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


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R1 = R2 
Solution: At ground the circuit is an inverting amplifier. 
AV = \u2013Rf/R1 
AV = \u20131 
When the wiper is 10% away from ground, so that the 
noninverting gain will be 10% of its maximum of 2. 
Av(non) = 10% (2) = 0.2 
AV = Av(in) + Av(non) 
AV = \u20131 + 0.2 
AV = \u20130.8 
Answer: The gain with the wiper at ground is \u20131, and 
10% away is \u20130.8. 
 20-13. Given: 
R = 5 k\u2126 
nR = 75 k\u2126 
nR/(n \u2013 1)R = 5.36 k\u2126 
Solution: 
AV = \u2013nR/R 
AV = \u201375 k\u2126/5 k\u2126 
AV = \u201315 
Answer: The maximum positive gain is 15, and the 
maximum negative gain is \u201315. 
 20-14. Given: 
R\u2032 = 10 k\u2126 
R = 22 k\u2126 
C = 0.02 \u3bcF 
fin = 100 Hz, 1 kHz, 10 kHz 
Solution: 
fC = 1/(2\u3c0 RC) 
fC = 1/[(2\u3c0RC)(0.02 \u3bcF)] 
fC = 362 Hz 
\u444 = \u20132 arctan (f/fC) 
\u444 = \u20132 arctan (100 Hz/362 Hz) 
\u444 = \u201330.9° 
\u444 = \u20132 arctan (f/fC) 
\u444 = \u20132 arctan (1 kHz/362 Hz) 
\u444 = \u2013140° 
\u444 = \u20132 arctan (f/fC) 
\u444 = \u20132 arctan (10 kHz/362 Hz) 
\u444 = \u2013176° 
Answer: The phase shift is \u201330.9° at 100 Hz, \u2013140° at 
1 kHz, and \u2013176° at 10 kHz. 
 20-15. Given: 
R1 = 1.5 k\u2126 
Rf = 30 k\u2126 
Solution: 
Av(inv) = \u2013Rf/R1 (Eq. 20-6) 
Av(inv) = \u201330 k\u2126/1.5 k\u2126 
Av(inv) = \u201320 
Av(non) = [(R2/R1) + 1][R 2
\u2032 /(R 1
\u2032 + R 2
\u2032 )] (Eq. 20-7) 
Av(non) = [(30 k\u2126/1.5 k\u2126) + 1][30 k\u2126/(1.5 k\u2126 + 30 k\u2126)] 
Av(non) = 20 
Av(CM) = ±4(0.1%) = ±4(0.001) = ±0.004 
Answer: The differential voltage gain is \u201320, and the 
common mode gain is ±0.004. 
1-101 
 20-16. Given: 
R1 = 1 k\u2126 
Rf = 20 k\u2126 
Solution: 
Av(inv) = \u2013Rf/R1 (Eq. 20-6) 
Av(inv) = \u201320 k\u2126/1 k\u2126 
Aiv(inv) = \u201320 
Av(CM) = ±4 \u394R/R (Eq. 20-5) 
Av(CM) = ±4 (1%) = ±4(0.01) 
Av(CM) = ±0.04 
Answer: The differential voltage gain is \u201320, and the 
common-mode gain is ±0.04. 
 20-17. Given: 
R1 = 10 k\u2126 
R2 = 20 k\u2126 
R3 = 20 k\u2126 
R4 = 10 k\u2126 
Solution: 
V2 = [R2/(R1 + R2)]VCC 
V2 = [20 k\u2126/(10 k\u2126 + 20 k\u2126)]15 V 
V2 = 10 V 
V4 = [R4/(R3 + R4)]VCC 
V4 = [10 k\u2126/(20 k\u2126 + 10 k\u2126)]15 V 
V4 = 5 
Answer: No, the bridge is not balanced. 
 20-18. Given: 
R1 = 1 k\u2126 
\u394R = 15 \u2126 
AV = \u2013100 
Solution: 
vin = (\u394R/4R)VCC 
vin = (15 \u2126/4 (1 k\u2126))15 V 
vin = 56.3 mV 
vout = A1(vin) 
vout = (\u2013100)(56.3 mV) 
vout = \u20135.63 V 
Answer: The output voltage is \u20135.63 V. 
 20-19. Given: 
R1 = 1 k\u2126 
Rf = 99 k\u2126 
R = 10 k\u2126 ± 0.5% 
vin = 2 mV 
Solution: 
AV = (Rf/R1) + 1 
AV = (99 k\u2126/1 k\u2126) + 1 
AV = 100 
vout = Avvin 
vout = 100(2 mV) 
vout = 200 mV 
AV(CM) = ±2(\u394R/R) 
AV(CM) = ±2(0.005) 
AV(CM) = ±0.01 
CMRR = |AV|/|AV(CM)| 
CMRR = 100/0.01 
CMRR = 10,000 
Answer: The output voltage is 200 mV, and the CMRR 
is 10,000. 
 20-20. Given: vin(CM) = 5 V 
Solution: Since the first stage has a common-mode gain 
of 1, both sides have the same voltage of 5 V. The guard 
voltage is 5 V. 
Answer: The guard voltage is 5 V. 
 20-21. Given: 
RG = 1008 \u2126 
vin = 20 mV 
Solution: 
AV = (49.4 k\u2126/RG) + 1 (Eq. 20-17) 
AV = (49.4 k\u2126/1008 \u2126) + 1 
AV = 50 
vout = Av(vin) 
vout = 50(20 mV) 
vout = 1 V 
Answer: The output voltage is 1 V. 
 20-22. Given: 
R = 10 k\u2126 
v1 = \u201350 mV 
v2 = \u201330 mV 
Solution: 
vout = v1 \u2013 v2 
vout = (\u201350 mV) \u2013 (\u201330 mV) 
vout = \u201320 mV 
Answer: The output voltage is \u201320 mV. 
 20-23. Given: 
R1 = 10 k\u2126 
R2 = 20 k\u2126 
R3 = 15 k\u2126 
R4 = 15 k\u2126 
R5 = 30 k\u2126 
RF = 75 k\u2126 
v1 = 1 mV 
v2 = 2 mV 
v3 = 3 mV 
v4 = 4 mV 
Solution: 
Av(1) = \u2013Rf/R1 
Av(1) = \u201375 k\u2126/10 k\u2126 
Av(1) = \u20137.5 
Av(2) = \u2013Rf/R2 
Av(2) = \u201375 k\u2126/20 k\u2126 
Av(2) = \u20133.75 
Av(3) = {[Rf/(R1||R2)] + 1}{(R4||R5)/[R3 + (R4||R5)]} 
Av(3) = {[75 k\u2126/(10 k\u2126||20 k\u2126)] + 1}{(15 k\u2126||30 
k\u2126)/[15 k\u2126 + (15 k\u2126||30 k\u2126)]} 
Av(3) = (12.25)(0.455) 
Av(3) = 5.57 
Av(4) = {[Rf/(R1||R2)] + 1}{(R3||R5)/[R4 + (R3||R5)]} 
Av(4) = {[75 k\u2126/(10 k\u2126||20 k\u2126)] + 1}{(15 k\u2126||30 
k\u2126)/[15 k\u2126 + (15 k\u2126||30 k\u2126)]} 
Av(4) = (12.25)(0.4) 
Av(4) = 4.9 
vout = Av(1)v1 + Av(2)v2 + Av(3)v3 + Av(4)v4 
vout = \u20137.5(1 mV) + \u20133.75(2 mV) + 4.9(3 mV) + 4.9 
(4 mV) 
vout = 19.3 mV 
Answer: The output voltage is 19.3 mV. 
1-104 
Solution: 
AV = (Rf/R1) + 1 
AV = (15 k\u2126/1.5 k\u2126) + 1 
AV = 11 
f1 = 1/[2\u3c0(R/2)C1] 
f1 = 1/[2\u3c0(68 k\u2126/2)(1 µF)] 
f1 = 4.68 Hz 
f2 = 1/(2\u3c0RLC2) 
f2 = 1/[2\u3c0(15 k\u2126)(2.2 µF)] 
f2 = 4.82 Hz 
f3 = 1/(2\u3c0R1C3) 
f3 = 1/[2\u3c0(1 k\u2126)(3.3 µF)] 
f3 = 32.2 Hz 
Answer: The gain is 11, and the cutoff frequencies are 
f1 = 4.68 Hz, f2 = 4.82 Hz, and f3 = 32.2 Hz. 
CRITICAL THINKING 
 20-40. Answer: Since the terminal is floating, the output would 
be saturated or VCC. To fix this problem, a large-value 
resistor could be connected to the noninverting terminal. 
This would keep it at ground potential during the 
transition and prevents a spike. 
 20-41. Given: 
R1(min) = 990 \u2126 
R1(max) = 1010 \u2126 
Rf(min) = 99 k\u2126 
Rf(max) = 101 k\u2126 
Solution: 
AV(min) = \u2013Rf(min)/R1(max) 
AV(min) = \u201399 k\u2126/1010 \u2126 
AV(min) = 98 
AV(max) = \u2013Rf(max)/R1(min) 
AV(max) = \u2013101 k\u2126/990 \u2126 
AV(max) = 102 
Answer: The minimum gain is 98, and the maximum 
gain is 102. 
 20-42. Given: 
Transistor: 
R1 = 22 k\u2126 
Rf = 10 k\u2126 
RS = 1 k\u2126 
RE = 5.6 k\u2126 
RC = 6.8 k\u2126 
VCC = 15V 
Op amp 
R1 = 1 k\u2126 
Rf = 47 k\u2126 
Solution: 
VBB = [Rf/(R1 + Rf + RS)]VCC (Eq. 8-1) 
VBB = [10 k\u2126/(22 k\u2126 + 10 k\u2126 + 1 k\u2126)]15 V 
VBB = 4.54 V 
VE = VBB \u2013 VBE (Eq. 8-2) 
VE = 4.54 V \u2013 0.7 V 
VE = 3.84 V 
IE = VE/RE (Eq. 8-3) 
IE = 3.84 V/5.6 k\u2126 
IE = 0.685 mA 
er\u2032 = 25 mV/IE (Eq. 9-10) 
er\u2032 = 25 mV/0.685 mA 
er\u2032 = 36.5 \u2126 
rc = RC 
rc = 6.8 k\u2126 
Av = rc/ er\u2032 (Eq. 10-7) 
Av = 6.8 k\u2126/36.5 \u2126 
Av = 186 
Op amp: 
AV = (Rf/R1) + 1 
AV = (47 k\u2126/1 k\u2126) + 1 
AV = 48 
AV = (48)(186) 
AV = 9114 
Answer: The voltage gain is 9114. 
 20-43. Given: 
R1 = 1 k\u2126 
Rf = 10 k\u2126 
RL = 100 \u2126 
\u3b2 = 50 
vin = 0.5 VCC 
Solution: 
AV = \u2013Rf/R1 
AV = \u201310 k\u2126/1 k\u2126 
AV = \u201310 
vout = AV(vin) 
vout = \u201310(0.5 V) 
vout = \u20135 V 
Iout = vout/RL 
Iout = \u20135 V/100 \u2126 
Iout = 50 mA 
IB = Iout/\u3b2 
IB = 50 mA/50 
IB = 1 mA 
Answer: The base current is 1 mA. 
 20-44. Answer: 
Trouble 1: Since there is voltage at E and not at F, there 
is an open between E and F. 
Trouble 2: Since the output is only 200 mV, which is the 
amplified output of A, R2 is open. 
Trouble 3: Since the input is 2 mV and the output is 
maximum, R1 is shorted. 
 20-45. Answer: 
Trouble 4: Since there is no voltage at B, there is an 
open between K and B. 
Trouble 5: Since the voltage at C is 3 mV and the 
voltage at D is zero, there is an open between C and D. 
Trouble 6: Since the voltage at A is zero, there is an 
open between J and A. 
 20-46. Answer: 
Trouble 7: Since the input voltage is 3 mV and the 
output is maximum, R3 is open. 
Trouble 8: Since the output is only 250 mV, which is the 
amplified output of B, R1 is open. 
Trouble 9: Since the output voltage is the same as the 
input voltage, R3 is shorted. 
Trouble 10: Since the input is 5 mV and the output is 
maximum, R2 is shorted. 
1-108 
 21-20. Given: 
R1 = 56 k\u2126 
Rf = 10 k\u2126 
C = 680 pF 
Solution: 
11/(2\u3c0 / )p ff C R R= 
1/[2\u3c0(680 pF) (10 k )(56 k )]pf = \u3a9 \u3a9 
fp = 9.89 kHz 
10.5 ( / )fQ R R= 
0.5 (56k )/(10k )Q = \u3a9 \u3a9 
Q = 1.18 
Kc = 1.04 (from Fig. 21-26) 
K3 = 1.30 (from Fig. 21-26) 
fc = fp/Kc (Eq. 21-31) 
fc = 9.89 kHz/1.04 
fc = 9.51 kHz 
f3 = fp/K3 
f3 = 9.89 kHz/1.30 
f3 = 7.61 kHz 
Answer: The pole frequency is 9.89 kHz, the cutoff 
frequency is 9.51 kHz, the 3-dB frequency is 7.61 kHz, 
and the Q is 1.18. 
 21-21. Given: 
R1 = 91 k\u2126 
Rf = 15 k\u2126 
C = 220 pF 
Solution: 
11/(2\u3c0 / )p ff C R R= 
1/[2\u3c0(220 pF) (15 k )(91 k )]pf = \u3a9 \u3a9 
fp = 19.6 kHz 
10.5 ( / )fQ R R= 
0.5 (91k )/(15k )Q = \u3a9 \u3a9 
Q = 1.23 
Kc = 1.06 (from Fig. 21-26) 
K3 = 1.32 (from Fig. 21-26) 
fc = fp/Kc (Eq. 21-31) 
fc = 19.6 kHz/1.06 
fc = 18.5 kHz 
f3 = fp/K3 
f3 = 19.6 kHz/1.32 
f3 = 14.8 kHz 
Answer: The pole frequency is 19.6 kHz, the cutoff 
frequency is 18.5 kHz, the 3-dB frequency is 14.8 kHz, 
and the Q is 1.23. 
 21-22. Given: 
R1 = 2 k\u2126 
Rf = 56 k\u2126 
C = 270 pF 
Solution: 
AV = \u2013Rf/2R1 (Eq. 21-35) 
AV = \u201356 k\u2126/2(2
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