Digital Electronics For Engineering and Diploma Courses

By Dr. S.K. Bhattacharya

This is a helpful book for the students of engineering courses it includes the following chapters :

Number Systems and Codes, Logic Gates, Boolean algebra and logic simplification, Design of Combinational Logic Circuits, Arithmetic Circuits, Decoder, Encoder, Multiplexer, Demultiplexer, Sequential Circuit Design, Shift Registers, Counters, A/D and D/A Converters and Logic Family

• Language: English
• Publisher: Abhishek Publications
• Release date: Jun 14, 2023
• ISBN: 9788182474482

Book preview

Dr. S.K. Bhattacharya Ms. Balaka Biswas

ABHISHEK PUBLICATIONS

CHANDIGARH / DELHI (INDIA)

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ISBN : 978-81-8247-404-8

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First Edition : 2012

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Abhishek Publications

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PREFACE 

Digital Electronics is one of the core subjects taught to students of Electronics and Communication Engineering, Electrical and Electronics Engineering, Electronics and Instrumentation Engineering, Computer Science and Engineering, Computer Applications and Information Technology.  To understand the  design and  working of all electronic gadgets and equipment ranging from consumer electronics to industrial electronics, from embedded systems to computers, to security systems, military equipments, to satellite communication technology. This book has been written to help students learn the basics of this subject with minimum effort. This is a basic book which will help students understand the basic concept and principles with ease and also do well in the examination. Although for detailed study a number of voluminous books on this subject are available, an attempt has been made to bring out a concise book which the students will find very useful.

The text has been systematically organized and the presentation has been kept

simple.

The text has been kept very concise but plenty of illustrations and solved examples

have been provided in each chapter.

This book will be most suitable for the students of B.Tech of all Indian Universities.

Suggestions for improvement of the book is most welcome.

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Dr. S.K. Bhattacharya Ms. Balaka Biswas

CONTENTS 

Preface

Number Systems and Codes 7

Logic Gates 49

Boolean algebra and logic simplification 65

Design of Combinational Logic Circuits 94

Arithmetic Circuits 111

Decoder, Encoder, Multiplexer, Demultiplexer 148

Sequential Circuit Design 179

Shift Registers 215

Counters 225

A/D and D/A Converters 262

Logic Family 284

1.1  Introduction

We are familiar with analog signals which are continuous with respect to time. For example, when we speak using a microphone, we create an analog electrical signal. The sound is converted into a time varying electrical signal. Quantities like sound, temperature, pressure, etc are examples of creating time varying analog signals. Sine wave, cosine wave, triangular wave, etc which we are familiar with are examples of analog signals.

Digital signals are discrete. These discrete time signals have values only at different instants of time. Square wave, binary pulses (on/off) are examples of digital signals. Examples of analog and digital signals

Fig 1.1 (a) Analog signal, (b) and (c)Digital signals

have been shown in fig 1.1 The study of digital electronics involves understanding of digital circuits and system which work on the principle of digital techniques. The operation of computers, calculators, digital watches, communication system, mobile, wi-fi network etc are examples of digital systems.

In digital electronics two discrete signal levels i.e. HIGH and LOW are representation by the binary digitals 1 and 0 respectively. One binary digit is called a bit. Binary number system i.e. 1 and 0 can be used to represent the two voltage levels i.e. HIGH and LOW respectively.

Like binary number system i.e. 1 and 0, there are other number systems used in digital electronics which will be discussed in this chapter.

1.2  Number Systems

We are familiar with 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. This number system is called decimal number system. Any number can be represented using these ten numbers. The base or radix of this number system is 10. Here, the maximum number of digital used is 10 and hence the radix is taken as 10. The other number systems that have been developed are binary, octal and hexadecimal number system. Any number system uses a set of rules of symbols to represents numbers. The four types of number systems and their base or radix are shown in table 1.1. The knowledge of number system is important for the design of digital circuits.

Table 1.1 Types of number systems

1.3  Decimal Number System

This number system uses ten digits and therefore has a base 10. The digits of number system are 0,1,2,3,4,5,6,7,8 and 9. For writing any member bigger than 9, we use two or more digits. In this member system any number is expressed as

Integer .   Fractional part

————10² 10¹ 10⁰ 10–1 10–2 10–3————

weighted values weighted values

Before and after the decimal point we use some multiplies of 10. These multiplies are called Weighted Values.

For example, decimal members 591 and 92359 are represented as 591 = 5 x 10² + 9 x 10¹ + 1 x 10⁰

923.59 = 9 x 10² + 2 x 10¹ + 3 x 10⁰ . 5 x 10–1 + 9 x 10–2

These decimal numbers can be represented as (591)10 , (923.59)10 etc. the suffix 10 indicates that these are decimal number with base 10.

1.4  Binary Number System

This number system uses only two digits, ie 0 and 1. Numbers are represented by zeroes and ones. The base or radix is 2. Representation of a number in binary code is done as

....... MSB LSB

....... 2⁴   2³   2²   2¹   2⁰ . 2–1 2–2 2–3 2–4 .........

weighted values weighted values

Before and after the decimal point we use some multiples of 2 having different weights. The weight of first bit is called the least significant bit (LSB) as its weight is minimum. The weight gets increased when we move towards left. The extreme left is called the most significant bit (MSB). After the decimal point, the weight of the first bit is 2-1. The weight decreases as we move from left to right. For example, the weight of binary members (1 0 0 11)2 and (10 10 111 . 011)2 are shown below. The decimal equivalent number also has been calculated as:

i. 1 0 0 1 1

24 23 22 21 20

N = 1 x 2⁴ + 0 x 2³ + 0 x 2² + 1 x 2¹ + 1 x 2⁰

= 16 + 0 + 0 + 2 + 1

= 19

ii. 1 0 1 0 1 1 1 . 0 1 1

26 25 24 23 22 21 20 . 2-1 2-2 2-3

N = 1 x 2⁶ + 0 x 2⁵ +1 x 2⁴ + 0 x 2³ +1 x 2² + 1 x 2¹ +1 x 2⁰ . 0 x 2-1 +1 x 2-2 + 1 x 2-3

= 64 + 0 + 16 + 0 + 4 + 2 + 1 . 0 + 0.25 + 0.125

= 87 . 375

1.5  Binary to Decimal Conversion

The procedure for conversion of binary number to decimal number is given below:

i)  Write the binary number which is to be converted into decimal number with spacing between the binary numbers.

ii)  Write the weights 2⁰, 2¹, 2², 2³ etc under each binary digit starting from the bit on the right hand side but before the decimal point. After the decimal point put weight of 2-1, 2-2, 2-3 etc.

iii)  Calculate the weighted numbers and add them ignoring the zeroes of the number.

The conversion of binary number to decimal number is illustrated through a number of examples.

Example 1.1 Convert the following binary numbers into their equivalent decimal numbers:

a) (1 1 1 0 0 1 1 0 1)2

b) (1 0 1 1 0 1)2

c) (1 0 0 1 1 0 1 . 1 0 1)2

Solution

a) 1   1   1   0 0 1   1   0   1

28 27 26 25   24   23   22   21   20

N = 1 x 2⁸ +1 x 2⁷ +1 x 2⁶ +0 x 2⁵ + 0 x 2⁴ +1 x 2³ +1 x 2² + 0 x 2¹ +1 x 2⁰

= 256 + 128 + 64 + 8 + 4 + 1

= 461

Thus, (1 1 1 0 0 1 1 0 1)2 = (461)10

b) 1   0   1 1   0 1

25   24   23   22   21   20

N = 1 x 2⁵ + 0 x 2⁴ +1 x 2³ +1 x 2² +0 x 2¹ +1 x 2⁰

= 32 + 8 + 4 + 1

= 45

Thus, (1 0 1 1 0 1)2 = (45)10

c) 1 0 0 1 1 0 1 . 1 0 1

26 25 24   23   22   21   20   . 2-1   2-2   2-3

N = 1 x 2⁶ +0 x 2⁵ + 0 x 2⁴ + 1 x 2³ + 1 x 2² + 0 x 2¹ + 1 x 2⁰ .1x2-1 +0x2-2 +1 x 2-3

= 64 + 0 + 0 + 8 + 4 + 0 + 1 . 1/2 + 0 + 1/8

= 77 . 625

Thus, (1 0 0 1 1 0 1 . 1 0 1)2 = (77.625)10

Example 1.2 Convert the following binary numbers into their equivalent decimal numbers:

a) (1 1 0 0 1 . 1 0 1)2

b) (1 0 0 0 1 1 . 0 0 1)2

c) (1 1 0 0 1 . 0 1 1)2

Solution

a) (1 1 0 0 1 . 1 0 1)2

N = 1 x 2⁴ + 1 x 2³ + 0 x 2² + 0 x 2¹ + 1 x 2⁰ . 1 x 2-1 + 0 x 2-2 +1 x 2-3

= 16 + 8 + 0 + 0 + 1 . ½ + 0 + 1/8

= 25.625

Thus, (1 1 0 0 1 . 1 0 1)2 = (25.625)10

b) (1 0 0 0 1 1 . 0 0 1)2

N = 1 x 2⁵ + 0 x 2⁴ + 0 x 2³ + 0 x 2² + 1 x 2¹ +1 x 2⁰ . 0 x 2-1 + 0 x 2-2 +1 x 2-3

= 32 + 0 + 0 + 0 + 2 + 1 . 0 + 0 + .125

= 35.125

Thus, (1 0 0 0 1 1 . 0 0 1)2 = (35.125)10

c) (1 1 0 0 1 . 0 1 1)2

N   = 1 x 2⁴ + 1 x 2³ + 0 x 2² + 0 x 2¹ +1 x 2⁰ . 0 x 2-1 + 1 x 2-2 +1 x 2-3

= 16 + 8 + 0 + 0 + 1 . 0 + .25 + .125

= 25.375

Thus, (1 1 0 0 1 . 0 1 1)2 = (25.375)10

In digital circuits binary digits are most commonly used because processing of information using electronic circuits in binary numbers is simpler than processing using decimal numbers.

Now we will take up conversion of decimal numbers into binary equivalent numbers.

1.6  Decimal to Binary Conversion

Like conversion of binary numbers into decimal numbers, we can convert a decimal number into its equivalent binary number. The procedure for conversion of a decimal number into its binary equivalent is as follows.

-  Write the decimal numbers and divide it successively by 2

-  Record the reminders of each division which will either be 0 or 1.

-  Stop the division till a 0 quotient is reached.

-  Read the reminders in the reverse order i.e. from down to upward direction. A few examples will clarify the procedure.

Example 1.3   Convert the following decimal numbers into binary numbers :

a) (135)10   b) (50)10 c) (32)10 d) (47)10   Cross-check by reverse conversion.

Solution

a) 135 2 = 67 Remainder 1

67 2 = 33 Remainder 1

33 2 = 16 Remainder 1

16 2 = 8 Remainder 0

8 2 = 4 Remainder 0

4 2 = 2 Remainder 0

2 2 = 1 Remainder 0 Read this

1 2 = 0 Remainder 1 | way Read the remainders in the reverse way.

By reading in the reverse way we get equivalent of decimal 135 as 1 0 0 0 0 1 1 1.

To cross-check we will convert (1 0 0 0 0 1 1 1) into decimal number

1 0   0   0   0   1   1 1

27 26 25   24   23   22   21   20

N = 1 x 2⁷ + 0 x 2⁶ + 0 x 2⁵ + 0 x 2⁴ + 0 x 2³ + 1 x 2² + 1 x 2¹ + 1 x 2⁰

= 128 + 4 + 2 + 1 = 135

Thus, (135)10 = (1 0 0 0 0 1 1 1)2

b) (50)10 = ( ? )2

Thus, (50)10 = (1 1 0 0 1 0)2

To cross-check, we write the binary number 1   1 0   0   1 0

25   24   23   22   21   20

N = 1 x 2⁵ + 1 x 2⁴ + 0 x 2³ + 0 x 2² + 1 x 2¹ + 0 x 2⁰

= 32 + 16 + 0 + 0 + 2 + 0

= 50

Hence the conversion is checked and found correct. c) (32)10 = ( ? )2

Thus, (32)10 = (1 0 0 0 0 0)2

To cross-check, we write the binary number as and convert it back to decimal number as

1   0   0 0 0   0

25 24 23   22   21   20

N = 1 x 2⁵ + 0 x 2⁴ + 0 x 2³ + 0 x 2² + 0 x 2¹ + 0 x 2⁰ = 32

Hence the conversion is found to be correct.

d) (47)10 = ( ? )2

Thus, (47)10 = (1 0 1 1 1 1)2

To cross-check, write the binary number and convert it back to decimal number

as

1   0   1 1 1   1

25 24 23   22   21   20

N = 1 x 2⁵ + 0 x 2⁴ + 1 x 2³ + 1 x 2² + 1 x 2¹ + 1 x 2⁰

= 32 + 0 + 8 + 4 + 2 + 1

= 47

Thus the conversion is correct.

Example 1.4 Convert the following decimal numbers into binary numbers and cross- check by reverse conversion : a) (291)10  b) (119)10

Solution

a)  291 2 = 145 Remainder 1

145 2 = 72 Remainder 1

72 2 = 36 Remainder 0

36 2 = 18 Remainder 0

18 2 = 9 Remainder 0

9 2 = 4 Remainder 1

4 2 = 2 Remainder 0

2 2 = 1 Remainder 0 Read

1 2 = 0 Remainder 1 | this way Thus, (291)10 = (1 0 0 1 0 0 0 1 1)2

To cross-check, we convert (1 0 0 1 0 0 0 1 1)2 into its decimal equivalent as

N = 1 x 2⁸ + 0 x 2⁷ + 0 x 2⁶ + 1 x 2⁵ + 0 x 2⁴ + 0 x 2³ + 0 x 2² + 1 x 2¹ + 1 x 2⁰

= 256 + 32 + 2 + 1

= 291

b)  119 2 = 59 Remainder 1

59 2 = 29 Remainder 1

29 2 = 14 Remainder 1

14 2 = 7 Remainder 0

7 2 = 3 Remainder 1

3 2 = 1 Remainder 1 Read

1 2 = 0 Remainder 1 | this way Thus, (119)10 = (1 1 1 0 1 1 1)2

To cross-check, we find decimal equivalent of (1 1 1 0 1 1 1)2 as

N = 1 x 2⁶ + 1 x 2⁵ + 1 x 2⁴ + 0 x 2³ + 1 x 2² + 1 x 2¹ + 1 x 2⁰

= 64 + 32 + 16 + 4 + 2 + 1

= 119

1.7  Conversion of Decimal Fraction into Binary Equivalent

Conversion of fractional number from one radix to another can be done using successive multiplication method. The number to be converted is multiplied by the radix of the new number system. For example, to convert a decimal fraction to binary equivalent, we will multiply the decimal fraction by 2 producing a product which will have integer portion and fractional portion. The integer part of the product becomes a numeral in the new radix number. Non existence of integer part of the product has to be considered as zero. The fractional part of the product is again multiplied by the radix i.e. 2 in this case, producing a product whose integer part becomes another numeral in the new system and the process is continued. This process is to be continued until the fractional part of the product reached zero or until the multiplication process has been carried out for sufficient times. The steps are as follows.

i)  Write the fractional decimal and multiply by the radix i.e. 2, in this case;

ii)  After each multiplication, the integer part of the product is noted separately. If the product does not contain any integer part a 0 is to be noted. The fractional part of the product is again multiplied by 2 and the process of separation of integer part is continued;

iii)  If the fractional part of the product does not becomes 0 after four or five multiplication, the process should be stopped.

iv)  The conversion result be recorded by writing the integers starting from top to bottom.

A few examples, will further clarify the process.

Example 1.5 Convert (0.75)10 into binary equivalent and check your result by reverse conversion.

Thus, (0.75)10 = (0.110)2

To check the result, we convert (.110) into decimal equivalent.

.110 = 1 x 2-1 + 1 x 2-2 + 0 x 2-3

= 0.5 + 0.25 + 0

= 0.75

Example 1.6   Convert the following decimal numbers into binary equivalent :

a)  (16.55)10 b) (85.0625)10 Cross-check the conversion by reverse conversion.

Solution

a) (16.55)10 = ( ? )2

First we will convert the integer part,

16 2 = 8 Remainder 0

For fractional part,

We close this