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Feedback
Amplifiers can be classified in one of four ways according to their input and output
parameters:
vo
e.g. op amp
v in
i
Current A i  o e.g. BJT
i in
i
Transconductance Gt  o e.g. FET
v in
v
Transresistance G r  o e.g. Norton op amp
i in
Voltage A v 
(1)
(2)
(3)
(4)
Feedback itself can be:
derived as a voltage or current and
applied in series or parallel with the input signal.
Figure 1 shows these four possibilities schematically. In each case the feedback term  is not
used. In practice, of course, it would modify the proportion of the term being fed back.
There now follows a set of four feedback circuits. These will be:

Series current (voltage current)

Shunt current (current current)

Shunt voltage (current voltage)

Series voltage (voltage voltage)
Current derived
Voltage derived
Amplifier
Amplifier
vo
io
iIN
iE
iF
Vin
VE
Amplifier
Amplifier
VF
1
For each of these, an expression will be derived for gain, input impedance and output
impedance. From these, you will be able to see that:




current derived feedback increases the output impedance;
voltage derived feedback decreases the output impedance;
series applied (voltage) feedback increases the input impedance;
parallel applied (current) feedback decreases the input impedance.
The trick is to identify from a given circuit diagram which type of feedback is which.
Series current
The feedback is proportional to the output current and applied in series with the input as
shown in Figure 2.
IIN
VIN
Vx
Rin
A
io
RL
Rf
Vo
The feedback is proportional to the output current and applied in series with the input as
shown in Figure 2.
βVo  i o R f 
Vo R f
R
β  f
RL
RL
Reduction In Gain
Vx = Vin - Vo ----------------------------------------------------------------(1)
Also Vo = AVx --------------------------------------------------------------(2)
Vx  Vin  βVo
Vx  Vin  Aβ Vx
Vx  Aβ Vx  Vin
Vin  Vx (1  Aβ )
now from (2)
Vo
Vo
 Vx Vin 
(1  Aβ ) and
A
A
Vo
A

the gain is reduced by a factor of (1  Aβ )
Vin (1  Aβ )
2
Input Impedance
Vin = Vx + Vo ----------------------------------------------------------------(1)
Also Vo = AVx -------------------------------------------------------------- (2)
Vin  Vx  βVo
Now Vin  Vx  Aβ Vx
Vin  Vx (1  Aβ )
Z in 
Vin
i in
Z in 
Vx (1  Aβ ) iin RIN (1  Aβ )

 RIN (1  Aβ )
i IN
i IN
Output Impedance
To measure the output impedance of a current series negative feedback amplifier input
terminals should be closed and in an impedance equal to the source impedance. E volts and
zero impedance should be connected across the output terminals. The output impedance is
E
given by R o ( f ) 
ohms
io
IIN
i
Ro
Vx
Rs
Rin
AVx
io
Vo
Vo
E  AVx  I o (R f  R o ) but Vx 
E A
E
Rf
R in
x IoR f
R in  R s
R in
x I o R f  I o (R f  R o )
R in  R s
E  I o (R f  R o )  A
E  I o (R f  R o  A
R in
x Io R f
R in  R s
R in R f
)
R in  R s
3
R R
E
 R o ( f )  (R f  R o  A in f )
Io
R in  R s
if R s is small in comparison to R in then the equation reduces to :
R o ( f )  (R f  R o  AR f )
R o ( f )  R f (1  A )  R o
Example Common Emitter Circuit
+ Vcc
RC
R1
C2
C1
Vin
Vs
R2
Vout
Iout
RE
Vout
0V
This is also known as series current feedback. The feedback signal is a voltage developed
from the output current, i.e. it is current derived and applied to act in series opposition to the
input signal, as shown in Fig. above.
4
Current Shunt Feedback
IIN
Rs
Ix
io
Iout
Vx
Rin
Vout
RL
Rf1
Es
Rf2
By current division Iout =
R s R in
IoR f
Rf
where R is the parallel sum of
hence  
Rf1  Rf 2  R
R s  R in
Rf1  Rf 2  R
also I x  I in  I o
I o  AI x  A(I in  I o )
I o  AI o  AI in
I o (1  A)  AI in
Io
A
 A i(f ) 
I in
1  A
Input Impedance
I in  I x  I o now I o 
(1  A)I o
A
x I in hence I in 
1  A
A
(1  A)I o
A
AI x  AI o  (1  A)I o
I x  I o 
AI x  (1  A)I o  AI o
AI x  I o
 I in  I x  AI x
I in  I x (1  A)
R in ( f ) 
Vin
Vin
V

and in  R in
I in
I x (1  A)
Ix
hence R in ( f ) 
R in
Thus the input impedance is reduced by the applicatio n of
(1  A)
current - shunt negative feedback
5
Output Impedance
For this calculation the input terminals should be closed with an input impedance equal to the
source impedance and a voltage generator of e.m.f. volts and zero internal impedance
connected across the output terminals.
Ix
I
I
Rs
Rin
AiIx
Ro
E
Rf1
Rf2
E  R o (I  I x )  IR f 1
I
Ix
where I x 
IR s
Rin
(R s  R in )
Rf2
I
I -I
Rf1
Rs
 1  AR s

R f1
  R f 1
R o ( f )  E  R o 
.
I
 R s  R in R f 1  R f 2  R 
If R f 2  R , R f 2  R f 1 and R s  R in then this expression reduces to
 AR f 1 
  R f 1 thus the outpuut current is increased using n.f.b.
R o ( f )  R o 1 
R f 2 

Example of Shunt Current Feedback
+ Vcc
R1
RC1
C2
R3
RC2
C3
C1
Vs
R2
R4
RE1
CE
Vout
Iout
RE2
0V
Iout
6
.The feedback signal is a current developed from the output current and applied to act in
parallel opposition to the input signal. The feedback current Io is derived from the output
current Io and acts in parallel to the input signal Vs. A practical example of this type of
feedback is shown at the bottom of page 6.
Shunt current feedback reduces the input impedance while increasing the output impedance
The Table on page indicates the effects of shunt voltage and shunt current feedback which
are both examples of current feedback. This type of feedback is often used to control the gain
over several stages in an amplifier system.
Voltage-Shunt Feedback
IIN
Rs
Ix
Iout
io
Vx
Rin
Ai
Vout
RL
Es
io
Iout
Rf
This type of feedback involves feeding back to the input circuit a current whose magnitude is
proportional to the output voltage. The block diagram of a voltage-shunt n.f.b. amplifier is
shown above.
I o 
Vo
where R  R s R in /( R s  R in ),
Rf  R
I o
Vo

Io
(R f  R )Vo / R L

RL
Rf  R
7
Input Impedance
IIN
Ix
Rs
Io
io
Vx
Rin
Vout
RL
Rf1
Now I x  I in   I o
I o  AI x  A (I in   I o )
I o  A I o  AI in
I o (1  A)  AI in
Io
A
 A i(f ) 
I in
1  A
(1  A)I
A
o
I  I  I now I 
x I hence I 
in
x
o
o 1  A in
in
A
(1  A)I
o
A
AI  AI  (1  A)I
x
o
o
AI  (1  A)I  AI
x
o
o
AI  I
x
o
 I  I  AI
in
x
x
 I  I 
x
o
I  I (1  A)
in
x
R
in (f )
V
V
in
 in 
and
I
I (1  A)
in
x
hence R
in (f )

V
in  R
in
I
x
R
in Thus the input impedance is reduced by the applicatio n of current - shunt negative feedback
(1  A)
8
Output Impedance
Ix
io
Iout
Rs
AiIx
Rin
Es
Ro
io
Iout
Rf
Ix 
Rs
E
.
where R  R s R in /( R s  R in ),
R f  R R s  R in
Io 
A i ER s
E
E


.
(R f  R )( R s  R in ) R o R f  R
Zo 
E

Io
Zo 
E
A i ER s
E
E


(R f  R )( R s  R in ) R o R f  R
1
AiR s
1
1


(R f  R )( R s  R in ) R o R f  R
Example of Current-voltage (shunt voltage) feedback
+ Vcc
RC
R1
Rf
Io
C1
Vs
C2
Vout
R2
RE
0V
Also called shunt voltage feedback. The feedback signal is a current signal developed output
voltage and applied to act in parallel (shunt) opposition to the input signal. Here the feedback
signal is the current Io which is derived from Vout and fed back to act in parallel to the input
signal Vs. This type of feedback can be used as shown by the circuit above.
9
Voltage Series Feedback
IIN
Vo
Rin Av
Vin
Rin
RL
N
E
Vo

E  I in (R s  R in )  Vo

Vin
(R s  R in )  Vo
R in
A vo( f ) 


Vo
is the overall gain of the amplifier
E
Vo
Vin
(R s  R in )  Vo
R in
Av
 R s  R in


 A v 
 R in


Vo
 R  R in

Vin  s
 A v 
 R in

where A v 
if R s  R in then the equation reduces to
Vo
is the open loop gain
Vin
Av
1  A v 
Input Impedance
IIN
Av
Vin
Vx
N
Rin
AvVx
RL
Rin(f)
AvVx

10
I in 
Vin
V
 x
R in ( f ) R in
but Vx  Vin   A v Vx or Vin  Vx (1   A v )
Hence R in ( f ) 
Vin Vx

(1   A v )
I in
I in
R in ( f )  R in (1   A v )
11
Output Impedance
IIN
i
Ro
Rs
Vx
E
Rin
AvVx
E

E  A v Vx  IR o
but Vx  ER in /( R s  R in ) so that

A v R in 
E 1 
  IR o
 R s  R in 
Ro
E

I 
A v R in 
1 

 R s  R in 
R o (R s  R in )
E
 R o(f ) 
I
R s  R in (1  A v )
The output impedance R o(f) of the amplifierw ith the load disconnect ed. R o is the outptu impedance
before feedback is applied and A v is the open circuit vo ltage gain of the amplifier.
Note if a load resistor is connected across the ouput of the amplifier then the the output impedance
of the amplifier is equal to R o ( f ) in parallel with the load resistance .
If R s  0 then the output impedance R o(f) 
Ro
(1  A v )
12
Example of series voltage feedback (Emitter Follower)
+ Vcc
R1
C1
C2
Vin
R2
Vs
RE Vout
Vout
0V
This is sometimes called series voltage feedback, where the feedback is a voltage signal
derived from the output voltage and applied to act in series opposition to the input signal as
shown in Fig. above. The feedback voltage (V0) is derived from the output voltage Vo and is
fed back to act in series with the input voltage signal Vs. A simple way of obtaining V0, is to
use a potential divider. A practical example of this is the emitter or source follower circuits
shown in Fig above.
Note In this circuit the output is taken across the emitter and source resistors.
Here the feedback voltage Vo = Vo this means that  = 100%, thus the emitter and source
follower will have unity voltage gain. That is, if Av =150, then
A v (f) 
Av(f) 
Av
1  A v
Av(f) 
150
1  (150 x 1)
150
 0.99
151
Av(f)  1
Feedback Voltage Feedback
Type
Series Voltage
Series Current
Voltage gain (Av)
reduced
reduced
Input impedance (Rin)
increased
increased
Output impedance (Ro)
decreased
increased
Input impedance (Rin)
decreased
decreased
Output impedance (Ro)
decreased
increased
Feedback Current Feedback
Type
Shunt Voltage
Shunt Current
Voltage gain (Ai)
decreased
decreased
13
E  I in (R s  R in )  Vo

Vin
(R s  R in )  Vo
R in
A vo( f ) 


Vo
is the overall gain of the amplifier
E
Vo
Vin
(R s  R in )  Vo
R in
Av
 R s  R in


 A v 
 R in


Vo
 R  R in

Vin  s
 A v 
 R in

where A v 
if R s  R in then the equation reduces to
Vo
is the open loop gain
Vin
Av
1  A v 
Input Impedance
IIN
Av
Vin
Vx
N
Rin
AvVx
RL
Rin(f)
AvVx
I in 

Vin
V
 x
R in ( f ) R in
but Vx  Vin   A v Vx or Vin  Vx (1   A v )
Hence R in ( f ) 
Vin Vx

(1   A v )
I in
I in
R in ( f )  R in (1   A v )
14
Output Impedance
IIN
i
Ro
Rs
Vx
E
Rin
AvVx
E

E  A v Vx  IR o
but Vx  ER in /( R s  R in ) so that

A v R in 
E 1 
  IR o
 R s  R in 
Ro
E

I 
A v R in 
1 

 R s  R in 
R o (R s  R in )
E
 R o(f ) 
I
R s  R in (1  A v )
The output impedance R o(f) of the amplifierw ith the load disconnect ed. R o is the outptu impedance
before feedback is applied and A v is the open circuit vo ltage gain of the amplifier.
Note if a load resistor is connected across the ouput of the amplifier then the the output impedance
of the amplifier is equal to R o ( f ) in parallel with the load resistance .
If R s  0 then the output impedance R o(f) 
Ro
(1  A v )
15