CHAPTER 10 FEEDBACK
Chapter Outline
10.1 The General Feedback Structure
10.2 Some Properties of Negative Feedback
10.3 The Four Basic Feedback Topologies
10.4 The Feedback Voltage Amplifier (Series-Shunt)
10.5 The Feedback Transconductance Amplifier (Series-Series)
10.6 The Feedback Transresistance Amplifier (Shunt-Shunt)
10.7 The Feedback Current Amplifier (Shunt-Series)
10.9 Determining the Loop Gain
10.10 The Stability Problem
10.11 Effect of Feedback on the Amplifier Poles
10.12 Stability Study Using Bode Plots
10.13 Frequency Compensation
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10.1 The General Feedback Structure
Feedback amplifier
Signal-flow diagram of a feedback amplifier
Open-loop gain: A
Feedback factor:
Loop gain: A
Amount of feedback: 1+A
𝑥 𝐴
Gain of the feedback amplifier (closed-loop gain): 𝐴 ≡ =
𝑥 1 + 𝐴𝛽
Negative feedback:
The feedback signal xf is subtracted from the source signal xs
Negative feedback reduces the signal that appears at the input of the basic amplifier
The gain of the feedback amplifier Af is smaller than open-loop gain A by a factor of (1+A)
The loop gain A is typically large (A >>1):
The gain of the feedback amplifier (closed-loop gain) Af 1/
Closed-loop gain is almost entirely determined by the feedback network better accuracy of Af
xf = xs(A)/(1+A) xs error signal xi = xs – xf
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10.2 Some Properties of Negative Feedback
Gain desensitivity
The negative feedback reduces the change in the closed-loop gain due to open-loop gain variation
𝑑𝐴 1 𝑑𝐴 1 𝑑𝐴
= → =
𝑑𝐴 1 + 𝐴𝛽 𝐴 1 + 𝐴𝛽 𝐴
Desensitivity factor: 1+A
Bandwidth extension
High-frequency response of a single-pole amplifier:
𝐴 𝐴 / 1+𝐴 𝛽
𝐴 𝑠 = →𝐴 𝑠 =
1 + 𝑠/𝜔 1 + 𝑠/𝜔 1 + 𝐴 𝛽
Low-frequency response of an amplifier with a dominant low-frequency pole:
𝑠𝐴 𝑠𝐴 / 1 + 𝐴 𝛽
𝐴 𝑠 = →𝐴 𝑠 =
𝑠+𝜔 𝑠+𝜔 / 1+𝐴 𝛽
Negative feedback:
Reduces the gain by a factor of (1+AM)
Extends the bandwidth by a factor of (1+AM)
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Interference reduction
The signal-to-noise ratio:
The amplifier suffers from interference introduced at the input of the amplifier
Signal-to-noise ratio: SNR = Vs/Vn
Enhancement of the signal-to-noise ratio:
Precede the original amplifier A1 by a clean amplifier A2
Use negative feedback to keep the overall gain constant
𝐴 𝐴 𝐴 𝑉
𝑉 =𝑉 +𝑉 → 𝑆𝑁𝑅 = 𝐴
1+𝐴 𝐴 𝛽 1+𝐴 𝐴 𝛽 𝑉
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Reduction in nonlinear distortion
The amplifier transfer characteristic is linearized through the application of negative feedback
= 0.01
𝐴
𝐴 = 1000 → 𝐴 = = 90.9
1+𝐴 𝛽
𝐴
𝐴 = 100 → 𝐴 = = 50
1+𝐴 𝛽
The gain decreases
The input linear range increases
Exercise 9.3
Exercise 9.4
Exercise 9.5
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10.3 The Four Basic Feedback Topologies
Voltage amplifiers
The most suitable feedback topologies is voltage-mixing and voltage-sampling one
Known as series-shunt feedback
Example:
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Current amplifiers
The most suitable feedback topologies is current-mixing and current-sampling one
Known as shunt-series feedback
Example:
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Transconductance amplifiers
The most suitable feedback topologies is voltage-mixing and current-sampling one
Known as series-series feedback
Example:
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Transresistance amplifiers
The most suitable feedback topologies is current-mixing and voltage-sampling one
Known as shunt-shunt feedback
Example:
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10.4 The Feedback Voltage Amplifier (Series-Shunt)
Ideal case for series-shunt feedback
Input resistance of the feedback amplifier: 𝑅 = (1 + 𝐴𝛽)𝑅
Output resistance of the feedback amplifier: 𝑅 = 𝑅 /(1 + 𝐴𝛽)
Voltage gain of the feedback amplifier (V/V): 𝐴 = 𝐴/(1 + 𝐴𝛽)
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The practical case for series-shunt feedback
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Analysis techniques for series-shunt feedback
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Example for series-shunt feedback
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Example for series-shunt feedback
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10.5 The Feedback Transconductance Amplifier (Series-Series)
Ideal case for series-series feedback
Input resistance of the feedback amplifier: 𝑅 = (1 + 𝐴𝛽)𝑅
Output resistance of the feedback amplifier: 𝑅 = (1 + 𝐴𝛽)𝑅
Transconductance gain of the feedback amplifier (-1): 𝐴 = 𝐴/(1 + 𝐴𝛽)
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The practical case for series-series feedback
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Analysis techniques for series-series feedback
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10.6 The Feedback Transresistance Amplifier (Shunt-Shunt)
Ideal case for shunt-shunt feedback
Input resistance of the feedback amplifier: 𝑅 = 𝑅 /(1 + 𝐴𝛽)
Output resistance of the feedback amplifier: 𝑅 = 𝑅 /(1 + 𝐴𝛽)
Transresistance gain of the feedback amplifier (): 𝐴 = 𝐴/(1 + 𝐴𝛽)
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10.7 The Feedback Current Amplifier (Shunt-Series)
Ideal case for shunt-series feedback
Input resistance of the feedback amplifier: 𝑅 = 𝑅 /(1 + 𝐴𝛽)
Output resistance of the feedback amplifier: 𝑅 = (1 + 𝐴𝛽)𝑅
Current gain of the feedback amplifier (A/A): 𝐴 = 𝐴/(1 + 𝐴𝛽)
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10.9 Determining the Loop Gain
Analysis of the feedback amplifier using the loop gain
In practical feedback amplifier, the feedback network may cause loading effect on the amplifier
And, sometimes, it is not easy to determine A and of the feedback amplifier
The loop-gain analysis method is introduced:
Identify the feedback network and use it to determine the value of
Determine the ideal value of the closed-loop gain Af as 1/, which is considered the upper-
bound value of Af to check the actual value in the calculation
Use open-loop analysis with equivalent loading to determine the loop gain A
Use the values of loop gain A and to determine the voltage gain A and Af
The value of loop gain determined using the method discussed here may differ somewhat from the
value determined by the approach studied in the previous session, but the difference is usually
limited to a few percent.
Open-loop analysis with equivalent loading:
Remove the external source
Break the loop with equivalent loading
Provide test signal Vt
Loop gain:
𝑉
𝐴𝛽 = −
𝑉
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Example
𝑅 || 𝑅 + 𝑅 || 𝑅 + 𝑅 𝑅 || 𝑅 + 𝑅 𝑅
𝐴𝛽 = 𝜇 × ×
𝑅 || 𝑅 + 𝑅 || 𝑅 + 𝑅 + 𝑟 𝑅 || 𝑅 + 𝑅 + 𝑅 𝑅 +𝑅
𝑅
→ 𝐴𝛽 = 𝜇 (for ideal op-amp)
𝑅 +𝑅
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Characteristic Equation
The gain of a feedback amplifier can be expressed as a transfer function (function of s) by taking
the frequency-dependent properties into consideration
The denominator determines the poles of the system and the numerator defines the zeros
From the study of circuit theory, the poles of a circuit are independent of the external excitation,
and the poles or the natural modes can be determined by setting the external excitation to zero
The characteristic equation and the poles are completely determined by the loop gain
𝑥 𝐴 𝑠 1 + 𝑎 𝑠 + ⋯+ 𝑎 𝑠
Transfer function: 𝐴 𝑠 ≡ = =
𝑥 1+𝐴 𝑠 𝛽 𝑠 1+ 𝑏 𝑠 +⋯+𝑏 𝑠
Characteristics equation: 1 + 𝑏 𝑠 + ⋯ + 𝑏 𝑠 = 0 → 1 + 𝐴 𝑠 𝛽 𝑠
Exercise 9.18
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10.10 The Stability Problem
Transfer function of the feedback amplifier
Transfer functions:
Open-loop transfer function: A(s)
Feedback transfer function: (s)
Closed-loop transfer function: Af (s)
𝐴 𝑠
𝐴 𝑠 =
1+𝐴 𝑠 𝛽 𝑠
For physical frequencies s = j
𝐴 𝑗𝜔
𝐴 𝑗𝜔 =
1 + 𝐴 𝑗𝜔 𝛽 𝑗𝜔
Loop gain: 𝐿 𝑗𝜔 ≡ 𝐴 𝑗𝜔 𝛽 𝑗𝜔 = |𝐴 𝜔 𝛽 𝜔 |𝑒 ( )
Evaluating the close-loop stability by the frequency response of the loop gain L(j):
For loop gain smaller than unity at 180:
Becomes positive feedback
Closed-loop gain becomes larger than open-loop gain
The feedback amplifier is still stable
For loop gain equal to unity at 180:
The amplifier will have an output for zero input (oscillation)
For loop gain larger than unity at 180:
Oscillation with a growing amplitude at the output
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The Nyquist plot
A plot used to evaluate the stability of a feedback amplifier
Plot the loop gain L(j) versus frequency on the complex plane
Magnitude decreases as frequency increases
Phase decreases as frequency increases due to the poles (final phase depends on number of poles)
Stability:
The plot does not encircle the point (-1, 0)
The magnitude of loop gain has to be less than unity when phase reaches -180
The system is more likely to become unstable as increases
Magnitude=A0 (increases with )
Exercise 9.20
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10.11 Effect of Feedback on the Amplifier Poles
Stability and pole location
The stability can be evaluated by the poles of the closed-loop transfer function
The poles have to be in the left half of the s-plane to ensure stability
Consider an amplifier with a pole pair at 𝑠 = 𝜎 ± 𝑗𝜔
The transient response contains the terms of the form 𝑣 𝑡 = 𝑒 𝑒 +𝑒 = 2𝑒 cos(𝜔 𝑡)
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Poles of the feedback amplifier
Characteristic equation: 1+A(s)(s) = 0
The feedback amplifier poles are obtained by solving the characteristic equation
Amplifier with single-pole response
𝐴 𝐴 / 1+𝐴 𝛽
𝐴 𝑠 = →𝐴 𝑠 =
1 + 𝑠/𝜔 1 + 𝑠/𝜔 1 + 𝐴 𝛽
𝜔 = 𝜔 1+𝐴 𝛽
The feedback amplifier is still a single-pole system
The pole moves away from origin in the s-plane as feedback ( ) increases
The bandwidth is extended by feedback at the cost of a reduction in gain
Unconditionally stable system (the pole never enters the right-half plane)
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Amplifier with two-pole response
Feedback amplifier
𝐴 𝐴(𝑠) 𝐴
𝐴 𝑠 = →𝐴 𝑠 = =
1 + 𝑠/𝜔 1 + 𝑠/𝜔 1 + 𝐴(𝑠)𝛽 1 + 𝑠/𝜔 1 + 𝑠/𝜔 +𝐴 𝛽
Still a two-pole system
Characteristic equation
𝑠 +𝑠 𝜔 +𝜔 + 1+𝐴 𝛽 𝜔 𝜔 =0
The closed-loop poles are given by
1 1
𝑠=− 𝜔 +𝜔 ± 𝜔 +𝜔 −4 1+𝐴 𝛽 𝜔 𝜔
2 2
The plot of poles versus is called a root-locus diagram
Unconditionally stable system (the pole never enters the right-half plane)
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Amplifier with three or more poles
Root-locus diagram:
As increases, the two poles become coincident and then become complex and conjugate
A value of exists at which this pair of complex-conjugate poles enters the right half of the s plane
The feedback amplifier is stable only if does not exceed a maximum value
Frequency compensation is adopted to ensure the stability
Exercise 9.22
Exercise 9.23
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10.12 Stability Study Using Bode Plots
Gain and phase margin
The stability of a feedback amplifier is determined by examining its loop gain as a function of
frequency L(j)= L(j)(j)
One of the simplest means is through the use of Bode plot for A
Stability is ensured if the magnitude of the loop gain is less than unity at a frequency shift of 180
Gain margin:
The difference between the value | A | of at 180 and unity
Gain margin represents the amount by which the loop gain can be increased while maintaining
stability
Phase margin:
A feedback amplifier is stable if the phase is
less than 180 at a frequency for which | A | =1
A feedback amplifier is unstable if the phase is
in excess of 180 at a frequency for which | A | =1
The difference between the phase at a frequency
(1) for which | A | =1 and -180
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Effect of phase margin on closed-loop response
Consider a feedback amplifier with a large low-frequency loop gain (A0 >> 1)
The closed-loop gain at low frequencies is approximately 1/
Denoting the frequency at which | A | = 1 by 1: A(j1) = 1e-j and - = 180 - phase margin
𝐴 1
𝐴 𝜔=0 = ≈
1+𝐴 𝛽 𝛽
𝐴(𝑗𝜔 ) 1 𝑒
𝐴 𝜔=𝜔 = = ×
1 + 𝐴(𝑗𝜔 )𝛽 𝛽 1 + 𝑒
1 1
|𝐴 𝜔 = 𝜔 | = ×
𝛽 |1 + 𝑒 |
Closed-loop gain at 1 peaks by a factor of 1.3 above the low-frequency gain for phase margin of 45
This peaking increases as the phase margin is reduced, eventually reaching infinite when the phase
margin is zero (sustained oscillations)
Closed-loop gain
(1) PM = 90 (- = -90) |Af(1)| = 0.707(1/)
(2) PM = 60 (- = -120) |Af(1)| = 1(1/)
(3) PM = 45 (- = -135) |Af(1)| = 1.3(1/)
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An alternative approach for investigating stability
In a Bode plot, the difference between 20log|A(j)| and 20log(1/) is 20log|A |
Example:
10
𝐴 𝑓 =
𝑓 𝑓 𝑓
1+𝑗 1+𝑗 1+𝑗
10 10 10
𝑓 𝑓 𝑓
𝜑 𝑓 = − 𝑡𝑎𝑛 + 𝑡𝑎𝑛 + 𝑡𝑎𝑛
10 10 10
(a)
= 0.000056
1/ = 17782 (85 dB)
f0-dB = 5.6105 Hz
(f0-dB) = -108
PM = 72
-180 = -90-tan-1(f180/106) -tan-1(f180/107)
f180 = 3.17106 Hz
|A(f180)| = 60 dB
GM = 25 dB
(b)
= 0.00316
1/ = 316 (50 dB)
f0-dB > f180 (Unstable)
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10.13 Frequency Compensation
Theory
Modify the open-loop transfer function A(s) so that the closed-loop amplifier is stable for a given
closed-loop gain
The simplest method for frequency compensation is to introduce a new pole at sufficiently low
frequency fD
The disadvantage of introducing a new pole at lower frequency is the significant bandwidth
reduction
Alternatively, the dominant pole can be shifted to a lower frequency f D such that the amplifier is
compensated without introducing a new pole
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Increase the time-constant of the dominant pole by adding additional capacitance
Add external capacitance CC at the node which contributes to a dominant pole
The required value of CC is usually quite large, making it unsuitable for IC implementation
1
𝑓 =
2𝜋𝑅 𝐶
1
𝑓 =
2𝜋𝑅 (𝐶 + 𝐶 )
Assume Rx = 1.6 M and Cx = 1 pF: fP1 = 105 Hz
For f D = 103 Hz, the required CC is 99 pF
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Miller compensation and pole splitting
Miller effect equivalently increase the capacitance by a factor of voltage gain
Use miller capacitance for compensation can reduce the need for large capacitance
1
𝑓 =
2𝜋𝑅 𝐶
1
𝑓 =
2𝜋𝑅 𝐶
𝑉 𝑠𝐶 − 𝑔 𝑅 𝑅
=
𝐼 1 + 𝑠 𝐶 𝑅 + 𝐶 𝑅 + 𝐶 𝑔 𝑅 𝑅 + 𝑅 +𝑅 + 𝑠 𝐶 𝐶 + 𝐶 𝐶 + 𝐶 𝑅 𝑅
1 1 1 1
𝑓 = ≈
2𝜋 𝑅 𝐶 + 𝑅 𝐶 + 𝐶 𝑔 𝑅 𝑅 + 𝑅 + 𝑅 2𝜋 𝑔 𝑅 𝑅 𝐶
1 𝑔 𝐶
𝑓 =
2𝜋 𝐶 𝐶 + 𝐶 𝐶 + 𝐶 𝐶
Pole splitting: as Cf increases, low-frequency pole reduces and high-frequency pole increases
It is desirable in terms of phase margin
The phase margin of an open-loop op amp defines the worst-case phase margin of a closed-loop
amplifier with = 1 (loop gain A = A)
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