University Syllabus
PART A
UNIT 1:
Diode Circuits: Diode Resistance, Diode equivalent circuits, Transition and diffusion capacitance, Reverse
recovery time, Load line analysis, Rectifiers, Clippers and clampers. (Chapter 1.6 to 1.14, 2.1 to 2.9)
6 Hours
UNIT 2:
Transistor Biasing: Operating point, Fixed bias circuits, Emitter stabilized biased circuits, Voltage divider
biased, DC bias with voltage feedback, Miscellaneous bias configurations, Design operations, Transistor
switching networks, PNP transistors, Bias stabilization. (Chapter 4.1 to 4.12)
7 Hours
UNIT 3:
Transistor at Low Frequencies: BJT transistor modeling, Hybrid equivalent model, CE Fixed bias
configuration, Voltage divider bias, Emitter follower, CB configuration, Collector feedback configuration,
Hybrid equivalent model. (Chapter 5.1 to 5.3, 5.5 to 5.17)
7 Hours
UNIT 4: Transistor Frequency Response: General frequency considerations, low frequency response, Miller effect
capacitance, High frequency response, multistage frequency effects. (Chapter 9.1 to 9.5, 9.6, 9.8,
9.9)
6 Hours
PART B
UNIT 5:
(a) General Amplifiers: Cascade connections, Cascode connections, Darlington connections. (Chapter 5.19
to 5.27)
3 Hours (b) Feedback Amplifier: Feedback concept, Feedback connections type, Practical feedback circuits.
(Chapter 14.1 to 14.4) 3
Hours
UNIT 6:
Sub Code 10ES32 IA Marks 25
Hrs/ Week 04 Exam Hours 03
Total Hrs. 52 Exam Marks 100
Analog Electronic Circuits 10ES32
SJBIT/ECE Page 2
Power Amplifiers: Definitions and amplifier types, series fed class A amplifier, Transformer coupled Class
A amplifiers, Class B amplifier operations, Class B amplifier circuits, Amplifier distortions. (Chapter 12.1 to
12.9)
7 Hours
UNIT 7: Oscillators: Oscillator operation, Phase shift Oscillator, Wienbridge Oscillator, Tuned Oscillator circuits,
Crystal Oscillator. (Chapter 14.5 to 14.11) (BJT version only)
6 Hours
UNIT 8: FET Amplifiers: FET small signal model, Biasing of FET, Common drain common gate configurations,
MOSFETs, FET amplifier networks. (Chapter 8.1 to 8.13)
7 Hours
TEXT BOOK:
1. Electronic Devices and Circuit Theory, Robert L. Boylestad and Louis Nashelsky, PHI/Pearson Eduication. 9TH Edition.
REFERENCE BOOKS:
1. Integrated Electronics, Jacob Millman & Christos C. Halkias, Tata - McGraw Hill, 1991 Edition 2. Electronic Devices and Circuits, David A. Bell, PHI, 4th Edition, 2004
3 Analog electronics circuits: A simplified approach U B mahadevaswamy, pearson education 9 th edition.
Question Paper Pattern: Student should answer FIVE full questions out of 8 questions to be set each
carrying 20 marks, selecting at least TWO questions from each part.
Analog Electronic Circuits 10ES32
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INDEX SHEET
SL.NO TOPIC PAGE NO.
1 University syllabus 04
PART A
UNIT 1: Diode Circuits
1.1 Diode Resistance 07
1.2 Diode equivalent circuits 08
1.3 Transition and diffusion capacitance 09
1.4 Reverse recovery time 11
1.5 Load line analysis 14
1.6 Clippers and clampers 20
UNIT - 2: Transistor Biasing
2.1 Bipolar Transistor 57
2.2 Bipolar Stability 59
2.3 Fixed with Emitter 63
2.4 Voltage divider biased 64
UNIT - 3: Transistor at Low Frequencies
3.1 AC Analysis BJT transistor modeling 87
3.2 Hybrid equivalent model 90
3.3 CE Fixed bias configuration 95
3.4 Voltage divider bias 96
3.5 Emitter follower 98
3.6 CB configuration 100
UNIT - 4: Transistor Frequency Response
4.1 General frequency Response 112
4.2 Low frequency response 113
4.3 Miller effect capacitance 118
4.4 High frequency response 120
PART B UNIT - 5: a) General Amplifiers
5.1 Amplifier Basics 132
5.2 Classification of Amplifier 133
5.3 Multistage Amplifier 134
5.4 RC couples Amplifier 140
UNIT - 5: b) Feedback Amplifiers
5.5 Feedback concept 140
5.6 Feedback connections type 142
UNIT - 6: Power Amplifiers
6.1 Definitions and amplifier types 180
6.2 Cass A amplifier 180
6.3 Transformer coupled Class A amplifiers 186
6.4 Class B amplifier operations 193
UNIT - 7: Oscillators
Analog Electronic Circuits 10ES32
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7.1 Oscillator operation 209
7.2 Phase shift Oscillator 209
7.3 Wienbridge Oscillator 210
7.4 Tuned Oscillator circuits 211
7.5 Crystal Oscillator 212
UNIT - 8: FET Amplifiers
8.1 FET small signal model 222
8.2 Biasing FET 226
Analog Electronic Circuits 10ES32
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Unit: 1 Hrs: 6
Diode Circuits: Diode Resistance, Diode equivalent circuits, Transition and diffusion capacitance, Reverse
recovery time, Load line analysis, Rectifiers, Clippers and clampers.
Recommended readings:
TEXT BOOK: 1. Electronic Devices and Circuit Theory, Robert L. Boylestad and Louis Nashelsky, PHI/Pearson
Eduication. 9TH Edition.
REFERENCE BOOKS: 1. Integrated Electronics, Jacob Millman & Christos C. Halkias, Tata - McGraw Hill, 1991 Edition
2. Electronic Devices and Circuits, David A. Bell, PHI, 4th Edition, 2004
Analog Electronic Circuits 10ES32
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1.1 DIODE RESITANCE
As the operating point of a diode moves from one region to another, the resistance of the diode will also
change due to the nonlinear shape of the diode characteristic curve.
The type of applied voltage or signal will define the resistance level of interest.
Three different levels will be introduced.
DC or Static Resistance
The application of a dc voltage to a circuit containing a semiconductor diode will result in an
operating point on the characteristic curve that will not change with time.
The resistance of the diode at the operating point is simply the quotient of the corresponding levels of
VD and ID.
The dc resistance levels at the knee and below will be greater than the resistance levels obtained for the
vertical rise section of the characteristics.
The resistance levels in the reverse bias region will be high.
In general, the lower the current through a diode the higher the dc resistance level
AC Resistance
It is used to find the diode resistance when the small signal ac input voltage is applied across the
diode.
VD
ID
DC resistance Rd= VD / ID
Analog Electronic Circuits 10ES32
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For small signal ac voltage, ID & VD changes around Q point which is fixed by large signal DC
voltage
The ac resistances is determined by
Drawing a tangent line at Q point
Then find the change in voltage and the current.
The ratio of this change in the voltage and the current is called ac resistance.
Average Resistance
It is used to find the diode resistance when the large signal ac input voltage is applied across the
diode.
For large signal, there is no Q point and limits of operation is large due large swing in current and
voltage.
Average resistance is ratio of change voltage to the change in current between two extreme points.
The average resistances is determined by
Drawing a straight line between two extreme voltages on characteristic curve
Then finding the difference in voltages and respective currents between the two
points.
1.2 Equivalent Circuits of Diode
Order of simplification:
Diode Characteristic Curve Non linear graph
Ac resistance rd= Vd / Id
Q Id
Vd
Analog Electronic Circuits 10ES32
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Piecewise Linear Equivalent Circuit approximate in to two lines, one horizontal and other with
slope 1/r
Simplified Equivalent Circuit approximate in to two lines, one horizontal and other one vertical
Ideal Diode with zero voltage across diode during forward bias and zero current through diode during
reverse bias
1.3 Diode Capacitance
Ideal Equivalent circuit
Ideal diode
rave = 0, Vk =0
Simplified Equivalent Circuit
Ideal diode
rave = 0
Analog Electronic Circuits 10ES32
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Two types of Diode Capacitance :
Transition Capacitance
Diffusion capacitance.
Effect of capacitance:
It is stray capacitance and has very low value
Diode becomes frequency sensitive, mainly at very high frequency.
At high frequency, Xc becomes low enough to introduce a low reactance shorting path.
Transition capacitance (CT):
Predominant effect in reverse bias condition.
Also called as Depletion region Capacitance or space charge capacitance
Basic capacitance eqn = A/d where
= permittivity of dielectric between the two plates
A =Area and d = distance between the plates.
Depletion region behaves like dielectric between two charged plates.
Depletion width d increases with increase in reverse bias.
So CT decreases as reverse bias increases.
Application Ex Schottky diode, varactor (Varicap) diodes.
CT is present in forward bias also, but is effect is neglected by the presence of larger CD
Diffusion capacitance (CD):
Predominant effect in forward bias condition.
Also called as Storage Capacitance.
Depends on rate at which charge is injected in to the PN region. (outside the depletion).
So as current increases, CD increases.
However increase in current reduces resistance. This helps in high frequency operation as T =
RC
Analog Electronic Circuits 10ES32
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1.4 Reverse Recovery Time:-
Denoted by trr.
In forward bias, large number of free electrons in P region and holes in N region during conduction.
This results in minority carriers in each region
Sudden changing to reverse bias results into large reverse current due to large minority carriers. ( I
reverse = I forward)
Stays for initial storage time ts
After movement of minority carriers top other region Ir decreases to Is within time tt.
trr = ts +tt
Important in high speed switching applications
Normal value few nanosec to 1us . Very low trr of picosecs are also available
-25 -20 -15 -10 -5 0 0.25 0.5
5
10
15
CT
Cpf
CT + CD ~ CD
CT
1uF
CD
1
D1
1N5402
Analog Electronic Circuits 10ES32
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DIODE SPECIFICATIONS
Data provided by manufactures.
Must be included data :-
VF at specified temp and IF
IF max at specified temp.
IR at specified voltage and temp.
PIV or BR or PRV at specified temp
PD max = VDID
Capacitance levels
Reverse recovery time .. trr
Operating temp Range
Additional Data depends on application :-
Frequency range
Noise Level.
Switching time.
Thermal resistance levels
Peak repetitive values.
ID
t
IF
t1
ts
tt
Diode is reverse biased
Ir
Is
Desired response.
trr = ts + tt
Analog Electronic Circuits 10ES32
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Diode Notation:- A & K Depends on application, manufactures, current/voltage rating
Testing can be done by using
DMM (Digital Multimeter)- with diode checking function
ohm meter
Curve Tracer
With diode checking :- Internal meter voltage is used
In one direction it shows 0.7 V as diode is forward biased
In other direction it is around 2.5V (depends on Vbattery)
With Ohm meter
One direction low resistance (RF)
Other direction it shows high resistance (RR)
With curve tracer
Characteristic curve of the diode is displayed.
Vertical axis is 1mA/div(Can be adjusted)
Horizontal axis is 100mV/div (can be adjusted)
Expensive and looks more complex.
Analog Electronic Circuits 10ES32
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Load line Analysis:-
Load line - defined by the network
Characteristic curve defined by device
V1 = VD+IDR
ID = V1/R at VD =0V
VD = V1/R at ID =0A
1. In any given circuit, check biasing of diode.
2. During forward bias (i.e diode is ON) replace diode by short for ideal diodes or with 0.7V
3. During reverse bias (i.e diode is OFF) replace diode with open circuit.
4. Do the ckt analysis and find the output voltage.
In a circuit, diode can be in Series, Parallel or Series and Parallel
Answers :-
Load line Analysis:-
R11k
+ V110V
D1
1N5402
10mA
V1
Q
Load line
Characteristic
Analog Electronic Circuits 10ES32
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5.Determine the current I for each of the configurations of fig 2.150 using the approximate equivalent model
for the diode.
a) I = 0 mA; diode reverse-biased.
(b)V in loop of 20 = 20 V 0.7 V = 19.3 V (Kirchhoffs voltage law)
I = 19.3/20 = 0.965 A
(c)I = 10v/10 = 1 A; center branch open is open as one diode is forward biased and the other one is reverse
biased.
Analog Electronic Circuits 10ES32
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6) Determine Vo and Id for the networks of fig.2.151
a)Diode forward-biased,
Kirchhoffs voltage law (CW):
5 V + 0.7 V Vo = 0
So Vo = 4.3 V
IR = ID = 4.3V/2.2K = 1.955 mA
(b)Diode forward-biased,
ID = (8-0.7)/ (1.2k+4.7k) = 1.24 mA
Vo = ID* 4.7 k + VD
= (1.24 mA)(4.7 k ) + 0.7 V = 6.53 V
7) Determine the level of Vo for each network of fig.2.152
a)Vo = (Vdc-VD1-VD2) = (20 V 1 V)
= (19 V) = 9.5 V
b) I = (10-(-2)-0.7)/ (1.2+4.7)k
= (11.3/5.9) = 1.915 mA
Analog Electronic Circuits 10ES32
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V = IR = (1.915 mA)(4.7 k ) = 9 V
Vo = V 2 V = 9 V 2 V = 7 V
8) Determine Vo and Id for the networks of fig.2.153.
a) Determine the Thevenin equivalent circuit for the 10mA source and 2.2 k resistor.
ETh = IR = (10 mA)(2.2 k ) = 22 V and RTh = 2.2k
So ID =22/(2.2+1.2) = 6.26 mA
Vo = ID(1.2 k ) = (6.26 mA)(1.2 k ) = 7.51 V
(b)Diode forward-biased,
ID = 20-(-5)-0.7 /6.8k= 3.57 mA
Kirchhoffs voltage law (CW):
Vo 0.7 V + 5 V = 0
Vo = 4.3 V
9) Determine Vo1 and Vo2 for the networks of fig.2.154.
(a)Vo1 = 12 V 0.7 V = 11.3 V
Vo2 = 0.3 V
Analog Electronic Circuits 10ES32
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(b) Vo1 = 10 V + 0.3 V + 0.7 V = 9 V
I = 9V/(1.2+3.3)k = 2 mA,
Vo2 = (2 mA)(3.3 k ) = 6.6 V
10) Determine Vo and Id for the netwoks of Fig.2.155.
(a) Both diodes forward-biased
(b) IR = (20-0.7)/4.7K = 4.106 mA
Assuming identical diodes:
ID = 4.106/2 = 2.05 mA,Vo = 20 V 0.7 V = 19.3 V
(b)Right diode forward-biased:
ID =15-(-5)-0.7 /2.2K =20/2.2 8.77 mA
Vo = 15 V 0.7 V = 14.3 V
Analog Electronic Circuits 10ES32
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11) Determine Vo and I for the networks of Fig 2.156.
(a)Both diodes forward-biased
IR = (20-0.7)/4.7K = 4.106 mA
Assuming identical diodes:
ID = 4.106/2 = 2.05 mA
Vo = 20 V 0.7 V = 19.3 V
(b)Right diode forward-biased:
ID =15-(-5)-0.7 /2.2K
=20-.7/2.2 = 8.77 mA
Vo = 15 V 0.7 V = 14.3 V
Analog Electronic Circuits 10ES32
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12) Determine Vo1 and Vo2 and I for the network of Fig.2.157.
Both diodes forward-biased:
Vsi = 0.7 V, Vge = 0.3 V
Vo1 = 20-0.7=19.3V Vo2 = 0.3V
Current through 1k resistor (I1)
= (20-0.7)I1 k 19.3/1k = 19.3 mA
Current through 0.7k resistor (I2)
13) Determine Vo and Id for the network of fig.2.158.
Both diodes are forward biased and parallel (in series with 2k ).
Thevinins eqt circuit for this is 2k//2k with 0.7 V in series = 1kohm in series with 0.7V
So current through load resistor
= (10-0.7)/ (1+2)k = 3.1mA
ID = 3.1/ 2 = 1.55 m
Vo = 3.1mA* 2 K = 6.2V
Analog Electronic Circuits 10ES32
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1.5 Clippers
A clipper is a circuit that is used to eliminate a portion of an input signal. There are two basic types of clippers: series
clippers and shunt/parallel clippers. As shown in Figure 4-1, the series clipper contains a diode that is in series with
the load. The shunt clipper contains a diode that is in parallel with the load.
FIGURE 4-1 Basic clippers.
The series clipper is a familiar circuit. The half-wave rectifier is nothing more than a series clipper. When the diode in
the series clipper is conducting, the load waveform follows the input waveform. When the diode is not conducting, the
output is approximately 0 V or fixed dc voltage which is connected in parallel. (Figure 4.2). The direction of the
diode determines the polarity of the output waveform. If the diode symbol (in the schematic diagram) points toward the
source, the circuit is a positive series clipper, meaning that it clips the positive alternation of the input. If the diode
symbol points toward the load, the circuit is a negative series clipper, meaning that it clips the negative alternation of
the input (Figure 4.11). With this di
Ideally, a series clipper has an output of when the diode is conducting (ignoring the voltage across the diode).
When the diode is not conducting, the input voltage is dropped across the diode, and .
Unlike a series clipper, a shunt clipper provides an output when the diode is not conducting. For example, refer to
Figure 4-1. When the diode is off (not conducting), the component acts as an open. When this is the case, and
form a voltage divider, and the output from the circuit is found using
When the diode in the circuit is on (conducting), it shorts out the load. In this case, the circuit ideally has an output of
. Again, this relationship ignores the voltage across the diode. In practice, the output from the circuit is
generally assumed to equal 0.7 V, depending upon whether the circuit is a positive shunt clipper or a negative shunt
clipper. The direction of the diode determines whether the circuit is a positive or negative shunt clipper. The series
current-limiting resistor ( ) is included to prevent the conducting diode from shorting out the source.
A biased clipper is a shunt clipper that uses a dc voltage source to bias the diode. A biased clipper is shown in Figure
4-2. (Several more are shown in Figures 4.9 and 4.10). The biasing voltage ( ) determines the voltage at which the
diode begins conducting. The diode in the biased clipper turns on when the load voltage reaches a value of .
In practice, the dc biasing voltage is usually set using a potentiometer and a dc supply voltage, as shown in Figure 4.10.
Analog Electronic Circuits 10ES32
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FIGURE 4-2 A biased clipper.
Clippers are used in a variety of systems, most commonly to perform one of two functions:
1. Altering the shape of a waveform
2. Protecting circuits from transients
The first application is apparent in the operation of half-wave rectifiers. As you know, these circuits are series clippers
that change an alternating voltage into a pulsating dc waveform. A transient is an abrupt current or voltage spike of
extremely short duration. Left unprotected, many circuits can be damaged by transients. Clippers can be used to protect
sensitive circuits from the effects of transients, as illustrated in Figure 4.12.
Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
1
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
OFF
ON,
Diode condition
Vo = VR =0V
Vo= Vin - Vd
Output voltage
IVinI < I0.7IVFor all values of Vin
Reverse Biased
IVinI > I0.7IVForward biased
Negative Cycle
Positive Cycle
Analog Electronic Circuits 10ES32
SJBIT/ECE Page 22
Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
4
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
Vo
Vin0V+5V
-4.3V
Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
5
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
OFF
ON,
Diode condition
Vo = VR =0V
Vo= Vin Vdc+ Vd
= 0 for +ive cycle
=-(Vin + 1.3V) in ive
cycle
Output voltage
Vin >2V -0.7Reverse Biased
For all values of Vin
Vin
Analog Electronic Circuits 10ES32
SJBIT/ECE Page 23
Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
6
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
OFF
ON,
Diode condition
Vo = VR =0V
Vo= Vin- (Vdc+0.7)
Output voltage
I Vin I<
Vdc+0.7
For all values
of Vin
Reverse
Biased
I Vin I > Vdc+0.7V
Forward biased
Negative Cycle
Positive Cycle
Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
7
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
Vo
Vin-5V
- 2.3V
Analog Electronic Circuits 10ES32
SJBIT/ECE Page 24
Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
8
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
OFF
ON,
Diode condition
Vo = Vdc =2V
Vo= Vin- 0.7V
Output voltage
Vin> Vdc-0.7Reverse
Biased
For all values of Vin
Vin< Vdc-0.7V
Forward biased
Negative Cycle
Positive Cycle
Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
9
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
Vo
Vin-5V
- 4.3V
2V
Analog Electronic Circuits 10ES32
SJBIT/ECE Page 25
Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
11
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
Vo
Vin
0V
+5V
4.93V
Staff: - KRS Session (Aug 08 Dec08)
TE Department PESIT, Bangalore
10
Syllabus: - Analog Electronic Circuit
Unit I /c: - Diode Applications - Clippers
OFF
ON,
Diode condition
Vo = VR =0V
Vo= Vin - Vd
Output voltage
For all values of Vin
Vin < 0.7V Reverse Biased
Vin > 0.7V Forward biased
Negative Cycle
Positive Cycle
Analog Electronic Circuits 10ES32
SJBIT/ECE Page 26
Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
12
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
OFF
ON,
Diode condition
Vo = Vin.
Vo= Vd =0.7V
Output voltage
For all
values of Vin
Vin 0.7VForward biased
Negative Cycle
Positive Cycle
Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
13
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
Vo
Vin
0.7V
-5V
-5V
Analog Electronic Circuits 10ES32
SJBIT/ECE Page 27
Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
15
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
Vo
Vin
- 0.7V
+5V
-5V
Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
16
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
OFF
ON,
Diode condition
Vo = Vin
Vo= Vdc + Vd== 2+0.7 =2.7V
Output voltage
For all values of Vin
Vin < Vdc +0.7VReverse Biased
Vin > Vdc +0.7VForward biased
Negative Cycle
Positive Cycle
Analog Electronic Circuits 10ES32
SJBIT/ECE Page 28
Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
17
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
Vo
Vin
2.7V
+5V
4.93V
Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
18
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
OFF
ON,
Diode
Vo = Vin.
Vo= -Vdc+Vd ==
-2 + 0.7 = -1.3V
Output voltage
IVinI > I(Vdc-0.7V)IReverse Biased
IVinI < I(Vdc-0. 7V)IFor all
values of Vin
Forw ard
biased
Negative CyclePositive
Cycle
Analog Electronic Circuits 10ES32
SJBIT/ECE Page 29
Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
19
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
Vo
Vin
-1.3V
-5V
5V
Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
20
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
OFF
ON,
Diode
Vo = Vin.
Vo= Vdc- Vd == 2 -0.7
=1.3V
Output voltage
For all values of Vin
Vin > Vdc -0.7VReverse Biased
Vin < Vdc -0.7VForward biased
Negative CyclePositive Cycle
Analog Electronic Circuits 10ES32
SJBIT/ECE Page 30
Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
21
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
Vo
Vin
+2.7V
-5V
-5V
Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
22
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
OFF
ON,
Diode
Vo = Vin.
Vo= -(Vdc+Vd)
== -2.7
Output voltage
IVinI< I(Vdc+0.7V) IFor all values of Vin
Reverse Biased
IVinI> I(Vdc+0.7V) IForward
biased
Negative CyclePositive Cycle
Analog Electronic Circuits 10ES32
SJBIT/ECE Page 31
Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
23
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
Vo
Vin
-2.7V+5V
5V
Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
24
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
OFF
ON,
Diode
Vo = Vin.
Vo= -(Vdc+Vd)
== -2.7
Output voltage
IVinI< I(Vdc+0.7V) IFor all values of Vin
Reverse Biased
IVinI> I(Vdc+0.7V) IForward
biased
Negative CyclePositive Cycle
Analog Electronic Circuits 10ES32
SJBIT/ECE Page 32
Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
25
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
Vo
Vin
-2.7V
2.7V
Clampers (DC Restorers)
A clamper is a circuit that is designed to shift a waveform above or below a dc reference voltage without altering the
shape of the waveform. This results in a change in the dc average of the waveform. Both of these statements are
illustrated in Figure 4-3. (The clamper has changed the dc average of the input waveform from 0 V to +5 V without
altering its shape.)
FIGURE 4-3 A clamper with its input and (ideal) output waveforms.
There are two basic types of clampers:
A positive clamper shifts its input waveform in a positive direction, so that it lies above a dc reference voltage.
For example, the positive clamper in Figure 4-3 shifts the input waveform so that it lies above 0 V (the dc
reference voltage).
A negative clamper shifts its input waveform in a negative direction, so that it lies below a dc reference
voltage.
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Both types of clampers, along with their input and output waveforms, are shown in Figure. The direction of the diode
determines whether the circuit is a positive or negative clamper.
Clamper operation is based on the concept of switching time constants. The capacitor charges through the diode and
discharges through the load. As a result, the circuit has two time constants:
For the charge cycle, and (where is the resistance of the diode)
For the discharge cycle, and (where is the resistance of the load)
Since is normally much greater than , the capacitor charges much more quickly than it discharges. As a result,
the input waveform is shifted as illustrated in Figure 4.16.
A biased clamper allows a waveform to be shifted above (or below) a dc reference other than 0 V. Several examples
of biased clampers are shown in Figure 4-4.
FIGURE Several biased clampers.
The circuit in Figure (a) uses a dc supply voltage (V) and a potentiometer to set the potential at the cathode of . By
varying the setting of , the dc reference voltage for the circuit can be varied between approximately 0 V and the
value of the dc supply voltage.
The zener clamper in Figure (b) uses a zener diode to set the dc reference voltage for the circuit. The dc reference
voltage for this circuit is approximately equal to . Note that zener clampers are limited to two varieties:
Negative clampers with positive dc reference voltages
Positive clampers with negative dc reference voltages
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Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
26
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clampers
First positive cycle:-
Diode is reverse biased and Vo= 0V.
First negative cycle:-
Diode is forward biased and capacitor is charging with very low time constant. At
negative peak, Vc=Vm.-Vdc After peak diode becomes reverse biased as
Vc>Vin.
Vo = Vin+Vc
Subsequent positive and negative cycles :- Time constant of Capacitor
discharge is very high.(=C*100k). In each negative cycle, Vc charges to max. value.
In both cycles Vo= Vin + Vc
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Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
27
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
Positive clamper with waveform in negative side. Swing level decreases with increase in voltage. Swing level is max at
Vdc =0V. Swing level can be varied from 0V to Vm
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Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
28
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clampers
First positive cycle:-
Diode is reverse biased and Vo= 0V.
First negative cycle:-
Diode is forward biased and capacitor is charging with very low time constant. At
negative peak, Vc=Vm.+Vdc After peak diode becomes reverse biased as
Vc>Vin.
Vo = Vin+Vc
Subsequent positive and negative cycles :- Time constant of Capacitor
discharge is very high.(=C*100k). In each negative cycle, Vc charges to max. value.
In both cycles Vo= Vin + Vc (Vin is +ivefor positve cycle and ive for ve cycle)
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Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
29
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
Positive clamper with waveform in positive side. Swing level increases with increase in voltage. Swing level is min at Vdc
=0V. Swing level can be varied from Vm to Vm+Vdc.
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Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
30
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clampers
First positive cycle:-
Diode is forward biased and capacitor is charging with very low time constant. At positive peak, Vc=Vm.-Vdc After peak,
diode becomes reverse biased as Vc>Vin.
Vo = Vin-Vc
Subsequent negative and positive cycles :- Time constant of Capacitor discharge is very high.(=C*100k). In each
positive cycle, Vc charges to max. value. In both cycles Vo= Vin Vc. (Vin is +ive
for postive cycle and ive for ve cycle)
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Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
31
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
Negative clamper with waveform in positive side. Swing level decreases with increase in voltage. Swing level is max
Vdc =0V. Swing level can be varied from 0 to Vm
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Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
32
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clampers
First positive cycle:-
Diode is forward biased and capacitor is charging with very low time constant. At positive peak, Vc=Vm.-Vdc After peak,
diode becomes reverse biased as Vc>Vin.
Vo = Vin-Vc
Subsequent negative and positive cycles :- Time constant of Capacitor discharge is very high.(=C*100k). In each
positive cycle, Vc charges to max. value. In both cycles Vo= Vin Vc. (Vin is +ive
for positive cycle and ive for ve cycle)
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Staff:- KRS
Session (Aug 08 Dec08)
TE Department
PESIT, Bangalore
33
Syllabus:- Analog Electronic Circuit
Unit I /c:- Diode Applications - Clippers
Negative clamper with waveform in negative side only. Swing level increases with increase in voltage. Swing level
is min Vdc =0V. Swing level can be varied from -Vm to -Vm +VDC).
Troubleshooting Diode Circuits
Because diodes are so common in the electronics industry, it is important to be able to troubleshoot and repair
systems that employ diodes.
Diode defects include:
Anode-to-cathode short.
Anode-to-cathode open.
Low front-to-back ratio.
Out-of-tolerance parameters.
Tests that can performed on diodes to check for their operation are:
Voltage measurements.
Ohmmeter tests.
Diode testers.
Instruments that used to measure the healthiness of diode are
Digital multimeter in diode mode
Ohm-meter ( multimeter in resistance mode)
Curve tracer
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Question paper with Solutions
Q. 1 What is the origin of diffusion capacitance. Obtain an expression for the diffusion capacitance in terms of current in a p-n diode.
(Jan 2004(6), July 2004 (6), Jan 2007 (7), July 2007 (5) ) Jan 2009 (7)
Sol.: In forward biased condition, the width of the depletion region decreases and holes from p side get diffused in 'n' side while
electrons from 'n' side move into the p-side the applied voltage increases, concentration of injected charged particles increases.
This rate of change of the injected charge with applied voltage is defined as capacitance called diffusion capaacitance.
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Q 2) Draw a double diode clipper which limits at two independent levels and explain its working. (Jan 2004(6), July 2004 (8), July 2005 (6), Jan 2007(6))
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Q. 3 Define the terms P.I.V and regulation as applied to rectifiers.
(July 2004 (4), Jan 2008 (4) , Jan 2009
Sol.: i) Peak Inverse Voltage (PIV) :
When the diode is not conducting, the reverse voltage gets applied across the diode. The peak value of such
voltage decides the peak universe voltage i.e. PIV rating of a diode.
Regulation of the output voltage:
As the load current changes, load voltage changes. Practically load voltage should remain constant So
concept of regulation is to study the effect of change in load current on the load voltage.
Q 4) Draw the piece-wise linear volt-ampere characteristics of a p-n diode. Give the circuit model for the
ON state and OFF state. Jan./Feb. 2005, July 2007 (10).
Another way to analyse the diode circuits is to approximate the V-I characteristics of a diode using only
straight lines i.e. linear relationships. In such approximation, the diode forward resistance is neglected and
the diode is assumed to conduct instantaneously when applied forward biased voltage Vo is equal to cut-in
voltage Vy' And then it is assumed that current increases instantaneously giving straight line nahlre of V-I
characteristics. While in reverse biased condition
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when Vo < 0, the diode does not conduct at all.
Hence when diode forward resistance is assumed zero, the circuit model of diode is as shown in the
Fig. 1 (a). In reverse biased, the diode is open circuit as shown in the Fig. 1 (b). As the diode conducts at Vo
=Vy' the V-I characteristics with straight lines is as shown in the Fig. 1 (c). As the method models the diode
with the pieces of straight lines, the name given to such approximation is piecewise-linear method. The
characteristics of diode shown in the Fig. 1 (c) are called the piecewise linear diode characteristics.
Open circuit
For the clipping circuit shown incharacteristic. Assume ideal diode.
150 volts.
the following figure, obtain its transfer
The input varies linearly from 0 to(7)
Q 5) Sketch and explain the circuit of a double ended clipper using ideal p-n diodes which limit the output
between 10 V. (6) (July 2005(6) July 2007(10),
July 2008 (10))
Vin = Vim sin w t
During positive half cycle, the diode D} becomes forward biased and conducts, only when Vin is greater
than battery voltage Vl' So as long as Vin is less that V1 both the diodes are reverse biased and output
follows input. When D1 conducts, D2 is OFF and hence the output is constant at V1 volts. This is shown in
the Fig. 2
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In case of negative half cycle, as long as Viis greater than V2' the diodes D1 and D2 both remain reverse
biased and the output follows input. Once input goes below V 2 then the diode D2 conducts and output
remains constant equal to V2' This is shown in the Fig. 3 (a) and (b).
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Q 6) Draw the bridge rectifier with capacitor filter and explain.
(July 2005(10), june 2008)
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Q 7) Explain the working of a full wove voltage doubler circuit. Jan./Feb. - 2004.Jan-
2006,July2008
Q 8) Design a power supply usinfS a FWR with capacitance filter to given an output voltage of 10V at 10mA from a 220
Hz, 50 Hz supply. The ripple factor must be less than 0.01. (Jan 2004(10))
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Q 9) For the clipping circuit shown in characteristic. Assume ideal diode.150 volts.the following figure, obtain its transferThe input
varies linearly from 0 to 7 Jan 2005 (10) July 2007 (10) Jan 2009 (10)
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Q 10) Design a full wave' rectifier with a capacitor filter to meet the following specifications.
DC output voltage = 15 volts, Load resistance = 1 kD. RMS ripple voltage on capacitor = < 1% of DC output
voltage. Assume the AC supply voltage as 230 Volts, 50 Hz. (8)
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Jan 2005(10)
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Recommended Questions
1. What do you understand by diffusion Capacitance? (Jan /Feb 2004, 6 marks) 2. Draw a doubl diode clipper, which limits at two independent levels and explain its operation
(Jan /Feb 2004, 6 marks)
3. what is the origin of diffusion capacitance? (July/ Aub 2004 6 marks) 4. Draw a double diode clipper which limits two independent levels and explain its workin?
(July/ Aub 2004 8 marks) 5. Draw a simple clamping circuit and explain its working?
(July/ Aub 2004 6 marks) 6. Define the terms P.I.V and regulation as applied to rectifiers
(July/ Aub 2004 4marks) 7. Explain the validity of the piecewise linear approximation of the diode model
(July/ Aub 2004 4 marks) 8. Draw the piece-wise linear volt-ampere characteristics of a p-n diode. Give the circuit model for the
ON state and OFF state.
9. Sketch and explain the circuit of a double ended clipper using ideal p-n diodes which limit the output between +/- 10V (July / Aug 2005 6 marks)
10. Draw the circuit diagram ofa bridge rectifier. Plot its input and output waveforms. (July / Aug 2005- 10 Marks)
11. Explain diffusion capacitance? (Jan/Feb 2007, 6 marks) 12. Draw and explain a double diode clipper circuit, which limits the output at two independent levels?
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Unit: 2 Hrs: 7
Transistor Biasing: Operating point, Fixed bias circuits, Emitter stabilized biased circuits, Voltage divider
biased, DC bias with voltage feedback, Miscellaneous bias configurations, Design operations, Transistor
switching networks, PNP transistors, Bias stabilization.
Recommended readings:
TEXT BOOK:
1. Electronic Devices and Circuit Theory, Robert L. Boylestad and Louis Nashelsky, PHI/Pearson Eduication. 9TH Edition.
REFERENCE BOOKS:
1. Integrated Electronics, Jacob Millman & Christos C. Halkias, Tata - McGraw Hill, 1991 Edition 2. Electronic Devices and Circuits, David A. Bell, PHI, 4th Edition, 2004
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2.1 Bipolar transistor biasing
DC Biasing is a static operation. It deals with setting a fixed level of the current which should flow through
the transistor with a devised fixed voltage drop across the transistor junctions.
Bias establishes the dc operating point for proper linear operation of an amplifier.
The proper flow of zero signal collector current and the maintenance of proper collector emitter voltage
during the passage of signal is known as transistor.
The purpose of dc biasing of a transistor is to obtain certain dc collector current at a certain dc collector
voltage. These values of current and voltage define the point at which transistor operates. This point is
known as operating point.
The transistor functions most linearly when it is constrained to operate in its active region. Once an operating
point Q is established, time-varying excursions of the input signal should cause an output signal of the same
wave form. If the output is not the one desired, i.e., the o/p does not suit the required conditions, the
operating point is unsatisfactory and should be relocated on the output characteristics.
Now, about choosing the operating point, we should note that the transistor cannot be operated everywhere in
the active region even if we have the liberty to choose the external circuit parameters. This is because of the
various transistor ratings which limit the range of operation. These ratings are maximum collector dissipation
Pcmax, maximum collector voltage V cmax, and maximum collector current Icmax & maximum emitter to base
voltage VEBmax.
Requirements upon biasing circuit
The operating point of a device, also known as bias point or quiescent point (or simply Q-point), is the DC
voltage and/or current which, when applied to a device, causes it to operate in a certain desired fashion.
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1. For analog circuit operation, the Q-point is placed so the transistor stays in active mode (does not shift to operation in the saturation region or cut-off region) when input is applied. For digital
operation, the Q-point is placed so the transistor does the contrary - switches from "on" to "off" state.
Often, Q-point is established near the center of active region of transistor characteristic to allow
similar signal swings in positive and negative directions.
2. Q-point should be stable. In particular, it should be insensitive to variations in transistor parameters (for example, should not shift if transistor is replaced by another of the same type), variations in
temperature, variations in power supply voltage and so forth.
3. The circuit must be practical: easily implemented and cost-effective.
Note on temperature dependence
At constant current, the voltage across the emitter-base junction VBE of a bipolar transistor decreases about 2
mV for each 1C rise in temperature. Oppositely, if VBE: is held constant and the temperature rises, IC
increases, also increasing the power consumed in the transistor, tending to further increase its temperature.
Unless steps are taken to control this positive feedback of increased temperature increased current increased temperature, thermal runaway ensues. An electrical approach to avoid thermal runaway is to use
negative feedback, as described in conjunction with some of the circuits below. A different approach is to use
heat sinks that carry away the extra heat.
Comparison between various configurations for a given transistor sample
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2.2 BIAS STABILITY-
There are two reasons for the operating point to shift. Firstly, the transistor parameters such as , VBE are not the same for every transistor, even of the same type. Secondly, the
transistor parameters (,IC0 , VBE ) are functions of temperature. It is therefore, very important that biasing network be so designed that operating point should be independent of
transistor parameter variations.
The techniques normally used to do so maybe classified into-
1.Stabilization techniques
2. Compensation techniques
STABILITY FACTOR-
As Ic is a function of ICO , VBE, & , it is convenient to introduce three partial derivatives of IC w.r.t these variables. These are called stability factors S,S&S and defined as follows: S = (Ic / ICO ) = (1+ )[ (1+(Rb/Re))/(1+ +(Rb/Re))] S = (Ic / VBE ) = -/Re [1+ +(Rb/Re)] S = (Ic / ) (Ic1/1) [ (1+(Rb/Re))/(1+ 2+(Rb/Re))]
Types of bias circuit
The following discussion treats five common biasing circuits used with bipolar transistors:
1. Fixed bias 2. Collector-to-base bias 3. Fixed bias with emitter resistor 4. Voltage divider bias 5. Emitter bias
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Fixed bias (base bias)
Fixed bias (Base bias)
This form of biasing is also called base bias. In the example image on the right, the single power source (for
example, a battery) is used for both collector and base of transistor, although separate batteries can also be
used.
In the given circuit,
VCC = IBRB + Vbe
Therefore,
IB = (VCC - Vbe)/RB
For a given transistor, Vbe does not vary significantly during use. As VCC is of fixed value, on selection of
RB, the base current IB is fixed. Therefore this type is called fixed bias type of circuit.
Also for given circuit,
VCC = ICRC + Vce
Therefore,
Vce = VCC - ICRC
From this equation we can obtain Vce. Since IC = IB, we can obtain IC as well. In this manner, operating point given as (VCE,IC) can be set for given transistor.
Merits:
It is simple to shift the operating point anywhere in the active region by merely changing the base
resistor (RB).
Very few number of components are required.
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Demerits:
The collector current does not remain constant with variation in temperature or power supply voltage.
Therefore the operating point is unstable.
When the transistor is replaced with another one, considerable change in the value of can be expected. Due to this change the operating point will shift.
Usage:
Due to the above inherent drawbacks, fixed bias is rarely used in linear circuits, ie. those circuits which use
the transistor as a current source. Instead it is often used in circuits where transistor is used as a switch.
Collector-to-base bias
Collector-to-base bias
In this form of biasing, the base resistor RB is connected to the collector instead of connecting it to the battery
VCC. That means this circuit employs negative feedback to stabilize the operating point.
From Kirchhoff's voltage law, the voltage across the base resistor is
VRb = VCC - (IC + Ib)RC - Vbe.
From Ohm's law, the base current is
Ib = VRb / Rb.
The way feedback controls the bias point is as follows. If Vbe is held constant and temperature increases,
collector current increases. However, a larger IC causes the voltage drop across resistor RC to increase, which
in turn reduces the voltage VRb across the base resistor. A lower base-resistor voltage drop reduces the base
current, which results in less collector current, so increase in collector current with temperature is opposed,
and operating point is kept stable.
For the given circuit,
IB = (VCC - Vbe) / (RB+RC).
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Merits:
Circuit stabilizes the operating point against variations in temperature and (ie. replacement of transistor)
Demerits:
In this circuit, to keep IC independent of the following condition must be met:
which is approximately the case if
RC >> RB.
As -value is fixed for a given transistor, this relation can be satisfied either by keeping RC fairly large, or making RB very low.
If RC is of large value, high VCC is necessary. This increases cost as well as precautions
necessary while handling.
If RB is low, the reverse bias of the collector-base is small, which limits the range of collector
voltage swing that leaves the transistor in active mode.
The resistor RB causes an ac feedback, reducing the voltage gain of the amplifier. This undesirable
effect is a trade-off for greater Q-point stability.
Usage: The feedback also decreases the input impedance of the amplifier as seen from the base, which can
be advantageous. Due to the gain reduction from feedback, this biasing form is used only when the trade-off
for stability is warranted.
2.3 Fixed bias with emitter resistor
Fixed bias with emitter resistor
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The fixed bias circuit is modified by attaching an external resistor to the emitter. This resistor introduces
negative feedback that stabilizes the Q-point. From Kirchhoff's voltage law, the voltage across the base
resistor is
VRb = VCC - IeRe - Vbe.
From Ohm's law, the base current is
Ib = VRb / Rb.
The way feedback controls the bias point is as follows. If Vbe is held constant and temperature increases,
emitter current increases. However, a larger Ie increases the emitter voltage Ve = IeRe, which in turn reduces
the voltage VRb across the base resistor. A lower base-resistor voltage drop reduces the base current, which
results in less collector current because Ic = IB. Collector current and emitter current are related by Ic = Ie with 1, so increase in emitter current with temperature is opposed, and operating point is kept stable.
Similarly, if the transistor is replaced by another, there may be a change in IC (corresponding to change in -value, for example). By similar process as above, the change is negated and operating point kept stable.
For the given circuit,
IB = (VCC - Vbe)/(RB + (+1)RE).
Merits:
The circuit has the tendency to stabilize operating point against changes in temperature and -value.
Demerits:
In this circuit, to keep IC independent of the following condition must be met:
which is approximately the case if
( + 1 )RE >> RB.
As -value is fixed for a given transistor, this relation can be satisfied either by keeping RE very large, or making RB very low.
If RE is of large value, high VCC is necessary. This increases cost as well as precautions
necessary while handling.
If RB is low, a separate low voltage supply should be used in the base circuit. Using two
supplies of different voltages is impractical.
In addition to the above, RE causes ac feedback which reduces the voltage gain of the amplifier.
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Usage:
The feedback also increases the input impedance of the amplifier as seen from the base, which can be
advantageous. Due to the above disadvantages, this type of biasing circuit is used only with careful
consideration of the trade-offs involved.
2.4 Voltage divider bias
Voltage divider bias
The voltage divider is formed using external resistors R1 and R2. The voltage across R2 forward biases the
emitter junction. By proper selection of resistors R1 and R2, the operating point of the transistor can be made
independent of . In this circuit, the voltage divider holds the base voltage fixed independent of base current provided the divider current is large compared to the base current. However, even with a fixed base voltage,
collector current varies with temperature (for example) so an emitter resistor is added to stabilize the Q-point,
similar to the above circuits with emitter resistor.
In this circuit the base voltage is given by:
voltage across
provided .
Also
For the given circuit,
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Merits:
Unlike above circuits, only one dc supply is necessary.
Operating point is almost independent of variation. Operating point stabilized against shift in temperature.
Demerits:
In this circuit, to keep IC independent of the following condition must be met:
which is approximately the case if
where R1 || R2 denotes the equivalent resistance of R1 and R2 connected in parallel.
As -value is fixed for a given transistor, this relation can be satisfied either by keeping RE fairly large, or making R1||R2 very low.
If RE is of large value, high VCC is necessary. This increases cost as well as precautions
necessary while handling.
If R1 || R2 is low, either R1 is low, or R2 is low, or both are low. A low R1 raises VB closer to
VC, reducing the available swing in collector voltage, and limiting how large RC can be made
without driving the transistor out of active mode. A low R2 lowers Vbe, reducing the allowed
collector current. Lowering both resistor values draws more current from the power supply
and lowers the input resistance of the amplifier as seen from the base.
AC as well as DC feedback is caused by RE, which reduces the AC voltage gain of the amplifier. A
method to avoid AC feedback while retaining DC feedback is discussed below.
Usage:
The circuit's stability and merits as above make it widely used for linear circuits.
Voltage divider with AC bypass capacitor
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Voltage divider with capacitor
The standard voltage divider circuit discussed above faces a drawback - AC feedback caused by resistor RE
reduces the gain. This can be avoided by placing a capacitor (CE) in parallel with RE, as shown in circuit
diagram.
This capacitor is usually chosen to have a low enough reactance at the signal frequencies of interest such that
RE is essentially shorted at AC, thus grounding the emitter. Feedback is therefore only present at DC to
stabilize the operating point. Of course, any AC advantages of feedback are lost.
Of course, this idea can be used to shunt only a portion of RE, thereby retaining some AC feedback.
Emitter bias
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When a split supply (dual power supply) is available, this biasing circuit is the most effective. The negative
supply VEE is used to forward-bias the emitter junction through RE. The positive supply VCC is used to
reverse-bias the collector junction. Only three resistors are necessary.
We know that,
VB - VE = Vbe
If RB is small enough, base voltage will be approximately zero. Therefore emitter current is,
IE = (VEE - Vbe)/RE
The operating point is independent of if RE >> RB/
Merit:
Good stability of operating point similar to voltage divider bias.
Demerit:
This type can only be used when a split (dual) power supply is available.
Stability factors
S (ICO) = IC / IC0
S (VBE) = IC / VBE
S () = IC /
Networks that are quite stable and relatively insensitive to temperature variations have
low stability factors.
The higher the stability factor, the more sensitive is the network to variations in that
parameter.
S( ICO)
Analyze S( ICO) for emitter bias configuration
fixed bias configuration Voltage divider configuration
For the emitter bias configuration, S( ICO) = ( + 1) [ 1 + RB / RE] / [( + 1) + RB / RE] If
RB / RE >> ( + 1) , then
S( ICO) = ( + 1)
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For RB / RE 10R2, the voltage VB will remain fairly constant for changing levels of IC. VBE = VB VE, as IC increases, VE increases, since VB is constant, VBE drops making IB to fall, which will try to offset the increases level of IC.
S(VBE)
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S(VBE) = IC / VBE
For an emitter bias circuit, S(VBE) = - / [ RB + ( + 1)RE]
If RE =0 in the above equation, we get S(VBE) for a fixed bias circuit as,
S(VBE) = - / RB. For an emitter bias,
S(VBE) = - / [ RB + ( + 1)RE] can be rewritten as,
S(VBE) = - (/RE )/ [RB/RE + ( + 1)]
If ( + 1)>> RB/RE, then
S(VBE) = - (/RE )/ ( + 1) = - 1/ RE The larger the RE, lower the S(VBE) and more stable is the system.
Total effect of all the three parameters on IC can be written as,
IC = S(ICO) ICO + S(VBE) VBE + S() General conclusion: The ratio RB / RE or Rth / RE should be as small as possible considering all aspects of
design.
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Question paper with Solutions
Q 1) Desigh a self bias transistor circuit for a stability of s< 5. The give date is as follows: Jan 2004(8)
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Q 2) For a self bias circuit, derive an expression for the stability factor s.
July 2004(8)
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Q 4.
Jan 2005 (10) JAN2009(10)
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Q 5) In the circuit of Fig. 9 given below, Vcc = 10V, Rc = 1.5 kn, ICQ = 2 mA, VCE = 5V, VBE = 0.7 V, 0 = 50 and stability factor S ~ 5. Find R] and R2.
July 2005 (9),July2009(9)
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Q. 6
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July 2006 (10)
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Recommended Questions
1. What are the causes of instability in a transistor? Explain them in brief.(Jan/Feb 2006, 5 marks) 2. Discuss the causes for bias instability in a transistor(July / Aug 2005, 5 marks) 3. What is meant by biasing of a transistor? List the different types of transistor biasing circuits. 4. What to do you mean by operating point of a transistor? Draw the output characteristic of transistor
with various limits of operation and explain it.
5. Differentiate the active region, saturating region and cut off region of a transistor with the requirement of biasing.
6. Analyze the fixed bias circuit operation and derive the expression for operating point (Iceq, Vceq) Vce max and Ic max
7. Analyse the Emitter bias circuit operation and derive the expression for operating point (Iceq, Vceq) Vce max and Ic max
8. What are the different areas of operation in the BJT Characteristic curve? And explain them. 9. Analyse the voltage divider bias circuit operation and derive the expression for operating point
(Iceq, Vceq) Vce max and Ic max (using both approximate and exact method)
10. List out the various types of biasing circuits and compare their merits and demerits. 11. What do you understand of designing the transistor bias circuit? List the parameters to be
calculated and list the parameters required to design.
12. What is meant by transistor switching circuit? Explain with the required biasing. 13. What do you mean by stabilization? 14. Give the essential requirements of stabilization 15. Differentiate between saturation, linear region & cutoff region of transistor operation & show this
in the characteristic curve
16. Explain the fixed bias of transistor with circuit diagram and output equations 17. What is meant by biasing of a transistor? List the different types of transistor biasing circuits 18. a) Draw the transistor amplifier with the fixed bias circuit using the given component values
Input coupling Capacitor C1 =10uF, RB= 240Kohm, RC = 22Kohm VCC= +12V Output coupling
capacitor C2 = 10uF, Beta = 50 Input signal is ac signal
b)Determining the following for the fixed bias transistor configuration
i)Ibq & Icq ii) Vceq iii) VB & Vc iv) VBC v)Ve
Recalculate for B =100 and compare the results
19. a) Draw the transistor amplifier with the Emitter bias circuit using the given component values Input coupling Capacitor C1 =10uF, RB= 510Kohm, RC = 2.4Kohm, RE= 1.5K ohm VCC= +20V
Output coupling capacitor C2 = 10uF, Beta = 100 Input signal is ac signal
b)Determining the following for the Emitter bias transistor configuration
i)Ibq & Icq ii) Vceq iii) VB & Vc iv) VBC v)Ve
Recalculate for B =50 and compare the results
20. a) Draw the transistor amplifier with the voltage divider bias circuit using the given component values
Input coupling Capacitor C1 =10uF, R1= 62Kohm, R2=9.1 kohm, RC = 3.9 k ohm, RE= 0.68 k
ohm VCC= +16V Output coupling capacitor C2 = 10uF, Beta = 80 Input signal is ac signal
b)Determining the following for the voltage divider bias transistor configuration
i)Ibq & Icq ii) Vceq iii) VB & Vc iv) VBC v)Ve
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Recalculate for B =50 and compare the results
21. Explain the concept of Load line in case of transistors and thus discuss the biasing techniques applied to NPN transistors
22. What do you mean by bias stabilization? 23. Define stability factor. Find the relationship between stability factor and Ib? What is its ideal
value?
24. Give the essential requirements of stabilization of transistor
25. Design the transistor inverter with Rb & RC , Vcc=5V to operate with saturation current of 8mA, B=100. Use level of Ib equal to 120% Ibmax and standard resistor values.
26. Write short notes on Relay driver circuit using transistor.
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Unit: 3 Hrs: 6
Transistor at Low Frequencies: BJT transistor modeling, Hybrid equivalent model, CE Fixed bias
configuration, Voltage divider bias, Emitter follower, CB configuration, Collector feedback configuration,
Hybrid equivalent model.
Recommended readings:
TEXT BOOK:
1. Electronic Devices and Circuit Theory, Robert L. Boylestad and Louis Nashelsky, PHI/Pearson Eduication. 9TH Edition.
REFERENCE BOOKS:
1. Integrated Electronics, Jacob Millman & Christos C. Halkias, Tata - McGraw Hill, 1991 Edition 2. Electronic Devices and Circuits, David A. Bell, PHI, 4th Edition, 2004
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3.1 AC Analysis of BJT transistors
Two types of analyses are usually used depending on the voltage and currents of the input ac signal
relative to the bias voltages and currents. They are small-signal analysis and large-signal analysis. In ac
analysis of BJT amplifier is done using small signal analysis,
Amplification in the AC domain
The transistor can be employed as an amplifying device. That is, the output sinusoidal signal is greater than
the input signal or the ac input power is greater than ac input power.
How the ac power output can be greater than the input ac power?
Conservation- output power of a system cannot be larger than its input and the efficiency cannot be greater
than 1.
The input dc plays an important role in the amplification and contributes in increasing its level to the ac
domain where the conversion will become as =Po(ac)/Pi(dc)
The superposition theorem is applicable for the analysis and design of the dc & ac components of a BJT
network. It permits the separation of the analysis of the dc & ac responses of the system.
In other words, one can make a complete dc analysis of a system before considering the ac response.
Once the dc analysis is complete, the ac response can be determined by doing a complete ac analysis.
Important Parameters for the ac analysis
Zi, Zo, Av, Ai are important parameters for the analysis of the AC characteristics of a
transistor circuit.
Using equivalent circuit
Zi = Vi/Ii where Ii= (Vs-Vi)/Rsense
Where Rsense is very low value resistor
used to measure input current
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Zo= Vo/Io where Io= (V-Vo)/Rsense
Where Rsense is very low value resistor
used to measure output current
Av = Vo/Vi
AVNL = Vo/Vi with RL = infinite
AVNL > AVLoad
Ai = Io/Ii
It also can be calculated as
Ai = -AvZi/RL
Phase Relationship
The phase relationship between input and output signal depends on the amplifier
Common Emitter : 180 degrees Common - Base : 0 degrees
Common Collector: 0 degrees
AC analysis using equivalent circuit:-
Schematic symbol for the device can be replaced by this equivalent circuit. Basic methods of circuit analysis
are applied.
DC levels are important to determine the Q-point. Once determined, the DC level can be ignored in the AC
analysis of the network.
Coupling capacitors & bypass capacitor are chosen, to have a very small reactance at the frequency of
applications.
The AC equivalent of a network is obtained by:
Setting all DC sources to zero & replacing them by a short-circuit equivalent.
Replacing all capacitors by a short-circuit equivalent.
Removing all elements bypassed by short-circuit equivalent.
Redrawing the network.
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The first step in modeling the ac behavior of the transistor is to determine its ac equivalent circuit and
use it to replace the transistor circuit symbol in the schematic. Normal circuit analysis is then performed.
To explain the transistor operation during small signal analysis, one of three models are usually used: the re
model, the hybrid model, and the hybrid equivalent model. The re model is a reduced version of the hybrid model which is exclusively used for high frequency analysis.
Disadvantage
Re model- It fails to account the output impedance level of device and feedback effect from output to input.
Hybrid equivalent model-It is limited to specified operating condition in order to obtain accurate result.
A device model is a combination of properly chosen circuit elements that best approximates the
actual behavior of the device under specific operating conditions.
The subsequent figures shows an example of how a typical CE circuit is usually converted to its ac
equivalent circuit. This is achieve by setting all DC sources as ground potential (or ac ground) and capacitors
as ac shorts and with small signal ac modeling of a transistor circuit
Short out capacitors
Set Vdc to ac gnd
Re-arranging
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3.2 The Hybrid Model The hybrid model is used for high frequency modeling of the transistor. We will apply this to frequency analysis discussions later on.
The re Model This model is more suitable for when transistor circuit is used at dc and low frequencies (e.g. audio). Its the same as the hybrid model except that the high frequency components are not included
Transistor Models
In this session, we will only be looking re model, and hybrid equivalent model..
The re transistor model
Common Base PNP Configuration
Transistor is replaced by a single diode between E & B, and control current source between B & C.
Collector current Ic is controlled by the level of emitter current Ie.
For the ac response the diode can be replaced by its equivalent ac resistance.
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The ac resistance of a diode can be determined by the equation;
where ID is the dc current through the diode at the Q-point.
Input impedance is relatively small and output impedance quite high.
Input impedance ranges from a few to max 50 . Typical values are in the M .
The common-base characteristics
Voltage Gain
EI
mVre
26
CBi reZ
CBZo
re
RA
re
R
rI
RI
V
VA
rI
ZI
ZIV
RI
RI
RIV
LV
L
ee
Le
i
O
V
ee
ie
iii
Le
LC
Loo
:gain voltage
: ageinput volt
)(
: tageoutput vol
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Current Gain
The fact that the polarity of the Vo as determined by the current IC is the same as defined by figure below.
It reveals that Vo and Vi are in phase for the common-base configuration.
Approximate model for a common-base npn transistor configuration
Example 1: For a common-base configuration in figure below with IE=4mA, =0.98 and AC signal of 2mV
is applied between the base and emitter terminal:
a) Determine the Zi b) Calculate Av if RL=0.56k
c) Find Zo and Ai
Solution:
1i
e
e
e
C
i
oi
A
I
I
I
I
I
IA
e
b b
c
ec I I
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common-base re equivalent cct
re
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ei
m
m
43.845.6
)56.0(98.0
r
R b)
e
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kA
98.0
Zc) o
i
oi
I
IA
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Example 2: For a common-base configuration in previous example with Ie=0.5mA, =0.98 and AC signal of
10mV is applied, determine:
a) Zi b) Vo if RL=1.2k c) Av d)Ai e) Ib
Common Emitter NPN Configuration
Base and emitter are input terminals.
Collector and emitter are output terminals.
Substitute re equivalent circuit
Current through diode
Input impedance
The output graph
205.0
10 Za)
:Solution
i
m
m
I
V
e
i
88mV5
(1.2k)0.98(0.5m)
RIRI b) LeLcoV8.58
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o
98.0A d) i
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ee
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bc II
bbe
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III
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Output impedance Zo
Voltage Gain
Current Gain
re model for common-emitter
re model for common-emitter
Example 3: Given =120 and IE(dc)=3.2mA for a common- emitter configuration with ro= , determine:
a) Zi b)Av if a load of 2 k is applied c) Ai with the 2 k load
Example 4: Using the npn common-emitter configuration, determine the following if =80, IE(dc)=2 mA and
ro=40 k .
a) Zi b) Ai if RL =1.2k c) Av if RL=1.2k
bI
c
e
bIi=I
b
re model for the C-E transistor configuration
re
ro
e
0AbI
c
e
bI
i=I
b
re
ro
e
Vs=0V
= 0A
impedance)high cct,(open Z
the thusignored is r if
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kk
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Common Collector Configuration
For the CC configuration, the model defined for the common-emitter configuration is normally applied rather
than defining a model for the common-collector configuration.
3.3 The re Transistor Model for the CE Fixed Biased Configuration
and ro are given in spec sheet; and re is determined from dc analysis
Ac analysis
Input impedance, Zi =Vi/Ii
From the figure, it is clear that, Ii = IRB+IB = Vi/RB+Vi/re = Vi(1/RB +1/re) Zi = Vi/Ii= (1/RB +1/re) i.e Zi = RB// re
6.8913
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e
oL
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Output impedance Zo=Vo/Io
By keeping Vi=0, Ii=0 So IB =0 i.e open in output side. Output current Io= Vo(1/ro+1/Rc)
So, Output impedance Zo = Vo/Io = Rc//ro Rc when ro>>Rc or ro>10Rc
Voltage gain Av=Vo/Vi
Vo= -IB(Rc//ro) = -(Vi/ re )(Rc//ro) by replacing IB= Vi/ re = -Vi Rc//ro /re
Av= Vo/Vi =-( Rc//ro)/re -Rc/re
-ive sign indicates the 1800 phase shift between input & output voltage signal
Current gain Ai =Io/Ii
Ii =Vi/Zi
Io=Vo/RL
Ai= Io/Ii = (Vo/RL)/(Vi/Zi)= -AvZi/RL
3.4 Ac analysis for Voltage divider circuit
Example
It is similar to that of fixed bias circuit with RB is replaced by R1//R2
So
Zi = R1//R2 //re here R1//R2 is comparitely smaller value than that of RB in fixed n\bias. So it may not be possible to ignore R1//R2 in calculation of Zi. So Zi with voltage divider is lesser than that of fixed bias
Zo =Rc//roRc Same as that of fixed bias. Av =-(Rc//ro)/re Rc/re Same as that of fixed bias
Hybrid Equivalent Model The hybrid parameters: hie, hre, hfe, hoe are developed and used to model the transistor.
These parameters can be found in a specification sheet for a transistor.
hi = input resistance hr = reverse transfer voltage ratio (Vi/Vo) hf = forward transfer current ratio (Io/Ii)
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ho = output conductance
General h-Parameters for any
Transistor Configuration
Vi =f1 (Ii, Vo) and Io =f2((Ii, Vo)
Vi = h11Ii+h12Vo
Io=h21Ii+h22Vo
Where
h11 = Vi/Ii with Vo=0 ie short circuit input resistance, unit , & designated as hi h12 = Vi/Vo with Ii=0 ie open circuit reverse transfer voltage ratio, unitless, & designated as hr
h21 = Io/Ii with Vo=0 ie short circuit forward transfer current ratio, unitless, & designated as hf
h22 = Vo/Io with Ii=0 ie open circuit output conductance, unit Siemens & designated as ho
So hi,hr,hf & ho are called hybrid parameters
By placing second subscript as b for CB, c for CC and e for CE, we can get hybrid parameters for each
configuration.
Simplified General h-Parameter Model
The above model can be simplified based on these approximations:
hr =0 therefore hrVo = 0 and 1/ho =
Common-Emitter h-Parameters
ac fe
hie =25mV/IBQ =hfeIBQ/IEQ
hfe =ac
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Common-Base h-Parameters
hib =25mV/IEQ
hib=-ac = -1
3.5 Common-Emitter (CE) Fixed-Bias Configuration
The input (Vi) is applied to the base and the output (Vo) is from the collector.
The Common-Emitter is characterized as having high input impedance and low output
impedance with a high voltage and current gain.
Determine hfe, hie, and hoe:
hfe and hoe: look in the specification sheet for the transistor or test the transistor using
a curve tracer.
hie: calculate hie using DC analysis:
hie =25mV/IBQ =hfe25mV/IEQ
Input impedance Zi = RB//hie hie if RB > 10hie
Output impedance Zo= Rc//(1/ho) Rc if 10Rc
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Voltage gain hie
hfeRc
hie
hohfeRcAv
)/1//(
if 1/ho>10Rc
Current gain Ai hfe if RB>10hie & 1/ho>10Rc
Or Ai = -AvZi/Rc
Phase Relationship
The phase relationship between input and output is 180 degrees. The negative sign used in
the voltage gain formulas indicates the inversion.
CE Voltage-Divider Bias Configuration
Input impedance Zi = R1//R2//hie
Output impedance Zo= Rc//(1/ho) Rc if 10Rc 10Rc
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Current gain Ai hfe if RB>10hie & 1/ho>10Rc
Or Ai = -AvZi/Rc Phase Relationship
A CE amplifier configuration will always have a phase relationship between input and
output is 180 degrees. This is independent of the DC bias.
CE Emitter-Bias Configuration
Unbypassed RE
Input impedance Zi = RB//hie+(1+hfeRE)
Output impedance Zo= Rc//(1/ho) Rc if 10Rc 10Rc, 1+hfe hfe
Current Gain Ai =
ieCoe
oefe
hRRRh
hRRh
IiIo)2//1(()1(
1)2//1(/ = -AvZi/Rc
For Emitter follower circuit
Zi = RB//(1+)(re+RE) = RB// RE if RE > 10re & 10RE
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3.6 For CB amplifier
Zi = RE//re =re ( hib in hybrid eq ckt)
Zo = Rc//ro=Rc
AV = Rc/re ( Rc/hib)
Effect of load resistance and source impedance:-
AvNL >AVL>AVs
Both load resistance & source impedance reduces the gain
If load resistance is very low compare to Zo, gain reduces drastically
If source impedance is very high compare to Zi gain reduces.
If we consider output equivalent circuit as voltage source with value AvNL , inseries with output
impedance Zo we can find the reduction factor as below
If AVNL is the no load gain, AVL is the gain with load RL, AVs is the gain with load and source resitance
rs
Then
RoR
R
rsZi
ZiAV
RoR
RAVAVs
rsZi
ZiAVAV
L
L
LN
L
LL
LNL
Analog Electronic Circuits 10