Solucionario Malvino Principios de Electronica 7o Edicicion

January 15, 2026 | Author: Anonymous | Category: N/A

Share Embed Donate

Report this link

Short Description

Download Solucionario Malvino Principios de Electronica 7o Edicicion...

Description

lOMoARcPSD|6455127

Solucionario. Malvino - Principios de Electronica 7º Edicicion Circuitos Amplificadores (Universidad Tecnológica del Perú)

Scan to open on Studocu

Studocu is not sponsored or endorsed by any college or university Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Instructor’s Manual to accompany

Electronic Principles Seventh Edition

Albert Malvino

David J. Bates Western Technical College

Prepared by

Patrick Hoppe

Boston Burr Ridge, IL Dubuque, IA Madison, WI New York San Francisco St. Louis Bangkok Bogot Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal New Delhi Santiago Seoul Singapore Sydney Taipei Toronto

Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Contents PREFACE PART 1

iv ELECTRONIC PRINCIPLES, SEVENTH EDITION SOLUTIONS TO TEXT PROBLEMS

PART 2

EXPERIMENTS MANUAL DATA FOR EXPERIMENTS AND LABORATORY QUIZZES

PART 3

1-1

2-1

TRANSPARENCY MASTERS TEXT FIGURES AND DATA SHEETS

3-1

iii Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Preface To best serve the needs of the instructor, the solutions or answers for the problems contained in the Malvino/Bates student text and experiments manual have been combined in this single volume. This instructor’s manual has been designed to provide you, the instructor, with a convenient reference source for answers to both even- and odd- numbered exercises. The sections of the Instructor’s Manual for Electronic Principles, Seventh Edition, are as follows: 1. Solutions to problems in Electronic Principles, Seventh Edition. Here you will find solutions for all the questions and problems at the end of the textbook chapters. In most cases, complete worked-out solutions are provided for your convenience. 2. Answers for the experiments manual. This part contains representative data for all experiments. Also included are the answers to the questions at the end of each experiment. 3. Transparency masters. Included in this section are 33 figures from the student text, along with manufacturers’ data sheets for popular devices for use as transparency masters or for duplication for student use. Albert Paul Malvino David J. Bates

iv Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Part

1

Electronic Principles Seventh Edition Chapter 1 Introduction SELF-TEST 1. a 2. c 3. a 4. b 5. d 6. d

7. b 8. c 9. b 10. a 11. a 12. a

13. c 14. d 15. b 16. b 17. a 18. b

19. b 20. c 21. b 22. b 23. c

give us more insight into how changes in load resistance affect the load voltage. 12. It is usually easy to measure open-circuit voltage and shorted-load current. By using a load resistor and measuring voltage under load, it is easy to calculate the Thevenin or Norton resistance.

PROBLEMS 1-1.

V = 12 V RS = 0.1 Ω

JOB INTERVIEW QUESTIONS

Solution:

Note: The text and illustrations cover many of the job interview questions in detail. An answer is given to job interview questions only when the text has insufficient information. 2. It depends on how accurate your calculations need to be. If an accuracy of 1 percent is adequate, you should include the source resistance whenever it is greater than 1 percent of the load resistance. 5. Measure the open-load voltage to get the Thevenin voltage VTH. To get the Thevenin resistance, reduce all sources to zero and measure the resistance between the AB terminals to get RTH. If this is not possible, measure the voltage VL across a load resistor and calculate the load current IL. Then divide VTH – VL by IL to get RTH. 6. The advantage of a 50 Ω voltage source over a 600 Ω voltage source is the ability to be a stiff voltage source to a lower value resistance load. The load must be 100 greater than the internal resistance in order for the voltage source to be considered stiff. 7. The expression cold-cranking amperes refers to the amount of current a car battery can deliver in freezing weather when it is needed most. What limits actual current is the Thevenin resistance caused by chemical and physical parameters inside the battery, not to mention the quality of the connections outside. 8. It means that the load resistance is not large compared to the Thevenin resistance, so that a large load current exists. 9. Ideal. Because troubles usually produce large changes in voltage and current, so that the ideal approximation is adequate for most troubles. 10. You should infer nothing from a reading that is only 5 percent from the ideal value. Actual circuit troubles will usually cause large changes in circuit voltages. Small changes can result from component variations that are still within the allowable tolerance. 11. Either may be able to simplify the analysis, save time when calculating load current for several load resistances, and

Given:

RL = 100RS RL = 100(0.1 Ω) RL = 10 Ω Answer: The voltage source will appear stiff for values of load resistance of g10 Ω. 1-2.

Given: RLmin = 270 Ω RLmax = 100 kΩ Solution: RS < 0.01 RL RS < 0.01(270 Ω) RS < 2.7 Ω

(Eq. 1-1)

Answer: The largest internal resistance the source can have is 2.7 Ω. 1-3.

Given: RS = 50 Ω Solution: RL = 100RS RL = 100(50 Ω) RL = 5 kΩ Answer: The function generator will appear stiff for values of load resistance of g5 kΩ.

1-4.

Given: RS = 0.04 Ω Solution: RL = 100RS RL = 100(0.04 Ω) RL = 4 Ω Answer: The car battery will appear stiff for values of load resistance of g 4 Ω.

1-1 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

1-5.

1-6.

Given:

Solution:

RS = 0.05 Ω I=2A Solution:

RL = 0.01RS RL = 0.01(250 kΩ) RL = 2.5 kΩ

V = IR (Ohm’s law) V = (2 A)(0.05 Ω) V = 0.1 V

IL = IT(RS)/(RS + RL)] (Current divider formula) IL = 5 mA[(250 kΩ)/(250 kΩ/(250 kΩ + 10 kΩ)] IL = 4.80 mA

Answer: The voltage drop across the internal resistance is 0.1 V.

Answer: The load current is 4.80 mA, and, no, the current source is not stiff since the load resistance is not less than or equal to 2.5 kΩ.

Given:

1-12. Solution:

V=9V RS = 0.4 Ω

VTH = VR2 VR2 = VS[(R2)/(R1 + R2)] (Voltage divider formula) VR2 = 36 V[(3 kΩ)/(6 kΩ + 3 kΩ)] VR2 = 12 V

Solution: I = V/R (Ohm’s law) I = (9 V)/(0.4 Ω) I = 22.5 A Answer: The load current is 22.5 A. 1-7.

(Eq. 1-4)

RTH = [R1R2/R1 + R2] (Parallel resistance formula) RTH = [(6 kΩ)(3 kΩ)/(6 kΩ + 3 kΩ)] RTH = 2 kΩ Answer: The Thevenin voltage is 12 V, and the Thevenin resistance is 2 kΩ.

Given: IS = 10 mA RS = 10 MΩ Solution: RL = 0.01 RS RL = 0.01(10 MΩ) RL = 100 kΩ Answer: The current source will appear stiff for load resistance of f 100 kΩ.

1-8.

Given: RLmin = 270 Ω RLmax = 100 kΩ Solution: RS > 100 RL RS > 100(100 kΩ) RS > 10 MΩ

(Eq. 1-3)

Answer: The internal resistance of the source is greater than 10 MΩ. 1-9.

Given: RS = 100 kΩ Solution: RL = 0.01RS RL = 0.01(100 kΩ) RL = 1 kΩ

(Eq. 1-4) (a) Circuit for finding VTH in Prob. 1-12. (b) Circuit for finding RTH in Prob. 1-12.

Answer: The maximum load resistance for the current source to appear stiff is 1 kΩ. IS = 20 mA RS = 200 kΩ RL = 0 Ω

Solution:

Solution: RL = 0.01RS RL = 0.01(200 kΩ) RL = 2 kΩ Answer: Since 0 Ω is less than the maximum load resistance of 2 kΩ, the current source appears stiff; thus the current is 20 mA. 1-11. Given: I = 5 mA RS = 250 kΩ RL = 10 kΩ

1-13. Given: VTH = 12 V RTH = 2 kΩ

1-10. Given:

I = V/R (Ohm’s law) I = VTH/(RTH + RL) I0Ω = 12 V/(2 kΩ + 0 Ω) = 6 mA I1kΩ = 12 V/(2 kΩ + 1 kΩ) = 4 mA I2kΩ = 12 V/(2 kΩ + 2 kΩ) = 3 mA I3kΩ = 12 V/(2 kΩ + 3 kΩ) = 2.4 mA I4kΩ = 12 V/(2 kΩ + 4 kΩ) = 2 mA I5kΩ = 12 V/(2 kΩ + 5 kΩ) = 1.7 mA I6kΩ = 12 V/(2 kΩ + 6 kΩ) = 1.5 mA Answers: 0 Ω 6 mA; 1 kΩ, 4 mA; 2 kΩ, 3mA; 3 kΩ, 2.4 mA; 4 kΩ, 2 mA; 5 kΩ, 1.7 mA; 6 kΩ, 1.5 mA.

1-2 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Solution: RN = RTH RTH = 10 kΩ

Thevenin equivalent circuit for Prob. 1-13.

(Eq. 1-10)

IN = VTH/RTH (Eq. 1-12) VTH = INRN VTH = (10 mA)(10 kΩ) VTH = 100 V Answer: RTH = 10 kΩ, and VTH = 100 V

1-14. Given: VS = 18 V R1 = 6 kΩ R2 = 3 kΩ Solution: VTH = VR2 (Voltage divider formula) VR2 = VS[(R2)/(R1 + R2)] VR2 = 18 V[(3 kΩ)/(6 kΩ + 3 kΩ)] VR2 = 6 V RTH = [(R1 × R2)/(R1 + R2)] (Parallel resistance formula) RTH = [(6 kΩ × 3 kΩ)/(6 kΩ + 3 kΩ)] RTH = 2 kΩ Answer: The Thevenin voltage decreases to 6V, and the Thevenin resistance is unchanged. 1-15. Given:

Thevenin circuit for Prob. 1-17.

1-18. Given (from Prob. 1-12): VTH = 12 V RTH = 2 kΩ Solution: (Eq. 1-10) RN = RTH RN = 2 kΩ IN = VTH/RTH (Eq. 1-12) IN = 12 V/2 kΩ IN = 6 mA

VS = 36 V R1 = 12 kΩ R2 = 6 kΩ

Answer: RN = 2 kΩ, and IN = 6 mA

Solution: VTH = VR2 VR2 = VS[(R2)/(R1 + R2)] (Voltage divider formula) VR2 = 36 V[(6 kΩ)/(12 kΩ + 6 kΩ)] VR2 = 12 V RTH = [(R1R2)/(R1 + R2)] (Parallel resistance formula) RTH = [(12 kΩ)(6 kΩ)/(12 kΩ + 6 kΩ)] RTH = 4 kΩ Answer: The Thevenin voltage is unchanged, and the Thevenin resistance doubles. 1-16. Given: VTH = 12 V RTH = 3 kΩ Solution: RN = RTH RN = 3 kΩ IN = VTH/RTH IN = 15 V/3 kΩ IN = 4 mA Answer: IN = 4 mA, and RN = 3 kΩ

IN 4mA

RN 3kΩ

Norton circuit for Prob. 1-18.

1-19. Shorted, which would cause load resistor to be connected across the voltage source seeing all of the voltage. 1-20. a. R1 is open, preventing any of the voltage from reaching the load resistor. b. R2 is shorted, making its voltage drop zero. Since the load resistor is in parallel with R2, its voltage drop would also be zero. 1-21. The battery or interconnecting wiring. 1-22. RTH = 2 kΩ Solution: RMeter = 100RTH RMeter = 100(2 kΩ) RMeter = 200 kΩ Answer: The meter will not load down the circuit if the meter impedance is g 200 kΩ.

CRITICAL THINKING 1-23. Given: Norton circuit for Prob. 1-16.

1-17. Given: IN = 10 mA RN = 10 kΩ

VS = 12 V IS = 150 A Solution: RS = (VS)/(IS) RS = (12 V)/(150 A) RS = 80 mΩ

1-3 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Answer: If an ideal 12 V voltage source is shorted and provides 150 A, the internal resistance is 80 mΩ. 1-24. Given: VS = 10 V VL = 9 V RL = 75 Ω

RTH = 500 Ω Answer: The value for R1 and R2 is 1 kΩ. Another possible solution is R1 = R2 = 4 kΩ. Note: The criteria will be satisfied for any resistance value up to 4 kΩ and when both resistors are the same value. 1-31. Given:

Solution:

VS = 30 V VL = 10 V RL > 1 MΩ RS < 0.01RL

VS = VRS + VL VRS = VS – VL VRS = 10 V – 9 V VRS = 1 V

(Kirchhoff’s law)

IRS = IL = VL/RL IRS = 9 V/75 Ω IRS = 120 mA

(Ohm’s law)

Solution:

RS = VRS/IRS RS = 8.33 Ω

(Ohm’s law)

RS < 0.01RL RS < 0.01(1 MΩ) RS < 10 kΩ Since the Thevenin equivalent resistance would be the series resistance, RTH < 10 kΩ.

RS < 0.01 RL (Eq. 1-1) 8.33 Ω < 0.01(75 Ω) 8.33 Ω 100(1 kΩ) RL > 100 kΩ V = IR V = (1 mA)(100 kΩ) V = 100 V Answer: A 100 V battery in series with a 100 kΩ resistor. 1-30. Given: VS = 30 V VL = 15 V RTH < 2 kΩ

1-33. Answer: First, measure the voltage across the terminals. This is the Thevenin voltage. Next, connect a resistor across the terminals. Next, measure the voltage across the resistor. Then, calculate the current through the load resistor. Then, subtract the load voltage from the Thevenin voltage. Then, divide the difference voltage by the current. The result is the Thevenin resistance. 1-34. Solution: Thevenize the circuit. There should be a Thevenin voltage of 0.148 V and a resistance of 6 kΩ. IL = VTH/(RTH + RL)

Solution: Assume a value for one of the resistors. Since the Thevenin resistance is limited to 2 kΩ, pick a value less than 2 kΩ. Assume R2 = 1 kΩ. VL = VS[R2/(R1 + R2)] (Voltage divider formula) R1 = [(VS)(R2)/VL] – R2 R1 = [(30 V)(1 kΩ)/(15 V)] – 1 kΩ R1 = 1 kΩ

IL = 0.148 V/(6 kΩ + 0) IL = 24.7 µA IL = 0.148 V/(6 kΩ + 1 kΩ) IL = 21.1 µA IL = 0.148 V/(6 kΩ + 2 kΩ) IL = 18.5 µA IL = 0.148 V/(6 kΩ + 3 kΩ) IL = 16.4 µA

RTH = (R1R2/R1 + R2) RTH = [(1 kΩ)(1 kΩ)]/(1 kΩ + 1 kΩ)

1-4 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

d. n-type e. p-type

IL = 0.148 V/(6 kΩ + 4 kΩ) IL = 14.8 µA IL = 0.148 V/(6 kΩ + 5 kΩ) IL = 13.5 µA

2-7.

IL = 0.148 V/(6 kΩ + 6 kΩ) IL = 12.3 µA

Solution:

Answer: 0, IL = 24.7 µA; 1 kΩ, IL = 21.1 µA; 2 kΩ, IL = 18.5 µA; 3 kΩ, IL = 16.4 µA; 4 kΩ, IL = 14.8 µA; 5 kΩ, IL = 13.5 µA; 6 kΩ, IL = 12.3 µA. 1-35. Trouble: 1: R1 shorted 2: R1 open or R2 shorted 3: R3 open 4: R3 shorted 5: R2 open or open at point C 6: R4 open or open at point D 7: Open at point E 8: R4 shorted

Chapter Two Semiconductors 2-8.

SELF-TEST 1. d 2. a 3. b 4. b 5. d 6. c 7. b 8. b 9. c 10. a 11. c 12. c 13. b 14. b

15. a 16. b 17. d 18. b 19. a 20. a 21. d 22. a 23. a 24. a 25. d 26. b 27. b 28. b

29. d 30. c 31. a 32. a 33. b 34. a 35. b 36. c 37. c 38. a 39. b 40. a 41. b

42. b 43. b 44. c 45. a 46. c 47. d 48. a 49. a 50. d 51. c 52. b 53. d 54. b

IS(new) = 2(ΔT/10)IS(old) (Eq. 2-5) IS(new) = 2[(75°C – 25°C)/10)] 10 nA IS(new) = 320 nA

2-9.

9. Holes do not flow in a conductor. Conductors allow current flow by virtue of their single outer-shell electron, which is loosely held. When holes reach the end of a semiconductor, they are filled by the conductor’s outer-shell electrons entering at that point. 11. Because the recombination at the junction allows holes and free electrons to flow continuously through the diode.

2-1.

–2

2-2.

–3

2-3.

a. b. c. d.

2-4.

500,000 free electrons

2-5.

a. 5 mA b. 5 mA c. 5 mA

2-6.

a. p-type b. n-type c. p-type

Semiconductor Conductor Semiconductor Conductor

ΔV = (–2 mV/°C)ΔT (Eq. 2-4) ΔV = (–2 mV/°C)(0°C – 25°C) ΔV = 50 mV Vnew = Vold + ΔV Vnew = 0.7 V + 0.05 V Vnew = 0.75 V ΔV = (–2 mV/°C)ΔT (Eq. 2-4) ΔV = (–2 mV/°C)(75°C – 25°C) ΔV = –100 mV Vnew = Vold + ΔV Vnew = 0.7 V – 0.1 V Vnew = 0.6 V Answer: The barrier potential is 0.75 V at 0°C and 0.6 V at 75°C. Given: IS = 10 nA at 25°C Tmin = 0°C – 75°C Tmax = 75°C Solution: IS(new) = 2(ΔT/10)IS(old) (Eq. 2-5) [(0°C – 25°C)/10] 10 nA IS(new) = 2 IS(new) = 1.77 nA

JOB INTERVIEW QUESTIONS

PROBLEMS

Given: Barrier potential at 25°C is 0.7 V Tmin = 25°C Tmin = 75°C

Answer: The saturation current is 1.77 nA at 0°C and 320 nA at 75°C. Given: ISL = 10 nA with a reverse voltage of 10 V New reverse voltage = 100 V Solution: RSL = VR/ISL RSL = 10 V/10 nA RSL = 1000 MΩ ISL = VR/RSL ISL = 100 V/1000 MΩ ISL = 100 nA

Answer: 100 nA. 2-10. Answer: Saturation current is 0.53 µA, and surfaceleakage current is 4.47 µA at 25°C. 2-11. Reduce the saturation current, and minimize the RC time constants.

Chapter 3 Diode Theory SELF-TEST 1. b 2. b 3. c 4. d 5. a 6. b

7. c 8. c 9. a 10. a 11. b

12. b 13. a 14. d 15. a 16. c

17. b 18. b 19. a 20. a 21. c

1-5 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Solution: The diode would be reversed-based and acting as an open. Thus the current would be zero, and the voltage would be source voltage.

Answer: IL = 19.3 mA VL = 19.3 V PL = 372 mW PD = 13.4 mW PT = 386 mW

Answer: VD = 12 V ID = 0 mA

3-15. Given:

3-19. Open

VS = 20 V VD = 0.7 V RL = 2 kΩ

3-20. The diode voltage will be 5 V, and it should burn open the diode. 3-21. The diode is shorted, or the resistor is open.

Solution:

3-22. The voltage of 3 V at the junction of R1 and R2 is normal if it is a voltage divider with nothing in parallel with R2. So, the problem is in the parallel branch. A reading of 0 V at the diode resistor junction indicates either a shorted resistor (not likely) or an open diode. A solder bridge could cause the resistor to appear to be shorted.

IL = VL/RL (Ohm’s law) IL = 19.3 V/2 kΩ IL = 9.65 mA Answer: 9.65 mA 3-16. Given:

3-23. A reverse diode test reading of 1.8 V indicates a leaky diode.

VS = 12 V VD = 0.7 V RL = 470 Ω

3-24. 1N4004

Solution: VS = VD + VL 12 V = 0.7 V + VL VL = 11.3 V

3-25. Cathode band. The arrow points toward the band. (Kirchhoff’s law)

IL = VL/RL (Ohm’s law) IL = 11.3 V/470 Ω IL = 24 mA

CRITICAL THINKING

PL = (VL)(IL) PL = (11.3 V)(24 mA) PL = 271.2 mW

3-27. Given: 1N914: forward 10 mA at 1 V; reverse 25 nA at 20 V 1N4001: forward 1 A at 1.1 V; reverse 10 µA at 50 V 1N1185: forward 10 A at 0.95 V; reverse 4.6 mA at 100 V

PD = (VD)(ID) PD = (0.7 V)(24 mA) PD = 29.2 mW

Solution:

PT = PD + PL PT = 29.2 mW + 271.2 mW PT = 300.4 mW

1N914 forward: R = V/I (Ohm’s law) R = 1 V/10 mA R = 100 Ω

Answer: VL = 11.3 V IL = 24 mA PL = 271.2 mW PD = 29.2 mW PT = 300.4 mW

1N914 reverse: R = V/I (Ohm’s law) R = 20 V/25 nA R = 800 MΩ 1N4001 forward: R = V/I (Ohm’s law) R = 1.1 V/1 A R = 1.1 Ω

3-17. Given: VS = 12 V VD = 0.7 V RL = 940 Ω Solution: VS = VD + VL (Kirchhoff’s law) 12 V = 0.7 V + VL VL = 11.3 V IL = VL/RL (Ohm’s law) IL = 11.3 V/940 Ω IL = 12 mA Answer: 12 mA 3-18. Given: VS = 12 V RL = 470 Ω

3-26. The temperature limit is 175°C, and the temperature of boiling water is 100°C. Therefore, the temperature of the boiling water is less than the maximum temperature and the diode will not be destroyed.

1N4001 reverse: R = V/I (Ohm’s law) R = 50 V/10 µA R = 5 MΩ 1N1185 forward: R = V/I (Ohm’s law) R = 0.95 V/10 A R = 0.095 Ω 1N1185 reverse: R = V/I (Ohm’s law) R = 100 V/4.6 mA R = 21.7 kΩ

1-8 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Answer:

Solution:

1N914: forward R = 100 Ω reverse R = 800 MΩ

(Eq. 3-7) rB = (V2 – V1)(I2 – I1) rB = (1 V – 0.7 V)/(500 mA – 0 mA) rB = 600 mΩ Answer: rB = 600 mΩ

1N4001: forward R = 1.1 Ω reverse R = 5 MΩ

3-31. 1. 2. 3. 4. 5.

1N1185: forward R = 0.095 Ω reverse R = 21.7 kΩ 3-28. Given: VS = 5 V VD = 0.7 V ID = 20 mA

6. 7.

Solution: VR = VS – VD (Kirchhoff’s law) VR = 5 V – 0.7 V VR = 4.3 V R = V/I (Ohm’s law) R = 4.3 V/20 mA R = 215 Ω

8. 9. 10. 11. 12.

Answer: R = 215 Ω 3-29. Given: VD = 0.7 V ID = 10 mA R1 = 30 kΩ R3 = 5 kΩ Solution: Find the voltage required on the parallel branch to achieve a diode current of 0.25 mA. VR = IR3 (Ohm’s law) VR = (0.25 mA)(5 kΩ) VR = 1.25 V V = VR + VD (Kirchhoff’s law) V = 1.25 V + 0.7 V V = 1.95 V

IR = ISL + IS 5 µA = ISL + IS(old) ISL = 5 µA – IS(old) 100 µA = ISL + IS(new) IS(new) = 2(ΔT/10) IS(old) (Eq. 2-6) Substitute formulas 2 and 5 into formula 4. 100 µA = 5 µA – IS(old) + 2(ΔT/10)IS(old) Put in the temperature values. 100 µA = 5 µA – IS(old) + 2 [(100ºC – 25ºC)/10]IS(old) Move the 5 µA to the left side, and simplify the exponent of 2. 95 µA = – IS(old) + 27.5 IS(old) Combine like terms. 95 µA = (27.5– 1)IS(old) 95 µA = (180.02) IS(old) Solve for the variable. IS(old) = 95 µA/(180.02) IS(old) = 0.53 µA

Using formula 3: 13. ISL = 5 µA – IS(old) 14. ISL = 5 µA – 0.53 µA 15. ISL = 4.47 µA Answer: The surface-leakage current is 4.47 µA at 25°C. 3-32. Given: R1 = 30 kΩ R2 = 10 kΩ R3 = 5 kΩ This condition will not occur if the diode is normal. It can be either opened or shorted. If it is shorted, the resistance would be 0 Ω. If it is open, it would be the resistance of the resistors.

This is the voltage at the junction of R1 and R2. Next find the voltage drop across R1.

Solution: The circuit would have R1 and R2 in parallel, and the parallel resistance in series with R3.

VR1 = VS – V (Kirchhoff’s law) VR1 = 12 V – 1.95 V VR1 = 10.05 V

R = [(R1)(R2)]/(R1 + R2) (Parallel resistance formula) R = [(30 kΩ)(10 kΩ)]/(30 kΩ + 10 kΩ) R = 7.5 kΩ RT = 5 kΩ + 7.5 kΩ RT = 12.5 kΩ

I = V/R (Ohm’s law) I = 10.05 V/30 kΩ I = 335 µA Now that the current through R1 is known, this is the total current for the parallel branches. The nest step is to find the current through R2. I 2 = I 1 – ID (Kirchhoff’s law) I2 = 335 µA – 0.25 mA I2 = 85 µA The next step is to use the voltage and current to calculate the resistance. R2 = V/I2 (Ohm’s law) R2 = 1.95 V/85 µA R2 = 23 kΩ Answer: R2 = 23 kΩ 3-30. Given: 500 mA at 1 V 0 mA at 0.7 V

Answer: The resistance would be 12.5 kΩ if the diode is open and 0 Ω if the diode is shorted. 3-33. During normal operation, the 15-V power supply is supplying power to the load. The left diode is forwardbiased and allows the 15-V power supply to supply current to the load. The right diode is reversed-based because 15 V is applied to the cathode and only 12 V is applied to the anode. This blocks the 12-V battery. Once the 15-V power supply is lost, the right diode is no longer reversed-biased, and the 12-V battery can supply current to the load. The left diode will become reverse-biased, preventing any current from going into the 15-V power supply. 3-34. It causes all of them to increase. 3-35. The source voltage does not change, but all other variables decrease. 3-36. I1, I2, and P1. Because the resistance of R2 increased, the total resistance of the voltage divider increases. This causes the current in the voltage divider to decrease. This explains the decreasing of the current. Since the voltage across R1 decreased and I1 decreased, P1 decreases.

1-9 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

3-37. VA, VB, VC, I1, I2, P1, P2. Since R is so large, it has no effect on the voltage divider; therefore, the variables associated with the voltage divider do not change. 3-38. VC, I3, P3. The voltage divider is unaffected. The increase in diode voltage drop will cause the voltage at point C to decrease, thus decreasing the current and power.

Answer: The peak voltage is –21.2 V, the average voltage is –6.74 V, and the dc voltage is –6.74 V. 4-3.

Solution: VP = 1.414 Vrms VP = 1.414 (50 V ac) VP = 70.7 V

Chapter 4 Diode Circuits

Vp(out) = Vp(in) – 0.7 V Vp(out) = 70.0 V

SELF-TEST 1. b 2. a 3. b 4. c 5. c 6. b 7. b

8. c 9. c 10. d 11. b 12. b 13. c

14. a 15. b 16. a 17. d 18. c 19. c

4-4.

7. The LC type is preferable when tighter regulation is required and (or) power cannot be wasted. Examples include transmitters, lab test equipment, and military gear when cost is not of primary concern. The LC filter ideally dissipates no power. In reality, the inductor losses result in some heal. The less costly RC filter consumes power in the resistor. 8. A full-wave rectifier is made up of two back-to-back halfwave rectifiers. 9. When you need a high dc output voltage from the power supply, but a step-up transformer is neither available nor practical in the design. 11. Because a transformer with a high turns ratio produces a few thousand volts, which means more insulation and expense. 13. There is probably a short in the circuit that caused excessive current through the resistor. You have to look at the schematic diagram and test the different components and wiring to try to locate the real trouble.

Answer: The peak voltage is –20.5 V, the average voltage is –6.52 V, and the dc voltage is –6.52 V.

Output waveform for Probs. 4-1 and 4-3. Waveform is negative for Probs. 4-2 and 4-4.

4-5.

4-2.

Vdc = 0.318 V (Eq. 4-2) Vdc = 0.318 (70.0 V) Vdc = 22.3 V Answer: The peak voltage is 70.0 V, the average voltage is 22.3 V, and the dc voltage is 22.3 V. Given: Vin = 15 V ac Solution: VP = 1.414 Vrms VP = 1.414 (15 V ac) VP = –21.2 V Vp(out) = Vp(in) – 0.7 V (Eq. 4-4) Vp(out) = –20.5 V Since the average and the DC values are the same: Vdc = 0.318 V (Eq. 4-2) Vdc = 0.318 (–20.5 V) Vdc = –6.52 V

PROBLEMS Given: Vin = 50 V ac Solution: VP = 1.414 Vrms VP = 1.414 (50 V ac) VP = 70.7 V Vp(out) = Vp(in) (Eq. 4-1) Vp(out) = 70.7 V Since the average and the dc values are the same: Vdc = 0.318 VP (Eq. 4-2) Vdc = 0.318 (70.7 V) Vdc = 22.5 V Answer: The peak voltage is 70.7 V, the average voltage is 22.5 V, and the dc voltage is 22.5 V. Given: Vin = 15 V ac Solution: VP = 1.414 Vrms VP = 1.414 (15 V ac) VP = –21.2 V (Eq. 4-1) Vp(out) = Vp(in) Vp(out) = –21.2 V Since the average and the dc values are the same: (Eq. 4-2) Vdc = 0.318 Vp Vdc = 0.318 (–21.2 V) Vdc = –6.74 V

(Eq. 4-4)

Since the average and the dc values are the same:

20. c 21. a 22. b 23. a 24. c 25. c

JOB INTERVIEW QUESTIONS

4-1.

Given: Vin = 50 V ac

4-6.

Given: Turns ratio = N1/N2 = 6:1 = 6 V1 = 120 Vrms Solution: V2 = V1/(N1/N2) (Eq. 4-5) V2 = 120 Vrms/6 V2 = 20 Vrms VP = (1.414) (Vrms) VP = (1.414) (20 Vrms) VP = 28.28 VP Answer: The secondary voltage is 20 Vrms or 28.28 VP. Given: Turns ratio = N1/N2 =1:12 = 0.083333 V1 = 120 V ac Solution: V2 = V1/(N1/N2) (Eq. 4-5) V2 = 120 V ac/0.083333 V2 = 1440 V ac VP = 1.414 Vrms VP = 1.414 (1440 V ac) VP = 2036.16 V Answer: The secondary rms voltage is 1440 V ac, and the peak voltage is 2036.16 V.

4-7.

Given: Turns ratio = N1/N2 = 8 : 1 = 8 V1 = 120 V ac (rms)

1-10 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Solution: V2 = V1/(N1/N2) V2 = 120 V ac/8 V2 = 15 V ac

(Eq. 4-5)

VP = 1.414 Vrms VP = 1.414 (17.14 V ac) VP = 24.24 V

VP = 1.414 Vrms VP = 1.414 (15 V ac) VP = 21.21 V Vp(out) = Vp(in) Vp(out) = 21.21 V

(Eq. 4-1)

Vdc = 0.318 VP (Eq. 4-2) Vdc = 0.318 (21.21 V) Vdc = 6.74 V Answer: The peak voltage is 21.21 V, and the dc voltage is 6.74 V. 4-8.

Given: Turns ratio = N1/N2 = 8:1 = 8 V1 = 120 V ac (rms) Solution: V2 = V1/(N1/N2) V2 = 120 V ac/8 V2 = 15 V ac

(Eq. 4-5)

VP(in) = 0.5 VP VP(in) = 0.5(24.24 V) VP(in) = 12.12 V Vp(out) = Vp(in) Vp(out) = 12.12 V

(Eq. 4-1)

Since the average and the dc values are the same: Vdc = 0.636 Vp (Eq. 4-6) Vdc = 0.636 (12.12 V) Vdc = 7.71 V Answer: The peak output voltage is 12.12 V, and the dc and average values are 7.71 V. 4-11. Given: Turns ratio = N1/N2 = 7:1 = 7 V1 = 120 V ac Solution: V2 = V1/(N1/N2) V2 = 120 V ac/7 V2 = 17.14 V ac

VP = 1.414 Vrms VP = 1.414 (15 V ac) VP = 21.21 V

4-9.

V2 = 120 V ac/7 V2 = 17.14 V ac

(Eq. 4-5)

Vp(out) = Vp(in) – 0.7 V (Eq. 4-4) Vp(out) = 20.51 V

VP = 1.414 Vrms VP = 1.414 (17.14 V ac) VP = 24.24 V

Vdc = 0.318 VP (Eq. 4-2) Vdc = 0.318 (20.51 V) Vdc = 6.52 V

VP(in) = 0.5 VP VP(in) = 0.5(24.24 V) VP(in) = 12.12 V

Answer: The peak voltage is 20.51 V, and the dc voltage is 6.52 V.

Vp(out) = Vp(in) – 0.7 V Vp(out) = 11.42 V

Given:

Since the average and the dc values are the same:

Turns ratio = N1/N2 = 4:1 = 4 V1 = 120 Vrms

Vdc = 0.636 VP (Eq. 4-6) Vdc = 0.636 (11.42 V) Vdc = 7.26 V

Solution: V2 = V1/(N1/N2) V2 = 120 Vrms/4 V2 = 30 Vrms

(Eq. 4-5)

(Eq. 4-4)

Answer: The peak output voltage is 11.42 V, and the dc and average values are 7.26 V. 4-12. Given:

Since it is a center-tapped transformer, each half of the secondary is half of the total secondary voltage.

Turns ratio = N1/N2 = 8:1 = 8 V1 = 120 V ac

Vupper = ½ V2 Vupper = ½ (30 Vrms) Vupper = 15 Vrms

Solution: V2 = V1/(N1/N2) V2 = 120 V ac/8 V2 = 15 V ac

Vlower = ½ VP Vlower = ½(30 Vrms) Vlower = 15 Vrms

VP(in) = 1.414 Vrms VP(in) = 1.414 (15 V ac) VP(in) = 21.21 V

VP = (1.414) (Vrms) VP = (1.414) (15 Vrms) VP = 21.21 VP

Vp(out) = Vp(in) Vp(out) = 21.21 V

Answer: Each half of the secondary has an rms voltage of 15 V and a peak voltage of 21.21 V. 4-10. Given: Turns ratio = N1/N2 = 7:1 = 7 V1 = 120 V ac Solution: V2 = V1/(N1/N2)

(Eq. 4-5)

(Eq. 4-1)

Since the average and the dc values are the same: Vdc = 0.636 Vp (Eq. 4-6) Vdc = 0.636 (21.21 V) Vdc = 13.49 V Answer: The peak output voltage is 21.21 V, and the dc and average values are 13.49 V.

(Eq. 4-5)

1-11 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Vout = 500 mV

4-13. Given: Turns ratio = N1/N2 = 8:1 = 8 V1 = 120 V ac

Answer: The ripple voltage would be 500 mV. 4-16. Given:

Solution: V2 = V1/(N1/N2) V2 = 120 V ac/8 V2 = 15 V ac

Vin = 14 V XL = 2 kΩ XC = 50 Ω

(Eq. 4-5)

Solution:

VP(in) = 1.414 Vrms VP(in) = 1.414 (15 V ac) VP(in) = 21.21 V Vp(out) = Vp(in) – 1.4 V Vp(out) = 19.81 V

Vout = (XC/XL)Vin (Eq. 4-9) Vout = (50 Ω/2 kΩ)(14 V) Vout = 350 mV (Eq. 4-8)

Since the average and the dc values are the same: Vdc = 0.636 Vp (Eq. 4-6) Vdc = 0.636 (19.81 V) Vdc = 12.60 V Answer: The peak output voltage is 19.81 V, and the dc and average values are 12.60 V.

Output waveform for Probs. 4-10 to 4-13.

Answer: The ripple voltage would be 350 mV. 4-17. Given: Turns ratio = N1/N2 = 8:1 = 8 V1 = 120 V ac RL = 10 kΩ C = 47 µF fin = 60 Hz Solution: V2 = V1/(N1/N2) V2 = 120 V ac/8 V2 = 15 V ac

VP = 1.414 V2 VP = 1.414 (15 V ac) VP = 21.2 V (This is the dc output voltage due to the capacitor input filter.)

4-14. Given: Turns ratio = N1/N2 = 8:1 = 8 V1(max) = 125 V ac V1(min) = 102 V ac Solution: V2(max) = V1(max)/(N1/N2) V2(max) = 125 V ac/8 V2(max) = 15.63 V ac

(Eq. 4-5)

V2(min) = V1(min)/(N1/N2) V2(min) = 105 V ac/8 V2(min) = 13.13 V ac

(Eq. 4-5)

I = V/R (Ohm’s law) I = 21.2 V/10 kΩ I = 2.12 mA fout = fin fout = 60 Hz

VP(in)max = 1.414 V2(max) VP(in)max = 1.414 (15.63 V ac) VP(in)max = 22.10 V

(Eq. 4-1)

Vp(out)min = Vp(in)min Vp(out)min = 18.57 V

(Eq. 4-1)

(Eq. 4-3)

VR = I/(fC) (Eq. 4-10) VR = (2.12 mA)/[(60 Hz)(47 µF)] VR = 752 mV Answer: The dc output voltage is 21.2 V with a 752 mV pp ripple.

VP(in)min = 1.414 V2(min) VP(in)min = 1.414 (13.13 V ac) VP(in)min = 18.57 V Vp(out)max = Vp(in)max Vp(out)max = 22.10 V

(Eq. 4-5)

Output waveform for Prob. 4-17.

4-18. Given:

Vdc(max) = 0.636 Vp(out)max Vdc = 0.636 (22.10 V) Vdc = 14.06 V

(Eq. 4-6)

Turns ratio = N1/N2 = 7:1 = 7 V1 = 120 V ac RL = 2.2 kΩ C = 68 µF fin = 60 Hz

Vdc(min) = 0.636 Vp(out)min Vdc = 0.636 (18.57 V) Vdc = 11.81 V

(Eq. 4-6)

Solution:

Answer: The maximum dc output voltage is 14.06 V, and the minimum is 11.81 V. 4-15. Given: Vin = 20 V XL = 1 kΩ XC = 25 Ω

V2 = V1/(N1/N2) V2 = 120 V ac/7 V2 = 17.14 V ac

(Eq. 4-5)

VP = 1.414 Vrms VP = 1.414 (17.14 V ac) VP = 24.24 V VP(in) = 0.5 VP VP(in) = 0.5(24.24 V) VP(in) = 12.12 V

Solution: Vout = (XC/XL)Vin (Eq. 4-9) Vout = (25 Ω/1 kΩ)(20 V)

Vp(out) = Vp(in) (Eq. 4-1) Vp(out) = 12.12 V (This is the dc output voltage due to the capacitor input filter.)

1-12 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

I = V/R (Ohm’s law) I = 12.12 V/2.2 kΩ I = 5.51 mA fout = 2fin (Eq. 4-7) fout = 2(60 Hz) fout = 120 Hz VR = I/(fC) (Eq. 4-10) VR = (5.51 mA)/[(120 Hz)(68 µF)] VR = 675 mV Answer: The dc output voltage is 12.12 V, with a 675 mV pp ripple. 4-19. Answer: (Eq. 4-10) VR = I/(fC) If the capacitance is cut in half, the denominator is cut in half and the ripple voltage will double. 4-20. Answer: VR = I/(fC) (Eq. 4-10) If the resistance is reduced to 500 Ω, the current increases by a factor of 20; thus the numerator is increased by a factor of 20 and the ripple voltage goes up by a factor of 20. 4-21. Given: Turns ratio = N1/N2 = 9:1 = 9 V1 = 120 V ac RL = 1 kΩ C = 470 µF fin = 60 Hz (Eq. 4-5)

VP = 1.414 Vrms VP = 1.414 (13.33 V ac) VP = 18.85 V Vp(out) = Vp (Eq. 4-1) Vp(out) = 18.85 V (This is the dc output voltage due to the capacitor input filter.) I = V/R (Ohm’s law) I = 18.85 V/1 kΩ I = 18.85 mA fout = 2fin fout = 2(60 Hz) fout = 120 Hz

(Eq. 4-7)

Vp(out) = VP (Eq. 4-1) Vp(out) = 16.50 V (This is the dc output voltage due to the capacitor input filter.) 4-23. Given: VP = 18.85 VP from Prob. 4-21 Solution: PIV = VP PIV = 18.85 V

(Eq. 4-13)

Answer: The peak inverse voltage is 18.85 V. 4-24. Given: Turns ratio = N1/N2 = 3:1 = 3 V1 = 120 Vrms Solution: V2 = V1/(N1/N2) V2 = 120 Vrms/3 V2 = 40 Vrms

(Eq. 4-5)

VP = (1.414) (Vrms) VP = (1.414) (40 Vrms) VP = 56.56 VP PIV = VP (Eq. 4-13) PIV = 56.56 V 4-25. Solution: From the information on p. 114. a. Secondary output is 12.6 V ac. VP = 1.414 Vrms VP = 1.414 (12.6 V ac) VP = 17.8 V b. Vdc = 17.8 V c. I = V/R (Ohm’s law) Idc = 17.8 V ac/1 kΩ Idc = 17.8 mA Rated current is 1.5 A. Answer: The peak output voltage is 17.8 V, and the dc output voltage is 17.8 V. It is not operating at rated current, and thus the secondary voltage will be higher. 4-26. Given:

VR = I/(fC) (Eq. 4-10) VR = (18.85 mA)/[(120 Hz)(470 µF)] VR = 334 mV Answer: The dc output voltage is 18.85 V, with a 334 mV pp ripple.

Output waveform for Probs. 4-18 and 4-21.

4-22. Given: Turns ratio = N1/N2 = 9:1 = 9 V1 = 105 V ac RL = 1 kΩ C = 470 µF fin = 60 Hz Solution: V2 = V1/(N1/N2)

VP = 1.414 Vrms VP = 1.414 (11.67 V ac) VP = 16.50 V

Answer: The peak inverse voltage is 56.56 V.

Solution: V2 = V1/(N1/N2) V2 = 120 V ac/9 V2 = 13.33 V ac

V2 = 105 V ac/9 V2 = 11.67 V ac

(Eq. 4-5)

Assume Pin = Pout Vdc = 17.8 V from Prob. 4-25 Idc = 17.8 mA from Prob. 4-25 Solution: Pout = Idc Vdc Pout = (17.8 mA)(17.8 V) Pout = 317 mW Pin = 317 mW Pin = V1Ipri Ipri = Pin/V1 Ipri = 317 mW/120 V Ipri = 2.64 mA Answer: The primary current would be 2.64 mA. 4-27. Given: VDC = 21.2 V from Prob. 4-17 VDC = 12.12 V from Prob. 4-18 Solution: Fig. 4-40(a)

1-13 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Idiode = V/R Idiode = (2.12 V)/(10 kΩ) Idiode = 2.12 mA Fig. 4-40(b) I = V/R I = (12.12 V)/(2.2 kΩ) I = 5.5 mA Idiode = 0.5 I Idiode = (0.5)/(5.5 mA) Idiode = 2.75 mA Answer: The average diode current in Fig. 4-40(a) is 212 ¿A and the current in Fig. 4-40(b) is 2.75 mA. 4-28 Given: Idc = 18.85 mA from Prob. 4-21 Solution: Idiode = (0.5)Idc Idiode = (0.5)(18.85 mA) Idiode = 9.43 mA 4-29 Given: Vp(out) = 18.85 V from Prob. 4-21 Solution: Without the filter capacitor to maintain the voltage at peak, the dc voltage is calculated the same way it would be done if the filter was not there. Vdc = 0.636 VP Vdc = 0.636(18.85 V) Vdc = 11.99 V Answer: The dc voltage is 11.99 V. 4-30. Answer: With one diode open, one path for current flow is unavailable. The output will look similar to a halfwave rectifier with a capacitor input filter. The dc voltage should not change from the original 18.85 V, but the ripple will increase to approximately double because the frequency drops from 120 to 60 Hz. 4-31. Answer: Since an electrolytic capacitor is polaritysensitive, if it is put in backward, it will be destroyed and the power supply will act as if it did not have a filter. 4-32. Answer: VP will remain the same, DC output equals VP, Vripple = 0 V. 4-33. Answer: Since this is a positive clipper, the maximum positive will be the diode’s forward voltage, and all the negative will be passed through. Maximum positive is 0.7 V, and maximum negative is –50 V.

Solution: Voltage at the cathode is found by using the voltage divider formula. (Eq. 4-18) Vbias = [R1/(R1 + R2)]Vdc Vbias = [1 kΩ/(1 kΩ + 6.8 kΩ)]15 V Vbias = 1.92 V The clipping voltage is the voltage at the cathode and the diode voltage drop. Vclip = 1.92 V + 0.7 V Vclip = 2.62 V Answer: Since it is a positive clipper, the positive voltage is limited to 2.62 V and the negative to –20 V.

Output waveform for Prob. 4-36.

4-37. Answer: The output will always be limited to 2.62 V. 4-38. Answer: Since this is a positive clamper, the maximum negative voltage will be –0.7 V and the maximum positive will be 29.3 V.

Output waveform for Prob. 4-38.

4-39. Answer: Since this is a negative clamper, the maximum positive voltage will be 0.7 V and the maximum negative will be –59.3 V.

Output waveform for Prob. 4-39.

4.40. Answer: The output will be 2VP or Vpp, which is 40 V. If the second approximation is used, the maximum for the clamp will be 39.3 V instead of 40 V, and since there is also a diode voltage drop, the output would be 38.6 V.

Output waveform for Prob. 4-40. Output waveform for Prob. 4-33.

4-34. Answer: Since this is a negative clipper, the maximum negative will be the diode’s forward voltage, and all the positive will be passed through. The maximum positive is 24 V, and the maximum negative is –0.7 V.

Output waveform for Prob. 4-34.

4-35. Answer: The limit in either direction is two diode voltage drops. Maximum positive is 1.4 V, and maximum negative is –1.4 V. 4-36. Given: DC voltage 15 V R1 = 1 kΩ R2 = 6.8 kΩ

4-41. Given: Turns ratio = N1/N2 = 1:10 = 0.1 V1 = 120 V ac Solution: (Eq. 4-5) V2 = V1/(N1/N2) V2 = 120 V ac/0.1 V2 = 1200 V ac VP = 1.414 Vrms VP = 1.414 (1200 V ac) VP = 1696.8 V Since it is a doubler, the output is 2VP. Vout = 2VP Vout = 2 (1696.8 V) Vout = 3393.6 V Answer: The output voltage will be 3393.6 V. 4-42. Given: Turns ratio = N1/N2 = 1:5 = 0.2 V1 = 120 V ac

1-14 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Solution: V2 = V1/(N1/N2) V2 = 120 V ac/0.2 V2 = 600 V ac

(Eq. 4-5)

I = V/R (Ohm’s law) I = 21.21 V/4.7 Ω I = 4.51 A Answer: The maximum surge current will be 4.51 A.

VP = 1.414 Vrms VP = 1.414 (600 V ac) VP = 848.4 V Since it is a tripler, the output is 3VP. Vout = 3VP Vout = 3 (848.4 V) Vout = 2545.2 V Answer: The output voltage will be 2545.2 V. 4-43. Given: Turns ratio = N1/N2 = 1:7 = 0.143 V1 = 120 V ac Solution: V2 = V1/(N1/N2) (Eq. 4-5) V2 = 120 V ac/0.143 V2 = 839.2 V ac VP = 1.414 Vrms VP = 1.414 (839.2 V ac) VP = 1186.6 V Since it is a quadrupler, the output is 4VP. Vout = 4VP Vout = 4(1186.6 V) Vout = 4746.4 V Answer: The output voltage will be 4746.4 V.

4-47. Answer: The signal is a sine wave, and thus the shape of the curve is a function of sine. The formula for the instantaneous voltage at any point on the curve is V = Vpsin¸. Using this formula, calculate the values for each point on the curve, add all 180 of the 1° points together and divide by 180. 4-48. Given: Turns ratio = N1/N2 = 8:1 = 8 V1 = 120 V ac Solution: V2 = V1/(N1/N2) V2 = 120 V ac/8 V2 = 15 V ac

(Eq. 4-5)

VP = 1.414 Vrms VP = 1.414 (15 V ac) VP = 21.21 V With the switch in the shown position, it is a bridge rectifier with a capacitor input filter. Thus the output voltage would be 21.21 V. With the switch in the other position, it is a full-wave rectifier with a capacitor input filter. Since it is a centertapped transformer, the peak voltage is half. VP = 10.6 VP The output would be 10.6 V. Answer: With the switch in the shown position, 21.21 V; with the switch in the other position, 10.6 V.

CRITICAL THINKING 4-44. Answer: If one of the diodes shorts, it will provide a low resistance path to either blow a fuse or damage the other diodes.

4-49. Answer: Both capacitors will charge to approximately 56 mV with opposite polarities. Vout will equal 56 mV – 56 mV. Vout will equal zero volts.

4-45. Given:

4-50. Fault 1—Since the load voltage is 0.636 of the peak voltage, the capacitor input filter is not doing its job; thus the capacitor is bad. Fault 2—Since the load voltage dropped a little and the ripple doubled, one of the diodes is open; this causes the frequence of the ripple to drop to half, which in turn causes the ripple to double. Fault 3—Since V1 is zero, the fuse must be blown. Since the load resistance is zero, the load resistor is shorted. This caused the excessive current in the secondary, which fed back to the primary and blew the fuse. Fault 4—Since V2 is good and all other voltages are bad, the transformer and fuse are good. R and C are good: thus either all four diodes opened (not likely) or there is an open in the ground circuit. Fault 5—Since V1 is zero, the fuse must be blown. Fault 6—The load resistor is open. No current is drawn, and thus there is no ripple. Fault 7—Since V1 is good and V2 is bad, the transformer is the problem. Fault 8—Since V1 is zero, the fuse must be blown. Since the capacitor reads zero, the capacitor is shorted. This caused the excessive current in the secondary, which fed back to the primary and blew the fuse. Fault 9—Since the load voltage is 0.636 of the peak voltage, the capacitor input filter is not doing its job and thus the capacitor is bad.

Turns ratio = N1/N2 = 8: 1 = 8 V1 = 120 V ac Solution: V2 = V1/(N1/N2) V2 = 120 V ac/8 V2 = 15 V ac

(Eq. 4-5)

VP = 1.414 Vrms VP = 1.414 (15 V ac) VP = 21.21 V Since each resistor is in the same current path and both have the same value, they equally divide the voltage. Since they both have a capacitor input filter, they divide the peak voltage. Answer: Each power supply has 10.6 V, but the load connected to the right side of the bridge is a positive 10.6 V and the load connected to the left side is a negative 10.6 V. 4-46. Given: VP = 21.21 VP from Prob. 4-1 R = 4.7 Ω Solution: The maximum surge current would be all of the peak voltage dropped across the resistor.

1-15 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Chapter 5 Special-Purpose Diodes

RS = 470 Ω RL = 15 kΩ

SELF-TEST

Solution:

1. d 2. b 3. b 4. a 5. a 6. c 7. c 8. a

9. c 10. b 11. c 12. a 13. b 14. d 15. d 16. a

17. c 18. c 19. b 20. b 21. a 22. c 23. c 24. c

25. b 26. d 27. a 28. c 29. b 30. b 31. d 32. a

VL = [RL/(RS + RL)]VS (Voltage divider formula) VL = [1.5 kΩ/(470 Ω + 1.5 kΩ)]24 V VL = 18.27 V Answer: The load voltage is 18.27 V. 5-5.

VS = 24 V VZ = 15 V RS = 470 Ω RL = 1.5 kΩ

JOB INTERVIEW QUESTIONS 3. The zener regulation is dropping out of regulation during worst-case conditions of low line voltage and high load current. 4. The LED is connected backward, or the LED current is excessive either because the series resistor is too small or the driving voltage is too high. 5. The basic idea is that a varactor is a voltage-controlled capacitance. By using a varactor as part of an LC tank circuit, we can control the resonant frequency with a dc voltage. 6. To provide a high degree of electrical isolation between input and output circuits. 7. The cathode lead is shorter than the anode lead. Also, the flat side of the dome package is the cathode.

Solution: IS = (VS – VZ)/RS (Eq. 5-3) IS = (24 V – 15 V)/470 Ω IS = 19.15 mA IL = VL/RL IL = 15 V/1.5 kΩ IL = 10 mA

5-2.

5-3.

5-4.

Given: VS = 24 V VZ = 15 V RS = 470 Ω Solution: (Eq. 5-3) IS = IZ = (VS – VZ)/RS IS = IZ = (24 V – 15 V)/)470 Ω) IS = IZ = 19.1 mA Answer: The zener current is 19.1 mA. Given: VS = 40 V VZ = 15 V RS = 470 Ω Solution: (Eq. 5-3) IS = IZ = (VS – VZ)/RS IZ = (40 V – 15 V)/470 Ω IZ = 53.2 mA Answer: The maximum zener current is 53.2 mA. Given: VS = 24 V VZ = 15 V RS = 470 Ω ± 5% RS(max) = 493.5 Ω RS(min) = 446.5 Ω Solution: IS = IZ(max) = (VS – VZ)/RS(min) (Eq. 5-3) IS = IZ = (24 V – 15 V)/(446.5 Ω) IS = IZ = 20.16 mA Answer: The maximum zener current is 20.16mA. Given: VS = 24 V VZ = 15 V

(Eq. 5-5, Ohm’s law)

IZ = I S – IL (Eq. 5-6, Kirchhoff’s current law) IZ = 19.15 mA – 10 mA IZ = 9.15 mA Answer: The series current is 19.15 mA, the zener current is 9.15 mA, and the load current is 10 mA. 5-6.

Given: VS = 24 V VZ = 15 V RS = 470 Ω ± 5% RS(max) = 493.5 Ω RS(min) = 446.5 Ω RL = 1.5 kΩ RL(max) = 1.575 kΩ RL(min) = 1.425 kΩ

PROBLEMS 5-1.

Given:

Solution: Looking at Eq. (5-6), the maximum zener current would occur at a maximum series current and a minimum load current. To achieve these conditions, the series resistance would have to be minimum and the load resistance would have to be maximum. IS = (VS – VZ)/RS(min) (Eq. 5-3) IS = (24 V – 15 V)/446.5 Ω IS = 20.16 mA IL = VL/RL(max) IL = 15 V/1.575 kΩ IL = 9.52 mA

(Eq. 5-5. Ohm’s law)

IZ = I S – IL (Eq. 5-6, Kirchhoff’s current law) IZ = 20.16 mA – 9.52 mA IZ = 10.64 mA Answer: The maximum zener current is 10.64 mA. 5-7.

Given: VS = 24 V to 40 V VZ = 15 V RS = 470 Ω Solution: Maximum current will occur at maximum voltage. IS = (VS – VZ)/RS (Eq. 5-3) IS = (40 V – 15 V)/470 Ω IS = 53.2 mA IL = VL/RL (Eq. 5-5, Ohm’s law) IL = 15 V/1.5 kΩ IL = 10 mA

1-16 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

IZ = I S – IL (Eq. 5.6, Kirchhoff’s current law) IZ = 53.2 mA – 10 mA IZ = 43.2 mA Given:

VS = 21.5 to 25 V RS = 470 Ω RZ = 14 Ω VZ = 15 V

VS = 24 V VZ = 12 V RS = 470 Ω RL = 1.5 kΩ

IS = (VS – VZ)/RS (Eq. 5-3) IS = (25 V – 15 V)/470 Ω IS = 21.28 mA

Answer: The maximum zener current is 43.2 mA. 5-8.

Solution:

Solution: VL = VZ = 12 V IS = (VS – VZ)/RS (Eq. 5-3) IS = (24 V – 12 V)/470 Ω IS = 25.5 mA IL = VL/RL (Eq. 5-5, Ohm’s law) IL = 12 V/1.5 kΩ IL = 8 mA IZ = I S – IL (Eq. 5-6, Kirchhoff’s current law) IZ = 25.5 mA – 8 mA IZ = 17.5 mA Answer: The load voltage is 12 V and the zener current is 17.5 mA. 5-9.

5-11. Given:

Given: VS = 20 V VZ = 12 V RS = 330 Ω RL = 1 kΩ

IL = VL/RL (Eq. 5-5, Ohm’s law) IL = 15 V/1.5 kΩ IL = 10 mA IZ = I S – IL (Eq. 5-6, Kirchhoff’s current law) IZ = 21.28 mA – 10 mA IZ = 11.28 mA ΔVL = IZRZ (Eq. 5-7) ΔVL = (11.28 mA)(14 Ω) ΔVL = 157.9 mV IS = (VS – VZ)/RS (Eq. 5-3) IS = (21.5 V – 15 V)/470 Ω IS = 13.83 mA IL = VL/RL (Eq. 5-5, Ohm’s law) IL = 15 V/1.5 kΩ IL = 10 mA IZ = I S – IL (Eq. 5-6, Kirchhoff’s current law) IZ = 13.83 mA – 10 mA IZ = 3.83 mA

VL = VZ = 12 V

ΔVL = IZRZ (Eq. 5-7) ΔVL = (3.83 mA)(14Ω) ΔVL = 53.6 mV

IS = (VS – VZ)/RS (Eq. 5-3) IS = (20 V – 12 V)/330 Ω IS = 24.24 mA

Answer: The load voltage changes from 15.054 V when the supply is 21.5 V, to 15.158 V when the supply is 25 V.

Solution:

IL = VL/RL (Eq. 5-5, Ohm’s law) IL = 12 V/1 kΩ IL = 12 mA IZ = I S – IL (Eq. 5-6, Kirchhoff’s current law) IZ = 24.24 mA – 12 mA IZ = 12.24 mA Answer: The load voltage is 12 V, and the zener current is 12.24 mA. RS 330 Ω 20 V

12 V

RL 1 kΩ

5-12. Given: VS = 24 V RS = 470 Ω RL = 1.5 kΩ VZ = 15 V Solution: The regulation is lost once the load voltage drops below 15 V. (Voltage divider formula) VL = [(RL)/(RS + RL)]VS VS = VL/[RL/(RS + RL)] VS = 15 V[(1.5 kΩ)/(470 Ω + 1.5 kΩ)] VS = 19.7 V Answer: The regulation will be lost when the source voltage drops below 19.7 V. 5-13. Given: VS = 20 to 26 V RS = 470 Ω RL = 500 to 1.5 kΩ VZ = 15 V

Zener regulator for Prob. 5-9.

5-10. Given: RS = 470 Ω RZ = 14 Ω VR(in) = 1 Vpp

Solution: The regulation is lost once the load voltage drops below 15 V.

Solution: VR (out) =

RZ RS

VR (in)

(Eq. 5-8)

VR(out) = (14 Ω/470 Ω)/1 Vpp VR(out) = 29.8 mVpp

(Eq. 5-5, Ohm’s law) IL = VL/RL IL = 15 V/1.5 kΩ RS(max) = [(VS(min)/VZ) – 1]RL(min) RS(max) = [(20 V/15 V) – 1]500 Ω RS(max) = 167 Ω

(Eq. 5-9)

Answer: The ripple voltage across the load resistor is 29.8 mVpp.

1-17 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Answer: The regulator will fail since the series resistor is greater than the maximum series resistance. For this regulator to work properly, the series resistor should be 167 Ω or less. 5-14. Given: VS = 18 to 25 V RS = 470 Ω IL = 1 to 25 mA VZ = 15 V

PZ = VI PZ = (15 V)(9.15 mA) PZ = 137.3 mW Answer: The power dissipation of the series resistor is 172.4 mW. The power dissipation of the load resistor is 150 mW. The power dissipation of the zener diode is 137.3 mW. 5-19. Given: VZ = 15 V ± 5%. The tolerance is in note 1 on the data sheet.

Solution: V −VZ ⎤R RS (max) = ⎡ S I(min) (Eq. 5-10) ⎣ L (max) ⎦ L (min) RS(max) = [(18 V – 15V)/25 mA RS(max) = 120 Ω

Solution:

Answer: Yes, the regulator will fail since the series resistance is greater than the maximum series resistance. For this regulator to work properly, the series resistor should be 120 Ω or less.

Answer: The minimum voltage of 14.25 V and the maximum voltage is 15.75 V.

5-15. Given:

Answer: P = 667 mW

RS(max) = [(VS(min)/VZ) – 1]RL(min) RL(min) = RS(max)/[(VS(min)/VZ) – 1] RL(min) = 470 Ω/[(24 V/15 V) – 1] RL(min) = 783 Ω

(Eq. 5-9)

Answer: The minimum load resistance is 783 Ω. 5-16. Given: VZ = 10 V IZ = 20 mA Solution: PZ = VZIZ (Eq. 5-11) PZ = (10 V)(20 mA) PZ = 0.2 W Answer: The power dissipation is 0.2 W. 5-17. Given: VZ = 20 V IZ = 5 mA

5-21. Given: VS = 24 V RS = 470 Ω RL = 1.5 kΩ RZ = 15 V Solutions: a. With the diode in parallel with the load, the load resistor is also effectively shorted and the output voltage would be 0 V. b. With the diode open, the load resistor and the series resistor form a voltage divider: VL = [RL/(RS + RL)]VS (Voltage divider formula) VL = [1.5 kΩ/(470 Ω + 1.5 kΩ)]24 V VL = 18.27 V c. With the series resistor open, no voltage reaches the load; thus the output voltage would be 0 V. d. The voltage drop across a short is 0 V. Answers:

Solution: P Z = V Z IZ (Eq. 5-11) PZ = (20 V)(5 mA) PZ = 0.1 W Answer: The power dissipation is 0.1 W. 5-18. Given: VS = 24 V RS = 470 Ω RL = 1.5 kΩ VZ = 15 V IS = 19.15 mA (from Prob. 5-5) IL = 10 mA (from Prob. 5-5) VZ = 9.15 mA (from Prob. 5-5)

P = I2R PL = (10 mA)2(1.5 kΩ) PL = 150 mW

T = 100°C 100°C – 50°C = 50°C Derating factor 6.67 mW/°C

Solution:

Solution:

5-20. Given: Solution:

VS = 24 V RS = 470 Ω VZ = 15 V

P = I2R PS = (19.15 mA)2(470Ω) PS = 172.4 mW

(15 V)(0.05) = 0.75 V 15 V + 0.75 V = 15.75 V 15 V – 0.75 V = 14.25 V

a. b. c. d.

0V 18.27 V 0V 0V

5-22. Answer: From the previous problem, the only trouble that caused this symptom is an open zener diode. 5-23. Answer: Check the series resistor. If it is shorted, it could damage the diode. If it had been operating correctly, the output voltage should have been 18.3 V. 5-24. Answers: a. If the V130LA2 is open, it will remove the over voltage protection and the LED will remain lit. b. If the ground is opened, there is no path for current and thus the LED will not be lit. c. If the filler capacitor is open, the voltage will have more ripple but the LED should remain lit. d. If the filter capacitor is shorted, the voltage across all devices in parallel with it will be zero; thus the LED will not be lit. e. If the 1N5314 is open, it will have no effect on the LED.

1-18 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

f. If the 1N5314 is shorted, the voltage across all devices in parallel with it will be zero; thus the LED will not be lit. 5-25. Given:

RZ = 14 Ω VZ = 15 V RL = 1 kΩ to 10 kΩ IS = 19.15 mA (from Prob. 5-29) Solution:

VS = 15 V VD = 2 V RS = 2.2 kΩ Solution:

IL(max) = VL/RL(min) IL(max) = 15 V/1 kΩ IL(max) = 15 mA

IS = (VS – VD)/RS (Eq. 5-13) IS = (15 V – 2 V)/2.2 kΩ IS = 5.91 mA

IL(min) = VL/RL(max) (Eq. 5-5, Ohm’s law) IL(min) = 15 V/10 kΩ IL(min) = 1.5 mA

Answer: The diode current is 5.91 mA.

IZ(min) = IS – IL(max) (Eq. 5-6, Kirchhoff’s current law) IZ(min) = 19.15 mA – 15 mA IZ(min) = 4.15 mA

5-26. Given:

(Eq. 5-5, Ohm’s law)

VS = 40 V. VD = 2 V RS = 2.2 kΩ

IZ(max) = IS – IL(min) (Eq. 5-6, Kirchhoff’s current law) IZ(max) = 19.15 mA – 1.5 mA IZ(max) = 17.65 mA

Solution:

ΔVL(min) = IZ(min)RZ ΔVL(min) = (4.15 mA)(14 Ω) ΔVL(min) = 58.1 mV

IS = (VS – VD)/RS (Eq. 5-13) IS = (40 V – 2 V)/2.2 kΩ IS = 17.27 mA Answer: The diode current is 17.27 mA. 5-27. Given:

ΔVL(max) = IZ(max)RZ ΔVL(max) = (17.65 mA)(14 Ω) ΔVL(max) = 247.1 mV VL(min) = 15.058 V VL(max) = 15.247 V

VS = 15 V VD = 2 V RS = 1 kΩ

Answer: The minimum load voltage would be 15.06 V and the maximum voltage would be 15.25 V.

Solution: IS = (VS – VD)/RS (Eq. 5-13) IS = (15 V – 2 V)/1 kΩ IS = 13 mA Answer: The diode current is 13 mA. 5-28. Answer: From Prob. 5-27, the resistor value will be 1 kΩ.

5-31. Given: VS = 20 V VZ = 6.8 V VL = 6.8 V IL = 30 mA Solution: RS(max) = [(VS(min) – VZ)/IL(max)] (Eq. 5-10) RS(max) = [(20 V – 6.8 V)/30 mA] RS(max) = 440 Ω

CRITICAL THINKING 5-29. Given:

RS(min) = [(VS – VZ)/IZM] RS(min) = [(20 V – 6.8 V)/55 mA] RS(min) = 240 Ω

VS = 24 V RS = 470 Ω RZ = 14 Ω VZ = 15 V Solution: IS = (VS – VZ)/RS (Eq. 5-3) IS = (24 – 15)/470 Ω IS = 19.15 mA IL = VL/RL (Eq. 5-5, Ohm’s law) IL = 15 V/1.5 kΩ IL = 10 mA IZ = I S – IL (Eq. 5-6, Kirchhoff’s current law) IZ = 19.15 mA – 10 mA IZ = 9.15 mA

Answer: Any similar design as long as the zener voltage is 6.8 V and the series resistance is less than 440 Ω, to provide the desired maximum output current, and greater than 240 Ω, if a 1N957B is used to prevent overcurrent if it becomes unloaded. The load resistance does not need to be specified because, as a power supply, the load resistance can vary. The only load parameter that is necessary is maximum current, and it is given.

ΔVL = IZRZ (Eq. 5-7) ΔVL = (9.15 mA)(14 Ω) ΔVL = 128.1 mV VL = 15.128 V or approximately 15.13 V

Zener regulator for Prob. 5-31.

Answer: The load voltage would be 15.13 V. 5-30. Given: VS = 24 V RS = 470 Ω

5-32. Given: VLED = 1.5 to 2 V ILED = 20 mA VS = 5 V Imax = 140 mA

1-19 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Solution: RS = [(VS – VLED(min))/ILED] RS = [(5 V – 1.5 V)/20 mA] RS = 175 Ω Answer: Same as Fig. 5-20 with resistor values of 175 Ω, which limits each branch to a maximum of 20 mA and a total of 140 mA. 5-33. Given: Vline = 115 V ac ± 10% VSec = 12.6 V ac R2 = 560 Ω ± 5% RZ = 7 Ω VZ = 5.1 V ± 5%

Answer: The maximum ripple voltage will be 0.438 V. 5-35. Given: VS = 6 V ac VD = 0.25 V Solution: VP(in) = 1.414 Vrms VP(in) = 1.414 (6 V ac) VP(in) = 8.48 V (Eq. 4-8; the 1.4 was changed VP(out) = VP(in) – 0.5 V to reflect using Schottky diodes) VP(out) = 7.98 V Answer: The voltage at the filter capacitor is 7.98 V. 5-36. Troubles:

Solution: To find the maximum zener current, the maximum secondary voltage, the minimum zener voltage, and the minimum resistance of R2 must be found. If the line voltage varies by 10 percent, the secondary voltage should also vary by 10 percent. VSec(max) = VSec + VSec (10%) VSec(max) = 12.6 V ac + 12.6 V ac (10%) VSec(max) = 13.86 V ac VP = 1.414 V ac VP = 1.414 (13.86 V ac) VP = 19.6 V VZ = 5.1 V ± 5% VZ = 5.1 V – [(5.1 V) (5%)] VZ = 4.85 V R2 = 560 Ω ± 5% R2 = 560 Ω – [(560 Ω) (5%)] R2 = 532 Ω The circuit can be visualized as a series circuit with a 19.6 V power supply, a 532 Ω R2, a 7 Ω RZ, and a 4.85 V zener diode.

1. Open RS, since there is voltage at A and no voltage at B; also could be a short from B or C to ground. 2. Open between B and D or an open at E. Since the voltages at B and C are 14.2 V, which is the voltage that would be present if the circuit were just a voltage divider with no zener diode, suspect something in the diode circuit. Since the diode is good, it is either an open between B and D or an open at E. 3. The zener is open. Since the voltages at B and C are 14.2 V, which is the voltage that would be present if the circuit were just a voltage divider with no zener diode, suspect something in the diode circuit. Since the diode reads an open, it is bad. 4. RS shorted, which caused the zener to open. With all the voltages at 18 V, the problem could be an open in the return path. But the zener is open, and the most likely cause of that is overcurrent. The only device that could short and cause the zener to burn open is RS. 5-37. Troubles: 5. Open at A. Since all the voltages are zero, the power must not be getting to the circuit. 6. Open RL, an open between B and C, or an open between RL and ground. To solve this problem, the second approximation must be used. With the load resistor operating normally, only part of the total current flows through the zener, which causes the 0.3V increase from its nominal voltage. But when the load resistor opens, all the total current flows through the diode, causing the voltage drop across the internal resistance to increase to 0.5 V. 7. Open at E. Since the voltages at B, C, and D are 14.2 V, which is the voltage that would be present if the circuit were just a voltage divider with no zener diode, suspect something in the diode circuit. Since the diode reads OK, that only leaves an open in the return path. 8. The zener is shorted or a short from B, C, or D to ground. Since the voltages at B, C, and D are 0, and A is 18 V, this could be caused by an open RS or a short from B, C, or D to ground. Since the diode reads 0 Ω, it confirms that the fault is a short.

IS = (VS – VZ)/RS + RZ) (Eq. 5-13) IS = (19.6 V – 4.85 V)/(532 Ω + 7 Ω) IS = 27.37 mA Answer: The maximum diode current is 27.37 mA. 5-34. Given: VSec = 12.6 V ac VD = 0.7 V I1N5314 = 4.7 mA ILED = 15.6 mA IZ = 21.7 mA C = 1000 ¿F ± 20% fin = 60 Hz Solution: The dc load current is the sum of all of the loads. I = I1N5314 + ILED + IZ I = 4.7 mA + 15.6 mA + 21.7 mA I = 42 mA fout = 2fin fout = 2(60 Hz) fout = 120 Hz

(Eq. 4-7)

Chapter 6 Bipolar Transistors SELF-TEST

The minimum capacitance will give the maximum ripple. C = 1000 ¿F ± 20% C = 1000 ¿F – 1000 ¿F (20%) C = 800 ¿F VR = I/(fC) (Eq. 4-10) VR = 42 mA/(120 Hz)(800 ¿F) VR = 0.438 V

1. b 2. a 3. d 4. a 5. c 6. a 7. b 8. b

9. b 10. a 11. a 12. b 13. d 14. b 15. a 16. b

1-20 Downloaded by Kevin Aguilar Arias ([email protected])

17. d 18. b 19. a 20. a 21. b 22. b 23. b 24. c

25. a 26. c 27. d 28. c 29. a 30. a

lOMoARcPSD|6455127

JOB INTERVIEW QUESTIONS 6. A transistor or semiconductor curve tracer. 7. Since there is almost zero power dissipation at saturation and cutoff, I would expect that the maximum power dissipation is in the middle of the load line. 10. Common emitter.

PROBLEMS 6-1.

Given: IE = 10 mA IC = 9.95 mA Solution: IE = IC + IB (Eq. 6-1) IB = IE – IC IB = 10 mA – 9.95 mA IB = 0.05 mA Answer: The base current is 0.05 mA.

6-2.

Given: IC = 10 mA IB = 0.1 mA Solution: βdc = IC/IB βdc = 10 mA/0.1 mA βdc = 100 Answer: The current gain is 100.

6-3.

Given: IB = 30 ¿A βdc = 150 Solution: IC = βdcIB IC = 150(30 ¿A) IC = 4.5 mA Answer: The collector current is 4.5 mA.

6-4.

Given: IC = 100 mA βdc = 65 Solution: IB = IC/βdc (Eq. 6-5) IB = 100 mA/65 IB = 1.54 mA IE = IB + I C IE = 1.54 mA + 100 mA Answer: The emitter current is 101.54 mA.

6-5.

Given: VBB = 10 V RB = 470 kΩ VBE = 0.7 V Solution: IB = [(VBB – VBE)/RB] (Eq. 6-6) IB = [(10 V – 0.7 V)/470 kΩ] IB = 19.8 ¿A Answer: The base current is 19.8 ¿A.

6-6.

Answer: The base current is unaffected by the current gain since 9.3 V/470 kΩ always equals 19.8 ¿A. The current gain will affect the collector current in this circuit.

Given: VBB = 10 V RB = 470 kΩ ± 5% VBE = 0.7 V Solution: The minimum resistance will yield the maximum current. RB = 470 kΩ ± 5% RB = 470 kΩ – 470 kΩ(5%) RB = 446.5 kΩ (Eq. 6-6) IB = [(VBB – VBE)/RB] IB = [(10 V – 0.7 V)/446.5 kΩ] IB = 20.83 ¿A Answer: The base current is 20.83 ¿A. 6-8. Given: IC = 6 mA RC = 1.5 kΩ VCC = 20 V Solution: VCE = VCC – ICRC VCE = 20 V – (6 mA)(1.5 kΩ) VCE = 11 V Answer: The collector to emitter voltage is 11 V. 6-9. Given: IC = 100 mA VCE = 3.5 V Solution: PD = VCEIC (Eq. 6-8) PD = (3.5 V)(100 mA) PD = 350 mW Answer: The power dissipation is 350 mW. 6-10. Given: VBB = 10 V RB = 470 kΩ VBE = 0.7 V (second approximation) VBE = 0 V (ideal) RC = 820 Ω VCC = 10 V βdc = 200 Solution: Ideal (Eq. 6-6) IB = [(VBB – VBE)/RB] IB = [(10 V – 0 V)/470 kΩ] IB = 21.28 ¿A IC = βdcIB IC = 200(21.28 ¿A) IC = 4.26 mA VCE = VCC – ICRC VCE = 10 V – (4.26 mA)(820 Ω) VCE = 6.5 V PD = VCEIC PD = (6.5 V)(4.26 mA) PD = 27.69 mW 2nd Approximation (Eq. 6-6) IB = [(VBB – VBE)/RB] IB = [(10V – 0.7 V)/470 kΩ] IB = 19.8 ¿A IC = βdcIB IC = 200(19.8 ¿A) IC = 3.96 mA 6-7.

1-21 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

VCE = VCC – ICRC VCE = 10 V – (3.96 mA)(820 Ω) VCE = 6.75 V PD = VCEIC PD = (6.75 V)(3.96 mA) PD = 26.73 mW Answer: The ideal collector-emitter voltage is 6.5 V and power dissipation is 27.69 mW. The second approximation collector-emitter voltage is 6.75 V and the power dissipation is 26.73 mW. 6-11. Given: VBB = 5 V RB = 330 kΩ VBE = 0.7 V (second approximation) VBE = 0 V (ideal) RC = 1.2 kΩ VCC = 15 V βdc = 150 Solution: Ideal (Eq. 6-6) IB = [(VBB – VBE)/RB] IB = [(5 V – 0 V)/330 kΩ] IB = 15.15 ¿A IC = βdcIB IC = 150(15.15 ¿A) IC = 2.27 mA

IB = [(12 V – 0)/680 kΩ] IB = 17.6 ¿A (ideal) IB = [(12 V – 0.7 V)/680 kΩ] IB = 16.6 ¿A (second approximation) IC = βdcIB (Eq. 6-4) IC = 175(17.6 ¿A) IC = 3.08 mA (ideal) IC = 175(16.6 ¿A) IC = 2.91 mA (second approximation) VCE = VCC – ICRC (Eq. 6-7) VCE = 12 V – (3.08 mA)(1.5 kΩ) VCE = 7.38 V (ideal) VCE = 12 V – (2.91 mA)(1.5 kΩ) VCE = 7.64 V (second approximation) PD = VCEIC (Eq. 6-8) PD = (7.38 V)(3.08 mA) PD = 22.73 mW (ideal) PD = (7.64 V)(2.91 mA) PD = 22.23 mW (second approximation) Answer: The ideal collector-emitter voltage is 7.38 V, and power dissipation is 22.73 mW. The second approximation collector-emitter voltage is 7.64 V, and power dissipation is 22.23 mW. 6-13. Answer: From the maximum ratings section, –55 to + 150°C.

VCE = VCC – ICRC VCE = 15 V – (2.27 mA)(1.2 kΩ) VCE = 12.28 V PD = VCEIC PD = (12.28 V)(2.27 mA) PD = 27.88 mW

6-14. Answer: From the on characteristics section, 70. 6-15. Given: PD(max) = 1 W IC = 120 mA VCE = 10 V Solution:

2nd Approximation IB = [(VBB – VBE)/RB] IB = [(5 V – 0.7 V)/330 kΩ] IB = 13.3 ¿A

(Eq. 6-6)

PD = VCEIC (Eq. 6-8) PD = (10 V)(120 mA) PD = 1.2 W Answer: The power dissipation has exceeded the maximum rating, and the transistor’s power rating is damaged and possibly destroyed.

IC = βdcIB IC = 150(13.03 ¿A) IC = 1.96 mA VCE = VCC – ICRC VCE = 15 V – (1.96 mA)(1.2 kΩ) VCE = 12.65 V

6-16. Given: PD = 625 mW Temperature = 65°C

PD = VCEIC PD = (12.65 V)(1.96 mA) PD = 24.79 mW

Solution:

Answer: The ideal collector-emitter voltage is 12.28 V and power dissipation is 27.88 mW. The second approximation collector-emitter voltage is 12.65 V, and power dissipation is 24.79 mW.

ΔP = ΔT (derating factor) ΔP = 40°C(2.8 mW/°C) ΔP = 112 mW

ΔT = 65°C – 25°C ΔT = 40°C

6-12. Given: VBB = 12 V RB = 680 kΩ VBE = 0.7 V (second approximation) VBE = 0 (ideal) RC = 1.5 kΩ VCC = 12 V βdc = 175 Solution: IB = [(VBB – VBE)/RB]

(Eq. 6-6)

PD(max) = 350 mW – 112 mW PD(max) = 238 mW Answer: The transistor is operating outside of its limits; the power rating is affected. 6-17. a. Increase: With the base resistor shorted, the baseemitter junction will have excessive current and will open, stopping all conduction. Thus source voltage is read from collector to emitter. b. Increase: With the base resistor open, the transistor goes into cutoff and source voltage is read from collector to emitter.

1-22 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

PD = VCEIC IC = PD/VCE IC = 280 mW/10 V IC = 28 mA

JOB INTERVIEW QUESTIONS

Answer: The maximum collector current is 28 mA. 6-23. Given: VBB = 10 V RB = 470 kΩ VBE = 0.7 V (second approximation) RC = 820 Ω VCC = 10 V βdc = 200

7. An increase in temperature almost always increases the current gain. 10. High leakage current, low breakdown voltage, and low current gain. 12. If in cutoff, VCE will be approximately equal to VCC. If in saturation, VCE is usually less than 1 V, typically 0.1 to 0.2 V for a small-signal transistor. 13. One with large RC.

PROBLEMS 7-1. 7-2. 7-3.

Solution: IB = [(VBB – VBE)/RB] (Eq. 6-6) IB = [(10 V – 0.7 V)/470 kΩ] IB = 19.8 ¿A IC = βdcIB IC = 200(19.8 ¿A) IC = 3.96 mA Answer: The LED current is 3.96 mA. 6-24. Answer: VCE(Sat) = 0.3 V 6-25. Answer: An increase in VBB causes the base current to increase, and, since the transistor is controlled by base current, all other dependent variables increase except VCE, which decreases because of the transistor being further into conduction. 6-26. Answer: The increase in VCC had no effect on the base circuit, which means that it also had no effect on IC and the voltage drop across the collector resistor. The increase did increase VCE and the power dissipation across the transistor.

VCE(cutoff) = VCC (Eq. 7-3) VCE(cutoff) = 20 V Answer: The collector current at saturation is 6.06 mA, and the collector-emitter voltage at cutoff is 20 V. The load line would connect these points. 7-4.

Given: VCC = 25 V VBB = 10 V RB = 1 MΩ RC = 3.3 kΩ Solution: IC(Sat) = VCC/RC (Eq. 7-2) IC(Sat) = 25 V/3.3 kΩ IC(Sat) = 7.58 mA Answer: The load line moves futher away from the origin.

7-5.

Given: VCC = 20 V VBB = 10 V RB = 1 MΩ RC = 4.7 kΩ Solution: IC(Sat) = VCC/RC (Eq. 7-2) IC(Sat) = 20 V/4.7 kΩ IC(Sat) = 4.25 mA

6-27. Answer: IC, IB, and all power dissipations decreased. The power dissipations decreased because of the drop in current (P = IV). The base current decreased because the voltage drop across it did not change and the resistance increased (I = V/R). The collector current decreased because the base current decreased (IC = IBβdc). 6-28. Answer: VA, VB, VD, IB, IC, and PB show no change. VA and VD do not change since the power supply voltages did not change. IB, VB, and PB do not change because the collector resistance does not affect the base circuit. IC does not change because IB did not change. 6-29. Answer: The only variable to decrease is VC. With an increase in βdc, the same base current will cause a greater collector current, which will create a greater voltage drop across the collector resistor. This leaves less voltage to drop across the transistor.

VCE(cutoff) = VCC VCE(cutoff) = 20 V Answer: The left side of the load line would move down while the right side remains at the same point.

Chapter 7 Transistor Fundamentals SELF-TEST 1. a 2. b 3. d 4. d 5. c 6. c 7. a 8. c

9. b 10. a 11. b 12. c 13. d 14. c 15. a 16. b

17. d 18. c 19. b 20. b 21. b 22. d 23. b 24. a

25. a 26. c 27. a 28. b 29. a 30. d 31. c

Answer: β = 30 Answer: β = 85 Given: VCC = 20 V VBB = 10 V RB = 1 MΩ RC = 3.3 kΩ Solution: IC(Sat) = VCC/RC (Eq. 7-2) IC(Sat) = 20 V/3.3 kΩ IC(Sat) = 6.06 mA

7-6.

Given: VCC = 20 V VBB = 10 V RB = 2 MΩ RC = 4.7 kΩ Solution: IC(Sat) = VCC/RC (Eq. 7-2) IC(Sat) = 20 V/3.3 kΩ IC(Sat) = 6.06 mA

1-24 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

VD = 6.8 V VBE = 0.7 V

7-46. Given: VCC = 10 V VBB = 5 V RE = 100 Ω VBE = 0.7 V

Solution: With the left transistor in cutoff, the base circuit for the right transistor would consist of the zener diode and the resistor, and the base voltage would be 6.8 V.

Solution: VE = VBB – VBE(1) – VBE(2) VE = 5 V – 0.7 V – 0.7 V VE = 3.6 V

(Eq. 7-7)

IE = VE/RE (Ohm’s law) IE = 3.6 V/100 Ω IE = 36 mA IE ≈ I C IC = 36 mA

IC = 36 mA (from the previous problem) βdc(1) = 100 βdc(2) = 50 Solution: The overall current gain is βdc(1)βdc(2); thus the overall current gain is (100)(50) = 5000. (Eq. 6-5)

7-48. Given:

VCC = 10 V RC = 2 kΩ Solution: IC(Sat) = VCC/RC IC(Sat) = 10 V/2 kΩ IC(Sat) = 5 mA

(Eq. 7-2)

Answer: The maximum possible current through the transistor is 5 mA. RC = 2 kΩ VRC = 2 V RE = 430 Ω Solution:

VBB(1) = 0 V VBB(2) = 10 V VCC = 10 V RC = 240 Ω RB = 2.4 kΩ RE = 270 Ω VD = 5 V VBE = 0.7 V

IC = VRC/RC IC = 2 V/2 kΩ IC = 1 mA

(Ohm’s law)

ILED = 1 mA (from the graph) ILED = IC = IE

Solution: With the left transistor in cutoff, it is essentially an open. Thus the base circuit for the right transistor would consist of the zener and the resistor, and the base voltage would be 5 V. (Eq. 7-7)

IE = 1 mA VE = IERE VE = (1 mA)(430 Ω) VE = 430 mV VBB = VE + VBE (Eq. 7-8) VBB = 0.43 V + 0.7 V VBB = 1.13 V Answer: The base voltage is 1.13 V.

IE = VE/RE (Ohm’s law) IE = 4.3 V/270 Ω IE = 15.93 mA IE ≈ I C = ID ID = 15.93 mA With the left transistor conducting, assume an ideal transistor. The collector voltage would be 0 V, which would cause the right transistor to cutoff. Therefore, there would be no collector current and no diode current. Answer: With VBB at 0 V, the diode current is 15.93 mA, and with VBB at 10 V, the diode current is 0 mA. VCC = 10 V VBB = 0 V RB = 2.4 kΩ RC = 240 Ω RE = 270 Ω

(Ohm’s law)

7-51. Given:

Answer: The base current of the first transistor is 7.2 ¿A.

7-49. Given:

IE = VE/RE IE = 6.1 V/270 Ω IE = 22.6 mA ID = IE = 22.6 mA 7-50. Given:

7-47. Given:

VE = VBB – VBE VE = 5 V – 0.7 V VE = 4.3 V

(Eq. 7-7)

Answer: The diode current is 22.6 mA.

Answer: The collector current is 36 mA.

IB = IC/βdc IB = 36 mA/5000 IB = 7.2 ¿A

VE = VBB – VBE VE = 6.8 V – 0.7 VE = 6.1 V

7-52. Given: VBB = 3 V VBE = 0.7 V Answer: Since the collector-base junction is a diode like the emitter-base junction, the internal impedance of the voltmeter will complete the circuit to ground and the collector-base diode will forward-bias. Thus the voltage will be a diode voltage drop less than the base voltage, or 2.3 V. 7-53. Given: RB = 1 MΩ VBB = 10 V RC = Open Solution: VCE = VBB VCE = 0.7 V Answer: The collector to ground voltage is approximately 0.7 V.

1-30 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

7-54. Given: VCC = 15 V VBB = 0 V and 15 V IC(Sat) = 5 mA Solution: IC(Sat) = VCC/RC (Eq. 7-2) RC = VCC/IC(Sat) RC = 15 V/5 mA RC = 3 kΩ RB = (VBB – VBE)/IB RB = (15 V – 0.7)/476 ¿A RB = 30 kΩ Answer: RC = 3 kΩ, RB = 30 kΩ 7-55. Given: VCC = 10 V VBB = 1.8 V VCE = 6.6 V RE = 1 kΩ Solution: IE = IC = (VBB – VBE)/RE IE = IC = (1.8 V – 0.7 V)/1 kΩ IE = IC = 1.1 mA VE = IERE VE = (1.1 mA)(1 kΩ) VE = 1.1 V VRC = VCC – VE – VCE VRC = 10 V – 1.1 V – 6.6V VRC = 2.3 V RC = VRC/IC RC = 2.3 V/1.1 mA RC = 2 kΩ Answer: RC = 2 kΩ 7-56. Answer: Since VBB increases, the transistor’s VB and VE values will increase along with an increase in all currents. VC will decrease because of an increased voltage drop across RC. 7-57. Answer: The increase in collector supply voltage causes an increase in collector voltage because the collector current remains constant. Thus the voltage drop across the collector resistor remains constant and the collector voltage has to increase. 7-58. Answer: All of the currents decrease. Since the emitter voltage remains constant, the increase in emitter resistance causes a decrease in emitter current. Since the collector current is approximately the same as the emitter current, it also decreases. The base current is related to the collector current by βdc and also decreases. 7-59. Answer: VB, VE, IE, IC and IB show no change. Since the base voltage did not change, VB and VE will not change. Since these voltages do not change, none of the currents change. VC will decrease since the voltage drop across RC will increase.

4. Emitter-feedback bias and collector-feedback bias. They were developed in an attempt to stabilize the Q point against transistor replacement and temperature changes. 6. No. Saturation and cutoff. 7. Changes in current gain will change the collector current. The base resistors should be made smaller to satisfy the condition described in the text. 10. The circuit will be highly sensitive to changes in current gain.

PROBLEMS 8-1.

8-2.

Chapter 8 Transistor Biasing SELF-TEST 1. d 2. a 3. a 4. d 5. b 6. b 7. b 8. a

9. c 10. a 11. b 12. a 13. c 14. c 15. c 16. a

17. b 18. a 19. d 20. a 21. c 22. a 23. d

JOB INTERVIEW QUESTIONS 2. The collector current changes only slightly, if at all.

24. b 25. b 26. c 27. b 28. c 29. a 30. d

8-3.

Given: R1 = 10 kΩ R2 = 2.2 kΩ RC = 3.6 kΩ RE = 1 kΩ VCC = 25 V VBE = 0.7 V Solution: (Eq. 8-1) VBB = [R2/(R1 + R2)]VCC VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]25 V VBB = 4.51 V VE = VBB – VBE (Eq. 8-2) VE = 4.51 V – 0.7 V VE = 3.81 V (Eq. 8-3) IE = VE/RE IE = 3.81 V/1 kΩ IE = 3.81 mA I C ≈ IE (Eq. 8-4) (Eq. 8-5) VC = VCC – ICRC VC = 25 V – (3.81 mA)(3.6 kΩ) VC = 11.28 V Answer: The emitter voltage is 3.81 V, and the collector voltage is 11.28V. Given: R1 = 10 kΩ R2 = 2.2 kΩ RC = 2.7 kΩ RE = 1 kΩ VCC = 15 V VBE = 0.7 V Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]15 V VBB = 2.7 V VE = VBB – VBE (Eq. 8-2) VE = 2.7 V – 0.7 V VE = 2.0 V (Eq. 8-3) IE = VE/R3 IE = 2.0 V/1 kΩ IE = 2 mA (Eq. 8-4) I C ≈ IE VC = VCC – ICRC (Eq. 8-5) VC = 15 V – (2 mA)(2.7 kΩ) VC = 9.59 V Answer: The emitter voltage is 2.0 V, and the collector voltage is 9.59 V. Given: R1 = 330 kΩ R2 = 100 kΩ RC = 150 kΩ RE = 51 kΩ VCC = 10 V VBE = 0.7 V

1-31 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

8-4.

8-5.

Solution: (Eq. 8-1) VBB = [R2/(R1 + R2)]VCC VBB = [100 kΩ/(330 kΩ + 100 kΩ)]10 V VBB = 2.33 V (Eq. 8-2) VE = VBB – VBE VE = 2.33 V – 0.7 V VE = 1.63 V (Eq. 8-3) IE = VE/RE IE = 1.63 V/51 kΩ IE = 31.96 ¿A (Eq. 8-4) I C ≈ IE (Eq. 8-5) VC = VCC – ICRC VC = 10 V – (31.96 ¿A)(150 kΩ) VC = 5.21 V Answer: The emitter voltage is 1.63 V, and the collector voltage is 5.21 V. Given: R1 = 150 Ω R2 = 33 Ω RC = 39 Ω RE = 10 Ω VCC = 12 V VBE = 0.7 V Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) VBB = [33 Ω/(150 Ω + 33 Ω)]12 V VBB = 2.16 V VE = VBB – VBE (Eq. 8-2) VE = 2.16 V – 0.7 V VE = 1.46 V IE = VE/RE (Eq. 8-3) IE = 1.46 V/10 Ω = 146 mA I C ≈ IE (Eq. 8-4) VC = VCC – ICRC (Eq. 8-5) VC = 12 V – (146 mA)(39 Ω) VC = 6.3 V Answer: The emitter voltage is 1.46 V. The collector voltage is 6.3 V. Given: R1 = 330 kΩ ± 5% R2 = 100 kΩ ± 5% RC = 150 kΩ ± 5% RE = 51 kΩ ± 5% VCC = 10 V VBE = 0.7 V Solution: VBB(max) = [R2(max)/(R1(min) + R2(max))]VCC (Eq. 8-1) VBB(max) = [105 kΩ/(313.5 kΩ + 105 kΩ)]10 V VBB(max) = 2.51 V VBB(min) = [R2(min)/(R1(max) + R2(min))]VCC (Eq. 8-1) VBB(min) = [95 kΩ/346.5 kΩ + 95 kΩ)]10 V VBB(min) = 2.15 V VE(max) = VBB(max) – VBB (Eq. 8-2) VE(max) = 2.51 V – 0.7 V VE(max) = 1.81 V VE(min) = VBB(min) – VBE (Eq. 8-2) VE(min) = 2.15 V – 0.7 V VE(min) = 1.45 V (Eq. 8-3) IE(max) = VE(max)/RE(min) IE(max) = 1.81 V/48.45 kΩ IE(max) = 37.36 ¿A

8-6.

8-7.

IE(min) = VE(min)/RE(max) (Eq. 8-3) IE(min) = 1.45 V/53.55 kΩ IE(min) = 27.08 ¿A I C ≈ IE (Eq. 8-4) (Eq. 8-5) VC(max) = VCC – IC(min)RC(min) VC(max) = 10 V – (27.08 ¿A)(142.5 kΩ) VC(max) = 6.14 V VC(min) = VCC – IC(max)RC(max) (Eq. 8-5) VC(min) = 10 V – (37.36 ¿A)(157.5 kΩ) VC(min) = 4.12 V Answer: The lowest collector voltage is 4.12 V, and the highest collector voltage is 6.14 V. Given: R1 = 150 Ω R2 = 33 Ω RC = 39 Ω RE = 10 Ω VCC = 12 V ± 10% VBE = 0.7 V Solution: VBB(max) = [R2/(R1 + R2)]VCC(max) (Eq. 8-1) VBB(max) = [33 Ω/(150 Ω + 33 Ω)]13.2 V VBB(max) = 2.38 V (Eq. 8-2) VE(max) = VBB(max) – VBE VE(max) = 2.38 V – 0.7 V VE(max) = 1.68 V IE(max) = VE(max)/RE (Eq. 8-3) IE(max) = 1.68 V/10 Ω IE(max) = 168 mA VBB(min) = [R2/(R1 + R2)]VCC(min) (Eq. 8-1) VBB(min) = [33 Ω/(150 Ω + 33 Ω)]10.8 V VBB(min) = 1.95 V VE(min) = VBB(min) – VBE (Eq. 8-2) VE(min) = 1.95 V – 0.7 V VE(min) = 1.25 V IE(min) = VE(min)/RE (Eq. 8-3) IE(min) = 1.25 V/10 Ω IE(min) = 125 mA I C ≈ IE (Eq. 8-4) VC(max) = VCC(max) – IC(min)RC (Eq. 8-5) VC(max) = 13.2 V – (125 mA)(39 Ω) VC(max) = 8.33 V VC(min) = VCC(min) – IC(max)RC (Eq. 8-5) VC(min) = 10.8 V – (168 mA)(39 Ω) VC(min) = 4.25 V Answer: The lowest collector voltage is 4.25 V and the highest collector voltage is 8.33 V. Given: R1 = 10 kΩ R2 = 2.2 kΩ RC = 3.6 kΩ RE = 1 kΩ VCC = 25 V VBE = 0.7 V VBB = 4.51 V (from Prob. 8-1) VE = 3.81 V (from Prob. 8-1) IE = IC = 3.81 mA (from Prob. 8-1) VC = 11.28 V (from Prob. 8-1) Solution: VCE = VC – VE (Eq. 8-6) VCE = 11.28 V – 3.81 V VCE = 7.47 V Answer: The Q point is IC = 3.81 mA, and VCE = 7.47 V.

1-32 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

8-8.

Given:

I C ≈ IE

R1 = 10 kΩ R2 = 2.2 kΩ RC = 2.7 kΩ RE = 1 kΩ VCC = 15 V VBE = 0.7 V

VC = VCC – ICRC (Eq. 8-5) VC = 12 V – (146 mA)(39 Ω) VC = 6.3 V VCE = VC – VE (Eq. 8-6) VCE = 6.3 V – 1.46 V VCE = 4.85 V

Solution:

Answer: The Q point is IC = 146 mA, and VCE = 4.85 V.

VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) VBB = [2.2 kΩ/(10 Ω + 2.2 kΩ)]15 V VBB = 2.7 V VE = VBB – VBE VE = 2.7 V – 0.7 V VE = 2.0 V

(Eq. 8-2)

IE = VE/RE (Eq. 8-3) IE = 2.0 V/1 Ω IE = 2 mA I C ≈ IE

(Eq. 8-4)

VCE = VC – VE (Eq. 8-6) VCE = 9.59 V – 2.0 V VCE = 7.59 V Answer: The Q point is IC = 2 mA, and VCE = 7.59 V. Given: R1 = 330 kΩ R2 = 100 kΩ RC = 150 kΩ RE = 51 kΩ VCC = 10 V VBE = 0.7 V VBB = 2.33 V (from Prob. 8-3) VE = 1.63 V (from Prob. 8-3) IE = IC = 31.96 µA (from Prob. 8-3) VC = 5.21 V (from Prob. 8-3) Solution: VCE = VC – VE (Eq. 8-6) VCE = 5.21 V – 1.63 V VCE = 3.58 V Answer: The Q point is IC = 31.96 µA, and VCE = 3.58 V. 8-10. Given: R1 = 150 Ω R2 = 33 Ω RC = 39 Ω RE = 10 Ω VCC = 12 V VBE = 0.7 V

R1 = 330 kΩ ± 5% R2 = 100 kΩ ± 5% RC = 150 kΩ ± 5% RE = 51 kΩ ± 5% VCC = 10 V VBE = 0.7 V VBB(max) = [R2(max)/(R1(min) + R2(max))]VCC (Eq. 8-1) VBB(max) = [105 kΩ/(313.5 kΩ + 105 kΩ)]10 V VBB(max) = 2.51 V VBB(min) = [R2(min)/(R1(max) + R2(min))]VCC (Eq. 8-1) VBB(min) = [95 kΩ/(346.5 kΩ + 95 kΩ)]10 V VBB(min) = 2.15 V VE(max) = VBB(max) – VBE VE(max) = 2.51 V – 0.7 V VE(max) = 1.81 V

(Eq. 8-2)

VE(min) = VBB(min) – VBE VE(min) = 2.15 V – 0.7 V VE(min) = 1.45 V

(Eq. 8-2)

IE(max) = VE(max)/RE(min) (Eq. 8-3) IE(max) = 1.81 V/48.45 kΩ IE(max) = 37.36 µA IE(min) = VE(min)/RE(max) (Eq. 8-3) IE(min) = 1.45 V/53.55 kΩ IE(min) = 27.08 µA I C ≈ IE

(Eq. 8-4)

Answer: The lowest collector current is 27.08 µA, and the highest collector current is 37.36 µA. 8-12. Given: R1 = 150 Ω R2 = 33 Ω RC = 39 Ω RE = 10 Ω VCC = 12 V ± 10% VBE = 0.7 V Solution: VBB(max) = [R2/(R1 + R2)]VCC(max) (Eq. 8-1) VBB(max) = [33 Ω/(150 Ω + 33 Ω)]13.2 V VBB(max) = 2.38 V

Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) VBB = [33 Ω/(150 Ω + 33 Ω)]12 V VBB = 2.16 V VE = VBB – VBE (Eq. 8-2) VE = 2.16 V – 0.7 V VE = 1.46 V IE = VE/RE (Eq. 8-3) IE = 1.46 V/10 Ω IE = 146 mA

8-11. Given:

Solution:

VC = VCC – ICRC (Eq. 8-5) VC = 15 V – (2 mA)(2.7 Ω) VC = 9.59 V

8-9.

(Eq. 8-4)

VE(max) = VBB(max) – VBE VE(max) = 2.38 V – 0.7 V VE(max) = 1.68 V

(Eq. 8-2)

IE(max) = VE(max)/RE (Eq. 8-3) IE(max) = 1.68 V/10 Ω IE(max) = 168 mA VBB(min) = [R2/(R1 + R2)]VCC(min) (Eq. 8-1) VBB(min) = [33 Ω/(150 Ω + 33 Ω)]10.8 V VBB(min) = 1.95 V

1-33 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

VE(min) = VBB(min) – VBE VE(min) = 1.95 V – 0.7 V VE(min) = 1.25 V

(Eq. 8-2)

IE(min) = VE(min)/RE (Eq. 8-3) IE(min) = 1.25 V/10 Ω IE(min) = 125 mA Answer: The lowest collector current 125 mA, and the highest collector current is 168 mA. 8-13. Given: RB = 10 kΩ RC = 4.7 kΩ RE = 10 kΩ VCC = 12 V VEE = –12 V Solution: IE = (–0.7 V – VEE)/RE IE = [–0.7 V – (–12 V)]/10 kΩ IE = 1.13 mA VC = VCC – ICRC VC = 12 V – (1.13 mA)(4.7 kΩ) VC = 6.69 V Answer: The emitter current is 1.13 mA, and the collector voltage is 6.69 V. 8-14. Given: RB = 20 kΩ RC = 9.4 kΩ RE = 20 kΩ VCC = 12 V VEE = –12 V Solution: IE = (–0.7 V – VEE)RE IE = [–0.7 V –(–12 V)]/20 kΩ IE = 565 µA VC = VCC – ICRC VC = 12 V – (565 µA)(9.4 kΩ) VC = 6.69 V Answer: The emitter current is 565 µA, and the collector voltage is 6.69 V. 8-15. Given: RB = 10 kΩ ± 5% RC = 4.7 kΩ ± 5% RE = 10 kΩ ± 5% VCC = 12 V VEE = –12 V Solution: IE(max) = (–0.7 V – VEE)/RE(min) IE(max) = [–0.7 V – (–12 V)]/9.5 kΩ IE(max) = 1.19 mA

(Eq. 8-14)

VC(max) = VCC – IC(min)RC(min) VC(max) = 12 V – (1.08 mA)(4465 kΩ) VC(max) = 7.18 V

(Eq. 8-15)

IE(min) = (–0.7 V – VEE)/RE(max) IE(min) = [–0.7 V – (–12 V)]/10.5 kΩ IE(min) = 1.08 mA

(Eq. 8-14)

VC(min) = VCC – IC(max)RC(max) VC(min) = 12 V – (1.19 mA)(4935 kΩ) VC(min) = 6.13 V

(Eq. 8-15)

Answer: The maximum collector voltage is 7.18 V. The minimum collector voltage is 6.13 V.

8-16. a. Increase: If R1 increases, VB decreases, VE decreases, IE decreases, IC decreases, the voltage drop across RC decreases, and VC increases. b. Increase: If R2 decreases, VB decreases, VE decreases, IE decreases, IC decreases, the voltage drop across RC decreases, and VC increases. c. Increase: RE increases, IE decreases, IC decreases, the voltage drop across RC decreases, and VC increases. d. Increases: RC decreases, the voltage drop across RC decreases, and VC increases. e. Increases: If VCC increases and the voltage drop across RC does not change, VC increases. f. Remain the same: βdc does not affect IC. Therefore the voltage drop across RC does not change, nor does VC. 8-17. a. Decreases: If R1 increases, VB increases, VE increases, IE decreases, IC decreases, the voltage drop across the collector resistor decreases, and VC decreases. b. Increases: If R2 increases, VB decreases, VE decreases, IE increases, IC increases, the voltage drop across the collector resistor increases, and VC increases. c. Decreases: RE increases, IE decreases, IC decreases, the voltage drop across the collector resistor decreases, and VC decreases. d. Increase: IC remains the same, RC increases, the voltage drop across the collector resistor increases, and VC increases. e. Increase: Since VBE does not increase in proportion to the increase in voltage supply, as do VB and VCC, the voltage drop across the emitter resistor increases, causing IE to increase. This causes the voltage drop across the collector resistor to increase and VC to increase. f. Remain the same: βdc does not affect IC. Therefore the voltage drop across RC does not change, nor does VC. 8-18. a. The approximate collector voltage is 12 V when R1 is open due to no collector current. b. The approximate collector voltage is 2.93 V when R2 is open, the transistor is in saturation. CEB can be approximated as a short. c. The approximate collector voltage is 12 V when RE is open due to no collector current. d. The approximate collector voltage is 0.39 V when RC is open. The collector current is zero, therefore the base current is equal to the emitter current. The circuit becomes a voltage divider of 150 Ω and 33 Ω driving 10 Ω through the base-emitter diode. Thevenize the base voltage divider to get a VTH = 2.16 V and a RTH = 27 V Ω. This Thevenin circuit has a load of 10 Ω and a diode. Now solve for a current of 39.57 mA, which leads to an emitter voltage of 395 mV. e. The approximate collector voltage is 12 V when the collector-emitter is open due to no collector current. 8-19. a. If R1 is open, the base voltage increases to 10 V and the transistor cuts off. Therefore, the collector voltage is zero. b. If R2 is open, the transistor goes into saturation, similar to the preceding problem. Again, you can approximate the saturated transistor as a CEB short; that is, all three terminals shorted. Then, 10 kΩ is in parallel with 3.6 kΩ, which is 2.65 kΩ. This is in series with 1 kΩ and 10 V. The series current is 10 V divided by 3.65 kΩ, or 2.74 mA. Multiply by 2.65 kΩ to get 7.57 V, the approximate value of collector voltage. c. With RE open, there is no collector current and the collector voltage is zero. d. With RC open, the transistor has no collector current. Similar to the preceding problem, the circuit becomes a voltage divider driving the emitter resistor through

1-34 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

the base-emitter diode. The Thevenin voltage and resistance facing the base-emitter diode are 1.8 V and 1.8 kΩ. The current through the emitter resistor is (1.8 V – 0.7 V) divided by (1.8 kΩ + 1 kΩ), or 0.393 mA. Multiply by 1 kΩ to get 0.393 V for the voltage across the emitter resistor. Subtract this from 10 V to gel 9.6 V at the emitter node. Subtract 0.7 V to get 8.9 V at the base node. Add 0.7 V to get the voltage at the collector node. The final answer is therefore 9.4 V at the collector when RC is open. If you don’t believe it, build the circuit and measure the collector voltage with the collector resistor open. e. When the collector-emitter terminals are open, there is no collector current and the collector voltage is zero. 8-20. Given:

V2 = [R2/(R1 + R2)]VCC V2 = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]10 V V2 = 1.8 V VRE = V2 – 0.7 V VRE = 1.8 V – 0.7 V VRE = 1.1 V

At cutoff, the maximum possible voltage across VCE is 8.9 V. Answer: The saturation current is 2.47 mA, and the collector-emitter cutoff voltage is 8.9 V.

Solution: VBB = [R2/(R1 + R2)]VCC VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)] – 10 V VBB = –1.8 V VE = V2 + 0.7 V VE = –1.8 V + 0.7 V VE = –1.1 V

IE = VRE/RE IE = 1.1 V/1 kΩ IE = 1.1 mA

IE = VE/RE IE = 1.1 V/1 kΩ IE = 1.1 mA

(Eq. 8-4)

I C ≈ IE

VC = ICRC VC = (1.1 mA)(3.6 kΩ) VC = 3.96 V Answer: The collector voltage is 3.96 V. 8-21. Given: R1 = 10 kΩ R2 = 2.2 kΩ RE = 1 kΩ RC = 3.6 kΩ VCC = 10 V VBE = 0.7 V V2 = 1.8 V (from Prob. 8-20) VRE = 1.1 V (from Prob. 8-20) IE = 1.1 mA (from Prob. 8-20) VC = 3.96 V (from Prob. 8-20) Solution: VCE = VCC – VC – VRE VCE = 10 V – 3.96 V – 1.1 V VCE = 4.94 V Answer: The collector-emitter voltage is –4.94 V since the collector is less positive than the emitter. R1 = 10 kΩ R2 = 2.2 kΩ RE = 1 kΩ RC = 3.6 kΩ VCC = 10 V VBE = 0.7 V

IC = 8.9 V/Rc IC = 8.9 V/3.6 kΩ IC = 2.47 mA

R1 = 10 kΩ R2 = 2.2 kΩ RE = 1 kΩ RC = 3.6 kΩ VCC = –10 V VBE = 0.7 V

Solution:

8-22. Given:

Solution: Because of the voltage divider, there will always be a 1.1-V drop across RE, and at saturation VCE = 0 V. This leaves 8.9 V across RC at saturation.

8-23. Given:

R1 = 10 kΩ R2 = 2.2 kΩ RE = 1 kΩ RC = 3.6 kΩ VCC = 10 V VBE = 0.7 V

I C ≈ IE

V2 = 1.8 V (from Prob. 8-20) VRE = 1.1 V (from Prob. 8-20) IE = 1.1 mA (from Prob. 8-20) VC = 3.96 V (from Prob. 8-20)

(Eq. 8-4)

VC = VCC + ICRC VC = –10 V + (1.1 mA)(3.6 kΩ) VC = –6.04 V Answer: The collector voltage is –6.04 V, and the emitter voltage is –1.1 V.

CRITICAL THINKING 8-24. The circuit is no longer considered stiff or independent of Beta. The base current is not small as compared to the voltage divider current. 8-25. The maximum power dissipation of the 2N3904 is 625 mW. The transistor is dissipating 705 mW. The transistor will probably overheat and fail. 8-26. As long as the voltmeter has a high enough input resistance, it should read approximately 4.83 V. 8-27. Increase the power supply value, short R1. 8-28. Connect an ammeter between the power supply and the circuit. Measure VR1 and VC, then calculate and add their respective currents. 8-29. Given: (for Q1): R1 = 1.8 kΩ R2 = 300 Ω RE = 240 Ω RC = 1 kΩ VCC = 15 V VBE = 0.7 V

1-35 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

8-30. Given:

Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) VBB = [300 Ω/(1.8 kΩ + 300 Ω)]15 V VBB = 2.14 V VE = VBB – 0.7 V (Eq. 8-2) VE = 2.14 V – 0.7 V VE = 1.44 V

R1 = 10 kΩ RE = 1 kΩ RC = 8.2 kΩ VCC = 20 V VD = 0.7 V Solution:

IE = VE/RE (Eq. 8-3) IE = 1.44 V/240 Ω IE = 6 mA

VBB = 3(VD) VBB = 3(0.7 V) VBB = 2.1 V

I C ≈ IE

VE = VBB – 0.7 V VE = 2.1 V – 0.7 V VE = 1.4 V

(Eq. 8-4)

VC = VCC – ICRC (Eq. 8-5) VC = 15 V – (6 mA)(1 kΩ) VC = 9.0 V Given (for Q2): R1 = 910 Ω R2 = 150 Ω RE = 120 Ω RC = 510 Ω VCC = 15 V VBE = 0.7 V

IE = VE/RE (Eq. 8-3) IE = 1.4 V/1 kΩ IE = 1.4 mA I C ≈ IE

(Eq. 8-4)

VC = VCC – ICRC (Eq. 8-5) VC = 20 V – (1.4 mA)(8.2 kΩ) VC = 8.52 V Answer: The emitter current is 1.4 mA, and the collector voltage is 8.52 V.

Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) VBB = [150 Ω/(910 Ω + 150 Ω)]15 V VBB = 2.12 V VE = VBB – 0.7 V (Eq. 8-2) VE = 2.12 V – 0.7 V VE = 1.42 V IE = VE/RE (Eq. 8-3) IE = 1.42 V/120 Ω IE = 11.83 mA I C ≈ IE

(Eq. 8-2)

8-31. Given: VBB(1) = 2 V RE(1) = 200 Ω RC(1) = 1 kΩ RE(2) = 1 kΩ VCC = 16 V Solution: VE(1) = VBB(1) – 0.7 V VE(1) = 2.0 V – 0.7 V VE(1) = 1.3 V

(Eq. 8-4)

(Eq. 8-2)

VC = VCC – ICRC (Eq. 8-5) VC = 15 V – (11.83 mA)(510 Ω) VC = 8.97 V

IE(1) = VE/RE IE(1) = 1.3 V/200 Ω IE(1) = 6.5 mA

Given (for Q3):

I C ≈ IE

R1 = 1 kΩ R2 = 180 Ω RE = 150 Ω RC = 620 Ω VCC = 15 V VBE = 0.7 V

VC(1) = VCC – ICRC (Eq. 8-5) VC(1) = 16 V – (6.5 mA)(1 kΩ) VC(1) = 9.5 V

(Eq. 8-4)

VC(1) = VBB(2) VE(2) = VBB(2) – 0.7 V VE(2) = 9.5 V – 0.7 V VE(2) = 8.8 V

Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) VBB = [180 Ω/(1 kΩ + 180 Ω)]15 V VBB = 2.29 V VE = VBB – 0.7 V (Eq. 8-2) VE = 2.29 V – 0.7 V VE = 1.59 V IE = VE/RE (Eq. 8-3) IE = 1.59 V/150 Ω IE = 10.6 mA I C ≈ IE

(Eq. 8-3)

(Eq. 8-2)

Answer: The output voltage is 8.8 V. 8-32. Given: R1 = 620 Ω R2 = 680 Ω RE = 200 Ω VCC = 12 V VBE = 0.7 V Solution:

(Eq. 8-4)

VC = VCC – ICRC (Eq. 8-5) VC = 15 V – (10.6 mA)(620 Ω) VC = 8.43 V Answer: The collector voltage for Q1 is 9.0 V, for Q2 is 8.97 V, and for Q3 is 8.43 V.

V2 = [R2/(R1 + R2)]VCC V2 = [680 Ω/(620 Ω + 680 Ω)]12 V V2 = 6.28 V VRE = V2 – 0.7 V VRE = 6.28 V – 0.7 V VRE = 5.58 V IE = VRE/RE IE = 5.58 V/200 Ω

1-36 Downloaded by Kevin Aguilar Arias ([email protected])

(Eq. 8-3)

lOMoARcPSD|6455127

8-40. Answer:

IE = 27.9 mA ILED ≈ IE

Trouble 9: Since the base voltage is 1.1 V, it appears that the voltage divider is working but not properly. The emitter voltage is 0.7 V less than the base, so the emitter-base junction is working. If RC is open, the meter would complete the circuit and give a low voltage reading. The trouble is an open RC.

Answer: The LED current is 27.9 mA. 8-33. Given: R1 = 620 Ω RE = 200 Ω VCC = 12 V VBE = 0.7 V VZ = 6.2 V

Trouble 10: This is very similar to trouble 9 except that the collector voltage is 10 V. Since source voltage is read above an open, the trouble is an open collector-base junction.

Solution:

8-41. Answer:

VRE = VZ – 0.7 V VRE = 6.2 V – 0.7 V VRE = 5.5 V IE = VRE/RE IE = 5.5 V/200 Ω IE = 27.5 mA

Trouble 11: Since all the voltages are 0 V, the power supply is not working. (Eq. 8-3)

Trouble 12: With the emitter voltage at 0 V and the base voltage at 1.83 V, the emitter-base diode of the transistor is open.

ILED ≈ IE Answer: The LED current is 27.5 mA. 8-34. Given: RE = 51 kΩ R1 = 3.3R2; this ratio is necessary to prevent moving the Q point. Assume βdc = 100 Solution: R1||R2 < 0.01 βdcRE (Eq. 8-9) R1||R2 = 0.01(100)(51 kΩ) R1||R2 = 51 kΩ Since R2 is the smaller of the two resistors, make it 51 kΩ. Then the parallel resistance will not be higher than 51 kΩ, which satisfies the requirement. R1 = 3.3R2 R1 = 3.3(51 kΩ) R1 = 168.3 kΩ Answer: R1 maximum of 168.3 kΩ, R2 maximum of 51 kΩ, and the ratio between them 3.3:1. 8-35. Answer: With VB at 10 V and R2 is good, the trouble is R1 shorted. 8-36. Answer: Since VB is 0.7 V and VE is 0 V, the trouble is RE is shorted.

Chapter 9 AC Models SELF-TEST 1. a 2. b 3. c 4. c 5. a 6. d

PROBLEMS 9-1.

8-38. Answer: Trouble 5: Since VB is 0 V, it is either R1 open or R2 shorted. R2 is OK, so the trouble is R1 open.

Trouble 8: R2 is shorted.

Given: C = 47 µF R = 10 kΩ Solution:

Trouble 4: Since all the voltages are the same, the trouble is that all the transistor terminals are shorted together.

9-2.

Trouble 7: Since VC is 10 V, there is an open below it or a short above it. A shorted RC would not affect VB; therefore there must be an open below it. If the transistor is open, VB would be 0 V; therefore the trouble is an open RE.

17. c 18. b 19. b 20. c 21. a

7. To permit the output voltage to swing over the largest possible voltage when the input signal is large enough to produce a maximum output. 8. Models provide mathematical and logical insight into the operation of a device. The two common transistor models are the T and the π. 11. It would become zero because there is no collector current.

Trouble 3: Since VC is 10 V and VE is 1.1 V, the transistor is good. Therefore the trouble is RC, which is shorted.

8-39. Answer:

12. d 13. b 14. b 15. d 16. b

JOB INTERVIEW QUESTIONS

8-37. Answer:

Trouble 6: R2 is open.

7. b 8. b 9. c 10. c 11. b

XC = 1/(2πfC) (Eq. 9-1) XC < 0.1R 1/(2πfC) = 0.1R 1/(2πC) = (0.1R)f f = 1/{[2π(47µF)][0.1(10 kΩ)]} f = 3.39 Hz Answer: The lowest frequency where good coupling exists is 3.39 Hz. Given: C = 47 µF R = 1 kΩ Solution: XC = 1/(2πfC) XC < 0.1R (Eq. 9-1) 1/(2πfC) = 0.1R 1/(2πC) = (0.1R)f f = 1/{[2π(47 µF)][0.1(1 kΩ)]} f = 33.9 Hz

1-37 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

VE = VBB – VBE VE = 2.7 V – 0.7 V VE = 2.0 V IE = VE/RE IE = 2.0 V/940 Ω IE = 2.128 mA

(Eq. 8-3)

ie(pp) < 0.1 IEQ (Eq. 9-6) ie(pp)max = 0.1 (2.13 mA) ie(pp)max = 213 µA Answer: The maximum ac emitter current for small signal operation is 213 µA. 9-11. Given: ic = 15 mA ib = 100 µA Solutions: β = ic/ib (Eq. 9-8) β = 15 mA/100 µA β = 150 Answer: The ac beta is 150. 9-12. Given: β = 200 ib = 12.5 µA

9-15. Given: (from Prob. 9-10) IE = 2.13 mA Solution: re′ = 25 mV/IE (Eq. 9-10) re′ = 25 mV/2.13 mA re′ = 11.7 Ω Answer: The ac resistance of the emitter diode is 11.7 Ω. 9-16. Given: re′ = 5.88 Ω (from Prob. 9-14) β = 200 Solution: Zin(base) = βre′ (Eq. 9-11) Zin(base) = 200 (5.88 Ω) Zin(base) = 1.18 kΩ Answer: The input impedance to the base is 1.18 kΩ. 9-17. Given: re′ = 11.7 Ω (from Prob. 9-15) β = 200 Solution: (Eq. 9-11) Zin(base) = βre′ Zin(base) = 200 (11.7 Ω) Zin(base) = 2.34 kΩ

Solutions:

Answer: The input impedance to the base is 2.34 kΩ.

β = ic/ib (Eq. 9-8) ic = βib ic = 200 (12.5 µA) ic = 2.5 mA

9-18. Given: Since the collector resistor does not affect the dc emitter current, the ac emitter resistance does not change. Since the beta did not change either, the input resistance remains the same as in problem 9-16.

Answer: The ac collector current is 2.5 mA. 9-13. Given:

Answer: The input impedance to the base is 1.18 kΩ. 9-19. Answer:

β = 100 ic = 4 mA

Zin(base) = 207 Ω

Zout = 1.02 kΩ

Solutions: β = ic/ib (Eq. 9-8) ib = ib/β ib = 4 mA/100 ib = 40 µA

+ –

1.5 kΩ

330 Ω

b re’

IC

1.2 kΩ

6.8 kΩ

Answer: The ac base current is 40 µA. b = 150

9-14. Given:

re’ = 5.86 Ω

9-20. Answer:

R1 = 1.5 kΩ R2 = 330 Ω RC = 1.2 kΩ RE = 470 Ω VCC = 15 V VBE = 0.7 V

+ –

3 kΩ

660 Ω

b re’

IC

2.4 kΩ

13.6 kΩ

Solution: VBB = [R2/(R1 + R2)]VCC (Voltage divider formula) VBB = [330 Ω/(1.5 kΩ + 330 Ω)]15 V VBB = 2.7 V VE = VBB – VBE VE = 2.7 V – 0.7 V VE = 2.0 V IE = VE/RE IE = 2.0 V/470 Ω IE = 4.26 mA

(Eq. 8-3)

re′ = 25 mV/IE (Eq. 9-10) re′ = 226.7 Ω mV/4.26 mA re′ = 5.88 Ω

Answer: The ac resistance of the emitter diode is 5.88 Ω.

9-21. Answer: min hfe = 50 max hfe = 200 Current is 1 mA Temperature 25°C 9-22. Given: IE = IC = 5 mA From Fig. 13 on the data sheet hie is 875 Ω at 5 mA; from Fig. 11 on the data sheet hfe is 150 Ω at 5 mA. Solution: re′ = (25 mV)/IE (Eq. 9-10) re′ = (25 mV)/5 mA re′ = 5 Ω

1-39 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

re′ = hic/hfe re′ = 875 Ω/150 re′ = 5.83

Answer: The value of re′ is 5.83 Ω. The calculated value is larger than the ideal.

CRITICAL THINKING

1/(2πfC) = 0.1R 1/(2πC) = (0.1R)(f) 1/[(2πC)(0.1R) = f f = 1/{[2π(2 µF)][0.1(800 Ω)]} f = 995 Hz Answer: The lowest frequency at which good bypassing exists is 995 Hz. 9-27. Given (for the first transistor):

9-23. Answer: The capacitor has a certain amount of leakage current, and this current will flow through the resistor and create a voltage drop across the resistor. 9-24. Answer: A wire has a very small inductance value. As the frequency increases, the inductive reactance starts to become significant. The wires connected to the capacitor and the leads will start to have an inductive reactance, causing the voltage to rise at the node. 9-25. Given:

R1 = 10 kΩ R2 = 2.2 kΩ RC = 3.6 kΩ RE = 1 kΩ VCC = 10 V VBE = 0.7 V β = 250 Solution (for the first transistor): VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]10 V VBB = 1.8 V

R1 = 10 kΩ R2 = 30 kΩ R3 = 20 kΩ R4 = 40 kΩ R5 = 40 kΩ C = 10 µF

VE = VBB – VBE VE = 1.8 V – 0.7 V VE = 1.1 V

Solution: R12(EQ) = 1/(1/R1 + 1/R2) R12(EQ) = 1/(1/(10 kΩ) + 1/(30 kΩ)) R12(EQ) = 7.5 kΩ R35(EQ) = 1/(1/R3 + 1/R4 + 1/R5) R35(EQ) = 1/[1/(20 kΩ) + 1/(40 kΩ) + 1/(40 kΩ)] R35(EQ) = 10 kΩ RT = R12(EQ) + R35(EQ) RT = 7.5 kΩ + 10 kΩ RT = 17.5 kΩ

(Eq. 8-2)

IE = VE/RE (Eq. 8-3) IE = 1.1 V/1 kΩ IE = 1.1 mA re′ = (25 mV)/IE (Eq. 9-10) re′ = (25 mV)/1.1 mA re′ = 22.7 Ω zin(base) = βre′ (Eq. 9-11) zin(base) = (250)(22.7 Ω) zin(base) = 5.68 kΩ Answer (for the first transistor): The input impedance of the base is 5.68 kΩ.

XC = 1/(2πfC) (Eq. 9-1) XC < 0.1R 1/(2πfC) = 0.1R 1/(2πC) = (0.1R)(f) 1/(2πC)(0.1R) = f f = 1/{[2π(10 µF)][0.1(17.5 kΩ)]} f = 9.09 Hz Answer: The lowest frequency at which good coupling exists is 9.09 Hz. 9-26. Given: R1 = 1 kΩ R2 = 4 kΩ C = 2 µF Answer: The Thevenin resistance is R1 in parallel with R2. R12(EQ) = 1/(1/R1 + 1/R2) R12(EQ) = 1/[1/(1 kΩ) + 1/(4 kΩ)] R12(EQ) = 800 Ω XC = 1/(2πfC) XC < 0.1R (Eq. 9-5)

Given (for the second transistor): R1 = 10 kΩ R2 = 2.2 kΩ RC = 3.6 kΩ RE = 1 kΩ VCC = 10 V VBE = 0.7 V β = 100 VBB = 1.8 V (from the first part of the problem) VE = 1.1 V (from the first part of the problem) IE = 1.1 mA (from the first part of the problem) re′ = 22.7 Ω (from the first part of the problem) Solution (for the second transistor): zin(base) = βre′ (Eq. 9-11) zin(base) = (100)(22.7 Ω) zin(base) = 2.27 kΩ Answer: The input impedance of the first base is 5.68 kΩ, and the input impedance of the second base is 2.27 kΩ. 9-28. Answer: See figure at foot of page.

AC equivalent circuit for Prob. 9-26.

1-40 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

9-29. Given: R = 30 Ω f = 20 Hz to 20 kHz Solution: XC = 1/(2πfC) (Eq. 9-5) XC < 0.1R 1/(2πfC) = 0.1R 1/(2πf) = (0.1R)(C) 1/(2πf)(0.1R) = C C = 1/{[2π(20 Hz)][0.1(30 Ω)]} C = 2653 µF Answer: The capacitor would have to be at least 2653 µF, or 2700 µF (standard value).

rc = 2.65 kΩ Av = rc / re′ (Eq. 10-3) Av = 2.65 kΩ/22.7 Ω Av = 117 vout = Avin vout = 117(2 mV) vout = 234 mV Answer: The output voltage is 234 mV. 10-2.

R1 = 10 kΩ R2 = 2.2 kΩ RC = 3.6 kΩ RE = 1 kΩ RL = 5 kΩ VCC = 10 V VBB = 1.8 V VE = 1.1 V IE = 1.1 V re′ = 22.7 Ω

Chapter 10 Voltage Amplifiers SELF-TEST 1. c 2. b 3. a 4. c 5. d 6. c 7. c

8. b 9. c 10. c 11. d 12. b 13. a 14. d

15. a 16. a 17. d 18. b 19. a 20. d 21. b

22. c 23. b 24. a 25. a 26. a 27. b 28. c

PROBLEMS 10-1.

Given: R1 = 10 kΩ R2 = 2.2 kΩ RC = 3.6 kΩ RE = 1 kΩ RL = 10 kΩ VCC = 10 V VBE = 0.7 V Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]10 V VBB = 1.8 V VE = VBB – VBE (Eq. 8-2) VE = 1.8 V – 0.7 V VE = 1.1 V IE = VE/RE (Eq. 8-3) IE = 1.1 V/1 kΩ IE = 1.1 mA re′ = (25 mV)/IE (Eq. 9-10) re′ = (25 mV)/1.1 mA re′ = 22.7 Ω rc = RC||RL (Eq. 10-2) rc = 3.6 kΩ||10 kΩ

(from Prob. 10-1) (from Prob. 10-1) (from Prob. 10-1) (from Prob. 10-1)

Solution: rc = RC||RL rc = 3.6 kΩ||5 kΩ rc = 2093 kΩ Av = rc / re′ Av = 2093 kΩ/22.7 Ω Av = 92.2

JOB INTERVIEW QUESTIONS 5. Coupling capacitors C1 and C2 must be replaced by wires. The ground reference has to be shifted to provide 0 V at the collector. Also, the bypass capacitor needs to be eliminated, and some of the resistors resized. 7. Very high input impedance to limit the current drawn from the preceding stage and to prevent distortion. Also, high current gain and low output impedance to provide a match to a speaker. 9. 100 is good choice for small-signal transistors.

Given:

Answer: The voltage gain is 92.2. 10-3.

Given: R1 = 10 kΩ R2 = 2.2 kΩ RC = 3.6 kΩ RE = 1 kΩ RL = 10 V VCC = 15 V Solution: VBB = [R2/(R1 + R2)]VCC VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]15 V VBB = 2.7 V VE = VBB – VBE VE = 2.7 V – 0.7 V VE = 2.0 V IE = VE/RE IE = 2.0 V/1 kΩ IE = 2 mA re′ = (25 mV)/IE re′ = (25 mV)/2 mA re′ = 12.5 Ω rc = RC||RL rc = 3.6 kΩ||10 kΩ rc = 2.65 kΩ Av = rc / re′ Av = 2.65 kΩ/12.5 Ω Av = 212 vout = Av(vin) vout = 212(1 mV) vout = 212 mV Answer: The voltage gain is 212, the output voltage is 212 mV.

1-41 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

10-4.

IE = 1.1 V/2 kΩ IE = 0.55 mA re′ = (25 mV)/IE (Eq. 9-10) re′ = (25 mV)/0.55 mA re′ = 45.5 Ω

Given: R1 = 10 kΩ R2 = 2.2 kΩ RC = 3.6 kΩ RE = 1 kΩ RL = 10 kΩ RG = 600 Ω VBE = 0.7 V VCC = 15 V Assume β = 100 VBB = [R2/(R1 + R2)]VCC VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]15 V VBB = 2.7 V

rc = RC||RL (Eq. 10-2) rc = 3.6 kΩ||10 kΩ rc = 2.65 kΩ Av = rc / re′ (Eq. 10-3) Av = 2.65 kΩ/45.5 Ω Av = 58 zin = R1||R2|| βre′ zin = 10 kΩ||2.2 kΩ||100(45.5 Ω) zin = 1.29 kΩ

VE = VBB – VBE VE = 2.7 V – 0.7 V VE = 2.0 V

vin = [zin(RG + zin)]vg (Eq. 10-4) vin = [1.29 kΩ/(600 Ω + 1.29 kΩ)]1 mV vin = 0.683 mV

IE = VE/RE IE = 2.0 V/1 kΩ IE = 2 mA re′ = (25 mV)/IE re′ = (25 mV)/2 mA re′ = 12.5 Ω

vout = Av(vin) vout = 58(0.683 mV) vout = 39.6 mV

Solution:

Answer: The output voltage is 39.6 mV. 10-6.

rc = RC||RL rc = 3.6 kΩ||10 kΩ rc = 2.65 kΩ Av = rc / re′ Av = 2.65 kΩ/12.5 Ω Av = 212 zin = R1||R2|| βre′ zin = 10 kΩ||2.2 kΩ||1.25 kΩ zin = 738 Ω

10-5.

Given: R1 = 10 kΩ R2 = 2.2 kΩ RC = 3.6 kΩ RE = 1 kΩ RL = 10 kΩ RG = 300 Ω VBE = 0.7 V VCC = 10 V Assume β = 100 Solution:

vin = [zin/(RG + zin)]vg vin = [738 Ω/(600 Ω + 738 Ω)]1 mV vin = 551.57 µV

VBB = [R2/(R1 + R2)]VCC VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]10 V VBB = 1.8 V

vout = Av(vin) vout = 212(551.57 µV) vout = 117 mV

VE = VBB – VBE VE = 1.8 V – 0.7 V VE = 1.1 V

Answer: The voltage gain is 212, the output voltage is 117 mV.

IE = VE/RE IE = 1.1 V/1 kΩ IE = 1.1 mA

Given:

re′ = (25 mV)/IE re′ = (25 mV)/1.1 mA re′ = 22.7 Ω

R1 = 10 kΩ R2 = 2.2 kΩ RC = 3.6 kΩ RE = 2 kΩ RL = 10 kΩ RG = 600 Ω VCC = 10 V VBE = 0.7 V Assume β = 100 Solution: VBB = [R1/(R1 + R2)]VCC (Eq. 8-1) VBB = [10 kΩ/(10 kΩ + 2.2 kΩ)]10 V VBB = 1.8 V VE = VBB – VBE VE = 1.8 V – 0.7 V VE = 1.1 V IE = VE/RE

(Eq. 8-2)

(Eq. 8-3)

rc = RC||RL rc = 3.6 kΩ||10 kΩ rc = 2.65 kΩ Av = rc / re′ Av = 2.65 kΩ/22.7 Ω Av = 117 zin = R1||R2|| βre′ zin = 10 kΩ||2.2 kΩ||2.27 kΩ zin = 1 k Ω vin = [zin/(RG + zin)]vg vin = [1 kΩ/(300 Ω + 1 kΩ)]1 mV vin = 769 ¿V vout = Av(vin)

1-42 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

vout = 117(769 µV) vout = 90 mV

10-8.

R1 = 10 kΩ R2 = 2.2 kΩ RC = 3.6 kΩ RE = 1 kΩ RL = 10 kΩ RG = 600 Ω VCC =12 V VBE = 0.7 V β = 100

Answer: The voltage gain is 117, the output voltage is 90 mV. 10-7.

Given: R1 = 10 kΩ R2 = 2.2 kΩ RC = 3.6 kΩ RE = 1 kΩ RL = 10 kΩ RG = 600 Ω VCC = 10 V VBE = 0.7 V β = 100

Solution: VBB = [R 2 /(R 1 + R2 )]VCC (Eq. 8-1) VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]12 V VBB = 2.16 V VE = VBB – VBE (Eq. 8-2) VE = 2.16 V – 0.7 V VE = 1.46 V

Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]10 V VBB = 1.8 V VE = VBB – VBE VE = 1.8 V – 0.7 V VE = 1.1 V

Given:

I E = VE / RE (Eq. 8-3) IE = 1.46 V/1 kΩ IE = 1.46 mA re′ = (25 mV) / I E (Eq. 9-10) re′ = (25 mV) /1.46 mA re′ = 17.1 Ω zin = R1||R2 ||βre′ zin = 10 kΩ||2.2 kΩ||100(17.1 Ω) zin = 878 Ω

(Eq. 8-2)

IE = VE/RE (Eq. 8-3) IE = 1.1 V/1 kΩ IE = 1.1 mA re′ = (25 mV) / I E (Eq. 9-10) re′ = (25 mV) /1.1 mA re′ = 22.7 Ω zin = R1 || R2 || βre′ zin = 10 kΩ||2.2 kΩ||100(22.7 Ω) zin = 1.0 kΩ

The input impedance for each stage is 878 Ω. vin(1) = [zin/(RG + zin)] vg (Eq. 10-4) vin(1) = (878 Ω/ (600 Ω + 878 Ω)]1 mV vin(1) = 0.594 mV

The input impedance for each stage is 1.0 kΩ.

The input impedance for the second stage is the load resistance for the first stage.

vin(1) = [zin/(RG + zin)]vg (Eq. 10-4) vin(1) = [1.0 kΩ/(600 Ω + 1.0 kΩ)]1 mV vin(1) = 0.625 mV

(Eq. 10-2) rc = RC||RL rc = 3.6 kΩ||878 Ω rc = 706 Ω Av = rc / re′ (Eq. 10-7) Av = 706 Ω/17.1 Ω Av = 41.3

The input impedance for the second stage is the load resistance for the first stage. rC = RC ||RL (Eq. 10-2) rc = 3.6 kΩ||1.0 kΩ rc = 783 Ω Av = re / re′ (Eq.10-3) Av = 783 Ω/22.7 Ω Av = 34.5

The output voltage of the first stage is the input voltage for the second stage. vout(1) = Avin vout(1) = 41.3(0.594 mV) vout(1) = 24.5 mV

The output voltage of the first stage is the input voltage for the second stage.

rc(2) = RC||RL (Eq. 10-2) rc(2) = 3.6 kΩ||10 kΩ rc(2) = 2.65 kΩ Av(2) = rc / re′ (Eq. 10-7) Av(2) = 2.65 kΩ/17.1 Ω Av(2) = 155

vout(1) = Av(vin) vout(1) = 34.5(0.625 mV) vout(1) = 21.6 mV rc (2) = RC || RL (Eq.10-2) rc(2) = 3.6 kΩ||10 kΩ rc(2) = 2.65 kΩ Av(2) = rc / re′ (Eq. 10-3) Av(2) = 2.65 kΩ/22.7Ω Av(2) = 117 vout(2) = Av(vin) vout(2) = 117(21.6 mV) vout(2) = 2.53 V

vout(2) = Av(vin) vout(2) = 155(24.5 mV) vout(2) = 3.80 V Answer: The output voltage is 3.80 V. 10-9.

Answer: The base voltage of the first stage is 0.625 mV, the base voltage of the second stage is 21.6 mV, and the voltage across the collector resistor is 2.53 V.

Given: R1 = 10 kΩ R2 = 2.2 kΩ RC = 3.6 kΩ RE = 1 kΩ RL = 10 kΩ RG = 600 Ω

1-43 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

rc = 2.65 kΩ

VCC = 10 V VBE = 0.7 V β = 300 Solution:

Av = rc/re Av = 2.65 kΩ/180 Ω Av = 14.7

VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]10 V VBB = 1.8 V

zin = R1||R2||βrc zin = 10 kΩ||2.2 kΩ||18 kΩ zin = 1.64 kΩ

VE = VBB – VBE VE = 1.8 V – 0.7 V VE = 1.1 V

vin = [zin/(RG + zin)]vg vin = [1.64 kΩ/(600 Ω + 1.64 kΩ)]25 mV vin = 18.3 mV

(Eq. 8-2)

IE = VE/RE (Eq. 8-3) IE = 1.1 V/1 kΩ IE = 1.1 mA re′ = (25 mV) / I E (Eq. 9-10) re′ = (25 mV) /1.1 mA re′ = 22.7 Ω zin = R1 || R2 || βre′ zin = 10 kΩ||2.2 kΩ||300(22.7 Ω) zin = 1.43 kΩ

The input impedance for each stage is 1.43 kΩ. vin(1) = [zin/(RG + zin)]vg (Eq. 10-4) vin(1) = [1.43 kΩ/(600 Ω + 1.43 kΩ)]1 mV vin(1) = 0.704 mV The input impedance for the second stage is the load resistance for the first stage. rc = RC||RL (Eq. 10-2) rc = 3.6 kΩ||1.43 kΩ rc = 1.02 kΩ Av = rc / re′ (Eq. 10-3) Av = 1.02 kΩ/22.7 Ω Av = 45 The output voltage of the first stage is the input voltage for the second stage. vout(1) = Av (vin) vout(1) = 45(0.704 mV) vout(1) = 31.7 mV rc(2) = RC||RL (Eq. 10-2) rc(2) = 3.6 kΩ||10 kΩ rc(2) = 2.65 kΩ Av(2) = rc / re′ (Eq. 10-3) Av(2) = 2.65 kΩ/22.7 Ω Av(2) = 117 vout(2) = Avin vout(2) = 117 (31.7 mV) vout(2) = 3.71 V

R1 = 10 kΩ R2 = 2.2 kΩ RC = 3.6 kΩ RE = 820 Ω re = 180 Ω RL = 10 kΩ RG = 600 Ω VBE = 0.7 V VCC = 10 V Assume β = 100 Solution:

Answer: The voltage gain is 14.7, the output voltage is 269 mV. 10-11. Given: R1 = 10 kΩ R2 = 2.2 kΩ RC = 3.6 kΩ RE = 820 Ω re = 180 Ω RL = 10 kΩ RG = 50 Ω VBE = 0.7 V VCC = 10 V Assume β = 100 Solution: rc = RC||RL rc = 3.6 kΩ||10 kΩ rc = 2.65 kΩ Av = rc/re Av = 2.65 kΩ/180 Ω Av = 14.7 zin = R1||R2||βre zin = 10 kΩ||2.2 kΩ||18 kΩ zin = 1.64 kΩ vin = [zin/(RG + zin)]vg vin = [1.64 kΩ/(50 Ω + 1.64 kΩ)]50 mV vin = 48.52 mV vout = Av(vin) vout = 14.7(48.52 mV) vout = 713 mV Answer: The voltage gain is 14.7, the output voltage is 713 mV. 10-12. Given:

Answer: The output voltage is 3.71 V. 10-10. Given:

vout = Av(vin) vout = 14.7(18.3 mV) vout = 269 mV

R1 = 10 kΩ R2 = 2.2 kΩ RC = 3.6 kΩ RE = 820 Ω re = 180 Ω RL = 3.6 kΩ RG = 600 Ω VBE = 0.7 V VCC = 10 V Assume β = 100 Solution: rc = RC||RL rc = 3.6 kΩ||3.6 kΩ rc = 1.8 kΩ

rc = RC||RL rc = 3.6 kΩ||10 kΩ

1-44 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

VCC = 10V VBE = 0.7 V

Av = rc/re Av = 1.8 kΩ/180 Ω Av = 10 Answer: The voltage gain is 10. 10-13. Given: R1 = 10 kΩ R2 = 2.2 kΩ RC = 3.6 kΩ RE = 820 Ω re = 180 Ω RL = 10 kΩ RG = 600 Ω VCC = 30 V VBE = 0.7 V Solution: rc = RC||RL (Eq. 10-2) rc = 3.6 kΩ||10 kΩ rc = 2.65 kΩ Av = rc / re (Eq. 10-7) Av = 2.65 kΩ/180 Ω Av = 14.7

Answer: The voltage gain is 14.7. 10-14. Given:

Solution: (Eq. 10-10)

Answer: The voltage gain is 100. 10-15. Given: re = 125 Ω Av = 100 Solution: Av = rf / re rf = 100(125 Ω) rf = 12.5 kΩ

(Eq. 10-10)

Answer: The feedback resistor would need to be 12.5 kΩ. 10-16. Answer: Since the capacitor is an open to direct current, the dc voltages do not change. The first stage is now swamped. Therefore the voltage gain is greatly reduced and the input impedance is increased so that more of the generator voltage is seen at the input. The overall effect is a reduced input voltage to the second stage. Since the gain of the second stage remains the same and the input voltage is reduced, the output voltage is also reduced. 10-17. Answer: Since there is a voltage at the second stage input, the cause is most likely in the second stage. Some of the possible causes are: open transistor, open emitter resistor, open collector resistor, or open output coupling capacitor.

CRITICAL THINKING 10-18. Given: R1 = 20 kΩ R2 = 4.4 kΩ RC = 7.2 kΩ RE = 2 kΩ RL = 20 kΩ

VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) VBB = [4.4 kΩ/(20 kΩ + 4.4 kΩ)]10 V VBB = 1.8V VE = VBB – VBE VE = 1.8V – 0.7V VE = 1.1 V

(Eq. 8-2)

IE = VE/RE IE = 1.1 V/2 kΩ IE = 0.55 mA

(Eq. 8-3)

re′ = (25 mV) / I E re′ = (25 mV) / 0.55 mA re′ = 45.5 Ω

(Eq. 9-10)

rc = RC||RL (Eq. 10-2) rc = 7.2 kΩ||20 kΩ rc = 5.3 kΩ Av = rc / re′ (Eq. 10-3) Av = 5.3 kΩ/45.5 Ω Av = 116 Answer: The voltage gain is 116. 10-19. Given:

rf = 5 kΩ re = 50 Ω Av = rf / re Av = 5 kΩ/50 Ω Av = 100

Solution:

R1 = 20 kΩ R2 = 4.4 kΩ RC = 7.2 kΩ RE = 2 kΩ RL = 20 kΩ RG = 1.2 kΩ VCC = 10 V VBE = 0.7 V Assume β = 100 VBB = 1.8 V (from Prob. 10-18) VE = 1.1 V (from Prob. 10-18) IE = 0.55 mA (from Prob. 10-18) re′ = 45.5 Ω (from Prob. 10-18) rc′ = 5.3 kΩ Av = 116 Solution: zin = R1 || R2 || βre′ zin = 20 kΩ||4.4 kΩ || 100(45.5 Ω) zin = 2.01 kΩ

vin = [zin/(RG + zin)]vg (Eq. 10-4) vin = [2.01 kΩ/(1.2 kΩ + 2.01 kΩ)]1 mV vin = 0.626 mV vout = Av(vin) vout = 116(0.626 mV) vout = 72.6 mV Answer: The output voltage is 72.6 mV. 10-20. Given: R1 = 20 kΩ R2 = 4.4 kΩ RC = 7.2 kΩ RE = 2 kΩ RL = 20 kΩ RG = 1.2 kΩ VCC = 10 V VBB = 0.7 V β = 100

1-45 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Trouble 4: Since the dc base voltage is 0 and there is an ac base voltage, the problem is an R1 open.

Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) VBB = [4.4 kΩ/(20 kΩ + 4.4 kΩ)] 10 V VBB = 1.8 V VE = VBB – VBE VE = 1.8 V – 0.7 V VE = 1.1 V

10-23. Answers: Trouble 5: Since there is no output ac voltage, the problem is C2 open.

(Eq. 8-2)

Trouble 6: Since there are no ac voltages and the base voltage has changed, the problem is in the input circuit. The voltage points to an open R2.

IE = VE/RE (Eq. 8-3) IE = 1.1 V/2 kΩ IE = 0.55 mA re′ = (25 mV) / I E (Eq. 9-10) re′ = (25 mV) / 0.55 mA re′ = 45.5 Ω zin = R1 || R2 || βre′ zin = 20 kΩ||4.4 kΩ||100(45.5 Ω) zin = 2.01 kΩ The input impedance for each stage is 2.01 kΩ.

Trouble 7: All the dc voltages are OK; thus the transistor and resistors are OK. Since the base and emitter ac voltages are the same, the problem appears to be an open bypass capacitor C3. Trouble 8: Since there are no ac voltages and the base voltage has changed, the problem is in the input circuit. Since the collector voltage is so low, the collector resistor is open. 10-24. Answers:

vin(1) = [zin/(RG + zin)]vg (Eq. 10-4) vin(1) = [2.01 kΩ/(1.2 kΩ + 2.01 kΩ)]1 mV vin(1) = 0.626 mV

Trouble 9: Since there are no dc voltages, the problem is no VCC. Trouble 10: Since the emitter voltage is 0 and the base voltage is near normal, the problem is an open BE diode.

The input impedance for the second stage is the load resistance for the first stage.

Trouble 11: With all the dc voltages the same, the problem is a shorted transistor in all three terminals.

(Eq. 10-2) rc = RC||RL rc = 3.6 kΩ||2.01 Ω rc = 1.29 kΩ Av = rc / re′ (Eq. 10-3) Av = 1.29 kΩ/45.5 Ω Av = 28.4

Chapter 11 CC and CB Amplifiers

The output voltage of the first stage is the input voltage for the second stage.

SELF-TEST

Trouble 12: Since all the ac voltages are 0, the problem could be the generator, RG open, or C1 open.

vout(1) = Av(vin) vout(1) = 28.4(0.626 mV) vout(1) = 17.8 mV rc(2) = RC||RL (Eq. 10-2) rc(2) = 7.2 kΩ||20 kΩ rc(2) = 5.3 kΩ Av(2) = rc / re′ (Eq. 10-3) Av(2) = 5.3 kΩ/45.5 Ω Av(2) = 116 vout(2) = Av(vin) vout(2) = 116(17.8 mV) vout(2) = 2.06 V Answer: The output voltage is 2.06 V. 10-21. Answer: The rc would be the collector resistance only: 3.6 kΩ. 10-22. Answers: Trouble 1: Since all the ac voltages are 0, the problem could be the generator, RG open, or C1 open. Trouble 2: Since the input voltage increased to 0.75 mV, the problem is an open RE. Trouble 3: Since there are no ac voltages and the base voltage has changed, the problem is in the input circuit. Since there is a 0.7-V drop across the BE diode, the transistor should be conducting and thus the collector voltage should be less than 10V. It appears that the BC diode is open, except the base voltages are not consistent with that problem. To make this problem correct for the BC diode open, return VB, VE, and vb to the OK values.

1. b 2. c 3. b 4. c 5. d 6. c 7. a 8. a

9. d 10. c 11. a 12. c 13. d 14. a 15. c 16. c

17. a 18. c 19. c 20. a 21. a 22. d 23. a 24. d

25. a 26. c 27. d 28. c 29. b 30. d

JOB INTERVIEW QUESTIONS 5. Voltage gain is always less than but usually near 1. The circuit is used as a current or power amplifier. Applications include stereo output stages, linear power-supply regulation, and drivers for relays, LEDs. 7. They allow excellent impedance matching and maximum power transfer to low-impedance loads. 11. None. 12. Power gain is the product of voltage gain and current gain. Although the voltage gain is slightly less than 1, the current gain is very large. Therefore, the power gain is very large.

PROBLEMS 11-1.

Given: R1 = 2.2 kΩ R2 = 2.2 kΩ RE = 1 kΩ RL = 3.3 kΩ RG = 50 Ω β = 200 VCC = 15 V VBE = 0.7 V

1-46 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

vin = [zin/(zin + RG)]VG vin = [1.09 kΩ/(1.09 kΩ + 50 Ω)]1 V vin = 0.956 V

Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) VBB = [2.2 kΩ/(2.2 kΩ + 2.2 kΩ)]15 V VBB = 7.5 V VE = VBB – VBE VE = 7.5 V – 0.7 V VE = 6.8 V

(Eq. 8-2)

IE = VE/RE IE = 6.8 V/1 kΩ IE = 6.8 mA

(Eq. 8-3)

re′ = 25 mV / I E re′ = 25 mV / 6.8 mA re′ = 3.68 Ω

(Eq. 9-10)

vout = Av(vin) (Eq. 9-3) vout = (0.995)(0.956 V) vout = 0.951 V Answer: The gain is 0.995, and the output voltage is 0.951 V. 11-4.

R1 = 2.2 kΩ R2 = 2.2 kΩ RE = 1 kΩ RL = 3.3 kΩ RG = 50 Ω β = 50 to 300 VCC =15 V VBE = 0.7 V re′ = 3.68 Ω (from Prob. 11-1) re = 767 Ω (from Prob. 11-1)

re = RE||RL (Eq. 11-1) re = 1 kΩ||3.3 kΩ re = 767 Ω

zin ( baSe) = β(re + re′) (Eq. 11-3) zin(base) = 200(767 Ω + 3.48 Ω) zin(base) = 154 kΩ zin(base) = 154 kΩ||2.2 kΩ||2.2 kΩ = 1.09 kΩ

Solution: zin(min) = R1 || R2 || β(re + re′) (Eq. 11-4) zin(min) = 2.2 kΩ||2.2 kΩ||50(767 Ω + 3.48 Ω) zin(min) = 1.07 kΩ zin(max ) = R1 || R2 || β(re + re′) (Eq. 11-4) zin(max) = 2.2 kΩ||2.2 kΩ||300(767 Ω + 3.48 Ω) zin(max) = 1.09 kΩ

Answer: The input impedance of the base is 154 kΩ, and the input impedance of the stage is 1.09 kΩ. 11-2.

Given: R1 = 2.2 kΩ R2 = 2.2 kΩ RE = 1 kΩ RL = 3.3 kΩ RG = 50 Ω β = 150 VCC = 15 V VBE = 0.7 V re′ = 3.68 Ω (from Prob. 11-1) re = 767 Ω (from Prob. 11-1)

vin(min) = [zin/(zin + RG)]VG vin(min) = [1.07 kΩ/(1.07 kΩ + 50 Ω)]1 V vin(min) = 0.955 V vin(min) = [zin/(zin + RG)]VG vin(min) = [1.09 kΩ/(1.09 kΩ + 50 Ω)]1 V vin(min) = 0.956 V Answer: The input voltage varies over the range of 0.955 to 0.956 V.

Solution: zin = R1 || R2 || β(re + re′) (Eq. 11-4) zin = 2.2 kΩ||2.2 kΩ||150(767 Ω + 3.48 Ω) zin = 1.09 kΩ

vin = [zin/(zin + RG)]VG vin = [1.09 kΩ/(1.09 kΩ + 50 Ω)]1 V vin = 0.956 V Answer: The input voltage is 0.956 V. 11-3.

Given: R1 = 2.2 kΩ R2 = 2.2 kΩ RE = 1 kΩ RL = 3.3 kΩ RG = 50 Ω β = 175 VCC = 15V VBE = 0.7 V re′ = 3.68 Ω (from Prob. 11-1) re = 767 Ω (from Prob. 11 − 1) Solution: Av = re /( re + re′) (Eq. 11-2) Av = 767/(767 + 3.48) Av = 0.995 zin = R1 || R2 || β(re + re′) (Eq. 11-4) zin = 2.2 kΩ||2.2 kΩ||175(767 Ω + 3.48 Ω) zin = 1.09 kΩ

Given:

11-5.

Given: R1 = 4.4 kΩ R2 = 4.4 kΩ RE = 2 kΩ RL = 6.6 kΩ RG = 100 Ω β = 150 VCC = 15 V VBE = 0.7 V Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) VBB = [4.4 kΩ/(4.4 kΩ + 4.4 kΩ)]15 V VBB = 7.5 V VE = VBB – VBE VE = 7.5 V – 0.7 V VE = 6.8 V

(Eq. 8-2)

IE = VE/RE (Eq. 8-3) IE = 6.8 V/2 kΩ IE = 3.4 mA re′ = 25 mV / I E (Eq. 9-10) re′ = 25 mV / 3.4 mA re′ = 7.35 Ω re = RE||RL (Eq. 11-1) re = 2 kΩ||6.6 kΩ re = 1.53 kΩ

1-47 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

zin = R1 || R2 || β(re + re′) (Eq. 11-4) zin = 4.4 kΩ||4.4 kΩ||150(1.53 kΩ + 7.35 kΩ) zin = 2.18 kΩ

11-8. Given:

R1 = 100 Ω R2 = 200 Ω RE = 30 Ω RL = 10 Ω RG = 50 Ω β = 175 VCC = 20 V VBE = 0.7 V re′ = 0.06 Ω (from Prob. 11-6) re = 7.5 Ω (from Prob. 11-6)

vin = [zin/(zin + RG)]VG vin = [2.18 kΩ/(2.18 kΩ + 100 Ω)]1 V vin = 0.956 V Answer: The input impedance doubles to 2.18 kΩ, and the input voltage remains the same at 0.956 V. 11-6.

Given: R1 = 100 Ω R2 = 200 Ω RE = 30 Ω RL = 10 Ω RG = 50 Ω β = 200 VCC = 20 V VBE = 0.7 V

Solution:

Solution:

Av = re /( re + re′) (Eq. 11-2) Av = 7.5/(7.5 + 0.06) Av = 0.992 zin = R1 || R2 || β(re + re′) (Eq. 11-4) zin = 100 Ω||200 Ω||175(7.5 Ω + 0.06 Ω) zin = 63.5 Ω

VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) VBB = [200 Ω/(100 Ω + 200 Ω)]20 V VBB = 13.3 V

vin = [zin/(zin + RG)]VG vin = [63.5 Ω/(63.5 Ω + 50 Ω)]1 V vin = 0.559 V

VE = VBB – VBE (Eq. 8-2) VE = 13.3 V – 0.7 V VE = 12.6 V

vout = Av(vin) (Eq. 9-3) vout = (0.992)(0.559 V) vout = 0.555 V

IE = VE/RE (Eq. 8-3) IE = 12.6 V/30 Ω IE = 420 mA re′ = 25 mV / I E (Eq. 9-10) re′ = 25 mV / 420 mA re′ = 0.06 Ω

Answer: The gain is 0.992, and the output voltage is 0.555 V.

re = RE||RL re = 30 Ω||10 Ω re = 7.5 Ω

(Eq. 11-1)

zin(base) = β(re + re′) (Eq. 11-3) zin(base) = 200(7.5 Ω + 0.06 Ω) zin(base) = 1.51 kΩ zin = R1 || R2 || β(re + re′) (Eq. 11-4) zin = 100 Ω||200 Ω||1.51 kΩ zin = 63.8 Ω Answer: The input impedance of the base is 1.51 kΩ, and the input impedance to the stage is 63.8 Ω. 11-7.

Given: R1 = 100 Ω R2 = 200 Ω RE = 30 Ω RL = 50 Ω RG = 50 Ω β = 150 VCC = 20 V VBE = 0.7 V re′ = 0.06 Ω (from Prob.11-6) re = 7.5 Ω (from Prob. 11-6) Solution: zin = R1 || R2 || β(re + re′) (Eq. 11-4) zin = 100 Ω||200 Ω||150(7.5 Ω + 0.06 Ω) zin = 63 Ω

vin = [zin/(zin + RG)]VG vin = [63 Ω/(63 Ω + 50 Ω)]1 V vin = 0.558 V

11-9.

Given: R1 = 2.2 kΩ R2 = 2.2 kΩ RE = 1 kΩ RL = 3.3 kΩ RG = 50 Ω β = 200 VCC = 15 V VBE = 0.7V re′ = 3.68 Ω (from Prob. 11-1) re = 767 Ω (from Prob. 11-1) Solution: zout = RE || [re′ + ( RG || R1 || R2 ) / β] (Eq. 11-5) zout = 1 kΩ||[3.68 Ω + (50 Ω||2.2 kΩ||2.2 kΩ)/200] zout = 3.9 Ω

Answer: The output impedance is 3.9 Ω. 11-10. Given:

R1 = 100 Ω R2 = 200 Ω RE = 30 Ω RL = 10 Ω RG = 50 Ω β = 100 VCC = 20 V VBE = 0.7 V re′ = 0.06 Ω (from Prob. 11-6) re = 7.5 Ω (from Prob. 11-6) Solution: zout = RE || [re′ + ( RG || R1 || R2 ) / β (Eq. 11-5) zout = 30 Ω||[0.06 Ω + (50 Ω||100 Ω||200 Ω)/100] zout = 0.342 Ω

Answer: The output impedance is 0.342 Ω.

Answer: The input voltage is 0.558 V.

1-48 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

re′ = (25 mV) / 5.85 mA re′ = 4.27 Ω

β = β1β2 (Eq. 11-7) β = 150(150) β = 22500 zin(base) = βre zin(base) = (22500)(4.44) zin(base) = 100 kΩ

rc = RC||RL rc = 1.5 kΩ||150 Ω rc = 136.5 Ω Av1 = rc / re′ Av1 = 136.5 Ω/4.27 Ω Av1 = 31.9

Answer: The input impedance of the base is 100 kΩ. 11-18. Given:

Answer: The voltage gain drops to 31.9. 11-15. Given:

R1 = 150 kΩ R2 = 150 kΩ RE = 470 Ω RL = 1 kΩ RG = 5.1 kΩ VCC = 15 V β = 5000

Solution: zin(base) = βre zin(base) = (2000)(4.44) zin(base) = 8.88 kΩ

Solution: re = RE||RL (Eq. 11-1) re = 470 Ω||1 kΩ re = 320 Ω

zin = R1||R2||zin(base) zin = 1 kΩ||2 kΩ||8.88 kΩ zin = 620 Ω

zin(base) = βre zin(base) = (5000)(320) zin(base) = 1.6 MΩ Answer: The input impedance of the base is 1.6 MΩ. 11-16. Given:

(Eq. 11-8)

vin = [zin/(zin + RG)]VG vin = [620 Ω/(620 Ω + 600 Ω)]1 V vin = 0.508 V Answer: The input voltage is 0.508 V.

R1 = 150 kΩ R2 = 150 kΩ RE = 470 Ω RL = 1 kΩ RG = 5.1 kΩ VCC = 15 V β = 7000 re = 320 Ω (from Prob. 11-15) Solution: zin(base) = βre zin(base) = (7000) (320) zin(base) = 2.24 MΩ zin = R1||R2||zin(base) (Eq. 11-4) zin = 150 kΩ|| 150 kΩ||2.24 MΩ zin = 72.6 kΩ vin = [zin/(zin + RG)]VG vin = [72.6 kΩ/(72.6 kΩ + 5.1 kΩ)]10 mV vin = 9.34 mV Answer: The input voltage is 9.34 mV. 11-17. Given:

11-19. Given:

VZ = 7.5 V VBE = 0.7 V R = 1 kΩ VCC = 15V Solution: Vout = VZ – VBE (Eq. 11-9) Vout = 7.5 V – 0.7 V Vout = 6.8 V IZ = (VCC – VZ)/R IZ = (15 – 7.5)/1 kΩ IZ = 7.5 mA Answer: The output voltage is 6.8 V, and the zener current is 7.5 mA. 11-20. Given:

VZ = 7.5 V VBE = 0.7 V R = 1 kΩ VCC = 25 V Solution:

R1 = 1 kΩ R2 = 2 kΩ RE = 10 Ω RL = 8 Ω RG = 600 Ω VCC = 20 V β1 = 150 β2 = 150

Vout = VZ – VBE (Eq. 11-9) Vout = 7.5 V – 0.7 V Vout = 6.8 V Take the base current into account. IZ = (VCC – VZ)/R – Iout/β IZ = (25 – 7.5)/1 kΩ – (6.8 V/33 Ω)/150 IZ = 17.5 mA – 1.37 mA IZ = 16.1 mA

Solution: re = RE||RL re = 10 Ω||8 Ω re = 4.44 Ω

R1 = 1 kΩ R2 = 2 kΩ RE = 10 Ω RL = 8 Ω RG = 600 Ω VCC = 20 V β = 2000 re = 4.44 Ω (from Prob. 11-17)

(Eq. 11-1)

Answer: The output voltage is 6.8 V, and the zener current is 16.1 mA. 11-21. Given: With the wiper in the middle, the voltage divider is effectively two resistors: each has a value of 1.5 kΩ.

1-50 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

re′ = (25 mV) / I E re′ = (25 mV) / 650 µA re′ = 38.46 Ω

VZ = 7.5 V VBE = 0.7 V Solution: Vout = [(R3 + R4)/R4](VZ + VBE) (Eq. 11-12) Vout = [(1.5 kΩ + 1.5 kΩ)/1.5 kΩ](7.5 V + 0.7 V) Vout = 16.4 V Answer: The output voltage is 16.4 V. 11-22. Given: With the wiper all the way up, the voltage divider is effectively two resistors: the top has a value of 1 kΩ, and the bottom has a value of 2 kΩ.

With the wiper all the way down, the voltage divider is effectively two resistors: the top has a value of 2 kΩ, and the bottom has a value of 1 kΩ. Solution: (Eq. 11-12) Vout(top) = [(R3 + R4)/R4](VZ + VBE) Vout(top) = [(1 kΩ + 2 kΩ)/1.5 kΩ](7.5 V + 0.7 V) Vout(top) = 12.3 V Vout(bottom) = [(R3 + R4)/R4](VZ + VBE) (Eq. 11-12) Vout(bottom) = [(2 kΩ + 1 kΩ)/1 kΩ](7.5 V + 0.7 V) Vout(bottom) = 24.6 V Answer: The output voltage with the wiper all the way up is 12.3 V, and all the way down is 24.6 V. 11-23. Given:

R1 = 10 kΩ R2 = 2 kΩ RC = 3.3 kΩ RE = 2 kΩ VCC =12 V

R1 = 10 kΩ R2 = 2 kΩ RC = 3.3 kΩ RE = 2 Ω VCC = 12 V Solution: VBB = [R2/(R1 + R2)]VCC VBB = [2 kΩ/(10 kΩ + 2 kΩ)]12 V VBB = 2 V VE = VBB – VBE VE = 2 V – 0.7 V VE = 1.3 V IE = VE/RE IE = 1.3 V/2 kΩ IE = 650 µA re′ = (25 mV) / I E re′ = (25 mV) / 650 µA re′ = 38.46 Ω = 38.5 Ω Z in (emitter ) = re′

Solution: VBB = [R2/(R1 + R2)]VCC VBB = [2 kΩ/(10 kΩ + 2 kΩ)]12 V VBB = 2 V VE = VBB – VBE VE = 2 V – 0.7 V VE = 1.3 V

Zin(emitter) = 38.5 Ω Z in(Stage) = RE || re′ Since RE >> re′

Z in(Stage) ≅ re′ = 38.5 Ω Zout ≈ RC Zout = 3.3 kΩ Answer: The Z in(emitter) = 38.5 Ω, the Z in(Stage) = re′ = 38.5 Ω, and Z out = 3.3kΩ.

IE = VE/RE IE = 1.3 V/2 kΩ IE = 650 µA

11-26. Given:

Answer: The emitter current is 650 µA. 11-24. Given:

R1 = 10 kΩ R2 = 2 kΩ RC = 3.3 kΩ RE = 2 Ω VCC = 12 V

R1 = 10 kΩ R2 = 2 kΩ RC = 3.3 kΩ RE = 2 kΩ RG = 50 Ω VCC = 12 V Vgen = 2 mV Solution:

Solution: VBB = [R2/(R1 + R2)]VCC VBB = [2 kΩ/(10 kΩ + 2 kΩ)]12 V VBB = 2 V

IE = VE/RE IE = 1.3 V/2 kΩ IE = 650 µA

Answer: The voltage gain at 64.4. 11-25. Given:

VZ = 7.5 V VBE = 0.7 V

VE = VBB – VBE VE = 2 V – 0.7 V VE = 1.3 V

rc = RC||RL rc = 3.3 kΩ||10 kΩ rc = 2.48 kΩ Av = rc / re′ Av = 2.48 kΩ/38.46 Ω Av = 64.4

VBB = [R2/(R1 + R2)]VCC VBB = [2 kΩ/(10 kΩ + 2 kΩ)]12 V VBB = 2V VE = VBB – VBE VE = 2 V – 0.7 V VE = 1.3 V IE = VE/RE IE = 1.3 V/2 kΩ IE = 650 µA re′ = (25 mV) / I E re′ = (25 mV) / 650 µA re′ = 38.46 Ω

1-51 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Vout = 6.8 V (from Prob. 11-19) RL = 33 Ω

re = RC||RL re = 3.3 kΩ||10 kΩ re = 2.48 kΩ Av = rc / re′ Av = 2.48 kΩ/38.46 Ω Av = 64.4 Z in (Stage) = RE || re′

Solution: VCE = VCC – Vout VCE = 15 V – 6.8 V VCE = 8.2 V IC = Iout = Vout/RL IC = 6.8 V/33 Ω IC = 206 mA

Since RE >> re′

Z in(Stage) = re′ = 38.5 Ω vin ≈ (zin(RG + zin)Vgen) vin = (38.5 kΩ (50 Ω + 38.5 Ω) 2 mV) vin = 870 µV

P = VCEIC P = (8.2 V)(206 mA) P = 1.69 W

vout = Av(vin) vout = 64.4(870 µV) vout = 56 mV

Answer: 1.69 W 11-29. Given:

Answer: The output voltage is 56 mV. 11-27. Given:

R1 = 10 kΩ R2 = 2 kΩ RC = 3.3 kΩ RE =2 kΩ RG = 50 Ω VCC = 15 V Vgen = 2 mV

Solution: VBB = [R2/(R1 + R2)] VCC (Eq. 8-1) VBB = [2 kΩ/(4.7 kΩ + 2 kΩ)]15 V VBB = 4.48 V

Solution: VBB = [R2/(R1 + R2)]VCC VBB = [2 kΩ/(10 kΩ + 2 kΩ)]15 V VBB = 2.5 V VE = VBB – VBE VE = 2.5 V – 0.7 V VE = 1.8 V

(Eq. 8-3)

IE = IC

(Eq. 8-4)

(Eq. 6-5)

Answer: The values are VB = 4.48 V, VE = 3.78 V, VC = 11.22 V, IE = 3.78 mA, IC = 3.78 mA, and IB = 25.2 µA. 11-30. Given:

vin ≈ (zin(RG + zin)Vgen) vin = (27.8 kΩ (50 Ω + 27.8 Ω)2 mV) vin = 715 µV

R1 = 4.7 kΩ R2 = 2 kΩ RC = 1 kΩ RE = 1 kΩ VCC = 15 V β = 150 vin = 5 mV vout(2) is an emitter follower that has a gain of 1. Solution:

vout = Av(vin) vout = 89.3(715 µV) vout = 63.8 mV

rc = 1 kΩ re = 1 kΩ

Answer: The output voltage is 63.8 mV.

VZ = 7.5 V VCC = 15 V

IE = VE/RE IE = 3.78 V/1 kΩ IE = 3.78 mA

IB = IC/β IB = 3.78 mA/150 IB = 25.2 µA

rc = RC||RL rc = 3.3 kΩ||10 kΩ rc = 2.48 kΩ Av = rc / re′ Av = 2.48 kΩ/27.8 Ω Av = 89.3 Z in (Stage) = RE || re′ Since RE >> re′ Z in(Stage) = re′ = 27.8 Ω

11-28. Given:

VE = VBB – VBE (Eq. 8-2) VE = 4.48 V – 0.7 V VE = 3.78 V

VC = VCC – ICRC (Eq. 8-15) VC = 15 V – 3.78 mA(1 kΩ) VC = 11.22 V

IE = VE/RE IE = 1.8 V/2 kΩ IE = 900 µA re′ = (25 mV) / I E re′ = (25 mV) / 900 µA re′ = 27.8 Ω

CRITICAL THINKING

R1 = 4.7 kΩ R2 = 2 kΩ RC = 1 kΩ RE = 1 kΩ VCC = 15V β = 150

Av = rc/re Av = 1 kΩ/1 kΩ Av = 1

(Eq. 10-7)

Answer: Both outputs are 5 mV; the top one is 180° out of phase. The purpose of this circuit is to produce two signals that are the same magnitude and 180° out of phase.

1-52 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

VE = 2.15 V

Trouble 7: Since there is voltage at G and none at H, the trouble is an open Q2.

IE = ICQ = VE/RE (Eq. 8-3) ICQ = 2.15 V/220 Ω ICQ = 9.77 mA

Chapter 12 Power Amplifiers

Since ICQ is the center of the load line, the load line is linear, the other end is zero, and the ac saturation current is double the Q point current. The ac saturation current is 19.5 mA.

SELF-TEST 1. b 2. b 3. d 4. a 5. c 6. d 7. d 8. b 9. b

10. d 11. c 12. d 13. b 14. b 15. b 16. b 17. c 18. a

19. a 20. c 21. b 22. d 23. a 24. a 25. b 26. c 27. c

28. a 29. d 30. d 31. b 32. c 33.d 34.c 35. a

Answer: The ac collector resistance is 543 Ω, and the ac saturation current is 10.9 mA. 12-3.

R1 = 2 kΩ R2 = 470 Ω RC = 680 Ω RE = 220 Ω RL = 2.7 kΩ VCC = 15 V VBE = 0.7 V RG = 50 Ω rc = 543 Ω (from Prob. 12-2) ICQ = 9.77 mA (from Prob. 12-2) VE = 2.15 V (from Prob. 12-2)

JOB INTERVIEW QUESTIONS 6. Tuned RF amplifier. It would be impractical to use a class C amplifier for an audio application because it would distort the signal. 8. The lower the duty cycle is, the less the current drain. 11. Thermal conductive paste used to create a low thermal resistance path between the case and the heat sink. 12. Class A. No signal is lost in a class A amplifier: 360° in, 360° out. With class C, over half the signal is lost. 13. Narrowband.

Solution: VC = VCC – RCICQ VC = 15 V – (680 Ω)(9.77 mA) VC = 8.36 V MP = ICQrc or VCEQ MP = (9.77 mA)(543 Ω) MP = 5.31 V

PROBLEMS 12-1.

Given:

(Eq. 12-8)

or

R1 = 2 kΩ R2 = 470 Ω RC = 680 Ω RE = 220 Ω RL = 2.7 kΩ VCC = 15V

VCEQ = VC – VE VCEQ = 8.36 V – 2.15 V VCEQ = 6.21 V MPP = 2MP MPP = 2(5.31 V) MPP = 10.62 V

Solution: RC = 680 Ω IC(Sat) = VCC/(RC + RE) (Eq. 12-1) IC(Sat) = 15 V/(680 Ω + 220 Ω) IC(Sat) = 16.67 mA

Answer: The maximum peak-to-peak voltage is 10.62 V. 12-4.

Given: R1 = 4 kΩ R2 = 940 Ω RC = 1.36 kΩ RE = 440 Ω RL = 5.4 kΩ VCC = 15V VBE = 0.7 V RG = 100 Ω

Answer: The dc collector resistance 680 Ω, and the dc saturation current is 16.67 mA. 12-2.

Given:

Given: R1 = 2 kΩ R2 = 470 Ω RC = 680 Ω RE = 220 Ω RL = 2.7 kΩ VCC = 15 V VBE = 0.7 V RG = 50 Ω

Solution: rc = RC||RL (Eq. 10-2) rc = 1.36 kΩ||5.4 kΩ rc = 1086 Ω Answer: The ac collector resistance is 1086 Ω.

Solution:

12-5.

rc = RC||RL (Eq. 10-2) rc = 680 Ω||2.7 kΩ rc = 543 Ω VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) VBB = [470 Ω/(2 kΩ + 470 Ω)]15 V VBB = 2.85 V VE = VBB – VBE (Eq. 8-2) VE = 2.85 V – 0.7 V

Given: R1 = 6 kΩ R2 =1.41 kΩ RC = 2.04 kΩ RE = 660 Ω RL = 8.1 kΩ VCC = 15V VBE = 0.7 V RG = 150 Ω

1-54 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

IE = ICQ = VE/RE ICQ = 9.3 V/68 Ω ICQ = 137 mA

Solution: rc = RC||RL (Eq. 10-2) rc = 2.04 kΩ||8.1 kΩ rc = 1.63 kΩ

Ic(Sat) = 2(ICQ) Ic(Sat) = 2(137 mA) Ic(Sat) = 274 mA

VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) VBB = [1.41 kΩ/(6 kΩ + 1.41 kΩ)]15 V VBB = 2.85 V VE = VBB – VBE (Eq. 8-2) VE = 2.84 V – 0.7 V VE = 2.15 V

Answer: The ac collector resistance is 50 Ω, and the ac saturation current is 274 mA. 12-8.

R1 = 200 Ω R2 = 100 Ω RC = 100 Ω RE = 68 Ω RL = 100 Ω VCC = 30 V VBE = 0.7 V rc = 50 Ω (from Prob. 12-7) ICQ = 137 mA (from Prob. 12-7) ic(Sat) = 174 mA (from Prob. 12-7) VE = 9.3 V (from Prob. 12-7)

IE = ICQ = VE/RE (Eq. 8-3) ICQ = 2.15 V/660 Ω ICQ = 3.26 mA VC = VCC – RCICQ VC = 15 V – (2.04 kΩ)(3.26 mA) VC = 8.35 V MP = ICQrc or VCEQ MP = (3.26 mA)(1.63 kΩ) MP = 5.31 V

(Eq. 12-8)

or

Solution:

VCEQ = VC – VE VCEQ = 8.35 V – 2.15 V VCEQ = 6.2 V

VC = VCC – RCICQ VC = 30 V – (100 Ω)(137 mA) VC = 16.3 V VCEQ = VC – VE VCEQ = 16.3V – 9.3V VCEQ = 7 V

MPP = 2MP MPP = 2(5.31 V) MPP = 10.62 V

MP = VCEQ = 7 V MPP = 2MP = 14 V

Answer: The maximum peak-to-peak voltage is 10.62 V. 12-6.

Given:

or

R1 = 200 Ω R2 = 100 Ω RC = 100 Ω RE = 68 Ω RL = 100 Ω VCC = 30 V

MP = ICQrc MP = (137 mA)(50 Ω) MP = 6.85 V MPP = 2MP = 13.7 V Answer: The maximum peak-to-peak voltage is 13.7 V.

Solution:

12-9.

RC = 100 Ω IC(Sat) = VCC/RC + RE IC(Sat) = 30 V/100 Ω + 68 Ω IC(Sat) = 179 mA Answer: The dc collector resistance is 100 Ω, and the saturation current is 179 mA. 12-7.

Given:

Given:

Solution:

R1 = 200 Ω R2 = 100 Ω RC = 100 Ω RE = 68 Ω RL = 100 Ω VCC = 30 V VBE = 0.7 V

rc = RC||RL rc = 200 Ω||200 Ω rc = 100 Ω Answer: The ac collector resistance is 100 Ω. 12-10. Given:

Solution: rc = RC||RL rc = 100 Ω||100 Ω rc = 50 Ω VBB = [R2/(R1 + R2)]VCC VBB = [100 Ω/(200 Ω + 100 Ω)]30 V VBB = 10 V VE = VBB – VBE VE = 10 V – 0.7 V VE = 9.3 V

Given: R1 = 400 Ω R2 = 200 Ω RC = 200 Ω RE = 136 Ω RL = 200 Ω VCC = 30 V VBE = 0.7 V

R1 = 600 Ω R2 = 300 Ω RC = 300 Ω RE = 204 Ω RL = 300 Ω VCC = 30 V VBE = 0.7 V Solution: VBB = [R2/(R1 + R2)]VCC

1-55 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

VBB = [300 Ω/(600 Ω + 300 Ω)]30 V VBB = 10 V

VE = 2.85 V – 0.7 V VE = 2.15 V

VE = VBB – VBE VE = 10 V – 0.7 V VE = 9.3 V

IE = VE/RE IE = 2.15 V/220 Ω IE = 9.77 mA

IE = ICQ = VE/RE ICQ = 9.3 V/204 Ω ICQ = 45.59 mA

Idc = Ibias + IE Idc = 6.07 mA + 9 77 mA Idc = 15.84 mA

VC = VCC – RCICQ VC = 30 V – (300 Ω)(45.59 mA) VC = 16.3 V

Answer: The current drain is 15.84 mA.

VCEQ = VC – VE VCEQ = 16.3 V – 9.3 V VCEQ = 7 V

(Eq. 8-3)

12-14. Given:

Idc = 15.84 mA (from Prob. 12-13) VCC = 15 V Solution: Pdc = IdcVCC Pdc = (15.84 mA)(15 V) Pdc = 237.6 mW

MP = VCEQ = 7 V MPP = 2MP = 14 V or

(Eq. 12-17)

Answer: The dc input power is 237.6 mW.

MP = ICQrc MP = (45.59 mA) (150 Ω) MP = 6.85 V MPP = 2MP = 13.7 V

12-15. Given:

Answer: The maximum peak-to-peak voltage is 13.7 V.

MPP = 10.62 V (from Prob. 12-3) RL = 2.7 kΩ Pdc = 237.6 mW (from Prob. 12-14) Solution:

12-11. Given:

Pout(max) = MPP2/8RL (Eq. 12-15) Pout = (10.62 V)2/8(2.7 kΩ) Pout = 5.22 mW

Pout = 2 W Pin = 4 mW Solution: AP = Pout/Pin AP = 2 W/4 mW AP = 500

(Eq. 12-12)

Answer: The power gain is 500. 12-12. Given:

Answer: The efficiency is 2.2%. 12-16. Given:

ICQ = 9.77 mA (from Prob. 12-2) VCEQ = 6.21 V (from Prob. 12-3)

Vout = 15 V pp RL = 1 kΩ Pin = 400 µW

Solution: PDQ = VCEQ ICQ (Eq. 12-16) PDQ = (6.21 V)(9.77 mA) PDQ = 60.7 mW

Solution: 2

Pout = V /8R Pout = (15 V)2/8 kΩ Pout = 28.1 mW

Answer: The quiescent power dissipation is 60.7 mW.

AP = Pout/Pin (Eq. 12-12) AP = 28.1 mW/400 µW AP = 70.3 Answer: The power gain is 70.3. 12-13. Given:

R1 = 2 kΩ R2 = 470 Ω RC = 680 Ω RE = 220 Ω RL = 2.7 kΩ VCC = 15 V VBE = 0.7 V RG = 50 Ω VBB = 2.85 V (from Prob. 12-2) Solution: Ibias = VCC/(R1 + R2) Ibias = 15 V/(2 kΩ + 470 Ω) Ibias = 6.07 mA VE = VBB – VBE

· = [Pout/Pin]100% · = [5.22 mW/237.6 mW]100% · = 2.2%

(Eq. 8-2)

12-17. Given:

R1 = 200 Ω R2 = 100 Ω RC = 100 Ω RE = 68 Ω RL = 100 Ω VCC = 30 V VBE = 0.7 V VBB = 10 V (from Prob. 12-7) Solution: Ibias = VCC/(R1 + R2) Ibias = 30 V/(200 Ω + 100 Ω) Ibias = 100 mA VE = VBB – VBE VE = 10 V – 0.7 V VE = 9.3 V IE = VE/RE IE = 9.3 V/68 Ω IE = 136.8 mA ≈ 137 mA

1-56 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Solution:

Idc = Ibias + IE Idc = 100 mA + 137 mA Idc = 237 mA Answer: The current drain is 237 mA. 12-18. Given:

Idc = 2.37 mA VCC = 30V

(from Prob. 12-17)

Pdc = IdcVCC Pdc = (237 mA)(30 V) Pdc = 7.11 W Answer: The dc input power is 7.11 W. 12-19. Given:

MPP = 2MP = 13.7 V (from Prob. 12-10) Pdc = 7.11 W (from Prob. 12-18) RL = 100 Ω Solution: Pout(max) = MPP2/8RL Pout = (13.7 V)2/8(100 Ω) Pout = 235 mW

Ibias = VCC/(R1 + R2) Ibias = 10 V/(10 Ω + 2.2 Ω) Ibias = 0.82 A

VE = VBB – VBE VE = 1.80 V – 0.7 V VE = 1.10 V

(Eq. 8-2)

IE = VE/RE IE = 1.10 V/1 Ω IE = 1.1 A

(Eq. 8-3)

Idc = Ibias + IE Idc = 0.82 A + 1.1 A Idc = 1.92 A Pdc = IdcVCC Pdc = (1.92 A)(10 V) Pdc = 19.2 W

· = [Pout/Pin]100% · = [235 mW/7.11 W]100% · = 3.3% Answer: The efficiency is 3.3%. 12-20. Given:

ICQ = 137 mA (from Prob. 12-7) VCEQ = 7 V (from Prob. 12-8) Solution:

(Eq. 12-17)

· = [Pout/Pin]100% · = [0.977 W/19.2 W]100% · = 5.1% Answer: The output power is 0.977 W, and the efficiency is 5.1%. 12-23. Given:

Vcut-off = 12 V Solution:

PDQ = VCEQICQ PDQ = (7 V)(137 mA) PDQ = 960 mW Answer: The quiescent power dissipation is 960 mW. 12-21. Given:

MPP = 12 Vcut-off MPP = 2(12 V) MPP = 24 V Answer: The maximum peak-to-peak voltage is 24 V. 12-24. Given:

R1 = 10 Ω R2 = 2.2 Ω RE = 1 Ω VCC = 10 V VBE = 0.7 V

VCC = MPP = 30 V RL = 16 Ω Solution:

Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) VBB = [2.2 Ω/(10 Ω + 2.2 Ω)]10 V VBB = 1.80 V VE = VBB – VBE VE = 1.80 V – 0.7 V VE = 1.10 V

(Eq. 8-2)

IE = VE/RE IE = 1.1 V/1 Ω IE = 1.1 A

(Eq. 8-3)

Answer: The dc emitter current is 1.1 A. R1 = 10 Ω R2 = 2.2 Ω RE = 1 Ω VCC = 10 V VBE = 0.7 V RC = 3.2 Ω Vout = 5 V pp

(Eq. 12-15)

VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) VBB = [2.2 Ω/(10 Ω + 3.2 Ω)] 10 V VBB = 1.80 V

Solution:

12-22. Given:

Pout = Àout2/8RL Pout = (5 V)2/8(3.2 Ω) Pout = 0.977 W

PD(max) = MPP2/40RL PD(max) = (30 V)2/40(16 Ω) PD(max) = 1.41 W Answer: The maximum power dissipation of each transistor is 1.41 W. 12-25. Given:

VCC = MPP = 30 V RL = 16 Ω Solution: Pout(max) = MPP2/8RL Pout(max) = (30 V)2/8(16 Ω) Pout(max) = 7.03 W Answer: The maximum output power is 7.03 W. 12-26. Given:

R1 = 100 Ω R2 = 100 Ω RL = 50 Ω VCC = 30 V VDiode = 0.7 V

1-57 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Solution: Ibias = (VCC – 2VDiode)/(R1 + R2) Ibias = 28.6 V/(100 Ω + 100 Ω) Ibias = 143 mA ICEQ ≈ Ibias = 143 mA Answer: The quiescent collector current is 143 mA. 12-27. Given:

VCC = MPP = 30 V RL = 50 Ω Solution: Ibias = (VCC – 2VDiode)/(R1 + R2) Ibias = 28.6 V/(100 Ω + 100 Ω) Ibias = 143 mA Idc = 238 mA

Pout(max) = MPP2/8RL Pout(max) = (30 V)2/8(50 Ω) Pout(max) = 2.25 W · = [Pout/Pin]100% · = [2.25 W/750 mW]100% · = 31.5% Answer: The efficiency is 31.5%. 12-28. Given:

VCC = MPP = 30 V RL = 50 Ω

R1 = 12 kΩ R2 = 1 kΩ RE = 100 kΩ β = 200 VCC = 30 V Solution: re′ = 25 mV/IE re′ = 25 mV/(10 V/1 kΩ) re′ = 2.5 Ω re = RE re = 100 Ω

(second stage)

Answer: The voltage gain of the first stage is 188. 12-31. Given for 2nd Stage:

R1 = 12 kΩ R2 = 1 kΩ RE = 100 Ω RC = 1 kΩ VE = 1.43 V 3rd Stage:

Solution: Ibias = (VCC – 2VDiode)/(R1 + R2) Ibias = 28.6 V/(1 kΩ + 1 kΩ) Ibias = ICQ = 14.3 mA Idc = 110 mA

β = 200 VCC = 30 V RE = 100 Ω Solution: IE = VE/RE IE = 1.43 V/100 Ω IE = 14.3 mA re′ = 25 mV/IE re′ = 25 mV/(14.3 mA) re′ = 1.75 Ω

Pdc = IdcVCC Pdc = 110 mA(30 V) Pdc = 3.3 W Pout(max) = MPP2/8RL Pout(max) = (30 V)2/8(50 Ω) Pout(max) = 2.25 W · = [Pout/Pin]100% · = [2.25 W/3.3 W]100% · = 68.3% Answer: The efficiency is 68.3% and the quiescent collector current is 14.3 mA. 12-29. Given:

MPP = 30 V RL = 100 Ω Solution: Pout(max) = MPP2/8RL Pout(max) = (30 V)2/8(100 Ω) Pout(max) = 1.13 W Answer: The maximum power output is 1.13 W. R1 = 10 kΩ R2 = 5.6 kΩ RC = 1 kΩ RE = 1 kΩ

Second Stage:

rc = RC||Zin(Stage 2) Zin(Stage 2) = 12 kΩ||910 Ω||β re′ rc = 1 kΩ||12 kΩ||910 Ω||200(100 Ω) rc = 496 Ω Av(Stage 1) = rc / re′ Av(Stage 1) = 496 Ω/2.5 Ω Av(Stage 1) = 188

Pdc = IdcVCC Pdc = 238 mA(30 V) Pdc = 7.14 W

12-30. Given for 1st stage:

VBB = 10.7 V VE = 10 V

re = RE re = 100 Ω Zin(base) = β re′ Zin(base) = 200(100Ω) Zin(base) = 20 kΩ

(Eq. 8-3)

(Eq. 9-10)

(second stage) (Eq. 10-9)

rc = RC||Zin(base) rc = 1 kΩ||20 kΩ rc = 952 Ω Av = rc/(re + re′ ) Av = 952 Ω/(100 Ω + 1.75 Ω) Av = 9.36 Answer: The gain of the second stage is 9.36. 12-32. Given: IE = 14.3 mA Solution: ICQ = Ibias = 14.3 mA Answer: The quiescent collector current is 14.3 mA. 12-33. Given: (from Prob. 12-30) Av1 = 188 Av2 = 9.36 (from Prob. 12-31)

1-58 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Solution:

Answer: The output power is 31.25 mW.

Av3 = 1 (Eq. 12-25) Av = Av1Av2Av3 Av = (188)(9.36)(1) Av = 1679

12-40. Given: VCC = 30 V.

Solution:

Answer: The total voltage gain is 1679. 12-34. Given: vin = 5 Vrms.

MPP = 2 VCC MPP = 2(30 V) MPP = 60 V

(Eq. 12-38)

Pout = MPP2/8RL (Eq. 12-15) Pout = (60 V pp)2/8(10 kΩ) Pout = 45 mW

Solution: Vpp = 2.828 Vrms. Vpp = 2.828(5 V) Vpp = 14.14 V pp

Answer: The maximum output power is 45 mW. 12-41. Given:

Since the input is clamped at 0.7 V, the negative peak is –13.44 V. The average value is –6.37 V, so the voltmeter will read –6.37 V. Answer: The input voltage is 14.14 V pp, and the base voltage is –6.37 V. 12-35. Given:

L = 1 µH C = 220 pF

Idc = 0.5 mA VCC = 30 V Solution: Pdc = VCCIdc Pdc = (30 V)(0.5 mA) Pdc = 15 mW

(Eq. 12-17)

Answer: The dc input power is 15 mW. 12-42. Given:

Solution: fr = 1/(2 π LC ) (Eq. 12-29) fr = 1/[2 π (1 μH)(220pF)] fr = 10.73 MHz Answer: The resonant frequency is 10.73 MHz. 12-36. Given:

Idc = 0.4 mA VCC = 30 V vout = 30 V pp RL = 10 kΩ Solution: Pdc = VCCIdc Pdc = (30 V)(0.4 mA) Pdc = 12 mW

L = 2 µH C = 220 pF Solution: (Eq. 12-29) fr = 1/(2 π LC ) fr = 1/[2 π (2¿H)(220 pF)] fr = 7.59 MHz Answer: The resonant frequency is 7.59 MHz. 12-37. Given:

(Eq. 12-17)

Pout = vout2/8RL (Eq. 12-14) Pout = (30 V pp)2/8(10 kΩ) Pout = 11.25 mW · = (Pout/Pin)100% (Eq. 12-18) · = (11.25 mW/12 mW)100% · = 93.75% Answer: The efficiency is 93.75%.

L = 1 µH C = 100 pF

12-43. Given:

Solution: fr = 1/(2 π LC ) fr = 1/[2 π (1¿H)(100pF)] fr = 15.92 MHz

(Eq. 12-29)

Answer: The resonant frequency is 15.92 MHz. 12-38. Given:

Q = 125 fr = 10.73 MHz

(from Prob. 12-35)

Solution: B = fr/Q B = 10.73 MHz/125 B = 85.84 kHz Answer: The bandwidth is 85.84 kHz.

Pout = 11 mW Pin = 50 µW

12-44. Given:

Ap = Pout/Pin (Eq. 12-12) Ap = 11 mW/50 µW Ap = 220

Q = 125 fr = 10.73 MHz (from Prob. 12-35) RL = 10 kΩ MPP = 60 V (from Prob. 12-40) L = 1 µH

Answer: The power gain is 220.

Solution:

Solution:

12-39. Given:

XL = 2πfL XL = 2(3.14)(10.73 MHz)(1 µH) XL = 67.38 Ω

vout = 50 V pp RL = 10 kΩ Solution: 2

Pout = vout /8RL (Eq. 12-14) Pout = (50 V pp)2/8(10 kΩ) Pout = 31.25 mW

RP = QXL RP = (125)(67.38 Ω) RP = 8.42 kΩ

(Eq. 12-33)

rC = RP||RL

(Eq. 12-34)

1-59 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Solution:

rC = 8.42 kΩ||10 kΩ rC = 4.57 kΩ PD = MPP2/40rC (Eq. 12-39) PD = (60 V)2/40(4.57 kΩ) PD = 19.7 mW Answer: The worst-case power dissipation is 19.7 mW. 12-45. Given:

PD = 625 mW D = 5 mW/°C TA = 100°C

The left side of the dc load line is IC(Sat), and the right side is VCC. The Q point is ICQ, VCEQ. The ac load line passes through the Q point. The right side of the ac load line is ICQ rc above the Q point, or 11.52 V. This gives the line a slope of ICQ/ICQ rc = 9.77 mA/5.31 V = 1.84 mA/V. To find the ac saturation current, take the ac voltage maximum multiplied by the slope = (11.52 V) (1.84 mA/V) = 21.2 mA. Answer: See the graph.

Solution: (Eq. 12-40) ΔP = D(TA – 25°C) ΔP = (5 mW/°C)(100°C – 25°C) ΔP = 375 mW PD(max) = PD – ΔP PD(max) = 625 mW – 375 mW PD(max) = 250 mW Answer: The worst-case power rating is 250 mW. 12-46. Given: Derating curve on Fig. 12-36.

Answer: The maximum dissipation at 100°C is 2 W. 12-47. Given:

PD = 115 W D = 0.657 W/°C TC = 90°C

12-53. Answer:

PL: The power across the load resistor should increase because the emitter current is increased, the gain is slightly increased, and the output voltage and the output power will increase. PD: Since MPP is reduced, the PD is also reduced (PD = MPP2/40rc).

Solution: ΔP = D(TC – 25°C) (Eq. 12-40) ΔP = (0.657 W/°C)(90°C – 25°C) ΔP = 42.7 W PD(max) = PD – ΔP PD(max) = 115 W – 42.7 W PD(max) = 72.3 W Answer: The power rating is 72.3 W with a case temperature of 90°C.

CRITICAL THINKING 12-48. Answer: The input is larger than the maximum allowed input for an undistorted output. The input is driving the output into saturation, clipping the wave off, and turning it into a square wave.

PS: A higher VCC will cause the input power to increase (P = V2/R). MPP: A higher VCC will cause the emitter voltage to increase and thus the emitter current to increase. This causes an increase in the collector resistor voltage drop, which reduces VCEQ and, in turn, the MPP. ·: Since the dc input power is increased and the output power is increased, the efficiency remains constant. 12-54. Answer:

PL: The power across the load resistor should decrease because the emitter current is decreased, the gain is slightly decreased, and the output voltage and the output power will increase. PD: Since MPP is increased, the PD is also increased (PD = MPP2/40rc).

12-49. Answer: Electrically, it would be safe to touch, but it may be hot and cause a burn.

PS: A higher R1 will cause the input power to decrease (P = V2/R).

12-50. Answer: No, the maximum efficiency of anything is 100 percent. It is impossible to get more power out of a device than is put into the device.

MPP: A higher R1 will cause the emitter voltage to decrease and thus the emitter current to decrease. This causes a decrease in the collector resistor voltage drop, which will increase VCEQ and, in turn, increase the MPP.

12-51. Answer: No, the ac load line is more vertical because the ac collector resistance is usually less than the dc collector resistance. If the collector had an inductor instead of a resistor, the ac resistance would be greater than the dc resistance and make the ac load line less vertical. 12-52. Given:

IC(Sat) = 16.67 mA (from Prob. 12-1) VCC = 15 V ICQ = 9.77 mA (from Prob. 12-2) MP = ICQ rc = 5.31 V (from Prob. 12-3) VCEQ = 6.21 V (from Prob. 12-3)

·: Since the dc input power is decreased and the output power is decreased, the efficiency is not changed. 12-55. Answer:

PL: The power across the load resistor should increase because the emitter current is increased, the gain is slightly increased, and the output voltage and the output power will increase. PD: Since MPP is reduced, the PD is also reduced (PD = MPP2/40rc). PS: A higher R2 will cause the input power to decrease (P = V2/R).

1-60 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Solution: (Eq. 13-2) VGS(off) = –VP VP = 6 V (Eq. 13-1) RDS = VP/IDSS RDS = 6 V/30 mA RDS = 200 Ω VD = [RDS/(RDS + RD)]VDD VD = [200 Ω/(200 Ω + 20 kΩ)]20 V VD = 0.198 V Answer: The drain voltage is 0.198 V. 13-12. Given: VDD = 20 V RD = 10 kΩ VGS(off) = –6 V IDSS = 30 mA Solution: ID(Sat) = VDD/RD ID(Sat) = 20 V/10 kΩ ID(Sat) = 2 mA VGS(off) = VP. VP = 6 V

(Eq. 13-2)

RDS = VP/IDSS RDS = 6 V/30 mA RDS = 200 Ω

(Eq. 13-1)

VD = [RDS/(RDS + RD)]VDD VD = [200 Ω/(200 Ω + 10 kΩ)]20 V VD = 0.392 V Answer: The drain saturation current is 2 mA, and the drain voltage is 0.392 V. 13-13. Given:

R1 = 1.5 MΩ R2 = 1 MΩ RS = 22 kΩ RD = 10 kΩ VDD = 25 V Solution: VG = [R2/(R1 + R2)]VDD VG = [1 MΩ/(1.5 MΩ + 1 MΩ)] 25 V VG = 10 V ID = VG/RS ID = 10 V/22 kΩ ID = 0.455 mA

(Eq. 13-10)

VD = VDD – IDRD (Eq. 13-4) VD = 25 V – (0.455 mA)(10 kΩ) VD = 20.45 V Answer: The drain voltage is 20.45 V. 13-14. Given: R1 = 1.5 MΩ R2 = 1 MΩ RS = 22 kΩ RD = 10 kΩ VDD = 25 V VG = 10 V (from Prob. 13-13) ID = 0.455 mA (from Prob. 13-13) VD = 20.45 V (from Prob. 13-13) Solution: ID(Sat) = VDD/(RD + RS) ID(Sat) = 25 V/(10 kΩ + 22 kΩ) ID(Sat) = 0.781 mA VS ≈ VG

VDSQ = VD – VS VDSQ = 20.45 V – 10 V VDSQ = 10.45 V

DC load line and Q point for Prob. 13-14.

13-15. Given:

VDD = 25 V VSS = –25 V RD = 7.5 kΩ RS = 18 kΩ Solution: (Eq. 13-12) ID = VSS/RS ID = –25 V/18 kΩ ID = 1.39 mA (Eq. 13-4) VD = VDD – IDRD VD = 25 V – (1.39 mA)(7.5 kΩ) VD = 14.58 V Answer: The drain voltage is 14.58 V. 13-16. Given: VDD = 25 V VSS = –25 V RD = 7.5 kΩ RS = 30 kΩ Solution: (Eq. 13-12) ID = VSS/RS ID = –25 V/30 kΩ ID = 0.833 mA VD = VDD – IDRD (Eq. 13-4) VD = 25 V – (0.833 mA)(7.5 kΩ) VD = 18.75 V Answer: The drain voltage is 18.75 V. 13-17. Given:

VDD = 15 V VEE = –9 V RD = 7.5 kΩ RE = 8.2 kΩ VBE = 0.7 V Solution: ID = (VEE – VBE)/RE ID = (9 V – 0.7 V)/8.2 kΩ ID = 1.01 mA

(Eq. 13-13)

(Eq. 13-4) VD = VDD – IDRD VD = 15 V – (1.01 mA)(7.5 kΩ) VD = 7.43 V Answer: The drain voltage is 7.43 V, and the drain current is 1.01 mA. 13-18. Given:

VDD = 15 V VEE = –9 V RD = 4.7 kΩ RE = 8.2 kΩ VBE = 0.7 V

1-63 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

13-19.

13-20.

13-21.

13-22.

13-23.

13-24.

Solution: (Eq. 13-13) ID = (VEE – VBE)/RE ID = (9 V – 0.7 V)/8.2 kΩ ID = 1.01 mA (Eq. 13-4) VD = VDD – IDRD VD = 15 V – (1.01 mA)(4.7 kΩ) VD = 10.25 V Answer: The drain voltage is 10.25 V, and the drain current is 1.01 mA. Given: VDD = 25 V RD = 8.2 kΩ RS = 1 kΩ ID = 1.5 mA Solution: (Eq. 13-7) VGS = –IDRS VGS = –(1.5 mA)(1 kΩ) VGS = –1.5 V VD = VDD – IDRD – IDRS VD = 25 V – (1.5 mA)(8.2 kΩ) – (1.5 mA)(1 kΩ) VD = 11.2 V Answer: The gate-source voltage is –1.5 V, and the drain-source voltage is 11.2 V. Given: VDD = 25 V RD = 8.2 kΩ RS = 1 kΩ VS = 1.5 V Solution: ID = VS/RS ID = 1.5 V/1 kΩ ID = 1.5 mA VD = VDD – IDRD VD = 25 V – (1.5 mA)(8.2 kΩ) VD = 12.7 V Answer: The drain voltage is 12.7 V. Given: VDD = 25 V RD = 10 kΩ RS = 22 kΩ R1 = 1.5 MΩ R2 = 1 MΩ Answer: The gate-source voltage is –2.5 V, and the drain current is 0.55 mA. Given: VDD = 15 V RG = 2.2 MΩ RE = 8.2 kΩ VEE = –9 V Answer: The gate-source voltage is –2.0 V, and the drain voltage is 7.5 V. Given: VDD = 25 V RG = 1.5 MΩ RS = 1 kΩ Answer: The gate-source voltage is –2.0 V, and the drain current is 1.5 mA. Given: VDD = 25 V RG = 1.5 MΩ RS = 2 kΩ

Answer: The gate-source voltage is –5.0 V, and the drain current is 1 mA and the drain-source voltage is 14.8 V. 13-25. Given:

gm0 = 4000 ¿S IDSS = 10 mA Solution: VGS(off) = –2I/DSS/gm0 VGS(off) = –2(10 mA)/4000 ¿S VGS(off) = –5 V

(Eq. 13-15)

gm = gm0 [1 – (VGS/VGS(off))] gm = 4000 ¿S[1 – (–1 V/–5V)] gm = 3200 ¿S

(Eq.13-16)

Answer: The gate-source cutoff voltage is –5 V, and the gm0 for VGS = –1 V is 3200 ¿S. 13-26. Given:

gm0 = 1500 ¿S IDSS = 2.5 mA VGS = –1 V Solution: (Eq. 13-15) VGS(off) = –2IDSS/gm0 VGS(off) = –2(2.5 mA)/1500 ¿S VGS(off) = –3.33 V gm = gm0 [1 – (VGS/VGS(off))] gm = 1500 ¿S[1 – (–1 V/–3.3 V)] gm = 1045 ¿S

(Eq. 13-16)

Answer: The gm for VGS = –1 V is 1045 ¿S. 13-27. Given:

gm0 = 6000 ¿S IDSS = 12 mA VGS = –2 V Solution: VGS(off) = –2IDSS/gm0 VGS(off) = –2(12 mA)/6000 ¿S VGS(off) = –4 V

(Eq. 13-15)

Since the ratio of VGS to VGS(off) is one-half, the following equation can be used: ID/IDSS = 1/4 ID = 1/4(IDSS) ID = 1/4(12 mA) ID = 3 mA gm = gm0 [1 – (VGS/VGS(off))] gm = 6000 ¿S[1 – (–2 V/–4 V)] gm = 3000 ¿S

(Eq. 13-16)

Answer: The drain current is 3 mA, and the transconductance is 3000 ¿S. 13-28. Given:

VDD = 30 V R1 = 20 MΩ R2 = 10 MΩ RD = 1 kΩ RS = 2 kΩ RL = 10 kΩ vin = 2 mV gm = 3000 ¿S Solution: rd = RD||RL rd = 1 kΩ||10 kΩ rd = 909 Ω

1-64 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Av = gmrd (Eq. 13-17) Av = (3000 ¿S)(909 Ω) Av = 2.73 vout = Av(vin) vout = (2.73)(2 mV) vout = 5.46 mV Answer: The output voltage is 5.46 mV. 13-29. Given:

VDD = 30 V R1 = 20 MΩ R2 = 10 MΩ RD = 1 kΩ RS = 2 kΩ RL = 10 kΩ vin = 2 mV IDSS = 12 mA (from the graph) VGS(off) = –4 V (from the graph) Solution: rd = RD||RL rd = 1 kΩ||10 kΩ rd = 909 Ω VGS(off) = –2IDSS/gm0 (Eq. 13-15) gm0 = -2IDSS/VGS(off) gm0 = –2(12 mA)/–4 V gm0 = 6000 ¿S VG = [R2/(R1 + R2)]VDD VG = [10 MΩ/(20 MΩ + 10 MΩ)] 30 V VG = 10 V ID = VG/RS ID = 10 V/2 kΩ ID = 5 mA

(Eq. 13-10)

13-31. Given: VDD = 30 V R1 = 20 MΩ R2 = 10 MΩ RS = 3.3 kΩ RL = 1 kΩ vin = 5 mV IDSS = 6 mA (from the graph) VGS(off) = –4 V (from the graph) rS = 767 Ω (from Prob. 13-30) Solution:

VGS(off) = –2IDSS/gm0 (Eq. 13-15) gm0 = –2IDSS/VGS(off) gm0 = –2(6 mA)/ –4 V gm0 = 3000 ¿S VG = [R2/(R1 + R2)]VDD VG = [10 MΩ/(20 MΩ + 10 MΩ)] 30 V VG = 10 V (Eq. 13-7) ID = VG/RS ID = 10 V/3.3 kΩ ID ≅ 3 mA From the graph, VGS is roughly –1.25 V when ID = 3 mA. With Eq. (13-16), gm = 2060 ¿S. With Eq. (13-18): gmrS = (2060 ¿S)(767 Ω) = 1.58 Av = 1.58/(1 + 1.58) = 0.612 vout = Avin vout = (0.612)(5 mV) vout = 3.06 mV Answer: The output voltage is 3.06 mV. 13-32. Given:

From the graph, VGS is approximately –1.4 V when ID is 5 mA. With Eq. 13-16, gm0 = 3900 ¿S. Then: Av = gmrd (Eq. 13-17) Av = (3900 ¿S)(909 Ω) Av = 3.54 vout = Av(vin) vout = 3.54(2 mV) vout = 7.09 mV Answer: The output voltage is 7.09 mV. 13-30. Given:

VDD = 30 V R1 = 20 MΩ R2 = 10 MΩ RS = 3.3 kΩ RL = 1 kΩ vin = 5 mV gm = 2000 ¿S

Rin = 22 kΩ vin = 50 mV pp IDSS = 10 mA VP = 2 V Solution: RDS = VP/IDSS (Eq. 13-1) RDS = 2 V/10 mA RDS = 200 Ω With VGS at –10 V the JFET is cut off and appears as an open; thus vout = vin = 50 mV pp. With VGS at 0 V, the JFET is conducting and a voltage divider is created with the input resistance. vout = [RDS/(RDS + Rin)]vin vout = [200 Ω/(200 Ω + 22 kΩ)]50 mV pp vout = 0.45 mV pp On-off ratio = vout(max)/vout(min) (Eq. 13-19) On-off ratio = 50 mV pp/0.45 mV pp On-off ratio =111 Answer: The output voltage at a VGS of –10 V is 50 mV pp, the output voltage at a VGS of 0 V is 0.45 mV pp, and the on-off ratio is 111.

Solution: rS = RS||RL rS = 3.3 kΩ||1 kΩ rS = 767 Ω

13-33. Given:

Av = (gmrs)/(1 + gmrs) (Eq. 13-18) Av = (2000 ¿S)(767 Ω)/[1 + (2000 ¿S)(767 Ω)] Av = 0.605 vout = Av(vin) vout = (0.605)(5 mV) vout = 3.03 mV

Rout = 33 kΩ vin = 25 mV pp IDSS = 5 mA VP = 3 V Solution: RDS = VP/IDSS

(Eq. 13-1)

Answer: The output voltage is 3.03 mV.

1-65 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

RDS = 3 V/5 mA RDS = 600 Ω With VGS at –10 V the JFET is cut off and appears as an open; thus vin = 0 mV pp. With VGS at 0 V, the JFET is conducting and a voltage divider is created with the output resistance. vout = [Rout/(RDS + Rout)]vin vout = [33 kΩ/(600 Ω + 33 kΩ)]25 mV pp vout = 24.55 mV pp On-off ratio = vout(max)/vout(min) (Eq. 13-19) On-off ratio = 24.55 mV pp/0 mV pp On-off ratio = ∞ Answer: The output voltage at a VGS of –10 V is 0 mV pp, the output voltage at a VGS of 0 V is 24.55 mV pp, and the on-off ratio is ∞.

CRITICAL THINKING 13-34. Answer:

IDSS = 20 mA VDS(max) = 5 V for the ohmic region VDS = 5 to 30 V in the active range 13-35. Given:

VGS(off) = –8 V (from the graph) IDSS = 32 mA (from the graph) VGS(1) = –4 V VGS(2) = –2 V Solution: VGS(off) = –2IDSS/gm0 (Eq. 13-15) gm0 = –2IDSS/VGS(off) gm0 = –2(32 mA)/–8 V gm0 = 8000 ¿S gm = gm0 [1 – (VGS/VGS(off))] (Eq. 13-16) gm = 8000 ¿S[1 – (VGS/–8 V)] ID = IDSS[1 – (VGS(1)/VGS(off))]2 ID = 32 mA[1 – (–4 V/–8V)]2 ID = 8 mA

(Eq. 13-3)

ID = IDSS[1 – (VGS(2)/VGS(off))]2 ID = 32 mA[1 – (–2 V/–8V)]2 ID = 8 mA

(Eq. 13-3)

Answer: The transconductance equation is gm = 8000 ¿S [1 – (VGS/–8 V)], the drain current at –4 V is 8 mA, and the drain current at –2 V is 18 mA. 13-36. Given:

VGS(off) = –5 V (from the graph) IDSS = 12 mA (from the graph) VGS(1) = –1 V Solution: ID = IDSS[1 – (VGS(1)/VGS(off))]2 ID = 12 mA[1 – (–1 V/–5V)]2 ID = 7.68 mA

(Eq. 13-3)

Answer: The drain current is 7.68 mA. 13-37. Given:

VDD = 15 V VEE = –10 V RD = 3.3 kΩ RE = 4.7 kΩ

VBE = 0.7 V gm = 2000 ¿S vin = 3 mV Solution: (Eq. 13-13) ID = (VEE – VBE)/RE ID = (10 V – 0.7 V)/4.7 kΩ ID = 2 mA (Eq. 13-4) VD = VDD – IDRD VD = 15 V – (2 mA)(3.3 kΩ) VD = 8.4 V rd = RD||RL rd = 3.3 kΩ||15 kΩ rd = 2.7 kΩ Av = gmrd (Eq. 13-17) Av = (2000 ¿S)(2.7 kΩ) Av = 5.4 vout = Av(vin) vout = (5.4)(3 mV) vout = 16.2 mV Answer: The drain voltage is 8.4 V, and the output voltage is 16.2 mV. 13-38. Answer: a. Multiply 4 mA and 510 Ω to get 2.04 V. b. It must equal 2.04 V. c. Because of the linearity of the circuit, the meter reads half of maximum, or 0.5 mA. 13-39. Given: IDSS = 16 mA RDS = 200 Ω RL = 10 kΩ VDD = 30 V Solution: Since VGS is 0 V, it is operating in the active region. The JFET appears to be a current source, but since the load is so large, the power supply cannot supply enough voltage to produce that current and it drops into the ohmic region and the JFET acts like resistor. I = VDD/(RDS + RL) I = 30 V/(200 Ω + 10 kΩ) I = 2.94 mA VDS = IRDS VDS = (2.94 mA)(200 Ω) VDS = 0.59 V If the load is shorted, RL = 0 Ω and the JFET operates in the active region. I = IDSS I = 16 mA VDS = VDD VDS = 30 V Answer: During normal operation, the current is 2.94 mA and the voltage across the JFET is 0.59 V. With the load shorted, the current is 16 mA and the voltage is 30 V. 13-40. Answer: a. The gm0 is 6000 ¿S. Multiply this by 1 kΩ to get a voltage gain of 6. b. At –1 V, the gm is 4500 ¿S and the voltage gain is 4.5. c. 3 d. 1.5 e. 0.75

1-66 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

ID = IDSS(1 – (VGS/VGS(off)))2 ID = 4 mA(1 – (–1 V/–2.0 V))2 ID = 1 mA ID = IDSS(1 – (VGS/VGS(off)))2 ID = 4 mA(1 – (–1.5 V/–2.0 V))2 ID = 0.25 mA

Answer: The drain current is 12 mA and the drainsource voltage is 6.36 V. 14-5.

gm0 = 4000 µS RD = 470 Ω RL = 2 kΩ Vin = 100 mV

Answer:

Solution:

VGS = –0.5 V, ID = 2.25 mA VGS = –1 V, ID = 1 mA VGS = –1.5 V, ID = 250 mA 14-2.

rd = RD||RL rd = 470 Ω||2 kΩ rd = 381 Ω

Given:

Av = gmrd Av = (4000 µS)(381 Ω) Av = 1.52

VGS = –0.5 V VGS = –1.0 V VGS = –1.5 V VGS = +0.5 V VGS = +1.0 V VGS = +1.5 V VGS(off) = –2 V IDSS = 4 mA

vout = VinAv vout = (100 mV)(1.52) vout = 152 mV Answer: The voltage gain is 1.52, the voltage out is 152 mV, and rd is 381 Ω.

Solution: ID = IDSS(1 – (VGS/VGS(off)))2 (Eq. 14-1) ID = 4 mA(1 – (+0.5 V/–2.0 V))2 ID = 6.25 mA

14-6.

Solution:

(Eq. 14-1) ID = IDSS(1 – (VGS/VGS(off))) ID = 4 mA(1 – (+1.5 V/–2.0 V))2 ID = 12.25 mA

rd = RD||RL rd = 680 Ω||10 kΩ rd = 637 Ω

Answer:

Av = gmrd Av = (4000 µS)(637 Ω) Av = 2.55

2

VGS = 0.5 V, ID = 6.25 mA VGS = 1 V, ID = 9 mA VGS = 1.5 V, ID = 12.25 mA

vout = VinAv vout = (100 mV)(2.55) vout = 255 mV

Given: VGS = –1.0 V VGS = –2.0 V VGS = 0 V VGS = +1.5 V VGS = +2.5 V VGS(off) = +3 V IDSS = 12 mA

Answer: The voltage gain is 2.55, the voltage out is 255 mV, and rd is 637 Ω. 14-7.

Solution: ID = IDSS(1 – (VGS/VGS(off)))2 (Eq. 14-1) ID = 4 mA(1 – (+1.5 V/+3.0V))2 ID = 3 mA ID = IDSS(1 – (VGS/VGS(off)))2 (Eq. 14-1) ID = 4 mA(1 – (2.5 V/+3.0 V))2 ID = 0.333 mA

Given: gm0 = 4000 µS RD = 680 Ω RL = 10 kΩ Vin = 100 mV RG = 1 MΩ Solution: Zin ≈ RG ≈ 1 MΩ Answer: The input impedance is approximately 1 MΩ.

14-8a. Given:

VDS(on) = 0.1 V ID(on) = 10 mA

Answer: VGS = 1.5 V, ID = 3 mA VGS = 2.5 V, ID = 333 µA 14-4.

Given: gm0 = 4000 µS RD = 680 Ω RL = 10 kΩ Vin = 100 mV

ID = IDSS(1 – (VGS/VGS(off)))2 (Eq. 14-1) ID = 4 mA(1 – (+1 V/–2.0 V))2 ID = 9 mA

14-3.

Given:

Solution: RDS(on) = VDS(on)/ID(on) RDS(on) = 0.1 V/10 mA RDS(on) = 10 Ω

Given: VGS(off) = –3 V IDSS = 12 mA

(Eq. 14-1)

Answer: The drain-source resistance is 10 Ω.

Solution: VDS = VDD – (IDSSRD) (Eq. 14-2) VDS = 12 V – ((12 mA)(470 Ω)) VDS = 6.36 V

14-8b. Given:

VDS(on) = 0.25 V ID(on) = 45 mA

ID = IDSS = 12 mA

1-68 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Solution:

Solution:

RDS(on) = VDS(on)/ID(on) (Eq. 14-1) RDS(on) = 0.25 V/45 mA RDS(on) = 5.56 Ω

VDS = ID(Sat)RDS(on) VDS = 200 mA (2 Ω) VDS = 0.4 V

Answer: The drain-source resistance is 5.56 Ω.

Answer: The voltage across the E-MOSFET is 0.4 V.

14-8c. Given:

14-10. Given:

VDS(on) = 0.75 V ID(on) = 100 mA Solution: RDS(on) = VDS(on)/ID(on) RDS(on) = 0.75 V/100 mA RDS(on) = 7.5 Ω

(Eq. 14-1)

Solution:

Answer: The drain-source resistance is 7.5 Ω. 14-8d. Given:

VDS(on) = 0.15 V ID(on) = 200 mA

Answer: The voltage across the E-MOSFET is 0.5 V.

(Eq. 14-1)

Answer: The drain-source resistance is 0.75 Ω.

VD = [RDS(on)/(RDS(on) + RD)]VDD VD = [28 Ω/(28 Ω + 1.8 kΩ)]15 V VD = 0.23 V

VGS(on) = 3 V ID(on) = 500 mA RDS(on) = 2 Ω ID(Sat) = 25 mA

Answer: The drain voltage is 0.23 V.

Solution:

14-12. Given:

VDS = ID(Sat)RDS(on) VDS = 25 mA (2 Ω) VDS = 0.05 V Answer: The voltage across the E-MOSFET is 0.05 V. 14-9b. Given:

VGS(on) = 5 V (from Table 14-1) ID(on) = 200 mA (from Table 14-1) RDS(on) = 7.5 Ω (from Table 14-1) VDD = 25 V RD = 150 Ω Solution:

VGS(on) = 3 V ID(on) = 500 mA RDS(on) = 2 Ω ID(Sat) = 50 mA

VD = [RD/(RDS(on) + RD)]VDD VD = [150 Ω/(7.5 Ω + 150 Ω)] 25 V VD = 23.8 V Answer: The drain voltage is 23.8 V.

Solution:

14-13. Given:

VDS = ID(Sat)RDS(on) VDS = 50 mA (2 Ω) VDS = 0.1 V Answer: The voltage across the E-MOSFET is 0.1 V. 14-9c. Given:

VGS(on) = 3 V ID(on) = 500 mA RDS(on) = 2 Ω ID(Sat) = 100 mA

VGS(on) = 10 V (from Table 14-1) ID(on) = 1 A (from Table 14-1) RDS(on) = 0.9 Ω (from Table 14-1) VDD = 12 V RD = 18 Ω Solution: VD = [RDS(on)/(RDS(on) + RD)]VDD VD = [0.9 Ω/(0.9 Ω + 18 Ω)]12 V VD = 0.57 V

Solution:

Answer: The drain voltage is 0.57 V.

VDS = ID(Sat)RDS(on) VDS = 100 mA (2 Ω) VDS = 0.2 V

14-14. Given:

Answer: The voltage across the E-MOSFET is 0.2 V. VGS(on) = 3 V ID(on) = 500 mA RDS(on) = 2 Ω ID(Sat) = 200 mA

VGS(on) = 2.6 V (from Table 14-1) ID(on) = 20 mA (from Table 14-1) RDS(on) = 28 Ω (from Table 14-1) VDD = 15 V RD = 1.8 kΩ Solution:

14-9a. Given:

14-9d. Given:

VD = [RDS(on)/(RDS(on) + RD)]VDD VD = [10 Ω/(10 Ω + 390 Ω)]20 V VD = 0.5 V 14-11. Given:

Solution: RDS(on) = VDS(on)/ID(on) RDS(on) = 0.15 V/200 mA RDS(on) = 0.75 Ω

VGS(on) = 2.5 V (from Table 14-1) ID(on) = 100 mA (from Table 14-1) RDS(on) = 10 Ω (from Table 14-1) VDD = 20 V RD = 390 Ω

VGS(on) = 5 V (from Table 14-1) ID(on) = 200 mA (from Table 14-1) RDS(on) = 7.5 Ω (from Table 14-1) VDD = 30 V RD = 1 kΩ VLED = 2 V

1-69 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Solution:

14-20. Answer: Inverted.

ID = (VDD – VLED)/(RDS(on) + RD) ID = (30 V – 2 V)/(7.5 Ω + 1 kΩ) ID = 27.8 mA

14-21. Answer: The on MOSFET has an RDS(on) of 10 V divided by 1 mA, which equals 10 kΩ. The off MOSFET has an RDS(off) of 10 V divided by 1 µA, which equals 10 MΩ. When the input voltage is high, the lower MOSFET is on, and the output voltage is given by:

Answer: The LED current is 27.8 mA. 14-15. Given:

10 kΩ 12 V ≅ 0.012 V 10.01 MΩ When the input voltage is low, the lower MOSFET is off, and the output voltage is given by: 10 MΩ Vout = 12 V ≅ 12 V 10.01 MΩ 14-22. Given: Vout =

VGS(on) = 2.6 V (from Table 14-1) ID(on) = 20 mA (from Table 14-1) RDS(on) = 28 Ω (from Table 14-1) VDD = 20 V RD = 1 kΩ Solution: ID = (VDD)/(RDS(on) + RD) ID = (20 V)/(28 Ω + 1 kΩ) ID = 19.5 mA

12-V peak square-wave input f = 1 kHz

IL = VDD/RL IL = 20 V/2 Ω IL = 10 A

Assume the same values from the previous problem.

Answer: The MOSFET current is 19.5 mA. The load current is 10 A.

14-23. Given:

VDD = 12 V RDS(on) = 5 kΩ

14-16. Given:

ID(active) = 1 mA VDS(active) = 10 V

Solution:

Solution: RD = VDS(active)/ID(active) RD = 10 V/1 mA RD = 10 kΩ

Answer: The signal will be 180° out of phase and have a maximum value of 12 V and a minimum value of 0 V.

(Eq. 14-6)

ID = VDD/2(RDS(on)) ID = 12 V/2(5 kΩ) ID = 1.2 mA Answer: The current is 1.2 mA.

Answer: The drain resistance is 10 kΩ.

14-24. Given:

VGS(on) = 10 V (from Table 14-2) ID(on) = 2 A (from Table 14-2) RDS(on) = 1.95 Ω (from Table 14-2) VDD = 12 V RD = 10 Ω

14-17. Given:

RDS(on) = 300 Ω VDD = 12 V RD = 8 kΩ Solution: When the input is low, the lower MOSFET is open and the output voltage is pulled up to the supply voltage. When the input is high, the lower MOSFET has a resistance of 300 Ω. vout = [RDS(on)/(RDS(on) + RD)]VDD vout = [300 Ω/(300 Ω + 8 kΩ)]12 V vout = 0.43 V Answer: When the input voltage is low, the output voltage is 12 V; when the input voltage is high, the output voltage is 0.43 V.

Solution: When the input is low, the MOSFET is open and no current flows. When the input is high, the MOSFET has a resistance of RDS(on) = 1.95 Ω. ID = (VDD)/(RDS(on) + RD) ID = (12 V)/(1.95 Ω + 10 Ω) ID = 1 A Answer: The current is 0 A when the input is low, and 1 A when the input is high. 14-25. Given:

VGS(on) = 10 V (from Table 14-2) ID(on) = 2 A (from Table 14-2) RDS(on) = 1.95 Ω (from Table 14-2) VDD = 12 V RD = 6 Ω

14-18. Given:

RDS(on) = 150 Ω VDD = 18 V RD = 2 kΩ Solution: When the input is low, the lower MOSFET is open and the output voltage is pulled up to the supply voltage. When the input is high, the lower MOSFET has a resistance of 150 Ω. vout = [RDS(on)/(RDS(on) + RD)]VDD vout = [150 Ω/(150 Ω + 2 kΩ)]18 V vout = 1.26 V Answer: When the input voltage is low, the output voltage is 18 V; when the input voltage is high, the output voltage is 1.26 V. 14-19. Answer: The output waveform is a square wave with an upper peak of +12 V and a lower peak of 0.43 V.

Solution: When the input is high, the MOSFET has a resistance of RDS(on) = 1.95 Ω. ID = (VDD)/(RDS(on) + RD) ID = (12 V)/(1.95 Ω + 6 Ω) ID = 1.51 A Answer: The current is 1.51 A when the input is high. 14-26. Given:

VGS(on) = 10 V (from Table 14-2) ID(on) = 5 A (from Table 14-2) RDS(on) = 1.07 Ω (from Table 14-2)

1-70 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

VB = 12 V V = 19 V R = 5 kΩ

Solution: As the temperature rises 100°C, the normalized resistance increases by a factor of 2.25. Thus 2.25/100°C = 0.0225/°C. The temperature increases 75°C. Thus the resistance increases by a factor of 75°C(0.0225/°C) = 1.69.

Solution: Just before breakover, the capacitor voltage is VB.

0.17(1.69) = 0.29 Ω

I = (V – VB)/R I = (19 V – 12 V)/5 kΩ I = 1.4 mA

Answer: The resistance at 100°C is 0.29 Ω. 14-41. Given:

Vin = 12 V Turns ratio = 4:1

While the diode is conducting, the voltage across it is 0.7 V.

Solution: The primary voltage will be 12 V.

I = (V – VD)/R I = (19 V – 0.7 V)/5 kΩ I = 3.66 mA

N1/N2 = 4 N1/N2 = V1/V2 V2 = V1/(N1/N2) V2 = 12 V/4 V2 = 3 V

(Eq. 4-14)

Answer: The current through the resistor just before breakover is 1.4 mA, and during conduction is 3.66 mA. 15-3.

Given: VD = 0.7 V VB = 12 V V = 19 V R = 5 kΩ C = 0.2 µF

Answer: The output voltage is 3 V.

Chapter 15 Thyristors SELF-TEST

Solution:

1. c 2. b 3. d 4. c 5. b 6. b 7. a 8. b

RC = (5 kΩ)(0.02 µF) RC = 0.1 ms T = 0.1 ms since the period equals the RC time

9. b 10. c 11. a 12. b 13. b 14. d 15. d 16. d

17. d 18. a 19. a 20. b 21. c 22. b 23. c 24. b

25. d 26. d 27. b 28. a 29. c 30. b 31. a

JOB INTERVIEW QUESTIONS

f = 1/T f = 1/10.1 ms f = 10 kHz Answer: The RC time is 0.1 ms, and the frequency is 10 kHz. 15-4.

VD = 0.7 V VB = 20 V IH = 3 mA R = 1 kΩ

5. The SCR remains latched once the initial stimulus is removed; the transistor does not. This prevents silencing the alarm by a clever burglar or destruction of the sending unit by fire or flood, etc. 6. In every section of the field. 7. Power-handling capability: The SCR can handle the most current, and the power FET the least current. Efficiency: The SCR is the most efficient since the control signal can be removed once SCR is conducting, and the power FET is the next-most efficient since its control current is low. Control input: The power FET and BJT are easier to control because they can be shut off using the control input. Maximum frequency: The power FET switches the fastest.

15-1.

Solution: Since the diode is open before breakover, no current flows before the device breaks over. Thus when the power supply reaches breakover voltage, the device will break over. V = IH R + 0.7 V (Eq. 15-2) V = (3 mA)(1 kΩ) + 0.7 V V = 3.7 V Answer: The power supply voltage will be 20 V at breakover and 3.7 V at dropout. 15-5.

PROBLEMS

Solution: The maximum voltage across the capacitor will be breakover voltage, because as soon as the device breaks over, the voltage drops to about 0.7 V.

Solution: V = IH R + 0.7 V (Eq. 15-2) V = (4 mA)(1 kΩ) + 0.7 V V = 4.7 V

RC = (10 kΩ)(0.06 µF) RC = 0.6 ms

Answer: The power supply voltage will be 4.7 V at dropout. Given: VD = 0.7 V

Given: VD = 0.7 V VB = 12 V V = 19 V R = 10 kΩ C = 0.06 µF

Given: VD = 0.7 V IH = 4 mA R = 1 kΩ

15-2.

Given:

Answer: The maximum voltage across the capacitor is 12 V, and the time constant is 0.6 ms. 15-6.

Given:

1-73 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

15-10. Given:

VGT = 1.0 V IGT = 2 mA IH = 12 mA VCC = 12 V RG = 2.2 kΩ R = 47 Ω Solution: When the SCR is off, no current flows. The output voltage when the SCR is off is the same as the power supply voltage. (Eq. 15-1) Vin = VGT + IGTRG Vin = 1 V + (2 mA)(2.2 kΩ) Vin = 5.4 V V = IHR + 0.7 V (Eq. 15-2) VCC = (12 mA)(47 Ω) + 0.7 V VCC = 1.26 V

15-7.

Vin = VGT + IGTRG (Eq. 15-1) Vin = 2 V + (8 mA)(6.6 kΩ) Vin = 54.8 V Answer: The input voltage required to turn on the SCR is 54.8 V. 15-11. Given:

R = 750 Ω R1 = 3.3 kΩ R2 = 6.8 kΩ C = 4.7 µF

Given:

Solution:

VGT = 0.7 V IGT = 1.5 mA IH = 2 mA VCC = 12 V RG = 4.4 kΩ R = 94 Ω

RC = Rth(cap)C RC = (2.54 kΩ)(4.7 ¿F) RC = 11.9 msec

Solution:

Answer: The charging time constant is 11.9 ms. and the Thevenin resistance is 611 Ω.

Rth = R||R1 Rth = 750||3.3 kΩ Rth = 611 Ω

15-12. Given:

Answer: The input voltage required to turn on the SCR is 7.3 V.

R1 = 1 kΩ R2 = 4.6 kΩ C = 0.47 µF

Answer: The highest output occurs when 0.8 V is across the 500-Ω resistor. The current through this resistor is 0.8 V divided by 500 Ω, which equals 1.6 mA. This 1.6 mA must flow through the 3.3-kΩ resistor. The 200 µA of gate current must also flow through the 3.3-kΩ resistor. If we ignore the 200 µA on the grounds that it is much smaller than 1.6 mA, we get an approximate answer of:

XC = 1/(2πfC) XC = 1/(2π(60 Hz)(0.47 µF) XC = 5644 Ω

V = 0.8 V + (1.6 mA)(3.3 kΩ) = 6.08 V If we include the 200 µA, we get a slightly larger output voltage: V = 0.8 V + (1.6 mA + 200 µA)(3.3 kΩ) = 6.74 V 15-9.

Solution:

Answer: The output voltage when the SCR is off is 12 V. The input voltage required to turn on the SCR is 5.4 V, and the supply voltage required to turn the SCR off is 1.26 V.

Vin = VGT + IGTRG (Eq. 15-1) Vin = 0.7 V + (1.5 mA)(4.4 kΩ) Vin = 7.3 V

15-8.

VGT = 2 V IGT = 8 mA IH = 2 mA VCC = 12 V RG = 6.6 kΩ R = 141 Ω

Given: VGT = 1.5 V IGT = 15 mA IH = 10 mA VCC = 12 V RG = 2.2 kΩ R = 47 Ω

Solution:

Z = R2 + X 2 Z = 5.6 k Ω 2 + 5.644 k Ω 2 Z = 7.95 kΩ

»X ¿ ¸ Z =∠ – arctan ¼ C À ½ R Á » 5.644 k Ω ¿ ¸ Z =∠ – arctan ¼ À ½ 5.6 k Ω Á ¸ Z = ∠ – 45°

IC ∠ ¸ =

Solution: Vin = VGT + IGTRG (Eq. 15-1) Vin = 1.5 V + (15 mA)(2.2 kΩ) Vin = 34.5 V VCC = IHR + 0.7 V (Eq. 15-2) VCC = (10 mA)(47 Ω) + 0.7 V VCC = 1.17 V Answer: The input voltage required to turn on the SCR is 34.5 V, and the supply voltage required to turn the SCR off is 1.17 V.

Vin

»X ¿ ZT ∠ − arctan ¼ C À ½ R Á 120 V ∠ 0° IC ∠ ¸ = 7.95 k Ω ∠ − 45° I C ∠ ¸ =15 mA ∠ – 45°

VC = ( I C ∠ ¸ )( X C ∠ – 90° ) VC = (15 mA ∠ – 45°)(5644 Ω ∠ – 90°) VC = 85 V ∠ – 45°)

¸°cond = 180° – ¸°firing ¸°cond = 180° – 45°

1-74 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

¸°cond = 135 Answer: The firing angle is 45°, the conduction angle is 135°, and the voltage across the capacitor is 85 Vac. 15-13. Given:

R1= 1 kΩ R2 = 50 kΩ pot C = 0.47 ¿F Solution: Perform the following calculations with an R value of 1 kΩ and 51 kΩ. XC = 1/(2πfC) XC = 1/(2π(60 Hz)(0.47 ¿F) XC = 5644 Ω

VC = (20.9 mA ∠ 80°)(5644 Ω ∠ − 90° ) VC = 118 V ∠ − 10°

¸°cond = 180° – ¸°firing ¸°cond = 180° – 10° ¸°cond = 170° Answer: The minimum conduction angle is 170°, and the maximum conduction angle is 96.3°. 15-15. Given:

VGT = 0.8 V IGT = 200 ¿A VZ = 10 V Solution:

Z = R2 + X 2 Z = 1 kΩ 2 + 5.644 kΩ 2 Z = 5.732 kΩ

VCC = VZ + VGT (Eq. 15-3) VCC = 10 V + 0.8 V VCC = 10.8 V

»X ¿ ¸ Z = ∠ − arctan ¼ C À ½ R Á

Answer: The voltage needed to trigger the crowbar is 10.8 V.

» 5.644 kΩ ¿ ¸ Z = ∠ − arctan ¼ À ½ 1 kΩ Á ¸ Z = ∠ − 80° Vin

IC ∠ ¸ =

VGT = 1.5 V IGT = 200 ¿A VZ = 10 V ± 10% Solution:

»X ¿ ZT ∠ − arctan ¼ C À ½ R Á 120 V ∠ 0° IC ∠ ¸ = 5.732 kΩ ∠ − 80° I C ∠ ¸ = 20.9 mA ∠ 80° VC = ( I C ∠ ¸)( X C ∠ − 90°) VC = (20.9 mA ∠ 80°)(5644 Ω ∠ − 90°) VC = 118 V ∠ − 10°

Answer: The minimum firing angle is 10°, and the maximum firing angle is 83.7°. 15-14. Given:

R1 = 1 kΩ R2 = 50 kΩ pot C = 0.47 ¿F

VZ(max) = VZ + 0.1(VZ) VZ(max) = 10 V + 0.1(10 V) VZ(max) = 11 V VCC = VZ + VGT (Eq. 15-3) VCC = 11 V + 1.5 V VCC = 12.5 V Answer: The voltage needed to trigger the crowbar is 12.5 V. 15-17. Given:

VGT = 0.8 V IGT = 200 ¿A VZ = 12 V Solution:

Solution: Perform the following calculations with an R value of 1 kΩ and 51 kΩ. XC = 1/(2πfC) XC = 1/(2π(60 Hz))(0.47 ¿F) XC = 5644 Ω Z = R2 + X 2 Z = 1 kΩ 2 + 5.644 kΩ 2 Z = 5.732 kΩ

Vtrig = VZ + VGT (Eq. 15-3) Vtrig = 12 V + 0.8 V Vtrig = 12.8 V Answer: The SCR will trigger at 12.8 V. 15-18. Given:

VGT = 0.8 V IGT = 200 ¿A VZ = 12 V Solution:

»X ¿ ¸ Z = ∠ − arctan ¼ C À ½ R Á » 5.644 k Ω ¿ ¸ Z = ∠ − arctan ¼ À ½ 1 kΩ Á ¸ Z = ∠ − 80° IC ∠ ¸ =

15-16. Given:

Vin »X ¿ ZT ∠ − arctan ¼ C À ½ R Á

120 V ∠ 0° 5.732 kΩ ∠ − 80° lC ∠ ¸ = 20.9 mA ∠ 80° IC ∠ ¸ =

VCC = VZ + VGT (Eq. 15-3) VCC = 12 V + 0.8 V VCC = 12.8 V Answer: The voltage needed to trigger the crowbar is 12.8 V. 15-19. Given:

VB = 20 V VGT = 2.5 V Solution: Ignore the gate current in the triac. Then VC = VB + VGT VC = 20 V + 2.5 V VC = 22.5 V

VC = ( I C ∠ ¸)( X C ∠ − 90°)

1-75 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Answer: The capacitor voltage required to turn on the triac is 22.5 V. 15-20. Given:

VCC = 100 V R = 15 Ω

Solution: Rmax = R2 + R1 (max) Rmax = 1 kΩ + 50 kΩ Rmax = 51 kΩ

Solution: Ideally, when the triac is conducting, the voltage drop across it is 0 V.

Rmin = R2 + R1 (min) Rmin = 1 kΩ + 0 kΩ Rmin = 1 kΩ

I = VCC/R I = 100 V/15 Ω I = 6.67 A

RCmax = RmaxC RCmax = (51 kΩ)(0.1 µF) RCmax = 5.1 ms

Answer: The load current is 6.67 A.

RCmin = RmaxC RCmin = (1 kΩ)(0.1 µF) RCmin = 0.1 msec

15-21. Given:

VB = 28 V VGT = 2.5 V Solution: Ignore the current through the diac and triac. Then VC = VB + VGT VC = 28 V + 2.5 V VC = 30.5 V Answer: The capacitor voltage required to turn on the triac is 30.5 V.

Tmax = 0.2(RCmax) Tmax = 0.2(5.1 ms) Tmax = 1.02 ms Tmin = 0.2(RCmin) Tmin = 0.2(0.1 ms) Tmin = 0.02 ms fmax = 1/Tmin fmax = 1/0.02 ms fmax = 50 kHz fmin = 1/Tmax fmin = 1/1.02 ms fmin = 980 Hz

15-22. Given:

VCC = 15 V R2 = 1 kΩ R3 = 2V

Answer: The maximum frequency is 50 kHz, and the minimum is 980 Hz.

Solution: Vgate = [R3/(R2 + R3)]VCC (Voltage divider formula) Vgate = [2 kΩ/(1 kΩ + 2 kΩ)]15 V Vgate = 10 V

15-28. Given:

R = 100 Ω VCC = 15 V

Vanode = VTrig + 0.7 V Vanode = 10 V + 0.7 Vanode = 10.7 V

Solution: In a dark room the SCR is off and the output voltage is 15 V. Once the SCR fires, its voltage drops to 0.7 V.

Answer: The gate trigger voltage is 10 V and the anode is 10.7 V.

I = (VCC – 0.7 V)/R I = (15 V – 0.7 V)/100 Ω I = 143 mA

15-23. Given:

Answer: The output voltage when it is dark is 20 V and when it is light is 0.7 V, and the current through the resistor is 143 mA when it is light.

VCC = 15 V Vgate = 10 V Vanode = 10.7 V

15-29. Answer:

Solution:

Trouble 1: Since there is voltage at D and not at E, the wire connecting the two is open.

VR4 = Vanode – 0.7 V VR4 = 10.7 V – 0.7 VR4 = 10 V

Trouble 2: No supply voltage.

Answer: The peak voltage across R4 = 10 V. 15-24. Answer: The output waveform will be a sawtooth waveform from 0 V to 10.7 V.

Trouble 3: Since there is voltage at B and not at C, the transformer is the problem. Trouble 4: Since there is voltage at A and not at B, the fuse is open. 15-30. Answer:

CRITICAL THINKING 15-25. Answer: The breakover voltage of the diode, which is 10 V.

Trouble 5: Since there is an overvoltage and the crowbar is off, the problem is the crowbar.

15-26. Answer: The breakover voltage of the diode, which is 10 V.

Trouble 6: Since there is voltage at C and not at D and the load resistor is not shorted, the rectifier is the problem.

15-27. Given:

R1 = 0 to 50 kΩ R2 = 1 kΩ C = 0.1 µF T = 20%(RC)

Trouble 7: Since there is voltage at E and not at F, the wire connecting the two is open. Trouble 8: Since there is voltage at A and not at B, the fuse is open.

1-76 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

16-3.

Chapter 16 Frequency Effects

Av(mid) = 200 f2 = 10 kHz f = 100 kHz, 200 kHz, 500 kHz, 1 MHz

SELF-TEST 1. a 2. b 3. c 4. c 5. b

6. c 7. b 8. c 9. c 10. d

11. c 12. c 13. d 14. a 15. c

16. a 17. d 18. b 19. c 20. a

Solution: Substitute in the appropriate value for f. A v = A v(mid) 2

» f ¿ 1+ ¼ À ½ f2 Á A v = 200

JOB INTERVIEW QUESTIONS 1. Too much stray-wiring capacitance. Shorten the leads as much as possible. 2. With a sine wave, find the frequency at which the voltage gain is down 3 dB. With a square wave, use the stepresponse method. 5. Some oscilloscopes (with plug-in vertical preamps) specify the risetime of the main frame. A risetime of 0.7 ns converts to a bandwidth of 50 MHz. 6. Use a step voltage and measure the risetime of the output signal. 8. Maximum power transfer. 9. dBm is referenced to 1 mW, whereas dB is not referenced to any standard. 10. Because it amplifies down to 0 Hz, which is the frequency of a dc signal. 11. Semilogarithmic 12. It is a computer program that provides electronic circuit simulation. It is used to build, test, and analyze simulated circuits.

» 100 kHz ¿ 1+ ¼ À ½ 10 kHz Á Av = 19.9

Given: Av(mid) = 1000 f1 = 100 Hz f2 = 100 kHz Solution: (Eq. 16-3) Av(20K) = Amid /[ 1+ ( f1 / f ) 2 ] Av(20K) = 1000 /[ 1+ (100 Hz/20 Hz) 2 ] Av(20K) = 196 Av(300K) = Amid /[ 1+ ( f / f 2 ) 2 ] (Eq. 16-3) Av(300K) = 1000 /[ 1+ (300 kHz/100 kHz) 2 ] Av(300K) = 316

2

Answer: Av = 19.9 at 100 kHz, Av = 9.98 at 200 kHz, Av = 4 at 500 kHz, Av = 2 at 1 MHz. 16-4.

Given: AP = 5, 10, 20, 40 Solution: AP(dB) = 10 logAP AP(dB) = 10 log(5) AP(dB) = 7 dB AP(dB) = 10 logAP AP(dB) = 10 log(10) AP(dB) = 10 dB AP(dB) = 10 logAP AP(dB) = 10 log(20) AP(dB) = 13 dB

PROBLEMS 16-1.

Given:

AP(dB) = 10 logAP AP(dB) = 10 log(40) AP(dB) = 16 dB Answer: The decibel power gain is 7 dB at a power gain of 5, 10 dB at 10, 13 dB at 20, and 16 dB at 40. 16-5.

Given: AP = 0.4, 0.2, 0.1, 0.05 Solution:

Answer: The frequency response looks like the figure below; the gain at 20 Hz is 196, and at 300 kHz is 316.

AP(dB) = 10 logAP AP(dB) = 10 log(0.4) AP(dB) = –3.98

(Eq. 16-8)

AP(dB) = 10 logAP AP(dB) = 10 log(0.2) AP(dB) = –6.99

(Eq. 16-8)

AP(dB) = 10 logAP AP(dB) = 10 log(0.1) AP(dB) = –10

(Eq. 16-8)

AP(dB) = 10 logAP AP(dB) = 10 log(0.05) AP(dB) = –13

Answer: The decibel power gain is –3.98 at a power gain of 0.4, –6.99 at 0.2, –10 at 0.1, and –13 at 0.05.

Frequency response for Prob. 16-1.

16-2.

Answer: See the figure below.

(Eq. 16-8)

16-6.

Given: AP = 2, 20, 200, 2000 Solution: AP(dB) = 10 logAP AP(dB) = 10 log(2) AP(dB) = 3 dB

(Eq. 16-8)

AP(dB) = 10 logAP AP(dB) = 10 log(20) AP(dB) = 13 dB

(Eq. 16-8)

Frequency response for Prob. 16-2.

1-77 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

AP(dB) = 10 logAP AP(dB) = 10 log(200) AP(dB) = 23 dB

(Eq. 16-8)

AP(dB) = 10 logAP AP(dB) = 10 log(2000) AP(dB) = 33 dB

(Eq. 16-8)

16-11. Given:

Answer: The decibel power gain is 3 dB at a power gain of 2, 13 dB at 20, 23 dB at 200, and 33 dB at 2000. 16-7.

Given: AP = 0.4, 0.04, 0.004 Solution: AP(dB) = 10 logAP AP(dB) = 10 log(0.4) AP(dB) = –3.98

(Eq. 16-8)

AP(dB) = 10 logAP AP(dB) = 10 log(0.04) AP(dB) = –13.98

(Eq. 16-8)

AP(dB) = 10 logAP AP(dB) = 10 log(0.004) AP(dB) = –23.98

(Eq. 16-8)

Solution: Av1 = antilog(Av1(dB)/20) Av1 = antilog(30 dB/20) Av1 = 31.6

(Eq. 16-15)

Av2 = antilog(Av2(dB)/20) Av2 = antilog(52 dB/20) Av2 = 398

(Eq. 16-15)

Answer: The voltage gain of the first stage is 31.6, and the second stage is 398. Solution: Av(dB) = 20 logAv (Eq. 16-9) Av(dB) = 20 log(100,000) Av(dB) = 100 Answer: The decibel voltage gain is 100 dB. 16-13. Given: AdB = 34 dB

Given:

Solution:

Av1 = 200 Av2 = 100

Av = antilog(Av(dB)/20) Av = antilog(34 dB/20) Av = 50.1

Solution: Av = Av1Av2 (Eq. 16-10) Av = (200)(100) Av = 20,000

(Eq. 16-15)

Answer: The voltage gain is 50.1. 16-14. Given:

Av(dB) = 20 logAv (Eq. 16-8) Av(dB) = 20 log(20,000) Av(dB) = 86 dB Answer: The voltage gain is 20,000, and the decibel voltage gain is 86 dB. 16-9.

Av1(dB) = 30 dB Av2(dB) = 52 dB

16-12. Given: Av = 100,000

Answer: The decibel power gain is –3.98 dB at a power gain of 0.4, –13.98 dB at 0.04, and –23.98 dB at 0.004. 16-8.

Answer: The decibel voltage gain is 82, and the voltage gain is 12,589.

Given: Av1 = 200 Av2 = 100 Solution:

Av1 = 25.8 Av2 = 117 Solution: Av1(dB) = 20 logAv1 Av1(dB) = 20 log(25.8) Av1(dB) = 28.2 dB

(Eq. 16-9)

Av2(dB) = 20 logAv2 Av2(dB) = 20 log(117) Av2(dB) = 41.4 dB

(Eq. 16-9)

Av1(dB) = 20 logAv1 Av1(dB) = 20 log(200) Av1(dB) = 46 dB

(Eq. 16-8)

Av(dB) = Av1(dB) + Av2(dB) (Eq. 16-11) Av(dB) = 28.2 dB + 41.4 dB Av(dB) = 69.6 dB

Av2(dB) = 20 logAv2 Av2(dB) = 20 log(100) Av2(dB) = 40 dB

(Eq. 16-8)

Answer: The decibel voltage gain for the first stage is 28.2 dB, the second stage is 41.4 dB, and overall is 69.6 dB.

Answer: The decibel voltage gain for stage 1 is 46 dB, and stage 2 is 40 dB. 16-10. Given:

16-15. Given:

AP1(dB) = 23 dB AP2(dB) = 18 dB Solution:

Av1(dB) = 30 dB Av2(dB) = 52 dB

AP1(dB) = Av1(dB) = 23 dB AP2(dB) = Av2(dB) = 18 dB

Solution: Av(dB) = Av1(dB) + Av2(dB) Av(dB) = 30 dB + 52 dB Av(dB) = 82 dB

(Eq. 16-11)

Av(dB) = Av1(dB) + Av2(dB) Av(dB) = 23 dB + 18 dB Av(dB) = 41 dB

Av = antilog(Av(dB)/20) Av = antilog(82 dB/20) Av = 12,589

(Eq. 16-15)

Answer: The total decibel voltage gain is 41 dB, the first-stage decibel voltage gain is 23 dB, and the second stage voltage gain is 18 dB.

1-78 Downloaded by Kevin Aguilar Arias ([email protected])

(Eq. 16-11)

lOMoARcPSD|6455127

16-23. Given:

C = 1000 pF R = 10 kΩ Solution: f2 = 1/(2πRC) f2 = 1/[2π(10 kΩ)(1000 pF)] f2 = 15.9 kHz Answer: See figure below.

Ideal Bode plot for Prob. 16-25.

16-26. Given:

C = 5 pF Av = 200,000 Solution: Cin = C(Av + 1) (Eq. 16-26) Cin = 5 pF(200,000 + 1) Cin = 1 ¿F

Ideal Bode plot for Prob. 16-23.

16-24. Given:

Answer: The Miller input capacitance is 1 ¿F.

R = 1 kΩ C = 50 pF

16-27. Given:

Solution: f2 = 1/(2πRC) f2 = 1/[2π(1 kΩ)(50 pF)] f2 = 3.18 MHz

C = 15 pF Av = 250,000 RL = 10 kΩ Rin = 1 kΩ

Answer: See figure below.

Solution: Cin = C(Av + 1) (Eq. 16-26) Cin = 15 pF(250,000 + 1) Cin = 3.75 ¿F f2 = 1/(2πRC) f2 = 1/[2π(1 kΩ)(3.75 ¿F)] f2 = 42 kHz (Eq. 16-9) Av(dB) = 20 log Av Av(dB) = 20 log(250,000) Av(dB) = 108 dB Answer: See figure below.

Ideal Bode plot for Prob. 16-24.

16-25. Given:

R = 15 kΩ C = 100 pF Av(mid) = 400 Solution: f2 = 1/(2πRC) f2 = 1/[2π(15 kΩ)(100 pF)] f2 = 106 kHz Av(dB) = 20 log Av Av(dB) = 20 log(400) Av(dB) = 52 dB

(Eq. 16-9)

Ideal Bode plot for Prob. 16-27.

16-28. Given:

Answer: See figure at top of next column.

C = 50 pF Av = 200,000 Solution: (Eq. 16-26) Cin = C(Av + 1) Cin = 50 pF(200,000 + 1) Cin = 10 ¿F Answer: The Miller input capacitance is 10 ¿F.

1-80 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

16-29. Given:

C = 100 pF Av = 150,000 RL = 10 kΩ Rin = 1 kΩ Solution: (Eq. 16-26) Cin = C(Av + 1) Cin = 100 pF(150,000 + 1) Cin = 15 µF f2 = 1/(2πRC) f2 = 1/[2π(1 kΩ)(15 µF)] f2 = 11 Hz (Eq. 16-9) Av(dB) = 20 log Av Av(dB) = 20 log(150,000) Av(dB) = 104 dB Answer: See figure below.

fC1 = 1/(2πRinC) fC1 = 1/[2π(717 Ω)(1 µF)] fC1 = 222 Hz Answer: The lower cutoff frequency for the base coupling circuit is 222 Hz. 16-34. Given:

C = 4.7 µF RC = 3.6 kΩ RL = 10 kΩ Solution: R = RC + RL R = 36 kΩ + 10 kΩ R = 13.6 kΩ fC1 = 1/(2πRC) fC1 = 1/[2π(13.6 kΩ)(4.7 µF)] fC1 = 2.49 Hz Answer: The lower cutoff frequency for the collector coupling circuit is 2.49 Hz. 16-35. Given:

C = 25 µF RC = 3.6 kΩ RL = 10 kΩ Solution: fC1 = 1/(2πZoutC) fC1 = 1/[2π(22.4 Ω)(25 µF)] fC1 = 284 Hz Answer: The lower cutoff frequency for the emitter bypass circuit is 284 Hz.

Ideal Bode plot for Prob. 16-29.

16-30. Given: TR = 10 µs Solution: (Eq. 16-29) f2 = 0.35/TR f2 = 0.35/10 µs f2 = 35 kHz Answer: The upper cutoff frequency is 35 kHz. 16-31. Given: TR = 0.25 µs Solution: (Eq. 16-29) f2 = 0.35/TR f2 = 0.35/0.25 µs f2 = 1.4 MHz Answer: The bandwidth is 1.4 MHz. 16-32. Given: f2 = 100 kHz Solution: (Eq. 16-29) f2 = 0.35/TR TR = 0.35/f2 TR = 0.35/100 kHz TR = 3.5 µs Answer: The risetime is 3.5 µs. 16-33. Given: C = 1 µF R = 50 Ω Solution: Rin = Zin(Stage) Rin = R1 || R2 || βre′ Rin = 717 Ω

16-36. Given: Cc′ = 2 pF Ce′ = 10 pF ′ = 5 pF CStray R1 = 10 kΩ R2 = 2.2 kΩ RC = 3.6 kΩ RL = 10 kΩ RG = 50 kΩ β = 200

Solution: rg = RG||R1||R2 rg = 50 Ω||10 kΩ||2.2 kΩ rg = 48 Ω Cin(m) = 236 pF C = Cin (m) + Ce′ = 246 pF Base f2 = 1/(2πrgC) f2 = 1/[2π(48 Ω)(246 pF)] f2 = 13.5 MHz Collector C = Ce′ + CStray = 7 pF Cout(m) = 2 pF R = RC + RL R = 3.6 kΩ + 10 kΩ R = 13.6 kΩ f2 = 1/(2πRC) f2 = 1/[2π(13.6 Ω)(1.3 pF)] f2 = 8.59 MHz Answer: The high cutoff frequency for the base is 13.5 MHz and the collector is 8.59 MHz.

1-81 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

C = CdS + Cout(m) C = 15 pF + 5.3 pF

16-37. Given:

gm = 16.5 mS CiSS = 30 pF CoSS = 20 pF CrSS = 5 pF

Drain f2 = 1/(2πRC) f2 = 1/[2π(909 Ω)(20.3 pF)] f2 = 8.61 MHz

Solution: Cgd = CrSS = 5 pF CgS = CiSS – CrSS CgS = 30 pF – 5 pF CgS = 25 pF

Answer: The high frequency cutoff for the gate is 30.3 MHz and the drain is 8.61 MHz.

CRITICAL THINKING

CdS = CoSS – CrSS CdS = 20 pF – 5 pF CdS = 15 pF

16-40. Given:

Answer: CgS = 25 pF, Cgd = CrSS = 5 pF, CdS = 15 pF. 16-38. Given:

Solution:

R1 = 2 MΩ R2 = 1 MΩ RD = 1 kΩ RL = 10 kΩ RG = 50 Ω Cin = 0.01 µF Cout = 1 µF

Av(mid) = antilog(Av(dB)/20) Av(mid) = antilog(80 dB/20) Av(mid) = 10,000

(Eq. 16-15)

Av(20K ) = Av(mid) /[ 1 + ( f/f 2 ) 2 ] ( Eq. 16-3) Av(20K ) = 10,000 /[ 1 + (20 kHz /100 Hz) 2 ] Av(20K) = 50 Av(dB) = 20 log A Av(dB) = 20 log(50) Av(dB) = 34 dB

Solution: Rin = Zin(Stage) Rin = R1||R2 Rin = 667 kΩ

(Eq. 16-9)

Av(44.4K ) = Av(mid) /[ 1 + ( f/f 2 ) 2 ] (Eq. 16-3) Av(44.4K ) = 10,000 /[ 1 + (44.4 kHz/100 Hz) 2 ] Av(44.4K) = 22.5

fC1 = 1/(2πRinC) fC1 = 1/[2π(667 kΩ)(0.01 µF)] fC1 = 23.9 Hz or 14.5 Hz Answer: The dominant low cutoff frequency is 23.9 Hz. 16-39. Given:

R1 = 2 MΩ R2 = 1 MΩ RD = 1 kΩ RL = 10 kΩ RG = 50 kΩ Cgd = 5 pF CgS = 25 pF CdS = 015 pF

f2 = 100 Hz Av(dB) = 80 dB

Av(dB) = 20 log Av Av(dB) = 20 log(22.5) Av(dB) = 27 dB

(Eq. 16-9)

Answer: The decibel voltage gain at 20 kHz is 34 dB, and at 44.4 kHz is 27 dB. 16-41. Given:

(from Prob. 16-37.) (from Prob. 16-37.) (from Prob. 16-37.)

Solution: Av = gmrd Av = (16.5 mS)(1 kΩ||10 kΩ) Av = 15 Cin(m) = Cgd(Av + 1) Cin(m) = 5 pF(15 + 1) Cin(m) = 80 pF C = CgS + Cin(m) C = 25 pF + 80 pF C = 105 pF R = RG||R1||R2 R = 50 Ω||2 MΩ||1 MΩ R = 50 Ω

(Eq. 16-40)

Solution: Since the roll-off is 20 dB/decade at a frequency of 1 kHz (one decade above the cutoff frequency), the gain is 100 dB (20 dB less than the midband), and at 10 kHz the gain is 80 dB. From this point the roll-off increases to 40 dB/decade; thus at 100 kHz, the gain will be 40 dB. Answer: The voltage gain at 100 kHz is 40 dB. 16-42. Given:

Vin = 20 mV Av(mid) = 100 Solution: Vout(max) = Av(mid)Vin Vout(max) = (100)(20 mV) Vout(max) = 2 V At the 10% point = 0.1 Vout(max) At the 10% point = 0.1(2 V) At the 10% point = 0.2 V

Gate f2 = 1/(2πRC) f2 = 1/[2π(50 Ω)(105 pF)] f2 = 30.3 MHz Collector Cout(M) = Cgd[(Av + 1)/Av] Cout(M) = 5 pF[(15 + 1)/15] Cout(M) = 5.3 pF

f2 = 100 Hz Second breakpoint is 10 kHz Av(mid) = 120 dB

(Eq. 16-41)

At the 90% point = 0.9 Vout(max) At the 90% point = 0.9(2 V) At the 90% point = 1.8 V Answer: The voltage at the 10% point is 0.2 V, and at the 90% point is 1.8 V.

1-82 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

16-43. Given:

17-2.

VCC = 15 V VEE = –15 V RE = 270 kΩ RC = 180 kΩ

R = 4 kΩ C = 50 pF Solution: f2 = 1/(2πRC) f2 = 1/[2π(4 kΩ)(50 pF)] f2 = 796 kHz

Solution: IT = ( – VEE – VBE)/RE IT = (–15 V – 0.7 V)/270 kΩ IT = 53 µA IE = 1/2 IT IE = 1/2(53 µA) IE = 26.5 µA VC = VCC – (26.5 µA)(180 kΩ) VC = 10.2 V

(Eq. 16-29) f2 = 0.35/TR TR = 0.35/f2 TR = 0.35/796 kHz TR = 0.44 µs Answer: The risetime is 0.44 µs. 16-44. Given:

Answer: The tail current is 53 µA, the emitter is 26.5 µA, and the quiescent voltage is 10.2 V.

f2 = 1 MHz TR = 1 µs 17-3.

Solution: f2 = 0.35/TR f2 = 0.35/1 µs f2 = 350 kHz

Solution: IT = (– VEE)/RE IT = (– 12 V)/200 kΩ IT = 60 µA

Chapter 17 Differential Amplifiers

IE = 1/2 IT IE = 1/2(60 µA) IE = 30 µA

SELF-TEST 7. b 8. a 9. d 10. a 11. c 12. b

13. c 14. a 15. a 16. b 17. d 18. c

Given: VCC = 12 V VEE = –12 V RE = 200 kΩ RC = 200 kΩ

(Eq. 16-29)

Answer: The amplifier with the cutoff frequency of 1 MHz has the larger bandwidth.

1. b 2. c 3. a 4. c 5. b 6. a

Given:

19. b 20. c 21. a 22. c 23. c

Right Side VC = VCC – (30 µA)(200 kΩ) VC = 6 V Left Side VC = 12 V Answer: The tail current is 60 µA, the emitter is 30 µA, and the quiescent voltage is 6 V on the right side and 12 V on the left side.

JOB INTERVIEW QUESTIONS 6. Use a transistor as a current source instead of a tail resistor. It could be a regulator configuration or a current source. 9. A transistor acting as a current source. 11. Current sources and active loads. 12. Increased voltage gain and higher CMRR. 13. Trick question. You can’t test a 741 with an ohmmeter.

PROBLEMS 17-1.

17-4.

Given: VCC = 12 V VEE = –12 V RE = 200 kΩ RC = 200 kΩ Solution: IT = (– VEE – VBE)/RE IT = (– 12 V – 0.7 V)/200 kΩ IT = 56.5 µA

Given: VCC = 15 V VEE = –15 V RE = 270 kΩ RC = 180 kΩ

IE = ½ IT IE = 1/2(56.5 µA) IE = 28.3 µA

Solution:

Right Side

IT = –VEE/RE IT = –15 V/270 kΩ IT = 55.6 µA

VC = VCC – (28.3 µA)(200 kΩ) VC = 6.35 V Left Side

IE = 1/2 IT IE = 1/2 (55.6 µA) IE = 27.8 µA

VC = 12 V

VC = VCC – (27.8 µA)(180 kΩ) VC = 10 V

Answer: The tail current is 56.5 µA, the emitter is 28.3 µA, and the quiescent voltage is 6.35 V on the right side and 12 V on the left side.

Answer: The tail current is 55.6 µA, the emitter is 27.8 µA, and the quiescent voltage is 10 V.

1-83 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

17-5.

RC = 47 kΩ β = 275 v1 = 0 mV v1 = 1 mV

Given: VCC = 15 V VEE = –15 V RE = 68 kΩ RC = 47 kΩ β = 275 v1 = 2.5 mV

Solution: IT = ( – VEE)/RE IT = (– 15 V)/68 kΩ IT = 220.6 µA

Solution:

IE = 1/2 IT (Eq. 17-6) IE = 1/2(220.6 µA) IE = 110.3 µA re′ = 25 mV/IE (Eq. 9-10) re′ = 25 mV/110.3 ¿A re′ = 226.7 Ω Av = RC / re′ (Eq. 17-10) Av = 47 kΩ/226.7 Ω Av = 207.3

(Eq. 17-5) IT = (– VEE/RE) IT = (– 15 V)/68 kΩ IT = 220.6 µA IE = 1/2 IT (Eq. 17-6) IE = 1/2 (220.6 µA) IE = 110.3 µA re′ = 25 mV/IE (Eq. 9-10) re′ = 25 mV/110.3 ¿A re′ = 226.7 Ω (Eq. 17-10) Av = RC/ re′ Av = 47 kΩ/226.7 Ω Av = 207.3 vout = Av(v1 – v2) (Eq. 17-2) vout = 207.3(2.5 mV – 0) vout = 518 mV zin = 2β re′ (Eq. 17-11) zin = 2(275)(226.7 Ω) zin = 125 kΩ

vout = Av(v1 – v2) (Eq. 17-2) vout = 207.3(0 V – 1 mV) vout = –207 mV zin = 2β re′ (Eq. 17-11) zin = 2(275)(226.7 Ω) zin = 125 kΩ Answer: The output voltage is –207 mV, and the input impedance is 125 kΩ. 17-8.

Answer: The output voltage is 518 mV, and the input impedance is 125 kΩ. 17-6.

Given: Av = 360 Iin(biaS) = 600 nA Iin(off) = 100 nA Vin(off) = 1 mV RB1 = 10 kΩ

Given: VCC = 15 V VEE = –15 V RE = 68 kΩ RC = 47 kΩ β = 275 v1 = 2.5 mV

Solution: V1err = (RBl – RB2)/Iin(biaS) V1err = (10 kΩ – 0)600 nA V1err = 6 mV

Solution:

(Eq. 17-16)

V2err = (RB1 + RB2)(Iin(off)/2) V2err = (10 kΩ + 0)(100 nA/2) V2err = 0.5 mV

IT = (– VEE – VBE)/RE (Eq. 17-5) IT = (– 15 V – 0.7 V)/68 kΩ IT = 210.3 µA

V3err = Vin(off) V3err = 1 mV

IE = 1/2 IT (Eq. 17-6) IE = 1/2(210.3 µA) IE = 105.2 µA re′ = 25 mV / I E (Eq. 9-10) re′ = 25 mV /105.2 µA re′ = 237.6 Ω Av = RC / re′ (Eq. 17-10) Av = 47 kΩ/237.6 Ω Av = 197.8

17-7.

(Eq. 17-5)

(Eq. 17-17)

(Eq. 17-18)

Verror = Av(V1err + V2err + V3err) (Eq. 17-19) Verror = 360(6 mV + 0.5 mV + 1 mV) Verror = 2.7 V

With base resistors equal. V1err = 0.

(Eq. 17-20)

V2err = RBIin(off) (Eq. 17-21) V2err = (10 kΩ)(100 nA) V2err = 1 mV

vout = Av(v1 – v2) (Eq. 17-2) vout = 197.8(2.5 mV – 0) vout = 494 mV zin = 2β re′ (Eq. 17-11) zin = 2(494)(237.6 Ω) zin = 131 kΩ

Verror = Av(V1err + V2err + V3err) (Eq. 17-19) Verror = 360(0 mV + 1 mV + 1 mV) Verror = 0.72 V

Answer: The output voltage is 494 mV, and the input impedance is 131 kΩ.

Answer: The output error voltage is 2.7 V. If the base resistors are equal, the output error voltage is 0.72 V.

Given: VCC = 15 V VEE = –15 V RE = 68 kΩ

V3err = Vin(off) V3err = 1 mV

17-9.

Given: Av = 250 Iin(biaS) = 1 µA

1-84 Downloaded by Kevin Aguilar Arias ([email protected])

(Eq. 17-18)

lOMoARcPSD|6455127

–Av3(CL)max = –100 kΩ/40 kΩ –Av3(CL)max = –2.5

Av(CL)max = (100 kΩ/2 kΩ) + 1 Av(CL)max = 51 Av(CL)min = (R2(min)/R1) + 1 Av(CL)min = (0 kΩ/2 kΩ) + 1 Av(CL)min = 1

(Eq. 18-12)

f2(CL)max = funity/Av(CL)min f2(CL)max = 20 MHz/1 f2(CL)max = 20 MHz

(Eq. 18-5)

f2(CL)min = funity/Av(CL)max f2(CL)min = 20 MHz/51 f2(CL)min = 392 kHz

(Eq. 18-5)

Answer: The maximum output voltage is 1.35 Vpp, and the minimum output voltage is zero. 18-21. Given:

Answer: The voltage gain has a range of 1 to 51 and a bandwidth of 392 kHz to 20 MHz. 18-16. Answer: The voltage across the closed-loop output impedance is the difference between the ideal 50 mV and the actual 49.98 mV. In other words, 0.02 mV is dropped across the closed-loop output impedance. The load current is 49.98 mV divided by 2 Ω, which is approximately 25 mA. Divide 0.02 mV by 25 mA to get 0.0008 Ω for the closed-loop output impedance. 18-17. Given:

f = 15 kHz VP = 2 V

R1 = 220 Ω RF1 = 47 kΩ RF2 = 18 kΩ RF3 = 39 kΩ Solution:

–Av1(CL) = –RF1/R1 (Eq. 18-3) –Av1(CL) = –47 kΩ/220 Ω –Av1(CL) = –214 –Av2(CL) = –RF2/R1 (Eq. 18-3) –Av2(CL) = –18 kΩ/220 Ω –Av2(CL) = –82 –Av3(CL) = –RF3/R1 (Eq. 18-3) –Av3(CL) = –39 kΩ/220 Ω –Av3(CL) = –177 Solution: The gain at position 1 is 214, at position 2 is 82, and at position 3 is 177.

Solution: SS = 2πfVP SS = 2π(15 kHz)(2 V) SS = 188 mV/¿s

18-22. Given:

R1 = 6 kΩ at position 2 R2 = 6 kΩ||3 kΩ at position 1 = 2 kΩ R2 = 120 kΩ funity = 1 MHz

SS = 2πfVP SS = 2π(30 kHz)(2 V) SS = 376 mV/¿s Answer: The initial slope is 188 mV/¿s, with a peak of 2 V, and 376 mV/¿s, with a frequency of 30 kHz. 18-18. Answer: a. b. c. d. e.

vout = Av1(CL)max(vin1) + Av2(CL)max(vin2) + Av3(CL)max(vin3) vout = –10(50 mVpp) + 5(90 mVpp) + 2.5(160 mVpp) vout = –1.35 Vpp

OP-07A TL082 and TL084 LM12 OP-64E OP-07A

Solution: Av1(CL) = (R2/R1) + 1 (Eq. 18-12) Av1(CL) = (120 kΩ/2 kΩ)+ 1 Av1(CL) = 61 Av2(CL) = (R2/R1) + 1 (Eq. 18-12) Av2(CL) = (120 kΩ/6 kΩ) + 1 Av2(CL) = 21 f2(CL)1 = funity/Av(CL1) f2(CL)1 = 1 MHz/61 f2(CL)1 = 16.4 kHz

18-19. Answer:

CMRR = 38 dB (from Fig. 18-6a) MPP = 21 V (from Fig. 18-6b) Av = 1000 (from Fig. 18-6c) 18-20. Given:

(Eq. 18-5)

f2(CL)2 = funity/Av(CL1)(max) f2(CL)2 = 1 MHz/21 f2(CL)2 = 47.6 kHz

(Eq. 18-5)

Answer: The voltage gain at position 1 is 61, with a bandwidth of 16.4 kHz, and at position 2 is 21, with a bandwidth of 47.6 kHz.

R1 = 10 kΩ R2 = 20 kΩ R3 = 40 kΩ Rf(max) = 100 kΩ Rf(min) = 100 kΩ v1 = 50 mVpp v1 = 90 mVpp v1 = 160 mVpp

18-23. Given:

Solution: When the resistance is zero, the voltage gains are zero and the output voltage is zero.

R1 = ∞ at position 2 R1 = 3 kΩ at position 1 R2 = 120 kΩ funity = 1 MHz AVOL = 100,000 Solution:

–Av1(CL)max = –Rf /R1 (Eq. 18-3) –Av1(CL)max = –100 kΩ/10 kΩ –Av1(CL)max = –10

AV1(CL) = (R2/R1) + 1 (Eq. 18-12) AV1(CL) = (120 kΩ/3 kΩ) + 1 Av1(CL) = 41

–Av2(CL)max = –Rf /R1 (Eq. 18-3) –Av2(CL)max = –100 kΩ/20 kΩ –Av2(CL)max = –5

At position 2, it becomes a voltage follower: AvCL2 = 1.

–Av3(CL)max = –Rf /R1

Answer: The voltage gain at position 1 is 41, and at position 2 is 1.

(Eq. 18-3)

1-91 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

18-24. Answer: The output will go to positive or negative saturation. 18-25. Answer:

Position 1: The input voltage is applied directly to the noninverting input. Because of the virtual short between the noninverting and inverting input terminals, there is no ac voltage across the left 10-kΩ resistor. Since there is no ac voltage across the resistor, it can be removed from the circuit without changing the operation. With the resistor removed, the circuit reduces to a voltage follower and Av(CL) = 1 and a closed-loop bandwidth of f unity

f 2( CL ) =

Av (CL )

1 MHz = = 1 MHz 1

Position 2: The circuit is an inverting amplifier. The magnitude of the voltage gain is Av(CL) = 1. Note that the closed-loop bandwidth is only half as much because f unity

1 MHz f 2( CL ) = = = 500 kHz AV ( CL ) +1 1+1

This was covered briefly in the chapter. See the equation at the top of p. 633 and the brief explanation that follows. Chapter 19 discusses the closed-loop bandwidths in more detail. 18-26. Answer:

Position 1: With the left resistor open, the circuit reduces to a voltage follower and AV(CL) = 1. Position 2: With the left resistor open, the voltage gain is zero. 18-27. Answer: Go to positive or negative saturation. 18-28. Given:

18-29. Given:

R1 = 2 kΩ R2 = 100 kΩ C = 1 ¿F vin = 50 mV pp f = 1 kHz Solution: XC = 1/2πfC XC = 1/[2π(1 kHz)(1 ¿F)] XC = 159 Ω

Since XC is less than one-tenth of 2 kΩ, the bottom of the 2 kΩ is approximately an ac ground. AV(CL) = (R2/ R1′ ) + 1 (Eq. 18-12) AV(CL) = (100 kΩ/2 kΩ) + 1 AV(CL) = 51 vout = AV(CL)vin vout = 51(50 mV pp) vout = 2.55 V pp Answer: The output voltage is 2.55 V. 18-30. Given:

Iin(biaS) = 500 nA Iin(off) = 200 nA Vin(off) = 6 mV R1 = 2 kΩ R2 = 100 kΩ Solution: R1′ = XC = R1 R1′ = 0 + 2 kΩ R1′ = 2 kΩ

Iin(biaS) = 500 nA Iin(off) = 200 nA Vin(off) = 6 mV R1 = 2 kΩ R2 = 100 kΩ C = 1 ¿F

RB2 = R1||R2 (Eq. 18-11) RB2 = 2 kΩ||100 kΩ RB2 = 1.96 kΩ

Solution: XC = 1/2πfC XC = 1/[2π(0)(1 ¿F)] XC = ∞ R1′ = X C + R1 R1′ = ∞ + 2 kΩ R1′ = ∞

V2err = (RB1 + RB2)(Iin(off)/2) (Eq. 18-8) V2err = (0 + 1.96 kΩ)(200 nA/2) V2err = 196 ¿V V3err = Vin(off) = 6 mV (Eq. 18-12) AV(CL) = ( R2 / R1′) + 1 AV(CL) = (100 kΩ/2 kΩ) + 1 AV(CL) = 51

RB2 = R1||R2 RB2 = ∞||100 kΩ RB2 = 100 kΩ

Verror = ± AV(CL)( ± V1err ± V2err ± V3err) Verror = 51(980 ¿V + 196 ¿V + 6 mV) Verror = 366 mV

V1err = (RB1 – RB2)Iin(biaS) V1err = (0 – 1.96 kΩ)(500 nA) V1err = 980 ¿V

(Eq. 18-11)

V1err = (RB1 – RB2)Iin(biaS) V1err = (0 – 100 kΩ)(500 nA) V1err = 50 mV

(Eq. 18-8)

V2err = (RB1 + RB2)(Iin(off)/2) (Eq. 18-8) V2err = (0 + 100 kΩ)(200 nA/2) V2err = 10 mV V3err = Vin(off) = 6 mV AV(CL) = (R2/ R1′ ) + 1 AV(CL) = (100 kΩ/∞) + 1 AV(CL) = 1

(Eq. 18-12)

Verror = ± ACL( ± V1err ± V2err ± V3err) Verror = 1(50 mV + 10 mV + 6 mV) Verror = 66 mV Answer: The output voltage is 66 mV.

(Eq. 18-8)

Answer: The output voltage is 366 mV. 18-31. Answer:

For IB1: V1—increase. Because of the increase in voltage drop across the resistor. V2—no change. Not affected. Vin—increase. Because of the increase in V1. Vout—increase. Because of the increase in input voltage. MPP—no change. Since the load resistance and VCC did not change. fmax—no change. Since slew rate did not change. For IB2: V1—no change. Not affected. V2—increase. Because of the increase in voltage drop across the resistor.

1-92 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Av = antilog(Av(dB)/20) Av = antilog(88 dB/20) Av = 25,119

Vin—increase. Because of the increase in V2. Vout—increase. Because of the increase in input voltage. MPP—no change. Since the load resistance and VCC did not change. fmax—no change. Since slew rate did not change.

%error = 100%(1 + AVOLB) (Eq 19-5) %error = 100%[1 + 25,119(0.038)] %error = 0.10%

18-32. Given:

For VCC: V1—no change. Not affected. V2—no change. Not affected. Vin—no change. Not affected. Vout—no change. Not affected. MPP—increase Since VCC is increased. fmax—no change. Since slew rate did not change.

(Eq. 16-15)

AV = AVOL/(1 + AVOLB) (Eq. 19-3) AV = 25,119/[l + 25,119(0.038)] AV = 26.29 Answer: The feedback fraction is 0.038, the ideal closed-loop voltage gain is 26.32, the percent error is 0.10%, and the exact voltage gain is 26.29. 19-2.

18-33. Given:

R1 = 2.7 kΩ Rf = 39 kΩ AVOL(dB) = 88 dB

V1—no change. Not affected. V2—no change. Not affected. Vin—no change. Not affected. Vout—no change. Not affected. MPP—no change. Since the load resistance and VCC did not change. fmax—increase. Since slew rate increased.

Solution: B = R1/(R1 + Rf) (Eq. 19-6) B = 2.7 kΩ/(2.7 kΩ + 39 kΩ) B = 0.065

18-34. Given:

V1—no change. Not affected. V2—no change. Not affected. Vin—no change. Not affected. Vout—no change. Not affected. MPP—no change. Since the load resistance and VCC did not change. fmax—decrease. Since the increase in voltage causes the rate of voltage rise to increase.

Given:

AV = 1/B AV = 1/0.065 AV = 15.44

(Eq. 19-4)

Answer: The feedback fraction is 0.065, and the closedloop voltage gain is 15.44. 19-3.

Given: R1 = 4.7 kΩ Rf = 68 kΩ AVOL(dB) = 88 dB Solution:

Chapter 19 Negative Feedback

B = R1/(R1 + Rf) (Eq. 19-6) B = 4.7 kΩ/(4.7 kΩ + 68 kΩ) B = 0.065

SELF-TEST 1. b 2. d 3. a 4. a 5. a 6. c 7. b

8. b 9. b 10. b 11. d 12. b 13. b 14. b

15. b 16. d 17. c 18. b 19. c 20. b 21. c

JOB INTERVIEW QUESTIONS 8. Increased voltage gain and possible oscillation. 12. Current amplifier and transconductance amplifier.

Given:

(Eq. 19-4)

Answer: The feedback fraction is 0.065, and the closedloop voltage gain is 15.47. 19-4.

Given: R1 = 2.7 kΩ Rf = 68 kΩ AVOL(dB) = 108 dB Solution:

AV = 1/B AV = 1/0.038 AV = 26.32

R1 = 2.7 kΩ Rf = 68 kΩ AVOL(dB) = 88 dB Solution: B = R1/(R1 + Rf) (Eq. 19-6) B = 2.7 kΩ/(2.7 kΩ + 68 kΩ) B = 0.038 AV = 1/B AV = 1/0.038 AV = 26.32

AV = 1/B AV = 1/0.065 AV = 15.47

B = R1/(R1 + Rf) (Eq. 19-6) B = 2.7 kΩ/(2.7 kΩ + 68 kΩ) B = 0.038

PROBLEMS 19-1.

22. d 23. d 24. b 25. a 26. b 27. d 28. a

(Eq. 19-4)

(Eq. 19-4)

AV = antilog(AVOL(dB)/20) AV = antilog(108 dB/20) AV = 251,189

(Eq. 16-15)

%error = 100%(1 + AVOLB) (Eq 19-5) %error = 100%[1 + 251,189(0.038)] %error = 0.01% AV = AVOL/(1 + AVOLB) (Eq. 19-3) AV = 251,189/[l + 251,189(0.038)] AV = 26.31

1-93 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Solution vout(1) = iinR3 (from Table 19-2) vout(1) = 1 ¿A(10 kΩ) vout(1) = 10 mV AV(2) = Rf/R1 + 1 AV(2) = 99 kΩ/1 kΩ + 1 AV(2) = 100

Rout = 75 Ω Solution: B(1) = R1(1)/(R1(1) + Rf) (Eq. 19-6) B(1) = 1 kΩ/(1 kΩ + 50 kΩ) B(1) = 0.0196 zin1(CL) = (1 + AVOLB(1))Rin (Eq. 19-8) zin1(CL) = (1 + (100,000)(0.0196))2 MΩ zin1(CL) = 3924 MΩ

vout(2) = AV(vin(2)) vout(2) = 100(10 mV) vout(2) = 1 V Answer: The output voltage is 1 V. 19-24. Given:

zout1(CL) = Rout/(1 + AVOLB(1)) (Eq. 19-10) zout1(CL) = 75 Ω/(1 + (100,000)(0.0196)) zout1(CL) = 38 mΩ B(2) = R1(2)/(R1(2) + Rf) (Eq. 19-6) B(2) = 25 kΩ/(25 kΩ + 50 kΩ) B(2) = 0.333

Rf = 50 kΩ R1(1) = 1 kΩ R1(2) = 25 kΩ R1(3) = 100 kΩ Solution:

zin2(CL) = (1 + AVOLB(2))Rin (Eq. 19-8) zin2(CL) = (1 + (100,000)(0.333))2 MΩ zin2(CL) = 66,669 MΩ

AV1(CL) = Rf/R1(1) + 1 AV1(CL) = 50 kΩ/1 kΩ + 1 AV1(CL) = 51

zout2(CL) = Rout/(1 + AVOLB(2)) (Eq. 19-10) zout2(CL) = 75 Ω/(1 + (100,000)(0.333)) zout2(CL) = 2.5 mΩ

AV2(CL) = Rf/R1(2) + 1 AV2(CL) = 50 kΩ/25 kΩ + 1 AV2(CL) = 3

B(3) = R1(3)/(R1(3) + Rf) (Eq. 19-6) B(3) = 100 kΩ/(100 kΩ + 50 kΩ) B(3) = 0.667

AV3(CL) = Rf/R1(3) + 1 AV3(CL) = 50 kΩ/100 kΩ + 1 AV3(CL) = 1.5

zin3(CL) = (1 + AVOLB(3))Rin (Eq. 19-8) zin3(CL) = (1 + (100,000)(0.667))2 MΩ zin3(CL) = 133,335 MΩ

Answer: The voltage gains are 51 at the 1-kΩ position, 3 at the 25-kΩ position, and 1.5 at the 100-kΩ position.

zout3(CL) = Rout/(1 + AVOLB(3)) (Eq. 19-10) zout3(CL) = 75 Ω/(1 + (100,000)(0.667)) zout3(CL) = 1.25 mΩ

19-25. Given:

Rf = 50 kΩ R1(1) = 1 kΩ R1(2) = 25 kΩ R1(3) = 100 kΩ AV1(CL) = 51 AV2(CL) = 3 AV3(CL) = 1.5 vin = 10 mV

(from Prob. 19-24) (from Prob. 19-24) (from Prob. 19-24)

19-27. Given:

Solution: vout(1) = AV1(CL)(vin) vout(1) = 51(10 mV) vout(1) = 510 mV vout(2) = AV2(CL)(vin) vout(2) = 3(10 mV) vout(2) = 30 mV vout(3) = AV3(CL)(vin) vout(3) = 1.5(10 mV) vout(3) = 15 mV Answer: The output voltages are 510 mV at the 1-kΩ position, 30 mV at the 25-kΩ position, and 15 mV at the 100-kΩ position. 19-26. Given:

Rf = 50 kΩ R1(1) = 1 kΩ R1(2) = 25 kΩ R1(3) = 100 kΩ AV1(CL) = 51 AV2(CL) = 3 AV3(CL) = 1.5 AVOL = 100,000 Rin = 2 MΩ

Answer: At the 1-kΩ position the input impedance is 3,924 MΩ and the output impedance is 38 mΩ. At the 25-kΩ position the input impedance is 66,669 MΩ and the output impedance is 2.5 mΩ. At the 100-kΩ position the input impedance is 133,335 MΩ and the output impedance is 1.25 mΩ. Note: The RCM of the op amp is not included in the calculations for input impedance. See Example 19-2. Iin(bias) = 80 nA Iin(off) = 20 nA Vin(off) = 1 mV AVOL = 100,000 Rf = 100 kΩ R1(1) = 1 kΩ R1(2) = 25 kΩ R1(3) = 100 kΩ AV1(CL) = 101 AV2(CL) = 5 AV3(CL) = 2 Solution: RB2(1) = R1(1)||Rf (Eq. 18-11) RB2(1) = 1 kΩ||100 kΩ RB2(1) = 990 Ω V1err(1) = (RB1 – RB2(1))Iin(biaS) V1err(1) = (0 – 990 Ω)(80 nA) V1err(1) = – 79.2 ¿V

(from Prob. 19-24) (from Prob. 19-24) (from Prob. 19-24)

V2err(1) = (RB1 + RB2(1))(Iin(off)/2) V2err(1) = (0 + 990 Ω)(20 nA/2) V2err(1) = 9.9 ¿V V3err(1) = Vin(off) = 1 mV

1-96 Downloaded by Kevin Aguilar Arias ([email protected])

(Eq. 18-8)

(Eq. 18-8)

lOMoARcPSD|6455127

Rmin = 130 Ω Rmax = 25.13 kΩ funity = 1 MHz

fC1 = 1/(2πR3C1) fC1 = 1/[2π(100 kΩ)(2.2 ¿F)] fC1 = 0.72 Hz

Solution:

fC2 = 1/(2πRLC2) fC2 = 1/[2π(25 kΩ)(4.7 ¿F)] fC2 = 1.35 Hz

Bmin = (10 kΩ||130 Ω)/(10 kΩ||130 Ω + 180 kΩ) Bmin = 0.000712

fC3 = 1/(2πR1C3) fC3 = 1/[2π(2 kΩ)(1 ¿F)] fC3 = 79.6 Hz

Bmax = (10 kΩ||25.13 kΩ)/(10 kΩ||25.13 kΩ + 180 kΩ) Bmax = 0.0382 f2(min) = Bfunity f2(min) = 0.000712(1 MHz) f2(min) = 712 Hz

Answer: The midband voltage gain is 42, the upper cutoff frequency is 71.4 kHz, and the lower cutoff frequency is 79.6 Hz.

f2(max) = Bfunity f2(max) = 0.0382(1 MHz) f2(max) = 38.2 kHz

20-6.

R1 = 3.3 kΩ R2 = 150 kΩ R3 = 100 kΩ RL = 10 kΩ C1 = 1 ¿F C2 = 10 ¿F C3 = 4.7 ¿F funity = 1 MHz

−R f

−180 kΩ = = 18 Av = 10 kΩ R1 Answer: The minimum bandwidth is 712 Hz and the maximum bandwidth is 38.2 kHz. 20-4.

Given: R1 = 1.5 kΩ Rf = 100 kΩ Rmin = 100 Ω Rmax = 5.1 kΩ funity = 1 MHz

Solution: AV = (R2/R1) + 1 AV = (150 kΩ/3.3 kΩ) + 1 AV = 46.5 f2 = funity/AV f2 = 1 MHz/46.5 f2 = 21.5 kHz

Solution: Bmin = (R1||Rmin)/(R1||Rmin + Rf) Bmin = (1.5 kΩ||100 Ω)/(1.5 kΩ||100 Ω + 100 kΩ) Bmin = 0.000937

fC1 = 1/(2πR3C1) fC1 = 1/[2π(100 kΩ)(1 ¿F)] fC1 = 1.59 Hz

Bmax = (R1||Rmax)/(R1||Rmax + Rf) Bmax = (1.5 kΩ||5.1 kΩ)/(1.5 kΩ||5.1 kΩ + 100 kΩ) Bmax = 0.01146

fC2 = 1/(2πRLC2) fC2 = 1/[2π(10 kΩ)(10 ¿F)] fC2 = 1.59 Hz

f2(min) = Bminfunity f2(min) = 0.000937(1 MHz) f2(min) = 937 Hz

fC3 = 1/(2πR1C3) fC3 = 1/[2π(3.3 kΩ)(4.7 ¿F)] fC3 = 10.3 Hz

f2(max) = Bmaxfunity f2(max) = 0.01146(1 MHz) f2(max) = 11.5 kHz

Answer: The midband voltage gain is 46.5, the upper cutoff frequency is 21.5 kHz, and the lower cutoff frequency is 10.3 Hz.

AV = –Rf/R1 AV = –100 kΩ/1.5 kΩ AV = –66.7

20-5.

Given:

20-7.

Given:

vout = Avvin vout = –66.7(4 mV) vout = 266.8 mV

R1 = 2 kΩ Rf = 100 kΩ vin = 10 mV

Answer: The minimum bandwidth is 937 Hz and the maximum bandwidth is 11.5 kHz. The output voltage is 266.8 mV.

Solution: AV = (Rf/R1) + 1 AV = (100 kΩ/2 kΩ) + 1 AV = 51

Given: R1 = 2 kΩ Rf = 82 kΩ RL = 25 kΩ C1 = 2.2 ¿F C2 = 4.7 ¿F funity = 3 MHz Solution: AV = (Rf/R1) + 1 AV = (82 kΩ/2 kΩ) + 1 AV = 42

vout = Avvin vout = 51(10 mV) vout = 510 mV Answer: The output voltage at A, B, and C is 510 mV. 20-8.

Given: R1 = 91 kΩ Rf = 12 kΩ R3 = 1 kΩ vin = 2 mV

f2 = funity/AV f2 = 3 MHz/42 f2 = 71.4 kHz

1-99 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Answer: The maximum gain is –10, and the maximum positive gain is 0.

Solution: Low gate: AV = (Rf/R1) + 1 AV = (12 kΩ/91 kΩ) + 1 AV = 1.13

20-12. Given: R1 = R2

Solution: At ground the circuit is an inverting amplifier. AV = –Rf/R1 AV = –1

vout = Avvin vout = 1.13(2 mV) vout = 2.26 mV High gate: AV = [Rf/(R1||R3)] + 1 AV = [12 kΩ/(91 kΩ||1 kΩ)] + 1 AV = 13.1 vout = Avvin vout = 13.1(2 mV) vout = 26.2 mV

Av(non) = 10% (2) = 0.2 AV = Av(in) + Av(non) AV = –1 + 0.2 AV = –0.8 Answer: The gain with the wiper at ground is –1, and 10% away is –0.8.

Answer: When the gate is low, the output is 2.26 mV; when the gate is high, the output is 26.2 mV. 20-9.

When the wiper is 10% away from ground, so that the noninverting gain will be 10% of its maximum of 2.

Given: R1 = 20 kΩ Rf = 68 kΩ R3 = 1 kΩ vin = 1 mV

20-13. Given:

R = 5 kΩ nR = 75 kΩ nR/(n – 1)R = 5.36 kΩ Solution: AV = –nR/R AV = –75 kΩ/5 kΩ AV = –15

Solution:

Answer: The maximum positive gain is 15, and the maximum negative gain is –15.

Low gate: AV = (Rf/R1) + 1 AV = (68 kΩ/20 kΩ) + 1 AV = 4.4

20-14. Given:

R′ = 10 kΩ R = 22 kΩ C = 0.02 ¿F fin = 100 Hz, 1 kHz, 10 kHz

vout = Avvin vout = 4.41(1 mV) vout = 4.4 mV High gate: AV = [Rf/(R1||R3)] + 1 AV = [68 kΩ/(20 kΩ||1 kΩ)] + 1 AV = 72.4 vout = Avvin vout = 72.4(1 mV) vout = 72.4 mV Answer: When the gate is low, the output is 4.4 mV, and when the gate is high, the output is 72.4 mV. 20-10. Given:

Solution: fC = 1/(2π RC) fC = 1/[(2πRC)(0.02 ¿F)] fC = 362 Hz ф = –2 arctan (f/fC) ф = –2 arctan (100 Hz/362 Hz) ф = –30.9° ф = –2 arctan (f/fC) ф = –2 arctan (1 kHz/362 Hz) ф = –140°

R1 = 10 kΩ Rf = 10 kΩ Vin = 2.5 V

ф = –2 arctan (f/fC) ф = –2 arctan (10 kHz/362 Hz) ф = –176°

Solution:

Answer: The phase shift is –30.9° at 100 Hz, –140° at 1 kHz, and –176° at 10 kHz.

AV = (Rf/R1) + 1 AV = (10 kΩ/10 kΩ) + 1 AV = 2

20-15. Given:

R1 = 1.5 kΩ Rf = 30 kΩ

Vout = AV(vin) Vout = 2(2.5 V) Vout = 5 V

Solution:

Answer: The new output reference voltage is 5 V. 20-11. Given:

R1 = 1 kΩ Rf = 10 kΩ Solution: –R2/R1 < Av < 0 –10 kΩ/1 kΩ < Av < 0 –10 < Av < 0

Av(inv) = –Rf/R1 (Eq. 20-6) Av(inv) = –30 kΩ/1.5 kΩ Av(inv) = –20 Av(non) = [(R2/R1) + 1][R ′2 /(R 1′ + R ′2 )] (Eq. 20-7) Av(non) = [(30 kΩ/1.5 kΩ) + 1][30 kΩ/(1.5 kΩ + 30 kΩ)] Av(non) = 20 Av(CM) = ±4(0.1%) = ±4(0.001) = ±0.004 Answer: The differential voltage gain is –20, and the common mode gain is ±0.004.

1-100 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

20-16. Given:

20-20. Given: vin(CM) = 5 V

R1 = 1 kΩ Rf = 20 kΩ

Solution: Since the first stage has a common-mode gain of 1, both sides have the same voltage of 5 V. The guard voltage is 5 V.

Solution: Av(inv) = –Rf/R1 (Eq. 20-6) Av(inv) = –20 kΩ/1 kΩ Aiv(inv) = –20 Av(CM) = ±4 ΔR/R (Eq. 20-5) Av(CM) = ±4 (1%) = ±4(0.01) Av(CM) = ±0.04 Answer: The differential voltage gain is –20, and the common-mode gain is ±0.04. 20-17. Given:

Answer: The guard voltage is 5 V. 20-21. Given:

RG = 1008 Ω vin = 20 mV Solution: AV = (49.4 kΩ/RG) + 1 (Eq. 20-17) AV = (49.4 kΩ/1008 Ω) + 1 AV = 50 vout = Av(vin) vout = 50(20 mV) vout = 1 V

R1 = 10 kΩ R2 = 20 kΩ R3 = 20 kΩ R4 = 10 kΩ

Answer: The output voltage is 1 V. 20-22. Given:

Solution: V2 = [R2/(R1 + R2)]VCC V2 = [20 kΩ/(10 kΩ + 20 kΩ)]15 V V2 = 10 V V4 = [R4/(R3 + R4)]VCC V4 = [10 kΩ/(20 kΩ + 10 kΩ)]15 V V4 = 5 Answer: No, the bridge is not balanced. 20-18. Given:

R = 10 kΩ v1 = –50 mV v2 = –30 mV Solution: vout = v1 – v2 vout = (–50 mV) – (–30 mV) vout = –20 mV Answer: The output voltage is –20 mV. 20-23. Given:

R1 = 1 kΩ ΔR = 15 Ω AV = –100

vout = A1(vin) vout = (–100)(56.3 mV) vout = –5.63 V

R1 = 10 kΩ R2 = 20 kΩ R3 = 15 kΩ R4 = 15 kΩ R5 = 30 kΩ RF = 75 kΩ v1 = 1 mV v2 = 2 mV v3 = 3 mV v4 = 4 mV

Answer: The output voltage is –5.63 V.

Solution:

Solution: vin = (ΔR/4R)VCC vin = (15 Ω/4 (1 kΩ))15 V vin = 56.3 mV

Av(1) = –Rf/R1 Av(1) = –75 kΩ/10 kΩ Av(1) = –7.5

20-19. Given: R1 = 1 kΩ Rf = 99 kΩ R = 10 kΩ ± 0.5% vin = 2 mV

Av(2) = –Rf/R2 Av(2) = –75 kΩ/20 kΩ Av(2) = –3.75

Solution:

Av(3) = {[Rf/(R1||R2)] + 1}{(R4||R5)/[R3 + (R4||R5)]} Av(3) = {[75 kΩ/(10 kΩ||20 kΩ)] + 1}{(15 kΩ||30 kΩ)/[15 kΩ + (15 kΩ||30 kΩ)]} Av(3) = (12.25)(0.455) Av(3) = 5.57

AV = (Rf/R1) + 1 AV = (99 kΩ/1 kΩ) + 1 AV = 100 vout = Avvin vout = 100(2 mV) vout = 200 mV

Av(4) = {[Rf/(R1||R2)] + 1}{(R3||R5)/[R4 + (R3||R5)]} Av(4) = {[75 kΩ/(10 kΩ||20 kΩ)] + 1}{(15 kΩ||30 kΩ)/[15 kΩ + (15 kΩ||30 kΩ)]} Av(4) = (12.25)(0.4) Av(4) = 4.9

AV(CM) = ±2(ΔR/R) AV(CM) = ±2(0.005) AV(CM) = ±0.01 CMRR = |AV|/|AV(CM)| CMRR = 100/0.01 CMRR = 10,000 Answer: The output voltage is 200 mV, and the CMRR is 10,000.

vout = Av(1)v1 + Av(2)v2 + Av(3)v3 + Av(4)v4 vout = –7.5(1 mV) + –3.75(2 mV) + 4.9(3 mV) + 4.9 (4 mV) vout = 19.3 mV Answer: The output voltage is 19.3 mV.

1-101 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

re′ = 25 mV/0.685 mA re′ = 36.5 Ω

Solution: AV = (Rf/R1) + 1 AV = (15 kΩ/1.5 kΩ) + 1 AV = 11

rc = RC rc = 6.8 kΩ Av = rc/ re′ (Eq. 10-7) Av = 6.8 kΩ/36.5 Ω Av = 186

f1 = 1/[2π(R/2)C1] f1 = 1/[2π(68 kΩ/2)(1 µF)] f1 = 4.68 Hz

Op amp: AV = (Rf/R1) + 1 AV = (47 kΩ/1 kΩ) + 1 AV = 48

f2 = 1/(2πRLC2) f2 = 1/[2π(15 kΩ)(2.2 µF)] f2 = 4.82 Hz f3 = 1/(2πR1C3) f3 = 1/[2π(1 kΩ)(3.3 µF)] f3 = 32.2 Hz

AV = (48)(186) AV = 9114

Answer: The gain is 11, and the cutoff frequencies are f1 = 4.68 Hz, f2 = 4.82 Hz, and f3 = 32.2 Hz.

CRITICAL THINKING 20-40. Answer: Since the terminal is floating, the output would be saturated or VCC. To fix this problem, a large-value resistor could be connected to the noninverting terminal. This would keep it at ground potential during the transition and prevents a spike. 20-41. Given:

R1(min) = 990 Ω R1(max) = 1010 Ω Rf(min) = 99 kΩ Rf(max) = 101 kΩ

Answer: The voltage gain is 9114. 20-43. Given: R1 = 1 kΩ Rf = 10 kΩ RL = 100 Ω β = 50 vin = 0.5 VCC

Solution: AV = –Rf/R1 AV = –10 kΩ/1 kΩ AV = –10 vout = AV(vin) vout = –10(0.5 V) vout = –5 V Iout = vout/RL Iout = –5 V/100 Ω Iout = 50 mA

Solution: AV(min) = –Rf(min)/R1(max) AV(min) = –99 kΩ/1010 Ω AV(min) = 98

IB = Iout/β IB = 50 mA/50 IB = 1 mA

AV(max) = –Rf(max)/R1(min) AV(max) = –101 kΩ/990 Ω AV(max) = 102

Answer: The base current is 1 mA.

Answer: The minimum gain is 98, and the maximum gain is 102. 20-42. Given:

20-44. Answer:

Trouble 1: Since there is voltage at E and not at F, there is an open between E and F. Trouble 2: Since the output is only 200 mV, which is the amplified output of A, R2 is open.

Transistor: R1 = 22 kΩ Rf = 10 kΩ RS = 1 kΩ RE = 5.6 kΩ RC = 6.8 kΩ VCC = 15V

Trouble 3: Since the input is 2 mV and the output is maximum, R1 is shorted. 20-45. Answer:

Trouble 4: Since there is no voltage at B, there is an open between K and B.

Op amp R1 = 1 kΩ Rf = 47 kΩ

Trouble 5: Since the voltage at C is 3 mV and the voltage at D is zero, there is an open between C and D.

Solution: VBB = [Rf/(R1 + Rf + RS)]VCC (Eq. 8-1) VBB = [10 kΩ/(22 kΩ + 10 kΩ + 1 kΩ)]15 V VBB = 4.54 V (Eq. 8-2) VE = VBB – VBE VE = 4.54 V – 0.7 V VE = 3.84 V IE = VE/RE (Eq. 8-3) IE = 3.84 V/5.6 kΩ IE = 0.685 mA re′ = 25 mV/IE (Eq. 9-10)

Trouble 6: Since the voltage at A is zero, there is an open between J and A. 20-46. Answer:

Trouble 7: Since the input voltage is 3 mV and the output is maximum, R3 is open. Trouble 8: Since the output is only 250 mV, which is the amplified output of B, R1 is open. Trouble 9: Since the output voltage is the same as the input voltage, R3 is shorted. Trouble 10: Since the input is 5 mV and the output is maximum, R2 is shorted.

1-104 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

f0 = 55.7 kHz

21-20. Given:

R1 = 56 kΩ Rf = 10 kΩ C = 680 pF

Answer: The Q is 2.65, the voltage gain is –14, and the center frequency is 55.7 kHz. 21-23. Given:

Solution: f p = 1/(2πC R1/R f ) f p = 1/[2π(680 pF) (10 kΩ)(56 kΩ)] fp = 9.89 kHz

R1 = 3.6 kΩ Rf = 7.5 kΩ R3 = 27 Ω C = 22 nF

Q = 0.5 ( R1 / R f )

Solution:

Q = 0.5 (56 kΩ)/(10 kΩ) Q = 1.18

AV = –Rf/2R1 (Eq. 21-35) AV = –7.5 kΩ/2(3.6 kΩ) AV = –1.04

Kc = 1.04 (from Fig. 21-26) K3 = 1.30 (from Fig. 21-26)

Q = 0.5 [ R f /( R1 || R3 )]

(Eq. 21-40)

Q = 0.5 [7.5kΩ /(2 kΩ || 27 Ω)] Q = 8.39

fc = fp/Kc (Eq. 21-31) fc = 9.89 kHz/1.04 fc = 9.51 kHz

f 0 = 1/(2πC ( R1 || R3 ) R f ]

(Eq. 21-41)

f 0 = 1/[2π(22 nF) (2kΩ || 27 Ω)(7.5kΩ)] f0 = 16.2 kHz

f3 = fp/K3 f3 = 9.89 kHz/1.30 f3 = 7.61 kHz Answer: The pole frequency is 9.89 kHz, the cutoff frequency is 9.51 kHz, the 3-dB frequency is 7.61 kHz, and the Q is 1.18. 21-21. Given:

R1 = 91 kΩ Rf = 15 kΩ C = 220 pF

Answer: The Q is 8.39, the voltage gain is –1.04, and the center frequency is 16.2 kHz. 21-24. Given:

R1 = 28 kΩ R3 = 1.8 kΩ C = 1.8 nF AV = –1 Solution:

Solution:

Q = 0.707 [( R1 + R3 ) / R3 ]

f p = 1/(2πC R1/R f ) f p = 1/[2π(220 pF) (15 kΩ)(91 kΩ)] fp = 19.6 kHz

Q = 0.5 ( R1/R f )

(Eq. 21-43)

Q = 0.707 [(28kΩ+1.8kΩ) /1.8kΩ ] Q = 2.88 (Eq. 21-44) f 0 = 1/(2πC [2 R1 ( R1 || R3 )]) f 0 = 1/{2π(1.8 nF) [2(28 kΩ)(28 kΩ ||1.8kΩ)]} f0 = 9.09 kHz

Q = 0.5 (91kΩ)/(15kΩ) Q = 1.23

Answer: The Q is 2.88, the voltage gain is –1, and the center frequency is 9.09 kHz.

Kc = 1.06 (from Fig. 21-26) K3 = 1.32 (from Fig. 21-26)

21-25. Given:

fc = fp/Kc (Eq. 21-31) fc = 19.6 kHz/1.06 fc = 18.5 kHz

R1 = 20 kΩ Rf = 10 kΩ R = 56 kΩ C = 180 nF

f3 = fp/K3 f3 = 19.6 kHz/1.32 f3 = 14.8 kHz

Solution:

Answer: The pole frequency is 19.6 kHz, the cutoff frequency is 18.5 kHz, the 3-dB frequency is 14.8 kHz, and the Q is 1.23. 21-22. Given:

R1 = 2 kΩ Rf = 56 kΩ C = 270 pF

AV = (Rf/R1) + 1 (Eq. 21-46) AV = (10 kΩ/20 kΩ) + 1 AV = 1.5 f0 = 1/(2πRC) (Eq. 21-47) f0 = 1/[2π(56 kΩ)(180 nF)] f0 = 15.8 Hz Q = 0.5/(2 – AV) Q = 0.5/(2 – 1.5) Q=1

Solution:

(Eq. 21-48)

AV = –Rf/2R1 (Eq. 21-35) AV = –56 kΩ/2(2 kΩ) AV = –14

BW = f0/Q (Eq. 21-34) BW = 15.8 Hz/1 BW = 15.8 Hz

Q = 0.5 ( R f /R1 )

Answer: The voltage gain is 1.5, the Q is 1, the resonant frequency is 15.8 Hz, and the bandwidth is 15.8 Hz.

(Eq. 21-36)

Q = 0.5 (56 kΩ) /(2 kΩ) Q = 2.65 f 0 = 1/(2πC ( R1 / R f )

21-26. Given:

(Eq. 21-38)

f 0 = 1/[2π ( 270 pF ) (56 kΩ)(2 kΩ)]

R = 3.3 kΩ C = 220 nF

1-108 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Roll-off = 20n dB/decade (Eq. 21-4a) Roll-off = 20(10) dB/decade Roll-off = 200 dB/decade 4 kHz is 1 octave above Attenuation = 60 dB 8 kHz is 2 octaves above Attenuation = 120 dB 20 kHz is 1 decade above Attenuation = 200 dB Answer: The attenuation is 60 dB at 4 kHz, 120 dB at 8 kHz, and 200 dB at 20 kHz. 21-34. Given: n=2 R = 10 kΩ fc = 5 kHz Solution: Q = 0.5 C2 / C1 (from Fig. 21-24) (Butterworth response) 0.707 = 0.5 C2 / C1 1.414 = C2 / C1 2 = C2/C1 C2 = 2C1 fp = 1/2πR C1 (2C1 ) fp = 1/2π RC1 2 C1 = 1/2πRfp 2 C1 = 1/2π(10 kΩ)(5 kHz) 2 C1 = 2.25 nF C2 = 4.5 nF 21-35. Given: n=2 R = 25 kΩ fc = 7.5 kHz Ap = 12 dB

JOB INTERVIEW QUESTIONS 5. It means using back-to-back zener diodes or other circuits to limit the output voltage swing. 8. An IC comparator does not have an internal compensating capacitor.

PROBLEMS 22-1.

Solution: AVOL = antilog(AV(dB)/20) AVOL = antilog(l06 dB/20) AVOL = 200,000 vin(min) = ±VSat/AVOL vin(min) = ±20 V/200,000 vin(min) = ±100 µV

22-2.

Given: vin = 50 V R = 10 kΩ Answer: ID = (vin – 0.7 V)/R ID = (50 V –0.7 V)/10 kΩ ID = 4.93 mA

22-3.

Given: VZ = 6.8 V VS = ±15 V Solution: Vout = ±(VZ + VD) (Eq. 22-1) Vout = ±(6.8 V + 0.7 V) Vout = ±7.5 V Answer: The output voltage will be limited to ±7.5 V.

22-4.

(Chebyshev response)

Given: VS = ±12 V Answer: The output voltage would vary between 0.7 V and –12 V.

8 = C2 / C1 64 = C2/C1 C2 = 64C1 fp = 1/2πR C1 (64C1 ) fp = 1/16πRC1 C1 = 1/16πRfp C1 = 1/16π(25 kΩ)(5.39 kHz) C1 = 148 pF C2 = 9.45 nF

22-5.

Given: VS = ±12 V Answer: When the strobe is high, the output is zero. When the strobe is low, the output will vary between 0.7 V and –9 V.

22-6.

Solution:

SELF-TEST 17. a 18. c 19. b 20. c 21. d 22. d 23. a 24. b

Given: VS = ±15 V R1 = 47 kΩ R2 = 12 kΩ C = 0.5 µF

Chapter 22 Nonlinear Op-Amp Circuits 9. c 10. b 11. c 12. b 13. b 14. b 15. a 16. b

(Eq. 22-1)

Solution: The diode current is 4.93 mA.

fc = Kcfp (Eq. 21-23) fp = fc/Kc fp = 7.5 kHz/1.391 fp = 5.39 kHz Q = 0.5 C2 / C1 (from Fig. 21-25)

1. d 2. a 3. c 4. b 5. c 6. a 7. a 8. b

(Eq. 16-15)

Answer: An input voltage of 100 µV will produce positive saturation, assuming rail-to-rail output.

Solution: Since Ap = 12 dB, Kc = 1.391 and Q = 4 (from Table 21-3)

4 = 0.5 C2 / C1

Given: AV(dB) = 106 dB VSat = ±20 V

25. a 26. a 27. d 28. b 29. c 30. a

vref = [R2/(R1 + R2)]VCC (Eq. 22-2) vref = [12 kΩ/(47 kΩ + 12 kΩ)]15 V vref = 3.05 V (Eq. 22-3) fC = 1/[2π(R1||R2)C] fC = 1/[2π(47 kΩ||12 kΩ) 0.5 µF] fC = 33.3 Hz Answer: The reference voltage is 3.05 V, and the cutoff frequency is 33.3 Hz.

1-110 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

TR ≈ 2.2(RC) (Eq. 16-28) TR ≈ 2.2(1 kΩ)(50 pF) TR ≈ 110 ns

f = 1/T f = 122.4 ms f = 45 Hz Answer: The lowest frequency is 45 Hz. 22-33. Answer: With the diode reversed, it becomes a negative peak detector and the output voltage is –106 mV. 22-34. Given: vin = 150 mV peak

Answer: The output voltage is 150 mV. 22-35. Given: vin = 100 mV peak

Answer: The risetime is 110 ns. 22-41. Given:

R1 = 33 kΩ Rf = 3.3 kΩ C = 47 µF Vripple = 1 V rms Solution:

Solution: vout = vin + V peak (Eq. 22-21) vout = 100 mV peak + 100 mV peak vout = 200 mV peak Answer: The output swings from 0 V to 200 mV peak. 22-36. Given:

RL = 10 kΩ C = 4.7 µF

fC = 1/[2π(R1||Rf)C] (from Fig. 22-11) fC = 1/[2π(33 kΩ||3.3 kΩ)47 µF] fC = 1.1 Hz The power supply ripple is 120 Hz because rectification. This is 2 decades above the cutoff frequency. Since there is one capacitor, the roll-off is 20 dB/decade. The input is attenuated by 40 dB, equivalent to 0.01. vtb =

Solution: RLC > 10T (Eq. 22-20) RLC = 10T for the highest period or lowest frequency [(10 kΩ)(4.7 µF)]/10 = T T = 4.7 ms

3.3 kΩ (1V) = 0.1V 33 kΩ + 3.3 kΩ

vout = (0.01)(0.1 V) = 0.001 V

Answer: The cutoff frequency is 1:1 Hz, and the ripple voltage at the inverting input is 0.001 V rms. 22-42. Given:

f = 1/T f = 1/4.7 ms f = 213 Hz Answer: The lowest frequency is 213 Hz.

VCC = 15 V R1 = 33 kΩ Rf = 3.3 kΩ vin(peak) = 5 V vref = 1.36 V (from Prob. 22-9)

22-37. Given: f = 10 kHz

Solution:

¸ = 16° and 164° (from Prob. 22-9) D = 41% (from Prob. 22-9)

1 Hz = 1 cycle/second 10 kHz = 10,000 cycles/second or 10,000 cycles in 1 second Each cycle has two transitions (low to high and high to low); thus there are 2 pulses per cycle. 10,000 cycles in 1 second × 2 pulses/cycle = 20,000 pulses in 1 second. Answer: There are 20,000 pulses in 1 second. 22-38. Given: f = 1 kHz

Solution: Since there are 2 pulses per cycle, a pulse occurs every T/2. T = 1/f T = 1/1 kHz T = 1 mS

Solution: Ihigh = V/R Ihigh = 5 V/1 kΩ Ihigh = 5 mA Since the output is high only 41% of the time, the average current is: Iave = Dihigh Iave = (0.41)(5 mA) Iave = 2.05 mA Answer: The average current is 2.05 mA. 22-43. Given:

R1 = 1.5 kΩ ± 5% Rf = 68 kΩ ± 5% VSat = 13.5 V

Answer: A pulse occurs every T/2 or 0.5 mS.

CRITICAL THINKING 22-39. Answer: Make the 3.3-kΩ resistor a variable so that it can be adjusted to any desired value. 22-40. Given:

R = 1 kΩ C = 50 pF Solution: Risetime is from the 10% point to the 90% point (discussed in Chap. 16). Using the universal time constant chart, it takes about 3 time constants to go from 10% to 90%.

Solution: R1(max) = 1.5 kΩ + 5%(1.5 kΩ) = 1575 Ω R1(min) = 1.5 kΩ – 5%(1.5 kΩ) = 1425 Ω Rf(max) = 68 kΩ + 5%(68 kΩ) = 71.4 kΩ Rf(min) = 68 kΩ – 5%(68 kΩ) = 64.6 kΩ B(min) = R1(min)/(R1(min) + Rf(max)) B(min) = 1425 Ω/(1425 kΩ + 71.4 kΩ) B(min) = 0.0196

(Eq. 22-4)

H(min) = 2B(min) VSat (Eq. 22-9) H(min) = 2(0.0196)(13.5 V) H(min) = 0.529 V Answer: The minimum hysteresis is 0.529 V.

1-114 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

8. There must be an unwanted positive feedback path between the output and the input of the three-stage amplifier. Lowfrequency oscillations may be caused by the high powersupply impedance. You can try using a large filter capacitor at the supply point for each stage. If this docs not work, then a power supply with better regulation is needed. For high-frequency oscillations, you can try shielding the stages, using a single ground point, filter capacitors on each stage supply, and ferrite beads on each base or gate lead.

Answer: The maximum frequency is 3.98 kHz, and the minimum frequency is 332 Hz. 23-3c. Given:

C = 0.002 µF Rmin = 2 kΩ Rmax = 24 kΩ Solution: fr(max)= 1/[2πRminC] (Eq. 23-4) fr(max) = 1/[2π(2 kΩ)(0.002 µF)] fr(max) = 39.8 kHz

PROBLEMS 23-1.

fr(min) = 1/[2πRmaxC] (Eq. 23-4) fr(min) = 1/[2π(24 kΩ)(0.002 µF)] fr(min) = 3.32 kHz

Given: RF = 1 kΩ Solution: The oscillator becomes stable with a lamp resistance of 500 Ω and from the graph a lamp voltage of 3 V rms. IL = 3 V/500 Ω IL = 6 mA

Answer: The maximum frequency is 39.8 kHz, and the minimum frequency is 3.32 kHz. 23-3d. Given:

Vout = IL(RF + RL) Vout = 6 mA(1 kΩ + 500 Ω) Vout = 9 V rms 23-2.

C = 200 pF Rmin = 2 kΩ Rmax = 24 kΩ

Answer: The output voltage is 9 V rms.

Solution:

Given:

fr(max) = 1/[2πRminC] (Eq. 23-4) fr(max) = 1/[2π(2 kΩ)(200 pF)] fr(max) = 398 kHz

C = 200 pF Rmin = 2 kΩ Rmax = 24 kΩ

fr(min) = 1/[2πRmaxC] (Eq. 23-4) fr(min) = 1/[2π(24 kΩ)(200 pF)] fr(min) = 33.2 kHz

Solution: fr(max) = 1/[2πRminC] (Eq. 23-4) fr(max) = 1/[2π(2.2 kΩ)(200 pF)] fr(max) = 398 kHz fr(min) = 1/[2πRmaxC] (Eq. 23-4) fr(min) = 1/[2π(2.4 kΩ)(200 pF)] fr(min) = 33.2 kHz

Answer: The maximum frequency is 398 kHz, and the minimum frequency is 33.2 kHz. 23-4.

Given: Vout = 6 V rms RF = 2Rlamp Solution: Since the lamp resistance is one-third of the total resistance, its voltage will be one-third of the total voltage, or 2 V rms. According to the graph, the lamp resistance would be 350 Ω, so the feedback resistor would need to be twice that, or 700 Ω.

Answer: The maximum frequency is 398 kHz, and the minimum frequency is 33.2 kHz. 23-3a. Given:

C = 0.2 µF Rmin = 2 kΩ Rmax = 24 kΩ

Answer: Change the feedback resistor to 700 Ω. 23-5.

Solution: fr(max) = 1/[2πRminC] (Eq. 23-4) fr(max) = 1/[2π(2 kΩ)(0.2 µF)] fr(max) = 398 Hz fr(min) = 1/[2πRmaxC] (Eq. 23-4) fr(min) = 1/[2π(24 kΩ)(0.2 µF)] fr(min) = 33.2 Hz

Solution: 1 decade above 398 kHz is 3.98 MHz. Answer: The cutoff frequency is 3.98 MHz. 23-6.

Given: R = 10 kΩ C = 0.01 µF

Answer: The maximum frequency is 398 Hz, and the minimum frequency is 33.2 Hz.

Solution: fr = 1/[2πRC] (Eq. 23-4) fr = 1/[2π(10 kΩ)(0.01 µF)] fr = 1.59 kHz

23-3b. Given:

C = 0.02 µF Rmin = 2 kΩ Rmax = 24 kΩ

Given: Maximum frequency is 398 kHz, from Prob. 23-3.

Answer: The resonant frequency is 1.59 kHz. 23-7.

Solution: fr(max) = 1/[2πRminC] (Eq. 23-4) fr(max) = 1/[2π(2 kΩ)(0.02 µF)] fr(max) = 3.98 kHz fr(min) = 1/[2πRmaxC] (Eq. 23-4) fr(min) = 1/[2π(24 kΩ)(0.02 µF)] fr(min) = 332 Hz

Given: R = 20 kΩ C = 0.02 µF Solution: fr = 1/[2πRC] (Eq. 23-4) fr = 1/[2π(20 kΩ)(0.02 µF)] fr = 398 Hz

1-116 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Answer: The resonant frequency is 398 Hz. 23-8.

Given:

f r = 1/(2π LC ) (Eq. 23-5) f r =1/(2π (20 µH)(909 pF)) fr = 1.18 MHz

R1 = 10 kΩ R2 = 5 kΩ RE = 1 kΩ VBE = 0.7 V VCC = 12 V

Answer: The frequency is 1.18 MHz. 23-12. Answer: Reduce the inductance by a factor of 4 (since there is a square root in the denominator).

Solution: VB = [R2/(R1 + R2)]VCC (Eq. 8-1) VB = [5 kΩ/(10 kΩ + 5 kΩ)]12 V VB = 4 V VE = VB – VBE VE = 4 V – 0.7 V VE = 3.3 V IE = VE/RE IE = 3.3 V/1 kΩ IE = 3.3 mA

(Eq. 8-2)

VCE = VC – VE VCE = 12 V – 3.3 V VCE = 8.7 V

(Eq. 8-6)

Answer: The emitter current is 3.3 mA, and the collector-to-emitter voltage is 8.7 V. Given:

C1 = 0.001 µF C2 = 0.01 µF C3 = 47 pF L = 10 µH f r = 1/(2π LC3 ) (Eq. 23-18) f r = 1/(2π (10 µH)(47 pF)) fr = 7.34 MHz

Answer: The frequency is 7.34 MHz. 23-14. Given:

L1 = 1 µH L2 = 0.2 µH C = 1000 pF Solution: B = L2L1 (Eq. 23-16) B = 0.2 µH/1 µH B = 0.2

C1 = 0.001 µF C2 = 0.01 µF L = 10 µH Solution: C = C1C2/(C1 + C2) (Eq. 23-6) C = (0.001 µF)(0.01 µF)/(0.001 µF + 0.01 µF) C = 909 pF

( (

23-13. Given:

Solution:

(Eq. 8-3)

Since the RF choke is a short to direct current, the collector voltage is 12 V.

23-9.

C = (0.001 µF)(0.01 µF)/(0.001 µF + 0.01 µF) C = 909 pF

)

f r = 1/ 2π LC (Eq. 23-5) f r = 1/ 2π (10 µH)(909 pF) fr = 1.67 MHz

)

B = C1/C2 (Eq. 23-7) B = 0.001 µF/0.01 µF B = 0.10 Av(min) = C2/C1 (Eq. 23-8) Av(min) = 0.001 µF/0.01 µF Av(min) = 10 Answer: The frequency is 1.67 MHz, the feedback fraction is 0.10, and the minimum gain is 10. 23-10. Given:

C1 = 0.001 µF C2 = 0.01 µF Solution: B = C1/(C1 + C2) B = 0.001 µF/(0.001 µF + 0.01 µF) B = 0.091 Answer: The feedback fraction is 0.091.

f r = 1/[2π LC ] (Eq. 23-5) f r = 1/[2π (1.2 µH)(1000 pF)] fr = 4.59 MHz Av(min) = L1/L2 Av(min) = 1 µH/0.2 µH Av(min) = 5 Answer: The frequency is 4.59 MHz, the feedback fraction is 0.2, and the minimum gain is 5. 23-15. Given:

M = 0.1 µH L = 3.3 µH Solution: B = M/L (Eq. 23-14) B = 0.1 µH/3.3 µH B = 0.030 Av(min) = L/M Av(min) = 3.3 µH/0.1 µH Av(min) = 33 Answer: The feedback fraction is 0.03, and the minimum gain is 33. 23-16. Given: f = 5 MHz

Answer: The first overtone is 10 MHz, the second overtone is 15 MHz, and the third overtone is 20 MHz.

23-11. Given:

C1 = 0.001 µF C2 = 0.01 µF L = 20 µH

23-17. Answer: Since the frequency is inversely proportional to thickness, if thickness is reduced by 1% the frequency will increase by 1%.

Solution: C = C1C2/(C1 + C2)

L = L1 + L2 L = 1 µH + 0.2 µH L = 1.2 µH

(Eq. 23-6)

1-117 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

R3 = 40 kΩ C = 0.1 µF

Answer: The period is 100 µs, the quiescent pulse width is 5.61 µs, the maximum pulse width is 8.66 µs, the minimum pulse width is 3.71 µs, the maximum duty cycle is 0.0866, and the minimum duty cycle is 0.0371.

Solution: The waveform is a sine wave. f = 1.1RC f = 1/(20 kΩ)(0.1 µF) f = 500 Hz

23-24. Given:

VCC = 10 V R1 = 1.2 kΩ R2 = 1.5 kΩ C = 4.7 nF vmod = 1.5 V

Amplitude = 2.4 Vp Amplitude = 4.8 Vpp Answer: The output is a sine wave at a frequency of 500 Hz and a peak voltage of 2.4 V.

Solution: W = 0.693(R1 + R2)C (Eq. 23-26) W = 0.693(1.2 kΩ + 1.5 kΩ)(4.7 nF) W = 8.79 µs

23-27. Given:

S1 = Open R = 10 kΩ R3 = 40 kΩ C = 0.01 µF

T = 0.693(R1 + 2R2)C (Eq. 23-27) T = 0.693[1.2 kΩ + 2(1.5 kΩ)](4.7 nF) T = 13.68 µs

Solution:

UTPmax = 2VCC/3 + vmod (Eq. 23-34) UTPmax = 2(10 V)/3 + 1.5 V UTPmax = 8.17 V UTPmin = 2VCC/3 – vmod UTPmin = 2(10 V)/3 – 1.5 V UTPmin = 5.17 V

The waveform is a triangle wave. f = 1.1RC f = 1/(10 kΩ)(0.01 µF) f = 10 kHz

(Eq. 23-34)

Wmax = – (R1 + R2)C1n[(VCC – UTPmax)/(VCC – 0.5UTPmax)] (Eq. 23-35) Wmax = – {[(1.2 kΩ + 1.5 kΩ)(4.7 nF)]ln[(10 – 8.17 V)/ (10 V – 0.5(8.17 V)]} Wmax = 14.89 µs

Amplitude = 5 Vp Amplitude = 10 Vpp Answer: The output is a triangle wave at a frequency of 10 kHz and a peak voltage of 5 V. 23-28. Given:

R1 = 2 kΩ R2 = 10 kΩ C = 0.01 µF

Wmin = – (R1 + R2)C1n[(VCC – UTPmin)/(VCC – 0.5UTPmin)] (Eq. 23-35) Wmin = – {[(1.2 kΩ + 1.5 kΩ)(4.7 nF)]ln[(10 – 5.17 V)/ (10 V – 0.5(5.17 V)]} Wmin = 5.44 µs

Solution: f =

Space = 0.693R2C Space = 0.693(1.5 kΩ)(4.7 nF) Space = 4.89 µs

2 » 1 ¿ ¼ À 0.1 µF ½ 2 kΩ + 10 kΩ Á f = 1.67 kHz f =

Answer: The quiescent pulse width is 8.79 µs, the quiescent period is 13.69 µs, the maximum pulse width is 14.89 µs, the minimum pulse width is 5.44 µs, and the space between pulses is 4.89 µs.

D = R1/(R1 + R2) D = 2 kΩ/(2 kΩ + 10 kΩ) D = 0.167

23-25. Given:

IC = 0.5 mA VCC = 10 V C = 47 nF

Answer: The frequency is 1.67 kHz, and the duty cycle is 0.167. Decrease. With the lamp open, there is no path for feedback current. Thus the voltage at the inverting terminal will equal the output voltage and it should be driven to 0 V. Increase. With the inverting input grounded, there is no feedback and the gain is open-loop gain and the output will be saturation. Same. The upper potentiometer affects frequency, not output voltage. Same. Unless the supply falls low enough for clipping. Same. Only a very small change at the output.

23-29a.

Solution: S = IC/C (Eq. 23-39) S = 0.5 mA/47 nF S = 10.6 V/mS V = 2VCC/3 V = 2(10 V)/3 V = 6.67 V

2» 1 ¿ ¼ À C ½ R1 + R2 Á

23-29b.

(Eq. 23-40) 23-29c.

T = 2VCC/3S (Eq. 23-41) T = 2(10 V)/3(10.6 V/mS) T = 0.629 mS

23-29d.

Answer: The slope is 10.6 V/mS, the peak value is 6.67 V, and the duration is 0.629 mS.

23-30. Answer:

23-26. Given:

S1 = Closed R = 20 kΩ

23-29e. 1. 2. 3. 4. 5. 6.

Shorted inductor Open inductor Shorted capacitors Open capacitors Open in the feedback path Loss of the power supply

1-119 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

TABLE 2-5

TABLE 4-2 DMM TESTING RTH = 600 Ω Forward Reverse

ANSWERS 1. 2. 3. 4. 5. 6.

b c d d a The calculated Thevenin resistance is 2.39 kΩ. A load of 100 kΩ implies the load voltage will be down approximately 2 percent from the ideal open-load or Thevenin voltage. This means the voltmeter reads slightly less than the ideal Thevenin voltage. 7. In Fig. 2-1a, a shorted 2.2-kΩ resistor means the voltage between point A and common is lower than it should be, which implies a Thevenin voltage that is less than before. Also, the shorted resistor means less Thevenin resistance. 8. In Fig. 2-1a, an open 2.2-kΩ resistor implies that all of the source voltage will appear across the AB terminals when the load is open. Furthermore, opening the resistor will increase the Thevenin resistance. 9. If I wanted to stay in business, I had better produce batteries with very low internal resistance because batteries are supposed to act like stiff voltage sources for most load resistances.

Measured 1

Measured 2

Measured 3

0.552 V OL

0.571 V OL

0.544 V OL

TABLE 4-3 DATA FOR FORWARD BIAS Calculated

Diode 1 Diode 2 Diode 3

Measured

VD

VL

VD

VL

0.7 V 0.7 V 0.7 V

9.3 V 9.3 V 9.3 V

0.68 V 0.72 V 0.67 V

9.32 V 9.28 V 9.33 V

TABLE 4-4 DATA FOR REVERSE BIAS Calculated

Diode 1 Diode 2 Diode 3

Measured

VD

VL

VD

VL

10 V 10 V 10 V

0 0 0

10 V 10 V 10 V

0 0 0

TABLE 4-5 DIODE CONDUCTION

Experiment 3 TABLE 3-1 TROUBLES AND VOLTAGES Trouble

VA

Circuit OK R1 open R2 open R3 open R4 open R1 shorted R2 shorted R3 shorted R4 shorted

VB

5.21 V 0 6.9 V 6.81 V 6.81 V 10 V 0 2.72 V 4.92 V

1.06 V 0 1.41 V 0 6.81 V 2.04 V 0 2.72 V 0

ANSWERS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

a c d d d a c a a d

D2

D3

D4

On Off

Off On

Off On

On Off

TABLE 4-6. DIODE CONDUCTION

Normal Reversed

D1

D2

D3

D4

On Off

On Off

Off On

Off On

TABLE 4-7 DIODE AND LOAD VOLTAGES VD1

VD2

VD3

VD4

VL

Calculated Measured

0.7 V 0.68 V

0.7 V 0.72 V

9.3 V 9.32 V

9.3 V 9.28 V

ANSWERS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Experiment 4 TABLE 4-1 OHMMETER TESTING

RF RR

Normal Reversed

D1

Expected

Measured 1

Measured 2

Measured 3

Low High

25 Ω Infinite

24 Ω Infinite

26 Ω Infinite

a b d c b b c b a c

2-2 Downloaded by Kevin Aguilar Arias ([email protected])

8.6 V 8.6 V

lOMoARcPSD|6455127

TABLE 7-5 CRITICAL THINKING

Experiment 7 TABLE 7-1 HALF-WAVE RECTIFIER

RMS secondary voltage Peak output voltage DC output voltage Ripple frequency

Calculated

Measured

12.6 V 17.8 V 5.67 V 60 Hz

15.6 V 20 V 6.6 V 60 Hz

RMS secondary voltage Peak output voltage DC load voltage DC load current Ripple frequency Load resistance

Calculated

Measured

12.6 V 8.9 V 5.67 V 120 Hz

15.6 V 9.5 V 6.3 V 120 Hz

Calculated

Measured

12.6 V 17.8 V 11.3 V 120 Hz

15.6 V 19 V 12.5 V 120 Hz

TABLE 7-4 TROUBLESHOOTING Calculated Vdc Diode open Half-secondary short

5.67 V 5.67 V

7.73 V 8.8 V 5.53 V 20 mA 120 Hz 267 Ω

Measured fout

60 Hz 120 Hz

Vdc 6.25 V 5.68 V

fout 60 Hz 120 Hz

TABLE 8-3 RL = 1 kΩ AND C = 470 µF

Experiment 8 TABLE 8-1 TRANSFORMER RESISTANCES

RMS secondary voltage Peak output voltage DC output voltage DC load current Ripple frequency Peak-to-peak ripple

Rpri = 33.6 Ω Rsec = 1.1 Ω

TABLE 8-2 RL = 1 kΩ AND C = 47 µF

RMS secondary voltage Peak output voltage DC output voltage DC load current Ripple frequency Peak-to-peak ripple

6.3 V 8.9 V 5.67 V 21 mA 120 Hz 270 Ω

1. d 2. c 3. c 4. c 5. d 6. For a given transformer, the bridge rectifier has the largest unfiltered dc output voltage, ideally 90 percent of the rms secondary voltage versus 45 percent for the others. 7. When any diode opens, the circuit reverts to a half-wave rectifier. For this reason, the unfiltered dc output voltage and the ripple frequency are half of their normal values. 8. If any diode is shorted, the other diode on the same side of the bridge is destroyed and the remaining diodes are unaffected. For instance, if D1 shorts in Fig. 7-2, D3 is destroyed because of excessive current. Diodes D2 and D4 are unaffected. 9. I used only half the secondary winding to drive the bridge rectifier. This reduced the dc output voltage to approximately 5.67 V. To get approximately 20 mA of dc load current, I selected a load resistance of 270 Ω.

TABLE 7-3 BRIDGE RECTIFIER

RMS secondary voltage Peak output voltage DC output voltage Ripple frequency

Measured

ANSWERS

TABLE 7-2 FULL-WAVE RECTIFIER

RMS secondary voltage Peak output voltage DC output voltage Ripple frequency

Calculated

Calculated

Measured

12.6 V 17.8 V 17.8 V 17.8 mA 120 Hz 3.16 V

15.4 V 21.5 V 19.1 V 19 mA 120 Hz 2.5 V

Calculated

Measured

12.6 V 17.8 V 17.8 V 17.8 mA 120 Hz 0.316 V

15.5 V 20 V 19.9 V 20 mA 120 Hz 0.28 V

TABLE 8-4 RL = 10 kΩ AND C = 470 µF RMS secondary voltage Peak output voltage DC output voltage DC load current Ripple frequency Peak-to-peak ripple

2-4 Downloaded by Kevin Aguilar Arias ([email protected])

Calculated

Measured

12.6 V 17.8 V 17.8 V 1.78 mA 120 Hz 31.6 mV

15.6 V 20.4 V 20.4 V 1.99 mA 120 Hz 25 mV

lOMoARcPSD|6455127

TABLE 14-3 OPTOCOUPLER

TABLE 15-2 CALCULATIONS

VS

Vout

Transistor

VCB

IE

αdc

βdc

2V 4V 6V 8V 10 V 12 V 14 V

13.8 V 13 V 11.9 V 10.7 V 9.6 V 9V 8.7 V

1 2 3

13.7 V 14.2 V 13.5 V

3.8 mA 1.9 mA 6.5 mA

0.996 0.992 0.998

253 127 433

TABLE 15-3 TROUBLESHOOTING Trouble

TABLE 14-4 TROUBLESHOOTING Estimated VLED Open LED Shorted LED

Measured VLED

15V 0V

15 V 0V

Estimated VC

Measured VC

+ 15 V + 15 V 0 0 + 15V

+ 15V + 15V + 0.04 V 0 + 15 V

Open 470 kΩ Shorted 1 kΩ Open 1 kΩ Shorted collector-emitter Open collector-emitter

TABLE 15-4 CRITICAL THINKING TABLE 14-5 CRITICAL THINKING Calculated R Red LED Green LED

680 Ω 680 Ω

ILED

Measured VLED

19.1 mA 2 V 19.1 mA 2 V

ILED

VLED

20.1 mA 1.7 V 19.6 mA 2 V

ANSWERS 1. 2. 3. 4. 5. 6.

c c c a a When the cathode of a LED is grounded, there is LED current and the LED segment lights up. By grounding one or more cathodes, we can display any digit between 0 and 9. With a LED voltage drop of 2 V, the current through the series resistor is (5 – 2)/270, or 11.1 mA. The LED current in any lit segment equals 11.1 mA/n, where n is the number of lit segments. Therefore, LED brightness decreases as more segments light up. 7. Kirchhoff’s voltage law says the voltage across the LED equals the source voltage minus the drop across the series resistor. When the LED is open, there is no drop across the series resistor and the entire source voltage appears across the open LED. 8. I subtracted 2 V (LED drop) from 15 V (source voltage) to get 13 V (voltage across the series resistor). Then I calculated R = 13 V/20 mA, or 650 Ω. The nearest standard sizes are 620 and 680 Ω. Arbitrarily, I selected 680 Ω. 9. By using nine identical resistors, one in series with each cathode. Then the positive supply voltage is connected directly to pin 3. By grounding the lower end of any resistor, we can set up LED current and tight the associated segment.

TABLE 15-1 TRANSISTOR VOLTAGES AND CURRENTS

1 2 3

RB

VC

360 kΩ 370 kΩ

7.45 V 7.76 V

ANSWERS 1. 2. 3. 4. 5. 6.

7.

8. 9. 10.

c b d c a The VBE drop was close to 0.7 V for all three transistors. It was also apparent that collector current is much greater than base current. My weakest transistor had a βdc of 197, which means collector current was approximately 200 times greater than base current. I measured the full supply voltage, + 15 V. With the base resistor open, the transistor goes into cutoff and operates at the lower end of the dc load line, which has a collector voltage of + 15 V. Stated another way, there was no collector current, which means no voltage drop across the collector resistor. Therefore, the full supply voltage appeared at the collector. A shorted transistor has either zero ohms or a very low value of resistance because of internal damage. Because of this, the voltage across a very low resistance approaches zero. Answers will vary. Because the base current and the beta current set the current.

Experiment 16 TABLE 16-1 FIRST CIRCUIT

Calculated Measured

Experiment 15

Transistor

Calculated Measured

VBE

VCE

Region

0.7 V 0.72 V

6.04 V 6.97 V

Active Active

TABLE 16-2 SECOND CIRCUIT

VBE

VCE

IB

IC

0.69 V 0.65 V 0.75 V

14.4 V 14.8 V 14.2 V

31.8 µA 31.8 µA 31.8 µA

6.4 mA 4.8 mA 3.2 mA

Calculated Measured

VBE

VCE

Region

0.7 V 0.75 V

0V 0.1 V

Saturation Saturation

2-9 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Figure 5-44

3-4 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Figure 8-30

3-7 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Figure 9-8

3-8 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Figure 12-6

3-11 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Figure 12-43

3-12 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Figure 11-30

3-13 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Figure 15-21

3-14 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Figure 18-4

3-15 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Figure 18-7

3-16 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Figure 18-13

3-17 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Figure 20-27

3-20 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Figure 20-45

3-21 Downloaded by Kevin Aguilar Arias ([email protected])

lOMoARcPSD|6455127

Figure 23-15

3-29 Downloaded by Kevin Aguilar Arias ([email protected])

Solucionario Malvino Principios de Electronica 7o Edicicion

Short Description

Description

Comments