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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.
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.
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.
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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