BPHCT-143

DIGITAL AND ANALOG

Indira Gandhi National
Open University
CIRCUITS AND
School of Sciences INSTRUMENTATION

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Volume – II Blocks 3 & 4

 Volume II
Block 3 ANALOG CIRCUITS

Block 4 OPERATIONAL AMPLIFIER

AND INSTRUMENTATION
 BPHET-143
DIGITAL AND ANALOG
Indira Gandhi National
Open University
CIRCUITS AND
School of Sciences INSTRUMENTATION

Block

3
ANALOG CIRCUITS
UNIT 10
Amplifiers 7
UNIT 11
Oscillators 37
UNIT 12
Regulated Power Supply 53
Course Design Committee
Prof. A. K. Ghatak, Retd. Dr. Parthasarathy Prof. Shubha Gokhale
IIT Delhi, New Delhi Dept. of Physics, School of Sciences
Maharaja Agrasen College, IGNOU, New Delhi
Prof. Suresh Garg, Retd.
University of Delhi, Delhi
School of Sciences Prof. Sanjay Gupta
IGNOU, New Delhi Prof. M.S. Nathawat School of Sciences
Vice Chancellor, Former Director, IGNOU, New Delhi
Usha Martin University School of Sciences,
Dr. Subhalakshmi Lamba
IGNOU, New Delhi
Prof. R.M. Mehra, Retd. School of Sciences
Dept. of Electronics, Prof. Vijayshri IGNOU, New Delhi
South Campus, School of Sciences
University of Delhi, Delhi IGNOU, New Delhi
Dr. Ashok Goyal, Retd. Prof. Sudip Ranjan Jha
Dept. of Physics, Hansraj College School of Sciences
University of Delhi, Delhi IGNOU, New Delhi

Block Preparation TeamProf. S. Gokhale

School of Sciences
IGNOU, New Delhi
Prof. Shubha Gokhale (Units 10-12)
School of Sciences
IGNOU, New Delhi
The material of this block is based on Block 2 of the earlier B.Sc. Physics elective
PHE-10 entitled Electrical Circuits and Electronics, edited by late Prof. Abhai Mansingh
of the University of Delhi and coordinated by Dr. Shubha Gupta.

Course Coordinators: Prof. Shubha Gokhale, Prof. Vijayshri

Block Production Team

AR (P), IGNOU
Acknowledgement: Shri Gopal Krishan Arora, EDP, SOS for CRC preparation.
May, 2022
© Indira Gandhi National Open University, 2022
ISBN:
Disclaimer: Any materials adapted from web-based resources in this module are being used for educational
purposes only and not for commercial purposes.
All rights reserved. No part of this work may be reproduced in any form, by mimeograph or any other means,
without permission in writing from the Copyright holder.
Further information on the Indira Gandhi National Open University courses may be obtained from the
University’s office at Maidan Garhi, New Delhi-110 068 or the official website of IGNOU at www.ignou.ac.in.
Printed and published on behalf of Indira Gandhi National Open University, New Delhi by Prof. Sujatha
Varma, Director, SOS, IGNOU.
Printed at
 CONTENTS
Block and Unit Titles 1
Credit page 2
Contents 3
BLOCK 3: ANALOG CIRCUITS 5

Unit 10 Amplifiers 7
10.1 Introduction 8
10.2 Classification of Amplifiers 9
10.3 Small Signal Low Frequency Amplifier 11
10.3.1 Coupling and Bypass Capacitors 13
10.3.2 Frequency Response of an RC-Coupled Amplifier 13
10.4 Multistage Amplifiers and Coupling 16
10.4.1 Gain of a Multistage Amplifier 16
10.4.2 Coupling of Amplifier Stages 18
10.5 Power Amplifiers 21
10.5.1 Class A Power Amplifier 22
10.5.2 Push-Pull Amplifier 26
10.6 Amplifier Performance under Negative Feedback 29
10.6.1 Negative Feedback and its Effect on Amplifier Performance 30
10.7 Summary 32
10.8 Terminal Questions 32
10.9 Solutions and Answers 33

Unit 11 Oscillators 37
11.1 Introduction 38
11.2 Positive Feedback and Oscillations 38
11.2.1 Oscillations in Tuned Circuits 40
11.2.2 Positive Feedback Amplifier as an Oscillator 42
11.3 LC Oscillators 43
11.3.1 Tuned-Collector Oscillator 43
11.3.2 Hartley Oscillator 45
11.3.3 Colpitts Oscillator 46
11.4 RC Oscillators 47
11.4.1 Phase-Shift Oscillator 47
11.4.2 Wien Bridge Oscillator 48
11.5 Summary 50
11.6 Terminal Questions 50
11.7 Solutions and Answers 51
3
 Unit 12 Regulated Power Supply 53
12.1 Introduction 54
12.2 DC Power Unit 55
12.2.1 The Transformer 55
12.2.2 Half-Wave Rectification 56
12.2.3 Full-Wave Rectification 60
12.3 Rectifier Performance 64
12.3.1 Performance of Half-Wave Rectifier 66
12.3.2 Performance of Full-Wave Rectifier 67
12.4 Filter Circuits 71
12.4.1 Capacitance Filter 71
12.4.2 Inductance Filter 72
12.4.3 LC Filter 73
12.5 Regulation of Output Voltage 75
12.5.1 Voltage Regulator Parameters 75
12.5.2 Principle of Voltage Regulation 76
12.5.3 Shunt Voltage Regulator 78
12.5.4 Series Pass Voltage Regulator 80
12.5.5 Variable Voltage Supply using Potential Divider 83
12.6 Summary 84
12.7 Terminal Questions 84
12.8 Solutions and Answers 85

Further Readings 169

Table of Physical Constants 170
List of Blocks and Units: BPHET-143 171
Syllabus: Digital and Analog Circuits and Instrumentation (BPHET-143) 172

4
 BLOCK 3 : ANALOG CIRCUITS

In the previous blocks of this course you learnt about various semiconductor devices like
diodes, transistors and their applications in digital circuits.

The most important property of the semiconductors is their ability to function as amplifiers of
electrical voltage, current or power. When voltage or current signals are applied to the input
terminals of an amplifier, larger voltage or current signals are available at the output
terminals. In Unit 10 we will introduce you to different types of amplifiers and their
classification based on their applications, frequency range of operation, the position of
operating point (Q) on the transistor output characteristics etc. The audio circuits require a
distortion-free amplification of input signal for satisfactory listening experience. Class A
amplifiers are the most commonly used amplifiers for such applications. We will analyse the
class A amplifier and its frequency response. In order to enhance the gain of the circuit, it is
common practice to cascade multiple amplifiers together. Various methods to couple different
stages of amplifiers will be described. You will also learn to estimate the performance of a
power amplifier and get familiar with push-pull amplifier.

Feedback in electronic circuits plays a very important role. The negative feedback in an
amplifier enhances its stability and bandwidth (range of operating frequency). On the other
hand a positive feedback gives rise to oscillations in the circuits.

Sometimes, it becomes necessary to generate alternating currents of high frequencies

(ranging upto millions and even billions of cycles per second). Transistors together with
associated components, may be used to generate these alternating signals using proper
feedback network. We call such generators as oscillators. Unit 11 discusses a variety of
sinusoidal oscillators include RC oscillators and LC oscillators.

In all previous units, the voltages and currents required to operate the analog and digital
circuits were assumed to be available. We will examine how necessary voltages are
generated and the means by which they are regulated. Most electronic circuits need a dc
voltage in order to work properly. Since line voltage is alternating, the first thing that has to be
done in any electronic equipment is to convert ac voltage to dc voltage. In Unit 12 we explain
how well-regulated dc power supplies for electronic circuits can be derived from the ac mains.

We wish you all the best in your studies.

5
6
Unit 10 Amplifiers

UNIT 10
Amplifiers are used to enhance the
AMPLIFIERS
signals. Class A amplifier shown in
this figure is used in audio circuits
because of its low distortion, as you
will learn in this unit.

Structure
https://pixabay.com/photos/power-plant-
industry-chimney-2411932/

10.1 Introduction 10.5 Power Amplifiers

Expected Learning Outcomes Class A Power Amplifier
10.2 Classification of Amplifiers Push-Pull Amplifier
10.3 Small Signal Low Frequency Amplifier 10.6 Amplifier Performance under Negative
Coupling and Bypass Capacitors Feedback
Frequency Response of an RC-Coupled Negative Feedback and its Effect on Amplifier
Amplifier Performance
10.4 Multistage Amplifiers and Coupling 10.7 Summary
Gain of a Multistage Amplifier 10.8 Terminal Questions
Coupling of Amplifier Stages 10.9 Solutions and Answers

STUDY GUIDE
In this course you have so far learnt about the semiconductor devices and their electrical
characteristics. Most of the electronics systems use these devices for their functioning. The most
prominent use of transistors is in amplifier circuits that enlarge the input voltage signal or amplify the
power (voltage  current product) for driving the loads like speakers or transmitter antenna in
communication circuits. In this unit you will learn the use of transistor in different types of amplifiers. To
understand these circuits, you should revise the transistor input-output characteristics discussed in
Unit 4 and the transistor equivalent circuits described in Unit 5. You should solve all the SAQs and
Terminal Questions on your own to have better understanding of the topic.

“Negative feedback effected amplifier performance Harold Stephen

significantly.” Black

7
Block 3 Analog Circuits
10.1 INTRODUCTION
You are familiar with an „audio system‟ which is used to play CDs, cassettes
and receive radio broadcasts. Fig. 10.1a shows a simple diagram for such a
system. It consists of various input devices connected to the input of an
amplifier whose output is connected to the loudspeaker. The microphone and
other input devices convert sound into electrical signals. The loudspeaker
carries out the reverse process by converting electrical signals into sound.
The electrical output from the input devices is too weak to provide an
adequate sound level in the loudspeaker. The amplifier is used to convert the
input signal into a sufficiently powerful electrical signal for the loudspeaker.

Fig. 10.1: A block diagram for an audio system.

In Fig. 10.1 we have not shown a power supply but you must remember that
all the electronic systems need such supply. It may either be provided with
batteries or plugged into the mains supply. In Fig. 10.2 we have shown a
signal from a microphone given to an amplifier. Here the amplifier controls the
flow of power from the power supply to the loudspeaker in response to the
electrical signal received from the microphone. In this way amplifier supplies
much stronger signal power to the loudspeaker than the microphone can
provide.

Fig. 10.2: Power flow in the audio system.

8
Unit 10 Amplifiers
The function of an amplifier is to increase (boost) the level of the signal. The
amplifier circuits use transistors or integrated circuits (ICs) as amplifying
devices.

The amplifier circuits can be classified in a wide variety of ways as discussed

in Sec. 10.2. The amplifiers used for handling small amplitude and low
frequency signals are discussed in Sec. 10.3. You will also learn about their
frequency response.

There are certain conditions when a single amplifier is not enough for
providing the necessary voltage/current/power output. In such cases the
amplifying action can be augmented by connecting more than one amplifier in
succession or cascade. You will learn about such multistage amplifiers in
Sec. 10.4. Many a times we need high power output from an amplifier to drive
certain loads like radio transmitters. You will learn about such high power
amplifiers in Sec. 10.5.

By sampling a part of output signal and providing it at the input of an amplifier,

we can improve the quality of amplifier performance significantly. This is
essentially done with the help of negative feedback. You will study the effects
of negative feedback on the amplifier performance in Sec. 10.6.

Expected Learning Outcomes

After studying this unit, you should be able to:
 state the classification of amplifiers on the basis of applications,
operating point, coupling, circuit configuration, bandwidth and frequency;
 explain the variation of Q-point in a small signal class A amplifier;
 describe the frequency response curve of an RC-coupled amplifier;
 calculate the overall gain of multistage amplifier;
 draw and explain the working of single ended and push pull amplifier
circuits;
 calculate the output ac power delivered by a single-ended power
amplifier; and
 explain the advantages of negative feedback in amplifiers.

10.2 CLASSIFICATION OF AMPLIFIERS

Amplifier circuits can be classified in a wide variety of ways. They can be
classified according to their use, the type of bias used, the frequency or
bandwidth of the signals they amplify, the type of coupling used to join two
stages and their circuit configuration.

Amplifiers according to use: They fall into two main groups – voltage
amplifiers and power amplifiers. Voltage amplifiers increase the voltage level
of an applied signal. Since the output voltage of an amplifier is determined by
the voltage drop across the output load, the impedance of the load is made as
large as is practical in most voltage amplifiers.
9
Block 3 Analog Circuits
Power amplifiers are also called current amplifiers. They deliver a large
amount of current to the output load so that large power can be delivered to
the loads like speaker, transmitters etc.

Amplifiers according to bias: Amplifiers are also classified according to their

biasing conditions, or, in other words, according to the portion of the input
signal voltage cycle during which output current flows. You have learnt about
the loadline and operating point of a transistor based on the biasing conditions
in Unit 4. There are four classes of amplifiers according to bias: class A,
class B, class AB and class C.

Class A amplifiers are biased such that the Q point is in the centre of their
operating curve so that output current flows during the entire cycle of the input
voltage (see Fig. 10.3a). This results in minimum distortion of the signal at the
output and as a result, class A amplifiers are widely used in audio system,
where low distortion is important. However, these amplifiers are not energy
efficient because, even for no input signal, the transistor always conducts
current at its quiescent value.

(a) (b)

(c) (d)

Fig. 10.3: Classification of amplifier according to bias.

Class B amplifiers are biased such that the operating point is at the cutoff, so
output current flows for approximately one-half of the input signal voltage
cycle as shown in Fig. 10.3b. In effect, a class B amplifier cuts off one half of
the ac input signal waveform. When no input signal is present, no output
current flows, hence it is very efficient with low power consumption.
Class AB amplifiers are biased so that output current flows for appreciably
more than one half of the input cycle, but for less than the entire cycle as
10 shown in Fig. 10.3c. Essentially, class AB amplifiers are a compromise
Unit 10 Amplifiers
between the low distortion of class A amplifiers and the high efficiency of
class B amplifiers.

Class C amplifiers are biased beyond the cutoff so that output current only
flows during the small part of positive going peak of the input cycle (see
Fig. 10.3d). Such, amplifiers have high power outputs but they have a high
degree of distortion, which prevents their use in audio applications.

Amplifiers according to coupling: When large amplification is required, it is

practical to use cascade (or multi-stage) amplifiers. Such amplifiers are often
classified according to the way in which they are coupled. The basic coupling
methods are: resistance-capacitance (RC) coupling, transformer coupling and
direct coupling. In RC coupling, the output of first stage is coupled to the input
of next stage through a capacitor. In transformer coupling, the output of one
circuit is coupled to the input of the next by means of a transformer. In direct
coupling, the output of one stage is applied directly to the input of the next
stage.

Amplifiers according to circuit configuration: Amplifiers are also classified

on the basis of transistor configuration used. These are:

i) grounded base or common base (CB) amplifier.

ii) grounded emitter or common emitter (CE) amplifier.

iii) grounded collector or common collector (CC) amplifier.

Amplifiers according to frequency: They are classified as direct current

(dc) amplifier, audio frequency (af) amplifier, radio frequency (rf) amplifier and
video frequency (vf) amplifier. As their names imply, dc amplifiers amplify
signals of very low frequency. Audio amplifiers operate in the audio frequency
range i.e. from 20 to 20,000 Hz. Video amplifiers amplify signals from the
lower audio frequencies to as high as 4 or 5 MHz.

You must have noticed that these classifications are somewhat overlapping.
For instance, audio amplifiers may also be voltage or power amplifiers.
Similarly, an rf amplifier at the same time may be a common emitter or
common base amplifier.

SAQ 1 – Classification of amplifiers

Why are class C amplifiers preferred for amplifying the oscillator outputs?

10.3 SMALL SIGNAL LOW FREQUENCY

AMPLIFIER
In Unit 4 of this course we discussed the biasing of transistor for linear
operation. After the transistor has been biased with a Q point near the middle
of the dc load line we can couple a small ac signal into the transistor. This
produces fluctuations in the collector current of the same frequency and
shape. For example, if the input is a sine wave with a frequency of 1 kHz, the
11
Block 3 Analog Circuits
output will be an enlarged sine wave with a frequency of 1 kHz. The amplifier
is called a linear amplifier if it does not change the shape of the signal. As long
as the amplitude of the input signal is small, the transistor will use only a small
portion of the load line and the operation will be linear. For this purpose, a
class A amplifier shown in Fig. 10.4a with universal bias is used.

Look at the output characteristics of a transistor shown in Fig. 10.4b. The

operating point Q is defined by ICQ ,VCEQ and IBQ .

(a)

(b)

Fig. 10.4: a) Class A amplifier circuit; b) swing of Q-point for a sinusoidal input.

When a sinusoidal signal is applied to the base, the base current changes to
IB1 during positive half-cycle of the signal and to I B2 during the negative half.
The Q-point swings from Q to A and Q to B during this time. You can observe
that under all circumstances the operating point remains on the linear portion
of the characteristics. So an amplifier in which the operation is linear for the
applied signal, is called a small signal amplifier.

Now we will get familiar with some concepts required to analyze small signal
amplifier. We will begin with coupling capacitors, devices that allow us to
12
Unit 10 Amplifiers
couple ac signals into and out of a transistor stage without changing the dc
bias voltages.

10.3.1 Coupling and Bypass Capacitors

In Fig. 10.5 we show a small signal (class A) amplifier. The output is taken
between point D and ground. The coupling capacitor (Cc ) passes (couples)
an ac signal from collector C to output D. For this to happen, the capacitive
reactance XC must be very small at the frequency of signal being amplified.

A bypass capacitor (CE ) is similar to a coupling capacitor, except that it

couples an ungrounded emitter (E) to a grounded point. The bypass capacitor
ideally looks like a short to an ac signal. Because of this, point E is shorted to
ground as far as the ac signal is concerned. This is why we have labelled the
emitter (point E) as ac ground. A bypass capacitor will not disturb the dc
voltage at point E because it looks open to dc current and keeps the dc
operating point Q in the middle of the loadline.

Fig. 10.5: Small signal amplifier with coupling and bypass capacitors.

10.3.2 Frequency Response of an RC-Coupled Amplifier

A practical amplifier circuit is meant to raise the voltage level of the input
signal. This signal may be obtained from the CD player or the microphone in
case of a public address (PA) system. Such a signal is not of a single
frequency, but it consists of a band of frequencies. For example, the electrical
signal produced by our speech or by a musical orchestra may contain
frequencies as low as 30 Hz and as high as 15 kHz. If the loudspeakers are to
reproduce the original sound faithfully, the amplifier used must amplify all the
frequency components of the signal equally well. If it does not do so, the
output of the loudspeaker will not be an exact replica of the original sound.

The performance of an amplifier is judged by observing whether all frequency

components of the signal are amplified equally well. This information is
provided by its frequency response curve. This curve illustrates how the
magnitude of the voltage gain of amplifier ( Av ) varies with the frequency of
the input signal.
13
Block 3 Analog Circuits

Fig. 10.6: Frequency response curve of an RC-coupled amplifier.

Fig. 10.6 shows a frequency response curve of a typical RC-coupled amplifier.

You will observe that the gain is constant only for a limited band of
frequencies. This range of frequencies is called the mid-frequency range and
the gain is called mid-band gain, Avm . On both sides of the mid-frequency
range, the gain decreases. For very low and for very high frequencies, the
gain of the amplifier reduces significantly.

Low Frequency Range

In Unit 5, we analysed an amplifier circuit to determine its voltage gain. This

was the mid-frequency gain. In mid-frequency range, the coupling and bypass
capacitors are as good as short circuits. But, when the frequency is low, these
capacitors can no longer be replaced by the short circuit approximation. The
lower the frequency, the greater is the value of reactance of these capacitors,
since

1
Xc 
2 fC

This causes a significant voltage drop across Cc . The lower the frequency of
the signal, higher will be the reactance of the capacitor Cc and more will be
the reduction in output voltage. At zero frequency (dc signals), the reactance
of capacitor Cc is infinitely large (an open circuit). The effective output voltage
then reduces to zero. Thus, we see that the output voltage (and hence the
voltage gain) decreases as the frequency of the signal decreases below the
mid-frequency range.

In practical circuits, the value of the bypass capacitor C E is very large

(= 100 F) hence its effect is not significant at low frequencies. Effectively, it
is the coupling capacitor that has more pronounced effect in reducing the gain
at low frequencies.

High Frequency Range

As the frequency of the input signal increases beyond mid frequency range,
the gain of the amplifier reduces. Several factors are responsible for this
14
Unit 10 Amplifiers
reduction in gain. Firstly, the beta () of the transistor is frequency dependent.
Its value decreases at high frequencies. Because of this, the voltage gain of
the amplifier reduces as the frequency increases. When depletion layer
is formed, it has a
Another important factor responsible for the reduction in gain of the amplifier barrier potential
across it due to ions
at high frequencies is the presence of the semiconductor device itself. In a
present there. This
transistor, there exist some capacitance due to the formation of a depletion potential acts as a
layers at the junctions. These inter-electrode capacitance Cbc , Cbe , Cce are parallel plate capacitor
shown in Fig. 10.7. Note that the connection for these capacitances are shown with charge
by dotted lines to indicate that these are not physically present in the circuits, accumulation across
but are inherently present within the device. the junction.

Fig. 10.7: RC-coupled amplifier. Capacitances that affect high-frequency

response are shown by dotted connections.

The higher the frequency, the lower is the impedance offered by these
capacitors and lower will be the output voltage.

Bandwidth of an Amplifier

Frequency response curve of an RC-coupled amplifier of Fig. 10.6, shows that

the gain remains constant only for a limited band of frequencies. On both the
low-frequency side as well as on the high frequency side, the gain falls. Now,
an important question arises – where exactly should we fix the frequency
limits (of input signal) within which the amplifier will have almost constant gain
and may be called a good amplifier? The limit is set at those frequencies at
which the voltage gain reduces to 70.7% of the maximum gain Avm. These
frequencies are known as the cut-off frequencies of the amplifier. These
frequencies are marked in Fig. 10.6. The frequency f1 is the lower cut off
frequency and frequency f2 is the upper cut-off frequency. The difference of
the two frequencies, that is (f2  f1), is called the bandwidth (BW) of the
amplifier. The mid-frequency range of the amplifier is from f1 to f2 .

You may like to attempt an SAQ.

15
Block 3 Analog Circuits

SAQ 2 – Class A amplifier

a) What should be the bandwidth of an amplifier used to amplify an audio

signal?
b) Design a class A amplifier in Fig. 10.4a to get VCEQ  10 V, ICQ  10 mA.
Consider VCC  20 V , a silicon transistor with dc  100 and VB  4 V .

Let us now discuss the arrangement of amplifiers used for increasing the gain.
Why are class C amplifiers preferred for amplifying the oscillator outputs?
10.4 MULTISTAGE AMPLIFIERS AND COUPLING
An amplifier is the basic building block of most electronic systems. Just as one
brick does not make a house, a single-stage amplifier is not sufficient to build
a practical electronic system. In section 10.3 we discussed the single-stage
amplifier. The gain of a single stage is not sufficient for some applications. The
voltage level of a signal can be raised to the desired level if we use more than
one stage. When a number of amplifier stages are used in succession (one
after the other) it is called a multi-stage amplifier or a cascaded amplifier.
Much higher gains can be obtained from the multi-stage amplifier.

Let us estimate the gain of a multistage amplifier.

10.4.1 Gain of a Multistage Amplifier

A multi-stage amplifier (n-stages) can be represented by a block diagram as
shown in Fig. 10.8.

Fig. 10.8: Block diagram of a multistage amplifier having n stages.

You should note that the output of the first stage makes the input of the
second stage: the output of the second stage makes the input of the third
stage, …, and so on. The signal voltage v s is applied to the input of the first
stage. The final output v o is then available at the output terminals of the last
stage. The output of the first (or the input to the second stage) is

v1  A1 v s

where A1 is the voltage gain of the first stage. Then the output of the second
stage (or the input to the third stage) is

v 2  A2 v1  A2( A1 vs )

Similarly, the final output v o is given as

v o  v n  An v n 1

where An is the voltage gain of the last (nth) stage.

16
Unit 10 Amplifiers
We may look upon this multi-stage amplifier as a single amplifier, whose input
is v s and output is v o . The overall gain A of the amplifier is then given as

v v v v v
A  o  1  2  ...  n 1  o
v s v s v1 v n 2 v n 1

or

A  A1  A2  ...  An 1  An (10.1)

The gain of an amplifier can also be expressed in another unit called decibel.

Decibel

The gains of amplifiers are usually very large. While handling such large
numbers it is convenient to compare two powers on a logarithmic scale rather
than on a linear scale. The number of bels by which a power P2 exceeds a
power P1 is defined as

P2
Number of bels = log10
P1

For practical purposes it has been found that the unit bel is quite large.
Another unit, one-tenth of bel, is more convenient. This smaller unit is called
decibel (abbreviated as dB). Since one decibel is one-tenth of a bel, we can
write
P2
Number of dB = 10  Number of bels = 10 log10 (10.2)
P1
where P1 represents the input power and P2 the output power of an amplifier:
If V1 and V2 are the input and output voltages of the amplifier, then

V2
P1  1
Ri
and
V2
P2  2
Ro

where, Ri and Ro are the input and output impedances of the amplifier
respectively. Then, Eq. (10.2) can be written as
V 2 / Ro
Number of dB = 10 log10 2
V12 / RI

In case the input and output impedances of the amplifier are equal, i.e.
Ri  Ro  R, then Eq. (10.2) simplifies to
2
V2 V 
Number of dB = 10 log10 2  10 log10  2 
V12  V1 

V V
 10  2 log10 2  20 log10 2 (10.3)
V1 V1
17
Block 3 Analog Circuits
However, in general, the input and output impedances are not always equal.
But the expression of Eq. (10.3) is adopted as a convenient definition of the
decibel voltage gain of an amplifier, regardless of the magnitudes of the input
and output impedances.

As an example, if the voltage gain of an amplifier is 100, it can be denoted on

the dB scale as

V
Gain in dB = 20 log10 2  20 log10 100
V1

 20  2  40 dB

Gain of Multi-Stage Amplifier in dB

The gain of a multi-stage amplifier can be easily computed if the gains of the
individual stages are known in dB. The overall voltage gain in dB of a multi-
stage amplifier is the sum of the decibel voltage gains of the individual stages.
That is

AdB  AdB1  AdB2  ...  AdBn (10.4)

SAQ 3 – Gain of a multistage amplifier

A multi-stage amplifier consists of three stages. The voltage gains of the

stages are 30, 50 and 80. Calculate the overall voltage gain. Also express the
overall voltage gain in dB.

10.4.2 Coupling of Amplifier Stages

In a multi-stage amplifier, the output of one stage makes the input of the next
stage (as shown in Fig. 10.8). Can we connect the output terminals of one
amplifier to the input terminals of the next amplifier directly? This may not
always be possible due to practical difficulties. We must use a suitable
coupling network between two stages so that a minimum loss of voltage
occurs when the signal passes on to the next stage. Also, the dc voltage at
the output of one stage should not be permitted to go to the input of the next.
If it does, the biasing conditions of the next stages are disturbed.

The coupling network not only couples two stages; it also forms a part of the
load impedance of the preceding stage. Thus, the performance of the amplifier
will also depend upon the type of coupling network used. The three generally
used coupling schemes are:

i) Resistance-capacitance (R-C) coupling

ii) Transformer coupling

18 iii) Direct coupling

Unit 10 Amplifiers

Resistance-Capacitance Coupling

Fig. 10.9 shows how to couple two stages of amplifier using resistance-
capacitance (RC) coupling scheme. This is the most widely used method. In
this scheme, the signal at the collector of the first stage is coupled to the base
of the second stage through the capacitor C c . The coupling capacitor C c
blocks the dc voltage of the first stage from reaching the base of the second
stage. In this way, the dc biasing of the next stage is not interfered with. For
this reason, the capacitor C c is also called a blocking capacitor.

Fig. 10.9: Two-stage RC-coupled amplifier using transistors.

Transformer Coupling

In this type of coupling, a transformer is used to transfer the ac output voltage

of the first stage to the input of the second stage as shown in Fig. 10.10.

Fig. 10.10: Two stages of amplifier are coupled by a transformer.

19
Block 3 Analog Circuits
Here, the resistor Rc is replaced by the primary winding of the transformer.
The secondary winding of the transformer replaces the wire between the
voltage divider (of the biasing network) and the base of the second stage.

Note that in this circuit there is no coupling capacitor. The dc isolation

between the two stages is provided by the transformer itself. There exists no
dc path between the primary and the secondary windings of a transformer.
However, the ac voltage across the primary winding is transferred (with a
multiplication factor depending upon the turns ratio of the transformer) to the
secondary winding.

The main advantage of the transformer coupling over RC coupling is that all
the dc voltage supplied by Vcc is available at the collector. There is no voltage
drop across the collector resistor Rc (of RC-coupled amplifier). The absence
of resistor Rc in the collector circuit also eliminates the unnecessary power
loss in the resistor.

The transformer coupling scheme is not used for amplifying low frequency
(audio) signals. However, these are widely used in radio-frequency (rf)
amplifiers. In radio receivers, the rf ranges from 550 kHz to 3 MHz, while in TV
receivers it ranges from 54 MHz to 216 MHz. The transformer coupling
provides high gain at the desired rf frequency.

The use of a transformer for coupling also helps in proper impedance

matching. By suitably selecting the turns ratio of the transformer, we can
match any load with the output impedance of the amplifier. This help in
transferring maximum power from the amplifier to the load. This is discussed
in more details in Section 10.5 on power amplifier.

Direct Coupling

In certain applications, the signal voltages are of very low frequency. The
amplifier used for the amplification of such slowly varying signals makes use
of direct coupling. In this type of coupling scheme, the output of one stage of
the amplifier is connected to the input of the next stage by means of a simple
connecting wire.

However, the direct coupling scheme has one serious drawback. The
transistor parameters like VBE and  vary with temperature. This causes the
collector current and voltage to change. Because of the direct coupling, this
voltage change appears at the final output.

For applications where the signal frequency is below 10 Hz, coupling

capacitors and bypass capacitors cannot be used. At low frequencies, these
capacitors can no longer be treated as short circuits, since they offer
sufficiently high impedance. Fig. 10.11 shows a two stage direct coupled
20 amplifier.
Unit 10 Amplifiers

Fig. 10.11: Direct-coupling of two stages of amplifier.

Before proceeding further, you should solve an SAQ.

SAQ 4 – Direct coupled amplifier

In the circuit given in Fig. 10.11, if VCC  20 V, R1  15 k, R2  5 k,

collector current in T1 is 10 mA, VCE1  8 V,VBE1  0.6 V and its dc current gain
is 80, calculate RC1, RE1 and the dc voltage at the base of T2 .

Now, let us consider same high power amplifiers.

10.5 POWER AMPLIFIERS

In almost all electronic systems, the last stage has to be a power amplifier. For
example, in a public address system, it is the power amplifier that drives the
loudspeakers. When a person speaks into a microphone, the sound waves
are converted by it into electrical signal. This electrical signal is of very low
voltage (a few mV). This signal, if fed directly, cannot drive the loudspeakers,
to give sound (audio) output. The voltage level of this signal is first raised to
sufficiently high values (a few V) by passing it through a multi-stage voltage
amplifier. This voltage is then used to drive (or excite) the power amplifier. The A voltage amplifier
increases the voltage
power amplifier amplifies the current and hence delivers higher power to the
level of the input
loudspeakers. The loudspeakers finally convert the electrical energy into signal, but does not
sound energy. Thus, a large audience can hear the speech (or music from the increase its current. A
orchestra). power amplifier may
not increase the
In this way, a power amplifier is an essential part of most of the electronic voltage level, but
systems. increase the current
output. Effectively
The primary function of the voltage amplifier is to raise the voltage level of the
power (i.e. current
signal. It is designed to achieve the largest possible voltage gain. Only very multiplied by voltage)
little power can be drawn from its output. On the other hand, a power amplifier is large, although
is meant to boost the power level of the input signal. This amplifier can feed a there is no increase in
large amount of power to the load. To obtain large power at the output of the the voltage level.
power amplifier, its input-signal voltage must be large. That is why, in an
electronic system, a voltage amplifier precedes the power amplifier, and the
power amplifiers are called large-signal amplifiers.
21
Block 3 Analog Circuits
Now the question arises that why a voltage amplifier cannot work as a power
amplifier or in other words what is the difference between a voltage amplifier
and a power amplifier. Refer to Fig. 10.5 in Section 10.3.1.
The total dc power drawn from the supply is VCC ICQ . Out of this, only
VCEQ ICQ  is the effective dc input power to the amplifier because, at best,
that is the power that could be converted into useful ac power. The difference
of power dissipated in RC and RE which is
2 (R  R )
ICQ C E

goes waste in unnecessarily heating the resistors.

We can attempt to reduce this wastage of power. The resistor RE has to be
there in the circuit, because it is a part of the biasing network. If RE is absent,
the stabilisation of the operating point becomes poor. However, we can do
something about the resistance RC .

We can replace RC by a component whose dc resistance is zero, but ac

impedance is very high. We can do this by replacing RC by a choke (an
inductor). Two things are achieved by doing this. First, no dc voltage drop
occurs across the choke (since the dc resistance is almost zero). So we can
use a lower voltage supply VCC for the same amplifier. Second, the dc power
loss in the choke is almost nil. Thus, this circuit is much efficient as compared
to the one in Fig. 10.9.
Still more improvements can be made in this circuit so that it works as a better
power amplifier. Let is discuss it now.

10.5.1 Class A Power Amplifier

Fig. 10.12 shows a typical Class A transistor power amplifier which operates
linearly throughout the full cycle of the signal. In many electronic systems,
such as radio, television, tape recorder, public address system, etc. the final
output is in the form of sound. In such systems, the loudspeaker is the load for
the power amplifier. The power amplifier makes the final stage and it drives
the loudspeaker. In place of choke we have used a transformer in the circuit
because it provides impedance matching.

Fig. 10.12: Class A power amplifier

22
Unit 10 Amplifiers

In a power amplifier, the ac output power is sufficiently high, whereas the

power in the base circuit is quite low. The question naturally arises where the
power comes from. The only source of power in a power amplifier is the dc
supply, VCC . A portion of this dc input power appears as useful ac power
across the load RL . The rest of it is lost in the circuit. That is

dc input power = ac output power + losses in devices

or
Pi (dc)  Po (ac)  Pow er losses (10.5)

In Eq. (10.5), the input dc power is obtained from the battery or a dc power
supply. It is given by the product of voltage VCC and the average current
drawn from the supply. If the amplifier is working in class A configuration, the
average collector current will be the same as the quiescent collector current
ICQ . Therefore, the dc input power is

Pi  VCC ICQ (10.6)

For the transformer coupled amplifier, the only power lost is PD which is
dissipated by the transistor (other losses are negligible). We can now write
Eq. (10.5) as

PD  Pi  Po (ac ) (10.7)

or
PD  VCC ICQ  Po (10.7a)

This equation is very significant in the operation of a power amplifier. The

maximum power dissipation in the transistor occurs when the ac output power
is zero, in which case

PD(max)  VCC ICQ (10.8)

Now let us understand this with an example. Fig. 10.13 shows output
characteristics of a power transistor. Assume that its power dissipation rating
is 3.5 W, then we must ensure that PD does not exceed 3.5 W. We first plot its
collector dissipation curve. For this, we take some arbitrary values of VCE and
calculate corresponding values of IC so that we always have
VCE IC  PD  3.5 W. The curve obtained from these values is a hyperbola, as
shown in Fig. 10.13. If this transistor is used in a power amplifier, its Q point
must lie below this curve.

A power amplifier is said to have high efficiency, if it can convert a greater

portion of the dc input power drawn from the supply into the useful ac output
power. We define circuit efficiency as the ratio of ac power to dc input power
supplied to the collector-emitter circuit.

P (ac) Po
 o  (10.9)
Pi (dc ) VCC ICQ
23
Block 3 Analog Circuits
Efficiency is a measure of how well an amplifier converts dc power from the
supply into useful ac output power.

Fig. 10.13: Characteristics of a power transistor with its collector dissipation

curve.

Let us now analyse the circuit shown in Fig. 10.14. The output characteristics
of the transistor are given in Fig. 10.13. We shall find, for this circuit, the rms
values of collector current and voltage, and the ac power developed at the
collector and the amplifier efficiency.

Fig. 10.14: A practical class A power amplifier.

From the graph in Fig. 10.13, we can find the maximum and minimum values
of the collector current and voltage between which the signal swings. The ac
power developed across the transformer primary can be calculated to be
24
Unit 10 Amplifiers
V (peak) IC (peak )
Po (ac)  VCE (rms )  IC (rms )  CE 
2 2

But

V (max )  VCE (min )

VCE (peak )  CE
2

And

I (max )  IC (min )
IC (peak )  C
2

(VCE (max)  VCE (min) ) (IC(max)  IC(min) )

 P0 (ac )  
2 2 2 2

or

[VCE (max)  VCE (min) ] [IC(max)  IC(min) ]

P0 (ac)  (10.10)
8

The same power appears across the load RL , if the transformer is 100%
efficient. Assuming the losses in the transformer to be negligible, we can now
calculate the ac power delivered to the loudspeaker RL from Eq. (10.10).

In this case, if we consider

VCE (max)  18.0 V; VCE (min)  2.0 V

IC(max)  245 mA; IC(min)  25 mA

then,

(18.0  2.0)  (245  25)  10 3

P0 (ac) 
8

16.0  0.220

8

 0.44 W

The quiescent collector current, from the graph, is ICQ  135 mA. The dc input
power to the amplifier is given by Eq. (10.6) and in this case

Pi (dc)  VCC ICQ

 20  0.135  2.7 W

The output circuit efficiency can now be calculated by using Eq. (10.9)

P (ac) 0.44
 o   100%  16.3%
Pi (dc) 2.70

We can use Eq. (10.7) to determine the power dissipated by the transistor.

PD  Pi (dc)  Po  2.70  0.44  2.26 W 25

Block 3 Analog Circuits
From the above calculations we see that, under the conditions specified, the
collector-circuit efficiency of the single-ended power amplifier circuit is 16.3%.
This, in fact, is quite low. A large part of the net input dc power is dissipated by
the transistor as heat. By increasing the signal at the input we could obtain
more output power. This way, the efficiency also goes up. But then the
amplifier is driven into saturation (or into cut-off, or both) during a part of the
ac cycle. This will cause a large distortion in the output. To keep the distortion
minimum, the amplifier has to work under class A operation. For this
operation, the maximum theoretical efficiency is 50%. In practical circuits, the
efficiency is much less than 30%.
It is possible to have higher efficiency by working the amplifier under class B,
class AB, or even class C operation; and yet keep the distortion low. This is
achieved by a circuit called a push-pull amplifier. We shall discuss this circuit
in the next sub-section.
10.5.2 Push-Pull Amplifier
A push-pull amplifier circuit uses two transistors as shown in Fig. 10.15. This
circuit can work in class B, class AB or class A operation. Because of the
special circuit connection, it provides a low distortion, and at the same time a
highly efficient operation (class AB or class B).

The dots marked on the

primary and secondary
winding of the
transformer are the
points with same phase
of signal.

Fig. 10.15: Push pull amplifier circuit using transistors.

The matched transistors Here we use two matched transistors, T1 and T2 , with their emitter terminals
have identical connected together. The circuit has two transformers – one at the input and
characteristics like , , the other at the output. The input transformer has a centre-tapped secondary
temperature drift etc.
winding. It provides opposite polarity signals to the two transistor inputs. The
primary of the output transformer is also centre-tapped. The collector
terminals of the two transistors are connected to the supply VCC through the
primary of this transformer, that is, this primary acts as the choke in the
collector circuits.
The load resistance (usually a loudspeaker) is connected across the
secondary of the output transformer. Note that the resistors R1, R2 and RE
26 form the biasing network of both the transistors.
Unit 10 Amplifiers
To understand how opposite polarity signals appear at the two transistor
inputs when we apply a signal Vi to the amplifier inputs, refer to Fig. 10.16.
Assume that when we apply a sinusoidal signal Vi at the input, it induced
voltage of 20 V (peak) across the secondary winding (i.e. across the terminals
AB). Point C is the centre-tap of the secondary. This point is at ground (0 V
potential) for ac.

Fig. 10.16: Opposite-polarity inputs to the two transistors obtained by using a

centre-tapped transformer.

The voltages across each half of the secondary are 10 V (peak). The two
voltages add upto a total of 20 V across the whole winding. Let us consider
the instant at which the voltage at point A is 20 V with respect to point B, i.e.
when VAB  20 V. At this instant, the voltage at point A with respect to point C
is 10 V, i.e. VAC  VA  10 V. At the same time, the voltage at point C with
respect to point B is also 10 V, i.e. VCB  10 V. That is VBC  VCB  10 V.
This is the voltage at point B with respect to the ground, and this voltage VB is
appearing at the input of the transistor T2. Thus, the signals appearing at the
base of the two transistors are of opposite polarity. It can also be said that,
they are in opposite phase (phase difference of  radians). The resulting base
current of the two transistors can then be written as

i b1  Ib sin t (10.11)

ib2  Ib sin(t  ) (10.12)

We shall now see what happens at the output of the amplifier. We first
consider class A operation (although push-pull connection can be more
efficiently used in class B or class AB operation). As shown in Fig. 10.17,
under no input signal, the quiescent collector currents (ICQ ) of the two
transistors flow in opposite directions through the two halves of the primary
winding. These currents produce opposite flux through the magnetic core of
the output transformer. If the two transistors are perfectly matched, the net flux
in the core is zero.
27
Block 3 Analog Circuits

Fig. 10.17: Details of push pull operation at the output.

When an ac signal is applied to the input, opposite phased, varying base

currents flow in the two transistors. As a result, the ac collector currents in the
two transistors are also in opposite phase. The total current i c1 in transistor
T1 and the total current i c 2 in transistor T2 varies as shown in Fig. 10.18a
and b, respectively. These currents flow in opposite directions in two halves of
the primary winding. The flux produced by these currents will also be in
opposite directions. The next flux in the core will be the same as that
produced by the difference of the currents i c1 and ic 2 . To find the difference
i c1 and i c 2 , we first find the negative of i c 2 . This is done in Fig. 10.18c. We
can now add the currents of Fig. 10.18a and c to get the difference, since
ic1  ic 2  ic1  (ic 2 )

Fig. 10.18: a) Collector current ic 1 in transistor T1 ; b) Collector current ic 2 in

transistor T2 ; c) The negative of ic 2 ; d) the difference ( ic 1  ic 2 ).

The difference of the two collector currents is obtained in Fig. 10.18d. Note
that during this process the quiescent currents (ICQ ) of the two transistors get
cancelled, but the ac currents get added up. The overall operation results in a
net ac current flow through the primary of the transformer. This results in a
varying flux in the core. An ac voltage is induced in the secondary, and the ac
power is delivered to the load resistor RL .
28
Unit 10 Amplifiers
From Fig. 10.18a and b, it may be seen that during the first half-cycle, the
current i c1 increases, but at the same time the current i c 2 decreases. In other
words, when one transistor is being driven into more conduction, the other is
driven into less conduction. The reverse happens in the next half-cycle. This
amounts to saying that when the current in one transistor is “pushed-up”, the
current in the other transistor is “pulled down”. Hence, the name push-pull
amplifier.

Before proceeding further solve an SAQ.

SAQ 5 – Class A power amplifier

a) The quiescent operating collector current for the circuit shown in

Fig. 10.12 is 25 mA. If VCC  15 V, what is the input dc power to the
circuit?

b) Calculate output ac power, if the efficiency of the amplifier is 15%

The amplifier performance can improve if a feedback from its output is used to
stabilize its parameters. Let us now discuss how a negative feedback
achieves this.

10.6 AMPLIFIER PERFORMANCE UNDER

NEGATIVE FEEDBACK
Feedback simply means transferring a portion of the energy from the output of
a device back to its input. In other words, feedback is the process of taking a
part of output signal and feeding it back to the input circuit. Look at the block
diagram shown in Fig. 10.19. Let A be the gain of the amplifier when there is
no feedback.

Fig. 10.19: Schematic diagram of feedback amplifier.

A portion Vo where   1, is then applied back to the amplifier input. The
actual input to the amplifier thus consists of the sum of the signal voltage Vi
plus feedback voltage, Vf  Vo . We call  as the feedback fraction defined
as

V
 f (10.13)
Vo
29
Block 3 Analog Circuits
So the total input voltage with feedback  Vi  Vo

The output voltage with this input is Vo  (Vi  Vo ) A

i.e. Vo  AVi  AVo

as Vo  AVo  AVi

Vo (1  A)  AVi

V A
 Gain with feedback, Af  o  (10.14)
Vi 1  A

10.6.1 Negative Feedback and its Effect on Amplifier

Performance

In Eq. (10.13) if  is negative, then the feedback signal is out of phase with the
applied signal. In such a case, the net input voltage to the amplifier becomes
the difference of the external input voltage and the feedback voltage. Since
the net input to the amplifier is reduced, the output of the amplifier also
decreases. In other words, the gain of the amplifier reduces because of the
feedback. Such a feedback is called negative or degenerative feedback. By
putting  as a negative quantity into Eq. (10.14) you will get the gain with
negative feedback as

A A
Af   (10.15)
1  A( ) 1  A

Since you are dividing A by a positive quantity greater than 1, obviously

Af  A. Thus, negative feedback reduces the gain of the amplifier.

Before moving further you may solve the following SAQ.

SAQ 6 – Negative feedback gain

Calculate the gain of a negative feedback amplifier with an internal gain,

A = 100, and feedback factor  = 1/10.

In Sec. 10.3.2, we discussed the frequency response of an amplifier. The

difference (f 2  f1 ) is called the bandwidth (BW) of the amplifier. For an
amplifier, the product of gain and BW, called gain-bandwidth product,
remains constant i.e. A  BW = constant. Since the gain of the amplifier is
decreased with negative feedback, to get the product Af  BW same as
before, BW has to increase. In other words, the bandwidth of the amplifier
increases with negative feedback. Fig. 10.20 shows the frequency response of
the amplifier both, with and without feedback.
30
Unit 10 Amplifiers

Fig. 10.20: Frequency response of negative feedback amplifier.

In Eq. (10.15), if A is very large compared to 1 in the denominator, then 1

can be neglected in comparison with A. So Eq. (10.15) reduces to

A 1
Af   (10.16)
A 

Since  does not depend upon the parameters of the active device that you
have used in the amplifier, such as a transistor, the gain with feedback, Af is
almost independent of the actual gain A. On the other hand, A is dependent
upon the transistor parameters. Thus, by introducing negative feedback we
can have the gain to be independent of transistor parameters. This is known
as stabilization of amplifier gain.

Likewise other effects of negative feedback are, reduction in noise,

modification of the input and output resistances of the amplifier etc.

Thus, we have seen that the gain of an amplifier is reduced when negative
feedback is used. However, the negative feedback improves the performance
of the amplifier from so many other points of view. The advantages of negative
feedback are as follows:

i) It increases the bandwidth

ii) It improves the stability of amplifier gain

iii) It reduces distortion

iv) It increases the input impedance

v) It decreases the output impedance

Now we will recapitulate the points discussed in this unit.

31
 Block 3 Analog Circuits
10.7 SUMMARY

Concept Description

Q-point  The operating point (Q-point) with fixed I B , IC and VCE values is provided to
the transistor through biasing when no signal is applied.

Small signal  In a small signal amplifier, the signal amplitude makes the Q-point to swing on
amplifier the linear portion of the characteristics.

Power amplifier  A large signal amplifiers is the one in which the signal amplitude swings the
Q-point even over the non-linear portion of the characteristics. This produces
distortion.

Push-pull  In push-pull amplifier, the signal to be amplified is converted to two 180

amplifier degree out of phase signals that are applied simultaneously to the inputs of
two matched transistors. The amplified output signal is obtained across a
transformer connected between the collectors of two transistors.

Negative  Negative feedback reduces the gain of the amplifier and increases its
feedback bandwidth.

10.8 TERMINAL QUESTIONS

1. Why do you select the operating point on the linear portion of the
characteristics?

2. What is meant by biasing? Which biasing is the most commonly used?

3. Fill in the gap by choosing the correct option.

a) If in an RC coupled amplifier the value of a coupling capacitor is

increased to twice the original value, the low frequency response of the
amplifiers would be ……………….. (worse/the same/better).

b) In the above case the high frequency response would be …………….

(worse/the same/better).

4. An RC-coupled amplifier has a voltage gain of 100 in the frequency range

of 400 Hz to 25 kHz. On either side of these frequencies, the gain falls so
that it is reduced by 3 dB at 80 Hz and 40 kHz. Calculate gain in dB at cut-
off frequencies and also construct a plot of frequency response curve.

5. In class B push pull amplifier battery supply is 5 volts, output power is

400 mW and maximum collector dissipation of each transistor is 100 mW.
Calculate peak collector current.
32
Unit 10 Amplifiers

10.9 SOLUTIONS AND ANSWERS

Self-Assessment Questions
1. Oscillator circuits are used for generation of a single frequency signal. It is
purely sinusoidal signal of fixed frequency and hence there is no particular
need of distortion-free amplification. Class C amplifier provides good
results for such applications, even though it uses only a small fraction of
input signal for operation.

2. a) Audio signal ranges between 20 Hz to 20 kHz. Hence audio amplifier

bandwidth should be 20 kHz.

b)

Fig. 10.21: Class A amplifier design.

IC  10 mA

I 10 mA
IB  C   100 A
 100

VB  4 V

 VE  VB  VBE  4 V  0.6V  3.4 V

Considering IC  IE  10 mA

3.4 V
RE   340 
10 mA

VCC  20 V VB  4 V

VR2  VB  4 V

 VR1  20  4  16 V

If we wish to draw 100 A base current, the current through the resistor
R1 should be ~ 10 times the value of IB . Hence current through R1 ,

IR1  1mA

VR1 16 V
 R1    16 k
IR1 1mA 33
Block 3 Analog Circuits
Now current through R2 , IR2  IR1  IB  1mA  100 A  0.9 mA

VR2 4V
 R2    4.4 k
IR2 0.9 mA

Now

VE  3.4 V

The supply voltage VCC is getting divided into voltage across RC , VCE ,
and voltage across RE . It is given that VE  3.4 V and VCE  10 V .

 Voltage drop across RC

= VCC  VCE  VE

= 20  10  3.4  6.6 V

Current through RC  IC  10 mA

6.6 V
 RC   660
10 mA

With these values of R1, R2, RC and RE , we obtain operating point at

VCEQ  10 V, ICQ  10 mA and IBQ  100 A

3. Using Eq. (10.1), the overall voltage gain is

A  A1  A2  A3

 30  50  80  120000

The overall voltage gain in dB of three-stage amplifier is given as

AdB  AdB1  AdB2  AdB3

We are given the voltage gain of the individual stages as ratios. So, we
would first find the gains of the individual stages in decibels. Thus

AdB1  20 log10 30  29.54 dB

AdB2  20 log10 50  33.98 dB

AdB3  20 log10 80  38.06 dB

Therefore

AdB  29.54  33.98  38.06  101.58 dB

From the overall voltage gain also you can verify that the gain in dB is

34 AdB  20 log10 120000  101.58 dB

Unit 10 Amplifiers
4. VCC  20 V

R2 5 k
VB1  VCC   20 V  5 V
R1  R1 (5  15) k

VE1  VB1  VBE1  5 V  0.6  4.4 V

Since IC1  IE1  10 mA,

4.4 V
RE1   440 
10 mA

VRC1  VCC  (VCE1  VE1 )

= 20  (8 + 4.4) = 7.6 V
7.6 V
 RC1   760 
10 mA
and voltage at the base of Q2

VC1  VCC  VRC1

= 20 V  7.6 V

= 12.4 V

5. a) Given that,
VCC  15 V, ICQ  25 mA

From Eq. (10.6)

Pi (dc)  VCC ICQ  15 V  25 mA

= 0.375 W

b) From Eq. (10.9),

P (ac )
 o
Pi (dc )

Po (ac )    Pi (dc )  (0.15  0.375 W)  0.056 W  56 mW

6. For negative feedback, we use Eq. (10.15)

A  100;   1 / 10.

A 100 100
 Af     9.09
1  A 1  (100  0.1) 1  10

Terminal Questions
1. So that the change in the operating point does not introduce any distortion
at the output.

2. Applying various voltages to different elements of a transistor or any

device for its proper operation is called biasing. The universal or potential
divider bias is the most commonly used biasing scheme. 35
Block 3 Analog Circuits
3. a) better b) the same

4. The gain in dB is

A dB  20 log10 A  20log10100  40 dB

This is the mid band gain. The gain at cut-off frequencies is 3 dB less than
the mid band gain, i.e.

( A dB ) (at cut-off frequencies) = 40  3 = 37 dB

The plot of the frequency response curve is given in Fig. 10.22.

Fig. 10.22: Frequency response of an RC-coupled amplifier.

5. Given Vmax  VCC  5 V.

P(out )  400 mW  400  103 W

The output power is given by

I V
P  max  max
2 2

or

I
400  10 3  max  5
2

 Imax  160 mA.

36
Unit 11 Oscillators

UNIT 11
OSCILLATORS
Oscillators are used for generating
sine waves. LC circuits are used for
high frequency generation as you will
learn in this unit.

Structure
https://pixabay.com/photos/power-plant-
industry-chimney-2411932/

11.1 Introduction 11.4 RC Oscillators

Expected Learning Outcomes Phase-Shift Oscillator
11.2 Positive Feedback and Oscillations Wien Bridge Oscillator
Oscillations in Tuned Circuits 11.5 Summary
Positive Feedback Amplifier as an Oscillator 11.6 Terminal Questions
11.3 LC Oscillators 11.7 Solutions and Answers
Tuned-Collector Oscillator
Hartley Oscillator
Colpitts Oscillator

STUDY GUIDE
In the last unit you learnt about the amplifiers required for increasing the power (or voltage / current) of
a signal. You also learnt about the concept of negative feedback, and how it provides stability to the
amplifier performance. In this unit, you will learn about the use of transistors for generating ac signals of
desired frequency. Such circuits are called the Oscillators. For this you should brush up your
knowledge about RC, LC and LCR resonant circuits you worked with in the second semester courses
BPHCT-133 entitled Electricity and Magnetism and BPHCL-134 entitled Electricity and Magnetism:
Laboratory.
You will also require the knowledge of transistor biasing (mostly in CE configuration) discussed in the
earlier units of this course.
You should try to work out the SAQs, TQs and design examples given in this unit on your own in order
to get complete knowledge of the topic.

“External interference, which can never be entirely eliminated

R.V.L. Hartley
in practice, always reduces the effectiveness of the system.”

37
Block 3 Analog Circuits
11.1 INTRODUCTION
Generation of high frequencies is essential in all communication systems. For
example, in radio and television broadcasting, the transmitter radiates the
signal using a carrier of very high frequency say from 550 kHz to 22 MHz in
radio broadcasting and from 47 MHz to few GHz in TV broadcasting. In radio
and TV receivers too there is an oscillator circuit which generates very high
frequencies. Even mobile phones require oscillator circuits. The oscillators are
usually transistor amplifier circuits with positive feedback. In the last unit you
learnt about the concept of feedback.

If the portion of the output that is fed back is in phase with respect to the input,
then the feedback is termed as positive feedback. With positive feedback a
circuit can be made to generate an output with no external input. In this unit
we will use positive feedback in building oscillators.

In Sec. 11.2 we will discuss about the positive feedback and the principle
governing the oscillations. You will be able to establish the criteria for
sustained oscillations in a circuit called Barkhausen Criteria.

Based on the principle used for generation of oscillations, the oscillators are
classified as tuned circuit (LC) oscillators and RC oscillators. In Sec. 11.3 you
will learn about two important LC oscillators viz. Hartley oscillator and Colpitts
oscillator.

In Sec. 11.4 we will discuss about the RC oscillators, namely, Wien bridge
oscillator and phase shift oscillator.

Expected Learning Outcomes

After studying this unit, you should be able to:

 workout the feedback gain in case of positive feedback;

 state the conditions under which a feedback amplifier works as an oscillator;

 state the classification of oscillators;

 draw the circuit and explain the working, state the frequency relation and design
the LC oscillators (Hartley and Colpitt’s oscillators);

 explain the working of phase shift oscillator and design its feedback R-C
network; and

 explain the working and calculate the frequency of a Wien bridge oscillator.

Now, we begin with understanding the role of positive feedback in sustaining

oscillations in an oscillator circuit.

11.2 POSITIVE FEEDBACK AND OSCILLATIONS

We had discussed the concept of feedback in the last unit. You also know the
effect of negative feedback on the amplifier performance in form of reduced
38
gain and increased bandwidth.
Unit 11 Oscillators

Now, when the feedback voltage is in phase with the input signal, then it adds
to the input signal. In this case  is positive and feedback is termed as
positive or regenerative feedback. You can observe that when  is positive,
the gain with feedback can be written from Eq. (10.14) as

A
Af  (11.1)
1  A

Since A is divided by a number smaller than unity, Af  A. So positive

feedback increases the gain of an amplifier. This in turn reduces the
bandwidth (BW) because you know that the product (Gain  BW) is constant.

In Eq. (11.1) if A = 1, then Af  . In other words you have an amplifier

which gives an output without an input! Such a circuit is called as an
oscillator. You should not be under the impression that you are getting an
output power without any input power. The circuit draws the power from the dc
supply connected to the transistor, and converts it into ac power. So we can
define an oscillator as that circuit which converts dc power into ac power.
Fig. 11.1 illustrates the difference between an oscillator and an amplifier.

Fig. 11.1: Comparison between an amplifier and an oscillator.

For a circuit to work as an oscillator certain conditions have to be satisfied.

They are as follows:

1. The feedback should be positive.

2. The product A should be unity.

A = 1 (11.2)
This condition is known as Barkhausen Criteria of oscillations.

3. The circuit must amplify and the amplification should be sufficient to

overcome the losses in the circuit.
The third condition is to be satisfied in order to sustain oscillations (see
Fig. 11.2a) otherwise you will get the damped oscillations as shown in
Fig. 11.2b.
39
Block 3 Analog Circuits

Fig. 11.2: a) Sustained oscillations; b) damped oscillations.

Oscillators are mainly divided into two types, namely, sinusoidal and
relaxation oscillators. Sinusoidal oscillators produce continuously varying
signals like sine waves with single frequency. Whereas relaxation oscillators
produce non-sinusoidal signals like square waves, triangular waves etc. In this
unit we will study about a few sinusoidal oscillators. Depending upon how
oscillations are produced, sinusoidal oscillators are of the following type:

i) Tuned circuit (LC) oscillators

ii) RC oscillators

These two classes require different conditions to be satisfied by the feedback

circuits. Let us discuss these in detail now.

11.2.1 Oscillations in Tuned Circuits

An inductor and a capacitor connected in parallel form a tuned or tank circuit.
In Fig. 11.3a, energy is introduced into this circuit by connecting the capacitor

Fig. 11.3: Damped oscillation in an LC circuit.

40
Unit 11 Oscillators
to a dc voltage source, that is with switch S in position 1. The negative
terminal of the battery supplies electrons to the lower plate of the capacitor.
Because of the accumulation of electrons, the capacitor gets charged and
there is a voltage across it. We say that energy is stored in the capacitor in the
form of electric potential energy. When the switch S is thrown to position 2 as
shown in Fig. 11.3b, current starts flowing in the circuit comprising the
capacitor and inductor. The capacitor now starts discharging through the
inductor. Since the inductor has the property of opposing any change in
current, the current builds up slowly. Maximum current flows in the circuit
when the capacitor is fully discharged. At this instant, the potential energy of
the system is zero. But the electron motion being greatest (maximum current),
the magnetic field energy around the coil is maximum.

Once the capacitor is fully discharged, the magnetic field begins to collapse.
The back emf in the inductor keeps the current flowing in the same direction.
The capacitor starts charging, but with opposite polarity this time, as shown in
Fig. 11.3c. As the charge builds up across the capacitor, the current
decreases and the magnetic field decreases. When the magnetic field energy
drops to zero, the capacitor charges to the value it had in condition (a) but with
opposite polarity. Once again all the energy is in the form of potential energy.
The capacitor now begins to discharge again. This time current flows in the
opposite direction. Fig. 11.3d shows the capacitor fully discharged, and also
shows maximum current flowing in the circuit. Again, all the energy is in the
magnetic field. The interchange or “oscillation” of energy between L and C is
repeated again and again. This situation is similar to an oscillating pendulum,
in which the energy keeps on interchanging between potential energy and
kinetic energy.

In a practical pendulum, because of the friction at the pivot and the air
resistance, some energy is lost during each swing. The amplitude of each
cycle goes on decreasing and eventually the pendulum comes to rest, though
it may take a long time. The oscillations of the pendulum are said to be
damped.

An ideal LC circuit is expected to perform the oscillations forever, however, a

practical LC circuit deviates from the ideal one. The inductor coil will have
some resistance, and the dielectric material of the capacitor will have some
finite resistance, and will have some leakage. Because of these factors, some
energy loss takes place during each cycle of the oscillation. As a result of this
loss, the amplitude of oscillation decreases continuously and ultimately the
oscillations die down just like in case of practical pendulum. Thus, we find that
a tank circuit by itself is capable of producing oscillations, but they are
damped as shown in Fig. 11.3e.

Frequency of Oscillations in an LC Circuit

In LC circuit, the constants of the system are the inductance and capacitance
values. The frequency of oscillation is the same as the resonant frequency of
the tank circuit. It is given by
1
f0  (11.3)
2 LC
41
Block 3 Analog Circuits
Sustained Oscillations
The oscillations of a pendulum can be maintained at a constant level, if we
supply additional energy to it from time to time, to overcome the effect of
frictional losses.
The oscillations of an LC circuit can also be maintained at a constant level in a
similar way. For this, we have to supply a pulse of energy at the right time in
each cycle. The resulting “undamped oscillations” are called sustained
oscillations, as shown in Fig. 11.2a. Such sustained oscillations (or continuous
waves) are generated by the electronic oscillator circuits.
There are many varieties of LC-oscillator circuits. All of them have following
two features in common:
i) They must contain an active device (transistor) that works as an amplifier
to overcome the losses.
ii) There must be positive feedback in the amplifier.
11.2.2 Positive Feedback Amplifier as an Oscillator
We discussed that an oscillator generates ac output signal without any input
ac signal. A part of the output is fed back to the input; and this feedback signal
is the only input to the internal amplifier.
To understand how an oscillator produces an output signal without an external
input signal, let us consider Fig. 11.4a. The voltage source v drives the input
terminals YZ of the internal amplifier (with voltage gain A). The amplified
signal Av drives the feedback network is to produce feedback voltage Av.
This voltage returns to the point X. If the phase shift due to the amplifier and
feedback network is correct, the signal at point X will be exactly in phase with
the signal driving the input terminals YZ of the internal amplifier.

Fig. 11.4: Proper positive feedback in an amplifier makes it an oscillator.

42
Unit 11 Oscillators
Let us assume that we connect points X and Y and remove voltage source v.
The feedback signal now drives the input terminals YZ of the amplifier (see
Fig. 11.4b). If A is less than unity, Av is less than v, and the output signal
will die out as shown in Fig. 11.4c. This happens because enough voltage is
not returned to the input of the amplifier. On the other hand, if A is greater
than unity, Av is greater than v, and the output voltage builds up as shown in
Fig. 11.4d. Such oscillations are called growing oscillations. Finally, if A
equals unity, no change occurs in the output and we get an output with
constant amplitude as shown in Fig. 11.4e.
You must be wondering that how do we get the initial voltage v, which is
required to start the oscillations, if no external source is connected. In
practice, this starting voltage required to initiate the oscillations is actually
generated within the circuit itself by the components like resistors used in it.
Every resistor has some free electrons in it. Due to the room temperature,
these electrons gain some energy to move around randomly in the resistor.
This causes generation of a noise voltage across the resistor. This noise can
contain many frequencies due to its random nature. So a resistor can be
considered as a source of voltage generated due to thermal noise.
When we turn the power to the circuit on, the only signal existing in the circuit
is this noise generated voltage. At the output of the amplifier, this noise
voltage appears as an amplified voltage containing many frequencies. This
voltage is fed to the input of the amplifier via the feedback network and this
starts the oscillations. At this stage A product is little more them unity. Due to
frequency selective nature of the feedback network, only single frequency
oscillations sustain in the oscillator and eventually the Barkhausen Criteria
(A = 1) condition is satisfied at this frequency.
You may now like to attempt an SAQ.

SAQ 1 – Barkhausen criteria

For an oscillator circuit if  factor is 5% of the output voltage, what should be
the gain of the amplifier in order to obtain sustained oscillations?

Now we will discuss about some tuned circuit (LC) oscillators.

11.3 LC OSCILLATORS
LC oscillators or resonant-circuit oscillators are widely used for generating
high frequencies. With practical values of inductors and capacitors, it is
possible to produce frequencies as high as 500 MHz. The oscillators used in
rf (radio frequency) generators, radio and TV receivers etc. are LC oscillators.
Such an oscillator has an amplifier, an LC resonant circuit and a feedback
arrangement. There is a large variety of LC-oscillator circuits. Here, we shall
discuss only a few important ones.
11.3.1 Tuned-Collector Oscillator
Fig. 11.5 shows a basic LC-oscillator circuit. It is called tuned-collector
oscillator, because the tuned (LC) circuit is connected to the collector. We use
a transformer here. The primary of the transformer and the capacitor form the
43
Block 3 Analog Circuits
tuned circuit (or tank circuit) which decides the frequency of oscillation. The
secondary winding is connected to the base. You are aware that the voltage in
primary and secondary winding of the transformer can have 0 or 180 phase
difference depending upon the direction of windings of the primary and
secondary coils. On the transformer symbol, the points of equal phase are
indicated by the dots near the ends of the two windings. In present case,
shown in Fig. 11.5, the primary and secondary voltages are out of phase
i.e. have 180 phase difference. Since a phase difference of 180 is provided
by the CE transistor amplifier, and an additional 180 by the transformer, the
type of feedback is positive. The transistor amplifier provides sufficient gain for
oscillator action to take place.

Fig. 11.5: Tuned-collector oscillator.

Resistors R1 R2 and RE provide dc bias to the transistor. The capacitors
CE and C2 bypass resistors RE and R2 , respectively for ac signal. It is for
this reason, the resistors RE and R2 have no effect on the ac operation of
the circuit. The dc bias voltage set by the potential divider R1 and R2 is
connected to the base through the low-resistance secondary winding of the
transformer. This secondary of the transformer provides ac feedback voltage
which is 180 out of phase than the output. This voltage appears across the
base-emitter junction, since the junction point of R1 and R2 is at ac ground
(due to bypass capacitor C2 ) .
The moment we switch on the supply, the current starts building up through
the primary winding of the transformer connected to the collector. This induces
a varying voltage in the secondary of the transformer. An amplified voltage
again appears in the tank circuit, which responds most to its resonant
frequency. Because of the sufficient gain provided by the transistor, and the
proper amount of feedback in the correct phase, the oscillations grow till a
certain voltage level is reached. Thus, sustained oscillations are obtained at
the resonant frequency
1
f0  (11.4)
2 LC

44 Now you may like to attempt an SAQ.

Unit 11 Oscillators

SAQ 2 – Tuned collector oscillator

What will be the effect on voltage fed back at the base, if capacitor C2 is not
connected in the circuit shown in Fig. 11.5?

11.3.2 Hartley Oscillator

The Hartley oscillator is one of the simplest types of tuned oscillator circuits. In
this circuit only one coil is used, which is tapped such that a portion L1 of the
coil is in the collector circuit, while L2 is in the base circuit with the tapped
point connected to the ground. The amplified energy in the collector section is
fed back to the base by means of inductive coupling and amount of coupling
will depend upon the number of turns in L1 and L2 . Fig. 11.6 shows a Hartley
oscillator circuit.

Fig. 11.6: Hartley oscillator.

An RFC (radio frequency choke) permits an easy flow of dc current. At the

same time, it offers very high impedance to high frequency currents. In other
words, an RFC ideally looks like a dc short and an ac open. Hence the high
frequency signal at the collector does not reach the dc power supply and
disturb the dc biasing of the circuit. The presence of the coupling capacitor Cc
in the output circuit of the Hartley oscillator does not permit the dc currents to
go to the tank circuit. The radio-frequency energy developed across the RFC
is capacitively coupled to the tank circuit through the capacitor Cc .

In the tank circuit, the inductors L1 and L2 are in series; so effective

inductance

L  L1  L2 (11.5)

Hence, the resonance frequency can be expressed as

1
f  (11.6)
2 (L1  L2 ) C
45
Block 3 Analog Circuits
11.3.3 Colpitts Oscillator

The Colpitts oscillator shown in Fig. 11.7 is a widely used circuit in commercial
signal generators above 1 MHz. The oscillator is similar to the Hartley
oscillator given in Fig. 11.6 The only difference is that the Colpitts oscillator
uses a split-tank capacitor instead of a split-tank inductor. The RFC has the
same function as in the Hartley oscillator. The voltage developed across the
capacitor C2 provides the regenerative feedback required for the sustained
oscillations to the base circuit.

Fig. 11.7: Colpitts oscillator using a transistor.

The values of L, C1 and C2 determine the frequency of oscillation. The

frequency of oscillation is given by

1
f  (11.7)
2 LC

where

C1C2
C (11.8)
C1  C2

since C1 and C2 are in series.

You may now like to attempt an SAQ.

SAQ 3 – LC oscillators
Considering the condition for starting the oscillations is A > 1, what should
be the condition on the value of gain A in terms of values of capacitors C1
and C2 in a Colpitts oscillator?

46
Unit 11 Oscillators

11.4 RC OSCILLATORS
Till now we have discussed only those oscillators which use an LC-tuned
circuit. These tuned circuit oscillators are good for generating high
frequencies. But for low frequencies (say, audio frequencies), the LC circuit
becomes impracticable (due to extremely high values of L and C). In such
cases, RC oscillators are more suitable. There are many types of RC
oscillators, but the following two are the most important:

i) Phase-shift oscillator

ii) Wien bridge oscillator

Before discussing the particular RC oscillators let us understand the working

principle of these oscillators.

Basic Principles of RC Oscillators

We know that a single stage of an CE amplifier not only amplifies the input
signal but also shifts its phase by 180. If we take a part of the output and
directly feed it back to the input, a negative feedback takes place. The net
output voltage then decreases. But for producing oscillations we must have
positive feedback (of sufficient amount). Positive feedback occurs only when
the feedback voltage is in phase with the original input signal. This condition
can be achieved in two ways. We can take a part of the output of a single
stage amplifier (giving a phase shift of 180) and then pass it through a phase-
shift network giving an additional phase shift of 180. Thus a total phase shift
of 180 + 180 = 360 (which is equivalent to a phase shift of 0) occurs, as
the signal passes through the amplifier and the phase-shift network. This is
the principle of phase-shift oscillator.

Another way of getting a phase shift of 360 is to use two stages of amplifiers
each giving a phase shift of 180. A part of this output is fed back to the input
through a feedback network without producing any further phase shift. This is
the principle of Wien bridge oscillator.

11.4.1 Phase-Shift Oscillator

Fig. 11.8 shows a phase-shift oscillator. It consists of amplifier with three

sections of RC sections as feedback network.

The phase of the signal at the input (B) gets reverse at output (C), when it is
amplified by the amplifier. The output of the amplifier goes to a feedback
network. The feedback network consists of three identical RC sections. Each
RC section provides a phase of 60 Thus a total of 60  3 = 180 phase shift
is provided by the feedback network. The output of this network is now in the
same phase as the originally assumed input to the amplifier, as shown in the
figure. If the condition A = 1 is satisfied, oscillations will be maintained.
47
Block 3 Analog Circuits

Fig. 11.8: Phase-shift oscillator.

It may be shown by a straightforward (but a little complicated) analysis that the

frequency at which this RC network provides exactly 180 phase-shift is given
by

1
f  (11.9)
2 RC 6

Hence, this must be the frequency of oscillation.

SAQ 4 – Phase-shift oscillator

A transistor phase-shift oscillator uses three identical RC sections in the
feedback network. The values of the components are R = 100 k and
C = 0.01 F. Calculate the frequency of oscillation.

11.4.2 Wien Bridge Oscillator

The Wien bridge oscillator is a standard circuit for generating low frequencies
in the range of 10 Hz to about 1 MHz. It is used in most of the commercial
audio generators. Basically, this oscillator consists of two stages of RC-
coupled amplifier and a feedback network. The block diagram of Fig. 11.9
explains the principle of working of this oscillator.
Here, the blocks A1 and A2 represent two amplifier stages with each
providing 180 phase shift. The output of the second stage goes to the
feedback network. The voltage across the parallel combination C2 R2 is fed to
the input of the first stage. The net phase shift through the two amplifiers is
zero. Therefore, it is evident that for the oscillation to be maintained, the
phase shift through the coupling network must be zero. It can be shown that
this condition occurs at a frequency given by

1
f  (11.10)
2 R1C1R2C2
48
Unit 11 Oscillators

Fig. 11.9: Block diagram of a basic Wien bridge oscillator.

To have a gain we add some amount of negative feedback. The addition of
negative feedback modifies the circuit in Fig. 11.9 to that shown in
Fig. 11.10a.

Here we are giving the

positive and negative
feedbacks to two
separate inputs of an
amplifier. You will
learn about such
amplifiers in Unit 13,
where we discuss the
operational amplifiers.

Fig. 11.10: Wien bridge oscillator with negative feedback; b) same circuit
redrawn to depict the “bridge” in the circuit. 49
 Block 3 Analog Circuits
The same circuit is redrawn in Fig. 11.10b. The two amplifier blocks in
Fig. 11.10a representing the two stages of the amplifier are replaced by a
single block in Fig. 11.10b. You may now see why this circuit is called a bridge
oscillator. In this circuit, the resistors R3 and R 4 provide the desired negative
feedback.

We can have a continuous variation of frequency in the oscillator by varying

the two capacitors C1 and C2 simultaneously. These capacitors are variable
air-gang capacitors. We can change the frequency range of the oscillator by
switching into the circuit with different values of resistors R1 and R2 .

SAQ 5 – Wien bridge oscillator

The RC network of a Wien bridge oscillator consists of resistors and

capacitors of values R1  R2  220 k and C1  C2  250 pF. Calculate the
frequency of oscillations.

11.5 SUMMARY

Concept Description

Positive  If the feedback signal is in phase with the applied signal and aids it, positive
feedback or regenerative feedback takes place.

Gain with  Gain increases with positive feedback, which may lead to oscillations. The
positive gain is given by
feedback
A
Af 
1  A

Oscillator  An oscillator acts as energy converter which changes direct current energy
into alternating current energy.

Components of  Essential parts of an oscillator are (i) the frequency determining network
oscillator circuit (ii) source of dc energy and (iii) a feedback circuit to provide positive
feedback.

LC oscillators  Hartley oscillator uses a tapped coil in the feedback circuit.

 Colpitt’s oscillator uses a tapped capacitance network in the feedback circuit.
It has better frequency stability than Hartley oscillator.

RC oscillators  Phase-shift oscillator has three RC sections to provide 180 phase shift in
feedback voltage.
 A Wien bridge oscillator is an RC oscillator whose frequency of oscillation can
be varied over a wide range.

11.6 TERMINAL QUESTIONS

1. What is meant by loop gain?
2. A Wien bridge oscillators uses 10 k resistors and 4.70 nF capacitors in
50 its bridge circuit. What is the frequency of oscillation?
Unit 11 Oscillators
3. In Hartley oscillator the input and output inductors are 0.1 mH each and
capacitor is 0.1 nF. Calculate the frequency of this oscillator.

11.7 SOLUTIONS AND ANSWERS

Self-Assessment Questions
1.  = 5% = 0.05
By Barkhausen criteria,
A = 1 for sustained oscillations
1 1
 A   20.
 0.05

2. If capacitor C2 is not connected across resistor R2 , the complete

feedback voltage induced in the secondary of the transformer would not
go the input of the transistor because some of this voltage will drop across
R2 .

3. To start the oscillations, initial condition is A > 1. Hence

1
A

In the circuit of Colpitts oscillator (Fig. 11.7), the output voltage appears
across C1 , whereas the feedback voltage across C2 . Since, same current
is flowing through these capacitors in the tank circuit made up of C1 and
C2 , we can write

vf i XC2 XC2
  
v out i XC1 XC1

1 (2 fC2 ) C1
 
1 (2 fC1) C2
In order to start the oscillations, we need
C2
A
C1
4. The frequency of oscillation of a phase-shift oscillator is given as
1
f 
2 RC 6

Here, R  100 k  10 5 ; C  0.01 F  10 8 F.

Therefore,
1
f   64.97 Hz
2  3.141  10  108  2.452
5

5. For a Wien bridge oscillator, the frequency of oscillation is given as

1 1
f0  
2 R1R2C1C2 2RC

where R1  R2  R and C1  C2  C.
51
Block 3 Analog Circuits
Here, R  220 k  2.2  105 , C  250 pF  2.5  10 10 F.

Therefore,

1
f0 
2  3.141  2.2  105  2.5  10 10

 2893.7 Hz

 2.89 kHz.

Terminal Questions
1. The product A is known as loop gain. The input signal is multiplied by A
times in passing through the amplifier and  times in the feedback network
before it arrives at the input. Hence the name loop gain.
1
2. f  ; R  10 k  10  10 3 
2 RC

C  4.7 nF  4700 pF.

1
f 
2  3.142  10  10 3  4700  10 12

 3386 Hz
3. For Hartley oscillator
1
f 
2 (L1  L2 ) C

1

2 (0.1  0.1)  10 3  0.1  10 9

1 107
   1.125 MHz.
2 0.2  0.1 10 12 2 2

52
Unit 12 Regulated Power Supply

UNIT 12
REGULATED
Power supplies provide dc voltage to
the system. While obtaining dc from
ac, rectifiers are used alongwith
filters. The figure shows output
obtained from full wave rectifier and
capacitance filter as discussed in this
POWER SUPPLY
unit.

Structure
12.1 Introduction 12.5 Regulation of Output Voltage
Expected Learning Outcomes Voltage Regulator Parameters
12.2 DC Power Unit Principle of Voltage Regulation
The Transformer Shunt Voltage Regulator
Half-Wave Rectification Series Pass Voltage Regulator
Full-Wave Rectification Variable Voltage Supply using Potential Divider
12.3 Rectifier Performance 12.6 Summary
Performance of Half-Wave Rectifier 12.7 Terminal Questions
Performance of Full-Wave Rectifier 12.8 Solutions and Answers
12.4 Filter Circuits
Capacitance Filter
Inductance Filter
LC Filter

STUDY GUIDE
The power required to drive the circuits discussed so far in this course is obtained from a circuit called
power supply. It provides a dc voltage of desired value. It requires conversion of ac mains voltage
received from the electric supply company into dc voltage using the circuits called rectifier followed by a
filter. Efficiency of rectification can be calculated using simple mathematics. You will require basic
knowledge of calculus for this. The filtered voltage is then given to a circuit that regulates the output
voltage to a set value. These regulators commonly use zener diode as a constant voltage reference
device. You should revise the zener characteristics you studied in Unit 3 of Block 1 of this course to
understand its use as a voltage reference.

You should work out the SAQs and TQs given in this unit on your own before turning to the solutions
given at the end.

“Power is a neutral energy source, like tofu. It’s what you do Elizabeth
with it gives it flavour.” Gilbert

53
Block 3 Analog Circuits
12.1 INTRODUCTION
Many electronic devices, such as pocket calculators and small radios operate
on batteries that provide steady (dc) voltage and currents. Usually, we use dry
cells, but sometimes a battery eliminator is used in place of dry cells. The
battery eliminator converts the ac mains voltage into dc voltage and thus
eliminates the need for dry cells. So far, in all the previous units, where you
learnt about various circuits like amplifiers, oscillators, digital circuits it was
assumed that voltage and currents required to operate them were available.
For the operation of most of the devices in electronic equipment, a dc voltage
is needed. Most common dc source used by us is the charger used for our
mobiles or laptops.
In our country, the electrical energy available in homes, laboratories and
industries is in the form of alternating voltage of 220 V (rms) at a frequency of
50 Hz. But that cannot be used as received. Hence almost all electronic
equipment include a circuit that converts the ac voltage of mains supply into
dc voltage. This part of the equipment is called power supply.
Every power supply can be thought of as comprising (a) a dc power unit
consisting of a transformer followed by a diode circuit called rectifier, which
creates a dc voltage from the ac mains; (b) some circuitry consisting of filter to
remove any ac component present in the output of the rectifier; and (c) a
voltage regulator circuit, which gives a precisely controlled dc voltage. A block
diagram of such a power supply is shown in Fig. 12.1

Fig. 12.1: The block diagram of power supply. It consists of a dc power unit,
which converts the ac supply to a dc voltage plus some ripple;
followed by circuitry which reduces the ripple and regulates the
output voltage.

The output from a dc power unit on its own is usually unsatisfactory for two
reasons:

i) It usually carries a small amount of ac ripple superimposed on the dc

voltage. If it is used to supply power to an audio amplifier, it would give
rise to some audible ‘hum’ from the loudspeaker.

ii) The internal resistance of the dc power unit is usually higher than
desirable. Because of this, the voltage output can be significantly affected
54
Unit 12 Regulated Power Supply

by variations in the current drawn from the unit. The change in output
voltage per unit change of output current is called the regulation of the dc
source.

The regulating circuitry added to the dc power unit in a regulated power supply
reduces the ripple and improves the regulation.
In Sec. 12.2 you will learn about the dc power unit where we describe the
transformer and the rectifier circuits using diodes. The main two types of
rectifiers are half wave rectifier and full wave rectifier. Along with circuits and
working, you will also learn about their important characteristics like output dc
voltage and peak inverse voltage (PIV). The rectification efficiency and the
amount of ac component in the rectified voltage are very important factors and
influence the choice of rectifier in our circuit. You will learn about the
performance of the half and full wave rectifiers in Sec. 12.3.

The output voltage of rectifiers always possesses some ac component (ripple)

in it. This ripple can be removed by the filter circuits comprising either inductor
or capacitor or both. In Sec. 12.4 some important filter circuits are described.
The output of the filter needs to be regulated to obtained a constant desired
value of output voltage. You will study about the regulator circuits in Sec. 12.5.

Expected Learning Outcomes

After studying this unit, you should be able to:
 explain the following terms: half-wave rectification, full-wave rectification,
bridge rectifier, ripple, load regulation, line regulation;
 draw the circuit diagram and explain the working of half-wave rectifier,
centre-tapped full-wave rectifier and bridge rectifier;
 derive the expressions for output dc voltage, average or dc current, rms
current, ripple factor and rectifier efficiency in case of half-wave and full-
wave rectifiers;
 explain with the help of suitable waveforms the working of filters using
shunt-capacitor, series inductor and LC filters; and
 design the shunt and series pass voltage regulator circuits.

12.2 DC POWER UNIT

DC power units consist of transformer and rectifier circuit.

12.2.1 The Transformer

The first component in any power supply circuit is a transformer. It serves two
purposes. Firstly, it allows us to step the voltage up or down. This way we can
get the desired level of dc voltage. For example, the charger used with
mobiles phone gives a dc voltage of about 5V. We can use a step down
transformer to get such a low ac voltage at the input of the rectifier. On the
other hand, the cathode-ray tube used in an oscilloscope needs a very high dc
voltage – of the order of a few kV. Here, we may use a step up transformer. 55
Block 3 Analog Circuits
The second advantage of the transformer is the isolation it provides from the
power line. It reduces the risk of electrical shock.
The output voltage from the transformer is determined by the number of turns
in the primary and secondary windings of the transformer.
If n1 is the number of turns in the primary winding and n2 is the number of
turns in the secondary winding, then the ratio n1 : n2 is called the turns ratio
of the transformer. If the input voltage at primary is v in and the output voltage
at the secondary is vout , then

v in n
 1 (12.1)
v out n2
Hence
n
v out  2 v in (12.1a)
n1
So you can easily understand that when we wish to step down the voltage, we
have n1  n2 .

The transformer in a dc power unit uses either a single secondary winding or a

centre-tapped one, depending on the design of the rectifier circuit, you will be
learning in the next sections. Before moving further, you may solve the
following SAQ which will serve as revision of your previous knowledge.

SAQ 1 – Transformer voltage

a) Recall from your earlier knowledge that 220 V of mains supply means that
rms (root mean square) voltage is 220 V. What is the maximum voltage
(also called peak voltage amplitude)?
b) You know that to obtain lower voltages, a step down transformer is
needed. If the turns ratio of such a transformer is 15:1 and if the input
mains rms voltage is 220 V, what will be the rms and maximum voltage
across the secondary?

12.2.2 Half-Wave Rectification

The simplest type of a rectifier circuit is the half-wave rectifier. Fig. 12.2 shows
the circuit of half-wave rectifier where the diode forms a series circuit with the
secondary of the transformer and the load resistor RL .

Fig. 12.2: A half-wave rectifier circuit consisting of a transformer, a diode and a

56 load resistor.
Unit 12 Regulated Power Supply
Let us see how this circuit rectifies ac into dc.

The primary of the transformer is connected to the power mains. An ac

voltage is induced across the secondary of the transformer. The voltage may
be less than, or equal to, or greater than the primary voltage depending upon
the turns ratio of the transformer. We can represent the voltage across the
secondary (input voltage to rectifier) by the equation.
v  Vm sin t (12.2)

where  (= 2f) is the angular frequency of the input ac voltage.

Fig. 12.3a shows how input voltage varies with time. It has alternate positive
and negative half-cycles. Voltage Vm is the peak value of this alternating
voltage.

Fig. 12.3: Half-wave rectifier: a) input voltage waveform; b) output voltage

waveform; c) the transformer output voltage (input voltage) waveform
(continuous line) together with the rectified voltage (output voltage)
waveform across the load (shown dashed) in practical circuit.

During the positive half-cycle of the input voltage, the polarity of the voltage
across the secondary is as shown in Fig. 12.4a. This polarity makes the diode
forward biased. The diode conducts, and current i L flows through the load
resistor RL . This current makes the terminal A positive with respect to
terminal B. Since a forward-biased diode offers a very low resistance, the
voltage drop across it is also very small (about 0.3 V for Ge diode and about
0.7 V for Si diode). Therefore, the voltage appearing across the load terminals
AB is practically the same as the input voltage v at every instant. But ideally
speaking the situation is slightly different. By solving SAQ 2 you can find it out
by yourself.
57
Block 3 Analog Circuits

Fig. 12.4: Half-wave rectifier circuit: a) during positive half-cycle; b) during

negative half-cycle.

During the negative half-cycle of the input voltage, the polarity gets reversed
as shown in Fig. 12.4b. The diode is now reverse biased and hence non-
conducting. Practically no current flows through the circuit. Therefore, almost
no voltage is developed across the load resistance. All the input voltage
appears across the diode itself. The output voltages in the two half cycles are
shown in Fig. 12.3b.

To sum up, when the input voltage is going through its positive half-cycle, the
voltage of the output is almost the same as the input voltage. During the
negative half-cycle, no voltage is available across the load. The complete
waveform of the output voltage v o across the load is shown in Fig. 12.3b.
This voltage, though not a perfect dc, is at least unidirectional.

SAQ 2 – Voltage drop across forward biased diode

Fig. 12.3c shows the transformer output (input voltage to half wave rectifier)
waveform as a continuous line and the voltage across the load resistor (output
voltage) as a dashed line. Why is the output voltage less than the input
voltage in the positive half cycle?

Peak Inverse Voltage

Let us again focus our attention on the diode in Fig. 12.4b. During the
negative half-cycle of the input, the diode is reverse biased. The whole of the
input voltage appears across the diode (as there is no voltage across the load
resistance). When the input reaches its peak value Vm in the negative half-
cycle, the voltage across the diode is also maximum. This maximum voltage is
known as the peak inverse voltage (PIV). It represents the maximum voltage
the diode must withstand during the negative half-cycle of the input. Thus, for
a half-wave rectifier,

PIV  Vm (12.3)

Output dc Voltage
The average value of a sine wave (such as that in Fig. 12.3a) over one
complete cycle is zero. If a dc ammeter (moving coil type) is connected in an
58 ac circuit, it will read zero. (The dc meter reads average value of current in a
Unit 12 Regulated Power Supply
circuit.) Now, if the dc ammeter is connected in the half-wave rectifier circuit
(Fig. 12.2), it will show some reading. This indicates that there is some dc
current flowing through the load RL . We can find out the value of this current
for a half-wave rectifier circuit.

Fig. 12.5: Waveform of the current flowing through load RL in a half-wave

rectifier.

In Fig. 12.3b, we had plotted the waveform of the voltage across the load
resistor RL . If we divide each ordinate of this curve by the value of resistance
RL , we get the current waveform. This is shown in Fig. 12.5. Note that the
two waveforms (for current and for voltage) are similar. Mathematically, we
can describe the current waveform as follows:
iL  Im sin t; for 0  t   (12.4a)

and iL  0; for   t  2 (12.4b)

Here, Im is the peak value of the current i L . It is obviously related to the peak
value of voltage Vm as

V
Im  m , (12.5)
RL
since the diode resistance in the conducting state is assumed to be zero. To
find the dc or average value of current, we add or integrate the instantaneous
values of the current over one complete cycle, i.e. from 0 to 2 (curve repeats
itself after the first cycle). Using Eq. (12.4a) and (12.4b) we find the dc current
as follows:
2
1
Idc 
2  iLd (t ) (12.6)
0

 2
1  

2  
 i m sin t d (t )  0 d (t )


0 


1
2

I m (  cos t ) 0  0 
1
 Im (  cos   (  cos 0))
2
1 I
 Im (1  1)  m
2 
I
 Idc  m (12.7)
 59
Block 3 Analog Circuits
The dc voltage developed across the load RL is
I V
Vdc  Idc  RL  m  RL  m (12.8)
 
Remember that while writing Eq. (12.5), we have assumed that
i) the diode resistance in forward bias is zero, and
ii) the secondary winding of transformer has zero resistance.
The second assumption is often very near to the truth. The winding resistance
is almost zero. But, the forward diode resistance rd is sometimes not so small.
If it is comparable to the load resistance RL , we must take it into
consideration. Eq. (12.5) for peak current then gets modified to
Vm
Im  (12.9)
(RL  rd )
The dc voltage across the load resistor RL , can now be written with the help
of Eq. (12.8) as
Vm RL Vm
Vdc  
 (RL  rd )  (1  rd / RL )
V
 m (if rd  RL ) (12.10)


XAMPLE 12.1: DC VOLTAGE AND PIV RATING OF DIODE

The turns ratio of a transformer used in a half-wave rectifier (shown in

Fig. 12.2) is 12:1. The primary is connected to the power mains: 220 V,
50 Hz. Assuming the diode resistance in forward bias to be zero, calculate
the dc voltage across the load. What is the PIV of the diode?
SOLUTION  The maximum (peak value) primary voltage is

Vp  2 Vrms  2  220  311 V.

Therefore, the maximum secondary voltage is

1
Vm   311  25.9 V
12
The dc load voltage is
V 25.9
Vdc  m   8.24 V
 
The peak inverse voltage is
PIV  Vm  25.9 V

12.2.3 Full-Wave Rectification

In a half-wave rectifier we utilize only one half-cycle of the input wave. In a full-
wave rectifier we utilize both the half cycles. Alternate half cycles are inverted
to give a unidirectional load current. For full-wave rectification we can use two
types of circuit. One is called centre-tap rectifier that uses two diodes. The
60
other is called bridge rectifier and it uses four diodes.
Unit 12 Regulated Power Supply
Centre-Tap Rectifier
The circuit of a centre-tap rectifier is shown in Fig. 12.6a. It uses two diodes
D1 and D2. During the positive half-cycles of secondary voltage, diode D1 is
forward biased and D2 is reverse biased. The current flows through the diode
D1, load resistor RL , and the upper half of the secondary winding, as shown
in Fig. 12.6b. During negative half-cycles diode D2 becomes forward biased
and D1 reverse biased. Now D2 conducts and D1 becomes open. The current
flows through diode D2, load resistor RL , and the lower half of the winding, as
shown in Fig. 12.6c. Note that the load current in both Figs. 12.6b and c flows
in the same direction. The waveform of the current i L , and hence of the load
voltage v o , is shown in Fig. 12.6d.

Fig. 12.6: Centre-tap full-wave rectifier: a) circuit; b) operating during positive

half cycle; c) operating during negative half cycle; d) voltage across
the load resistor.
Peak Inverse Voltage
Fig. 12.7 shows the centre-tap rectifier circuit at the instant the secondary
voltage reaches its positive maximum value.

Fig. 12.7: The PIV across the non-conducting diode D2 in a centre-tap rectifier is
2Vm . 61
Block 3 Analog Circuits
The voltage Vm is the maximum (peak) voltage across half of the secondary
winding. At this instant, the diode D1 is conducting and it offers almost zero
resistance. The whole of the voltage Vm across the upper half winding
appears across the load resistor RL . Therefore, the reverse voltage that
appears across the non-conducting diode is the summation of the voltage
across the lower half winding and the voltage across the load resistor RL .
From the figure this voltage is Vm  Vm  2Vm . Thus,

PIV  2Vm (12.11)

Bridge Rectifier

A more widely used full-wave rectifier circuit is the bridge rectifier, shown in
Fig. 12.8a. It uses four diodes instead of two, but avoids the need for a centre-
tapped transformer. During the positive half-cycle of the secondary voltage,
diodes D2 and D4 are conducting and diodes D1 and D3 are non-conducting.
Therefore, current flows through the secondary winding, diode D2, load
resistor RL and diode D4, as shown in Fig. 12.8b. During negative half-cycles
of the secondary voltage, diodes D1 and D3 conduct, and diodes D2 and D4
do not conduct. The current flows through the secondary winding, diode D3,
load resistor RL and the diode D1 as shown in Fig. 12.8c. In both the cases,
the current flows through the load resistor in the same direction. Therefore, a
fluctuating, unidirectional voltage is developed across the load. The load
voltage waveform is shown in Fig. 12.8d.

Fig. 12.8: Bridge rectifier.

Peak Inverse Voltage

Let us now find the peak inverse voltage that appears across a non-
conducting diode in a bridge rectifier. Fig. 12.9 shows the bridge rectifier
62
Unit 12 Regulated Power Supply
circuit at the instant the secondary voltage reaches its positive peak value,
Vm . The diodes D2 and D4 are conducting whereas diodes D1 and D3 are
reverse biased and are non-conducting. The conducting diodes D2 and D4
have almost zero resistance (and hence zero voltage drops across them).
Point B is at the same potential as the point A. Similarly, point D is at the
same potential as the point C. The entire voltage Vm across the secondary
winding appears across the load resistor RL . The reverse voltage across the
non-conducting diode D1 (or D3) is also Vm . Thus,

PIV  Vm (12.12)

Fig. 12.9: The PIV across the non-conducting diode D1 or D3 is Vm .

Output dc Voltage in Full-Wave Rectifiers

The voltage waveform in Fig. 12.8d is exactly the same as that in Fig. 12.6d.
In both the rectifier circuits the load voltage is the same. However, there is one
difference. In the bridge rectifier, Vm is the maximum voltage across the
secondary winding. But in the centre-tap rectifier, Vm represents the maximum
voltage across half the secondary winding.

Now, let us compare the full-wave rectifier voltage waveform (of Fig. 12.6d or
Fig. 12.8d) with the half-wave rectifier voltage waveform (of Fig. 12.3b). In a
half-wave rectifier, only positive half-cycles are utilized for the dc output. But a
full-wave rectifier utilizes both the half-cycles. Therefore, the dc or average
voltage available in a full-wave rectifier will be double the dc voltage available
in a half-wave rectifier. If the resistance of a forward biased diode is assumed
to be zero, the dc voltage of a full-wave rectifier (refer Eq. 12.10) is

2Vm
Vdc  (12.13)


We can mathematically derive Eq. 12.13, on the same lines as we derived

Eq. (12.8) in the previous sub-section. Try to derive it by solving the following
SAQ.
63
Block 3 Analog Circuits

SAQ 3 – DC output from a full-wave rectifier

The output voltage of a full-wave rectifier (see Fig. 12.8b) is described as

Vo  Vm sin t 0  t  

Vo  Vm sin t   t  2

A minus sign appears in the second equation because during the second half-
cycle the wave is still sinusoidal, but inverted. The average or the dc value of
voltage is
2
1
Vdc 
2  Vo d (t )
0

Prove that

2Vm
Vdc 


So, now you know that a dc power unit can be of any one of the three forms: it
can contain a half-wave rectifier or a centre-tap full-wave rectifier or a bridge
rectifier circuit.

Let us now compare the performances of half and full-wave rectifiers.

12.3 RECTIFIER PERFORMANCE

If we connect a load resistor RL directly across the secondary of the
transformer, the current flowing through it will be purely ac (sinusoidal having
zero average value). This current is as shown in Fig. 12.10a.

In some applications, we require a dc current to flow through the load. The dc

current is unidirectional and, ideally, has no fluctuations with time. The ideal
dc current is shown in Fig. 12.10b. To see how effectively a rectifier converts
ac into dc, we compare its output current waveshape with the ideal dc current.

If the load takes current from a half-wave rectifier, the current waveform will be
as in Fig. 12.10c. It is unidirectional, but fluctuates greatly with time. The
waveform of the load current, when the load is connected to a full-wave
rectifier, is shown in Fig. 12.10d. This too is unidirectional and fluctuates with
time. A unidirectional, fluctuating waveform may be considered as consisting
of a number of components. It has an average or dc value over which a
number of ac (sinusoidal) components of different frequencies are
superimposed. These undesired ac components are called ripples. The
lowest ripple frequency in case of a half-wave rectifier is the same as the
power-mains frequency. But, for full-wave rectifier it is not so. As can be seen
from Figs. 12.10d and a, the period of the output wave of a full-wave rectifier
is half the period of the input wave. The variation in current (or voltage)
64
Unit 12 Regulated Power Supply

Fig. 12.10: Comparison of half-wave and full-wave rectifiers with an ideal ac-to-
dc converter.
repeats itself after each angle  of the input wave. Therefore, the lowest
frequency of the ripple in the output of a full-wave rectifier is twice the input
frequency. That is, the ripple frequency
f r  f i  50 Hz (half-wave rectifier)
and
f r  2f i  100 Hz (full-wave rectifier)
Ripple Factor
In order to have an assessment of the ac component in the output of a rectifier
circuit, we use a parameter called Ripple factor. It is defined as
rms value of ac component of load voltage
Ripple factor (  )  (12.14)
dc component of load voltage
Now, the rms voltage across the load comprises both, a dc component and an
ac component. We can express it as
2
v Lrms  (VLdc )2  (v Lacrms )2
(12.15)
 (v Lacrms ) 2  (v Lrms ) 2  (VLdc ) 2

(v Lacrms ) (v Lrms )2  (VLdc )2

  Ripple factor  
(VLdc ) (VLdc )2
2 2
v  i 
   Lrms   1   Lrms   1 (12.16)
 Ldc 
V  Ldc 
I
65
Block 3 Analog Circuits
Form Factor (F): It is defined as the ratio of rms value of load voltage to the
dc component.
v i
Form factor (F)  Lrms  Lrms (12.17a)
VLdc ILdc

Clearly,  (Ripple factor) = F 2  1. (12.17b)

Rectification Efficiency ():
It tells us what percentage of total input ac power is converted into useful dc
output power. Thus, rectification efficiency is defined as
dc power delivered to load

ac input power from transformer secondary
or
P
Wattmeter or power   dc (12.18)
meter is the instrument Pac
to measure the ac Here, Pac is the power that would be indicated by a wattmeter connected in
power. You can
calculate dc power by the rectifying circuit with its voltage terminals placed across the secondary
multiplying dc current winding and Pdc is the dc output power.
and dc voltage at the We shall now analyse half-wave and full-wave rectifiers to find their ripple
output of the circuit at
factor and rectification efficiency.
any instant of time.
12.3.1 Performance of Half-Wave Rectifier
The half-wave rectified current wave is plotted in Fig. 12.11 and is described
mathematically as
i L  Im sin t for 0  t   (12.19a)
and iL  0 for   t  2 (12.19b)
For determining the ripple factor or rectification efficiency, we first find the rms
value of the current.

Fig. 12.11: Half-wave rectified current waveform. (The instantaneous ac

component of current is the difference between instantaneous total
current and dc current, i.e., i   i L  I dc ) .

RMS Value of Current

The rms or effective value of the current flowing through the load is given as

2
1
i rms 
2  i L2 (t ) (12.20)
66 0
Unit 12 Regulated Power Supply
where current i L is described by Eqs. (12.19a) and (12.19b). Therefore,
 2
1  2 
i rms 
2   2

 Im sin t d (t )  0 d (t )

0 


2 (1  cos 2 t )
Im

2  2
d (t )
0


2
Im  t  sin 2t 
 
2  2  2 0
I
 i rms  m (12.21)
2
Form Factor
Using the values of I dc from Eq. (12.7) and i rms from Eq. (12.21), the form
factor is given by
i I /2 
F  Lrms  m   1.57 (12.22)
i Ldc Im /  2
Ripple Factor
From Eq. (12.17b), the ripple factor is given as
  F 2  1  (1.57) 2  1  1.21 (12.23)
Here, we see that the ripple current (or voltage) exceeds the dc current (or
voltage). This shows that the half-wave rectifier is a poor converter of ac into
dc.
Rectification Efficiency
For a half-wave rectifier, the dc power delivered to the load is
2
2 R   Im  R
Pdc  I dc L   L
  
and the total input ac power is
2
2 (r  R )   I m  (r  R )
Pac  i rms d L   d L
 2 
Therefore, the rectification efficiency is
P (I m / ) 2 R L
  dc   100%
Pac (I m / 2) 2 (rd  R L )
40.5
 % (12.24)
1  rd / RL 
If rd  RL ,   40.5% . It means that under the best conditions (i.e. no diode
loss), only 40.5% of the ac input power is converted into dc power. The rest
remains as ac power in the load.
12.3.2 Performance of Full-Wave Rectifier
Fig. 12.12 shows a full-wave rectified current waveform. You will notice that its
period is . The wave repeats itself after each . Therefore, while computing
the average or rms values, we should take the integration between the limits 0
to , instead of 0 to 2. The waveshape between 0 to  is described as
i L  I m sin t (12.25) 67
Block 3 Analog Circuits

Fig. 12.12: Full-wave rectified current waveform.

RMS Value of Current

Effective or rms value of current is given as
 
1 2 1 2
i rms  
I d (t ) 
 L  
I m sin 2 t d (t )
0 0


2
Im  1  cos 2t d (t ) 
2 t
Im sin 2t 

  
 2

  2

4 0
0

2
Im 
 
 2
or
I
i rms  m (12.26)
2
Note that this is the same as the rms value of the full sinusoidal ac wave.
The dc or average value of the current is
 
1 1
I dc 
 
i L d (t ) 
 
I m sin t d (t )
0 0
2I m
 (12.27)

This current is obviously double the dc current of a half-wave rectifier.
Form Factor
i I / 2 
F  Lrms  m   1.11 (12.28)
i dc 2I m /  2 2
Ripple Factor
From Eq. (12.17b)

  F 2  1  (1.11) 2  1  0.48 (12.29)

Rectification Efficiency
For a full-wave rectifier, the dc power delivered to the load is
2
2 R   2I m  R
Pdc  I dc L   L
68
  
Unit 12 Regulated Power Supply
and the total input ac power is
2
2 (r  R )   I m  (r  R )
Pac  i rms d L   d L
 2
Therefore, the rectification efficiency is
P (2I m / ) 2 R L
  dc   100%
Pac (I m / 2 ) 2 (rd  R L )
For rd  RL
8
  100  81.1% (12.30)
2
This shows that the rectification efficiency of a full wave rectifier is twice that of
a half-wave rectifier under identical conditions. The maximum possible
efficiency can be 81.1% (when r d  R L ).

XAMPLE 12.2: RIPPLE FACTOR AND RECTIFICATION

EFFICIENCY

In a centre-tap full-wave rectifier, the load resistance RL  1k. Each

diode has a forward-bias dynamic resistance rd of 10 . The voltage
across half the secondary winding is (220 sin 314 t). Find a) the peak value
of current, b) the dc or average value of current, c) the rms value of current,
d) the ripple factor, and e) the rectification efficiency.
SOLUTION  The voltage across half the secondary winding is given as
v  220 sin 314 t

a) The peak value of voltage is

Vm  220 V

Therefore, peak value of current is

Vm 220
Im    0.2178 A
rd  RL 10  1000

 217.8 mA

b) The dc or average value of current for full wave rectifier is

2I m 2  217.8
I dc    138.7 mA
 
c) The rms value of current for full wave rectifier is
I
I rms  m  154 mA
2

d) The ripple factor is given by Eq. (12.16)

2
I  154  2
   rms   1     1  0.482
 Idc   138.7 
69
Block 3 Analog Circuits

e) The rectification efficiency is given

P
  dc
Pac

But, Pdc  Idc

2 R  (0.1387 )2  1000  19.23 W
L

Pac  I rms
2 (r  R )
d L

 (0.154)2 (10  1000)

 23.96 W

P 19.23
   dc   0.8026  100%  80.26%
Pac 23.96

A full-wave rectifier is preferred to a half-wave rectifier, because its

rectification efficiency is double and its ripple factor is low. Table 12.1 gives
the comparison between different rectifiers discussed so far. Unless otherwise
indicated, all rectifiers we will use in our further discussion are full-wave
rectifiers (either centre-tap or bridge).
Table 12.1: Comparison between different rectifiers

Half-wave Full-wave
Centre-tap Bridge
Number of 1 2 4
diodes
Transformer Single secondary Centre tapped Single secondary
necessary winding secondary winding winding
Peak secondary Vm Vm Vm
voltage
Peak inverse Vm 2Vm Vm
voltage
Peak load Vm /(rd  RL ) Vm /(2rd  RL ) Vm /(rd  RL )
current, I m

RMS current, Im / 2 Im / 2 Im / 2
i rms

DC current, I dc Im /  2I m /  2I m / 

Forms factor 1.57 1.1 1.1

Ripple factor, r 1.21 0.48 0.48
Rectification 40.5% 81.1% 81.1%
efficiency (max)
Lowest ripple fi 2f i 2f i
frequency, f r
70
Unit 12 Regulated Power Supply

Most electronic equipment require smooth dc operating voltages. The output

of a rectifier cannot be applied directly to such equipment because of the
ripple. Consequently, the ripple must be eliminated. Circuits for accomplishing
this are called filter circuits. Let us now discuss about the filter circuits.

12.4 FILTER CIRCUITS

The aim of rectification is to provide a steady dc voltage, similar to the voltage
from a battery. We have seen that a full-wave rectifier provides a better dc
than a half-wave rectifier. But, even a full-wave rectifier does not provide
ripple-free dc voltage. Actually, the rectifiers provide what we may call “a
pulsating dc”. We can filter or smooth out the ac variations from the rectified
voltage. For this we use a filter or smoothing circuit (see Fig. 12.1). In this
section, we shall discuss different types of filter circuits.

12.4.1 Capacitance Filter

The ripple output of a rectifier represents energy being supplied to the load in
pulses. The ripple fluctuations can be reduced considerably if some of the
output is stored while the rectifier is delivering a pulse and then released to
the load between output pulses. This is the basic operating principle of the
capacitance filter.

Such a filter consists of a large value capacitor C in shunt (parallel) with the
load resistor RL , as shown in Fig. 12.13a. The capacitor offers a low-
resistance path to the ac components of current. To dc (with zero frequency),
this is an open circuit. All the dc current passes through the load. But due to
capacitor, most of the ac component part flows to the ground and only a small
part of the ac component passes through the load producing a small ripple
voltage.

(a)

Fig. 12.13: a) Circuit of full-wave rectifier with shunt capacitor filter; b) filtering
action by capacitor charging-discharging. 71
Block 3 Analog Circuits
The waveforms of filter output are shown in Fig. 12.13b. When the rectifier
output voltage is increasing, the capacitor charges to the peak voltage Vm .
Just past the positive peak (point B), the rectifier output voltage starts to fall
(see the dotted curve in Fig. 12.13b). But at point B, the capacitor has + Vm
volts across it. Since the source voltage becomes slightly less than Vm the
capacitor will try to send current back through the diode (of the rectifier). This
reverse-biases the diode, i.e. it becomes open-circuited. Thus, the capacitor
changes the conditions under which the diodes (of the rectifier) conduct.

The diode (open-circuit) disconnects or separates the source from the load.
The capacitor starts to discharge through the load. This prevents the load
voltage from falling to zero. The capacitor continues to discharge until the
source voltage (the dotted curve) becomes more than the capacitor voltage (at
point C). The diode again starts conducting, and the capacitor is again
charged to peak value Vm . During the time the capacitor is charging (from
point C to point D) the rectifier supplies the charging current i c through the
capacitor branch as well as the load current i L . When the capacitor
discharges (from point B to point C), the rectifier does not supply any current,
the capacitor sends current i L through the load. In this way, the current is
maintained through the load all the time.

The rate at which the capacitor discharges between points B and C (in
Fig. 12.13b) depends upon the time constant CRL . The longer this time
constant is, the steadier is the output voltage. If the load current is fairly small
(i.e., RL is sufficiently large) the capacitor does not discharge very much, and
the average load voltage Vdc is slightly less than the peak value Vm .

Any increase in the load current (i.e. decrease in the value of RL ) makes the
time constant of the discharge path smaller. The capacitor then discharges
more rapidly, and the load voltage does not remain constant. The ripple
increases with increase in load current. Also, the dc output voltage, Vdc
decreases.

12.4.2 Inductance Filter

An inductor is another device that can store and release electrical energy. It
does this by extracting energy from a flowing current and storing it in the form
of magnetic field when the current is increasing. Then it releases the energy to
keep the current flowing when the current begins to decrease. This ability of
an inductor to store and release energy can be used to help prevent the
abrupt changes in the output of a rectifier. This property is used in the
inductance filter of Fig. 12.14a. Whenever the current through an inductor
tends to change, a “back emf” is induced in the inductor. This induced back
emf prevents the current from changing its value. Any sudden change in the
current that might have occurred in the circuit without an inductor is smoothed
out by the presence of the inductor. Its effect on the output waveform is shown
in Fig. 12.14b.
72
Unit 12 Regulated Power Supply

Fig. 12.14: a) Full-wave rectifier with inductance filter; b) the filtering action.

The inductance filter prevents the current, and, therefore, the output voltage
from ever reaching the peak value that would be obtained if the inductor were
not in the circuit.

Therefore, a rectifier that has an inductance filter will not produce as high an
output voltage as the one that has a capacitance filter. However, a large load
current can be drawn from the inductance filter without changing the output
voltage.

The operation of a series inductor filter depends upon the current flowing
through it. Therefore this filter (and also the choke (inductor) input LC filter
discussed in the next sub-section) can only be used together with a full-wave
rectifier (since it requires current to flow at all times). Furthermore, the higher
the current flowing through it, the better is its filtering action. Therefore, an
increase in load current results in reduced ripple.

12.4.3 LC Filter
We have seen that an inductance filter has a feature of decreasing the ripples
when the load current is increased. Reverse is the case with a capacitor filter.
In this case, as the load current is increased, the ripples also increase. An LC
filter combines the features of both the inductor filter and shunt capacitor filter.
Therefore, the ripples remain fairly the same even when the load current
changes. There are two types of LC filter: capacitor input filter and choke
(inductor) input filter. Let us describe each one of them.

Capacitor input filter

This type of LC filter is shown in Fig. 12.15. This is called a capacitor input
filter because capacitor (C1) is the first filtering element directly after the
rectifier. It is also called the “pi” filter because its schematic arrangement of
C’s and L resembles the Greek letter .
73
Block 3 Analog Circuits

Fig. 12.15: Capacitor input filter.

In this type of filter, capacitor C1 performs the same function as the simple
capacitance filter described earlier. It charges to the peak voltage of the
rectified output pulses and then discharges through the load, when the rectifier
output falls. Capacitor C2 provides similar filtering action but to a lesser
degree. Inductor L1 adds to the overall filtering action by opposing changes in
both, the output current filtered by C2 and the current drawn by the load.

The output of such a filter contains only small amount of ripple. However, the
voltage regulation of such a filter is relatively poor. This is because of the
decrease in the voltage across C1 as it discharges between rectified pulses.

Choke (inductor) input filter

When an inductor is used as primary filtering element in an LC filter, the

network is called a choke input filter. The term choke is used because of the
inductor’s ability to stop, or choke, the passage of ripple voltage to the load.
Fig. 12.16 shows a simple choke input filter.

Fig. 12.16: Choke input filter.

74
Unit 12 Regulated Power Supply

Due to the arrangement of components, this filter is also known as “L” shaped
filter.
The inductor opposes current changes while the capacitor charges and
discharges in standard filter-capacitor fashion. Since the inductor reduces the
peak rectifier current and this in turn reduces the maximum voltage to which
the capacitor charges, the output voltage of the choke input filter is lower than
that of the capacitor input filter. This output voltage, though, is affected less by
changes in the load current.

12.5 REGULATION OF OUTPUT VOLTAGE

So far we have discussed the rectifier and filter system. The filters are used to
reduce the ripple and help in regulation of voltage. A very important
characteristics of any power supply is the degree to which its output voltage
remains constant despite changes in the amount of current drawn from the
supply by the load. This characteristic is called the regulation of the supply. A
supply whose output voltage changes little over a wide variation in load
current is said to have good regulation. If output voltage falls sharply as the
load current increases such power supply circuit has poor regulation. To
obtain good regulation, additional circuits are added so that the output voltage
remain constant. These additional circuits are called voltage regular circuits.

12.5.1 Voltage Regulator Parameters

Basically, the output voltage tends to vary because of two things:
(a) fluctuating line (input ac) voltage, and (b) fluctuating load current. Every
time the line voltage goes up, the output voltage will go up. But when the load
current goes up, the output voltage will go down. This is because the output
voltage is actually what is left after the voltage drops across the internal
components in the circuit are taken away from the input voltage. As the load
current increases, the internal drop increases and less is left for the output.
Thus, we define two terms: (i) source regulation, and (ii) load regulation.
Source regulation
It is also called source effect or line regulation. It is the change in regulated
load (output) voltage for the specified range of line voltage and is defined as:

SR
% Source Regulation   100% (12.31)
Vnom

where
SR = change in load voltage for full line voltage change
Vnom = nominal load voltage

For instance, if the change in load voltage is 5 mV and the nominal load
voltage is 10V then

5 mV
% Source Regulation   100%  0.05%
10 V
75
Block 3 Analog Circuits
Load regulation

It is also called load effect. It is defined as the change in regulated output

voltage when the load current changes from minimum to maximum.

LR  VNL  VFL (12.32)

where

LR = load regulation

VNL = load voltage with no load current

VFL = load voltage with full load current

Load regulation is often expressed as a percent by dividing the change in load

voltage by the no load voltage

V  VFL
%Load Regulation  NL  100% (12.33)
VNL

Before proceeding further, you may like to solve an SAQ.

SAQ 4 – Load regulation

If no load voltage is 10V and the full load voltage is 9.9V, then what is the
percent load regulation?

12.5.2 Principle of Voltage Regulation

There are different types of voltage regulator. One is the linear regulator
which is operating throughout the cycle of ac, and so there is no ripple
problem. But its efficiency is less.

The other kind of regulator is called a switching regulator, where we use a

device to charge and discharge and supply this shunt dc voltage to the load.
But since we are going to switch on and off the charging of that storage
device, we are going to get a ripple which will be riding on the output dc. In
this unit we will be restricting our discussion to the linear voltage regulators
only.

The principle of linear regulator is quite simple as shown in Fig. 12.17. The
rectified and filtered input (unregulated) (VUR ) is given it to the linear regulator
circuit, and we get a regulated output (VOR ) which is smaller than the input
that we have given.

Fig. 12.17: Principle of linear regulation

76
Unit 12 Regulated Power Supply

In the regulator circuit we compare the input unregulated voltage with a fixed
reference voltage and that fixed reference voltage determines the output
regulated voltage. The excess voltage, that is the difference between the input
and the output voltage is dissipated within the circuit as an ohmic loss i.e. in
the form of heat. It is not a very efficient circuit because we are losing on a
power here; but we get excellent regulation in this method.

The linear regulators are of two types:

i) shunt voltage regulator; and

ii) series voltage regulator or series pass regulator.

In the shunt voltage regulator, we connect the load in parallel with the
reference voltage. The reference voltage is derived through a special device
which ensures constant voltage across it. Since the load is directly parallel to
the reference voltage, it will always have a fixed voltage (equal to the
reference voltage) across it.

In series voltage regulator or series pass regulator, we connect the load in

series with a variable element that drops the excess voltage between the input
and the output across it.

Now for these regulator circuits, we need the reference voltage device which
will give a rock steady reference voltage irrespective of any conditions like
change the input voltage, change the current flowing through the load, or
change in ambient conditions like temperature, humidity etc. One of the best
candidate used for generating a reference voltage is a zener diode. You have
learnt about in Unit 2 of this course. A zener diode is a p-n junction diode with
very high doping and so the depletion region between the p and n junctions is
very thin. When we apply a forward bias to this diode, it just operates like a
normal rectifier diode and has same forward bias characteristics. But in case
of the reverse bias, when we give certain voltage across the depletion layer,
due to its small thickness, it allows a lot of charge generation in the form of
avalanche and zener breakdown. Then a current gets sustained in the device
and the voltage across the device remains constant. It is called breakdown
voltage or the zener voltage. In this particular case, the breakdown is not a
permanent deformity and does not damage the device. Whenever we reduce
the reverse voltage it comes back to its normal operation.

Before discussing the types of regulators you may solve one SAQ.

SAQ 5 – Zener reference voltage

Why can normal p-n junction diode not be used as reference voltage source?
77
Block 3 Analog Circuits
12.5.3 Shunt Voltage Regulator

Use of zener diode in a shunt voltage regulator is shown in Fig. 12.18. Here,
we connect the zener diode cathode to the positive terminal and anode to the
ground potential. In this way, we operate it in the reverse bias mode with a
resistance R connected between the supply and the diode. Due to reverse
bias condition, the voltage across the zener diode remains constant. Since the
load is attached in parallel to the zener diode, it always has the same voltage
(the zener voltage) across it. In this way, we have a constant voltage delivered
to the load.

Fig. 12.18: Shunt voltage regulator using zener diode.

Now the zener is going to draw Vz voltage across it. But the input is the
rectified, filtered unregulated voltage Vin . The balance of (Vin  Vz ) voltage
gets dropped across R. So R is the dissipative element in shunt voltage
regulator. For the zener to get operational; minimum zener current should
always pass through it and that is denoted by I z .

The load current IL  Vz / RL . Hence the resistance value of the load

determines I L . Here the input current I in drawn from filter output gets divided
in two parts: one goes to the zener, and other to the load. Minimum I z is
determined from the specification of the zener diode, but the load current is
determined by the value of load resistance. For lower values of RL , I L is larger.
Now, in order to keep the zener working, we need minimum I z flowing
through it irrespective of I L value. So when we design a regulator, we should
have provision to draw the input current

I in  I z  I L(max)

Let us understand the design of shunt regulator with an example.

78
Unit 12 Regulated Power Supply

XAMPLE 12.3: DESIGN OF SHUNT REGULATOR FOR

VARIABLE LOAD

Design a shunt regulator using zener diode to provide 5V dc voltage for a

variable load of 100  to 500 . The minimum zener current for its proper
operation is 5 mA. The unregulated input voltage = 6 V.

SOLUTION  Refer to Fig. 12.18. For obtaining 5 V across the load, we

choose the zener with breakdown voltage = 5 V. When the load resistance
is maximum, current though the load

5V
I Lmin   10 mA
500 

When the load resistance is minimum, maximum load current flows in the
load,

5V
I Lmax   50 m
100 

The input current sourced from the filter output is the sum of zener current
and maximum load current

 I in  I Z  I Lmax

 5 mA  50 mA  55 mA

This current always flows from the filter output. It gets divided among load
and zener diode. For lowest load current (10 mA),

zener current = 55 – 10 = 45 mA

Hence the power rating of zener

= 5 V  45 mA

= 225 mW

Since the input voltage = 6 V and, zener voltage = 5 V.

Hence 1 V is dropped across R. Current flowing through R  I in  55 mA \

1V
 R  18.2 
55 mA

And the power rating of R is

 2  R  (55 mA) 2  (18.2 )  0.055 W

I in

So the circuit of voltage regulator will have zener diode with 5 V and
250 mW rating. The resistance is 18.2  with 55 mW rating.

From this example you can see that 55 mA current is always going to flow
through 18.2  resistor. Hence it is a dissipative kind of configuration;
79
Block 3 Analog Circuits
because whether we require just 10 mA current or 50 mA current through the
load, R is always going to draw 55 mA.
So far we considered about the load regulation where we get regulated
voltage even when load is changing. But now, the rectified input that we are
getting is also unregulated. So its value may also vary. We considered in our
example a fixed voltage drop of 1 V (= 6V – 5V) across the dissipative
element R. But this drop will change, if input voltage varies. This will
effectively change the current flowing through resistor R, as you will find out
from the next SAQ.

SAQ 6 – Regulation again input voltage regulation

Consider the regulator discussed in Example 12.1. Find out the range of
current flowing through the resistor R when the input voltage Vin varies
between 6 V and 8 V; if R = 18.2 .

From this SAQ you will realize that the current in R depends on the voltage
drop across it. So this I in is going to change with Vin but I L will remain
constant for given load. Now, the extra variation in the current will have to be
borne by current through zener.
Hence the power rating of the zener has to be calculated not just on the basis
of load variation, but also based on the expected changes in the input supply.
The shunt regulator is not very energy efficient, since the excess voltage will
result into excess power dissipation in R. Now, we will discuss another
relatively efficient configuration of linear regulator called the series pass
voltage regulator.

12.5.4 Series Pass Voltage Regulator

Refer to Fig. 12.19 which shows a schematic of series pass voltage regulator.

Fig. 12.19: Schematic of series pass voltage regulator.

In this regulator a variable element (like transistor) is placed in series with the
load. The reference voltage is given to the series element which controls the
current passing through the element.
In order to supply regulated voltage at the output, we control the voltage drop
80 across the series pass element by changing its resistance. That is, the
Unit 12 Regulated Power Supply
difference between Vin and Vout is dropped by the series pass element
which, in turn, will be determined by the current passing through it. For this
purpose, the series pass element should be controllable electronically
depending on the reference voltage. Hence, we need to have an active device
as our series pass element.

The first choice is a transistor, because we can always control the voltage
across the transistor which depends on the current flowing through it as well
as the base current supplied to it. Here, we use emitter follower (common
collector) configuration of an n-p-n transistor as the series pass element as
shown in Fig. 12.20.

Fig. 12.20: Transistor as series pass element.

We apply the reference voltage to the base of the transistor in the form of a
reverse biased zener diode. The transistor is biased to conduct in its active
region. Thus, the base emitter voltage (Vbe ) of the transistor will be constant
(of the order of 0.6 V, if we use a silicon transistor). So the voltage appearing
at the output of this series regulator is Vo  Vz  Vbe , i.e. Vo is less than the
reference voltage by 0.6 V.

In this case, the load current is nothing but the emitter current flowing through
the transistor and the voltage drop across the transistor, that is the collector-
emitter voltage Vce  Vin  Vo .

Now, the current given to the base of this transistor is going to be determined
by the load current, because the load current is nothing but the emitter
current. Hence we can estimate

I
Ib  L ,


where  is the current gain of the transistor. Hence, the zener current is now
restricted only to the minimum current that is required for zener to operate (I z )
plus the base current required for the transistor to operate.
81
Block 3 Analog Circuits
Hence this is an efficient circuit, than the shunt regulator because the
requirement of current is quite less as far as the base current is concerned,
and so there are less losses in the resistor. An added advantage is that with
the series pass element regulator we can also have a possibility of feedback
configuration. Before studying the advantages of feedback configuration, let
us discuss an example on designing of a series pass regulator.

XAMPLE 12.4: DESIGN OF SERIES PASS REGULATOR

Design a series pass regulator to obtain 10 V dc output with maximum load

current of 100 mA. Consider the nominal input voltage to be 15 V.

SOLUTION  Refer to Fig. 12.20. Let us consider that we are using a

silicon n-p-n transistor with  = 50. We need output voltage = 10 V. For
silicon transistor, VBE  0.6 V.

Hence Vz  Vo  VBE  10 V  0.6 V  10.6 V.

Maximum output current I L  I E  100 mA.

100
For   50, I B   2 mA .
50

VCE  Vin  Vout  15  10  5 V

Current through the resistor R

I R  I Z  I BE

If we consider, I Z  5 mA.

I R  5  2  7 mA

Voltage across R,

VR  Vin  VZ  15 V  10.6 V  4.4 V

4. 4 V
 R  628 
7 mA

In this case, since very small currents are flowing through R and zener diode,
their power ratings will be quite low; as you will calculate in the following SAQ.

SAQ 7 – Regulator components

Calculate the power ratings of the zener diode and resistor R used in
Example 12.4.
82
Unit 12 Regulated Power Supply
Now let us discuss how series pass regulator with feedback can be used as a
variable voltage supply.

12.5.5 Variable Voltage Supply using Potential Divider

So far we discussed the power supply with fixed output voltage, determined by
the reference voltage. But, we can have a feedback, sampled from the output
voltage and feed it back to a comparator which compares the sampled voltage
with the reference voltage and generates an error signal to control the series
element. In such case, the portion of output voltage sampled decides the
value of output voltage.
You will learn about
Typically for a comparator, we use operational amplifier because it has high the operational
gain and so with small difference between two input voltages it generates amplifier in Unit 13.
adequate output signal to control the series pass element. The feedback
circuit can be a potential divider consisting of two resistors R1 and R 2 as
shown in Fig. 12.21.

Fig. 12.21: Voltage supply with potential divider.

This configuration of the regulator circuit allows us to have higher output

voltage than the zener voltage rating. Here, we are using the sampling circuit
in the form of a potential divider R1 and R 2 and the voltage across R 2 is fed
back to the negative terminal of the comparator.

R2
The feedback voltage  Vfb  Vo (12.33)
R1  R 2

The second input of the comparator is connected to the zener diode. Hence,
the feedback voltage is compared to the reference voltage and the error signal
is generated to control the series pass element to provide output voltage Vo .
So we are able to generate the output voltage which is greater than the
reference voltage given by the relation:

R1  R2
Vo   Vref (12.34)
R2
83
 Block 3 Analog Circuits
Now, if we replace the resistor R 2 by a variable resistor, by changing its value,
we can obtain variable output voltage from a fixed reference voltage.

You will appreciate the working of potential divider regulator by solving the
following SAQ.

SAQ 8 – Potential divider regulator

Design a potential divider circuit to obtain 10 V output voltage using a

reference voltage of 5 V.

Now, we summarise the points discussed in this unit.

12.6 SUMMARY

Concept Description

DC power unit  The ac mains voltage is first reduced by means of a transformer and then
rectified by using half wave or full wave rectifier.
Half-wave  A half-wave rectifier consists of a diode and a load resistor. It delivers an
rectifier output during only half of the input voltage cycle.
Full-wave  Full-wave rectifier delivers an output during both halves of the input voltage
rectifier cycle.
 Full-wave rectifier with centre tapped transformer circuit consists of two
diodes having a common load.
 Bridge rectifier consists of four diodes and a common load.
Filters  Fluctuation of the dc output from a rectifier above and below its average value
is called ripple, which can be removed by filter circuit.
 Inductance filter consists of an inductor connector in series with the load.
 Capacitance filter is connected in shunt with the load.
 Combination of L and C improves filtering action.

Shunt regulator  A reference voltage derived usually from a reverse biased zener diode is
applied to the load in parallel.
Series pass  As series pass element in the form of a transistor is used for dissipating the
regulator difference voltage between input and output. The load current is delivered by
the emitter current in CC configuration. Very small (base) current is required
to control the series pass element.
Potential divider  Variable voltage can be obtained using single reference voltage with the help
regulator of a feedback network.

12.7 TERMINAL QUESTIONS

1. If the peak voltage across a transformer secondary is 100 V, what is the dc
voltage value of a) a half-wave rectifier, and b) full-wave rectifier?

2. Which has greater ripple frequency: a half-wave rectifier whose input is

84 500 Hz or a full-wave rectifier whose input is 300 Hz?
Unit 12 Regulated Power Supply
3. The turns ratio of the transformer used in a bridge rectifier is
n1 : n2  12 : 1. The primary is connected to 220 V, 50 Hz power mains.
Assuming that the diode voltage drop to be zero, find the dc voltage
across the load. What is the PIV of each diode? If the same dc voltage is
obtained by using a centre tap rectifier, what is the PIV?
4. In a full-wave rectifier without filter, the load resistance is of 4000 .
Each diode has a resistance of 800 , voltage applied to each diode is
(240 sin 100 t). Calculate
a) peak, average and rms values of current,
b) dc power output and total power input
c) rectifier efficiency
d) form factor
e) ripple factor
5. Nominal supply voltage of a regulator is 10 V and over the entire variable
range of input, the output changes from 9.97 V to 10.02 V. Calculate
percentage source regulation.
6. Design a potential divider voltage regulator circuit to give 5 V to 10 V dc
with maximum load current of 100 mA using an unregulated dc input of 15
V and a zener diode with VZ  5 V.

12.8 SOLUTIONS AND ANSWERS

Self-Assessment Questions
1. a) The rms value of a sinusoidal waveform is the amplitude divided by
2 , so in this case the peak amplitude, or maximum voltage, is
220  2  311 V.
1
b) rms voltage across secondary = 220   14.7 V.
15
311
Maximum voltage across secondary =  20.7 V.
15
2. The difference between the output voltage and input voltage is caused by
the presence of a silicon (or Germanium) diode. When the transformer
output voltage is positive the diode conducts, so the voltage applied to the
load resistance is the ac voltage minus the voltage dropped across the
diode. When the transformer output voltage is negative no current flows so
there is zero voltage drop across the load, as shown in Fig. 12.3c.
2
1
3. Vdc 
2  v o d ( t )
0

 2
1

2 
(Vm sin t ) d (t )   (Vm sin t ) d (t )
0 


1
2

 Vm cos t 0  Vm cos t 2 
85
Block 3 Analog Circuits
V
 m  cos   cos 0  cos 2  cos 
2
2Vm
 .

10 V  9.9 V
4. %LR   100%  1% .
10 V
5. In case of normal p-n junction diode the reverse bias breakdown voltage is
very large, and beyond this voltage the diode gets damaged. Hence it
cannot act as a constant reference voltage device in reverse bias
condition.
6. Voltage across R,
VR  Vin  VZ
We have taken VZ  5 V.

Hence for
1V
Vin  6 V, VR  6  5  1 V; I in   55 mA
18.2 

and for
3V
Vin  8 V, VR  8  5  3 V; I in   165 mA
182 

Hence the current in the range of 55 mA to 165 mA may flow in the

resistance R for variation in the input voltage.
7. Voltage across zener diode = 10.6 V
and current through zener = 5 mA
Hence the power rating of zener diode = 53 mW.
Power rating of the resistor R
 4.4 V  7 mA  31 mW.

These power ratings are quite low as compared to the components used in
the shunt regulator.
8. We want Vo  10 V and Vref  5 V.

Now
R1  R 2
Vo  Vref
R2

R1  R 2
 10 V  5V
R2

Hence,
10
 R1  R2  R2  2R2
5
 R1  R2

86 If R1  10 k, then R2  10 k.

Unit 12 Regulated Power Supply

Terminal Questions
100
1. a) V

200
b) V

2. Full-wave rectifier.
For half-wave rectifier, ripple frequency is 500 Hz.
For full-wave rectifier, ripple frequency is 300  2 = 600 Hz.
3. The maximum primary voltage is
Vp  2 Vrms  2  220  311 V

Therefore, the maximum secondary voltage is

n 1
Vm  2 Vp   311  25.9 V
n1 12
The dc voltage across the load is
2Vm 2  25.9
Vdc    16.49 V
 
The PIV (for bridge rectifier) is
PIV  Vm  25.9 V
For the centre-tap rectifier, the PIV is
PIV  2Vm  2  25.9  51.8 V
Vm
4. a) Peak value of current, Im 
rd  RL
240
 Im   50  103 A  50 mA
800  4000
2Im 50  2
Average value of current, Idc    31.83 mA
 
I 50
The rms value of current output, Irms  m   35.36 mA
2 1.41
b) The dc power output
Pdc  Idc
2 R
L

 (31.83)2  106  4000

 4.05 W
Total power input
Pin  (Irms )2  (rd  RL )

 (35.36  103 )2  (800  4000 )

 6.0 W
c) Rectifier efficiency
P 4.05 W
  dc   100%  67.5%
Pin 6.001 W
87
Block 3 Analog Circuits
d) Form factor,
I rms 35.36 mA
  1.11
I dc 31.84 mA
2
I 
e) Ripple factor   rms   1
 Idc 

 (1.112 )  1  0.48
V  Vmin
5. % Source Regulation  max  100
Vnom
10.02  9.97
  100
10
0.05
  100  0.5%
10
6 Vref  Vz  5 V
R1  R 2  V 
Range of  i.e. o  is 1 to 2. So potential divider R1,R2 can be
R1  Vref 
designed by using a potentiometer of 100 k as shown in Fig. 12.22.

Fig. 12.22: Variable dc supply

Transistor emitter current ~ 100 mA for maximum load condition. With

assumed current gain of 50, base current  2 mA. By changing the
position of pointer on the potentiometer, ratio of R1 and R 2 can be
changed to obtain desired output voltage.
For V0  5 V , the pointer is at the top of the potentiometer, so that,
R1  0, and
R1  R2 0  R2
V0  Vref  5 V  5 V
R2 R2
For V0  10 V , the pointer is kept in the middle of the potentiometer, so
that R1  R2 and

R1  R2
V0  Vref  2Vref  10 V
R2
88