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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.
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.
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.
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.
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.
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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