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


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of 
2 V, and 376 mV/\u3bcs, with a frequency of 30 kHz. 
 18-18. Answer: 
 a. OP-07A 
 b. TL082 and TL084 
 c. LM12 
 d. OP-64E 
 e. OP-07A 
 18-19. Answer: 
CMRR = 38 dB (from Fig. 18-6a) 
MPP = 21 V (from Fig. 18-6b) 
Av = 1000 (from Fig. 18-6c) 
 18-20. Given: 
R1 = 10 k\u2126 
R2 = 20 k\u2126 
R3 = 40 k\u2126 
Rf(max) = 100 k\u2126 
Rf(min) = 100 k\u2126 
v1 = 50 mVpp 
v1 = 90 mVpp 
v1 = 160 mVpp 
Solution: When the resistance is zero, the voltage gains 
are zero and the output voltage is zero. 
\u2013Av1(CL)max = \u2013Rf /R1 (Eq. 18-3) 
\u2013Av1(CL)max = \u2013100 k\u2126/10 k\u2126 
\u2013Av1(CL)max = \u201310 
\u2013Av2(CL)max = \u2013Rf /R1 (Eq. 18-3) 
\u2013Av2(CL)max = \u2013100 k\u2126/20 k\u2126 
\u2013Av2(CL)max = \u20135 
\u2013Av3(CL)max = \u2013Rf /R1 (Eq. 18-3) 
\u2013Av3(CL)max = \u2013100 k\u2126/40 k\u2126 
\u2013Av3(CL)max = \u20132.5 
vout = Av1(CL)max(vin1) + Av2(CL)max(vin2) + Av3(CL)max(vin3) 
vout = \u201310(50 mVpp) + 5(90 mVpp) + 2.5(160 mVpp) 
vout = \u20131.35 Vpp 
Answer: The maximum output voltage is 1.35 Vpp, and 
the minimum output voltage is zero. 
 18-21. Given: 
R1 = 220 \u2126 
RF1 = 47 k\u2126 
RF2 = 18 k\u2126 
RF3 = 39 k\u2126 
Solution: 
\u2013Av1(CL) = \u2013RF1/R1 (Eq. 18-3) 
\u2013Av1(CL) = \u201347 k\u2126/220 \u2126 
\u2013Av1(CL) = \u2013214 
\u2013Av2(CL) = \u2013RF2/R1 (Eq. 18-3) 
\u2013Av2(CL) = \u201318 k\u2126/220 \u2126 
\u2013Av2(CL) = \u201382 
\u2013Av3(CL) = \u2013RF3/R1 (Eq. 18-3) 
\u2013Av3(CL) = \u201339 k\u2126/220 \u2126 
\u2013Av3(CL) = \u2013177 
Solution: The gain at position 1 is 214, at position 2 is 
82, and at position 3 is 177. 
 18-22. Given: 
R1 = 6 k\u2126 at position 2 
R2 = 6 k\u2126||3 k\u2126 at position 1 = 2 k\u2126 
R2 = 120 k\u2126 
funity = 1 MHz 
Solution: 
Av1(CL) = (R2/R1) + 1 (Eq. 18-12) 
Av1(CL) = (120 k\u2126/2 k\u2126)+ 1 
Av1(CL) = 61 
Av2(CL) = (R2/R1) + 1 (Eq. 18-12) 
Av2(CL) = (120 k\u2126/6 k\u2126) + 1 
Av2(CL) = 21 
f2(CL)1 = funity/Av(CL1) (Eq. 18-5) 
f2(CL)1 = 1 MHz/61 
f2(CL)1 = 16.4 kHz 
f2(CL)2 = funity/Av(CL1)(max) (Eq. 18-5) 
f2(CL)2 = 1 MHz/21 
f2(CL)2 = 47.6 kHz 
Answer: The voltage gain at position 1 is 61, with a 
bandwidth of 16.4 kHz, and at position 2 is 21, with a 
bandwidth of 47.6 kHz. 
 18-23. Given: 
R1 = \u221e at position 2 
R1 = 3 k\u2126 at position 1 
R2 = 120 k\u2126 
funity = 1 MHz 
AVOL = 100,000 
Solution: 
AV1(CL) = (R2/R1) + 1 (Eq. 18-12) 
AV1(CL) = (120 k\u2126/3 k\u2126) + 1 
Av1(CL) = 41 
At position 2, it becomes a voltage follower: AvCL2 = 1. 
Answer: The voltage gain at position 1 is 41, and at 
position 2 is 1. 
1-92 
 18-24. Answer: The output will go to positive or negative 
saturation. 
 18-25. Answer: 
Position 1: The input voltage is applied directly to the 
noninverting input. Because of the virtual short between 
the noninverting and inverting input terminals, there is 
no ac voltage across the left 10-k\u2126 resistor. Since there 
is no ac voltage across the resistor, it can be removed 
from the circuit without changing the operation. With 
the resistor removed, the circuit reduces to a voltage 
follower and Av(CL) = 1 and a closed-loop bandwidth of 
 unity2( )
( )
1 MHz 1 MHz
1CL v CL
f
f
A
= = = 
Position 2: The circuit is an inverting amplifier. The 
magnitude of the voltage gain is Av(CL) = 1. Note that the 
closed-loop bandwidth is only half as much because 
 unity2( )
( )
1 MHz 500 kHz
1 1 1CL V CL
f
f
A
= = =+ + 
This was covered briefly in the chapter. See the equation 
at the top of p. 633 and the brief explanation that 
follows. Chapter 19 discusses the closed-loop 
bandwidths in more detail. 
 18-26. Answer: 
Position 1: With the left resistor open, the circuit 
reduces to a voltage follower and AV(CL) = 1. 
Position 2: With the left resistor open, the voltage gain 
is zero. 
 18-27. Answer: Go to positive or negative saturation. 
 18-28. Given: 
Iin(biaS) = 500 nA 
Iin(off) = 200 nA 
Vin(off) = 6 mV 
R1 = 2 k\u2126 
R2 = 100 k\u2126 
C = 1 \u3bcF 
Solution: 
XC = 1/2\u3c0fC 
XC = 1/[2\u3c0(0)(1 \u3bcF)] 
XC = \u221e 
1 1CR X R\u2032 = + 
1R
\u2032 = \u221e + 2 k\u2126 
1R
\u2032 = \u221e 
RB2 = R1||R2 (Eq. 18-11) 
RB2 = \u221e||100 k\u2126 
RB2 = 100 k\u2126 
V1err = (RB1 \u2013 RB2)Iin(biaS) (Eq. 18-8) 
V1err = (0 \u2013 100 k\u2126)(500 nA) 
V1err = 50 mV 
V2err = (RB1 + RB2)(Iin(off)/2) (Eq. 18-8) 
V2err = (0 + 100 k\u2126)(200 nA/2) 
V2err = 10 mV 
V3err = Vin(off) = 6 mV 
AV(CL) = (R2/ 1R\u2032 ) + 1 (Eq. 18-12) 
AV(CL) = (100 k\u2126/\u221e) + 1 
AV(CL) = 1 
Verror = ± ACL( ± V1err ± V2err ± V3err) 
Verror = 1(50 mV + 10 mV + 6 mV) 
Verror = 66 mV 
Answer: The output voltage is 66 mV. 
 18-29. Given: 
R1 = 2 k\u2126 
R2 = 100 k\u2126 
C = 1 \u3bcF 
vin = 50 mV pp 
f = 1 kHz 
Solution: 
XC = 1/2\u3c0fC 
XC = 1/[2\u3c0(1 kHz)(1 \u3bcF)] 
XC = 159 \u2126 
Since XC is less than one-tenth of 2 k\u2126, the bottom of 
the 2 k\u2126 is approximately an ac ground. 
AV(CL) = (R2/ 1R
\u2032 ) + 1 (Eq. 18-12) 
AV(CL) = (100 k\u2126/2 k\u2126) + 1 
AV(CL) = 51 
vout = AV(CL)vin 
vout = 51(50 mV pp) 
vout = 2.55 V pp 
Answer: The output voltage is 2.55 V. 
 18-30. Given: 
Iin(biaS) = 500 nA 
Iin(off) = 200 nA 
Vin(off) = 6 mV 
R1 = 2 k\u2126 
R2 = 100 k\u2126 
Solution: 
1R\u2032 = XC = R1 
1R\u2032 = 0 + 2 k\u2126 
1R\u2032 = 2 k\u2126 
RB2 = R1||R2 (Eq. 18-11) 
RB2 = 2 k\u2126||100 k\u2126 
RB2 = 1.96 k\u2126 
V1err = (RB1 \u2013 RB2)Iin(biaS) (Eq. 18-8) 
V1err = (0 \u2013 1.96 k\u2126)(500 nA) 
V1err = 980 \u3bcV 
V2err = (RB1 + RB2)(Iin(off)/2) (Eq. 18-8) 
V2err = (0 + 1.96 k\u2126)(200 nA/2) 
V2err = 196 \u3bcV 
V3err = Vin(off) = 6 mV 
AV(CL) = 2 1( / )R R\u2032 + 1 (Eq. 18-12) 
AV(CL) = (100 k\u2126/2 k\u2126) + 1 
AV(CL) = 51 
Verror = ± AV(CL)( ± V1err ± V2err ± V3err) 
Verror = 51(980 \u3bcV + 196 \u3bcV + 6 mV) 
Verror = 366 mV 
Answer: The output voltage is 366 mV. 
 18-31. Answer: 
For IB1: 
V1\u2014increase. Because of the increase in voltage drop 
across the resistor. 
V2\u2014no change. Not affected. 
Vin\u2014increase. Because of the increase in V1. 
Vout\u2014increase. Because of the increase in input voltage. 
MPP\u2014no change. Since the load resistance and VCC did 
not change. 
fmax\u2014no change. Since slew rate did not change. 
For IB2: 
V1\u2014no change. Not affected. 
V2\u2014increase. Because of the increase in voltage drop 
across the resistor. 
1-93 
Vin\u2014increase. Because of the increase in V2. 
Vout\u2014increase. Because of the increase in input voltage. 
MPP\u2014no change. Since the load resistance and VCC did 
not change. 
fmax\u2014no change. Since slew rate did not change. 
 18-32. Given: 
For VCC: 
V1\u2014no change. Not affected. 
V2\u2014no change. Not affected. 
Vin\u2014no change. Not affected. 
Vout\u2014no change. Not affected. 
MPP\u2014increase Since VCC is increased. 
fmax\u2014no change. Since slew rate did not change. 
 18-33. Given: 
V1\u2014no change. Not affected. 
V2\u2014no change. Not affected. 
Vin\u2014no change. Not affected. 
Vout\u2014no change. Not affected. 
MPP\u2014no change. Since the load resistance and VCC did 
not change. 
fmax\u2014increase. Since slew rate increased. 
 18-34. Given: 
V1\u2014no change. Not affected. 
V2\u2014no change. Not affected. 
Vin\u2014no change. Not affected. 
Vout\u2014no change. Not affected. 
MPP\u2014no change. Since the load resistance and VCC did 
not change. 
fmax\u2014decrease. Since the increase in voltage causes the 
rate of voltage rise to increase. 
Chapter 19 Negative Feedback 
SELF-TEST 
1. b 8. b 15. b 22. d 
2. d 9. b 16. d 23. d 
3. a 10. b 17. c 24. b 
4. a 11. d 18. b 25. a 
5. a 12. b 19. c 26. b 
6. c 13. b 20. b 27. d 
7. b 14. b 21. c 28. a 
JOB INTERVIEW QUESTIONS 
 8. Increased voltage gain and possible oscillation. 
 12. Current amplifier and transconductance amplifier. 
PROBLEMS 
 19-1. Given: 
R1 = 2.7 k\u2126 
Rf = 68 k\u2126 
AVOL(dB) = 88 dB 
Solution: 
B = R1/(R1 + Rf) (Eq. 19-6) 
B = 2.7 k\u2126/(2.7 k\u2126 + 68 k\u2126) 
B = 0.038 
AV = 1/B (Eq. 19-4) 
AV = 1/0.038 
AV = 26.32 
Av = antilog(Av(dB)/20) (Eq. 16-15) 
Av = antilog(88 dB/20) 
Av = 25,119 
%error = 100%(1 + AVOLB) (Eq 19-5) 
%error = 100%[1 + 25,119(0.038)] 
%error = 0.10% 
AV = AVOL/(1 + AVOLB) (Eq. 19-3) 
AV = 25,119/[l + 25,119(0.038)] 
AV = 26.29 
Answer: The feedback fraction is 0.038, the ideal 
closed-loop voltage gain is 26.32, the percent error is 
0.10%, and the exact voltage gain is 26.29. 
 19-2. Given: 
R1 = 2.7 k\u2126 
Rf = 39 k\u2126 
AVOL(dB) = 88 dB 
Solution:
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