feedback amplifier 1

8/3/2019 Feedback Amplifier 1

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Feedback Amplifier Topologies : As expected, there are four

basic single-loop feedback amplifier topologies, all of which have

the structure given in figure below.

The basic structure of a single-loop feedback amplifier.



The symbol indicated by the circle with the summation sign §

enclosed represents the summing network whose output is thealgebraic sum of the inputs. Thus Xi = Xs + Xf

The signal Xi, representing the output of the summing network, is

the amplifier input .

The ideal feedback amplifier :

iXÖ

If the feedback signal Xf is 1800 out of phase with the input Xs, as

is true in negative-feedback system, then Xi is a difference signal.

That is, Xi decreases as |Xf | increases.

The reverse transmission of the feedback network F is defined by

o

f

X

X| F



The transfer ratio F is often a real number but, in general, is a

function of frequency. The gain of the amplifier A is defined by

i

o

i

o

X

X

X

XA !|

The gain with feedback AF is obtained by substituting the above

equations is

F!| X1

A

X

X

As

o

F

The gain A in the above equations represents the transfer function

without feedback. If F = 0, eliminating the fed-back signal, no

feedback exists. Frequently, A is referred to as the open-loop gain

( F = 0) and designated by AOL. When F { 0, a feedback loop existsand AF is open called the closed-loop gain.

If |AF| < |A|, the feedback is termed negative; if |AF| > |A|, the

feedback is positive (regenerative). We see that in the case of

negative feedback |1 - A F| > 1.



FeedbackAmplifier :

Series-Shunt Feedback :

Series-shunt feedback configuration

From the above figure

fbi

fb

vvv

fvv

avv

!

!

!

I

I

0

0



Shunt-Shunt Feedback :

Ii

va

o!

where a is a transresistance o

fb

v

i

f !

where f is a transconductance

Shunt-shunt feedback configuration.



Iaiv !

0

fbi iii !I

Substitution of i fb we get oi fvii !I

oi

o fvia

v!

R earranging terms we find

A

a f

a

i

v

i

o

!! 1

The input impedance can be calculated asa f

ii i

!1

I

The input impedance ZI with feedback isi

i

i

i

v Z !

Substituting givesT

z

a f i

v Z ii

i

!

!

11

1

I

It is easily shown that the output impedance in this case isT

z Z o

o

!1



Shunt-Series Feedback :

If zL « zo it can be shown that

a f

a

i

i

i

o

!

1 T

z Z i

i

!1

(a) Equivalent circuit of a shunt-shunt feedback amplifier. ( b) Equivalent circuit

of a shunt-shunt feedback amplifier for apw.

)1( T z Z oo !



Shunt-series feedback configuration.



Series-series feedback configuration



Series-Series Feedback :

If zL « zo it can be shown that

a f

a

v

i

i

o

!1

)1( T z Z ii !

)1( T z Z oo

!



The number of poles in a transfer function is equal to the number of

independent energy-storing elements in the network.

Often, the high-frequency response of an amplifier has no finite

zeros; that is, AH(S) contains only poles. For this situation an

amplifier with three real poles has a transfer function

321 111 p s p s p s

A s A

o H

!

Alternatively, equation can, by performing the indicated

multiplication, be rewritten as

3

3

2

211 sa sa sa

A s A

o H

!



where

321

1111

p p pa !

323121

2

111

p p p p p pa !

3213

1

p p pa !



Consider the situation where p1<< p2 < p3; in other words, where

p1 is the dominant pole. Then

1

1

1

pa $

2

1

21

21

pa

p pa !$

3

2

3213

1

p

a

p p pa !$

or

or

or

1

1

1

a p $

2

12

aa p $

3

23

a

a p $



The importance of previous equation is that we can approximate

the pole locations by knowing the coefficients a1, a2, and a3 inAH(s). Furthermore, the dominant-pole approximation gives the

value of the 3-dB frequency f H as

1

1

21

2 a

p f H

TT!$

The asymptotic Bode diagram is extremely useful for the pencil-and- paper calculations made by circuit designers. Again,

computer simulations are used to obtain the accuracy required for

the component values in the final design.



Example

The return ratio of a two-pole amplifier is

76 10/110/1

100

s s sT

!

(a) Determine the phase margin. ( b) Is the amplifier stable?

Solution:

(a) The asymptotic Bode diagram is diagram is displayed in the

figure, from which [G

= 107.5 = 3.16 v 107 rad/s. On the phase

curve, we see that T= -157.50, and use of equation gives

05.221805.157 !! M J

as indicated on figure.



( b) As JM > 0, the amplifier is stable. Calculation of [G

and JM

using the actual Bode diagram, and verified by MICR OCAP II,

gives [G

= 3.09 v 107 rad/s and JM = 20.20. These are in good

agreement with the values obtained from the asymptotic Bode

diagram.



Figure. Asymptotic Bode diagram .



Example

The return ratio of a three-pole amplifier is

87

1 1011011 s s s

T sT

O

!

[

(a) Determine the gain and phase margins for TO = 104 when (1)

[1 = 106 rad/s and (2) [1 = 100 rad/s. ( b) Is the closed-loop

amplifier stable for each case in part a? ( c) R epeat parts a and b

for [1

= 106 rad/s, but when TO

is reduced to 10.



Figure. The asymptotic Bode diagram for the three-pole amplifier.



Figure. Asymptotic Bode diagram for a three-pole amplifier

illustrating compensation by narrow-banding the amplifier.



Example

The compensated return ratio of a single-loop amplifier is

? A ? A ? A

87

1

4

10/110/1/1

10

s s s

sT

!

[

Determine [1 so that the phase margin is approximately 67.50.



Figure. Bode diagrams used to compensate the amplifier.

(b)



Input Resistance : The KVL for the loop gives Vi = Vs + Vf . The

feedback signal is Vf = FXo is the output signal and Xo = AVi.

Combination of these relationships gives

F!!

A1

VIR V s

ii

from which the resistance with feedback R IF is

)T1(R )A1(R I

V

R ii

s

IF ! F!|

For this circuit

ioof f si AIX;XIIII ! F!!

Combination of these equations and use of V = IiR i yields

T1

R

A1

R

I

VR ii

s

IF

! F

!|



Output Impedance :

Pertaining to the calculation of the output resistance to a feedback

amplifier : (a) shunt connections; ( b) series connection.



so XA1

AV

F!

o

s

o

isc

R

AX

R

AXI !!

With Vo = 0 (short circuit), no signal is fed back, Xf = 0, and

Xi = Xs. Formation of the ratio Vo/Isc yields

T1

R

A1

R

I

VR oo

sc

oOF

!

F

!|

The output impedance is increased when a negative-feedback

amplifier employs a series-connection output.

We obtainssco X

A1

AII

F

!!

With terminals 1 and 2 open-circuited, no signal is fed back

(Xf = 0) and Xi = Xs and the open-circuit voltage Voc = -AXiR o.

Combining these relations and forming Voc/Isc, we obtain

R OF

= R o(1 - A F) = R

o(1 + T)



Example

Determine AF, T, and R OF for the common-source stage with

source resistance in the figure below.

(a) Common-source amplifier with source resistance. ( b) Schematic

diagram and (c) equivalent circuit of the amplifier without feedback.

(a) ( b)

©



Solution : The input circuit is analogous to that for the emitter

follower and is thus series-connected. Making Vo = 0 does not

eliminate the feedback because Io and, hence, Vf do not become zero.When Io = 0, Vf = 0 and the output is also series connected. (This

amplifier is a series-series type.)

To determine the input circuit of the amplifier without feedback,

open-circuit the output (Io = 0). The feedback resistance R s appearsin series with Vs. To find the output circuit, set Ii =0. Again, as

indicated in the figure R s appears in the outer loop. The equivalent

circuit in Figure b is depicted in Figure c.

In Figure c, Vo = -IoR D and Vf = -IoR s; hence F = Vf /Io = -R s.

Since no current exists in the gate loop, Vgs = Vs. Use of KVL for

the drain loop gives Io = QVs/(r d + R D + R S). It follows that



SDds

o

OLR R r V

IA

Q!!

The return ratio isSDd

SOL

R R r R AT

Q! F!

Combination of these equations and clearing fractions yields

DDd

OL

F R )1(R r T1

A

A Q

Q

!!

Since Vo = -IoR D, Vo/Vs is the same result.

Inspection of Figure shows that R OD, the output resistance with the

controlled source suppressed, is R S

+ r d

. The return ratios TOC

and TSC

are

obtained by setting R D = w and R D = 0, respectively. Thus TOC = T |R D pw =

0 and TSC = T | R D p0 = QR S/(r d + R S). The output resistance with feedback

R OF = R OD(1 + TSC) = (R S + r d) [1 + Q R S/(r d + R S)]. Clearing fractions and

rearranging terms, we obtain R OF = r d + R S (1 + Q) } r d(1+gmR S) for Q » 1.



Example

(a) Determine the input resistance R IF of an inverting Op-Amp

stage. Include the Op-Amp input resistance R i in the model for the stage. ( b) Evaluate R IF for R i p w.

Solution :

Equivalent cir cuit f or the inverting O p- Am p stage.



(a) The equivalent circuit of the stage is displayed in figure. The

dead-system input resistance R ID obtained by setting = 0, isiVÖ

21 || R R R R i ID !With the input terminals open-circuited, no current exists in R 1,

and consequently

2i

vi

i

iOC

R R

AR

VÖ

VT

!

!

The resistance R 1 and R i are in parallel when the input terminals are

short-circuited. Hencev

i

i

SC A R R R

R RT

21

1

)||(

||

!

Observe that neither TOC not TSC is zero in this amplifier.

Substitution of these values into Equation gives

2vi

i21

2ivi

2i1v1i2i1IF

R )A1(R

R R R

)]R R /(AR [1

]R )R ||R /[(A)R ||R (1]R ||R R [R

!

!



In figure, it is clear that R IF =R 1 + R xx; therefore

)]1/([

)]1/([

)1( 2

2

2

2

vi

vi

vi

i xx A R R

A R R

R A R

R R

R

!

!

The resistance R xx can be recognized as R i||R 2/(1 + AV); the

resistance R 2/(1 + Av) is exactly that which is reflected across the

X-X terminals by using Miller¶s theorem.

( b) From part a, allowingR i pw, gives

xx1

v

21IF R R

A1

R R R !

!

For a large value of Av (Av p w), the input resistance is simply

R 1 as R xx p 0. However, this is the expected result, for when

Av p w, the inverting terminal is a virtual ground.



Series-shunt configuration fed from a finite source impedance.

From the previous equations

a f a

vv

i

!1

0

s

si

i

iv

z Z

Z v

! where Zi is the input impedance seen by



Series-shunt configuration with finite impedances in the basic amplifier.

Iavv !

0

oi fvvv ! I

Substituting the above equations gives

)1( a f va fvvvi !! III

i

i

z

vi

I

!Also

Substituting we geta f z

vi

i

i

i

!

1

1



Thus, from the previous equation input impedance Zi with feedback applied is

i

i

ii zT

i

v Z )1( !!

Series feedback at the input always raises the input impedance by (1+T).

0! ve f v

o

e

z

avvi

!

Circuit for the calculation of the output impedance of the series-shunt

feedback configuration.



(a) Equivalent circuit of a series-shunt feedback amplifier.( b) Equivalent circuit of a series-shunt feedback amplifier for apw.

Substituting we get

o z

a fvvi

!

From the above equation the output impedance Zo with feedback applied is

T

z

i

v Z

o

o

!!

1

(a) ( b)

feedback amplifier 1

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