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1. Feedback Amplifiers
Dr. Monir Hossen
ECE, KUET
Department of Electronics and Communication Engineering, KUET

2. Department of Electronics and Communication Engineering, KUET 2
Feedback Concepts
 The input signal Vs and
a feedback signal Vf is
applied to the mixer.
 For negative feedback
the difference between
Vs and Vf equal to Vi is
applied to the
amplifier.
 Vo is the output and a
portion of output β
connected to the
feedback network.
 The process of injecting a fraction of output energy of
some device back to the input is known as feedback.

3. Department of Electronics and Communication Engineering, KUET 3
Positive Feedback
 When the feedback signal is in phase with the input signal and
thus aids it, it is called positive feedback.
 Both amplifier and feedback network introduce a phase shift
of 180°.
 The result is a 360° phase shift around the loop, causing the
feedback voltage Vf to be in phase with the input signal Vin.
 The positive feedback increases the gain of the amplifier.
However, it increased distortion and instability.

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Negative Feedback
o When the feedback signal is out of phase with the input signal
and thus opposes it, it is called negative feedback.
o The amplifier introduces a phase shift of 180° into the circuit
while the feedback network is so designed that it introduces 0°
phase shift.
o The result is that the feedback voltage Vf is 180° out of phase
with the input signal Vin.
o Negative feedback reduces the gain of the amplifier.
o However, it provides low distortion, stability in gain, increased
bandwidth and improved input and output impedances.

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Advantages of Negative Feedback
 Higher input impedance.
 Better stabilized voltage gain.
 Improved frequency response.
 Lower output impedance.
 Reduced noise.
 More linear operation.
 Reduce non-linear distortion.

6. Department of Electronics and Communication Engineering, KUET 6
Feedback Connection Types
 Feedback can be connected in four basic ways.
 Both voltage and current can be feed back to the
input either in series or parallel.
1) Voltage – series feedback.
2) Voltage – shunt feedback.
3) Current – series feedback.
4) Current – shunt feedback.
oSeries feedback connections increase the input resistance.
oShunt feedback connections decrease the input resistance.
oVoltage feedback decrease the output impedance.
oCurrent feedback increase the output impedance.

7. Department of Electronics and Communication Engineering, KUET 7
Voltage Series Feedback
 Here, a portion of output voltage
feedback in series to the input.
 Negative feedback results overall
gain reduction.
 If there is no feedback (vf = 0).
The voltage gain of the amplifier
is:
i
o
s
o
v
v
v
v
A 
 If vf is connected in series with the input, then => vi = vs – vf
Since vo = Avi = A(vs – vf) = Avs – Avf
= Avs – A(βvo)
=> vo(1+ βA) = Avs
A
A
v
v
s
o


1 A
A
v
v
A
s
o
f


1
So overall voltage
gain with feedback is

8. Department of Electronics and Communication Engineering, KUET 8
Example
Ex. 01) When negative voltage feedback is applied to an amplifier of
gain 100, the overall gain falls to 50. (i) Calculate the fraction of the
output voltage fedback. (ii) If this fraction is maintained, calculate the
value of the amplifier gain required if the overall stage gain is to be 75.

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Example

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Ex. 2) The gain of an amplifier without feedback is 50 whereas
with negative voltage feedback, it falls to 25. If due to ageing, the
amplifier gain falls to 40, find the percentage reduction in stage
gain (i) without feedback and (ii) with negative feedback.
Example

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Ex. 03) An amplifier is required with a voltage gain of 100 which does
not vary by more than 1%. If it is to use negative feedback with a basic
amplifier the voltage gain of which can vary by 20%, determine the
minimum voltage gain required and the feedback factor.
Example

12. Department of Electronics and Communication Engineering, KUET 12
The gain with feedback
Voltage Shunt Feedback
fi
i
s
o
f
II
AI
I
v
A


Here,
Vo = AIi
Is = Ii+ If
oi
i
f
vI
AI
A


ii
i
f
AII
AI
A


A
A
Af


1
Ii
Is
i
o
I
v
A 
o
f
v
I

of vI 
vo RL

13. Department of Electronics and Communication Engineering, KUET 13
Input Impedance with Feedback (1/3)
 Voltage Series Feedback:
The input impedance of the network can be determined as:
i
is
i
os
i
fs
i
i
i
z
Avv
z
vv
z
vv
z
v
I
 






14. Department of Electronics and Communication Engineering, KUET 14
Input Impedance with Feedback (2/3)
isii AvvZI 
iiis AvZIv 
iiiis ZAIZIv 
i
iiii
i
s
if
I
ZAIZI
I
v
Z


We know,
)1(So, AZZ iif 

15. Department of Electronics and Communication Engineering, KUET 15
Input Impedance with Feedback (3/3)
 Voltage Shunt Feedback:
Ii
Is
If = βvo
Ri
Ro
AIi
vo RL
o
f
v
I

oi
i
fi
i
s
i
if
vI
v
II
v
I
v
Z




Here,
Zi = vi /Ii
A=vo / Ii
ioii
ii
if
IvII
Iv
Z
//
/


A
Z
Z i
if


1

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 Voltage Series Feedback:
o The output impedance is determined by applying a voltage v
resulting in a current I with vs shorted out ( vs = 0)
o The voltage v is then –
v = IZ0 + Avi
Output Impedance with Feedback (1/3)
for vs = 0, vi = - vf

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Output Impedance with Feedback (2/3)
fo AvIZv ,thatSo
)( vAIZv o 
Rewriting the equation as:
oIZAvv  
A
Z
I
v
Z o
of


1

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Current Series Feedback:
IA
Z
v
Av
Z
v
Av
Z
v
I
o
f
o
i
o

Output Impedance with Feedback (3/3)
By applying a signal
voltage v to the output
with vs shorted out
resulting a current I, the
ratio of v to I being the
output impedance.
At, vs = 0, vi = - vf
IAZvIZ oo 
vIAZo  )1(  )1( AZ
I
v
Z oof 

19. Department of Electronics and Communication Engineering, KUET 19
Effect of Negative Feedback on Gain and
Bandwidth (1/2)
 We know overall gain with negative feedback is:
 For a practical amplifier the open-loop gain drops off at high
frequencies due to the active device and circuit capacitance.
 Gain also drop off at low frequencies for capacitively coupled
amplifier stages.
 When open-loop gain A drops enough low then the βA is no
longer larger than 1, so Af = 1/ β is not true.
A
A
A
A
Af




1

1
So, fA
1for A

20. Department of Electronics and Communication Engineering, KUET 20
 The figure shows that the amplifier with negative feedback has
more bandwidth Bf than the amplifier without feedback B.
 The feedback amplifier also has a higher upper 3-dB frequency
and smaller lower 3-dB frequency.
 However, the product of gain and frequency for both cases are
same value.
 Ao*f1 = Afo*f1f and Ao*f2 = Afo*f2f
Effect of Negative Feedback on Gain and
Bandwidth (2/2)

21. Department of Electronics and Communication Engineering, KUET 21
Gain Stability with Feedback
 In addition to the β factor setting a precise gain value, we are
also interested in how stable the feedback amplifier is
compared to an amplifier without feedback.
 This shows that the magnitude of the relative change in gain
is reduced by the factor compared to that without
feedback
f
f
A
dA
A
A
dA

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Example
The improvement is 100 times. Thus, while the amplifier gain
changes by 20%, the gain with feedback changes by only 0.2%.
Ex. 04) If an amplifier with gain of 1000 and feedback of 0.1 has a
gain change of 20% due to temperature, calculate the change in
gain of the feedback amplifier.
Soln:
We know

23. Department of Electronics and Communication Engineering, KUET 23
Practical Feedback Circuits (1/2)
 Here, R1 and R2
resistors are used as a
feedback network.
 A part of output
signal is obtained
from R2 to ground.
 vf is connected in
series with the source
signal vs.
 Without feedback the
amplifier gain is:
A = vo/vi = -gmRL
 where RL is the
parallel combination
of RD, Ro and (R1+R2).
 Voltage series feedback:

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 The feedback factor:
21
2
RR
R
v
v
o
f



Practical Feedback Circuits (2/2)
 We know the gain with negative feedback is:
mL
Lm
f
gR
RR
R
Rg
A
A
A
21
2
11







 If βA>>1 we have:
2
211
R
RR
Af




25. Department of Electronics and Communication Engineering, KUET 25
Example
Exam: 05) Calculate the gain without and with feedback for the
FET amplifier circuit of figure bellow and the following values: R1
= 80 kΩ, R2 = 20 kΩ, Ro = 10 kΩ, RD = 10 kΩ, and gm = 4000 µS,

26. Department of Electronics and Communication Engineering, KUET 26
Example
Soln:

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Exam. 06) Calculate the amplifier gain of the circuit of Fig. bellow
for op-amp gain A 100,000 and resistances R1 1.8 k and R2 200 .
Example

28. Department of Electronics and Communication Engineering, KUET 28
Barkhausen Criterion (1/3)
 Typically, the feedback network is composed of
passive lumped components that determine the
frequency of oscillation.
 So, the feedback is complex-transfer function,
hence denoted as β.
 We can derive the requirements for oscillation as
follows:
o Initially, sinusoidal source is xin
o The condition of oscillation is even xin is zero
the output xout can be non-zero.

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β
 A
xin +
-
xin + βxout
βxout
xout
Barkhausen Criterion (2/3)
 The output of the amplifier block can be written as
][ outinfout xxAx 
infoutfout xAxAx  
inffout xAAx  )1( 
f
inf
out
A
xA
x


1

30. Department of Electronics and Communication Engineering, KUET 30
Barkhausen Criterion (3/3)
 If xin is zero, the only way the output can be
nonzero is to have βAf = 1.
 This condition is known as Barkhausen
criterion.

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Example/ Problem
 Ex/Pb 07): An amplifier has an open-loop gain A=100,000. A
negative feedback of 10 dB is applied. Find (i) voltage gain with
feedback, (ii) value of feedback fraction β.
 Soln: (i) open-loop gain without feedback in dB
= 20log10 100,000
= 20log10105
= 100 dB
voltage gain with feedback = 100-10 = 90 dB
Now, 20log10(Af) = 90
=> log10 (Af) = 90/20 =4.5
so, Af = antilog 4.5 = 31622
(ii)
A
A
Af


1 *000,1001
000,100
31622


5
10*17.2 


32. Department of Electronics and Communication Engineering, KUET 32
Nonlinear Distortion in Feedback
Amplifier (1/3)
Nonlinear
distortion
Fundamental
& Harmonics
In nonlinear distortion, elongate +ve half cycle and
compress –ve half cycle.

33. Department of Electronics and Communication Engineering, KUET 33
Nonlinear Distortion in Feedback
Amplifier(2/3)
 Nonlinear distortion produces harmonics of the
input signal.
 If a input signal has a 1KHz frequency then the
output signals frequencies are 1, 2, 3 … KHz.
 Here, 1KHz is fundamental frequency, and all others
are harmonics.
 Sometime nonlinear distortion is called harmonic
distortion.
 Total Harmonic Distortion (THD) without feedback
is:
%100
VoltagelFundamenta
VoltageHarmonicTotal
THD

34.  Total Harmonic Distortion (THD) with feedback is:
 So THDf <THD
Department of Electronics and Communication Engineering, KUET 34
Nonlinear Distortion in Feedback
Amplifier (3/3)
v
f
A
THD
THD


1
 Here, Av is gain
without feedback.
 β is the feedback
fraction

35. Thanks for Your Kind
Attention
Department of Electronics and Communication Engineering, KUET