.34 A iC max = β (β + 1) Rb β (β + 1) 40 ⋅ 41 So, Rb = Vin − 2Vγ = (5 − 1.2) = 18,329Ω 0.34 0.34 Since 18.33 kΩ is a standard resistor value, we should select Rb = 18.33 kΩ, which will result in a slightly lower maximum current. ( ) 10.18 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 Section 10.4: Selecting an Operating Point for a BJT Problem 10.28 Solution: Known quantities: The circuit of Figure of 10.22 in the text, RC = 1KΩ , VBB = 5V, β min = 50, VCC = 10V. Find: The range of RB to make the transistor in the saturation state. Analysis: Assuming VCEsat = 0.2 V , the current I C is: IC = VCC − VCEsat = 9.8 mA RC I Therefore, I B = C = 0.196 mA b Assuming Vγ = VBEsat = 0.6 V , we have RB = VBB − VBE = 22.45 kΩ IB 0 < RB < 22.45 kΩ That is Problem 10.29 Solution: Known quantities: The circuit of Figure of 10.22 in the text, RC = 1KΩ , RB = 10KΩ, β min = 50, VCC = 5V. Find: The range of VBB to make the transistor in the saturation state. Analysis: Assume VCEsat = 0.2 V , the current I C can be found as V − VCEsat I C = CC = 4.8 mA RC I Therefore, I B = C = 0.096 mA = 96 µA b Assuming Vγ = VBEsat = 0.6 V , we have VBB = I B RB + VBEsat = 1.56 V That is VBB > 1.56 V 10.19 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 Problem 10.30 Solution: Known quantities: The circuit of Figure 10.20 in the text, RC = 2kΩ , I BB = 20 µA, β = 100, VCC = 10V. Find: I C , I E , VCE , VCB . Analysis: I C = βI B = 100 × 20 × 10 −6 = 2 mA I E = I B + I C = ( β + 1) I B = 101 × 20 × 10 − 6 = 2.02 mA VCE = VCC − I C RC = 10 − 2 × 2 = 6 V Assume VBE = 0.6 V Then VCB = VCE − VBE = 6 − 0.6 = 5.4 V Problem 10.31 Solution: Known quantities: For the circuit shown in Figure P10.31: VCC = 20 V β = 130 R1 = 1.8 M R2 = 300 k RC = 3 k RL = 1k RE = 1 k RS = 0 .6 k vS = 1 cos(6.28 × 103 t ) mV . Find: The Thèvenin equivalent of the part of the circuit containing R1 , R2 , and VCC with respect to the terminals of R2 . Redraw the schematic using the Thèvenin equivalent. Analysis: Extracting the part of the circuit specified, the Thèvenin equivalent voltage is the open circuit voltage. The equivalent resistance is obtained by suppressing the ideal independent voltage source: Note that VCC must remain in the circuit because it supplies current to other parts of the circuit: 10.20 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 Problem 10.32 Solution: Known quantities: For the circuit shown in Figure P10.32: VCC = 12 V β = 130 R1 = 82 k R2 = 22 k RE = 0.5 k RL = 16 . Find: VCEQ at the DC operating point. Analysis: Simplify the circuit by obtaining the Thèvenin equivalent of the biasing network (R1,, R2, VCC) in the base circuit: 12 ⋅ 22 V CC R 2 = = 2.538 V R1 + R 2 82 + 22 R 1 R 2 = 82 ⋅ 22 = 17.35 k Suppress V CC : R B = R eq = 82 + 22 R1 + R 2 Redraw the circuit using the Thèvenin equivalent. The "DC blocking" or "AC coupling" capacitors act as open circuits for DC; therefore, the signal source and load can be neglected since this is a DC problem. Specify directions of current and polarities of voltages. VD : V BB = V TH = V OC = Assume the transistor is operating in its active region. Then, the base-emitter junction is forward biased. V BEQ ≈ 700 mV [Si] I EQ = [ β + 1 ] I BQ KVL : - V BB + I BQ R B + V BEQ + I EQ R E = 0 - V BB + I BQ R B + V BEQ + [ β + 1 ] I BQ R E = 0 I BQ = V BB - V BEQ 2.538 − 0.7 = = 22.18 µA + (β + 1) ⋅ R E 17350 + (130 + 1) ⋅ 500 RB −6 I EQ = (β + 1 ) I BQ = (130 + 1 ) ⋅ 22.18 ⋅ 10 = 2.906 mA KVL : - I EQ R E - V CEQ + V CC = 0 V CEQ = V CC - I EQ R E = 12 − 2.906 ⋅ 0.5 = 10.55 V The collector-emitter voltage is greater than its saturation value (0.3 V for Silicon). Therefore the initial assumption (operation in the active region) was correct and the solution is valid. 10.21 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 Problem 10.33 Solution: Known quantities: For the circuit shown in Figure P10.33: VCC = 12 V β = 100 VEE = 4 V RB = 100 k RC = 3 k RL = 6 k RE = 3 k RS = 0.6 k vS = 1 cos(6.28 × 103 t ) mV . Find: VCEQ and the region of operation. Analysis: The "DC blocking" or "AC coupling" capacitors act as open circuits for DC; therefore, the signal source and load can be neglected since this is a DC problem. Specify directions of current and polarities of voltages. Assume the transistor is operating in its active region; then, the baseemitter junction is forward biased and: VBEQ ≈ 700 mV [ Si] I CQ = β ⋅ I BQ I EQ = ( β + 1) I BQ ⇒ VEE − VBEQ 4 − 0.7 = 8.189µA 100000 + (100 + 1)(3000) KVL : − VEE + I BQ RB + VBEQ + I EQ RE = 0 I BQ = RB + [β + 1]RE = I CQ = β ⋅ I BQ = (100) ⋅ 8.189 ⋅ 10 − 6 = 818.9µA I EQ = ( β + 1) ⋅ I BQ = (100 + 1) ⋅ 8.189 ⋅ 10 − 6 = 827.0 µA KVL : + VEE − I EQ RE − VCEQ − I CQ RC + VCC = 0 = 11.06 V ⇒ VCEQ = VEE + VCC − I CQ RC − I EQ RE = 4 + 12 − 818.9 ⋅ 10 − 6 ⋅ 3000 − 827.0 ⋅ 10 ⋅ 10 − 6 ⋅ 3000 The collector-emitter voltage is greater (more positive) than its saturation value (+ 0.3 V for Silicon). Therefore the initial assumption (operation in the active region) was correct and the solution is valid. Notes: 1. DC power may be supplied to an npn BJT circuit by connecting the positive terminal of a DC source to the collector circuit, or, by connecting the negative terminal of a DC source to the emitter circuit, or, as was done here, both. 2. In a pnp BJT circuit the polarities of the sources must be reversed. Negative to collector and positive to emitter. 10.22 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 Problem 10.34 Solution: Known quantities: For the circuit shown in Figure P10.34: VCC = 12 V RB = 325 k β = 130 RC = 1.9 k RE = 2.3 k RL = 10 k RS = 0.5 k vS = 1 cos(6.28 × 103 t ) mV . Find: VCEQ and the region of operation. Analysis: The "DC blocking" or "AC coupling" capacitors act as open circuits for DC; therefore, the signal source and load can be neglected since this is a DC problem. Specify directions of current and polarities of voltages. Assume the transistor is operating in its active region; then, the base-emitter junction is forward biased. The base and collector currents both flow through the collector resistor in this circuit. VBEQ ≈ 700 mV [ Si ] I CQ = β ⋅ I BQ I EQ = ( β + 1) I BQ ⇒ 12 − 0.7 (325 + (130 + 1) ⋅ (2.3 + 1.9)) ⋅ 103 KCL : I BQ + I CQ − I RC = 0 VCC − VBEQ RB + ( β + 1)( RE + RC ) I RC = I CQ + I BQ = ( β + 1) I BQ I BQ = = = 12.91µA I RC = I EQ = ( β + 1) ⋅ I BQ = (130 + 1) ⋅ 12.96 ⋅ 10 − 6 = 1.691 mA KVL : − I EQ RE − VCEQ − I RC RC + VCC = 0 ⇒ VCEQ = VCC − I RC RC − I EQ RE = 12 − 1.691 ⋅ 1.9 − 1.691 ⋅ 2.3 = 4.896 V KVL : − I EQ RE − VBEQ − I BQ RB − I RC RC + VCC = 0 ⇒ - ( β + 1) I BQ ( RE + RC ) − VBEQ − I BQ RB + VCC = 0 The collector-emitter voltage is greater than its saturation value (0.3 V for Silicon). Therefore the initial assumption (operation in the active region) was correct and the solution is valid. 10.23 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 Problem 10.35 Solution: Known quantities: For the circuit shown in Figure P10.35: vS = 3 V RB = 60 k β = 100 Find: The value of RE so that I E is 1 mA. b) RC so that VC is 5 V. c) The small-signal equivalent circuit of the amplifier for RL = 5kΩ d) The voltage gain. a) Analysis: (a) With RB = 60 kΩ and VB = 3 V , applying KVL, we have 3 = I B RB + 0.6 + (1 + β ) I B RE 2 .4 IB = 60kΩ + 101RE I E = 101 2 .4 = 1mA 60kΩ + 101RE Therefore, 101 ⋅ 2.4 − 60 RE = = 1.81 kΩ 101 (b) VCE = 15 − I C RC − I E RE From (a), we have I C = I E Therefore, RC = β = 0.99 mA β +1 RB + 15 − 5 − 1.81 = 8.27 kΩ 0.99 B ∆ IB C ∆ IC RC 1 RL h oe (c) The small signal equivalent circuit is shown below + v OUT - (d) VS ∆I B = RB + hiw vS h ie - h fe ∆ I B 1 V vout = −∆I C RL ∆I C = out + h fe ∆I B 1 hoe hoe ∂V 0.6 hie = BE I BQ = = 60 .6 kΩ ∂I B 0.0099 × 10 − 3 Since hoe is not given, we can reasonably assume that 1/hoe is very large. Therefore, v 100 ⋅ RL AV = out = − = −4.15 vs RB + hie E 10.24 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 Problem 10.36 Solution: Known quantities: For the circuit shown in Figure P10.36: RC = 200 k Find: e) f) The operating point of the transistor. Voltage gain vout vin ; current gain iout iin g) Input resistance ri h) Output resistance ro Analysis: (a) VB = VCC R2 = 6 .1 V R1 + R2 RB = R1 || R2 = 3749.87 Ω Assuming VBE = 0.6 V , we have VEV = VB − VBE = 5.5 V V I E = E = 22 mA RE I I B = E = 0.088 mA b +1 and VCE = VC − VE = (V.CC − RC I C ) − 5.5 = 15 - 200 ⋅ 21.912 ⋅ 10 -3 − 5.5 = 5.12 V (b) The AC equivalent circuit is shown on the right: hie = ∂VBE ∂I B I BQ ≈ 0.6 0.088 × 10 − 3 = 6.82kΩ vout = RE ( I B + I C ) = 250(250 + 1) I B vin = I B hie + vout = I B hie + 250 ⋅ 251 ⋅ I B Therefore, the voltage gain is v AV = out = 0.902 and vin iout = I B + I C + ( β + 1) ⋅ I B v iin = I B + in = I B + ( I B hie + 250 ⋅ 251 ⋅ I B ) RB RB and the current gain is iout ( β + 1) I B = = 12.84 iin I B + ( I B hie + 250 ⋅ 251 ⋅ I B ) RB 10.25 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10 (c) To find the input resistance we compute: vin = I B hie + 250 ⋅ 251 ⋅ I B iin = I B + ( I B hie + 250 ⋅ 251 ⋅ I B ) RB Therefore. the input resistance is v ri = in = 3558Ω iin (d) To find the output resistance we compute vout = RE ( I B + I C ) = 250(250 + 1) I B iout = I B + I C + ( β + 1) ⋅ I B Therefore, the output resistance is v ro = out = 250Ω iout Problem 10.37 Solution: Known quantities: The circuit shown in Figure P10.37(a), P10.37(b): Find: The duration of the fuel injector pulse. Analysis: (a) With VCE = 0.3 V, VBE = 0.9 V and VBATT = 13 V, TC = 100°, from Figure P9.6(d), we have KC = 0, VCIT = 16/13= 1.23 ms. The signal duration is: τ = 1×10-3×0 + 1.23×10-3 = 1.23 ms When Vsignal is applied, the base current is IB = VBATT /80= 0.1625 A Thus, the transistor will be in the saturation region. Therefore, Vinj = VBATT - VCE = 13 - 0.3 = 12.7 V The time constant of the injector circuit is: τ' = L/R= 0.1 ms
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