100 pág.

# Eletronica 1 Malvino 7ed Respostas (em ingles)

DisciplinaEletrônica I6.544 materiais31.730 seguidores
Pré-visualização32 páginas
```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:```
Estudante fez um comentário
falatando pagina vergonha
0 aprovações
Marcus fez um comentário
tá faltando páginas véi....
0 aprovações
naruto fez um comentário
alguien me podria enviar el archivo a mi email por favor 5542sp@gmail.com
0 aprovações
Carregar mais