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SOLUTION MANUAL Electronic Principles 9th Edition by Albert Paul Malvino, Patrick Hoppe Part

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 Solution: RL = 100RS RL = 100(0.1 Ω) RL = 10 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:

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.

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.

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

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

1-9.

6 kΩ

3 kΩ

6 kΩ

3 kΩ

36 V

Given:

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)

(Voltage divider formula)

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

36 V

VTH

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.

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

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 Solution: VTH = VR2

VTH 100 V

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

(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. 1-15. Given: VS = 36 V

R1 = 12 kΩ

R2 = 6 kΩ Solution: VTH = VR2 VR2 = VS[(R2)/(R1 + R2)]

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

3 kΩ

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

(Voltage divider formula)

VR2 = 36 V[(6 kΩ)/(12 kΩ + 6 kΩ)] VR2 = 12 V

10 kΩ

IN 6 mA

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 Norton circuit for Prob. 1-16.

1-17. Given: IN = 10 mA RN = 10 kΩ

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

RS = 80 mΩ

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 ≮ 0.75 Ω Answer: a. The internal resistance (RS) is 8.33 Ω. b. The source is not stiff since RS ≮ 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

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) [(0°C – 25°C)/10]

IS(new) = 2 10 nA IS(new) = 1.77 nA (ΔT/10) IS(new) = 2 IS(old) [(75°C – 25°C)/10)]

IS(new) = 2 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

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

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

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

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.

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) 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.

VD = 0 V

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

Given:

R = 220 Ω V=6V

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

Given:

VS = 20 V

PROBLEMS

Solution: I = V/R I = 6 V/220 Ω I = 27.27 mA

Given:

VS = 20 V

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

JOB INTERVIEW QUESTIONS

3-1.

Given:

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

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

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 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 3-16. 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

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

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.

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

I2 = I1 – ID 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

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

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

(Kirchhoff’s law)

I2 = 335 μA – 0.25 mA

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

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) (Eq. 4-1) Vp(out) = 70.7 V Since the average and the dc values are the same: Vdc = 0.318 Vp (Eq. 4-2) Vdc = 0.318 (70.7 V)

Vdc = 22.5 V Answer: The peak voltage is 70.7 V, the average voltage is 22.5 V, and the dc voltage is 22.5 V.

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

3-38. D1 is reverse biased 3-39. 1N4001 silicon rectifier diode 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 (Eq. 4-4) Vp(out) = 70.0 V Since the average and the dc values are the same: Vdc = 0.318 V (Eq. 4-2)

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.

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

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

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 (Eq. 4-4) Vp(out) = 11.42 V Since the average and the dc values are the same: Vdc = 0.636 Vp (Eq. 4-6) 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 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) – 1.4 V (Eq. 4-8) Vp(out) = 19.81 V 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

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. 4-5)

Vp(in)max = 1.414 V2(max) 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.)

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) (Eq. 4-5) V2 = 120 V ac/7 V2 = 17.14 V ac Vp = 1.414 Vrms

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)

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.

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.

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

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

V2 = 839.2 V ac Vp = 1.414 Vrms Vp = 1.414 (839.2 V ac)

Vp = 1186.6 V Since it is a quadrupler, the output is 4Vp. Vout = 4Vp Vout = 4(1186.6 V) Vout = 4746.4 V Answer: The output voltage will be 4746.4 V.

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.

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

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.

5-5.

Given:

IS = (40 V – 15 V)/470 Ω

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

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. 5-8.

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 Solution: VL = VZ = 12 V

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

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

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: VR (out) =

(Voltage divider formula)

RZ VR (in) RS (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) (Eq. 5-9) RL(min) = RS(max) /[(VS(min) /VZ) – 1] RL(min) = 470 Ω /[(24 V/15 V) – 1] RL(min) = 783 Ω 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

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.

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

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

IL = 30 mA

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.

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 Answer: The maximum ripple voltage will be 0.438 V. 5-35. Given: VS = 6 V ac VD = 0.25 V Solution: Vp(in) = 1.414 Vrms Vp(in) = 1.414 (6 V ac) Vp(in) = 8.48 V 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

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 5-39. Power supply failed (0 V) 5-40. Zener diode open 5-41. RL is shorted 5-42. Zener diode is forward biased 5-43. Common anode 5-44. Biased off 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.

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 Solution: VCE = VCC – ICRC VCE = 20 V – (6 mA)(1.5 kΩ) VCE = 11 V Answer: The collector to emitter voltage is 11 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:

6-9.

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

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]

(Eq. 6-6)

IB = 19.8 μA Answer: The base current is 19.8 μA.

6-7.

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)

IB = [(10 V – 0.7 V)/470 kΩ]

6-6.

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.

Given:

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.

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)

Given:

PD = 27.69 mW 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