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G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Chapter 10 Instructor Notes Chapter 10 introduces bipolar junction transistors. The material on transistors has been reorganized in this 4th Edition, and is now 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. 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. Example 10.5 introduces the use of Electronics Workbench (EWB) as a tool for analyzing electronics circuits; the CDROM contains an introduction to the software package and a number of solved problems. Section 10.3 introduces large-signal models of the BJT, and also includes the box Focus on Methodology: Using device data sheets (pp. 535-537). Example 10.7 (LED Driver) and the box Focus on Measurements: Large Signal Amplifier for Diode Thermometer (pp. 539-541) provide two application examples that include EWB solutions. Section 8.4 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. Learning Objectives 1. Understand the basic principles of amplification and switching. Section 1. 2. Understand the physical operation of bipolar transistors; determine and select the operating point of a bipolar transistor circuit; understand the principle of small signal amplifiers. Section 2. 3. Understand the large-signal model of the bipolar transistor, and apply it to simple amplifier circuits. Section 3. 4. Understand the operation of bipolar transistor as a switch and analyze basic analog and digital gate circuits. Section 4. 10.1 G. Rizzoni, Principles and Applications of Electrical Engineering 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 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 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 b) saturation region. = - 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. d) With VBE ______________________________________________________________________________________ Problem 10.3 Solution: Known quantities: The circuit of Figure P10.3. β= IC = 100 . IB Find: The operating point and the state of the transistor. Analysis: VBE = 0.6 V and the BE junction is forward biased. V − VBE 12 − 0.6 = = 13.9µA I B = CC R1 820 I C = β ⋅ I B = 1.39 mA 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 B + I C )RE VBC = VBE VCE > VBE = 12 − (1.39)(2.2) − (1.39 + 0.0139)(0.910) = 7.664 V + VCE = 0.6 + 7.664 = 8.264 V The transistor is in the active region. ______________________________________________________________________________________ 10.3 G. Rizzoni, Principles and Applications of Electrical Engineering 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 emitter-base 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 ≈ VCEIC = 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: Applying KVL to the right-hand side of the circuit, Then, on the left-hand side, assuming β >> 1 : (0.6 + 15) ) = −520µA § V + 15 · I E = −¨ BE ¸=− 30000 © 30000 ¹ VCB = 10 − I C ⋅ RC − 10 + I C RC + VCB = 0 = 10 − (−520) ⋅ (15) ⋅ 10 −3 = 17.8 V ______________________________________________________________________________________ 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. 10.4 G. Rizzoni, Principles and Applications of Electrical Engineering Analysis: Define RC Problem solutions, Chapter 10 = 3.3kΩ, RE = 1.2kΩ, R1 = 62kΩ, R2 = 15kΩ, VCC = 18 V By applying Thevenin’s theorem from base and mass, we have RB = R1 || R2 = 12.078 kΩ 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 VCC − VEB 20 − 0.6 = = 497.4µA . Since β >> 1 , I C ≈ I E = 497.4µA RE 39 ⋅10 3 Applying KVL to the left-hand side: VCB + I C RC − VDD = 0 IE = VCB = VDD − I C RC = 20 − 497.4 ⋅ 20 ⋅10 −3 = 10.05 V ______________________________________________________________________________________ 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 = VCC − VEB 20 − 0.6 = = 881.8µ A , I C ≈ I E = 881.8µA RE 22 ⋅10 3 10.5 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 VCB = VDD − I C RC = 20 − 881.8 ⋅ 20 ⋅10 − 3 = 2.364 V ______________________________________________________________________________________ Problem 10.9 Solution: Known quantities: The collector characteristics for a certain transistor, as shown in Figure P10.9. Find: a) The ratio IC / IB for VCE = 10 V and IB = 100 µA, 200 µA, and 600 µA b) VCE, assuming the maximum allowable collector power dissipation is 0.5 W for IB = 500 µA. Analysis: = 100 µA and VCE = 10 V, from the characteristics, we have IC = 17 mA. The ratio IC / IB is 170. For IB = 200 µA and VCE = 10 V, from the characteristics, we have IC = 33 mA. The ratio IC / IB is 165. For IB = 600 µA and VCE = 10 V, from the characteristics, we have IC = 86 mA. The ratio IC / IB is 143. -3 b) For IB = 500 µA, and if we consider an average β from a., we have IC = 159·500 10 = 79.5 mA. The power dissipated by the transistor is P = VCE I C + VBE I B ≈ VCE I C , therefore: 0 .5 P = = 6.29 V . VCE ≈ I C 79.5 ⋅ 10 −3 a) For IB ______________________________________________________________________________________ Problem 10.10 Solution: Known quantities: Figure P10.10, assuming both transistors are silicon-based with Find: a) IC1, b) IC2, β = 100 . VC1, VCE1. VC2, VCE2. Analysis: a) From KVL: − 30 + I B1 RB1 + VBE1 = 0 I B1 = I C1 = β ⋅ I B1 = 3.907 mA VC1 = 30 − RC1 I C1 = 30 − 3.907 ⋅ 6.2 = 5.779 V VCE1 = VC1 = 5.779 V . b) Again, from KVL: and I C 2 − 5.779 + VBE 2 + I E 2 RE 2 = 0 § β · § 100 · ¸¸ = 1.081 ⋅ ¨ = I E 2 ¨¨ ¸ = 1.07 mA . © 101 ¹ © β +1¹ 10.6 30 − 0.7 = 39.07µA 750 ⋅ 10 3 I E2 = 5.779 − 0.7 = 1.081 mA 4.7 ⋅ 10 3 G. Rizzoni, Principles and Applications of Electrical Engineering Also, Problem solutions, Chapter 10 − 30 + I C 2 ( RC 2 + RE 2 ) + VCE 2 = 0 VCE 2 = 30 − (1.07) ⋅ (20 + 4.7) = 3.574 V . Finally, I C 2 = 30 − VC 2 RC 2 VC 2 = 30 − (1.07) ⋅ (20) = 8.603 V . ______________________________________________________________________________________ Problem 10.11 Solution: Known quantities: Collector characteristics of the 2N3904 npn transistor, see data sheet pg. 536. 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: load line The operating point is at the intersection of the load line and the Therefore, I B = 20µA line of the characteristic. 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 C 1 ⋅ 10 −3 β≈ = = 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. 10.7 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 _____________________________________________________________________________________ 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: 10 − 0.2 = 9.8 mA RC 5 .7 − 0 .6 = Vγ = 0.6 V , therefore I B = = 102µA RB VCE = Vsat = 0.2 V , therefore I C = VBE I C 9.8 ⋅10 −3 = = 96.08 << β I B 102 ⋅ 10 −6 ______________________________________________________________________________________ Problem 10.13 Solution: Known quantities: For the circuit shown in Figure P10.13, Find: a) VB b) c) IB IE Analysis: a) VEB = VE VE = 1V and Vγ = 0.6 V . d) IC e) β f) α . − VB = Vγ = 0.6 V VB = VE − VEB = 1 − 0.6 = 0.4 V 10.8 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 VB 0 .4 = = 20µA RB 20 ⋅ 10 3 5 − VE 5 − 1 c) I E = = = 800µA RE 5000 d) I C = I E − I B = 800 − 20 = 780µA b) e) f) IB = I C 780 = = 39 IB 20 I 780 α= C = = 0.975 . I E 800 β= ______________________________________________________________________________________ Problem 10.14 Solution: Known quantities: For the circuit shown in Figure P10.14: VCC = 20 V β = 130 R1 = 1.8 MΩ RL = 1kΩ RS = 0.6 kΩ vS = cos 6.28 ×103 ⋅ t mV . [ R2 = 300 kΩ ] Find: The Thèvenin equivalent of the part of the circuit containing terminals of RC = 3 kΩ RE = 1kΩ R1 , R2 , and VCC with respect to the 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: ______________________________________________________________________________________ 10.9 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Problem 10.15 Solution: Known quantities: For the circuit shown in Figure P10.14: VCC = 15 V RL = 1.5 kΩ β = 100 R1 = 68 kΩ RS = 0.9 kΩ [ R2 = 11.7 kΩ RC = 200Ω 3 ] RE = 200Ω vS = cos 6.28 ×10 ⋅ t mV . Find: VCEQ and the region of operation. Analysis: Simplify the circuit by obtaining the Thèvenin equivalent of the biasing network in the base circuit: 15 ⋅11.7 V CC R 2 = = 2.202 V 68 + 11.7 R1 + R2 R1 R 2 = 68 ⋅11.7 = 9.982 kΩ R B = Req = R1 + R 2 68 + 11.7 VD : V BB = V TH = Suppress V CC : 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. Assume the transistor is operating in its active region; then, the base-emitter junction is forward biased. V BEQ ≈ 700 mV [Si] I CQ = β ⋅ I BQ I EQ = (β + 1) I BQ Then: 10.10 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 KVL: - V BB + I BQ RB +V BEQ+ I EQ RE = 0 - V BB + I BQ RB +V BEQ+ [ β +1 ] I BQ RE = 0 2.202- 0.7 V BB - V BEQ = 49.76 µA = I BQ = RB + [ β +1 ] RE 9982+ (100 + 1)(200) −6 I CQ = β I BQ = 100 ⋅ 49.76 ⋅10 = 4.976 mA −6 I EQ = (β + 1)⋅ I BQ = (100 + 1) ⋅ 49.76 ⋅ 10 = 5.026 mA KVL : - I EQ R E - V CEQ - I CQ RC + V CC = 0 V CEQ = V CC - I CQ RC - I EQ R E = 15 − 4.976 ⋅ 0.2 − 5.026 ⋅ 0.2 = 13.00 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. Notes: 1. If the collector-emitter voltage were less than its saturation value, the transistor would be operating in its "saturation" or "ohmic" region. In this case, the collector-emitter voltage is equal to its saturation value (0.3 V for Silicon). The solution for the base current remains valid. However, the parameter β and the solution for the collector and emitter curents become invalid. In the saturation region, the base-emitter junction is still forward biased and its voltage remains the same. A solution for the collector and emitter currents is possible using the collector-emitter saturation voltage (0.3 V for Silicon) in a KVL. In the saturation region, the base current has no control over the collector or emitter currents (since β is invalid). Therefore, amplification is impossible. ______________________________________________________________________________________ 2. Problem 10.16 Solution: Known quantities: For the circuit shown in Figure P10.14: VCC = 15 V β = 100 R1 = 68 kΩ RL = 1.5 kΩ RS = 0.9 kΩ vS = cos 6.28 ×103 ⋅ t mV . [ R2 = 11.7 kΩ RC = 4 kΩ ] Find: VCEQ and the region of operation. Analysis: 10.11 RE = 200Ω G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Simplify the circuit by obtaining the Thèvenin equivalent of the biasing network in the base circuit: 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. Assume the transistor is operating in its active region. Then, the base-emitter junction is forward biased. V BEQ ≈ 700 mV [Si] I CQ = β ⋅ I BQ 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 = 2.202 - 0.7 V BB - V BEQ = 49.76 µA = 9982 + (100 + 1)(200) RB + [ β +1 ] RE −6 I CQ = β I BQ = 100 ⋅ 49.76 ⋅10 = 4.976 mA −6 I EQ = (β + 1)⋅ I BQ = (100 + 1) ⋅ 49.76 ⋅10 = 5.026 mA KVL : - I EQ R E - V CEQ - I CQ RC + V CC = 0 V CEQ = V CC - I CQ RC - I EQ R E = 15 − 4.976 ⋅ 4 − 5.026 ⋅ 0.2 = − 5.91 V The collector-emitter voltage is less (more negative) than its saturation value (+ 0.3 V for Silicon). Therefore the initial assumption (operation in the active region) was incorrect and the solution is not valid. The device is operating in the saturation region with a saturation collector-emitter voltage equal to 0.3 V. The solutions for the collector and emitter currents are invalid. THE VALID SOLUTION: KVL : - I EQ R E - vCE -SAT - I CQ RC + V CC = 0 I EQ = I BQ + I CQ - [ I BQ + I CQ ] R E - vCE -SAT - I CQ RC + V CC = 0 I CQ = 15 − 49.76 ⋅ 0.2 ⋅10 −3 − 0.3 V CC - I BQ R E - vCE - SAT = = 3.498 mA 200 + 4000 R E + RC The solution for the base current is valid. The value of beta given is not valid. ______________________________________________________________________________________ 10.12 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Problem 10.17 Solution: Known quantities: For the circuit shown in Figure P10.17: VCC = 12 V β = 130 R1 = 82 kΩ R2 = 22 kΩ RE = 0.5 kΩ RL = 16Ω . 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 R2 = = 2.538 V R1 + R2 82 + 22 R1 R2 = 82 ⋅ 22 = 17.35 kΩ R B = Req = 82 + 22 R1 + R2 VD : V BB = V TH = V OC = Suppress V CC : 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. 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 = 2.538 − 0.7 V BB - V BEQ = = 22.18µA 17350 + (130 + 1) ⋅ 500 R B + (β + 1) ⋅ R E −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. ______________________________________________________________________________________ Problem 10.18 Solution: Known quantities: For the circuit shown in Figure P10.18: 10.13 G. Rizzoni, Principles and Applications of Electrical Engineering VCC = 12 V RL = 6 kΩ VEE = 4 V Rs = 0.6 kΩ β = 100 [ RB = 100 kΩ 3 ] Problem solutions, Chapter 10 RC = 3 kΩ RE = 3 kΩ vS = cos 6.28 ×10 ⋅ 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 and: V BEQ ≈ 700 mV [Si] I CQ = β ⋅ I BQ I EQ = (β + 1) I BQ KVL : - V EE + I BQ R B + V BEQ + I EQ R E = 0 4 - 0.7 V EE - V BEQ = 8.189 µ A = I BQ = 100000 + (100 + 1)(3000) RB + [ β +1 ] RE I CQ = β ⋅ I BQ = (100 ) ⋅ 8.189 ⋅10 −6 = 818.9 mA I EQ = (β + 1) ⋅ I BQ = (100 + 1) ⋅ 8.189 ⋅ 10 −6 = 827.0 mA KVL : + V EE - I EQ R E - V CEQ - I CQ RC + V CC = 0 V CEQ = V EE + V CC - I CQ RC - I EQ R E = 4 + 12 − 818.9 ⋅ 3 ⋅ 10 −3 − 827.0 ⋅ 3 ⋅10 −3 = 11.06 V 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. ______________________________________________________________________________________ Problem 10.19 Solution: Known quantities: For the circuit shown in Figure P10.19: VCC = 12 V β = 130 RB = 325 kΩ RC = 1.9 kΩ 10.14 RE = 2.3 kΩ G. Rizzoni, Principles and Applications of Electrical Engineering RL = 10 kΩ Rs = 0.5 kΩ [ Problem solutions, Chapter 10 ] vS = 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. V BEQ ≈ 700 mV [Si] I CQ = β I BQ I EQ = [ β + 1 ] I BQ KCL : I BQ + I CQ - I RC = 0 I RC = I CQ + I BQ = β I BQ + I BQ = [ β + 1 ] I BQ KVL : - I EQ R E - V BEQ - I BQ R B - I RC RC + V CC = 0 - [ β + 1 ] I BQ [ R E + RC ] - V BEQ - I BQ R B + V CC = 0 12 − 0.7 V CC - V BEQ = = 12.91µA [325 + (130 + 1) ⋅ (2.3 + 1.9) ]⋅103 R B + [ β + 1 ] [ R E + RC ] = I EQ = [ β + 1 ] I BQ = (130 + 1) ⋅12.91 ⋅10 −6 = 1.691mA I BQ = I RC KVL : - I EQ R E - V CEQ - I RC RC + V CC = 0 V CEQ = V CC - I RC RC - I EQ R E = 12 − 1.691 ⋅1.9 − 1.691 ⋅ 2.3 = 4.896 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. ______________________________________________________________________________________ Problem 10.20 Solution: Known quantities: For the circuit shown in Figure P10.19: VCC = 15 V C = 0.5µF β = 170 RL = 1.7 kΩ Rs = 70Ω vS = cos 6.28 ×103 ⋅ t mV . [ RB = 22 kΩ ] 10.15 RC = 3.3 kΩ RE = 3.3 kΩ G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Find: VCEQ and the region of operation. Analysis: When β is very high (say, greater than 50) the "high beta approximation" can be used, i.e.: IC ≈ I E ≈ β I B Using this approximation in this problem: I BQ ≈ 12.5µA I CQ ≈ I EQ ≈ 2.125 mA V CEQ ≈ 0.975 V The collector-emitter voltage is greater than the saturation value (but not by much) so the original assumption (operation in the active region) and the solution are valid. However, when a signal is introduced into the circuit, the collector-emitter voltage will vary about its Q point value. Since the Q point is close to saturation, saturation and severe distortion due to clipping is likely. ______________________________________________________________________________________ Problem 10.21 Solution: Known quantities: • For the circuit shown in Figure P10.14: VCC = 15 V RE = 710Ω RL = 3 kΩ • • C = 0.47µ F R1 = 220 kΩ R2 = 55 kΩ Rs = 0.6 kΩ vS = Vi 0 ⋅ sin (ωt ) RC = 3 kΩ Vi 0 = 10 mV . DC operating point: I BQ = 19.9µ A V CEQ = 7.61 V Transfer characteristic and β of the npn silicon transistor: v BE V BEQ + vbe IC ≈ I s eVT = I s e VT β = 100 • Device i-v characteristic plotted in Figure P9.21. Find: a) The no-load large signal gain [v0/vi]. b) Sketch the waveform of the output voltage as a function of time. c) How the output voltage is distorted compared with the input waveform. Analysis: a) Simplify the circuit by determining the Thèvenin equivalent of the biasing network in the base circuit. VD : V BB = V TH = 15 ⋅ 55 V CC R2 = = 3.00 V R1 + R2 220 + 55 Suppress V CC : 3 R1 R2 = 220 ⋅ 55 ⋅10 = 44.0 kΩ R B = Req = 220 + 55 R1 + R2 First, determine the collector and emitter Q point: 10.16 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 KVL : - I EQ R E - V CEQ - I CQ RC + V CC = 0 V CEQ = − 2.010 ⋅ 0.710 − 1.990 ⋅ 3 + 15 = 7.603 V −6 I CQ = β I BQ = 100 ⋅19.9 ⋅10 = 1.99 mA −6 I EQ = [ β + 1 ] I BQ = 101 ⋅19.9 ⋅10 = 2.01mA Now, determine the DC load line (circuit characteristic). KVL : - i E R E - vCE - iC RC + V CC = 0 iC = iE = β +1 iC β V CC - vCE V CC - vCE = β +1 3717 R E + RC β If : vCE = V CC = 15 V [Intercept 1] = 0 ° =® 15 °̄ = 3717 = 4.035 mA If : vCE = 0 [Intercept 2] Plot the DC load line on the same plot as the transistor i-v characteristic. Note that the collector current is a linear function of the collector-emitter voltage. Only two points (the two intercepts shown as Points 1 and 2 on the plot) are required to plot a linear function. The DC load line should pass through the Q point. The AC load line can now be plotted. It will pass through the Q point as does the DC load line; however, its intercepts cannot be directly determined. Instead, its slope will be determined. The slope and the Q point can then be used to plot the AC load line. Slope AC = - 1 = - RC 1 mA 2mA ∆ iC = - 0.3333 = = 3000 V 6V ∆ vCE For AC the capacitors act as short circuits. The bypass capacitor in parallel with the emitter resistor shorts out any AC voltage across it. The collector coupling capacitor acts as a short and connects the load resistor to the circuit. However, no-load conditions are specified, i.e., the load current is zero or the load resistor is replaced by an open circuit. With no signal, the circuit operates at its Q point. When a signal is introduced, the operating point changes along the AC load line as a function of the signal voltage. Unfortunately, the device i-v characteristic is plotted as a function of input (base) current instead of voltage. But, there is the static characteristic that relates collector current to base-emitter voltage: v BE iC = I s e V T = I s e V BEQ + v be VT = [ Is e V BEQ VT v be v be ] eV T = I CQ eV T v be iC = I CQ vbe = iB = eV T I BQ eV T β β The base-emitter signal voltage is related to the input signal voltage. At sufficiently high frequencies (the "mid"-frequency range) the capacitors can be modeled as short circuits for AC. This means that there is no AC voltage across either the AC coupling (DC blocking) capacitor or the capacitor bypassing the emitter resistor. Using superposition to sum only AC voltages around the loop: 10.17 G. Rizzoni, Principles and Applications of Electrical Engineering KVL : - vi + vc + vbe + ve = 0 iB = vbe I BQ eV T vbe = vi vc = ve = 0 = [ 19.9 µA ] Problem solutions, Chapter 10 vi e 26 mV This transfer function can now be used to give the base current as a function of the input voltage. The intersection of the base current curves with the AC load line give the corresponding values of the collector current and collector-emitter voltage (points 3, Q and 4 on the plot): ωt [rad] 0,π,2π π/2 3π/2 vi [mV] 0 +10 -10 iB [µ µA] 19.9 29.23 13.54 iC [mA] 1.99 2.92 1.35 vCE [V] 7.61 4.55 9.15 vo [V] 0 -3.06 +1.54 The capacitor connecting the load to the circuit is a DC blocking [and AC coupling] capacitor. Remember that for no load conditions, the load resistor is assumed to be open. Again using superposition to sum only AC voltages around the loop and modeling the capacitors as shorts: - ve + vce + vc + vo = 0 ve = vc = 0 vo = vce = vCE - V CEQ The no-load voltage gain can be determined using the maximum excursion in the input and output voltages: Avo = ∆ vo (− 1.54) − (+ 3.06) = −230 = ∆ vi 10 ⋅10 −3 − − 10 ⋅10 −3 ( ) ( ) When the input voltage becomes more positive, the output voltage becomes more negative and vice versa. This is called "phase inversion" and is why the gain is negative. In normal amplification this is not important. If the loading effect of the load resistance is included, the AC load line will be steeper and the gain will be reduced. b) A plot is a large number of calculated points plotted and connected by a smooth curve. In a sketch, only the most significant points [maxima, minima, intercepts, etc] are calculated, plotted, and connected by a carefully drawn smooth curve. c) The output waveform is very distorted, i.e., it is not a true linearly amplified copy of the input waveform. This is most obvious when the positive and negative peaks of the two curves are compared: Input voltage +10 mV -10 mV Output voltage -3.06 V +1.94 V The two peaks of the input voltage are equal; however, the positive peak is amplified less than the negative peak. The nonlinear (exponential) behavior of the transistor causes the amplification to be nonlinear and the output waveform to be very distorted. This is characteristic of large signal amplifiers. Even more serious distortion could occur if the input voltage (also called the "excitation" or "drive") were increased so that saturation or cutoff occurs. This causes "clipping" and a severely distorted output. ______________________________________________________________________________________ 10.18 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Section 10.3: BJT Large-Signal Model Problem 10.22 Solution: Known quantities: For the circuit shown in Figure 10.24: Voff = 0 V, Von = 5 V, I B = 5 mA, RB = 1 kΩ, 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 RC. Analysis: RC = VCC − Vγ LED − VCEsat I LED ≤ 5 − 1.4 − 0.2 = 340 Ω 0.01 From the maximum power I LED max = RC > Pmax 0.1 = = 71 mA Vγ LED 1.4 VCC − Vγ LED − VCEsat I LED max = 47 Ω Therefore, RC ∈[47, 340] Ω ______________________________________________________________________________________ Problem 10.23 Solution: Known quantities: For the circuit shown in Figure 10.26: VD = 1.1 V, RB = 33 kΩ, 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 IB = VD − VBEQ RB = 1.1 − 0.75 = 10.6 µA 33000 The current through RS is IS = VCEQ − VD RS = 6 − 1.1 = 9.8 mA 500 It follows that the current through the resistance RC is I CQ = β I B + I S = 11.8 mA Finally, RC = VCC − VCEQ I CQ = 12 − 6 = 508.5 Ω 0.0118 ______________________________________________________________________________________ 10.19 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Problem 10.24 Solution: Known quantities: For the circuit shown in Figure 10.24: Voff = 0 V, Von = 5 V, I B max = 5 mA, RC = 340 Ω, 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 IC = VCC − Vγ LED − VCEsat RC = 10 mA In order to guarantee that the BJT is in saturation RB ≤ RB ≥ Von − Vγ IC / β Von − Vγ I B max = 5 − 0.7 = 40.85 kΩ 0.01 95 = 860 Ω ______________________________________________________________________________________ Problem 10.25 Solution: Known quantities: For the circuit shown in Figure 10.24: Voff = 0 V, Von = 5 V, I B max = 5 mA, RB = 10 kΩ, RC = 340Ω, VCC = 5 V, Vγ = 0.7V, VCEsat = 0.2 V, Vγ LED = 1.4 V, I LED ≥ 10 mA, Pmax = 100 mW Find: Minimum value of β that will ensure the correct operation of the LED. Analysis: IB = β min Von − Vγ RB I = LED min IB 4.3 = 0.43 mA 10000 0.01 = = 23.25 0.43 ⋅ 10 −3 = ______________________________________________________________________________________ Problem 10.26 Solution: Known quantities: For the circuit shown in Figure 10.24: Voff = 0 V, Von = 3.3 V, I B max = 5 mA, RB = 10 kΩ, RC = 340Ω, VCC = 5 V, Vγ = 0.7V, VCEsat = 0.2 V, Vγ LED = 1.4 V, I LED ≥ 10mA, Pmax = 100mW 10.20 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Find: Minimum value of β that will ensure the correct operation of the LED. Analysis: IB = β min Von − Vγ RB I = LED min IB 3.3 − 0.7 = 0.26 mA 10000 0.01 = = 38.5 0.26 ⋅ 10 −3 = ______________________________________________________________________________________ Problem 10.27 Solution: Known quantities: For the circuit shown in Figure 10.24: Voff = 0 V, Von = 5 V, I B max = 1 mA, RB = 1 kΩ, R = 12 Ω, 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 = β min VCC − VCEsat 13 − 1 = = 1A 12 R I 1 = C = = 1000 I B max 1 ⋅ 10 −3 ______________________________________________________________________________ Problem 10.28 Solution: Known quantities: For the circuit shown in Figure 10.24: Voff = 0 V, Von = 5 V, I B max = 1 mA, β = 2000, R = 12 Ω, 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. 10.21 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Analysis: If the BJT is in saturation IC = VCC − VCEsat = 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 RB ≤ RB ≥ Von − Vγ IC / β Von − Vγ I B max = 5 − 0.7 = 8.6 kΩ 1 2000 = 4.3 kΩ ______________________________________________________________________________________ Problem 10.29 Solution: Known quantities: For the circuit shown in Figure 10.24: Voff = 0 V, Von = 3.3 V, I B max = 1 mA, β = 2000, R = 12 Ω, 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 IC = VCC − VCEsat = 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 RB ≤ RB ≥ Von − Vγ IC / β Von − Vγ I B max = 3.3 − 0.7 = 5.2 kΩ 1 2000 = 2.6 kΩ ______________________________________________________________________________________ 10.22 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Section 10.4: BJT Switches and Gates Problem 10.30 Solution: Known quantities: The circuit given in Figure P10.30. Find: Show that the given circuit functions as an OR gate if the output is taken at v01. Analysis: Construct a state table. This table clearly describes an AND gate when the output is taken at vo1 . v1 v2 Q1 Q2 Q3 vo1 vo 2 0 0 off off on 0 5V 0 5V off on off 5V 0 5V 0 on off off 5V 0 5V 5V on on off 5V 0 ______________________________________________________________________________________ Problem 10.31 Solution: Known quantities: The circuit given in Figure P10.30. Find: Show that the given circuit functions as a NOR gate if the output is taken at v02. Analysis: See the state table constructed for Problem 10.30. This table clearly describes a NOR gate when the output is taken at vo 2 . ______________________________________________________________________________________ Problem 10.32 Solution: Known quantities: The circuit given in Figure P10.32. Find: Show that the given circuit functions as an AND gate if the output is taken at v01. Analysis: Construct a state table. This table clearly describes an AND gate when the output is taken at vo1 . 10.23 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 v1 v2 Q1 Q2 Q3 vo1 vo 2 0 0 off off on 0 5V 0 5V off on on 0 5V 5V 0 on off on 0 5V 5V 5V on on off 5V 0 ______________________________________________________________________________________ Problem 10.33 Solution: Known quantities: The circuit given in Figure P10.32. Find: Show that the given circuit functions as a NAND gate if the output is taken at v02. Analysis: See the state table constructed for Problem 10.32. This table clearly describes a NAND gate when the output is taken at vo 2 . ______________________________________________________________________________________ Problem 10.34 Solution: Known quantities: In the circuit given in Figure P10.34 the minimum value of vin for a high input is 2.0 transistor Q1 has a β of at least 10. V. Assume that the Find: The range for resistor RB that can guarantee that the transistor is on. Analysis: 5 − 0 .2 = 2.4 mA , therefore, iB = iC/β = 0.24 mA. 2000 (vin)min = 2.0 V and (vin)max = 5.0 V, therefore, applying KVL: -vin +RB iB + 0.6 = 0 v − 0 .6 or . Substituting for (vin)min and (vin)max , we find the following range for RB: RB = in iB ic = 5.833 kΩ ≤ RB ≤ 18.333k Ω ______________________________________________________________________________________ Problem 10.35 Solution: Known quantities: For the circuit given in Figure P10.35: R1C = R2C = 10 kΩ , R1B = R2 B = 27 kΩ . Find: a) vB, vout, and the state of the transistor Q1 when vin is low. b) vB, vout, and the state of the transistor Q1 when vin is high. 10.24 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Analysis: vin is low Q1 is cutoff vB = 5 V Q2 is in saturation vout = low = 0.2 V. b) vin is high Q1 is in saturation vB = 0.2 V Q2 is cutoff vout = high = 5 V. a) ______________________________________________________________________________________ Problem 10.36 Solution: Known quantities: For the inverter given in Figure P10.36: RC1 = RC 2 = 2 kΩ , RB = 5 kΩ . Find: The minimum values of β1 and β2 to ensure that Q1 and Q2 saturate when vin is high. Analysis: 5 − 0 .2 2.5 = 2.4 mA , therefore, ic = mA . Applying KVL: 2000 β − 5 + RB i B1 + 0.6 + 0.6 + 0.6 = 0 + i or 0.64 ⋅ β1 = 1.2 + 2.5 Therefore, iB1= 0.64 mA. i E1 = β1 ⋅ i B1 = 600 500 B 2 β2 ic = Choose β2 = 10 β1 = 2.27. ______________________________________________________________________________________ Problem 10.37 Solution: Known quantities: For the inverter given in Figure P10.36: RC1 = 2.5 kΩ , Find: Show that Q1 saturates when vin is high. Find a condition for RC 2 = 2 kΩ , β1 = β 2 = 4 . RC 2 to ensure that Q2 also saturates. Analysis: 3 .2 = 0.8 mA iC1 = 3.2 mA 4000 600 Applying KCL: + i B 2 = 3.2 i B 2 = 2 mA ; iC 2 = β ⋅ i B 2 = 8 mA 500 Applying KVL: 5 − 0.2 = 0.008 ⋅ RC 2 RC 2 = 600Ω i B1 = ______________________________________________________________________________________ Problem 10.38 Solution: Known quantities: The basic circuit of a TTL gate, shown in Figure P10.38. Find: The logic function performed by this circuit. Analysis: The circuit performs the function of a 2-input NAND gate. The analysis is similar to Example 10.8. 10.25 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 ______________________________________________________________________________________ Problem 10.39 Solution: Known quantities: The circuit diagram of a three-input TTL NAND gate, given in Figure P10.39. Find: vB1, vB2, vB3, vC2, and vout, assuming that all the input voltages are high. Analysis: Q2 and Q3 conduct, while Q4 is cutoff. vB1 = 1.8 V, vB2 = 1.2 V, vB3 = 0.6 V, and vC2 = vout = 0.2 V. ______________________________________________________________________________________ Problem 10.40 Solution: Known quantities: Figure P10.40. Find: Show that when two or more emitter-follower outputs are connected to a common load, as shown in Figure P10.54, the OR operation results; that is, v0 = v1 OR v2. Analysis: v2 v1 Q1 Q2 v0 L L L L L L H H L H H L L H H H H H H H L : Low; H : High. ______________________________________________________________________________________ 10.26

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