Contemporary Electronics: Fundamentals. Experiments in. Fundamentals First Edition

Transcription

1 Instructor Solutions Manual to accompany Contemporary Electronics: Fundamentals First Edition and Experiments in Contemporary Electronics: Fundamentals First Edition Louis Frenzel

2 CONTENTS Preface PART A iii Solutions to Questions and Problems in Frenzel s Contemporary Electronics: Fundamentals, 1/e PART B Results and Data for Experiments in Frenzel s Contemporary Electronics: Fundamentals, 1/e - 2 -

3 PREFACE This manual provides the answers to all of the textbook questions and problems as well as the solutions to all of the experiments from the companion experiments manual. All of the answers have been checked several times for accuracy. However, if you do find any errors or questionable solutions, please let McGraw Hill know so corrections can be made in any future editions. All of the lab manual experiments have been built and tested multiple times. Most experiments have also been tested in a community college AAS degree electronics program. However, because of differences in components, test equipment, measurement techniques and other factors, your results may not be exact. After extensive testing, most experiments have a repeatability error of 10% or less. If the exact components specified cannot be found, a web search can usually turn up an equivalent. Most of the parts used in the experiments manual were obtained through online supplier Jameco. Other sources for selected parts were All Electronics and Ramsey Electronics. Should you find any problems or discrepancies please let us know through McGraw Hill so that issues can be resolved and corrections made if necessary. Consideration was given to making one or more prewired boards containing all the circuitry associated with the experiments. Such boards (DC, AC and Semiconductor/Linear) would save considerable wiring and troubleshooting in the lab. If any of you are interested in such an accessory, let me know. I would like to thank Bill Hessmiller who compiled and edited this manual. Your comments, suggestions and other feedback are always appreciated. Lou Frenzel lfrenzel@austin.rr.com - 3 -

4 PART A Chapter 1 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. a 2. b 3. b 4. d 5. a 6. c 7. c 8. b 9. a 10. c 11. b 12. d 13. a 14. c 15. b 16. d 17. d 18. b 19. a 20. c ANSWERS TO CHAPTER ESSAY QUESTIONS - 4 -

5 Chapter 2 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. b 2. a 3. c 4. a 5. a 6. d 7. b 8. a 9. c 10. d 11. b 12. a 13. b 14. d 15. c 16. a 17. c 18. d 19. b 20. d 21. c 22. b 23. a 24. d 25. a ANSWERS TO PROBLEMS Section 2.5 The Volt Unit of Potential Difference 2.1 V = 8V J Section 2.6 Charge in Motion is Current 2.3 I = 4A 2.4 I = 10A - 5 -

6 Section 2.7 Resistance is Opposition to Current 2.5 (a.) R = 500Ω (b.) R = 250Ω (c.) R = 120Ω (d.) R = 4Ω 2.6 (a.) G = 5mS (b.) G = 10mS (c.) G = 20mS (d.) G = 40mS - 6 -

7 Chapter 3 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. b 2. d 3. a 4. c 5. c 6. a 7. a 8. d 9. c 10. c 11. c 12. b 13. d 14. a 15. b 16. b 17. c ANSWERS TO PROBLEMS Section 3.2 Resistor Color Coding 3.1 a. 1.5kΩ, ±10% b. 27Ω, ±5% c. 470kΩ, ±5% d. 6.2Ω, ±5% e. 91kΩ, ±5% f. 10Ω, ±5% g. 1.8MΩ, ±10% h. 1.5kΩ, ±20% i. 330Ω, ±10% j. 560kΩ, ±5% k. 2.2kΩ, ±5% l. 8.2Ω, ±5% m. 51kΩ, ±5% n. 680Ω, ±5% o. 0.12Ω, ±5% p. 1kΩ, ±5% q. 10kΩ, ±10% r. 4.7kΩ, ±5% 3.2 a. 470kΩ - 7 -

8 b. 1.2kΩ c. 330Ω d. 10kΩ - 8 -

9 3.3 a. ±195Ω or 3.705kΩ to 4.095kΩ b. ±10Ω or 90Ω to 110Ω c. ±2.4kΩ or 117.6kΩ to 122.4kΩ d. ±0.11Ω or 2.09Ω to 2.31Ω e. ±0.75Ω or 74.25Ω to 75.75Ω 3.4 Reading from left to right the colors are: a. brown, black, orange and gold b. red, violet, gold and gold c. green, blue, red and silver d. brown, green, green and gold e. red, red, silver and gold 3.5 Reading from left to right the colors are: a. brown, brown, black, black and brown b. orange, yellow, black, red and green c. gray, red, green, red and red d. blue, red, blue, gold and brown e. brown, black, green, orange and violet Section 3.4 Rheostats and Potentiometers - 9 -

10 3.6 See Figure 3-1 Figure 3-1 Chapter 4 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. c 2. d 3. a 4. b 5. d 6. b 7. d 8. c 9. a 10. c 11. b 12. c 13. b 14. a 15. d 16. c 17. a 18. b 19. d ANSWERS TO PROBLEMS Section 4.1 The Current V I = R

11 4.1 a. I = 2A b. I = 3A c. I = 4A 4.2 a. I = 0.005A b. I = 0.02A c. I = 0.015A 4.3 I = 0.12A 4.4 Yes, because the current, I, is only 15A. Section 4.2 The Voltage V = I x R 4.5 a. V = 180V b. V = 60V 4.6 a. V = 10V b. V = 7.5V 4.7 a. V = 60V b. V = 13.2V 4.8 V = 10V 4.9 V = 120V

12 Section 4.3 The Resistance 4.10 a. R = 4Ω b. R = 6Ω 4.11 a. R = 4Ω b. R = 12Ω 4.12 a. R = 6,000Ω b. R = 5,000Ω 4.13 R = 60Ω 4.14 R = 8.5Ω V R = I

13 Section 4.5 Multiple and Submultiple Units 4.15 a. I = 100μA b. V = 660V 4.16 I = 50μA 4.17 V =10V Section 4.6 The Linear Proportion between V and I 4.18 See Figure 4-11 Figure

14 Section 4.7 Electric Power 4.19 a. P = 75W b. I = 10A 4.20 a. I = 500mA b. V = 100V 4.21 a. P = 150μW b. V = 200V 4.22 a. I = 500mA b. I = 833.3mA c. I = 2.5A 4.23 V = 15V 4.24 P = 1.8W 4.25 Cost = 5 Dollars and 4 Cents 4.26 Cost = 64 Dollars and 80 Cents Section 4.8 Power Dissipation in Resistance 4.27 P = 500mW 4.28 P = 200W 4.29 P = 2.16W 4.30 P = 320mW

15 Section 4.9 Power Formulas 4.31 a. R = 312.5Ω b. V = 223.6V 4.32 I = 31.62mA 4.33 I = 2.38mA 4.34 R = 240Ω 4.35 V = 100V 4.36 V = 13.96V 4.37 I = 8A 4.38 V = 7.75V 4.39 R = 7.2Ω

16 Chapter 5 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. c 2. b 3. a 4. a 5. c 6. c 7. a 8. b 9. a 10. c 11. b 12. c 13. d 14. b 15. a 16. c 17. b 18. d 19. b 20. c ANSWERS TO PROBLEMS Section 5.1 Why I is the Same in All Parts of a Series Circuit 5.1 a. I = 100mA b. I = 100mA c. I = 100mA d. I = 100mA e. I = 100mA f. I = 100mA 5.2 I = 100mA

17 Section 5.2 Total R Equals the Sum of All Series Resistances 5.3 R T = 300Ω I = 30mA 5.4 R T = 4kΩ I = 6mA 5.5 R T = 6kΩ I = 4mA 5.6 R T = 1.2MΩ I = 200μA

18 Section 5.3 Series IR Voltage Drops 5.7 V 1 = 3.6V V 2 = 5.4V 5.8 V 1 = 6V V 2 = 7.2V V 3 = 10.8V 5.9 V 1 = 4.4V V 2 = 13.6V V 3 = 2V V 4 = 200V V 5 = 20V 5.10 R T = 2kΩ I = 10mA V 1 = 3.3V V 2 = 4.7V V 3 = 12V 5.11 R T = 16kΩ I = 1.5mA V 1 = 2.7V V 2 = 4.05V V 3 = 12.3V V 4 = 4.95V

19 Section 5.4 Kirchhoff s Law (KVL) 5.12 V T = 100V 5.13 V T = 15V 5.14 V 2 = 11V 5.15 V 1 = 7.2V V 2 = 8.8V V 3 = 4V V 4 = 60V V 5 = 40V V T = 120V 5.16 V 1 = 12V

20 Section 5.5 Polarity of IR voltage Drops 5.17 a. R T = 100Ω, I = 500mA, V 1 = 5V, V 2 = 19.5V, V 3 = 25.5V b. See Figure 5-1 c. See Figure 5-1 d. See Figure See Figure 5-1. Whether electron flow or conventional current is used, the polarity of the resistor voltage drops is the same The polarity of the individual resistor voltage drops is opposite to that shown in Figure 5-1. The reason is that the polarity of a resistor s voltage drop depends on the direction of current flow and reversing the polarity of V T reverses the direction of current. Figure

21 Section 5.6 Total Power in a Series Circuit 5.20 P 1 = 108mW P 2 = 162mW P T = 270mW 5.21 P 1 = 36mW P 2 = 43.2mW P 3 = 64.8mW P T = 144mW

22 Section 5.7 Series-Aiding and Series-Opposing Voltages 5.22 a. V T = 27V b. I = 10mA c. Electrons flow up through R a. V T = 14V b. I = 14mA c. Electrons flow down through R a. V T = 6V b. I = 6mA c. Electrons flow up through R a. V T = 12V b. I = 400mA c. Electrons flow down through R 1 and R 2. d. V 1 = 4.8V and V 2 = 7.2V

23 Section 5.8 Analyzing Series Circuits with Random Unknowns 5.26 R S = 120Ω 5.27 R T = 30kΩ V 3 = 14.4V V 4 = 5.64V R 3 = 12kΩ P T = 43.2mW P 1 = 4.752mW P 2 = 14.4mW P 3 = 17.28mW P 4 = 6.768mW 5.28 I = 20mA R T = 6kΩ V T = 120V V 2 = 36V P 1 = 400mW

24 5.29 R 3 = 800Ω I = 50mA V T = 100V P T = 5W 5.30 R = 1kΩ 5.31 R = 38Ω 5.32 V T = 15.3V

25 Section 5.9 Ground Connections in Electrical and Electronic Systems 5.33 V AG = 18V V BG = 7.2V V CG = 1.2V 5.34 V AG = 16.4V V BG = -13.6V V CG = -43.6V 5.35 V AG = 20V V BG = 16.4V V CG = -9.4V V DG = -16V 5.36 V AG = 22.5V V BG = 7.5V V CG = -7.5V

26 Chapter 6 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. a 2. c 3. b 4. d 5. b 6. d 7. c 8. a 9. d 10. a 11. d 12. b 13. c 14. b 15. c 16. b 17. a 18. c 19. d 20. b ANSWERS TO PROBLEMS Section 6.1 The Applied Voltage, V A is the Same Across Parallel Branches 6.1 a. 12V b. 12V c. 12V d. 12V V

27 Section 6.2 Each Branch I Equals R V A 6.3 I 1 = 100mA I 2 = 200mA 6.4 I 2 is double I 1 because R 2 is one-half the value of R a. I 3 = 1.2A b. I 1 and I 2 are not affected by the addition of R 3 because R 1 and R 2 still have the applied voltage of 12V across them. 6.6 I 1 = 600mA I 2 = 900mA I 3 = 300mA 6.7 Yes, I 1 and I 3 remain the same because R 1 and R 3 still have the applied voltage of 18V across them. 6.8 I 1 = 200mA I 2 = 15mA I 3 = 85mA I 4 = 20mA

28 Section 6.3 Kirchhoff s Current Law (KCL) 6.9 I T = 300mA 6.10 I T = 320mA 6.11 I T = 160mA 6.12 I 1 = 24mA I 2 = 20mA I 3 = 16mA I T = 60mA 6.13 a. 60mA b. 36mA c. 16mA d. 16mA e. 36mA f. 60mA 6.14 I 1 = 200mA I 2 = 300mA I 3 = 2A I T = 2.5A

29 6.15 a. 2.5A b. 500mA c. 300mA d. 300mA e. 500mA f. 2.5A g. 2A h. 2A 6.16 I 3 = 80mA 6.17 I 2 = 90mA

30 Section 6.4 Resistances in Parallel 6.18 R EQ = 40Ω 6.19 R EQ = 10Ω 6.20 R EQ = Ω 6.21 R EQ = 400Ω 6.22 R EQ = 20Ω 6.23 R EQ = 600kΩ 6.24 R EQ = 200Ω 6.25 R EQ = 5kΩ 6.26 R 2 = 1.5kΩ 6.27 R EQ = 112.5Ω 6.28 R EQ = 550Ω

31 6.29 a. R EQ = 269.9Ω (Ohmmeter will read 270Ω approximately.) b. R EQ = 256.6kΩ (Ohmmeter will read 256kΩ approximately.) c. R EQ = 559.3kΩ (Ohmmeter will read 559kΩ approximately.) d. R EQ = 1.497kΩ (Ohmmeter will read 1.5kΩ approximately.) e. R EQ = 9.868kΩ (Ohmmeter will read 9.87kΩ approximately.) Section 6.5 Conductances in Parallel 6.30 G 1 = 1mS G 2 = 250µS G 3 = 5mS G T = 6.25mS R EQ = 160Ω 6.31 G T = 500mS R EQ = 2Ω

32 Section 6.6 Total Power in Parallel Circuits 6.32 P 1 = 10.8W P 2 = 16.2W P 3 = 5.4W P T = 32.4W 6.33 P 1 = 20.4W P 2 = 1.53W P 3 = 8.67W P 4 = 2.04W P T = 32.64W

33 Section 6.7 Analyzing Parallel Circuits with Random Unknowns 6.34 V A = 18V R 1 = 360Ω I 2 = 150mA R EQ = 90Ω P 1 = 900mW P 2 = 2.7W P T = 3.6W 6.35 V A = 15V I 1 = 150mA I 2 = 50mA R 2 = 300Ω I T = 200mA P 2 = 750mW P T = 3W 6.36 R 3 = 500Ω V A = 75V I 1 = 150mA I 2 = 300mA I T = 600mA P 1 = 11.25W P 2 = 22.5W P 3 = 11.25W P T = 45W 6.37 I T = 1.2A I 1 = 100mA I 2 = 900mA R 1 = 1.08kΩ R 2 = 120Ω R 3 = 540Ω P 2 = 97.2W P 3 = 21.6W P T = 129.6W

34 Chapter 7 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. c 2. b 3. a 4. c 5. d 6. c 7. a 8. a 9. b 10. d 11. b 12. d 13. a 14. c 15. d 16. b 17. d 18. c 19. a 20. b ANSWERS TO PROBLEMS Section 7.1 Finding R T for Series-Parallel Resistances 7.1 Resistors R 1 and R 2 are in series and resistors R 3 and R 4 are in parallel. It should also be noted that the applied voltage, V T, is in series with R 1 and R 2 because they all have the same current. 7.2 a. 900Ω b. 600Ω c. 1.5kΩ d. 10mA e. 10mA f. 10mA 7.3 I 1 = 10mA I 2 = 10mA V 1 = 2.2V V 2 = 6.8V V 3 = 6V V 4 = 6V I 3 = 6mA I 4 = 4mA

35 7.4 Resistors R 1 and R 4 are in series and resistors R 2 and R 3 are in parallel. Also, the applied voltage, V T, is in series with R 1 and R 4 because they all have the same current. 7.5 a. 250Ω b. 200Ω c. 450Ω d. 40mA e. 40mA 7.6 I 1 = 40mA V 1 = 6V V 2 = 8V V 3 = 8V I 2 = 26.7mA I 3 = 13.3mA I 4 = 40mA V 4 = 4V P 1 = 240mW P 2 = 213.3mW P 3 = 106.7mW P 4 = 160mW P T = 720mW

36 Section 7.2 Resistance Strings in Parallel 7.7 a. 3kΩ b. 1kΩ c. I 1 = 5mA and I 2 = 15mA d. I T = 20mA e. R T = 750Ω f. V 1 = 6V, V 2 = 9V, V 3 = 2.7V and V 4 = 12.3V 7.8 a. I 1 = 8mA, I 2 = 24mA, I 3 = 16mA, I T = 48mA b. R T = 500Ω c. V 1 = 8V, V 2 = 16V, V 3 = 24V, V 4 = 8V and V 5 = 16V Section 7.3 Resistance Banks in Series 7.9 a. 1.32kΩ b. R T = 2kΩ c. I T = 7.5mA d. V AB = 9.9V e. V 3 = 5.1V f. I 1 = 4.5mA and I 2 = 3mA g. 7.5mA 7.10 a. 150Ω b. R T = 250Ω c. I T = 100mA d. V AB = 15V e. V 1 = 10V f. I 2 = 25mA and I 3 = 75mA g. 100mA

37 Section 7.4 Resistance Banks and Strings in Series-Parallel 7.11 R T = 240Ω I T = 50mA V 1 = 3.75V V 2 = 8.25V V 3 = 3.75V V 4 = 4.5V 7.12 R T = 300Ω I T = 250mA V 1 = 30V V 2 = 28V V 3 = 13.6V V 4 = 14.4V V 5 = 14.4V I 1 = 50mA I 2 = 25mA I 3 = 25mA I 4 = 25mA V 6 = 17V I 1 = 250mA I 2 = 50mA I 3 = 200mA I 4 = 120mA I 5 = 80mA I 6 = 250mA

38 7.13 R T = 1kΩ I 1 = 60mA I 2 = 44mA I 3 = 16mA V 1 = 7.2V V 2 = 52.8V V 3 = 52.8V V AB = 52.8V 7.14 R T = 200Ω I T = 120mA V 1 = 6V V 2 = 6V V 3 = 12V V 4 = 12V V 5 = 18V V 6 = 24V I 1 = 40mA I 2 = 30mA I 3 = 10mA I 4 = 20mA I 5 = 10mA I 6 = 80mA Section 7.5 Analyzing Series-Parallel Circuits with Random Unknowns 7.15 V 1 = 36V V 2 = 18V V 3 = 18V I 2 = 180mA R 3 = 150Ω R T = 180Ω V T = 54V 7.16 R 2 = 100Ω V 1 = 1.2V V 2 = 12V V 3 = 13.2V I 3 = 40mA I T = 160mA V T = 13.2V 7.17 R T = 90Ω I T = 200mA V 1 = 3V V 2 = 15V V 3 = 5V V 4 = 10V V 5 = 6V I 1 = 200mA I 2 = 150mA I 3 = 50mA I 4 = 10mA I 5 = 40mA I 6 = 40mA

39 V 6 = 4V

40 Section 7.6 The Wheatstone Bridge 7.18 a. When the bridge is balanced, the current in M 1 is zero. b. Zero volts a. R X = 6,816Ω b. V CB = V DB = 8.33V c. I T = 1.91mA 7.20 a. C to D b. D to C 7.21 a. R X(max) = Ω b. R X(max) = 9,999,900Ω 7.22 a. R X = 123Ω b. R X = 123.2Ω c. R X = Ω R 1 The ratio arm fraction 1 = provides a placement accuracy of R2 100 ±0.01Ω, and therefore the greatest accuracy R 3 must be adjusted to 1kΩ a. V CD becomes negative. b. V CD becomes positive a. The thermistor resistance is 4,250Ω. b. T A has increased above 25 C

41 Section 7.7 Series Voltage Dividers 7.26 V 1 = 3V V 2 = 6V V 3 = 9V 7.27 V 1 = 12V V 2 = 14.4V V 3 = 5.64V V 4 = 3.96V 7.28 V 1 = 4V V 2 = 6V V 3 = 8V 7.29 V 1 = 2.5V V 2 = 7.5V V 3 = 15V 7.30 a. V 1 = 20V; V 2 = 36V; V 3 = 44V b. V AG = 100V; V BG = 80V; V CG = 44V

42 Section 7.4 Series Voltage Divider with Parallel Load Current 7.31 a. I 1 = 150mA I 2 = 150mA I L = 0mA V BG = 18V V AG = 24V b. I 1 = 300mA I 2 = 100mA I L = 200mA V BG = 12V V AG = 24V 7.32 The voltage, V BG, decreases when S 1 is closed because R L in parallel with R 2 reduces the resistance from points B to G. This lowering of resistance changes the voltage division in the circuit. With S 1 closed, the resistance from B to G is a smaller fraction of the total resistance, which in turn means the voltage, V BG must also be less a. I 1 = 200mA I 2 = 200mA I L = 0mA V BG = 20V V AG = 32V b. I 1 = mA I 2 = 160mA I L = mA V BG = 16V V AG = 32V 7.34 Resistor R a. I 1 = 3mA I 2 = 3mA I L = 0mA V BG = 6V V AG = 12V b. I 1 = 4mA I 2 = 2mA I L = 2mA V BG = 4V V AG = 12V

43 Chapter 8 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. b 2. a 3. c 4. d 5. c 6. a 7. c 8. b 9. d 10. b 11. c 12. b 13. a 14. c 15. a 16. d 17. c 18. c 19. b 20. c 21. d 22. b ANSWERS TO PROBLEMS Section 8.1 Kirchhoff s Current Law (KCL) 8.1 I 3 = 15A 8.2 I 1 + I 2 I 3 = 0 or 5A + 10A 15A = I 3 = 6mA 8.4 I 3 = 8A, I 5 = 16A and I 8 = 11A 8.5 Point X: 6A + 11A + 8A 25A = 0 Point Y: 25A 2A 16A 7A = 0 Point Z: 16A 5A 11A =

44 Section 8.2 Kirchhoff s Voltage Law (KVL) 8.6 a. 4.5V + 5.4V + 8.1V 18V = 0 b. 6V + 18V + 12V 36V = 0 c. 6V + 4.5V + 5.4V + 8.1V + 12V 36V = a. 4.5V + 5.4V + 8.1V = 18V. This voltage is the same as the voltage V R3. b. 6V + 18V + 12V = 36V. This voltage equals the applied voltage, V T. c. 6V + 4.5V + 5.4V + 8.1V + 12V = 36V. This voltage equals the applied voltage, V T. d. 18V - 8.1V 5.4V = 4.5V. This voltage is the same as the voltage, V R V AG = +5V V BG = -10V 8.9 V AG = +8V V BG = 0V Section 8.3 Thevenin s Theorem 8.10 See Figure 8-1 Figure When R L = 3Ω, I L = 2.5A and V L = 7.5V When R L = 6Ω, I L = 1.67A and V L = 10V When R L = 12Ω, I L = 1A and V L = 12V 8.12 When R L = 100Ω, I L = 48mA and V L = 4.8V

45 When R L = 1kΩ, I L = 30mA and V L = 30V When R L = 5.6kΩ, I L = 10.29mA and V L = 57.6V 8.13 I L = 20mA and V L = 24V Section 8.4 Thevenizing a Bridge Circuit 8.14 See Figure 8-8 Figure I L = 40mA V L = 4V 8.16 See Figure 8-9 Figure

46 8.17 I L = 30mA V L = 3V Section 8.5 Voltage Sources Ω to 10 MΩ Ω kω mv

47 Chapter 9 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. b 2. a 3. c 4. b 5. d 6. b 7. a 8. c 9. c 10. d 11. b 12. c 13. d 14. a 15. b 16. a 17. b 18. c 19. b ANSWERS TO PROBLEMS Section 9.1 Function of the Conductor 9.1 a. 100 ft. b. R T = 8.16Ω c. I = 14.71A d. 1.18V e V f W g kW h kW i. 98.1%

48 Section 9.2 Standard Wire Gage Sizes 9.2 a. 25 cmils b. 441 cmils c cmils d cmils e. 10,000 cmils f. 40,000 cmils 9.3 No. 13 gage 9.4 a. R = 1.018Ω b. R = 2.042Ω c. R = 4.094Ω d. R = 26.17Ω 9.5 A 1,000 ft. length of No. 14 gage copper wire. 9.6 A 1,000 ft. length of No. 23 gage copper wire. Section 9.3 Types of Wire Conductors 9.7 No. 10 Gage 9.8 No. 19 Gage

49 Section 9.6 Switches 9.9 a. 0V b. 6.3V c. Yes d. 150mA 9.10 a. 6.3V b. 0V c. No d. 0A 9.11 See Figure 11-1 Figure (a) See Figure 9-2 (b) See Figure

50 Figure 9-2 Figure Yes, because the 1,000W appliance will only draw 8.33A of current. Section 9.8 Wire Resistance 9.14 a. R = 2.54Ω b. R = 4.16Ω 9.15 R = 0.127Ω (approx.) 9.16 V LOAD = V (approx.)

51 Section 9.9 Temperature Coefficient of Resistance 9.17 R = 25Ω 9.18 R = 1,040Ω

52 Chapter 10 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. d 2. b 3. d 4. a 5. c 6. b 7. a 8. c 9. b 10. c 11. d 12. a 13. a 14. d 15. d ANSWERS TO PROBLEMS Section 10.6 Series and Parallel Connected Cells 10.1 V L = 3V, I L = 30mA, the current in each cell equals 30mA V L = 8.4V, I L = 350mA, the current in each cell equals 350mA. V L = 1.25V, I L = 50mA, the current in each cell equals 25mA V L = 6.30V, I L = 630mA, the current in each cell equals 315mA V L = 3V, I L = 300mA, the current in each cell equals 100mA

53 Section 10.8 Internal Resistance of a Generator 10.6 r i = 10Ω 10.7 r i = 2Ω 10.8 I = 15A 10.9 r i = 6Ω r i = 25Ω r i = 15Ω

54 Section 10.9 Constant Voltage and Constant Current Sources a. I L = 9.9A; V L = 9.9V b. I L = 1.996A; V L = 9.98V c. I L = 999mA; V L = 9.99V d. I L = 99.99mA; V L 10V a. I L = 1μA; V L = 0V b. I L 1μA; V L 100μV c. I L 1μA; V L 1mV d. I L = 0.99μA; V L = 99mV See Figure Figure

55 Chapter 11 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. c 2. b 3. a 4. d 5. a 6. b 7. d 8. c 9. c 10. b 11. a 12. d 13. c 14. b 15. d 16. a 17. c 18. d 19. d 20. c 21. a 22. c 23. b 24. b 25. d 26. c 27. d 28. a 29. b 30. a 31. c 32. b 33. c 34. d 35. d 36. b 37. a 38. d 39. d 40. c 41. b 42. a 43. b 44. b 45. c ANSWERS TO ESSAY QUESTIONS

56 Chapter 12 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. a 2. c 3. b 4. c 5. d 6. a 7. b 8. a 9. b 10. d 11. c 12. b 13. c 14. a 15. d 16. b 17. c 18. a 19. b 20. d ANSWERS TO PROBLEMS Section 12.2 Meter Shunts 12.1 a. R S = 50Ω b. R S = 5.56Ω c. R S = 2.08Ω d. R S = 0.505Ω 12.2 a. R S = 1kΩ b. R S = 52.63Ω c. R S = 10.1Ω d. R S = 5.03Ω e. R S = 1Ω f. R S = 0.5Ω

57 12.3 So that the current in the circuit is approximately the same with or without the meter present. If the current meter s resistance is too high, the measured value of current could be significantly less than the current without the meter present. Section 12.3 Voltmeters 12.4 a. R mult = 9.95kΩ b. R mult = kΩ 12.5 Ω 1kΩ Rating = V V 12.6 a. R mult = 147.5kΩ b. R mult = 497.5kΩ c. R mult = MΩ d. R mult = MΩ e. R mult = MΩ 12.7 Ω Rating = V 50kΩ V

58 Section 12.4 Loading Effect of a Voltmeter 12.8 a. V = 7.2V b. V = 4.5V c. V = 7.16V 12.9 a. V = 4V b. V = 2.4V c. V = 3.75V The analog voltmeter with an R V of 1MΩ produced a greater loading effect. The reason is that its resistance is less than that of the DMM whose R V is 10MΩ. Section 12.5 Ohmmeters a. R X = 0Ω b. R X = 250Ω c. R X = 750Ω d. R X = 2.25kΩ e. R X = Ω No, because the internal battery voltage determines the direction of current through the meter movement On any range the zero-ohms adjustment control is adjusted for zero ohms with the ohmmeter leads shorted. The zero ohms control is adjusted to compensate for the slight changes in battery voltage, V b, when changing ohmmeter ranges. Without a zero ohms adjustment control, the scale of the ohmmeter would not be properly calibrated

59 Section 12.8 Meter Applications When measuring an unknown value of voltage or current always start on the highest range and work your way down. This practice will avoid excessive current in the meter which could damage it The ohmmeter could be damaged or the meter will read an incorrect value of resistance. When measuring resistance, power must be off in the circuit being tested! To remove any parallel paths which could produce an incorrect meter reading A current meter is connected in series to measure the current at some point in a circuit. Connecting a current meter in parallel could possibly ruin the meter due to excessive current. Remember, a current meter has a very low resistance and connecting it in parallel can effectively short-out a component By making voltage measurements and calculating I as R V a. 0Ω b. infinite ( )Ω a. V AG = 24V V BG = 20V V CG = 12V V DG = 7.2V b. V AG = 24V V BG = 0V V CG = 0V V DG = 0V c. V AG = 24V V BG = 24V V CG = 0V V DG = 0V d. V AG = 24V V BG = 24V V CG = 24V V DG = 0V

60 e. V AG = 24V V BG = 24V V CG = 24V V DG = 24V Chapter 13 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. b 2. d 3. c 4. b 5. d 6. a 7. d 8. c 9. b 10. c 11. d 12. a 13. d 14. d 15. c ANSWERS TO PROBLEMS Section 13.4 Troubleshooting: Opens and Shorts in Series Circuits 13.1 a. infinite b. zero c. V R1 = 24 V; V R2 = 0 V; V R3 = 0 V 13.2 a. 2 kω b. 8 ma c. V R1 = 8 V; V R2 = 16 V; V R3 = 0 V

61 13.3 a. R T increases b. I decreases c. V R2 decreases d. V R1 and V R3 decrease 13.4 Trouble 1 Trouble 2 Trouble 3 Trouble 4 Trouble 5 Trouble 6 Trouble 7 Trouble 8 Trouble 9 R 2 open R 3 shorted R 4 shorted R 1 open R 1 shorted R 4 open R 2 shorted R 3 open R 5 open Trouble 10 R 5 increased in value Trouble 11 R 3 decreased in value Trouble 12 R 5 shorted Trouble 13 R 1 increased in value Trouble 14 R 4 decreased in value Section 13.5 Troubleshooting: Opens and Shorts in Parallel Circuits

62 13.5 a. fuse blows b. total current decreases c. total current decreases d. fuse blows 13.6 R 1 open 13.7 Connection CD or GH open. Measure voltage across R 3 or R R 2 open 13.9 a. Zero for M 1 and M 3 b. Zero c. 36 V d. A short across any resistor e. Disconnect each resistor at one end and measure it. Look for a visual short Connection between AB or IJ open Measure voltage across R

63 13.11 Zero Disconnect one end of each resistor and measure it. Look for a visual short a. M 3 will read 600 ma b. Zero c. Zero d. 36 V Zero volts across the fuse Zero volts across the switch a. Zero b. M 1 = 3.7 A and M 3 = 1.2 A a. M 1 = 2.5 A and M 3 = zero b. 36 V

64 c. Zero Section 13.6 Troubleshooting: Opens and Shorts in Series-Parallel Circuits a. 25 V b V c V Trouble 1 R 6 open Trouble 2 R 2 shorted Trouble 3 R 4 open Trouble 4 R 5 open Trouble 5 R 6 shorted Trouble 6 R 2 increased in value Trouble 7 R 1 open Trouble 8 R 4 shorted Trouble 9 R 1 shorted Trouble 10 R 3 open Trouble 11 R 3 shorted Trouble 12 R 5 shorted

65 Figure Trouble 1 R 2 open Trouble 2 R 1 open Trouble 3 R 3 shorted Trouble 4 Ground on R 4 open Trouble 5 R 2 shorted Trouble 6 R 1 shorted Trouble 7 R 4 open Trouble 8 R 4 shorted Figure Trouble 1 R 2 open Trouble 2 Point C shorted to ground Trouble 3 R 3 open Trouble 4 Point B shorted to ground Trouble 5 R 1 open Trouble 6 Load B open or disconnected Trouble 7 R 4 or point D shorted to ground Trouble 8 R 1 shorted Chapter

66 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. d 2. c 3. a 4. d 5. b 6. c 7. a 8. b 9. c 10. a 11. a 12. c 13. b 14. c 15. b 16. d 17. a 18. b 19. a 20. b 21. a 22. d 23. c 24. d 25. c 26. b ANSWERS TO PROBLEMS Section 14.2 Alternating Voltage Generator 14.1 a b c d a rad or 2 π rad b rad or π rad 3π c rad or rad 2 d rad or 2π rad 14.3 a. at 90 0 b. at c. 0 0, and rad = approx. Section 14.3 The Sine Wave 14.5 a. v = 14.14V b. v = -10V

67 14.6 V pk = 51.96V

68 Section 14.4 Alternating Current 14.7 a. i = 2.5mA b. i = -4.33mA 14.8 a. counterclockwise b. clockwise Section 14.5 Voltage and Current Values for a Sine Wave 14.9 a. 30V peak-to-peak b. 10.6V rms c. 9.56V average a V peak b V peak-to-peak c. 36V average Section 14.6 Frequency a. f = 10Hz b. f = 500Hz c. f = 50,000Hz or 50kHz d. f = 2,000,000Hz or 2MHz a. 2,000 cps b. 15,000,000 cps c. 10,000 cps d. 5,000,000,000 cps

69 Section 14.7 Period a. T = 500μs b. T = 250μs c. T = 5μs d. T = 0.5μs a. f = 200Hz b. f = 100kHz c. f = 2MHz d. f = 30kHz a. 50μs b. 100μs c. 150μs d. 200μs Section 14.8 Wavelength a. 186,000 mi/s b. 3 x cm/s c. 3 x 10 8 m/s ,130 ft/s m a ft. b ft a. f = 1.875MHz b. f = MHz

70 Section 14.9 Phase Angle A sine wave has its maximum values at 90 o and whereas a cosine wave has its maximum values at 0 0 and a. See Figure 14-1 b. See Figure 14-2 Figure 14-1 Figure

71 Section The Time Factor in Frequency and Phase a. t = 83.3μs b. t = 250μs Section Alternating Current Circuits with Resistance R T = 250Ω I = 40mA V 1 = 4V V 2 = 6V P 1 = 160mW P 2 = 240mW P T = 400mW

72 Section Nonsinusoidal AC Waveforms a. V = 100V peak-to-peak f = 20kHz b. V = 30V peak-to-peak f = 500Hz c. V = 100V peak-to-peak f = 2.5kHz Any complex nonsinusoidal signal can be represented as a fundamental sine wave to which has been added harmonic sine waves GHz MHz a c d a b a c Average dc voltage = 1.5 Amplitude of third harmonic = V Chapter 15 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. b 2. a 3. d 4. c 5. c 6. d 7. a 8. b 9. c 10. d 11. c 12. a

73 13. d 14. d 15. a 16. d 17. a 18. c 19. a 20. b 21. a 22. d 23. b 24. c 25. d ANSWERS TO PROBLEMS Section 15.3 The Farad Unit of Capacitance 15.1 a. Q = 50µC b. Q = 136nC 15.2 a. Q = 500nC b. Q = 25µC 15.3 a. V = 2.5V b. V = 75V 15.4 a. V = 250V b. V = 150V 15.5 a. C = 15µF b. C = 0.5µF or 500nF 15.6 Area of the plates, distance between the plates, and type of dielectric a. C = 1.77pF b. C = 2.213nF Section 15.6 Ultracapacitors 15.8 Supercapacitors and electrochemical double-layer capacitor (EDLC) 15.9 True a

74 - 74 -

75 Section 15.7 Parallel Capacitances C T = 20μF C T = 0.38μF C T = 0.148μF a. V = 10V b. Q 1 = 1mC c. Q 2 = 2.2mC d. Q 3 = 6.8mC e. Q T = 10mC f. C T = 1,000μF Section 15.9 Series Capacitances C EQ = 0.08μF C EQ = 600pF C EQ = 8,561pF a. C EQ = 5μF b. Q 1 = Q 2 = Q 3 = 180μC c. V C1 = 18V, V C2 = 6V, V C3 = 12V d. Q T = 180μC Chapter 16 ANSWERS TO REVIEW QUESTIONS 1. a 2. c 3. b 4. d 5. b 6. c 7. a 8. d 9. b 10. c 11. b 12. d 13. a 14. b 15. a ANSWERS TO PROBLEMS Section 16.1 Alternating Current in a Capacitive Circuit

76 16.1 a. I = 0A b. V lamp = 0V c. V C = 12V 16.2 When S 1 is initially closed, a charging current will flow for a short period of time until the 100µF capacitor, C 1 is fully charged. This charging current flows through the filament of the 12V lamp and lights it for just an instant a. I = 400mA b. I = 400mA c. I = 400mA d. I = 400mA e. I = 0A 16.4 a. X C = 500Ω b. X C = 3kΩ c. X C = 50kΩ d. X C = 2MΩ 16.5 The amplitude of the applied voltage, the frequency of the applied voltage and the amount of capacitance

77 Section 16.2 The Amount of X C Equals 1 2πfC 16.6 a. X C = 31.83kΩ b. X C = 159.1Ω 16.7 a. X C = Ω b. X C = Ω 16.8 a. C = 0.5µF b. C = 0.37pF 16.9 a. f = 33.86Hz b. f = 67.73kHZ C = 0.01µF F = kHz a. I = 100mA b. I = 25mA c. I = 100mA d. I = 25mA a. X C = 20kΩ b. X C = 5kΩ c. X C = 2.5kΩ d. X C = 1kΩ a. X C = Ω b. X C = 636.6Ω

78 16.15 a. C = 0.05µF b. C = µF a. f = 795.8Hz b. f = 1kHz Section 16.3 Series or Parallel Capacitive Reactances a. X CT = 5kΩ b. X CT = 3kΩ a. X CEQ = 80Ω b. X CEQ = 720Ω Section 16.4 Ohm s Law Applied to X C I = 50mA a. I decreases b. I increases C 1 = C 2 = C 3 = C T = 0.2μF 0.42μF 0.63μF 1.25μF Chapter 17 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. d 2. c 3. b 4. a 5. c 6. b 7. d 8. a 9. c 10. a 11. c 12. a 13. d 14. c 15. b

79 ANSWERS TO PROBLEMS Section 17.1 Sine Wave V C Lags I C by a. 10V b. 10mA c. 10kHz d (i C leads V C by 90 0 ) 17.2 a. See Figure 17-1 b. See Figure 17-2 Figure 17-1 Figure

80 Section 17.2 X C and R in Series 17.3 a. 2A b. 2A c. 2A 17.4 a. I and V R are in phase b. V C lags I by 90 0 c. V C lags V R by V T = 100V 17.6 See Figure 17-3 Figure a. V R = 40V b. V C = 30V c. V T = 50V

81 Section 17.3 Impedance Z Triangle 17.8 Z T = 25Ω I = 4A V C = 80V V R = 60V θ Z = See Figure 17-4 Figure a. X C decreases b. Z T decreases c. I increases d. V C decreases e. V R increases f. θ Z decreases (becomes less negative)

82 Section 17.4 RC Phase-Shifter Circuit X C = 26.53kΩ Z T = 56.6kΩ I = 2.12mA V R = 106V V C = 56.24V θ Z = a. V R leads V T by 28 0 b. V C lags V T by See Figure 17-5 Figure

83 Section 17.5 X C and R in Parallel a. 120V b. 120V a. V A and I R are in phase b. I C leads V A by 90 0 c. I C leads I R by I R = 3A I C = 4A I T = 5A Z EQ = 24Ω θ I = See Figure 17-6 Figure I R = 200mA I C = 200mA I T = 282.8mA Z EQ = 63.65Ω θ I =

84 17.19 X C = 500Ω I R = 20mA I C = 48mA I T = 52mA Z EQ = Ω θ I = a. I R stays the same b. I C increases c. I T increases d. Z EQ decreases e. θ I increases

85 Section 17.6 RF and AF Coupling Capacitors C C = 1μF C C = 0.1μF f = 33.86kHz θ Z = Section 17.7 Capacitive Voltage Dividers a. X C1 = 10kΩ X C2 = 2kΩ X C3 = Ω X C4 = 45.46Ω X CT = 12.5kΩ b. I = 2.4mA c. V C1 = 24V V C2 = 4.8V V C3 = 1.091V V C4 = 109.1mV V C1 = 50V V C2 = 20V V C3 = 10V Chapter 18 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. a 2. d 3. b 4. c 5. b 6. a 7. b 8. d 9. c 10. a 11. b 12. d 13. a 14. c 15. d 16. c 17. b 18. a 19. b 20. d ANSWERS TO PROBLEMS Section 18.1 Induction by Alternating Current 18.1 A small current change of 1 to 2mA

86 18.2 a. Points A, E and I b. Points C and G 18.3 A high frequency alternating current. Section 18.2 Self-Inductance L 18.4 di a. dt di b. dt c. di dt di d. dt = 1.5A/s = 10,000A/s = 10A/s = 2,500A/s e. f. g. di dt di dt di dt = 5,000A/s = 40,000A/s = 1A/s 18.5 a. L = 10H b. L = 1.5mH c. L = 1.5H d. L = 6mH e. L = 3mH f. L = 375µH g. L = 15H 18.6 L = 30mH 18.7 a. L = 0.633µH b. L = 3.17mH c. L = 63.3µH d. L = 0.316µH e. L = 126mH 18.8 Choke

87 Section 18.3 Self-Induced Voltage, V L 18.9 V L = 500V V L = 49.5V a. V L = 10V b. V L = 100V Section 18.5 Mutual Inductance L M k = L M = 61.24mH The assumed value of k is 1. Section 18.6 Transformers a. V S = 24Vac b. I S = 2A c. P sec = 48W d. P pri = 48W e. I P = 400mA N P 3 a. = NS 1 b. I S = 2.5A c. I P = 833.3mA a. b. N P NS 1 = 5 N P N 5 = 1 S

88 18.18 % Efficiency = 80% There are two advantages of a transformer with an isolated secondary. a. Any dc current in the primary is not present in the secondary. b. The isolated secondary reduces the possibility of an electric shock because it is isolated from the earth ground connection of the 120Vac power line. Section 18.7 Transformer Ratings The power rating of a transformer is specified in volt-amperes (VA) which is the unit of apparent power a. Never apply more than the rated voltage to the primary. b. Never draw more than the rated current from the secondary To identify those transformer leads with the same instantaneous polarity a. V AB = 12.6Vac b. V AC = 6.3Vac c. V BC = 6.3Vac I P = 210mA

89 Section 18.8 Impedance Transformation a. Z P = 200Ω b. Z P = 5Ω a. b. N P = 1:3.46 NS N P = 4:47:1 N S N P = 11.18:1 N S a. Z P = 500Ω b. P RL = 5W c. P pri = 5W Section Inductances in Series or Parallel a. L T = 20mH b. L T = 1mH c. L T = 10mH a. L EQ = 3.75mH b. L EQ = 4mH c. L EQ = 102µH d. L EQ = 280.4µH L T = 660mH a. L T = 82.63mH b. L T = 37.37mH Chapter 19 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. b 2. c 3. a 4. d 5. a 6. c 7. b 8. a 9. d 10. b ANSWERS TO PROBLEMS

90 Section 19.1 How X L Reduces the Amount of I 19.1 X L = 0Ω at dc 19.2 The frequency, f, and the inductance, L I dc = 2.5A 19.4 X L = 0Ω with S 1 in position 1. This is because the inductor has no inductive reactance for direct current a. Because with S 1 in position 2 the inductor has an inductive reactance, X L, in addition to the dc resistance, r i, to limit the circuit s current flow. With S 1 in position 1 only the dc resistance of the coil limits current flow since there is no X L for direct current. b. X L = 4kΩ 19.6 a. X L = 5kΩ b. X L = 2.5MΩ c. X L = 800Ω d. X L = 150kΩ e. X L = 300Ω

91 Section 19.2 X L = 2πfL 19.7 a. X L = 37.7Ω b. X L = 6.28kΩ 19.8 a. X L = 18.85mΩ b. X L = 1.19kΩ 19.9 a. L = 100mH b. L = 31.83mH a. f = Hz b. f = kHz L = mH F = 63.66MHz a. I = 1mA b. I = 4mA a. X L = 5kΩ b. X L = 100kΩ a. X L = 2.64kΩ b. X L = 1kΩ

92 19.16 a. L = 63.66mH b. L = 2µH a. f = kHz b. f = 7.96kHz Section 19.3 Series or Parallel Inductive Reactances a. X LT = 1.75kΩ b. X LT = 5kΩ a. X LEQ = 720Ω b. X LEQ = 600Ω Section 19.4 Ohm s Law Applied to X L I = 80mA a. I increases b. I decreases a. X LT = 400Ω b. I = 60mA c. V L1 = 6V, V L2 = 7.2V and V L3 = 10.8V Section 19.5 Applications of X L for Different Frequencies a. L = 318.3mH b. L = 10mH Section 19.6 Waveshape of V L Induced by Sine Wave Current V L leads i L by a phase angle of This 90 0 phase relationship exists because V L depends on the rate of current change rather than on the actual value of current itself a. 0 0, and b and Chapter

93 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. b 2. a 3. d 4. c 5. c 6. a 7. b 8. d 9. c 10. a 11. c 12. d 13. a 14. d 15. b 16. a 17. d 18. b 19. c 20. b ANSWERS TO PROBLEMS Section 20.1 Sine Wave i L Lags V L by a. 10V b. 10mA c. 10kHz d a. Either +10V or 10V depending on the direction of i L. b. 0V c. 0V 20.3 a. See Figure 20-1 b. See Figure 20-2 Figure

94 Figure 20-2 Section 20.2 X L and R in Series 20.4 a. I = 4A b. I = 4A c. I = 4A 20.5 a. 0 0 b c V T = 100V 20.7 See Figure 20-3 Figure a. V R = 48V b. V L = 36V c. V T = 60V

95 Section 20.3 Impedance Z Triangle 20.9 Z T = 125Ω I = 288mA V L = 21.6V V R = 28.8V θ Z = See Figure 20-4 Figure

96 20.11 X L = 1.8kΩ Z T = 3.25kΩ I = 30.77mA V R = 83.1V V L = 55.4V θ Z = Section 20.4 X L and R in Parallel a. 12Vac b. 12Vac a. 0 0 b. I L lags V A by 90 0 c. I L lags I R by See Figure Figure I R = 18mA I L = 36mA I T = 40.25mA Z EQ = 894.4Ω θ I =

97 20.16 a. I R stays the same b. I L decreases c. I T decreases d. Z EQ increases e. θ I becomes less negative

98 Section 20.5 Q of a Coil a. Q = 6.28 b. Q = Because as the frequency increases the ac effective resistance of the coil also increases. The ac effective resistance of a coil represents the losses which are associated with a coil at higher frequencies which include skin effect, eddy current losses and hysteresis losses R e = 94.25Ω

99 Section 20.6 AF and RF Chokes a. L = 4.78H b. L = 2.39mH f = 47.75kHz Section 20.7 Power in AC Circuits Real power = 240 W Reactive power = 320 VAR Apparent power = 400 VA Power factor = Real power = 360 W Reactive power = 240 VAR Apparent power = 432 VA Power factor = 0.83 Section 20.8 Complex Impedances Z = j Z = 55 j90 Chapter

100 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. c 2. d 3. b 4. a 5. a 6. c 7. b 8. d 9. c 10. b 11. d 12. a 13. d 14. a 15. c 16. b 17. b 18. d 19. c 20. d 21. c 22. d 23. a 24. b 25. a ANSWERS TO PROBLEMS Section 21.1 Response of Resistance Alone 21.1 The current, I, reaches its steady-state value immediately because a resistor does not provide any reaction to a change in either voltage or current I = 6A Section 22.2 R L Time Constant 21.3 a. T = 40μs b. 80mA c. 0mA d. approximately 50.6mA e. 200μs 21.4 a. T = 250μs b. T = 5μs 21.5 a. either increase L or decrease R b. either decrease L or increase R

101 Section 21.3 High Voltage Produced by Opening an RL Circuit 21.6 a. T = 60ns b. V L = 80kV c. 300ns or 0.3μs 21.7 Without a resistor across S 1 there is no way to determine the time constant of the circuit with S 1 open. This is because there is no way of knowing what the resistance of the open switch is. We do know, however, that the time constant will be very short with S 1 open. This di short time constant will result in a very large value which in turn will dt produce a very large induced voltage across the open contacts of the switch. This will most likely produce internal arcing across the open switch contacts

102 Section 21.4 RC Time Constant 21.8 a. T = 20ms b. T = 20ms 21.9 a. V C = 31.6V b. V C = 50V c. V C = 50V a. V C = 18.4V b. V C = 0V c. V C = 0V a. T = 1s b. T = 200ms a. Increase the value of either R or C. b. Decrease the value of either R or C

103 Section 21.5 RC Charge and Discharge Curves a. 500μA b. zero c. V R = 18.4V d. 184μA a. 500μA b. 184μA c. zero Section 21.6 High Current Produced by Short-Circuiting RC Circuit a. T = 100ms b. T = 250μs a. 500ms b. 1.25ms Section 21.7 RC Waveshapes a. T = 200μs b. See Figure 21-1 c. See Figure 21-1 d. See Figure

104 Figure

105 Section 21.8 Long and Short Time Constants a. Long b. Short a. the output is taken across the resistor b. short a. the output is taken across the capacitor b. long S ection 21.9 Charge and Discharge with Short RC Time Constants a. T = 10μs b. See Figure 21-2 c. See Figure 21-2 tp d. = 10 RC

106 Figure 21-2 Section Long Time Constant for RC Coupling Circuit a. T = 1ms tp 1 b. = RC 10 c. See Figure

107 Figure

108 Section Advanced Time Constant Analysis T = 1s a. V C = 151V b. V C = 233.1V c. V C = 290.9V a. V R = 149V b. V R = 24.63V T = 7.5ms a. t = 2.16ms b. t = 7.36ms a. V R = 13.17V b. V R = 3.25V

109 21.29 a. t = 13.44ms b. t = 2.16ms Chapter b 2. c 3. a 4. d 5. b 6. a 7. c 8. c 9. b 10. c 11. a 12. a 13. d 14. b 15. d 16. c 17. a 18. d 19. b 20. a ANSWERS TO PROBLEMS Section 22.1 The Resonance Effect 22.1 The condition of equal and opposite reactances in an LC circuit. Resonance occurs at only one particular frequency, known as the resonant frequency Tuning is the main application of resonance in radio frequency (RF) circuits. Resonant circuits are tuned to the desired resonant frequency to select only the desired signal while rejecting all others X L = X C = 1kΩ Section 22.2 Series Resonance 22.4 a. X = 0Ω b. Z T = 40Ω c. I = 25μA d. θ = 0 0 e. V L = 50mV f. V C = 50mV g. V rs = 1mV

110 22.5 Z T increases as the frequency of the applied voltage increases or decreases from f r. This is because above or below f r, X L and X C are no longer equal and therefore the net reactance, X, is no longer zero. As Z T increases above and below f r, therefore I decreases from its maximum value at f r Because at f r the total impedance, Z T is purely resistive. Section 22.3 Parallel Resonance 22.7 a. I L = 5mA b. I C = 5mA c. I X = 0mA d. Z EQ = 100kΩ 22.8 As the frequency of the applied voltage increases or decreases from f r, X L and X C are no longer equal. As a result, the branch currents I L and I C will not completely cancel since they are not equal. Therefore, I T increases above and below f r which causes Z EQ to decrease from its maximum value at resonance. Section 22.4 Resonant Frequency f r = 2π 1 LC 22.9 a. f r = 400kHz b. f r = 5MHz L = 35.1μH C = 70.17pF From 5.63MHz to 1.78MHz C = pF With C set to 360pF, f r = 1.875MHz. To double f r C must be reduced to 90pF

111 Section 22.5 Q Magnification Factor of Resonant Circuit a. Q = 25 b. Q = a. f r = 1MHz b. Q = 100 c. V L = V C = 1V Q = Because as the frequency increases the ac resistance of the coil will also increase to some degree. This limits the Q of both the coil and the resonant circuit. The increase in the ac resistance represents the losses that occur in a coil at higher frequencies. These losses include skin effect, hysteresis and eddy current losses. Section 22.6 Bandwidth of a Resonant Circuit a. Δ f = 100kHz b. f 1 = 4.95MHz and f 2 = 5.05MHz c. I at f r = 796.2μA, I at f 1 and f 2 = 562.9μA Since Δ f = f r a higher Q corresponds to a narrower bandwidth. Q a. f r = 3MHz b. X L = X C = 942.5Ω c. Z T = 18.85Ω d. I = 2.65μA e. Q = 50 f. V L = V C = 2.5mV g. θ = 0 0 h. Δ f = 60kHz, f 1 = 2.97MHz and f 2 = 3.03MHz i. I = 1.87μA a. f r = 1MHz b. X L = X C = 1.571kΩ c. I L = I C = 1.273mA d. Q = 125 e. Z EQ = kΩ f. I T = 10.18μA g. θ I =

112 h. Δ f = 8kHz, f 1 = 996kHz and f 2 = 1.004MHz

113 Section 22.7 Tuning a. 3.16:1 Section 22.8 Damping of Parallel Resonant Circuits a. Q = 56 b. Δ f = 22.3kHz a. Q = 114 b. Δ f = 8.77kHz R P = 78.54kΩ R P = 196.4kΩ Chapter 23 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. c 2. c 3. a 4. c 5. d 6. b 7. a 8. d 9. c 10. b 11. d 12. a 13. d 14. a 15. b 16. a 17. b 18. c 19. d 20. b 21. b 22. d 23. a 24. b 25. b ANSWERS TO PROBLEMS Section 23.1 Ideal Responses 23.1 b 23.2 d

114 Section 23.2 Filter Circuits 23.3 a. Low-Pass b. High-Pass mV Section 23.3 Low-Pass Filters 23.5 a. The term passband refers to those frequencies below the cutoff frequency of a low-pass filter. Signal frequencies in the passband are allowed to pass from the input to the output of the filter with little or no attenuation. b. The term stopband refers to those frequencies above the cutoff frequency of a low-pass filter. Signal frequencies in the stopband are severely attenuated as they pass through the filter from input to output Because the π-type filter uses an additional inductor and capacitor, it provides a much sharper transition between the stopband and passband. Section 23.4 High-Pass Filters 23.7 Yes, except that for a high-pass filter the passband is above the cutoff frequency and the stopband is below the cutoff frequency Yes, because the T-type filter uses an additional inductor and capacitor

115 Section 23.5 Analyzing Filter Circuits 23.9 a. Low-Pass b. High-Pass c. Low-Pass d. High-Pass a. f C = 3.29kHz. b. f C = 339Hz. c. f C = 5.31kHz. d. f C = 1.15kHz a. V OUT = 49.99mV and θ = b. V OUT = 49.9mV and θ = c. V OUT = 47.84mV and θ = d. V OUT = 35.35mV and θ = -45 e. V OUT = 15.63mV and θ = f. V OUT = 8.12mV and θ = g. V OUT = 1.64mV and θ = a. V OUT = 58.97mV and θ = 88.3 b. V OUT = 291.8mV and θ = 81.6 c. V OUT = 565.9mV and θ = d. V OUT = 1.41V and θ = 45 e. V OUT = 1.89V and θ = f. V OUT = 2V and θ = g. V OUT = 2V and θ =

116 o a. 0 o b. -90 o f N = 4.42kHz Section 23.6 Decibels and Frequency Response Curves a. N db = 3dB b. N db = 10dB c. N db = 13dB d. N db = 16dB a. N db = -3dB b. N db = -10dB c. N db = -60dB d. N db = -20dB The rate of rolloff well above the cutoff frequency of an RC low-pass filter is 6dB/octave or 20dB/decade. At f = 10kHz, N db = -10.1dB At f = 20kHz, N db = dB At f = 100kHz, N db = dB As shown, between 10kHz and 20kHz the filter attenuation increases by 5.69dB or approximately 6dB/octave. Between 10kHz and 100kHz the filter attenuation increases by 19.58dB or approximately 20dB/decade

117 Section 23.7 Resonant Filters The Circuit Q a. Bandpass b. Bandstop c. Bandpass d. Bandstop Chapter 24 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. c 2. d 3. a 4. b 5. a 6. c 7. d 8. d 9. a 10. b 11. a 12. c 13. a 14. c 15. b 16. c 17. d 18. a 19. b 20. d 21. d 22. c 23. d 24. b 25. d 26. c 27. a 28. b 29. a 30. b 31. d 32. c 33. b 34. d 35. a 36. b 37. d 38. c 39. a 40. a ANSWERS TO PROBLEMS Section 24.1 Power Generation rpm 24.2 Increase its rotational speed Section 24.2 Power Distribution

118 1200 watts lost Efficiency = 50% watts lost Section 24.3 Home Electric Wiring watt-hours kwh $ Yes Section 24.4 Three-Phase AC Power V V V V Section 24.6 Electrical Safety

119 ma 100 ma If an accidental short occurs from hot wire to a metal enclosure that is grounded by the green or bare ground wire, the circuit breaker or fuse will blow preventing a shock, if the metal enclosure is touched. Section 24.7 Alternative Energy V 150 ma High cost; large surface area required for sufficient power; need for DC to AC conversion; no power without light so storage is needed High cost; not enough transmission lines to the grid; noise, high maintenance; no power without wind so storage needed. Chapter 25 ANSWERS TO CHAPTER REVIEW QUESTIONS

120 1. e 2. b 3. a 4. d 5. a 6. c 7. b 8. c 9. b 10. c 11. a 12. d 13. d 14. c 15. d 16. b 17. a 18. c 19. b 20. b 21. c 22. a 23. d 24. a 25. b ANSWERS TO PROBLEMS Section 25.2 Principles of Transmission Lines ns MHz meter MHz meters, 6.67 nanoseconds 25.6 Yes MHz Section 25.3 Transmission Line Specifications and Characteristics ohms ns/ft nh db/ft, 0.31/dB/ft watt db watt Section 25.4 How a Transmission Line Works ohms, 1.57 ns/ft μs ohms Infinity degrees per foot Chapter 26 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. a 2. c 3. b 4. d 5. a 6. b 7. c 8. b

121 9. d 10. a 11. c 12. b 13. b 14. d 15. c 16. a ANSWERS TO PROBLEMS Section 26.5 Voltage and Time Measurements Vpp 30.1 μs 33.2 khz 12 mv 50 Hz T = 100 ns t p = 27 ns DC = 27% 26.4 t r = 12 ns degrees ns MHz ns Section 26.8 AC Testing and Troubleshooting Examples 26.9 One possible scenario is to: 1. Do a visual inspection of the wires, connectors and physical connections. 2. Use an ohmmeter to do a continuity check of the cables. 3. Disconnect the speaker and check coil for continuity. 4. Use an oscilloscope to check the power amplifier with a signal at its input, One possible scenario is to: 1. Check to see that AC power is applied. 2. Check all cables, connectors and connections. 3. Use a DMM to check all DC power supply voltages. 4. Apply a voice or tone input to the microphone. Use a scope to observe the amp output. 5. Use the scope to look at the amplifier input/preamp output

122 6. Use the scope to observe the microphone output. Chapter 27 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. b 2. a 3. a 4. c 5. b 6. a 7. d 8. a 9. d 10. c 11. b 12. a 13. d 14. c 15. d 16. b 17. c 18. b 19. c 20. a 21. d 22. d 23. a 24. d 25. b 26. a 27. c 28. b 29. a 30. d 31. a 32. b 33. b 34. c 35. c ANSWERS TO PROBLEMS Section 27.3 Linear, or Analog, Circuits or 34 db mw db db Section 27.6 System Examples Drawing of a basic block diagram representing a system Chapter 28 ANSWERS TO CHAPTER REVIEW QUESTIONS

123 1. d 2. a 3. b 4. b 5. d 6. c 7. a 8. b 9. c 10. a 11. c 12. c 13. b 14. b 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. b 29. d 30. c 31. a 32. a 33. b 34. a 35. b 36. c 37. c 38. a 39. b 40. a 41. a 42. b 43. b 44. c 45. a 46. c 47. d 48. b 49. a 50. c 51. a 52. d ANSWERS TO PROBLEMS a. Semiconductor b. Conductor c. Semiconductor d. Conductor , a. 5 ma b. 5 ma c. 5 ma

124 Chapter 29 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. b 2. b 3. c 4. d 5. a 6. b 7. c 8. c 9. a 10. a 11. d 12. b 13. b 14. d 15. a 16. c 17. b 18. b 19. a 20. c 21. a 22. d 23. a 24. c 25. a 26. b 27. b 28. a 29. c 30. c 31. c 32. b 33. d 34. b 35. b 36. d 37. a ANSWERS TO PROBLEMS Section 29.1 Basic Diode Characteristics ma mw mw Section 29.2 The Ideal Diode 29.4 I L = 20 ma V L = 20 V P L = 400 mw

125 P D = 0 mw P T = 400 mw ma 29.6 V L = 12 V I L = 25.5 ma P L = 306 mw P D = 0 mw P T = 306 mw 29.7 V D = 12 V I D = 0 ma Section 29.3 The Second and Third Approximations 29.8 I L = 19.3 ma V L = 19.3 V P L = 372 mw P D = 13.5 mw

126 P T = 386 mw 29.9 V L = 11.3 V I L = 24 ma P L = mw P D = 29.2 mw P T = mw V D = 12 V I D = 0 ma Section 29.7 The Zener Diode I D = 31 ma I D = 31 ma I D = 33.5 ma Section 29.8 The Loaded Zener Regulator V L = 18 V

127 I S = 41 ma I D = 31 ma I L = 10 ma I D = 31 ma Section 29.9 Zener Specifications P D = 200 mw P Rs = 369 mw P RL = 150 mw P D = 465 mw Section Optoelectronic Devices I D = 5 ma I D = 18 ma I D = 15 ma R = 153 Ω Chapter 30 ANSWERS TO CHAPTER REVIEW QUESTIONS

128 1. b 2. a 3. b 4. c 5. b 6. a 7. b 8. c 9. c 10. b 11. b 12. d 13. c 14. d 15. b 16. a 17. c 18. a 19. a 20. c 21. a 22. c 23. b 24. a 25. d 26. d 27. c 28. c 29. c 30. a 31. c ANSWERS TO PROBLEMS Section 30.2 The Half-Wave Rectifier 30.1 Peak voltage is 70.7 V Average voltage is 22.5 V Dc voltage is 22.5 V 30.2 Peak voltage is 21.2 V Average voltage is 6.74 V Dc voltage is 6.74 V Section 30.3 The Transformer

129 Secondary rms voltage is 1440 V ac Peak voltage is V 30.4 Peak voltage is V Dc voltage is 6.74 V Section 30.4 The Full-Wave Rectifier 30.5 Each half of the secondary has an rms voltage of 15 V and a peak voltage of V Peak output voltage is V Average voltage is 7.71 V Dc voltage is 7.71 V Section 30.5 The Bridge Rectifier 30.7 Peak output voltage is V Average voltage is V Dc voltage is V

130 30.8 Maximum dc output voltage is V Minimum dc output voltage is V Section 30.6 Power Supply Filtering 30.9 Dc output voltage is V Peak-to-peak ripple is 675 mv If the capacitance is cut in half, the denominator is cut in half and the ripple voltage will double 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 goes up by a factor of Dc output voltage is V Peak-to-peak ripple is 334 mv V P(out) = V (This is the dc output voltage due to the capacitor input filter

131 30.14 Peak-to-peak ripple voltage across the capacitor is 25 mv. Section 30.7 Peak Inverse Voltage Peak inverse voltage is V. Section 30.8 Other Power Supply Topics Peak voltage across the secondary winding is 12.6 Vac Dc output voltage is 12.4 V Transformer is being operated at its rated output current. Dc output voltage will be lower than normal The primary current is 150 ma. Section 30.9 Troubleshooting Dc voltage is V 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 from the original V, but the ripple will increase to approximately double because the frequency drops from 120 to 60 Hz

132 30.20 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 V P will remain the same. Dc output equals V P. V ripple = 0 V Section Voltage Multipliers Output voltage is V Chapter 31 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. a 2. c 3. b 4. c 5. b 6. b 7. b 8. b 9. c 10. b 11. c 12. c 13. a 14. b 15. a 16. d 17. c 18. a Chapter 32 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. a 2. d 3. c 4. d 5. b 6. d 7. c 8. a 9. c 10. b

133 11. a 12. c 13. d 14. c 15. a 16. d 17. b 18. d 19. d 20. b 21. a 22. b 23. d 24. c 25. c 26. c 27. b 28. d 29. a 30. b 31. d 32. d 33. b 34. d 35. a 36. c 37. d 38. a 39. b 40. d 41. d 42. b 43. a *Question 36: Reference text is found in the seventh edition of Electronic Principles by Malvino/Bates in Chapter 14 on page 400 listed under, Lack of Thermal Runaway. ANSWERS TO PROBLEMS Section 32.2 Junction FET Basics 32.1 Input resistance is 15 GΩ Section 32.3 Drain Curves 32.2 Maximum drain current is 20 ma Gate-source cutoff voltage is 4 V R DS is 200 Ω

134 32.3 Pinch-off voltage is 2 V R DS is 125 Ω Section 32.4 The Transconductance Curve 32.4 Gate voltage at the ½ cutoff point is 3 V Drain current is 4mA 32.5 Gate voltage at the ½ cutoff point is 2 V Drain current is 2.5 ma 32.6 Drain current is 7.88 ma when the gate voltage is 1 V Drain current is 0.88 ma when the gate voltage is 3 V Section 32.5 Biasing the JFET 32.7 Drain voltage is V 32.8 Gate-source voltage is 1.5 V

135 Drain-source voltage is 11.2 V 32.9 The drain voltage is 12.7 V Gate-source voltage is 2.0 V Drain current is 1.5 ma Gate-source voltage is 5.0 V Drain current is 1 ma Drain-source voltage is 14.8 V Section 32.6 Transconductance Gate-source cutoff voltage is 5 V The g m0 for V GS = 1 V is 3200 μs Drain current is 3 ma Transconductance is 3000 μs Section JFET Amplifiers

136 32.14 Output voltage is 5.46 mv Output voltage is 3.03 mv Section D-MOSFET Curves V GS = 0.5 V, I D = 2.25 ma V GS = 1 V, I D = 1 ma V GS = 1.5 V, I D = 250 ma V GS = 0.5 V, I D = 6.25 ma V GS = 1 V, I D = 9 ma V GS = 1.5 V, ma Section Depletion-Mode MOSFET Drain current is 12 ma Drain-source voltage is 6.36 V Voltage gain is

137 Voltage out is 152 mv R d is 381 Ω Approximate impedance is 1 MΩ Section The Ohmic Region a Drain-source resistance is 10 Ω b Drain-source resistance is 7.5 Ω a Approximate drain current is 157 ma b Approximate drain current is 635 ma Chapter 33 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. b 2. a 3. d 4. a 5. c 6. a 7. b 8. b 9. a 10. b 11. b 12. b 13. b 14. b 15. b 16. b 17. a 18. b 19. a 20. b 21. c 22. a 23. a 24. a

138 25. b 26. a 27. a 28. c 29. c 30. d 31. a 32. b 33. b 34. b 35. c 36. b 37. b 38. c 39. a 40. a 41. c 42. a 43. a 44. d 45. a 46. a 47. c 48. c 49. d 50. c 51. c 52. a 53. c 54. a 55. c 56. d 57. b 58. c 59. a 60. b 61. a 62. b 63. b 64. b 65. a 66. a 67. a 68. c 69. a 70. d 71. a 72. c 73. c 74. a 75. a 76. c 77. d 78. d 79. a 80. d 81. a 82. d 83. c 84. a 85. a 86. b 87. c 88. a 89. a 90. b 91. d 92. b 93. b 94. d 95. c 96. a *Question 52: Reference text is found in the seventh edition of Electronic Principles by Malvino/Bates in Chapter 8 on Page 265 under, Stiff Voltage Divider. ANSWERS TO PROBLEMS Section 33.3 Transistor Currents ma A ma

139 Section 33.5 The Base Curve μa Section 33.6 Collector Curves mw Section 33.7 Reading Data Sheets 33.7 The power dissipation has exceeded the transistor s maximum power rating, and the transistor is damaged and possibly destroyed. Section 33.9 Basic Biasing 33.8 Ideal collector-emitter voltage is 7.38 V Power dissipation is mw Second approximation collector-emitter voltage is 7.64 V Power dissipation is mw Section Emitter Bias 33.9 Collector voltage is 10 V Emitter voltage is 1.8 V

140 V ma Section Voltage Divider Bias Emitter voltage is 3.81 V Collector voltage is V Section Troubleshooting Transistor Circuits a. The approximate collector voltage is 12 V when R 1 is open due to no collector current. b. The approximate collector voltage is 2.93 V when R 2 is open; the transistor is in saturation. CEB can be approximated as a short. c. The approximate collector voltage is 12 V when R E is open due to no collector current. d. The approximate collector voltage is 0.39 V when R C 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 V TH = 2.16 V and a R TH = 27 Ω. This Thevenin circuit has a load of 10 Ω and a diode. Now, solve for a current of

141 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. Section PNP Transistors V Collector-emitter voltage is 4.94 V since the collector is less positive than the emitter. Section Typical Bipolar Amplifiers Ω mv Section Emitter Follower Input impedance of the base is 154 kω Input impedance of the stage is 1.09 kω V

142 Gain is Output voltage is V Impedance of the base is 1.6 MΩ Output voltage is 50 V Section The Differential Amplifier Tail current is 55.6 μa Emitter current is 27.8 μa Quiescent voltage is 10 V Tail current is 60 μa Emitter current is 30 μa Quiescent voltage is 6 V (on the right side) Quiescent voltage is 12 V (on the left side) Output voltage is 518 mv Input impedance is 125 kω

143 33.26 Common-mode voltage gain is 0.5 Common-mode output voltage is 10 μv Output voltage is 286 mv Common-mode voltage is 2.5 mv Common-mode voltage gain is 31.6 Output voltage is 158 μv Section The Transistor Switch a. The transistor is in saturation because the calculated collector current is greater than the saturation current b. The transistor is not in saturation because the calculated collector current is less than the saturation current. c. The transistor is not in saturation because the collector current is less than the saturation current. d. The transistor is not in saturation because the calculated collector current is less than the saturation current

144 33.30 With the switch open, the collector voltage is 5 V. With the switch closed, the collector voltage is 0 V ma ma Chapter 34 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. a 2. b 3. c 4. c 5. b 6. c 7. b 8. c 9. c 10. d 11. c 12. c 13. d 14. a 15. c 16. c 17. b 18. d 19. a 20. c ANSWERS TO PROBLEMS Section 34.2 Frequency Response of an Amplifier 34.1 The frequency response looks like the figure below; the gain at 20 Hz is 196, and at 300 khz is

145 Frequency response for Problem See figure below Frequency response for problem db at a power gain of 5 10 db at a power gain of db at a power gain of db at a power gain of

146 db at a power gain of db at a power gain of db at a power gain of db at a power gain of db at a power gain of 2 13 db at a power gain of db at a power gain of db at a power gain of db at a power gain of db at a power gain of db at a power gain of Section 34.4 Decibel Voltage Gain db 12,

147 34.9 First stage: 31.6 Second stage: a. 41 db b a b db First stage: 28.2 db Second stage: 41.4 db Overall gain: 69.6 db Section 34.5 Impedance Matching

148 Total db voltage gain: 41 db First stage db voltage gain: 23 db Second stage db voltage gain: 18 db Load voltage is 0.56 mv Load power is 1.05 nw mv Section 34.6 Decibels above a Reference mw mv mw is 14 dbm 93.5 mw is 19.7 dbm 4.87 W is 36.9 dbm μv is 120 dbv

149 34.8 mv is 29.2 dbv 12.9 V is 22.2 dbv 345 V is 50.8 dbv Section 34.7 Bode Plots See Figure below Figure for Problem Section 34.8 More Bode Plots See figure below

150 Figure for Problem See Figure below Figure for Problem See Figure below

151 Figure for Problem Section 34.9 Rise-Time-Bandwidth khz MHz μs Voltage gain at 100 khz is 40 db At the 10% point: 0.2 V At the 90% point: 1.8 V The amplifier with the cutoff frequency of 1 MHz has the larger bandwidth. Chapter

152 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. c 2. d 3. a 4. b 5. d 6. a 7. b 8. a 9. b 10. c 11. d 12. d 13. d 14. d 15. d 16. c 17. c 18. c 19. b 20. a 21. c 22. b 23. c 24. d 25. b 26. b 27. d 28. c 29. c 30. a 31. b 32. c 33. d 34. d 35. d 36. b 37. d 38. b 39. a 40. a ANSWERS TO PROBLEMS Section 35.2 The 741 Op Amp μv db or 100, Voltage gain at 1 khz is 19,900 Voltage gain at 10 khz is 2000 Voltage gain at 100 khz is V/μS MHz

153 Section 35.3 The Inverting Amplifier 35.6 Closed-loop voltage gain is 10 Bandwidth is 2 MHz Output voltage at 1 khz is 250 mvpp Output voltage at 10 MHz is 49 mvpp Section 35.4 The Noninverting Amplifier 35.7 Closed-loop gain is 11 Bandwidth is 1.82 khz Output voltage at 100 khz is 275 mvpp Section 35.5 Common Op-Amp Applications 35.8 Output voltage is 740 mvpp and the compensating resistor should be 5 kω 35.9 Output voltage is 50 mvpp Bandwidth is 1 MHz Section 35.7 Differential Amplifiers

154 35.10 Differential voltage gain is 20 Common-mode gain is ± Differential voltage gain is 20 Common-mode gain is ± No, the bridge is not balanced V Section 35.8 Instrumentation Amplifiers Output voltage is 200 mv CMRR is 10, V Chapter 36 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. a 2. c 3. b 4. c 5. a 6. a 7. b 8. c 9. c 10. b

155 11. b 12. a 13. b 14. a 15. c 16. b 17. c 18. d 19. d 20. b 21. d 22. c 23. c 24. a 25. c 26. b 27. b 28. c 29. c 30. b ANSWERS TO PROBLEMS Section 36.1 Comparators with Zero Reference 36.1 An input voltage of 100 μv will produce positive saturation, assuming rail-to-rail output ma 36.3 The output voltage would vary between 0.7 V and 12 V Section 36.2 Comparators with Nonzero References 36.4 Reference voltage is 3.05 V Cutoff frequency is 33.3 Hz % % Section 36.3 Comparators with Hysteresis

156 36.7 Upper trip point is 1.52 V Lower trip point is 1.52 V Hysteresis is 3.05 V 36.8 Upper trip point is V Lower trip point is V Hysteresis is V 36.9 Upper trip point is V Lower trip point is V Hysteresis is 906 mv Section 36.4 The Integrator ma V Section 36.5 Waveform Conversion The output voltage is a triangular wave with a peak-to-peak voltage of 7.8 mv

157 36.13 The output voltage is a triangular waveform with a peak-to-peak voltage of V The duty cycle is 0.5 at the top and 0 at the bottom. Section 36.6 Waveform Generation Hz Section 36.7 Active Diode Circuits The output will be a half-wave signal with a peak voltage of 100 mv mv With the diode reversed, it becomes a negative peak detector and the output voltage is 106 mv mv Section 36.9 First-Order Stages Hz Voltage gain is 5.4 Cutoff frequency is 11.7 khz

158 Voltage gain is 21.4 Cutoff frequency is 10.3 khz Section Bandpass Filters Q is 2.65 Voltage gain is 14 Center frequency is 55.7 khz Section Bandstop Filters Voltage gain is 1.5 Q is 1 Resonant frequency is 15.8 Hz Bandwidth is 15.8 Hz Chapter 37 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. b 2. a 3. c 4. a 5. c 6. b 7. c 8. c 9. b 10. d 11. c 12. d 13. b 14. b 15. b 16. b 17. a 18. c 19. a 20. c

159 21. a 22. d 23. a 24. a 25. b 26. c 27. c 28. q 29. d 30. c 31. b 32. b 33. d 34. c 35. a Chapter 38 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. a 2. b 3. a 4. c 5. d 6. a 7. b 8. d 9. c 10. b 11. a 12. b 13. b 14. d 15. b 16. d 17. b 18. a 19. d 20. b 21. a 22. d 23. c 24. b 25. d 26. a 27. c 28. b a 31. c 32. b 33. a 34. d 35. d 36. a 37. c 38. c 39. a 40. d ANSWERS TO PROBLEMS Section 38.2 The Wien-Bridge Oscillator ,190 Hz Section 38.3 The Colpitts Oscillator

160 Frequency is 1.67 MHz Feedback fraction is 0.10 Minimum gain is Reduce the inductance by a factor of 4 (since there is a square root in the denominator. Section 38.4 Other LC Oscillators MHz Section First overtone is 10 MHz Second overtone is 15 MHz Third overtone is 20 MHz 38.6 Since the frequency is inversely proportional to thickness, if thickness is reduced by 1%, the frequency will increase by 1%. Section 38.6 Crystal Oscillators Hz

161 % Section 38.7 Square-Wave, Pulse, and Function Generators μs 38.9 Frequency is 46.8 khz Duty cycle is Hz Period is 100 μs Quiescent pulse width is 5.61 μs Maximum pulse width is 8.66 μs Minimum pulse width is 3.71 μs Maximum duty cycle is Minimum duty cycle is μh

162 Section 38.9 Frequency Synthesizers khz steps GHz output Chapter 39 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. a 2. b 3. b 4. b 5. c 6. c 7. b 8. a 9. c 10. b 11. b 12. b 13. a 14. c 15. a 16. c 17. b 18. a 19. a 20. c 21. b 22. d 23. a 24. c 25. a 26. c 27. b 28. d 29. a 30. b 31. a 32. d 33. a 34. b 35. c 36. b 37. a 38. d 39. c 40. b ANSWERS TO PROBLEMS Section 39.2 Supply Characteristics e.45% % % %

163 Section 39.4 Series Regulators 39.5 Output voltage is 16.9 V Power dissipation is 1.05 W % V 11.3 V Section 39.5 Linear IC Regulators 39.8 Load current is 750 ma Headroom voltage is 5 V Power dissipation is 3.75 W μv V Maximum efficiency is 83.3% Minimum efficiency is 60% Section 39.7 Switching Regulators

164 V V V V Section 39.8 DC-DC Converters % A V V Ω Chapter 40 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. c 2. b 3. d 4. c 5. b 6. b 7. a 8. b 9. b 10. c 11. a 12. b 13. d 14. d 15. d 16. a 17. b 18. c 19. b 20. c 21. d 22. c 23. a 24. b 25. b ANSWERS TO PROBLEMS Section 40.2 The Silicon-Controlled Rectifier

165 40.1 Output voltage when the SCR is off is 12 V. Input voltage required to turn on the SCR is 5.4 V. Supply voltage required to turn the SCR off is 1.26 V V 40.3 Input voltage required to turn on the SCR is 34.5 V. Supply voltage required to turn the SCR off is 1.17 V Charging time constant is 11.9 ms. Thevenin resistance is 611 Ω Firing angle is 45 Conduction angle is 135 Voltage across the capacitor is 85 Vac 40.6 Minimum firing angle is

166 Maximum firing angle is Minimum conduction angle is 170 Maximum conduction angle is 96.3 Section 40.3 The SCR Crowbar V V Section 40.5 Bidirectional Thyristors V A Chapter 41 ANSWERS TO CHAPTER REVIEW QUESTIONS 1. d 2. a 3. b 4. c 5. c 6. b 7. d 8. a 9. b 10. b 11. c 12. d 13. a 14. c 15. a 16. b 17. c 18. d 19. a 20. d 21. a 22. b 23. c 24. a 25. a

167 ANSWERS TO PROBLEMS Section 41.3 Circuit Troubleshooting 41.1 Transistor defective or R 1 open 41.2 Desired output is 1.8 V The 4 kω feedback resistor may be open. Section 41.4 Power Supply Troubleshooting 41.3 Desired output is +12 V DC C 1 open 41.4 C 1 or C 2 open or must be larger. Add C across 8 V output. Section 41.5 Alternative Troubleshooting Techniques 41.5 Buy the replacement board. It is faster and cheaper than the repair Search the Internet Section 41.6 System Troubleshooting

168 PART B SOLUTIONS TO LAB EXPERIMENTS The material in Part B contains representative data for the experiments in the first 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 using both digital and analog test equipment. Experiment 1 Resistor Code Identification and Ohmmeters (Reference Chapter 3) Steps 1-7 The instructor is responsible for assembling a package of standard axial lead and chip resistors so he or she will know the values and other details. Step 8 Common tolerances for chip resistors are ± 5% and ± 1%. Step 10 When the ohmmeter leads are touched together, the meter reads zero ohms or a short. The resistance of this short is a fraction of an ohm. The resistance of a short piece of wire is near zero or a fraction of an ohm which is too low for the meter to register. Step 11 When the ohmmeter leads are open, the DMM display is a one (1) indicating an infinite resistance or open circuit. Step 12 In this step, the body resistance between the two hands is being measured. The resistance will vary from person to person but will normally be in the 100k ohm to 1 megohm range. When the fingers are wet, there is less resistance. So the reading could be lower than several 100k ohms. ANSWERS TO QUESTIONS 1. The power rating of chip resistors is generally 1/8-watt or ¼-watt. 2. The resistance of a short is typically less than one ohm and near zero ohms in some cases. 3. The theoretical resistance of an open circuit is infinity or in reality, greater than 100 megohms. Experiment 2 Soldering In this experiment, the student was asked to build a kit to practice soldering. Successful construction and demonstration of the kit determines the student s grade. NOTE: A soldering practice kit is also available for surface mount devices (SMD). It can be acquired from Jameco, part number ANSWERS TO QUESTIONS 1. A cold solder joint is one that did not get enough heat to melt the solder so that it adheres to the parts being connected. It is also one that moved and did not

169 solidify completely. Usually, just applying more heat to re-melt the solder is all that is needed to fix a cold joint. 2. Flux 3. The main disadvantage of lead-free solder is that it takes higher temperatures to melt the solder. These temperatures can damage the components being soldered if care is not taken. Experiment 3 Ohm s Law and Multimeters (Reference Chapters 4, 12) Step 2 Step 3 Step 4 Step 5 Step 6 I = 5/1.5k = 3.33 ma Measured value should be very close to calculated value. Resistor tolerance and meter accuracy could account for differences. Current with the 4.7k resistor should be: 5/4.7k = 1.06 ma. Measurement should be close. Increasing the resistance causes the current to decrease according to Ohm s law. Increasing the voltage to 15 volts on the 4.7 k resistor will produce a current of: 15/4.7k = 3.2 ma. Measured value should be close. Increasing the voltage on a resistance increases the current. Decreasing the voltage decreases the current and increasing the voltage increases the current as predicted by Ohm s law. ANSWERS TO QUESTIONS 1. Doubling the voltage and halving the resistance in a circuit should cause current to be four times than previously. 2. To measure current with a DMM, the meter must be in series with the source voltage and the load resistance so that the same current flows through all of them. 3. If the resistance is zero, theoretical current is infinite. This is not possible in reality as very high current will cause components to burn up or a fuse to blow or circuit breaker to trip. Experiment 4 DC Power in Resistive Circuits (Reference Chapter 4) Step 3 Step 4 The resistor may be warm but not hot. The power being dissipated in a 330 ohm, ¼ -watt resistor is mw. So, the rating of the resistor is not being exceeded. Step 5 With 15 volts applied to the 330 resistor, the power dissipated is 0.68 W. So, the power rating is exceeded. The resistor will heat up fast and will probably burn

170 Steps 8 and 9 See table below: Voltage I (ma) 1k P (mw) 1k I (ma) 2.2k P (mw) 2.2k Step 10 See Figure 4-1. Figure 4-1 Step 11 See Figure

171 Figure 4-2 ANSWERS TO QUESTIONS 1. The curves are linear. 2. The curves have a square law function. 3. Three ways to cool: (1) Blow air over the component. (2) Attach a heat sink. (3) Immerse the component in an insulating liquid. Experiment 5 Cells and Batteries (Reference Chapter 10) Step 3 Step 4 Step 5 The type and specifications of the cells will vary depending upon what the instructor supplies. The open circuit voltages will depend upon the batteries provided. See textbook for typical values. Both the potato and lemon should produce a voltage of several tenths of a volt. The copper wire is positive. ANSWERS TO QUESTIONS 1. The output voltage depends on the chemical make-up of the cell. 2. The current capability depends on the volume of chemicals. 3. A dime-cola cell should produce an output of about 0.5 volts

172 Experiment 6 Series Circuits (Reference Chapter 5) Step 2 The total resistance is 97k ohms. Step 3 The measured resistance should be 97k ohms. Step 4 Any measured variation is caused by resistor tolerance and/or ohmmeter error. Step 6 The calculated current with 15 volts applied is ma. Step 7 The measured current should be the same as the calculated value within 10% or less. Step 8 Any difference between calculated and measured values is due to resistor tolerance and/or DMM error. Step 9 The voltage drops are: V, V, 5.12 V and 8.68 V. Step 10 The measured voltage drops should be the same as the calculated values within 10 %. Any differences are due to resistor tolerance and/or DMM error. Step 11 The sum of the voltage drops should equal the applied voltage within 10% or less. Step 12 Power dissipated is 2.32 mw. Step 14 The calculated series resistor is 476 ohms. Step 15 The closest standard value is 470 ohms. Step 17 The voltage across the LED may be less than 2 volts or slightly higher depending upon the LED used. The current should be approximately 20 ma. Step 19 Varying the potentiometer (pot) used as a variable resistor changes the circuit current and therefore varies the LED brightness. ANSWERS TO QUESTIONS volts. An open component in series circuit will show the source voltage. 2. Zero. Since no current flows in the circuit, there are no voltage drops. 3. Rheostat. Experiment 7 Parallel Circuits (Reference Chapter 6) Step 2 R T = 1.182k ohms. Measured value should be within ± 5%. Error may be due to resistor tolerance and/or DMM accuracy. Step 3 I T = 12.7 ma. Measured value should be within ± 5%. Error may be due to resistor tolerance and/or DMM accuracy. Step 4 Calculating the resistance from voltage and current should yield 1.182k ohms as before. Measured value should be within ± 5%. Error may be due to resistor tolerance and/or DMM accuracy. Step 5 Branch currents are: 6.8 ma, 2.73 ma, 2.2 ma, 1 ma. Sum should equal 12.7 ma which validates Kirchhoff s current law. Measured value should be within ± 5%. Error may be due to resistor tolerance and/or DMM accuracy

173 ANSWERS TO QUESTIONS 1. You could use 10kΩ in series with 1kΩ or two 22kΩ resistors in parallel. Measured value should be within ± 5%. Error may be due to resistor tolerance and/or DMM accuracy ohms. Closest standard values are 110 or 120 ohms. With 120 ohms, the error is 6.2 ohms and with 110 ohms, the error is 3.8 ohms or 3.3%. 3. If one parallel resistor opens, the total resistance increases. This causes the total current to decrease. Experiment 8 Maximum Power Transfer (Reference Chapter 8) Steps 2 and 3 Load Resistance (Ω) Current (ma) Voltage Out (V) Power Out (mw) k k k k Step 4 See graph in Figure 8-1. Figure

174 The highest power output was achieved with the 1k ohm resistor which matches the internal resistance of the generator. The highest output voltage occurred with the 10k ohm resistor. ANSWERS TO QUESTIONS 1. B. False 2. Greatest efficiency occurs with the 10k ohm resistor. 3. The power dissipated in the internal resistance is released as heat. Experiment 9 Series-Parallel Circuits (Reference Chapter 7) Step 2 R T = 2192 ohms. Measured value should be within 10% of this value. Step 4 I T = 5.5 ma. Measured value should be within 10% of this value. Step 5 I R2 = 1.71 ma Step 6 V R3 = 2.1 V Step 7 V O = V. Measured value should be within 10% of this value. A = ANSWERS TO QUESTIONS 1. See Figure 9-1. Figure Four groups of four series-connected 100 ohm resistors in parallel ohms 4. 8 ohms Experiment 10 Voltage Dividers (Reference Chapter 7) Step 2 Step 4 V O = 2.64 V. To decrease the output, increase the 22k ohm resistance or decrease the 4.7k ohm resistance. The two resistor values are 300 ohms and 200 ohms. The 2 V output is taken from the 200 ohm resistor

175 Step 5 Closest standard values are 220 and 330 ohms. The output across the 220 ohm resistor would be 2 V. Step 6 The output voltage should be 2 V within 5 %. Step 9 With a supply voltage of 15 volts, the potentiometer (pot) output will vary from zero to 15 volts. Step 11 The calculated values are 20k and 30k which are standard values. Step 12 See Figure Figure 10-1 Step V Step 15 Load Resistance (Ω) Output Voltage (V) 10k k M None, open, infinity Step 16 As the load resistance decreases, the output voltage decreases. Step 17 The voltage division ratio is the same but the resistor values are smaller. Therefore, the output voltage is less affected by the loads. Step 18 There is little or no output voltage variation with any load because the output resistor value (100 ohms) is so much lower than any load resistance. The output remains at about V. Step 19 See Figure

176 Figure 10-2 ANSWERS TO QUESTIONS 1. With no load, the output voltage is 5 V. With the load, the output is 3.9 V. 2. The output should be about half the 17 V or about 7.86 V. The internal resistance of most DMMs is 11M ohm. So, the 15 V divides between the 10M and 11M ohm resistances V Experiment 11 Thermistors (Reference Chapter 3) Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 The thermistor should have a value of 10k ohms or close to it at room temperature. As the thermistor gets cold, its resistance should increase. The actual amount of the increase will depend on how cold it gets in the cold environment you created. It probably did not double in resistance but perhaps increased to 12k to 15 k ohms. As the thermistor gets hot, its resistance decreases. Again, the actual amount of decrease depends on how hot the resistor became. It may have decreased below 7k ohms The voltage across the resistor and thermistor should be about 7.5 V each. Total circuit current should be 0.75 ma. Calculated resistance should be near 10 k ohms. Differences are due to temperature variations. With the 22k ohm resistor, the voltage across it should be 10.3 volts with 4.7 volts across the thermistor. With the thermistor in cold water, the thermistor voltage should rise to 6 or 7 volts depending upon the actual temperature of the water. ANSWERS TO QUESTIONS 1. b. Decrease 2. a. Increase 3. c. Fuse

177 Experiment 12 Photocells (Reference Chapter 3) Step 4 Step 5 Step 6 Step 8 With no light, the resistance of the cell should be near maximum which is about 300 k ohms. With full light intensity, the cell resistance should be near its minimum value of 4k ohms. As the light intensity increases, the resistance decreases and vice versa. If your dark resistance is 250k ohms and your light resistance is 5k ohms, the ratio is 50 to 1. At full light intensity, the output should be about 5 volts. At full dark resistance, the output should be about 14 volts. ANSWERS TO QUESTIONS 1. d 2. a 3. a Experiment 13 Thevenin s Theorem (Reference Chapter 8) Step 1 Calculated values are V TH = V, R TH = ohms, V o = 7.63 V Step 2 Measured V TH should be within 5% percent of the calculated value V. Step 3 Measured V O should be about 7.63 V within 5%. Step 4 Measured R TH should be ohms within 5%. Step 5 Closest standard value is either 1.1k ohm or 1.2k ohm, 1.2k ohm is the most common. See Figure Figure 13-1 Step 7 Output should be about 7.5 to 7.6 V. ANSWERS TO QUESTIONS ohms or 1200 ohms standard value. 2. The Thevenin s resistance changes from about 75 ohms at maximum resistance setting and approaches zero resistance at minimum setting. The Thevenin s resistance peaks at about 267 ohms at mid-point of the potentiometer setting. 3. Zero

178 Experiment 14 Bridge Circuits (Reference Chapter 8) Step 1 R x should be 2.62k ohms. Step 3 Zero Step 4 The 10k potentiometer should measure about 2.62k ohms or be equal to the calculated value within 5%. Steps 6 and 7 Varying the 10k potentiometer (pot) should produce a voltage variation of about 4.8 volts from 0 to 10k ohms. At about the midpoint of the pot, the output will be zero. The polarity depends upon which lead of the DMM is connected to points X and Y. The voltage polarity will reverse at about the mid-point of the 10k pot. Step 9 If the ambient lighting is too dark, the photocell resistance may be too high to balance the circuit. Move it to a brighter area but not too bright. Step 10 Blocking the light increases the photocell resistance so the bridge will go out of balance and the voltage will increase. (The voltage could be negative depending upon how you connected the DMM leads to X and Y.) Step 11 Increasing the light decreases the photocell resistance. This should cause the bridge voltage to switch polarities and increase in the opposite direction. ANSWERS TO QUESTIONS 1. Zero 2. At a higher temperature, the thermistor resistance will decrease and the bridge voltage will go from zero, at balance, to some negative value. 3. If the temperature decreases, the thermistor resistance will increase and the bridge voltage will go to some positive value. NOTE: Answers assume DMM negative or ground lead is on X. Experiment 15 Solar Cells and Batteries (Reference Chapter 10) Step 1 The solar cell module used here had an operating voltage of 3 volts and current of 50 ma. Step 2 6 cells Step 3 About 2 volts or so depending on the light level. Step 4 In bright light, output should be over 3 V and as high as 3.6 V. Step 5 The output voltage varies with the light level, low voltage at low light. Step 6 Output drops to near zero. Some leakage from incomplete coverage will produce a few tenths of a volt. Step 7 Output is highest with maximum brightness and lowest at minimum brightness. Step 8 Output current should match the operating current level in brightest light, in this case 50 ma. The actual internal resistance of the ammeter section of the DMM may decrease the level below maximum operating current

179 ANSWERS TO QUESTIONS 1. See Figure cells in series for 9 volts, four strings in parallel for 120 ma. Figure To charge a battery. 3. High cost and the need to store energy in a battery during no light conditions. Experiment 16 Capacitors and Supercapacitors in DC Circuits (Reference Chapters 15, 21) Step 2 When the wire is touched to the power supply, the capacitor charges to 15 volts. When the wire is touched to the 100 ohm resistor and the LED, the LED lights momentarily because the capacitor discharges into it. Step 4 The time constant is 7.05 seconds. Time for full charge is seconds. In one time constant, the capacitor should charge to 9.5 volts. Step 6 The measured time constants should match the calculated values. However, because of component tolerances, DMM accuracy and error in observation of the timing device, measured values could be off as much as 20%. Step 7 With a 10 µf capacitor, the time constant is 150 ms and the amount of time for full charge is 750 ms. The voltage after one time constant should

180 be 9.5 volts. These times are too short to measure with a stop watch or other timing device. Step 8 A decrease in resistance would decrease the charge time. Step 10 The time constant is seconds. The time for full discharge is 16 seconds. The voltage after one time constant is 5.5 volts. Step 11 The measured time constants should match the calculated values. However, because of component tolerances, DMM accuracy and error in observation of the timing device, measured values could be off as much as 20%. Step 12 Increasing the capacitance would increase the discharge time. Step 14 The time constant is 1000 seconds or minutes. Time for full discharge is 5000 seconds or minutes (1 hour and minutes). Voltage after one time constant is 1.85 volts. Step 15 The measured time constants should match the calculated values. However, because of component tolerances, DMM accuracy and error in observation of the timing device, measured values could be off as much as 20%. ANSWERS TO QUESTIONS 1. Battery F at 11 volts. 3. V c = V(1 e -t/rc ) Charge, V c = Ve -t/rc Discharge Experiment 17 Reed Relays (Reference Chapter 11) Step 2 Step 3 Step 4 Step 6 Step 7 Step 8 With no bar magnet near the relay, the contacts should be open giving a DMM reading of 1 or infinity. Moving the magnet in parallel to the relay should cause the contacts to close giving a DMM reading of zero ohms. The open or closed state of the relay depends on its distance from the relay and its orientation. Parallel orientation is best. The contacts may not close if the magnet is perpendicular to the relay. The farther the distance from the relay, the less effect it has on the contacts. This is also a function of the magnet s strength. The magnet magnetizes the reeds and causes them to become bar magnets. These magnetized reeds attract one another and close the contacts. The pull-in voltage varies with different relays. The one used here has a pull-in voltage of about 9 volts. The drop-out voltage should be about one volt. When the bar magnet is close to the relay, its contacts should be closed. When 12 volts is applied, the relay contacts should open. This effect is present depending upon the strength of the bar magnet, its orientation and whether the relay coil magnetism can overcome the bar magnet

181 ANSWERS TO QUESTIONS 1. b. Closed 2. This depends on the relay coil resistance. The one used here had a resistance of 1050 ohms. The current, with 12 volts applied therefore, is 11.4 ma. 3. b. Experiment 18 DC Motor Demonstration (Reference Chapter 11) Step 1 The resistance of the motor winding will vary with the motor type and size. The one used here had a resistance of 28 ohms. Steps 3 and 4 As you increase the voltage on the motor, the shaft will rotate slowly at the low voltages and faster with increased voltage. At 12 volts, the motor will be at maximum speed. For this motor, the speed is about 5800 rpm. Step 5 Reversing the voltage polarity will change the direction of shaft rotation. Step 6 The operating current will be a function of the motor type and size. The one used here had a current of 0.11 A at 12 V. Step 7 The calculated motor current based on winding resistance should be 0.42 A. This is higher than the actual current. ANSWERS TO QUESTIONS 1. The voltage polarity will only change the direction of rotation and should not affect the speed, if the applied voltage is the same with both polarities. 2. The motor speed is proportional to applied voltage. 3. The operating motor current is less than the current based on the winding resistance alone. The reason for this is that when the motor is rotating, the magnetic field induces a counter voltage into the winding that opposes the applied voltage thereby decreasing the current. Experiment 19 DC Feedback Control Demonstration Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 8 Step 11 Step 14 P = 2.88 W 10k ohms at room temperature (about 72 degrees) The thermistor resistance with decrease with an increase of temperature V. If the thermistor resistance is actually 10k. The voltage could be higher or lower depending on the actual ambient temperature. As temperature increases, the voltage at pin 2 should decrease V The comparator output could be zero or 5 volts depending on the temperature of the thermistor. The voltage divider voltages will be close to one another, about 3.75 volts. The fan is probably off if the ambient temperature is less than 72 degrees. As the thermistor heats up, its resistance decreases thereby decreasing the voltage output of the divider

182 ANSWERS TO QUESTIONS 1. The set point voltage comes from the divider made up of R 1 and R The monitored physical characteristic is heat from the heating element as sensed by the thermistor. 3. The process implemented is comparison of the set point to the measured temperature. This is the function of the comparator IC. Experiment 20 Sine Wave Measurements (Reference Chapter 14) Step V rms. Calculated and measured values should be within 5%. Step 4 1 ms Step 5 1 khz Step mvpp = 177 mv rms, T = ms. 3 Vpp square wave. 15 khz, T = µs. The rms value of a square wave is different than the rms value of a sine wave. The rms calculation measures the area under the sine curve which is different from a square wave. Step 7 10 Vpp = V rms Step 8 I = ma Step 9 Measured and calculated values should have been within 5%. Any error is scope and/or DMM accuracy and measurement reading of the scope. Step 10 With the scope trace set to the center line for zero volts, applying +6 volts DC caused the horizontal trace to move up 6 volts. The -12 V input caused the horizontal trace to move down -12 volts. Step 11 See Figure Figure 20-1 ANSWERS TO QUESTIONS 1. Error sources include resistor tolerance, DMM and scope accuracy and visual measurement inaccuracy on the scope screen. 2. c. rms 3. a. True Experiment 21 Series RC Circuits (Reference Chapters 16, 17)

183 Step 3 X C = 3979 ohms Z = 4547 ohms V C = 7 Vpp V R = 3.87 Vpp θ = Φ = 28.4 Step 4 See Figure 21-1 Figure 21-1 Step 5 V R = 3.87 Vpp Step 6 I = 622 µa rms = 1.76 ma pp Step lead Step 9 V R = 5.8 Vpp Doubling the capacitance halved the reactance to about 1990 ohms so the voltage distribution changes. The impedance decreased making the current greater thereby making the output higher. Step 11 Vc = 7 Vpp Step 12 The values of V C and V R are the same as in Figure 21-1 The component values are the same and so are voltages and frequency. Only the physical configuration of the circuit changed. Step 13 Phase angle should measure The output lags the input. Step 14 The output voltage decreases with an increase of frequency. As frequency increases, capacitive reactance decreases causing less voltage drop. Step Hz. At this frequency, R = X L and the phase shift is 45. ANSWERS TO QUESTIONS 1. 0 and Capacitive reactance decreases with an increase in frequency. Impedance decreases. 3. a. Increase Experiment 22 Parallel RC Circuits (Reference Chapters 16, 17) Step 2 Z = 2541 ohms Step 3 See Figure

184 Figure 22-1 Step 4 I = 1.11 ma, Z = 2.828/.0011 = 2.54 k ohms Step 5 θ = Step 6 Z = 3k ohms, I = 0.93 ma rms, θ = 22.7 Step 7 Decreasing the capacitance, increases the reactance causing current to be less. ANSWERS TO QUESTIONS 1. Component tolerance will produce some error. DMM current measurement may introduce some error as most DMMs do not have a good AC frequency response. 2. No 3. As frequency increases, reactance decreases, and impedance decreases. Therefore, current increases. Experiment 23 Inductance Measurement and RL Circuits (Reference Chapters 18, 19) Step 2 R w = 32 ohms. (Varies with manufacturer) Step 3 R w = 33 mh within 10% Step 5 4 Vpp = V rms Step 6 V L = 0.28 V rms or 0.8 Vpp Step 7 V R = 1.35 V rms or 3.82 Vpp Step 8 R T = 1032 ohms, (1k plus 32 ohms of resistance in the inductor), Z = 1050 ohms, I = 1.35 ma rms, θ = Measured and calculated values will vary because of component tolerances, scope and DMM accuracy and accuracy of setting the frequency. Step 9 See Figure Figure

185 Step 10 L = 35 mh. The coil resistance introduces some error as the voltage across the coil is actually the voltage across the inductance and the winding resistance. Tolerance of the inductor is usually 20%. Step 11 Q of coil = Q of circuit = 0.2 Step 12 R T = 1032 ohms, X L = 1036 ohms, Z = 1.46k ohms, I = ma rms, V R = V rms, V L = V rms, θ = 46. See Figure Figure 23-2 Steps 13 and 14 X L increases with frequency. Therefore, Z increases with frequency making the circuit current less. Step 16 With 5 volts applied, the neon bulb does not light. It has a 60 to 90 volt threshold before it glows. Step 17 When the connection is removed, the magnetic field in the inductor collapses quickly. This action induces a high voltage that is sufficient to light the bulb momentarily. ANSWERS TO QUESTIONS 1. b. Decrease 2. The coil resistance makes the inductor voltage a bit higher than the calculations indicate. 3. About 156 ma Experiment 24 Transformers (Reference Chapter 18) Step 1 Step 3 Step 4 Step 5 Some transformers will have red secondary wires. Primary: 27.8 ohms, Secondary: 2.6 ohms, half of Secondary: 1.3 ohms. Actual values will vary with the manufacturer but these are representative. V s = 30.9 volts. V s (half) = volts. V p = 120 V ± 5 V This is a step down transformer. The red-yellow wire is the secondary center tap. Step 6 Turns ratio is Step 8 V S = 3.88 Vpp Step 9 Turns ratio = 1 : Step 10 Primary and secondary voltages should either be in phase (0 shift) or inverted (180 shift). Assume they are in phase as shown in the Figure

186 Figure 24-1 Step 11 Reversing the scope probe connections causes the secondary voltage to change from 0 to 180 or vice versa. This will depend on how they were initially connected in step 10. See Figure 24-1 (above). ANSWERS TO QUESTIONS ma 2. a. True ohms Experiment 25 RLC Circuits and Resonance (Reference Chapter 22) Step 1 R W = 32.5 ohms (Varies somewhat by manufacturer) Step 3 In order for this step to be valid, the frequency response of the DMM should be greater than about 5 khz. Analog multimeters are usually more accurate when making AC measurements at audio frequencies. V R = 1.29 V, V L = V, V C = 1.02 V Step 4 I = 1.32 ma, Z = 1101 ohms, θ = Circuit is capacitive as X C is greater than X L Step 6 V R = 1.36 V, V L = V, V C = V. I = 1.36 ma, Z = 1036 ohms, θ = 5.2. Circuit is inductive as X L is greater than X C Step 7 See Figure

187 Figure 25-1 Step 8 Step 11 Step 12 Step 13 Step 15 Step 16 Step 17 Step 18 Step 19 f r = khz As the frequency varies, the voltage across the resistor will rise and fall and peak at some point. T = 361 µs, f r = 2.70 khz Component tolerances, frequency accuracy, and error when reading graticule measurement on scope screen. V R = 1.36 V, V L = V, V C = V. Voltage across the inductor is greater because of its resistance. Q L = 12.75, Q circuit = 0.4 at 2kHz, Q L = 19.3, Q circuit = 0.6 at 3kHz I = 1.36 ma, Z = 1036 ohms At resonance, X L = X C. So, the circuit is resistive and no phase shift occurs. There is 0 phase shift at resonance. Below resonance, circuit is capacitive; so current leads voltage. Above resonance, the circuit is inductive; so current lags voltage. ANSWERS TO QUESTIONS 1. a. Inductive 2. a. Maximum 3. c. Band pass filter Experiment 26 Filters (Reference Chapter 23) Step 2 Step 3 Step 5 Low pass filter f c = 3981 Hz As frequency increases, output voltage decreases. Yes, low pass. Steps 6 and 7 Frequency (khz) Output voltage (pp) db 100 Hz

188 See plotted curve in Figure Figure

189 Step 8 f c = 4 khz. This is close to the calculated value. Errors caused by inaccurate scope readings and component tolerances. Step 9 6 db/octave, 20 db/decade Step 10 High pass filter Step 12 Bandpass filter Step Hz Step k ohms (Based on coil resistance of 80 ohms. This varies with coil manufacturer.) Q = 28 Step 15 See curve in Figure 26-2 Figure 26-2 Step 16 BW = 127 Hz ANSWERS TO QUESTIONS 1. Add more RC sections. 2. a. True 3. Reverse positions of the resistor and parallel LC circuits. Experiment 27 Ceramic Band-Pass Filter (Reference Chapter 23) Step 3 Step 4 Step 5 The peak should occur at about 455 khz ± 3kHz. Some filters show three peaks, a major one at the center and minor peaks on either side of center. The output will be about 2 volts. Voltage db attenuation is 6 db. These filters have broad characteristics; so attenuation could be in the 5 to 7 db range. Power db differs for different load resistance values. The output varies sharply and quickly above and below the center frequency indicating a very selective filter

190 Step 6 Step 7 Step 8 At the -3 db points, the frequencies should have indicated approximately 452 khz and 458 khz. The bandwidth is about 6 khz. Most of these filters measure the bandwidth at the 6 db points which should produce a bandwidth of about 8 khz. Some have a bandwidth of up to 13 khz. What you should observe is a curve that looks like the selectivity plot of a band pass filter. This filter should cover a frequency range so that the bandwidth is shown. ANSWERS TO QUESTIONS 1. This filter has an attenuation of about 5 to 6 db. 2. This filter is much more selective than an LC filter. 3. The bandwidth is fixed. Experiment 28 Fourier Theory (Reference Chapter 14) Step 4 Output may be about 2.5 Vpp. Step 5 Attenuation factor would be about 0.5. The actual output will depend on the filter attenuation and this varies with different filters and manufacturers. Step 7 The DC output is zero as the ceramic filter does not pass DC. Step 8 The output may be about 2 Vpp. This depends on filter attenuation. Step 10 Output at the third harmonic may be about 0.7 volts Steps 11 and 12 5 th harmonic output should be about 0.4 volts and the 7 th harmonic should be about 0.28 volts. Step 13 See Figure Figure 28-1 ANSWERS TO QUESTIONS 1. Zero output on even harmonics. 2. Measurement errors in scope and filter

191 3. Triangle waves have all harmonics both odd and even. Experiment 29 Pulse Circuits (Reference Chapter 21) Step 2 High pass filter, differentiator circuit Step 3 T = 20 µs, full charge time = 100 µs Step 5 Most function generators produce a square wave as identified by a 50% duty cycle. Function generators usually generate both AC and DC based square waves. Select DC. Step 6 Rise and fall times will vary with the specific function generator but should be less than 20 ns. Step 8 Period of 100 khz square wave is 10 µs. Half period is 5 µs. Step 9 See Figure Figure 29-1 Step 10 A 10kHz period is 100 µs, 1 khz period is 1 ms or 1000 µs. See Figures 29-2 and Figure

192 Figure 29-3 Step 11 25kHz Step 14 Low pass filter, integrator circuit Step 15 T= 4.7 µs, 23.5 µs Step 17 See Figure Figure 29-4 Step khz period is 100 µs, 100 khz period is 10 µs. See Figures 29-5 and Figure

193 Figure 29-6 Step 20 Changing the resistor and/or capacitor values will change the time constant. The goal in most differentiators is a very short time constant compared to the period of the input to produce short spikes at the leading and trailing edges of the input square wave. For the integrator, the goal is a long time constant compared to the input period. ANSWERS TO QUESTIONS 1. A differentiator filters out all or most of the lower frequencies leaving only the higher harmonics to produce the sharp pulses. The integrator filters out all or most of the higher frequencies leaving only the lower harmonics and the DC response to create the output. 2. AC 3. The goal in most differentiators is a very short time constant compared to the period of the input to produce short spikes at the leading and trailing edges of the input square wave. For the integrator, the goal is a long time constant compared to the input period. Experiment 30 Pulse Averaging with Pulse Width Modulation (Reference Chapter 21) Step 2 Step 4 Step 6 Step 10 Step 11 The measured period should be about 8 ms corresponding to a 125 Hz oscillation frequency. Pulse amplitude should be about 12 volts. The average output should be about 6 volts DC with a 50 % duty cycle. The measured average DC should be about 6 volts as calculated. The average DC with the filter should be a bit less than 6 volts but close. The output waveform should be a low amplitude triangle wave riding on the average DC voltage value. The low pass filter is acting as an integrator to smooth the pulses into a more constant DC. ANSWERS TO QUESTIONS volts µs 3. Filter output variations could be reduced in amplitude by making the low pass filter time constant longer compared to the pulse period

194 Experiment 31 Solar Powered AC System Step 1 Open circuit voltage will vary with light intensity but should be in the 15 to 22 volt range. Step 4 Charge controller input will depend on light level but should be 17 or 18 volts. Step 5 Battery voltage should be greater than 12.6 volts or about 13 to 15 volts. Step 6 AC inverter output should be about 115 to 120 volts. Step 7 Step 8 You would normally expect a sine wave output. Inverters have different waveforms. Some are a square wave others are a pulse wave. It depends on the manufacturer. In any case, the period should be about ms. ANSWERS TO QUESTIONS 1. The open circuit output is much larger than the loaded output. The voltage loss occurs across the internal resistance of the solar cells. 2. No. 3. The battery voltage is greater than the 12.6 volts normal output in order to reverse the current flow in the battery to charge it. Experiment 32 Semiconductor Diodes (Reference Chapters 28, 29) Steps 2 and 6 In this step, the diode is forward biased. The approximate forward voltage drops and circuit currents are: 1N60 (1N34) volts 4.7 ma 1N4148 (1N914) volts 4.3 ma 1N volts 4.2 ma 1N volts 4.6 ma Step 3 The diodes are reverse biased and you should read the supply voltage of 5 V. Steps 10 and 11 See Figure

195 Figure 32-1 Step 12 Step 13 The clipped negative part of the signal should be equal to the forward biased values from Steps 2 and 6. The forward and reverse resistance values of diodes vary widely and with different ohmmeters. The lowest forward resistance was probably from the germanium or Schottky diodes (1N60 and 1N5819). The highest reverse resistances were probably from the 1N4148 or 1N5819. ANSWERS TO QUESTIONS 1. When reversed biased, the voltage across the diode is equal to the applied voltage. 2. The lowest forward voltage is probably from the germanium or Schottky diodes. The highest forward voltage was probably from the silicon diodes (1N4001 or 1N4148) 3. The most unusual waveform was probably from the 1N4001 rectifier diode. It has a poor high frequency response. Experiment 33 Zener Diodes (Reference Chapter 29) Step 2 Step 4 The voltage across the diode should be about 0.7 volts and the resistor voltage about 4.3 volts. The low voltage across the diode indicates forward bias. In this connection, the diode is reversed biased. You will measure the supply voltage from zero up to the breakdown of the zener at its rated voltage of 5.1volts. After that, the diode voltage remains almost constant at about 5.1volts

196 Supply Voltage Diode Voltage Step 5 Step 6 Step 7 Step 8 Step 9 The estimated zener value is 5.1 volts. The zener voltage is 5.1 volts and the resistor voltage is 9.9 volts. The zener current is 2.1 ma. As you reduce the supply voltage from 15 volts, the zener voltage remains the same at about 5 volts. Just below 5 volts the zener no longer breaks down and you simply measure the supply voltage down to zero. Decreasing the supply voltage from 15 to 12 volts had minimal change in the zener voltage which should have remained at about 5.1 volts. Increasing the voltage to 18 volts may have caused a minor zener voltage increase to 5.2 or 5.3 volts. With 15 volts supply voltage, the resistor voltage is 9.9 volts producing a zener current of 2.1 ma. The zener power dissipation is about 10.7 mw, which is much lower than the zener rating of 400 mw. The zener should not be hot. ANSWERS TO QUESTIONS 1. With a 400 mw rating, the maximum current for this zener diode is 78.4 ma. 2. The zener voltage remains essentially constant as the supply voltage varies over a wide range. 3. When the supply voltage is less than the zener breakdown voltage, the voltage across the zener is the supply voltage because the zener is not conducting. Experiment 34 Light Emitting Diodes (Reference Chapter 29) Step 1 Step 3 Step 4 Step 5 Step ohms The LED voltage is about 1.8 volts and the resistor voltage is 3.2 volts. The LED current therefore is about 21.3 ma. The calculated and actual current values are very close. The main difference is the actual LED forward voltage which may vary from about 1.6 to 2.2 volts. Increasing the resistor value reduced the LED current; so its brightness decreased significantly

197 ANSWERS TO QUESTIONS 1. See Figure The resistor limits the current. The standard silicon diode conducts during the negative half cycles of the AC wave protecting the LED from excessive reverse bias. Figure Insert a potentiometer connected as a rheostat, or variable resistor in series with the LED to vary the current which will change the brightness. 3. False Experiment 35 Rectifier Circuits (Reference Chapter 30) Step 2 With no load the rms value will be greater than the 9 volt value under maximum load. The value depends on the specific transformer but is probably 11 to 12 volts. This translates to a peak-to-peak value of 15 to 16 Vpp and about 7.5 to 8 Vp. Step 3 The peak value expected is about 8 volts and an average of about 2.5 volts. Step 4 The measured values should be very close to your calculated values. The main difference is the actual transformer voltage values which can vary from unit to unit. See Figure Figure 35-1 Step 5 Reversing the diode changes the output from positive pulses to negative pulses. Voltage values should be the same but just with a negative polarity

198 Step 8 With the 47 µf capacitor DC output should be about 7.5 volts. The ripple should be in the 3 mv peak-to-peak range. Step 9 With the 1000 µf capacitor the DC output should have been about 8 volts with a ripple of about 130 to 150 µv peak-to-peak. The ripple with the larger capacitor is significantly less. Values vary considerable because of actual capacitor values. Step 11 Assuming a peak voltage of 8 volts, the DC average should be about 5 volts. Step 12 Measured values should be close to your calculated values. Step 13 With the 1000 µf capacitor, the DC output should be near 8 volts and the ripple should be about 60 to 70 µv peak-to-peak. Again, the actual values depend on the value of the capacitor which can have a wide tolerance value. Step 14 The half and full wave rectifiers produce about the same DC output but the half wave has a higher ripple value. The higher the capacitor value the lower the ripple voltage. ANSWERS TO QUESITONS 1. Assuming a load current of 8 ma, the ripple with the 47 µf capacitor on the half wave rectifier, should be about 2.8 mvpp and about 133 µvpp with the 1000 µf capacitor. In the full wave rectifier the ripple should be about 66 µv with the 1000 µf capacitor. Capacitor tolerance will cause variations. 2. Simply reverse the ground connections on the load. 3. If the diode is shorted, AC would be applied to the electrolytic capacitor reversing its polarity and causing it to short or otherwise be damaged. Experiment 36 Bipolar Transistor Operation (Reference Chapters 31, 33) Step 4 I C = 12.8 ma Step 5 I B = 128 µa Step 6 β = 100 Note: The beta of a 2N3904 varies from a low of about 100 to more than 400; so the results here will vary considerably. For example, if you measured a base resistance of 88k ohms, the base current would be about 128 µa. This will give a beta of 100. Step 8 V CE will be either higher or lower than 6 volts since the new transistor has a different beta. Step 9 Again, you measured the base resistance and calculated the base current. You can find the collector current by measuring the voltage across R C and dividing by the 470 ohm collector resistor. Then you can calculate the new beta. It could be higher or lower than before. Step 10 The beta values you calculated should be in the range of 100 to 400. Step 11 As you vary the base potentiometer resistance, you vary I B and that simulates a small input signal. You should see a rather large variation in V CE as I C changes with I B

199 ANSWERS TO QUESTIONS 1. b. Decreases 2. a. Large 3. I E = 12.8 ma (about the same as I C ) Experiment 37 Bipolar Transistor Amplifier (Reference Chapter 33) Step 2 V b = 2.32 V V e = 1.62 V I c = I e = 1.62 ma V Rc = 7.6 V V c = 7.4 V I b = 16.2 µa (assuming β = 100) Step 3 r e = 15.4 ohms r c = 3.9k ohms A (with emitter capacitor) = 253 A (without emitter capacitor) = 3.84 Step 5 The actual measured DC bias conditions should be within 10% of your calculated values. Calculation approximations, rounding and resistor tolerance will account for any differences. Step 7 The peak-to-peak output should be about volts before distorting. The input voltage should measure about volts peak-to-peak with 12 V peak-to-peak output. Step 8 See Figure Figure 37-1 Step 11 Step 17 Step 23 The peak-to-peak output should be about volts before distorting. Input voltage is about 50 mv peak-to-peak. This gives a gain A of about 240 to 260. Measured values should be within 20% of calculated values k ohms (this is the approximate value based on a β or 100. For higher β, this value will be higher.) 4.7k ohms

200 ANSWERS TO QUESTIONS 1. The emitter capacitor effectively eliminates the AC emitter voltage. The AC emitter voltage is negative feedback that lowers the effective input voltage and reduces the gain. 2. The output voltage clip as the output swings from cut-off to saturation. 3. When the variable potentiometer produces half the open circuit output voltage, its value is equal to the internal impedance of the source which is the output impedance. Or the potentiometer resistance becomes the internal resistance of the generator to produce half the output across the input impedance. Experiment 38 Transistor Switch Inverter (Reference Chapters 31, 32) Step 2 Output is +5 V. Step 3 V CE = 0.3 V, V BE = 0.7 V and V CB = 0.4 V. Step 4 Yes, the transistor is saturated as the gain of the transistor is sufficient to produce saturation with the given base current. Step 5 Minimum h FE is 100. Step 6 I C = 10 ma. I B = 0.43 ma Step 7 Yes, with the minimum h FE the transistor is saturated. Step 8 The waveforms are the opposite of one another or inverted. Step 9 Rise time should be in the 40 ns range and fall time in the 50 ns range. Step 11 With zero gate voltage, the drain voltage should be +12 V. Step 12 Threshold should be about 2 V for the 2N7000 and about 3.5 V with the IRFD110. The drain voltage with V TH applied should be less than 1 V. Step 13 The output is a distorted square wave. ANSWERS TO QUESTIONS 1. The output should be the supply voltage with no load connected. 2. The BJT 3. With a load connected, it would form a voltage divider with the collector or drain resistor so the output would be less than the supply voltage. Experiment 39 Integrated Circuit Amplifier (Reference Chapter 34) Step 3 The actual values depend on the specific version of this IC. The minimal specifications are given here. a. 4 V minimum, 12 V maximum b. 250 mw minimum, 325 mw maximum c. 50 k ohms d. Yes, differential e. Gain 26 db or 46 db with capacitor from pins 1 to 8. f. Pins 4 and 6. g. 8 ohms

201 h. 300 khz Step 5 6 V DC. This is half the supply voltage indicating a linear amplifier. Step 7 Output voltage is about 5 Vpp with input voltage about 250 mvpp. Step 8 Gain = 20 Step mw Step 13 The output waveform clips and appears to be more rectangular. ANSWERS TO QUESTIONS 1. This network acts as a mild low pass filter to reduce the upper frequency limit of the amplifier. 2. This capacitor blocks the DC output from the speaker. 3. Connect a large capacitor across pins 1 and 8. Experiment 40 Op Amp Circuits (Reference Chapter 35) Step 2 a. ± 18 V b. ± 15 V c. 3 MHz d ohms e. 13 V/µs typical, 5 V/µs minimum Step 4 Figure 40-2, Gain = 14.67, V O = +7.33V. V O = 2.93 Vpp, inverted Figure 40-3, Gain = 6.66, V O = +10 V, 6.66 Vpp, non-inverted Figure 40-4, Gain 1, V O = 6 V, 6 Vpp, non-inverted Step 5 Your measured values should match the calculated values within 5%. Step 6 See Figures 40-2, 40-3 and Figure

202 Figure 40-3 Figure 40-4 ANSWERS TO QUESTIONS 1. a. True 2. a. True 3. Mostly bipolar but JFETs at the input and output. Experiment 41 Differential Amplifiers (Reference Chapter 35) Step 2 8 V Step 3 8 V Measured value should be very close to calculated value. Step 4 Gain = 1 Step 5 Measured output = 3.16 mvpp Step 6 A v (cm) =.00158, CMRR = 56 db Note: The actual (measured) output will vary widely depending on resistor values. The results in steps 5 and 6 are representative only. Step 7 Yes, output voltage was reduced. New value of CMRR = 68 db Note: Again, results will vary. ANSWERS TO QUESTIONS db

203 2. Buy 1% resistors. More precision resistors (e.g. 0.1% etc.) are also available at a higher cost. 3. Yes, circuit is more balanced as the variable value for R 4 minimizes some but not all of the tolerance variations. Experiment 42 Active Low Pass Filter (Reference Chapter 36) Step Hz Step 4 See Figure

204 Figure 42-1 Step 5 Roll off rate is approximately 40 db/decade. Step µs Step Hz Step 9 The cut-off is about the same as measured