Solutions Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patri

Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

Solution Manual For Electronic Principles 9th Edition by Albert Paul Malvino, Patrick Hoppe Part

1

Electronic Principles 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

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 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 give

Given:

Solution: RL = 100RS RL = 100(0.1 Ω) RL = 10 Ω Answer: The voltage source will appear stiff for values of load resistance of ≥10 Ω.

1-2.

Given:

RLmin = 270 Ω RLmax = 100 kΩ Solution: RS < 0.01 RL (Eq. 1-1) RS < 0.01(270 Ω)

RS < 2.7 Ω

1-3.

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

Given: RS = 50 Ω Solution: RL = 100RS RL = 100(50 Ω)

RL = 5 kΩ

1-4.

Answer: The function generator will appear stiff for values of load resistance of ≥5 kΩ.

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 ≥ 4 Ω.

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1-1

Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

1-5.

Given: RS = 0.05 Ω I=2A

Solution: RL = 0.01RS (Eq. 1-4) RL = 0.01(250 kΩ) RL = 2.5 kΩ IL = IT [(RS)/(RS + RL)] (Current divider formula) IL = 5 mA [(250 kΩ)/(250 kΩ + 10 kΩ)] IL = 4.80 mA

Solution: V = IR (Ohm’s law) V = (2 A)(0.05 Ω) V = 0.1 V Answer: The voltage drop across the internal resistance is 0.1 V. 1-6.

Given: V=9V RS = 0.4 Ω Solution: I = V/R (Ohm’s law) I = (9 V)/(0.4 Ω) I = 22.5 A

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.

1-12. Solution: VTH = VR2 VR2 = VS[(R2)/(R1 + R2)]

RTH = [(6 kΩ)(3 kΩ)/(6 kΩ + 3 kΩ)]

Answer: The load current is 22.5 A. 1-7.

1-8.

RTH = 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 ≤100 kΩ.

(Voltage divider formula)

VR2 = 36 V[(3 kΩ)/(6 kΩ + 3 kΩ)] VR2 = 12 V RTH = [R1R2/R1 + R2] (Parallel resistance formula)

Answer: The Thevenin voltage is 12 V, and the Thevenin resistance is 2 kΩ.

6 kΩ

R1

3 kΩ

R2

6 kΩ

R1

3 kΩ

R2

36 V

Given:

VTH

RLmin = 270 Ω RLmax = 100 kΩ Solution: RS > 100 RL (Eq. 1-3) RS > 100(100 kΩ)

1-9.

RS > 10 MΩ Answer: The internal resistance of the source is greater than 10 MΩ.

Given: RS = 100 kΩ Solution: RL = 0.01RS (Eq. 1-4)

36 V

RTH

RL = 0.01(100 kΩ)

RL = 1 kΩ Answer: The maximum load resistance for the current source to appear stiff is 1 kΩ.

1-10. Given: IS = 20 mA RS = 200 kΩ

RL = 0 Ω 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 appea rs stiff; thus the current is 20 mA. 1-11. Given: I = 5 mA RS = 250 kΩ RL = 10 kΩ

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

1-13. Given: VTH = 12 V RTH = 2 kΩ Solution: 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.

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1-2

R

Solution: RN = RTH (Eq. 1-10) RTH = 10 kΩ IN = VTH/RTH (Eq. 1-12) VTH = INRN

TH

V

R

TH

L

VTH = (10 mA)(10 kΩ)

VTH = 100 V Answer: RTH = 10 kΩ, and VTH = 100 V

Thevenin equivalent circuit for Prob. 1-13.

1-14. Given: VS = 18 V R1 = 6 kΩ R2 = 3 kΩ

R VTH

Solution: VTH = VR2 (Voltage divider formula)

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 6 V, and the Thevenin resistance is unchanged.

Thevenin circuit for Prob. 1-17.

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

1-15. Given: VS = 36 V R1 = 12 kΩ

R2 = 6 kΩ 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 = 12 V/3 kΩ IN = 4 mA Answer: IN = 4 mA, and RN = 3 kΩ

4 mA

10 kΩ

100 V

VR2 = VS[(R2)/(R1 + R2)]

IN

TH

RN

3 kΩ

Norton circuit for Prob. 1-16.

1-17. Given: IN = 10 mA RN = 10 kΩ

IN 6 mA

R N

2 kΩ

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 ≥ 200 kΩ.

CRITICAL THINKING 1-23. Given: VS = 12 V IS = 150 A Solution: RS = (VS)/(IS) RS = (12 V)/(150 A)

RS = 80 mΩ

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

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 Solution: VS = VRS + VL (Kirchhoff’s law) VRS = VS – VL VRS = 10 V – 9 V VRS = 1 V IRS = IL = VL/RL (Ohm’s law) IRS = 9 V/75 Ω

IRS = 120 mA RS = VRS/IRS (Ohm’s law) RS = 8.33 Ω RS < 0.01 RL (Eq. 1-1) 8.33 Ω < 0.01(75 Ω) 8.33 Ω 𝕍 0.75 Ω Answer: a. The internal resistance (RS) is 8.33 Ω. b. The source is not stiff since RS 𝕍 0.01 RL. 1-25. Answer: Disconnect the resistor and measure the voltage.

1-26. Answer: Disconnect the load resistor, turn the internal voltage and current sources to zero, and measure the resistance.

1-27. Answer: Thevenin’s theorem makes it much easier to solve problems where there could be many values of a resistor.

1-28. Answer: To find the Thevenin voltage, disconnect the load resistor and measure the voltage. To find the Thevenin resistance, disconnect the battery and the load resistor, short the battery terminals, and measure the resistance at the load terminals.

1-29. Given: RL = 1 kΩ I = 1 mA Solution: RS > 100RL RS > 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 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 RTH = (R1R2/R1 + R2) RTH = [(1 kΩ)(1 kΩ)]/(1 kΩ + 1 kΩ) 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: VS = 30 V VL = 10 V

RL > 1 MΩ RS < 0.01RL (since the voltage source must be stiff) (Eq. 1-1) Solution: RS < 0.01RL RS < 0.01(1 MΩ) RS < 10 kΩ Since the Thevenin equivalent resistance would be the series resistance, RTH < 10 kΩ.

Assume a value for one of the resistors. Since the Thevenin resistance is limited to 1 kΩ, pick a value less than 10 kΩ. Assume R2 = 5 kΩ.

VL = VS[R2/(R1 + R2)] (Voltage divider formula) R1 = [(VS)(R2)/VL] – R2 R1 = [(30 V)(5 kΩ)/(10 V)] – 5 kΩ R1 = 10 kΩ RTH = R1R2/(R1 + R2) RTH = [(10 kΩ)(5 kΩ)]/(10 kΩ + 5 kΩ) RTH = 3.33 kΩ Since RTH is one-third of 10 kΩ, we can use R1 and R2 values that are three times larger.

Answer: R1 = 30 kΩ R2 = 15 kΩ Note: The criteria will be satisfied as long as R1 is twice R2 and R2 is not greater than 15 kΩ.

1-32. Answer: First, measure the voltage across the terminals.

This is the Thevenin voltage. Next, connect the ammeter to the battery terminals—measure the current. Next, use the values above to find the total resistance. Finally, subtract the internal resistance of the ammeter from this result. This is the Thevenin resistance.

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) 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

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1-4

IL = 0.148 V/(6 kΩ + 4 kΩ) IL = 14.8 μA IL = 0.148 V/(6 kΩ + 5 kΩ) IL = 13.5 μA IL = 0.148 V/(6 kΩ + 6 kΩ) IL = 12.3 μA 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 1-36. R1 shorted

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 d. n-type e. p-type

2-7.

Given:

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

1-37. R2 open 1-38. No supply voltage 1-39. R4 open

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.

1-40. R2 shorted

Chapter 2

Semiconductors

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

2-8.

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

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

JOB INTERVIEW QUESTIONS 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.

PROBLEMS 2-1.

–2

2-2.

–3

2-3.

a. Semiconductor b. Conductor c. Semiconductor d. Conductor

Given:

IS = 10 nA at 25°C Tmin = 0°C – 75°C Tmax = 75°C Solution: (ΔT/10) IS(new) = 2 IS(old) IS(new) = 2[(0°C – 25°C)/10]10 nA IS(new) = 1.77 nA (ΔT/10) IS(new) = 2 IS(old) IS(new) = 2[(75°C – 25°C)/10)] 10 nA IS(new) = 320 nA Answer: The saturation current is 1.77 nA at 0°C and 320 nA at 75°C. 2-9.

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.

2-12. R1 = 25 Ω

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1-5

Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

2-13. R1 open

3-3.

VD1 = 0.75 V VD2 = 0.8 V ID1 = 400 mA Solution: Since the diodes are in series, the current through each is the same.

2-14. D1 shorted 2-15. D1 open 2-16. V1 = 0 V

Chapter 3

Answer: 400 mA

Diode Theory 3-4.

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

Given:

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

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

VS = 20 V

18. b 19. a 20. b 21. a 22. c

VD = 0 V

RL = 1 kΩ Solution: VS = VD + VL (Kirchhoff’s law) 20 V = 0 V + VL VL = 20 V IL = VL/RL (Ohm’s law)

JOB INTERVIEW QUESTIONS 4. If you have a data sheet, look up the maximum current rating and the breakdown voltage. Then, check the schematic diagram to see whether the ratings are adequate. If they are, check the circuit wiring.

7. Measure the voltage across a resistor in series with the diode. Then, divide the voltage by the resistance.

8. With the power off, check the back-to-front ratio of the diode with an ohmmeter or use the diode test function on a DMM. If it is high, the diode is OK. If it is not high, disconnect one end of the diode and recheck the back-to-front ratio. If the ratio is now high, the diode is probably OK. If you are still suspicious of the diode for any reason, the ultimate test is to replace it with a known good one.

10. Connect a diode in series between the alternator and the battery for the recreational vehicle. The diode arrow points from the alternator to the RV battery. This way, the alternator can charge the vehicle battery. When the engine is off, the diode is open, preventing the RV battery from discharging.

11. Use a voltmeter or oscilloscope for a diode in the circuit. Use an ohmmeter, DMM or curve tracer when the diode is out of the circuit.

Given:

IL = 20 V/1 kΩ

IL = 20 mA PL = (IL)(VL) PL = (20 mA)(20 V) PL = 400 mW PD = (ID)(VD) PD = (20 mA)(0 V) PD = 0 mW PT = PD + PL PT = 0 mW + 400 mW PT = 400 mW Answer: IL = 20 mA VL = 20 V PL = 400 mW

PD = 0 mW PT = 400 mW 3-5.

Given:

VS = 20 V VD = 0 V

RL = 2 kΩ Solution: IL = VL/RL (Ohm’s law) IL = 20 V/2 kΩ IL = 10 mA Answer: 10 mA

PROBLEMS 3-1.

Given:

R = 220 Ω V=6V Solution: I = V/R I = 6 V/220 Ω I = 27.27 mA Since it is a series circuit, the current flowing through the diode is the same as the current through the resistor.

Answer: 27.27 mA 3-2.

Given:

VD = 0.7 V ID = 100 mA Solution: P = VI P = (0.7 V)(100 mA) P = 70 mW Answer: 70 mW

3-6.

Given:

VS = 12 V VD = 0 V RL = 470 Ω Solution: VS = VD + VL (Kirchhoff’s law) 12 V = 0 V + VL VL = 12 V IL = VL/RL (Ohm’s law) IL = 12 V/470 Ω

IL = 25.5 mA PL = (VL)(IL) PL = (12 V) (25.5 mA) PL = 306 mW

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1-6

PD = (VD)(ID) PD = (0 V) (25.5 mA)

PD = 0 mW PT = PD + PL PT = 0 mW + 306 mW PT = 306 mW Answer: VL = 12 V IL = 25.5 mA

PL = 306 mW PD = 0 mW PT = 306 mW 3-7.

Given:

VS = 12 V VD = 0 V RL = 940 Ω Solution: IL = VL/RL (Ohm’s law) IL = 12 V/940 Ω IL = 12.77 mA Answer: 12.77 mA 3-8.

Given:

VS = 12 V RL = 470 Ω Solution: The diode would be reversed-biased and acting as an open. Thus the current would be zero and the voltage would be source voltage.

Answer: VD = 12 V ID = 0 mA 3-9.

Given:

VS = 20 V VD = 0.7 V

RL = 1 kΩ Solution: VS = VD + VL (Kirchhoff’s law) 20 V = 0.7 V + VL VL = 19.3 V IL = VL/RL (Ohm’s law) IL = 19.3 V/1 kΩ

IL = 19.3 mA PL = (IL)(VL) PL = (19.3 mA)(19.3 V) PL = 372 mW PD = (ID)(VD) PD = (19.3 mA)(0.7 V) PD = 13.5 mW PT = PD + PL PT = 13.5 mW + 372 mW

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

PD = 13.5 mW PT = 386 mW

3-10. Given: VS = 20 V VD = 0.7 V

RL = 2 kΩ Solution: IL = VL/RL (Ohm’s law) IL = 19.3 V/2 kΩ IL = 9.65 mA Answer: 9.65 mA 3-11. Given: VS = 12 V VD = 0.7 V RL = 470 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/470 Ω

IL = 24 mA PL = (VL)(IL) PL = (11.3 V)(24 mA) PL = 271.2 mW PD = (VD)(ID) PD = (0.7 V)(24 mA) PD = 29.2 mW PT = PD + PL PT = 29.2 mW + 271.2 mW

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

PD = 29.2 mW PT = 300.4 mW 3-12. 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 3-13. Given: VS = 12 V RL = 470 Ω Solution: The diode would be reversed-biased and acting as an open. Thus the current would be zero and the voltage would be source voltage.

Answer: VD = 12 V ID = 0 mA

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

Note for Probs. 3-14 to 3-18: Since the bulk resistance of a 1N4001 is 0.23 Ω and meets the criteria of Eq. (3-6) (rB < 0.01 RTH), it can be ignored for these calculations.

3-14. Given: VS = 20 V VD = 0.7 V

RL = 1 kΩ Solution: VS = VD + VL (Kirchhoff’s law) 20 V = 0.7 V + VL VL = 19.3 V IL = VL /RL (Ohm’s law) IL = 19.3 V/1 kΩ

IL = 19.3 mA PL = (IL)(VL) PL = (19.3 mA)(19.3 V) PL = 372 mW PD = (ID)(VD) PD = (19.3 mA)(0.7 V) PD = 13.5 mW PT = PD + PL PT = 13.5 mW + 372 mW

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

PD = 13.4 mW PT = 386 mW 3-15. Given: VS = 20 V

Answer: VL = 11.3 V IL = 24 mA PL = 271.2 mW

PD = 29.2 mW PT = 300.4 mW 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 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: VD = 12 V ID = 0 mA 3-19. Open 3-20. The diode voltage will be 5 V, and it should burn open the diode.

VD = 0.7 V

RL = 2 kΩ

3-21. The diode is shorted, or the resistor is open.

Solution: IL = VL/RL (Ohm’s law) IL = 19.3 V/2 kΩ IL = 9.65 mA

3-22. The voltage of 3 V at the junction of R1 and R 2 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.

Answer: 9.65 mA 3-16. Given: VS = 12 V VD = 0.7 V RL = 470 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/470 Ω

IL = 24 mA PL = (VL)(IL) PL = (11.3 V)(24 mA) PL = 271.2 mW PD = (VD)(ID) PD = (0.7 V)(24 mA) PD = 29.2 mW PT = PD + PL

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

3-24. 1N4004 3-25. Cathode band. The arrow points toward the band.

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.

CRITICAL THINKING 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

PT = 29.2 mW + 271.2 mW

PT = 300.4 mW

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1-8

V = VR + VD (Kirchhoff’s law) V = 1.25 V + 0.7 V V = 1.95 V

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

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

VR1 = VS – V (Kirchhoff’s law) VR1 = 12 V – 1.95 V VR1 = 10.05 V I = V/R (Ohm’s law) I = 10.05 V/30 kΩ I = 335 μA

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

Now that the current through R1 is known, this is the total current for the parallel branches. The next step is to find the current through R2.

I2 = I1 – ID (Kirchhoff’s law)

1N4001 reverse: R = V/I (Ohm’s law) R = 50 V/10 μA R = 5 MΩ

I2 = 335 μA – 0.25 mA

I2 = 85 μA

1N1185 forward: R = V/I (Ohm’s law) R = 0.95 V/10 A R = 0.095 Ω

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

1N1185 reverse: R = V/I (Ohm’s law) R = 100 V/4.6 mA R = 21.7 kΩ Answer: 1N914: forward R = 100 Ω reverse R = 800 MΩ 1N4001: forward R = 1.1 Ω reverse R = 5 MΩ 1N1185: forward R = 0.095 Ω reverse R = 21.7 kΩ 3-28. Given: VS = 5 V VD = 0.7 V ID = 20 mA 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 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

3-30. Given: 500 mA at 1 V 0 mA at 0.7 V Solution: rB = (V2 – V1)(I2 – I1) (Eq. 3-7) rB = (1 V – 0.7 V)/(500 mA – 0 mA) rB = 600 mΩ Answer: rB = 600 mΩ 3-31.

1. IR = ISL + IS 2. 5 μA = ISL + IS(old) 3. ISL = 5 μA – IS(old) 4. 100 μA = ISL + IS(new) (∆T/10) 5. IS(new) = 2 IS(old) Substitute formulas 2 and 5 into formula 4. (∆T/10) 6. 100 μA = 5 μA –IS(old) + 2 IS(old) Put in the temperature values. [(100ºC – 25ºC)/10] 7. 100 μA = 5 μA –IS(old) + 2 IS(old) Move the 5 μA to the left side, and simplify the exponent of 2.

7.5 8. 95 μA = – IS(old) + 2 IS(old) Combine like terms.

7.5 9. 95 μA = (2 – 1)IS(old) 10. 95 μA = (180.02) IS(old) Solve for the variable.

11. IS(old) = 95 μA/(180.02) 12. 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

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1-9

Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

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.

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

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 Answer: The resistance would be 12.5 kΩ if the diode is open and 0 Ω if the diode is shorted.

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.

PROBLEMS 4-1.

Solution: Vp = 1.414 Vrms

3-33. During normal operation, the 15-V power supply is sup-plying power to the load. The left diode is forward-biased 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, prevent-ing any current from going into the 15-V power supply.

Vp = 1.414 (50 V ac)

Vp = 70.7 V Vp(out) = Vp(in) Vp(out) = 70.7 V

(Eq. 4-1)

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.

3-34. D1 is shorted 4-2.

3-35. D1 is open

Given: vin = 50 V ac

Given: vin = 15 V ac

3-36. Power supply has failed and is 0 V

Solution: Vp = 1.414 Vrms

3-37. R3 is shorted

Vp = 1.414 (15 V ac)

Vp = –21.2 V Vp(out) = Vp(in) (Eq. 4-1) Vp(out) = –21.2 V

3-38. D1 is reverse biased 3-39. 1N4001 silicon rectifier diode

Since the average and the dc values are the same: Vdc = 0.318 Vp (Eq. 4-2)

3-40. Yes, the 1N4002 has the same forward current rating and a higher reverse breakdown rating.

Vdc = 0.318 (–21.2 V)

Vdc = –6.74 V Answer: The peak voltage is –21.2 V, the average voltage is –6.74 V, and the dc voltage is –6.74 V.

3-41. Cathodes 3-42. Normally reverse biased 4-3.

3-43. Normally reverse biased

Chapter 4

Solution: Vp = 1.414 Vrms

Diode Circuits

Vp = 1.414 (50 V ac)

Vp = 70.7 V 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

Given: vin = 50 V ac

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

JOB INTERVIEW QUESTIONS 7. The LC type is preferable when tighter regulation is required and (or) power cannot be wasted. Examples include transmit-ters, lab test equipment, and military gear when cost is not of primary concern. The LC filter ideally dissipates no power. 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.

(Eq. 4-4)

Since the average and the dc values are the same: 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. 4-4.

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

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1-10

Since the average and the DC values are the same: Vdc = 0.318 V (Eq. 4-2)

VP = 1.414 Vrms

Vdc = 0.318 (–20.5 V)

VP = 21.21 V Vp(out) = Vp(in) – 0.7 V (Eq. 4-4) Vp(out) = 20.51 V Vdc = 0.318 Vp (Eq. 4-2)

VP = 1.414 (15 V ac)

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

Vdc = 0.318 (20.51 V)

Vdc = 6.52 V Answer: The peak voltage is 20.51 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.

4-9.

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. 4-6.

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) Solution: V2 = V1/(N1/N2) (Eq. 4-5) V2 = 120 V ac/8 V2 = 15 V ac Vp = 1.414 Vrms Vp= 1.414 (15 V ac)

Vp = 21.21 V Vp(out) = Vp(in) (Eq. 4-1) Vp(out) = 21.21 V 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) (Eq. 4-5) V2 = 120 V ac/8 V2 = 15 V ac

Given:

Turns ratio = N1/N2 = 4:1 = 4 V1 = 120 Vrms Solution: V2 = V1/(N1/N2) (Eq. 4-5) V2 = 120 Vrms/4 V2 = 30 Vrms Since it is a center-tapped transformer, each half of the secondary is half of the total secondary voltage.

Vupper = ½ V2 Vupper = ½ (30 Vrms) Vupper = 15 Vrms Vlower = ½ Vp Vlower = ½(30 Vrms) Vlower = 15 Vrms Vp = (1.414) (Vrms) Vp = (1.414) (15 Vrms) Vp = 21.21 Vp 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) V2 = 120 V ac/7 V2 = 17.14 V ac 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 Vp(out) = Vp(in) (Eq. 4-1) Vp(out) = 12.12 V 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 Vac Solution: V2 = V1/(N1/N2) (Eq. 4-5) V2 = 120 V ac/7 V2 = 17.14 V ac

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1-11

Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

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

(Eq. 4-4)

Vdc = 0.636 (11.42 V)

Vdc = 7.26 V Answer: The peak output voltage is 11.42 V, and the dc and average values are 7.26 V. 4-12. Given: Turns ratio = N1/N2 = 8:1 = 8 V1 = 120 V ac Solution: V2 = V1/(N1/N2) (Eq. 4-5) V2 = 120 V ac/8 V2 = 15 V ac Vp(in) = 1.414 Vrms Vp(in) = 1.414 (15 V ac) Vp(in) = 21.21 V Vp(out) = Vp(in) (Eq. 4-1) Vp(out) = 21.21 V 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. 4-13. Given: Turns ratio = N1/N2 = 8:1 = 8 V1 = 120 V ac (Eq. 4-5)

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

(Eq. 4-5)

(Eq. 4-5)

Vp(in)max = 1.414 V2(max)

Since the average and the dc values are the same: Vdc = 0.636 Vp (Eq. 4-6)

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

Solution: V2(max) = V1(max)/(N1/N2) V2(max) = 125 V ac/8 V2(max) = 15.63 V ac V2(min) = V1(min)/(N1/N2) V2(min) = 105 V ac/8 V2(min) = 13.13 V ac

(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.

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

Vp(in)max = 1.414 (15.63 V ac) Vp(in)max = 22.10 V 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

(Eq. 4-1)

Vp(out)max = 22.10 V Vp(out)min = Vp(in)min

(Eq. 4-1)

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

(Eq. 4-6)

Vdc = 0.636 (22.10 V)

Vdc = 14.06 V Vdc(min) = 0.636 Vp(out)min (Eq. 4-6) Vdc = 0.636 (18.57 V) Vdc = 11.81 V 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 Solution: vout = (XC/XL)vin (Eq. 4-9) vout = (25 Ω/1 Ω)(20 V) vout = 500 mV Answer: The ripple voltage would be 500 mV. 4-16. Given: vin = 14 V XL = 2 kΩ XC = 50 Ω Solution: vout = (XC/XL)vin (Eq. 4-9) vout = (50 Ω/2 kΩ)(14 V) vout = 350 mV 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)

(Eq. 4-5)

V2 = 120 Vac/8

V2 = 15 Vac 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.)

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1-12

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

C = 470 μF fin = 60 Hz

fout = fin (Eq. 4-3) fout = 60 Hz VR = I/(fC) (Eq. 4-10)

V2 = 120 V ac/9

VR = (2.12 mA)/[(60 Hz)(47 μF)]

VR = 752 mV Answer: The dc output voltage is 21.2 V with a 752 mVp-p ripple.

Output waveform for Prob. 4-17.

4-18. Given: Turns ratio = N1/N2 = 7:1 = 7 V1 = 120 V ac RL = 2.2 kΩ C = 68 μF fin = 60 Hz Solution: V2 = V1/(N1/N2) V2 = 120 V ac/7 V2 = 17.14 V ac Vp = 1.414 Vrms

(Eq. 4-5)

Solution: V2 = V1/(N1/N2)

(Eq. 4-5)

V2 = 13.33 V ac Vp = 1.414 Vrms Vp = 1.414 (13.33 Vac) 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

(Eq. 4-7)

fout = 2(60 Hz)

fout = 120 Hz 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 mVp-p ripple.

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

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 mVp-p ripple.

4-19. Answer: VR = I/(fC) (Eq. 4-10) 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

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)

(Eq. 4-5)

V2 = 105 V ac/9

V2 = 11.67 V ac Vp = 1.414 Vrms Vp = 1.414 (11.67 V ac) Vp = 16.50 V 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 (Eq. 4-13) PIV = 18.85 V 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)

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

Vp = (1.414) (Vrms) Vp = (1.414) (40 Vrms) Vp = 56.56 Vp PIV = Vp (Eq. 4-13) PIV = 56.56 V Answer: The peak inverse voltage is 56.56 V. 4-25. Solution: From the information in Section 4-8. 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: 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) Idiode = V/R Idiode = (2.12 V)/(10 kΩ) Idiode = 212 μA 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 μA 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 half-wave rectifier with a capacitor input filter. The dc voltage should not change much 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 polarity-sensitive, 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.

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 Solution: Voltage at the cathode is found by using the voltage divider formula.

Vbias = [R1/(R1 + R2)]Vdc (Eq. 4-18) 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.

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1-14

Vp = 1.414 Vrms Vp = 1.414 (600 V ac) 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.

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)

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 Vp-p, 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.

4-41. Given: Turns ratio = N1/N2 = 1:10 = 0.1 V1 = 120 V ac Solution: V2 = V1/(N1/N2)

(Eq. 4-5)

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 Solution: V2 = V1/(N1/N2) (Eq. 4-5) V2 = 120 V ac/0.2

V2 = 600 V ac

(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.

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.

4-45. Given: Turns ratio = N1/N2 = 8:1 = 8 V1 = 120 V ac Solution: V2 = V1/(N1/N2)

(Eq. 4-5)

V2 = 120 V ac/8

V2 = 15 V ac 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.

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.

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1-15

Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

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 = Vp sin θ. 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) (Eq. 4-5)

4-54. V1 failed 4-55. XMFR secondary winding is open 4-56. Full-wave bridge rectifier 4-57. 15 V ac 4-58. Full-wave rectifier 4-59. Approximately 8.1 V 4-60. 14 V

V2 = 120 V ac/8

V2 = 15 V ac 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. 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.

4-50. 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.

4-51. C1 is open 4-52. Full-wave bridge is open

Chapter 5

Special-Purpose Diodes

SELF-TEST 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

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.

PROBLEMS 5-1.

Given:

VS = 24 V VZ = 15 V RS = 470 Ω Solution: IS = IZ = (VS – VZ)/RS (Eq. 5-3) IS = IZ = (24 V – 15 V)/(470 Ω) IS = IZ = 19.1 mA Answer: The zener current is 19.1 mA. 5-2.

Given:

VS = 40 V VZ = 15 V RS = 470 Ω Solution: IS = IZ = (VS – VZ)/RS

(Eq. 5-3)

IZ = (40 V – 15 V)/470 Ω

IZ = 53.2 mA Answer: The maximum zener current is 53.2 mA.

4-53. C1 is shorted

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1-16

5-3.

Given:

IZ = IS – IL

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)

IZ = 20.16 mA – 9.52 mA

IZ = 10.64 mA Answer: The maximum zener current is 10.64 mA. 5-7.

VZ = 15 V

(Eq. 5-3)

RS = 470 Ω Solution: Maximum current will occur at maximum voltage.

IS = (VS – VZ)/RS (Eq. 5-3)

IS = IZ = 20.16 mA Answer: The maximum zener current is 20.16 mA. Given:

IS = (40 V – 15 V)/470 Ω

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

IS = 53.2 mA IL = VL/RL (Eq. 5-5, Ohm’s law) IL = 15 V/1.5 kΩ IL = 10 mA IZ = IS – IL (Eq. 5.6, Kirchhoff’s current law) IZ = 53.2 mA – 10 mA IZ = 43.2 mA Answer: The maximum zener current is 43.2 mA.

Solution: 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.

5-8.

Given:

Solution: VL = VZ = 12 V

Solution: IS = (VS – VZ)/RS (Eq. 5-3) IS = (24 V – 15 V)/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 = IS – 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. 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 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

Given:

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

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

5-6.

Given:

VS = 24 V to 40 V

IS = IZ = (24 V – 15 V)/(446.5 Ω)

5-4.

(Eq. 5-6, Kirchhoff’s current law)

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 = IS – 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.

Given:

VS = 20 V VZ = 12 V RS = 330 Ω

RL = 1 kΩ Solution: VL = VZ = 12 V IS = (VS – VZ)/RS

(Eq. 5-3)

IS = (20 V – 12 V)/330 Ω

IS = 24.24 mA IL = VL/RL (Eq. 5-5, Ohm’s law) IL = 12 V/1 kΩ

IL = 12 mA IZ = IS – IL

(Eq. 5-6, Kirchhoff’s current law)

IZ = 24.24 mA – 12 mA

(Eq. 5-5. Ohm’s law)

IZ = 12.24 mA Answer: The load voltage is 12 V, and the zener current is 12.24 mA.

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1-17

Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

VL = [(RL)/(RS + RL)]VS VS = VL/[RL/(RS + RL)]

RS

VS = 15 V[(1.5 kΩ)/(470 Ω + 1.5 kΩ)]

330 Ω 12 V

20 V

RL 1 kΩ

Zener regulator for Prob. 5-9.

5-10. Given: RS = 470 Ω RZ = 1 4Ω VR(in) = 1 Vp-p Solution: V R(out)

(Voltage divider formula)

=

RZ RS

V R(in)

(Eq. 5-8) VR(out) = (14 Ω /470 Ω )/1 Vp-p VR(out) = 29.8 mVp-p Answer: The ripple voltage across the load resistor is 29.8 mVp-p.

5-11. Given: VS = 21.5 to 25 RS = 470 Ω RZ = 14 Ω VZ = 15 V Solution: IS = (VS – VZ)/RS (Eq. 5-3) IS = (25 V – 15 V)/470 Ω IS = 21.28 mA IL = VL/RL (Eq. 5-5, Ohm’s law) IL = 15 V/1.5 kΩ IL = 10 mA IZ = IS – 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 = IS – IL (Eq. 5-6, Kirchhoff’s current law) IZ = 13.83 mA – 10 mA

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

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 Solution: The regulation is lost once the load voltage drops below 15 V. IL = VL/RL (Eq. 5-5, Ohm’s law) IL = 15 V/1.5 kΩ RS(max) = [(VS(min) /VZ ) – 1]RL(min) RS(max) = [(20 V/15 V) – 1]500 Ω

(Eq. 5-9)

RS(max) = 167 Ω 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 Solution: RS (max) =

VS (min) − VZ I L(max)

(Eq. 5-10)

RS(max) = (18 V – 15V)/25 mA RS(max) = 120 Ω 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.

5-15. Given: VS = 24 V RS = 470 Ω VZ = 15 V Solution: 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

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1-18

Solution: PZ = VZIZ

(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

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: a. 0 V b. 18.27 V c. 0 V d. 0 V 5-22. Answer: From the previous problem, the only trouble that caused this symptom is an open zener diode.

(from Prob. 5-5)

IL = 10 mA (from Prob. 5-5) VZ = 9.15 mA (from Prob. 5-5) Solution: 2 P=I R 2 PS = (19.15 mA) (470Ω) PS = 172.4 mW 2 P=I R 2 PL = (10 mA) (1.5 kΩ) PL = 150 mW 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 determined from the data sheet.

Solution: (15 V)(0.05) = 0.75 V 15 V + 0.75 V = 15.75 V 15 V – 0.75 V = 14.25 V Answer: The minimum voltage of 14.25 V and the maximum voltage is 15.75 V. 5-20. Given: T = 100°C Solution: 100°C – 50°C = 50°C Derating factor 6.67 mW/°C Answer: P = 667 mW 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

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 filter 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. 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: VS = 15 V VD = 2 V RS = 2.2 kΩ Solution: IS = (VS – VD)/RS (Eq. 5-13) IS = (15 V – 2 V)/2.2 kΩ IS = 5.91 mA Answer: The diode current is 5.91 mA. 5-26. Given: VS = 40 V VD = 2 V RS = 2.2 kΩ Solution: 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: VS = 15 V VD = 2 V

RS = 1 kΩ 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.

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1-19

Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

CRITICAL THINKING 5-29. Given: 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 = IS – IL

(Eq. 5-6, Kirchhoff’s current law)

IZ = 19.15 mA – 10 mA

Solution: RS(max) = [(VS(min) – VZ)/IL(max)] RS(max) = [(20 V – 6.8 V)/30 mA] RS(max) = 440 Ω RS(min) = [(VS – VZ)/IZM] RS(min) = [(20 V – 6.8 V)/147 mA] RS(min) = 90 Ω

(Eq. 5-10)

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 90 Ω, if a 1N4736A 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.

IZ = 9.15 mA ΔVL = IZRZ (Eq. 5-7) ΔVL = (9.15 mA)(14 Ω ) ΔVL = 128.1 mV VL = 15.128 V or approximately 15.13 V Answer: The load voltage would be 15.13 V. 5-30. Given: VS = 24 V RS = 470 Ω RZ = 14 Ω VZ = 15 V RL = 1 kΩ to 10 kΩ IS = 19.15 mA (from Prob. 5-29) Solution: IL(max) = VL/RL(min) (Eq. 5-5, Ohm’s law) IL(max) = 15 V/1 kΩ IL(max) = 15 mA IL(min) = VL/RL(max) IL(min) = 15 V/10 kΩ

(Eq. 5-5, Ohm’s law)

IL(min) = 1.5 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 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 ΔVL(min) = IZ(min)RZ ΔVL(min) = (4.15 mA)(14 Ω ) ΔVL(min) = 58.1 mV 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 Answer: The minimum load voltage would be 15.06 V and the maximum voltage would be 15.25 V. 5-31. Given: VS = 20 V VZ = 6.8 V VL = 6.8 V IL = 30 mA

Zener regulator for Prob. 5-31.

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

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1-20

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.

IS = (VS – VZ)/(RS + RZ) 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 μF ± 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 (Eq. 4-7) fout = 2(60 Hz)

fout = 120 Hz The minimum capacitance will give the maximum ripple.

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.

5-38. RS is open

C = 1000 μF ± 20% C = 1000 μF – 1000 μF (20%) C = 800 μF VR = I/(fC) (Eq. 4-10) VR = 42 mA/(120 Hz)(800 μF) VR = 0.438 V

5-39. Power supply failed (0 V)

Answer: The maximum ripple voltage will be 0.438 V.

5-44. Biased off

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 Vp(out) = Vp(in) – 0.5 V (Eq. 4-8; the 1.4 was changed to reflect using Schottky diodes) Vp(out) = 7.98 V Answer: The voltage at the filter capacitor is 7.98 V. 5-36. Troubles: 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

5-40. Zener diode open 5-41. RL is shorted 5-42. Zener diode is forward biased 5-43. Common anode 5-45. 25 mA 5-46. 200 Ω 5-47. 470 Ω, 1/2 watt

Chapter 6

BJT Fundamentals

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

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

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

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

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.

14. An increase in temperature almost always increases the current gain.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

PROBLEMS 6-1.

Solution: The minimum resistance will yield the maximum current.

RB = 470 kΩ ± 5% RB = 470 kΩ – 470 kΩ(5%) RB = 446.5 kΩ IB = [(VBB – VBE)/RB] (Eq. 6-6)

Given:

IE = 10 mA IC = 9.95 mA Solution: IE = IC + IB (Eq. 6-1) IB = IE – IC

IB = [(10 V – 0.7 V)/446.5 kΩ]

IB = 10 mA – 9.95 mA

IB = 20.83 μA Answer: The base current is 20.83 μA.

IB = 0.05 mA Answer: The base current is 0.05 mA. 6-2.

6-8.

Given:

IC = 6 mA RC = 1.5 kΩ VCC = 20 V

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:

Solution: PD = VCEIC

IC = 4.5 mA Answer: The collector current is 4.5 mA. Given:

VBE = 0 V (ideal) RC = 820 Ω

IB = 1.54 mA IE = IB + IC IE = 1.54 mA + 100 mA Answer: The emitter current is 101.54 mA. Given:

VBB = 10 V RB = 470 kΩ VBE = 0.7 V Solution: IB = [(VBB – VBE)/RB]

VCC = 10 V

βdc = 200 Solution: Ideal IB = [(VBB – VBE)/RB] (Eq. 6-6) IB = [(10 V – 0 V)/470 kΩ] IB = 21.28 μA IC = βdcIB IC = 200(21.28 μA)

(Eq. 6-6)

IB = [(10 V – 0.7 V)/470 kΩ]

IB = 19.8 μA Answer: The base current is 19.8 μA. 6-6. Answer: The base current is unaffected by the current gain since 9.3 V/470 kΩ always equals 19.8 μA. The current gain will affect the collector current in this circuit.

6-7.

PD = 350 mW Answer: The power dissipation is 350 mW.

VBE = 0.7 V (second approximation)

(Eq. 6-5)

IB = 100 mA/65

6-5.

(Eq. 6-8)

PD = (3.5 V)(100 mA)

6-10. Given: VBB = 10 V RB = 470 kΩ

IC = 100 mA βdc = 65 Solution: IB = IC/βdc

Given:

IC = 100 mA VCE = 3.5 V

IC = 150(30 μA)

6-4.

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.

IB = 30 μA βdc = 150 Solution: IC = βdcIB

Given:

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

Given:

2nd Approximation IB = [(VBB – VBE)/RB]

VBB = 10 V RB = 470 kΩ ± 5% VBE = 0.7 V

IB = 19.8 μA IC = βdcIB

(Eq. 6-6)

IB = [(10V – 0.7 V)/470 kΩ]

IC = 200(19.8 μA)

IC = 3.96 mA

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1-22

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 IB = [(VBB – VBE)/RB] IB = [(5 V – 0 V)/330 kΩ] IB = 15.15 μA IC = βdcIB

Solution: IB = [(VBB – VBE)/RB] (Eq. 6-6) IB = [(12 V – 0)/680 kΩ] IB = 17.6 μA (ideal) IB = [(12 V – 0.7 V)/680 kΩ] IB = 16.6 μA (second approximation) IC = βdcIB (Eq. 6-4) IC = 175(17.6 μA)

IC = 3.08 mA (ideal) IC = 175(16.6 μA) 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)

(Eq. 6-6)

IC = 150(15.15 μA)

IC = 2.27 mA 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 2nd Approximation IB = [(VBB – VBE)/RB] (Eq. 6-6) IB = [(5 V – 0.7 V)/330 kΩ] IB = 13.3 μA IC = βdcIB

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. 6-14. Answer: From the on characteristics section, 70.

6-15. Given: PD(max) = 1 W IC = 120 mA

VCE = 10 V Solution: PD = VCEIC

(Eq. 6-8)

PD = (10 V)(120 mA)

IC = 150(13.03 μA)

PD = 1.2 W

IC = 1.96 mA VCE = VCC – ICRC VCE = 15 V – (1.96 mA)(1.2 kΩ) VCE = 12.65 V PD = VCEIC PD = (12.65 V)(1.96 mA) PD = 24.79 mW 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.

Answer: The power dissipation has exceeded the maximum rating, and the transistor’s power rating is damaged and possibly destroyed.

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

6-16. Given: PD = 625 mW Temperature = 65°C Solution: ΔT = 65°C – 25°C ΔT = 40°C ΔP =ΔT (derating factor) ΔP = 40°C(2.8 mW/°C) ΔP = 112 mW 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. Answer: β = 30 6-18. Answer: β = 85

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

6-19. Given: VCC = 20 V VBB = 10 V RB = 1 MΩ RC = 3.3 kΩ Solution: IC(sat) = VCC/RC IC(sat) = 20 V/3.3 kΩ

Solution: IC(sat) = VCC/RC IC(sat) = 5 V/470 Ω IC(sat) = 10.64 mA VCE(cutoff) = VCC VCE(cutoff) = 5 V (Eq. 6-11)

IC(sat) = 6.06 mA VCE(cutoff) = VCC

(Eq. 6-12) 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.

6-20. Given: VCC = 25 V VBB = 10 V

RB = 1 MΩ RC = 3.3 kΩ Solution: IC(sat) = VCC/RC

6-24. Given: VCC = 10 V VBB = 5 V RB = 680 kΩ RC = 470 Ω Solution: IC(sat) = VCC/RC (Eq. 6-11) IC(sat) = 10 V/470 Ω IC(sat) = 21.28 mA VCE(cutoff) = VCC (Eq. 6-12) VCE(cutoff) = 10 V

(Eq. 6-11)

Answer: The load line moves futher away from the origin.

6-21. Given: VCC = 20 V

Answer: Load line moved farther from the origin on the graph.

6-25. Given: VCC = 5 V VBB = 5 V

RB = 680 kΩ

RC = 1 kΩ Solution: IC(sat) = VCC/RC IC(sat) = 5 V/1 kΩ IC(sat) = 5 mA

VBB = 10 V

RB = 1 MΩ RC = 4.7 kΩ Solution: IC(sat) = VCC/RC (Eq. 6-11) IC(sat) = 20 V/4.7 kΩ IC(sat) = 4.25 mA 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.

6-22. Given: VCC = 20 V VBB = 10 V RB = 500 kΩ RC = 3.3 kΩ Solution: IC(sat) = VCC/RC (Eq. 6-11) IC(sat) = 20 V/3.3 kΩ IC(sat) = 6.06 mA VCE(cutoff) =VCC VCE(cutoff) = 20 V Answer: The load line does not change.

6-23. Given: VCC = 5 V VBB = 5 V RB = 680 kΩ RC = 470 Ω

(Eq. 6-12)

Answer: The collector current at saturation is 10.64 mA, and the collector-emitter voltage at cutoff is 5 V. The load line would connect these points.

IC(sat) = 25 V/3.3 kΩ

IC(sat) = 7.58 mA

(Eq. 6-11)

VCE(cutoff) = VCC VCE(cutoff) = 5 V

(Eq. 6-11)

(Eq. 6-12)

Answer: The left side of the load line will decrease by half, and the right will not move.

Ic

(b) (a) (c)

V CE

Load lines for (a) Prob. 6-23, (b) Prob. 6-24, and (c) Prob. 6-25

6-26. Given: VCC = 20 V VBB = 10 V

RB = 1 MΩ RC = 3.3 kΩ

β = 200 Solution: IB = (VBB − VBE)/RB (Eq. 6-13) IB = (10 V − 0.7 V)/1 MΩ IB = 9.3 μA IC = β IB (Eq. 6-3) IC = 200(9.3 μA)

IC = 1.86 mA VCE = VCC – ICRC

(Eq. 6-15)

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VCE = 20 V – (1.86 mA)(3.3 kΩ) VCE = 13.86 V Answer: The voltage between the collector and ground is 13.86 V. 6-27. Given: VCC = 20 V VBB =10 V

RB =1 MΩ

RC = 470 Ω

β =150 Solution: IB = (VBB − VBE)/RB (Eq. 6-13) IB = (5 V − 0.7 V)/680 kΩ IB = 6.32 μA IC = βIB (Eq. 6-3) IC = 150(6.32 μA)

RC = 3.3 kΩ

βdc(min) = 25 βdc(max) = 300 IB = 9.3 μA (from the previous problem) Solution: IC(min) = βdc(min)IB (Eq. 6-3) IC(min) = 25(9.3 μA) IC(min) = 232.5 μA IC(max) = βdc(max)IB (Eq. 6-3) IC(max) = 300(9.3 μA) IC(max) = 2.79 mA VCE(min) = VCC − IC(max)RC (Eq. 6-15) VCE(min) = 20 V − (2.79 mA)(3.3 kΩ) VCE(min) = 10.79 V VCE(max) = VCC − IC(min)RC (Eq. 6-15) VCE(max) = 20 V − (232.5 μA)(3.3 kΩ ) VCE(max) = 19.23 V Answer: The maximum collector to ground voltage is 19.23 V, and the minimum is 10.79 V. 6-28. Given: VCC = 20 V ± 10% VBB = 10 V ± 10%

RB = 1 MΩ ± 5% RC = 3.3 kΩ ± 5% βdc(min) = 50 βdc(max) = 150 Solution: IB(min) = (VBB(min) – VBE)/RB(max) IB(min) = (9 V – 0.7 V)/1.05 MΩ IB(min) = 7.90 μA

(Eq. 6-13)

IB(max) = (VBB(max) – VBE)/RB(min) IB(max) = (11 V – 0.7 V)/0.95 MΩ IB(max) = 10.84 μA

(Eq. 6-13)

IC(min) = βdc(min)IB(min) IC(min) = 50(7.90 μA) IC(min) = 395 μA

(Eq. 6-3)

IC(max) = βdc(max)IB(max) IC(max) = 150(10.84 μA) IC(max) =1.63 mA

(Eq. 6-3)

VCE(min) =VCC(max) – IC(max)RC(max) VCE(min) =18 V – (1.63 mA)(3.47 kΩ) VCE(min) =12.34 V

IC = 948 μA VCE = VCC − ICRC (Eq. 6-15) VCE = 5 V − (948 μA)(470 Ω) VCE = 4.55 V Answer: The voltage between the collector and ground is 4.55 V. 6-30. Given: VCC = 5 V VBB =10 V RB = 680 Ω RC = 470 Ω β(min) = 100 β(max) = 300 IB = 6.32 μA (from the previous problem) Solution: IC(min) = β(min) IB (Eq. 6-3) IC(min) =100(6.32 μA) IC(min) = 632 μA IC(max) = β(max) IB (Eq. 6-3) IC (max) = 300(6.32 μA) IC (max) = 1.90 mA VCE(min) = VCC − IC(max) RC (Eq. 6-15) VCE(min) = 5 V − (1.9 mA)(470 Ω) VCE(min) = 4.1 V VCE (max) = VCC − IC (min) RC (Eq. 6-15) VCE(max) = 5 V − (6.32 μA)(470 Ω) VCE(max) = 4.99 V Answer: The maximum collector to ground voltage is 4.99 V, and the minimum is 4.1 V. 6-31. Given: VCC = 5 V ± 10% VBB = 5 V ±10% RB = 680 kΩ ± 5% RC = 470 Ω ± 5% βdc(min) = 50 βdc(max) = 150 Solution:

(Eq. 6-15)

VCE(min) =VCC(max) – IC(max)RC(max) (Eq. 6-15) VCE(max) =22 V – (395 μA)(3.14 kΩ) VCE(max) = 20.76 V Answer: The maximum collector to ground voltage is 20.76 V, and the minimum is 12.34 V. 6-29. Given: VCC = 5V VBB = 10 V

RB = 680 kΩ

IB(min) = (VBB(min) − VBE)/RB(max) IB(min) = (4.5 V − 0.7 V)/714 kΩ IB(min) = 5.32 μA

(Eq. 6-13)

IB(min) = (VBB(min) − VBE)/RB(max) IB(max) = (5.5 V − 0.7 V)/646 kΩ IB(max) = 7.43 μA

(Eq. 6-13)

IC(min) = βdc(min)IB(min) IC(min) = 50(5.32 μA) IC(min) = 266 μA

(Eq. 6-3)

IC(max) = βdc(max)IB(max) IC(max) = 150(7.43 μA) IC(max) = 1.11 mA

(Eq. 6-3)

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

VCE(min) = VCC (max) − IC(max)RC(max) VCE(min) = 4.5 V − (1.11 mA)(493.5 Ω) VCE(min) = 3.95 V

(Eq. 6-15)

VCE(max) = VCC(max) − IC(min)RC(min) (Eq. 6-15) VCE(max) = 5.5 V − (266 μA)(446.5 Ω) VCE(max) = 5.38 V Answer: The maximum collector to ground voltage is 5.38 V, and the minimum is 3.95 V. 6-32a. Given: VCC = 20 V VBB = 10 V RB = 33 kΩ RC = 3.3 kΩ hFE = βdc = 100 Solution: IC(sat) = VCC/RC (Eq. 6-11) IC(sat) = 20 V/3.3 kΩ IC(sat) = 6.06 mA IB = (VBB − VBE)/RB (Eq. 6-13) IB = (10 − 0.7)/33 kΩ

IB = 281.8 μA IC = βdcIB (Eq. 6-3) IC = 100(281.8 μA)

IC = 28.18 mA Answer: The transistor is in saturation because the calculated collector current is greater than the saturation current.

6-32b. Given: VCC = 20 V VBB = 5 V

RB = 1 MΩ RC = 3.3 kΩ hFE = βdc = 100 Solution: IC(sat) = VCC/RC (Eq. 6-11) IC(sat) = 20 V/3.3 kΩ IC(sat) = 6.06 mA IB = (VBB − VBE)/RB (Eq. 6-13) IB = (5 − 0.7)/1 MΩ IB = 4.3 μA IC = βdcIB (Eq. 6-3) IC = 200(4.3 μA)

IC = 860 μA Answer: The transistor is not in saturation because the calculated collector current is less than the saturation current.

6-32c. Given: VCC = 20 V VBB = 10 V

RB = 1 MΩ RC = 10 kΩ hFE = βdc = 50 Solution: IC(sat) = VCC/RC (Eq. 6-11) IC(sat) = 20 V/10 kΩ IC(sat) = 2 mA IB = (VBB − VBE)/RB (Eq. 6-13) IB = (10 − 0.7)/1 MΩ IB = 9.3 μA

IC = βdcIB

(Eq. 6-3)

IC = 50(9.3 μA)

IC = 465 μA Answer: The transistor is not in saturation because the calculated collector current is less than the saturation current.

6-32d. Given: VCC = 10 V VBB = 10 V

RB = 1 MΩ RC = 3.3 kΩ hFE = βdc = 100 Solution: IC(sat) = VCC/RC (Eq. 6-11) IC(sat) = 10 V/3.3 kΩ IC(sat) = 3.03 mA IB = (VBB − VBE)/RB (Eq. 6-13) IB = (10 V − 0.7 V)/1 MΩ IB = 9.3 μA IC = βdcIB (Eq. 6-3) IC = 100(9.3 μA)

IC = 930 μA Answer: The transistor is not in saturation because the calculated collector current is less than the saturation current.

6-33a. Given: VCC = 5 VBB = 5 V RB = 51 kΩ RC = 470 Ω hFE = βdc = 100 Solution: IC(sat) = VCC/RC (Eq. 6-11) IC(sat) = 5 V/470 Ω IC(sat) = 10.64 mA IB = (VBB − VBE)/RB (Eq. 6-13) IB = (5 V − 0.7 V)/51 kΩ IB = 84.3 μA IC = βdcIB (Eq. 6-3) IC = 100(84.3 μA)

IC = 8.43 mA Answer: The transistor is not in saturation because the calculated current is less than the saturation current.

6-33b. Given: VCC = 5 V VBB = 10 V RB = 680 kΩ RC = 470 Ω hFE = βdc = 500 Solution: IC(sat) = VCC/RC (Eq. 6-11) IC(sat) = 5 V/470 Ω IC(sat) = 10.64 mA IB = (VBB − VBE)/RB (Eq. 6-13) IB = (10 V − 0.7 V)/680 kΩ IB = 13.68 μA IC = βdcIB (Eq. 6-3) IC = 500(13.68 μA)

IC = 6.84 mA

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Answer: The transistor is not in saturation because the calculated collector current is less than the saturation current.

6-33c. Given: VCC = 5 V VBB = 5 V RB = 680 kΩ RC = 10 kΩ hFE = βdc = 100 Solution: IC(sat) = VCC/RC (Eq. 6-11) IC(sat) = 5 V/10 kΩ IC(sat) = 0.5 mA IB = (VBB − VBE)/RB (Eq. 6-13) IB = (5 − 0.7)/680 kΩ

IB = 6.32 μA IC = βdcIB

(Eq. 6-3)

CRITICAL THINKING 6-36. Given: βdc = 200 Solution: αdc = IC /IE

(Eq. 6-2)

IC = αdcIE βdc = IC /IB

(Eq. 6-3)

IE = IC + IB (Eq. 6-1) Substitute Eq. (6-2) for IC: βdc = αdcIE /IB Substitute Eq. (6-1) for IE: βdc = αdc(IC + IB)/IB Distribute αdc: βdc = (αdcIC + αdcIB)/IB βdc = αdcIC /IB + αdcIB /IB Simplify: βdc = αdcIC /IB + αdc Substitute Eq. (6-3) for IC /IB: βdc = αdcβdc + αdc

IC = 100(6.32 μA)

Factor out the αdc: βdc = αdc( βdc + 1)

IC = 632 μA

Solve for αdc: αdc = βdc/(βdc + 1) αdc = 200/(200 + 1) αdc = 0.995 Answer: The αdc is 0.995.

Answer: The transistor is in saturation because the calculated collector is greater than the saturation current.

6-33d. Given: VCC = 10 V VBB = 5 V RB = 680 kΩ RC = 470Ω hFE = βdc = 100 Solution: IC(sat) = VCC/RC (Eq. 6-11) IC(sat) = 10 V/470 Ω IC(sat) = 21.28 mA IB = (VBB − VBE)/RB (Eq. 6-13) IB = (5 V − 0.7 V)/680 kΩ

IB = 6.32 μA IC = βdcIB

(Eq. 6-3)

IC = 100(6.32 μA)

IC = 632 μA Answer: The transistor is not in saturation because the calculated current is less than the saturation current.

6-34. Answer: With the switch open, the collector voltage is 5 V, and with the switch closed, the collector voltage is 0 V. 6-35. 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. c. Increase: With the collector resistor shorted, it is the only thing in the circuit with the source and will read source voltage at all times. d. Decrease: There will be no voltage present at the collector or emitter. e. Increase: With the base supply gone, the transistor goes into cutoff and source voltage is read from collector to emitter. f. Decrease: There will be no voltage present at the collector or emitter.

6-37. Given: αdc = 0.994 Solution: From the previous solution: βdc = αdcβdc + αdc βdc – αdcβdc = αdc Factor out the βdc: βdc(1 – αdc) = αdc Solve for βdc: βdc = αdc/(1 – αdc) βdc = 0.994/(1 – 0.994) βdc = 165.67 Answer: The βdc is 165.67.

6-38. Given: VBB = 5 V VCC = 15 V hFE = βdc = 120 IC = 10 mA VCE = 7.5 V VBE = 0.7 V Solution: βdc = IC /IB (Eq. 6-3) IB = IC/βdc IB = 10 mA/120 IB = 83.33 μA IB = (VBB – VBE)RB (Eq. 6-6) RB = (VBB – VBE)/IB RB = (5 V – 0.7 V)/83.33 μA

RB = 51.6 kΩ VCE = VCC – ICRC (Eq. 6-7) RC = (VCC – VCE)/IC RC = (15 V – 7.5 V)/10 mA RC = 750 Ω Answer: The base resistor needs to be 51.6 Ω, and the collector resistor needs to be 750 Ω. Note: These may not be standard values, so it may take more than one resistor, or a potentiometer may be used.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

VBE = 0.7 V (second approximation) RC = 820 Ω VCC = 10 V βdc = 200

6-39. Given: VBB = 10 V VCE = 6.7 V VBE = 0.7 V (second approximation)

VBE = 0 V (ideal) RC = 820 Ω

Solution: IB = [(VBB – VBE)/RB] (Eq. 6-6) IB = [(10 V – 0.7 V)/470 kΩ] IB = 19.8 μA IC = βdcIB

VCC = 10 V

βdc = 200 Solution: Ideal VCC = VRC + VCE VRC = VCC – VCE VRC = 10 V – 6.7 V VRC = 3.3 V IC = VRC/RC IC = 3.3 V/820 Ω IC = 4 mA

IB = IC/βdc IB = 4 mA/200 IB = 20.1 μA IB = [(VBB – VBE)/RB] RB = [(VBB – VBE)/IB]

RB = [(10 V – 0 V)/20.1 μA]

RB = 497.5 kΩ 2nd Approximation VCC = VRC + VCE VRC = VCC – VCE VRC = 10 V – 6.7 V VRC = 3.3 V IC = VRC/RC IC = 3.3 V/820 Ω IC = 4 mA

IB = IC/βdc IB = 4 mA/200 IB = 20.1 μA IB = [(VBB – VBE)/RB] RB = [(VBB – VBE)/IB] RB = [(10 V – 0.7 V)/20.1 μA] RB = 462.69 kΩ Answer: (Ideal) RB = 497.5 kΩ, (2nd Approximation) RB = 462.69 kΩ.

6-40. Given: PD = 350 mW @25°C T = 50°C VCE = 10 V Solution: ΔT = 50°C – 25°C ΔT = 25°C ΔP = ΔT (derating factor) ΔP = 25°C (2.8 mW/°C) ΔP = 70 mW

IC = 200(19.8 μA)

IC = 3.96 mA Answer: The LED current is 3.96 mA. 6-42. Answer: VCE(sat) = 0.3 V 6-43. No collector supply 6-44. RC is shorted 6-45. RB is shorted 6-46. No base supply 6-47. RB is open 6-48. Base bias; common emitter 6-49. Synch input; clock output 6-50. 5 mA 6-51. Saturated 6-52. +5 V

Chapter 7 SELF-TEST 1. b 2. b 3. d 4. b 5. a 6. a 7. c 8. a 9. c 10. d

11. a 12. a 13. d 14. b 15. b 16. b 17. a 18. c 19. a 20. b

21. a 22. c 23. c 24. c 25. a 26. b 27. a 28. d 29. a 30. c

31. a 32. d 33. b 34. b 35. c 36. b 37. c 38. a 39. d

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

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.

PD(max) = 350 mW – 70 mW PD(max) = 280 mW PD = VCEIC IC = PD/VCE IC = 280 mW/10 V IC = 28 mA

PROBLEMS

Answer: The maximum collector current is 28 mA.

7-1.

6-41. Given: VBB = 10 V RB = 470 kΩ

BJT Biasing

10. The circuit will be highly sensitive to changes in current gain.

Given:

VBB = 2.5 V VCC = 20 V

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RC = 10 kΩ RE = 1.8 kΩ VBE = 0.7 V Solution: VE = VBB − VBE (Eq. 7-1) VE = 2.5 V − 0.7 V VE = 1.8 V IE = VE/RE (Ohm’s law) IE = 1.8 V/1.8 kΩ IE = 1 mA IE ≈ IC VC = VCC − ICRC (Kirchhoff’s law)

RC = 910 Ω RE = 180 Ω VBE = 0.7 V Solution: VE = VBB − VBE (Eq. 7-1) VE = 2 V − 0.7 V VE = 1.3 V IE = VE/RE (Ohm’s law) IE = 1.3 V/180 Ω IE = 7.22 mA

IE ≈ IC VC = VCC − ICRC (Kirchhoff’s law) VC = 10 V − (7.22 mA)(910 Ω) VC = 3.43 V Answer: The collector voltage is 3.43 V.

VC = 20 V − (1 mA)(10 kΩ)

VC = 10 V Answer: The collector voltage is 10 V, and the emitter voltage is 1.8 V. 7-2.

7-5.

Given:

VBB = 2.3 V VCC = 10 V RC = 910 Ω RE = 360 Ω VBE = 0.7 V

VBB = 2.5 V VCC = 20 V RC = 10 kΩ RE = 3.6 kΩ VBE = 0.7 V Solution: VE = VBB − VBE (Eq. 7-1) VE = 2.5 V − 0.7 V VE = 1.8 V IE = VE /RE (Ohm’s law) IE = 1.8 V/3.6 kΩ

Solution: VE = VBB − VBE (Eq. 7-1) VE = 2.3 V − 0.7 V VE = 1.6 V IE = VE /RE (Ohm’s law) IE = 1.6 V/360 Ω IE = 4.44 mA

IE = 0.5 mA

IE ≈ IC VC = VCC − ICRC

IE ≈ IC VC = VCC − ICRC

(Kirchhoff’s law)

VC = 5.96 V VCE = VC −VE

VCE = 15 V − 1.8 V

7-3.

Given:

VBB = 2.5 V VCC = 15 V RC = 10 kΩ RE = 1.8 kΩ VBE = 0.7 V Solution: VE = VBB − VBE (Eq. 7-1) VE = 2.5 V −0.7 V VE = 1.8 V IE = VE/RE (Ohm’s law) IE = 1.8 V/1.8 kΩ IE = 1 mA IE ≈ IC VC = VCC − ICRC (Kirchhoff’s law) VC = 15 V − (1 mA)(10 kΩ) VC = 5 V

7-4.

(Eq. 7-2)

VCE = 5.96 V − 1.6 V

(Eq. 7-2)

VCE = 13.2 V Answer: The collector-emitter voltage is 13.2 V.

(Kirchhoff’s law)

VC = 10 V − (4.44 mA)(910 Ω)

VC = 20 V − (0.5 mA)(10 kΩ)

VC = 15 V VCE = VC − VE

Given:

VCE = 4.36 V Answer: The collector-emitter voltage is 4.36 V.

7-6.

Given:

VBB = 1.8 V VCC = 15 V RC = 910 Ω RE = 180 Ω VBE = 0.7 V Solution: VE = VBB −VBE (Eq. 7-1) VE = 1.8 V −0.7 V VE = 1.1 V IE = VE/RE (Ohm’s law) IE = 1.1 V/180 Ω IE = 6.11 mA

IE ≈ IC VC = VCC − ICRC

(Kirchhoff’s law)

VC = 15 V − (6.11 mA)(910 Ω)

VC = 9.44 V VCE = VC − VE

(Eq. 7-2)

VCE = 9.44 V − 1.1 V

Answer: The collector voltage is 5 V.

VCE = 8.34 V

Given:

Answer: The collector-emitter voltage is 8.34 V.

VBB = 2 V VCC = 10 V Copyright 2021 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

7-7.

Given:

VCC = 5 V VBB = 2 V RE = 100 Ω Solution: VE = VBB −VBE (Eq. 7-1) VE = 2 V − 0.7 V VE = 1.3 V IE = VE /RE (Ohm’s law) IE = 1.3 V/100 Ω IE = 13 mA IE ≈ IC = ID Answer: The diode current is 13 mA. 7-8.

Given:

VCC = 5 V VBB = 1.8 V RE = 100 Ω VD ≈ 2 V Solution: VE = VBB − VBE (Eq. 7-1) VE = 1.8 V− 0.7 V VE = 1.1 V IE = VE /RE (Ohm’s law) IE = 1.1 V/100 Ω IE = 11 mA IE ≈ IC = ID VC = VCC −VD VC = 5 V −2 V VC = 3 V Answer: The diode current is 11 mA, and the collector voltage is 3 V. 7-9.

Answer: RC could be shorted; the transistor could be open collector-emitter; RB could be open, keeping the transistor in cutoff; open in the base circuit; open in the emitter circuit.

7-10. Answer: With the ground open, the base would read VBB and the collector would read VCC because source voltage is read above an open.

7-11. Answer: Shorted transistor; RB value very low; VBB too high.

7-12. Answer: RC could be shorted; the transistor could be open collector-emitter; RB could be open, keeping the transistor in cutoff; RE could be open; open in the base circuit; open in the emitter circuit.

7-13. Answer: With the emitter resistor open, the base would read VBB and the collector would read VCC because source voltage is read above an open.

7-14. Answer: Shorted transistor collector-emitter because the emitter voltage should be 1.1 V; open collector resistor; loss of VCC. 7-15. Given: R1 = 10 kΩ R2 = 2.2 kΩ RC = 3.6 kΩ

RE = 1 kΩ VCC = 25 V VBE = 0.7 V

Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 7-4) VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]25 V VBB = 4.51 V VE = VBB – VBE (Eq. 7-5) VE = 4.51 V – 0.7 V VE = 3.81 V IE = VE/RE (Eq. 7-6) IE = 3.81 V/1 kΩ IE = 3.81 mA IC ≈ IE (Eq. 7-7) VC = VCC – ICRC

(Eq. 7-8)

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. 7-16. 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. 7-4) VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]15 V VBB = 2.7 V VE = VBB – VBE (Eq. 7-5) VE = 2.7 V – 0.7 V VE = 2.0 V IE = VE/R3 (Eq. 7-6) IE = 2.0 V/1 kΩ IE = 2 mA IC ≈ IE (Eq. 7-7) VC = VCC – ICRC (Eq. 7-8) 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. 7-17. Given: R1 = 330 kΩ R2 = 100 kΩ RC = 150 kΩ RE = 51 kΩ VCC = 10 V VBE = 0.7 V Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 7-4) VBB = [100 kΩ/(330 kΩ + 100 kΩ)]10 V VBB = 2.33 V VE = VBB – VBE (Eq. 7-5) VE = 2.33 V – 0.7 V VE = 1.63 V IE = VE/RE (Eq. 7-6) IE = 1.63 V/51 kΩ IE = 31.96 μA IC ≈ IE (Eq. 7-7)

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VC = VCC – ICRC (Eq. 7-8) VC = 10 V – (31.96 μA)(150 kΩ) VC = 5.21 V Answer: The emitter voltage is 1.63 V, and the collector voltage is 5.21 V. 7-18. Given: R1 = 150 Ω R2 = 33 Ω RC = 39 Ω RE = 10 Ω VCC = 12 V VBE = 0.7 V

7-20. Given: R1 = 150 Ω R2 = 33 Ω RC = 39 Ω RE = 10 Ω VCC = 12 V ± 10% VBE = 0.7 V

Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 7-4) VBB = [33 Ω/(150 Ω + 33 Ω)]12 V VBB = 2.16 V VE = VBB – VBE (Eq. 7-5) VE = 2.16 V – 0.7 V VE = 1.46 V IE = VE/RE (Eq. 7-6) IE = 1.46 V/10 Ω = 146 mA IC ≈ IE (Eq. 7-7) VC = VCC – ICRC

VC(min) = VCC – IC(max)RC(max) (Eq. 7-8) VC(min) = 10 V – (37.36 μA)(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.

(Eq. 7-8)

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. 7-19. 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(min) = [R2(min)/(R1(max) + R2(min))]VCC (Eq. 7-4) VBB(min) = [95 kΩ/(346.5 kΩ + 95 kΩ)]10 V VBB(min) = 2.15 V VE(max) = VBB(max) – VBB (Eq. 7-5) VE(max) = 2.51 V – 0.7 V VE(max) = 1.81 V (Eq. 7-5)

VC(min) = VCC(min) – IC(max)RC (Eq. 7-8) 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. 7-21. 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

IE(max) = VE(max)/RE(min) (Eq. 7-6) IE(max) = 1.81 V/48.45 kΩ IE(max) = 37.36 μA IE(max) = VE(max)/RE(min) IE(min) = 1.45 V/53.55 kΩ IE(min) = 27.08 μA IC ≈ IE (Eq. 7-7)

VE(min) = VBB(min) – VBE (Eq. 7-5) VE(min) = 1.95 V – 0.7 V VE(min) = 1.25 V IE(min) = VE(min)/RE (Eq. 7-6) IE(min) = 1.25 V/10 Ω IE(min) = 125 mA IC ≈ IE (Eq. 7-7) VC(max) = VCC(max) – IC(min)RC (Eq. 7-8) VC(max) = 13.2 V – (125 mA)(39 Ω) VC(max) = 8.33 V

VBB(max) = [R2(max)/(R1(min) + R2(max))]VCC (Eq. 7-4) VBB(max) = [105 kΩ/(313.5 kΩ + 105 kΩ)]10 V VBB(max) = 2.51 V

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

Solution: VBB(max) = [R2/(R1 + R2)]VCC(max) (Eq. 7-4) VBB(max) = [33 Ω/(150 Ω + 33 Ω)]13.2 V VBB(max) = 2.38 V VE(max) = VBB(max) – VBE (Eq. 7-5) VE(max) = 2.38 V – 0.7 V VE(max) = 1.68 V IE(max) = VE(max)/RE (Eq. 7-6) IE(max) = 1.68 V/10 Ω IE(max) = 168 mA VBB(min) = [R2/(R1 + R2)]VCC(min) (Eq. 7-4) VBB(min) = [33 Ω/(150 Ω + 33 Ω)]10.8 V VBB(min) = 1.95 V

(Eq. 7-6)

(from Prob. 7-15)

VE = 3.81 V (from Prob. 7-15) IE = IC = 3.81 mA (from Prob. 7-15) VC = 11.28 V (from Prob. 7-15) Solution: VCE = VC – VE (Eq. 7-9) 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.

VC(max) = VCC – IC(min)RC(min) (Eq. 7-8) VC(max) = 10 V – (27.08 μA)(142.5 kΩ) VC(max) = 6.14 V

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

7-22. Given: R1 = 10 kΩ R2 = 2.2 kΩ

IC ≈ IE

VC = VCC – ICRC

RE = 1 kΩ VCC = 15 V VBE = 0.7 V Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 7-4) VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]15 V VBB = 2.7 V VE = VBB – VBE (Eq. 7-5) VE = 2.7 V – 0.7 V VE = 2.0 V IE = VE/RE (Eq. 7-6) IE = 2.0 V/1 kΩ IE = 2 mA IC ≈ IE (Eq. 7-7) VC = VCC – ICRC

VC = 6.3 V VCE = VC – VE

VCE = 4.85 V Answer: The Q point is IC = 146 mA, and VCE = 4.85 V. 7-25. 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. 7-4) VBB(max) = [105 kΩ/(313.5 kΩ + 105 kΩ)]10 V VBB(max) = 2.51 V

(Eq. 7-8)

VC = 9.59 V VCE = VC – VE

VBB(min) = [R2(min)/(R1(max) + R2(min))]VCC (Eq. 7-4) VBB(min) = [95 kΩ/(346.5 kΩ + 95 kΩ)]10 V VBB(min) = 2.15 V

(Eq. 7-9)

VCE = 9.59 V – 2.0 V

VCE = 7.59 V Answer: The Q point is IC = 2 mA, and VCE = 7.59 V.

VE(max) = VBB(max) – VBE VE(max) = 2.51 V – 0.7 V VE(max) = 1.81 V VE(min) = VBB(min) – VBE VE(min) = 2.15 V – 0.7 V VE(min) = 1.45 V

(from Prob. 7-17)

(Eq. 7-9)

7-24. 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. 7-4) VBB = [33 Ω/(150 Ω + 33 Ω)]12 V VBB = 2.16 V VE = VBB – VBE (Eq. 7-5) VE = 2.16 V – 0.7 V VE = 1.46 V IE = VE/RE (Eq. 7-6) IE = 1.46 V/10 Ω IE = 146 mA

(Eq. 7-5)

IE(min) = VE(min)/RE(max) (Eq. 7-6) IE(min) = 1.45 V/53.55 kΩ IE(min) = 27.08 μA IC ≈ IE (Eq. 7-7) Answer: The lowest collector current is 27.08 μA, and the highest collector current is 37.36 μA.

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.

(Eq. 7-5)

IE(max) = VE(max)/RE(min) (Eq. 7-6) IE(max) = 1.81 V/48.45 kΩ IE(max) = 37.36 μA

VE = 1.63 V (from Prob. 7-17) IE = IC = 31.96 μA (from Prob. 7-17) VC = 5.21 V (from Prob. 7-17) Solution: VCE = VC – VE

(Eq. 7-9)

VCE = 6.3 V – 1.46 V

VC = 15 V – (2 mA)(2.7 kΩ)

VBB = 2.33 V

(Eq. 7-8)

VC = 12 V – (146 mA)(39 Ω)

RC = 2.7 kΩ

7-23. Given: R1 = 330 kΩ R2 = 100 kΩ RC = 150 kΩ RE = 51 kΩ VCC = 10 V VBE = 0.7 V

(Eq. 7-7)

7-26.

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. 7-4) VBB(max) = [33 Ω/(150 Ω + 33 Ω)]13.2 V VBB(max) = 2.38 V VE(max) = VBB(max) – VBE (Eq. 7-5) VE(max) = 2.38 V – 0.7 V VE(max) = 1.68 V IE(max) = VE(max)/RE (Eq. 7-6) IE(max) = 1.68 V/10 Ω IE(max) = 168 mA VBB(min) = [R2/(R1 + R2)]VCC(min) (Eq. 7-4) VBB(min) = [33 Ω/(150 Ω + 33 Ω)]10.8 V VBB(min) = 1.95 V

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VE(min) = VBB(min) – VBE (Eq. 7-5) VE(min) = 1.95 V – 0.7 V VE(min) = 1.25 V IE(min) = VE(min)/RE (Eq. 7-6) 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. 7-27. 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. 7-28. 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. 7-29. 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) (Eq. 7-11) IE(max) = [–0.7 V – (–12 V)]/9.5 kΩ IE(max) = 1.19 mA VC(max) = VCC – IC(min)RC(min) (Eq. 7-12) VC(max) = 12 V – (1.08 mA)(4465 Ω) VC(max) = 7.18 V IE(min) = (–0.7 V – VEE)/RE(max) (Eq. 7-11) IE(min) = [–0.7 V – (–12 V)]/10.5 kΩ IE(min) = 1.08 mA VC(min) = VCC – IC(max)RC(max) (Eq. 7-12) VC(min) = 12 V – (1.19 mA)(4935 Ω) VC(min) = 6.13 V Answer: The maximum collector voltage is 7.18 V. The minimum collector voltage is 6.13 V.

77.30. 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.

77.31. 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 VEE, 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.

77.32. 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.

77.33. 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 approxi-mate 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 7.26 V, the approximate value of collector voltage.

c. With RE open, there is no collector current and the col-lector voltage is zero.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

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 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.

7-34. Given: R1 = 10 kΩ R2 = 2.2 kΩ RE = 1 kΩ RC = 3.6 kΩ VEE = 10 V VBE = 0.7 V Solution: V2 = [R2/(R1 + R2)]VEE 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 IE = VRE/RE IE = 1.1 V/1 kΩ IE = 1.1 mA IC ≈ IE VC = ICRC

(Eq. 7-7)

VC = (1.1 mA)(3.6 kΩ)

VC = 3.96 V Answer: The collector voltage is 3.96 V. 7-35. Given: R1 = 10 kΩ R2 = 2.2 kΩ RE = 1 kΩ RC = 3.6 kΩ VEE = 10 V VBE = 0.7 V V2 = 1.8 V (from Prob. 7-34) VRE = 1.1 V (from Prob. 7-34) IE = 1.1 mA (from Prob. 7-34) VC = 3.96 V (from Prob. 7-34) 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.

7-36. Given: R1 = 10 kΩ R2 = 2.2 kΩ

RE = 1 kΩ

RC = 3.6 kΩ VEE = 10 V VBE = 0.7 V V2 = 1.8 V

(from Prob. 7-34)

VRE = 1.1 V (from Prob. 7-34) IE = 1.1 mA (from Prob. 7-34) VC = 3.96 V (from Prob. 7-34) 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.

IC = 8.9 V/RC IC = 8.9 V/3.6 kΩ IC = 2.47 mA 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. 7-37. Given: R1 = 10 kΩ R2 = 2.2 kΩ

RE = 1 kΩ RC = 3.6 kΩ VCC = –10 V VBE = 0.7 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 = VE/RE IE = 1.1 V/1 kΩ IE = 1.1 mA IC ≈ IE (Eq. 7-7) 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 7-38. The circuit is no longer considered stiff or independent of Beta. The base current is not small as compared to the voltage divider current.

7-39. The maximum power dissipation of the 2N3904 is 625 mW. The transistor is dissipating 705 mW. The transistor will probably overheat and fail.

7-40. As long as the voltmeter has a high enough input resistance, it should read approximately 4.83 V. 7-41. Increase the power supply value, short R1.

7-42. Connect an ammeter between the power supply and the circuit. Measure VR1 and VC, then calculate and add their respective currents.

7-43. Given: (for Q1): R1 = 1.8 kΩ R2 = 300 Ω

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RE = 240 Ω RC = 1 kΩ VCC = 15 V VBE = 0.7 V

VC = VCC – ICRC

Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 7-4) VBB = [300 Ω/(1.8 kΩ + 300 Ω)]15 V VBB = 2.14 V VE = VBB – 0.7 V (Eq. 7-5) VE = 2.14 V – 0.7 V

VE = 1.44 V IE = VE/RE

(Eq. 7-6)

IE = 1.44 V/240 Ω

IE = 6 mA IC ≈ IE

(Eq. 7-7)

VC = VCC – ICRC

(Eq. 7-8)

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

7-44. Given: R1 = 10 kΩ RE = 1 kΩ RC = 8.2 kΩ VCC = 20 V VD = 0.7 V Solution: VBB = 3(VD) VBB = 3(0.7 V) VBB = 2.1 V VE = VBB – 0.7 V (Eq. 7-5) VE = 2.1 V – 0.7 V VE = 1.4 V IE = VE/RE (Eq. 7-6) IE = 1.4 V/1 kΩ IE = 1.4 mA IC ≈ IE (Eq. 7-7) (Eq. 7-8)

VC = 20 V – (1.4 mA)(8.2 kΩ)

VE = 2.12 V – 0.7 V

VE = 1.42 V IE = VE/RE

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.

VC = VCC – ICRC

Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 7-4) VBB = [150 Ω/(910 Ω + 150 Ω)]15 V VBB = 2.12 V VE = VBB – 0.7 V (Eq. 7-5)

(Eq. 7-6)

IE = 1.42 V/120 Ω

IE = 11.83 mA IC ≈ IE (Eq. 7-7) VC = VCC – ICRC

(Eq. 7-8)

VC = 15 V – (11.83 mA)(510 Ω)

VC = 8.97 V Given (for Q3): R1 = 1 kΩ R2 = 180 Ω RE = 150 Ω RC = 620 Ω VCC = 15 V VBE = 0.7 V Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 7-4) VBB = [180 Ω/(1 kΩ + 180 Ω)]15 V VBB = 2.29 V VE = VBB – 0.7 V (Eq. 7-5) VE = 2.29 V – 0.7 V

VE = 1.59 V IE = VE/RE

(Eq. 7-8)

VC = 15 V – (10.6 mA)(620 Ω)

(Eq. 7-6)

IE = 1.59 V/150 Ω

IE = 10.6 mA IC ≈ IE (Eq. 7-7)

VC = 8.52 V Answer: The emitter current is 1.4 mA, and the collector voltage is 8.52 V. 7-45. 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 (Eq. 7-5) VE(1) = 2.0 V – 0.7 V VE(1) = 1.3 V IE(1) = VE/RE (Eq. 7-6) IE(1) = 1.3 V/200 Ω IE(1) = 6.5 mA IC ≈ IE (Eq. 7-7) VC(1) = VCC – ICRC (Eq. 7-8) VC(1) = 16 V – (6.5 mA)(1 kΩ) VC(1) = 9.5 V VC(1) = VBB(2) VE(2) = VBB(2) – 0.7 V (Eq. 7-5) VE(2) = 9.5 V – 0.7 V VE(2) = 8.8 V Answer: The output voltage is 8.8 V. 7-46. Given: R1 = 620 Ω R2 = 680 Ω RE = 200 Ω VEE = 12 V VBE = 0.7 V Solution: V2 = [R2/(R1 + R2)]VEE V2 = [680 Ω/(620 Ω + 680 Ω)]12 V V2 = 6.28 V

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

VRE = V2 – 0.7 V VRE = 6.28 V – 0.7 V VRE = 5.58 V IE = VRE/RE (Eq. 7-6) IE = 5.58 V/200 Ω IE = 27.9 mA ILED ≈ IE Answer: The LED current is 27.9 mA. 7-47. Given: R1 = 620 Ω RE = 200 Ω VEE = 12 V VBE = 0.7 V VZ = 6.2 V Solution: VRE = VZ – 0.7 V VRE = 6.2 V – 0.7 V VRE = 5.5 V IE = VRE/RE (Eq. 7-6) IE = 5.5 V/200 Ω IE = 27.5 mA ILED ≈ IE Answer: The LED current is 27.5 mA. 7-48. 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. 7-12) 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.

7-49. Answer: With VB at 10 V and R2 is good, the trouble is R1 shorted.

7-50. Answer: Since VB is 0.7 V and VE is 0 V, the trouble is RE is shorted.

7-51. Answer: 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.

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

7-52. 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 6: R2 is open.

7-53. Answer: 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. Trouble 8: R2 is shorted.

7-54. Answer: 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 emitterbase 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. 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.

7-55. Answer: Trouble 11: Since all the voltages are 0 V, the power supply is not working.

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.

7-56. R2 is shorted 7-57. RC is shorted 7-58. RE is shorted 7-59. No VCC 7-60. Transistor B-E open

Chapter 8

Basic BJT Amplifiers

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

11. b 12. d 13. b 14. b 15. d 16. b 17. c 18. b 19. b 20. c

21. a 22. c 23. b 24. c 25. d 26. c 27. c 28. b 29. c 30. c

31. b 32. a 33. a 34. b 35. c 36. b 37. a

JOB INTERVIEW QUESTIONS 4. To permit the output voltage to swing over the largest possible voltage when the input signal is large enough to produce a maximum output.

10. 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.

PROBLEMS 8-1.

Given:

C = 47 μF R = 10 kΩ Solution: XC = 1/(2πfC) XC < 0.1R (Eq. 8-1) 1/(2πfC) = 0.1R 1/(2πC) = (0.1R)f

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f = 1/{[2π(47 μF)][0.1(10 kΩ)]} f = 3.39 Hz

Solution: REQ = (R1 × R2)/(R1 + R2)

8-2.

Given:

C = 47 μF R = 1 kΩ Solution: XC = 1/(2πfC) XC < 0.1R

Answer: The lowest frequency where good coupling exists is 1.45 Hz.

(Eq. 8-1)

1/(2πfC) = 0.1R 1/(2πC) = (0.1R)f f = 1/{[2π(47 μF)][0.1(1 kΩ)]} f = 33.9 Hz

8-7.

Given:

REQ = (2.2 kΩ × 10 kΩ)/(2.2 kΩ + 10 kΩ)

C = 100 μF R = 10 kΩ

REQ = 1.8 kΩ XC = 1/(2πfC) XC < 0.1REQ (Eq. 8-1) 1/(2 πfC) = 0.1REQ f = 1/{[2π(47 μF)][0.1(1.8 kΩ)]} f = 18.8 Hz

Solution: XC = 1/(2πfC) XC < 0.1R

(Eq. 8-1)

1/(2πfC) = 0.1R 1/(2πC) = (0.1R)f f = 1/{[2π(100 μF)][0.1(10 kΩ)]} f = 1.59 Hz Answer: The lowest frequency where good coupling exists is 1.59 Hz. 8-4.

Answer: The lowest frequency where good coupling exists is 18.8 Hz. 8-8.

f = 100 Hz R = 10 kΩ Solution: XC = 1/(2πfC)

REQ = (2.2 kΩ × 10 kΩ)/(2.2 kΩ + 10 kΩ)

REQ = 1.8 kΩ XC = 1/(2πfC) XC < 0.1REQ (Eq. 8-1) 1/(2πfC) = 0.1R C = 1/[2π(1 kHz)(0.1)(1.8 kΩ)] C = 0.88 μF Answer: A capacitor value of 0.88 μF is required for good coupling.

(Eq. 8-1)

1/(2πfC) = 0.1R C = 1/[2π(100 Hz)(0.1)(10 kΩ)] C = 1.59 μF Answer: A capacitor value of 1.59 μF is required for good coupling.

8-5.

Given:

C = 220 μF R1 = 2.2 kΩ R2 = 10 kΩ Solution: REQ = (R1 × R2)/(R1 + R2)

8-9.

(Parallel resistance formula)

REQ = (2.2 kΩ × 10 kΩ)/(2.2 kΩ + 10 kΩ) REQ = 1.8 kΩ XC = 1/(2 πfC) XC < 0.1REQ (Eq. 8-1) 1/(2πfC) = 0.1REQ f = 1/{[2π(220 μF)][0.1(1.8 kΩ)]} f = 4 Hz Answer: The lowest frequency where good coupling exists is 4 Hz. 8-6.

Given:

C = 220 μF R1 = 10 kΩ R2 = 10 kΩ

Given:

f = 1 kHz R1 = 2.2 kΩ R2 = 10 kΩ Solution: REQ = (R1 × R2)/(R1 + R2) (Parallel resistance formula)

Given:

XC < 0.1R

Given:

C = 47 μF R1 = 2.2 kΩ R2 = 10 kΩ Solution: REQ = (R1 × R2)/(R1 + R2) (Parallel resistance formula)

Answer: The lowest frequency where good coupling exists is 33.9 Hz. 8-3.

(Parallel resistance formula)

REQ = (10 kΩ × 10 kΩ)/(10 kΩ + 10 kΩ) REQ = 5 kΩ XC = 1/(2πfC) XC < 0.1REQ (Eq. 8-1) 1/(2πfC) = 0.1REQ f =1/{[2π(220 μF)][0.1(5 kΩ)]} f = 1.45 Hz

Answer: The lowest frequency where good coupling exists is 3.39 Hz.

Given:

R1 = 1.5 kΩ R2 = 330 Ω RC = 1.2 kΩ RE = 470 Ω VCC = 15 V VBE = 0.7 V Solution: VBB = [R2/(R1 + R2)]VCC 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

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

ie(p-p) < 0.1 IEQ (Eq. 8-6) ie(p-p)max = 0.1 (4.26 mA) ie(p-p)max = 426 μA Answer: The maximum ac emiter current for small signal operation is 426 μA. 8-10. Given: R1 = 1.5 kΩR2 = 330 Ω RC = 1.2 kΩ RE = 940 Ω VCC = 15 V VBE = 0.7 V 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 (Eq. 7-6) IE = 2.0 V/940 Ω

IE = 2.128 mA ie(p-p) < 0.1 IEQ

RE = 470 Ω VCC = 15 V VBE = 0.7 V Solution: VBB = [R2/(R1 + R2)]VCC

r'e =25 mV/4.26 mA

r'e = 5.88 Ω Answer: The ac resistance of the emitter diode is 5.88 Ω.

8-15. Given: IE = 2.13 mA Solution: r'e = 25 mV/IE

(Eq. 8-6)

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

8-12. Given: β = 200 ib = 12.5 μA Solutions: β = ic/ib (Eq. 8-8) ic = βib ic = 200 (12.5μA) ic = 2.5 mA Answer: The ac collector current is 2.5 mA. 8-13. Given: β = 100 ic = 4 mA Solutions: β = ic/ib (Eq. 8-8) ib = ib ⁄β

(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 r'e = 25 mV/IE (Eq. 8-10)

(from Prob. 8-10) (Eq. 8-10)

r'e = 25 mV/2.13 mA

r'e = 11.7 Ω Answer: The ac resistance of the emitter diode is 11.7 Ω.

8-16. Given: r'e = 5.88 Ω β = 200 Solution: Zin(base) = βr'e

(from Prob. 8-14)

(Eq. 8-12)

Zin(base) = 200 (5.88 Ω)

Zin(base) = 1.18 kΩ Answer: The input impedance to the base is 1.18 kΩ.

8-17. Given: r'e = 11.7 Ω β = 200 Solution: Zin(base) = βr'e

(from Prob. 8-15)

(Eq. 8-12)

Zin(base) = 200 (11.7 Ω)

Zin(base) = 2.34 kΩ Answer: The input impedance to the base is 2.34 kΩ.

8-18. 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 8-16.

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

8-19. Answer:

ib = 4 mA/100 ib = 40 μA

Answer: The ac base current is 40 μA. 8-14. Given: R1 = 1.5 kΩ R2 = 330 Ω RC = 1.2 kΩ

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8-20. Answer:

8-24. Given: 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 8-21. Answer: min hfe = 50 max hfe = 200 Current is 1 mA Temperature 25°C 8-22. 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: r'e = (25 mV)/IE (Eq. 8-10) r'e = (25 mV)/5 mA r'e = 5 Ω r'e = hie/hfe r'e = 875 Ω/150

r'e = 5.83 Answer: The value of r'e is 5.83 Ω. The calculated value is larger than the ideal.

8-23. Given: R1 = 10 kΩ = 2.2 R2 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. 7-4) VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]10 V VBB = 1.8 V VE = VBB – VBE (Eq. 7-5) VE = 1.8 V – 0.7 V VE = 1.1 V IE = VE/RE (Eq. 7-6) IE = 1.1 V/1 kΩ = 1.1 IE mA r'e = (25 mV)/IE (Eq. 8-10) r'e = (25 mV)/1.1 mA

r'e = 22.7 Ω rc = RC || RL (Eq. 8-15) rc = 3.6 kΩ || 10 kΩ rc = 2.65 kΩ AV = rc/r'e (Eq. 8-16) AV = 2.65 kΩ/22.7 Ω AV = 117

vout = AV (vin) vout = 117(2 mV) vout = 234 mV Answer: The output voltage is 234 mV.

(from Prob. 8-23) (from Prob. 8-23) (from Prob. 8-23)

r'e = 22.7 Ω (from Prob. 8-23) Solution: rc = RC || RL rc = 3.6 kΩ || 5 kΩ rc = 2093 kΩ Av = rc/r'e Av = 2093 kΩ/22.7 Ω Av = 92.2 Answer: The voltage gain is 92.2.

8-25. 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 r'e = (25 mV)/IE r'e = (25 mV)/2 mA r'e = 12.5 Ω rc = RC || RL rc = 3.6 kΩ || 10 kΩ rc = 2.65 kΩ AV = rc/r'e 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. 8-26. 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

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

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 r'e = (25 mV)/IE r'e = (25 mV)/2 mA r'e = 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 Zin = R1 || R2 || βr'e Zin = 10 kΩ || 2.2 kΩ || 1.25 kΩ Zin = 738 Ω vin = [Zin/(RG + Zin)]vg vin = [738 Ω/(600 Ω + 738 Ω)]1 mV vin = 551.57 μV vout = AV(vin) vout = 212(551.57 μV) vout = 117 mV Answer: The voltage gain is 212, the output voltage is 117 mV. 8-27. Given: 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 = [R2/(R1 + R2)]VCC (Eq. 7-4) VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]10 V VBB = 1.8 V VE = VBB – VBE (Eq. 7-5) VE = 1.8 V – 0.7 V VE = 1.1 V IE = VE/RE (Eq. 7-6) IE = 1.1 V/2 kΩ IE = 0.55 mA r'e = (25 mV)/IE

(Eq. 8-10)

r'e = (25 mV)/0.55 mA

r'e = 45.5 Ω rc = RC || RL (Eq. 8-15) rc = 3.6 kΩ || 10 kΩ rc = 2.65 kΩ AV = rc / r'e (Eq. 8-16) AV = 2.65 kΩ/45.5 Ω AV = 58

Zin = R1 || R2 || βr'e Zin = 10 kΩ || 2.2 kΩ || 100(45.5 Ω) Zin = 1.29 kΩ vin = [Zin/(RG + Zin)]vg (Eq. 8-17) vin = [1.29 kΩ/(600 Ω + 1.29 kΩ)]1 mV vin = 0.683 mV vout = AV(vin) vout = 58(0.683 mV) vout = 39.6 mV Answer: The output voltage is 39.6 mV. 8-28. 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: VBB = [R2/(R1 + R2)]VCC 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 IE = VE/RE IE = 1.1 V/1 kΩ IE = 1.1 mA r'e = (25 mV)/IE r'e = (25 mV)/1.1 mA r'e = 22.7 Ω rc = RC || RL rc = 3.6 kΩ || 10 kΩ rc = 2.65 kΩ AV = rc/r'e AV = 2.65 kΩ/22.7 Ω AV = 117 Zin = R1 || R2 || βr'e Zin = 10 kΩ || 2.2 kΩ || 2.27 kΩ Zin = 1 kΩ vin = [Zin/(RG + Zin)]vg vin = [1 kΩ/(300 Ω + 1 kΩ)]1 mV vin = 769 μV vout = AV(vin) vout = 117(769 μV) vout = 90 mV Answer: The voltage gain is 117, the output voltage is 90 mV. 8-29. Given: 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

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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/(600 Ω + 1.64 kΩ)]25 mV vin = 18.3 mV vout = AV(vin) vout = 14.7(18.3 mV) vout = 269 mV Answer: The voltage gain is 14.7, the output voltage is 269 mV. 8-30. 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. 8-31. Given: 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 AV = rc/re AV = 1.8 kΩ/180 Ω AV = 10 Answer: The voltage gain is 10.

8-32. 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. 8-15) rc = 3.6 kΩ || 10 kΩ rc = 2.65 kΩ AV = rc/re (Eq. 8-19) AV = 2.65 kΩ/180 Ω AV = 14.7 Answer: The voltage gain is 14.7.

8-33. Answer: Since the capacitor is an open to direct current, the dc voltages do not change and the ac voltage gain will be drastically reduced.

8-34. Answer: Some of the possible causes are: open transistor, open emitter resistor, or open output coupling capacitor.

CRITICAL THINKING 8-35. 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.

8-36. 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.

8-37. Given: R = 30 Ω f = 20 Hz to 20 kHz Solution: XC = 1/(2πfC) XC < 0.1R

(Eq. 8-5)

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).

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

8-38. Given: R1 = 20 kΩ R2 = 4.4 kΩ

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 10 V. 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.

RC = 7.2 kΩ

RE = 2 kΩ RL = 20 kΩ VCC = 10 V VBE = 0.7 V Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 7-4) VBB = [4.4 kΩ/(20 kΩ + 4.4 kΩ)]10 V VBB = 1.8 V VE = VBB – VBE (Eq. 7-5) VE = 1.8 V – 0.7 V VE = 1.1 V IE = VE/RE (Eq. 7-6) IE = 1.1 V/2 kΩ IE = 0.55 mA re' = (25 mV)/IE (Eq. 8-10) re' = (25 mV)/0.55 mA

re' = 45.5 Ω rc = RC || RL (Eq. 8-15) rc = 7.2 kΩ || 20 kΩ rc = 5.3 kΩ AV = rc/r'e (Eq. 8-16) AV = 5.3 kΩ/45.5 Ω AV = 116

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

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

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.

8-41. Answers: 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.

Trouble 9: Since there are no dc voltages, the problem is no VCC.

Answer: The voltage gain is 116.

Trouble 10: Since the emitter voltage is 0 and the base voltage is near normal, the problem is an open BE diode.

8-39. Given: R1 = 20 kΩ R2 = 4.4 kΩ

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

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. 8-38) VE = 1.1 V (from Prob. 8-38) IE = 0.55 mA (from Prob. 8-38) re' = 45.5 Ω

(from Prob. 8-38)

re' = 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. 8-40. Answers: Trouble 1: Since all the ac voltages are 0, the problem could be the generator, RG open, or C1 open.

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

8-42. C3 is open 8-43. C1 is open 8-44. RE is shorted 8-45. Transistor is upside down.

8-46. No VCC

Chapter 9 Multistage, CC, and CB Amplifiers SELF-TEST 1.a 2. b 3. c 4. b 5. c 6. b 7. c 8. d 9. c

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

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

28. a 29. c 30. d 31. c 32. b 33. d 34. b 35. d 36. b

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

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1-42

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 9-1. 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

VE = 1.1 V IE = VE/RE (Eq. 7-6) IE = 1.1 V/1 kΩ

9-2. Given: 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

(Eq. 8-10)

r'e = 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 1.0 kΩ. = [Zin/(RG + Zin)]vg

(Eq. 8-17)

= [1.0 kΩ/(600 Ω + 1.0 kΩ)]1 mV

vin(1) = 0.625 mV The input impedance for the second stage is the load resistance for the first stage. rc = RC || RL (Eq. 8-15) rc = 3.6 kΩ || 1.0kΩ

rc = 783 Ω AV = rc/r'e (Eq. 8-16) 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) = AV(vin) vout(1) = 34.5(0.625 mV) vout(1) = 21.6 mV rc(2) = Rc || RL (Eq. 8-15) rc(2) = 3.6 kΩ || 10 kΩ rc(2) = 2.65 kΩ AV(2) = rc/r'e (Eq. 8-16)

AV(2) = 2.65 kΩ/22.7Ω AV(2) = 117

(Eq. 7-4) VBB = [R2/(R1 + R2)]VCC VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]12 V VBB = 2.16 V VE = VBB − VBE (Eq. 7-5) VE = 2.16 V – 0.7 V

r'e = (25 mV)/1.46 mA

r'e = (25 mV) /1.46 mA

vin(1)

voltage across the collector resistor is 2.53 V.

VE = 1.46 V IE = VE/RE (Eq. 7-6) IE = 1.46 V/1 kΩ IE = 1.46 mA r'e = (25 mV)/IE (Eq. 8-10)

VBB = 1.8 V VE = VBB – VBE (Eq. 7-5) VE = 1.8 V – 0.7 V

vin(1)

the base voltage of the second stage is 21.6 mV, and the

Solution:

Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 7-4) VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]10 V

IE = 1.1 mA r'e = (25 mV)/IE

vout(2) = AV(vin) vout(2) = 117(21.6 mV) vout(2) = 2.53 V Answer: The base voltage of the first stage is 0.625 mV,

r'e = 17.1 Ω Zin = R1 || R2 || βr'e Zin = 10 kΩ || 2.2 kΩ || 100(17.1 Ω) Zin = 878 Ω The input impedance for each stage is 878 Ω.

(Eq. 8-17) vin(1) = [Zin/(RG + Zin)]vg vin(1) = [878 Ω/ (600 Ω + 878 Ω)]1 mV vin(1) = 0.594 mV The input impedance for the second stage is the load resistance for the first stage.

(Eq. 8-15) rc = RC || RL rc = 3.6 kΩ || 878 kΩ rc = 706 Ω AV = rc/r'e

(Eq. 8-19)

AV = 706 Ω/17.1 Ω

AV = 41.3 The output voltage of the first stage is the input voltage for the second stage. vout(1) = Av(vin) vout(1) = 41.3(0.594 mV) vout(1) = 24.5 mV rc(2) = RC || RL (Eq. 8-15) rc(2) = 3.6 kΩ || 10 kΩ

rc(2) = 2.65 kΩ AV(2) = rc/r'e (Eq. 8-19) AV(2) = 2.65 kΩ/17.1 Ω AV(2) = 155 vout(2) = AV(vin) vout(2) = 155(24.5 mV) vout(2) = 3.80 V Answer: The output voltage is 3.80 V.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

9-3.

Given:

Solution: AV = rf /re (Eq. 9-2) AV = 5 kΩ/50 Ω AV = 100 Answer: The voltage gain is 100.

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 = 300

9-5.

re = 125 Ω AV = 100

Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 7-4) VBB = [2.2 kΩ/(10 kΩ + 2.2 kΩ)]10 V VBB = 1.8 V VE = VBB – VBE (Eq. 7-5) VE = 1.8 V – 0.7 V

Solution: AV = rf /re rf = 12.5 kΩ

Answer: The feedback resistor would need to be 12.5 kΩ.

9-6.

Given:

VE = 1.1 V IE = VE/RE (Eq. 7-6) IE = 1.1 V/1 kΩ

R1 = 2.2 kΩ

IE = 1.1 mA r'e = (25 mV)/IE

RG = 50 Ω

R2 = 2.2 kΩ

RE = 1 kΩ RL = 3.3 kΩ β = 200 VCC = 15 V VBE = 0.7 V

(Eq. 8-10)

r'e = 22.7 Ω Zin = R1 || R2 || βr'e 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. 8-17) 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. 8-15) rc = 3.6 kΩ || 1.43 kΩ rc = 1.02 kΩ AV = rc/r'e (Eq. 8-16) AV = 1.02 kΩ/22.7 Ω AV = 45

Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 7-4) VBB = [2.2 kΩ/(2.2 kΩ + 2.2 kΩ)]15 V VBB = 7.5 V VE = VBB – VBE (Eq. 7-5) VE = 7.5 V – 0.7 V VE = 6.8 V IE = VE/RE (Eq. 7-6) IE = 6.8 V/1 kΩ IE = 6.8 mA r'e = 25 mV/IE (Eq. 8-10) r'e = 25 mV/6.8 mA

r'e = 3.68 Ω re = RE || RL (Eq. 9-3) re = 1 kΩ || 3.3 kΩ re = 767 Ω Zin(base) = β(re + r'e) (Eq. 9-5)

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

Zin(base) = 200(767 Ω + 3.48 Ω)

vout(1) = AV(vin) vout(1) = 45(0.704 mV) vout(1) = 31.7 mV rc(2) = RC || RL (Eq. 8-15) rc(2) = 3.6 kΩ || 10 kΩ rc(2) = 2.65 kΩ AV(2) = rc/r'e (Eq. 8-16) AV(2) = 2.65 kΩ/22.7 Ω AV(2) = 117

vout(2) = AV(vin) vout(2) = 117(31.7 mV) vout(2) = 3.71 V Answer: The output voltage is 3.71 V. Given:

rf = 5 kΩ re = 50 Ω

(Eq. 9-2)

rf = 100(125 Ω)

r'e = (25 mV) /1.1 mA

9-4.

Given:

Zin(base) = 154 kΩ Zin(stage) = 154 kΩ || 2.2 kΩ || 2.2 kΩ = 1.09 kΩ Answer: The input impedance of the base is 154 kΩ, and the input impedance of the stage is 1.09 kΩ.

9-7.

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 r'e = 3.68 Ω (from Prob. 9-6) re = 767 Ω (from Prob. 9-6) Solution: Zin = R1 || R2 || β(re + r'e) (Eq. 9-6) Zin = 2.2 kΩ || 2.2 kΩ || 150(767 Ω + 3.48 Ω) Zin = 1.09 kΩ

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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. 9-8.

Given:

R1 = 2.2 kΩ R2 = 2.2 kΩ

RE = 1 kΩ RL = 3.3 kΩ RG = 50 Ω

β = 175 VCC = 15 V VBE = 0.7 V r'e = 3.68 Ω (from Prob. 9-6) re = 767 Ω (from Prob. 9-6) Solution: AV = re /(re + r'e) (Eq. 11-2) AV = 767 Ω/(767 Ω + 3.48 Ω) AV = 0.995 Zin = R1 || R2 || β(re + r'e) (Eq. 9-6) Zin = 2.2 kΩ || 2.2 kΩ || 175(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 vout = AV(vin) (Eq. 8-3) vout = (0.995)(0.956 V) vout = 0.951 V Answer: The gain is 0.995, and the output voltage is 0.951 V. 9-9.

Given: 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

β = 150 VCC = 15 V VBE = 0.7 V Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 7-4) VBB = [4.4 kΩ/(4.4 kΩ + 4.4 kΩ)]15 V VBB = 7.5 V VE = VBB – VBE (Eq. 7-5) VE = 7.5 V – 0.7 V VE = 6.8 V IE = VE/RE (Eq. 7-6) IE = 6.8 V/2 kΩ IE = 3.4 mA r'e = 25 mV/IE (Eq. 8-10) r'e = 25 mV/3.4 mA

r'e = 7.35 Ω re = RE || RL (Eq. 9-3) re = 2 kΩ || 6.6 kΩ re = 1.53 kΩ Zin = R1 || R2 || β(re + r'e) (Eq. 9-6) Zin = 4.4 kΩ || 4.4 kΩ || 150(1.53 kΩ + 7.35 kΩ) Zin = 2.18 kΩ 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.

RG = 50 Ω

(from Prob. 9-6)

re = 767 Ω (from Prob. 9-6) Solution: Zin(min) = R1 || R2 || β(re + r'e) (Eq. 9-6) 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 + r'e) (Eq. 9-6) Zin(max) = 2.2 kΩ || 2.2 kΩ || 300(767 Ω + 3.48 Ω) Zin(max) = 1.09 kΩ 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. 9-10. Given: R1 = 4.4 kΩ R2 = 4.4 kΩ

RG = 100 Ω

9-11. Given: R1 = 100 Ω R2 = 200 Ω RE = 30 Ω RL = 10 Ω

R1 = 2.2 kΩ

r'e = 3.68 Ω

RE = 2 kΩ RL = 6.6 kΩ

β = 200 VCC = 20 V VBE = 0.7 V Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 7-4) VBB = [200 Ω/(100 Ω + 200 Ω)]20 V VBB = 13.3 V VE = VBB – VBE (Eq. 7-5) VE = 13.3 V – 0.7 V VE = 12.6 V IE = VE/RE (Eq. 7-6) IE = 12.6 V/30 Ω IE = 420 mA r'e = 25 mV/IE (Eq. 8-10) r'e = 25 mV/420 mA

r'e = 0.06 Ω re = RE || RL (Eq. 9-3) re = 30 Ω || 10 Ω re = 7.5 Ω Zin(base) = β(re + r'e) (Eq. 9-5) Zin(base) = 200(7.5 Ω + 0.06 Ω) Zin(base) = 1.51 kΩ

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

Zin (stage) = R1 || R2 || β(re + r'e)

(Eq. 9-6)

Zin (stage) = 100 Ω || 200 Ω || 1.51 kΩ

Zin (stage) = 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.

9-12. 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 r'e = 0.06 Ω (from Prob. 9-11) re = 7.5 Ω (from Prob. 9-11) Solution: Zin = R1 || R2 || β(re + r'e) (Eq. 9-6) 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 Answer: The input voltage is 0.558 V. 9-13. 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 r'e = 0.06 Ω (from Prob. 9-11) re = 7.5 Ω (from Prob. 9-11) Solution: AV = re/(re + r'e) (Eq. 9-4) AV = 7.5/(7.5 + 0.06)

AV = 0.992 Zin = R1 || R2 || β(re + r'e) (Eq. 9-6) Zin = 100 Ω || 200 Ω || 175(7.5 Ω + 0.06 Ω) Zin = 63.5 Ω vin = [Zin/(Zin + RG)]VG vin = [63.5 Ω/(63.5 Ω + 50 Ω)]1 V vin = 0.559 V vout = Av(vin) (Eq. 8-3) vout = (0.992)(0.559 V) vout = 0.555 V Answer: The gain is 0.992, and the output voltage is 0.555 V. 9-14. Given: R1 = 2.2 kΩ R2 = 2.2 kΩ

(Eq. 9-7)

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.

9-15. 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 r'e = 0.06 Ω

(from Prob. 9-11)

re = 7.5 Ω (from Prob. 9-11) Solution: (Eq. 9-7) Zout = RE || [r'e + (RG || R1 || R2)/β] 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.

9-16. Given: Q2 β = 200 (ac & dc) R1 = 4.7 kΩ

R2= 1 kΩ RC = 1.5 kΩ RE = 330 Ω RL = 150 Ω VCC = 15 V Solution: VBB = [R2/(R1 + R2)]VCC VBB = [1 kΩ/(4.7 kΩ + 1 kΩ)]15 V VBB = 2.63 V VE = VBB − VBE VE = 2.63 V − 0.7 V VE = 1.93 V IE = VE/RE IE = 1.93 V/330 Ω IE = 5.85 mA r'e = (25 mV)/IE r'e = (25 mV)/5.85 mA r'e = 4.28 Ω ZinQ2 = βre2 ZinQ2 = 200 (150 Ω) ZinQ2 = 30 kΩ rc = RC || zinQ2 rc = 1.5 kΩ || 30 kΩ rc = 1.429 kΩ AV = rc/r'e AV = 1.429 kΩ/4.28 Ω AV = 335 Answer: The voltage gain is 335.

RE = 1 kΩ RL = 3.3 kΩ RG = 50 Ω

β = 200 VCC = 15 V VBE = 0.7V r'e = 3.68 Ω re = 767 Ω

Solution: Zout = RE || [r'e + (RG || R1 || R2)/β]

(from Prob. 9-6) (from Prob. 9-6)

9-17. Given: β = 150 (ac & dc) R1 = 4.7 kΩ R2 = 1 kΩ RC = 1.5 kΩ RE = 330 Ω RL = 150 Ω

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VCC = 15 V vg = 10 mV Solution: VBB = [R2/(R1 + R2)]VCC VBB = [1 kΩ/(4.7 kΩ + 1 kΩ)]15 V VBB = 2.63 V VE = VBB − VBE VE = 2.63 V − 0.7 V VE = 1.93 V IE = VE/RE IE = 1.93 V/330 Ω IE = 5.85 mA r'e = (25 mV)/IE r'e = (25 mV)/5.85 mA r'e = 4.28 Ω ZinQ2 = βRL ZinQ2 = 150 (150 Ω) ZinQ2 = 22.5 kΩ rc = RC || ZinQ2 rc = 1.5 kΩ || 22.5 kΩ rc = 1.406 kΩ AV1 = rc/r'e AV1 = 1.4 kΩ/4.28 Ω AV1 = 327 vout = AV1(vg) = 327(10 mV) = 3.27 V Answer: The 1st stage voltage gain is 327, the 2nd stage voltage gain is 1, vout = 3.27 V. 9-18. Given: β = 200 (ac & dc) R1 = 4.7 kΩ R2 = 1 kΩ RC = 1.5 kΩ RE = 330 Ω RL = 125 Ω VCC = 15 V Solution: VBB = [R2/(R1 + R2)]VCC VBB = [1 kΩ/(4.7 kΩ + 1 kΩ)]15 V VBB = 2.63 V VE = VBB − VBE VE = 2.63 V − 0.7 V VE = 1.93 V IE = VE/RE IE = 1.93 V/330 kΩ IE = 5.85 mA r'e = (25 mV)/IE r'e = (25 mV)/5.85 mA r'e = 4.28 Ω ZinQ2 = βRL ZinQ2 = 200 (125 Ω) ZinQ2 = 25 kΩ rc = RC || ZinQ2 rc = 1.5 kΩ || 25 kΩ rc = 1.415 kΩ AV1 = rc/r'e AV1 = 1.415 kΩ/4.28 Ω AV1 = 331

9-19. Given: β = 200 (ac & dc) R1 = 4.7 kΩ R2 = 1 kΩ RC = 1.5 kΩ RE = 330 Ω RL = 150 Ω VCC = 15 V Solution: VBB = [R2/(R1 + R2)]VCC VBB = [1 kΩ/(4.7 kΩ + 1 kΩ)]15 V VBB = 2.63 V VE = VBB − VBE VE = 2.63 V − 0.7 V VE = 1.93 V IE = VE /RE IE = 1.93 V/330 Ω IE = 5.85 mA r'e = (25 mV)/IE r'e = (25 mV)/5.85 mA r'e = 4.27 Ω rc = RC || RL rc = 1.5 kΩ || 150 Ω rc = 136.5 Ω AV1 = rc/r'e AV1 = 136.5 Ω/4.27 Ω AV1 = 31.9 Answer: The voltage gain drops to 31.9.

9-20. Given: R1 = 150 kΩ R2 = 150 kΩ RE = 470 Ω

RL = 1 kΩ RG = 5.1 kΩ VCC = 15 V β = 5000 Solution: re = RE || RL (Eq. 9-3) re = 470 Ω || 1 kΩ re = 320 Ω Zin(base) = βre Zin(base) = (5000)(320) Zin(base) = 1.6 MΩ Answer: The input impedance of the base is1.6 MΩ.

9-21. Given: 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. 9-20)

Solution: Zin(base) = βre Zin(base) = (7000) (320) Zin(base) = 2.24 MΩ

Answer: The voltage gain remains at 331.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

Zin = R1 || R2 || Zin(base) (Eq. 9-6) 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. 9-22. Given: R1 = 1 kΩ R2 = 2 kΩ RE = 10 Ω

RL = 8 Ω RG = 600 Ω VCC = 20 V β1 = 150 β2 = 150 Solution: re = RE || RL (Eq. 9-3) re = 10 Ω || 8 Ω re = 4.44 Ω β = β1β2 (Eq. 9-9) β = 150(150) β = 22500 Zin(base) = βre Zin(base) = (22500)(4.44 Ω) Zin(base) = 100 kΩ Answer: The input impedance of the base is 100 kΩ.

9-23. Given: R1 = 1 kΩ R2 = 2 kΩ

VBE = 0.7 V

RS = 1 kΩ VCC = 25 V Solution: Vout = VZ – VBE (Eq. 9-11) Vout = 7.5 V – 0.7 V Vout = 6.8 V Take the base current into account.

IZ = (VCC – VZ)/RS – 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 Answer: The output voltage is 6.8 V, and the zener current is 16.1 mA. 9-26. Given: With the wiper in the middle, the voltage divider is effectively two resistors: each has a value of 1.5 kΩ.

Vz = 7.5 V VBE = 0.7 V Solution: Vout = [(R3 + R4)/R4](VZ + VBE) (Eq. 9-14) Vout = [(1.5 kΩ + 1.5 kΩ)/1.5 kΩ](7.5 V + 0.7 V) Vout = 16.4 V 9-27. 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.

VZ = 7.5 V VBE = 0.7 V

(from Prob. 9-22)

Zin(base) = βre Zin(base) = (2000)(4.44 Ω) Zin(base) = 8.88 kΩ Zin = R1 || R2 || Zin(base) (Eq. 9-6) Zin = 1 kΩ || 2 kΩ || 8.88 kΩ Zin = 620 Ω 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. 9-24. Given: VZ = 7.5 V VBE = 0.7 V

9-25. Given: VZ = 7.5 V

Answer: The output voltage is 16.4 V.

RE = 10 Ω

RL = 8 Ω RG = 600 Ω VCC = 20 V β = 2000 re = 4.44 Ω Solution:

Answer: The output voltage is 6.8 V, and the zener current is 7.5 mA.

RS = 1 kΩ VCC = 15 V Solution: Vout = VZ – VBE (Eq. 9-11) Vout = 7.5 V – 0.7 V Vout = 6.8 V IZ = (VCC – VZ)/RS IZ = (15 V– 7.5 V)/1 kΩ IZ = 7.5 mA

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: Vout(top) = [(R3 + R4)/R4](VZ + VBE) (Eq. 9-14) 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. 9-14)

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. 9-28. Given: R1 = 10 kΩ R2 = 2 kΩ

RC = 3.3 kΩ

RE = 2 kΩ 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

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IE = VE/RE IE = 1.3 V/2 kΩ IE = 650 μA Answer: The emitter current is 650 μA. 9-29. Given: 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 r'e = (25 mV)/I E r'e = (25 mV)/650 μA r'e = 38.46 Ω rc = RC || RL rc = 3.3 kΩ || 10 kΩ rc = 2.48 kΩ AV = rc/r'e AV = 2.48 kΩ/38.46 Ω AV = 64.4 Answer: The voltage gain at 64.4.

9-30. Given: 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 r'e = (25 mV)/IE r'e = (25 mV)/650 μA r'e = 38.46 Ω = 38.5 Ω Zin(emitter) = r'e Zin(emitter) = 38.5 Ω Zin(stage) = RE || r'e Since RE >> r'e Zin(stage) ≅ r'e = 38.5 Ω Zout ≈ RC Zout = 3.3 kΩ Answer: The Zin(emitter) = 38.5 Ω, the Zin(stage) = r'e = 38.5 Ω.

9-31. Given: R1 = 10 kΩ R2 = 2 kΩ RC = 3.3 kΩ

RE = 2 kΩ RG = 50 Ω VCC = 12 V vg = 2 mV 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 r'e = (25 mV)/I E r'e = (25 mV)/650 μA r'e = 38.46 Ω re = RC || RL re = 3.3 kΩ || 10 kΩ re = 2.48 kΩ AV = rc/r'e AV = 2.48 kΩ/38.46 Ω AV = 64.4 Zin(stage) = RE || r'e Since RE >> r'e Zin(stage) = r'e = 38.5 Ω vin ≈ [Zin/(RG + Zin)]vg vin = [38.5 Ω/(50 Ω + 38.5 Ω)] 2 mV vin = 870 μV vout = AV(vin) vout = 64.4(870 μV) vout = 56 mV Answer: The output voltage is 56 mV.

9-32. Given: R1 = 10 kΩ R2 = 2 kΩ RC = 3.3 kΩ

RE = 2 kΩ RG = 50 Ω VCC = 15 V vg = 2 mV 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 IE = VE/RE IE = 1.8 V/2 kΩ IE = 900 μA r'e = (25 mV)/I E r'e = (25 mV) /900 μA r'e = 27.8 Ω

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

rc = RC || RL rc = 3.3 kΩ || 10 kΩ rc = 2.48 kΩ AV = rc/r'e AV = 2.48 kΩ/27.8 Ω AV = 89.3 Zin(stage) = RE || r'e Since RE >> r'e Zin(stage) = r'e = 27.8 Ω vin ≈ [Zin/(RG + Zin)]vg vin = [27.8 kΩ/(50 Ω + 27.8 Ω)]2 mV vin = 715 μV vout = AV(vin) vout = 89.3(715 μV) vout = 63.8 mV Answer: The output voltage is 63.8 mV.

CRITICAL THINKING 9-33. Given: VZ = 7.5 V VCC = 15 V Vout = 6.8 V RL = 33 Ω

(from Prob. 9-24)

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

VE = 2.79 V – 0.7 V

VE = 2.09 V IE = VE/RE

(Eq. 7-6)

r'e = 25 mV/IE

(Eq. 8-10)

r'e = 25 mV/0.95 mA

VE = 4.48 V – 0.7 V

(Eq. 7-6)

IE = 3.78 V/1 kΩ IE = 3.78 mA

IE = IC (Eq. 7-7) VC = VCC – ICRC (Eq. 7-18) VC = 15 V – 3.78 mA(1 kΩ) VC = 11.22 V IB = IC/β IB = 25.2 μA

9-36. Given: R1 = 33 kΩ R2 = 10 kΩ RC = 4.7 kΩ RE = 2.2 kΩ VCC = 12 V vin = 10 mV Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 7-4) VBB = [10 kΩ/(33 kΩ + 10 kΩ)]12 V VBB = 2.79 V VE = VBB – VBE (Eq. 7-5)

IE = 0.95 mA rc = 4.7 kΩ

Solution: VBB = [R2/(R1 + R2)] VCC (Eq. 7-4) VBB = [2 kΩ/(4.7 kΩ + 2 kΩ)]15 V VBB = 4.48 V VE = VBB – VBE (Eq. 7-5)

IB = 3.78 mA/150

Solution: rc = 1 kΩ re = 1 kΩ AV = rc/re (Eq. 8-19) AV = 1 kΩ/1 kΩ AV = 1 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.

IE = 2.09 V/2.2 kΩ

9-34. Given: R1 = 4.7 kΩ R2 = 2 kΩ RC = 1 kΩ RE = 1 kΩ VCC = 15 V β = 150

VE = 3.78 V IE = VE/RE

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. 9-35. Given: 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.

r'e = 26.3 Ω AV = rc/re'

(Eq. 8-16)

AV = 4.7 kΩ /26.3 Ω

AV = 179 vout = AV(vin) vout = 179(10 mV) vout = 1.79 V When the control voltage is 5 V, the control transistor is saturated and grounds the input; thus the output is zero.

Answer: With the control voltage at 0 V, the output is 1.79 V. With the control voltage at 5 V, the output is 0 V. This circuit could be a mute circuit.

9-37. Given: RL = 33 Ω RS = 1 kΩ VCC = 15 V βdc = 200

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vin = [815 Ω/(270 Ω + 815 Ω)](100 mVp-p) vin = 75 mVp-p vout = (75 mVp-p)(4.49) vout = 337 mVp-p Answer: The output voltage would be 337 mVp-p.

Solution: IE = (VCC – VBE)/(RL + (RS/βdc)) IE = (15 V – 0.7 V)/(33 Ω + (1 kΩ/200))

IE = 376 mA Vout = VE = IERE = (376 mA)(33 Ω) Vout = 12.4 V Answer: The output voltage is 12.4 V. 9-38. Given: RL = 33 Ω RS = 1 kΩ VCC = 15 V βdc = 100 Solution: IB = (VCC – VBE)/RS IB = (15 V – 0.7 V)/1 kΩ IB = 14.3 mA IE = βdcIB

9-42. Answer: The output would decrease to zero volts.

9-43. Given: Trouble 1: Since there is voltage at H and none at I, the trouble is an open C4.

Trouble 2: Since there is voltage at F and none at G, the trouble is an open between F and G. Trouble 3: Since there is voltage at A and none at B, the trouble is an open C1.

9-44. Answers: Trouble 4: Since E is not a ground, potential trouble is an open C3.

IE = (100)(14.3 mA)

IE = 1.43 A PD = ICVCE PD = (1.43 A)(15 V) PD = 21.5 W Answer: The transistor will dissipate 21.5 W and be destroyed.

9-39. Given: Wiper is at 50% RL = 100 Ω Solution: Vout = [(1.5 kΩ + 1.5 kΩ)/1.5 kΩ](8.2 V) Vout = 16.4 V Iout = 16.4 V/100 Ω Iout = 164 mA PD = (164 mA)(25 V – 16.4 V) PD = 1.41 W Answer: The power dissipation of Q2 is 1.41 W when the wiper is at 50%.

9-40. Given: βdc = 100 for both transistors Solution: IE = (7.5 V – 0.7 V – 0.7 V)/470 Ω IE = 13 mA r'e = 25 mV/13 mA r'e = 1.93 Ω Zout= r'e + [((5.1 kΩ || 150 kΩ || 150 kΩ)/10,000) || 470 Ω] Zout = (1.93 Ω + 0.478 Ω) || 470 Ω Zout = 2.4 Ω Answer: The output impedance is 2.4 Ω.

9-41. Given: vg = 100 mV Solution: AV1 = 1.5 kΩ/(4.27 Ω + 330 Ω) AV1 = 4.49 Zin = 4.7 kΩ || 1 kΩ || (200)(334) Zin = 815 Ω

Trouble 5: Since there is a voltage at B and none at C, the trouble is an open between B and C. Trouble 6: Since there is voltage at D and none at F, the trouble is an open C2.

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

9-45. C3 is shorted 9-46. C2 is open 9-47. Q1 B-E short 9-48. Q2 is open 9-49. R2 is shorted

Chapter 10

Power Amplifiers

SELF-TEST 1. b 2. b 3. c 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

JOB INTERVIEW QUESTIONS 6. Tuned RF amplifier. It would be impractical to use a ClassC 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.

PROBLEMS 10-1.

Given:

R1 = 2 kΩ R2 = 470 Ω RC = 680 Ω

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

RE = 220 Ω RL = 2.7 kΩ VCC = 15 V Solution: RC = 680 Ω IC(sat) = VCC/(RC + RE)

or 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

(Eq. 10-1)

IC(sat) = 15 V/(680 Ω + 220 Ω)

IC(sat) = 16.67 mA Answer: The dc collector resistance 680 Ω, and the dc saturation current is 16.67 mA. 10-2.

10-4.

VE = 2.85 V – 0.7 V

VE = 2.15 V IE = ICQ = VE/RE

(Eq. 7-6)

ICQ = 2.15 V/220 Ω

ICQ = 9.77 mA 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. Answer: The ac collector resistance is 543 Ω, and the ac saturation current is 19.5 mA. Given:

Solution: rc = RC || RL (Eq. 8-15) rc = 1.36 kΩ || 5.4 kΩ rc = 1086 Ω Answer: The ac collector resistance is 1086 Ω.

10-5.

Given:

R1 = 6 kΩ R2 =1.41 kΩ RC = 2.04 kΩ RE = 660 Ω RL = 8.1 kΩ VCC = 15 V VBE = 0.7 V RG = 150 Ω Solution: rc = RC || RL (Eq. 8-15) rc = 2.04 kΩ || 8.1 kΩ rc = 1.63 kΩ VBB = [R2/(R1 + R2)]VCC (Eq. 7-4) VBB = [1.41 kΩ/(6 kΩ + 1.41 kΩ)]15 V VBB = 2.85 V VE = VBB – VBE (Eq. 7-5) VE = 2.84 V – 0.7 V

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

Given:

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

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. 8-15) rc = 680 Ω || 2.7 kΩ rc = 543 Ω VBB = [R2/(R1 + R2)]VCC (Eq. 7-4) VBB = [470 Ω/(2 kΩ + 470 Ω)]15 V VBB = 2.85 V VE = VBB – VBE (Eq. 7-5)

10-3.

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

VE = 2.15 V IE = ICQ = VE/RE

(Eq. 7-6)

ICQ = 2.15 V/660 Ω

(from Prob. 10-2)

ICQ = 9.77 mA (from Prob. 10-2) VE = 2.15 V (from Prob. 10-2) Solution: VC = VCC – RCICQ VC = 15 V – (680 Ω)(9.77 mA)

VC = 8.36 V MP = ICQrc or VCEQ (Eq. 10-8) MP = (9.77 mA)(543 Ω) MP = 5.31 V

ICQ = 3.26 mA VC = VCC – RCICQ VC = 15 V – (2.04 kΩ)(3.26 mA) VC = 8.35 V MP = ICQrc or VCEQ (Eq. 10-8) MP = (3.26 mA)(1.63 kΩ) MP = 5.31 V or VCEQ = VC – VE VCEQ = 8.35 V – 2.15 V VCEQ = 6.2 V MPP = 2MP MPP = 2(5.31 V) MPP = 10.62 V Answer: The maximum peak-to-peak voltage is 10.62 V.

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10-6.

Given:

Solution: VC = VCC – RCICQ

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

10-7.

VC = 30 V – (100 Ω)(137 mA)

Solution: RC = 100 Ω

VC = 16.3 V VCEQ = VC – VE VCEQ = 16.3 V – 9.3 V VCEQ = 7 V MP = VCEQ = 7 V MPP = 2MP = 14 V

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.

or MP = ICQrc MP = (137 mA)(50 Ω) MP = 6.85 V MPP = 2MP = 13.7 V

Given:

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

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

10-9.

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 IE = ICQ = VE/RE ICQ = 9.3 V/68 Ω ICQ = 137 mA VC = VCC – ICQRC VC = 30 V – (137 mA)(100 Ω)

VC = 16.3 V VCEQ = VC – VE VCEQ = 16.3 V – 9.3 V VCEQ = 7 V ic(sat) = ICQ + VCEQ/rc ic(sat) = 137 mA + 7 V/50 Ω ic(sat) = 277 mA Answer: The ac collector resistance is 50 Ω, and the ac saturation current is 277 mA. 10-8.

Given:

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

Given:

R1 = 400 Ω R2 = 200 Ω RC = 200 Ω RE = 136 Ω RL = 200 Ω VCC = 30 V VBE = 0.7 V Solution: rc = RC || RL rc = 200 Ω || 200 Ω rc = 100 Ω Answer: The ac collector resistance is 100 Ω.

10-10. Given: 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 VBB = [300 Ω/(600 Ω + 300 Ω)]30 V VBB = 10 V VE = VBB – VBE VE = 10 V – 0.7 V VE = 9.3 V IE = ICQ = VE/RE ICQ = 9.3 V/204 Ω ICQ = 45.59 mA VC = VCC – RCICQ VC = 30 V – (300 Ω)(45.59 mA)

(from Prob. 10-7)

ICQ = 137 mA (from Prob. 10-7) ic(sat) = 277 mA (from Prob. 10-7) VE = 9.3 V (from Prob. 10-7)

VC = 16.3 V VCEQ = VC – VE VCEQ = 16.3 V – 9.3 V VCEQ = 7 V MP = VCEQ = 7 V MPP = 2MP = 14 V or MP = ICQrc MP = (45.59 mA) (150 Ω)

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

MP = 6.85 V MPP = 2MP = 13.7 V Answer: The maximum peak-to-peak voltage is 13.7 V.

10-11. Given: Pout = 2 W Pin = 4 mW Solution: Ap = Pout/Pin (Eq. 10-12) Ap = 2 W/4 mW Ap = 500 Answer: The power gain is 500.

10-12. Given: vout = 15 Vp-p RL = 1 kΩ Pin = 400 μW Solution: 2 Pout = V /8RL 2 Pout = (15 V) /8 kΩ Pout = 28.1 mW Ap = Pout/Pin (Eq. 10-12) Ap = 28.1 mW/400 μW

Ap = 70.3 Answer: The power gain is 70.3.

10-13. 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. 10-2) Solution: Ibias = VCC/(R1 + R2) Ibias = 15 V/(2 kΩ + 470 Ω) Ibias = 6.07 mA VE = VBB – VBE (Eq. 7-5) VE = 2.85 V – 0.7 V VE = 2.15 V IE = VE/RE (Eq. 7-6) IE = 2.15 V/220 Ω IE = 9.77 mA Idc = Ibias + IE Idc = 6.07 mA + 9.77 mA Idc = 15.84 mA Answer: The current drain is 15.84 mA. 10-14. Given: Idc = 15.84 mA VCC = 15 V

(from Prob. 10-13)

Solution: Pdc = IdcVCC (Eq. 10-17) Pdc = (15.84 mA)(15 V) Pdc = 237.6 mW Answer: The dc input power is 237.6 mW.

10-15. Given: MPP = 10.62 V (from Prob. 10-3) RL = 2.7 kΩ Pdc = 237.6 mW (from Prob. 10-14) Solution: 2 Pout(max) = MPP /8RL (Eq. 10-15) 2 Pout = (10.62 V) /8(2.7 kΩ) Pout = 5.22 mW η = [Pout/Pin]100% η = [5.22 mW/237.6 mW]100% η = 2.2% Answer: The efficiency is 2.2%.

10-16. Given: ICQ = 9.77 mA VCEQ = 6.21 V

(from Prob. 10-2) (from Prob. 10-3)

Solution: PDQ = VCEQ ICQ

(Eq. 10-16)

PDQ = (6.21 V)(9.77 mA)

PDQ = 60.7 mW Answer: The quiescent power dissipation is 60.7 mW.

10-17. 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. 10-7) 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 Idc = Ibias + IE Idc = 100 mA + 137 mA Idc = 237 mA Answer: The current drain is 237 mA. 10-18. Given: Idc = 237 mA VCC = 30 V

(from Prob. 10-17)

Solution: Pdc = IdcVCC Pdc = (237 mA)(30 V)

Pdc = 7.11 W Answer: The dc input power is 7.11 W. 10-19. Given: MPP = 2MP = 13.7 V (from Prob. 10-10) Pdc = 7.11 W (from Prob. 10-18) RL = 100 Ω

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Solution: 2 Pout(max) = MPP /8RL 2 Pout = (13.7 V) /8(100 Ω) Pout = 235 mW η = [Pout/Pin]100% η = [235 mW/7.11 W]100% η = 3.3% Answer: The efficiency is 3.3%.

10-20. Given: ICQ = 137 mA (from Prob. 10-7) VCEQ = 7 V (from Prob. 10-8) Solution: PDQ = VCEQICQ PDQ = (7 V)(137 mA) PDQ = 960 mW Answer: The quiescent power dissipation is 960 mW.

10-21. Given: R1 = 10 Ω

R2 = 2.2 Ω

RE = 1 Ω VCC = 10 V VBE = 0.7 V Solution: VBB = [R2/(R1 + R2)]VCC (Eq. 7-4) VBB = [2.2 Ω/(10 Ω + 2.2 Ω)] 10 V VBB = 1.80 V VE = VBB – VBE (Eq. 7-5) VE = 1.80 V – 0.7 V VE = 1.10 V IE = VE/RE (Eq. 7-6) IE = 1.1 V/1 Ω IE = 1.1 A Answer: The dc emitter current is 1.1 A. 10-22. Given: R1 = 10 Ω

R2 = 2.2 Ω

RE = 1 Ω VCC = 10 V VBE = 0.7 V RC = 3.2 Ω Vout = 5 Vp-p Solution: 2 Pout = v /8RL (Eq. 10-14) out 2 Pout = (5 V) /8(3.2 Ω) Pout = 0.977 W Ibias = VCC/(R1 + R2) Ibias = 10 V/(10 Ω + 2.2 Ω) Ibias = 0.82 A VBB = [R2/(R1 + R2)]VCC (Eq. 7-4) VBB = [2.2 Ω/(10 Ω + 3.2 Ω)]10 V VBB = 1.80 V VE = VBB – VBE (Eq. 7-5) VE = 1.80 V – 0.7 V VE = 1.10 V IE = VE/RE (Eq. 7-6) IE = 1.10 V/1 Ω IE = 1.1 A

Idc = Ibias + IE Idc = 0.82 A + 1.1 A Idc = 1.92 A Pdc = IdcVCC (Eq. 10-17) Pdc = (1.92 A)(10 V) Pdc = 19.2 W η = [Pout/Pin]100% η = [0.977 W/19.2 W]100% η = 5.1% Answer: The output power is 0.977 W, and the efficiency is 5.1%.

10-23. Given: VCE(cutoff) = 12 V Solution: MPP = 12 VCE(cutoff) MPP = 2(12 V) MPP = 24 V Answer: The maximum peak-to-peak voltage is 24 V. 10-24. Given: VCC = MPP = 30 V RL = 16 Ω Solution: 2 PD(max) = MPP /40RL 2 PD(max) = (30 V) /40(16 Ω) PD(max) = 1.41 W Answer: The maximum power dissipation of each transistor is 1.41 W. 10-25. Given: VCC = MPP = 30 V RL = 16 Ω Solution: 2 Pout(max) = MPP /8RL 2 Pout(max) = (30 V) /8(16 Ω) Pout(max) = 7.03 W Answer: The maximum output power is 7.03 W. 10-26. Given: R1 = 100 Ω R2 = 100 Ω RL = 50 Ω VCC = 30 V VDiode = 0.7 V 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. 10-27. 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

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

Pdc = IdcVCC

rc = R3 || Zin(stage 2)

Pdc = 238 mA(30 V)

Zin(stage 2) = 12 kΩ || 910 Ω || β r'e

Pdc = 7.14 W 2 Pout(max) = MPP /8RL 2 Pout(max) = (30 V) /8(50 Ω) Pout(max) = 2.25 W η = [Pout/Pin]100% η = [2.25 W/7.14 W]100% η = 31.5%

rc = 1 kΩ || 12 kΩ || 910 Ω || 200(100 Ω) rc = 496 Ω AVstage 1) = rc / r'e AV(stage 1) = 496 Ω/2.5 Ω AV(stage 1) = 188

Answer: The efficiency is 31.5%.

10-28. Given: VCC = MPP = 30 V RL = 50 Ω Solution: Ibias = (VCC – 2VDiode)/(R1 + R2) Ibias = 28.6 V/(1 kΩ + 1 kΩ) Ibias = ICQ = 14.3 mA Idc = 110 mA Pdc = IdcVCC Pdc = 110 mA(30 V)

Pdc = 3.3 W 2 Pout(max) = MPP /8RL Pout(max) = (30 V)2/8(50 Ω) Pout(max) = 2.25 W η = [Pout/Pin]100% η = [2.25 W/3.3 W]100% η = 68.3% Answer: The efficiency is 68.3% and the quiescent collector current is 14.3 mA. 10-29. Given: MPP = 30 V RL = 100 Ω Solution: 2 Pout(max) = MPP /8RL 2 Pout(max) = (30 V) /8(100 Ω) Pout(max) = 1.13 W Answer: The maximum power output is 1.13 W. 10-30. Given for 1st stage: R1 = 10 kΩ R2 = 5.6 kΩ

R3 = 1 kΩ R4 = 1 kΩ VBB = 10.7 V VE = 10 V Second Stage: R5 = 12 kΩ R6 = 1 kΩ (variable) R7 = 1 kΩ R8 = 100 Ω

β = 200 VCC = 30 V Solution: r'e = 25 mV/IE r'e = 25 mV/(10 V/1 kΩ) r'e = 2.5 Ω re = R8 (second stage) re = 100 Ω

Answer: The voltage gain of the first stage is 188.

10-31. Given for 2nd Stage: R5 = 12 kΩ R6 = 1 kΩ (variable) R8 = 100 Ω

R7 = 1 kΩ VE = 1.43 V 3rd Stage: β = 200 VCC = 30 V RL = 100 Ω Solution: IE = VE/RE (Eq. 7-6) IE = 1.43 V/100 Ω IE = 14.3 mA r'e = 25 mV/IE (Eq. 8-10) r'e = 25 mV/(14.3 mA)

r'e = 1.75 Ω re = R8 (second stage) re = 100 Ω Zin(base) = β re (Eq. 8-21) Zin(base) = 200(100 Ω) Zin(base) = 20 kΩ rc = R7 || Zin(base) rc = 1 kΩ || 20 kΩ rc = 952 Ω AV = rc/(re + r'e) AV = 952 Ω/(100 Ω + 1.75 Ω) AV = 9.36 Answer: The gain of the second stage is 9.36.

10-32. Given: IE = 14.3 mA Solution: ICQ = Ibias = 14.3 mA Answer: The quiescent collector current is 14.3 mA.

10-33. Given: AV1 = 188 AV2 = 9.36

(from Prob. 10-30) (from Prob. 10-31)

Solution: AV3 = 1 (Eq. 10-25) AV = AV1AV2AV3 AV = (188)(9.36)(1)

AV = 1679 Answer: The total voltage gain is 1679.

10-34. Given: νin = 5 Vrms.

Solution: Vp-p = 2.828 Vrms.

Vp-p = 2.828(5 V)

Vp-p = 14.14 Vp-p

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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 DMM will read –6.37 V. Answer: The input voltage is 14.14 Vp-p, and the base voltage is –6.37 V. 10-35. Given: L = 1 μH C = 220 pF Solution: fr = 1 / (2

LC )

(Eq. 10-29)

fr = 1 / [2 (1 μH)(220 pF)] ]

fr = 10.73 MHz Answer: The resonant frequency is 10.73 MHz. 10-36. Given: L = 2 μH C = 220 pF Solution:

10-41. Given: Idc = 0.5 mA VCC = 30 V Solution: Pdc = VCCIdc

(Eq. 10-17)

Pdc = (30 V)(0.5 mA)

Pdc = 15 mW Answer: The dc input power is 15 mW. 10-42. Given: Idc = 0.4 mA VCC = 30 V vout = 30 Vp-p RL = 10 kΩ Solution: Pdc = VCCIdc

(Eq. 10-17)

Pdc = (30 V)(0.4 mA)

fr = 1 / (2 LC )

(Eq. 10-29)

fr = 1 / [2 (2 μH)(220 pF)]]

fr = 7.59 MHz Answer: The resonant frequency is 7.59 MHz. 10-37. Given: L = 1 μH C = 100 pF Solution:

Pdc = 12 mW 2 Pout = v /8RL (Eq. 10-14) 2 out P out = (30 Vp-p) /8(10 kΩ) Pout = 11.25 mW η = (Pout/Pin)100% (Eq. 10-18) η = (11.25 mW/12 mW)100% η = 93.75% Answer: The efficiency is 93.75%.

fr = 1 / (2π LC )

(Eq. 10-29)

fr = 1 / [2π (1 μH)(100 pF)]

fr = 15.92 MHz Answer: The resonant frequency is 15.92 MHz. 10-38. Given: Pout = 11 mW Pin = 50 μW Solution: Ap = Pout/Pin (Eq. 10-12) Ap = 11 mW/50 μW Ap = 220 Answer: The power gain is 220.

10-39. Given: νout = 50 Vp-p RL = 10 kΩ Solution: Pout = vout2/8RL (Eq. 10-14) 2 Pout = (50 Vp-p) /8(10 kΩ) Pout = 31.25 mW Answer: The output power is 31.25 mW. 10-40. Given: VCC = 30 V. Solution: MPP = 2 VCC MPP = 2(30 V) MPP = 60 V 2 Pout = MPP /8RL

10-43. Given: Q = 125 fr = 10.73 MHz

(from Prob. 10-35)

Solution: B = fr/Q B = 10.73 MHz/125 B = 85.84 kHz Answer: The bandwidth is 85.84 kHz. 10-44. Given: Q = 125 fr = 10.73 MHz RL = 10 kΩ MPP = 60 V L = 1 μH

(from Prob. 10-35) (from Prob. 10-40)

Solution: XL = 2πfL XL = 2(3.14)(10.73 MHz)(1 μH)

XL = 67.38 Ω RP = QXL (Eq. 10-33) RP = (125)(67.38 Ω)

RP = 8.42 kΩ rC = RP || RL (Eq. 10-34) rC = 8.42 kΩ || 10 kΩ rC = 4.57 kΩ 2 PD = MPP /40rC (Eq. 10-39) PD = (60 V)2/40(4.57 kΩ)

(Eq. 10-38)

PD = 19.7 mW Answer: The worst-case power dissipation is 19.7 mW.

(Eq. 10-15)

2

Pout = (60 Vp-p) /8(10 kΩ) Pout = 45 mW

10-45. Given: P = 625 mW D

D = 5 mW/°C TA = 100°C

Answer: The maximum output power is 45 mW.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

Solution: P = D(TA – 25°C) (Eq. 10-40) P = (5 mW/°C)(100°C – 25°C) P = 375 mW PD(max) = PD – P PD(max) = 625 mW – 375 mW PD(max) = 250 mW Answer: The worst-case power rating is 250 mW. 10-46. Given: Derating curve on Fig. 10-34.

Answer: The maximum dissipation at 100°C is 2 W. 10-47. Given: PD = 115 W D = 0.657 W/°C TC = 90°C

Answer: See the graph.

IC 20 mA 16 mA

Alternating current

12 mA

Q

8 mA Direct 4 mA current

V

0 mA

CE

0V

11.5 V 15 V

10-53. C2 is shorted 10-54. D1 is open 10-55. VCC is now 20 V

Solution: ∆P = D(TC – 25°C) (Eq. 10-40) ∆P = (0.657 W/°C)(90°C – 25°C) ∆P = 42.7 W

10-56. Q1 B-E shorted

PD(max) = PD – ∆P PD(max) = 115 W – 42.7 W PD(max) = 72.3 W

10-59. Approximately 24 Vp-p

Answer: The power rating is 72.3 W with a case temperature of 90°C.

CRITICAL THINKING 10-48. 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.

10-49. Answer: Electrically, it would be safe to touch, but it may be hot and cause a burn.

10-50. 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.

10-51. 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.

10-52. Given: IC(sat) = 16.67 mA VCC = 15 V ICQ = 9.77 mA

(from Prob. 10-1) (from Prob. 10-2)

MP = ICQ rc = 5.31 V (from Prob. 10-3) VCEQ = 6.21 V (from Prob. 10-3) Solution: 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.

10-57. R6 is shorted 10-58. Class-B/AB push-pull power amplifier

10-60. Compensation diodes used for temperature stability 10-61. 511 microamps 10-62. Approximately zero volts dc

Chapter 11

JFETs

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

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

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

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

JOB INTERVIEW QUESTIONS 7. The gate can be triggered using the static electricity of the human hand to put the device into saturation briefly, enough to trigger another circuit, such as a one-shot multivibrator.

10. It has low input capacitance that allows it to amplify higher frequencies (VHF and UHF) than are possible with a CS amplifier.

11. Although they do not have as much voltage gain as bipolar transistors, they have a high input impedance and very low noise. This is preferred in applications in which the incoming signal may be a few microvolts to be followed by an amplification of a million or more.

PROBLEMS 11-1.

Given:

IG = 1 nA Reverse voltage = –15 V Solution: Rin = Reverse voltage/IG Rin = 15 V/1 nA Rin = 15 GΩ Answer: The input resistance is 15 GΩ.

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11-2.

Given:

Solution: VGS/VGS(off) = 1/2 VGS = 1/2(VGS(off)) VGS = 1/2(–6 V) VGS = –3 V ID /IDSS = 1/4 ID = 1/4 16 mA ID = 4 mA Answer: The gate voltage at the 1/2 cutoff point is –3 V, and the drain current is 4 mA.

IG = 1 μA Reverse voltage = –20 V Ambient Temperature = 100°C Solution: Rin = Reverse voltage/IG Rin = 20 V/1 nA

Rin = 20 MΩ Answer: The input resistance of the gate is 20 MΩ at 100°C. 11-3.

11-7.

Given:

IDSS = 10 mA VGS(off) = –4 V

Solution: IDSS = Maximum drain current = 20 mA VGS(off) = –VP (Eq. 11-2) VGS(off) = –4 V

Solution: VGS/VGS(off) = 1/2 VGS = 1/2(VGS(off)) VGS = 1/2(–4 V) VGS = –2 V ID/IDSS = 1/4 ID = 1/4 10 mA ID = 2.5 mA Answer: The gate voltage at the 1/2 cutoff point is –2 V, and the drain current is 2.5 mA.

RDS = VP /IDSS

(Eq. 11-1)

RDS = 4 V/20 mA

RDS = 200 Ω Answer: The maximum drain current is 20 mA, the gatesource cutoff voltage is –4 V, and the value of RDS is 200 Ω.

11-4.

11-8.

Given:

Solution: VGS(off) = –VP VP = 2 V RDS = VP /IDSS

(Eq. 11-2)

Solution: ID = IDSS[1 – (VGS(1)/VGS(off))]2 2 ID = 14 mA[1 – (–1 V/–4V)] ID = 7.88 mA ID = IDSS[1 – (VGS(2)/VGS(off))]2 ID = 14 mA[1 – (–3 V/–4V)]2 ID = 0.88 mA

(Eq. 11-1)

RDS = 2 V/16 mA

RDS = 125 Ω Answer: The pinch-off voltage is 2 V, and the value of RDS is 125 Ω.

Given:

11-9.

(Eq. 11-2)

RDS(min) = VP(min)/IDSS(min) RDS(min) = 0.5 V/1 mA RDS(min) = 500 Ω

(Eq. 11-1)

RDS(max) = VP(max)/IDSS(max) (Eq. 11-1) RDS(max) = 6 V/5 mA RDS(max) = 1.1 kΩ Answer: The minimum value of RDS is 500 Ω, and the maximum value is 1.1 kΩ.

11-6.

Given:

IDSS = 16 mA VGS(off) = –6 V

(Eq. 11-3)

(Eq. 11-3)

Answer: The drain current is 7.88 mA when the gate voltage is –1 V, and 0.88 mA when the gate voltage is –3 V.

IDSS(min) = 1 mA IDSS(max) = 5 mA VGS(off)min = –0.5 V VGS(off)max = –6 V Solution: VGS(off) = –VP VP(min) = 0.5 V VP(max) = 6 V

Given:

IDSS = 14 mA VGS(off) = –4 V VGS(1) = –1 V VGS(2) = –3 V

IDSS = 16 mA VGS(off) = –2 V

11-5.

Given:

IDSS = 20 mA VP = 4 V

Given:

VDD = 15 V RD = 10 kΩ VGS(off) = –3 V IDSS = 5 mA Solution: ID(sat) = VDD /RD ID(sat) = 15 V/10 kΩ ID(sat) = 1.5 mA VGS(off) = –VP (Eq. 11-2) VP = 3 V RDS = VP /IDSS (Eq. 11-1) RDS = 3 V/5 mA RDS = 600 Ω VD = [RDS/(RDS + RD)]VDD VD = [600 Ω/(600 Ω + 10 kΩ)]15 V VD = 0.849 V Answer: The drain saturation current is 1.5 mA, and the drain voltage is 0.849 V.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

11-10. Given: VDD = 15 V RD = 20 kΩ IDSS = 5 mA VGS(off) = –3 V Solution: VGS(off) = –VP (Eq. 11-2) VP = 3 V RDS = VP /IDSS (Eq. 11-1) RDS = 3 V/5 mA RDS = 600 Ω VD = [RDS/(RDS + RD)]VDD VD = [600 Ω/(600 Ω + 20 kΩ)]15 V VD = 0.437 V Answer: The drain voltage is 0.437 V.

11-11. Given: VDD = 20 V RD = 20 kΩ VGS(off) = –6 V IDSS = 30 mA Solution: VGS(off) = –VP (Eq. 11-2) VP = 6 V RDS = VP/IDSS (Eq. 11-1) 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.

11-12. 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 (Eq. 11-2) VP = 6 V RDS = VP/IDSS (Eq. 11-1) RDS = 6 V/30 mA RDS = 200 Ω 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. 11-13. 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 (Eq. 11-10) ID = 10 V/22 kΩ ID = 0.455 mA VD = VDD – IDRD (Eq. 11-4) VD = 25 V – (0.455 mA)(10 kΩ)

VD = 20.45 V Answer: The drain voltage is 20.45 V. 11-14. Given: R1 = 1.5 MΩ R2 = 1 MΩ RS = 22 kΩ RD = 10 kΩ VDD = 25 V

VG = 10 V

(from Prob. 11-13)

ID = 0.455 mA

(from Prob. 11-13)

VD = 20.45 V (from Prob. 11-13) 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 ID (mA) 0.781 Q

0.455

10.45

VDS (V)

25

DC load line and Q point for Prob. 11–14.

11-15. Given: VDD = 25 V VSS = –25 V RD = 7.5 kΩ RS = 18 kΩ Solution: ID = VSS/RS (Eq. 11-12) ID = –25 V/18 kΩ ID = 1.39 mA VD = VDD – IDRD (Eq. 11-4) VD = 25 V – (1.39 mA)(7.5 kΩ)

VD = 14.58 V Answer: The drain voltage is 14.58 V.

11-16. Given: VDD = 25 V VSS = –25 V RD = 7.5 kΩ RS = 30 kΩ

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Solution: ID = VSS/RS (Eq. 11-12) ID = –25 V/30 kΩ ID = 0.833 mA VD = VDD – IDRD (Eq.11-4) VD = 25 V – (0.833 mA)(7.5 kΩ)

VD = 18.75 V Answer: The drain voltage is 18.75 V. 11-17. Given: VDD = 15 V VEE = –9 V RD = 7.5 kΩ RE = 8.2 kΩ VBE = 0.7 V Solution: ID = (VEE – VBE)/RE (Eq.11-13) ID = (9 V – 0.7 V)/8.2 kΩ ID = 1.01 mA VD = VDD – IDRD (Eq.11-4) 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. 11-18. Given: VDD = 15 V VEE = –9 V RD = 4.7 kΩ RE = 8.2 kΩ VBE = 0.7 V Solution: ID = (VEE – VBE)/RE (Eq.11-13) ID = (9 V – 0.7 V)/8.2 kΩ ID = 1.01 mA VD = VDD – IDRD (Eq.11-4) 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. 11-19. Given: VDD = 25 V

RD = 8.2 kΩ

RS = 1 kΩ ID = 1.5 mA Solution: VGS = –IDRS (Eq.11-7) 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. 11-20. 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. 11-21. 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 from the transconductance curve.

11-22. 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 from the transconductance curve.

11-23. Given: VDD = 25 V RG = 1.5 MΩ

RS = 1 kΩ Answer: The gate-source voltage is –1.5 V, and the drain current is 1.5 mA. 11-24. Given: VDD = 25 V RG = 1.5 MΩ

RS = 2 kΩ Answer: The gate-source voltage is –2.0 V, and the drain current is 1 mA and the drain-source voltage is 14.8 V.

11-25. Given: gm0 = 4000 μS IDSS = 10 mA Solution: VGS(off) = –2IDSS/gm0 (Eq.11-15) VGS(off) = –2(10 mA)/4000 μS VGS(off) = –5 V gm = gm0 [1 – (VGS/VGS(off))] (Eq.11-16) gm = 4000 μS[1 – (–1 V/–5V)] gm = 3200 μS Answer: The gate-source cutoff voltage is –5 V, and the gm0 for VGS = –1 V is 3200 μS. 11-26. Given: gm0 = 1500 μS IDSS = 2.5 mA

VGS = –1 V Solution: VGS(off) = –2IDSS/gm0 (Eq.11-15) VGS(off) = –2(2.5 mA)/1500 μS VGS(off) = –3.33 V

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

gm = gm0 [1 – (VGS/VGS(off))] (Eq.11-16) gm = 1500 μS[1 – (–1 V/–3.3 V)] gm = 1045 μS Answer: The gm for VGS = –1 V is 1045 μS. 11-27. Given: gm0 = 6000 μs IDSS = 12 mA

VGS = –2 V Solution: VGS(off) = –2IDSS/gm0 (Eq.11-15) VGS(off) = –2(12 mA)/6000 μS VGS(off) = –4 V 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))] (Eq.11-16) gm = 6000 μS[1 – (–2 V/–4 V)] gm = 3000 μS Answer: The drain current is 3 mA, and the transconductance is 3000 μs.

11-28. Given: VDD = 30 V R1 = 20 MΩ R2 = 10 MΩ

RD = 1 kΩ RS = 2 kΩ RL = 10 kΩ vg = 2 mV gm = 3000 μS Solution: rd = RD || RL rd = 1 kΩ || 10 kΩ rd = 909 Ω Zin = R1 || R2 Zin = 20 MΩ || 10 MΩ Zin = 6.67 MΩ AV = gmrd (Eq.11-17) AV = (3000 μS)(909 Ω)

AV = 2.73 vout = AV(vin) vout = (2.73)(2 mV) vout = 5.46 mV Answer: zin = 6.67 MΩ and the output voltage is 5.46 mV.

11-29. Given: VDD = 30 V R1 = 20 MΩ

ID = 10 V/2 kΩ

ID = 5 mA From the graph, VGS is approximately –1.4 V when ID is 5 mA. With Eq.11-16, gm0 = 3900 μs. Then: AV = gmrd

(Eq.11-17)

AV = (3900 μS)(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. 11-30. Given: VDD = 30 V R1 = 20 MΩ R2 = 10 MΩ RS = 3.3 kΩ

RL = 1 kΩ vin = 5 mV gm = 2000 μS Solution: rS = RS || RL rS = 3.3 kΩ || 1 kΩ rS = 767 Ω AV = (gmrs)/(1 + gmrs)

(Eq.11-21)

AV = (2000 μS)(767 Ω)/[1 + (2000 μS)(767 Ω)]

AV = 0.605 vout = AV(vin) vout = (0.605)(5 mV) vout = 3.03 mV Zout = RS || 1/gm Zout = 3.3 kΩ || 1/2000 μs Zout = 434 Ω Answer: The output voltage is 3.03 mV and zout is 434 Ω.

11-31. Given: VDD = 30 V R1 = 20 MΩ R2 = 10 MΩ

R2 = 10 MΩ

RD = 1 kΩ RS = 2 kΩ RL = 10 kΩ

RS = 3.3 kΩ

vg = 2 mV

IDSS = 12 mA VGS(off) = –4 V

Solution: rd = RD || RL rd = 1 kΩ || 10 kΩ rd = 909 Ω VGS(off) = –2IDSS/gm0 (Eq.11-15) gm0 = –2IDSS/VGS(off) gm0 = –2(12 mA)/–4 V gm0 = 6000 μS VG = [R2/(R1 + R2)]VDD VG = [10 MΩ/(20 MΩ + 10 MΩ)] 30 V VG = 10 V ID = VG/RS (Eq.11-10)

(from the graph) (from the graph)

RL = 1 kΩ vin = 5 mV IDSS = 6 mA VGS(off) = –4 V rS = 767 Ω

(from the graph) (from the graph) (from Prob.11-30)

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Solution: VGS(off) = –2IDSS/gm0

(Eq.11-15)

gm0 = –2IDSS/VGS(off) gm0 = –2(6 mA)/ –4 V gm0 = 3000 μS VG = [R2/(R1 + R2)]VDD VG = [10 MΩ/(20 MΩ + 10 MΩ)] 30 V VG = 10 V ID = VG/RS (Eq.11-7) ID = 10 V/3.3 kΩ

From the graph, VGS is roughly –1.25 V when ID = 3 mA.

With Eq. (11-16), gm = 2060 μs.

With Eq. (11-21): gmrS = (2060 μS)(767 Ω) = 1.58 AV = 1.58/(1 + 1.58) = 0.612 vout = AV(vin) vout = (0.612)(5 mV) vout = 3.06 mV Answer: The output voltage is 3.06 mV.

RDS = 2 V/10 mA

(Eq.11-1)

RDS = 200 Ω With VGS at –10 V the JFET is cut off and appears as an open; thus vout = vin = 50 mVp-p.

With VGS at 0 V, the JFET is conducting and a voltage divider is created with RD. vout = [RDS/(RDS + Rin)]vin vout = [200 Ω/(200 Ω + 22 kΩ)]50 mVp-p vout = 0.45 mVp-p On-off ratio = vout(max)/vout(min) (Eq.11-23) On-off ratio = 50 mVp-p/0.45 mVp-p

On-off ratio = 111 Answer: The output voltage at a VGS of –10 V is 50 mVpp, the output voltage at a VGS of 0 V is 0.45 mVp-p, and the on-off ratio is 111.

11-33. Given: RD = 33 kΩ vin = 25 mVp-p

On-off ratio = ∞ Answer: The output voltage at a VGS of –10 V is 0 mVp-p, the output voltage at a VGS of 0 V is 24.55 mVp-p, and the

on-off ratio is ∞.

11-34. Answer: IDSS = 20 mA VDS(max) = 5 V for the ohmic region VDS = 5 to 30 V in the active range 11-35. 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.11-15)

gm0 = –2IDSS/VGS(off) gm0 = –2(32 mA)/–8 V gm0 = 8000 μS gm = gm0 [1 – (VGS /VGS(off))] gm = 8000 μS[1 – (VGS/–8 V)] 2 ID = IDSS[1 – (VGS(1)/VGS(off))] 2 ID = 32 mA[1 – (–4 V/–8 V)] ID = 8 mA 2 ID = IDSS[1 – (VGS(2)/VGS(off))] 2 ID = 32 mA[1 – (–2 V/–8 V)] ID = 18 mA

(Eq.11-16) (Eq.11-3)

(Eq.11-3)

Answer: The transconductance equation is gm = 8000 μs [1 – (VGS/–8 V)], the drain current at –4 V is 8 mA, and the drain current at –2 V is 18 mA. 11-36. Given: VGS(off) = –5 V (from the graph) IDSS = 12 mA (from the graph) VGS = –1 V Solution: 2 ID = IDSS[1 – (VGS/VGS(off))] (Eq.11-3) 2 ID = 12 mA[1 – (–1 V/–5 V)] ID = 7.68 mA Answer: The drain current is 7.68 mA.

IDSS = 5 mA

Vp = 3 V Solution: RDS = VP/IDSS

On-off ratio = 24.55 mVp-p/0 mVp-p

CRITICAL THINKING

ID ≅ 3 mA

11-32. Given: RD = 22 kΩ vin = 50 mVp-p IDSS = 10 mA VP = 2 V Solution: RDS = VP/IDSS

vout = [RD/(RDS + Rout)]vin vout = [33 kΩ/(600 Ω + 33 kΩ)]25 mVp-p vout = 24.55 mVp-p On-off ratio = vout(max)/vout(min) (Eq.11-23)

(Eq.11-1)

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 mVp-p.

With VGS at 0 V, the JFET is conducting and a voltage divider is created with the output resistance.

11-37. Given: VDD = 15 V VEE = –10 V RD = 3.3 kΩ RE = 4.7 kΩ VBE = 0.7 V gm = 2000 μS vg = vin = 3 mV Solution: ID = (VEE – VBE)/RE (Eq.11-13) ID = (10 V – 0.7 V)/4.7 kΩ ID = 2 mA

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

VD = VDD – IDRD

(Eq.11-4)

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.11-17) AV = (2000 μS)(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. 11-38. 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. 11-39. 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.

11-40. Answer: a. The gm0 is 6000 μS. Multiply this by 1 kΩ to get a voltage gain of 6.

b. At –1 V, the gm is 4500 μS and the voltage gain is 4.5.

c. 3 d. 1.5 e. 0.75 11-41. Given: Normal Operation VGS = –1.6 V vg = 100 mV

vS = 0 V

vS = 0 V

VDS = 9.6 V

vd = 200 mV

vout = 357 mV

vout = 200 mV

Answer: Open R1.

11-42. Given: Normal Operation VGS = –1.6 V

With Trouble VGS = –0.6 V

ID = 4.8 mA

ID = 7.58 mA

VDS = 9.6 V

VDS = 1.25 V

vg = 100 mV

vg = 100 mV

vS = 0 V vd = 357 mV

vS = 0 V vd = 29 mV

vout = 357 mV

vout = 29 mV

Answer: Open R2.

11-43. Given: Normal Operation VGS = –1.6 V ID = 4.8 mA VDS = 9.6 V

With Trouble VGS = –0.56 V ID = 0 mA VDS = 0 V

vg = 100 mV

vg = 100 mV

vs = 0 V vd = 357 mV

vs = 0 V vd = 0 mV

vout = 357 mV

vout = 0 mV

Answer: Open RD. 11-44. Given: Normal Operation VGS = –1.6 V ID = 4.8 mA VDS = 9.6 V

With Trouble VGS = –8 V ID = 0 mA VDS = 8 V

vg = 100 mV

vg = 100 mV

vs = 0 V vd = 357 mV

vs = 0 V vd = 0 mV

vout = 357 mV

vout = 0 mV

Answer: Open RS. 11-45. Given: Normal Operation VGS = –1.6 V ID = 4.8 mA VDS = 9.6 V

With Trouble VGS = +8 V ID = 0 mA VDS = 24 V

vg = 100 mV

vg = 100 mV

vs = 0 V vd = 357 mV

vs = 0 V vd = 0 mV

vout = 357 mV Answer: Open G-S.

vout = 0 mV

11-46. Given: Normal Operation VGS = –1.6 V

With Trouble VGS = –1.61 V

ID = 4.8 mA

ID = 4.8 mA

VDS = 9.6 V

VDS = 9.6 V

vg = 100 mV

vg = 100 mV

vs = 0 V vd = 357 mV

vs = 87 V vd = 40 mV

vout = 357 mV vout = 40 mV Answer: Open Bypass Capacitor.

With Trouble VGS = –2.75 V ID = 1.38 mA VDS = 19.9 V vg = 100 mV

ID = 4.8 mA

vd = 357 mV

11-47. Given: Normal Operation VGS = –1.6 V

With Trouble VGS = –1.6 V

ID = 4.8 mA

ID = 4.8 mA

VDS = 9.6 V

VDS = 9.6 V

vg = 100 mV

vg = 100 mV

vs = 0 V

vs = 0 V

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vd = 357 mV vout = 357 mV

vd = 397 mV

Solution: 2 ID = IDSS(1 – (VGS/VGS(off))) 2 ID = 4 mA(1 – (–0.5 V/–2.0 V)) ID = 2.25 mA 2 ID = IDSS(1 – (VGS/VGS(off))) 2 ID = 4 mA(1 – (–1 V/–2.0 V)) ID = 1 mA 2 ID = IDSS(1 – (VGS/VGS(off))) 2 ID = 4 mA(1 – (–1.5 V/–2.0 V)) ID = 0.25 mA

vout = 0 mV

Answer: Open Drain Coupling Capacitor.

11-48. Given: Normal Operation VGS = –1.6 V ID = 4.8 mA

VDS = 9.6 V vg = 100 mV vs = 0 V vd = 357 mV vout = 357 mV

With Trouble VGS = 0 V ID = 7.58 mA VDS = 1.5 V

vg = 1 mV vs = 0 V vd = 0 V

Answer: VGS = –0.5 V, ID = 2.25 mA VGS = –1 V, ID = 1 mA VGS = –1.5 V, ID = 0.25 mA

vout = 0 V

Answer: Open Drain Coupling Capacitor.

11-49. R2 is shorted 12-2.

11-50. C2 is open

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 Solution:

11-51. C3 is shorted 11-52. vg is 1 mVp-p 11-53. Q1 is shorted D-S

Chapter 12

MOSFETs

ID = IDSS(1 – (VGS/VGS(off)))2

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

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

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

22. b 23. d 24. d 25. c 26. d

5. MOS technology, especially CMOS, consumes low power and requires a small space. The result is complex circuits that are lightweight, will last a long time on batteries, and are suitable for solar power.

9. Because the thin insulating layer within the device is easily destroyed by static electricity 10. Ship or store the devices in antistatic foam material or wire wrap around leads. Also, technicians should be grounded by using an antistatic wrist strap, touching the chassis, or standing on antistatic (grounding) mats. Finally, use grounded sol-dering irons and test equipment.

11. MOSFETs have faster switching times, resulting in less time spent in the active region, which produces higher efficiency and reduced heat-sink requirements. MOSFETs are also immune to thermal runaway and are easily connected in parallel for greater power dissipation.

12-1.

Given:

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

(Eq. 12-1) 2

ID = 4 mA(1 – (+0.5 V/–2.0 V)) ID = 6.25 mA 2 ID = IDSS(1 – (VGS/VGS(off)))

ID = 4 mA(1 – (+1 V/–2.0 V)) ID = 9 mA

2

ID = IDSS(1 – (VGS/VGS(off)))2

(Eq. 12-1)

(Eq. 12-1) 2

ID = 4 mA(1 – (+1.5 V/–2.0 V)) ID = 12.25 mA

JOB INTERVIEW QUESTIONS

PROBLEMS

Given:

Answer: VGS = 0.5 V, ID = 6.25 mA VGS = 1 V, ID = 9 mA VGS = 1.5 V, ID = 12.25 mA 12-3.

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 Solution:

ID = IDSS(1 – (VGS/VGS(off)))2

(Eq. 12-1) 2

ID = 4 mA(1 – (+1.5 V/+3.0V)) ID = 3 mA ID = IDSS(1 – (VGS/VGS(off)))2

(Eq. 12-1) 2

ID = 4 mA(1 – (2.5 V/+3.0 V)) ID = 0.333 mA Answer: VGS = 1.5 V, ID = 3 mA VGS = 2.5 V, ID = 333 μA

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

12-4.

Given:

VGS(off) = –3 V IDSS = 12 mA Solution: VDS = VDD – (IDSSRD) (Eq. 12-2) VDS = 12 V – ((12 mA)(470 Ω)) VDS = 6.36 V ID = IDSS = 12 mA Answer: The drain current is 12 mA and the drainsource voltage is 6.36 V. 12-5.

Given:

gm0 = 4000 μS RD = 470 Ω

RL = 2 kΩ vin = 100 mV Solution: rd = RD || RL rd = 470 Ω || 2 kΩ rd = 381 Ω AV = gmrd AV = (4000 μS)(381 Ω)

AV = 1.52 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.

12-6.

Given:

gm0 = 4000 μS RD = 680 Ω RL = 10 kΩ vin = 100 mV Solution: rd = RD || RL rd = 680 Ω || 10 kΩ rd = 637 Ω AV = gmrd AV = (4000 μS)(637 Ω)

AV = 2.55 vout = vinAV vout = (100 mV)(2.55) vout = 255 mV Answer: The voltage gain is 2.55, the voltage out is 255 mV, and rd is 637 Ω.

12-7.

Given:

gm0 = 4000 μS RD = 680 Ω

RL = 2 kΩ Vin = 100 mV

RG = 1 MΩ Solution: Zin ≈ RG ≈ 1MΩ Answer: The input impedance is approximately 1 MΩ.

12-8a. Given: VDS(on) = 0.1 V ID(on) = 10 mA

Solution: RDS(on) = VDS(on)/ID(on) (Eq. 12-1) RDS(on) = 0.1 V/10 mA RDS(on) = 10 Ω Answer: The drain-source resistance is 10 Ω.

12-8b. Given: VDS(on) = 0.25 V ID(on) = 45 mA Solution: RDS(on) = VDS(on)/ID(on) (Eq. 12-1) RDS(on) = 0.25 V/45 mA RDS(on) = 5.56 Ω Answer: The drain-source resistance is 5.56 Ω.

12-8c. Given: VDS(on) = 0.75 V ID(on) = 100 mA Solution: RDS(on) = VDS(on)/ID(on) (Eq. 12-1) RDS(on) = 0.75 V/100 mA RDS(on) = 7.5 Ω Answer: The drain-source resistance is 7.5 Ω.

12-8d. Given: VDS(on) = 0.15 V ID(on) = 200 mA Solution: RDS(on) = VDS(on)/ID(on) (Eq. 12-1) RDS(on) = 0.15 V/200 mA RDS(on) = 0.75 Ω Answer: The drain-source resistance is 0.75 Ω.

12-9a. Given: VGS(on) = 3 V ID(on) = 500 mA RDS(on) = 2 Ω ID(sat) = 25 mA Solution: 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. 12-9b. Given: VGS(on) = 3 V ID(on) = 500 mA RDS(on) = 2 Ω ID(sat) = 50 mA Solution: 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. 12-9c. Given: VGS(on) = 3 V ID(on) = 500 mA RDS(on) = 2 Ω ID(sat) = 100 mA

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Solution: VDS = ID(sat)RDS(on) VDS = 100 mA (2 Ω) VDS = 0.2 V Answer: The voltage across the E-MOSFET is 0.2 V. 12-9d. Given: VGS(on) = 3 V ID(on) = 500 mA RDS(on) = 2 Ω ID(sat) = 200 mA Solution: VDS = ID(sat)RDS(on) VDS = 200 mA (2 Ω) VDS = 0.4 V Answer: The voltage across the E-MOSFET is 0.4 V. 12-10. Given: VGS(on) = 2.5 V (from Table 12-1) ID(on) = 100 mA (from Table 12-1) RDS(on) = 10 Ω (from Table 12-1) VDD = 20 V RD = 390 Ω Solution: VD = [RDS(on)/(RDS(on) + RD)]VDD VD = [10 Ω/(10 Ω + 390 Ω)]20 V VD = 0.5 V Answer: The voltage across the E-MOSFET is 0.5 V. 12-11. Given: VGS(on) = 2.6 V (from Table 12-1) ID(on) = 20 mA (from Table 12-1) RDS(on) = 28 Ω (from Table 12-1) VDD = 15 V RD = 1.8 kΩ Solution: VD = [RDS(on)/(RDS(on) + RD)]VDD VD = [28 Ω/(28 Ω + 1.8 kΩ)]15 V VD = 0.23 V Answer: The drain voltage is 0.23 V. 12-12. Given: VGS(on) = 5 V (from Table 12-1) ID(on) = 200 mA (from Table 12-1) RDS(on) = 7.5 Ω (from Table 12-1) VDD = 25 V RD = 150 Ω Solution: 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. 12-13. Given: VGS(on) = 10 V (from Table 12-1) ID(on) = 1 A (from Table 12-1) RDS(on) = 0.9 Ω (from Table 12-1) 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

Answer: The drain voltage is 0.57 V. 12-14. Given: VGS(on) = 5 V ID(on) = 200 mA RDS(on) = 7.5 Ω

(from Table 12-1) (from Table 12-1) (from Table 12-1)

VDD = 30 V

RD = 1 kΩ VLED = 2 V Solution: ID = (VDD – VLED)/(RDS(on) + RD) ID = (30 V – 2 V)/(7.5 Ω + 1 kΩ) ID = 27.8 mA Answer: The LED current is 27.8 mA. 12-15. Given: VGS(on) = 2.6 V ID(on) = 20 mA RDS(on) = 28 Ω

(from Table 12-1) (from Table 12-1) (from Table 12-1)

VDD = 20 V

RD = 1 kΩ Solution: ID = (VDD)/(RDS(on) + RD) ID = (20 V)/(28 Ω + 1 kΩ) ID = 19.5 mA IL = VDD/RL IL = 20 V/2 Ω IL = 10 A Answer: The MOSFET current is 19.5 mA. The load current is 10 A. 12-16. Given: ID(active) = 1 mA VDS(active) = 10 V Solution: RD = VDS(active) /ID(active) (Eq. 12-6) RD = 10 V/1 mA RD = 10 kΩ Answer: The drain resistance is 10 kΩ.

12-17. 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. 12-18. 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.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

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. 12-19. Answer: The output waveform is a square wave with an upper peak of +12 V and a lower peak of 0.43 V. 12-20. Answer: Inverted square wave from +12 V to 0 V. 12-21. 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:

vout =

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: vout = 10 MΩ 12 V ≅ 12 V 10.01 MΩ

12-22. Given: 12-V peak square-wave input f = 1 kHz Assume the same values from the previous problem.

Answer: The signal will be 180° out of phase and have a maximum value of 12 V and a minimum value of 0 V. 12-23. Given: VDD = 12 V RDS(on) = 5 kΩ Solution: ID = VDD/2(RDS(on)) ID = 12 V/2(5 kΩ) ID = 1.2 mA Answer: The current is 1.2 mA. 12-24. Given: VGS(on) = 10 V (from Table 12-2) ID(on) = 2 A (from Table 12-2) RDS(on) = 1.95 Ω (from Table 12-2) VDD = 12 V RD = 10 Ω 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.

12-25. Given: VGS(on) = 10 V (from Table 12-2) ID(on) = 2 A (from Table 12-2) RDS(on) = 1.95 Ω (from Table 12-2) VDD = 12 V

RD = 6 Ω 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.

12-26. Given: VGS(on) = 10 V (from Table 12-2) ID(on) = 5 A (from Table 12-2) RDS(on) = 1.07 Ω (from Table 12-2) VDD = 15 V

RD = 3 Ω 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) = 1.07 Ω.

ID = (VDD)/(RDS(on) + RD) ID = (15 V)/(1.07 Ω + 3 Ω) ID = 3.69 A Answer: The current is 0 A when the input is low, and 3.69 A when the input is high.

12-27. Given: VGS(on) = 10 V (from Table 12-2) ID(on) = 5 A (from Table 12-2) RDS(on) = 1.07 Ω (from Table 12-2) VDD = 15 V

RD = 5 Ω Solution: When it is dark, the photodiode acts like an open and the gate voltage is 10 V. When the input is 10 V, the MOSFET has a resistance of RDS(on) = 1.07 Ω.

ID = (VDD)/(RDS(on) + RD) ID = (15 V)/(1.07 Ω + 5 Ω)

ID = 2.47 A 2 P = (2.47 A) (5 Ω) = 30.5 W Answer: The power is 30.5 W when it is dark.

12.28. Given: VGS(on) = 10 V (from Table 12-2) ID(on) = 2 A (from Table 12-2) RDS(on) = 1.95 Ω (from Table 12-2) VDD = 24 V RD = 12 Ω 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 = (24 V)/(1.95 Ω +12 Ω) ID = 1.72 A Answer: The current is 0 A when the input is low, and 1.72 A when the input is high.

12-29. Given: VGS(on) = 10 V (from Table 12-2) ID(on) = 2 A (from Table 12-2) RDS(on) = 1.95 Ω (from Table 12-2) VDD = 12 V RD = 18 Ω Solution: When the probes are underwater, their resistance is low and the gate voltage is also low. When the input is low, the MOSFET is open and no current flows. When the probes are out of the water, their resistance is high and the gate voltage is also high. 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 + 18 Ω)

ID = 0.6 A

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Answer: The current is 0 A when the probes are underwater, and 0.6 A when the probes are above the water.

12-30. Given: VGS(on) = 10 V (from Table 12-2) ID(on) = 5 A (from Table 12-2) RDS(on) = 1.07 Ω (from Table 12-2) VDD = 20 V

RD = 4 Ω R1 = R2 = 1 MΩ C = 20 μF Solution: τ = RC τ = (1 MΩ || 1 MΩ)(20 μF) τ = 10 s At full brightness, the FET appears to have a resistance of RDS(on) = 1.07 Ω.

ID = (VDD)/(RDS(on) + RD) ID = (20 V)/(1.07 Ω + 4 Ω)

ID = 3.94 A 2 P=I R 2 P = (3.94 A) (4 Ω) P = 62.1 W Answer: The time constant is 10 s, and the lamp power dissipation at full brightness is 62.1 W. 12-31. Given: VGS(on) = 10 V (from Table 12-2) ID(on) = 5 A (from Table 12-2) RDS(on) = 1.07 Ω (from Table 12-2) VDD = 20 V

RD = 6 Ω R1 = R2 = 2 MΩ C = 20 μF Solution: τ = RC τ = (2 MΩ || 2 MΩ)(20 μF) τ = 20 s At full brightness, the FET appears to have a resistance of RDS(on) = 1.07 Ω.

ID = (VDD)/(RDS(on) + RD) ID = (20 V)/(1.07 Ω + 6 Ω) ID = 2.83 A Answer: The time constant is 20 s, and the lamp current at full brightness is 2.83 A. 12-32. Given: Vin = 15 V Q1 RDS(on) = 1 Ω R1 = 10 kΩ EN signal = 0 V and +5 V Solution: When the input EN is low (0 V), Q1 is off and ID = 0 A. When the EN signal is high (+5 V), Q1 is on.

ID = Vin/(R1 + RDS(on)) ID = 15 V/(10 kΩ + 1 Ω) ID = 1.5 mA Answer: The current of Q1 is 0 A when the EN signal is low and 1.5 mA when the enable signal is high.

12-33. Given: Vin = 15 V Q2 RDS(on) = 0.1 Ω

RL = 5 Ω EN signal = +5 V Solution: When the enable signal is high, the output voltage across the load is divided between RDS(on) and RL.

VL = [RL/(RL + RDS(on))]Vin VL = [5 Ω/(5 Ω + 0.1 Ω)]15 V VL = 14.7 V Answer: When the enable signal is high, the output voltage is 14.7 V. 12-34. Given: Vin = 15 V

Q2 RDS(on) = 0.1 Ω

RL = 5 Ω EN signal = +5 V

Solution: IL = Vin/(RDS(on) + RL) IL = 15 V/(0.1 Ω + 5 Ω) IL = 2.94 A 2 Q2 PLOSS = (IL) (RDS(on)) 2 Q2 PLOSS = (2.94 A) (0.1 Ω) Q2 PLOSS = 865 mW 2 PL = (IL) (RL) 2 PL = (2.94 A) (5 Ω) PL = 43.2 W Answer: When the enable signal is +5.0 V, Q2's power loss is 865 mW and the output load power is 43.2 W. 12-35. Given: ID(on) = 75 mA VGS(on) = 4.5 V VGS(th) = 0.8 Solution: 2 k = IDS(on)/(VGS(on) – VGS(th)) k = 75 mA/(4.5 V – 0.8 V)2 –3

2

k = 5.48 × 10 A/V 2 ID = k[VGS – VGS(th)] –3 2 2 ID = 5.48 × 10 A/V [3 – 0.8] ID = 26 mA –3

Answer: The value for k is 5.48 × 10 drain current is 26 mA. 12-36. Given: RD = 150 Ω RL = 1 kΩ vin = 50 mV Solution: rd = RD || RL rd = 150 Ω || 1 kΩ rd = 130 Ω –3 2 k = 5.48 × 10 A/V gm = 2k[VGS – VGS(th)] gm = 24 mS AV = gmrd

2

A/V and the

(from Problem 12-35)

AV = (24 mS)(130 Ω)

AV = 3.14 vout = vgAV vout = (50 mV)(3.14) vout = 157 mV Answer: The voltage gain is 3.14, the voltage out is 157 mV, and gm is 24 mS.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

RDS(on) = 0.9 Ω VDD = 12 V RD = 18 Ω

12-37. Given: RD = 50 Ω ID(on) = 600 mA VGS(on) = 4.5 V VGS(th) = 2.1 V Solution: 2 k = IDS(on)/(VGS(on) – VGS(th))

Solution: ID = (VDD)/(RDS(on) + RD) ID = (12 V)/(0.9 Ω + 18 Ω)

ID = 0.635 A 2 P = ID RD 2 P = (0.635 A) (18 Ω) P = 7.26 W

k = 600 mA/(4.5 V – 2.1 V)2 –3

2

k = 104 × 10 A/V 2 ID = k[VGS – VGS(th)] -3 2 2 ID = 104 × 10 A/V [3 – 2.1] ID = 84.4 mA –3

Answer: The value for k is 104 × 10 drain current is 84.4 mA.

2

A/V and the

12-38. Given: RD = 15 Ω RL = 1 kΩ vin = 50 mV VDD = 12 V Solution: rd = RD || RL rd = 15 Ω || 1 kΩ rd = 14.8 Ω AV = gmrd AV = (395.8 mS)(14.8 Ω)

AV = 5.85 vout = VinAV vout = (50 mV)(5.85) vout = 292 mV Answer: The voltage gain is 5.85, the voltage out is 292 mV, and gm is 395.8 mS.

CRITICAL THINKING 12-39. Given: VGS(on) = 5 V ID(on) = 200 mA RDS(on) = 7.5 Ω VDD = 25 V RD = 150 Ω f = 1 kHz

(from Table 12-1) (from Table 12-1) (from Table 12-1)

Pave = P (duty cycle) Pave = 7.26 W(0.25) Pave = 1.81 W Answer: The average power dissipation in the load resistor is 1.81 W. 12-41. Given: VDD = 12 V RDS(on) = 100 Ω RDS(off) = 10 MΩ Iave = 50 μA Solution: Since the FETs are complementary, one is off and the other is on. Thus the current drawn from the power supply is going to be controlled by the off device.

ID = VDD /RDS(off) ID = 12 V/10 MΩ ID = 1.2 μA P = VDDID P = (12 V)(1.2 μA) P = 14.4 μW P = VDDIave P = (12 V)(50 μA) P = 600 μW Answer: The quiescent power drain is 14.4 μW, and the average power drain is 600 μW. 12-42. Given: VG = 3 V VDD = 15 V

R1 = 1 MΩ

Solution: ID = (VDD)/(RDS(on) + RD) ID = (25 V)/(7.5 Ω + 150 Ω) ID = 158.7 mA Since the signal is a square wave, the off time equals the on time, which gives it a duty cycle of 0.5.

Duty cycle = on time/total time 2 P = ID RD P = (158.7 mA) 2(150 Ω) P = 3.78 W Pave = P (duty cycle) Pave = 3.78 W(0.5) Pave = 1.89 W Answer: The average power dissipation in the load resistor is 1.89 W. 12-40. Given: VGS(on) = 10 V ID(on) = 1 A

(from Table 12-1)

Solution: VR = VDD – VG VR = 15 V – 3 V VR = 12 V I = VR/R1 I = 12 V/1 MΩ I = 12 μA The current through R2 is 3 V divided by 2 MΩ, which equals 1.5 μA. Therefore, the photodiode current is 12 μA minus 1.5 μA, or 10.5 μA. Answer: The diode current is 10.5 μA. 12-43. Given: RDS(on) = 0.17 Ω at 25°C. 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.

0.17(1.69) = 0.29 Ω (from Table 12-1) (from Table 12-1)

Answer: The resistance at 100°C is 0.29 Ω.

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12-44. Given: Vin = 12 V Turns ratio = 4:1

13-2.

VD = 0.7 V VB = 12 V V = 19 V R1 = 5 kΩ Solution: Just before breakover, the capacitor voltage is VB. I = (V – VB)/R1 I = (19 V – 12 V)/5 kΩ I = 1.4 mA

Solution: The primary voltage will be 12 V. N1/N2 = 4 N1/N2 = V1/V2 V2 = V1/(N1/N2) V2 = 12 V/4 V2 = 3 V Answer: The output voltage is 3 V.

While the diode is conducting, the voltage across it is 0.7 V.

12-45. C1 is open

I = (V – VD)/R1 I = (19 V – 0.7 V)/5 kΩ I = 3.66 mA

12-46. VCC has failed and is now 0 V 12-47. Q1 has failed 12-48. VG is set to 5 Vp, not 50 mVp

Answer: The current through the resistor just before breakover is 1.4 mA, and during conduction is 3.66 mA.

12-49. R1 is shorted

Chapter 13

13-3.

Thyristors 9. b 10. c 11. a 12. b 13. d 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

RC = 0.1 ms

T = 0.1 ms since the period equals the RC time constant f = 1/T f = 1/0.1 ms f = 10 kHz

JOB INTERVIEW QUESTIONS 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.

Answer: The RC time constant is 0.1 ms, and the frequency is 10 kHz.

5.

13-4.

IH = 3 mA

RS = 1 kΩ 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 = IHRS + 0.7 V (Eq. 13-2) 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.

Given:

VD = 0.7 V IH = 4 mA

RS = 1 kΩ Solution: V = IH RS + 0.7 V (Eq. 13-2) V = (4 mA)(1 kΩ) + 0.7 V V = 4.7 V Answer: The power supply voltage will be 4.7 V at dropout.

Given:

VD = 0.7 V VB = 20 V

PROBLEMS 13-1.

Given:

VD = 0.7 V VB = 12 V V = 19 V R1 = 5 kΩ C1 = 0.02 μF Solution: RC = (5 kΩ)(0.02 μF)

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

Given:

13-5.

Given:

VD = 0.7 V VB = 12 V V = 19 V R1 = 10 kΩ C1 = 0.06 μF 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. RC = (10 kΩ)(0.06 μF) RC = 0.6 ms Answer: The maximum voltage across the capacitor is 12 V, and the time constant is 0.6 ms.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

13-6.

Given:

VGT = 1.0 V IGT = 2 mA IH = 12 mA VCC = 12 V RG = 2.2 kΩ RL = 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.

Vin = VGT + IGTRG (Eq. 13-1) Vin = 1 V + (2 mA)(2.2 kΩ)

Vin = 5.4 V V = IHRL + 0.7 V (Eq. 13-2) VCC = (12 mA)(47 Ω) + 0.7 V VCC = 1.26 V 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. 13-7. Given: VGT = 0.7 V IGT = 1.5 mA IH = 2 mA VCC = 12 V RG = 4.4 kΩ RL = 94 Ω Solution: Vin = VGT + IGTRG (Eq. 13-1) Vin = 0.7 V + (1.5 mA)(4.4 kΩ) Vin = 7.3 V Answer: The input voltage required to turn on the SCR is 7.3 V. 13-8. 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: 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 13-9. Given: VGT = 1.5 V IGT = 15 mA IH = 10 mA VCC = 12 V RG = 2.2 kΩ RL = 47 Ω Solution: Vin = VGT + IGTRG

(Eq. 13-1)

Vin = 1.5 V + (15 mA)(2.2 kΩ)

Vin = 34.5 V VCC = IHRL + 0.7 V

(Eq. 13-2)

VCC = (10 mA)(47 Ω) + 0.7 V

VCC = 1.17 V

13-10. Given: VGT = 2 V IGT = 8 mA IH = 2 mA VCC = 12 V RG = 6.6 kΩ RL = 141 Ω Solution: Vin = VGT + IGTRG

(Eq. 13-1)

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. 13-11. Given: R1 = 3.3 kΩ R2 = 6.8 kΩ R3 = 750 Ω C1 = 4.7 μF Solution: RC = RTH(cap)C1 RC = (2.54 kΩ)(4.7 μF) RC = 11.9 msec RTH = R || R1 RTH = 750 Ω || 3.3 kΩ RTH = 611 Ω Answer: The charging time constant is 11.9 ms. and the Thevenin resistance is 611 Ω.

13-12. Given: R1 = 1 kΩ R2 = 4.6 kΩ C = 0.47 μF Solution: XC = 1/(2πfC)

XC = 1/2π(60 Hz)(0.47 μF)

XC = 5644 Ω Z = R2 + X 2 Z= Z = 7.95 kΩ X   =  − arctan C Z  R     5.644 k  Z =  − arctan   5.6 k   Z =  − 45 Vin I C = X  ZT − arctan C   R  IC ∠ θ = 120 V ∠ 0º 7.95 kΩ ∠ – 45º IC ∠ θ = 15 mA ∠ 45º VC = (IC ∠ θ) (XC ∠ – 90º) VC = (15 mA ∠ 45º)(5644 Ω ∠ – 90º) VC = 85 V ∠ – 45º θºcond =180º– θºfiring θºcond =180º – 45º θºcond =135 Answer: The firing angle is 45º, the conduction angle is 135º, and the voltage across the capacitor is 85 Vac.

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. Copyright 2021 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

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IGT = 200 μA VZ = 10 V Solution: VCC = VZ + VGT

13-13. Given: R1 = 1 kΩ R2 = 50 kΩ pot C = 0.47 μF 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 μF) XC = 5644 Ω

Z = R2 + X 2 Z= Z = 5.732 kΩ  XC   R     5.644 k   =  − arctan Z  1 k   

 =  − arctan Z

Z =  − 80 I  = C

 XC  ZT − arctan   R 

120 V 0 IC = 5.732 k − 80

IC ∠ θ = 20.9 mA ∠ 80º VC = (IC ∠ θ) (XC ∠ – 90º) VC = (20.9 mA ∠ 80º)(5644 Ω ∠ – 90º) VC = 118 V ∠ – 10º Answer: The minimum firing angle is 10º, and the maximum firing angle is 83.7º.

13-14. Given: R1 = 1 kΩ R2 = 50 kΩ pot C = 0.47 μF 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 μF)

XC = 5644 Ω Z = R2 + X 2 Z= Z = 5.732 kΩ

 XC    R   5.644 k   =  − arctan Z  1 k    I  = C

Vin

Z − arctan

 XC 

   R  120 V 0 IC = 5.732 k − 80 T

IC ∠ θ = 20.9 mA ∠ 80º VC = (IC ∠ θ) (XC ∠ – 90º) VC = (20.9 mA ∠ 80º)(5644 Ω ∠ – 90º) VC = 118 V ∠ – 10º θºcond = 180º – θºfiring θºcond = 180º – 10º θºcond = 170º Answer: The minimum conduction angle is 96.3º, and the maximum conduction angle is 170º.

13-15. Given: VGT = 0.8 V

13-16. Given: VGT = 1.5 V IGT = 200 μA VZ = 10 V ± 10% Solution: VZ(max) = VZ + 0.1(VZ) VZ(max) = 10 V + 0.1(10 V) VZ(max) = 11 V VCC = 11 V + 1.5 V

(Eq. 13-3)

VCC = 12.5 V Answer: The voltage needed to trigger the crowbar is 12.5 V. 13-17. Given: VGT = 0.8 V IGT = 200 μA VZ = 12 V Solution: Vtrig = VZ + VGT (Eq. 13-3) Vtrig = 12 V + 0.8 V Vtrig = 12.8 V Answer: The SCR will trigger at 12.8 V. 13-18. Given: VGT = 0.8 V IGT = 200 μA VZ = 11 V Solution: VCC = VZ + VGT

(Eq. 13-3)

VCC = 11 V + 0.8 V

Z =  − arctan 

Z =  − 80

VCC = 10.8 V Answer: The voltage needed to trigger the crowbar is 10.8 V.

VCC = VZ + VGT

Vin

(Eq. 13-3)

VCC = 10 V + 0.8 V

VCC = 11.8 V Answer: The voltage needed to trigger the crowbar is 11.8 V. 13-19. Given: V = 20 V B

VGT = 2.5 V Solution: Ignore the gate current in the triac. Then V =V +V C

B

GT

VC = 20 V + 2.5 V VC = 22.5 V Answer: The capacitor voltage required to turn on the triac is 22.5 V. 13-20. Given: Vin = 100 V RL = 15 Ω Solution: Ideally, when the triac is conducting, the voltage drop across it is 0 V. I = Vin/RL I = 100 V/15 Ω I = 6.67 A Answer: The load current is 6.67 A.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

13-21. 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. 13-22. Given: VCC = 15 V R2 = 1 kΩ R3 = 2 kΩ Solution: Vgate = [R3/(R2 + R3)]VCC (Voltage divider formula) Vgate = [2 kΩ/(1 kΩ + 2 kΩ)]15 V Vgate = 10 V Vanode = VTrig + 0.7 V Vanode = 10 V + 0.7 Vanode = 10.7 V Answer: The gate trigger voltage is 10 V and the anode is 10.7 V. 13-23. Given: VCC = 15 V Vgate = 10 V Vanode = 10.7 V Solution: VR4 = Vanode – 0.7 V VR4 = 10.7 V – 0.7 VR4 = 10 V Answer: The peak voltage across R4 = 10 V.

RCmin = RmaxC1 RCmin = (1 kΩ)(0.1 μF) RCmin = 0.1 ms Tmax = 0.2(RC1 max) Tmax = 0.2(5.1 ms) Tmax = 1.02 ms Tmin = 0.2(RC1 min) 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 Answer: The maximum frequency is 50 kHz, and the minimum is 980 Hz. 13-28. Given: RL = 100 Ω VCC = 15 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. I = (VCC – 0.7 V)/RL I = (15 V – 0.7 V)/100 Ω I = 143 mA Answer: The output voltage when it is dark is 15 V and when it is light is 0.7 V, and the current through the resistor is 143 mA when it is light.

13-29. Answer: Trouble 1: Since there is voltage at D and not at E, the wire connecting the two is open.

13-24. Answer: The output waveform will be a sawtooth waveform from 0 V to 10.7 V.

Trouble 2: No supply voltage.

CRITICAL THINKING

Trouble 4: Since there is voltage at A and not at B, the fuse is open.

13-25. Answer: The breakover voltage of the diode, which is 10 V.

Trouble 3: Since there is voltage at B and not at C, the transformer is the problem.

13-30. Answer:

13-26. 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.

13-27. Given: R1 = 0 to 50 kΩ R2 = 1 kΩ C1 = 0.1 μF T = 20%(RC)

Trouble 6: Since there is voltage at C and not at D and the load resistor is not shorted, the rectifier is the problem.

Solution: Rmax = R2 + R1(max) Rmax = 1 kΩ + 50 kΩ Rmax = 51 kΩ Rmin = R2 + R1(min) Rmin = 1 kΩ + 0 kΩ Rmin = 1 kΩ RCmax = RmaxC1 RCmax = (51 kΩ)(0.1 μF) RCmax = 5.1 ms

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.

13-31. The fuse is open 13-32 The SCR D3 has been triggered on 13-33. Open transformer secondary 13-34. C1 is open 13-35. VS is 0 V

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Chapter 14

Frequency Effects

14-2. Answer: See the figure below.

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

Av

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

500,000 353,000

JOB INTERVIEW QUESTIONS

f

Too much stray-wiring capacitance. Shorten the leads as much as possible.

2. With a sine wave, find the frequency at which the voltage

15 Hz

1.

gain is down 3 dB. With a square wave, use the step-response

Frequency response for Prob. 14-2.

14-3.

f2 = 10 kHz f = 100 kHz, 200 kHz, 500 kHz, 1 MHz Solution: Substitute in the appropriate value for f.

𝐴𝑉 = 𝐴𝑉(𝑚𝑖𝑑) 𝑓 2 √1 +( ) 𝑓2 𝐴𝑉 = 200 100 kHz 2 √ 1+( ) 10 kHz

12. It is a computer program that provides electronic circuit simulation. It is used to build, test, and analyze simulated circuits.

PROBLEMS 14-1.

AV=19.9 Answer: AV = 19.9 at 100 kHz, AV = 9.98 at 200 kHz, AV = 4 at 500 kHz, AV = 2 at 1 MHz. 14-4.

Given:

AV(mid) = 1000 f1 = 100 Hz f2 = 100 kHz Solution: AV(20K) = Av(mid)/ [ 1 + ( f1 / f )2 ] (Eq. 14-3) AV(20K) = 1000/ [ 1 + (100 Hz / 20 Hz)2 ] AV(20K) = 196 AV(300K) = Av(mid)/ [ 1 + ( 1f / f )2 ]

(Eq. 14-3) AV(300K) = 1000]/ [ 1 + (300 kHz / 20 kHz)3 ] AV(300K) = 316 Answer: The frequency response looks like the figure below; the gain at 20 Hz is 196, and at 300 kHz is 316.

Av

Given:

AV(mid) = 200

method.

5. Some oscilloscopes (with plug-in vertical preamps) specify the risetime of the main frame. A risetime of 7.0 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

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 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.

14-5. Given: Ap = 0.4, 0.2, 0.1, 0.05

1000 707

100 Hz

300 kHz

Frequency response for Prob. 14-1.

Solution: Ap(dB) = 10 logAp (Eq. 14-8) Ap(dB) = 10 log(0.4) Ap(dB) = –3.98 Ap(dB) = 10 logAp (Eq. 14-8) Ap(dB) = 10 log(0. 2) Ap(dB) = –6.99 Ap(dB) = 10 logAp (Eq. 14-8) Ap(dB) = 10 log(0.1) Ap(dB) = –10 Ap(dB) = 10 logAp (Eq. 14-8) Ap(dB) = 10 log(0.05) Ap(dB) = –13

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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.

14-6. Given: Ap = 2, 20, 200, 2000 Solution: Ap(dB) = 10 logAp (Eq. 14-8) Ap(dB) = 10 log(2) Ap(dB) = 3 dB Ap(dB) = 10 logAp (Eq. 14-8) Ap(dB) = 10 log(20) Ap(dB) = 13 dB Ap(dB) = 10 logAp (Eq. 14-8) Ap(dB) = 10 log(200) Ap(dB) = 23 dB Ap(dB) = 10 logAp (Eq. 14-8) Ap(dB) = 10 log(2000) Ap(dB) = 33 dB 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.

14-7.

Given: AP = 0.4, 0.04, 0.004

Solution: Ap(dB) = 10 logAp (Eq. 14-8) Ap(dB) = 10 log(0.4) Ap(dB) = –3.98 Ap(dB) = 10 logAp (Eq. 14-8) AP(dB) = 10 log(0.04)

Ap(dB) = –13.98 Ap(dB) = 10 logAp

(Eq. 14-8)

Ap(dB) = –23.98 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.

AV = 12,589 Answer: The decibel voltage gain is 82, and the voltage gain is 12,589.

14-11. Given: AV1(dB) = 30 dB AV2(dB) = 52 dB Solution: AV1 = antilog(AV1(dB)/20)

(Eq. 14-15)

AV1 = antilog(30 dB/20)

AV1 = 31.6 AV2 = antilog(AV2(dB)/20)

(Eq. 14-15)

AV2 = antilog(52 dB/20)

AV2 = 398 Answer: The voltage gain of the first stage is 31.6, and the second stage is 398.

Solution: AV(dB) = 20 logAV (Eq. 14-9) AV(dB) = 20 log(100,000) AV(dB) = 100 Answer: The decibel voltage gain is 100 dB. Solution: AV = antilog(AV(dB)/20)

(Eq. 14-15)

AV = antilog(34 dB/20)

(Eq. 14-10)

AV = (200)(100)

AV = 20,000 AV(dB) = 20 logAV (Eq. 14-8) 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. 14-9.

AV = antilog(82 dB/20)

14-13. Given: AdB = 34 dB

Given:

AV1 = 200 AV2 = 100 Solution: AV = AV1AV2

Solution: AV(dB) = AV1(dB) + AV2(dB) (Eq. 14-11) AV(dB) = 30 dB + 52 dB AV(dB) = 82 dB AV = antilog(AV(dB)/20) (Eq. 14-15)

14-12. Given: AV = 100,000

AP(dB) = 10 log(0.004)

14-8.

14-10. Given: AV1(dB) = 30 dB AV2(dB) = 52 dB

Given:

AV1 = 200 AV2 = 100 Solution: AV1(dB) = 20 logAV1 (Eq. 14-8) AV1(dB) = 20 log(200) AV1(dB) = 46 dB AV2(dB) = 20 logAV2 (Eq. 14-8) AV2(dB) = 20 log(100) AV2(dB) = 40 dB Answer: The decibel voltage gain for stage 1 is 46 dB, and stage 2 is 40 dB.

AV = 50.1 Answer: The voltage gain is 50.1.

14-14. Given: AV1 = 25.8 AV2 = 117 Solution: AV1(dB) = 20 logAV1 (Eq. 14-9) AV1(dB) = 20 log(25.8) AV1(dB) = 28.2 dB AV2(dB) = 20 logAV2 (Eq. 14-9) AV2(dB) = 20 log(117) AV2(dB) = 41.4 dB AV(dB) = AV1(dB) + AV2(dB) (Eq. 14-11) AV(dB) = 28.2 dB + 41.4 dB AV(dB) = 69.6 dB 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. 14-15. Given: Ap1(dB) = 23 dB Ap2(dB) = 18 dB

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Solution: Ap1(dB) = AV1(dB) = 23 dB Ap2(dB) = AV2(dB) = 18 dB AV(dB) = AV1(dB) + AV2(dB) (Eq. 14-11) AV(dB) = 23 dB + 18 dB AV(dB) = 41 dB Answer: The total decibel voltage gain is 41 dB, the firststage decibel voltage gain is 23 dB, and the second stage voltage gain is 18 dB.

PdBm = 10 log(P/1 mW)

PdBm = 36.9 dBm Answer: 25 mW is 14 dBm, 93.5 mW is 19.7 dBm, and 4.87 W is 36.9 dBm. 14-20. Given: V = 1 μV, 34.8 mV, 12.9 V, and 345 V Solution: VdBV = 20 logV

14-16. Given: vin = 10 μV RG = 300 Ω AP1(dB) = 23 dB AP2(dB) = 18 dB Ap(dB) = 41 dB (from Prob. 14-15) Solution: Ap = antilog(ApdB/10) (Eq. 14-15) Ap = antilog(41 dB/10) Ap = 12,589

VdBV = –120 dBV VdBV = 20 logV

(Eq. 14-18)

VdBV = 20 log(34.8 mV)

VdBV = –29.2 dBV VdBV = 20 logV

(Eq. 14-18)

VdBV = 20 log(12.9 V)

VdBV = 22.2 dBV VdBV = 20 logV

(Eq. 14-18)

VdBV = 20 log(345 V)

VdBV = 50.8 dBV Answer: 1 μV is –120 dBV, 34.8 mV is –29.2 dBV, 12.9 V is 22.2 dBV, and 345 V is 50.8 dBV. 14-21. Given: AV(mid) = 200,000 f2 = 10 Hz Roll-off = 20 dB/decade Answer: AV(dB) = 20 log AV (Eq. 14-9) AV(dB) = 20 log(200,000) AV(dB) = 106 dB

14-17. Given: PdBm = 20 dBm Solution: P = antilog(PdBm/10) (Eq. 14-17) P = antilog(20 dBm/10) P = 100 mW

(Eq. 14-18)

VdBV = 20 log(1 μV)

2

Pin = V /R 2 Pin = (5 μV) /300 Ω (because half of the source voltage appears at the input) Pin = 0.0833 pW Pout = ApPin Pout = (12,589)(0.0833 pW) Pout = 1.05 nW vout = vout = (1.05 nW)(300) vout = 0.56 mV Answer: The load voltage is 0.56 mV, and the load power is 1.05 nW.

(Eq. 14-16)

PdBm = 10 log(4.87 W/1 mW)

From the graph below, the gain at 1 MHz is 6 dB. AV = antilog(AV(dB)/20) (Eq. 14-15) AV = antilog(6 dB/20) AV = 2 Answer: The voltage gain at 1 MHz is 2.

A V(dB)

Answer: The output power is 100 mW. 106 dB

14-18. Given: VdBV = –45 Solution: V = antilog(VdBV/20) (Eq. 14-19) V = antilog(–45 dBV/20) V = 5.6 mV

86 dB 66 dB 46 dB 26 dB

Answer: The output voltage is 5.6 mV. 14-19. Given: P = 25 mW, 93.5 mW, and 4.87 W Solution: PdBm = 10 log(P/1 mW) (Eq. 14-16) PdBm = 10 log(25 mW/1 mW) PdBm = 14 dBm PdBm = 10 log(P/1 mW) (Eq. 14-16) PdBm = 10 log(93.5 mW/1 mW) PdBm = 19.7 dBm

6 dB 10 Hz 100 Hz 1 kHz 10 kHz 100 kHz 1 MHz

Ideal Bode plot for Prob. 14-21.

14-22. Given: AV(mid) = 316,000 f2 = 40 Hz Roll-off = 20 dB/decade Answer: AV(dB) = 20 log AV (Eq. 14-9) AV(dB) = 20 log(316,000) AV(dB) = 110 dB

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

Answer: See figure below.

14-25. Given: R = 15 kΩ C = 100 pF AV(mid) = 400

A V(dB)

110 dB

Solution: f2 = 1/(2πRC) f2 = 1/[2π(15 kΩ)(100 pF)] f2 = 106 kHz AV(dB) = 20 log AV (Eq. 14-9)

90 dB 70 dB 50 dB

AV(dB) = 20 log(400)

30 dB

AV(dB) = 52 dB 10 dB

Answer: See figure at top of next column.

40 Hz

4 kHz

400 kHz

A

Ideal Bode plot for Prob. 14-22.

V(dB)

14-23. 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.

52 dB 32 dB 12 dB 106 kHz

A

1.06 MHz

10.6 MHz

V(dB)

Ideal Bode plot for Prob. 14-25.

15.9 kHz 0 dB

f

20 dB/decade

14-26. Given: C = 5 pF AV = 200,000 Solution: Cin = C(AV + 1) (Eq. 14-26) Cin = 5 pF(200,000 + 1) Cin = 1 μF Answer: The Miller input capacitance is 1 μF.

14-27. Given: Cin(M) = 15 pF

Ideal Bode plot for Prob. 14-23.

14-24. Given: R = 1 kΩ C = 50 pF

AV = 250,000

RL = 10 kΩ RG = 1 kΩ

Solution: f2 = 1/(2πRC) f2 = 1/[2π(1 kΩ)(50 pF)] f2 = 3.18 MHz

Solution: Cin(M) = C(AV + 1) (Eq. 14-26) Cin(M) = 15 pF(250,000 + 1) Cin(M) = 3.75 μF f2 = 1/(2πRC) f2 = 1/[2π (1 kΩ)(3.75 μF)] f2 = 42 Hz AV(dB) = 20 log AV (Eq. 14-9)

Answer: See figure below.

A V(dB)

3.18 MHz

AV(dB) = 20 log(250,000)

0 dB

f

20 dB/decade

AV(dB) = 108 dB Answer: See figure below.

Ideal Bode plot for Prob. 14-24.

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A V(dB)

108 dB

14-31. Given: TR = 0.25 μs Solution: f2 = 0.35/TR (Eq. 14-29) f2 = 0.35/0.25 μs f2 = 1.4 MHz Answer: The bandwidth is 1.4 MHz.

88 dB 68 dB 48 dB 28 dB

14-32. Given: f2 = 100 kHz

8 dB 42 Hz 420 Hz 4.2 kHz 42 kHz 420 kHz 4.2 MHz

Ideal Bode plot for Prob. 14-27.

14-28. Given: C = 50 pF AV = 200,000 Solution: Cin(M) = C(AV + 1) (Eq. 14-26) Cin(M) = 50 pF(200,000 + 1) Cin(M) = 10 μF Answer: The Miller input capacitance is 10 μF. 14-29. Given: C = 100 pF AV = 150,000 RL = 10 kΩ RG = 1 kΩ Solution: Cin(M) = C(AV + 1) (Eq. 14-26) Cin(M) = 100 pF(150,000 + 1) Cin(M) = 15 μF f2 = 1/(2πRC) f2 = 1/[2π(1 kΩ)(15 μF)] f2 = 11 Hz AV(dB) = 20 log AV (Eq. 14-9) AV(dB) = 20 log(150,000)

AV(dB) = 104 dB Answer: See figure below.

A V(dB)

104 dB 84 dB

Solution: f2 = 0.35/ TR (Eq. 14-29) TR = 0.35/f2 TR = 0.35/100 kHz TR = 3.5 μs Answer: The risetime is 3.5 μs.

14-33. Given: Cin = 1 μF RG = 50 Ω Solution: R = RG + Zin(stage) R = RG + R1 || R2 || βr'e R = 1.34 kΩ fC1 = 1/(2πRCin) fC1 = 1/[2π(1.34 kΩ)(1 μF)] fC1 = 119 Hz Answer: The lower cutoff frequency for the base coupling circuit is 119 Hz. 14-34. Given: Cout = 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/(2πRCout) 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.

64 dB

14-35. Given: CE = 25 μF

44 dB

RG = 50 Ω

24 dB 4 dB 11 Hz 110 Hz 1.1 kHz 11 kHz 110 kHz 1.1 MHz

Ideal Bode plot for Prob. 14-29.

14-30. Given: TR = 10 μs Solution: f2 = 0.35/TR (Eq. 14-29) f2 = 0.35/10 μs f2 = 35 kHz Answer: The upper cutoff frequency is 35 kHz.

RE = 1 kΩ Solution: fC1 = 1/(2πzoutCE) fC1 = 1/[2π(22.4 Ω)(25 μF)] fC1 = 284 Hz Answer: The lower cutoff frequency for the emitter bypass circuit is 284 Hz. 14-36. Given: C'c = 2 pF C'e = 10 pF

C'Stray = 5 pF R1 = 10 kΩ R2 = 2.2 kΩ

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RC = 3.6 kΩ RL = 10 kΩ

14-39. Given: R1 = 2 MΩ R2 = 1 MΩ RD = 1 kΩ RL = 10 kΩ RG = 50 Ω

RG = 50 Ω

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) + C'e = 246 pF Base f2 = 1/(2πrgC) f2 = 1/[2π(48 Ω)(246 pF)] f2 = 13.5 MHz

Cgd = 5 pF

AV = (16.5 ms)(1 kΩ || 10 kΩ)

AV = 15 Cin(M) = Cgd(AV + 1) (Eq. 14-40) 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

Collector C = Cout(M) + Cstray C = 2 pF + 5 pF = 7 pF R = RC || RL R = 3.6 kΩ || 10 kΩ R = 2.65 kΩ f2 = 1/(2πRC) f2 = 1/[2π(2.65 kΩ)(7 pF)] f2 = 8.59 MHz Answer: The high cutoff frequency for the base is 13.5 MHz and the collector is 8.59 MHz.

Gate f2 = 1/(2πRC) f2 = 1/[2π(50 Ω)(105 pF)] f2 = 30.3 MHz Collector Cout(M) = Cgd[(AV + 1)/AV]

14-37. Given: gm = 16.5 mS Ciss = 30 pF

(Eq. 14-41)

Cout(M) = 5 pF[(15 + 1)/15]

Coss = 20 pF

Crss = 5 pF Solution: Cgd = Crss = 5 pF Cgs = Ciss – Crss Cgs = 30 pF – 5 pF Cgs = 25 pF Cds = Coss – Crss Cds = 20 pF – 5 pF Cds = 15 pF Answer: Cgs = 25 pF, Cgd = Crss = 5 pF, Cds = 15 pF. 14-38. Given: R1 = 2 MΩ R2 = 1 MΩ RD = 1 kΩ RL = 10 kΩ RG = 50 Ω

Cout(M) = 5.3 pF C = Cds + Cout(M) C = 15 pF + 5.3 pF C = 20.3 pF Drain R = RD || RL

f2 = 1/(2πRC)

f2 = 1/[2π(909 Ω)(20.3 pF)] f2 = 8.61 MHz Answer: The high frequency cutoff for the gate is 30.3 MHz and the drain is 8.61 MHz.

CRITICAL THINKING 14-40. Given: f2 = 100 Hz AV(dB) = 80 dB

C1 = 0.01 μF

C2 = 1 μF Solution: Rin = Zin(stage) Rin = R1 || R2 Rin = 667 kΩ (RG is considered insignificant) fC1 = 1/(2πRinC1) fC1 = 1/[2π(667 kΩ)(0.01 μF)]

fC1 = 23.9 Hz or 14.5 Hz capacitor)

(from Prob. 14-37)

Cgs = 25 pF (from Prob. 14-37) Cds = 15 pF (from Prob. 14-37) Solution: AV = gmrd

(from the output coupling

Answer: The dominant low cutoff frequency is 23.9 Hz.

Solution: AV(mid) = antilog(AV(dB)/20) AV(mid) = antilog(80 dB/20)

(Eq. 14-15)

AV(mid) = 10,000 AV(20K) = Av(mid)/[ 1 + ( f / f 2 )2 ] (Eq. 14-3) AV20K) = 10,000/[ 1 + (20 kHz/100 Hz)2 ] AV(20K) = 50 AV(dB) = 20 log A (Eq. 14-9) AV(dB) = 20 log(50) AV(dB) = 34 dB AV(44.4K) = AV(mid)/[ 1 + ( f / f 2 )2 ] (Eq. 14-3) AV(44.4K) = 10,000/[ 1 + (44.4 kHz/100 Hz)2 ] AV(44.4K) = 22.5

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AV(dB) = 20 log AV (Eq. 14-9) AV(dB) = 20 log(22.5) AV(dB) = 27 dB Answer: The decibel voltage gain at 20 kHz is 34 dB, and at 44.4 kHz is 27 dB. 14-41. Given: f2 = 100 Hz Second breakpoint is 10 kHz AV(mid) = 120 dB 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. 14-42. 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

14-48. CE is open 14-49. VCC is at 15 V, not 10 V

Chapter 15 SELF-TEST 1. b 2. c 3. a 4. c 5. b 6. a

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

PROBLEMS 15-1.

Given:

VCC = 15 V VEE = –15 V RE = 270 kΩ RC = 180 kΩ Solution: IT = VEE/RE (Eq. 15-5) IT = 15 V/270 kΩ IT = 55.6 μA IE = 1/2 IT IE = 1/2 (55.6 μA) IE = 27.8 μA VC = VCC – (27.8 μA)(180 kΩ) VC = 10 V Answer: The tail current is 55.6 μA, the emitter is 27.8 μA, and the quiescent voltage is 10 V. 15-2.

Given:

VCC = 15 V VEE = –15 V RE = 270 kΩ RC = 180 kΩ Solution: IT = (VEE – VBE)/RE (Eq. 15-8) 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 Answer: The tail current is 53 μA, the emitter current is 26.5 μA, and the quiescent voltage is 10.2 V.

14-44. Given: f2 = 1 MHz TR = 1 μs Solution: f2 = 0.35/TR (Eq. 14-29) f2 = 0.35/1 μs f2 = 350 kHz Answer: The amplifier with the cutoff frequency of 1 MHz has the larger bandwidth.

14-45. RG is 500 Ω instead of 50 Ω

14-47. Cin has changed to 0.1 μF

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

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.

Answer: The voltage at the 10% point is 0.2 V, and at the 90% point is 1.8 V. 14-43. Given: R = 4 kΩ C = 50 pF

14-46. RE is 2 kΩ instead of 1 kΩ

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

JOB INTERVIEW QUESTIONS

At the 90% point = 0.9 vout(max) At the 90% point = 0.9(2 V) At the 90% point = 1.8 V

Solution: f2 = 1/(2πRC) f2 = 1/[2π (4 kΩ)(50 pF)] f2 = 796 kHz f2 = 0.35/TR (Eq. 14-29) TR = 0.35/f2 TR = 0.35/796 kHz TR = 0.44 μs Answer: The risetime is 0.44 μs.

Differential Amplifiers

15-3.

Given:

VCC = 12 V VEE = –12 V

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RE = 200 kΩ RC = 200 kΩ Solution: IT = (VEE)/RE IT = (12 V)/200 kΩ IT = 60 μA IE = 1/2 IT IE = 1/2(60 μA) IE = 30 μA

15-4.

vout = AV(v1 – v2)

vout = 518 mV Zin = 2β r'e (Eq. 15-11) Zin = 2(275)(226.7 Ω)

Zin = 125 kΩ Answer: The output voltage is 518 mV, and the input impedance is 125 kΩ.

15-6.

Given:

Right Side VC = VCC – (30 μA)(200 kΩ) VC = 6 V Left Side VC = 12 V

VCC = 15 V VEE = –15 V RE = 68 kΩ

Answer: The tail current is 60 μA, the emitter current is 30 μA, and the quiescent voltage is 6 V on the right side and 12 V on the left side.

Solution: IT = (VEE – VBE)/RE (Eq. 15-5) IT = (15 V – 0.7 V)/68 kΩ IT = 210.3 μA IE = 1/2 IT (Eq. 15-6)

RC = 47 kΩ

β = 275 v1 = 2.5 mV

Given:

VCC = 12 V VEE = –12 V RE = 200 kΩ RC = 200 kΩ

IE = 1/2(210.3 μA)

IE = 105.2 μA r'e = 25 mV/ IE r'e = 237.6 Ω AV = RC/r'e

Given:

VCC = 15 V VEE = –15 V RE = 68 kΩ RC = 47 kΩ

β = 275 v1 = 2.5 mV Solution: IT = (VEE/RE) (Eq. 15-5) IT = (15 V)/68 kΩ IT = 220.6 μA IE = 1/2 IT (Eq. 15-6) IE = 1/2 (220.6 μA)

IE = 110.3 μA r'e = 25 mV/IE r'e = 25 mV/110.3 μA r'e = 226.7 Ω AV = RC/r'e (Eq. 15-10) AV = 47 kΩ/226.7 Ω

(Eq. 15-10)

AV = 47 kΩ/237.6 Ω

AV = 197.8 vout = AV(v1 – v2)

(Eq. 15-2)

vout = 197.8(2.5 mV – 0)

vout = 494 mV Zin = 2β r'e (Eq. 15-11) Zin = 2(275)(237.6 Ω)

Zin = 131 kΩ

Left Side VC = 12 V Answer: The tail current is 56.5 μA, the emitter current is 28.3 μA, and the quiescent voltage is 6.35 V on the right side and 12 V on the left side.

(Eq. 8-10)

r'e = 25 mV/105.2 μA

Solution: IT = (VEE – VBE)/RE IT = (12 V – 0.7 V)/200 kΩ IT = 56.5 μA IE = ½ IT IE = 1/2(56.5 μA) IE = 28.3 μA Right Side VC = VCC – (28.3 μA)(200 kΩ) VC = 6.35 V

15-5.

(Eq. 15-2)

vout = 207.3(2.5 mV – 0)

Answer: The output voltage is 494 mV, and the input impedance is 131 kΩ.

15-7.

Given:

VCC = 15 V VEE = –15 V RE = 68 kΩ RC = 47 kΩ

β = 275 v1 = 0 mV v1 = 1 mV Solution: IT = (VEE)/RE (Eq. 15-5) IT = (15 V)/68 kΩ IT = 220.6 μA IE = 1/2 IT (Eq. 15-6) IE = 1/2(220.6 μA)

IE = 110.3 μA r'e = 25 mV/IE

(Eq. 8-10)

r'e = 25 mV/110.3 μA

r'e = 226.7 Ω AV = RC/r'e

(Eq. 15-10)

AV = 47 kΩ /226.7 Ω

AV = 207.3

AV = 207.3

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vout = AV(v1 – v2)

(Eq. 15-2)

vout = 207.3(0 V – 1 mV)

vout = –207 mV Zin = 2β r′e (Eq. 15-11) Zin = 2(275)(226.7 Ω)

Zin = 125 kΩ Answer: The output voltage is –207 mV, and the input impedance is 125 kΩ.

15-8.

Given:

AV = 360 Iin(bias) = 600 nA Iin(off) = 100 nA Vin(off) = 1 mV RB1 = 10 kΩ Solution: V1 err = (RBl – RB2)/Iin(bias) (Eq. 15-16) V1 err = (10 kΩ – 0)600 nA V1 err = 6 mV V2 err = (RB1 + RB2)(Iin(off)/2) (Eq. 15-17) V2 err = (10 kΩ + 0)(100 nA/2) V2 err = 0.5 mV V3 err = Vin(off)

(Eq. 15-18)

V3 err = 1 mV Verror = AV(V1 err + V2 err + V3 err) (Eq. 15-19) Verror = 360(6 mV + 0.5 mV + 1 mV) Verror = 2.7 V With base resistors equal.

V1 err = 0.

V2 err = RBIin(off) V2 err = (10 kΩ)(100 nA) V2 err = 1 mV V3 err = Vin(off) (Eq. 15-18) V3 err = 1 mV Verror = AV(V1 err + V2 err + V3 err) (Eq. 15-19) Verror = 360(0 mV + 1 mV + 1 mV) Verror = 0.72 μV Answer: The output error voltage is 2.7 V. If the base resistors are equal, the output error voltage is 0.72 V. 15-9.

V2 err = RBIin(off) V2 err = (10 kΩ)(200 nA) V2 err = 2 mV V3 err = Vin(off) (Eq. 15-18) V3 err = 5 mV Verror = AV(V1 err + V2 err + V3 err) (Eq. 15-19) Verror = 250(0 mV + 2 mV + 5 mV) Verror = 1.75 V Answer: The output error voltage is 4 V. If the base resistors are equal, the output error voltage is 1.75 V. 15-10. Given: RC = 500 kΩ RE = 500 kΩ vin(CM) = 20 μV Solution: AV(CM) = RC /2RE (Eq. 15-20) AV(CM) = 500 kΩ/2(500 kΩ) AV(CM) = 0.5 vout(CM) = AV(CM)(vin(CM)) vout(CM) = 0.5(20 μV) vout(CM) = 10 μV Answer: The common-mode voltage gain is 0.5, and the common-mode output voltage is 10 μV. 15-11. Given: RC = 500 kΩ RE = 500 kΩ VCC = 15 V VEE = −15 V vin(CM) = 5 mV vin = 2 mV AV(CM) = 0.5 (from Prob. 15-10) Solution: vout(CM) = AV(CM)(vin(CM)) vout(CM) = 0.5(5 mV) vout(CM) = 2.5 mV IT = (VEE − VBE)RE (Eq. 15-8) IT = (15 V − 0.7)/500 kΩ

Given

IT = 28.6 μA IE = IC = IT /2

AV = 250 Iin(bias) = 1 μA Iin(off) = 200 nA Vin(off) = 5 mV

(Eq. 15-6)

IE = IC = 28.6 μA/2

RB1 = 10 kΩ Solution: V1 err = (RB1 − RB2)Iin(bias) V1 err = (10 kΩ − 0)1 μA

(Eq. 15-16)

V1 err = 10 mV V2 err = (RB1 + RB2)(Iin(off)/2) (Eq. 15-17) V2 err = (10 kΩ + 0)(200 nA/2) V2 err = 1 mV V3 err = Vin(off)

With base resistors equal, V1 err = 0

(Eq. 15-18)

V3 err = 5 mV Verror = AV(V1 err + V2 err + V3 err) (Eq. 15-19) Verror = 250(10 mV + 1 mV + 5 mV)

IE = IC = 14.3 μA r'e = 25 mV/IE r'e = 25 mV/14.3 μA r'e = 1.75 kΩ Since it is a single-ended output: AV = Rc/2r'e (Eq. 15-9) AV = 500 kΩ/2(1.75 kΩ) AV = 143 vout1 = AV(v1 − v2) (Eq. 15-2) vout1 = 143(2 mV − 0)

vout1 = 286 mV Answer: The output voltage is 286 mV (desired) and 2.5 mV (common mode).

Verror = 4 V

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

15-12. Given: AV = 100,000 CMRRdB = 70 dB vin(CM) = 5 μV Solution: CMRR = antilog(CMRRdB/20) CMRR = antilog(70 dB/20) CMRR = 3162 AV(CM) = AV/CMRR

IT = 221.6 μA IE = IC = IT /2

AV(CM) = 31.6 vout(CM) = AV(CM)(vin(CM)) vout(CM) = 31.6(5 μV) vout(CM) = 158 μV vout(desired) = AV(vin(desired)) vout = (100,000)(5 μV) vout = 500 mV

(Eq. 15-8)

(Eq. 15-6)

IE = IC = 221.6 μA/2

IE = IC = 110.8 μA r'e = 25 mV/IE r'e = 25 mV/110.8 μA r'e = 225.6 Ω

Answer: The common-mode voltage gain is 31.6, and the output voltage is 158 μV. The desired output is 500 mV.

Since it is differential output: AV = RC/r'e (Eq. 15-10) AV = 51 kΩ/225.6 Ω

AV = 226 vout = AV(v1 − v2)

(Eq. 15-2)

vout = 226(5 mV − 0 V)

(from Prob. 15-10)

Solution: IT = (VEE − VBE)RE

(Eq. 15-8)

IT = (10 V − 0.7)/500 kΩ

IT = 18.6 μA IE = IC = IT /2

Solution: IT = (VEE − VBE)/RE IT = (12 V − 0.7 V)/51 kΩ

AV(CM) = 100,000/3,162

15-13. Given: RC = 500 kΩ RE = 500 kΩ VCC = 10 V VEE = −10 V AV(CM) = 0.5

15-15. Given: VCC = 12 V VEE = −12 V RC = 51 kΩ RE = 51 kΩ RL = 27 kΩ v1 = 5 mV

(Eq. 15-6)

IE = IC = 18.6 μA/2

IE = IC = 9.3 μA r'e = 25 mV/IE r'e = 25 mV/9.3 μA r'e = 2.69 kΩ Since it is a single-ended output: AV = RC /2r'e (Eq. 15-9) AV = 500 kΩ/2(2.69 kΩ) AV = 93 CMRR = AV/AV(CM) CMRR = 93/0.5 CMRR = 186 CMRRdB = 20 logCMRR CMRRdB = 20 log186 CMRRdB = 45.4 dB Answer: The common-mode rejection ratio is 45.4 dB. 15-14. Given: AV = 150,000 CMRRdB = 85 dB Solution: CMRR = antilog(CMRRdB/20) CMRR = antilog(85 dB/20) CMRR = 17,783 AV(CM) = AV/CMRR AV(CM) = 150,000/17,783 AV(CM) = 8.4

vout = 1.13 V RTH = 2RC RTH = 2(51 kΩ) RTH = 102 kΩ The output voltage is divided between the Thevenin resistance and the load resistance. vL = [RL/(RTH + RL)]vout vL = [27 kΩ/(102 kΩ + 27 kΩ)]vout vL = 237 mV Answer: The load voltage is 237 mV. 15-16. Given: VCC = 12 V VEE = −12 V RC = 51 kΩ RE = 51 kΩ RL = 27 kΩ v1 = 5 mV Solution: IT = (VEE − VBE)/RE

(Eq. 15-8)

IT = (12 V − 0.7 V)/51 kΩ

IT = 221.6 μA IE = IC = IT /2

(Eq. 15-6)

IE = IC = 221.6 μA/2

IE = IC = 110.8 μA r'e = 25 mV/IE r'e = 25 mV/110.8 μA r'e = 225.6 Ω Since it is differential output: AV = RC/r'e (Eq. 15-10) AV = 51 kΩ/225.6 Ω

AV = 226 vout = AV(v1 − v2)

(Eq. 15-2)

vout = 226(5 mV − 0 V)

vout = 1.13 V

Answer: The common-mode voltage gain is 8.4.

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RTH = 2RC RTH = 2(51 kΩ) RTH = 102 kΩ IL = vout/RTH IL = 1.13 V/102 kΩ IL = 11.1 μA Answer: The load current is 11.1 μA. 15-17. Answer: With the open base, the right transistor will go into cutoff and its collector voltage will go high. Since there is still a path for current through the left transistor, it will conduct hard and its collector voltage will go low.

Thus the output will be high and the input will have little effect. A diff amp or an op amp needs a current path to ground for both bases.

15-18. Given: VCC = 12 V VEE = −12 V RC = 20 kΩ RE = 200 kΩ Solution: IT = (VEE − VBE)/RE IT = (12 V − 0.7 V)/200 kΩ IT = 56.5 μA IE = 1/2 IT IE = 1/2 (56.5 μA)

IE = 28.3 μA Vout = VCC − ICRC Vout = 12 V − (28.3 μA)(20 kΩ) Vout = 11.4 V Answer: The output voltage is 11.4 V. 15-19. Answer: C. When the left base is open, all the tail current must flow through the right transistor. This will pull the output voltage down to almost zero.

CRITICAL THINKING 15-20. Given: VCC = 12 V VEE = −12 V RC = 200 kΩ RE = 200 kΩ Solution: IT = (VEE − VBE)/RE IT = (12 V − 0.7 V)/200 kΩ IT = 56.5 μA IE = 1/2 IT IE = 1/2 (56.5 μA)

IE = 28.3 μA Vout = VCC − ICRC Vout = 12 V − (28.3 μA)(20 kΩ) Vout = 11.4 V Answer: The output voltage is 11.4 V. 15-21. Given: Both bases are connected to ground.

Answer: Both bases are at 0 V. 15-22. Answer: This is a current mirror. With Eq. (15-25), you can calculate an ideal tail current of 2 mA and a secondapproximation current of 1.95 mA. The current

through the active load is half the tail current, ideally, 1 mA. 15-23. Answer: The resistor has to be changed to an ideal value of 30 V/15 μA, which equals 2 MΩ.

15-24. Given: VCC = 12 V VEE = −12 V RC = 200 kΩ RE = 200 kΩ VBE = 0.7 V Room temperature = 25°C Left transistor: −2 mV/°C Right transistor: −2.1 mV/°C Solution: ∆T = 75°C − 25°C ∆T = 50°C VBE(L) = 0.7 V + (−2 mV/°C) T VBE(L) = 0.7 V + (−2 mV/°C)50°C VBE(L) = 0.6 V VBE(R) = 0.7 V + (−2.1 mV/°C) T VBE(R) = 0.7 V + (−2.1 mV/°C)50°C VBE(R) = 0.595 V Since each transistor has half the tail current through it, the tail resistor would appear twice as large to each transistor.

IE = (VEE − VBE(R))RE (Eq. 15-8) IE = (12 V − 0.595 V)/400 kΩ IE = 28.5 μA Vout = VCC − ICRC Vout = 12 V − (28.5 μA)(200 kΩ) Vout = 6.3 V Answer: The output voltage is 6.0 V ideally, and 6.3 V using the 2nd approximation.

15-25. Given: R1 = 5 kΩ R2 = 10 kΩ RC1 = RC2 = RE = 2 kΩ VCC = 15 V VEE = −15 V Solution: VB(3) = [R1/(R1 + R2)]VEE VB(3) = [5 kΩ/(5 kΩ + 10 kΩ)] −15 V VB(3) = −5 V VE(3) = VB + 0.7 VE(3) = −5 V − 0.7 V VE(3) = −5.7 V IE(3) = IT = (VEE − VE)/RE IE(3) = IT = (15 V − 5.7 V)/2 kΩ IE(3) = IT = 4.65 mA IE(1) = IC(1) = IT /2 (Eq. 15-6) IE(1) = IC(1) = 4.65 mA/2 IE(1) = IC(1) = 2.33 mA r'e = 25 mV/IE r'e = 25 mV/2.33 mA r'e = 10.7 Ω

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

AV = RC/r'e AV = 2 kΩ/10.7 Ω AV = 187 Answer: The r'e is 10.7 Ω, and the gain is 187.

PROBLEMS 16-1.

VCC = ±18 V Negative saturation voltage is −17 V

15-26. Given: RC1 = RC2 = RE = 5 kΩ R2 = R4 = 10 kΩ R1 = R3 = 20 kΩ VCC = 30 V Solution: VB = [R2/(R2 + R1)]VCC VB = [10 kΩ/(10 kΩ + 20 kΩ)]30 V VB = 10 V VE = VB − 0.7 VE = 10 V − 0.7 V VE = 9.3 V IE = IT = VE/RE IE = IT = 9.3 V/5 kΩ IE = IT = 1.86 mA IE = IC = IT /2 (Eq. 15-6) IE = IC = 1.86 mA/2 IE = IC = 0.93 mA vout = VCC − ICRC vout = 30 V − (930 μA)(5 kΩ) vout = 25.35 V

AV = 100,000 (from Table 16-1) Solution: v2 = 17 V/100,000 v2 = 170 μV

16-2.

AV = 100,000 Answer: The common-mode rejection ratio is 100 dB or 100,000.

16-3.

(from Table 16-1)

Solution:

15-31. Q2 is open from collector to emitter

SELF-TEST

1 = 200,000/[ 1 + (20 MHz/f2

]

AV(100k) = 200,000/[ AV(100k) = 200

(Eq. 14-3)

1 + (100 kHz/100 Hz)

2

]

Answer: The voltage gain at 1 kHz is 19,900, at 10 kHz is 2000, and at 100 kHz is 200.

16-4.

Given:

∆vout = 2 V ∆t = 0.4 μs

JOB INTERVIEW QUESTIONS 9. The LM318 is preferable when slew-rate distortion is too high with a 741C. Applications include high-frequency (wideband) video signals, in which a low slew rate produces a smeared picture, and digital interface circuits, in which timing slew would be objectionable. A disadvantage is that the LM318 is more likely to break into oscillation unless good design and construction practices are used.

11. Audio amps, video amps, IF amps, RF amps, and voltage regulators.

12. Open feedback loop.

]

AV(1k) = 19,900 AV(10k) = AV(mid)/[ 1 + ( f /f 2 )2 ] (Eq. 14-3) AV(10k) = 200,000/[ 1 + (10 kHz/100 Hz)2 ] AV(10k) = 2000 AV(100k) = AV(mid)/[ 1 + ( f /f 2 )2 ] (Eq. 14-3)

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

AV(unity) = AV(mid)/[

[ 1 + (20 MHz/f2 ] = 200,000 2 1 + (20 MHz/f2) = 40,000,000,000 2 (20 MHz/f2) = 40,000,000,001 20 MHz/f2 = 200,000 f2 = 20 MHz/200,000 f2 = 100 Hz AV(1k) = AV(mid)/[ 1 + ( f /f 2 )2 ] (Eq. 14-3) AV(1k) = 200,000/[ 1 + (1 kHz/100 Hz)2 ]

15-30. RE has changed to 33.8 kΩ

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

Given:

AV(mid) = 200,000 funity = 20 MHz AV(unity) = 1

15-29. VCC is at 25 V, not 15 V

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

(Eq. 14-15)

AV = antilog(100 dB/20)

15-28. VEE is at 0 V, not −15 V

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

Answer: The input voltage required to drive the 741C op amp into negative saturation is 170 μV. Given: CMRR = 100 dB (from Table 16-1) Solution: AV = antilog(AdB/20)

Answer: The tail current is 1.86 mA, and the output voltages are 25.35 V. 15-27. Q1 is open from collector to emitter

Chapter 16

Given:

Solution: SR = ∆vout/Δt

SR = 2 V/0.4 μs

SR = 5 V/μs Answer: The slew rate is 5 V/μs.

16-5.

Given:

SR = 70 V/μs Vp = 7 V

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Solution: fmax = SR/2π (Vp) (Eq. 16-2) fmax = 70 V/μs/2π (7 V) fmax = 1.59 MHz

AV(dB) = 20 logAV(CL) AV(dB) = 20 log(10) AV(dB) = 20 Answer: The closed-loop voltage gain is 10, the bandwidth is 2 MHz, the output voltage at 1 kHz is 250 mVpp, and the output voltage at 10 MHz is 49 mV p-p.

Answer: The power bandwidth is 1.59 MHz. 16-6a. Given: SR = 0.5 V/μs VP = 1 V Solution: fmax = SR/2π(Vp) (Eq. 16-2) fmax = 0.5 V/μs/2π (1 V) fmax = 79.6 kHz Answer: The power bandwidth is 79.6 kHz.

A

V(dB)

20 dB

20 dB/decade

16-6b. Given: SR = 3 V/μs VP = 5 V Solution: fmax = SR/2π (Vp) (Eq. 16-2) fmax = 3 V/μs/2π(5 V) fmax = 95.5 kHz Answer: The power bandwidth is 95.5 kHz.

2 MHz Ideal Bode plot for Prob. 16-7.

16-8.

Solution: AV(CL) = −Rf /R1 (Eq. 16-3) AV(CL) = −300 kΩ/15 kΩ AV(CL) = −20 RB2 = Rf || R1 (Eq. 16-11) RB2 = 300 kΩ || 15 kΩ RB2 = 14.29 kΩ V1 err = (RB1 − RB2)Iin(bias) (Eq. 16-8) V1 err = (0 − 14.29 kΩ)(30 pA) V1 err = −429 nV V2 err = (RB1 + RB2)(Iin(off)/2) (Eq. 16-9) V2 err = (0 + 14.29 kΩ)(3 pA/2) V2 err = 21.4 nV V3 err = Vin(off) = 1 mV

Solution: fmax = SR/2π (Vp) (Eq. 16-2) fmax = 15 V/μs/2π(10 V) fmax = 239 kHz Answer: The power bandwidth is 239 kHz. Given:

R1 = 180 Ω Rf = 1.8 kΩ vin = 25 mVp-p funity = 20 MHz Solution: AV(CL) = −Rf/R1 (Eq. 16-3) AV(CL) = −1.8 kΩ/180 Ω AV(CL) = −10 f2(CL) = funity/AVCL

Given:

Iin(bias) = 30 pA Iin(off ) = 3 pA Vin(off ) = 1 mV

16-6c. Given: SR = 15 V/μs VP = 10 V

16-7.

f 20 MHz

Verror = ± AV(CL)(±V1 err ± V2 err ± V3 err) Verror = 20(429 nV + 21.4 nV + 1 mV) Verror = 20 mV

(Eq. 16-5)

f2(CL) = 20 MHz/10

Answer: The output voltage is 20 mV.

f2(CL) = 2 MHz At 1 kHz the voltage gain is the closed loop gain.

vout = AV(CL)(vin) vout = 10(25 mVp-p) vout = 250 mVp-p At 10 MHz the voltage gain is reduced.

𝐴𝑉(10 MHx) =

𝐴𝑉(𝐶𝐿)

√1 + (𝑓/𝑓2)2 10

𝐴𝑉(10 MHx) =

√1 + (10 MHz/2 MHz)2 AV(10 MHz) = 1.96 vout = AV(10 MHz)(vin) vout = 1.96(25 mVp-p) vout = 49 mVp-p

16-9.

Given:

Iin(bias) = 50 pA Iin(off) = 10 pA Vin(off) = 2 mV Solution: AV(CL) = −Rf /R1 (Eq. 16-3) AV(CL) = −300 kΩ/15 kΩ AV(CL) = −20 RB2 = Rf || R1 (Eq. 16-11) RB2 = 300 kΩ || 15 kΩ RB2 = 14.29 kΩ V1 err = (RB1 − RB2)Iin(bias) (Eq. 16-8) V1 err = (0 − 14.29 kΩ)(50 pA) V1 err = −714.3 nV

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

V2 err = (RB1 + RB2)(Iin(off)/2) (Eq. 16-9) V2 err = (0 + 14.29 kΩ)(10 pA/2) V2 err = 71.4 nV V3 err = Vin(off) = 2 mV Verror = ± AV(CL)(± V1 err ± V2 err ± V3 err) Verror = 20(714.3 nV + 71.4 nV + 2 mV) Verror = 40 mV Answer: The output voltage is 40 mV. 16-10. Given: R1 = 150 Ω Rf = 3 kΩ vin = 25 mVp-p funity = 20 MHz Solution: AV(CL) = (Rf /R1) + 1 (Eq. 16-12) AV(CL) = (3 kΩ/150 Ω) + 1 AV(CL) = 21 f2(CL) = funity/AV(CL) (Eq. 16-5) f2(CL) = 20 MHz/21 f2(CL) = 952 kHz At 100 kHz the voltage gain is the closed loop gain. vout = AV(CL)(vin) vout = 21(25 mVp-p) vout = 525 mVp-p Answer: The closed-loop gain is 21, the bandwidth is 952 kHz, and the output voltage at 100 kHz is 525 mVp-p.

16-11. Given: Iin(bias) = 50 pA Iin(off) = 10 pA Vin(off) = 2 mV AV(CL) = 21 (from Prob. 16-10) Solution: (Eq. 16-11) RB2 = R1 || Rf RB2 = 150 Ω || 3 kΩ RB2 = 142.9 Ω V1 err = (RB1 − RB2)Iin(bias) (Eq. 16-8) V1 err = (0 − 142.9 Ω)(50 pA) V1 err = −7.15 nV V2 err = (RB1 + RB2)(Iin(off)/2) (Eq. 16-9) V2 err = (0 + 142.9 Ω)(10 pA/2) V2 err = 715 nV V3 err = Vin(off) = 2 mV Verror = ± AV(CL)(±V1 err ± V2 err ± V3 err) Verror = 21(7.15 nV + 715 pV + 2 mV) Verror = 42 mV Answer: The output voltage is 42 mV. 16-12. Given: R1 = 10 kΩ R2 = 20 kΩ R3 = 40 kΩ Rf = 40 kΩ v1 = 50 mVp-p v2 = 90 mVp-p v3 = 160 mVp-p

Solution: AV1(CL) = −Rf /R1 (Eq. 16-3) AV1(CL) = −40 kΩ/10 kΩ AV1(CL) = −4 AV2(CL) = −Rf /R2 (Eq. 16-3) AV2(CL) = −40 kΩ/20 kΩ AV2(CL) = −2 AV3(CL) = −Rf /R3 (Eq. 16-3) AV3(CL) = −40 kΩ/40 kΩ AV3(CL) = −1 vout = AV1(CL)(vin1) + AV2(CL)(vin2) + AV3(CL)(vin3) vout = −4(50 mVp-p) + −2(90 mVp-p) + −1(160 mVp-p) vout = −540 mVp-p RB2 = R1 || R2 || R3 || Rf (Eq. 16-14) RB2 = 10 kΩ || 20 kΩ || 40 kΩ || 40 kΩ RB2 = 5 kΩ Answer: The output voltage is 540 mVp-p, and the compensating resistor should be 5 kΩ.

16-13. Given: It is a voltage follower. funity = 1 MHz (from Table 16-1) vin = 50 mVp-p Solution: AV(CL) = 1 (Eq. 16-15) f2(CL) = funity = 1 MHz vout = AV(CL)(vin) vout = 1(50 mVp-p) vout = 50 mVp-p Answer: The output voltage is 50 mVp-p, and the bandwidth is 1 MHz.

CRITICAL THINKING 16-14. Given: R1 = 1 kΩ Rf(max) = 101 kΩ Rf(min) = 1 kΩ funity = 20 MHz Solution: AV(CL)max = –Rf(max)/R1 (Eq. 16-3) AV(CL)max = –101 kΩ/1 kΩ AV(CL)max = 101 (the minus sign for phase inversion is ignored here) AV(CL)min = –Rf(min)/R1 (Eq. 16-3) AV(CL)min = –1 kΩ/1 kΩ AV(CL)min = 1 f2(CL)max = funity/(AV(CL)min + 1) f2(CL)max = 20 MHz/2 f2(CL)max = 10 MHz f2(CL)min = funity/AV(CL)max (Eq. 16-5) f2(CL)min = 20 MHz/101 f2(CL)min = 198 kHz Answer: The voltage gain has a range of 1 to 101 and a bandwidth of 198 kHz to 10 MHz. 16-15. Given: R1 = 2 kΩ Rf(max) = 100 kΩ Rf(min) = 0 kΩ funity = 20 MHz

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Solution: AV(CL)max = (Rf(max)/R1) + 1 AV(CL)max = (100 kΩ/2 kΩ) + 1 AV(CL)max = 51 AV(CL)min = (Rf(min)/R1) + 1 AV(CL)min = (0 kΩ/2 kΩ) + 1 AV(CL)min = 1 f2(CL)max = funity/AV(CL)min f2(CL)max = 20 MHz/1 f2(CL)max = 20 MHz

(Eq. 16-12)

–AV2(CL)max = –Rf/R2 (Eq. 16-3) –AV2(CL)max = –100 kΩ/20 kΩ –AV2(CL)max = –5

(Eq. 16-12)

–AV3(CL)max = –Rf /R3 (Eq. 16-3) –AV3(CL)max = –100 kΩ/40 kΩ –AV3(CL)max = –2.5

(Eq. 16-5)

f2(CL)min = funity/AV(CL)max (Eq. 16-5) f2(CL)min = 20 MHz/51 f2(CL)min = 392 kHz Answer: The voltage gain has a range of 1 to 51 and a bandwidth of 392 kHz to 20 MHz. 16-16. 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.

16-17. Given: f = 15 kHz Vp = 2 V Solution: SS = 2πfVp SS = 2π(15 kHz)(2 V) SS = 188 mV/μs SS = 2πfVp SS = 2π(30 kHz)(2 V) SS = 376 mV/μs Answer: The initial slope is 188 mV/μs, with a peak of 2 V, and 376 mV/μs, with a frequency of 30 kHz. 16-18. Answer: a. OP-07A b. TL082 and TL084 c. LM3876 d. LM7171 e. OP-07A 16-19. Answer: CMRR = 38 dB (from Fig. 16-7a) MPP = 21 V (from Fig. 16-7b) AV = 1000 (from Fig. 16-7c) 16-20. Given: R1 = 10 kΩ R2 = 20 kΩ R3 = 40 kΩ Rf(max) = 100 kΩ Rf(min) = 0 Ω v1 = 50 mVp-p v2 = 90 mVp-p

v3 = 160 mVp-p Solution: When the resistance is zero, the voltage gains are zero and the output voltage is zero.

–AV1(CL)max = – Rf /R1 (Eq. 16-3) –AV1(CL)max = –100 kΩ/10 kΩ –AV1(CL)max = –10

vout = AV1(CL)max(vin1) + AV2(CL)max(vin2) + AV3(CL)max(vin3) vout = 10(50 mVp-p) + 5(90 mVp-p) + 2.5(160 mVp-p) vout = 1.35 Vp-p Answer: The maximum output voltage is 1.35 V p-p, and the minimum output voltage is zero.

16-21. Given: R1 = 220 Ω Rf1 = 47 kΩ Rf2 = 18 kΩ Rf3 = 39 kΩ Solution: –AV1(CL) = –Rf1/R1 (Eq. 16-3) –AV1(CL) = –47 kΩ/220 Ω –AV1(CL) = –214 –AV2(CL) = –Rf2/R1 (Eq. 16-3) –AV2(CL) = –18 kΩ/220 Ω –AV2(CL) = –82 –AV3(CL) = –Rf3/R1 (Eq. 16-3) –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.

16-22. Given: R1 = 6 kΩ at position 2 R1 = 6 kΩ || 3 kΩ at position 1 = 2 kΩ Rf = 120 kΩ funity = 1 MHz Solution: AV1(CL) = (Rf/R1) + 1 (Eq. 16-12) AV1(CL) = (120 kΩ/2 kΩ) + 1 AV1(CL) = 61 AV2(CL) = (Rf/R1) + 1 (Eq. 16-12) 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

(Eq. 16-5)

f2(CL)2 = funity/AV(CL1)(max) (Eq. 16-5) f2(CL)2 = 1 MHz/21 f2(CL)2 = 47.6 kHz 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. 16-23. Given: R1 = ∞ at position 2 R1 = 3 kΩ at position 1 Rf = 120 kΩ funity = 1 MHz AVOL = 100,000

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

Solution: AV1(CL) = (Rf/R1) + 1 (Eq. 16-12) AV1(CL) = (120 kΩ/3 kΩ) + 1 AV1(CL) = 41

V2 err = (RB1 + RB2)(Iin(off)/2) (Eq. 16-9) V2 err = (0 + 100 kΩ)(200 nA/2) V2 err = 10 mV V3 err = Vin(off) = 6 mV

At position 2, it becomes a voltage follower: AV2(CL) = 1.

AV(CL) = (Rf/R'1) + 1 (Eq. 16-12) AV(CL) = (100 kΩ/∞) + 1 AV(CL) = 1

Answer: The voltage gain at position 1 is 41, and at position 2 is 1.

16-24. Answer: The output will go to positive or negative saturation.

16-25. 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 resis-tor removed, the circuit reduces to a voltage follower and AV(CL) = 1 and a closed-loop bandwidth of f f2(CL) = unity = 1 MHz = 1 MHz AV (CL)

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 f2(CL) =

funity AV (CL) + 1

=

1 MHz 1+1

= 500 kHz

This was covered briefly in the chapter. See the graphs at the top of p. 680 and the brief explanation that follows.

Chapter 17 discusses the closed-loop bandwidths in more detail.

16-26. 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.

16-27. Answer: Go to positive or negative saturation.

16-28. Given: Iin(bias) = 500 nA Iin(off) = 200 nA Vin(off) = 6 mV R1 = 2 kΩ Rf = 100 kΩ

C = 1 µF Solution: XC = 1/2πfC XC = 1/[2π(0)(1 µF)] XC = ∞ R'1 = XC + R1 R'1= ∞ + 2 kΩ R'1= ∞ RB2 = R1 || Rf (Eq. 16-11) RB2 = ∞ || 100 kΩ RB2 = 100 kΩ V1 err = (RB1 – RB2)Iin(bias) (Eq. 16-8) V1 err = (0 – 100 kΩ)(500 nA) V1 err = 50 mV

Verror = ± AV(CL)(±V1 err ±V2 err ±V3 err) Verror = 1(50 mV + 10 mV + 6 mV) Verror = 66 mV Answer: The output error voltage is 66 mV. 16-29. Given: R1 = 2 kΩ

Rf = 100 kΩ

C = 1 μF vin = 50 mVp-p f = 1 kHz Solution: XC = 1/2πfC XC = 1/[2π(1 kHz)(1 μF)] 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) = (Rf/R'1) + 1 (Eq. 16-12) AV(CL) = (100 kΩ/2 kΩ) + 1 AV(CL) = 51 vout = AV(CL)vin vout = 51(50 mVp-p) vout = 2.55 Vp-p Answer: The output voltage is 2.55 Vp-p.

16-30. Given: Iin(bias) = 500 nA Iin(off) = 200 nA Vin(off) = 6 mV R1 = 2 kΩ Rf = 100 kΩ Solution: R'1 = XC = R1 R'1 = 0 + 2 kΩ R'1 = 2 kΩ RB2 = R1 || Rf (Eq. 16-11) RB2 = 2 kΩ || 100 kΩ RB2 = 1.96 kΩ V1 err = (RB1 – RB2)Iin(bias) (Eq. 16-8) V1 err = (0 – 1.96 kΩ)(500 nA) V1 err = 980 μV V2 err = (RB1 + RB2)(Iin(off)/2) (Eq. 16-9) V2 err = (0 + 1.96 kΩ)(200 nA/2) V2 err = 196 μV V3 err = Vin(off) = 6 mV AV(CL) = (Rf /R'1) + 1 (Eq. 16-12) AV(CL) = (100 kΩ/2 kΩ) + 1 AV(CL) = 51 Verror = ±AV(CL)(±V1 err ± V2 err ± V3 err) Verror = 51(980 μV + 196 μV + 6 m∆V) Verror = 366 mV Answer: The output voltage is 366 mV.

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16-31. Rf changed to 9 kΩ

Solution: B = R1/(R1 + Rf) (Eq. 17-6) B = 2.7 kΩ/(2.7 kΩ + 39 kΩ) B = 0.065

16-32. Rf changed to 94 kΩ 16-33. R1 4.7 kΩ, not 470 Ω

AV = 1/B

16-34. Rf is 3.9 kΩ instead of 39 kΩ

AV= 15.44 Answer: The feedback fraction is 0.065, and the closedloop voltage gain is 15.44.

16-35. Op amp has failed 16-36. Output of the XR2206 (U6) 16-37. Push-pull Class-B/AB power amp

17-3.

16-39. 10

Solution: B = R1/(R1 + Rf) (Eq. 17-6) B = 4.7 kΩ/(4.7 kΩ + 68 kΩ) B = 0.065

16-40. Output would latch at plus and minus Vmax

Negative Feedback

AV = 1/B

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

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

AV = 15.47 Answer: The feedback fraction is 0.065, and the closedloop voltage gain is 15.47.

17-4.

PROBLEMS

AV = 1/B (Eq. 17-4) AV = 1/0.038 AV = 26.32 AV = antilog(AVOL(dB)/20)

Given:

R1 = 2.7 kΩ Rf = 68 kΩ AVOL(dB) = 88 dB

(Eq. 17-3)

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.

Given:

R1 = 2.7 kΩ Rf = 39 kΩ AVOL(dB) = 88 dB

(Eq. 14-15)

AV= antilog(108 dB/20)

Solution: B = R1/(R1 + Rf) (Eq. 17-6) B = 2.7 kΩ/(2.7 kΩ + 68 kΩ) B = 0.038 AV = 1/B (Eq. 17-4) AV = 1/0.038 AV = 26.32 AV = antilog(AV(dB)/20) (Eq. 14-15) AV = antilog (88 dB/20) AV = 25,119 %error = 100%(1 + AVOLB) (Eq 17-5) %error = 100%[1 + 25,119(0.038)] %error = 0.10%

17-2.

Given:

R1 = 2.7 kΩ Rf = 68 kΩ AVOL(dB) = 108 dB Solution: B = R1/(R1 + Rf) (Eq. 17-6) B = 2.7 kΩ/(2.7 kΩ + 68 kΩ) B = 0.038

8. Increased voltage gain and possible oscillation.

12. Current amplifier and transconductance amplifier.

AV = AVOL/(1 + AVOLB)

(Eq. 17-4)

AV = 1/0.065

JOB INTERVIEW QUESTIONS

17-1.

Given:

R1 = 4.7 kΩ Rf = 68 kΩ AVOL(dB) = 88 dB

16-38. 24 Vp-p

Chapter 17

(Eq. 17-4)

AV = 1/0.065

AV = 251,189 % = 100%/(1 + AVOLB) (Eq 17-5) %error = 100%/[1 + 251,189(0.038)] %error = 0.01% AV = AVOL/(1 + AVOLB) (Eq. 17-3) AV = 251,189/[l + 251,189(0.038)] AV = 26.31 Answer: The feedback fraction is 0.038, the ideal closedloop voltage gain is 26.32, the percent error is 0.01%, and the exact voltage gain is 26.31.

17-5.

Given:

R1 = 100 Ω Rf = 7.5 kΩ Rin = 3 MΩ RCM = 500 MΩ AVOL = 200,000 Solution: B = R1/(R1 + Rf ) (Eq. 17-6) B = 100 Ω/(100 Ω + 7.5 kΩ) B = 0.013 Zin(CL) = (1 + AVOLB)Rin || RCM (Eq. 17-8) Zin(CL) = [1 + 200,000(0.013)]3 MΩ || 500 MΩ

Zin(CL) = 470 MΩ

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

Answer: The closed-loop input impedance is 470 MΩ.

17-6.

Given:

Solution: Zout(CL) = Rout/(1 + AVOLB) (Eq. 17-10) Zout(CL) = 50 Ω/[1 + (75,000)(0.013)] Zout(CL) = 0.051 Ω Answer: The closed-loop output impedance is 0.051 Ω.

Given:

AVOL = 200,000 B = 0.013 (from Prob. 17-5) THDOL = 10% Solution: THDCL = THDVOL/(1 + AVOLB)

(Eq. 17-12)

THDCL = 0.0038% Answer: The closed-loop total harmonic distortion is 0.0038%.

Given:

iin = 20 mApeak Rf = 51 kΩ f = 1 kHz Solution: vout = –(iinR2) (Eq. 17-14) vout = –(20 μApeak)(51 kΩ) vout = –1.02 Vpeak at 1 kHz Answer: The output voltage is –1.02 Vpeak at 1 kHz. 17-9.

Given:

iin = 20 μApeak Rf = 33 kΩ f = 1 kHz Solution: vout = –(iinRf) vout = (–20 μApeak)(33 kΩ) vout = –0.660 Vpeak Answer: The output voltage is –0.660 Vpeak.

17-10. Given: iin = –10 μArms Rf = 51 kΩ f = 1 kHz Solution: vout = –(iinRf) vout = (–10 μA)(51 kΩ) vout = –510 Vrms Vp-p = –1.44 Vp-p Answer: The peak-to-peak output voltage is –1.44 V at 1 kHz. 17-11. Given: R1 = 2.7 Ω RL = 1 Ω vin = 0.5 Vrms

iout = 185 mArms 2 PL = (iout) RL 2 PL = (185 mA) (1 Ω) PL = 34.2 mW Answer: The output current is 185 mArms, and the load power is 34.2 mW. 17-12. Given: R1 = 2.7 Ω RL = 3 Ω vin = 0.5 Vrms Solution: iout = vin/R1

(Eq. 17-19)

iout = 0.5 Vrms/2.7 Ω

THDCL = 10%/[1 + 200,000(0.013)]

17-8.

(Eq. 17-19)

iout = 0.5 V/2.7 Ω

AVOL = 75,000 Rout = 50 Ω B = 0.013 (from Prob. 17-5)

17-7.

Solution: iout = vin/R1

iout = 185 mArms 2 PL = (iout) RL 2 PL = (185 mArms) (3 Ω) PL = 277 mW Answer: The output current is 185 mArms, and the load power is 277 mW. 17-13. Given: R1 = 4.7 Ω RL = 1 Ω vin = 0.5 Vrms Solution: iout = vin/R1

(Eq. 17-19)

iout = 0.5 V/4.7 Ω

iout = 106 mArms 2 PL = (iout) RL 2 PL = (106 mA) (1 Ω) PL = 11.2 mW Answer: The output current is 106 mArms, and the load power is 11.2 mW. 17-14. Given: R1 = 1.8 Ω

R2 = 1.5 kΩ

RL = 1 Ω iin = 1 mAp-p Solution: Ai = R2/R1 + 1 (Eq. 17-23) Ai = 1.5 kΩ/1.8 Ω + 1 Ai = 834 iout = Aiiin iout = 834(1 mAp-p) iout = 834 mAp-p irms = ip-p /2.828 irms = 834 mAp-p/2.828 irms = 295 mArms 2 PL = (iout) RL 2 PL = (295 mA) (1 Ω) PL = 87 mW Answer: The current gain is 834, and the load power is 87 mW.

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17-15. Given: R1 = 1.8 Ω

R2 = 1.5 kΩ

RL = 2 Ω iin = 1 mAp-p Ai = 834 (from Prob. 17-14) Solution: iout = Aiiin iout = 834(1 mAp-p) iout = 834 mAp-p irms = ip-p /2.828 irms = 834 mAp-p/2.828 irms = 295 mArms 2 PL = (iout) RL 2 PL = (295 mA) (2 Ω) PL = 174 mW Answer: The output current is 834 mAp-p, and the load power is 174 mW. 17-16. Given: R1 = 7.5 Ω

R2 = 1.5 kΩ

RL = 1 Ω iin = 1 mAp-p Solution: Ai = R2/R1 + 1 (Eq. 17-23) Ai = (1.5 kΩ/7.5 Ω) + 1 Ai = 201 iout = AiIin iout = (201)(1 mAp-p) iout = 201 mAp-p irms = ip-p/(2.828) irms = 201 mAp-p/(2.828) irms = 71.1 mArms 2 PL = (iout) RL 2 PL = (71.1 mArms) (1 Ω) PL = 5 mW Answer: The current gain is 201, and the load power is 5 mW. 17-17. Given: (1 + AVOLB) = 1,000 f2(OL) = 2 Hz Solution: f2(CL) = (1 + AVOLB)f2(OL) f2(CL) = (1000)(2 Hz) f2(CL) = 2 kHz Answer: The closed-loop bandwidth is 2 kHz. 17-18. Given: AVOL = 316,000 f2(OL) = 4.5 Hz AV(CL) = 75 Solution: funity = AVOLf2(OL) = 316,000(4.5 Hz) = 1.42 MHz f2(CL) = funity/AV(CL) (Eq. 17-27) f2(CL) = 1.42 MHz/75 f2(CL) = 18.9 kHz Answer: The closed-loop bandwidth is 18.9 kHz.

17-19. Given: AVOL = 20,000 f2(OL) = 750 Hz Solution: f2(CL) = (1 + AVOL)f2(OL)

(from Table 17-2)

f2(CL) = (1 + 20,000)(750 Hz)

f2(CL) = 15 MHz Answer: The closed-loop bandwidth is 15 MHz. 17-20. Given: (1 + AVOLB) = 5000 f2(OL) = 120 Hz Solution: f2(CL) = (1 + AVOLB) f2(OL)

(from Table 17-2)

f2(CL) = 5000(120 Hz)

f2(CL) = 600 kHz Answer : The closed-loop bandwidth is 600 kHz. 17-21. Given: funity = 1 MHz SR = 0.5 V/µs AV(CL) = 10 Solution: f2(CL) = funity/AV(CL)

(Eq. 17-27)

f2(CL) = 1 MHz/10

f2(CL) = 100 kHz Vp(max) = SR/(2πf2(CL))

(Eq. 17-31) Vp(max) = 0.5 V/μs/(2π100 kHz)

Vp(max) = 796 mVp Answer: The closed-loop bandwidth is 100 kHz, and the maximum peak voltage is 796 mV.

CRITICAL THINKING 17-22. Given: Rf = 150 kΩ iin = 4 μA Solution: vout = iinRf vout = 4 μA(150 kΩ) vout = 600 mV Answer: The output voltmeter reads 600 mV. 17-23. Given: Rf1 = 10 kΩ iin = 1 μA

R1 = 1 kΩ Rf2 = 99 kΩ Solution: vout(1) = iinRf

(from Table 17-2)

vout(1) = 1 μA(10 kΩ)

vout(1) = 10 mV AV(2) = Rf2/R1 + 1 AV(2) = 99 kΩ/1 kΩ + 1 AV(2) = 100 vout(2) = AV(vin(2)) vout(2) = 100(10 mV) vout(2) = 1 V Answer: The output voltage is 1 V.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

17-24. Given: Rf = 50 kΩ R1 = 1 kΩ R2 = 25 kΩ R3 = 100 kΩ Solution: AV1(CL) = Rf/R1 + 1 AV1(CL) = 50 kΩ/1 kΩ + 1 AV1(CL) = 51 AV2(CL) = Rf/R2 + 1 AV2(CL) = 50 kΩ/25 kΩ + 1 AV2(CL) = 3 AV3(CL) = Rf/R3 + 1 AV3(CL) = 50 kΩ/100 kΩ + 1 AV3(CL) = 1.5 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.

17-25. Given: Rf = 50 kΩ R1 = 1 kΩ R2 = 25 kΩ R3 = 100 kΩ

AV1(CL) = 51 AV2(CL) = 3 AV3(CL) = 1.5 vin = 10 mV

(from Prob. 17-24) (from Prob. 17-24) (from Prob. 17-24)

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.

17-26. Given: Rf = 50 kΩ R1 = 1 kΩ R2 = 25 kΩ R3 = 100 kΩ AV1(CL) = 51 (from Prob. 17-24) AV2(CL) = 3 (from Prob. 17-24) AV3(CL) = 1.5 (from Prob. 17-24) AVOL = 100,000 Rin = 2 MΩ Rout = 75 Ω Solution: B(1) = R1/(R1 + Rf) (Eq. 17-6) B(1) = 1 kΩ/(1 kΩ + 50 kΩ) B(1) = 0.0196 Zin1(CL) = (1 + AVOLB(1))Rin (Eq. 17-8) Zin1(CL) = (1 + (100,000)(0.0196))2 MΩ Zin1(CL) = 3924 MΩ Zout1(CL) = Rout/(1 + AVOLB(1)) (Eq. 17-10)

Zout1(CL) = 75 Ω/(1 + (100,000)(0.0196)) Zout1(CL) = 38 mΩ (Eq. 17-6) B(2) = R2/(R2 + Rf) B(2) = 25 kΩ/(25 kΩ + 50 kΩ) B(2) = 0.333 Zin2(CL) = (1 + AVOLB(2))Rin (Eq. 17-8) Zin2(CL) = (1 + (100,000)(0.333))2 MΩ Zin2(CL) = 66,669 MΩ Zout2(CL) = Rout/(1 + AVOLB(2)) (Eq. 17-10) Zout2(CL) = 75 Ω/(1 + (100,000)(0.333)) Zout2(CL) = 2.5 mΩ (Eq. 17-6) B(3) = R3/(R3 + Rf) B(3) = 100 kΩ/(100 kΩ + 50 kΩ) B(3) = 0.667 Zin3(CL) = (1 + AVOLB(3))Rin (Eq. 17-8) Zin3(CL) = (1 + (100,000)(0.667))2 MΩ zin 3(CL) = 133,335 MΩ Zout3(CL) = Rout/(1 + AVOLB(3)) (Eq. 17-10) Zout3(CL) = 75 Ω/(1 + (100,000)(0.667)) Zout3(CL) = 1.25 mΩ Answer: At the 1-kΩ position the input impedance is 3,924 MΩ and the output impedance is 38 mΩ. At the 25kΩ 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 17-2.

17-27. Given: Iin(bias) = 80 nA Iin(off) = 20 nA Vin(off) = 1 mV AVOL = 100,000 Rf = 100 kΩ

R1= 1 kΩ R2 = 25 kΩ R3 = 100 kΩ AV1(CL) = 101 AV2(CL) = 5 AV3(CL) = 2 Solution: RB2(1) = R1 || Rf RB2(1) = 1 kΩ || 100 kΩ RB2(1) = 990 Ω V1 err(1) = (RB1 – RB2(1))Iin(bias) V1 err(1) = (0 – 990 Ω)(80 nA) V1 err(1) = – 79.2 µV V2 err(2) = (RB1 + RB2(1))(Iin(off)/2) V2 err(2) = (0 + 990 Ω)(20 nA/2) V2 err(2) = 9.9 µV V3 err(1) = Vin(off) = 1 mV

(Eq. 16-11)

(Eq. 16-8)

(Eq. 16-9)

Verror(1) = ±AV(CL)(±V1 err(1) ± V2 err(1) ± V3 err(1)) Verror(1) = 101(79.2 µV + 9.9 µV + 1 mV) Verror(1) = 110 mV RB2(2) = R2 || Rf (Eq. 16-11) RB2(2) = 25 kΩ || 100 kΩ RB2(2) = 20 kΩ V1 err(2) = (RB1 – RB2(2))Iin(bias) (Eq. 16-8) V1 err(2) = (0 – 20 kΩ)(80 nA) V1 err(2) = –1.6 µV

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V2 err(2) = (RB1 + RB2(2))(Iin(off)/2) V2 err(2) = (0 + 20 kΩ)(20 nA/2) V2 err(2) = 200 µV

(Eq. 16-9)

V3 err(2) = Vin(off) = 1 mV Verror(2) = ±Av(CL) (±V1 err(2) ± V2 err(2) ± V3 err(2)) Verror(2) = 5(1.6 mV + 200 µV + 1 mV) Verror(2) = 14 mV (Eq. 16-11) RB2(3) = R3 || Rf RB2(3) = 100 kΩ || 100 kΩ RB2(3) = 50 kΩ V1 err(3) = (RB1 – RB2(3))Iin(bias) (Eq. 16-8) V1 err(3) = (0 – 50 kΩ)(80 nA) V1 err(3) = – 4 mV V2 err(3) = (RB1 + RB2(3))(Iin(off)/2) (Eq. 16-9) V2 err(3) = (0 + 50 kΩ)(20 nA/2) V2 err(3) = 500 µV V3 err(3) = Vin(off) = 1 mV Verror(3) = ±AV(CL) (±V1 err(3) ± V2 err(3) ± V3 err(3)) Verror(3) = 2(4 mV + 500 µV + 1 mV) Verror(3) = 11 mV Answer: The output offset voltage is 110 mV at the 1-kΩ position, 14 mV at the 25-kΩ position, and 11 mV at the 100-kΩ position.

17-28. Given: Rf(1) = 100 Ω Rf(2) = 1 kΩ Rf(3) = 10 kΩ iin = 1 mA Solution: vout(1) = iinRf (1) (from Table 17-2) vout(1) = 1 mA(100 Ω) vout(1) = 100 mV vout(2) = iinRf (2) (from Table 17-2) vout(2) = 1 mA(1 kΩ) vout(2) = 1 V vout(3) = iinRf (1) (from Table 17-2) vout(3) = 1 mA(10 kΩ) vout(3) = 10 V Answer: The output offset voltage is 110 mV at position A, 1 V at position B, and 10 V at position C.

Rf = 2 V/1 mA

Rf = 2 kΩ Answer: The unknown resistor is 2 kΩ.

17-32. Given: R = 100 kΩ V = 10 V Rf (max) = 11 kΩ Rf (min) = 9 kΩ Solution: iin = V/R iin = 10 V/100 kΩ iin = 0.1 mA vout(max) = iinRf (max) (from Table 17-2) vout(max) = 0.1 mA(11 kΩ) vout(max) = 1.1 V vout(min) = iin Rf (min) (from Table 17-2) vout(min) = 0.1 mA(9 kΩ) vout(min) = 0.9 V Answer: The output voltage varies between 0.9 V and 1.1 V. 17-33. Given: R = 100 kΩ V = 10 V Rf (max) = 10 kΩ Rf (min) = 1 kΩ Solution: iin = V/R iin = 10 V/100 kΩ iin = 0.1 mA vout(max) = iinRf(max)

(from Table 17-2)

vout(max) = 0.1 mA(10 kΩ)

vout(max) = 1 V vout(min) = iinRf(min)

(from Table 17-2)

vout(min) = 0.1 mA(1 kΩ)

Answer: The output voltage varies between 0.1 V and 1 V. (from Table 17-2)

vout = 2 µA(100 kΩ)

vout = 200 mV Answer: The output voltage is 200 mV. 17-30. Given: Rf = 3.3 kΩ iin = 1 mA Solution: vout = iinRf

(from Table 17-2)

vout(min) = 0.1 V

17-29. Given: Rf = 100 kΩ iin = 2 µA Solution: vout = iinRf

17-31. Given: vout = 2 V iin = 1 mA Solution: vout = iinRf Rf = vout/iin

17-34. Given: R1 = 10 Ω R2 = 100 Ω

R3 = 1 kΩ R4 = 10 kΩ R5 = 100 kΩ

iout = 100 µA full scale Solution: iout = vin /R1 v = ioutR (from Table 17-2)

vout = 1 mA(3.3 kΩ)

vout = 3.3 V Answer: The output voltage is 3.3 V.

vin(1) = ioutR1 vin(1) = 100 µA(10 Ω) vin(1) = 1 mV

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

vin(2) = ioutR2 vin(2) = 100 µA(100 Ω) vin(2) = 10 mV vin(3) = ioutR3 vin(3) = 100 µA(1 kΩ) vin(3) = 100 mV vin(4) = ioutR4 vin(4) = 100 µA(10 kΩ) vin(4) = 1 V vin(5) = ioutR5 vin(5) = 100 µA(100 kΩ) vin(5) = 10 V Answer: The input voltages are 1 mV at the 10-Ω position, 10 mV at the 100-Ω position, 100 mV at the 1-kΩ position, 1 V at the 10-kΩ position, and 10 V at the 100kΩ position.

17-35. Answer: Trouble 1: Since there is voltage at C and not at D, the trouble is an open between C and D.

5. The problem is that the output is drawing too much current from the op amp. The circuit can be redesigned using a higher-power op amp or a current booster.

8. The output is weighted, that is, proportional to the weighted sum of the inputs.

9. Refer to Fig. 18-35. To modify the circuit for a dc response, accept a dc offset and correct it later, or use two batteries as a split supply.

10. Use an emitter follower on the output or employ a Class-B stage for bidirectional amplification.

11. Because of the high open-loop voltage gain of the op amp, the slightest input voltage immediately biases one of the output transistors. In effect, the knee voltage is divided by the open-loop gain.

PROBLEMS 18-1.

R1 = 10 MΩ R2 = 20 MΩ R3 = 15 kΩ R4 = 15 kΩ R5 = 75 kΩ Solution: First stage: AV = –R2/R1 AV = –20 MΩ/10 MΩ AV = –2

Trouble 2: Since all the voltages are zero, the trouble is a shorted R2.

Trouble 3: The trouble is a shorted R4.

17-36. Answer: Trouble 4: Since the output of the first stage is very high, the trouble is an open R2.

Trouble 5: Since there is voltage at F and not at G, the trouble is an open between F and G.

Second stage: AV = –R4/R3 AV = –15 kΩ/15 kΩ AV = –1 AV = –R5/R3 AV = –75 kΩ/15 kΩ AV = –5

Trouble 6: Since there is voltage at F and not at E, the trouble is an open R3.

17-37. Answer: Trouble 7: Since there is voltage at A and not at B, the trouble is an open between A and B. Trouble 8: Since the second stage has no gain, the trouble is a shorted R3.

Total: AV = (–2)(–1) AV = 2 for the switch position pointing to R4 AV = (–5)(–2) AV = 10 for the switch position pointing to R5 Answer: The gain is 2 in the switch position pointing to R4, and the gain is 10 for the switch position pointing to R5.

Trouble 9: The trouble is R4 open.

17-38. R1 is shorted 17-39. R2 is actually 500 Ω, not 1 kΩ 17-40. R3 is actually 51 kΩ, not 100 kΩ 17-41. R2 is actually 10 kΩ, not 1 kΩ 17-42. The op amp U2 has failed

Chapter 18 Applications

Linear Op-Amp Circuit

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

18-2.

Given:

R1 = 1.5 kΩ Rf = 75 kΩ RL = 15 kΩ

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

Given:

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

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

JOB INTERVIEW QUESTIONS 4. The first stage provides a high input impedance and voltage gain, and the second stage produces a high CMRR.

C1 = 1 µF C2 = 4.7 µF funity = 1 MHz Solution: AV = –Rf /R1 AV = –75 kΩ/1.5 kΩ AV = –50 fC1 = 1/(2πR1C1) fC1 = 1/[2π (1.5 kΩ)(1 µF)] fC1 = 106 Hz fC2 = 1/(2π RLC2) fC2 = 1/[2π (15 kΩ)(4.7 µF)] fC2 = 2.26 Hz

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Solution: AV = (Rf/R1) + 1 AV = (82 kΩ/2 kΩ) + 1 AV = 42 f2 = funity/AV f2 = 3 MHz/42 f2 = 71.4 kHz fC1 = 1/(2πR3C1) fC1 = 1/[2π(100 kΩ)(2.2 µF)] fC1 = 0.72 Hz fC2 = 1/(2πRLC2) fC2 = 1/[2π(25 kΩ)(4.7 µF)] fC2 = 1.35 Hz fC3 = 1/(2πR1C3) fC3 = 1/[2π(2 kΩ)(1 µF)] fC3 = 79.6 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.

Answer: The gain is 50 (inverted), and the cutoff frequen-cies are 2.26 Hz and 106 Hz. 18-3.

Given:

R1 = 10 kΩ Rf = 180 kΩ Rmin = 130 Ω Rmax = 25.13 kΩ funity = 1 MHz Solution: Bmin = (10 kΩ || 130 Ω)/(10 kΩ || 130 Ω + 180 kΩ) Bmin = 0.000712 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 f2(max) = Bfunity f2(max) = 0.0382(1 MHz) f2(max) = 38.2 kHz

AV =

−Rf R1

=

−180 k

= –18

18-6.

Answer: The voltage gain is 18 with an inverted output.

The minimum bandwidth is 712 Hz and the maximum bandwidth is 38.2 kHz. 18-4.

RL = 10 kΩ

C1 = 1 µF C2 = 10 µF C3 = 4.7 µF funity = 1 MHz Solution: AV = (Rf/R1) + 1 AV = (150 kΩ/3.3 kΩ) + 1 AV = 46.5 f2 = funity/AV f2 = 1 MHz/46.5 f2 = 21.5 kHz fC1 = 1/(2πR2C1) fC1 = 1/[2π(100 kΩ)(1 µF)] fC1 = 1.59 Hz fC2 = 1/(2πRLC2) fC2 = 1/[2π(10 kΩ)(10 µF)] fC2 = 1.59 Hz fC3 = 1/(2πR1C3) fC3 = 1/[2π(3.3 kΩ)(4.7 µF)] fC3 = 10.3 Hz 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.

Given:

R1 = 1.5 kΩ Rf = 100 kΩ Rmin = 100 Ω Rmax = 5.1 kΩ funity = 1 MHz Solution: Bmin = (R1 || Rmin)/(R1 || Rmin + Rf) Bmin = (1.5 kΩ || 100 Ω)/(1.5 kΩ || 100 Ω + 100 kΩ) Bmin = 0.000937 Bmax = (R1 || Rmax)/(R1 || Rmax + Rf) Bmax = (1.5 kΩ || 5.1 kΩ)/(1.5 kΩ || 5.1 kΩ + 100 kΩ) Bmax = 0.01146 f2(min) = Bminfunity f2(min) = 0.000937(1 MHz) f2(min) = 937 Hz f2(max) = Bmaxfunity f2(max) = 0.01146(1 MHz) f2(max) = 11.5 kHz AV = –Rf/R1 AV = –100 kΩ/1.5 kΩ AV = –66.7 vout = AVvin vout = –66.7(4 mV) vout = –266.8 mV Answer: The minimum bandwidth is 937 Hz and the maximum bandwidth is 11.5 kHz. The output voltage is –266.8 mV. 18-5.

Given:

R1 = 2 kΩ Rf = 82 kΩ RL = 25 kΩ C1 = 2.2 µF C2 = 4.7 µF funity = 3 MHz

Given:

R1 = 3.3 kΩ Rf = 150 kΩ R2 = 100 kΩ

10 k

18-7.

Given:

R1 = 2 kΩ Rf = 100 kΩ vin = 10 mV Solution: AV = (Rf/R1) + 1 AV = (100 kΩ/2 kΩ) + 1 AV = 51 vout = AV(vin) vout = 51(10 mV) vout = 510 mV Answer: The output voltage at A, B, and C is 510 mV.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

18-8.

Given:

R1 = 91 kΩ Rf = 12 kΩ

R2 = 1 kΩ vin = 2 mV Solution: Low gate: AV = (Rf/R1) + 1 AV = (12 kΩ/91 kΩ) + 1 AV = 1.13 vout = AV(vin) vout = 1.13(2 mV) vout = 2.26 mV High gate: AV = [Rf/(R1 || R2)] + 1 AV = [12 kΩ/(91 kΩ || 1 kΩ)] + 1 AV = 13.1 vout = AV(vin) vout = 13.1(2 mV) vout = 26.2 mV Answer: When the gate is low, the output is 2.26 mV; when the gate is high, the output is 26.2 mV. 18-9.

Given:

R1 = 20 kΩ Rf = 68 kΩ

R2 = 1 kΩ vin = 1 mV Solution: Low gate: AV = (Rf/R1) + 1 AV = (68 kΩ/20 kΩ) + 1 AV = 4.4 vout = AV(vin) vout = 4.41(1 mV) vout = 4.4 mV High gate: AV = [Rf/(R1 || R2)] + 1 AV = [68 kΩ/(20 kΩ || 1 kΩ)] + 1 AV = 72.4 vout = AV(vin) 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. 18-10. Given: R1 = 10 kΩ Rf = 10 kΩ Vin = 2.5 V Solution: AV = (Rf/R1) + 1 AV = (10 kΩ/10 kΩ) + 1 AV = 2 Vout = AV(vin) Vout = 2(2.5 V) Vout = 5 V

18-11. Given: R1 = 1 kΩ R2 = 10 kΩ Solution: –R2/R1 < AV < 0 –10 kΩ/1 kΩ < AV < 0 –10 < AV < 0 Answer: The maximum inverting gain is –10, and the maximum positive gain is 0.

18-12. Given: R1 = R2 Solution: At ground the circuit is an inverting amplifier.

AV = –Rf/R1 AV = –1 When the wiper is 10% away from ground, so that the noninverting gain will be 10% of its maximum of 2.

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.

18-13. 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 Answer: The maximum positive gain is 15, and the maximum negative gain is –15.

18-14. Given: R' = 10 kΩ R = 22 kΩ C = 0.02 µF fin = 100 Hz, 1 kHz, 10 kHz Solution: fC = 1/(2π RC) fC = 1/[(2π22 kΩ)(0.02 µF)] fC = 362 Hz ϕ = –2 arctan (f/fC) ϕ = –2 arctan (100 Hz/362 Hz) ϕ = –30.9° ϕ = –2 arctan (f/fC) ϕ = –2 arctan (1 kHz/362 Hz) ϕ = –140° ϕ = –2 arctan (f/fC) ϕ = –2 arctan (10 kHz/362 Hz) ϕ = –176° Answer: The phase shift is –30.9° at 100 Hz, –140° at 1 kHz, and –176° at 10 kHz. 18-15. Given: R1 = 1.5 kΩ R2 = 30 kΩ

Answer: The new output reference voltage is 5 V.

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Solution: AV(inv) = –R2/R1 (Eq. 18-6) AV(inv) = –30 kΩ/1.5 kΩ AV(inv) = –20 AV(non) = [(R2/R1) + 1][R'2/(R'1 + R'2)] (Eq. 18-7) 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.

18-16. Given: R1 = 1 kΩ R2 = 20 kΩ Solution: AV(inv) = –R2/R1 (Eq. 18-6) AV(inv) = –20 kΩ/1 kΩ AV(inv) = –20 AV(CM) = ±4 ∆R/R (Eq. 18-5) 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.

18-17. Given: R1 = 10 kΩ R2 = 20 kΩ R3 = 20 kΩ R4 = 10 kΩ 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.

18-18. Given: R1 = 1 kΩ ∆R = 15 Ω AV = –100 Solution: vin = (∆R/4R)VCC vin = (15 Ω/4 (1 kΩ))15 V vin = 56.3 mV vout = AV(vin) vout = (–100)(56.3 mV) vout = –5.63 V Answer: The output voltage is –5.63 V. 18-19. Given: R1 = 1 kΩ R2 = 99 kΩ

R = 10 kΩ ± 0.5% vin = 2 mV Solution: AV = (R2/R1) + 1 AV = (99 kΩ/1 kΩ) + 1 AV = 100

vout = AV(vin) vout = 100(2 mV) vout = 200 mV at preamp output and –200 mV at diff amp output.

AV(CM) = ±2(∆R/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.

18-20. Given: vin(CM) = 5 V 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. Answer: The guard voltage is 5 V. 18-21. Given: RG = 1008 Ω vin = 20 mV Solution: AV = (49.4 kΩ/RG) + 1 (Eq. 18-17) AV = (49.4 kΩ/1008 Ω) + 1 AV = 50 vout = AV(vin) vout = 50(20 mV) vout = 1 V Answer: The output voltage is 1 V. 18-22. 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. 18-23. Given: 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 Solution: AV(1) = –Rf/R1 AV(1) = –75 kΩ/10 kΩ AV(1) = –7.5 AV(2) = –Rf/R2 AV(2) = –75 kΩ/20 kΩ AV(2) = –3.75 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)]}

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

AV(3) = (12.25)(0.455) AV(3) = 5.57 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 vout = AV(1)v1 + AV(2)v2 + AV3)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.

18-28. Given: D3 – D0 = 0001 Solution: 0 1 2 3 BIN = (D0 × 2 ) + (D1 × 2 ) + (D2 × 2 ) + (D3 × 2 ) BIN = (1 × 20) + (0 × 21) + (0 × 22) + (0 × 23)

+ (0 × 24)

BIN = 1

 BIN

 V  2N ref    102  V =−  (2.5 V) out  24    (

V

=−

out

(Eq. 18-19)

Vout = –312.5 mV Answer: The smallest output voltage is –312.5 mV.

18-24. Given: R = 10 kΩ v1 = 1.5 V v2 = 2.5 V

18-29. Given: R1 = 2 kΩ

R2 = 47 kΩ

β = 100 ISC = 25 mA

v3 = 4 V Solution: vout = –(v1 + v2 + v3)/3 vout = –(1.5 V + 2.5 V + 4 V)/3 vout = –2.67 V

Solution: AV = (R2/R1) + 1 AV = (47 kΩ/2 kΩ) + 1 AV = 24.5 Imax = βISC Imax = (100) 25 mA Imax = 2.5 A Answer: The voltage gain is 24.5, and the maximum current is 2.5 A (assumes 25 mA output maximum of IC).

Answer: The output voltage is –2.67 V. 18-25. Given: v0 = 5 V v1 = 0 V v2 = 5 V v3 = 0 V Solution: vout = –(v3 + 0.5v2 + 0.25v1 + 0.125v0) vout = –(0 + 0.5(5 V) + 0 + 0.125(5 V)) vout = –3.125 V

18-30. Given: R1 = 1 kΩ

R2 = 10 kΩ

β = 125 ISC = 25 mA

Answer: The output voltage is –3.125 V. 18-26. Given: D7 – D0 = 10100101 Solution: 0 1 2 3 BIN = (D0 × 2 ) + (D1 × 2 ) + (D2 × 2 ) + (D3 × 2 ) 4 5 6 7 + (D4 × 2 ) + (D5 × 2 ) + (D6 × 2 ) + (D7 × 2 ) BIN = (1 × 20) + (0 × 21)

+ (1 × 22) + (0 × 23) + (0 × 24)

5

6

7

+ (1 × 2 ) + (0 × 2 ) + (1 × 2 ) BIN = 165 Answer: The decimal equivalent value is 165.

18-27. Given: D7 – D0 = 01100110 Vref = +5 V Solution: 0 1 2 3 BIN = (D0 × 2 ) 4+ (D1 × 2 ) + 5(D2 × 2 ) +6(D3 × 2 ) 7 + (D × 2 ) + (D × 2 ) + (D × 2 ) + (D × 2 ) 4

BIN = (0 × 20) + (1 × 21) 5

5

6

6

7

+ (1 × 2 ) + (1 × 2 ) + (0 × 2 ) BIN = 102  BIN  Vout = −  N  Vref   2   102  V =−  2(5 V) out  28   

7

+ (1 × 22) + (0 × 23) + (0 × 24)

(Eq. 18-19)

Solution: AV = –(R2/R1) AV = –(10 kΩ/1 kΩ) AV = –10 Imax = βIsc Imax = 125 (25 mA) Imax = 3.125 A Answer: The voltage gain is –10, and the maximum current is 3.125 A. 18-31. Given: R = 2 kΩ RL = 75 Ω vin = 1 V VCC = 15 V Solution: iout = vin/R iout = 1 V/2 kΩ iout = 0.5 mA RL(max) = R[(VCC/vin) – 1] RL(max) = 2 kΩ[(15 V/1 V) – 1] RL(max) = 28 kΩ Answer: The output current is 0.5 mA, and the maximum load resistance is 28 kΩ.

Vout = –3.98 V Answer: The output voltage is –3.98 V.

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18-32. Given: R = 3.3 kΩ RL = 150 Ω vin = 5 V VCC = 15 V Solution: iout = (VCC – vin)/R iout = (15 V – 5 V)/3.3 kΩ iout = 3.03 mA RL(max) = R/[(VCC/vin) – 1] RL(max) = 3.3 kΩ/[(15 V/5 V) – 1] RL(max) = 1.65 kΩ

18-36. Given: R1 = 5.1 kΩ R2 = 51 kΩ

Answer: The output current is 3.03 mA, and the maximum load resistance is 1.65 kΩ.

18-33. Given: R = 10 kΩ vin = 3 V VCC = 15 V Solution: iout = vin/R iout = 3 V/10 kΩ iout = 0.3 mA RL(max) = R[(VCC/vin) – 1] RL(max) = 10 kΩ[(15 V/3 V) – 1] RL(max) = 40 kΩ

AV(max) = [–(51 kΩ/5.1 kΩ)][(1 kΩ + 5 MΩ)/ (68 kΩ + 1 kΩ + 5 MΩ)] AV(max) = –9.87 Answer: The maximum voltage gain is –9.87, and the minimum voltage gain is –0.16.

18-37. Given: R1 = 10 kΩ R2 = 10 kΩ R5 = 75 kΩ R6 = 1.2 kΩ R7(min) = 180 Ω R7(max) = 10 MΩ

Answer: The output current is 0.3 mA, and the maximum load resistance is 40 kΩ.

18-34. Given: R = 2 kΩ RL = 500 Ω vin = 6 V vin(max) = 7.5 V

AV(min) = [–(R2/R1)][(R6 + R7(min))/(R5 + R6 + R7 (min))]

Solution: iout = –vin/R iout = 6 V/2 kΩ iout = 3 mA RL(max) = (R/2)[(VCC/vin) – 1] RL(max) = (2 kΩ/2)[(15 V/7.5 V) – 1] RL(max) = 1 kΩ Answer: The output current is 3 mA, and the maximum load resistance is 1 kΩ.

18-35. Given: R1 = 10 kΩ R2 = 100 kΩ R3 = 100 kΩ R4 = 10 kΩ rds(min) = 200 Ω rds(max) = 1 MΩ Solution: AV(min) = [(R2/R1) + 1][rds(min)/(rds(min) + R3)] AV(min) = [(100 kΩ/10 kΩ) + 1][200/(200 + 100 kΩ)] AV(min) = 0.02 AV(max) = [(R2/R1) + 1][(rds(max)/rds(max) + R3)] AV(max) = [(100 kΩ/10 kΩ) + 1][1 MΩ/(1 MΩ + 100 kΩ)]

AV(max) = 10 Answer: The maximum voltage gain is 10, and the minimum voltage gain is 0.02.

R5 = 68 kΩ

R6 = 1 kΩ rds(min) = 120 Ω rds(max) = 5 MΩ Solution: AV(min) = [–(R2/R1)][(R6 + rds(min))/(R5 + R6 + rds(min))]

AV(min) = [–(51 kΩ/5.1 kΩ)][(1 kΩ + 120 Ω)/ (68 kΩ + 1 kΩ + 120 Ω)] AV(min) = –0.16 AV(max) = [–(R2/R1)][(R6 + rds(max))/(R5 + R6 + rds(max))]

Solution: AV(min) = [–(10 kΩ/10 kΩ)][(1.2 kΩ + 180 Ω)/(75 kΩ + 1.2 kΩ + 180 Ω)] AV(min) = –0.018 AV(max) = [–( R2/R1 )][(R6 + R7(max))/(R5 + R6 + R7(max))]

AV(max) = [–(10 kΩ/10 kΩ)][(1.2 kΩ + 10 MΩ)/(75 kΩ + 1.2 kΩ + 10 MΩ)] AV(max) = –0.99 Answer: The maximum voltage gain is –0.99, and the minimum voltage gain is –0.018.

18-38. Given: R1 = 3.3 kΩ R2 = 82 kΩ RL = 10 kΩ R = 91 kΩ C1 = 4.7 µF C2 = 10 µF C3 = 4.7 µF Solution: AV = –R2/R1 AV = –82 kΩ/3.3 kΩ AV = –24.8 f1 = 1/(2π R1C1) f1 = 1/[2π (3.3 kΩ)(4.7 µF)] f1 = 10.26 Hz f2 = 1/(2π RLC2) f2 = 1/[2π (10 kΩ)(10 µF)] f2 = 1.59 Hz f3 = 1/[2π (R/2)C3] f3 = 1/[2π (91 kΩ/2)(4.7 µF)] f3 = 0.74 Hz Answer: The gain is –24.8, and the cutoff frequencies are f1 = 10.26 Hz, f2 = 1.59 Hz, and f3 = 0.74 Hz.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

18-39. Given: R1 = 1.5 kΩ R2 = 15 kΩ RL = 15 kΩ R = 68 kΩ C1 = 1 µF C2 = 2.2 µF C3 = 3.3 µF Solution: AV = (R2/R1) + 1 AV = (15 kΩ/1.5 kΩ) + 1 AV = 11 f1 = 1/[2π(R/2)C1] f1 = 1/[2π(68 kΩ/2)(1 µF)] f1 = 4.68 Hz 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 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 18-40. 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.

18-41. Given: R1(min) = 990 Ω R1(max) = 1010 Ω Rf(min) = 99 kΩ Rf(max) = 101 kΩ Solution: AV(min) = –Rf(min)/R1(max) AV(min) = –99 kΩ/1010 Ω AV(min) = –98 AV(max) = –Rf(max)/R1(min) AV(max) = –101 kΩ/990 Ω AV(max) = –102 Answer: The minimum gain is 98, and the maximum gain is 102.

18-42. Given: Transistor: R1 = 22 kΩ R2 = 10 kΩ

RS = 1 kΩ RE = 5.6 kΩ RC = 6.8 kΩ VCC = 15 V Op amp R3 = 1 kΩ Rf = 47 kΩ Solution: VBB = [R2/(R1 + R2 + RS)]VCC VBB = [10 kΩ/(22 kΩ + 10 kΩ + 1 kΩ)]15 V VBB = 4.54 V

VE = VBB – VBE VE = 4.54 V – 0.7 V VE = 3.84 V IE = VE/RE IE = 3.84 V/5.6 kΩ IE = 0.685 mA r'e = 25 mV/IE r'e = 25 mV/0.685 mA r'e = 36.5 Ω rc = Rc rc = 6.8 kΩ AV = rc/r'e AV = 6.8 kΩ/36.5 Ω AV = 186 Op amp: AV = (Rf/R3) + 1 AV = (47 kΩ/1 kΩ) + 1 AV = 48 AV= (186)(48)

AV = 9114 Answer: The voltage gain is 9114.

18-43. Given: R1 = 1 kΩ R2 = 10 kΩ RL = 100 Ω

β = 50 vin = 0.5 V Solution: AV = –R2/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 IB = Iout/β IB = 50 mA/50 IB = 1 mA Answer: The base current is 1 mA. 18-44. 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.

Trouble 3: Since the input is 2 mV and the output is maximum, R1 is shorted.

18-45. Answer: Trouble 4: Since there is no voltage at B, there is an open between K and B. 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. Trouble 6: Since the voltage at A is zero, there is an open between J and A.

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18-46. Answer: Trouble 7: Since the input voltage is 3 mV and the output is maximum, R3 is open.

Q = f0/BW (Eq. 19-3) Q = 1.86 kHz/7.36 kHz Q = 0.25

Trouble 8: Since the output is only 250 mV, which is the amplified output of B, R1 is open.

Since Q < 1, it is wideband.

Trouble 9: Since there is input voltage and no voltage reading a points C and D, the op amp is shorted at the input.

Answer: The bandwidth is 7.36 kHz, the center frequency is 1.86 kHz, the Q is 0.25, and it is wideband.

19-2.

Given:

18-48. Rf is 10 kΩ not 100 kΩ

f1 = 20 kHz f2 = 22.5 kHz Solution: BW = f2 – f1 (Eq. 19-1) BW = 22.5 kHz – 20 kHz

18-49. Rf is open

f0 =

18-50. Rf is shorted

f0 = (20 kHz)(22.5 kHz)

Trouble 10: Since the input is 5 mV and the output is maximum, R2 is shorted.

18-47. R1 is actually 10 kΩ not 1 kΩ

BW = 2.5 kHz

18-51. The closed loop feedback for the op amp U2 has opened 18-52. Noninverting voltage amplifier 18-53. Adjusts the op amp output to zero volts when the input signal is zero.

18-55. AV = AVOL; The output would be clipped at (+) and (–) 12 V 18-56. 0 Hz to 1.36 MHz

Active Filters

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

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

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

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

JOB INTERVIEW QUESTIONS 6. Low attenuation and the edge frequency. High attenuation and the edge frequency.

7. A filter designed to control the phase of a signal rather than its amplitude.

8. It compares the voltage gain to the frequency.

PROBLEMS 19-1.

Given:

f0 = 21.2 kHz Q = f0/BW (Eq. 19-3) Q = 21.2 kHz/2.5 kHz Q = 8.48 Since Q > 1, it is narrowband.

Answer: The bandwidth is 2.5 kHz, the center frequency is 21.2 kHz, the Q is 8.48, and it is narrowband.

18-54. Vout = (AV)(vin) = (11)(0.5 Vp-p) = 5.5 Vp-p

Chapter 19

(Eq. 19-2)

19-3a. Given: f1 = 2.3 kHz f2 = 4.5 kHz Solution: BW = f2 – f1 (Eq. 19-1) BW = 4.5 kHz – 2.3 kHz BW = 2.2 kHz f0 = (Eq. 19-2) f0 = (2.3 kHz)(4.5 kHz) f0 = 3.2 kHz Q = f0/BW (Eq. 19-3) Q = 3.2 kHz/2.2 kHz Q = 1.45 Since Q > 1, it is narrowband.

Answer: Narrowband.

19-3b. Given: f1 = 47 kHz f2 = 75 kHz Solution: BW = f2 – f1 (Eq. 19-1) BW = 75 kHz – 47 kHz BW = 28 kHz f0 = (Eq. 19-2)

f1 = 445 Hz

f0 = (75 kHz)(47 kHz)

f2 = 7800 Hz

f0 = 59.4 kHz

Solution: (Eq. 19-1) BW = f2 – f1 BW = 7800 Hz – 445 Hz BW = 7355 f0 = f f (Eq. 19-2) f0 = (445 Hz)(7800 Hz)

Q = f0/BW (Eq. 19-3) Q = 59.4 kHz/28 kHz Q = 2.12 Since Q > 1, it is narrowband.

1 2

Answer: Narrowband.

f0 = 1.86 kHz

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

19-3c. Given: f1 = 2 Hz f2 = 5 Hz Solution: BW = f2 – f1 (Eq. 19-1) BW = 5 Hz – 2 Hz BW = 3 Hz f0 = (Eq. 19-2)

19-7.

f0 = 3.16 Hz Q = f0 /BW (Eq. 19-3) Q = 3.16 Hz/3 Hz Q = 1.05

L = 20 mH C = 5 μF R = 600 Ω Solution: f0 = 1/(2π LC ) (Eq. 19-6) f0 = 1/[2π (20 mH)(5μF) ] f0 = 503 Hz XL = 2πf0L XL = 2π(503 Hz)(20 mH) XL = 63.2 Ω Q = R/XL (Eq. 19-7) Q = 600/63.2 Q = 9.5

Since Q > 1, it is narrowband.

Answer: Narrowband.

Answer: The resonant frequency is 503 Hz, and the Q is 9.5.

f0 = (2 Hz)(5 Hz)

19-3d. Given: f1 = 80 Hz f2 = 160 Hz Solution: BW = f2 – f1

19-8.

(Eq. 19-1)

f0 = (160 Hz)(80 Hz) f0 = 113 Hz Q = f0/BW (Eq. 19-3) Q = 113 Hz/80 Hz Q = 1.4 Since Q > 1, it is narrowband.

Given: Seven capacitors

19-9.

Given:

R1 = 15 kΩ C = 270 nF

Answer: Seventh order Given: 10 capacitors

(Eq. 19-4)

Solution: fC = 1/(2πR1C1)

Roll-off = 20n dB/decade Roll-off = 20(10) dB/decade Roll-off = 200 dB/decade

(Eq. 19-4a)

fC = 39.3 Hz

Roll-off = 6n dB/octave Roll-off = 6(10) dB/octave Roll-off = 60 dB/octave

(Eq. 19-4a)

Solution: n = number of capacitors n = 10

(Eq. 19-6)

Answer: The resonant frequency is 712 Hz, and the Q is 13.4.

Answer: Narrowband.

(Eq. 19-9)

fC = 1/[2π(15 kΩ)(270 nF)]

Answer: The cutoff frequency is 39.3 Hz.

Answer: The roll-off rate is roll-off = 200 dB/decade or roll-off = 60 dB/octave.

19-6.

)

mH)(5µF) ]

f0 = 712 Hz XL = 2πf0L XL = 2π(712 Hz)(10 mH) XL = 44.7 Ω Q = R/XL (Eq. 19-7) Q = 600/44.7 Q = 13.4

1 2

19-5.

Given:

L = 10 mH C = 5 μF R = 600 Ω Solution: f0 = 1/(2π f0 = 1/[2π

BW = 160 Hz – 80 Hz BW = 80 Hz f0 = f f (Eq. 19-2)

19-4.

Given:

Given: 14 capacitors

Solution: n = number of capacitors n = 14

(Eq. 19-4)

Number of ripples = n/2 Number of ripples = 14/2 Number of ripples = 7

(Eq. 19-5)

19-10. Given: R1 = 7.5 kΩ R2 = 33 kΩ R3 = 20 kΩ C = 680 pF Solution: AV = (R2 /R1) + 1 (Eq. 19-10) AV = (33 kΩ/7.5 kΩ) + 1 AV = 5.4 fC = 1/(2πR3C1) (Eq. 19-11) fC = 1/[2π (20 kΩ)(680 pF)]

Answer: There are seven ripples.

fC = 11.7 kHz Answer: The voltage gain is 5.4, and the cutoff frequency is 11.7 kHz.

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Q = 0.5 19-11. Given: R1 = 2.2 kΩ R2 = 47 kΩ C = 330 pF

Q = 0.5

Solution: AV = –R2/R1 (Eq. 19-12) AV = –47 kΩ/2.2 kΩ AV = –21.4 fC = 1/(2πR2C1) (Eq. 19-13) fC = 1/[2π(47 kΩ)(330 pF)]

fC = 10.3 kHz Answer: The voltage gain is –21.4, and the cutoff frequency is 10.3 kHz. 19-12. Given: R1 = 10 kΩ C1 = 15 nF

(Eq. 19-19)

(200 pF)/ (100 pF)

Q = 0.707 Since it is a Butterworth response, the cutoff and 3-dB frequencies are the same as the pole frequency.

Answer: The frequencies are 15 kHz, and the Q is 0.707.

19-16. Given: R = 51 kΩ C1 = 100 pF C2 = 680 pF Solution: fp = 1/(2πR

(Eq. 19-20) fp = 1/[2π(51 kΩ) (100 pF)(680 pF) ] fp = 12 kHz

Q = 0.5 (C / C ) (Eq. 19-19) Q = 0.5 (680 pF)/ (100 pF) 2

Solution: fC = 1/(2πR1C1)

(Eq. 19-14)

1

Q = 1.3

fC = 1/[2π(10 kΩ)(15 nF)]

fc = Kc fp

fC = 1.06 kHz

fc = 1.12(12 kHz)

Answer: The cutoff frequency is 1.06 kHz.

fc = 13.44 kHz f3dB = K3 fp

19-13. Given: R1 = 12 kΩ R2 = 24 kΩ R3 = 20 kΩ C = 220 pF

fC = 1/[2π(20 kΩ)(220 pF)]

fC = 36.2 kHz Answer: The voltage gain is 3, and the cutoff frequency is 36.2 kHz. 19-14. Given: R1 = 8.2 kΩ C1 = 560 pF

C2 = 680 pF (Eq. 19-17)

AV = –560 pF/680 pF

AV = –0.824 fC = 1/(2πR1C2)

(Eq. 19-18)

fC = 1/[2π(8.2 kΩ)(680 pF)]

fC = 28.5 kHz Answer: The voltage gain is –0.824, and the cutoff frequency is 28.5 kHz. 19-15. Given: R = 75 kΩ C1 = 100 pF C2 = 200 pF

19-17. Given: R1 = 51 kΩ R2 = 30 kΩ R = 33 kΩ C = 220 pF Solution: AV = (R2/R1) + 1 (Eq. 19-29) AV = (30 kΩ/51 kΩ) + 1 AV = 1.58 Q = 1/(3 – AV) (Eq. 19-30) Q = 1/(3 – 1.58) Q = 0.707 fp = 1/(2πRC) fp = 1/[2π(33 kΩ) (220 pF)] fp = 21.9 kHz Since it is a Butterworth response, the cutoff and 3-dB frequencies are the same as the pole frequency.

Answer: The frequencies are 21.9 kHz, and the Q is 0.707.

19-18. Given: R1 = 33 kΩ R2 = 33 kΩ R = 75 kΩ C = 100 pF

V

)

(Eq. 19-20)

fp = 1/[2π(75 kΩ) (100 pF)(200 pF) ] fp = 15 kHz

f3dB = 16.32 kHz Answer: The pole frequency is 12 kHz, the cutoff frequency is 13.44 kHz, the 3-dB frequency is 16.32 kHz, and the Q is 1.3

Solution: A 1) + 1kΩ) + (Eq. AV = = (R (332/R kΩ/33 1 19-29)

Solution:

fp = 1/(2π R

(Eq. 19-24)

f3dB = 1.36 (12 kHz)

Solution: AV = (R2/R1) + 1 (Eq. 19-15) AV = (24 kΩ/12 kΩ) + 1 AV = 3 fC = 1/(2πR3C1) (Eq. 19-16)

Solution: AV = –C1/C2

(Eq. 19-23)

AV = 2 Q = 1/(3 – AV) Q = 1/(3 – 2) Q=1

(Eq. 19-30)

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

fp = 1/(2πRC) fp = 1/[2π(75 kΩ)(100 pF)] fp = 21.2 kHz Kc = 1.000 (from Table 19-3) K3 = 1.272 fc = Kc fp

(from Table 19-3) (Eq. 19-23)

fc = 1.000(21.2 kHz)

fc = 21.2 kHz f3 = K3 fp (Eq. 19-24) f3 = 1.272(21.2 kHz)

f3 = 27 kHz

Answer: The pole frequency is 21.2 kHz, the cutoff frequency is 21.2 kHz, the 3-dB frequency is 27 kHz, and the Q is 1.

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.

19-21. Given: R1 = 91 kΩ R2 = 15 kΩ C = 220 pF Solution: fp = 1/(2πC

fp = 1/[2π(220 pF) (15 kΩ)(91 kΩ) ] fp = 19.6 kHz Q = 0.5 (R /R ) 1

19-19. Given: R1 = 75 kΩ R2 = 56 kΩ R = 68 kΩ C = 120 pF

R1 R 2 )

Q = 0.5

2

(91 kΩ)/ (15 kΩ)

Q = 1.23

Solution: AV = (R2/R1) + 1 (Eq. 19-29) AV = (56 kΩ/75 kΩ) + 1 AV = 1.75 Q = 1/(3 – Av) (Eq. 19-30) Q = 1/(3 – 1.75) Q = 0.8 fp = 1/(2πRC) fp = 1/[2π (68 kΩ)(120 pF)] fp = 19.5 kHz Kc = 0.661 (from Table 19-3) K3 = 1.115 (from Table 19-3) fc = Kc fp (Eq. 19-23) fc = 0.661(19.5 kHz)

fc = 12.89 kHz f3 = K3 fp (Eq. 19-24) f3 = 1.115(19.5 kHz)

Kc = 1.06 K3 = 1.32 fc = fp/Kc

(from Fig. 19-26) (from Fig. 19-26) (Eq. 19-31)

fc = 19.6 kHz/1.06

fc = 18.5 kHz f3 = fp/K3 f3 = 19.6 kHz/1.32 f3 = 14.8 kHz 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.

19-22. Given: R1 = 2 kΩ R2 = 56 kΩ C = 270 pF Solution: AV = –R2/2R1 (Eq. 19-32) AV = –56 kΩ/2(2 kΩ)

f3 = 21.74 kHz

AV = –14

Answer: The pole frequency is 19.5 kHz, the cutoff frequency is 12.89 kHz, the 3-dB frequency is 21.74 kHz, and the Q is 0.8.

Q = 0.5 Q = 0.5 (56 kΩ)/ (2 kΩ) Q = 2.65 f0 = 1/[2πC (R1R2 ) ]

19-20. Given: R1 = 56 kΩ R2 = 10 kΩ C = 680 pF Solution: fp = 1/[2π (680 pF) (10 kΩ)(56 kΩ) ] fp = 9.89 kHz Q = 0.5 Q = 0.5 (56 k) / (10 k) Q = 1.18 (from Fig. 19-26) (from Fig. 19-26) (Eq. 19-31)

fc = 9.89 kHz/1.04

fc = 9.51 kHz

(Eq. 19-36)

f0 = 1/[2π(270 pF) (2 kΩ)(56 kΩ) ]f0

fp = 1/(2πC R1 R2 )

Kc = 1.04 K3 = 1.30 fc = fp/Kc

(Eq. 19-33)

f0 = 55.7 kHz Answer: The Q is 2.65, the voltage gain is –14, and the center frequency is 55.7 kHz. 19-23. Given: R1 = 3.6 kΩ R2 = 7.5 kΩ R3 = 27 Ω C = 22 nF Solution: AV = –R2/2R1

(Eq. 19-32)

AV = –7.5 kΩ/2(3.6 kΩ)

AV = –1.04 Q = 0.5 Q = 0.5 Q = 8.39

[(R2 / (R1 || R3 )]

(Eq. 19-37)

7.5 kΩ/ (2 kΩ || 27 Ω)

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f0 = 1/[2πC

(R1 || R3 )R2 ]

(Eq. 19-38)

f0 = 1/[2π(22 nF) (2 kΩ ||27Ω) (7.5 kΩ)] f0 = 16.2 kHz Answer: The Q is 8.39, the voltage gain is –1.04, and the center frequency is 16.2 kHz. 19-24. Given: R1 = 28 kΩ R3 = 1.8 kΩ C = 1.8 nF AV = –1 Solution: Q = 0.707 Q = 0.707 Q = 2.88

[(R1 + R3 ) / R3 ]

(Eq. 19-40)

[(28 kΩ +1.8 kΩ)/1.8 kΩ]

f0 =1/(2πC [2R (R || R )] ) (Eq. 19-41) f0 =1/{2π(1.8 nF) 2(28 kΩ)(28 kΩ || 1.8 kΩ) } f0 = 9.09 kHz Answer: The Q is 2.88, the voltage gain is –1, and the center frequency is 9.09 kHz 1

1

3

Solution: AV = (R2/R1) + 1 (Eq. 19-43) AV = (10 kΩ/20 kΩ) + 1 AV = 1.5 f0 = 1/(2πRC) (Eq. 19-44) f0 = 1/[2π(56 kΩ)(180 nF)]

(Eq. 19-45)

BW = f0/Q BW = 15.8 Hz/1 BW = 15.8 Hz 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. 19-26. Given: R = 3.3 kΩ C = 220 nF Solution: f0 = 1/(2πRC) f0 = 1/[2π(3.3 kΩ)(220 nF)] f0 = 219 Hz f = 2(219 Hz) f = 438 Hz ϕ = –2 arctan f/f0 ϕ = –2 arctan (438 Hz/219 Hz) ϕ = –127° 19-27. Given: R = 47 kΩ C = 6.8 nF Solution: f0 = 1/(2πRC)

f0 = 498 Hz f = 0.5(498 Hz) f = 249 Hz ϕ = 2 arctan f/f0 ϕ = 2 arctan (498 Hz/249 Hz) ϕ = 127° 19-28. Given: R1 = 24 kΩ R2 = 100 kΩ R3 = 10 kΩ R4 = 15 kΩ C = 3.3 nF Solution: AV = –R2/R1 AV = –100 kΩ/24 kΩ AV = –4.17 Q = R2/R3 Q = 100 kΩ/10 kΩ Q = 10 f0 = 1/(2πR3C ) f0 = 1/[2π(10 kΩ)(3.3 nF)] f0 = 4.82 kHz BW = f0/Q (Eq. 19-34) BW = 4.82 kHz/10 BW = 482 Hz

19-25. Given: R1 = 20 kΩ R2 = 10 kΩ R = 56 kΩ C = 180 nF

f0 = 15.8 Hz Q = 0.5/(2 – Av) Q = 0.5/(2 – 1.5) Q=1

f0 = 1/[2π(47 kΩ)(6.8 nF)]

Answer: The voltage gain is – 4.17, the Q is 10, the center frequency is 4.82 kHz, and the bandwidth is 482 Hz.

19-29. Given: R1 = 24 kΩ R2 = 100 kΩ R3(min) = 2 kΩ R3(max) = 10 kΩ R4 = 15 kΩ C = 3.3 nF Solution: Q(min) = R2/R3(max) Q(min) = 100 kΩ/10 kΩ Q(min) = 10 f0(min) = 1/(2πR3(max)C) f0(min) = 1/[2π(10 kΩ)(3.3 nF)] f0(min) = 4.82 kHz Q(max) = R2/R3(min) Q(max) = 100 kΩ/2 kΩ Q(max) = 50 f0(max) = 1/(2πR3(min)C) f0(max) = 1/[2π(2 kΩ)(3.3 nF)] f0(max) = 24.1 kHz BW =

1 2 R2C

=

1 = 482 Hz 2π (100 kΩ)(3.3 nF)

Answer: The maximum center frequency is 24.1 kHz, the maximum Q is 50, the minimum bandwidth is 482 Hz, and the maximum bandwidth 482 Hz. 19-30. Given: R = 6.8 kΩ R1 = 6.8 kΩ R2 = 100 kΩ C = 5.6 nF

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

Solution: AV = 1/3[(R2/R1) + 1] AV = 1/3[(100 kΩ/6.8 kΩ) + 1] AV = 5.23 Q = Av Q = 5.23

19-34. Given: n=2 R = 10 kΩ fc = 5 kHz Solution:

Q = 0.5 C / C 0.707 = 0.5 C / C 1.414 = C / C 2 = C2/C1 C2= 2C1 fp = 1/2πR C (2C ) fp = 1/2πRC1 2 C1 = 1/2πRfp 2 C1 = 1/2π(10 kΩ) C1 = 2.25 nF C2 = 4.5 nF 2

f0 = 1/(2πRC) f0 = 1/[2π(6.8 kΩ)(5.6 nF)] f0 = 4.18 kHz Answer: The voltage gain and Q are 5.23, and the center frequency is 4.18 kHz.

2

2

19-31. Given: f0 = 50 kHz Q = 20

f1 = f0 – ½ BW f1 = 50 kHz – ½ (2.5 kHz) f1 = 48.75 kHz f2 = f0 + ½ BW f2 = 50 kHz + ½ (2.5 kHz) f2 = 51.25 kHz Answer: The cutoff frequencies are 48.75 kHz and 51.25 kHz.

(5 kHz)

2

Ap = 12 dB Solution: Since Ap = 12 dB, Kc = 1.391 and Q = 4 (from Table 19-3) fc = Kcfp (Eq. 19-23) fp = fc/Kc fp = 7.5 kHz/1.391 fp = 5.39 kHz Q = 0.5 (from Fig. 19-25) 4 = 0.5 (Chebyshev response) 8 = C /C 64 = C2/C1 C2 = 64C1 fp = 1/2πR C (64C ) fp = 1/16πRC1 C1= 1/16πRfp C1= 1/16π (25 kΩ) (5.39 kHz) C1= 148 pF

19-32. Given: f2 = 84.7 kHz BW = 12.3 kHz

2

Solution: f1 = f2 – BW f1 = 84.7 kHz – 12.3 kHz f1 = 72.4 kHz Answer: The lower cutoff frequency is 72.4 kHz. 19-33. Given: n = 10 fc = 2 kHz

Roll-off = 20n dB/decade Roll-off = 20(10) dB/decade Roll-off = 200 dB/decade

1

19-35. Given: n=2 R = 25 kΩ fc = 7.5 kHz

Solution: BW = f0/Q BW = 50 kHz/20 BW = 2.5 kHz

Solution: Roll-off = 6n dB/octave Roll-off = 6(10) dB/octave Roll-off = 60 dB/octave

1

1

1

CRITICAL THINKING

(from Fig. 19-24) (Butterworth response)

1

1

1

1

C2= 9.47 nF 19-36. R3 is open 19-37. U1 has failed 19-38. C4 is shorted

(Eq. 19-4b)

19-39. C3 is open 19-40. U2 has failed

(Eq. 19-4a)

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.

Chapter 20 Nonlinear Op-Amp Circuit Applications SELF-TEST 1. d 2. a 3. c 4. b 5. c 6. a 7. a 8. b

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

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

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

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1-108

JOB INTERVIEW QUESTIONS

20-7.

VS = ±12 V R1 = 15 kΩ

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.

R2 = 7.5 kΩ

C = 1 μF

Solution: vref = [R2/(R1 + R2)]VEE (Eq. 20-2) vref = [7.5 kΩ/(15 kΩ + 7.5 kΩ)]−12 V vref = −4 V fc = 1/[2π (R1 || R2)C] (Eq. 20-3) fc = 1/[2π (15 kΩ || 7.5 kΩ)1 μF] fc = 31.8 Hz Answer: The reference voltage is −4 V, and the cutoff frequency is 31.8 Hz.

PROBLEMS 20-1.

Given:

AV(dB) = 106 dB Vsat = ±20 V Solution: AVOL = antilog(Av(dB)/20) AVOL = antilog(l06 dB/20) AVOL = 200,000 vin(min) = ±Vsat/AVOL (Eq. 20-1)

20-8.

vin(min) = ±100 µV Answer: An input voltage of 100 µV will produce positive saturation, assuming rail-to-rail output.

Solution: vref = [R2/(R1 + R2)]VCC (Eq. 20-2) vref = [4.7 kΩ/(22 kΩ + 4.7 kΩ)]9 V vref = 1.58 V vin = vin(peak)sinθ θ = arcsin(vin/vin(peak)) θ = arcsin(1.58 V/7.5 V) θ = 12° and 168° D = conduction angle/360° D = (168° − 12°)/360° D = 43.3%

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 Solution: The diode current is 4.93 mA. 20-3.

Given:

VZ = 6.8 V VS = ±15 V Solution: vout = ±(VZ + VD) vout = ±(6.8 V + 0.7 V) vout = ±7.5 V Answer: The output voltage will be limited to ±7.5 V. 20-4. Given: VS = ±12 V Answer: The output voltage would vary between 0.7 V and –12 V. 20-5. 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. 20-6.

Given:

VS = ±15 V R1 = 47 kΩ R2 = 12 kΩ CBY = 0.5 µF Solution: vref = [R2/(R1 + R2)]VCC (Eq. 20-2) vref = [12 kΩ /(47 kΩ + 12 kΩ)]15 V vref = 3.05 V fc = 1/[2π(R1 || R2)CBY] (Eq. 20-3) 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.

Given:

VCC = 9 V R1 = 22 kΩ R2 = 4.7 kΩ vin(peak) = 7.5 V

vin(min) = ±20 V/200,000

20-2.

Given:

Answer: The duty cycle is 43.3%.

20-9.

Given:

VCC = 15 V R1 = 33 kΩ R2 = 3.3 kΩ vin(peak) = 5 V Solution: vref = [R2/(R1= R2)]VCC (Eq. 20-2) vref = [3.3 kΩ/(33 kΩ + 3.3 kΩ)]15 V vref = 1.36 V vin = vin(peak)sinθ θ = arcsin(vin/vin(peak)) θ = arcsin(1.36 V/5 V) θ = 16° and 164° D = conduction angle/360° D = (164° − 16°)/360° D = 41% Answer: The duty cycle is 41%.

20-10. Given: Vsat = 14 V R1 = 2.2 kΩ R2 =18 kΩ Solution: B = R1/(R1 + R2) (Eq. 20-4) B = 2.2 kΩ/(2.2 kΩ + 18 kΩ) B = 0.1089

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

UTP = BVsat UTP = 0.1089(14 V) UTP = 1.52 V

(Eq. 20-6)

LTP = −BVsat (Eq. 20-7) LTP = −0.1089(14 V) LTP = −1.52V H = 2BVsat H = 2(0.1089)(14 V) H = 3.05 V

(Eq. 20-9)

Answer: The upper trip point is 1.52 V, the lower trip point is −1.52 V. The hysteresis is 3.05 V. 20-11. Given: R1 = 1 kΩ R2 = 20 kΩ Vsat = 15 V Solution: UTP = (R1/R2)Vsat (Eq. 20-10) UTP = (1 kΩ/20 kΩ)15 V UTP = 0.75 V LTP = −(R1/R2)Vsat (Eq. 20-11) LTP = −(1 kΩ/20 kΩ)15 V LTP = −0.75 V H = UTP − LTP (Eq. 20-8) H = 0.75 V − (−0.75 V) H = 1.5 V Answer: The maximum peak-to-peak noise allowed is 1.5 V. 20-12. Given: R1 = 1 kΩ R2 = 18 kΩ C1 = 3 pF Solution: C2 = (R1/R2)C1 (Eq. 20-12) C2 = (1 kΩ/18 kΩ)3 pF C2 = 0.167 pF Answer: The speed-up capacitor should be at least 0.167 pF. 20-13. Given: R1 = 1.5 kΩ R2 = 68 kΩ Vsat = 13.5 V Solution: B = R1/(R1 + R2) (Eq. 20-4) B = 1.5 kΩ/(1.5 kΩ + 68 kΩ) B = 0.0216 UTP = BVsat (Eq. 20-6) UTP = 0.0216(13.5 V) UTP = 0.292 V LTP = −BVsat (Eq. 20-7) LTP = −0.0216(13.5 V) LTP = −0.292 V H = 2BVsat (Eq. 20-9) H = 2(0.0216)(13.5 V) H = 0.584 V Answer: The upper trip point is 0.292 V, the lower trip point is −0.292, and the hysteresis is 0.584 V.

20-14. Given: R1 = 2.2 kΩ R2 = 68 kΩ Vsat = 14 V Solution: UTP = (R1/R2)Vsat (Eq. 20-10) UTP = (2.2 kΩ/68 kΩ)14 V UTP = 0.453 V LTP = −(R1/R2)Vsat (Eq. 20-11) LTP = −(2.2 kΩ/68 kΩ)14 V LTP = −0.453 V H = UTP − LTP (Eq. 20-8) H = 0.453 V − (−0.453 V) H = 906 mV Answer: The upper trip point is 0.453 V, the lower trip point is −0.453 V, and the hysteresis is 906 mV. 20-15. Given: LTP = 3.5 V UTP = 4.75 V vin(peak) = 10 V Vsat = 12 V Solution: The output voltage will be low when the input voltage is between 3.5 and 4.75 V. 20-16. Given: VCC = 12 V Solution: UTP = [R/(4R + R)]12 V UTP = 2.4 V LTP = [R/(6R + R)]12 V LTP = 1.71 V Answer: The lower trip point is 1.71 V, and the upper trip point is 2.4 V. 20-17. Given: Vin = 5 V R =1 kΩ Solution: Iin = Vin /R Iin = 5 V/1 kΩ Iin = 5 mA Answer: The charging current is 5 mA. 20-18. Given: Vin = 5 V R = 1 kΩ C = 10 μF T = 1 ms Solution: V = [T/(RC)]Vin (Eq. 20-13) V = [1 ms/(1 kΩ)(10 μF)]5 V V = 0.5 V Answer: The output voltage is 0.5 V. 20-19. Given: Vin = 0.1 V R = 1 kΩ C(1) = 0.1 μF C(2) = 1 μF C(3) = 10 μF C(4) = 100 μF T = 1 ms

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Solution: V(1) = [T/(RC(1))]Vin (Eq. 20-13) V(1) = [1 ms/(1 kΩ)(0.1 μF)](0.1 V) V(1) = 1 V V(2) = [T/(RC(2))]Vin (Eq. 20-13) V(2) = [1 ms/(1 kΩ)(1 μF)](0.1 V) V(2) = 0.1 V V(3) = [T/(RC(3))]Vin (Eq. 20-13) V(3) = [1 ms/(1 kΩ)(10 μF)](0.1 V) V(3) = 10 mV V(4) = [T/(RC(4))]Vin (Eq. 20-13) V(4) = [1 ms/(1 kΩ)(100 μF)](0.1 V) V(4) = 1 mV Answer: The output voltages are 1 V with a 0.1-μF capacitor, 0.1 V with a 1-μF capacitor, 10 mV with a 10-μF capacitor, and 1 mV with a 100-μF capacitor.

20-20. Given: f = 10 kHz R = 4.7 kHz C = 6.8 μF VP = 5 V Solution: Vout(p-p) = VP/[2fRC] (Eq. 20-17) Vout(p-p) = 5 V/[2(10 kHz)(4.7 kΩ)(6.8 ΩF)] Vout(p-p) = 7.8 mV Answer: The output voltage is a triangular wave with a peak-to-peak voltage of 7.8 mV. 20-21. Given: f = 10 kHz R = 4.7 kHz C = 0.068 μF Vp = 5 V Solution: Vout(p-p) = Vp/[2fRC] (Eq. 20-17) Vout(p-p) = Vp/[2(10 kHz)(4.7 kΩ)(0.068 μF)] Vout(p-p) = 0.782 Vp-p Answer: The output voltage is a triangular waveform with a peak-to-peak voltage of 0.782 V. 20-22. Given: f(1) = 5 kHz f(2) = 20 kHz R = 4.7 kHz C = 6.8 μF VP = 5 V Solution: Vout(p-p)1 = Vp /[2fRC] (Eq. 20-17) Vout(p-p)1 = 5 V/[2(5 kHz)(4.7 kΩ)(6.8 μF)] Vout(p-p)1 = 15.6 mV Vout(p-p)2 = Vp /[2fRC] (Eq. 20-17) Vout(p-p)2 = 5 V/[2(20 kHz)(4.7 kΩ)(6.8 μF)] Vout(p-p)2 = 3.9 mV Answer: The output voltage is 15.6 mV at 5 kHz, and 3.9 mV at 20 kHz. 20-23. Given: R1 = 5 kΩ

R2 = 0 Ω (top) and 10 kΩ (bottom)

VCC = 15 V VP = 10 V

Solution: vref = [R2/(R2 +R1)]VCC vref = [10 kΩ/(10 kΩ + 5 kΩ)]15 V vref = 10 V Since the reference voltage is the same as the maximum voltage, the output does not go high and the duty cycle is 0%.

With the wiper at the top, the reference voltage is zero and the output is high half of the time. Therefore, the duty cycle is 0.50.

Answer: The duty cycle is 0.5 at the top and 0 at the bottom.

20-24. Given: R1 = 5 kΩ R2 = 5 kΩ VCC = 15 V VP = 10 V Solution: vref = [R2/(R2 + R1)]VCC vref = [5 kΩ/(5 kΩ + 5 kΩ)]15 V vref = 7.5 V D = (VP − vref)/(VP − (−VP)) D =(10 V − 7.5 V)/(10 V − (−10 V)) D = 0.125 Answer: The output is high for 1/8 the cycle, therefore the duty cycle is 0.125.

20-25. Given: R1 = 33 kΩ

R2 = 4.7 kΩ

R = 2 kΩ C = 0.1 μF

Solution: B = R1/(R1 + R2) (Eq. 20-4) B = 33 kΩ/(33 kΩ + 4.7 kΩ) B = 0.875 T = 2RCln[(1 + B)/(1 − B)] (Eq. 20-18) T = 2(2 kΩ)(0.1 μF)ln[(1+ 0.875)/(1 − 0.875)] T = 1.08 ms f = 1/T f = 1/1.08 ms f = 923 Hz Answer: The frequency is 923 Hz. 20-26. Given: R1 = 66 kΩ

R2 = 9.4 kΩ

R = 4 kΩ C = 0.1 μF

Solution: B = R1/(R1 + R2) B = 66 kΩ/(66 kΩ + 9.4 kΩ) B = 0.875

(Eq. 20-4)

T = 2RCln[(1 + B)/(1 − B)] (Eq. 20-18) T = 2(4 kΩ)(0.1 μF)ln[(1 + 0.875)/(1 − 0.875)] T = 2.16 ms f = 1/T f = 1/2.16 ms f = 463 Hz Answer: The frequency is reduced by half.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

20-27. Given: R1 = 33 kΩ R2 = 4.7 kΩ

R = 2 kΩ C = 0.47 μF Solution: B = R1/(R1 + R2) (Eq. 20-4) B = 33 kΩ/(4.7 kΩ + 33 kΩ) B = 0.875 T = 2RCln[(1 + B)/(1 − B)] (Eq. 20-18) T = 2(2 kΩ)(0.47 μF)ln[(1 + 0.875)/(1 − 0.875)] T = 5.09 ms f = 1/T f = 1/5.09 ms f = 196 Hz Answer: The frequency is 196 Hz. 20-28. Given: R1 = 2.2 kΩ R2 = 22 kΩ Vsat = 12 V Solution: UTP = (R1/R2)Vsat (Eq. 20-10) UTP = (2.2 kΩ/22 kΩ)12 V UTP = 1.2 V LTP = −(R1/R2)Vsat (Eq. 20-11) LTP = −(2.2 kΩ/22 kΩ)12 V LTP = −1.2 V H = UTP – LTP (Eq. 20-8) H = 1.2 V − (−1.2 V) H = 2.4 V Answer: The upper trip point is 1.2 V, and the lower trip point is −1.2 V. The hysteresis is 2.4 V. 20-29. Given: f = 5 kHz R3 = 2.2 kΩ R4 = 22 kΩ C = 4.7 μF Vp-p = 28 V Vp = 14 V Solution: Vout(p-p) = Vp/[2fR3C] (Eq. 20-17) Vout(p-p) = 14 V/[2(5 kHz)(2.2 kΩ)(4.7 μF)] Vout(p-p) = 135 mVp-p Answer: The output voltage is 135 mVp-p.

20-30. Given: vin = 100 mVP Answer: The output will be a half-wave signal with a peak voltage of 100 mV. 20-31. Given: vin = 75 mVrms f = 20 kHz Solution: vin(p) = 1.414(vin(rms)) vin(p) = 1.414(75 mVrms) vin(p) = 106 mV peak Answer: The output voltage is 106 mV.

20-32. Given: RL = 33 kΩ C = 6.8 μF Solution: RLC > 10T (Eq. 20-20) RLC = 10T for the highest period or lowest frequency [(33 kΩ)(6.8 μF)]/10 = T T = 22.4 ms f = 1/T f = 122.4 ms f = 45 Hz Answer: The lowest frequency is 45 Hz. 20-33. Answer: With the diode reversed, it becomes a negative peak detector and the output voltage is –106 mV. 20-34. Given: vin = 150 mVp-p Answer: The output voltage is 75 mV. 20-35. Given: vin = 100 mVp Solution: vout = vin + Vp (Eq. 20-21) vout = 100 mVp + 100 mVp vout = 200 mVp Answer: The output swings from 0 V to 200 mV peak.

20-36. Given: RL = 10 kΩ C = 4.7 μF Solution: RLC > 10T (Eq. 20-20) RLC = 10T for the highest period or lowest frequency [(10 kΩ)(4.7 μF)]/10 = T T = 4.7 ms f = 1/T f = 1/4.7 ms f = 213 Hz Answer: The lowest frequency is 213 Hz. 20-37. Given: f = 10 kHz Solution: 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.

20-38. 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 Answer: A pulse occurs every T/2 or 0.5 ms.

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CRITICAL THINKING 20-39. Answer: Make the 3.3-kΩ resistor a variable so that it can be adjusted to any desired value.

20-40. Given: R = 1 kΩ C = 50 pF Solution: Risetime is from the 10% point to the 90% point (discussed in Chap. 14). Using the universal time constant chart, it takes about 3 time constants to go from 10% to 90%.

TR ≈ 2.2(RC) (Eq. 14-28) TR ≈ 2.2(1 kΩ)(50 pF) TR ≈ 110 ns Answer: The risetime is 110 ns.

20-41. Given: R1 = 33 kΩ R2 = 3.3 kΩ C = 47 μF Vripple = 1 Vrms Solution: fC = 1/[2π(R1 || R2)C] (from Fig. 20-11) 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.

R2(max) = 68 kΩ + 5%(68 kΩ) = 71.4 kΩ R2(min) = 68 kΩ – 5%(68 kΩ) = 64.6 kΩ B(min) = R1(min)/(R1(min) + R2(max)) (Eq. 20-4) B(min) = 1425 Ω/(1425 Ω + 71.4 kΩ) B(min) = 0.0196 H(min) = 2B(min) Vsat (Eq. 20-9) H(min) = 2(0.0196)(13.5 V) H(min) = 0.529 V Answer: The minimum hysteresis is 0.529 V. 20-44. Given: LTP = 3.5 V UTP = 4.75 V vin(peak) = 10 V Vsat = 12 V Solution: vin = vin(peak) sinθ θ = arcsin(vin/vin(peak)) θ = arcsin(3.5 V/10 V) θ = 21° and 159° vin = vin(peak) sinθ θ = arcsin(vin/vin(peak)) θ = arcsin(4.75 V/10 V) θ = 28° and 152° The output will be high from 21° to 28° and from 152° to 159°, or a total of 14° D = conduction angle/360° D = 14°/360° D = 3.9% Answer: The duty cycle is 0.039 or 3.9%.

Answer: The cutoff frequency is 1.1 Hz, and the ripple voltage at the inverting input is 0.001 V rms.

20-42. Given: VCC = 15 V R1 = 33 kΩ R2 = 3.3 kΩ vin(peak) = 5 V vref = 1.36 V (from Prob. 20-9) θ = 16° and 164° (from Prob. 20-9) D = 41% (from Prob. 20-9) 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. 20-43. Given: R1 = 1.5 kΩ ± 5% R2 = 68 kΩ ± 5% Vsat = 13.5 V Solution: R1(max) = 1.5 kΩ + 5%(1.5 kΩ) = 1575 Ω R1(min) = 1.5 kΩ – 5%(1.5 kΩ) = 1425 Ω

20-45. Given: With Eq. (20-13), the output voltage at the end of the pulse is V = [(1 ms)/(1 kΩ)(10 μF)](5 V) = 0.5 V C = (T/RV)Vin C = [(1 ms)/(1 kΩ)(10 V)](5 V) = 0.5 μF The foregoing is for T = 1 ms. Therefore, we need 0.05 μF for T = 0.1 ms and 5 μF for T = 10 ms.

Answer: Switch in different capacitors of 0.05, 0.5, and 5 μF. Also, cascade an inverter to get a positive-going output.

20-46. Given: f = 20 kHz R1 = 33 kΩ R2 = 4.7 kΩ Solution: B = R1/(R1 + R2) (Eq. 19-6) B = 33 kΩ/(33 kΩ + 4.7 kΩ) B = 0.875 T = 1/f T = 1/20 kHz T = 50 μs T = 2RCln[(1 + B)/(1 − B)] (Eq. 20-18) C = T/2Rln[(1 + B)/(1 − B)] C = 50 μs/2(2 kΩ)ln[(1 + 0.875)/(1 − 0.875)] C = 4.6 nF Answer: Change the capacitor to 4.7 nF.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

20-47. Given: R1 = 2.2 kΩ R2 = 82 kΩ Vsat = 14 V Solution: UTP = (R1/R2) Vsat (Eq. 20-10) UTP = (2.2 kΩ/82 kΩ)14 V UTP = 0.376 V

20-59. U1 has failed 20-60. The trip point has changed from −10 V to −7.3 V

Chapter 21 SELF-TEST 1. a 2. b 3. a 4. c 5. b 6. a 7. c 8. d

H = 2 UTP = 0.751 V R1 = 1/0.751 (2.2 kΩ) = 2.93 kΩ Use nearest standard value of 3 kΩ.

To increase hysteresis to 1 V, increase R1 by a factor of 1 V/0.751 V, to get R1 ≅ 3 kΩ. For a safety factor, use R1 = 3.3 kΩ.

Answer: Increase R1 to 3.3 kΩ.

20-48. Answer: Use a window comparator with an upper trip point of 5 V and a lower trip point of −5 V. 20-49. Answer: Use a comparator with hysteresis and a lightdependent resistor in a voltage divider as the input.

20-50. Answer: Rectify the incoming signal and send it to a comparator. When the input drops below the reference output, the alarm sounds.

5.

8.

(Eq. 20-13)

T = (V/Vin)RC

T = (1.23 V/5 V)(1 M Ω)(10 μF) T = 2.46 s

21-1.

Therefore, the distance is half this amount: 228,780 miles.

Answer: The moon is 228,780 miles away.

20-52. Answer: Trouble 1: The positive clamper circuit is the trouble.

Trouble 2: The integrator circuit is the trouble.

20-53. Answer: Trouble 3: The relaxation oscillator circuit is the trouble.

Trouble 5: The positive clamper circuit is the trouble.

20-54. Answer: Trouble 6: The comparator circuit is the trouble.

Trouble 7: The integrator circuit is the trouble.

20-55. Answer: Trouble 8: The peak detector circuit is the trouble.

Trouble 9: The integrator circuit is the trouble.

Trouble 10: The comparator circuit is the trouble.

20-56. C4 is open 20-57. D1 is shorted 20-58. D2 is open

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

25. a 26. d 27. c 28. b 29. d 30. d 31. a 32. d 33. c

The monostable timer has pins 6 and 7 connected together and is externally triggered, whereas the astable timer has a resistor between pins 6 and 7 and requires no external trigger.

There must be an unwanted positive feedback path between the output and the input of the three-stage amplifier. Low-frequency oscillations may be caused by the high power-supply 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.

PROBLEMS

(2.46 s)186,000 miles/s = 457,560 miles for a round trip.

Trouble 4: The peak detector circuit is the trouble.

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

JOB INTERVIEW QUESTIONS

20-51. Given: Vin = 5 V R = 1 MΩ C = 10 μF V = 1.23 V Solution: V = [T/(RC)]Vin

Oscillators

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 Vrms.

IL = 3 V/500 Ω IL = 6 mA vout = IL(Rf + RL) vout = 6 mA(1 kΩ + 500 Ω) vout = 9 Vrms Answer: The output voltage is 9 Vrms.

21-2.

Given:

C = 200 pF Rmin = 2 kΩ Rmax = 24 kΩ Solution: fr(max) = 1/[2πRminC] (Eq. 21-4) fr(max) = 1/[2π(2 kΩ)(200 pF)] fr(max) = 398 kHz fr(min) = 1/[2πRmaxC] (Eq. 21-4) fr(min) = 1/[2π(24 kΩ)(200 pF)] fr(min) = 33.2 kHz Answer: The maximum frequency is 398 kHz, and the minimum frequency is 33.2 kHz. 21-3a. Given: C = 0.2 μF Rmin = 2 kΩ Rmax = 24 kΩ

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Solution: fr(max) = 1/[2πRminC]

(Eq. 21-4)

fr(max) = 1/[2π(2 kΩ)(0.2 μF)]

fr(max) = 398 Hz fr(min) = 1/[2πRmaxC] (Eq. 21-4) fr(min) = 1/[2π(24 kΩ)(0.2 μF)] fr(min) = 33.2 Hz Answer: The maximum frequency is 398 Hz, and the minimum frequency is 33.2 Hz. 21-3b. Given: C = 0.02 μF Rmin = 2 kΩ Rmax = 24 kΩ Solution: fr(max) = 1/[2πRminC] (Eq. 21-4) fr(max) = 1/[2π (2 kΩ)(0.02 μF)] fr(max) = 3.98 kHz fr(min) = 1/[2πRmaxC] (Eq. 21-4)

21-5. Given: Maximum frequency is 398 kHz, from Prob. 21-3.

Solution: 1 decade above 398 kHz is 3.98 MHz. Answer: The cutoff frequency is 3.98 MHz. 21-6.

R = 10 kΩ C = 0.01 μF Solution: fr = 1/[2πRC] (Eq. 21-4) fr = 1/[2π(10 kΩ)(0.01 μF)] fr = 1.59 kHz Answer: The resonant frequency is 1.59 kHz. 21-7.

Solution: fr = 1/[2πRC] (Eq. 21-4) fr = 1/[2π(20 kΩ)(0.02 μF)] fr = 398 Hz Answer: The resonant frequency is 398 Hz. 21-8.

21-4.

Given:

Vout = 6 Vrms 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 Vrms. 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: Change the feedback resistor to 700 Ω.

Given:

R1 = 10 kΩ R2 = 5 kΩ RE = 1 kΩ VBE = 0.7 V VCC = 12 V Solution: VB = [R2/(R1 + R2)]VCC 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 Since the RF choke is a short to direct current, the collector voltage is 12 V. VCE = VC – VE VCE = 12 V – 3.3 V VCE = 8.7 V Answer: The emitter current is 3.3 mA, and the collector-to-emitter voltage is 8.7 V.

21-3c. Given: C = 0.002 μF Rmin = 2 kΩ Rmax = 24 kΩ Solution: fr(max) = 1/[2πRminC] (Eq. 21-4) fr(max) = 1/[2π(2 kΩ)(0.002 μF)] fr(max) = 39.8 kHz fr(min) = 1/[2πRmaxC] (Eq. 21-4) fr(min) = 1/[2π(24 kΩ)(0.002 μF)] fr(min) = 3.32 kHz Answer: The maximum frequency is 39.8 kHz, and the minimum frequency is 3.32 kHz. 21-3d. Given: C = 200 pF Rmin = 2 kΩ Rmax = 24 kΩ Solution: fr(max) = 1/[2πRminC] (Eq. 21-4) fr(max) = 1/[2π(2 kΩ)(200 pF)] fr(max) = 398 kHz fr(min) = 1/[2πRmaxC] (Eq. 21-4) fr(min) = 1/[2π(24 kΩ)(200 pF)] fr(min) = 33.2 kHz Answer: The maximum frequency is 398 kHz, and the minimum frequency is 33.2 kHz.

Given:

R = 20 kΩ C = 0.02 μF

fr(min) = 1/[2π (24 kΩ)(0.02 μF)]

fr(min) = 332 Hz Answer: The maximum frequency is 3.98 kHz, and the minimum frequency is 332 kHz.

Given:

21-9.

Given:

C1 = 0.001 μF C2 = 0.01 μF L = 10 μH Solution: C = C1C2/(C1 + C2) (Eq. 21-6) C = (0.001 μF)(0.01 μF)/(0.001 μF + 0.01 μF) C = 909 pF fr = 1/(2π LC ) (Eq. 21-5) fr = 1/(2π (10 μH)(909 pF) ) fr = 1.67 MHz B = C1/C2 (Eq. 21-7) B = 0.001 μF/0.01 μF B = 0.10

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

AV(min) = C2/C1 (Eq. 21-8) AV(min) = 0.01 μF/0.001 μF AV(min) = 10 Answer: The frequency is 1.67 MHz, the feedback fraction is 0.10, and the minimum gain is 10.

21-10. 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.

Solution: B = M/L (Eq. 21-14) 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.

21-16. Given: f = 5 MHz Answer: The first overtone is 10 MHz, the second overtone is 15 MHz, and the third overtone is 20 MHz.

21-11. Given: C1 = 0.001 μF C2 = 0.01 μF L = 20 μH Solution: C = C1C2/(C1 + C2) (Eq. 21-6) C = (0.001 μF)(0.01 μF)/(0.001 μF + 0.01 μF) C = 909 pF

fr = 1/(2π LC ) (Eq.21-5) fr = 1/(2π (20 μH)(909 pF) ) fr = 1.18 MHz Answer: The frequency is 1.18 MHz. 21-12. Answer: Reduce the inductance by a factor of 4 (since there is a square root in the denominator).

21-13. Given: C1 = 0.001 μF C2 = 0.01 μF C3 = 47 pF L = 10 μH Solution: fr = 1/(2π )

21-17. Answer: Since the frequency is inversely proportional to thickness, if thickness is reduced by 1% the frequency will increase by 1%.

21-18. Given: L=1H Cs = 0.01 pF R = 1 kΩ Cm = 20 pF Solution: fs = 1/[2π

]

(Eq.21-20)

fs = 1/[2π (1 H)(0.01 pF) ] fs = 1.5915 MHz Cp = CmCs/(Cm + Cs)

(Eq. 21-21)

Cp = (20 pF)(0.01 pF)/(20 pF + 0.01 pF)

Cp = 0.009995 pF fp = 1/[2π

]

(Eq.21-22)

fp = 1/[2π (1 H)(0.009995 pF) ] (Eq.21-18)

fr = 1/(2π (10 μH)(47 pF) ) fr = 7.34 MHz Answer: The frequency is 7.34 MHz. 21-14. Given: L1 = 1 μH L2 = 0.2 μH C = 1000 pF

fp = 1.5919 MHz XL = 2πfL XL = 2π (1.5915 MHz)(1 H)

XL = 10 MΩ Q = XL/R Q = 10 MΩ/1 kΩ Q = 10,000 Answer: The series frequency is 1.5915 MHz, the parallel frequency is 1.5919 MHz, and the Q is 10,000.

Solution: B = L2/L1 (Eq. 21-16) B = 0.2 μH/1 μH B = 0.2 L = L1 + L2 L = 1 μH + 0.2 μH L = 1.2 μH

fr = 1/[2π

21-15. Given: M = 0.1 μH L = 3.3 μH

]

(Eq. 21-5)

fr =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.

21-19. Given: R = 10 kΩ C = 0.047 μF Solution: W = 1.1RC (Eq. 21-25) W = 1.1(10 kΩ)(0.047 μF) W = 517 μs Answer: The pulse width is 517 μs.

21-20. Given: VCC = 10 V R = 2.2 kΩ C = 0.2 μF

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Solution: LTP = VCC/3 LTP = 10 V/3 LTP = 3.33 V

(Eq. 21-24)

UTP = 2VCC/3 UTP = 2(10 V)/3 UTP = 6.67 V

(Eq. 21-23)

W = 1.1RC (Eq. 21-25) W = 1.1(2.2 kΩ)(0.2 μF) W = 484 μs Answer: The minimum trigger voltage is 3.33 V, the maximum capacitor voltage is 6.67V, and the pulse width is 484 μs.

21-21. Given: R1 = 10 kΩ R2 = 2 kΩ C = 0.0022 μF Solution: f =

1.44

(Eq. 21-28)

(R1 + 2R2 )C

f =

1.44 10 kΩ + 2(2 kΩ) (0.0022 μF)

f = 46.8 kHz Answer: The frequency is 46.8 kHz. 21-22. Given: R1 = 20 kΩ R2 = 10 kΩ C = 0.047 μF Solution: f = f =

1.44 (R! + 2R2 )C

(Eq.21-18) 1.44

[20 k + 2(10 k)](0.047 μF)

f = 766 Hz D = (R1 + R2)/(R1 + 2R2) (Eq. 21-29) D = (20 kΩ + 10 kΩ)/[20 kΩ + 2(10 kΩ)] D = 0.75 Answer: The frequency is 766 Hz, and the duty cycle is 0.750.

21-23. Given: VCC = 10 V R = 5.1 kΩ C = 1 nF f = 10 kHz vmod = 1.5 V Solution: T = 1/f T = 1/10 kHz T = 100 μs W = 1.1RC (Eq. 21-25) W = 1.1(5.1 kΩ)(1 nF) W = 5.61 μs UTPmax = 2VCC /3 + vmod (Eq. 21-35) UTPmax = 2(10 V)/3 + 1.5 V UTPmax = 8.17 V UTPmin = 2VCC /3 − vmod (Eq. 21-35) UTPmin = 2(10 V)/3 − 1.5 V UTPmin = 5.17 V

Wmax = −RCln(1 − UTPmax/VCC) (Eq. 21-34) Wmax = −(5.1 kΩ)(1 nF)ln[1 − (8.17 V/10 V)] Wmax = 8.66 μs Wmin = −RCln(1 − UTPmin/VCC) (Eq. 21-34) Wmin = −(5.1 kΩ)(1 nF)ln[1 − (5.17 V/10 V)] Wmin = 3.71 μs Dmax = Wmax/T Dmax = 8.66 μs/100 μs Dmax = 0.0866 Dmin = Wmin/T Dmin = 3.71 μs/100 μs Dmin = 0.0371 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.

21-24. Given: VCC = 10 V R1 = 1.2 kΩ R2 = 1.5 kΩ C = 4.7 nF vmod = 1.5 V Solution: W = 0.693(R1 + R2)C (Eq. 21-26) W = 0.693(1.2 kΩ + 1.5 kΩ)(4.7 nF) W = 8.79 μs T = 0.693(R1 + 2R2)C (Eq. 21-27) T = 0.693[1.2 kΩ + 2(1.5 kΩ)](4.7 nF) T = 13.68 μs UTPmax = 2VCC/3 + vmod (Eq. 21-35) UTPmax = 2(10 V)/3 + 1.5 V

UTPmax = 8.17 V UTPmin = 2VCC/3 − vmod

(Eq. 21-35)

UTPmin = 2(10 V)/3 – 1.5 V

UTPmin = 5.17 V Wmax = – (R1 + R2)C1n[(VCC − UTPmax)/(VCC – 0.5 UTPmax)]

(Eq. 21-36) 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 Wmin = – (R1 + R2)C1n[(VCC – UTPmin)/(VCC – 0.5 UTPmin)] (Eq. 21-36)

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 Space = 0.693R2C Space = 0.693(1.5 kΩ)(4.7 nF) Space = 4.89 μs Answer: The quiescent pulse width is 8.79 μs, the quiescent period is 13.68 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.

21-25. Given: IC = 0.5 mA VCC = 10 V C = 47 nF Solution: S = IC/C (Eq. 21-40) S = 0.5 mA/47 nF S = 10.6 V/ms

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

V = 2VCC/3 V = 2(10 V)/3 V = 6.67 V

(Eq. 21-41)

T = 2VCC/3S (Eq. 21-42) T = 2(10 V)/3(10.6 V/ms) T = 0.629 ms Answer: The slope is 10.6 V/ms, the peak value is 6.67 V, and the duration is 0.629 ms.

21-26. Given: S1 = Closed R = 20 kΩ R3 = 40 kΩ C = 0.1 μF Solution: The waveform is a sine wave.

21-29c. Same. The upper potentiometer affects frequency, not output voltage.

21-29d. Same. Unless the supply falls low enough for clipping.

23-29e. Same. Only a very small change at the output.

21-30. Answer: 1. Shorted inductor 2. Open inductor 3. Shorted capacitors 4. Open capacitors 5. Open in the feedback path 6. Loss of the power supply 21-31. Answer: The fuzz is probably oscillations. To correct this, make sure that the leads are short and are not running really close to each other. Also, a ferrite bead in the feedback path may dampen them out.

f0 = 1/RC (Eq. 21-31)

f0 = 1/(20 kΩ)(0.1 μF) f0 = 500 Hz Amplitude = 2.4 Vp (from Fig. 21-53c) Amplitude = 4.8 Vp-p Answer: The output is a sine wave at a frequency of 500 Hz and a peak voltage of 2.4 V. 21-27. Given: S1 = Open R = 10 kΩ R3 = 40 kΩ C = 0.01 μF Solution: The waveform is a triangle wave. f0 = 1/RC

(Eq. 21-31)

f0 = 1/(10 kΩ)(0.01 μF) f0 = 10 kHz Amplitude = 5 Vp (from Fig. 21-53c) Amplitude = 10 Vp-p Answer: The output is a triangle wave at a frequency of 10 kHz and a peak voltage of 5 V. 21-28. Given: R1 = 2 kΩ R2 = 10 kΩ C = 0.1 μF Solution: 2 1  f =   C  R1 + R2  2  1  f = 0.1 μF  2 k + 10 k 

f = 1.67 kHz D = R1/(R1 + R2) D = 2 kΩ/(2 kΩ + 10 kΩ) D = 0.167 Answer: The frequency is 1.67 kHz, and the duty cycle is 0.167.

21-29a. 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. 21-29b. Increase. With the inverting input grounded, there is no feedback and the gain is open-loop gain and the output will be saturation.

CRITICAL THINKING 21-32. Given: f = 20 Hz to 20 kHz vout = 5 Vrms Rmin = 2.2 kΩ Solution: One of many possible designs is C = 0.22 μF, 0.022 μF, and 0.0022 μF. Change 2 kΩ in Fig. 21-53a to 3.3 kΩ and use a 50-kΩ potentiometer. Use a 1-kΩ potentiometer in the place of the 1 kΩ in series with the lamp.

Adjust 1 kΩ to get an output of 5 V. 21-33. Given: f = 2.5 MHz C1 = 0.001 μF C2 = 0.01 μF C = 909 pF (from Prob. 21-9) Solution: fr = 1/(2π ) (Eq. 21-5) 2 L = [1/(2πfr)] /C 2 L = {1/[2π (2.5 MHz)] /(909 pF)} L = 4.46 μH Answer: The inductor would have to be 4.46 μH. 21-34. Given: C = 0.001 μF R1 = 10 kΩ R2 = 10 kΩ

R3 = 5 kΩ f2(CL) = 15.9 kHz Solution: XC = 1/(2πfC) XC = 1/[2π (15.9 kHz)(0.001 μF)]

XC = 10 kΩ ϕ = −arctan(R/XC) ϕ = −arctan(10 kΩ/10 kΩ) ϕ = −45° Below its critical frequency, the op amp has 180° of phase shift. At 15.9 kHz, the op amp has an additional phase shift of 90° because it is well above its critical fre-quency.

This means that the op amp has approximately 270° of phase shift at 15.9 kHz. Each lag circuit has a phase shift of 45° at 15.9 kHz. Therefore, the total or loop phase shift is 270° plus 45° plus 45°, or 360°.

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21-35. Given: f = 1 kHz D = 0.75 Solution: Pick a value for R1 = 10 kΩ.

D = (R1 + R2)/(R1 + 2R2) (Eq. 21-29) D(R1 + 2R2) = (R1 + R2) DR1 + 2DR2 = R1 + R2 2DR2 − R2 = R1 − DR1 R2(2D − 1) = (R1 − DR1) R2 = (R1 − DR1)/(2D − 1) R2 = [10 kΩ − 0.75(10 kΩ)]/[2(0.75) − 1]

PROBLEMS 22-1.

Given:

VNL = 15 V VFL = 14.5 V Solution: Load regulation = (VNL − VFL)/VFL × 100% Load regulation = (15 V − 14.5 V)/14.5 V × 100% Load regulation = 3.45% Answer: The load regulation is 3.45%.

22-2.

VHL = 20 V VLL = 19 V Solution: Line regulation = (VHL − VLL)/VLL × 100% Line regulation = (20 V − 19 V)/19 V × 100% Line regulation = 5.26%

R2 = 5 kΩ f = 1.44/[(R1 + 2R2)C] (Eq. 21-28) C = 1.44/[(R1 + 2R2)f] C = 1.44/{[(10 kΩ + (2)(5 kΩ)(1 kHz)]} C = 72 nF Answer: R1 is 10 kΩ , R2 is 5 kΩ , and the capacitor is 72 nF. 21-36. C1 is shorted

Given:

Answer: The line regulation is 5.26%.

22-3.

Given:

VHL = 12.3 V VLL = 12 V

21-37. VCC has failed 21-38. U1 has failed

Solution: Line regulation = (VHL − VLL)/VLL × 100% (Eq. 22-2) Line regulation = (12.3 V − 12 V)/12 V × 100% Line regulation = 2.5% Answer: The line regulation is 2.5%.

21-39. R2 is shorted 21-40. R1 is 1 kΩ instead of 75 kΩ 21-41. Sinewave 21-42. 10 Hz 21-43. VR7 = 1.8 kΩ

22-4.

21-44. VR5 and VR8

Given:

21-45. Output of pin 11 to Q3

RTH = 2 Ω RL(min) = 50 Ω

Chapter 22 Supplies

Solution: Load regulation = RTH/RL(min) × 100% Load regulation = (2 Ω/50 Ω) × 100% Load regulation = 4%

Regulated Power

Answer: The load regulation is 4%.

SELF-TEST 1. a 2. b 3. b 4. b 5. c 6. a 7. c 8. c 9. b 10. a

22-5.

11. c 12. b 13. b 14. b 15. a 16. c 17. b 18. d 19. c 20. a

21. c 22. b 23. a 24. a 25. b 26. a 27. c 28. d 29. b

30. d 31. c 32. a 33. a 34. c 35. a 36. c 37. b 38. d

Given:

Vin = 25 V RS = 22 Ω VZ = 18 V VBE = 0.75 V RL = 100 Ω Solution: Vout = VZ + VBE (Eq. 22-5) Vout = 18 V + 0.75 V Vout = 18.75 V IS = (Vin − Vout)/RS (Fig. 22-4)

JOB INTERVIEW QUESTIONS

IS = (25 V − 18.75 V)/22 Ω

6. The first is a member of the LM78XX family and is a pos-itive voltage regulator with an output voltage of 6 V. The sec-ond is a member of the LM79XX family and is a negative voltage regulator with an output voltage of −12 V.

IS = 484 mA IL = Vout/RL

10. Thermal shutdown means that the chip automatically shuts itself off when it overheats.

IL = 18.75 V/100 Ω

IL = 187.5 mA IC = IS − IL IC = 284 mA − 187.5 mA IC = 96.5 mA Answer: The output voltage is 18.75 V, the input current is 484 mA, the load current is 187.5 mA, and the collector current is 96.5 mA.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

22-6.

Vin = 25 V RS = 15 Ω VZ = 5.6 V VBE = 0.77 V RL = 80 Ω R1 = 330 Ω R2 = 680 Ω Solution: Vout = [(R1 + R2)/R1](VZ + VBE) (Eq. 22-6) Vout = [(330 Ω + 680 Ω)/330 Ω](5.6 V + 0.77 V) Vout = 19.5 V IS = (Vin − Vout)/RS (Fig. 22-4) IS = (25 V – 19.5 V)/15 Ω

IS = 367 mA IL = Vout/RL IL = 19.5 V/80 Ω IL = 244 mA IC = IS − IL IC = 367 mA − 244 mA IC = 123 mA Answer: The output voltage is 19.5 V, the input current is 367 mA, the load current is 244 mA, and the collector current is 123 mA. 22-7.

Given:

Vin = 25 V RS = 8.2 Ω VZ = 5.6 V RL = 50 Ω R1 = 2.7 kΩ R2 = 6.2 kΩ Solution: Vout = [(R1 + R2)/R1]VZ (Eq. 22-7) Vout = [(2.7 kΩ + 6.2 kΩ)/2.7 kΩ]5.6 V Vout = 18.46 V IS = (Vin − Vout)/RS (Fig. 22-6) IS = (25 V − 18.46 V)/8.2 Ω

IS = 798 mA IL = Vout/RL

(Fig. 22-6)

IL = 18.46 V/50 Ω

IL = 369 mA IC = IS − IL (Fig. 22-6) IC = 798 mA − 369 mA

IC = 429 mA Answer: The output voltage is 18.46 V, the input current is 798 mA, the load current is 369 mA, and the collector current is 429 mA. 22-8.

IL = Vout/RL IL = 16.9 V/50 Ω IL = 339 mA PD = (Vin − Vout)IL

Given:

Given:

Vin = 20 V VZ = 4.7 V VBE = 0.77 V RL = 50 Ω R1 = 2.2 kΩ R2 = 4.7 kΩ R3 = 1.5 kΩ R4 = 2.7 kΩ Solution: Vout = [(R1 + R2)/R1](VZ + VBE) (Fig. 22-8) Vout = [(2.2 kΩ + 4.7 kΩ)/2.2 kΩ](4.7 V + 0.7 V) Vout = 16.9 V

(Fig. 22-8)

PD = (20 V − 16.9 V) 336 mA

PD = 1.05 W Answer: The output voltage is 16.9 V, and the power dissipation is 1.05 W. 22-9.

Given:

Vin = 20 V Vout = 16.9 V (from Prob. 22-8) Solution: Efficiency = Vout /Vin × 100% (Eq. 22-13) Efficiency = 16.9 V/20 V × 100% Efficiency = 84.5% Answer: The efficiency is 84.5%.

22-10. Given: VZ = 6.2 V RL = 4 Ω R1 = 2.7 kΩ R2 = 2.2 kΩ Solution: Vout = [(R1 + R2)/R1](VZ) (Eq. 22-14) Vout = [(2.7 kΩ + 2.2 kΩ)/2.7 kΩ](6.2 V) Vout = 11.25 V≈ 11.3 V Answer: The output voltage is 11.25 V≈ 11.3 V. 22-11. Given: VZ = 4.7 V Vin(max) = 30 V R3 = 820 Ω Solution: IZ(max) = (Vin(max) − VZ)/R3 IZ(max) = (30 V − 4.7 V)820 Ω IZ(max) = 30.9 mA Answer: The maximum zener current is 30.9 mA. 22-12. Given: VZ = 4.7 V Radj = 1.5 kΩ R1(min) = 750 Ω R1(max) = 2.25 kΩ R2(min) = 750 Ω R2(max) = 2.25 kΩ Solution: Vout(max) = [(R1(min) + R2(max))/R1(min)]VZ (Eq. 22-14) Vout(max) = [(750 Ω + 2.25 kΩ)/750 Ω](4.7 V) Vout(max) =18.8 V Vout(min) = [(R1(max) + R2(min))/R1(max)]VZ (Eq. 22-14) Vout(min) = [(2.2 kΩ + 750 Ω)/2.25 k Ω](4.7 V) Vout(min) = 6.27 V Answer: The maximum output voltage is 18.8 V, and the minimum is 6.27 V. 22-13. Given: Vout(reg) = 10 V R4 = 3 Ω

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Solution: The current limiting starts at a VBE of 0.6 V. IL = 0.6 V/R4 IL = 0.6 V/3 Ω IL = 200 mA RL = Vout(reg)/IL RL = 10 V/200 mA RL = 50 Ω ISL = 0.7 V/R4 ISL = 0.7 V/3 Ω ISL = 233 mA Answer: Current limiting starts at a load resistance of 50 Ω and the short-circuit current is 233 mA. 22-14. Given: Vout = 15 V Vin = 20 V RL = 20 Ω Solution: IL = Vout/RL IL = 15 V/20 Ω IL = 750 mA Headroom voltage = Vin − Vout (Eq. 22-11) Headroom voltage = 20 V − 15 V Headroom voltage = 5 V PD = (Headroom voltage)(IL)

(Eq. 22-12)

PD = (5 V)(750 mA)

PD = 3.75 W Answer: The load current is 750 mA, the headroom voltage is 5 V, and the power dissipation is 3.75 W. 22-15. Given: f = 120 Hz C = 4700 μF IL = 750 mA (from Prob. 22-14) RRdB = 70 dB (from Table 22-1) Solution: VR(in) = IL/fC (Eq. 4-10) VR(in) = 750 mA/[(120 Hz)(4700 μF)] VR(in) = 1.33 V RR = antilog(RRdB/20) RR = antilog(70 dB/20) RR = 3162 VR(out) = VR(in)/RR VR(out) = 1.33 V/3162 VR(out) = 421 μV Answer: The output ripple voltage is 421 μV. 22-16. Given: R1 = 2.7 kΩ R2 = 20 kΩ Solution: Vout = [(R1 + R2)/R1](1.25 V) (Eq. 22-19) Vout = [(2.7 kΩ + 20 kΩ)/2.7 kΩ](1.25 V) Vout = 10.5 V Answer: The output voltage is 10.5 V. 22-17. Given: Vout = 15 V Vin(min) = 18 V Vin(min) = 25 V

Solution: Maximum efficiency = (Vout/Vin(min)) × 100% (Eq. 22-13) Maximum efficiency = (15 V/18 V) × 100% Maximum efficiency = 83.3% Minimum efficiency = (Vout/Vin(max)) × 100% (Eq. 22-13) Minimum efficiency = (15 V/25 V) × 100% Minimum efficiency = 60% Answer: The maximum efficiency is 83.3%, and the minimum efficiency is 60%.

22-18. Given: Vout = 12 V Vin = 5 V Iout = 0.25 A Iin = 1 A Solution: Pout = VoutIout Pout = (12 V)(0.25 A) Pout = 3 W Pin = VinIin Pin = (5 V)(1 A) Pin = 5 W Efficiency = Pout/Pin × 100% Efficiency = 3 W/5 W × 100% Efficiency = 60% Answer: The efficiency is 60%.

22-19. Given: Vout = 5 V Vin = 12 V Iin = 2 A Efficiency is 80% Solution: Pin = VinIin Pin = 12 V(2 A) Pin = 24 W Efficiency = Pout/Pin × 100% (Efficiency/100%)Pin = Pout

Pout = (80%/100%)24 W Pout = 19.2 W Pout = VoutIout Iout = Pout/Vout Iout = 19.2 W/5 V Iout = 3.84 A Answer: The output current is 3.84 A. 22-20. Given: R1 = 1.5 kΩ R2 = 10 kΩ Vref = 2.5 V Solution: Vout = [(R1 +R2)/R1](Vref) (Eq. 22-22) Vout = [(1.5 kΩ + 10 kΩ)/1.5 kΩ](2.5 V) Vout = 19.2 V Answer: The output voltage is 19.2 V. 22-21. Given: D = 30% Vin = 20 V

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Solution: Vout = DVin (Eq. 22-21) Vout = (0.3)(20 V) Vout = 6 V Answer: The output voltage is 6 V. 22-22. Given: R1 = 1.2 kΩ R2 = 15 kΩ Vref = 1.25 V Solution: Vout = [(R1 + R2)/R1](Vref) (Eq. 22-22) Vout = [(1.2 kΩ + 15 kΩ)/1.2 kΩ](1.25 V) Vout = 16.9 V Answer: The output voltage is 16.9 V. 22-23. Given: R1 = 2.1 kΩ R2 = 12 kΩ Vref = 2.1 V Solution: Vout = [(R1 + R2)/R1](Vref) (Eq. 22-22) Vout = [(2.1 kΩ + 12 kΩ)/2.1 kΩ](2.1 V) Vout = 14.1 V Answer: The output voltage is 14.1 V.

CRITICAL THINKING 22-24. Given: R1 = 240 Ω R2 = 0 to 5 kΩ Vin = 30 V Solution: Vout(min) = [(R1 + R2)/R1]1.25 (Eq. 22-19) Vout(min) = [(240 Ω + 5 kΩ)/240 Ω]1.25 V Vout(min) = 27.3 V As R2 approaches 0 Ω, the voltage approaches 1.25 V. Since the LM317 has a dropout of 2 V, the highest regulated output is 28 V with an input of 30 V. With the transistor saturated, the output is 1.25 V. Answer: The output voltage range with the shutdown signal low is 1.25 to 27.3 V and is 1.25 V when the shutdown signal is high.

22-25. Given: R1 = 240 Ω R2 = 0 to 5 kΩ Vin = 30 V Vout = 18 V Solution: Using “Vout = [(R1 + R2)/R1]1.25” solve for R2 Vout = [(R1 + R2)/R1]1.25 Vout/1.25 V = [(R1 + R2)/R1] R1(Vout/1.25 V) = R1 + R2 R1(Vout/1.25 V) − R1 = R2 R1[(Vout/1.25 V) − 1] = R2 R2 = R1 [(Vout/1.25 V) − 1] R2 = 240 Ω[(18 V/1.25 V) − 1] R2 = 3.22 kΩ Answer: The adjustable resistor needs to be 3.22 kΩ.

22-26. Answer: Since the output current and voltage are essentially constant, it is logical to assume (since the input voltage is essentially constant) that the input current is essentially constant. If the capacitor is supplying a constant current, it should discharge at a linear rate.

22-27. Given: Load regulation is 5% VNL = 12.5 V Solution: Load regulation = (VNL − VFL)/VFL × 100%

(Eq. 22-1)

Load regulation/100% = (VNL − VFL)/VFL

VFL(load regulation/100%) = VNL − VFL VFL(load regulation/100%) + VFL = VNL VFL[(load regulation/100%) + 1] = VNL VFL = VNL/[(load regulation/100%) + 1] VFL = 12.5 V/[(5%/100%) + 1] VFL = 11.9 V Answer: The full load voltage is 11.9 V. 22-28. Given: Line regulation is 3% VLL = 16 V Solution: Line regulation = (VHL − VLL)/VLL × 100% (Eq. 22-2) Line regulation/100% = (VHL − VLL)/VLL VLL(line regulation/100%) = VHL − VLL VLL(line regulation/100%) + VLL = VHL

VLL[(line regulation/100%) + 1] = VHL VHL = 16 V/[(3%/100%) + 1] VHL = 16.48 V Answer: The high line voltage is 16.48 V. 22-29. Given: Load regulation = 1% RL(min) = 10 Ω Solution: Load regulation = RTH/RL(min) × 100% Load regulation/100% = RTH/RL(min)

RTH = RL(min)[load regulation/100%] RTH = 10 Ω[1%/100%] RTH = 0.1 Ω Answer: The power supply output resistance is 0.1 Ω.

22-30. Given: Vin = 35 V IC = 60 mA IL = 140 mA

RS = 100 Ω Solution: IC = IS − IL IS = IC + IL IS = 60 mA + 140 mA

IS = 200 mA Vout = Vin − ISRS Vout = 35 V − (200 mA)(100 Ω) Vout = 15 V RL = Vout/IL RL = 15 V/140 mA RL =107 Ω Answer: The load resistance is 107 Ω.

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22-31. Given: Start current limiting at 250 mA VBE where current limiting starts is 0.6 V Solution: R4 = VBE/I R4 = 0.6 V/250 mA

R4 = 2.4 Ω Answer: R4 needs to be 2.4 Ω.

22-32. Given: Vout = 10 V VBE = 0.7 V

R4 = 1 Ω K = 0.70

Solution: ISL = VBE/KR4 ISL = 0.7 V/0.70 (1 Ω) ISL = 1 A Imax = ISL + [(1 − K)Vout]/KR4 Imax = 1 A + [(1 − 0.70)10 V]/0.70(1 Ω) Imax = 5.29 A Answer: The shorted load current is 1 A, and the maximum current is 5.29 A. 22-33. Given: R5 = 7.5 kΩ R6 = 1 kΩ R7 = 9 kΩ C3 = 0.001 μF

22-35. Answer: Since the voltage at D is stuck low, the trouble is the triangle-to-pulse converter.

22-36. Answer: Since the voltage at A is stuck low, the trouble is the relaxation oscillator.

22-37. Answer: Since the voltage at D is high, the voltage at E should be 12.8 V, but since it is 0 V, the trouble is Q1.

22-38. Answer: Since the voltage at C is stuck low, the trouble is the integrator.

22-39. Answer: Since the voltage at A is stuck high, the trouble is the relaxation oscillator.

22-40. Answer: Since the voltage at E is higher than the desired output of 5 V, the transistor should be off more of the time. Thus the output of the comparator should be high.

But because it is low, the problem is the comparator.

22-41. Answer: Since the voltage at D is stuck at 0 V, the trouble is the triangle-to-pulse converter.

22-42. Answer: Since the voltage at E is low, the output of the comparator should be low. But because it is high, the trouble is the comparator.

22-43. Answer: The voltage at D is stuck high, since the output of the comparator is high, as it should be, given the voltage at E. The trouble is the triangle-to-pulse converter.

22-44. Q1 is open 22-45. R4 is open 22-46. R2 is open 22-47. D1 is shorted

Solution: 6.= R7/(R6 + R7) B = 9 kΩ/(1 kΩ + 9 kΩ) B = 0.9

22-48. U1 has failed

T = 2RCln[(1 + B)/(1 − B)] (Eq. 20-18) T = 2(7.5 kΩ)(0.001 μF)ln[(1 + 0.9)/(1 − 0.9)] T = 44.16 μs f = 1/T f = 1/44.16 μs f = 22.6 kHz

22-51. Vin = 14−15 V

Answer: The switching frequency is 22.6 kHz. 22-34. Given: VZ = 4.7 V R1(mid) = 1250 Ω R2(mid) = 1250 Ω Solution: Vout(mid) = [(R1(mid) + R2(mid))/R1(mid)]VZ (Eq. 22-14) Vout(mid) = [(1250 Ω + 1250 Ω)/1250 Ω]4.7 V Vout(mid) = 9.2 V Answer: The output voltage is 9.2 V.

22-49. U1 0 to 20 V; U2 +12 V; U3 +5 V; U4 −12 V; U5 0 to −20 V

22-50. 417 mVp-p 22-52. 20 μVp-p 22-53. VR1 = 840 Ω

Chapter 23 SELF-TEST 1. a 2. d 3. c 4. d 5. a 6. d 7. b 8. a 9. b 10. a 11. b 12. c

Industry 4.0

13. a 14. d 15. a 16. a 17. d 18. c 19. c 20. a 21. b 22. d 23. a 24. d

25. a 26. c 27. a 28. c 29. d 30. c 31. b 32. b 33. a 34. d 35. b 36. c

37.

38.

39.

40.

41.

42.

43.

44.

45.

b a a c d b c b a

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

1-1 JOB INTERVIEW QUESTIONS 23-49. Industry 4.0 is the interconnection of and the communication between robotic manufacturing cells and a centralized computer system. Data from the entire network is used to improve efficiency of the overall manufacturing process.

23-50. A passive sensor does not require a voltage source to function.

An RTD varies in resistance based on temperature. Whereas an active sensor requires a voltage source to function. A Photoelectric sensor emits a light of a specific wavelength. A corresponding detector, sensitive to the wavelength of light emitted, detects the presence or absence of the light.

23-12. Bi-directional 23-13. Transducer, Analog Interface, Microprocessor, Data Exchange 23-14. Active, passive 23-15. Voltage source 23-16. Voltage source

23-51. There are three methods in which the photoelectric sensor is used; Diffuse (this is sometimes referred to as the Reflective method), Through-beam, and Retroreflective.

23-17. Resistance

23-52. An ultrasonic sensor uses sound waves to measure distance.

The ultrasonic sensor contains both a transmitter and a receiver.

The transmitter emits a short bust of high-frequency sound waves and then detects them when they bounce back to the sensor. The sensor uses the time between the transmission and detection of the soundwave to determine the distance to an object.

23-19. Wheatstone bridge

23-18. Platinum, Nickel, and Copper

23-20. Given: V1 = 10 V, R1 = 100 Ω, R2 = 100 Ω, R3 = 100 Ω, the RTD will vary from 100 Ω at 0oC to 138.5 Ω at 100oC. The resistance of the cable is ignored.

Solution:

 1 RTD  −V 1  R1+ RTD   2     119.25  1 VAB = 10V   − 10V   2   100  +119.25   VAB = 439 mV

23-53. In smart sensors, a second transmitter and receiver are incorporated and the results of both sets are constantly monitored by the programmable logic controller. If the data from one transmitter/receiver set begins to differ from the other, the sensor will generate a signal to alert a technician that a sensor window may be dirty.

23-54. In the conversion of a DC voltage to a digital signal, the DC voltage is divided into a specific number of steps. Each step representing a specific increase in voltage is assigned a binary number from zero to (2# of bits - 1). The number of steps is based on the number of bits the ADC circuit provides.

23-55. The data is typically stored in an internal cloud in an IIoT system to keep the network and associated data secure from hackers.

=V

V

AB

1

(Eq. 23-4)

Answer: The voltage VAB is 439 mV. 23-21. Given: V1 = 10 V, R1 = 100 Ω, R2 = 100 Ω, R3 = 100 Ω, the RTD will vary from 100 Ω at 0oC to 138.5 Ω at 100oC. The resistance of the cable is 25 Ω.

Solution:

 RTD + R V AB = V 1  R1+ RTD +Cable R

  1  − V1  2     Cable     1  (Eq. 23-5) VAB = 10V  119.25  + 25  − 10V   100  +119.25  + 25 2     VAB = 905.8 mV

PROBLEMS 23-1. Steam engine by Thomas Newcomen in 1712 23-2. Factories

Answer: The voltage VAB is 905.8 mV 23-3. Electricity 23-22. Voltage 23-4. Assembly line 23-23. Piezoelectric transducer 23-5. Industrial automation 23-24.

23-6. Computer Numerical Control (CNC) machines 23-7. Programmable Logic Controller (PLC) 23-8. Forth 23-9. Temperature, pressure, relative location, motion, and ambient light 23-10. Resistance, voltage, current 23-11. Stiff voltage source Copyright 2021 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

1-124

23-25. Given: speed of sound travels 1126 feet per second at 20⁰C, troundtrip = 7 ms Solution: (Eq. 23-6) Distance to object = 1126 ft x sec

7 ms

= 3.941 feet

23-30. The anti-aliasing filter is typically a low-pass filter with a corner frequency of half the Nyquist frequency. The anti-aliasing filter removes any signals above the frequency of the original signal.

2

Convert feet to inches: 3.941 feet x

12 inches

= 47.29 inches

1 ft Distance to object = 3.941 feet or 47.29 inches

Answer: The distance to the liquid is 3.941 feet or 47.29 inches.

23-26. Given: The tank is 12 feet in diameter and 16 feet tall. 1500 ft3 of solution is required in the tank. Both tank and solution are at room temperature.

Solution:

V =  r 2h V h= r 2 1500 ft 3 h=  (6 ft )2 h =13.263 ft tRoundtrip =

23-29. 50 or 60 Hz, because it is power line noise.

23-31. RFID systems typically operate at frequencies between 900 MHz and 915 MHz.

23-32. The passive RFID tag collects energy from the radio waves and uses it to power a tiny circuit inside the tag.

23-33. The active RFID tag typically has a battery.

23-34. Programmable logic controllers, smart sensors, robots, conveyor lines and machining centers are programmed to act on the information received from the other devices.

23-35. The centralized process management system of remote devices and automated systems provides a wealth of data and facilitated informed decision making in the manufacturing process.

Distance to object = time of one way trip x 2 Speedof ultrasonic waves

23-36. The data is commonly stored on an internal cloud.

16 ft − 13.263 ft = 2.4307 ms x 2 ft 1126 sec tRoundtrip = 4.861ms t Roundtrip =

Answer: The required height of the solution is 13.263 ft and the round trip time for the ultrasonic waves is 4.861 ms.

23-27. Given: DC voltage varies from 0 V to 3 V, 8 bit ADC Solution: 2# of bits = number of steps, 28 = 256 steps

CRITICAL THINKING 23-37. The black case would absorb the light beam, not reflect it. Either the Through-beam method or Retroreflective method would detect the black plastic case.

23-38. The ultrasonic sensor is the best choice to measure the fluid level in a tank. Using sound waves, it can continuously measure the fluid level in a tank.

23-39. A capacitive proximity sensor can be mounted on the side of a tank. When the fluid in the tank reaches the sensor, the fluid changes the capacitance of the sensor and this change is detected.

Maximum DC Voltage = Voltage Step (Eq. 23-10) 2# of Bits 3 V 3V = = 11.7mV 28 256

23-40. The input impedance of the Differential amplifier is very high. According to Ohm’s law, I = V/Z. If “Z” is very large, the current “I” approaches zero.

Answer: The number of steps is 256. The step size is 11.7 mV

TROUBLESHOOTING

23-28. Given: Analog signal frequency = 2 kHz Solution: Nyquist rate = 2 x original analog signal frequency.

2 x 2 kHz = 4 kHz

23.41. Trouble 1: The source decreased to a 0V to 2 mV pulse.

The pulse entering U3 is 2 V. The reference voltage is 3 V (Point C). The output remains at -13.3 V. Trouble 2: R3 is shorted Trouble 3: R3 is open

Answer: The Nyquist rate is 4 kHz.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

23-42. Trouble 4: No supply voltage Trouble 5: Rf is open Trouble 6: U2 failed 23-43. Trouble 7: U3 failed Trouble 8: U1 failed Trouble 9: The feedback loop of U1 is open MULTISIM TROUBLESHOOTING 23-44. The op amp U1 failed.

23-45. Resistor R3 is open.

23-46. The feedback loop of U2 is open 23-47. The op amp U3 failed.

23-48. The signal source (vin) failed.

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Part

2

Experiments Manual Note: The tables in this part contain representative data for the experiments in the ninth edition. Variations from the data may be expected, depending on the accuracy of the equipment used and the tolerance of the components. Measurements were made with both digital and analog test equipment, along with Multisim simulation software.

Experiment 1 TABLE 1-1 VOLTAGESOURCE

R

Estimated VL

Measured VL

10 V 10 V 9.9 V 9.5 V

10 V 9.99 V 9.91 V 9.53 V

0Ω 10 Ω 100 Ω 470 Ω

has an internal resistance that is at least 100 times the load resistance.

7. A shorted load has zero resistance. Ohm’s law says that the load voltage equals the load current times the load resistance. The product of load current times zero gives zero load voltage.

8. With the load resistor open, the current through the series resistor R of Fig. 1-1 is zero. Therefore, Ohm’s law tells us that no voltage is dropped across R. Kirchhoff’s voltage law tells us the sum of voltages around the loop is zero; therefore, the load voltage equals the source voltage.

9. 1 MΩ: For the current source to appear stiff, its internal resistance must be at least 100 times the maximum load resistance. In this case, 100(10 kΩ) is 1 MΩ.

Experiment 2 TABLE 2-1 THEVENIN VALUES

TABLE 1-2 CURRENT SOURCE

RL 0Ω 10 Ω 47 Ω 100 Ω

Estimated IL

Measured IL

10 mA 9.9 mA 9.5 mA 9 mA

9.36 mA 9.28 mA 8.94 mA 8.56 mA

TABLE 1-3 TROUBLESHOOTING

Trouble

0V 10 V

RTH

IN

5.43 V 5.67 V

2.39 kΩ 2.36 kΩ

2.23 mA 2.40 mA

TABLE 2-2 LOAD VOLTAGES

VL for 1 kΩ

VL for 4.7 kΩ

1.6 V 1.76 V

3.6 V 3.83 V

Calculated Measured

Measured VL

Shorted load Open load

Calculated Measured

VTH

TABLE 2-3 TROUBLESHOOTING

Estimated

Measured

TABLE 1-4 CRITICAL THINKING

Type Voltage source Current source

R

Measured Quantity

100 Ω 10 kΩ

9.91 V 0.98 mA

ANSWERS

Trouble

VTH

RTH

VTH

RTH

Shorted 2.2 kΩ Open 2.2 kΩ

1.5 V 15 V

1 kΩ 5.4 kΩ

1.52 V 15 V

1.13 kΩ 5.13 kΩ

TABLE 2-4 CRITICAL THINKING

1. d 2. b 3. b 4. b 5. a 6. A stiff voltage source has an internal resistance that is less than 1⁄100 of the load resistance, whereas a stiff current source

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2-1

Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

TABLE 2-5

RTH =600 Ω

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

9. 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.

10. 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.

Experiment 3 TABLE 3-1 TROUBLES AND VOLTAGES

Trouble

VA

VB

Circuit OK R1 open R2 open R3 open R4 open R1 shorted R2 shorted R3 shorted R4 shorted

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

TABLE 3-2 TROUBLES AND VOLTAGES

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2-2

Experiment 5

Experiment 6

TABLE 5-1 FORWARD BIAS

TABLE 6-1 TWO POINTS ON THE FORWARD CURVE

I

VS

V

I

R

0 0.5 V 1V 2V 4V 6V 8V 10 V 15 V

0 0.45 V 0.55 V 0.6 V 0.64 V 0.66 V 0.68 V 0.69 V 0.71 V

0 0.03 mA 0.4 mA 1.35 mA 3.4 mA 5.5 mA 7.6 mA 9.7 mA 14.2 mA

(no entry) 15 kΩ 1.38 kΩ 444 Ω 188 Ω 120 Ω 89 Ω 71 Ω 50 Ω

−1 V −5 V −10 V −15 V

V

I

R

−1 V −5 V −10 V −15 V

0 0 0 0

Infinite Infinite Infinite Infinite

TABLE 5-3 DIODE CONDUCTION

Estimated I Fig. 4-3a Fig. 4-3b

45 mA 0

Measured I 37.7 mA 0

Normal diode Shorted diode Open diode

Measured VL

0.7 V 0 15 V

0.71 V 0 14.9 V

0.7 V 0.7 V

0.69 V 0.78 V

TABLE 6-2 DIODE VALUES

Vknee = 0.69 V rB = 2.25 Ω

TABLE 6-3 DIODE CURRENT

Ideal I = 27.6 mA Second I = 25.7 mA Third I = 25.6 mA Measured I = 24.1 mA TABLE 6-4 TROUBLESHOOTING

Shorted 470 Ω Open 470 Ω

Estimated I

Measured I

0 30 mA

0 30.4 mA

TABLE 6-5 CRITICAL THINKING

TABLE 5-4 TROUBLESHOOTING

Estimated VL

Measured V

10 mA 50 mA

TABLE 5-2 REVERSE BIAS

VS

Estimated V

VS R1 R2 R3 I

Design 1

Design 2

Design 3

12.5 V 470 Ω 220 Ω 220 Ω 7.87 mA

5.86 V 220 Ω 470 Ω 220 Ω 8.2 mA

11.7 V 220 Ω 220 Ω 470 Ω 8.59 mA

TABLE 5-5 CRITICAL THINKING

VS = 10.7 V; RS = 1 kΩ; I = 10.2 mA

ANSWERS 1. b 2. a 3. c 4. c 5. d 6. An ordinary resistor has the same resistance in either direction. A diode has a low resistance when forward-biased and a high resistance when reverse-biased.

7. Because the load is in parallel with a forward-biased diode whose typical voltage drop is 0.7 V. 8. Because the circuit becomes a voltage divider with 1 kΩ in series with 100 kΩ. This implies that approximately 99 percent of the source voltage appears across the load resistor.

9. The higher the source voltage the better, because it swamps out the effect of the diode offset voltage.

ANSWERS 1. c 2. b 3. c 4. c 5. c 6. The steeper the diode curve, the larger the changes in current for a given voltage change. This implies the bulk resistance decreases.

7. With the diode shorted, there is no voltage to drive the diode.

8. Two things happen. The Thevenin voltage increases from 10.2 to 15 V, an increase of approximately 47 percent. The Thevenin resistance increases from 370 to 440 Ω, an increase of approximately 19 percent. Therefore, the short-circuit current is greater, which implies more diode current.

9. c

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2-3

Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

Experiment 7 TABLE 7-5 CRITICAL THINKING

TABLE 7-1 HALF-WAVERECTIFIER

RMS secondary voltage Peak output voltage DC output voltage Ripple frequency

Calculated

Multisim

Measured

12.6 V 16.6 V 5.44 V 60 Hz

12.6 V 17.1 V 5.36 V 60 Hz

14.64 V 19.7 V 6.15 V 60 Hz

Note: The 2nd approximation for the diode was used.

RMS secondary voltage Peak output voltage DC load voltage DC load current Ripple frequency Load resistance

Calculated

Measured

6.3 V 8.9 V 5.67 V 21 mA 120 Hz 270 Ω

7.73 V 8.8 V 5.53 V 20 mA 120 Hz 267 Ω

TABLE 7-2 FULL-WAVE RECTIFIER

RMS secondary voltage Peak output voltage DC output voltage Ripple frequency

Calculated

Multisim

Measured

6.3 V 8.4 V 5.67 V 120 Hz

6.32 V 8.28 V 5.06 V 120 Hz

7.28 V 10.29 V 6.15 V 120 Hz

Note: The 2nd approximation for the diodes was used.

TABLE 7-3 BRIDGE RECTIFIER

RMS secondary voltage Peak output voltage DC output voltage Ripple frequency

Calculated

Multisim

Measured

12.6 V 16.6 V 11.2 V 120 Hz

12.78 V 15.2 V 9.3 V 120 Hz

14.64 V 19.3 V 12.28 V 120 Hz

Note: The 2nd approximation for the diodes was used.

TABLE 7-4 TROUBLESHOOTING

Calculated

Diode open Half-secondary short

ANSWERS 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 voltageand 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.

Measured

Vdc

fout

Vdc

fout

5.67 V 5.67 V

60 Hz 120 Hz

6.25 V 5.68 V

60 Hz 120 Hz

Experiment 8 TABLE 8-3 RL = 1 KΩ AND C = 470 μF

TABLE 8-1 TRANSFORMER RESISTANCES

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 Peal-to-peak ripple

Calculated

Multisim

Measured

12.6 V 16.6 V 16.6 V 16.6 mA 120 Hz 2.94 V

12.6 V 17.1 V 15.36 V 15.36 mA 120 Hz 2.17 V

14.5 V 20.5 V 17.7 V 17.7 mA 120 Hz 2.8 V

Note: The 2nd approximation for the diodes was used.

RMS secondary voltage Peak output voltage DC output voltage DC load current Ripple frequency Peal-to-peak ripple

Calculated

Multisim

Measured

12.6 V 16.6 V 16.6 V 16.6 mA 120 Hz 0.24 V

12.6 V 17.1 V 15.93 V 15.93 mA 120 Hz 0.236 V

14.5 V 20.5 V 18.3 V 18.3 mA 120 Hz 0.276 V

Note: The 2nd approximation for the diodes was used.

TABLE 8-4 RL = 10 KΩ AND C = 470 μF

RMS secondary voltage Peak output voltage DC output voltage DC load current Ripple frequency Peal-to-peak ripple

Calculated

Multisim

Measured

12.6 V 16.6 V 16.6 V 1.66 mA 120 Hz 24 mV

12.6 V 17.1 V 16.75 V 1.65 mA 120 Hz 24 mV

14.5 V 20.5 V 18.71 V 1.9 mA 120 Hz 28 mV

Note: The 2nd approximation for the diodes was used.

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2-4

TABLE 8-5 TROUBLESHOOTING

Calculated

Open diode Open capacitor

Measured

Vdc

fout

Vrip

Vdc

fout

Vrip

17.8 V 11.3 V

60 Hz 120 Hz

0.63 V 17.8 V

19.6 V 12.5 V

60 Hz 120 Hz

0.48 V 20 V

TABLE 8-6 CRITICAL THINKING

Calculated

Measured

12.6 V 17.8 V 17.8 V 4.56 mA 120 Hz 1.73 V

15.5 V 20.1 V 20.1 V 5.2 mA 120 Hz 1.5 V

RMS secondary voltage Peak output voltage DC output voltage DC load current Ripple frequency Peal-to-peak ripple

through the load. Because of the long time constant, the capacitor voltage decreases only slightly. When the secondary voltage again approaches the peak, the diode turns on briefly and recharges the capacitor.

7. a 8. When the filter capacitor is open, the circuit reverts to an average rectifier. It is still a bridge rectifier; therefore, the output voltage is ideally a full-wave signal with a dc value of 0.636 times the peak secondary voltage.

9. I used 220 μF. I calculated the new load current, approximately 17.8 V/3.9 kΩ, which equals 4.56 mA. Ten percent of the peak secondary voltage is 1.78 V. With a ripple frequency of 120 Hz, I used Eq. (4-8) to calculate an ideal capacitance of 214 μF. The nearest standard size is 220 μF. 10. Primary impedance is stepped down by the square of the turns ratio. Since 92 is 81:

ANSWERS 1. b 2. d 3. d 4. c 5. a 6. The capacitor ideally charges to the peak secondary voltage.

Then the diode stops conducting. The capacitor discharges

RTH = Rpri/81 + Rsec = 33.6/81 + 1.1 = 1.51 Ω

Experiment 9 TABLE 9-1 LIMITERS

Estimated

Positive limiter Negative limiter Combination limiter Biased limiter

Measured

Pos Peak

Neg Peak

Pos Peak

Neg Peak

+0.7 V –0.7 V +0.7 V Variable

−4.5 V +4.5 V −0.7 V −0.7 V

+0.68 V −0.68 V +0.66 V Variable

−4.66 V +4.66 V −0.66 V −0.66 V

TABLE 9-2 PEAK DETECTOR

TABLE 9-4 CRITICAL THINKING

Estimated Vdc fout Vrip

5V 1 kHz 100 mV

Measured 9.12 V 1 kHz 90 mV

Capacitance DC output voltage Ripple frequency Peak-to-peak ripple

Calculated

Measured

0.033 μF +4.3 V 5 kHz 0.563 V

(No entry) 4.1 V 5 kHz 0.47 V

TABLE 9-3 TROUBLESHOOTING

Estimated

Open diode Shorted diode

Measured

Pos Peak

Neg Peak

Pos Peak

Neg Peak

+5 V 0

−0.7 V 0

+4.9 V 0

−0.64 V 0

ANSWERS 1. a 2. b 3. b 4. b 5. a 6. The diode on the right clips off the negative portions of the signal; the clipping level is approximately –0.7 V. The diode on the left clips off positive portions of the signal. Because

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

this diode is biased by the positive supply, the positive clipping level is adjustable between 0 and +10 V. 7. When the left diode is open, the circuit becomes a negative clipper. When the left diode is shorted, no voltage can appear across the output because it is shorted.

8. The filter capacitor is open. The circuit is then a half-wave rectifier, equivalent to a negative limiter.

9. The second approximation looks like the best compromise. If we subtract 0.7 V from the peak input voltage to get the dc output voltage, our calculated dc output will be fairly close to the measured values.

10. The knee voltage is only 0.3 V. Because less signal is wasted.

Experiment 10 TABLE 10-1 CLAMPERS

Positive clamper Negative clamper

Estimated

Measured

Pos Peak Neg Peak

Pos Peak Neg Peak

+10 V 0V

+9.8 V +0.5 V

0V −10 V

−0.8V −9.8V

6. During the first few negative half cycles, the diode conducts and charges the input capacitor. After several cycles, the capacitor voltage approximately equals the peak input voltage. If you apply Kirchhoff’s voltage law, you find that the output voltage is the sum of the ac source voltage plus the capacitor voltage. Since the capacitor is charged to approximately VP, the output voltage swings from approximately 0 to 2 VP. 7. I got about 8.9 V. The positive dc clamper was still working; therefore, the input to the right diode was a sine wave swinging between approximately 0 and +10 V. This meant that the right diode was forward-biased for most of the cycle. In other words, the output voltage was swinging from approximately 0 to +10 V. The average of this is +5 V. If we allow a drop of 0.7 V across the right diode, the dc output voltage will be very close to +8.9 V, the measured value.

8. More than likely, it is a classic dc-return problem. The generator that works is dc-coupled. The one that does not is accoupled, meaning an internal coupling capacitor. The proof is to connect the questionable generator and to insert a dc return.

Experiment 11 TABLE 11-1 TRANSFORMER RESISTANCES

TABLE 10-2 PEAK-TO-PEAK DETECTOR

Estimated

Measured

10 V 0.1 V

8.9 V 0.09 V

Vdc Vrip TABLE 10-3 DC RETURN

Estimated VP

Measured VP

5V 0

4.1 V 0

With dc return Without dc return

TABLE 11-2 HALF-WAVE DOUBLER

ANSWERS 1. b 2. d 3. d 4. b 5. d

Measured

6.3 V 17.8 V 60 Hz 0.632 V

7.72 V 19.1 V 60 Hz 0.5 V

Calculated

Measured

6.3 V 17.8 V 120 Hz 0.632 V

7.73 V 19.5 V 120 Hz 0.5 V

Half rms secondary voltage DC output voltage Ripple frequency Peak-to-peak ripple

Estimated Vdc

Measured Vdc

0 5V 5V 0 0 0 0 5V

0 4.9 V 4.9 V 0 0 0 0 4.8 V

Half rms secondary voltage DC output voltage Ripple frequency Peak-to-peak ripple TABLE 11-4 TROUBLESHOOTING

Estimated

TABLE 10-5 CRITICAL THINKING

Capacitance DC output voltage Ripple frequency Peak-to-peak ripple

Calculated

TABLE 11-3 FULL-WAVEDOUBLER

TABLE 10-4 TROUBLESHOOTING

Open C1 Short C1 Open C2 Short C2 Open D1 Short D1 Open D2 Short D2

Rpri = 33.6 Ω Rsec = 1.1 Ω

Calculated

Measured

1.5 μF 9.8 V 2.5 kHz 0.107 V

(No entry) 9.8 V 2.5 kHz 0.1 V

Open C1 Open D2 Open C2

Measured

Vdc

fout

Vrip

Vdc

fout

Vrip

0 0 9V

0 0 60 Hz

0 0 18 V

0 0 9.4 V

0 0 60 Hz

0 0 20 V

TABLE 11-5 CRITICAL THINKING

Half rms secondary voltage DC output voltage Ripple frequency Peak-to-peak ripple

Calculated

Measured

6.3 V 17.8 V 60 Hz 1.78 V

7.73 V 18.5 V 60 Hz 1.4 V

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2-6

TABLE 12-3 TROUBLESHOOTING

ANSWERS 1. d 2. a 3. b 4. b 5. a 6. On the positive half cycle the upper filter capacitor charges to

Trouble

Estimated Vdc

Measured Vdc

0 15 V 0 0.7 V

0 15 V 0 0.82 V

Shorted diode Open diode Open resistor Fwd. Biased Diode

the peak input voltage (approximately). On the negative half cycle the lower filter capacitor charges to the input Vp with the same polarity. The two capacitor voltages add to produce the load voltage, which is approximately double the peak input voltage.

7. Filter capacitor C1 still charges to the peak voltage (approximately). Since C2 is open, diode D2 conducts for most of the cycle. The load voltage is the sum of the input ac voltage plus the dc voltage across C1. The output waveform then swings from approximately 0 to twice the peak input voltage. (As discussed later in Sec. 4-17, the circuit becomes a dc clamper.)

8. If either filter capacitor is shorted, the nearest diode is connected across half the secondary winding. The excessive forward current will destroy the diode because the Thevenin resistance facing the diode is extremely small.

9. I calculated an ideal peak voltage of 17.8 V. Ten percent of this is 1.78 V. For a load resistance of 3.9 kΩ, the dc load current is ideally 17.8 V/3.9 kΩ, which equals 4.56 mA. Using a ripple frequency of 60 Hz in Eq. (4-8) gives an ideal capacitance of 42.7 μF. The nearest standard size is 47 μF. 10. The turns ratio is 115/12.6, or 9.13. The primary impedance is stepped down by the square of 9.13, which equals 83.4.

The Thevenin resistance facing the capacitor is RTH = Rpri/83.4 + Rsec = 30/83.4 + 1 = 1.36 Ω

Experiment 12

Experiment 13

TABLE 12-1 DATA FOR ZENER DIODE

TABLE 13-1 SPLIT SUPPLY

Vin

Estimated Vout

Measured Vout

0 2V 4V 6V 8V 10 V 12 V 14 V

0 2V 4V 6V 6.2 V 6.2 V 6.2 V 6.2 V

0 2V 4V 5.99 V 6.18 V 6.20 V 6.22 V 6.23 V

Calculated

Positive regulator Negative regulator

Measured

Vin

Vout

Vin

Vout

8.9 V −8.9 V

6.2 V −6.2 V

9.78 V −9.76 V

6.15 V −5.96 V

TABLE 13-2 VOLTAGE REGULATION

Estimated TABLE 12-2 ZENER RESISTANCE

Vin

Calculated IZ

Calculated RZ

0 2V 4V 8V 10 V 12 V 14 V

0 0 12 µA 10.1 mA 21.1 mA 32.1 mA 43.2 mA

(No entry) (No entry) (No entry) (No entry) 1.8 Ω 1.4 Ω (No entry)

Measured

RL

+Vout

−Vout

+Vout

−Vout

470 Ω 4.7 kΩ 47 kΩ

6.2 V 6.2 V 6.2 V

−6.2V −6.2V −6.2V

6.15 V 6.26 V 6.28 V

−5.96 V −6.05 V −6.07 V

TABLE 13-3 RIPPLE

RL 470 Ω 47 Ω

Calculated Vrip In Out 0.319 V 0.319 V

14.9 mV 14.9 mV

Measured Vrip In Out 0.34 V 0.34 V

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10 mV 10 mV

Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

Design for two-stage voltage regulator (for Experiment 13)

TABLE 13-4 TROUBLESHOOTING

Estimated

Open CT Open D1 Open D6

Measured

+Vout

–Vout

+Vout

–Vout

6.2 V 6.2 V 6.2 V

−6.2 V −6.2 V −6.75 V

6.14 V 6.15 V 6.16 V

−5.94 V −5.92 V −7.58 V

TABLE 13-5 CRITICAL THINKING

Calculated

Preregulator input Regulator input Regulator output

Measured

Vdc

Vrip

Vdc

Vrip

17.8 V 12.4 V 6.2 V

0.709 V 76.4 mV 1.78 mV

19.5 V 12.4 V 6.11 V

0.6 V 30 mV 1.5 mV

ANSWERS

4. b 5. c 6. Ideally, the dc voltage across the filter capacitor is 8.9 V, half of the secondary peak voltage. The zener diode breaks down at approximately 6.2 V. This implies a current of 18 mA through the series resistor. As long as the load cur- rent is less than 18 mA, the zener diode operates in the breakdown region and holds the output voltage approximately constant.

7. With the center tap open, the entire secondary voltage drives the bridge diodes, which in turn drive the two filter capacitors in series. If the filter capacitances are identical, each has the same voltage, which ideally is half the second- ary peak voltage. Usually, the filter capacitances are not equal because of tolerances; this is why the center tap is normally used.

8. D4 still conducts during the negative half cycles and the circuit becomes a half-wave rectifier working into a capacitorinput filter. As a result, the capacitor still charges to the peak input voltage. Even though the circuit works, it has twice as much ripple, which is undesirable.

9. I used 7 Ω for the zener resistance of each diode. Since my measured ripple was less than calculated ripple, the zener diodes probably had less than 7 Ω of zener resistance.

1. d 2. b 3. c

Experiment 14

TABLE 14-2 SEVEN-DIGIT INDICATOR

TABLE 14-1 LED DATA

I

Vred

Vgreen

10 mA 20 mA 30 mA 40 mA

1.6 V 1.69 V 1.74 V 1.78 V

1.9 V 2.02 V 2.12 V 2.2 V

Display

Pins Grounded

0 1 2 3 4 5 6 7 8 9 LDP RDP

1, 2, 7, 8, 10, 13 10, 13 1, 7, 8, 11, 13 1, 8, 10, 11, 13 2, 10, 11, 13 1, 2, 8, 10, 11 1, 2, 7, 8, 10, 11 1, 10, 13 1, 2, 7, 8, 10, 11, 13 1, 2, 8, 10, 13 6 9

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2-8

TABLE 14-5 TROUBLESHOOTING

TABLE 14-3 OPTOCOUPLER

VS

Vout

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

TABLE 14-4 PHOTDIODE

Open LED Shorted LED

Estimated VLED

Measured VLED

15 V 0V

15 V 0V

TABLE 14-6 CRITICAL THINKING

Calculated Red LED Green LED

Measured

R

ILED

VLED

ILED

VLED

680 Ω 680 Ω

19.1 mA 19.1 mA

2V 2V

20.1 mA 19.6 mA

1.7 V 2V

Experiment 15

TABLE 15-1 TRANSISTOR VOLTAGES AND CURRENTS

ANSWERS

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

Experiment 16

Experiment 17

TABLE 16-1 FIRST CIRCUIT

TABLE 17-1 βdc VALUES

Calculated Measured

VBE

VCE

Region

0.7 V 0.69 V

6.04 V 6.79 V

Active Active

R

RB

βdc

IC

203 kΩ 338 kΩ 474 kΩ

225 kΩ 360 kΩ 506 kΩ

100 160 225

6.36 mA 6.36 mA 6.36 mA

Transistor 1 2 3

TABLE 16-2 SECOND CIRCUIT TABLE 17-2 CALCULATIONS

Calculated Measured

VBE

VCE

Region

0.7 V 0.69 V

0V 89 mV

Saturation Saturation

Test

IE(ideal)

IE(second)

1 2 3

6.73 mA 6.71 mA 6.7 mA

6.42 mA 6.4 mA 6.39 mA

TABLE 16-3 THIRD CIRCUIT TABLE 17-3 CALCULATIONS

VBE

VCE

Region

Calculated Measured

10 V 10 V

Cutoff Cutoff

VBE

IC(sat)

VCE(sat)

VCE(cutoff)

0.7 V

6.82 mA

0

15 V

TABLE 17-4 MEASUREMENTS TABLE 16-4 FOURTH CIRCUIT

VBE

VCE

Region

Calculated Measured

10 V 10 V

Cutoff Cutoff

TABLE 16-5 FIFTH CIRCUIT

Q1 Q2 Q3 Q4

VBE

IC

VCE

Operating Region

R=0 22 kΩ open VCE ≅ 7.5 V

0.71 V 0V 0.7 V

6.75 mA 0 3.4 mA

0.1 V 15 V 7.5 V

Saturation Cutoff Active

ANSWERS

VBE

VCE

Region

0.7 V 0V 0V 0.7 V

0V 10 V 10 V 6V

Saturation Cutoff Cutoff Active

VBE

VCE

Region

0V 0V 0.7 V 0.7 V

10 V 10 V 0V 10 V

Cutoff Cutoff Saturation Active

TABLE 16-6 TROUBLESHOOTING

RBO RBG RCO RCS

Condition

1. a 2. d 3. d 4. d 5. a 6. c 7. c 8. c 9. Answers will vary.

10. Answers will vary.

11. Answers will vary.

Experiment 18 TABLE 18-1 TRANSISTOR SWITCH

Calculated

Measured

ANSWERS 1. b 2. d 3. a 4. d 5. b 6. a 7. d 8. a 9. Answers will vary.

Transistor 1 2 3

IB

IC

1.43 mA 13 mA -

VCE 0 -

IB

IC

VCE

1.3 mA 11.9 mA 0.07 V 1.3 mA 11.9 mA 0.05 V 1.3 mA 11.9 mA 0.08 V

TABLE 18-2 TRANSISTOR CURRENT SOURCE

Calculated Transistor 1 2 3

VE

IC

Measured VCE

VE

IC

VCE

4.3 V 19.5 mA 3.7 V 4.37 V 20.5 mA 3.97 V 4.29 V 20.5 mA 4 V 4.25 V 20.2 mA 3.98 V -

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2-10

ANSWERS

TABLE 18-3 TROUBLESHOOTING THE TRANSISTOR SWITCH

Trouble

Estimated VC

Measured VC

13 V 0 0 13 V

13.8 V 0.05 V 0 13.8 V

Open 10 kΩ Open 1 kΩ Shorted collector-emitter Open collector-emitter

TABLE 18-4 TROUBLESHOOTING THE TRANSISTOR CURRENT SOURCE

Estimated

Measured

Trouble

VC

VE

VC

VE

Open 220 Ω Shorted collector-emitter Open collector-emitter

10 V 8V 10 V

5V 8V 0

8.87 V 8.32 V 8.87 V

4.57 V 8.32 V 0

TABLE 18-5 CRITICAL THINKING

Calculated Transistor Switch Current source

Measured

R

VE

IC

VE

IC

430 Ω 150 Ω

0 4.3 V

30.2 mA 28.7 mA

0 4.29 V

28.4 mA 28 mA

Experiment 19

TABLE 19-4 DESIGN VALUES: R1 = 9.1 KΩ; R2 = 1.5 KΩ; RC = 3 KΩ; RE = 750 Ω

TABLE 19-1 VOLTAGE-DIVIDER BIAS

Calculated Transistor 1 2 3

VB

Measured

1 2 3

8.63 V 8.62 V 8.48 V

1.82 V

1.16 V 5.95 V

— —

1.82 V 1.83 V

1.17 V 5.92 V 1.2 V 5.82 V

ANSWERS

VC

Measured VC

VB

VE

VC

TABLE 19-3 TROUBLESHOOTING

Open R1 Shorted R1 Open R2 Shorted R2 Open RC Shorted RC Open RE Shorted RE Open collector-emitter Shorted collector-emitter

9.31 V -

5.71 V

VE

0 − 0.7 V 5.58 V −0.015 V −0.66 V 5.74 V — — — −0.022 V −0.66 V 5.76 V — — — −0.009 V −0.63 V 5.73 V

Trouble

Measured VC

1 2 3

Calculated VE

Calculated VC

VB

TABLE 19-2 EMITTER BIAS

Transistor VB

Transistor

VC

VE

1.8 1.1 V V — — — —

1. b 2. a 3. b 4. a 5. a 6. A transistor switch operates at saturation, while a transistor current source operates in the active region. Also, the transistor switch appears shorted when conducting, while the transistor current source appears like a current source with a high internal impedance.

7. The most likely trouble is a collector-emitter short. This would ground the cathode of the LED, producing a collector voltage of approximately zero and keeping the LED lit.

8. With the emitter resistor open, no collector current exists. When I connected the voltmeter to the collector, it pro- duced a small current through the LED. This was not enough current to light the LED, but it was enough to produce a LED drop of approximately 1.13 V because I measured 8.87 V at the collector. Similarly, the measured emitter voltage was 4.57 V, slightly less than 5 V because of the VBE drop produced by the current through the voltmeter.

9. The major problem is the variation in βdc with transistor replacement and temperature change. Most small-signal transistors have a βdc greater than 10 under all conditions. Hard saturation means using a 10:1 ratio for collector to base current. This guarantees transistor saturation under all operatingconditions.

Estimated VC

Measured VC

10 V 9.3 V 2.2 V 10 V 1.1 V 10 V 10 V 0 10 V 2.2 V

10 V 9.2 V 2.89 V 10 V 0.45 V 10 V 10 V 0.05 V 10 V 2.34 V

1. c 2. b 3. d 4. c 5. a 6. I learned that the Q point is almost constant, even though transistors are changed. The collector voltage in either circuit had a variation of 0.13 V or less when I changed transistors.

7. Open R1, shorted R2, shorted RC, open RE, and open collectoremitter.

8. I measured 2.34 V. A shorted collector-emitter is equivalent to transistor saturation. The collector resistor (3.6 kΩ) and the emitter resistor (1 kΩ) form a voltage divider driven by a supply of 10 V. Theoretically, this means approximately 2.2 V will appear at the collector, close enough to the measured value.

9. The measured VC ranged from 8.48 to 8.63 V. The calculated VC was 9.31 V. The main cause of differences is the tolerance of the resistors. Some of my resistors had a tolerance of 5 percent and others a tolerance of 10 percent. The small variation in measured values from 8.48 to 8.63 V was caused mainly by variations in hFE.

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2-11

Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

Experiment 20 TABLE 20-1 NEGATIVE POWER SUPPLY

Calculated

Measured

Transistor

VB

VE

VC

VB

VE

VC

1 2 3

−1.8 V −1.8 V −1.8 V

−1.1 V −1.1 V −1.1 V

−5.71 V −5.71 V −5.71 V

−1.79 V −1.79 V −1.8 V

−1.14 V −1.19 V −1.14 V

−5.94 V −5.76 V −5.94 V

TABLE 20-2 POSITIVE POWER SUPPLY

Calculated

Measured

Transistor

VB

VE

VC

VB

VE

VC

1 2 3

8.2 V 8.2 V 8.2 V

8.9 V 8.9 V 8.9 V

4.28 V 4.28 V 4.28 V

8.21 V 8.21 V 8.22 V

8.85 V 8.81 V 8.88 V

4.08 V 4.23 V 4.04 V

TABLE 20-3 TROUBLESHOOTING

Estimated

Measured

Trouble

VB

VE

VC

VB

VE

VC

Open R1 Shorted R1 Open R2 Shorted R2 Open RC Shorted RC Open RE Shorted RE Open CE Shorted CE

7.3 V 10 V 10 V 0 8.2 V 8.2 V 8.2 V 9.3 V 8.2 V 8V

8V 10 V 10 V 0.7 V 8.9 V 8.9 V 0 10 V 10 V 8V

8V 0 0 0.7 V 8.9 V 0 0 10 V 0 8V

6.45 V 10 V 10 V 0 8.93 V 8.23 V 8.21 V 9.31 V 8.22 V 7.77 V

7.17 V 10 V 10 V 0.83 V 9.61 V 8.89 V 1.98 V 10 V 10 V 7.7 V

7.14 V 0 0 0.82 V 9.61 V 0 0 9.96 V 0 7.7 V

TABLE 20-4 DESIGN VALUES: R1 = 100 Ω; R2 = 200 Ω; RE = 51 Ω

Transistor

Calculated IC

Measured IC

1 2 3

19 mA 19 mA 19 mA

18 mA 17.9 mA 18.5 mA

3. b 4. d 5. b 6. Conventional flow: R1 down, R2 down, RE down, RC down, transistor down; electron flow: all currents are up.

7. Shorted R1, open R2, shorted RC, open RE, open CE. 8. Shorted R1 or open R2.

9. Many designs are possible. Arbitrarily, I selected VE = +4 V. Then I calculated an RE of 1 V/20 mA, or 50 Ω. The nearest standard size is 51 Ω.

ANSWERS 1. c 2. b

Experiment 21 TABLE 21-1 BASE BIAS

Calculated

Measured

Transistor

IB

IC

VC

IB

IC

VC

1 2 3

53 μA 53 μA 53 μA

10.6 mA 10.6 mA 10.6 mA

7.79 V 7.79 V 7.79 V

53.1 μA 53.1 μA 53.1 μA

9.8 mA 8.3 mA 16 mA

8.8 V 9.91 V 3.05 V

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2-12

TABLE 21-2 EMITTER-FEEDBACK BIAS

Calculated

Measured

Transistor

IC

VC

VE

IC

VC

VE

1 2 3

6.36 mA 6.36 mA 6.36 mA

9.21 V 9.21 V 9.21 V

0.636 V 0.636 V 0.636 V

5.8 mA 4.85 mA 10.4 mA

9.89 V 10.8 V 5.98 V

0.59 V 0.48 V 1.03 V

TABLE 21-3 COLLECTOR-FEEDBACK BIAS

Calculated

Measured

Transistor

VB

IC

VC

VB

IC

VC

1 2 3

0.7 V 0.7 V 0.7 V

7.15 mA 7.15 mA 7.15 mA

7.85 V 7.85 V 7.85 V

0.69 V 0.69 V 0.69 V

6.3 mA 5.8 mA 8.4 mA

8.47 V 9.05 V 6.27 V

ANSWERS

TABLE 21-4 TROUBLESHOOTING

Trouble

Estimated VC

Measured VC

15 V 0.7 V 0 15 V 15 V 0

15 V 0.74 V 0 15 V 15 V 0

Open 200 kΩ Shorted 200 kΩ Open 1 kΩ Shorted 1 kΩ Open collector-emitter Shorted collector-emitter

TABLE 21-5 CRITICAL THINKING: RB = 39 KΩ; RC = 240 Ω

Transistor

Calculated VC

Measured VC

1 2 3

5.25 V 5.25 V 5.25 V

7.56 V 7.25 V 5.4 V

Experiment 22

1. d 2. c 3. c 4. c 5. c 6. Base bias had the worst variation. Emitter-feedback biaswas almost as bad. Collector-feedback bias has the most stable Q point.

7. Three possible troubles are an open base resistor, a shorted collector resistor, and a transistor with an open between the collector-emitter terminals.

8. Two possible troubles are an open collector resistor and a transistor with a short between the collector-emitter terminals.

9. To get a Q point near the middle of the load line, I calculated an RC of 5 V/2 mA, or 2.5 kΩ. The nearest standard value was 2.4 kΩ. Figure 15 of the 2N3904 data sheet indicates that hFE is approximately 0.85 of its value at IC = 10 mA. This implies a typical hFE of about 0.85 × 200, or 170. Then I used Eq. (8-37) in the textbook to calculate an RB of 408 kΩ . The nearest standard value is 390 kΩ .

TABLE 22-2 BYPASS CAPACITOR

TABLE 22-1 COUPLING CAPACITOR

fc 10fc 0.1fc

ANSWERS 1. b 2. b 3. b

f

vout

179 Hz 1.79 kHz 17.9 Hz

0.707 V 0.995 V 0.1 V

fc 10fc 0.1fc

f

vA

658 Hz 6.58 kHz 65.8 Hz

0.707 V 0.1 V 0.995 V

4. c 5. b 6. a 7. c 8. d

Copyright 2021 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

2-13

Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

TABLE 23-4 CRITICAL THINKING

Experiment 23 TABLE 23-1 CE AMPLIFIER

Calculated

Calculated

dc ac

Measured

B

E

C

B

E

C

1.8 V 10 mV

1.1 V 0

5.71 V 1.71 V

1.8 V 10 mV

1.2 V 0

5.76 V 1.6 V

Measured

Transistor

r'e

RC

AV

vin

vout

AV

1 2 3

22.7 Ω 22.7 Ω 22.7 Ω

2.2 kΩ 2.2 kΩ 2.2 kΩ

97 97 97

10 mV 10 mV 10 mV

0.95 V 1V 0.95 V

95 100 95

ANSWERS

TABLE 23-2 VOLTAGE GAIN

Calculated

Measured

Experimental

Transistor

r'e

AV

vin

vout

AV

r'e

1 2 3

22.7 22.7 22.7

171 171 171

10 mV 10 mV 10 mV

1.6 V 1.7 V 1.6 V

160 170 160

24.4 22.9 24.4

TABLE 23-3 TROUBLESHOOTING

Estimated

Measured

Trouble

vb

ve

vc

vb

ve

vc

Open C1 Open R2 Open RE

0 10 mV 10 mV

0 0 0

0 10 mV 0

0 3 mV 10 mV

0 0 0

0 0 0

Experiment 24

1. a 2. d 3. c 4. d 5. a 6. The small ac base voltage appears across r'e, producing an ac collector current of vb/r'e. This flow is through RC, producing an ac collector voltage of vbRC/r'e. Since RC/r'e is ideally 159, the ac collector voltage is approximately 159 times larger than the ac base voltage.

7. The dc voltages are unaffected by an open coupling capacitor. The ac voltages disappear except for hum and noise.

8. With RE open, the transistor is cut off; so there is no collector current and no ac collector voltage. The emitter is still bypassed to ground; therefore, there is no ac emitter voltage.

The ac base voltage is unaffected by the open RE. 9. I assumed an r'e of 22.7 Ω. I multiplied this by 100 to get 2.27 kΩ, or 2.2 kΩ for the nearest standard size.

TABLE 24-4 CRITICAL THINKING

TABLE24-1 CE AMPLIFIERWITHSOURCEANDLOADRESISTANCES

Calculated

Measured

Transistor

Zin

vb

vc

vb

vc

1 2 3

1.18 kΩ 1.18 kΩ 1.18 kΩ

10.8 mV 10.8 mV 10.8 mV

0.5 V 0.5 V 0.5 V

12 mV 11 mV 12 mV

0.52 V 0.5 V 0.52 V

Calculated

Measured

Transistor

rE

vb

vc

vb

vc

1 2 3

300 300 300

63.5 mV 63.5 mV 63.5 mV

662 mV 662 mV 662 mV

64 mV 64 mV 64 mV

650 mV 650 mV 650 mV

ANSWERS TABLE 24-2 SWAMPED AMPLIFIER

Calculated

Measured

Transistor

Zin

vb

vc

vb

vc

1 2 3

1.7 kΩ 1.7 kΩ 1.7 kΩ

63 mV 63 mV 63 mV

1.05 V 1.05 V 1.05 V

64 mV 64 mV 64 mV

1V 1V 1V

1. a 2. d 3. a 4. d 5. c 6. The swamping resistor is in series with r'e, which can vary with temperature change and transistor replacement. Since the voltage gain equals RC/(rE + r'e) the changes in r'e have less effect on the denominator.

TABLE 24-3 TROUBLESHOOTING

Estimated Trouble

vb

Open CE 66 mV Shorted CE 20 mV Open collector-emitter 66 mV Shorted collector-emitter 20 mV Open C2 63 mV Shorted C2 63 mV

Measured

ve

vc

vb

ve

vc

66 mV 20 mV 0 20 mV 63 mV 63 mV

0.2 V 20 mV 0 20 mV 1V 1V

64 mV 15 mV 64 mV 14 mV 64 mV 63 mV

60 mV 15 mV 0 14 mV 60 mV 60 mV

200 mV 5 mV 0 14 mV 1.05 V 1.05 V

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2-14

impedance of the base drops to approximately 180 Ω. This is why the base voltage decreases significantly.

9. I calculated 360 Ω for the sum of rE + r'e. Since r'e is approximately 22.7 Ω, I selected 330 Ω for the swamping resistor.

7. The voltage gain decreases significantly because the bottom of the swamping resistor is no longer at ac ground. The effective swamping resistance equals RE, the dc emitter resistance.

8. The transistor is driven into saturation because the dc emitter increases to the value of IC(sat). Because of this, the input

Experiment 25 TABLE 25-2 AC VOLTAGES: hfe = 120

TABLE 25-1 DC VOLTAGES

Calculated

Measured

Calculated

Stage

VB

VE

VC

VB

VE

VC

Stage

1 2

1.8 V 1.8 V

1.1 V 1.1 V

5.97 V 5.97 V

1.79 V 1.73 V

1.13 V 1.07 V

6.25 V 6.59 V

1 2

vb

Measured

ve

vc

vb

ve

vc

12.2 mV 9 mV 143 mV 13 mV 10 mV 150 mV 143 mV 105 mV 1.4 V 150 mV 100 mV 1.4 V

TABLE 25-3 TROUBLESHOOTING

DC voltages Trouble Open C4 Shorted R4 Shorted R10 Open R3 Open C5

AC voltages

Stage

VB

VE

VC

vb

ve

vc

1 1 2 1 2

1.8 V 1.8 V 1.75 V 3.58 V 1.75 V

1.1 V 1.1 V 1.11 V 2.88 V 1.09 V

6.05 V 10 V 6.06 V 2.92 V 6.12 V

14 mV 12 mV 130 mV 2 mV 150 mV

14 mV 10 mV 10 mV 2 mV 150 mV

15 mV 10 mV 3V 1 mV 140 mV

6. With the coupling capacitor open, we are measuring the unloaded, or Thevenin, ac voltage at the collector. When the coupling capacitor is reconnected, the input impedance of the second stage loads down the Thevenin circuit. The voltagedivider action implies less ac collector voltage.

7. The effective swamping resistance increases to the value of dc emitter resistance. This reduces the voltage gain significantly.

8. It should be in the vicinity of 68 Ω because the base-emitter terminals are ideally shorted.

9. First, replace the npn transistors by pnp transistors. Second, transpose the plus and minus sides of the power supply. Third, reverse the polarity of all electrolytic capacitors. The amplifier should now work similar to the one with npn transistors.

TABLE 25-4 CRITICAL THINKING: rE = 120 Ω

AV1 = 12.2 AV2 = 6.19 AV = 75.5

ANSWERS 1. b 2. c 3. b 4. c 5. b

Experiment 26

TABLE 26-3 TROUBLESHOOTING

TABLE 26-1 EMITTER FOLLOWER: hfe = 120

Calculated

dc ac

Estimated Measured

B

E

C

B

E

C

5.00 V 0.58 V

4.30 V 0.58 V

10 V 0

5.08 V 0.58 V

4.43 V 0.58

10 V 0

TABLE 26-2 OUTPUT IMPEDANCE

Calculated rout = 42.2 Ω Unloaded vout = 58 mV Loaded vout = 30 mV Experimental rout = 47.6 Ω

Measured

Trouble

VE

ve

VE

ve

Open R1 Shorted R1 Open R2 Shorted R2 Open RE Shorted RE

0 9.3 V 9.3 V 0 4.3 V 0

0 0 0 0 0 0

0 9.31 V 9.15 V 0 4.82 V 0

0 0 0 0 0.2 V 0

TABLE 26-4 COMMON-BASE (CB)

Calculated

dc ac

Measured

B

E

C

B

E

C

1.67 V 0

0.97 V 20 mV

8.3 V 1.03

1.66 V 0

1.02 V 19.8 mV

8.19 V 983

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2-15

Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

TABLE 27-5 RIPPLE

TABLE 26-5 (CB) INPUT IMPEDANCE

Calculated

Measured

50.4 Ω

51 Ω

Zin

TABLE 26-6 CRITICAL THINKING

Calculated

Measured

% Error

2 Vp-p 100

1.87 Vp-p 93.5

6.5 6.5

vout AV

RL 100 Ω 1 kΩ 10 kΩ

Calculated Vrip

Measured Vrip

In

Out

In

Out

1.82 V 0.477 V 0.343 V

38.3 mV 10.2 mV 7.3 mV

1.25 V 0.4 V 0.3 V

10 mV 5 mV 4 mV

TABLE 27-6 TROUBLESHOOTING

Estimated

ANSWERS 1. b 2. a 3. d 4. d 5. d 6. Measured a voltage drop across the inserted 51 Ω input resistor as 9.94 mV. 9.94 mV divided by 51 Ω equals 195 μA for iin. Zin = vin/iin = 9.96 mV/195 μA which results in 51 Ω.

Open D1 Open R1 Shorted D5 Collector-emitter open

Measured

Vin

Vout

Vin

Vout

17.8 V 17.8 V 17.8 V 17.8 V

1.34 V 0 0 0

5.5 V 20.5 V 19.4 V 19.7 V

20 mV 0 0 0

Experiment 28 TABLE 28-1 CE AMPLIFIER

Experiment 27 TABLE 27-1 SPEAKER DRIVER

VB VE VC Vspeaker VV

Calculated

Measured

6.88 V 6.18 V 10 V 270 mVp-p 0.9

6.75 V 5.94 V 10 V 249 mVp-p 0.83

TABLE 27-2 DARLINGTON PAIR

VB Multisim Actual

IB1

IB2

Imotor

β1

β2

β

3 V 80 μA 13 mA 681 mA 121.5 49.5 6019 3 V 17 μA 2.68 mA 136 mA 158 51 8000

ICQ VCEQ PP IS

100 Ω 1 kΩ 10 kΩ

1.11 mA 9V 2.35 V 2.34 mA

1.05 mA 10.8 V 2.5 V 2.3 mA

PD(max) PL(max) PS η

Calculated

Measured

9.99 mW 0.46 mW 35.1 mW 1.3%

11.3 mW 0.52 mW 34.5 mW 1.5%

TABLE 28-3 TROUBLESHOOTING

Calculated

Measured

Calculated

Measured

Trouble

PP

IS

PP

IS

17.8 V 9.1 V 8.4 V

18.6 V 9.06 V 8.43 V

Shorted R2 Open CE Open RL C-E Open

0 2.35 V 7.99 V 0

1.5 mA 2.34 mA 2.34 mA 1.23 mA

0 2.5 V 9V 0

1.41 mA 2.3 mA 2.3 mA 1.2 mA

TABLE 27-4 VOLTAGE REGULATION

RL

Measured

TABLE 28-2 POWER AND EFFICIENCY

TABLE 27-3 REGULATED POWER SUPPLY

Vin VZ Vout

Calculated

Estimated Vout

Measured Vout

8.4 V 8.4 V 8.4 V

8.39 V 8.45 V 8.32 V

TABLE 28-4 CRITICAL THINKING: RE = 910 Ω

PP IS PL(max) PS η

Calculated

Measured

4.66 V 3.43 mA 1.8 mW 51.5 mW 3.5%

5V 3.4 mA 2.08 mW 51 mW 4.08%

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2-16

ANSWERS

Experiment 30

1. b 2. b 3. b 4. b 5. c 6. When the signal swing approaches cutoff, r'e increases, this decreases the voltage gain and produces a squashing effect on the positive half of the cycle. When the signal swing approaches saturation, r'e decreases; this increases the voltage gain and produces an elongation effect on the negative half of the cycle.

7. The ac output compliance does not change when the bypass capacitor opens. The voltage gain decreases significantly.

TABLE 30-1 DC VOLTAGES

Calculated

Q1 Q2 Q3 Q4

Measured

B

E

C

B

E

C

1.8 V 1.61 V 5.7 V 4.3 V

1.1 V 0.92 V 5V 5V

6.1 V 4.3 V 10 V 0

1.7 V 1.56 V 5.65 V 4.48 V

1.05 V 0.92 V 5.03 V 5.03 V

6.4 V 4.48 V 10 V 0

TABLE 30-2 AC VOLTAGES

8. When RL opened, the effective ac load increased to RC. This meant that the product ICQrC had increased to ICQRC. The ac output compliance increased to 2ICQRC. 9. Using trial and error, 910 Ω produced and ICQof 2.2 mA and a VCEQ of 3.12 V. This meant the ac output compliance was equal to approximately 2ICQrC or 4.66 V.

Experiment 29 TABLE 29-1 SINGLE-SUPPLY CIRCUIT

Transistor

Idiode

IC

1 2 3

9.2 mA 9.1 mA 9.4 mA

8 mA 10 mA 11 mA

vin

Measured

Calculated

Q1 Q2 Q3 Q4

B

E

C

B

1.42 mV 1.25 V 4V 4V

114 mV 1.25 V 4V 4V

1.25 V 4V 0 0

170 mV 1.16 V 4.4 V 4.4 V

E

C

130 mV 1.16 V 1.16 V 4.4 V 4V 0 4V 0

TABLE 30-3 CRITICAL THINKING

vout

2 Vp-p 1.8 Vp-p 2 Vp-p 1.82 Vp-p 2 Vp-p 1.78 Vp-p

MPP 8.5 Vp-p 8.6 Vp-p 8.4 Vp-p

TABLE 29-2 DUAL-SUPPLY CIRCUIT

Initial MPP = 5.8 V Final MPP = 7 V Changes were as follows: R8 was changed to 1 kΩ, R9 was changed to 100 Ω, and R7 was readjusted for a maximum unclipped output signal.

ANSWERS

Transistor

Idiode

IC

vin

vout

MPP

1 2 3

9.1 mA 9.2 mA 9.4 mA

10 mA 8 mA 11 mA

2 Vp-p 2 Vp-p 2 Vp-p

1.82 Vp-p 1.8 Vp-p 1.78 Vp-p

8.6 Vp-p 8.5 Vp-p 8.4 Vp-p

ANSWERS 1. b 2. c 3. a 4. c 5. a 6. c 7. The diode and transistor currents are different because the match between the diode curve and the transistor’s VBE curve is different.

8. They reduced VBE.

1. c 2. a 3. b 4. b 5. c 6. The input stage operated class A with a voltage gain of approximately 7. The amplified collector signal drives a CE driver that has a voltage gain of approximately 4. The final stage is a push-pull emitter follower with a gain of approximately 0.9.

7. This will result in crossover distortion because the emitter diode has a quiescent voltage of only 0.35 V. 8. The overall voltage gain will increase significantly. Also, nonlinear distortion will become noticeable because the swamping resistor of the first stage has been shorted.

9. To begin with, the VBE drop of each output transistor will reduce the ac output compliance of 2VBE. Furthermore, onthe positive output peak, Q2 approaches cutoff. Base current for Q3 has to flow through R8. The drop across R8 reduces the positive output peak. Similarly, at the negative output peak, Q2 saturates. The drop across R9 is approximately 1.5 V. This further reduces the ac compliance. In summary, the MPP is reduced on the positive output peak by VBE + IBR8 and on the negative peak by VBE + IER9.

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2-17

Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

9. The problem is the Q of the coil. It is too low at 20 kHz. If you assume a coil Q of 15, the equivalent resistance of the coil was only 26.9 kΩ, compared with a load resistance of 100 kΩ. With the same ac voltage across the coil and the load, the coil dissipated almost four times as much power as the load. The solution is to use a better coil or to operate at a higher frequency where the coil Q is higher.

Experiment 31 TABLE 31-1 WAVEFORMS

W =0.12 ms T = 50 μs D = 12.5% +peak =+0.7 V −peak =−2.6 V

Experiment 32 TABLE 32-1 JFET DATA

TABLE 31-2 CALCULATIONS FORTUNED AMPLIFIER

fr = 40.4 kHz XL = 8.28 kΩ RS = 558 Ω RP = 126 kΩ rc = 5.58 kΩ Q = 6.67 B = 6.06 kHz MPP = 20 V PL(max) = 0.5 mW

JFET

IDSS

VGS(off)

1 2 3

9.25 mA 12.3 mA 17.2 mA

−2.7 V −3.35 V −4.8 V

TABLE 32-2 GATE BIAS: VGG = −1.5 V

TABLE 31-3 MEASUREMENTS FOR TUNED AMPLIFIER

Q=8 B = 4.9 kHz MPP = 18.1 V PL(max) = 0.335 mW

JFET

VGS

ID

VD

1 2 3

−1.5 V −1.5 V −1.5 V

2.33 mA 4.66 mA 9.82 mA

13.6 V 12.4 V 10 V

TABLE 32-3 SELF-BIAS

Calculated TABLE 31-4 FREQUENCY MULTIPLIER

fin 20.2 kHz 13.5 kHz

fout

n

40.4 kHz 40.4 kHz

2 3

Measured

JFET

ID

VD

VS

ID

VD

VS

1 2 3

2.94 mA 2.94 mA 2.94 mA

13.6 V 13.6 V 13.6 V

2V 2V 2V

2.26 mA 2.89 mA 4.18 mA

13.7 V 13.3 V 12.7 V

1.55 V 1.92 V 2.73 V

TABLE 32-4 VOLTAGE-DIVIDERBIAS

TABLE 31-5 TROUBLESHOOTING

Trouble

Estimated Vout

Measured Vout

R1 Open

0

0

Q1 Collector-emitter short C2 Short C3 Open

0 0 0

0 0 0

Calculated JFET

ID

1 2 3

2.6 mA 2.6 mA 2.6 mA

VD

Measured VS

13.8 V 5.72 V 13.8 V 5.72 V 13.8 V 5.72 V

ID

VD

VS

2.41 mA 2.66 mA 3.17 mA

13.8 V 13.7 V 13.4 V

5.25 V 5.75 V 6.85 V

TABLE 32-5 CURRENT-SOURCE BIAS

Calculated

Measured

JFET

ID

VD

VS

ID

VD

VS

1 2 3

2.1 mA 2.1 mA 2.1 mA

10.4 V 10.4 V 10.4 V

2V 2V 2V

2.15 mA 2.17 mA 2.19 mA

13.6 V 13.6 V 13.4 V

1.58 V 2.19 V 3.54 V

TABLE 32-6 TROUBLESHOOTING

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2-18

ANSWERS

TABLE 32-7 CRITICAL THINKING: RS = 270 Ω

JFET

ID

VD

VS

1 2 3

3.81 mA 5.04 mA 7.24 mA

13.1 V 12.5 V 11.4 V

1.08 V 1.43 V 2.04 V

ANSWERS 1. d 2. d 3. a 4. a 5. b 6. Gate bias was the worst; the drain current varied from 2.33 to 9.82 mA. Self-bias was better; drain current varied from 2.26 to 4.18 mA. Voltage-divider bias was still better; drain current varied from 2.41 to 3.17 mA. Current-source bias was best; drain current varied from 2.15 to 2.19 mA. 7. Shorted R4, open MPF102, open R3, and no supply voltage.

8. Open MPF102, open R3, and shorted R2.

9. I used 270 Ω. I used VGS(off)/IDSS and the values of Table 32-1 calculated 292 Ω, 272 Ω, and 279 Ω. The average value was 281 Ω. The nearest standard value is 270 Ω.

Experiment 33 TABLE 33-1 CS AMPLIFIER

Calculated

1. c 2. b 3. b 4. d 5. c 6. The JFET amplifier has much less voltage gain than abipolar amplifier. In this experiment, the CS amplifier had a voltage gain of only 4.4. A bipolar amplifier with a similar Q point and load resistance would have a voltage gain in the hundreds.

7. An open coupling capacitor C2 is the most likely trouble.

8. If all ac voltages are zero, the most likely trouble is a shorted R1. Another almost equally likely trouble is no signal from the audio generator.

9. I subtracted 2 V from 7.5 V to get 5.5 V at the gate. Then I used Ohm’s law to work out the current through the 220-kΩ resistor. Finally, I divided (15 V − 5.5 V) by the calculated current to get approximately 390 kΩ.

Experiment 34 TABLE 34-1 JFET ANALOG SWITCH

Measured vout JFET

Measured rDS(on)

Calculated vout

S1 open

S1 closed

1 2 3

177 145 117

1.74 mV 1.43 mV 1.16 mV

1.8 mV 1.5 mV 1.2 mV

91 mV 91 mV 91 mV

Measured

JFET

vout

AV

rout

vout

AV

rout

1 2 3

0.44 V 0.44 V 0.44 V

4.4 4.4 4.4

2.2 kΩ 2.2 kΩ 2.2 kΩ

0.48 V 0.44 V 0.46 V

4.8 4.4 4.6

2.21 kΩ 2.1 kΩ 1.93 kΩ

TABLE 34-2 JFET CHOPPER

TABLE 33-2 SOURCE FOLLOWER

Calculated

Measured

JFET

vout

AV

rout

vout

AV

rout

1 2 3

0.815 V 0.815 V 0.815 V

0.815 0.815 0.815

407 Ω 407 Ω 407 Ω

0.82 V 0.82 V 0.83 V

0.82 0.82 0.83

383 Ω 380 Ω 325 Ω

TABLE 33-3 TROUBLESHOOTING

DC Symptoms

AC Symptoms

TABLE 34-3 VOLTAGE-VARIABLE RESISTANCE

VG

VD

VS

vg

vd

vs

vout

Trouble

0 0 0 0

15 15 15 0

3.7 3.7 3.7 0

1 1 0 1

0 0 0 0

0 0.82 0 0

0 0 0 0

RL shorted C2 open R1 shorted No supply voltage

TABLE 33-4 CRITICAL THINKING: R = 390 KΩ; R = 220 KΩ 1

2

JFET

VS

AV

1 2 3

7.98 V 6.57 V 8.2 V

8.8 9 8.8

VGG

Vout

rds(on)

0 −0.5 V −1 V −1.5 V −2 V −2.5 V −3 V −3.5 V −4 V −4.5 V −5 V

1.2 mV 1.3 mV 1.5 mV 1.7 mV 1.9 mV 2.2 mV 2.8 mV 4 mV 5.8 mV 23 mV 96 mV

120 Ω 130 Ω 150 Ω 170 Ω 190 Ω 220 Ω 280 Ω 400 Ω 616 Ω 2.98 kΩ 240 kΩ

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2-19

Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

TABLE 35-3 VOLTAGE-DIVIDER BIAS (VN10KM)

TABLE 34-4 AGC CIRCUIT

Condition

vout

VGG

Maximum output Max/2 Max/4 Max/8

0.7 V 0.35 V 0.18 V 0.09 V

−2.67 V −4.14 V −4.57 V −4.70 V

TABLE 34-5 TROUBLESHOOTING

Calculated

Measured

VMOS

ID

VG

VS

ID

VG

1 2 3

9.82 mA 9.82 mA 9.82 mA

7.5 V 7.5 V 7.5 V

5.5 V 10.1 mA 5.5 V 10.7 mA 5.5 V 10.9 mA

VS

7.39 V 5.57 V 7.4 V 6 V 7.4 V 6.04 V

TABLE 35-4 SOURCE FOLLOWER (VN10KM)

DC Symptoms

AC Symptoms

VG

VD

vg

vd

Trouble

OK OK 0

OK 0 0

0 OK 0

0 0 0

R1 short R2 open Q1 short

Calculated gm = 30,000 μS Calculated AV = 0.944 Calculated rout = 33.3 Ω Measured AV = 0.95 Measured PP = 12 V Measured rout = 27.4 Ω

ANSWERS 1. b 2. c 3. a 4. c 5. When S1 is open, VGS is zero and rds(on) is low. The voltagedivider action between R1 and rds(on) produces a low output signal. On the other hand, when S1 is closed, VGS is more than VGS(off) and rds(on) approaches infinity. In this case, the output voltage is almost as large as the input voltage.

6. When VGG varies from zero to more negative values, rds(on) increases. The 10-kΩ resistor and rds(on) form a voltage divider. Because rds(on) is variable and controlled by VGG, we can control the output voltage with VGG. 7. No supply voltage, open R2, Q1 short.

8. R1 open, R3 short, MPF102short.

9. Low rds(on). Because we can get more attenuation with reasonable sizes of R1.

Experiment 35 TABLE 35-1 THRESHOLD VOLTAGE (IRF510)

VMOS

VGS(off)

1

0.85 V

TABLE 35-2 TRANSCONDUCTANCE CURVE (IRF510)

ID

VGS

10 μA 0.5 mA 1 mA 2 mA 3 mA 4 mA 5 mA 6 mA 7 mA 8 mA 9 mA 10 mA

0.85 V 1.25 V 1.35 V 1.52 V 1.57 V 1.63 V 1.67 V 1.7 V 1.73 V 1.77 V 1.8 V 1.83 V

Experiment 36 TABLE 36-1 TRANSISTOR LATCH

Calculated

Measured

LED

VA

ILED

VA

ILED

Off

+15 V

0

+14.1 V

0

On

+0.8 V

12.2 mA

+0.79 V

12.6 mA

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2-20

TABLE 36-2 SCR CIRCUIT

LED off: Calculated V2 = +15 V Calculated ILED = 0 Measured V2 = +14 V Measured ILED = 0 LED on: Calculated V2 = +0.8 V Calculated ILED = 37 mA Measured V2 = +0.73 V Measured ILED = 41.2 mA Triggering: Calculated V1 = +3.06 V Measured V1 = +1.9 V Holding: Calculated VCC = +4.78 V Measured VCC = +1.9 V

Experiment 38 TABLE 38-1 TAIL AND BASE CURRENTS

Calculated

Measured

IT = 9.53 mA IB1 = 23.8 μA IB2 = 23.8 μA

IT = 9.51 mA IB1 = 21.2 μA IB2 = 32.5 μA

TABLE 38-2 INPUT OFFSET AND BIAS CURRENTS

Theoretical

Experimental

Iin(off) = 0 Iin(bias) = 23.8 μA

Iin(off) = 11.3 μA Iin(bias) = 26.9 μA

TABLE 38-3 INPUT AND OUTPUT OFFSET VOLTAGES

Vout(off) = 0.65 V Vin(off) = 21.7 mA

ANSWERS

TABLE 38-4 VOLTAGE GAINS AND CMRR

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

9.

TABLE 38-5 TROUBLESHOOTING

Reduce power supply to zero.

Trouble

Experiment 37

Q1 CE short Q1 CE open Q2 CE short Q2 CE open

Estimated Vout

Measured Vout

+15 V 0 0 +15 V

+14.6 V 0.8 V 0 +15 V

TABLE 38-6 CRITICAL THINKING

RE = 4.7 kΩ RC = 5.1 kΩ VC2 = 7.38 V

ANSWERS 1. d 2. b 3. b 4. c 5. c 6. Because it means the differential signal is amplified much differential signal is amplified approximately 55 times as much as a common-mode signal.

7. The dc output voltage will be near zero because Q2 will go into saturation. Q1 will probably be destroyed by excessive power dissipation because the tail current is almost 100 mA. There will be no ac signal at the output.

8. I expect to see an ac voltage of about 5 mV pp, half of the input signal.

9. I used RE = 4.7 Ω and RC = 5.1 kΩ. The new value of A is 5100/ (22 + 16.4) = 133. The new value of ACM is 5100/(2 × 4700) = 0.542. The new value of CMRR is 133/0.542, or 245.

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2-21

Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

Experiment 39 TABLE 39-1 DC VOLTAGES

VC1

VC2

VE

VEE

VB1

VB2

4.9 V

5.2 V

−0.71 V

−10 V

−6 mV

−5 mV

TABLE 39-2 NONINVERTING INPUT

Magnitude Phase

v1

vC1

vC2

ve

50 mV 0°

4.75 V 180°

5.15 V 0°

25 mV 0°

TABLE 39-3 INVERTING INPUT

Magnitude Phase

6. With a common-mode signal, the output of each collector has the same magnitude and phase. Therefore, the differential output is ideally zero. Any deviation from zero is caused by tolerances in components. Even with tolerances, the differential output signal with a common-mode input signal is extremely small. On the other hand, a differential input signal produces equal-magnitude out-of-phase signals on the collectors. The differential output is approximately twice as much as the v2 signal.

Experiment 40 TABLE 40-1 FINE-ADJUSTMENT CIRCUIT

V1 Vout

v2

vC1

vC2

ve

50 mV 0°

4.75 V 0°

5.15 V 180°

25 mV 0°

Positive saturation

Negative saturation

0V 14.2 V

0V −14.2 V

TABLE 40-2 COARSE-ADJUSTMENT CIRCUIT

Positive saturation

Negative saturation

0V 13.5 V

0V −13.5 V

TABLE 39-4 DIFFERENTIAL INPUT

Magnitude Phase

v1

vC1

v2

vC2

ve

vout

V1 Vout

50 mV 0°

4.75 V 180°

50 mV 180°

5.15 V 0°

0 0°

10 V 0°

TABLE 40-3 NONINVERTING COMPARATOR

TABLE 39-5 COMMON-MODE INPUT

v1

vC1

v2

vC2

ve

vout

Magnitude 50 mV 25 mV 50 mV 25 mV 50 mV 50 mV Phase 0° 180° 0° 0° 0° 0°

V2

V1 (calculated)

V1 (measured)

0 1 2 5 10

0 1 2 5 10

0 1 2 5 10

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2-22

8. Because the peak voltage was much higher.

9. Because the load resistor was much smaller.

Experiment 41 TABLE 41-1 INVERTING AMPLIFIER (POSITIVE INPUTS)

vin

vout (calculated)

vout (measured)

−1 V −2 V −5 V −10 V

−1.02 V −2.04 V −5.1 V −10.2 V

0.1 V 0.2 V 0.5 V 1V

TABLE 41-2 INVERTING AMPLIFIER (NEGATIVE INPUTS)

vin

vout (calculated)

vout (measured)

1V 2V 5V 10 V

1.02 V 2.04 V 5.1 V 10.2 V

−0.1 V −0.2 V −0.5 V −1 V

TABLE 41-3 INVERTING AMPLIFIER (SINUSOIDAL INPUTS)

vin

vout (calculated)

vout (measured)

1 Vp-p 2 Vp-p 5 Vp-p 10 Vp-p 20 Vp-p

1.02 Vp-p 2.04 Vp-p 5.1 Vp-p 10.2 Vp-p 20.4 Vp-p

0.1 Vp-p 0.2 Vp-p 0.5 Vp-p 1 Vp-p 2 Vp-p

Experiment 42 TABLE 42-1 INPUT VOLTAGES

Calculated

Measured

Op amp

v1

v2

v1

v2

1 2 3

17.6 mV 17.6 mV 17.6 mV

17.6 mV 17.6 mV 17.6 mV

14.4 mV 9.3 mV 13.9 mV

12.9 mV 11.2 mV 14.6 mV

TABLE 42-2 INPUT OFFSET AND BIAS CURRENTS

Op amp

Iin(off)

Iin(bias)

1 2 3

6.8 nA 6.6 nA 3.2 nA

62 nA 47.6 nA 64.8 nA

TABLE 42-3 INPUT AND OUTPUT OFFSET VOLTAGES

Op Amp

Vout(off)

Vin(off)

Imax

1 2 3

0.7 V 0.77 V 0.59 V

0.7 mV 0.77 mV 0.59 mV

14 mA 30.3 mA 18 mA

TABLE 41-4 NONINVERTING AMPLIFIER (SINUSOIDAL INPUTS)

TABLE 42-4 SLEW RATE, POWER BANDWIDTH, AND MPP VALUE

vin

vout (calculated)

vout (measured)

Op amp

SR

fmax

MPP

1.1 Vp-p 2.2 Vp-p 5.5 Vp-p 11 Vp-p 22 Vp-p

1.12 Vp-p 2.24 Vp-p 5.6 Vp-p 11.2 Vp-p 22.4 Vp-p

1 2 3

0.637 V/μs 0.583 V/μs 0.528 V/μs

9 kHz 9 kHz 9 kHz

25 V 25.2 V 25 V

0.1 Vp-p 0.2 Vp-p 0.5 Vp-p 1 Vp-p 2 Vp-p

TABLE 42-5 TROUBLESHOOTING

Trouble

DC Output Voltage

AC Output Voltage

No +15-V supply No −15-V supply Pin 2 shorted to GND

−13.2 V +11.7 V +14.2 V

0 0 0

TABLE 41-5 BANDWIDTH OF NONINVERTING AMPLIFIER

vout 1 Vp-p 20 Vp-p

BW

SR Distortion

100 kHz 8 kHz

No Yes

TABLE 41-6 MAXIMUM PEAK-TO-PEAK OUTPUT

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

RL

MPP

10 kΩ 100 Ω

27 V 7V

TABLE 42-6 CRITICAL THINKING

Rf = 68 kΩ AV = 70.4

ANSWERS 1. d 2. b 3. d 4. b 5. c 6. The input offset current is the difference between the two base currents that flow into the noninverting and inverting inputs of the op amp. The input bias current is the average of the two base currents; in design, it can be used as an approximation for either base current.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

7. When very small return resistors are used at the inputs, the input offset current and input bias current have negligible effect on the output. All that remains is the differences in the VBE values of the input transistors. The input offset voltage is approximately equal to this VBE difference. It is the input voltage needed to zero the output voltage.

8. I overdrove the op-amp circuit to get an output with slew-rate limiting. Then I measured the voltage and time change. The ratio of these two quantities gave me the slew rate, which was near the typical value of 0.5 V/μs for a 741C. 9. I used 68 kΩ. Because the ratio of R2 to R1 has to equal 68, the voltage gain.

8. a. Positive half sine wave. b. Adjust the size of a positive cycle. c. Adjust the size of total waveform. d. 1 e. Zero if 10 kΩ is set to zero, or 0.0909 if 10 kΩ is set to maximum.

11. The output voltage is a negatively clipped sine wave. The upper 100 kΩ controls the size of the positive half cycle.

13. The output voltage is a positively clipped sine wave. The upper 100 kΩ controls the size of the negative half cycle.

15. The output is a sine wave with a peak-to-peak value of 10 V. 16. The 10 kΩ controls the amplitude of the sine wave.

17. The upper 100 kΩ changes the positive half cycle of theoutput, and the lower 100 kΩ changes the negative halfcycle.

Table 44-1. AV1 = 1 and AV2 = 1; vout = 0, 1 Vp-p, 1 Vp-p, 2 Vp-p

Experiment 43

Table 44-2. AV1 = 1 and AV2 = 1; vout = 0, 1 Vp-p 1 Vp-p, 2 Vp-p

TABLE 43-1 ERROR VOLTAGES WITHOUT COMPENSATION

Table 44-3. Calculated vout = 2.7 Vp-p, 5.4 Vp-p, 8.1 Vp-p, and 10.8 Vp-p; measured vout ≅ 2.7 Vp-p, 5.4 Vp-p, 8.1 Vp-p, and 10.8 Vp-p

V1err V2err V3err Verror

Calculated

Measured

0.728 mV 0.091 mV 2 mV 28.2 mV

DNA DNA DNA 32.6 mV

TABLE 43-2 ERROR VOLTAGES WITH COMPENSATION

V1 err V2 err V3 err Verror

Calculated

Measured

0 0.182 mV 2 mV 21.8 mV

DNA DNA DNA 25.2 mV

ANSWERS 1. b 2. b 3. c 4. d 5. b 6. a 7. The tolerance of the resistors. It has the most effect because the voltage gain of each channel depends on the ratio of resistances.

8. Use a summing amplifier with three input channels.

Experiment 45

TABLE 43-3 SMALL-SIGNAL OUTPUT IMPEDANCE

TABLE 45-1 CLOSED-LOOP VOLTAGE GAIN

RL VL

10 kΩ 1 Vp-p

1 kΩ 1 Vp-p

100 Ω 1 Vp-p

10 Ω 1 Vp-p

TABLE 43-4 LARGE-SIGNAL OUTPUT IMPEDANCE

RL VL

10 kΩ 27 Vp-p

1 kΩ 25 Vp-p

100 Ω 7 Vp-p

10 Ω 0.5 Vp-p

R1 10 kΩ 22 kΩ 47 kΩ 68 kΩ 100 kΩ

Calculated AV(CL)

Measured AV(CL)

11 23 48 69 101

11.4 22 49.5 71 100

ANSWERS 1. b 2. b 3. b 4. a 5. a 6. c 7. c 8. a 9. Because the load voltage drops in half when the load resistance equals the output impedance.

Experiment 44 3. 1.45 Vp-p 4. 1.3 Vp-p 5. 5.4 Vp-p

TABLE 45-2 STABLE VOLTAGE GAIN

Op amp

Calculated AV(CL)

Measured AV(CL)

1 2 3

34 34 34

35 35 35

TABLE 45-3 CLOSED-LOOP OUTPUT OFFSET VOLTAGE

R 10 kΩ 22 kΩ 47 kΩ 68 kΩ 100 kΩ

Calculated Voo(CL)

Measured Voo(CL)

22 mV 46 mV 96 mV 138 mV 202 mV

7.5 mV 32.6 mV 32.6 mV 46.8 mV 68.6 mV

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2-24

TABLE 46-3 ICVS FEEDBACK

TABLE 45-4 TROUBLESHOOTING

Estimated

Iin

Measured

Trouble

Vout

vout

Vout

vout

R1 short R1 open R2 short R2 open

0 0 0 0

100 mV 28 V 28 V 100 mV

1 mV 1V 0.9 V 7.8 mV

100 mV 27 V 27 V 100 mV

TABLE 45-5 CRITICAL THINKING

1 mA 2 mA 3 mA 4 mA 6 mA 8 mA 10 mA

Calculated Vout

Measured Vout

1V 2V 3V 4V 6V 8V 10 V

1.03 V 2.06 V 3.08 V 4.10 V 6.12 V 8.16 V 10.2 V

TABLE 46-4 ICIS FEEDBACK

Rf = 39 kΩ AV(CL) = 40.4

Iin

ANSWERS 1. c 2. c 3. b 4. d 5. a 6. 100 mV. Because the inverting input is bootstrapped to the noninverting input. Another way of saying this is the error voltage approaches zero because the open-loop voltage gain is very large.

7. There is no feedback; so the op amp is operating open-loop.

This means the 100-mV input signal will overdrive the op amp so that the output will be clipped. The peak-to-peak output is approximately equal to the ac output compliance. With 15-V supplies, this means around 27 to 28 V. 8. Unity. Voltagefollower.

0.1 mA 0.2 mA 0.3 mA 0.4 mA 0.6 mA 0.8 mA 1 mA

Calculated Iout

Measured Iout

1 mA 2 mA 3 mA 4 mA 6 mA 8 mA 10 mA

1.01 mA 2.02 mA 3.04 mA 4.05 mA 6.07 mA 8.08 mA 10.1 mA

TABLE 46-5 TROUBLESHOOTING

Trouble

Description

1 2 3

All troubles depend on what the instructor inserted into the circuits.

TABLE 46-6 CRITICAL THINKING

9. I would specify feedback resistors, R1 and R2, to have a tolerance of 1 percent. In the worst case, R1 can be 1 percent high and R2 can be 1 percent low. The ratio produces a 2 percent error.

Experiment 46 TABLE 46-1 VCVS FEEDBACK

Vin

Calculated Vout

Measured Vout

0.1 V 0.2 V 0.3 V 0.4 V 0.6 V 0.8 V 1V

1V 2V 3V 4V 6V 8V 10 V

1.04 V 2.05 V 3.06 V 4.07 V 6.12 V 8.16 V 10.2 V

TABLE 46-2 VCIS FEEDBACK

Vin

Calculated Iout

Measured Iout

1V 2V 3V 4V 6V 8V 10 V

1 mA 2 mA 3 mA 4 mA 6 mA 8 mA 10 mA

1.01 mA 2.02 mA 3.02 mA 4.03 mA 6.06 mA 8.07 mA 10.1 mA

ANSWERS 1. c 2. d 3. b 4. b 5. d 6. The answers here will depend on what impressed the student.

It’s up to the instructor to decide if the ideas are worthwhile.

7. The transconductance drops by a factor of 10, meaning that an input of 1 V will only produce 0.1 mA instead of 1 mA. 8. The output voltage will go into positive saturation. I learned this in Exp. 42 in the troubleshooting section.

9. To measure voltage, I’d use noninverting current feedback (Fig. 46-2). This will give an extremely high input impedance; the scale of the output meter movement would be calibrated in volts. To measure current, I’d use inverting current feedback (Fig. 46-4). This will give a very low input impedance; the scale of the output meter movement would be calibrated in amperes (or μA and mA).

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

Experiment 47

Experiment 48

TABLE 47-1 GAIN AND CRITICAL FREQUENCY

TABLE 48-1 NONINVERTING AMPLIFIER

Calculated

Calculated: AV(CL) = 48 f2(CL) = 20.8 kHz fin = 46.8 Hz fout = 3.39 Hz fby = 33.9 Hz MPP = 13 V Measured: AV(CL) = 47 f2(CL) = 17.3 kHz f1(CL) = 63 Hz MPP = 12.5 V

Measured

R1

AV(CL)

f2(CL)

AV(CL)

f2(CL)

4.7 kΩ 6.8 kΩ 10 kΩ 22 kΩ 33 kΩ 47 kΩ

48 69 101 221 331 471

20.8 kHz 14.5 kHz 9.9 kHz 4.52 kHz 3.02 kHz 2.12 kHz

48.1 70 105 215 320 46

17.1 kHz 11 kHz 7.5 kHz 3.9 kHz 2.5 kHz 1.88 kHz

TABLE 47-2 RISETIME

R1 4.7 kΩ 6.8 kΩ 10 kΩ 22 kΩ 33 kΩ 47 kΩ

Measured TR

Experimental f2(CL)

20 μs 29 μs 42 μs 94 μs 140 μs 200 μs

17.5 kHz 12.1 kHz 8.33 kHz 3.72 kHz 2.5 kHz 1.75 kHz

TABLE 47-3 TROUBLESHOOTING

Trouble

Estimated TR

Measured TR

R1 shorted No +15-V supply 100 Ω open

0.35 μs No output 0.35 μs

0.5 μs No output 0.5 μs

TABLE 48-2 INVERTING AMPLIFIER

Calculated

Measured

R3

AV(CL)

f2(CL)

AV(CL)

f2(CL)

47 kΩ 68 kΩ 100 kΩ 220 kΩ 330 kΩ 470 kΩ

47 68 100 220 330 470

21.3 kHz 14.7 kHz 10 kHz 4.55 kHz 3.03 kHz 2.12 kHz

49 70 103 210 305 490

19.8 kHz 13.9 kHz 9.4 kHz 4.62 kHz 3.18 kHz 1.98 kHz

TABLE 48-3 NONINVERTER/INVERTER

Calculated: AV(non) = 10 AV(inv) = 10 Measured: AV(non) = 10.3 AV(inv) = 10.3

TABLE 47-4 CRITICAL THINKING

R1 = 2.7 kΩ ACL = 28.5 TR = 11 μs f2(CL) = 31.8 kHz

TABLE 48-4 TROUBLESHOOTING

ANSWERS

Trouble

DC Voltage at Pin 6

1. b 2. b 3. a 4. b 5. a 6. Because you can then calculate the bandwidth for any specified voltage gain. This is essential in troubleshooting and design.

R1 open R1 short R2 open R2 short C1 open

+0.660 V +14.5 V +14.5 V +0.660 V +7.6 V

7. This will remove supply voltage from the op amp. The symptoms include a saturated dc output voltage and no ac signal.

8. Possible troubles are both supply voltages are zero, pin 6 is unconnected, the output is shorted to ground, and the op amp

TABLE 48-5 CRITICAL THINKING

C1 = 1.0 μF C3 = 1.5 μF f1(CL) = 12 Hz

is defective.

9. The higher the funity, the better. Because bandwidth increases with funity. This implies a faster risetime.

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TABLE 49-4 VOLTAGE-CONTROLLED CURRENT SOURCE: RL = 2.2 KΩ

ANSWERS 1. c 2. a 3. b 4. b 5. b 6. Because of the virtual ground, all of the input current through the 1-kΩ source resistor flows through feedback resistor R. This produces an output voltage across R. Since the current through R is the same as through the 1-kΩ resistor, the voltage gain equals the ratio of R to the source resistance.

The voltage increases to +14.5 V. Since +15 V is applied to the noninverting input, the output runs out of MPP. In other words, it saturates.

7. The dc voltages will be all right, but the ac output voltage will be much lower than it should be because the circuit becomes a voltage follower.

8. I selected 10 μF for C1 to get a cutoff frequency much lower than 20 Hz. I selected 1.5 μF for C3 to get a cutoff frequency lower than 20 Hz. The bypass capacitor dominates this design.

Vin

Calculated IL

Measured IL

0V 1V 2V 3V 4V 5V

0 mA 1 mA 2 mA 3 mA 4 mA 4.7 mA (saturation)

0 mA 0.99 mA 1.98 mA 2.98 mA 3.97 mA 4.6 mA

TABLE 49-5 TROUBLESHOOTING

Trouble

Estimated IL

Measured IL

R3 short R4 open Q1 open Q2 open

0 0 0 0

0 16 μA 16 μA 0

TABLE 49-6 CRITICAL THINKING

Experiment 49 IL = 0.2mA IL = 0.2 mA for 10 Ω, 100 Ω, and 1 kΩ IL = 0.6 mA for 10 Ω, 100 Ω, and 1 kΩ IL = 0.6 mA for 10 Ω, 100 Ω, and 1 kΩ (current reverses direction) 23. The Howland current source produces a constant load current. It is more useful than the current source of Fig. 49-3 because the Howland can produce a load current in either direction by reversing the polarity of the input voltage.

R = 2.2 kΩ IL = 2.31 mA

18.

19.

21.

22.

TABLE 49-1 Vin =1.36 V

RL VL IL

1 kΩ 1.36 V 1.36 mA

330 Ω 1.36 V 4.13 mA

100 Ω 1.36 V 13.6 mA

33 Ω 0.833 V 25 mA

10 Ω 0.25 V 25 mA

100 Ω 1.36 V 13.6 mA

33 Ω 1.36 V 41.3 mA

10 Ω 1.36 V 136 mA

TABLE 49-2 Vin =1.36 V

RL VL IL

1 kΩ 1.36 V 1.36 mA

330 Ω 1.36 V 4.13 mA

TABLE 49-3 VOLTAGE-CONTROLLED CURRENT SOURCE: RL = 100 Ω

Vin

Calculated IL

Measured IL

0V 1V 2V 3V 4V 5V

0 mA 1 mA 2 mA 3 mA 4 mA 5 mA

0 mA 0.99 mA 1.98 mA 2.98 mA 3.97 mA 4.96 mA

ANSWERS 1. b 2. d 3. b 4. b 5. a 6. It cannot produce 5 mA through 2.2 kΩ (11 V) because the transistor saturates when the load voltage is 10 V. 7. A shorted R3 means that there is 15 V at pin 2 of the second 741C. This implies 0 V across the emitter resistor, which means zero emitter and collector current.

8. The transistor dissipation is greater than the power rating of 2N3904. When the input voltage is 1.36 V, the transistor voltage is 15 V − 1.36 V, or 13.6 V. The load current is 136 mA. Therefore, PD is 1.85 W.

Experiment 50

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

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2-28

TABLE 53-2 RELAXATION OSCILLATOR AND INTEGRATOR

TABLE 52-3 ZERO-CROSSING DETECTOR

Calculated: f = 796 Hz vout(1) = 28 V vout(2) = 8.79 V Measured: f = 733 Hz vout(1) = 27 V

Step 22: vin = +100 mV Color = green Step 23: vin = −100 mV Color = red

vout(2) = 8.5 V

TABLE 52-4 LIMIT DETECTOR

TABLE 53-3 TROUBLESHOOTING

Calculated trip point = +7.5 V Measured trip point = +7.65 V

Calculated

TABLE 52-5 TROUBLESHOOTING

Symptoms

Trouble

f

vout

f

vout

28 V 7V 4V

81 Hz

27 V

2.1 kHz 734 Hz

2.8 V 4.2 V

Trouble

R1 is 22 kΩ

79 Hz

1. Red LED always on 2. Trip point equals zero

R2 short R2 open or R3 short

3. Red LED goes on and off, green LED is always off 4. Neither LED comes on

Green LED open

R2 is 1.8 kΩ R4 is 22 kΩ

2.34 kHz 796 Hz

Open op amp

TABLE 53-4 CRITICAL THINKING

R1 = 1.8 kΩ f =980 Hz

TABLE 52-6 CRITICAL THINKING

R3 = 5.1 kΩ Trip point = +5.12 V

ANSWERS 1. a 2. a 3. b 4. b 5. c 6. The voltage divider sets up +7.5 V at the inverting input. This is the reference voltage or trip point. When the input voltage is less than +7.5 V, the op-amp output is negative and the red LED is on. When vin is greater than +7.5 V, the output is positive and the green LED is on.

7. If either R2 is open or R3 is shorted, the inverting input has zero voltage. This produces a trip point of zero.

8. We will get a half-wave rectified output instead of a peak detected output.

9. Since +5 V is approximately one-third of +15 V, I knew that I needed an R3 that was approximately one-half of R2. The nearest standard value is 5.1 kΩ.

Experiment 53 TABLE 53-1 SCHMITT TRIGGER

Calculated: Shape = square MPP = 20 V UTP = +0.1 V LTP = −0.1 V Measured: Shape = square MPP = 21.5 V UTP = +0.11 V LTP = −0.1 V

Measured

ANSWERS 1. c 2. b 3. c 4. b 5. c 6. When the output is in positive saturation, a reference voltage of +0.1 V appears at the noninverting input. This means that the inverting input voltage must be slightly greater than +0.1 V to force the output to change to the negative state.

Similarly, a negative output gives a reference voltage of −0.1 V; so the inverting input must be slightly more negative than −0.1 V to switch the output to the positive state.

7. A positive output produces a positive reference voltage at the noninverting input. Therefore, the output cannot change states until the capacitor voltage has become slightly more positive than the reference voltage. A similar statement applies to a negative output. The exponential charging and discharging of the capacitor control the frequency of theoutput signal.

8. The frequency decreases to 79.6 Hz, and the triangular output becomes saturated at both peaks. The frequency decreases by a factor of 10 because the RC time constant has increased by a factor of 10. Because of the greatly increased period, the integrator runs out of MPP, so that both peaks get clipped.

9. A Schmitt trigger uses the entire output range of an op amp. For this reason, slew rate becomes important in the choice of an op amp for use in a Schmitt trigger. A 741C has a slew rate of only 0.5 V/μs, which severely limits the switching time. The 318C is much better suited for use in a Schmitt trigger because its slew rate is 70 V/μs. In other words, its vertical transitions are 140 times faster than a 741 C.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

Experiment 54

TABLE 55-2 TROUBLESHOOTING

TABLE 54-1 OSCILLATOR

Trouble

Output Symptoms

R1 short R1 open R2 short R2 open R3 short R3 open C1 short C1 open C2 short C2 open C3 short C3 open

Small half-wave output No output No output Slightly less than maximum output No output No output No output No output No output Very small high-frequency output No output Smaller output, about half

Calculated R 10 kΩ 4.7 kΩ 2.2 kΩ

Measured

f

MPP

f

MPP

1.59 kHz 3.32 kHz 7.23 kHz

27 V 27 V 27 V

1.57 kHz 3.32 kHz 6.93 kHz

26 V 26 V 26 V

TABLE 54-2 TROUBLESHOOTING

Trouble

Symptoms

R1 short R1 open R2 short R2 open R3 short R3 open

No output Clipped but can be adjusted to normal Heavily clipped, unsymmetrical, lower frequency No output Small distorted output about 1.5 to 2 V pp Heavily clipped output, almost square

TABLE 55-3 CRITICAL THINKING

C1 = 47 pF C2 = 470 pF f =134 kHz

TABLE 54-3 CRITICAL THINKING

ANSWERS

R =6.8 kΩ f =2.3 kHz

1. d 2. a 3. c 4. a 5. a 6. A high-Q tank circuit is resonant at the oscillation frequency.

The transistor delivers power to this tank circuit. Part of the power in the tank circuit is fed back to the emitter or base of the transistor. In this way, we get positive feedback and the

ANSWERS 1. a 2. d 3. c 4. d 5. b 6. A lead-lag network connects to the noninverting input. This network is resonant at 1/2πRC; therefore, oscillations build up at this frequency. By returning a signal to the inverting input, we can reduce the loop gain to unity and prevent clipping.

7. When R1 is shorted, the circuit becomes a voltage follower and the circuit cannot oscillate because of the bootstrap effect. Furthermore, since the dc voltage at the noninverting input is approximately zero, the output voltage goes to zero rather than saturation.

8. When R3 is open, there is too little negative feedback. The only way to reduce loop gain to unity is to drive the op amp into saturation on both peaks. This is why we see heavy clipping.

9. By gang-tuning either the R resistors or the C capacitors.

Experiment 56

Experiment 55 TABLE 55-1 COLPITTS OSCILLATOR

Calculated

Measured

f = 91.8 kHz MPP = 20 V B = 0.091

f = 88.6 kHz MPP = 20 V B = 0.0825

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2-30

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

Experiment 57 TABLE 57-1 ASTABLE MULTIVIBRATOR

Calculated

Measured

RA

RB

f

D

f

D

10 kΩ 100 kΩ 10 kΩ

100 kΩ 10 kΩ 10 kΩ

686 Hz 1.2 kHz 4.8 kHz

52.4% 91.7% 66.7%

725 kHz 1.24 kHz 4.98 kHz

53% 92% 68%

6. The input voltage must have a positive peak of at least 0.15 V. The reason is this: The inverting input voltage being fed back by the voltage divider is approximately 1/100 of +Vsat, or roughly +0.15 V. This is the positive trip point of the Schmitt trigger. When the input voltage exceeds the UTP of +0.15 V, the output can switch states.

7. The Schmitt trigger produces a narrow trigger for the 555 timer, which is being used as a monostable multivibrator.

Instead of charging the capacitor through a resistor, we are using a transistor current source. The charging current is set

by R4, R5, and R. Because this charging current is fixed, the capacitor voltage increases linearly when the output is not grounded.

8. Trouble: C3 equals 0.001 μF instead of 0.01 μF. 9. I used a proportion as follows: Table 57-4 indicates that the slope is 41 V/ms for an R of 10 kΩ and a C of 0.01 μF. To get a slope of 15 V/ms, we need a C of 41 ms/15 ms, or 0.0273 μF. The nearest standard value is 0.027 μF.

Experiment 58

TABLE 57-2 VCO OPERATION

fmin = 691 Hz fmax = 1.13 kHz TABLE 57-3 MONOSTABLEMULTIVIBRATOR

R

Calculated W

Measured W

33 kΩ 47 kΩ 68 kΩ

363 μs 517 μs 748 μs

360 μs 510 μs 730 μs

TABLE 57-4 RAMP GENERATOR

R

Calculated Slope

Measured Slope

10 kΩ 22 kΩ 33 kΩ

41 V/ms 18.6 V/ms 12.4 V/ms

40.8 V/ms 18.5 V/ms 12.1 V/ms

TABLE 57-5 TROUBLESHOOTING

Trouble

Description

1 2 3 4

R open 2N3906 open C3 open C3 shorted

TABLE 57-6 CRITICAL THINKING

C3 = 0.027 μF Slope = 15.3 V/ms

ANSWERS 1. c 2. a 3. a 4. b 5. c

Experiment 59 3. VFL = 9.81 V 4. V = 9.86 V 5. V = 9.89 V 6. VNL = 9.91 V 7. 1.02 percent 9. VHL = 9.86 V 10. VLL = 9.28 V 12. 6.25 percent 14. VZ = 6.2 V 15. VL(max) = 12.14 V, VL(max) = 8.56 V 16. VFL = 10 V 17. VNL = 10.28 V 18. 2.8 percent 20. VHL = 10.19 V 21. VLL = 9.83 V 23. 3.66 percent 26. VL(max) = 10.96 V, VL(min) = 9.3 V 27. VFL = 10 V 28. VNL = 10.01 V 29. 0.1 percent 31. VHL = 10.02 V 32. VLL = 9.98 V 33. 0.4 percent

ANSWERS 1. b 2. c

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

TABLE 60-5 CRITICAL THINKING

3. c 4. b 5. b 6. a 7. a 8. c 9. b

R4 = 1 kΩ VL(min) = 8.9 V VL(max) = 13.9 V

ANSWERS

Experiment 60 23. VZ = 6.185 V 24. VL(max) = 11.81 V, VL(min) = 8.118 V 25. VFL = 10 V 26. VNL = 10.01 V 27. 0.1 percent 29. VHL = 10.01 V 30. VLL = 9.99 V 31. 0.2 percent 33. ISL = 34.9mA 34. PD = 317 mW (Note: Approximately 22 mA through the pass transistor and 13 mA through the sensing transistor).

35. Use a power transistor instead of a small-signal transistorfor the past transistor. Also, reduce the current-sensing resistor from 33 to 0.7 Ω. This assumes a maximum load current of 1 A. TABLE 60-1 MINIMUM AND MAXIMUM LOAD VOLTAGE

Calculated

Measured

VZ = 6.2 V VL(min) = 8.37 V VL(max) = 12.2 V

VZ = 6.18 V VL(min) = 8.14 V VL(max) =12.95 V

TABLE 60-2 REGULATION

Load Regulation

Source Regulation

VNL = 10 V VFL = 9.99 V %LR = 0.1%

VL(max) = 9.99 V VL(max) = 9.84 V %SR = 1.5%

TABLE 60-3 CURRENT LIMITING: ISL = 18.6 mA

RL

VL

10 kΩ 4.7 kΩ 1 kΩ 470 Ω 330 Ω 220 Ω 100 Ω 0

10 V 10 V 9.98 V 8.63 V 5.52 V 4.28 V 2.21 V 0

R2 open Zener open Zener short Q1 open

Experiment 61 15. Vdc = 17.8 V, Vrip = 2.77 Vp-p 17. Vdc = 17.4 V, Vrip = 2.5 Vp-p 18. Vdc = 5.1 V, Vrip = 200 μVp-p 19. The input voltage to the regulator is 17.5 V. This is the voltage out of the capacitor-input filter. The IC regulator produces an output voltage of 5 V. Since the ripple rejection is 80 dB, the ripple is reduced by a factor of 10,000.

22. 15 Vp-p 23. 75.3 V 24. 1.1 Vp-p 25. The circuit is a dc-to-dc converter. The transformer steps the 15-V pp output of the 555 timer to a higher voltage, whichis rectified to get a dc output voltage of approximately 75 V. TABLE 61-1 VOLTAGE REGULATOR

TABLE 60-4 TROUBLESHOOTING

Trouble

1. b 2. c 3. d 4. c 5. a 6. The circuit is a voltage amplifier because it uses noninverting voltage feedback. The zener voltage is the input voltage, and the voltage from the wiper is the feedback voltage. The circuit has a closed-loop voltage gain, producing an output voltage that is AV(CL) times VZ + VBE. 7. The shorted-load current is around 21.2 mA. Since the entire +15 V is across the pass transistor, the power dissipation is 318 mW. 8. When the zener diode opens, Q1 opens and there is no current through R2. For this reason, the base of Q2 goes high, pulling the emitter up to almost the input voltage.

9. First, change R3 to approximately 7 Ω. Second, a power transistor will be needed for the pass transistor because the required power dissipation under shorted-load conditions is approximately 1.5 W (15 V times 100 mA). I would connect this power transistor to Q2 to get a Darlington pair.

Estimated Vout

Measured Vout

0 +15 V +1 V +15 V

0 +14.2 V +1.06 V +14.2 V

Vin

Estimated Vout

Measured Vout

1V 5V 10 V 11 V 12 V 13 V 14 V 15 V

0 3V 5V 5V 5V 5V 5V 5V

0.06 V 3.8 V 4.8 V 4.8 V 4.81 V 4.81 V 4.81 V 4.82 V

TABLE 61-2 RIPPLE REJECTION

Calculated Vrip (minimum rejection) = 1.59 mV Calculated Vrip (typical rejection) = 0.2 mV Measured Vrip = 1mV Measured ripple rejection = 66 dB

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2-32

ANSWERS

TABLE 61-3 VOLTAGEAND CURRENT REGULATION

Measured

Calculated R2

Vout

Iout

Vout

Iout

10 Ω 22 Ω 33 Ω 47 Ω 68 Ω

5.43 V 5.95 V 6.43 V 7.04 V 7.94 V

41.3 mA 41.3 mA 41.3 mA 41.3 mA 41.3 mA

5.28 V 5.78 V 6.26 V 6.84 V 7.74 V

40.4 mA 40.2 mA 40.2 mA 40.2 mA 40.1 mA

TABLE 61-4 TROUBLESHOOTING

Trouble

Estimated Vout

Measured Vout

R1 short R1 open R2 short R2 open

6.8 V 5.54 V 5V 15 V

7.8 V 5.32 V 4.8 V 14.4 V

1. d 2. a 3. c 4. d 5. a 6. As discussed in your textbook, these IC regulators have dropout voltages between 2 and 2.5 V. Therefore, the input voltage of an LM340-5 is expected to have a minimum value of 7 to 7.5 V. 7. The input bypass capacitor prevents oscillations, and the output bypass capacitor improves the transient response.

8. I estimated 5.54 V and measured 5.32 V. With R1 open, the voltage across it is 5 V. With only IQ through an R2 of 68 Ω, the total output voltage is the sum of 5 V plus 8 mA times 68 Ω, which is 5.54 V. 9. Theoretically, I wanted 96.8 Ωbecause 41.3 mA times 96.8 Ω gives 4V.This added to 5V gives aregulated output of 9V.The measured output was 8.76V because I used astandard valueof 91 Ω (nearest standard value) and because of the tolerance in the LM340-5.

TABLE 61-5 CRITICAL THINKING

R2 = 91 Ω Vout = 8.76 V

System Applications 1

TABLE SA1-2 VOLTAGE OUTPUT

TABLE SA1-1 VOLTAGE OUTPUT

Vin (Vp)

Vin (Vp)

100 mV 300 mV 500 mV 600 mV 700 mV 800 mV 900 mV 1V 1.1 V 1.2 V 1.3 V 2V 5V

Expected Results

Multisim

Actual

VA (Vp)

VA (Vp)

VA (Vp)

100 mV 300 mV 500 mV 600 mV 700 mV 700 mV 700 mV 700 mV 700 mV 700 mV 700 mV 700 mV 700 mV

100 mV 300 mV 500 mV 550 mV 553 mV 595 mV 600 mV 610 mV 620 mV 625 mV 625 mV 650 mV 700 mV

100 mV 300 mV 500 mV 600 mV 700 mV 800 mV 900 mV 950 mV 1V 1.1 V 1.11 V 1.23 V 1.36 V

TROUBLESHOOTING OBSERVATIONS With D1 Removed: With D1 removed, D2 limits the negative peak’s voltage to its barrier potential of approximately 700 mV. The positive peak’s voltage will match that of the ac source.

With D2 Removed: With D2 removed, D2 limits the positive peak’s voltage to its barrier potential of approximately 700 mV. The negative peak’s voltage will match that of the ac source.

800 mV 900 mV 1V 1.1 V 1.2 V 1.3 V 1.4 V 1.5 V 1.8 V 2V 2.2 V 5V

Expected Results

Multisim

Actual

VA (Vp)

VA (Vp)

VA (Vp)

800 mV 900 mV 1V 1.1 V 1.2 V 1.3 V 1.4 V 1.4 V 1.4 V 1.4 V 1.4 V 1.4 V

800 mV 900 mV 1V 1.1 V 1.13 V 1.16 V 1.19 V 1.21 V 1.25 V 1.27 V 1.3 V 1.4 V

800 mV 900 mV 1V 1.1 V 1.2 V 1.3 V 1.4 V 1.5 V 1.8 V 1.9 V 2V 2.6 V

ANSWERS 1. b 2. a 3. a 4. b 5. b 6. c 7. c 8. d 9. None, the diode is open when VS =400 mVp 10. Five diodes in series, connected cathode to anode

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

TABLE SA2-4 TROUBLESHOOTING

System Applications 2

Measured

TABLE SA2-1 VOLTAGE MEASUREMENTS WITH NO REGULATION

Multisim

Measured Calculated

Multisim

Actual

B

B

B

11.2 V 795 mV

11.0 V 756 mV

11.3 V 910 mV

DC load voltage Peak-to-peak ripple

C1 = 4.7 µF DC load voltage Peak-to-peak ripple C1 = 1 µF DC load voltage Peak-to-peak ripple

TABLE SA2-2 VOLTAGE MEASUREMENTS WITH A ZENER DIODE FOR REGULATION

Actual

B

C

B

C

8.4 V 5V

6.2 V 218 mV

8.8 V 5V

6.2 V 220 mV

7.9 V 6.4 V

5.9 V 1.5 V

8.1 V 6.5 V

6.1 V 1.6 V

ANSWERS Measured Multisim

DC load voltage Peak-to-peak ripple

1. The voltage across the load would have a larger peak-to-peak ripple.

2. R1 satisfies Kirchhoff’s Voltage law by dropping the remaining voltage that wasn’t dropped across the zener diode. R1 also limits the current provided to the zener diode and the load resistor.

3. The waveform at point B had an ac ripple riding on a dc level whereas the waveform at point C was smooth.

4. The capacitor could be smaller. Instead of completely removing the ac ripple, it just needs to keep the dc voltage minus the Vp-p ripple voltage above the required input voltage of 7.5 V for the LM7805C. 5. a 6. a 7. d 8. c 9. b 10. c

Actual

B

C

B

C

8.7 V 4.6 V

6.2 V 77.6 mV

8.8 V 4.9 V

6.2 V 78 mV

TABLESA2-3 VOLTAGEMEASUREMENTS WITH AVOLTAGE REGULATOR

Measured Multisim

DC load voltage Peak-to-peak ripple

Actual

B

C

B

C

10.8 V 1V

5V 0V

10.9 V 1V

5V 0V

System Application 3 TABLE SA3-1 DC VOLTAGE MEASUREMENTS

Measured Multisim

Calculated

Q1 Q2 Q3 Q4

Actual

B

E

C

B

E

C

B

E

C

3.04 V 2.07 V 5.21 V 3.81 V

2.34 V 1.38 V 5.21 V 3.81 V

4.32 V 3.81 V 9V 0V

3.04 V 2.07 V 5.21 V 3.81 V

2.34 V 1.38 V 5.21 V 3.81 V

4.32 V 3.81 V 9V 0V

2.9 V 2.0 V 5.8 V 4.6 V

2.2 V 1.3 V 5.18 V 5.18 V

4.5 V 4.6 V 9V 0V

TABLE SA3-2 AC VOLTAGE MEASUREMENTS

Measured Calculated

Q1 Q2 Q3 Q4

Multisim

Actual

B

E

C

B

E

C

B

E

C

140 mV 1.4 V 3.84 V 3.84 V

114 mV 1.15 V 3.84 V 3.84 V

1.4 V 3.84 V 0V 0V

140 mV 1.4 V 3.84 V 3.84 V

114 mV 1.15 V 3.84 V 3.84 V

1.4 V 3.84 V 0V 0V

140 mV 1.4 V 3.98 V 3.98 V

118 mV 1.17 V 3.98 V 3.98 V

1.4 V 3.84 V 0V 0V

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TABLE SA3-3 VOLTAGE MEASUREMENTS

Predicted Value Test Point A B C E G

DC 13 V 9V 0V 2V 0V

AC 0V 0V 140 mV 980 mV 4V

Multisim DC 13 V 9.51 V 0V 2V 0V

Actual

AC 0V 0V 140 mV 980 mV 4V

DC 13 V 9V 0V 2V 0V

AC 0V 0V 140 mV 980 mV 4V

TABLE SA3-4 VOLTAGE MEASUREMENTS

Test Point C1 = 0.1 µF C D C1 Rmvd C D

Predicted Value DC 0V 3.04 V DC 0V 3.04 V

AC 140 mV 14 mV AC 140 mV 0V

Multisim DC 0V 3.04 V DC 0V 3.04 V

AC 140 mV 14 mV AC 140 mV 0V

Actual DC 0V 3.04 V DC 0V 3.04 V

AC 140 mV 14.1 mV AC 140 mV 0V

When C1 is decreased from 1 µF to 0.1 µF, additional voltage is dropped across it as it’s XC increases from 159 Ω to 1591 Ω.

When C1 is removed, the signal never reaches test point D. TABLE SA3-5 VOLTAGE GAIN MEASUREMENTS

C5 Removed

Calculated

Multisim

Actual

AV

AV

AV

1

1

0.98

QUESTIONS FOR SYSTEM APPLICATION 3 1. c 2. b 3. a 4. d 5. b 6. a 7. b 8. a 9. d 10. b

AC Model of Figure SA4-2

System Application 4 TABLE SA4-1 OUTPUT SIGNAL MEASUREMENTS

VS, 1 VP sine wave at

Multisim

Actual

10 Hz 100 Hz 500 Hz 1 kHz 1.2 kHz 1.3 kHz 1.4 kHz 1.5 kHz 1550 Hz 1585 Hz 1600 Hz 1.7 kHz 5 kHz 10 kHz 100 kHz

6.3 mVp 62.5 mVp 300 mVp 500 mVp 600 mVp 635 mVp 660 mVp 685 mVp 700 mVp 705 mVp 710 mVp 730 mVp 955 mVp 1 Vp 1 Vp

6.3 mVp 62.5 mVp 300 mVp 500 mVp 600 mVp 635 mVp 660 mVp 685 mVp 700 mVp 705 mVp 710 mVp 730 mVp 955 mVp 1 Vp 1 Vp

AC model redrawn with just VS, C1, and Zinstage as a single resistance.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

TABLE SA4-2 OUTPUT SIGNAL MEASUREMENTS

Corner Frequency

Calculated

Fig. SA4-1 Fig. SA4-3 Fig. SA4-4

1585 Hz 1.32 Hz 406 Hz

QUESTIONS FOR SYSTEM APPLICATION 4 1. a 2. c 3. d 4. c 5. c 6. a 7. d 8. b 9. a 10. c

System Application 5 TABLE SA5-1 DC VOLTAGE MEASUREMENTS

Test Points

Calculated

Multisim

Actual

A B D

13 V 9V 4.5 V

13 V 9.51 V 4.75 V

13 V 9V 4.49 V

TABLE SA5-2 VOLTAGE GAIN MEASUREMENTS

Calculated

Multisim

AV 29.3

Voltages

Actual AV

Voltages

AV

TP C 139 mVp-p 28.85 TP C 140 mVp-p 28.57 TP F 4.01 Vp-p TP F 4.00 Vp-p

TABLE SA5-3 TROUBLESHOOTING

Trouble

TP B

TP C

TP D

TP F

C5 is Shorted R2 Shorted LM7809 Fails D1 Shorts R1 Opens

9V 9V 9V 9V 9V

140 mVp-p 140 mVp-p 140 mVp-p 0V 140 mVp-p

0V 4.51 V 4.51 V 4.51 V 4.51 V

0V 0V 0V 0V 0V

QUESTIONS FOR SYSTEM APPLICATION 5 1. a 2. a 3. b 4. c 5. b 6. c 7. d 8. a 9. c 10. a

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System Application 6 TABLE SA6-1 DC VOLTAGE MEASUREMENTS AND OBSERVATIONS

TABLE SA6-2 LED DRIVER VOLTAGES

Test Points

LED Off

LED On

D E

840 mA 58 mV

−100 mV 2.09 V

TABLE SA6-3 TROUBLESHOOTING

Trouble R3 is Shorted R4 Shorted 2N3904 fails D1 Opens R6 Opens

Observations of how the circuit’s operation changed LTP doesn’t work UTP doesn’t work LED always on LTP doesn’t work LED doesn’t light

QUESTIONS FOR SYSTEM APPLICATION 6

1. d 2. b 3. c 4. b 5. d 6. c 7. a 8. a 9. b 10. b

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

Troubleshooting 1 TABLE TS1-1 TEST POINT VOLTAGES

TP A TP B TP C TP D

Calculated 12 V 11.3 V 6.2 V 2V

Multisim 12 V 11.3 V 6.2 V 2V

Actual 12 V 11.28 V 6.2 V 1.9 V

TABLE TS1-2 IMPROPERLY PLACED AND FAULTY COMPONENTS

D1 reverse biased D2 forward biased R1 and R2 swapped

D2 removed D3 removed V1 Polarities reversed

Expected Measured Expected Measured Expected Measured

TP A 12 V 12 V 12 V 12 V 12 V 12 V

TP B 0V 136 mV 11.2 V 11.2 V 11.2 V 11.33 V

TP C 0V 136 mV 700 mV 868 mV 4.8 4.72 V

TP D 0V 136 mV 700 mV 868 mV 2V 1.94 V

Expected Measured Expected Measured Expected Measured

12 V 12 V 12 V 12 V -12 V -12 V

11.2 V 11.32 V 11.8 V 11.27 V 0V 112 mV

8.5 V 8.55 V 6.2 V 6.24 V 0V 112 mV

2V 1.94 V 6.2 V 6.24 V 0V 112 mV

Conclusion drawn from results D1 is blocking the current flow in the circuit.

D2 is functioning as a standard diode; ≈0.7 V across it.

Total circuit current is below amount required for 6.2 V across D2 or (R2 and D3).

With zener removed, it becomes a series circuit (V1, D1, R1, R2, D3).

The zener D2 has no load.

D1 is blocking the current flow in the circuit.

QUESTIONS FOR TROUBLESHOOTING 1

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

b d c c d b b c d a

Troubleshooting 2 TABLE TS2-1 TEST POINT VOLTAGES

TP A TP B

Calculated DC AC 17.82 V 5V

Multisim DC AC 15.2 V 1.18 Vp 5V 0

Actual DC AC 17.88 V 155 mVp 5V 0

TABLE TS2-2 FAULTY COMPONENTS

D1 Open C1 Open

LM7805 Open

DC AC DC AC

TP A 16.6 V 2.68 Vp 11.95 Very Noisy

TP B 5V 0 4 2.5Vp

DC AC

17.88 V 155 mVp

0 0

Conclusion drawn from comparisons C1 smooths out the ac ripple enough that the LM7805 still functions properly.

The input to the LM7805 requires C1 to smooth out the ac ripple, w/o it the LM7805 does not function properly because the dc input voltage drops below the required value.

The DC voltage is present on the input of the LM7805 but is absent on the output. The signal is lost at the LM7805.

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2-38

TABLE TS2-3 TEST POINT CURRENTS AND VOLTAGES

Calculated 18.84 mA 9 mA 3.74 mA 6.8 V 9V

TP A TP B TP C TP D Vout

Multisim 18.7 mA 9.03 mA 3.56 mA 6.8 V 9V

Actual 19.5 mA 8.95 mA 4.6 mA 6.75 mA 8.86 V

TABLE TS2-4 FAULTY COMPONENTS

D1 Open

TP A 10.75 mA

TP B 9.77 mA

TP C 1.8 uA

TP D 12 V

Vout 9.68 V

Q1 Open

16 mA

0A

6.75 V

9.68 V

RL Shorted RL Open

60.6 mA 19 mA

9.77 mA 55.7 mA 0A

1.7 uA

6.75 V

0V

12.8 mA

6.75 V

8.98 V

Multisim AC 50 mVp 50 mVp 640 mVp 640 mVp

DC 0V 1.8 V 5.98 V 0V

Conclusion drawn from comparison D1 sets the output voltage, without it Vout is based on a voltage divider of RS and ((R2+R3)||RL) Open Q1 causes additional current to flow through RL, increasing the Vout.

Vout goes to 0 V and the current at TP A is equal to VS/RS. Vout remains the same, no increase in current draw from voltage source.

QUESTIONS FOR TROUBLESHOOTING 2

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

d d c a d a d d b d

Troubleshooting 3 TABLE TS3-1 TEST POINT VOLTAGES

TP A TP B TP C TP D

Calculated DC AC 0V 50 mVp 1.8 V 50 mVp 6.03 V 653 mVp 0V 653 mVp

DC 0V 1.8 V 6.03 V 0V

Actual AC 50 mVp 50 mVp 600 mVp 600 mVp

TABLE TS3-2 IMPROPERLY PLACED AND FAULTY COMPONENTS

C1 Open C2 Open C3 Open Q1 Open Q1 C-E Shorted C1 10 nF

DC AC DC AC DC AC DC AC DC AC

TP A 0V 50mVp 0V 50mVp 0V 50mVp 0V 50mVp 0V 50mVp

TP B 1.82 V 0V 1.82 V 50mVp 1.82 V 50mVp 1.82 V 50mVp 1.82 V 50mVp

TP C 5.98 V 0V 5.98 V 690mVp 5.98 V 109mVp 10V 0V 2.2 V 0V

TP D 0V 0V 0V 0V 0V 109mVp 0V 0V 0V 0V

DC AC

0V 50mVp

1.82 V 37.5 mVp

5.98 V 450 mVp

0V 450 mVp

Conclusion drawn from results The ac signal never reaches the base of Q1. The dc voltages remain the same.

The Av increased because rc = R3 not R3||R6 With C3 open, the calculated Av decreases to 2.59, the measured to 2.18 The ac signal stops at the base of Q1. The dc voltage at R2 is the same. The dc voltage at TP C = VCC. The ac signal is present at the base of Q1 but not the collector. R3, R4, and R5 form a dc voltage divider. The dc voltage at VC is the voltage across R4 and R5.

Some of the ac signal is dropped across C1 because it’s XC is 1.6 kΩ. The dc voltages remain the same.

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Solution Manual for Electronic Principles 9th Edition by Albert Paul Malvino, David J. Bates, Patrick E. Hoppe

TABLE TS3-3 TEST POINT VOLTAGES

Calculated DC AC 0V 40 mVp 1.8 V 40 mVp 6.03 V 224 mVp 1.8 V 224 mVp 6.03V 2.92 V 0V 2.92 V

TP A TP B TP C TP D TP E TP F

Multisim DC AC 0V 40 mVp 1.79 V 40 mVp 5.99 V 224 mVp 1.79 V 224 mVp 5.96 2.73 V 0V 2.73 V

Actual DC 0V 1.76 V 6.12 V 1.79 V 5.98 V 0V

AC 40 mVp 40 mVp 225 mVp 225 mVp 2.7 Vp 2.7 Vp

TABLE TS3-4 IMPROPERLY PLACED AND FAULTY COMPONENTS

DC AC DC AC

TP A 0V 40 mVp 0V 40 mVp

TP B 1.76 V 40 mVp 1.76 V 40 mVp

TP C 6.12 V 680 mVp 6.12 V 250 mVp

TP D 1.79 V 0V 1.79 V 250 mVp

TP E 6.03 V 0V 6.03 V 600 mVp

TPF 0V 0V 0V 600 mVp

Q1 Open

DC AC

0V 40 mVp

1.76 V 40 mVp

10 V 0V

1.79 V 0V

6.03 V 0V

0V 0V

Q2 Shorted

DC AC

0V 40 mVp

1.76 V 40 mVp

6.12 V 232 mVp

1.81 V 232 mVp

2.17 V 0V

0V 0V

Q1 PNP

DC AC

0V 40 mVp

2.59 V 40 mVp

3.27 V 40 mVp

1.79 V 40 mVp

6.03 V 106 mVp

0V 106 mVp

C2 Open C4 Open

Conclusion drawn from results The ac signal is interrupted at C2.

The dc voltages are not affected.

The voltage gain of Q2 is greatly reduced because R10 is not bypassed by C4. The dc voltages are not affected.

The ac signal is interrupted at the base of Q1. The dc collector voltage of Q1 is equal to Vcc. The ac signal is interrupted at the collector of Q2. Resistors R8, R9, and R10 form a dc voltage divider.

The voltage at TP E is the voltage across R9 & R10.

Q1 has no voltage gain because it is a PNP xstr. The dc voltages of Q1 change.

QUESTIONS FOR TROUBLESHOOTING 3

1.

2.

3.

4.

5.

6.

B A A A A A

Troubleshooting 4 TABLE TS4-1 TEST POINT CURRENTS AND VOLTAGES

TP A TP B TP C TP D Vout

Calculated 18.84 mA 9 mA 3.74 mA 6.8 V 9V

Multisim 18.7 mA 9.03 mA 3.56 mA 6.8 V 9V

Actual 19.5 mA 8.95 mA 4.6 mA 6.75 mA 8.86 V

TABLE TS4-2 FAULTY COMPONENTS

D1 Open

TP A 10.75 mA

TP B 9.77 mA

TP C 1.8 uA

TP D 12 V

Vout 9.68 V

Q1 Open

16 mA

0A

6.75 V

9.68 V

RL Shorted RL Open

60.6 mA 19 mA

9.77 mA 55.7 mA 0A

1.7 uA

6.75 V

0V

12.8 mA

6.75 V

8.98 V

Conclusion drawn from comparison D1 sets the output voltage, without it Vout is based on a voltage divider of RS and ((R1+R2)||RL) Open Q1 causes additional current to flow through RL, increasing the Vout.

Vout goes to 0 V and the current at TP A is equal to VS/RS. Vout remains the same, no increase in current draw from voltage source.

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