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V ac/7 V2 = 17.14 V ac VP = 1.414 Vrms VP = 1.414 (17.14 V ac) VP = 24.24 V VP(in) = 0.5 VP VP(in) = 0.5(24.24 V) VP(in) = 12.12 V Vp(out) = Vp(in) (Eq. 4-1) Vp(out) = 12.12 V (This is the dc output voltage due to the capacitor input filter.) 1-13 I = V/R (Ohm\u2019s law) I = 12.12 V/2.2 k\u2126 I = 5.51 mA fout = 2fin (Eq. 4-7) fout = 2(60 Hz) fout = 120 Hz VR = I/(fC) (Eq. 4-10) VR = (5.51 mA)/[(120 Hz)(68 µF)] VR = 675 mV Answer: The dc output voltage is 12.12 V, with a 675 mV pp ripple. 4-19. Answer: VR = I/(fC) (Eq. 4-10) If the capacitance is cut in half, the denominator is cut in half and the ripple voltage will double. 4-20. Answer: VR = I/(fC) (Eq. 4-10) If the resistance is reduced to 500 \u2126, the current increases by a factor of 20; thus the numerator is increased by a factor of 20 and the ripple voltage goes up by a factor of 20. 4-21. Given: Turns ratio = N1/N2 = 9:1 = 9 V1 = 120 V ac RL = 1 k\u2126 C = 470 µF fin = 60 Hz Solution: V2 = V1/(N1/N2) (Eq. 4-5) V2 = 120 V ac/9 V2 = 13.33 V ac VP = 1.414 Vrms VP = 1.414 (13.33 V ac) VP = 18.85 V Vp(out) = Vp (Eq. 4-1) Vp(out) = 18.85 V (This is the dc output voltage due to the capacitor input filter.) I = V/R (Ohm\u2019s law) I = 18.85 V/1 k\u2126 I = 18.85 mA fout = 2fin (Eq. 4-7) fout = 2(60 Hz) fout = 120 Hz VR = I/(fC) (Eq. 4-10) VR = (18.85 mA)/[(120 Hz)(470 µF)] VR = 334 mV Answer: The dc output voltage is 18.85 V, with a 334 mV pp ripple. Output waveform for Probs. 4-18 and 4-21. 4-22. Given: Turns ratio = N1/N2 = 9:1 = 9 V1 = 105 V ac RL = 1 k\u2126 C = 470 µF fin = 60 Hz Solution: V2 = V1/(N1/N2) (Eq. 4-5) V2 = 105 V ac/9 V2 = 11.67 V ac VP = 1.414 Vrms VP = 1.414 (11.67 V ac) VP = 16.50 V Vp(out) = VP (Eq. 4-1) Vp(out) = 16.50 V (This is the dc output voltage due to the capacitor input filter.) 4-23. Given: VP = 18.85 VP from Prob. 4-21 Solution: PIV = VP (Eq. 4-13) PIV = 18.85 V Answer: The peak inverse voltage is 18.85 V. 4-24. Given: Turns ratio = N1/N2 = 3:1 = 3 V1 = 120 Vrms Solution: V2 = V1/(N1/N2) (Eq. 4-5) V2 = 120 Vrms/3 V2 = 40 Vrms VP = (1.414) (Vrms) VP = (1.414) (40 Vrms) VP = 56.56 VP PIV = VP (Eq. 4-13) PIV = 56.56 V Answer: The peak inverse voltage is 56.56 V. 4-25. Solution: From the information on p. 114. a. Secondary output is 12.6 V ac. VP = 1.414 Vrms VP = 1.414 (12.6 V ac) VP = 17.8 V b. Vdc = 17.8 V c. I = V/R (Ohm\u2019s law) Idc = 17.8 V ac/1 k\u2126 Idc = 17.8 mA Rated current is 1.5 A. Answer: The peak output voltage is 17.8 V, and the dc output voltage is 17.8 V. It is not operating at rated current, and thus the secondary voltage will be higher. 4-26. Given: Assume Pin = Pout Vdc = 17.8 V from Prob. 4-25 Idc = 17.8 mA from Prob. 4-25 Solution: Pout = Idc Vdc Pout = (17.8 mA)(17.8 V) Pout = 317 mW Pin = 317 mW Pin = V1Ipri Ipri = Pin/V1 Ipri = 317 mW/120 V Ipri = 2.64 mA Answer: The primary current would be 2.64 mA. 4-27. Given: VDC = 21.2 V from Prob. 4-17 VDC = 12.12 V from Prob. 4-18 Solution: Fig. 4-40(a) 1-14 Idiode = V/R Idiode = (2.12 V)/(10 k\u2126) Idiode = 2.12 mA Fig. 4-40(b) I = V/R I = (12.12 V)/(2.2 k\u2126) I = 5.5 mA Idiode = 0.5 I Idiode = (0.5)/(5.5 mA) Idiode = 2.75 mA Answer: The average diode current in Fig. 4-40(a) is 212 \u3bcA and the current in Fig. 4-40(b) is 2.75 mA. 4-28 Given: Idc = 18.85 mA from Prob. 4-21 Solution: Idiode = (0.5)Idc Idiode = (0.5)(18.85 mA) Idiode = 9.43 mA 4-29 Given: Vp(out) = 18.85 V from Prob. 4-21 Solution: Without the filter capacitor to maintain the voltage at peak, the dc voltage is calculated the same way it would be done if the filter was not there. Vdc = 0.636 VP Vdc = 0.636(18.85 V) Vdc = 11.99 V Answer: The dc voltage is 11.99 V. 4-30. Answer: With one diode open, one path for current flow is unavailable. The output will look similar to a half- wave rectifier with a capacitor input filter. The dc voltage should not change from the original 18.85 V, but the ripple will increase to approximately double because the frequency drops from 120 to 60 Hz. 4-31. Answer: Since an electrolytic capacitor is polarity- sensitive, if it is put in backward, it will be destroyed and the power supply will act as if it did not have a filter. 4-32. Answer: VP will remain the same, DC output equals VP, Vripple = 0 V. 4-33. Answer: Since this is a positive clipper, the maximum positive will be the diode\u2019s forward voltage, and all the negative will be passed through. Maximum positive is 0.7 V, and maximum negative is \u201350 V. Output waveform for Prob. 4-33. 4-34. Answer: Since this is a negative clipper, the maximum negative will be the diode\u2019s forward voltage, and all the positive will be passed through. The maximum positive is 24 V, and the maximum negative is \u20130.7 V. Output waveform for Prob. 4-34. 4-35. Answer: The limit in either direction is two diode voltage drops. Maximum positive is 1.4 V, and maximum negative is \u20131.4 V. 4-36. Given: DC voltage 15 V R1 = 1 k\u2126 R2 = 6.8 k\u2126 Solution: Voltage at the cathode is found by using the voltage divider formula. Vbias = [R1/(R1 + R2)]Vdc (Eq. 4-18) Vbias = [1 k\u2126/(1 k\u2126 + 6.8 k\u2126)]15 V Vbias = 1.92 V The clipping voltage is the voltage at the cathode and the diode voltage drop. Vclip = 1.92 V + 0.7 V Vclip = 2.62 V Answer: Since it is a positive clipper, the positive voltage is limited to 2.62 V and the negative to \u201320 V. Output waveform for Prob. 4-36. 4-37. Answer: The output will always be limited to 2.62 V. 4-38. Answer: Since this is a positive clamper, the maximum negative voltage will be \u20130.7 V and the maximum positive will be 29.3 V. Output waveform for Prob. 4-38. 4-39. Answer: Since this is a negative clamper, the maximum positive voltage will be 0.7 V and the maximum negative will be \u201359.3 V. Output waveform for Prob. 4-39. 4.40. Answer: The output will be 2VP or Vpp, which is 40 V. If the second approximation is used, the maximum for the clamp will be 39.3 V instead of 40 V, and since there is also a diode voltage drop, the output would be 38.6 V. Output waveform for Prob. 4-40. 4-41. Given: Turns ratio = N1/N2 = 1:10 = 0.1 V1 = 120 V ac Solution: V2 = V1/(N1/N2) (Eq. 4-5) V2 = 120 V ac/0.1 V2 = 1200 V ac VP = 1.414 Vrms VP = 1.414 (1200 V ac) VP = 1696.8 V Since it is a doubler, the output is 2VP. Vout = 2VP Vout = 2 (1696.8 V) Vout = 3393.6 V Answer: The output voltage will be 3393.6 V. 4-42. Given: Turns ratio = N1/N2 = 1:5 = 0.2 V1 = 120 V ac 1-15 Solution: V2 = V1/(N1/N2) (Eq. 4-5) V2 = 120 V ac/0.2 V2 = 600 V ac VP = 1.414 Vrms VP = 1.414 (600 V ac) VP = 848.4 V Since it is a tripler, the output is 3VP. Vout = 3VP Vout = 3 (848.4 V) Vout = 2545.2 V Answer: The output voltage will be 2545.2 V. 4-43. Given: Turns ratio = N1/N2 = 1:7 = 0.143 V1 = 120 V ac Solution: V2 = V1/(N1/N2) (Eq. 4-5) V2 = 120 V ac/0.143 V2 = 839.2 V ac VP = 1.414 Vrms VP = 1.414 (839.2 V ac) VP = 1186.6 V Since it is a quadrupler, the output is 4VP. Vout = 4VP Vout = 4(1186.6 V) Vout = 4746.4 V Answer: The output voltage will be 4746.4 V. CRITICAL THINKING 4-44. Answer: If one of the diodes shorts, it will provide a low resistance path to either blow a fuse or damage the other diodes. 4-45. Given: Turns ratio = N1/N2 = 8: 1 = 8 V1 = 120 V ac Solution: V2 = V1/(N1/N2) (Eq. 4-5) V2 = 120 V ac/8 V2 = 15 V ac VP = 1.414 Vrms VP = 1.414 (15 V ac) VP = 21.21 V Since each resistor is in the same current path and both have the same value, they equally divide the voltage. Since they both have

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