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


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= 4.94 V 
Answer: The collector-emitter voltage is \u20134.94 V since 
the collector is less positive than the emitter. 
 8-22. Given: 
R1 = 10 k\u2126 
R2 = 2.2 k\u2126 
RE = 1 k\u2126 
RC = 3.6 k\u2126 
VCC = 10 V 
VBE = 0.7 V 
V2 = 1.8 V (from Prob. 8-20) 
VRE = 1.1 V (from Prob. 8-20) 
IE = 1.1 mA (from Prob. 8-20) 
VC = 3.96 V (from Prob. 8-20) 
Solution: Because of the voltage divider, there will always 
be a 1.1-V drop across RE, and at saturation VCE = 0 V. 
This leaves 8.9 V across RC at saturation. 
IC = 8.9 V/Rc 
IC = 8.9 V/3.6 k\u2126 
IC = 2.47 mA 
At cutoff, the maximum possible voltage across VCE is 
8.9 V. 
Answer: The saturation current is 2.47 mA, and the 
collector-emitter cutoff voltage is 8.9 V. 
 8-23. Given: 
R1 = 10 k\u2126 
R2 = 2.2 k\u2126 
RE = 1 k\u2126 
RC = 3.6 k\u2126 
VCC = \u201310 V 
VBE = 0.7 V 
Solution: 
VBB = [R2/(R1 + R2)]VCC 
VBB = [2.2 k\u2126/(10 k\u2126 + 2.2 k\u2126)] \u2013 10 V 
VBB = \u20131.8 V 
VE = V2 + 0.7 V 
VE = \u20131.8 V + 0.7 V 
VE = \u20131.1 V 
IE = VE/RE 
IE = 1.1 V/1 k\u2126 
IE = 1.1 mA 
IC \u2248 IE (Eq. 8-4) 
VC = VCC + ICRC 
VC = \u201310 V + (1.1 mA)(3.6 k\u2126) 
VC = \u20136.04 V 
Answer: The collector voltage is \u20136.04 V, and the 
emitter voltage is \u20131.1 V. 
CRITICAL THINKING 
 8-24. The circuit is no longer considered stiff or independent 
of Beta. The base current is not small as compared to the 
voltage divider current. 
 8-25. The maximum power dissipation of the 2N3904 is 
625 mW. The transistor is dissipating 705 mW. The 
transistor will probably overheat and fail. 
 8-26. As long as the voltmeter has a high enough input 
resistance, it should read approximately 4.83 V. 
 8-27. Increase the power supply value, short R1. 
 8-28. Connect an ammeter between the power supply and the 
circuit. Measure VR1 and VC, then calculate and add their 
respective currents. 
 8-29. Given: (for Q1): 
R1 = 1.8 k\u2126 
R2 = 300 \u2126 
RE = 240 \u2126 
RC = 1 k\u2126 
VCC = 15 V 
VBE = 0.7 V 
1-36 
Solution: 
VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) 
VBB = [300 \u2126/(1.8 k\u2126 + 300 \u2126)]15 V 
VBB = 2.14 V 
VE = VBB \u2013 0.7 V (Eq. 8-2) 
VE = 2.14 V \u2013 0.7 V 
VE = 1.44 V 
IE = VE/RE (Eq. 8-3) 
IE = 1.44 V/240 \u2126 
IE = 6 mA 
IC \u2248 IE (Eq. 8-4) 
VC = VCC \u2013 ICRC (Eq. 8-5) 
VC = 15 V \u2013 (6 mA)(1 k\u2126) 
VC = 9.0 V 
Given (for Q2): 
R1 = 910 \u2126 
R2 = 150 \u2126 
RE = 120 \u2126 
RC = 510 \u2126 
VCC = 15 V 
VBE = 0.7 V 
Solution: 
VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) 
VBB = [150 \u2126/(910 \u2126 + 150 \u2126)]15 V 
VBB = 2.12 V 
VE = VBB \u2013 0.7 V (Eq. 8-2) 
VE = 2.12 V \u2013 0.7 V 
VE = 1.42 V 
IE = VE/RE (Eq. 8-3) 
IE = 1.42 V/120 \u2126 
IE = 11.83 mA 
IC \u2248 IE (Eq. 8-4) 
VC = VCC \u2013 ICRC (Eq. 8-5) 
VC = 15 V \u2013 (11.83 mA)(510 \u2126) 
VC = 8.97 V 
Given (for Q3): 
R1 = 1 k\u2126 
R2 = 180 \u2126 
RE = 150 \u2126 
RC = 620 \u2126 
VCC = 15 V 
VBE = 0.7 V 
Solution: 
VBB = [R2/(R1 + R2)]VCC (Eq. 8-1) 
VBB = [180 \u2126/(1 k\u2126 + 180 \u2126)]15 V 
VBB = 2.29 V 
VE = VBB \u2013 0.7 V (Eq. 8-2) 
VE = 2.29 V \u2013 0.7 V 
VE = 1.59 V 
IE = VE/RE (Eq. 8-3) 
IE = 1.59 V/150 \u2126 
IE = 10.6 mA 
IC \u2248 IE (Eq. 8-4) 
VC = VCC \u2013 ICRC (Eq. 8-5) 
VC = 15 V \u2013 (10.6 mA)(620 \u2126) 
VC = 8.43 V 
Answer: The collector voltage for Q1 is 9.0 V, for Q2 is 
8.97 V, and for Q3 is 8.43 V. 
 8-30. Given: 
R1 = 10 k\u2126 
RE = 1 k\u2126 
RC = 8.2 k\u2126 
VCC = 20 V 
VD = 0.7 V 
Solution: 
VBB = 3(VD) 
VBB = 3(0.7 V) 
VBB = 2.1 V 
VE = VBB \u2013 0.7 V (Eq. 8-2) 
VE = 2.1 V \u2013 0.7 V 
VE = 1.4 V 
IE = VE/RE (Eq. 8-3) 
IE = 1.4 V/1 k\u2126 
IE = 1.4 mA 
IC \u2248 IE (Eq. 8-4) 
VC = VCC \u2013 ICRC (Eq. 8-5) 
VC = 20 V \u2013 (1.4 mA)(8.2 k\u2126) 
VC = 8.52 V 
Answer: The emitter current is 1.4 mA, and the collector 
voltage is 8.52 V. 
 8-31. Given: 
VBB(1) = 2 V 
RE(1) = 200 \u2126 
RC(1) = 1 k\u2126 
RE(2) = 1 k\u2126 
VCC = 16 V 
Solution: 
VE(1) = VBB(1) \u2013 0.7 V (Eq. 8-2) 
VE(1) = 2.0 V \u2013 0.7 V 
VE(1) = 1.3 V 
IE(1) = VE/RE (Eq. 8-3) 
IE(1) = 1.3 V/200 \u2126 
IE(1) = 6.5 mA 
IC \u2248 IE (Eq. 8-4) 
VC(1) = VCC \u2013 ICRC (Eq. 8-5) 
VC(1) = 16 V \u2013 (6.5 mA)(1 k\u2126) 
VC(1) = 9.5 V 
VC(1) = VBB(2) 
VE(2) = VBB(2) \u2013 0.7 V (Eq. 8-2) 
VE(2) = 9.5 V \u2013 0.7 V 
VE(2) = 8.8 V 
Answer: The output voltage is 8.8 V. 
 8-32. Given: 
R1 = 620 \u2126 
R2 = 680 \u2126 
RE = 200 \u2126 
VCC = 12 V 
VBE = 0.7 V 
Solution: 
V2 = [R2/(R1 + R2)]VCC 
V2 = [680 \u2126/(620 \u2126 + 680 \u2126)]12 V 
V2 = 6.28 V 
VRE = V2 \u2013 0.7 V 
VRE = 6.28 V \u2013 0.7 V 
VRE = 5.58 V 
IE = VRE/RE (Eq. 8-3) 
IE = 5.58 V/200 \u2126 
1-37 
IE = 27.9 mA 
ILED \u2248 IE 
Answer: The LED current is 27.9 mA. 
 8-33. Given: 
R1 = 620 \u2126 
RE = 200 \u2126 
VCC = 12 V 
VBE = 0.7 V 
VZ = 6.2 V 
Solution: 
VRE = VZ \u2013 0.7 V 
VRE = 6.2 V \u2013 0.7 V 
VRE = 5.5 V 
IE = VRE/RE (Eq. 8-3) 
IE = 5.5 V/200 \u2126 
IE = 27.5 mA 
ILED \u2248 IE 
Answer: The LED current is 27.5 mA. 
 8-34. Given: 
RE = 51 k\u2126 
R1 = 3.3R2; this ratio is necessary to prevent moving the 
Q point. Assume \u3b2dc = 100 
Solution: 
R1||R2 < 0.01 \u3b2dcRE (Eq. 8-9) 
R1||R2 = 0.01(100)(51 k\u2126) 
R1||R2 = 51 k\u2126 
Since R2 is the smaller of the two resistors, make it 51 k\u2126. 
Then the parallel resistance will not be higher than 51 
k\u2126, which satisfies the requirement. 
R1 = 3.3R2 
R1 = 3.3(51 k\u2126) 
R1 = 168.3 k\u2126 
Answer: R1 maximum of 168.3 k\u2126, R2 maximum of 51 k\u2126, 
and the ratio between them 3.3:1. 
 8-35. Answer: With VB at 10 V and R2 is good, the trouble is 
R1 shorted. 
 8-36. Answer: Since VB is 0.7 V and VE is 0 V, the trouble is 
RE is shorted. 
 8-37. Answer: 
Trouble 3: Since VC is 10 V and VE is 1.1 V, the 
transistor is good. Therefore the trouble is RC, which is 
shorted. 
Trouble 4: Since all the voltages are the same, the 
trouble is that all the transistor terminals are shorted 
together. 
 8-38. Answer: 
Trouble 5: Since VB is 0 V, it is either R1 open or R2 
shorted. R2 is OK, so the trouble is R1 open. 
Trouble 6: R2 is open. 
 8-39. Answer: 
Trouble 7: Since VC is 10 V, there is an open below it or 
a short above it. A shorted RC would not affect VB; 
therefore there must be an open below it. If the transistor 
is open, VB would be 0 V; therefore the trouble is an 
open RE. 
Trouble 8: R2 is shorted. 
 8-40. Answer: 
Trouble 9: Since the base voltage is 1.1 V, it appears 
that the voltage divider is working but not properly. The 
emitter voltage is 0.7 V less than the base, so the 
emitter-base junction is working. If RC is open, the meter 
would complete the circuit and give a low voltage 
reading. The trouble is an open RC. 
Trouble 10: This is very similar to trouble 9 except that 
the collector voltage is 10 V. Since source voltage is 
read above an open, the trouble is an open collector-base 
junction. 
 8-41. Answer: 
Trouble 11: Since all the voltages are 0 V, the power 
supply is not working. 
Trouble 12: With the emitter voltage at 0 V and the base 
voltage at 1.83 V, the emitter-base diode of the transistor 
is open. 
Chapter 9 AC Models 
SELF-TEST 
1. a 7. b 12. d 17. c 
2. b 8. b 13. b 18. b 
3. c 9. c 14. b 19. b 
4. c 10. c 15. d 20. c 
5. a 11. b 16. b 21. a 
6. d 
JOB INTERVIEW QUESTIONS 
 7. To permit the output voltage to swing over the largest 
possible voltage when the input signal is large enough to 
produce a maximum output. 
 8. Models provide mathematical and logical insight into the 
operation of a device. The two common transistor models 
are the T and the \u3c0. 
 11. It would become zero because there is no collector current. 
PROBLEMS 
 9-1. Given: 
C = 47 µF 
R = 10 k\u2126 
Solution: 
XC = 1/(2\u3c0fC) 
XC < 0.1R (Eq. 9-1) 
1/(2\u3c0fC) = 0.1R 
1/(2\u3c0C) = (0.1R)f 
f = 1/{[2\u3c0(47µF)][0.1(10 k\u2126)]} 
f = 3.39 Hz 
Answer: The lowest frequency where good coupling 
exists is 3.39 Hz. 
 9-2. Given: 
C = 47 µF 
R = 1 k\u2126 
Solution: 
XC = 1/(2\u3c0fC) 
XC < 0.1R (Eq. 9-1) 
1/(2\u3c0fC) = 0.1R 
1/(2\u3c0C) = (0.1R)f 
f = 1/{[2\u3c0(47 µF)][0.1(1 k\u2126)]} 
f = 33.9 Hz
1-39 
VE = VBB \u2013 VBE 
VE = 2.7 V \u2013 0.7 V 
VE = 2.0 V 
IE = VE/RE (Eq. 8-3) 
IE = 2.0 V/940 \u2126 
IE = 2.128 mA 
ie(pp) < 0.1 IEQ (Eq. 9-6) 
ie(pp)max = 0.1 (2.13 mA) 
ie(pp)max = 213 µA 
Answer: The maximum ac emitter
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