Chapter 19 Negative Feedback
12. Current amplifier and transconductance amplifier
PROBLEMS 19-1. Given:
R1 = 2.7 kΩ Rf = 68 kΩ AVOL(dB) = 88 dB Solution:
B = R1/(R1 + Rf) (Eq. 19-6) B = 2.7 kΩ/(2.7 kΩ + 68 kΩ)
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 kΩ Rf = 39 kΩ AVOL(dB) = 88 dB Solution:
B = R1/(R1 + Rf) (Eq. 19-6) B = 2.7 kΩ/(2.7 kΩ + 39 kΩ)
B = 0.065
AV = 1/B (Eq. 19-4) AV = 1/0.065
AV = 15.44
Answer: The feedback fraction is 0.065, and the closed-loop voltage gain is 15.44.
19-3. Given:
R1 = 4.7 kΩ Rf = 68 kΩ AVOL(dB) = 88 dB Solution:
B = R1/(R1 + Rf) (Eq. 19-6) B = 4.7 kΩ/(4.7 kΩ + 68 kΩ)
B = 0.065
AV = 1/B (Eq. 19-4) AV = 1/0.065
AV = 15.47
Answer: The feedback fraction is 0.065, and the closed-loop voltage gain is 15.47.
19-4. Given:
R1 = 2.7 kΩ Rf = 68 kΩ AVOL(dB) = 108 dB Solution:
B = R1/(R1 + Rf) (Eq. 19-6) B = 2.7 kΩ/(2.7 kΩ + 68 kΩ)
B = 0.038
AV = 1/B (Eq. 19-4) AV = 1/0.038
AV = 26.32
AV = antilog(AVOL(dB)/20) (Eq. 16-15) AV = antilog(108 dB/20)
AV = 251,189
%error = 100%(1 + AVOLB) (Eq 19-5)
%error = 100%[1 + 251,189(0.038)]
%error = 0.01%
AV = AVOL/(1 + AVOLB) (Eq. 19-3) AV = 251,189/[l + 251,189(0.038)]
AV = 26.31
Solution
Answer: The output voltage is 1 V.
19-24. Given:
Answer: The voltage gains are 51 at the 1-kΩ position, 3 at the 25-kΩ position, and 1.5 at the 100-kΩ position.
19-25. Given:
Answer: The output voltages are 510 mV at the 1-kΩ position, 30 mV at the 25-kΩ position, and 15 mV at the 100-kΩ position.
Answer: At the 1-kΩ position the input impedance is 3,924 MΩ and the output impedance is 38 mΩ. At the 25-kΩ position the input impedance is 66,669 MΩ and the output impedance is 2.5 mΩ. At the 100-kΩ position the input impedance is 133,335 MΩ and the output impedance is 1.25 mΩ. Note: The RCM of the op amp is not included in the calculations for input impedance. See Example 19-2.
Rmin = 130 Ω
Answer: The minimum bandwidth is 712 Hz and the maximum bandwidth is 38.2 kHz.
20-4. Given:
Answer: The minimum bandwidth is 937 Hz and the maximum bandwidth is 11.5 kHz. The output voltage is 266.8 mV.
Answer: The midband voltage gain is 42, the upper cutoff frequency is 71.4 kHz, and the lower cutoff frequency is 79.6 Hz.
20-6. Given:
Answer: The midband voltage gain is 46.5, the upper cutoff frequency is 21.5 kHz, and the lower cutoff frequency is 10.3 Hz.
20-7. Given:
Solution:
Answer: When the gate is low, the output is 2.26 mV;
when the gate is high, the output is 26.2 mV.
20-9. Given:
Answer: When the gate is low, the output is 4.4 mV, and when the gate is high, the output is 72.4 mV.
20-10. Given:
Answer: The new output reference voltage is 5 V.
20-11. Given:
Answer: The maximum gain is –10, and the maximum positive gain is 0.
20-12. Given: R1 = R2
Solution: At ground the circuit is an inverting amplifier.
AV = –Rf/R1
AV = –1
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 = –1 + 0.2 AV = –0.8
Answer: The gain with the wiper at ground is –1, and 10% away is –0.8.
Answer: The maximum positive gain is 15, and the maximum negative gain is –15.
20-14. Given:
Answer: The phase shift is –30.9° at 100 Hz, –140° at 1 kHz, and –176° at 10 kHz.
Answer: The differential voltage gain is –20, and the common mode gain is ±0.004.
20-16. Given:
Answer: The differential voltage gain is –20, and the common-mode gain is ±0.04.
20-17. Given:
Answer: No, the bridge is not balanced.
20-18. Given:
Answer: The output voltage is –5.63 V.
20-19. Given:
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:
Answer: The output voltage is 1 V.
20-22. Given:
Answer: The output voltage is –20 mV.
20-23. Given:
Answer: The output voltage is 19.3 mV.
Solution:
AV = (Rf/R1) + 1 AV = (15 kΩ/1.5 kΩ) + 1 AV = 11
f1 = 1/[2π(R/2)C1] f1 = 1/[2π(68 kΩ/2)(1 µF)]
f1 = 4.68 Hz f2 = 1/(2πRLC2)
f2 = 1/[2π(15 kΩ)(2.2 µF)]
f2 = 4.82 Hz f3 = 1/(2πR1C3)
f3 = 1/[2π(1 kΩ)(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 Ω R1(max) = 1010 Ω Rf(min) = 99 kΩ Rf(max) = 101 kΩ Solution:
AV(min) = –Rf(min)/R1(max)
AV(min) = –99 kΩ/1010 Ω AV(min) = 98
AV(max) = –Rf(max)/R1(min)
AV(max) = –101 kΩ/990 Ω AV(max) = 102
Answer: The minimum gain is 98, and the maximum gain is 102.
20-42. Given:
Transistor:
R1 = 22 kΩ Rf = 10 kΩ RS = 1 kΩ RE = 5.6 kΩ RC = 6.8 kΩ VCC = 15V Op amp R1 = 1 kΩ Rf = 47 kΩ Solution:
VBB = [Rf/(R1 + Rf + RS)]VCC (Eq. 8-1) VBB = [10 kΩ/(22 kΩ + 10 kΩ + 1 kΩ)]15 V VBB = 4.54 V
VE = VBB – VBE (Eq. 8-2) VE = 4.54 V – 0.7 V
VE = 3.84 V
IE = VE/RE (Eq. 8-3) IE = 3.84 V/5.6 kΩ
IE = 0.685 mA
r′ = 25 mV/Ie E (Eq. 9-10)
r′ = 25 mV/0.685 mA e
r′ = 36.5 Ω e
rc = RC
rc = 6.8 kΩ
Av = rc/r′ (Eq. e 10-7) Av = 6.8 kΩ/36.5 Ω
Av = 186 Op amp:
AV = (Rf/R1) + 1 AV = (47 kΩ/1 kΩ) + 1 AV = 48
AV = (48)(186) AV = 9114
Answer: The voltage gain is 9114.
20-43. Given:
R1 = 1 kΩ Rf = 10 kΩ RL = 100 Ω β = 50 vin = 0.5 VCC
Solution:
AV = –Rf/R1 AV = –10 kΩ/1 kΩ AV = –10
vout = AV(vin) vout = –10(0.5 V) vout = –5 V Iout = vout/RL
Iout = –5 V/100 Ω Iout = 50 mA IB = Iout/β 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.
21-20. Given:
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.
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.
Answer: The Q is 2.65, the voltage gain is –14, and the center frequency is 55.7 kHz.
21-23. Given:
Answer: The Q is 8.39, the voltage gain is –1.04, and the center frequency is 16.2 kHz.
21-24. Given: center frequency is 9.09 kHz.
21-25. Given:
Answer: The voltage gain is 1.5, the Q is 1, the resonant frequency is 15.8 Hz, and the bandwidth is 15.8 Hz.
21-26. Given:
R = 3.3 kΩ C = 220 nF
Roll-off = 20n dB/decade (Eq. 21-4a) Roll-off = 20(10) dB/decade
Roll-off = 200 dB/decade 4 kHz is 1 octave above Attenuation = 60 dB 8 kHz is 2 octaves above Attenuation = 120 dB 20 kHz is 1 decade above Attenuation = 200 dB
Answer: The attenuation is 60 dB at 4 kHz, 120 dB at