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Search Home Saved Top Charts Books Audiobooks Magazines News Documents Sheet Music 0 342 views Upload Sign In Join RELATED TITLES 0 Solution-Manual-for-ElectronicPrinciples-8th-Edition-By-MalvinoChapters-1-22.pdf Uploaded by a894881867 Save Embed Share Print 890 Resistor CI (Obsolete) 1 Download of 6 Electronic Devices and laboratorio 8 circuitos Search document Part 1 Electronic Principles Electronic Eighth Edition Chapter 1 Introduction SELF-TEST 12. 1. a 7. b 13. c 19. b 2. c 8. c 14. d 20. c 3. a 9. b 15. b 21. b 4. b 10. a 16. b 22. b 5. d 11. a 17. a 23. c 6. d 12. a 18. b us more insight into how changes in load resis load voltage. It is usually easy to measure open-circuit volta load current. By using a load resistor and mea under load, it is easy to calculate the Theve resistance. PROBLEMS 1-1. V = 12 V RS = 0.1 Ω JOB INTERVIEW QUESTIONS Note: The text and illustrations cover many of the job interview questions in detail. An answer is given to job interview questions only when the text has insufficient information. Solution: RL = 100RS RL = 100(0.1 Ω) RL = 10 Ω 2. It depends on how accurate your calculations need to be. If an accuracy of 1 percent is adequate, you should include the source resistance whenever it is greater than 1 percent of the load resistance. 5. Measure the open-load voltage to get the Thevenin voltage VTH. To get the Thevenin resistance, reduce all sources to zero and measure the resistance between the AB terminals to get RTH. If this is not possible, measure the voltage VL across a load resistor and calculate the load current IL. Then divide VTH – VL by IL to get RTH. 6. The advantage of a 50 Ω voltage source over a 600 Ω voltage source is the ability to be a stiff voltage source to a lower value resistance load. The load must be 100 greater than the internal resistance in order for the voltage source to be considered stiff. Special oﬀer for students: Only $4.99/month. 7. The expression cold-cranking amperes refers to the amount of current a car battery can deliver in freezing weather Master your semester with Scribd & The New York Times Given: Answer: The voltage source will appear sti load resistance of ≥10 Ω. 1-2. Given: RLmin = 270 Ω RLmax = 100 kΩ Solution: (Eq. 1-1) RS < 0.01 RL < Ω 0.01(270 ) RS <Free ΩForon 2.7 RSup Read 30this Days Sign to vote title 1-3. largest internal resistance Answer: Useful The Not useful Ω. haveCancel is 2.7anytime. Given: RS = 50 Ω Home Saved Top Charts Books Audiobooks Magazines News Documents Sheet Music Master your semester with Scribd & The New York Times Special oﬀer for students: Only $4.99/month. Upload Sign In Read Free For 30 Days Cancel anytime. Join Search Home Saved Top Charts Books Audiobooks Magazines News Documents Sheet Music 0 342 views Sign In Upload Join RELATED TITLES 0 Solution-Manual-for-ElectronicPrinciples-8th-Edition-By-MalvinoChapters-1-22.pdf Uploaded by a894881867 Save Embed Share Print 1 Download 1-5. 890 Resistor CI (Obsolete) of 6 Given: Solution: RS = 0.05 Ω I=2A (Eq. 1-4) RL = 0.01RS RL = 0.01(250 kΩ) RL = 2.5 kΩ Answer: The load current is 4.80 mA, and, no, th source is not stiff since the load resistance is not l or equal to 2.5 kΩ. Given: 1-12. V=9V RS = 0.4 Ω Solution: VTH = VR2 (Voltage divider VR2 = VS[(R2)/(R1 + R2)] VR2 = 36 V[(3 kΩ)/(6 kΩ + 3 kΩ)] VR2 = 12 V Solution: (Ohm’s law) I = V /R I = (9 V)/(0.4 Ω) I = 22.5 A (Parallel resistance RTH = [R1R2/R1 + R2] RTH = [(6 kΩ)(3 kΩ)/(6 kΩ + 3 kΩ)] RTH = 2 kΩ Answer: The load current is 22.5 A. Given: Answer: The Thevenin voltage is 12 V, and the T resistance is 2 kΩ. IS = 10 mA RS = 10 MΩ Solution: RL = 0.01 RS RL = 0.01(10 MΩ) RL = 100 kΩ Answer: The current source will appear stiff for load resistance of ≤100 kΩ. 1-8. (Current divider form IL = IT [(RS)/(RS + RL)] IL = 5 mA [(250 kΩ)/(250 kΩ + 10 kΩ)] IL = 4.80 mA Answer: The voltage drop across the internal resistance is 0.1 V. 1-7. laboratorio 8 circuitos Search document Solution: (Ohm’s law) V = IR V = (2 A)(0.05 Ω) V = 0.1 V 1-6. Electronic Devices and 6 kV R1 3 kV R2 6 kV R1 3 kV R2 36 V Given: VTH RLmin = 270 Ω RLmax = 100 kΩ Solution: (Eq. 1-3) RS > 100 RL RS > 100(100 kΩ) RS > 10 MΩ 36 V Answer: The internal resistance of the source is greater than 10 MΩ. 1-9. Given: RS = 100 kΩ Solution: = 0.01R (Eq. 1-4) Master RRyour semester with Scribd = 0.01(100 kΩ) R = 1 kΩ & The New Times The maximum load resistance for the current Answer: York L S L L source to appear is 1 k Ω. Special oﬀer for students: Onlystiff $4.99/month. 1-10. Given: RTH Read Free Foron 30this Days Sign up to vote title (a) Circuit for finding fo VTH in Prob. 1-12. ( b) Circuit finding RTH Prob. 1-12. inUseful Not useful Cancel anytime. 1-13. Given: V 12 V Home Saved Top Charts Books Audiobooks Magazines News Documents Sheet Music Master your semester with Scribd & The New York Times Special oﬀer for students: Only $4.99/month. Upload Sign In Read Free For 30 Days Cancel anytime. Join Search Home Saved Top Charts Books Audiobooks Magazines News Documents Sheet Music 0 342 views Sign In Upload Join RELATED TITLES 0 Solution-Manual-for-ElectronicPrinciples-8th-Edition-By-MalvinoChapters-1-22.pdf Uploaded by a894881867 890 Resistor CI (Obsolete) Save Embed Share Print 1 Download of 6 Electronic Devices and laboratorio 8 circuitos Search document Solution: RTH (Eq. 1-10) RN = RTH RTH = 10 kΩ VTH RL (Eq. 1-12) IN = VTH/RTH VTH = INRN VTH = (10 mA)(10 kΩ) VTH = 100 V Thevenin equivalent circuit for Prob. 1-13. Answer: RTH = 10 kΩ, and VTH = 100 V 1-14. Given: VS = 18 V R1 = 6 kΩ R2 = 3 kΩ RTH VTH 10 kV Solution: VTH = VR2 (Voltage divider formula) VR2 = VS[(R2)/(R1 + R2)] VR2 = 18 V[(3 kΩ)/(6 kΩ + 3 kΩ)] VR2 = 6 V (Parallel resistance formula) RTH = [[((R1 × R2)/(R1 + R2)] RTH = [(6 kΩ × 3 kΩ)/(6 kΩ + 3 kΩ)] RTH = 2 kΩ 100 V Thevenin circuit for Prob. 1-17. 1-18. VTH = 12 V RTH = 2 kΩ Answer: The Thevenin voltage decreases to 6 V, and the Thevenin resistance is unchanged. 1-15. Solution: (Eq. 1-10) Given: RN = RTH RN = 2 kΩ VS = 36 V R1 = 12 kΩ R2 = 6 kΩ IN = VTH/RTH IN = 12 V/2 kΩ IN = 6 mA Solution: Answer: RN = 2 kΩ, and IN = 6 mA VTH = VR2 (Voltage divider formula) VR2 = VS[(R2)/(R1 + R2)] VR2 = 36 V[(6 kΩ)/(12 kΩ + 6 kΩ)] VR2 = 12 V (Parallel resistance formula) RTH = [(R1R2)/(R1 + R2)] RTH = [(12 k Ω)(6 kΩ)/(12 kΩ + 6 kΩ)] RTH = 4 kΩ Answer: The Thevenin voltage is unchanged, and the Thevenin resistance doubles. 1-16. Given (from Prob. 1-12): IN RN 6 mA 2 kV Norton circuit for Prob. 1-18. 1-19. Given: VTH = 12 V RTH = 3 kΩ Master your semester with Scribd Solution: =R & The NewRR York = 3 kΩ Times N TH N Special oﬀer for students: $4.99/month. IN = VOnly TH/RTH IN = 12 V/3 kΩ (Eq. 1-12) Shorted, which would cause load resistor t across the voltage source seeing all of the 1-20. a. R1 is open, preventing any of the voltage the load resistor. b. R2 is shorted, making i zero. Since the load resistor is in parallel w age drop would also be zero. Read Free Foron 30this Days Sign up to vote title 1-21. The battery or interconnecting wiring. 1-22. Useful Not useful anytime. = 2 kΩ RTH Cancel Solution: Home Saved Top Charts Books Audiobooks Magazines News Documents Sheet Music Master your semester with Scribd & The New York Times Special oﬀer for students: Only $4.99/month. Upload Sign In Read Free For 30 Days Cancel anytime. Join Search Home Saved Top Charts Books Audiobooks Magazines News Documents Sheet Music 0 342 views Upload Sign In Join RELATED TITLES 0 Solution-Manual-for-ElectronicPrinciples-8th-Edition-By-MalvinoChapters-1-22.pdf Uploaded by a894881867 Save Embed Share Print 890 Resistor CI (Obsolete) 1 Download of 6 1-31. (Kirchhoff’s law) VS = VRS + VL VRS = VS – VL VRS = 10 V – 9 V VRS = 1 V RS = VRS/IRS RS = 8.33 Ω Given: VS = 30 V VL = 10 V RL > 1 MΩ RS < 0.01RL Solution: IRS = IL = VL/RL IRS = 9 V/75 Ω IRS = 120 mA Answer: The value for R1 and R2 is 1 kΩ. Anot sible solution is R1 = R2 = 4 kΩ. Note: The crit be satisfied for any resistance value up to 4 kΩ both resistors are the same value. Given: VS = 10 V VL = 9 V RL = 75 Ω laboratorio 8 circuitos Search document Answer: If an ideal 12 V voltage source is shorted and provides 150 A, the internal resistance is 80 mΩ. 1-24. Electronic Devices and (since the voltage source must (Eq. 1-1) Solution: RS < 0.01RL RS < 0.01(1 MΩ) RS < 10 kΩ (Ohm’s law) Since the Thevenin equivalent resistance woul series resistance, RTH < 10 kΩ. (Ohm’s law) Assume a value for one of the resistors. Si Thevenin resistance is limited to 1 kΩ, pick a v than 10 kΩ. Assume R2 = 5 kΩ. (Eq. 1-1) RS < 0.01 RL 8.33 Ω < 0.01(75 Ω) 8.33 Ω ≮ 0.75 Ω 1-25. Answer: Disconnect the resistor and measure the voltage. (Voltage divider form VL = VS[R2/(R1 + R2)] R1 = [(VS)(R2)/VL] – R2 R1 = [(30 V)(5 kΩ)/(10 V)] – 5 kΩ R1 = 10 kΩ 1-26. Answer: Disconnect the load resistor, turn the internal voltage and current sources to zero, and measure the resistance. RTH = R1R2/(R1 + R2) RTH = [(10 k Ω)(5 kΩ)]/(10 kΩ + 5 kΩ) RTH = 3.33 kΩ 1-27. Answer: Thevenin’s theorem makes it much easier to solve problems where there could be many values of a resistor. Since RTH is one-third of 10 kΩ, we can use values that are three times larger. Answer: To find the Thevenin voltage, disconnect the load resistor and measure the voltage. To find the Thevenin resistance, disconnect the battery and the load resistor, short the battery terminals, and measure the resistance at the load terminals. R1 = 30 kΩ R2 = 15 kΩ Answer: a. The internal resistance (RS) is 8.33 Ω. b. The source is not stiff since RS ≮ 0.01 RL. 1-28. 1-29. Answer: Note: The criteria will be satisfied as long as R2 and R2 is not greater than 15 kΩ. 1-32. Answer: First, measure the voltage across the te This is the Thevenin voltage. Next, connect the ter to the battery terminals—measure the curren use the values above to find the total resistance. subtract the internal resistance of the ammeter f result. This is the Thevenin resistance. 1-33. Answer: First, measure the voltage across the te This is the Thevenin voltage. Next, connect a across terminals. Next, measure the voltag Read Free Foron 30 Days Signthe up to vote this title the resistor. Then, calculate the current through Not useful Useful resistor. Then, subtract the load voltage from the Cancel anytime. nin voltage. Then, divide the difference voltag current. The result is the Thevenin resistance. Given: RL = 1 kΩ I = 1 mA Solution: RS > 100RL RS > 100(1 kΩ) RL > 100 kΩ Master your semester with Scribd V = IR kΩ)Times V = (1 mA)(100 & The New York = 100 V V Special oﬀer for students: Only $4.99/month. Answer: A 100 V battery in series with a 100 kΩ resistor. Home Saved Top Charts Books Audiobooks Magazines News Documents Sheet Music Master your semester with Scribd & The New York Times Special oﬀer for students: Only $4.99/month. Upload Sign In Read Free For 30 Days Cancel anytime. Join Search Home Saved Top Charts Books Audiobooks Magazines News Documents Sheet Music 0 342 views Sign In Upload Join RELATED TITLES 0 Solution-Manual-for-ElectronicPrinciples-8th-Edition-By-MalvinoChapters-1-22.pdf Uploaded by a894881867 Save Embed Share Print 890 Resistor CI (Obsolete) 1 Download of 6 Electronic Devices and 2-4. 2-5. 2-6. 2-7. Trouble: 1-38. No supply voltage 1-39. R4 open 1-40. R2 shorted a. p-type n-type c. p-type d. n-type e. p-type Given: Barrier potential at 25°C is 0.7 V Tmin = 25°C Tmin = 75°C 1: R1 shorted 2: R1 open or R2 shorted 3: R3 open 4: R3 shorted 5: R2 open or open at point C 6: R4 open or open at point D 7: Open at point E 8: R4 shorted R2 open 5 mA 5 mA 5 mA b. Answer: 0, IL = 24.7 μA; 1 kΩ, IL = 21.1 μA; 2 kΩ, IL = 18.5 μA; 3 kΩ, IL = 16.4 μA; 4 kΩ, IL = 14.8 μA; 5 kΩ, IL = 13.5 μA; 6 kΩ, IL = 12.3 μA. 1-37. a. c. IL = 0.148 V/(6 kΩ + 6 kΩ) IL = 12.3 μA R1 shorted 500,000 free electrons b. IL = 0.148 V/(6 kΩ + 5 kΩ) IL = 13.5 μA 1-36. Search document IL = 0.148 V/(6 kΩ + 4 kΩ) IL = 14.8 μA 1-35. laboratorio 8 circuitos Solution: ΔV = (–2 mV/°C) ΔT (Eq. 2-4) ΔV = (–2 mV/°C)(0°C – 25°C) ΔV = 50 mV Vnew = Vold + ΔV Vnew = 0.7 V + 0.05 V Vnew = 0.75 V ΔV = (–2 mV/°C) ΔT (Eq. 2-4) ΔV = (–2 mV/°C)(75°C – 25°C) ΔV = –100 mV ΔV Vnew = Vold + ΔV Vnew = 0.7 V – 0.1 V Vnew = 0.6 V Chapter 2 Semiconductor Semiconductors s Answer: The barrier potential is 0.75 0.6 V at 75°C. SELF-TEST 1. d 15. a 29. d 42. b 2. a 16. b 30. c 43. b 3. b 17. d 31. a 44. c 4. b 18. d 32. a 45. a 5. d 19. a 33. b 46. c 6. c 20. a 34. a 47. d 7. b 21. d 35. b 48. a 8. b 22. a 36. c 49. a 9. c 23. a 37. c 50. d 10. a 24. a 38. a 51. c 11. c 25. d 39. b 52. b 12. c 26. b 40. a 53. d 13. b 27. b 41. b 54. b 14. b 28. a Master your semester with Scribd & The New York QUESTIONS Times JOB INTERVIEW Special oﬀer for students: Only $4.99/month. 9. Holes do not flow in a conductor. Conductors allow current flow by virtue of their single outer-shell electron, which is 2-8. Given: IS = 10 nA at 25°C Tmin = 0°C – 75°C Tmax = 75°C Solution: IS(new) = 2(ΔT/10)IS(old) (Eq. 2-5) [(0°C – 25°C)/10] IS(new) = 2 IS(new) = 1.77 nA 10 nA IS(new) = 2(∆T/10) IS(old) [(75°C – 25°C) /10)] IS(new) = 2 IS(new) = 320 nA (Eq. 2-5) 10 nA is 1.77 n Answer: The saturation current Read Free Foron 30this Days Sign up to vote title 320 nA at 75°C. 2-9. Useful Given: Not useful Cancel anytime. ISL = 10 nA with a reverse voltage of 10 V New reverse voltage = 100 V Home Saved Top Charts Books Audiobooks Magazines News Documents Sheet Music Master your semester with Scribd & The New York Times Special oﬀer for students: Only $4.99/month. Upload Sign In Read Free For 30 Days Cancel anytime. Join Search Home Saved Top Charts Books Audiobooks Magazines News Documents Sheet Music 0 342 views Sign In Upload Join RELATED TITLES 0 Solution-Manual-for-ElectronicPrinciples-8th-Edition-By-MalvinoChapters-1-22.pdf Uploaded by a894881867 Save Embed Share Print 1 Download 2-13. R1 open 2-14. D1 shorted 2-15. D1 open 2-16. V1 = 0 V 890 Resistor CI (Obsolete) of 6 laboratorio 8 circuitos Search document 3-3. Given: Solution: Since the diodes are in series, the through each is the same. Answer: 400 mA 3-4. SELF-TEST Given: VS = 20 V VD = 0 V RL = 1 kΩ 7. c 13. a 18. b 2. b 8. c 14. d 19. a 3. c 9. a 15. a 20. b 4. d 10. a 16. c 21. a Solution: 5. a 11. b 17. b 22. c 6. b 12. b (Kirchhoff’s law) VS = VD + VL 20 V = 0 V + VL VL = 20 V JOB INTERVIEW QUESTIONS 4. 7. 8. 10. 11. If you have a data sheet, look up the maximum current rating and the breakdown voltage. Then, check the schematic diagram to see whether the ratings are adequate. If they are, check the circuit wiring. Measure the voltage across a resistor in series with the diode. Then, divide the voltage by the resistance. With the power off, check the back-to-front ratio of the diode with an ohmmeter or use the diode test function on a DMM. If it is high, the diode is OK. If it is not high, disconnect one end of the diode and recheck the back-to-front ratio. If the ratio is now high, the diode is probably OK. If you are still suspicious of the diode for any reason, the ultimate test is to replace it with a known good one. Connect a diode in series between the alternator and the battery for the recreational vehicle. The diode arrow points from the alternator to the RV battery. This way, the alternator can charge the vehicle battery. When the engine is off, the diode is open, preventing the RV battery from discharging. Use a voltmeter or oscilloscope for a diode in the circuit. Use an ohmmeter, DMM or curve tracer when the diode is out of the circuit. (Ohm’s law) IL = VL/RL IL = 20 V/1 kΩ IL = 20 mA PL = (IL)(VL) PL = (20 mA)(20 V) PL = 400 mW PD = (ID)(VD) PD = (20 mA)(0 V) PD = 0 mW PT = PD + PL PT = 0 mW + 400 mW PT = 400 mW Answer: IL = 20 mA VL = 20 V PL = 400 mW PD = 0 mW PT = 400 mW 3-5. Solution: Master Ryour = 220 Ω semester with Scribd V=6V & The New Solution:York Times Given: (Ohm’s law) IL = VL/RL / Ω 20 V 2 k IL =Read Free Foron 30this Days Sign up to vote title IL = 10 mA Useful Not useful mA anytime. Answer: 10Cancel I= V /R Special oﬀer for students: Only $4.99/month. I = 6 V/220 Ω Given: VS = 20 V VD = 0 V RL = 2 kΩ PROBLEMS 3-1. VD1 = 0.75 V VD2 = 0.8 V ID1 = 400 mA Chapter 3 Diode Theory 1. b Electronic Devices and 3-6. Given: Home Saved Top Charts Books Audiobooks Magazines News Documents Sheet Music Master your semester with Scribd & The New York Times Special oﬀer for students: Only $4.99/month. Upload Sign In Read Free For 30 Days Cancel anytime. Join