Analog Electronic Circuits (PC231EC)

UNIT 2
FEEDBACK AMPLIFIERS

Introduction
Feedback plays an important role in electronic circuits where some part of the output is fed
back to the input through a feedback network. So at the input side there is source and
feedback signals.
When the source signal is in phase with feedback signal, it is called as positive feedback.
When the source signal is out of phase with feedback signal, it is called as negative feedback.
Negative feedback employed in amplifying the signals and positive feedback is employed in
oscillator circuits.

Classification of Amplifiers
Amplifiers are classified based on input, output, load and source resistances.
1) Voltage Amplifier

Fig.3.1: Voltage Amplifier

Here the output voltage is proportional to input voltage and at both i/p and o/p of circuis we
have thevenin circuits
Vo = AVVi , RL>> Ro, Ri>>RS
1) Current Amplifier

In current amplifier, o/p current is directly proportional to input current. It maintains Norton
equivalent circuit at both i/p and o/p.

Fig.3.2: Current Amplifier

Io = AIII , RL<<Ro, Ri<<RS

2) Transconductance amplifier

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In transconductance amplifier the o/p current depends on i/p voltage, it has thevenin
equivalent at i/p and Norton equivalent to o/p.

Fig.3.3: Trans conductance amplifier

IL = GMVi , RL<<Ro, Ri>>RS

4) Transresistance Amplifier
In this amplifier, o/p voltage depends on i/p current, at i/p it has Norton equivalent and o/p it
has thevenin equivalent.

Fig.3.4: Trans resistance Amplifier

VO=RMII RL>>Ro, Ri>>RS

Block Diagram of Feedback Amplifier

Fig.3.5: Block Diagram of Feedback Amplifier

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Sampling Network:

A sampling network samples some part of the output signal and connected to input through
feedback network. Depending on type sampling quantity, we have two types, they are
a) Voltage sampling
b) Current Sampling

Fig.3.6: Voltage sampling

b) Current Sampling

Fig.3.7: Current sampling

Mixer or Comparator :
It is used to connect the feedback signal at the input circuit. There are two types of mixing
circuits.
a) Series Mixing
b) Shunt Mixing

Fig.3.8: Series Mixing

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 Fig.3.9: Shunt Mixing

Based on Sampling network and mixer, negative feedback amplifiers are classified into

1) Voltage Series Feedback Amplifier

2) Voltage Shunt Feedback Amplifier
3) Current Series Feedback Amplifier
4) Current Shunt Feedback Amplifier

1. Voltage Series Feedback Amplifier

In the voltage series feedback circuit, a fraction of the output voltage is applied in series
with the input voltage through the feedback circuit. This is also known as shunt-driven
series-fed feedback, i.e., a parallel-series circuit. Fig.3.9 shows the block diagram of
voltage series feedback, by which it is evident that the feedback circuit is placed in shunt
with the output but in series with the input. As the feedback circuit is connected in shunt with
the output, the output impedance is decreased and due to the series connection with the input, the
input impedance is increased.

Fig.3.10: Voltage Series Feedback Amplifier

2. Voltage Shunt Feedbcak Amplifier

In the voltage shunt feedback circuit, a fraction of the output voltage is applied in
parallel with the input voltage through the feedback network. This is also known
as shunt-driven shunt-fed feedback i.e., a parallel-parallel proto type. The below
figure 3.10 shows the block diagram of voltage shunt feedback, by which it is evident
that the feedback circuit is placed in shunt with the output and also with the input. As
the feedback circuit is connected in shunt with the output and the input as well, both the
output impedance and the input impedance are decreased.
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 Fig.3.11: Voltage Shunt Feedback Amplifier

3. Current Series Feedback Amplifier

In the current series feedback circuit, a fraction of the output voltage is applied in
series with the input voltage through the feedback circuit. This is also known
as series-driven series-fed feedback i.e., a series-series circuit.The following figure
shows the block diagram of current series feedback, by which it is evident that the
feedback circuit is placed in series with the output and also with the input. As the
feedback circuit is connected in series with the output and the input as well, both the
output impedance and the input impedance are increased.

Fig.3.12: Current Series Feedback Amplifier

4. Current Shunt Feedback Amplifier

In the current shunt feedback circuit, a fraction of the output voltage is applied in series
with the input voltage through the feedback circuit. This is also known as series-driven
shunt-fed feedback i.e., a series-parallel circuit. The below figure shows the block
diagram of current shunt feedback, by which it is evident that the feedback circuit is
placed in series with the output but in parallel with the input. As the feedback circuit is
connected in series with the output, the output impedance is increased and due to the
parallel connection with the input, the input impedance is decreased.

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 Fig.3.13: Current Shunt Feedback Amplifier

Feedback Network
A feedback network can be a simple resistor or capacitor or combination of R and C. but
more often a resistor is used where a received part of the output signal is fed back to the
input.

General Characteristics of Negative Feedback Amplifiers

1) Transfer Gain

Fig.3.14: Negative Feedback Amplifiers

Gain of an amplifier without feedback is A = Xo/Xi

Gain of an amplifier without feedback is Af = Xo/Xs
The output of mixer is Xi = Xs – Xf
Xs = Xi + Xf
𝑋𝑜
Af = 𝑋 +𝑋
𝑖 𝑓
𝑋𝑜
𝑋𝑖 𝐴
= 𝑋𝑓 = 𝑋𝑓 𝑋
1+ 1+ ∗ 𝑜
𝑋𝑖 𝑋𝑜 𝑋𝑖

𝐴
Af =
1+𝐴𝛽

Gain with feedback (Af) decreases by a factor of 1+Aβ

2) Stability of Gain
𝐴
Af = 1+𝐴𝛽
Differentiating w.r.t A we get
𝑑𝐴𝑓 (1+𝐴𝛽)−𝐴𝛽
=
𝑑𝐴 (1+𝐴𝛽)2
𝑑𝐴𝑓 1
=
𝑑𝐴 (1+𝐴𝛽)2

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 𝑑𝐴
dAf = (1+𝐴𝛽)2
𝑑𝐴𝑓 𝑑𝐴 1
= *𝐴
𝑑𝐴 (1+𝐴𝛽)2 𝑓
𝑑𝐴
𝑑𝐴𝑓 𝐴
=
𝑑𝐴 1+𝐴𝛽
𝑑𝐴𝑓
= 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑔𝑎𝑖𝑛 𝑤𝑖𝑡ℎ 𝑓𝑒𝑒𝑑𝑏𝑎𝑐𝑘
𝑑𝐴
𝑑𝐴
= 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑔𝑎𝑖𝑛 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑓𝑒𝑒𝑑𝑏𝑎𝑐𝑘
𝐴
1
𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑆𝑒𝑛𝑠𝑖𝑡𝑖𝑣𝑖𝑡𝑦 and 1+Aβ is Desensitivity
1+𝐴𝛽
The Ratio of fractional change in gain with feedback to fractional change in gain without
feedback is called “Sensitivity”

3) Frequency Response & Bandwidth

We know that AF = A/(1+Aβ)
For Low freq. range (AF)low = Alow /(1+Alowβ)
For mid freq. range (AF)mid = Amid /(1+Amidβ)
For high freq. range (AF)high = Ahigh /(1+Ahighβ)

a) Effect on lower cut-off freq. with negative feedback.

From two stage RC coupled BJT Amplifier

Alow = Amid / (1-j(fL/f))

(AF)low = Alow /(1+Alowβ)

𝐴𝑚𝑖𝑑
(AF)low = 𝑓𝐿
1−𝑗( )+ 𝐴𝑚𝑖𝑑 𝛽
𝑓
𝐴𝑚𝑖𝑑
(AF)low = 𝑓
𝑗( 𝐿 )
𝑓
(1+𝐴𝑚𝑖𝑑 𝛽)(1 − )
1+𝐴𝛽

(𝐴𝑓 )𝑚𝑖𝑑
(AF)Low = 𝑓𝐿
1−𝑗
𝑓(1+𝐴𝑚𝑖𝑑 𝛽)

(𝐴𝑓 )𝑚𝑖𝑑 𝑓𝐿
(AF)Low = 𝑓 𝑓𝐿𝑓 =
1−𝑗( 𝐿 ) 1+ 𝐴𝑚𝑖𝑑 𝛽
𝑓
Lower cut-off frequency with feedback decreases by a factor of 1+Amidβ

b) Effect on upper cut-off freq. with negative feedback.

𝐴𝑚𝑖𝑑
Ahigh = 𝑓
1+𝑗( )
𝑓𝐻

𝐴𝑚𝑖𝑑
(Af)high = 𝑓
1+𝑗( )+𝐴𝑚𝑖𝑑 𝛽
𝑓𝐻

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 (𝐴)𝑚𝑖𝑑
(Af)high = 𝑓
(1+𝐴𝑚𝑖𝑑 𝛽)[1+𝑗 ]
𝑓𝐻 (1+𝐴𝑚𝑖𝑑 𝛽)

(𝐴𝑓 )𝑚𝑖𝑑
(Af)high = 𝑓 where 𝑓𝐻𝑓 = 𝑓𝐻 (1 + 𝐴𝑚𝑖𝑑 𝛽)
1+𝑗( )
𝑓𝐻𝑓

Higher cutoff frequency with feedback increases by a factor of (1+A midβ)

(BW)f = fHf – fLf

𝑓𝐿
= 𝑓𝐻𝑓 (1 + 𝐴𝑚𝑖𝑑 𝛽) - 1+ 𝐴𝑚𝑖𝑑 𝛽

Fig 3.15: Effect of negative feedback on Bandwidth

Input & Output Resistance of negative feedback amplifier

1) Voltage Series Feedback Amplifier

Fig 3.16: Block diagram of Voltage Series Feedback Amplifier

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 Fig 3.17 : Equivalent circuit of Voltage Series Feedback Amplifier

Input Resistance of Voltage Series feedback amplifier

Fig 3.18 : Equivalent circuit of Voltage Series Feedback Amplifier for

calculating input resistance

Applying KVL at input

VS = Vi + Vf
VS = IiRi + βVo
Vo = ILRL
𝑅𝐿
Vo = AvVi [𝑅 +𝑅 ]
𝑜 𝐿
Let AV = Av RL / (Ro+ RL)
Vo = AV Vi
VS = IiRi + β (AVVi)
VS = IiRi (1+ βAV)
VS/Ii = Ri (1+ βAV)
Rif = Ri (1+ βAV)
Input resistance with feedback (Rif) increases by factor of (1+Avβ)

Output Resistance of Voltage Series feedback amplifier

To find output resistance short Vs , disconnect RL

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 Fig 3.18 : Equivalent circuit of Voltage Series Feedback Amplifier for
Calculating output resistance

Rof = V/I
At output side
V= IRo + AVVi
𝑉−𝐴𝑣 𝑉𝑖
I = 𝑅 𝑜
At input side
Vi + Vf =0
Vi = -βV
Now
𝑉+𝐴𝑣 𝛽𝑉 𝑉(1+𝐴𝑣 𝛽)
I = =
𝑅 𝑜 𝑅𝑜
𝑹𝒐
Rof = 𝟏+𝑨𝒗 𝜷
Output resistance decreases with factor of (1+Avβ)

2) Current Series Feedback Amplifier

Fig 3.19: Block diagram of Current Series Feedback Amplifier

Fig 3.20: Equivalent circuit of Current Series Feedback Amplifier

Input Resistance Current Series Amplifier with feedback

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 Fig 3.21: Equivalent circuit of Current Series Feedback Amplifier for
Calculating Input resistance

Applying KVL at input

Vs = Vi + Vf
Vs = IiRi + βIL
IL = gmvi[Ro/Ro+ RL]
𝑔 𝑅
IL = GMVi GM = 𝑅 𝑚+𝑅𝑜
𝑜 𝐿
Vs = IiRi + βGmVi
Vs = IiRi(1 + βGm)
Vs/Ii = Ri(1 + βGm)
Rif = Ri(1 + βGm)
Input resistance with feedback (Rif) increases by factor of (1+Gmβ)

Output Resistance Current Series Amplifier with feedback

Fig 3.22: Equivalent circuit of Current Series Feedback Amplifier for

Calculating output resistance

Io = -I
At input
Vi = Vf
Vi = -β Io
Vi = β I
At output
GmVi + I – V/Ro = 0
I = V/Ro - Gm β I
I(1+ Gm β) = V/Ro
Rof = Ro (1+ Gm β)

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 3) Current Shunt Amplifier

Fig 3.23: Block diagram of Current Shunt Feedback Amplifier

Fig 3.24: Equivalent circuit of Current Shunt Feedback Amplifier

Input Resistance Current Shunt Amplifier with feedback

Fig 3.25: Equivalent circuit of Current Shunt Feedback Amplifier for

Calculating input resistance

Applying KCL to output

IS = If + Ii
IS = βIL + Ii
IL = AiIi[Ro/Ro+ RL]
AI = Ai Ro/( Ro+ RL)
IL = AIIi
IS = β(AIIi) + Ii
IS = Ii(1+βAI)

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 Rif = Vi/IS
𝐼𝑖 𝑅𝑖
𝑅𝑖𝑓 =
𝐼𝑖 [1 + 𝐴𝐼 𝛽]
𝐑𝐢
𝐑 𝐢𝐟 =
[𝟏 + 𝐀𝐈 𝛃]
Input resistance with feedback (Rif) decreases by factor of (1+Aiβ)

Output Resistance Current Shunt Amplifier with feedback

Fig 3.26: Equivalent circuit of Current Shunt Feedback Amplifier for

Calculating output resistance
Io = -I
Apply KCL to o/p loop
AVIi + I – V/Ro = 0
I = V/Ro - AVIi
If = -I
-βVo = Ii I = V/Ro – AV βI
Ii = βI I(1+Aiβ) = V/Ro
Rof = V/I = Ro(1+Aiβ)
Output resistance increases with factor of (1+Aiβ)

4) Voltage Shunt Amplifier

Fig 3.27: Block diagram of voltage Shunt Feedback Amplifier

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 Fig 3.28: Equivalent circuit of Fig 3.27: Block diagram of voltage Shunt Feedback
Amplifier

Input Resistance Voltage Shunt Amplifier with feedback

Fig 3.29: Equivalent circuit of Fig 3.27: Block diagram of voltage Shunt Feedback
Amplifier for Calculating input resistance

Applying KCL at output

Is = If + Ii
= βVo + Ii
Vo = RmIi[RL/Ro+RL]
RM = RmRL/Ro+RL
Vo = RMIi
Is = βRMIi + Ii
Is = Ii(1 + βRM)
Rif = IiRi/Ii(1 + βRM)
Rif = Ri/(1 + βRM)
Input resistance with feedback (Rif) decreases by factor of (1+Rmβ)

Output Resistance of Voltage Shunt negative feedback amplifier

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 Fig 3.30: Equivalent circuit of voltage Shunt Feedback Amplifier for
Calculating output resistance

Rof = V/I
At o/p side
V = IRo + RmIi
𝑉−𝑅𝑚 𝐼𝑖
I = If = -Ii => βV = - Ii
𝑅𝑜
𝑉+𝛽𝑅𝑚 𝑉
I = 𝑅𝑜
𝑉(1+𝛽𝑅𝑚 )
I = 𝑅𝑜

𝑹𝒐
Rof = 𝟏+𝜷𝑹𝒎
Output resistance decreases with factor of (1+R mβ)

Methodologies to analyze negative feedback amplifiers

Step-I : Identify the topology

a) To find the type of sampling network
i) By shorting the o/p i.e. Vo = 0, if the feedback signal becomes 0, then it is voltage
sampling.
ii) By opening the o/p i.e. io = 0, if the feedback signal becomes 0, then it is current
sampling.
b) Find the type of mixing network
i) If the feedback signal is subtracted from external applied signal as a voltage in i/p
loop, then it is series mixing.
ii) If the feedback signal is subtracted from external applied signal as a current in i/p
loop, then it is shunt mixing.
Step-II: To find input circuit.
a) For voltage sampling make Vo = 0 by shorting output loop.
b) For current sampling make Io = 0 by opening output loop.
Step-III: To find output circuit
a) For series mixing make Ii = 0 by opening the input loop.
b) For shunt mixing make Vi = 0 by shorting the input loop.
Step-IV: Replace the transistor with appropriate h-parameter model and find out the
parameters like Ri, Ro, Rif, Rof and gain.
Problems :
1) For the given circuit identify the topology and find the required parameters

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Solution:
Step-I ---> Identify topology
By shorting output voltage feedback signal becomes zero and hence it is voltage
sampling.
From circuit, feedback signal Vf is subtracted from applied signal, so it is series
mixing
So the circuit is voltage series feedback amplifier.
Step 2 and 3 --------Find input and output circuits.
To find the input ckt, set Vo = 0, and hence Vs in series with Rs appears between
B and E. To find the output ckt, set Ii=Ib=0 and hence Re appears only in outer
loop as shown.

Step-4 : Replace transistor by its h-parameter equivalent circuit.

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Step 5: Find open loop voltage gain
ℎ𝑓𝑒 𝐼𝑏 𝑅𝑒
Av = Vo/Vs =
𝑉𝑠

Apply KVL to input loop

Vs = Ib(Rs+hie)
Substitute value of Vs we get
ℎ𝑓𝑒 𝑅𝑒 50∗100
Av = = = 2.38
𝑅𝑠 +ℎ𝑖𝑒 1𝐾+1.1𝐾
𝑉𝑓
We have β =
𝑉𝑜

D = 1+ βAV = 1+1*2.38 =2,38

𝐴𝑣 𝐴𝑣 2.38
Avf = = = = 0.7
1+𝛽𝐴𝑣 𝐷 3.38

Ri = Rs + hie = 1K + 1.1K= 2.1K

Rif = Ri D = 2.1K*3.38= 7.09K
R00 = ꭃ
Rꭃf = ꭃ

2) For the given circuit identify the topology and find the required parameters

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Solution:
Step 1: Identify topology
By opening output loop, feedback signal becomes zero and hence it is current
sampling. From the ckt, feedback signal Vf is subtracted from applied signal Vs
so it is a series mixing. So it is a current series feedback amplifier.
Step 2 and 3: Find i/p and o/p circuits.
To find i/p ckt set Io =0, then Re appears at i/p side
To find o/p ckt, set Ii =0 then Re appears in the o/p ckt.

Step 4: Replace transistor with its approproate h-parameter circuit

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 −ℎ𝑓𝑒 𝐼𝑏
GM = Io/Vi = = −0.015
𝐼𝑏 (𝑅𝑠 +ℎ𝑓𝑒 +𝑅𝑒 )

−𝐼𝑜 𝑅𝑒
β = Vf/Io = = −1.2𝐾
𝐼𝑜

D = 1+ β GM = -0.782*10^-3
Avf = Vo/Vs = -1.72
Ri = RS + hie + Re = 3.3K
Rif = Ri D = 63.294K
Ro = ꭃ
Rof = RoD = ꭃ

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 QUESTION BANK
 Short answer questions.
1) What are the characteristics of negative feedback amplifiers
2) Write about stability of negative feedback amplifiers
3) What is effect of negative feedback on i/p & o/p resistances of trans conductance
amplifier
4) Differentiate between local & global feedback
5) Explain a feedback amplifier with help of block diagram.
6) Write about stability of feedback amplifiers.
7) Draw the small signal equivalent circuit of FET amplifier in CS connection let RD=
4K Ω, b=40,rd=40KΩ.evaluate voltage gain.
8) Negative feedback improves gain stability of the amplifier, justify the statement.
9) An amplifier requires an input signal 60m Volts to produce certain output with a
negative feedback to set the same output. The voltage gain with feedback is 90. Find
open loop voltage gain and feedback factor.
10) An amplifier has a voltage gain of 200befor negative feedback is applied, when
negative feedback is applied with β=0.25 the nominal gain changes by 10% find the
percentage change in the overall gain.
11) How does negative feedback reduce distortion in amplifier?
12) Ri=1KΩ, R0=5KΩ, A=100, β=0.04 calculate Rif, R0f, of voltage shunt feedback
amplifier.
13) Draw the equivalent circuit of trans resistance amplifier and mention idle values for
Ri and R0

 Long answer questions.

1) Evaluate the effect of negative feedback on i/p & o/p impedances of voltage shunt
amplifier.
2) For a single stage voltage shunt feedback amplifier Rc=2K Ω,
Re=1KΩ,Rf=100KΩ,Rs=1KΩ, and hfe=50,calculate Rif and Rvsf.
3) Give an example of voltage series negative feedback amplifier.Analyse the amplifier
and find Avf, Rif and Rof if Re=4KΩ,Rs=1KΩ,R1=30KΩ, R2=20KΩ,
hfe=100,hie=1.1kΩ and hoe= hre=0.
4) Draw current series feedback amplifier if RC=1kΩ , RE=100Ω, R2=20kΩ,
R1=30KΩ, HFE=100, calculate A, Ri, Rif, Af.

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 ASSIGNMENT QUESTIONS

1. What are the characteristics of negative feedback amplifiers?

2. What is the effect of negative feedback on i/p & o/p resistances of transconductance
amplifier?
3. Differentiate local & global feedbacks.
4. If an amplifier has BW-200kHz & voltage gain =80, what will be new BW & gain if
5% of feedback.
5. What is effect of negative feedback on i/p, o/p resistances on current shunt amplifier?
6. Evaluate the effect of negative feedback on i/p & o/p impedances of voltage shunt
amplifier.
7. For single stage voltage shunt feedback amplifier if Rc-2k, Re=1K, Rf=100K, Rs=1K
hfe=50, find Rif, Rof.
8. Briefly explain the effect of negative feedback on stability. For single stage current
series feedback amplifier if Rc-2k, Re=1K, Rf=100K, Rs=1K hfe=50, find Rif, Rof.
9. An amplifier has midband gain of 100 and bandwidth of 250khz .
a) If 5% negative feedback is introduced find new bandwidth and gain
b) If bandwidth is to be restricted to 1MHz, find feedback ratio.

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PREVIOUS UNIVERSITY QUESTION PAPERS

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