55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Homework Assignment 13
Question 1
Short Takes – 2 points each.
1.
Classify the type of feedback uses in the circuit below (i.e., shunt-shunt, series-shunt, …)
2.
Answer: Series-shunt.
True or false: an engineer uses series-shunt negative feedback to extend the bandwidth of a
voltage amplifier—this will also increase the input resistance.
3.
Answer: True
What type of negative feedback (series-shunt, series-series,…) is used in the following
amplifier?
4.
Answer: Shunt-shunt
True or false: voltage regulators use negative feedback to stabilize/regulate their output
voltages: a side effect is that their output resistances are high.
5.
Answer: False
An amplifier has gain of 800. After adding negative feedback, the gain is measured as 25.
Find the loop gain.
Answer. 𝐴𝑓 = 𝐴𝑂𝐿 ⁄(1 + 𝛽𝐴𝑂𝐿 ) so that 25 = 800⁄(1 + 800𝛽). Solving for 𝑇 = 800𝛽
yields the loop gain 𝑇 = 31.
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
6.
An amplifier has gain of 800. After adding negative feedback, the gain is measured as 25.
Find the feedback factor.
Answer 𝐴𝑓 = 𝐴𝑂𝐿 ⁄(1 + 𝛽𝐴𝑂𝐿 ) so that 25 = 800⁄(1 + 800𝛽). Solving for 𝛽 yields
𝛽 = 0.0388
7.
8.
9.
An amplifier with gain of 200 has a 10% variation in gain over a certain frequency range.
Using negative feedback, what value of 𝛽 should one use to reduce the gain variation to
1%?
Answer. The improvement factor we want from the negative feedback is
Δ𝐴𝑂𝐿 )⁄Δ𝐴𝑓 = 10%⁄1% = 10. Therefore, (1 + 𝛽𝐴𝑂𝐿 ) = 10 ⇒ (1 + 200𝛽) = 10 ⇒
𝛽 = 0.045
An amplifier has gain of 100,000, and a 20% variation in gain over a certain temperature
range. Negative feedback is used to reduce the gain to 10. What is the variation in gain
with temperature of the feedback amplifier?
Answer. The gain is reduced by 1 + 𝛽𝐴𝑂𝐿 = 100,000⁄10 = 10,000. The temperature
variations are reduced by the same factor, so the feedback amplifier’s gain varies by
20% ⁄104 = 0.002%
An op-amp has an open-loop gain of 120 dB and an input resistance of 50 MΩ. An
engineer wants to use negative feedback to obtain an amplifier with input resistance of
5 GΩ. What is the gain (in dB) of the feedback amplifier? (2 points)
Answer. Negative feedback increases the resistance by (5 × 109 )⁄(50 × 106 ) = 100 (or
40 dB) and reduces the gain by the same factor, so the feedback amplifier’s gain is 80 dB.
10. A single-pole op-amp has an open-loop low-frequency gain of 𝐴 = 105 and an open
loop, 3-dB frequency of 4 Hz. If an inverting amplifier with closed-loop low-frequency
gain of �𝐴𝑓 � = 50 uses this op-amp, determine the closed-loop bandwidth.
11.
Answer. The gain-bandwidth product is 4 × 105 Hz. The bandwidth of the closed-loop
amplifier is then is 4 × 105 /50 = 8 kHz.
A single-pole op-amp has an open-loop gain of 100 dB and a unity-gain bandwidth
frequency 5 MHz. What is the open-loop bandwidth of the amplifier? The amplifier is
used as a voltage follower. What is the bandwidth of the follower?
Answer: A gain of 100 dB corresponds to 105 and the gain-bandwidth product is 5 MHz.
Thus, the open-loop bandwidth is (5 MHz)⁄105 = 50 Hz. A unity follower will have a
bandwidth of 5 MHz.
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 2 A certain audio power amplifier with a signal gain of 10 V/V is found to produce a
2-V peak-to-peak 60-Hz hum. We wish to reduce the output hum to less than 1 mV peak-topeak without changing the signal gain. To this end, we precede the power stage with a
preamplifier stage with gain 𝑎1 and then apply negative feedback around the composite
amplifier. What are the required values of 𝑎1 and 𝛽? Provide 𝛽 to four significant digits.
(6 points)
Original amplifier with 60-Hz hum
problem.
Preamplifier and negative feedback to fix hum
problem.
Solution
We need to reduce the hum by a factor (2 V)⁄(1 mV) = 2,000. Thus, the “improvement factor”
is
⋯ (1)
1 + 𝛽𝐴 = 2,000
where 𝐴 is the open-loop gain of the composite amplifier—the original × 10 amplifier and the
preamplifier, namely 𝐴 = 10𝑎1 . The closed-loop gain of the composite amplifier must be 10, so
𝐴𝑓 =
𝐴
𝐴
⇒ 10 =
1 + 𝛽𝐴𝑜𝑙
1 + 𝛽𝐴
⋯ (2)
We have two unknowns 𝛽 and 𝐴, and two equations. Solving yields 𝛽 = 0.09995, and 𝐴 =
20,000. Thus, 𝑎1 = 𝐴⁄10 = 2,000.
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 3 The amplifier below has an open-loop gain 𝐴𝑂𝐿 = 80 dB. What is 𝛽, the loop gain
𝑇, and the closed-loop gain 𝐴𝑓 ? (6 points)
𝑅1 = 1K
𝑅2 = 47K
𝑅𝐿 = 4.7K
Solution
𝑅1 , 𝑅2 form a voltage divider and feeds back a fraction 𝛽 = 1⁄(1 + 47) = 0.02083 of the
output voltage. The loop gain is 𝑇 = 𝛽𝐴𝑂𝐿 = (0.02083)(104 ) = 208.3. The closed-loop gain
is 𝐴𝑂𝐿 ⁄(1 + 𝛽𝐴𝑂𝐿 ) = 104 ⁄(1 + 208.3) = 47.78.
Question 4 For the non-inverting op-amp circuit below, the parameters are 𝐴 = 105 , 𝐴𝑣𝑓 =
20, 𝑅𝑖 = 100K, and 𝑅𝑜 = 100 Ω. Determine 𝑅𝑖𝑓 and 𝑅𝑜𝑓 respectively (6 points)
Solution
𝐴𝑓 =
𝐴𝑂𝐿
105
⇒ 20 =
⇒ 1 + 𝛽𝐴𝑂𝐿 = 5,000
1 + 𝛽𝐴𝑂𝐿
1 + 𝛽𝐴𝑂𝐿
𝑅𝑖𝑓 = (1 + 𝛽𝐴𝑂𝐿 )(100K) = 500.1M
100
𝑅𝑜𝑓 =
= 20 mΩ
1 + 𝛽𝐴𝑂𝐿
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 5 The parameters of the ideal shunt-series amplifier below are 𝐼𝑖 = 20 𝜇A, 𝐼𝑓𝑏 =
19 𝜇A, 𝑅𝑖 = 500 Ω, 𝑅𝑜 = 20K, and 𝛽𝑖 = 0.0095 A/A.. The open-loop gain is 𝐴𝑖 = 2,000 A⁄A.
Determine the values and units for 𝐼𝜀 , 𝐼𝑜 , 𝐴𝑓 , 𝑅𝑖𝑓 , and 𝑅𝑜𝑓 . (8 points)
Solution
𝐼𝜖 = 20 𝜇A − 19 𝜇A = 1 𝜇A
𝐼𝑓𝑏 19 × 10−6
𝐼𝑓𝑏 = 𝛽𝑖 𝐼𝑜 ⇒ 𝐼𝑜 =
=
= 2 mA
𝛽𝑖
0.0095
For the overall (feedback) amplifier
𝐴𝑓 =
𝐴𝑖
2,000
=
= 100 A/A
1 + 𝛽𝑖 𝐴𝑖 1 + (0.0095)(2,000)
Alternatively, use the approximation 𝐴𝑓 ≅ 1⁄𝛽𝑖 = 1⁄0.0095 = 105.3 A/A. One could also
determine 𝐼𝑂 as follows
𝑅𝑖𝑓 =
𝐼𝑂 = 𝐴𝑓 𝐼𝑖 = (20 × 10−6 )(100) = 2 mA
𝑅𝑖
500
=
= 25 Ω
1 + 𝛽𝑖 𝐴𝑖 1 + (0.0095)(2,000)
𝑅𝑜𝑓 = (1 + 𝛽𝑖 𝐴𝑖 )𝑅𝑜 = �1 + (0.0095)(2,000)�(20K) = 400K
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 6 The open-loop gain and input resistance of the
op-amp below is 106 and 1 MΩ respectively. Further,
𝑅1 = 99K, 𝑅2 = 1K. What is the closed-loop gain and input
resistance? (5 points)
Solution This is series-shunt (voltage-voltage) feedback, with 𝛽 = 𝑅2 ⁄(𝑅1 + 𝑅2 ) = 0.01.
Further, 1 + 𝛽𝐴𝑂𝐿 = 1 + (0.01)(106 ) = 104 . Thus
𝐴𝑓 =
𝐴𝑂𝐿
106
= 4 = 100
1 + 𝛽𝐴𝑂𝐿 10
𝑅𝑖𝑓 = 𝑅𝑖 (1 + 𝛽𝐴𝑂𝐿 ) = (106 )(104 ) = 104 MΩ
Question 7 An op-amp having a single-pole at 100 Hz, and a low-frequency gain of 105 is
operated is a feedback loop with 𝛽 = 0.01.
(a) What is the factor which feedback shifts the pole? (2 points)
(b) To what frequency? (2 points)
(c) If 𝛽 is changed to a value that results in a closed loop gain of +1, to what frequency does
the pole shift? (2 points)
Solution
Part (a). Feedback shifts the pole by a factor (1 + 𝛽𝐴) = 1 + 0.01 × 105 =1001
Part (b). The pole is shifted to 1001 × 100 = 100.1 kHz
Part (c)
𝐴𝑓 = 1 =
𝐴𝑂𝐿
105
=
1 + 𝛽𝐴𝑂𝐿
1 + 𝛽105
Thus (1 + 𝛽𝐴𝑂𝐿 ) = 105 . Feedback scales the bandwidth/shifts the pole by this factor to 100 ×
106 = 10 MHz.
Alternate solution. The gain-bandwidth product is 100 × 105 MHz , so that when the gain is 1,
the bandwidth is 10 MHz.
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 8 Part (a) of the figure below shows a non-feedback amplifier with gain A that
delivers 5 W into a 5 Ω speaker when the amplifier input is 50 mV rms. The nonlinear distortion
in the amplifier output is 1% of the total signal. In part (b) negative feedback is employed to
reduce the nonlinear distortion. A preamplifier is used to compensate for changes in gain the
feedback introduces
(a) Find the numerical value of the voltage gain A. (3 points).
(b) Find the value of β required to reduce the distortion to 0.1% with the same output signal
amplitude. Find the value of the preamplifier voltage gain 𝐴𝑃𝑅 . You can assume the
preamplifier nonlinear distortion is negligible. (5 points)
Solution
Part (a)
𝑣𝐿(rms) = �𝑃𝑅𝐿 = √5 × 5 = 5 V
𝐴 = 5/(50 × 10−3 ) = 100 (open loop)
Part (b) We need to reduce the nonlinear distortion by a factor 10. Thus
1 + 𝛽𝐴 = 10
The closed-loop gain is
𝛽 = 0.09
𝐴𝑓 =
10
= 10
1 + 𝛽𝐴
To provide the same output voltage, the pre-amplifier must have a voltage gain of 10.
7
55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 9 Consider an op-amp having a single-pole open-loop response with 𝐴𝑜 = 105 and an
open-loop 3-dB bandwidth of 10 Hz. The amplifier is ideal otherwise. The amplifier is
connected in the non-inverting configuration with a nominal low-frequency closed-loop gain of
100. (a) Find the feedback factor 𝛽. (2 points) (b) Make neat Bode plots showing the open-loop
gain and the phase of the open-loop amplifier. (6 points) (c) Add a plot of the loop gain 𝑇 to
you figure and find the frequency at which 𝑇 = 1. (3 points) (e) Find the phase margin of the
closed loop amplifier. (3 points) (f) Is the amplifier stable? (1 point)
Solution
𝐴𝑜
1 + 𝛽𝐴𝑜
105
100 =
1 + 𝛽105
𝛽 ≅ 0.01
𝐴𝑓 =
A plot of |𝛽𝐴𝑜 | is identical to that of 𝐴𝑜 , except that the magnitude is scaled by 𝛽. The phase is
the same.
From the plot, 𝑇 = |𝛽𝐴𝑜 | = 1 at f = 104. The phase at f = 104 is −90°, leaving another is −90°
before the phase shift becomes is −180° and making the amplifier unstable. Thus, the phase
margin is 90°. The feedback amplifier is stable.
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 10 The open-loop voltage gain of an amplifier is given by
𝐴𝑣 =
�1 + 𝑗
105
𝑓
𝑓
� �1 + 𝑗 5 �
103
10
An engineer used the amplifier to design a feedback amplifier with closed-loop gain 𝐴𝑓𝑣 = 100.
Will the amplifier be stable? If so, what is the phase margin? (15 points)
Solution
Determine 𝛽 for a closed-loop gain of 100:
𝐴𝑣
1 + 𝛽𝐴𝑣
105
100 =
1 + 𝛽(105 )
𝛽 = 9.99 × 10−3
𝐴𝑣𝑓 =
Find the frequency where the magnitude of the loop gain function is 1:
|𝑇(𝑓)| =
9.99 × 10−3 (105 )
2
2
�1 + � 𝑓 3 � �1 + � 𝑓 5 �
10
10
=1
Program this value into a programmable calculator and try different values for 𝑓 to find 𝑓 ≅
3.08 × 105 Hz. The phase of the loop gain function 𝑇(𝑓) = 𝛽𝐴𝑣 at this frequency is
𝑓
𝑓
−1
−
tan
103
105
5
3.08 × 10
3.08 × 105
−1
= tan−1
−
tan
103
105
= −161.8o
𝜙 = − tan−1
The phase margin is then 180 − 161.8 = 18.2o
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 11 Consider a feedback amplifier with loop gain transfer function
𝑇(𝑠) =
𝛽(100)
3
𝑠
�1 +
�
3
5 × 10
Determine the stability with 𝛽 = 0.2 (12 points)
Solution Substitute 𝑠 = 𝑗𝜔 = 𝑗2𝜋𝑓 to find the loop gain and 𝛽 = 0.2
𝑇(𝑓) =
𝛽(100)
�1 + 𝑗
2𝜋𝑓 3
�
5 × 103
=
20
2𝜋𝑓 2
�1 + �
� �
5 × 103
3/2
∠ −3 tan−1 �
The frequency where the phase become −180° is
−3 tan−1 �
𝑓180
� = −180° ⇒ 𝑓180 = 173 kHz
105
The magnitude of loop gain at this frequency is
|𝑇(𝑓)| =
20
3/2
2𝜋𝑓180 2
�1 + �
� �
5 × 103
Thus, the amplifier is unstable.
10
= 2.5
𝑓
�
105