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


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12-9. Given: 
R1 = 400 \u2126 
R2 = 200 \u2126 
RC = 200 \u2126 
RE = 136 \u2126 
RL = 200 \u2126 
VCC = 30 V 
VBE = 0.7 V 
Solution: 
rc = RC||RL 
rc = 200 \u2126||200 \u2126 
rc = 100 \u2126 
Answer: The ac collector resistance is 100 \u2126. 
 12-10. Given: 
R1 = 600 \u2126 
R2 = 300 \u2126 
RC = 300 \u2126 
RE = 204 \u2126 
RL = 300 \u2126 
VCC = 30 V 
VBE = 0.7 V 
Solution: 
VBB = [R2/(R1 + R2)]VCC 
1-56 
VBB = [300 \u2126/(600 \u2126 + 300 \u2126)]30 V 
VBB = 10 V 
VE = VBB \u2013 VBE 
VE = 10 V \u2013 0.7 V 
VE = 9.3 V 
IE = ICQ = VE/RE 
ICQ = 9.3 V/204 \u2126 
ICQ = 45.59 mA 
VC = VCC \u2013 RCICQ 
VC = 30 V \u2013 (300 \u2126)(45.59 mA) 
VC = 16.3 V 
VCEQ = VC \u2013 VE 
VCEQ = 16.3 V \u2013 9.3 V 
VCEQ = 7 V 
MP = VCEQ = 7 V 
MPP = 2MP = 14 V 
or 
MP = ICQrc 
MP = (45.59 mA) (150 \u2126) 
MP = 6.85 V 
MPP = 2MP = 13.7 V 
Answer: The maximum peak-to-peak voltage is 13.7 V. 
 12-11. Given: 
Pout = 2 W 
Pin = 4 mW 
Solution: 
AP = Pout/Pin (Eq. 12-12) 
AP = 2 W/4 mW 
AP = 500 
Answer: The power gain is 500. 
 12-12. Given: 
Vout = 15 V pp 
RL = 1 k\u2126 
Pin = 400 µW 
Solution: 
Pout = V2/8R 
Pout = (15 V)2/8 k\u2126 
Pout = 28.1 mW 
AP = Pout/Pin (Eq. 12-12) 
AP = 28.1 mW/400 µW 
AP = 70.3 
Answer: The power gain is 70.3. 
 12-13. Given: 
R1 = 2 k\u2126 
R2 = 470 \u2126 
RC = 680 \u2126 
RE = 220 \u2126 
RL = 2.7 k\u2126 
VCC = 15 V 
VBE = 0.7 V 
RG = 50 \u2126 
VBB = 2.85 V (from Prob. 12-2) 
Solution: 
Ibias = VCC/(R1 + R2) 
Ibias = 15 V/(2 k\u2126 + 470 \u2126) 
Ibias = 6.07 mA 
VE = VBB \u2013 VBE (Eq. 8-2) 
VE = 2.85 V \u2013 0.7 V 
VE = 2.15 V 
IE = VE/RE (Eq. 8-3) 
IE = 2.15 V/220 \u2126 
IE = 9.77 mA 
Idc = Ibias + IE 
Idc = 6.07 mA + 9 77 mA 
Idc = 15.84 mA 
Answer: The current drain is 15.84 mA. 
 12-14. Given: 
Idc = 15.84 mA (from Prob. 12-13) 
VCC = 15 V 
Solution: 
Pdc = IdcVCC (Eq. 12-17) 
Pdc = (15.84 mA)(15 V) 
Pdc = 237.6 mW 
Answer: The dc input power is 237.6 mW. 
 12-15. Given: 
MPP = 10.62 V (from Prob. 12-3) 
RL = 2.7 k\u2126 
Pdc = 237.6 mW (from Prob. 12-14) 
Solution: 
Pout(max) = MPP2/8RL (Eq. 12-15) 
Pout = (10.62 V)2/8(2.7 k\u2126) 
Pout = 5.22 mW 
\u3b7 = [Pout/Pin]100% 
\u3b7 = [5.22 mW/237.6 mW]100% 
\u3b7 = 2.2% 
Answer: The efficiency is 2.2%. 
 12-16. Given: 
ICQ = 9.77 mA (from Prob. 12-2) 
VCEQ = 6.21 V (from Prob. 12-3) 
Solution: 
PDQ = VCEQ ICQ (Eq. 12-16) 
PDQ = (6.21 V)(9.77 mA) 
PDQ = 60.7 mW 
Answer: The quiescent power dissipation is 60.7 mW. 
 12-17. Given: 
R1 = 200 \u2126 
R2 = 100 \u2126 
RC = 100 \u2126 
RE = 68 \u2126 
RL = 100 \u2126 
VCC = 30 V 
VBE = 0.7 V 
VBB = 10 V (from Prob. 12-7) 
Solution: 
Ibias = VCC/(R1 + R2) 
Ibias = 30 V/(200 \u2126 + 100 \u2126) 
Ibias = 100 mA 
VE = VBB \u2013 VBE 
VE = 10 V \u2013 0.7 V 
VE = 9.3 V 
IE = VE/RE 
IE = 9.3 V/68 \u2126 
IE = 136.8 mA \u2248 137 mA 
1-57 
Idc = Ibias + IE 
Idc = 100 mA + 137 mA 
Idc = 237 mA 
Answer: The current drain is 237 mA. 
 12-18. Given: 
Idc = 2.37 mA (from Prob. 12-17) 
VCC = 30V 
Solution: 
Pdc = IdcVCC 
Pdc = (237 mA)(30 V) 
Pdc = 7.11 W 
Answer: The dc input power is 7.11 W. 
 12-19. Given: 
MPP = 2MP = 13.7 V (from Prob. 12-10) 
Pdc = 7.11 W (from Prob. 12-18) 
RL = 100 \u2126 
Solution: 
Pout(max) = MPP2/8RL 
Pout = (13.7 V)2/8(100 \u2126) 
Pout = 235 mW 
\u3b7 = [Pout/Pin]100% 
\u3b7 = [235 mW/7.11 W]100% 
\u3b7 = 3.3% 
Answer: The efficiency is 3.3%. 
 12-20. Given: 
ICQ = 137 mA (from Prob. 12-7) 
VCEQ = 7 V (from Prob. 12-8) 
Solution: 
PDQ = VCEQICQ 
PDQ = (7 V)(137 mA) 
PDQ = 960 mW 
Answer: The quiescent power dissipation is 960 mW. 
 12-21. Given: 
R1 = 10 \u2126 
R2 = 2.2 \u2126 
RE = 1 \u2126 
VCC = 10 V 
VBE = 0.7 V 
Solution: 
VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) 
VBB = [2.2 \u2126/(10 \u2126 + 2.2 \u2126)]10 V 
VBB = 1.80 V 
VE = VBB \u2013 VBE (Eq. 8-2) 
VE = 1.80 V \u2013 0.7 V 
VE = 1.10 V 
IE = VE/RE (Eq. 8-3) 
IE = 1.1 V/1 \u2126 
IE = 1.1 A 
Answer: The dc emitter current is 1.1 A. 
 12-22. Given: 
R1 = 10 \u2126 
R2 = 2.2 \u2126 
RE = 1 \u2126 
VCC = 10 V 
VBE = 0.7 V 
RC = 3.2 \u2126 
Vout = 5 V pp 
Solution: 
Pout = \u3bdout2/8RL (Eq. 12-15) 
Pout = (5 V)2/8(3.2 \u2126) 
Pout = 0.977 W 
Ibias = VCC/(R1 + R2) 
Ibias = 10 V/(10 \u2126 + 2.2 \u2126) 
Ibias = 0.82 A 
VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) 
VBB = [2.2 \u2126/(10 \u2126 + 3.2 \u2126)] 10 V 
VBB = 1.80 V 
VE = VBB \u2013 VBE (Eq. 8-2) 
VE = 1.80 V \u2013 0.7 V 
VE = 1.10 V 
IE = VE/RE (Eq. 8-3) 
IE = 1.10 V/1 \u2126 
IE = 1.1 A 
Idc = Ibias + IE 
Idc = 0.82 A + 1.1 A 
Idc = 1.92 A 
Pdc = IdcVCC (Eq. 12-17) 
Pdc = (1.92 A)(10 V) 
Pdc = 19.2 W 
\u3b7 = [Pout/Pin]100% 
\u3b7 = [0.977 W/19.2 W]100% 
\u3b7 = 5.1% 
Answer: The output power is 0.977 W, and the 
efficiency is 5.1%. 
 12-23. Given: 
Vcut-off = 12 V 
Solution: 
MPP = 12 Vcut-off 
MPP = 2(12 V) 
MPP = 24 V 
Answer: The maximum peak-to-peak voltage is 24 V. 
 12-24. Given: 
VCC = MPP = 30 V 
RL = 16 \u2126 
Solution: 
PD(max) = MPP2/40RL 
PD(max) = (30 V)2/40(16 \u2126) 
PD(max) = 1.41 W 
Answer: The maximum power dissipation of each 
transistor is 1.41 W. 
 12-25. Given: 
VCC = MPP = 30 V 
RL = 16 \u2126 
Solution: 
Pout(max) = MPP2/8RL 
Pout(max) = (30 V)2/8(16 \u2126) 
Pout(max) = 7.03 W 
Answer: The maximum output power is 7.03 W. 
 12-26. Given: 
R1 = 100 \u2126 
R2 = 100 \u2126 
RL = 50 \u2126 
VCC = 30 V 
VDiode = 0.7 V 
1-58 
Solution: 
Ibias = (VCC \u2013 2VDiode)/(R1 + R2) 
Ibias = 28.6 V/(100 \u2126 + 100 \u2126) 
Ibias = 143 mA 
ICEQ \u2248 Ibias = 143 mA 
Answer: The quiescent collector current is 143 mA. 
 12-27. Given: 
VCC = MPP = 30 V 
RL = 50 \u2126 
Solution: 
Ibias = (VCC \u2013 2VDiode)/(R1 + R2) 
Ibias = 28.6 V/(100 \u2126 + 100 \u2126) 
Ibias = 143 mA 
Idc = 238 mA 
Pdc = IdcVCC 
Pdc = 238 mA(30 V) 
Pdc = 7.14 W 
Pout(max) = MPP2/8RL 
Pout(max) = (30 V)2/8(50 \u2126) 
Pout(max) = 2.25 W 
\u3b7 = [Pout/Pin]100% 
\u3b7 = [2.25 W/750 mW]100% 
\u3b7 = 31.5% 
Answer: The efficiency is 31.5%. 
 12-28. Given: 
VCC = MPP = 30 V 
RL = 50 \u2126 
Solution: 
Ibias = (VCC \u2013 2VDiode)/(R1 + R2) 
Ibias = 28.6 V/(1 k\u2126 + 1 k\u2126) 
Ibias = ICQ = 14.3 mA 
Idc = 110 mA 
Pdc = IdcVCC 
Pdc = 110 mA(30 V) 
Pdc = 3.3 W 
Pout(max) = MPP2/8RL 
Pout(max) = (30 V)2/8(50 \u2126) 
Pout(max) = 2.25 W 
\u3b7 = [Pout/Pin]100% 
\u3b7 = [2.25 W/3.3 W]100% 
\u3b7 = 68.3% 
Answer: The efficiency is 68.3% and the quiescent 
collector current is 14.3 mA. 
 12-29. Given: 
MPP = 30 V 
RL = 100 \u2126 
Solution: 
Pout(max) = MPP2/8RL 
Pout(max) = (30 V)2/8(100 \u2126) 
Pout(max) = 1.13 W 
Answer: The maximum power output is 1.13 W. 
 12-30. Given for 1st stage: 
R1 = 10 k\u2126 
R2 = 5.6 k\u2126 
RC = 1 k\u2126 
RE = 1 k\u2126 
VBB = 10.7 V 
VE = 10 V 
Second Stage: 
R1 = 12 k\u2126 
R2 = 1 k\u2126 
RE = 100 k\u2126 
\u3b2 = 200 
VCC = 30 V 
Solution: 
er\u2032 = 25 mV/IE 
er\u2032 = 25 mV/(10 V/1 k\u2126) 
er\u2032 = 2.5 \u2126 
er = RE (second stage) 
er = 100 \u2126 
rc = RC||Zin(Stage 2) 
Zin(Stage 2) = 12 k\u2126||910 \u2126||\u3b2 er\u2032 
rc = 1 k\u2126||12 k\u2126||910 \u2126||200(100 \u2126) 
rc = 496 \u2126 
Av(Stage 1) = /c er r\u2032 
Av(Stage 1) = 496 \u2126/2.5 \u2126 
Av(Stage 1) = 188 
Answer: The voltage gain of the first stage is 188. 
 12-31. Given for 2nd Stage: 
R1 = 12 k\u2126 
R2 = 1 k\u2126 
RE = 100 \u2126 
RC = 1 k\u2126 
VE = 1.43 V 
 3rd Stage: 
\u3b2 = 200 
VCC = 30 V 
RE = 100 \u2126 
Solution: 
IE = VE/RE (Eq. 8-3) 
IE = 1.43 V/100 \u2126 
IE = 14.3 mA 
er\u2032 = 25 mV/IE (Eq. 9-10) 
er\u2032 = 25 mV/(14.3 mA) 
er\u2032 = 1.75 \u2126 
re = RE (second stage) 
re = 100 \u2126 
Zin(base) = \u3b2 er\u2032 (Eq. 10-9) 
Zin(base) = 200(100\u2126) 
Zin(base) = 20 k\u2126 
rc = RC||Zin(base) 
rc = 1 k\u2126||20 k\u2126 
rc = 952 \u2126 
Av = rc/(re + er\u2032 ) 
Av = 952 \u2126/(100 \u2126 + 1.75 \u2126) 
Av = 9.36 
Answer: The gain of the second stage is 9.36. 
 12-32. Given: 
IE = 14.3 mA 
Solution: 
ICQ = Ibias = 14.3 mA 
Answer: The quiescent collector current is 14.3 mA. 
 12-33. Given: 
Av1 = 188 (from Prob. 12-30) 
Av2 = 9.36 (from Prob. 12-31) 
1-59 
Solution: 
Av3 = 1 (Eq. 12-25) 
Av = Av1Av2Av3 
Av = (188)(9.36)(1) 
Av = 1679 
Answer: The total voltage gain is 1679. 
 12-34. Given: vin = 5 Vrms. 
Solution: 
Vpp = 2.828 Vrms. 
Vpp = 2.828(5 V) 
Vpp = 14.14 V pp 
Since the input is clamped at 0.7 V, the negative peak is 
\u201313.44 V. The average value is \u20136.37 V, so the 
voltmeter will read \u20136.37 V. 
Answer:
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