Negative feedback Amplifiers

Prof. Yeshudas Muttu
FEEDBACK AMPLIFIERS
Assistant Professor
Don Bosco College of Engineering, Goa

INTRODUCTION
• AC analysis of an amplifier
• Zi, Zo, Av, Ai  more or less constant for an amplifier
• Most of the applications needs control of above
parameters.
• This can be achieved by feedback.
Prof. Yeshudas Muttu 2

What is Feedback?
• Part of output is combined with the input.
• It is a process of combining fraction of output energy
back to the input.
• Output energy can be voltage or current.

Types of Feedback
Positive Feedback
• If the feedback signal
increases the input signal
• i.e. when feedback signal is
in phase with input signal
Negative Feedback
• If the feedback signal
decreases the input signal
• i.e. when feedback signal is
out of phase w.r.t input signal.

Types of Feedback
Positive Feedback
• Causes excessive distortion
& instability, hence not used
in amplifiers.
• It has capability of
increasing the power of
original signal, hence used
in oscillators.
Negative Feedback
• Reduces but stabilizes the
amplifier gain & offers many
other advantages.
• This can be compensated by
increasing the number of
stages.

Classification of Amplifiers
1. Voltage Amplifiers
2. Current Amplifiers
3. Transconductance Amplifiers
4. Transresistance Amplifiers
Note:
• Voltage drop will
be maximum where
there is maximum
resistance
• Current flow will be
maximum through
the least resistance.
𝑉 = 𝐼𝑅
𝐼 =
𝑉
𝑅

1. Voltage Amplifiers
• Input signal  Voltage, Output signal Voltage.
• Hence, its convenient to depict the equivalent circuit
with thevenin’s equivalent circuit.
• Conditions For Voltage gain Av = Vout/ Vs,
1. Rin>>Rs  Vin = Vs
2. RL >> Rout  Vout = Av Vin
• Hence Voltage gain
• Av = Vout/Vin
• = Vout/Vs
• For an ideal Voltage amp
• Rin =∞ , Rout = 0
• Hence o/p v/g ∝ i/p v/g

2. Current Amplifiers
• Input signal  Current, Output signal current.
• Hence, its convenient to depict the equivalent circuit
with Norton's equivalent circuit.
• Conditions For Current gain Ai = Iout/ Is,
1. Rin<<Rs  Iin = Is
2. RL << Rout  Iout = Ai Iin
• Hence Current gain
• Ai = Iout/Iin
• = IL/Is
• For an ideal current amp
• Rin = 0, Rout = ∞
• Hence o/p current ∝ i/p current

What is Transconductance?
• The ratio of variation of the current at the output terminal
to the voltage at the input terminal of an active device.
• It is given as:
𝑮 𝒎 =
𝑰
𝑽
What is Transresistance?
• The ratio of the variation of output voltage to input
current .
• It is given as:
𝑹 𝒎 =
𝑽
𝑰

3. Transconductance Amplifiers
• Input signal  Voltage, Output signal current.
• Conditions For Output Current in terms of
Transconductance(Gm)
1. Rin>>Rs  Vin = Vs
2. RL << Rout  Iout = Gm Vin
• Hence Transconductance,
• Gm = Iout/Vin
• = IL/Vs
• For an ideal
Transconductance amp,
• Rin = ∞, Rout = ∞
• Hence o/p current ∝ i/p voltage

4. Transresistance Amplifiers
• Input signal  Current, Output signal  Voltage.
• Conditions For Output Current in terms of
Transresistance(Rm)
1. Rin<<Rs  Iin = Is
2. RL >> Rout  Vout = Rm Iin
• Hence Transresistance,
• Rm = Vout/Iin
• = Vout/Is
• For an ideal
Transresistance amp,
• Rin = 𝟎, Rout = 𝟎
• Hence o/p Voltage ∝ i/p current

Characteristics of Ideal Amplifier

NUMERICAL

𝑰𝒊𝒏 = 𝑰 𝑺 ×
𝑹 𝑺
𝑹 𝑺 + 𝑹𝒊𝒏
𝑰 𝑺 = 𝑰𝒊𝒏 ×
(𝑹 𝑺 + 𝑹𝒊𝒏)
𝑹 𝑺

Feedback Concept
• Below is the block diagram of an amplifier with
Feedback.

Signal Source
• It can be either a Voltage source, Vs in series with a
resistor Rs or
• A Current source Is, in parallel with Rs
Feedback Network
• It is a two port network which may be formed with R, L,
C (most often R).
• Its function is to return the part of output energy(V or I)
to the input of the amplifier.
Explanation of each block

Voltage/Node
Sampling Network
Current/Loop
Sampling Network
Sampling Network
It can be divided into two types:
• Output signal is sampled by
connecting feedback
network in shunt across the
output.
• Output signal is sampled by
connecting feedback
network in series across the
output.

Comparator/Mixer Network
• It can be divided into two types: Series Mixer & Shunt
Mixer.
• Differential amplifier (Dual i/p, single output) provides
Output proportional to difference i/p signal is used as
a mixer.

Types of Feedback Connection/Topologies

Explanation for Feedback Connections
• The four types are:
1. Voltage Series Feedback
2. Voltage Shunt Feedback
3. Current Series Feedback
4. Current Shunt Feedback
• Voltage refers to Output voltage is fed to the
feedback network. It is taken in parallel.
• Current refers to Output current is fed to the feedback
network. It is measured in series.
• Series refers to connecting the feedback signal in
series with input signal.
• Shunt refers to connecting the feedback signal in
parallel with an input current source.

Explanation for Feedback Connections
Series Feedback Connection
• It opposes the applied
voltage causing input
current to fall that makes
Zi to increase.
Shunt Feedback Connection
• Current drawn from the
signal source is increased
by amount = If, hence Zi
falls.
Voltage Feedback Connection
• Reduces o/p impedance
Current Feedback Connection
• Increases o/p impedance
In cascaded amplifiers, mostly high Zi & low Zo are desired.
This property is provided by Voltage series feedback, hence
they are commonly used.

23
Series Feedback Connection Shunt Feedback Connection
Rin = 0,
Rout = ∞
Rin = ∞,
Rout = ∞
Rin = 𝟎,
Rout = 𝟎
Rin =∞ ,
Rout = 0
Current
Shunt
Feedback
Voltage
Series
Feedback
Voltage
Shunt
Feedback
Current
Series
Feedback
VFC
VFC
CFC
CFC

Types of Feedback Connection

Parameters of Amplifier without Feedback
• Transfer gain of basic amplifier without feedback = A
(Output/ Input of basic amplifier).
1. Transfer ratio or Gain, Av = Vout/Vin
2. Current Gain, Ai = Iout/Iin
3. Transconductance, Gm = Iout/Vin
4. Transresistance, Rm = Vout/Iin

Parameters of Amplifier with Feedback
• Transfer gain of the amplifier with feedback = Af
(Output/ Input of the amplifier).
1. Transfer ratio or Gain with feedback, Avf = Vout/Vs
2. Current Gain with feedback, Aif = Iout/Is
3. Transconductance with feedback, Gmf = Iout/Vs
4. Transresistance with feedback, Rmf = Vout/Is

To derive relation between A & Af
• Above diagram is generalized feedback amplifier.
• The basic amplifier may be an ideal voltage,
Transresistance, Transconductance or current
amplifier.
• Rs is considered to be part of the amplifier.

• Transfer gain A includes effect of Loading from
feedback network β, as well as load resistance RL.
• The input signal Xs, the output signal Xout, the
feedback signal Xf & the difference signal Xd, each
represents either voltage or a current.

• From figure, mixing network’s output is the sum of the
inputs considering the sign indicated at each input.
• Hence Difference signal can be given as:
Xd = Xs – Xf = Xin
Xd is also known as error/comparison signal.
• Hence, Xs = 𝑿𝒊𝒏 + 𝑿 𝒇 -----(1)

• Reverse Transmission factor β, is defined as:
𝜷 =
𝑿 𝒇
𝑿 𝒐𝒖𝒕
------(2)
• It is usually a positive or negative real number.
• In general, β is a complex function of the signal
frequency.
• Transfer gain A, is defined as:
𝑨 =
𝑿 𝒐𝒖𝒕
𝑿 𝒊𝒏
-------(3)

• The expression for the transfer gain with feedback
can be given as:
𝑨 𝒇 =
𝑿 𝒐𝒖𝒕
𝑿 𝒔
-------(4)
• Hence, from equations 1 & 4,
𝑨 𝒇 =
𝑿 𝒐𝒖𝒕
𝑿𝒊𝒏 + 𝑿 𝒇
=
𝑿 𝒐𝒖𝒕
𝑿 𝒊𝒏(𝟏+
𝑿 𝒇
𝑿 𝒊𝒏
)
=
𝑿 𝒐𝒖𝒕
𝑿 𝒊𝒏
𝟏+
𝑿 𝒇
𝑿 𝒊𝒏
×
𝑿 𝒐𝒖𝒕
𝑿 𝒐𝒖𝒕

𝑨 𝒇 =
𝑿 𝒐𝒖𝒕
𝑿 𝒊𝒏
𝟏+
𝑿 𝒇
𝑿 𝒊𝒏
×
𝑿 𝒐𝒖𝒕
𝑿 𝒐𝒖𝒕
=
𝑿 𝒐𝒖𝒕
𝑿 𝒊𝒏
𝟏+
𝑿 𝒇
𝑿 𝒐𝒖𝒕
×
𝑿 𝒐𝒖𝒕
𝑿 𝒊𝒏
• From equations 2 & 3,
𝑨 𝒇 =
𝑨
𝟏 + 𝛃𝑨
• Af < A if the feedback is negative/degenerative/
Inverse.

• For Negative feedback,
𝑨 𝒇 =
𝑨
𝟏 + 𝛃𝑨
• For Positive feedback,
TPT: Af > A if the feedback is positive/regenerative/ direct.

Xin = Xs + Xf
Hence, Xs = Xin - X 𝒇
𝑨 𝒇 =
𝑿 𝒐𝒖𝒕
𝑿𝒊𝒏 − 𝑿 𝒇
𝑨 𝒇 =
𝑿 𝒐𝒖𝒕
𝑿 𝒔

𝑨 𝒇 =
𝑨
𝟏 − 𝛃𝑨
𝑨 𝒇 =
𝑿 𝒐𝒖𝒕
𝑿 𝒊𝒏
𝟏−
𝑿 𝒇
𝑿 𝒊𝒏
×
𝑿 𝒐𝒖𝒕
𝑿 𝒐𝒖𝒕
=
𝑿 𝒐𝒖𝒕
𝑿𝒊𝒏
𝟏 −
𝑿 𝒇
𝑿 𝒐𝒖𝒕
×
𝑿 𝒐𝒖𝒕
𝑿𝒊𝒏
Hence, For Positive feedback,
𝑨 𝒇 =
𝑨
𝟏 − 𝛃𝑨
Af > A if the feedback is positive/regenerative/ direct.

36
Series Feedback Connection Shunt Feedback Connection
Rin = 0,
Rout = ∞
Rin = ∞,
Rout = ∞
Rin = 𝟎,
Rout = 𝟎
Rin =∞ ,
Rout = 0
Current
Shunt
Feedback
Voltage
Series
Feedback
Voltage
Shunt
Feedback
Current
Series
Feedback
VFC
VFC
CFC
CFC

• For Negative feedback,
𝑨 𝒇 =
𝑨
𝟏 + 𝛃𝑨
• For Positive feedback,
𝑨 𝒇 =
𝑨
𝟏 − 𝛃𝑨

LOOP GAIN/RETURN RATIO
• Difference signal Xd is multiplied with basic amplifier
gain A which is later multiplied with feedback gain β,
that is later multiplied with ‘ – 1’ that results in:
Xd x A x β x (– 1) = –βA Xd
• Where (–βA) is the Loop gain/ Return ratio
• Return difference, D = 1 – (–βA) = 1 + βA
• Amount of feedback introduced in the amplifier is
expressed in decibels.
N = dB of feedback = 20 log10
𝑨 𝒇
𝑨
= 20 log10
𝑨
𝟏+𝛃𝑨
𝑨
N = 20 log10
𝟏
𝟏+𝛃𝑨
Where,
N will be
negative
number for
Negative
feedback

• Generally, A & β are phasor quantities. i.e. they
have magnitude & phase.
• Every amplifier stage introduces some phase shift.
Eg: Common Emitter amplifier o/p produces 180o
phase shift.
• For 3 – stage amplifier, total phase shift will be:
180 x 3 = 5400
• β usually has phase value of 00 or 1800. i.e. either in
phase with the i/p or in phase opposition to it.
• Hence, (1 – βA) is a complex quantity.

NUMERICALS

PROBLEMS FOR PRACTICE
1.
2.
3.

5.
47
4.

General Characteristics of Negative
Feedback Amplifiers
1. Provides gain stability
2. Reduces Non Linear distortion
3. Reduces Noise
4. Increases Bandwidth / Improved Frequency response.
5. Increased Input Impedance
6. Reduced Output Impedance

1. Stabilization of Gain with Negative Feedback
OR
Desensitivity of Gain
• The transfer gain of the amplifier is not constant as it
depends on the factors such as Operating point,
temperature, etc.
• This lack of stability in amplifiers can be reduced by
introducing negative feedback.
Differentiating with respect to A, (using u/v rule)
U
V

………………….(1)
From equation 1, we can conclude that the change in
the gain with feedback is less than the change in the
gain without feedback by a factor of (1+βA).

• The above ratio is called Sensitivity of the Transfer gain.
• Reciprocal of Sensitivity is called Desensitivity, D = 1 + βA

From Equation 2,

2. a) Reduction in Frequency Distortion
• For negative feedback, 𝑨 𝒇 =
𝑨
𝟏+𝛃𝑨
• If 𝛃𝑨>>1, 𝑨f = 𝑨/ 𝛃𝑨 = 1/ 𝛃
• Hence, Gain here, doesn't depend on A, but
depends on 𝛃.
• If feedback network is purely resistive, the overall
gain will be frequency independent. (even though
A is frequency dependent as its eliminated from the
equation)
• Practically, frequency distortion that arises due to
variations in amplifier gain with frequency, is
reduced considerably due to negative feedback.

2. b)Reduction in Non-linear Distortion with Feedback
• Consider an amplifier with open loop gain ‘A’ &
closed loop gain ‘Af’ with negative feedback.
• Let this amplifier produce a distortion ‘D’ without
any feedback.
• Let this amplifier produce a distortion ‘Df’ with
feedback.

• With feedback, a part of distortion 𝛃𝐃𝐟 is fed back & is
amplified to get A𝛃𝐃𝐟
• Hence net distortion, Df = D - A𝛃𝐃𝐟 (since –ve feedback
is in phase opposition)

D = Df + A𝛃𝐃𝐟
= Df (1+𝛃A)
Hence, Df =
𝑫
𝟏+𝛃A
Therefore, Distortion with feedback is reduced by a
factor of (1+ 𝛃A)
• Note:
This distortion can be cancelled only if amplifier
introduces it by itself.
Suppose if the original input signal itself is distorted,
no distortion cancellation of that shall take place.

3. Reduction in Noise with Feedback
• Noise in an amplifier is reduced in the same way as that
of distortion.
• i.e. Noise with feedback is reduced by a factor of
(1+ 𝛃A)
Nf =
𝑵
𝟏+𝜷A
• Note:
This Noise can be cancelled only if amplifier introduces it
by itself. Suppose if the original input signal itself is noisy,
no noise cancellation of that shall take place.

NUMERICALS

Feedback Amplifiers
3. Reduces Noise

4. Frequency Response & Bandwidth
• Frequency Response is a variation in the amplitude
level of its output signal when the frequency is changed.

…………(4)
…………(5)
• Hence Lower cutoff frequency with
feedback is less than that of
without feedback by factor of
(1+Amidβ).
• Hence with negative feedback
low frequency response of an
amplifier is improved.

Do the same as done for Lower cutoff
frequency & finally get the equation
below
…………(6)
…………(7)
• Hence Upper cutoff frequency with
feedback is more than that of
without feedback by factor of
(1+Amidβ).
• Hence with negative feedback
high frequency response of an
amplifier is improved.

Clearly,

NUMERICALS

Feedback Amplifiers
3. Reduces Noise

5. Input Resistance
• If the feedback signal is added to the input in series
with the applied voltage irrespective of which
sampling is used, it increases the input resistance.
• Since feedback Vf opposes Vs, the input current Ii is
less than what it would have been if Vf was absent.
Hence, Input
resistance with
feedback,
is greater than
the input
resistance (Ri)
without
feedback.

• If the feedback signal is added to the input in shunt
with the applied voltage irrespective of which
sampling is used, it decreases the input resistance.
• Since Is = Ii + If, current Is drawn from the source is
increased, than what it would have been if If was
absent
Hence, Input
resistance with
feedback,
is less than the
input resistance
(Ri) without
feedback for this
circuit.

VOLTAGE – SERIES FEEDBACK
• Voltage series feedback topology is replaced by its
Thevenin’s equivalent circuit.

…………(1)
…………(2)

Hence, Rif is increased
by (1+βA) times more
as compared to Ri

CURRENT – SHUNT FEEDBACK
• Current Shunt feedback topology is replaced by its
Nortons’s equivalent circuit.

…………(1)
…………(2)

Hence, Rif is
decreased
by a factor
of (1+βA) as
compared
to Ri

TASK 1: CURRENT SERIES FEEDBACK
=

TASK 2: VOLTAGE SHUNT FEEDBACK
=

Feedback Amplifiers
3. Reduces Noise

6. Output Resistance
• Negative feedback that samples the output voltage
irrespective of which mixer network is used, tends to
decrease the output resistance.

• Negative feedback that samples the output current
irrespective of which mixer network is used, tends to
increase the output resistance.

VOLTAGE – SERIES FEEDBACK
• Here, output resistance can be measured by shorting
the input source, i.e. Vs = 0.
• Now, looking into the terminals with RL disconnected,
+
-

+
-
………………….(1)
………………….(2)
+
-

• To find Rof’
&

CURRENT – SHUNT FEEDBACK
• Here, output resistance can be measured by Open
circuiting the input source. i.e. Is = 0.
• Now, looking into the terminals with RL disconnected,
97Prof. Yeshudas Muttu

……………(1)
……………(2) Prof. Yeshudas Muttu 98

• To find Rof’

TASK 1: VOLTAGE SHUNT FEEDBACK

TASK 2: CURRENT SERIES FEEDBACK

REFERENCES
• Electronic Devices & Circuits – J.B. Gupta
• Analog Electronic Circuits – A.P. Godse, U.A. Bakshi
• https://www.elprocus.com/
• https://electronicscoach.com/

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