Electronics Ch9

(1)

N

NTTUUEEE E EElelectronctronics ics –– L.L.HH. . LuLu 9-19-1

CHAPTER 9 FEEDBACK

Chapter Outline Chapter Outline

9.1 9.1 The

The General

General Feedback

Feedback Structure

Structure

9.2 9.2 Some

Some Properties

Properties of

of Negative

Negative Feedback

Feedback

9.3 9.3 The

The Four

Four Basic

Basic Feedback

Feedback Topologies

Topologies

9.4 9.4 The

The Feedback

Feedback Voltage

Voltage Amplifie

Amplifier

r (Series-Shunt

(Series-Shunt))

9.5 9.5 The

The Feedback

Feedback Transconductan

Transconductance

ce Amplifi

Amplifier

er (Series-Ser

(Series-Series)

ies)

9.6 9.6 The

The Feedback

Feedback Transresistan

Transresistance

ce Amplifi

Amplifier

er (Shunt-Shunt)

(Shunt-Shunt)

9.7 9.7 The

The Feedback

Feedback Current

Current Amplifie

Amplifier (

r (Shunt-Series)

Shunt-Series)

9.9 9.9 Determin

Determining

ing the

the Loop

Loop Gain

Gain

9.10 9.10 The

The Stability

Stability Problem

Problem

9.11 9.11 Effect of

Effect of Feedback on

Feedback on the

the Amplifier Poles

Amplifier Poles

9.12 9.12 Stability

Stability Study

Study Using Bo

Using Bode Pl

de Plots

ots

9.13

(2)

N

9.1 The General Feedback Structure

Feedback amplifier Feedback amplifier



Signal-flow diagram of a feedback amplifier Signal-flow diagram of a feedback amplifier



Open-loop gain:Open-loop gain: A A 

Feedback factor:Feedback factor:   

Loop gain:Loop gain: A A   

Amount of feedback: 1 +Amount of feedback: 1 + A A   

Gain of the feedback amplifier (closed-loop gain):Gain of the feedback amplifier (closed-loop gain):



 Negative feedback: Negative feedback:



The feedback signalThe feedback signal x x_f_fis subtracted from the source signalis subtracted from the source signal x x_ss 

 Negative feedback reduces Negative feedback reduces the signal that appears at the the signal that appears at the input of the basic amplifinput of the basic amplifier ier



The gain of the feedback amplifierThe gain of the feedback amplifier A A_f_fis smaller than open-loop gainis smaller than open-loop gain A A by a factor of (1+ by a factor of (1+ A A   ))



The loop gainThe loop gain A A   is typically large (is typically large ( A A   >>1):>>1):



The gain of the feedback amplifier (closed-loop gain)The gain of the feedback amplifier (closed-loop gain) A A_f_f1/1/   

The closed-loop gain is almost entirely determined by the feedback networkThe closed-loop gain is almost entirely determined by the feedback network  better accuracy of better accuracy of A A_f_f 

(3)

N

9.1 The General Feedback Structure

Feedback amplifier Feedback amplifier



Signal-flow diagram of a feedback amplifier Signal-flow diagram of a feedback amplifier



Open-loop gain:Open-loop gain: A A 

Feedback factor:Feedback factor:   

Loop gain:Loop gain: A A   

Amount of feedback: 1 +Amount of feedback: 1 + A A   

Gain of the feedback amplifier (closed-loop gain):Gain of the feedback amplifier (closed-loop gain):



 Negative feedback: Negative feedback:



The feedback signalThe feedback signal x x_f_fis subtracted from the source signalis subtracted from the source signal x x_ss 

 Negative feedback reduces Negative feedback reduces the signal that appears at the the signal that appears at the input of the basic amplifinput of the basic amplifier ier



The gain of the feedback amplifierThe gain of the feedback amplifier A A_f_fis smaller than open-loop gainis smaller than open-loop gain A A by a factor of (1+ by a factor of (1+ A A   ))



The loop gainThe loop gain A A   is typically large (is typically large ( A A   >>1):>>1):



The gain of the feedback amplifier (closed-loop gain)The gain of the feedback amplifier (closed-loop gain) A A_f_f1/1/   

The closed-loop gain is almost entirely determined by the feedback networkThe closed-loop gain is almost entirely determined by the feedback network  better accuracy of better accuracy of A A_f_f 

(4)



ExampleExample

The feedback ampl

The feedback amplifier is based on an opampifier is based on an opamp with infinite input resistance with infinite input resistance and zero output resistanceand zero output resistance (a

(a)) FiFind and an exn exprpressessioion fon for thr the fee feededbaback fck facactotor.r. (b)

(b) FinFind thd the coe condinditiotion unn under der whiwhich tch the che closlosed-ed-looloop gap gainin A A_f_fis almost entirely determined by the feedback network.is almost entirely determined by the feedback network. (c

(c)) If If ththe oe opepen-n-lloooop p gagainin A A= 10000 V/V, find= 10000 V/V, find R R₂₂// R R₁₁to obtain a closed-loop gainto obtain a closed-loop gain A A_f_fof 10 V/V.of 10 V/V. (d

(d)) WhWhat iat is ths the ame amouount of nt of fefeededbaback ick in decn decibibelel?? ((ee)) IIffV V _ss= 1 V, find= 1 V, findV V _o_o,,V V _f_fandandV V _ii..

((ff)) IIff A Adecreases by 20%, what is the decreases by 20%, what is the corresponding decrease incorresponding decrease in A A_f_f??

N

(5)

9.2 Some Properties of Negative Feedback

Gain desensitivity

The negative reduces the change in the closed-loop gain due to open-loop gain variation

Desensitivity factor: 1+ A 

Bandwidth extension

High-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+ A_M  )

Extends the bandwidth by a factor of (1+ A_M  )

(6)

Interference reduction

The signal-to-noise ratio:

The amplifier suffers from interference introduced at the input of the amplifier

Signal-to-noise ratio:S / I=V _s/V _n 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

(7)

Reduction in nonlinear distortion

The amplifier transfer characteristic is linearized through the application of negative feedback

  = 0.01

 Achanges from 1000 to 100

(8)

9.3 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:

(9)

Current amplifiers

The most suitable feedback topologies is current-mixing and current-sampling one

Known as shunt-series feedback

Example:

(10)

Transconductance amplifiers

The most suitable feedback topologies is voltage-mixing and current-sampling one

Known as series-series feedback

Example:

(11)

Transresistance amplifiers

The most suitable feedback topologies is current-mixing and voltage-sampling one

Known as shunt-shunt feedback

Example:

(12)

9.4 The Feedback Voltage Amplifier (Series-Shunt)

Ideal case

Input resistance of the feedback amplifier

Output resistance of the feedback amplifier

Voltage gain of the feedback amplifier

(13)

The practical case

(14)

(15)

Analysis techniques

(16)

Example

(17)

Example

(18)

9.5 The Feedback Transconductance Amplifier (Series-Series)

Ideal case

(19)

The practical case

(20)

Analysis techniques

(21)

9.6 The Feedback Transresistance Amplifier (Shunt-Shunt)

Ideal case

(22)

9.7 The Feedback Current Amplifier (Shunt-Series)

Ideal case

(23)

9.9 Determining the Loop Gain

An Alternative Approach for Finding Loop Gain

Open-loop analysis with equivalent loading:

Remove the external source

Break the loop with equivalent loading

Provide test signalV _t Loop gain:

Equivalent method for determining loop gain:

Usually convenient to employ in simulation

Remove the external source

Break the loop

Provide test signalV _t

Find the open-circuit voltage transfer function T _oc Find the short-circuit current transfer function T_sc

Loop gain:

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.

(24)

Example

(25)

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 that the poles of a circuit are independent of the external excitation.

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.

A given feedback loop may be used to general a number of circuits having the same poles but different transmission zeros.

the closed-loop gain and the transmission zeros depend on how and where the input signal is injected into the loop.

(26)

9.10 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: A_f(s)

For physical frequenciess= j

Loop gain:

Stability of the closed-loop transfer function:

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

(27)

The Nyquist plot

A plot used to evaluate the stability of a feedback amplifier

Plot the loop gain versus frequency on the complex plane

Stability:

The plot does not encircle the point (-1, 0)

(28)

9.11 Effect of Feedback on the Amplifier Poles

Stability and pole location

Consider an amplifier with a pole pair at

The transient response contains the terms of the form

(29)

NTUEE Electronics – L.H. Lu 9-28

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

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

(30)

Amplifier with two-pole response

Feedback amplifier

Still a two-pole system

Characteristic equation

The closed-loop poles are given by

The plot of poles versus  is called aroot-locus diagram

(31)

Amplifier with three or more poles

Root-locus diagram

As A₀  increases, the two poles become coincident and then become complex and conjugate.

A value of A₀  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.

(32)

9.12 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.

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 ₁₈₀and 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 a frequency for which | A  | =1 and 180

.

(33)

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:

The closed-loop gain at ₁ peaks by a factor of 1.3 above the low-frequency gain for a phase margin of 45.

This peaking increases as the phase margin is reduced, eventually reaching infinite when the phase margin is zero (sustained oscillations).

(34)

An alternative approach for investigating stability

In a Bode plot, the difference between 20 log| A(j )| and 20 log(1/  ) is 20 log| A  |.

Example:

(35)

9.13 Frequency Compensation

Theory

Modify the open-loop transfer function A(s) having three or more poles so that the closed-loop amplifier is stable for a given desirable value of closed-loop gain.

The simplest method for frequency compensation is to introduce a new pole at sufficiently low frequency f _D.

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 ’_Dsuch that the amplifier is compensated without introducing a new pole.

(36)

Increase the time-constant of the dominant pole by adding additional capacitance

Add external capacitance C _Cat the node which contributes to a dominant pole.

The required value of C _Cis usually quite large, making it unsuitable for IC implementation.