CHAPTER ONE

FEEDBACK AMPLIFIER

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FEEDBACK AMPLIFIER
What is feedback?
• Taking a portion of the signal arriving at the load and feeding it
back to the input.
• Shows how to modify the characteristics of amplifiers by combining
a portion of the output signal with the external signal.
• System can be a feed forward (Net gain ↑) or feedback (Net gain ↓)!
• Feedback can be –ve feedback or +ve feedback!
• Positive feedback will cause the amplifier oscillation.
• Many advantages are to be gained from the use of negative
feedback.
• What is negative feedback? Adding the feedback signal to the input
so as to partially cancel the input signal to the amplifier.
• Doesn’t this reduce the gain? Yes, this is the price we pay for using
negative feedback.

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 Why we use -ve feedback?
• Many desirable characteristics are obtained for the price of gain
reduction.
• It provides a series of benefits, such as improved bandwidth,
that outweigh the costs in lost gain and increased complexity in
amplifier design.

This is a signal-flow diagram, and the quantities x represent either voltage or current
signals.

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X f   f Xo where  f is called the feedback factor

X o  AX i where A is the amplifier' s gain, e.g . voltage gain

Xi  X s  X f where X i is the net input signal to the basic amplifier,

X s  the signal from the source

The amplifier' s gain with feedback is given by
Xo AX i A A A
Af      A
Xs Xi  X f Xf  f Xo 1  f A
1 1
Xi Xi
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 The General Feedback Equation

• Closed loop and open loop

• Closed loop gain
xo A
Af  
xs 1  A
• Feedback factor β
• Loop gain Aβ
• Amount of feedback (1+ Aβ)

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 Advantages of Negative Feedback
Gain desensitivity
• less variation in amplifier gain with changes in 
(current gain) of transistors due to dc bias, aging,
temperature, fabrication process, etc.
Defn:-
• The fractional change in amplification with feedback
divided by the fractional change without feedback is
called sensitivity of the transfer gain of an amplifier.

Desensitivity = = 1+βA

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Feedback can be used to desensitize the closed-loop gain to
variations in the basic amplifier. Let’s see how.
Assume beta is constant. Taking differentials of the closed-loop
gain equation gives…
A dA
Af  dA f 
1  A 1  A 2

Divide by Af
dA f dA 1  A 1 dA
 
Af 1  A 2 A 1  A A

This result shows the effects of variations in A on Af is mitigated

by the feedback amount. 1+Aβ is also called the desensitivity
amount.
Feedback also affects the input and output resistance of the
amplifier (increases Ri and decreases Ro by 1+A.β) 7
 D=1+βA
• With introduction of negative feedback the feed forward
transfer gain A is reduced by the factor D.
=

If Aβ>>1 then D → Aβ

≈ =
This implies the transfer gain with feedback depends
entirely on the feedback gain β.
So if the feedback network is designed using stable passive
elements then the stability of the system can be improved
substantially.

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 EXAMPLE

If an amplifier with gain of -1000 and feedback of β = -0.1 has a

gain change of 20% due to temperature, calculate the change in
gain of the feedback amplifier.
Solution:

The improvement is 100 times. Thus, whereas the amplifier gain

changes from |A| = 1000 by 20%, the gain with feedback changes
from || = 100 by only 0.2%.

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 Frequency distortion (Bandwidth extension)
• Extends dominant high and low frequency poles to
higher and lower frequencies, respectively.
From the equation
≈ = for Aβ>>1
If the feedback network does not contain reactive elements,
the overall gain is not a function of frequency.
Figure below shows that the amplifier with negative feedback has
more bandwidth ( B f )than the amplifier without feedback ( B ).
-

-
-

To simply state
becomes immune to any
frequency changes.
With FB gain is almost
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constant!
Nonlinear distortion
• In power amplifiers the input signal is large thus mostly
exceeding the linear region of operation of the
amplifier and as a consequence the output signal is
distorted.
• With negative feedback introduced, the input signal is
increased by the same amount () by which the gain is
reduced. So that the output signal amplitude remains
the same.
Improves amplifier linearity –
• reduces distortion in signal due to gain variations due
to transistors.

Cost of these advantages:

• Loss of gain, may require an added gain stage to compensate
added complexity in design 11
 Noise reduction
• improves signal-to-noise ratio
• Employing the same reasoning as that in the discussion of
non-linear distortion it can be shown that noise
introduced in the amplifier with negative feed back gets
divided by D at the output.

Xi
Xs
A Xo
 (1+βA)= +
D.= +
Xf 

• This necessary explains that noise at the output gets

reduced by a factor of D when negative feed back is
incorporated in to the system. 12
 Disadvantages of Negative Feedback

1. Circuit Gain – overall amplifier gain is

reduced compared to that of basic amplifier.
2. Stability – possibility that feedback circuit
will become unstable and oscillate at high
frequencies.

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 Classifications of amplifiers
Four broad categories of amplifiers are:-
1. Voltage amplifier (series-shunt)
2. Current amplifier (shunt series)
3. Trans conductance amplifier (series-series)
4. Trans resistance amplifier (shunt-shunt )
• The classifications of feedback amplifier is based on the
magnitude of the input and output impedances of an
amplifier relative to the source and load impedances
respectively.
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 1. Voltage amplifier

Vi Vo
if Ri  Rs
then Vi  Vs
and if Ro  R L
then,
Vo  AvVi  AvVs
Vo
hence Av 
Vi
with RL  
represent the open circuit voltage gain.

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• If the amplifier input resistance is large
compared with the source resistance , then .
• If the external load resistance is large compared
with the output resistance of the amplifier, then
≈.
• This amplifier provides a voltage output
proportional to the voltage input, and the
proportionality factor is independent of the
magnitudes of the source and load resistances.
Such a circuit is called a voltage amplifier.
• An ideal voltage amplifier must have infinite
input resistance and zero output resistance
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 2. Current amplifier
Current amplifier provides an output current proportional to the signal
current and the proportionality factor is independent of and .
if Ri  Rs
Ii Io
then Ii  I s
and if Ro  RL
then,
I o  Ai I i  Ai I s
Io
hence Ai 
Ii
with RL  0
represent the short circuit current gain.
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• An ideal current amplifier must have zero input
resistance and infinite output resistance
• In practice, the amplifier has low input resistance
and high output resistance.
• It drives a low resistance load () and is driven by
a high resistance source ().
= ≈
• Hence, the output current is proportional to the
signal current.

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 3.Transconductance amplifier

Vi Io
if Ri  Rs
then Vi  Vs
and if Ro  RL
then,

I o  GmVi  GmVs
Io
hence Gm 
Vi
with RL  0
represent the short circuit mutual or
transfer conductance
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• Trans conductance amplifier must have an infinite input
resistance and infinite output resistance
• A practical trans conductance amplifier has a large input
resistance () and hence must be driven by low resistance
source.
• It presents a high output resistance () and hence drives a
low resistance load.

I o  GmVi  GmVs

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4. Trans resistance amplifier

Ii Vo
if Ri  Rs

then Ii  I s
and if Ro  R L
then ,

Vo  Rm I i  Rmis
Vo
hence Rm 
Ii
represent the open circuit mutual or with RL  
transfer resistance.

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• Trans resistance amplifier ideally supplies an output
voltage in proportion to the signal current independently
of and .
• For practical trans resistance amplifier we must have ()
and ().
• Hence, the input and output resistances are low relative
to the source and load resistances.
Vo  Rm I i  Rmis
Vo
Rm  with RL  
Ii

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Paramete input output i/p o/p Gain
rs condition condition
Current
amplifier
Voltage
amplifier
Trans
conducta
nce
Trans
resistance

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Basic structure of feedback amplifier

Fig. Block diagram of feed back amplifier

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 Feedback Network

• This block is usually a passive two-port network.

• contain resistors, capacitors, and inductors.
• Usually it is simply a resistive network.

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 Sampling Network
Voltage sampler (shunt sampler)
Voltage sampler
can be a voltage
amplifier or
trans resistance
amplifier

• The output voltage is sampled by connecting the

feedback network in shunt across the output.
Type of connection is referred to as voltage or shunt or
node sampling.
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 Current sampler (series sampler)

Current sampler can be

current amplifier or
trans conductance
amplifier.

• The output current is sampled by connecting the

feedback network in series with the output
•Type of connection is referred to as current or series or
loop sampling.
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 Comparator or Mixer Network
Series mixer

series mixer can be trans

conductance or voltage
amplifier.

• voltage - applied feedback .

• identified as voltage or series or loop mixing.
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 Shunt mixer

Shunt mixer can be

current amplifier or
trans resistance.

• current - applied feedback

• identified as current or shunt or node mixing.

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 Feedback Topologies
There are four types of feedback amplifiers. Why?
 Output sampled can be a current or a voltage
 Quantity fed back to input can be a current or a
voltage
Four possible combinations of the type of output
sampling and input feedback
o One particular type of amplifier, e.g. voltage amplifier,
current amplifier, etc. is used for each one of the four
types of feedback amplifiers.
o Feedback factor βf is a different type of quantity, e.g.
voltage ratio, resistance, current ratio or conductance, for
each feedback configuration.
o Before analyzing the feedback amplifier’s performance,
need to start by recognizing the type of configuration.

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Terminology used to name types of feedback amplifier, e.g.
Series-shunt
• First term refers to nature of feedback connection at the
input.
• Second term refers to nature of sampling connection at
the output.
The four feedback circuit can be described by the types
of connections at the input and output of circuit.

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 fig. Feedback amplifier types: (a) voltage-series feedback, = ; (b) voltage-shunt feedback, =
; (c) current-series feedback, = ; (d) current-shunt feedback, = ;
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 1. Series-shunt topology

Voltage-mixing voltage-sampling
Voltage amplifier – voltage-controlled voltage source
Requires high input impedance, low output impedance
Use series-shunt feedback (voltage-voltage feedback)
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 2.Shunt-series topology

Current-mixing current-sampling
Current amplifier – current-controlled current source
Use shunt-series feedback (current-current feedback)
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 3.Series-series topology

Voltage-mixing current-sampling
Trans conductance amplifier – voltage-controlled current source
Use series-series feedback (current-voltage feedback)
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 4.Shunt-shunt topology

Current-mixing voltage-sampling
Trans impedance amplifier – current-controlled voltage source
Use shunt-shunt feedback (voltage-current feedback)
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Parameters input output sampler mixer FB Condn

Current Series/current shunt Current

amplifier series FB
Voltage Shunt/voltage series Voltage
amplifier series FB

Trans Shunt/voltage shunt Voltage

resistance shunt FB
Trans Series/current series Current
conductance series FB

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 General Feedback Structure
To make it general, the figure shows signal flow as opposed to
voltages or currents (i.e., signals can be either current or
voltage).

xi xo
Source  A Load
xs

xf


Fig. Basic structure of a feedback

amplifier.

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 Voltage and current signals in feed back amplifiers
Signal or Types of feedback
Ratio Voltage Current Current shunt Voltage shunt
series series (CA) (TR)
(VA) (TC)

voltage Current current voltage

voltage voltage current current

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 Method of Feedback Amplifier Analysis

• Recognize the feedback amplifier’s configuration, e.g. Series-shunt

• Calculate the appropriate gain A for the amplifier, e.g. voltage gain.
– This includes the loading effects of the feedback circuit (some combination
of resistors) on the amplifier input and output.
• Calculate the feedback factor βf A
Af 
• Calculate the factor βf A and make sure that it is: 1  f A

1) positive and 2) dimensionless

• Calculate the feedback amplifier’s gain with feedback Af using
• Calculate the final gain of interest if different from the gain calculated,
e.g. Current gain if voltage gain originally determined.
• Determine the dominant low and high frequency poles for the original
amplifier, but taking into account the loading effects of the feedback
network.
• Determine the final dominant low and high frequency poles of the
amplifier with feedback using
 
 Hf  1   f A  H
L
 Lf 
1   f A
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• Assumes feedback circuit does not load down the basic
amplifier A, i.e. doesn’t change its characteristics
• Doesn’t change gain A
• Doesn’t change pole frequencies of basic
amplifier A
• Doesn’t change Ri and Ro

• For the feedback amplifier as a whole, feedback does

change the midband voltage gain from A to Af
A
Af 
1  f A
• Does change input resistance from Ri to Rif

Rif  Ri 1   f A 
• Does change output resistance from Ro to Rof
Ro
Rof 
1  f A
• Does change low and high frequency 3dB frequencies


 Hf  1   f A  H  Lf 

L
1  f A 

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 Midband Gain
V A V AV AV AV
AVf  o  V i   
Vs Vi  V f Vf  f Vo 1   f AV
1 1
Vi Vi

Input Resistance
Vi  V f Vi   f Vo
V
Rif  s 
Ii Ii

Vi 

 Ri 1   f AV 
 R 
 i

Output Resistance
V  AV Vi
It  t
Ro
But Vs  0 so Vi  V f

and V f   f Vo   f Vt so

It 

Vt  AV  V f


Vt  AV  f Vt  
Ro Ro



Vt 1  AV  f 
Ro
V Ro
so Rof  t 
It 
1  AV  f 

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 Series-Shunt Feedback Amplifier - Ideal Case
Low Frequency Pole 
 A


 o   Ao 

 1 L   
Ao A    Ao  1   f Ao  A fo
For A   then A f      s      
1
L 1   f A     L   L  1    Lf 
s  Ao  1  s   f Ao  1    1 



1   f     1   f Ao  s   s 
  
1 L
 s 
Ao L
where A fo 
1   f Ao
 Lf 
1   f Ao
Low 3dB frequency lowered by feedback.

High Frequency Pole  

 
 Ao 
 s   Ao 
 
 1   1   f Ao 
Ao A   H  Ao   A fo
For A   then A f       
s 1   f A     s   s   
1 1    f Ao  1  s 
H 
1   f
Ao 
  H 
1 
 
  H 1   f Ao     Hf



 s 
1
 H 
 

where A fo 
Ao
1   f Ao

 Hf   H 1   f Ao  Upper 3dB frequency raised by feedback.

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• Feedback networks consist of a set of resistors
– Simplest case (only case considered here)
– In general, can include C’s and L’s (not considered
here)
– Transistors sometimes used (gives variable
amount of feedback) (not considered here)
• Feedback network needed to create Vf feedback signal
at input (desirable)

• Feedback network has parasitic (loading) effects

including:
• Feedback network loads down amplifier input
– Adds a finite series resistance
– Part of input signal Vs lost across this series
resistance (undesirable), so Vi reduced
• Feedback network loads down amplifier output
– Adds a finite shunt resistance
– Part of output current lost through this shunt
resistance so not all output current delivered to
load RL (undesirable)

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