-PARAMETER ODEL · evaluating unknown parameters. 2. That these parameters are not alike dimensionally. 3. That the unit of these parameters are different e.g. Ohms Ω, Mhos ℧,

P a g e | 1 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier

5th Edition for Session 2017-18

Syllabus: h-parameter model, Analysis of Transistor Amplifier circuits using h-parameters, CB, CE and CC

Amplifier configurations and performance factors.

Principle of Negative feedback in electronic circuits, Voltage series, Voltage shunt, Current series,

Current shunt types of Negative feedback, Typical transistor circuits effects of Negative feedback

on Input and Output impedance, Voltage and Current gains, Bandwidth, Noise and Distortion.

Why hybrid name is given to h-parameters? Why h-parameters are used for the analysis of small signal BJT circuits? (W-13/6m) What is small signal condition? Why hybrid parameters are used in analyzing low frequency network? (W-15/4m)

-PARAMETER MODEL ℎ refers to “Hybrid”. The name “hybrid” came because of few facts

1. That here both types of terminal conditions (i.e. open circuit and short circuit) are used for

evaluating unknown parameters.

2. That these parameters are not alike dimensionally.

3. That the unit of these parameters are different e.g. Ohms Ω, Mhos , constant, etc.

As V-I characteristics of BJT is non-linear, its analysis is complex. To simplify the analysis of the BJT,

its operation is restricted to the linear active region. This approximation is possible only with small

input signals. With small input signals the transistor can be replaced with small signal linear model.

This model is also called small signal equivalent circuit.

Generally, ℎ-parameters are used for the analysis of small signal BJT circuits, because they are easy

to measure, can be easily calculated from the transistor static characteristics curve and

manufacturers usually specify ranges for the various parameters for a type of transistor.

Two-Port Network or h-parameter model A transistor can be treated as a two-port network. The terminal behavior of any two port network

can be specified by the terminal voltages and at ports 1 and 2 respectively, and currents

and entering ports 1 and 2 respectively, as shown in following figure:

Of these four variables i.e. , , &, two can be selected as independent variables and

remaining two can be expressed in terms of these independent variables. This leads to various two-

port parameters, out of which, three are more important.

1. Z−Parameters or Impedance Parameters

2. Y−Parameters or Admiance Parameters

3. h−Parameters or Hybrid Parameters



Z−PARAMETERS OR IMPEDANCE PARAMETERS Let & be the independent variables. The voltages and will be

= +

= +

These four impedance parameters, i.e. , , & are defined as;

Z11 → Input impedance with output port open circuited

=

ℎ = 0

Z22 → Output impedance with input port open circuited

=

ℎ = 0

Z12 → Reverse transfer impedance with port 1 open circuited

=

ℎ = 0

Z21 → Forward transfer impedance with port 2 open circuited

=

ℎ = 0

Y−PARAMETERS OR ADMITTANCE PARAMETERS Let v1 and v2 are taken as independent variables. The currents i1 and i2 will be

= +

= +

These four admittance parameters i.e. , , & are defined as:

→ Input admiance with port 2 short circuited

=

ℎ = 0

→ Output admiance with port 1 short circuited

=

ℎ = 0

→ Reverse transfer admiance with port 1 short circuited

=

ℎ = 0

→ Forward transfer admiance with port 2 short circuited

=

ℎ = 0



HYBRID PARAMETERS OR −PARAMETERS [hint: we generally keep input current and output voltage constant i.e. i1 & v2 constant means

independent]

Let the input current i1 and the output voltage v2 be independent variables, the input voltage v1

and the output current i2 will be,

= ℎ + ℎ

= ℎ + ℎ

These four hybrid parameters i.e. ℎ, ℎ, ℎ&ℎ are defined as

ℎ→ Input impedance with output port short circuited

ℎ =

ℎ = 0

ℎ→ Output admiance with input port open circuited

ℎ =

ℎ = 0

ℎ→ Reverse voltage rao with input port open circuited

ℎ =

ℎ = 0

ℎ→ Forward current gain with output port short circuited

ℎ =

ℎ = 0

An alternative notation recommended by IEEE is commonly used today,

= 11 = ;

= 22 = ;

= 21 = ;

= 12 = ;

Notations of Transistor Circuits: Hybrid Parameter

Common Emitter

Common Collector

Common Base

Definition

ℎ = ℎ ℎ ℎ ℎ Short circuit input impedance

ℎ = ℎ ℎ ℎ ℎ Open circuit output admittance

ℎ = ℎ ℎ ℎ ℎ Open circuit reverse voltage transfer ratio

ℎ = ℎ ℎ ℎ ℎ Short circuit forward current gain



Hybrid Model of Two-Port Network:

ℎ-parameters for the hybrid model are given by:

= ℎ + ℎ

= ℎ + ℎ

Compare CB, CE and CC configuration of transistor. (4M/W-16) (5M/S-16) (S-15/5m)

[Add following point of comparison i.e hybrid model and equations to the answer of this question

in unit-2]

HYBRID MODEL OF THE TRANSISTOR IN CE CONFIGURATION:

= ℎ + ℎ

= ℎ + ℎ

HYBRID MODEL OF THE TRANSISTOR IN CB CONFIGURATION:

= ℎ + ℎ

= ℎ + ℎ



HYBRID MODEL OF THE TRANSISTOR IN CC CONFIGURATION:

= ℎ + ℎ

= ℎ + ℎ

TYPICAL VALUES OF HYBRID PARAMETERS OF A TRANSISTOR Parameter CE CC CB

1, 100 Ω 1, 100 Ω 22 Ω

2.5 × 10 1 3 × 10

50 −51 −0.98

25μ/ 25μ/ 0.49μ/

ANALYSIS OF TRANSISTOR AMPLIFIER CIRCUITS USING -

PARAMETERS

Current Gain (Ai): Current gain is defined as the ratio of output current to the input current.

=

=

−

= ℎ + ℎ = ℎ + ℎ[] = ℎ + ℎ[−] = ℎ − ℎ

+ ℎ = ℎ

(1 + ℎ) = ℎ



=ℎ

1 + ℎ

=−

ℎ

1 + ℎ

∴ =−ℎ

1 + ℎ

Voltage Gain (AV): Voltage gain is defined as the ration of output voltage to the input voltage.

=

=

∵ =

=

=

=

/

∵ =

∴ =

Input Impedance (Zi): Input impedance is defined as the ratio of input voltage to input current.

=

= ℎ + ℎ

=ℎ + ℎ

= ℎ + ℎ

= ℎ + ℎ ×

= ℎ + ℎ ×

()

= ℎ + ℎ ×

∵ =−ℎ

1 + ℎ

= ℎ + ℎ −ℎ

1 + ℎ = ℎ −

ℎℎ

1 + ℎ

Taking out ZL common from denominator

= ℎ −ℎℎ

1

+ ℎ

= ℎ −ℎℎ

1

+ ℎ

∵1

=

∴ = ℎ −ℎℎ

+ ℎ



Output Admittance (YO): Output admittance is defined as the ratio of output current to output voltage.

=

= ℎ + ℎ

= ℎ ×

+ ℎ

= ℎ ×

+ ℎ

Applying KVL in input circuit by short circuiting input voltage source VS.

+ ℎ + ℎ = 0

( + ℎ) + ℎ = 0

( + ℎ) = −ℎ

= −

ℎ

+ ℎ

= ℎ × −ℎ

+ ℎ + ℎ

∴ = ℎ −ℎℎ

+ ℎ

Voltage Amplification (AVS): Voltage amplification is defined as the ratio of output voltage to the input voltage.

=

=

×

= ×

= ×

= ( + )

=

+

=

+

=

+



= ×

+

=

=

×

+

=

+

Ideally RS is Zero.

=

=

AV is the voltage gain with ideal voltage source and AVS is the voltage gain with practical voltage

source.

Current Amplification (AIS): Current amplification is defined as the ratio of output current to the input current.

=

=

×

= ×

Norton’s equivalent circuit for the source is

=

=

+

=( + )

=

×

( + )

∵ =

=( + )

=

+

= ×

+

Ideally = ∞ for current source,

∴ =



AI is the current gain with ideal current source and AIS is the current gain with practical current

source.

Relation between Voltage Amplification (AVS) and Current Amplification

(AIS):

∵ =

+ =

+ ×

= ×

+ ×

= ×

Operating Power Gain (AP): It is defined as the ratio of output power to the input power.

=

=

=

=

=

×

= ×

∵ =

∴ =

×

= ×

=

×

LIST OF IMPORTANT FORMULAE:

=−ℎ

1 + ℎ

=

= ℎ −ℎℎ

+ ℎ

=1

= ℎ −

ℎℎ

+ ℎ

=

+

= ×

+

= ×



= ×

State Miller’s theorem with the help of a circuit and explain it with a suitable example. (W-16/6M) Write a short note on 'Miller's theorem'. (S-17/6M)

MILLER’S THEOREM: Miller’s theorem states that if an impedance Z is connected between input and output terminals of

the network, which provides a voltage gain AV, an equivalent circuit that gives the same effect can

be drawn by removing Z and connecting an impedance =

across the input and =

/

across the output as shown in following figure.

MILLER’S THEOREM: Dual of Miller’s Theorem states that if an impedance Z is connected as shunt element between

input and output terminals, it can be replaced by an impedance = (1 − ) at the input side

and =

at the output side as shown in following figure.



For the circuit shown in figure find . Given = . , = . ×, = & = / (W-16/8m)

Solution:

We know,

=−ℎ

1 + ℎ=

−ℎ

1 + ℎ=

−50

1 + 25 ×

As resistance 1K is connected between input and output, it can be simplified using Miller’s theorem.

10K

R1

10K

VS

R2

To find R1 and R2

We know,

=

1 − & =

1 − 1/≈ = 1

= ||10 =1 × 10

1 + 10= 909 =

=−50

1 + 25 × 909

= −48.88

Let Z be the input impedance of transistor, & R be the input impedance of the circuit



VS

RS

Zi VB

E

IB

R1

Ri

= ℎ −ℎℎ

+ ℎ= ℎ −

ℎℎ

(1/) + ℎ= 1

=

=

−48.88 × 909

1= −44.43192

=

1 − =

1

1 + 44.43192= 22

= || = ×

+ =

22 × 1

22 + 1= 21.52

=

+ =

−48.88 × 909

10 + 21.52= −.

The transistor amplifier shown in figure uses a transistor whose h-parameters are as follows: = Ω, = . × , = , = /,Calculate =

, , , (S-15/8m)

Solution:



VS

RS hie

hreVCE hfeIB hoe RL

5KR=

R1||R2

IO

IBIC

=

We Know,

=−ℎ

1 + ℎ=

−ℎ

1 + ℎ=

−50

1 + 24 × 5

= −44.64

We know,

= ℎ −ℎℎ

+ ℎ= ℎ −

ℎℎ

(1/) + ℎ= 1100 −

2.5 × 10 × 50

(1/5 ) + 24

= 1

=

=

(−44.64) × 5

1

= −223.2

VS

RS

Zi VB

E

IB

R=R1||R2

Ri

= ||

= || =

+ =

100 × 10

100 + 10= 9.09

=

+ =

9.09 ×

9.09 + =

9.09 × 1

9.09 + 1

= 900.89

We know,

=

+ =

+ =

−44.64 × 5

10 + 900.89

= −20.47



hfeIB hoeRL V

CEI

L

IC

RO

= ℎ −ℎℎ

+ ℎ= 24 −

50 × 2.5 × 10

10 + 1100= 1.126

=1

= 888

For the circuit shown find , , . Assume = . , = . × , =

= (S-14/10m)

Solution:

VS

RS hie

hreVCE hfeIB hoe RL

5KR=

R1||R2

IO

IBIC

=

We Know,

=−ℎ

1 + ℎ=

−ℎ

1 + ℎ=

−50

1 + 1

40 × 5

= −44.44



We know,

= ℎ −ℎℎ

+ ℎ= ℎ −

ℎℎ

(1/) + ℎ= 1100 −

2.5 × 10 × 50

(1/5 ) + 1

40

= 1.04

=

=

(−44.44) × 5

1.04= −213.65

VS

RS

Zi VB

E

IB

R=R1||R2

Ri

= ||

= || =

+ =

100 × 10

100 + 10= 9.09

=

+ =

9.09 ×

9.09 + =

9.09 × 1

9.09 + 1

= 933.23

We know,

=

+ =

+ =

−44.44 × 5

5 + 933.23

= −37.45

hfeIB hoeRL V

CEI

L

IC

RO

= ℎ −ℎℎ

+ ℎ=

1

40−

50 × 2.5 × 10

5 + 1.1k= 2.05

=1

= 488

Find , , for the circuit shown in figure. Given. = . , = . ×

, = ,

= (W-15/9m)



Solution:

We know,

=−ℎ

1 + ℎ=

−ℎ

1 + ℎ=

−50

1 + 1

40 ×

As resistance 200K is connected between input and output, it can be simplified using Miller’s

theorem.

10K

R1

10K

VS

R2

To find R1 and R2

We know,

=

1 − & =

1 − 1/≈ = 200

= ||10 =200 × 10

200 + 10= . =

=−50

1 + 1

40 × 9.52K

= −40.38

Let Z be the input impedance of transistor, & R be the input impedance of the circuit

VS

RS

Zi VB

E

IB

R1

Ri



= ℎ −ℎℎ

(1/) + ℎ= 1.1K −

2.5 × 10 × 50

(1/9.52K) + 1

40

= 1Ω

=

=

−40.38 × 9.52K

1= −.

=

1 − =

200

1 + 384.4= 518.94Ω

= || = ×

+ =

518.94 × 1

518.94 + 1= .

=

+ =

−40.38 × 9.52K

10 + 341.6= −37.17

CE AMPLIFIER WITH EMITTER RESISTOR:

Fig. AC Equivalent of the given circuit

The emitter resistor can be simplified using dual of Miller’s theorem as follows:

ℎ, = (1 − )& = − 1



Fig. h-parameter equivalent model of the given circuit

After splitting the emitter resistor , the total load impedance will be

= + = +

− 1

Current Gain ()

=−ℎ

1 + ℎ′=

−ℎ

1 + ℎ + − 1

=−ℎ

1 +ℎ( + ( − 1))

=−ℎ ×

+ ℎ( + ( − 1))

1 =−ℎ × 1

+ ℎ( + ( − 1))

+ ℎ( + ( − 1)) = −ℎ

+ ℎ + ℎ − ℎ = −ℎ

(1 + ℎ + ℎ) − ℎ = −ℎ

(1 + ℎ + ℎ) = −ℎ + ℎ

=ℎ − ℎ

1 + ℎ( + )

Input Impedance/Resistance ( )

=

=

ℎ + ℎ

= ℎ +

ℎ

= ℎ +

ℎ( )

Since, = /

= ℎ +ℎ()

= ℎ + ℎ



Voltage Gain ()

= ×

In this case,

= ×

Output Impedance () It is given by

=1

ℎ

1 + ℎ + ( + ℎ)(1 + ℎ)

+ + ℎ − ℎℎ/ℎ

Note: The a.c. analysis of CE amplifier with emitter resistor becomes very difficult. Therefore, to

simplify the analysis, h-parameter model is approximated to its reduced form as follows.

Approximated h-parameter model

Fig. Approximated hybrid model

Current Gain ()

=

= −

=

−ℎ

= −ℎ

Input Resistance

= ℎ + 1 + ℎ

Voltage Gain

=

=

−ℎ

ℎ + 1 + ℎ

Output Impedance The approximated circuit has infinite output impedance because with = 0 and external voltage

source applied at the output, it is found that = 0 and hence, = 0. With zero current output

impedance becomes infinite.



Consider a single stage CE amplifier with = , = , = , = , = . , = . , = /, = , = . × as shown in

the figure. Find , , , , & (S-17/8M)

Solution:

Note: This circuit has the emitter resistor which makes it of different type. But in AC analysis,

the capacitor shunted with it nullifies the resistance , that’s why, its value is not given in

question.

=

+ =

1 × 1.2

1 + 1.2= 545.45

We Know,

=−ℎ

1 + ℎ =

−50

1 + (25)545.45

= −49.33

We know,

= ℎ −ℎℎ

+ ℎ= ℎ −

ℎℎ

(1/ ) + ℎ

= 1.1K −2.5 × 10 × 50

(1/545.45) + (25μ)

= 1.09

=

=

(−49.33) × 545.45

1.09= −24.6



VS

RS

Zi VB

E

IB

R=R1||R2

Ri

= ||

= || =

+ =

50 × 2

50K + 2= 1.923

=

+ =

1.923 × 1.09

1.923 + 1.09

= 697

We know,

=

+ =

−49.33 × 545.45

1 + 697

= −15.85

= ×

+ = −49.33 ×

1

1 + 697

= −29

hfeIB hoeRL V

CEI

L

IC

RO

= ℎ −ℎℎ

+ ℎ= 25μ −

50 × 2.5 × 10

1 + 1.1= 19

=1

= 52.5



For the emitter follower with = and = as shown in the circuit diagram. Calculate , , , & (S-16/8m)

Solution:

Approximate hybrid model of the given circuit is

Let ℎ = 1100, ℎ = 50

Given: = 500Ω and = 5Ω

= −ℎ = −

= ℎ + 1 + ℎ = 1100 + (1 + 50)5 = .

= ×

=

×

=

−50 × 5

256.1= −.

= ×

+ =

×

+ =

−50 × 5

500 + 256.1= −.

= ∞



For an emitter follower circuit with = & = . Calculate , , , , & . Assume = , = & = /. (W-13/8m)

Solution:

The emitter follower circuit and its hybrid model are as follows

= −ℎ = −

= ℎ + 1 + ℎ = 1100 + (1 + 50)10 = .

= ×

+ =

−50 × 1

1 + 511.1= −.

= ×

=

×

=

−50 × 10

511.1= −.

= ×

+ =

×

+ =

−50 × 10

1 + 511.1= −.

= ∞



FEEDBACK AMPLIFIER: Feedback amplifier is a closed loop network in which some part of output signal is fed back to the

input. If A is the gain of amplifier and β is the gain of feedback network, then the feedback amplifier

will be as shown below:

Therefore ≠

CLOSED LOOP GAIN () WITH NEGATIVE FEEDBACK:

In negative feedback amplifier, feedback signal is subtracted from input.

− =

= + --- (1)

Gain of feedback network: = /

Gain of Amplifier: = /

Overall closed loop gain of feedback amplifier is,

=

=

+ =

/

/ + /=

1 +

=

1 +

×

=

1 +

The feedback signal can be either added to the input or subtracted from it. Depending upon

whether it is added or subtracted, it is classified into:

1. Positive Feedback Amplifier (Unit - 4)

2. Negative Feedback Amplifier

PRINCIPLE OF NEGATIVE FEEDBACK IN ELECTRONIC CIRCUITS: The principle of negative feedback is that a portion of the output signal is fed back to the input and

combined with the input signal in such a way as to reduce it. This reduces the overall gain of the



amplifier but also introduces a number of benefits, such as reducing distortion and noise, and

widening the amplifier’s bandwidth.

Explain different types of negative feedback connections. (6M/S-16) Show block schematically the different feedback connections in an amplifier. Explain the effect of each type of feedback on input and output impedance. (W-14/7m) Explain the following feedback topologies and draw the practical circuits. (i) Current series feedback (ii) Voltage shunt feedback (W-15/6m)

FEEDBACK TOPOLOGIES: Feedbacks are classified into following four topologies:

1. Voltage shunt feedback

2. Voltage series feedback

3. Current shunt feedback

4. Current series feedback

VOLTAGE SHUNT FEEDBACK

A

β

IS

Iin

IF=βVO

RL

V0

In such type of network, output voltage is fed back to the input in parallel. This type of amplifier is

also called as transresistance amplifier.

Equivalent circuit of transresistance amplifier is,

IS RS Ri

Ro

RL

Ii→

VO≈RmIS

Ri≪RS

RO≪RL

RmIi



VOLTAGE SERIES FEEDBACK

A

β

VS RL

V0Vin

Vf

Vf=βVO

In such type of network, output voltage is fed back to the input in series. This type of amplifier is

also called as voltage amplifier.

Equivalent circuit of voltage amplifier is,

+VS

-

Ri

Ro

RL

VO≈AVVS

RS≪Ri

RO≪RL

AVVi

RS

Vi

CURRENT SHUNT FEEDBACK

A

β

IS

Iin

IF=βIO

RL

Io

V0

In such type of network, output current is fed back to the input in shunt. As it affects the input

current signal, it can be used to control the current signal. Thus, it is also called as current amplifier.

Equivalent circuit of current amplifier is,

IS RS Ri

RL

Ii→

IL≈AiIS

Ri≪RS

RL≪RO

AiIi

Ro



CURRENT SERIES FEEDBACK

A

β

VS RL

IO→ V0Vin

Vf

In such type of network, a part of output current is made to develop a voltage proportional to the

output current and supplied back in series with the input. It is also called as transconductance

amplifier.

Equivalent circuit of transconductance amplifier is,

+VS

-

Ri

IO≈GmVS

RS≪Ri

RL≪RO

RS

Vi RLGmVi

Ro

Explain the effect of negative feedback on (i) Voltage gain (ii) Gain Stability (iii) Input resistance (iv) Bandwidth (8M/W-16). Explain the effect of negative feedback on various circuit parameters. (W-13/8m) Explain the effect of negative feedback on (i) Voltage gain (ii) Bandwidth and (iii) Noise (W-14/6m)

CHARACTERISTICS OF NEGATIVE FEEDBACK

(I) Effect of negative feedback on Input Impedance:

(A) Voltage Series Amplifier:

+VS

-

Ri

Ro

RLAVVi

Ii→

Vi

-Vf+

Rif

VO

Vf=βVO

=

=

+

=

+

Using voltage division theorem, can be obtained from output circuit as,



= ×

+ =

Where, is the open loop gain without considering and is the open loop gain considering

.

= + ()

=

+ ()

=

(1 + )

∴ = (1 + )

Thus, with negative feedback in voltage series amplifier, input impedance is increased by(1 +

) times.

(B) Current Series Amplifier:

RLGmVi

Ro

+VS

-

Ri

Ii→

Vi

-Vf+

RifVf=βIO

From figure,

=

=

+

=

+

Using Current division theorem, can be obtained from output circuit as,

= ×

+ =

Where, is the transconductance without considering and is the transconductance

considering .

= +

=

+ ×

=

(1 + )

= (1 + )

Thus, with negative feedback in current series amplifier, input impedance is increased

by(1 + ) times.



(C) Current Shunt Amplifier:

IS RiRL

Ii→

AiIi

Ro

Rif

If

If=βIO

From figure,

=

=

+ =

+

Using Current division theorem, can be obtained from output circuit as,

= ×

+ =

Where, is the open loop current gain without considering and is the open loop current

gain considering .

=

+ ()=

(1 + )

Where AI is the current gain including RL.

=

1 +

Thus, with negative feedback in current shunt amplifier, input impedance is decreased by(1 +

) times.

(D) Voltage Shunt Amplifier:

IS Ri

Ii→

Rif

If

Ro

RLRmIi

VOVi

If=βVO

=

=

+ =

+

Using voltage division theorem, can be obtained from output circuit as,



= ×

+ =

Where, is the transresistance without considering and is the transresistance considering

.

=

+ =

(1 + )

=

1 +

Thus, with negative feedback in voltage shunt amplifier, input impedance is decreased by(1 +

) times.

(II) Effect of negative feedback on Output Impedance:

(A) Voltage Series Amplifier:

+VS

-

Ri

Ro

RLAVVi

Ii→

Vi

-Vf+

VO

Rif ROf R’of

IOI

Applying KVL in output loop,

− − = 0

− (−) − = 0

= −

Applying KVL in input loop,

− − = 0

− − = 0

But for calculating output impedance, input voltage source is to be short circuited i.e. = 0

= −

= − (−)

=

(1 + )

=

=

(1 + )

=

1 +

Thus, with negative feedback in voltage series amplifier, output impedance is decreased

by(1 + ) times.



= ||

=

+ =

1 +

1 + +

=

+ (1 + )=

( + ) +

Dividing numerator and denominator by ( + )

=

+

1 +

+

ℎ =

+ &

=

+

=

1 +

Where is voltage gain including .

(B) Voltage Shunt Amplifier:

IS Ri

Ii→

Rif

If

Ro

RLRMIi

VOVi

If=βVO

R’ofRof

IO

I

Applying KVL in output side,

− − = 0

− (−) − = 0

= −

Applying KCL at input side,

= + = +

But for calculating output impedance, input current source is to be open circuited i.e. = 0

= −

= − (−)

=

(1 + )

The output resistance with feedback is given by,



=

=

(1 + )

=

1 +

Thus, with negative feedback in voltage shunt amplifier, output impedance is decreased

by(1 + ) times.

The output resistance with feedback including RL is,

= ||

=

+ =

1 +

1 + +

=

+ (1 + )=

( + ) +


=

+

1 + +

ℎ, =

+ &

=

+

=

1 +

Where, is the transresistance including .

(C) Current Shunt Amplifier:

IS RiRL

Ii→

AIIi

Ro

Rif

If Vi

If=βIO ROf R’of

IO

I

Applying KCL at output side,

− −

= 0

− (−) −

= 0

=

−

Applying KCL at input side,



= + = +

But for calculating output impedance, input current source is to be open circuited i.e. = 0

0 = +

= − = −(−) =

=

−

(1 + ) =

= (1 + )

= (1 + )

Thus, with negative feedback in current shunt amplifier, output impedance is increased by(1 +

) times.

Output resistance with feedback including RL is,

= ||

=

+ =

[(1 + )]

[(1 + )] + =

(1 + )

+ +


=

+

(1 + )

1 +

+

ℎ, =

+ &

=

+

=

(1 + )

(1 + )

Where, is the current gain including .

(D) Current Series Amplifier:

RLGMVi

Ro

+VS

-

Ri

Ii→

Vi

-Vf+

Rif ROf R’of

IO

I

VO

Applying KCL at the output side,



− −

= 0

− (−) −

= 0

=

−

Applying KVL at input side,

− − = 0

− − = 0

But for calculating output impedance, input voltage source is to be short circuited i.e. = 0

= −

=

− (−) =

+ (−) =

−

(1 + ) =

= (1 + )

= (1 + )

Thus, with negative feedback in current series amplifier, output impedance is increased

by(1 + ) times.

The output resistance with feedback including RL is,

= ||

=

+ =

[(1 + )]

[(1 + )] + =

(1 + )

+ +


=

+

(1 + )

1 + +

ℎ, =

+ &

=

+

=

(1 + )

(1 + )

Where, is the transconductance including .



(III) Effect of negative feedback on Gain:

In negative feedback amplifier, feedback signal is subtracted from input.

− =

= + --- (1)

Gain of feedback network: = /

Gain of Amplifier: = /

Overall closed loop gain of feedback amplifier is,

=

=

+ =

/

/ + /=

1 +

=

1 +

×

=

1 + = ×

1

1 +

∴ <

The closed loop gain of negative feedback amplifier is less than the open loop gain A.

Where, /( + ) is known as feedback factor.

DESENSITIZATION OR STABILIZATION OF GAIN: As we know, the transfer gain of amplifier changes with external parameters like temperature,

replacement of electronic component, etc. The closed loop gain of negative feedback amplifier is

given by,

=

1 +

Differentiating above equation with respect to A,

=

(1 + )1 −

(1 + )

=

1 + −

(1 + )=

1

(1 + )



=

(1 + )

Dividing both sides by

=

(1 + )×

1

=

(1 + )×

1

1 +

=

(1 + )×

1 +

=

×

1

1 +

=

/

1 +

The term / represents the fractional change in amplifier voltage gain with feedback and

/ represents the fractional change in amplifier voltage gain without feedback. And the term

1/(1 + ) is known as sensitivity.

Therefore, sensitivity is defined as the ratio of percentage change in voltage gain with feedback to

the percentage change in voltage gain without feedback.

=

=

1

1 +

The reciprocal of the term sensitivity is called as desensitivity D, which is given by,

= (1 + )

For the gain to be stable, ideally, sensitivity must be zero and desensitivity must be infinity.

Maximum stability can be achieved with large value of β.

If ||>> 1, then

=

1 + ≈

=

1

Hence gain entirely depends on the feedback network. If the feedback network contains only stable

passive elements, the improvement in stability will be high.

In above equation, A represents either , , and represents corresponding transfer

gain with feedback , , .

1. For Voltage series feedback ≈ 1/ shows stabilization of voltage gain,

2. For Current series feedback ≈ 1/ shows stabilization of transconductance gain,

3. For Current shunt feedback ≈ 1/ shows stabilization of current gain,

4. For Voltage shunt feedback ≈ 1/ shows stabilization of transresistance gain.

(IV) EFFECT OF NEGATIVE FEEDBACK ON LOWER & HIGHER CUTOFF

FREQUENCY:

EFFECT OF NEGATIVE FEEDBACK ON LOWER CUT-OFF FREQUENCY

We know,



=

1

1 − (/)− − − −(1)

=

1 − (/)

We know,

=

1 + =

1 − (/)

1 +

1 − (/)

=

1 − (/) 1 +

1 − (/)

=

1 − (/) + =

(1 + ) − (/)

Dividing numerator and denominator by (1 + )

=

1 +

1 + 1 +

−

×1

1 +

=

1 − 1

×

1 +

=

1

1 − 1

×

1 +

− − − −(2)

Comparing equation (1) & (2),

=

1 +

Therefore,

=

1

1 −

Where, is the lower cutoff frequency with feedback. From the above equation we can say that,

the lower cutoff frequency is reduced by a factor (1 + ) after applying negative feedback.

EFFECT OF NEGATIVE FEEDBACK ON HIGHER CUT-OFF FREQUENCY

We know,

=

1

1 − (/)− − − − − (1)

=

1 − (/)

We know,

=

1 + =

1 − (/)

1 +

1 − (/)

=

1 − (/) +



=

(1 + ) − (/)

Divide and multiply by (1 + ),

=

1 +

(1 + )(1 + )

−

×1

(1 + )

=

1 − ×

(1 + )

=

1

1 − ×

(1 + )

− − − −(2)

Comparing equation (1) & (2),

= (1 + )

Therefore,

=

1 − /

Where, is the higher cutoff frequency with feedback. From the above equation we can say that,

the higher cutoff frequency is increased by a factor (1 + ) after applying negative feedback.

(V) EFFECT OF NEGATIVE FEEDBACK ON BANDWIDTH: The bandwidth () of an amplifier is defined as the difference between upper/higher cutoff

frequency and lower cutoff frequency of a given frequency response.

= − --- (1)

The bandwidth of an amplifier with feedback is given by,

= − --- (2)

The upper cutoff frequency with feedback is given by,

= (1 + )

∴ > --- (3)

The lower cutoff frequency with feedback is given by,

=

1 +

∴ < --- (4)

From equations (1), (2), (3) & (4),

− > ( − )



>

Since bandwidth with negative increases by the factor (1 + ) and gain decreases by the same

factor, the gain bandwidth product of an amplifier does not altered, when negative feedback is

applied.

(VI) EFFECT OF NEGATIVE FEEDBACK ON NOISE: Negative feedback reduces the noise by increasing the signal to noise ratio. Consider the following

amplifier block with input signal , noise signal and gain .

Vn

A1

+VS

-VO

Assume that the noise is introduced at the input of the amplifier and the signal to noise ratio for

this amplifier is given by

=

Consider following negative feedback amplifier with another noise-free amplifier stage with gain

.

A1A2 VO

+VS

-

β

Vn

The output voltage will be due to &

To find output voltage due to only , noise source is to be short circuited.



=

1 +

To find output voltage due to only , source voltage is to be short circuited.

=

1 +

Thus, overall output voltage will be,

=

1 +

+

1 +

Therefore, signal to noise ratio with negative feedback will be,

=

1 +

1 +

=

×

=

×

Signal to noise ratio with negative feedback is A2 times larger than signal to noise ratio without

feedback. Thus, there is an improvement in SNR (signal to noise ratio).

(VII) EFFECT OF NEGATIVE FEEDBACK ON HARMONIC DISTORTION:

AmplifierA

FeedbackNetwork

β

---

βDf

InputSignal Output

βADf Df

Let an amplifier with open loop gain A produces a distortion D in the output signal without

feedback. When negative feedback is applied, output along with distortion is fed back to the input.

Let the gain with feedback be and the distortion at the output be . A part of this distortion

is fed back to the input. It gets amplified by a factor A and becomes .

Change in distortion due to application of feedback will be

= ℎ − ℎ

= − → (1 + ) =

=

1 +

Thus after applying feedback, distortion is reduced by (1 + ) times.

Documents

-PARAMETER ODEL · evaluating unknown parameters. 2. That these parameters are not alike dimensionally. 3. That the unit of these parameters are different e.g. Ohms Ω, Mhos ℧,