CHAPTER 10 OPERATIONAL-AMPLIFIER CIRCUITS

NTUEE Electronics – L.H. Lu 10-1

CHAPTER 10 FEEDBACK

Chapter Outline10.1 The General Feedback Structure10.2 Some Properties of Negative Feedback10.3 The Four Basic Feedback Topologies10.4 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 Gain10.10 The Stability Problem10.11 Effect of Feedback on the Amplifier Poles10.12 Stability Study Using Bode Plots10.13 Frequency Compensation


10.1 The General Feedback Structure

Feedback amplifierSignal-flow diagram of a feedback amplifier

Open-loop gain: A Feedback factor: Loop gain: A Amount of feedback: 1+A Gain of the feedback amplifier (closed-loop gain):

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

𝐴 ≡𝑥

𝑥=

𝐴

1 + 𝐴𝛽

10.2 Some Properties of Negative Feedback

Gain desensitivityThe negative feedback reduces the change in the closed-loop gain due to open-loop gain variation

Desensitivity factor: 1+ABandwidth extensionHigh-frequency response of a single-pole amplifier:

Low-frequency response of an amplifier with a dominant low-frequency pole:

Negative feedback: Reduces the gain by a factor of (1+AM) Extends the bandwidth by a factor of (1+AM)


𝑑𝐴

𝑑𝐴=

1

1 + 𝐴𝛽→

𝑑𝐴

𝐴=

1

1 + 𝐴𝛽

𝑑𝐴

𝐴

𝐴 𝑠 =𝐴

1 + 𝑠/𝜔→ 𝐴 𝑠 =

𝐴 / 1 + 𝐴 𝛽

1 + 𝑠/𝜔 1 + 𝐴 𝛽

𝐴 𝑠 =𝑠𝐴

𝑠 + 𝜔→ 𝐴 𝑠 =

𝑠𝐴 / 1 + 𝐴 𝛽

𝑠 + 𝜔 / 1 + 𝐴 𝛽

Interference reductionThe 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


𝑉 = 𝑉𝐴 𝐴

1 + 𝐴 𝐴 𝛽+ 𝑉

𝐴

1 + 𝐴 𝐴 𝛽→ 𝑆𝑁𝑅 =

𝑉

𝑉𝐴

Reduction in nonlinear distortionThe amplifier transfer characteristic is linearized through the application of negative feedback = 0.01

The gain decreases The input linear range increases

Exercise 9.3Exercise 9.4 Exercise 9.5


𝐴 = 1000 → 𝐴 =𝐴

1 + 𝐴 𝛽= 90.9

𝐴 = 100 → 𝐴 =𝐴

1 + 𝐴 𝛽= 50

10.3 The Four Basic Feedback Topologies

Voltage amplifiersThe most suitable feedback topologies is voltage-mixing and voltage-sampling oneKnown as series-shunt feedback

Example:


Current amplifiersThe most suitable feedback topologies is current-mixing and current-sampling oneKnown as shunt-series feedback

Example:


Transconductance amplifiersThe most suitable feedback topologies is voltage-mixing and current-sampling oneKnown as series-series feedback

Example:


Transresistance amplifiersThe most suitable feedback topologies is current-mixing and voltage-sampling oneKnown as shunt-shunt feedback

Example:


10.4 The Feedback Voltage Amplifier (Series-Shunt)

Ideal case for series-shunt feedback

Input resistance of the feedback amplifier:Output resistance of the feedback amplifier:Voltage gain of the feedback amplifier (V/V):


𝑅 = (1 + 𝐴𝛽)𝑅

𝑅 = 𝑅 /(1 + 𝐴𝛽)

𝐴 = 𝐴/(1 + 𝐴𝛽)

The practical case for series-shunt feedback



Analysis techniques for series-shunt feedback


Example for series-shunt feedback


Example for series-shunt feedback


10.5 The Feedback Transconductance Amplifier (Series-Series)

Ideal case for series-series feedback

Input resistance of the feedback amplifier:Output resistance of the feedback amplifier:Transconductance gain of the feedback amplifier (-1):


𝑅 = (1 + 𝐴𝛽)𝑅

𝑅 = (1 + 𝐴𝛽)𝑅

𝐴 = 𝐴/(1 + 𝐴𝛽)

The practical case for series-series feedback


Analysis techniques for series-series feedback


10.6 The Feedback Transresistance Amplifier (Shunt-Shunt)

Ideal case for shunt-shunt feedback

Input resistance of the feedback amplifier:Output resistance of the feedback amplifier:Transresistance gain of the feedback amplifier ():


𝑅 = 𝑅 /(1 + 𝐴𝛽)

𝑅 = 𝑅 /(1 + 𝐴𝛽)

𝐴 = 𝐴/(1 + 𝐴𝛽)

10.7 The Feedback Current Amplifier (Shunt-Series)

Ideal case for shunt-series feedback

Input resistance of the feedback amplifier:Output resistance of the feedback amplifier:Current gain of the feedback amplifier (A/A):


𝑅 = 𝑅 /(1 + 𝐴𝛽)

𝑅 = (1 + 𝐴𝛽)𝑅

𝐴 = 𝐴/(1 + 𝐴𝛽)

10.9 Determining the Loop Gain

Analysis of the feedback amplifier using the loop gainIn practical feedback amplifier, the feedback network may cause loading effect on the amplifierAnd, sometimes, it is not easy to determine A and of the feedback amplifierThe 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 sourceBreak the loop with equivalent loadingProvide test signal Vt

Loop gain:


𝐴𝛽 = −𝑉

𝑉

Example


𝐴𝛽 = 𝜇𝑅 || 𝑅 + 𝑅 || 𝑅 + 𝑅

𝑅 || 𝑅 + 𝑅 || 𝑅 + 𝑅 + 𝑟×

𝑅 || 𝑅 + 𝑅

𝑅 || 𝑅 + 𝑅 + 𝑅×

𝑅

𝑅 + 𝑅

→ 𝐴𝛽 = 𝜇𝑅

𝑅 + 𝑅 (for ideal op-amp)

Characteristic EquationThe gain of a feedback amplifier can be expressed as a transfer function (function of s) by taking

the frequency-dependent properties into considerationThe denominator determines the poles of the system and the numerator defines the zerosFrom 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 zeroThe characteristic equation and the poles are completely determined by the loop gain

Exercise 9.18


Transfer function:

Characteristics equation:

𝐴 𝑠 ≡𝑥

𝑥=

𝐴 𝑠

1 + 𝐴 𝑠 𝛽 𝑠=

1 + 𝑎 𝑠 + ⋯ + 𝑎 𝑠

1 + 𝑏 𝑠 + ⋯ + 𝑏 𝑠

1 + 𝑏 𝑠 + ⋯ + 𝑏 𝑠 = 0 → 1 + 𝐴 𝑠 𝛽 𝑠

10.10 The Stability Problem

Transfer function of the feedback amplifierTransfer functions: Open-loop transfer function: A(s) Feedback transfer function: (s) Closed-loop transfer function: Af (s)

For physical frequencies s = j

Loop gain: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 feedbackClosed-loop gain becomes larger than open-loop gainThe 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


𝐴 𝑠 =𝐴 𝑠

1 + 𝐴 𝑠 𝛽 𝑠

𝐴 𝑗𝜔 =𝐴 𝑗𝜔

1 + 𝐴 𝑗𝜔 𝛽 𝑗𝜔

𝐿 𝑗𝜔 ≡ 𝐴 𝑗𝜔 𝛽 𝑗𝜔 = |𝐴 𝜔 𝛽 𝜔 |𝑒 ( )

The Nyquist plotA plot used to evaluate the stability of a feedback amplifierPlot the loop gain L(j) versus frequency on the complex planeMagnitude decreases as frequency increasesPhase 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

Exercise 9.20


Magnitude=A0 (increases with )

10.11 Effect of Feedback on the Amplifier Poles

Stability and pole locationThe stability can be evaluated by the poles of the closed-loop transfer functionThe poles have to be in the left half of the s-plane to ensure stabilityConsider an amplifier with a pole pair at The transient response contains the terms of the form


𝑣 𝑡 = 𝑒 𝑒 + 𝑒 = 2𝑒 cos (𝜔 𝑡)

𝑠 = 𝜎 ± 𝑗𝜔


Poles of the feedback amplifierCharacteristic equation: 1+A(s)(s) = 0The feedback amplifier poles are obtained by solving the characteristic equation

Amplifier with single-pole response

The feedback amplifier is still a single-pole systemThe pole moves away from origin in the s-plane as feedback ( ) increasesThe bandwidth is extended by feedback at the cost of a reduction in gainUnconditionally stable system (the pole never enters the right-half plane)

𝐴 𝑠 =𝐴

1 + 𝑠/𝜔→ 𝐴 𝑠 =

𝐴 / 1 + 𝐴 𝛽

1 + 𝑠/𝜔 1 + 𝐴 𝛽

𝜔 = 𝜔 1 + 𝐴 𝛽


Amplifier with two-pole responseFeedback amplifier

Still a two-pole systemCharacteristic equation

The closed-loop poles are given by

The plot of poles versus is called a root-locus diagramUnconditionally stable system (the pole never enters the right-half plane)

𝐴 𝑠 =𝐴

1 + 𝑠/𝜔 1 + 𝑠/𝜔→ 𝐴 𝑠 =

𝐴(𝑠)

1 + 𝐴(𝑠)𝛽=

𝐴

1 + 𝑠/𝜔 1 + 𝑠/𝜔 + 𝐴 𝛽

𝑠 + 𝑠 𝜔 + 𝜔 + 1 + 𝐴 𝛽 𝜔 𝜔 = 0

𝑠 = −1

2𝜔 + 𝜔 ±

1

2𝜔 + 𝜔 − 4 1 + 𝐴 𝛽 𝜔 𝜔


Amplifier with three or more polesRoot-locus diagram:

As increases, the two poles become coincident and then become complex and conjugateA value of exists at which this pair of complex-conjugate poles enters the right half of the s planeThe feedback amplifier is stable only if does not exceed a maximum valueFrequency compensation is adopted to ensure the stability

Exercise 9.22Exercise 9.23

10.12 Stability Study Using Bode Plots

Gain and phase marginThe 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 AStability is ensured if the magnitude of the loop gain is less than unity at a frequency shift of 180Gain 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

stabilityPhase 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


Effect of phase margin on closed-loop responseConsider 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

Closed-loop gain at 1 peaks by a factor of 1.3 above the low-frequency gain for phase margin of 45This 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/)

𝐴 𝜔 = 0 =𝐴

1 + 𝐴 𝛽≈

1

𝛽

𝐴 𝜔 = 𝜔 =𝐴(𝑗𝜔 )

1 + 𝐴(𝑗𝜔 )𝛽=

1

𝛽×

𝑒

1 + 𝑒

|𝐴 𝜔 = 𝜔 | =1

𝛽×

1

|1 + 𝑒 |

An alternative approach for investigating stabilityIn a Bode plot, the difference between 20log|A(j)| and 20log(1/) is 20log|A |Example:


(a) = 0.0000561/ = 17782 (85 dB)f0-dB = 5.6105 Hz(f0-dB) = -108PM = 72-180 = -90-tan-1(f180/106) -tan-1(f180/107)f180 = 3.17106 Hz|A(f180)| = 60 dBGM = 25 dB

(b) = 0.003161/ = 316 (50 dB)f0-dB > f180 (Unstable)

𝐴 𝑓 =10

1 + 𝑗𝑓

101 + 𝑗

𝑓10

1 + 𝑗𝑓

10

𝜑 𝑓 = − 𝑡𝑎𝑛𝑓

10+ 𝑡𝑎𝑛

𝑓

10+ 𝑡𝑎𝑛

𝑓

10

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 gainThe 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


Increase the time-constant of the dominant pole by adding additional capacitanceAdd external capacitance CC at the node which contributes to a dominant poleThe required value of CC is usually quite large, making it unsuitable for IC implementation


Assume Rx = 1.6 M and Cx = 1 pF: fP1 = 105 HzFor f D = 103 Hz, the required CC is 99 pF

𝑓 =1

2𝜋𝑅 𝐶

𝑓 =1

2𝜋𝑅 (𝐶 + 𝐶 )

Miller compensation and pole splittingMiller effect equivalently increase the capacitance by a factor of voltage gainUse miller capacitance for compensation can reduce the need for large capacitance

Pole splitting: as Cf increases, low-frequency pole reduces and high-frequency pole increasesIt 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)


𝑓 =1

2𝜋𝑅 𝐶

𝑉

𝐼=

𝑠𝐶 − 𝑔 𝑅 𝑅

1 + 𝑠 𝐶 𝑅 + 𝐶 𝑅 + 𝐶 𝑔 𝑅 𝑅 + 𝑅 +𝑅 + 𝑠 𝐶 𝐶 + 𝐶 𝐶 + 𝐶 𝑅 𝑅

𝑓 =1

2𝜋𝑅 𝐶

𝑓 =1

2𝜋

1

𝑅 𝐶 + 𝑅 𝐶 + 𝐶 𝑔 𝑅 𝑅 + 𝑅 + 𝑅≈

1

2𝜋

1

𝑔 𝑅 𝑅 𝐶

𝑓 =1

2𝜋

𝑔 𝐶

𝐶 𝐶 + 𝐶 𝐶 + 𝐶 𝐶

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CHAPTER 10 OPERATIONAL-AMPLIFIER CIRCUITS