Feedback - 國立中興大學cc.ee.nchu.edu.tw/~aiclab/teaching/Electronics3/lect08.pdf · zThe Shunt-Shunt and Shunt-Series Feedback Amplifier zDetermining the Loop Gain zThe Stability

Ching-Yuan Yang

National Chung-Hsing UniversityDepartment of Electrical Engineering

Feedback

Microelectronic Circuits

8-1 Ching-Yuan Yang / EE, NCHUMicroelectrics (III)

Outline

The General Feedback Structure

Some Properties of Negative Feedback

The Four Basic Feedback Topologies

The Series-Shunt Feedback Amplifier

The Series-Series Feedback Amplifier

The Shunt-Shunt and Shunt-Series Feedback Amplifier

Determining the Loop Gain

The Stability Problem

Effect of Feedback on the Amplifier Poles

Stability study Using Bode Plots

Frequency Compensation


The General Feedback Structure


General structure of the feedback amplifier

xo = Axi , xf = βxo , xi = xs − xf

β : feedback factor

Aβ : loop gain

1 + Aβ : amount of feedback

If Aβ >> 1, then Af ≈ 1/β.

The gain of the feedback amplifier is almost entirely determined by the feedback network.

Feedback signal:

For Aβ >> 1 we see that xf ≈ xs , which implies that xi 0. (xi : error signal)

βAA

xx

As

of +

=≡1

sf xA

Ax

ββ

+=

1





Desensitize the gain

Extend the bandwidth of the amplifier

Reduce the effect of noise

Reduce nonlinear distortion


Gain Desensitivity

Assume that β is constant.

The percentage change in Af (due to variations in some circuit parameter) is

smaller than the percentage change in A by amount of feedback.

The amount of feedback, 1 + Aβ, is also known as the desensitivity factor.

βAA

xx

As

of +

=≡1

2)1(1βAdA

dAf

+=

2)1( βAdA

dAf +=

AdA

AA

dA

f

f

)1(1β+

=


Bandwidth Extension

Consider an amplifier whose high-frequency response is characterized by a single pole.

Apply the negative feedback with a frequency-independent factor β :

The feedback amplifier: (high-frequency response)

Midband gain

Upper 3-dB frequency ωHf = ωH (1 + AM β )

The amplifier band width is increased by the same factor, (1 + AM β ), by which

its midband gain is decreased, maintain the gain-bandwidth product constant.

Similarly, the low-frequency response of the feedback amplifier has a lower 3-dB

frequency

)1(1

1)(1

)()(

βω

ββ

MH

M

M

f

As

AA

sAsA

sA

++

+=+

=gain loop-Close

H

M

sA

sA

ω+

=1

)( AM : the midband gain

ωH : the upper 3-dB frequency

βM

MMf A

AA

+=

1

βωω

M

LLf A+

=1


Noise Reduction

n

s

VV

NS

= ratio noise-to-Singal

ββ 21

1

21

21

11 AAA

VAA

AAVV nso +

++

=

Vs : input signal Vo : output signalVn : noise or interference A1 : gain

The amplifier suffers from noise and the noise can be assumed to be introduced at the input of the amplifier.

Negative feedback is applied to improve SNR:

(clean amp.)

2 AVV

NS

n

s=∴


Reduction in Nonlinear Distortion

The open-loop voltage transfer characteristic of the amplifier is change from 1000 to 100 and then to 0. [curve (a)]Apply negative feedback with β = 0.01 to the amplifier [curve (b)] :

the slope of the steepest segment: the slope of the next segment:

The order-of-magnitude change in slop is considerably reduced.The price paid is a reduction in voltage gain.

9.9001.010001

10001 =

×+=fA

Curve (a) shows the amplifier transfer characteristic without feedback.

Curve (b) shows the characteristic with negative feedback (β = 0.01) applied.

5001.01001

1002 =

×+=fA


The Four Basic Feedback Topologies


Four basic feedback topologies

Base on the quantity to be amplified (voltage or current) and on the desired

form of the output (voltage or current), amplifiers can be classified into four

categories.

Voltage amplifiers

Current amplifiers

Transconductance amplifiers

Transresistance amplifiers

Four basic feedback topologies:

Voltage-sampling series-mixing (series-shunt) topology

Current-sampling shunt-mixing (shunt-series) topology

Current-sampling series-mixing (series-series) topology

Voltage-sampling shunt-mixing (shunt-shunt) topology


Voltage Amplifiers

Characteristics:

Input signal: voltage output signal: voltage

Voltage-controlled voltage source

High input impedance

Low output impedance

Feedback topology


The feedback network samples the output voltage, and the feedback signal xfis a voltage that can be mixed with the source voltage in series.

“Series” refers to the connection at the input and “shunt” refers to the connection at the output.

Characteristics:

⌦ Stabilize the voltage gain

⌦ Higher input resistance

(series connection at the input)

⌦ Lower output resistance

(parallel connection at the output)



Example

Vs vi Vo V− vi negative feedback

vi = V+ − V−


Current Amplifiers

Characteristics:

Input signal: current output signal: current

Current-controlled current source

Low input impedance

High output impedance

Feedback topology


The feedback network samples the output current, and the feedback signal xf is a current that can be mixed in shunt with the source current.

“Shunt” refers to the connection at the input and “series” refers to the connection at the output.

Characteristics:

⌦ Stabilize the current gain

⌦ Lower input resistance

(shunt connection at the input)

⌦ Higher output resistance

(series connection at the output)



Example

Is ii (Ib1) Ic1 Vc1 Ic2 (Io) Ie2 (Io /α) If ii (Ib1) negative feedback


Transconductance Amplifiers

Characteristics:

Input signal: voltage output signal: current

Voltage-controlled current source

High input impedance

High output impedance

Feedback topology


The feedback network samples the output current, and the feedback signal xf is a voltage that can be mixed with the source voltage in series.

“Series” refers to the connection at the input and “series” refers to the connection at the output.

Characteristics:

⌦ Stabilize the transconductance gain

⌦ Higher input resistance

(series connection at the input)

⌦ Higher output resistance

(series connection at the output)



Example

Vs vi (vbe) Ic1 vc1 Ic2 (Io) vc2 Io Io /α Vf vi negative feedback


Transresistance Amplifiers

Characteristics:

Input signal: current output signal: voltage

Current-controlled voltage source

Low input impedance

Low output impedance

Feedback topology


The feedback network samples the output voltage, and the feedback signal xf is a current that can be mixed in with shunt the source current.

“Shunt” refers to the connection at the input and “shunt” refers to the connection at the output.

Characteristics:

⌦ Stabilize the transresistance gain

⌦ Lower input resistance

(shunt connection at the input)

⌦ Lower output resistance

(shunt connection at the output)



Example

Norton’stheorem

Is ii Vo If ii negative feedback


Two-Port Network Parameters

y parameters

Equivalent circuit:


z parameters

Equivalent circuit:


h parameters

Equivalent circuit:


g parameters

Equivalent circuit:


The Series-Shunt Feedback Amplifier


The Ideal Situation of the Series-Shunt Feedback Amplifier

Ideal structure

A circuit: a unilateral open-loop amplifier

Ri : input resistance A : voltage gain Ro : output resistance

β circuit: an ideal voltage-sampling series-mixing feedback network

The source and load resistances are included inside the A circuit.

The β circuit does not load the A circuit. Do not change A ≈ Vo /Vi.


Equivalent circuit

The close-loop gain (A and β have reciprocal units)

Input resistance

The series-mixing increases the input resistance by (1 + Aβ).

General form Zif (s) = Zif (s)[1 + A(s)β(s)]

βAA

VV

As

of +

=≡1

)1(/

βββAR

VAVV

RV

VVR

VV

RRV

VIV

R ii

iii

i

oii

i

si

ii

s

i

sif +=

+=

+===≡


Output resistance (Vs = 0)

Output resistance

where

The voltage-sampling feedback reduces the output resistance by (1 + Aβ).

General form

IV

R tof ≡

0) ( =−=−=−=−

= stofio

it VVVVVR

AVVI forand ββ

o

tt

RVAV

Iβ+

=

βAR

IV

R otof +

=≡1

)()(1)(

)(ssA

sZsZ o

of β+=


The Practical Situation

Derivation of the A circuit and β circuit for the series-shunt feedback amplifier.

Block diagram of a practical series-shunt feedback amplifier

Rif and Rof are the input and output resistances, respectively, of the feedback amplifier, including Rs and RL.

The actual input and output resistances of the feedback amplifier usually exclude Rs and RL :

Lof

outsifin

RR

RRRR11

1

−=−= and


The circuit in with the feedback network represented by its h parameters.

The source and load resistances should be lumped with the basic amplifier.

In addition, the resistances in the feedback should be also lumped with the basic amplifier.

h-parameter two-port feedback network:

Neglect h21, the forward transmission effect of the feedback network, and thus omit the controlled source h21I1.

(h parameters)


The circuit in after neglecting h21.

Determine β : (port 1 open-circuited)012

112

=≡=

IVV

hβ

h21I1 is neglected.


Summary of the Series-Shunt Feedback Amplifier


Example 8.1 An op amp connected in the noninverting configuration

The op amp has an open-loop gain μ, a differential input resistance Rid, and an output resistance ro. Find expressions for A, β, the close-loop gain Vo /Vs, the input resistance Rin, and the output resistance Rout. (μ = 104, Rid = 100kΩ, ro = 1kΩ, RL = 2kΩ, R1 = 1kΩ, R2 = 1MΩ, and Rs = 10kΩ.)

The feedback network consists

R2 and R1. This network

samples the output voltage Vo

and and provides a voltage

signal (across R1) that mixed in

series with the input source Vs.


A circuit:

β circuit:

)(])([])([

''

2121

21

RRRRR

rRRRRRR

VV

Asid

id

oL

L

i

o

+++++

=≡ μ A ≈ 6000 V/V

V/V 10'

' 3

21

1 −≈+

=≡RR

RV

V

o

fβ


Voltage gain

Input resistance Rif = Ri(1 + Aβ ) where Ri is the input resistance of the A circuit.

Ri = Rs + Rid + (R1 ||R2) ≈ 111kΩ

Rif = 111×7 = 777kΩ

Rin = Rif − Rs = 739kΩ

Output resistance where Ro is the output resistance of the A circuit.

Ro = ro || RL || (R1 +R2) ≈ 667 Ω

∵ Rof = Rout || RL

Rout ≈ 100 Ω

V/V 8577

60001

==+

=≡βA

AVV

As

of

βAR

R oof +

=1

Ω== 3.957

667ofR


The Series-Series Feedback Amplifier


The Ideal Situation of the Series-Series Feedback Amplifier

Ideal structure


Ri : input resistance A : transconductance gain Ro : output resistance

β circuit: an ideal current-sampling series-mixing feedback network


The β circuit does not load the A circuit. Do not change A ≈ Io /Vi.


Equivalent circuit


Input resistance

The series-mixing increases the input resistance by (1 + Aβ ).

General form Zif (s) = Zi (s)[1 + A(s)β(s)]

βAA

VI

As

of +

=≡1

)1(/

βββAR

VAVV

RV

VVR

VV

RRV

VIV

R ii

iii

i

oii

i

si

ii

s

i

sif +=

+=

+===≡



Output resistance

In this case,

The voltage-sampling feedback increases the output resistance by (1 + Aβ ).

General form

tof I

VR ≡

0) ( =−=−=−= stofi VIIVV forββ

ottoit RIAIRAVIV )()( β+=−=

)1( βARIV

R ot

of +=≡

)]()(1)[()( ssAsZsZ oof β+=


Derivation of the A circuit and β circuit for the series-series feedback amplifier.

Block diagram of a practical series-shunt feedback amplifier




Lofoutsifin RRRRRR −=−= and


The circuit in with the feedback network represented by its z parameters.



z-parameter two-port feedback network:

Neglect z21, the forward transmission effect of the feedback network, and thus omit the controlled source z21I1.

(z parameters)


The circuit in after neglecting z21.

Determine β : (port 1 open-circuited)012

112

=≡=

IIV

zβ

z21I1 is neglected.


Summary of the Series-Series Feedback Amplifier


Example 8.2 A feedback triple is composed of three gain stages with series-series feedback

Assume that the bias circuit causes IC1 = 0.6mA, IC2 = 1mA, and IC3 = 4mA. Using these values and assuming hfe = 100 and ro = ∝, find the open-loop gain A, the feedback factor β, the closed loop gain Af = Io /Vs, the voltage gain Vo /Vs, the input resistance Rin = Rif, and the output resistance Rof (between nodes Y and Y’). Now, if ro of Q3 is 25kΩ, estimate an approximate value of the output resistance Rout.


A circuit:

Determine A:

V/V 92.14)]([

)(

211

21'1 −=

++−

=EFEe

C

i

c

RRRrrR

VV πα

{ } V/V 2.131))](()[1( 123221

2 −=+++−= EFEefeCmc

c RRRrhRgVV

mA/V 6.10)((

1

1233

3

2

'

=++

==EFEeb

e

c

o

RRRrVI

VI

A/V7.20'

'

1

2'1

'

'

=××=≡⇒i

o

c

c

i

c

i

o

VI

VV

VV

VI

A


β circuit:

Closed-loop gain:

Voltage gain:

Input resistance of the A circuit:

Input resistance of the feedback amplifier:

Output resistance Ro of the A circuit: The resistance looking between Y and Y’.

Output resistance of the feedback amplifier:

Ω=×++

== 9.11112

2'

'

EEFE

E

o

f RRRR

R

I

Vβ

mA/V 7.839.117.201

7.201

=×+

=+

=≡βA

AVI

As

of

V/V 2.50600107.83 33

3 −=××−=−≈−

= −Cf

s

Cc

s

o RAVRI

VV

Ω=+++= k 65.13))](()[1( 211 EFEefei RRRrhR

Determine β :

Ω=+= M 34.3)1( βARR iif

Ω=+

+++= 9.1431

])([ 2312

fe

CeEFEo h

RrRRRR

Ω=+= k 6.35)1( βARR oof


Find the output resistance from the equivalent circuit.

Ω=

+=

+≈

++=

M 5.2

)16.35(1001[25

)](1[

))(1(

333

3333

π

π

rRgr

rRrgrR

ofmo

ofomoout


Homework #1

Problems 6, 14, 24, 25, 27, 37


The Shunt-Shunt Feedback Amplifier


Ideal structure


Ri : input resistance A : transresistance gain Ro : output resistance

β circuit (transconductance): an ideal voltage-sampling shunt-mixing feedback network


The β circuit does not load the A circuit. Do not change A ≈ Vo /Ii.

The Ideal Situation of the Shunt-Shunt Feedback Amplifier


Equivalent circuit


Input resistance

The shunt-mixing reduces the input resistance by (1 + Aβ ).

General form

βAA

IV

As

of +

=≡1

βββ AR

AIII

RVI

IR

III

RII

RIRI

IV

R i

ii

ii

oi

ii

fi

ii

s

ii

s

ii

s

iif +

=+

=+

=+

===≡1

)()(1)(

)(ssA

sZsZ i

if β+=



Output resistance

where

The voltage-sampling feedback reduces the output resistance by (1 + Aβ).

General form

IV

R tof ≡

0) ( =−=−=−=−

= stofio

it IVVIIR

AIVI forand ββ

o

tt

RVAV

Iβ+

=

βAR

IV

R otof +

=≡1

)()(1)(

)(ssA

sZsZ o

of β+=




Block diagram of a practical shunt-shunt feedback amplifier



Lof

out

sif

in

RR

R

RR

R11

111

1

−=

−= and


The circuit in with the feedback network represented by its y parameters.



y-parameter two-port feedback network:

Neglect y21, the forward transmission effect of the feedback network, and thus omit the controlled source y21I1.

(y parameters)


The circuit in after neglecting y21.

Determine β : (port 1 short-circuited)012

112

=≡=

VVI

yβ

y21I1 is neglected.


Summary of the Shunt-Shunt Feedback Amplifier


Example 8.3 A feedback circuit of the shunt-shunt type

Determine the small-signal voltage gain Vo /Vs, the input resistance Rin , and the output resistance Rout = Rof . The transistor has β = 100.

dc analysis:

VVII

IV

IIV

CCB

BC

BBC

5.4 and 1.5mA, mA,015.0

07.0)1(

7.412

4799.347)07.0(7.0

==≈

⎪⎩

⎪⎨⎧

++=−

+=++=

β


Small-signal analysis

A circuit:

Input resistance of the A circuit

Ri = Rs || Rf || rπ = 1.4 kΩOutput resistance of the A circuit

Ro = RC || Rf = 4.27 kΩ

Transfer gain:

( )( )Cfmo

fsi

RRVgV

rRRIV

π

ππ

−=

=

'

'

( )( )Ω−=

−=

=

k 7.358

''

πrRRRRg

IV

A

fsCfm

i

o


Determine β :

Determine the closed-loop characteristics:

Ω−=−=≡

k4711

'

'

fo

f

RV

Iβ

Ω==+

=

Ω==+

=

−≈−

===

Ω−=+−

=+

=≡

49563.827.4

1

2.16263.84.1

1

V/V 16.410

6.41 ,

k 6.4147/7.3581

7.3581

β

β

β

AR

R

AR

R

RIV

VV

RIV

AA

IV

A

oof

iif

ss

o

s

osss

s

of

feedback withresistance Output

feedback withresistance Input

gain voltage the thusSince

gain loop-Closed


The Shunt-Series Feedback Amplifier


Ideal structure


Ri : input resistance A : current gain Ro : output resistance

β circuit : an ideal current-sampling shunt-mixing feedback network


The β circuit does not load the A circuit. Do not change A ≈ Io /Ii.

The Ideal Situation of the Shunt-Series Feedback Amplifier


Equivalent circuit


Input resistance

The series-mixing reduces the input resistance by (1 + Aβ ).

General form

βAA

II

As

of +

=≡1

βββ AR

AIII

RII

IR

III

RII

RIRI

IV

R i

ii

ii

oi

ii

fi

ii

s

ii

s

ii

s

iif +

=+

=+

=+

===≡1

)()(1)(

)(ssA

sZsZ i

if β+=



Output resistance

In this case,

The current-sampling feedback increases the output resistance by (1 + Aβ ).

General form

tof I

VR ≡

0) ( =−=−=−= stofi IIIII forββ

ottoit RIAIRAIIV )()( β+=−=

)1( βARIV

R ot

of +=≡

)]()(1)[()( ssAsZsZ oof β+=




Block diagram of a practical shunt-series feedback amplifier



Lofout

sif

in RRR

RR

R −=−

= and 11

1


The circuit in with the feedback network represented by its g parameters.



g-parameter two-port feedback network:

Neglect g21, the forward transmission effect of the feedback network, and thus omit the controlled source g21V1.

(g parameters)


The circuit in after neglecting g21.

Determine β : (port 1 short-circuited)012

112

=≡=

VII

gβ

g21V1 is neglected.


Summary of the Shunt-Series Feedback Amplifier


Example 8.4 A feedback circuit of the shunt-series type

Find the current gain Iout /Iin, the input resistance Rin , and the output resistance Rout. The transistor has β = 100 and VA = 75V.

dc analysis: (neglect the effect of finite transistor β and VA )

mA 187.087.0

V 87.07.057.1

V 57.115100

1512

1

1

1

=≈

=−≈

=+

≈

E

E

B

I

V

V

V 8.884.012

mA 4.04.33.1

V 3.17.02

V 211012

2

1

2

1

=×−≈

=≈

=−≈=×−≈

C

E

E

C

V

I

V

V


Small-signal equivalent circuit

A circuit:

])(['121 ππ rRRRRIV BfEsi +=

{ }))(1([ 221112 fEComb RRrRrVgV ++−= βππ

)('

22

2

fEe

bo RRr

VI

+≈ A/A45.201

''

−≈≡⇒i

o

II

A


Find input/output impedance:

Find the required current gain:

Determine β :

Determine the closed-loop characteristics:

254.0'

'

2

2 −=+

−=≡fE

E

o

f

RRR

I

Iβ

Ω=+= k 535.1)( 12 πrRRRRR BfEsi

Ω=+

=⇒ 5.291 βA

RR i

if

Ω≈−

=⇒ 5.29/1/1

1

sifin RR

R

Ω=+

++= k 69.21

)( 1122 β

oCefEo

rRrRRR

Ω≈+=⇒ k 1.140)1( βARR oof

Ω≈+=⇒ M 1.18)(1[ 22 ofmoout RrgrR π

A/A44.32

2

2

2 −=+

≈+

=≈s

o

CL

C

s

c

CL

C

s

out

in

out

II

RRR

II

RRR

II

II


Summary of Relationships for the Four Feedback_Amplifier Topologies


Determining The Loop Gain


Determining the Loop Gain

Loop gain Aβ is a very important quantity that characterizes a feedback loop.

Find the loop gain:

Let the external source xs = 0 .

Open the feedback loop by breaking the connection of xo to the feedback network and apply a test signal xt .

Determine the loop gain:

xf = βxt xi = −xf = −βxt xo = Axi = −Aβxt Loop gain Aβ = −xo /xt

The loop gain Aβ is given by the negative of the ratio of the returned signal to the applied test signal.


A conceptual feedback loop is broken at XX’ and a test voltage xt is applied. The impedance Zt is equal to that previously seen looking to the left of XX’. The loop gain Aβ = −Vr /Vt , where Vr is the returned voltage.

As an alternative, Aβ can be determined by finding the open-circuit transfer function Tocand the short-circuit transfer function Tsc and combining them as indicated.

open-circuit transfer function Toc short-circuit transfer function Tsc

t

r

VV

A −=β

scoc TT

A11

1

+−=β


Closed-loop circuit:

Break XX’ to determine the loop gain Aβ :

Vx

+

−

AxVx

Rx

Rt

X

X’

Y

Y’

Vx

+

−

AxVx

Rx

RtVr

+

−

Vt

tx

tx

t

r

RRR

AVV

A+

−=−=β

tx

ttxr RR

RVAV

+= ∵


Break YY’ to determine Toc and Tsc :

Open-circuit transfer function Toc

Short-circuit transfer function Tsc

Thus,

Vx

+

−

AxVx

Rx

RtVoc

+

−

Vt

Vx

+

−

AxVx

Rx

RtIscIt

xt

ococ

txoc

AVV

T

VAV

=≡∴

=

∵

x

tx

t

scsc

x

ttx

x

xxsc

ttx

RR

AII

T

RRIA

RVA

I

RIV

=≡∴

==

=

and

∵

βA

RAR

ATT tx

x

xscoc

=+

−=+

−1

111

1


Equivalence of Circuits from a Feedback-Loop Point of View

Break XX’ to determine the loop gain Aβ :

RRR

RRRRRRR

rRRRRR

RRRRRVV

id

id

id

id

oidL

idLr +++

++++

++−=

21

1

12

121 ])([

)(}])([{

}])([{μ

RRR

RRRRRRR

rRRRRR

RRRRR

VV

VV

ALid

id

id

id

oidL

idLr

t

r

++++

+++++

=−=−==21

1

12

12

1 ])([)(

}])([{

}])([{μβ


The Stability Problem


Stability Problem of a Feedback Amplifier

The closed-loop transfer function of a feedback amplifier

For physical frequencies s = jω ,

If for s = jω0, L(jω0) = −1, then the closed-loop gain approaches infinity at ω0. Under this condition, the circuit amplifies its own noise components at ω0 indefinely.

Barkhausen criteria: If a negative-feedback circuit has a loop gain that satisfies two conditions:

|L(jω0)| ≥ 1 and ∠L(jω0) = 180o,

then the circuit may oscillate at ω0.

In order to ensure oscillation in the presence of temperature and process variations, we typically choose the loop gain to be at least twice or three times the required value.

Oscillations could occur in a negative-feedback amplifier, we wish to find methods to prevent their occurrence.

)()(1)(

)(ssA

sAsAf β+

= where A(s) : the open-loop transfer function

β (s) : the feedback transfer function

)()(1)(

)(ωβω

ωωjjA

jAjAf +

= L( jω) ≡ A( jω)β ( jω) = |A( jω)β ( jω)|e−jφ(ω )


Nyquist Plot

Nyquist criterion:

If the intersection occurs to the left of (−1, 0), the magnitude of loop gain at this frequency is greater than unity and the amplifier will be unstable.

If the intersection occurs to the right of (−1, 0), the amplifier will be stable.


Effect Of Feedback On The Amplifier Poles


Stability and Pole Location

Consider an amplifier with a pole pair at s = σ0 ± jωn .

v(t) = eσ 0t [e +jωnt + e −jωnt ] = 2 eσ 0t cos (ωnt )

Relationship between pole location and transient response:

The existence of any

right-half-plane poles

results in instability.


Poles of the Feedback Amplifier

The closed-loop transfer function of a feedback amplifier

Find poles: The poles of the feedback amplifier are the zeros of 1 + A(s)β (s).

1 + A(s)β (s) = 0 (the characteristic equation of the feedback loop.)

It should therefore be apparent that applying feedback to an amplifier changes its poles.

)()(1)(

)(ssA

sAsAf β+

= where A(s) : the open-loop transfer function

β (s) : the feedback transfer function


Amplifier with Single-Pole Response

An amplifier whose open-loop transfer function is characterized by a single pole:

The closed-loop transfer function

Effect of feedback on the pole location and the frequency response:

psA

sAω/1

)( 0

+=

)1(/1)1/(

)(0

00

βωβAs

AAsA

pf ++

+=

)1( Pole

1 Gain Midband

loop-closed loop-open

00

βωωβ

oPPfP

oof

Aω

AA

AA

+=→+

=→


For frequencies ω >> ωP(1 + A0β ),

At such high frequencies the loop gain is much smaller than unity and the feedback is ineffective.

Effect of feedback on the frequency response:

Applying negative feedback to an amplifier results in extending its bandwidth at the expense of a reduction in gain.

Since the pole of the closed-loop amplifier never enters the right half of the s plane, the single-pole amplifier is stable for any value of β.

Unconditionally stable

)()( 0 sAs

AsA P

f ≈≈ω


Amplifier with Two-Pole Response

An amplifier whose open-loop transfer function is characterized by two poles:

Close-loop poles

Root-locus diagram

)/1)(/1()(

21 PP

o

ssA

sAωω ++

=

0)1()(0)(1 210212 =++++⇒=+ PPPP AsssA ωωβωωβ

2102

2121 )1(4)(

21

2 PPPPPP As ωωβωωωω

+−+±+

−=⇒

The feedback amplifier also is unconditionally stable.

The maximum phase shift of A(s) is 180o (90o per

pole), but this value is reached at ω = ∝.

The open-loop amplifier might have a dominant pole,

but this is not necessarily the case for the closed-

loop amplifier.

As is the case with second-order responses

generally, the closed-loop response can show a peak.


The characteristic equation of a second-order network:

Definition of ω0 and Q of a pair Determine Q factor: of complex conjugate poles:

Normalized magnitude response:

020

02 =++ ωωQ

ss ω0 : pole frequency Q : pole Q factor

21

210 )1(

PP

PPAQ

ωωωωβ

++

=

Q > 0.7, peaking. Q ≤ 0.7, no peaking.Q = 0.7 (poles at 45o), maximally flat.


Example 8.5 Positive-feedback circuitFind the loop transmission L(s) and the characteristic equation. Sketch a root-locus diagram for varying K, and find the value of K that results in a maximally flat response, and the value of K that makes the circuit oscillate.(Assume that the amplifier has infinite input impedance and output impedance.)SolutionThe loop transmission:

where T(s) is the transfer function of the two-port RC network.

Thus

The characteristic equation: 1 + L(s) = 0

∴ Pole frequency: Q factor:

Open-loop circuit:

)()()()( sKTVV

ssAsLt

r −=−=≡ β

221 )/1()/3(

)/1()(

CRCRssCRs

VV

sT r

++=≡

22 )/1()/3()/(

)(CRCRss

CRKssL

++−

=

013

013 2

22

2 =⎟⎠⎞

⎜⎝⎛+

−+=−⎟

⎠⎞

⎜⎝⎛++

CRCRK

ssCRK

sCRCR

ss

CR1

0 =ωK

Q−

=3

1


Root-locus diagram:

CR1

0 =ω

KQ

−=

31

The maximally flat response:Q = 0.707 K = 1.586

The poles are at 45o.K ≥ 3, the circuit is unstable.

The feedback is in fact positive,

and the circuit will oscillate at the

frequency for which the phase of T (jω) is zero.


Amplifier with Three or More Poles

A value of A0β exits at which this pair of complex-conjugate poles enters the right half of the s plane, thus causing the amplifier to become unstable.

An amplifier with three poles: Root-locus diagram

Since the amplifier with three poles has a phase shift that reach −270o as ω reaches ∝, there exists a finite frequency ω180o, at which the loop gain has 180o phase shift.

Frequency compensation

There exists a maximum value for β above which the feedback amplifier becomes unstable.Alternatively, there exists a minimum value value for the closed-loop gain Af0 below which the amplifier becomes unstable.

To obtain lower values of the closed-loop gain one needs to alter the loop transfer function L(s). This is the process of frequency compensation.


Stability Study Using Bode Plots


Gain and Phase Margins

Gain margin:

The difference between the value of |Aβ | at ω180o and unity (in dB).

Phase margin:

The difference between the phase angle at the 0-dB frequency and 180o.

The gain margin represents the amount by which the loop gain can be increased while stability is maintained.

If at the frequency of unity loop-gain magnitude, the phase lag is in excess of 180o, the amplifier will be unstable.

Feedback amplifiers are usually designed to have sufficient gain margin and phase margin to allow for the inevitable changes in loop gain with temperature, time, and so on.


Effect of Phase Margin on Closed-Loop Response

margin phase180 where

1)( o

1

−=

×= −

θ

βω θjejA

Close-loop gain (at ω1):

θ

θββω

ωω j

j

f ee

jAjA

jA −

−

+=

+=

1)/1(

)(1)(

)(1

11

θ

βωjf e

jA−+

=⇒1

/1)( 1

⌦For a phase margin of 45o, θ = 135o,

⌦The peaking increase as the phase margin is reduced, eventually reaching ∞ when the phase margin is zero.

⌦The closed-loop gain at low frequencies is approximately 1/β .

βω 1

3.1)( 1 =jAf Gain peaking


An Alternative Approach for Stability Analysis Using Bode Plot of |A|

Assume that β is independent of frequency, we can plot 20log(1/β )as a horizontal straight line on the same plane used for 20log|A|.

Study stability by examining the difference between the two plots.

If we wish to evaluate stability for a different feedback factor we simply draw another horizontal straight line at the level 20log(1/β ).

ββ

ω AjA log201

log20)(log20 =−

(expressed in dB)


Example: Stability Analysis Using Bode Plot of |A|

The open-loop gain of the amplifier

We can find

and

3.2×106 Hz

ββ

ω AjA log201

log20)(log20 =−

⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ +

=

765

5

101

101

101

10)(

fj

fj

fj

fA

⎥⎦⎤

⎢⎣⎡

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−=

⎟⎠⎞

⎜⎝⎛+−⎟

⎠⎞

⎜⎝⎛+−

⎟⎠⎞

⎜⎝⎛+−=

−−−7

16

15

1

2

7

2

6

2

5

10tan

10tan

10tan

101log20

101log20

101log20100

fff

ff

fA

φ

|A|


Frequency Compensation


Theory of Frequency Compensation

A’ :Introduce an additional pole at fD .poles: fD, fP1, fP2, fP3, ⋅⋅⋅

A” :Move the original low-frequency pole to f ’D .(i.e., fP1 f ’D )poles: f ’D, fP2, fP3, ⋅⋅⋅

β = 10−2

|A|


Implementation of Frequency Compensation

Two-Cascaded gain stages of a multistage amplifier

Equivalent circuit for the interface The circuit with a compensatingbetween the two stages capacitor CC added

xxP RC

fπ2

11 =

xCxD RCC

f)(2

1'

+=

π


Miller Compensation and Pole SplittingA gain stage in a multistage amplifier before compensation

A gain stage in a multistage amplifier with a compensating capacitor in the feedback loop


2121212

21212211

21

)]([)]([1

)(

RRCCCCCsRRRRgCRCRCs

RRgsC

IV

fmf

mf

i

o

++++++++−

=

21

2

1

21

2

2121

'''1

'''1

'1

1'

1'

1)(

PPP

PPPPPP

ss

ss

sssD

ωωω

ωωωωωω

++≈

+⎟⎟⎠

⎞⎜⎜⎝

⎛++=⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

(ω’P1 << ω’P2 )

12212122111

1)(

1'

RCRgRRRRgCRCRC fmmfP ≈

++++=⇒ ω

)('

21212 CCCCC

Cg

f

fmP ++≈ω

Pole splitting

(8.87)

(8.88)

ρ

jωBefore Compensation

ρ

jωAfter Compensation


Example 8.6: Frequency Compensation

The open-loop transfer function of the operational amplifier:

We wish to compensate the op amp so that the closed-loop amplifier with resistive feedback is stable for any gain (that is, for β up to unity) and PM (phase margin) ≥ 45o. Assume that the op amp circuit includes a stage such as that of Fig.a with C1 = 100pF, C2 = 5pF, and gm = 40mA/V, that the pole at fP1 is caused by by the input circuit of that stage, and that the pole is connected either between the input node B and ground or

in the feedback path of the transistor.

⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ +

=

765

5

101

101

101

10)(

fj

fj

fj

fA

Fig.a:


Find the compensating capacitor CC connected across the input terminals of the transistor stage.

(a) Determine R1 and R2 :

(b) Determine f ’D :

1st pole from fP1 to f ’D : and the 2nd pole remains unchanged.

PM ≥ 45o : The required value for f ’D is determined by drawing a −20dB/dec line from the 1-MHz

point on the 20log(1/β ) = 20log1 = 0dB line. This line will intersect the 100-dB dc gain line at 10Hz.

Thus

Ω===

Ω===

10

2

1MHz1

210

2

1MHz1.0

5

222

2

5

111

1

ππ

ππ

RRC

f

RRC

f

P

P

11 )(21

'RCC

fC

D +=

π

F1 )(2

1Hz10'

11

μπ

=∴+

== CC

D CRCC

f


Find the compensating capacitor Cf connected in the feedback path of the transistor.

By Eqs. (8.58) and (8.59):

Determine fP2 : Assume Cf >> C2 , then

The pole f ’P2 moves to a frequency higher than fP3 (= 10MHz).

The new 2nd pole is at fP3. This requires that the 1st pole be located at 100Hz for PM ≥ 45o :

Conclusion: Using Miller compensation not only results in a much smaller compensating capacitor but, owing to pole splitting, also enables us to place the dominant pole a decade higher in frequency.

)]([2' and

21

'2121

212

1 CCCCC

Cgf

RCRgf

f

fmP

fmP ++

≈≈ππ

MHz 6.60)(2

'21

2 =+

≈CC

gf m

P π

pF5.78 2

1Hz100'

121 =∴== f

fmP C

RCRgf

π



Homework #2

Problems 42, 58, 61, 67, 70, 80

Documents

Feedback - 國立中興大學cc.ee.nchu.edu.tw/~aiclab/teaching/Electronics3/lect08.pdf · zThe Shunt-Shunt and Shunt-Series Feedback Amplifier zDetermining the Loop Gain zThe Stability