-parameter odel · evaluating unknown parameters. 2. that these parameters are not alike...
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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
P a g e | 2 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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
P a g e | 3 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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
P a g e | 4 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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
P a g e | 5 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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
P a g e | 6 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
=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
P a g e | 7 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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.
=
=
×
= ×
= ×
= ( + )
=
+
=
+
=
+
P a g e | 8 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
= ×
+
=
=
×
+
=
+
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,
∴ =
P a g e | 9 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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
=
+
= ×
+
= ×
P a g e | 10 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
= ×
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.
P a g e | 11 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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
P a g e | 12 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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:
P a g e | 13 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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
P a g e | 14 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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
P a g e | 15 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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)
P a g e | 16 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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
P a g e | 17 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
= 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
P a g e | 18 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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
P a g e | 19 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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.
P a g e | 20 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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
P a g e | 21 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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
P a g e | 22 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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= −.
= ∞
P a g e | 23 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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= −.
= ∞
P a g e | 24 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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
P a g e | 25 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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
P a g e | 26 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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
P a g e | 27 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
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,
P a g e | 28 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
= ×
+ =
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.
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(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,
P a g e | 30 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
= ×
+ =
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.
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= ||
=
+ =
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,
P a g e | 32 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
=
=
(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 + )=
( + ) +
Dividing numerator and denominator by ( + )
=
+
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,
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5th Edition for Session 2017-18
= + = +
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 + )
+ +
Dividing numerator and denominator by ( + )
=
+
(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,
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5th Edition for Session 2017-18
− −
= 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 + )
+ +
Dividing numerator and denominator by ( + )
=
+
(1 + )
1 + +
ℎ, =
+ &
=
+
=
(1 + )
(1 + )
Where, is the transconductance including .
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(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 + )
P a g e | 36 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
=
(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,
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5th Edition for Session 2017-18
=
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 − (/) +
P a g e | 38 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
5th Edition for Session 2017-18
=
(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),
− > ( − )
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>
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.
P a g e | 40 Unit 3| Transistor Small Signal Analysis & Negative feedback amplifier
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=
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.
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