Eletronica 1 Malvino 7ed Respostas (em ingles)
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Eletronica 1 Malvino 7ed Respostas (em ingles)


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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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