(Course Code: EE222) Lecture 0 3: Digital Electronics ...home.iitj.ac.in/~sptiwari/DLD/Lecture0_3_DLD.pdf · i) Familiarization with logic gates and logic building. ii) Encoders and

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Digital Logic and Design (Course Code: EE222)

Lecture 0‐3: Digital Electronics Fundamentals

Indian Institute of Technology Jodhpur, Year 2018‐2019

Course Instructor: Shree Prakash TiwariEmail: [email protected]

Office: 210, Phone: 1356

Webpage: http://home.iitj.ac.in/~sptiwari/

Course related documents will be uploaded on http://home.iitj.ac.in/~sptiwari/DLD/

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Note: The information provided in the slides are taken form text books Digital Electronics (including Mano & Ciletti), and various other resources from internet, for teaching/academic use only

Digital Logic and Design

Objectives:

• To introduce the basic concepts of digital system and the use of Boolean algebra in logic analysis

d d iand design

• Understand the principles and methodology of digital logic design at the gate and switch level, including both combinational and sequential logic elements.

T i t d b i t l f l i d i d• To introduce basic tools of logic design and provide hands‐on experience designing digital circuits and components through simple logic circuits to hardware description language

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Learning Outcomes:

• Apply Boolean algebra and other techniques to express and simplify logic expressionsto express and simplify logic expressions.

• Analyze and design combinational and sequential digital systems.

• Use different techniques among them a hardware description language and a programming language, to design digital systems.

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Digital Logic and DesignCourse Contents: • Number system: binary numbers, 1s and 2s complement, arithmetic

operations in integer and floating point systems; ASCII, binary and gray codes;

• Boolean algebra: Boolean Equations, Minimization of Boolean functions; D i i bi ti l Ci it i t d/ M lti lDesigning combinational Circuits using gates and/or Multiplexers

• Combinational circuit: Adder, decoder, multiplexers, code converters (binary, gray and BCD);

• Sequential circuit: Bistable, Monostable, latches and flip‐flops, counters (binary, ring and Johnson), shift register, timer circuits;

• Hardware Description Languages: Combinational Logic, Structural Modeling, Sequential Logic, More Combinational Logic, Finite State Machines, Parameterized Modules, Testbenches

• Digital IC families: DTL TTL ECL MOS CMOS and their interfacingDigital IC families: DTL, TTL, ECL, MOS, CMOS and their interfacing.• ADC and DAC: Sample and hold circuits, ADCs, DACs.• Memories: semiconductor memories, PALs, PLAs and FPGAs; Pipelining

and timing issues, PROMs;

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Digital Logic and DesignLaboratory: Laboratory will contain experiments from following topics:

i) Familiarization with logic gates and logic building.

ii) Encoders and decoders

iii) Adder circuits: half adder and full adder

iv) Flip flops and countersiv) Flip‐flops and counters

v) Latches and memories

vi) Seven segment display

vii) Arithmetic logic circuits

viii) Digital to analog converters

ix) Analog to digital converters

x) Serial communication

xi) AND OR and EX OR gates using Nand (7400) gates; BCD to 6 3 1 1 Code converter;xi) AND, OR and EX_ OR gates using Nand (7400) gates; BCD to 6‐3‐1‐1 Code converter; 6‐3‐1‐1 to Gray Code converter; full adder circuit using AND, OR and XOR gates; a 4 ‐ Bit comparator using logic gates; Pseudo‐random bit generator; 4 ‐ bit ripple carry adder; Master ‐ Slave J‐K Flip‐Flop using Logic gates; Bi ‐ directional counter using J‐K Flip‐flops;

Priority encoder, multiplexer and decoder; VHDL code for simulation of a 4‐ Bit fast look ahead carry adder; VHDL code for simulation of an 8‐bit signed integer multiplier.

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Tentative Assessment: First Mid Semester Exam = 20%,

Second Mid Semester Exam = 20%,

End Semester Exam = 35%,

Laboratory: 20%, Quizzes: 5%

Reference Books

• M. Morris Mano, Michael D. Ciletti, "Digital Design: With an Introduction to the Verilog HDL", 5th edition, Prentice Hall of India, 2012

• Ronald J. Tocci, Neal Widmer, Greg Moss, Digital Systems: Principles and Applications, 10th edition, Pearson, 2009.

• D. M. Harris and S. L. Harris, Digital Design and Computer Architecture, Second Edition, Morgan Kuffmann

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Evolution of Electronic Devices

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Integrated Circuits

22 nm CMOS

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Before Electronics Era

9http://www.leapsecond.com/notes/cartoons.htm

After Electronics Revolution

10http://www.leapsecond.com/notes/cartoons.htm

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There are two way of representing the

numerical value of quantities:

analog

digital

Analog representation:

A quantity is represented by a voltage, current or meter movement that is proportional to the value of that quantityof that quantity

Example :

Automobile speedometer

Audio microphone

Analog quantities can vary over a continuous rangeof values

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Digital representation:The quantities are represented not by proportional quantities but by symbols called digitsExample :….Digital watch

It provides the time of day in the form of decimal digits which represent hours and minutes (and sometimes seconds)

The digital representation of the time of day changes in discrete steps

Advantages of Digital Techniques

1. Digital systems are easier to design as circuits usedare only switching circuits having only HIGH and LOWrange

2 I f ti t i2. Information storage is easy.

3. Accuracy and precision are greater

4. Operation can be programmed.

5. Digital circuits are less affected by noise, as thespurious fluctuation in voltage (noise) are not ascritical in digital systems became the exact value of avoltage is not important.

Limitation of Digital Techniques

The real world is mainly analog

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Design Hierarchy

SYSTEM

+

CIRCUIT

GATE

MODULE

n+n+S

GD

DEVICE

CIRCUIT

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Digital Number System

Many number systems are in use in digitaltechnologytechnology

The most common are

DecimalBinaryBinaryOctalHexadecimal

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Numbers Every number system is associated with a base or radix

The decimal system has a base of 10 and uses symbols (0,1,2,3,4,5,6,7,8,9) to represent

5 4 3 2 1 05 4 3 2 1 0 5 4 3 2 1 0( )ra a a a a a a r a r a r a r a r a r

A positional notation is commonly used to express numbers

numbers

3 2 1 010(2009) 2 10 0 10 0 10 9 10

2 1 0 1 210(123.24) 1 10 2 10 3 10 2 10 4 10

An octal number system has a base 8 and uses symbols (0,1,2,3,4,5,6,7)

3 2 1 08(2007) 2 8 0 8 0 8 7 8

What decimal number does it represent?

1 08(2007) 2 512 0 64 0 8 7 8 1033

A hexadecimal system has a base of 16

3 2 1 0(2 9) 2 16 16 16 9 16BC B C

Number Symbol

0 0

1 1

2 2

10(2 9) 2 16 16 16 9 16BC B C 3 3

4 4

5 5

6 6

7 7

8 8

9 9

10 A

How do we convert it into decimal number?

1 010(2 9) 2 4096 11 256 12 16 9 16 11209BC

10 A

11 B

12 C

13 D

14 E

15 F

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Example : the number 1011.1012

23 22 21 20 2‐1 2‐2 2‐3

1 0 1 1 1 0 1

Most Significant Bit(MSB) Binary Point

.Least Significant Bit(LSB)

A Binary system has a base 2 and uses only two symbols 0, 1 to represent all the numbers

3 2 1 02(1101) 1 2 1 2 0 2 1 2

20 1

21 2

22 4

23 8Which decimal number does this correspond to ?

1 02(1101) 1 8 1 4 0 2 1 2 13

1 1 0 1 . 1 0 0 1

2‐4

24 16

25 32

26 64

27 128

28 256

29 51223 22 21 20

2‐1 2‐32‐22‐4 29 512

210 1024(K)

220 1048576(M)

2‐1 2‐2 2‐3 2‐4 2‐5 2‐6

0.5 0.25 0.125 0.0625 0.03125 0.015625

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Converting decimal to binary number

Convert 45 to binary number

10 1 0(45) .......n nb b b

1 11 1 045 2 2 ....... 2n n

n nb b b b

Divide both sides by 2

1 2 01 1 0

4522.5 2 2 ....... 2 0.5n n

n nb b b b 1 1 02 n n

1 2 01 1 022 0.5 2 2 ....... 2 0.5n n

n nb b b b

0 1b

1 2 01 1 022 0.5 2 2 ....... 2 0.5n n

n nb b b b 0 1b

1 2 1 01 2 122 2 2 ....... 2 2n n

n nb b b b

Divide both sides by 2

2 3 01 2 1

2211 2 2 ....... 2 0.5

2n n

n nb b b b 1 0b

2 3 1 011 2 2 2 2n nb b b b 1 3 211 2 2 ...... 2 2n nb b b b

3 4 01 3 25.5 2 2 ...... 2 0.5n n

n nb b b b 2 1b

3 4 1 01 4 35 2 2 ...... 2 2n n

n nb b b b

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3 4 1 01 4 35 2 2 ...... 2 2n n

n nb b b b

4 5 01 4 32.5 2 2 ...... 2 0.5n n

n nb b b b 3 1b

4 5 1 01 5 42 2 2 ...... 2 2n n

n nb b b b

5 6 01 5 41 2 2 ...... 2 0.5n n

n nb b b b 4 0b

5 1b

10 5 4 3 2 1 0(45) 101101b b b b b b


Method of successive division by 2

45 remainder

22 1

11 0

5 1

2 1 45 = 1 0 1 1 0 1

1 0

0 1

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Convert (153)10 to octal number system

10 1 0 8(153) ( ....... )n nb b b

1 110 1 1 0(153) 8 8 ....... 8n n

n nb b b b

Divide both sides by 8Divide both sides by 8

1 2 0 01 1

15319.125 8 8 ....... 8

8 8n n

n n

bb b b

0 0.1258

b 0 1b

153

19 1

remainder

19 1

2 3

0 2153 = (231)8


Convert (0.35)10 to binary number

10 1 2 3(0.35) 0. ....... nb b b b

1 21 20.35 0 2 2 ....... 2 n

nb b b

How do we find the b‐1 b‐2 …coefficients?

Multiply both sides by 2

1 11 20.7 2 ....... 2 n

nb b b 1 0b

1 2 12 30.7 2 2 ....... 2 n

nb b b

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1 2 12 30.7 2 2 ....... 2 n

nb b b

Multiply both sides by 2

1 22 31.4 2 ....... 2 n

nb b b

2 1b Note that ½+1/4+1/8+……1

1 2 23 40.4 2 2 ....... 2 n

nb b b

1 30 8 2 2 nb b b 3 0b 3 40.8 2 ....... 2nb b b 3

0 .

0 . 25

125

x2

x2

0.125 = ?


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1. 00.125 = (.001)2

x2

0 .x2

0.8125 = ?0 . 8125

x21 . 625

25

0. 5

x2

1 .x2

1. 0

x2

0.8125 = (.1101)2

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Binary numbers

1011000111

Binary digit = bit

Least significant bit or LSBMost significant bit or MSB

This is a 10 bit number

decimal 2bit 3bit 4bit 5bit

0 00 000 0000 00000

1 01 001 0001 00001

2 10 010 0010 00010

3 11 011 0011 00011

4 100 0100 00100

5 101 0101 00101

6 110 0110 00110

7 111 0111 00111

8 1000 01000

N‐bit binary number can represent numbers from 0 to 2N ‐1

9 1001 01001

10 1010 01010

11 1011 01011

12 1100 01100

13 1101 01101

14 1110 01110

15 1111 01111

Converting Binary to Hex and Hex to BinaryNumber Symbol

0(0000) 0

1(0001) 1

2(0010) 2

3(0011) 3

4(0100) 4

5(0101) 5

7 6 5 4 3 2 1 0 1 0( ) ( , )b Hexb b b b b b b b h h

7 6 5 4 3 2 1 17 6 5 4 3 2 1 0 1 02 2 2 2 2 2 2 16b b b b b b b b h h

5(0101) 5

6(0110) 6

7(0111) 7

8(1000) 8

9(1001) 9

10(1010) A

11(1011) B

12(1100) C

3 2 1 4 3 2 1 17 6 5 4 3 2 1 0 1 0( 2 2 2 )2 ( 2 2 2 ) 16b b b b b b b b h h

h1 h0

(10110011) (1011)(0011) ( 3)b HB 13(1101) D

14(1110) E

15(1111) F

(10110011) (1011)(0011) ( 3)b HexB

(110011) (11)(0011) (33)b Hex

( ) (1110)(1100) (11101100)Hex bEC

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Binary Addition/Subtraction

0

0

0

0

1

1 1

1

1

1

0

1 01

1

1

1 1

1 0 1

1 1 0

1 0 1 1

1 1 0 1

1 1 0 1 1

+ 1 1 1 0

Complement of a number

Decimal system:

9’s complement

10’s complement

9’s complement of n‐digit number x is 10n ‐1 ‐x

10’s complement of n‐digit number x is 10n ‐x

9’s complement of 85 ? 210 1 85 99 85 14

9's complement of 123 = 999 123 876

10's complement of 123 = 9's complement of 123+1 877

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Complement of a binary number

Binary system:

1’s complement

2’s complement

1’s complement of n‐bit number x is 2n ‐1 ‐x

2’s complement of n‐bit number x is 2n ‐x

1’s complement of 1011 ? 42 1 1011 1111 1011 0100

1's complement of 1001101 = ?

0110010

1’s complement is simply obtained by flipping a bit (changing 1 to 0 and 0 to 1)

2's complement of 1010 = 1's complement of 1010+1 0110

2's complement of 110010 =

Leave all least significant 0’s as they are leave first 1 unchanged and then flip all subsequent bits

001110

Leave all least significant 0 s as they are, leave first 1 unchanged and then flip all subsequent bits

1011 0101

101101100 010010100

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Advantages of using 2’s complement

Adderx1

x2

S

CY

Can we carry out Y = X1 – X2 using such an adder?1 2

x

Adder

2's Complement

Y = S if Sign = 0Y = 2's Complement of S if Sign = 1

SYx1

x1,x2: N bit numbers

SignCYx2

Sign = 0 for psotive numbers = 1 for negative numbers

22N x1 2( , ) 2NCY S x x

Note that carry will be there only if x1 – x2 is positive as 2N is N+1 bits (1 followed by N zeros)

Sign = 0 for positive numbers= 1 for negative numbers

Advantages of using 2’s complement

Adder

2' C l t


SYx1


SiCYx2

2's Complement Sign


CY

22N x1 2( , ) 2NCY S x x

Note that carry will be there only if x1 – x2 is positive as 2N is N+1 bits (1 followed by N zeros)


A zero carry implies a negative number whose magnitude (x2 – x1) can be found as follows:

1 22NS x x

1 2 2 12'scomplement of 2 ( 2 )N NS x x x x

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Adder


SYx1=1010

Example

10 01000100

Adder

2's Complement Sign


CYx2=0110

61010

1 0 1 0+ 1 0 1 0

1 0Sign = 0 for positive numbers= 1 for negative numbers

1 0 1 0 0

Adder


SYx1=0110

Example

6 11000100

dde

2's Complement Sign


CYx2=1010

100110

0 1 1 0+ 0 1 1 0

0 1Sign = 0 for positive numbers= 1 for negative numbers

1 1 0 0

It makes sense to use adder as a subtractor as well provided additional circuit required for carrying out 2’s complement is simple

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Subtraction using 10’s complement

SYx1

x1,x2: N digit numbers


8

55

x2

Adder

Sign


CY10's Complement

8

310‐3=7

10Sign = 0 for positive numbers

= 1 for negative numbers

This way of subtraction would make sense only if subtracting a number x2 from 10N is much simpler than directly subtracting it directly from x1

Representing positive and negative binary numbers

One extra bit is required to carry sign information. Sign bit = 0 represents positive number and Sign bit = 1 represents negative number

decimal Signed Magnitude

0 0000

1 0001

decimal Signed 1’s complement

0 0000

1 0001

decimal Signed 2’s complement

0 0000

1 0001

2 0010

3 0011

4 0100

5 0101

6 0110

7 0111

‐0 1000

2 0010

3 0011

4 0100

5 0101

6 0110

7 0111

‐0 1111

1 0001

2 0010

3 0011

4 0100

5 0101

6 0110

7 0111

‐1 1111‐1 1001

‐2 1010

‐3 1011

‐4 1100

‐5 1101

‐6 1110

‐7 1111

‐1 1110

‐2 1101

‐3 1100

‐4 1011

‐5 1010

‐6 1001

‐7 1000

1 1111

‐2 1110

‐3 1101

‐4 1100

‐5 1011

‐6 1010

‐7 1001

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If we represent numbers in 2’s complement form carrying out subtraction is same as addition

Adder


SYx1


CYx2

2's Complement Sign


CY

S

x1

Answer is in 2’s complement form


Adder

x2

x1,x2: N bit numbers in 2's complement

CY

Example

Adder

S

x1

x2

x1,x2: N bit numbers in 2's complement

CY

1, 2 p

0 1 0 1+ 0 0 1 0

0 1 1 1

+ 5+ 2+7

+ 5- 2+3

0 1 0 1+ 1 1 1 0

0 0 1 1

- 5+ 2- 3

1 0 1 1+ 0 0 1 0

1 1 0 1

2’s complement is 0011 = 3

- 5- 2- 7

1 0 1 1+ 1 1 1 0

1 0 0 1

2’s complement is 0111 = 7

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CODES

When numbers, letters or words arerepresented by a special group of symbols, wesay that they are being encoded and they y ggroup of symbols is called code.

Binary‐ Coded‐ Decimal Code (BCD)

If each digit of a decimal number is represented by its binary equivalent the result is a code called binarybinary equivalent, the result is a code called binary‐coded‐ decimal. Since a decimal digit can be as large as 9, 4 bits are required to code each digit.

Ex. 874 = 1000 0111 0100 (BCD)

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Gray code

It belongs to a class of codes called minimum‐change codes, in which only one bit in the code groupscodes, in which only one bit in the code groups changes when going from one stage to the next. The Gray code is an unweighted code.

So, this code is not suited for arithmetic operation but finds application in input/output devices and

f A l Di i l C (ADC )some types of Analog to Digital Converters (ADCs).

Binary Equivalent Gray Code0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 10 0 1 0 0 0 1 10 0 1 1 0 0 1 0 0 1 0 0 0 1 1 00 1 0 0 0 1 1 00 1 0 1 0 1 1 10 1 1 0 0 1 0 10 1 1 1 0 1 0 01 0 0 0 1 1 0 01 0 0 1 1 1 0 11 0 1 0 1 1 1 11 0 1 1 1 1 1 01 1 0 0 1 0 1 01 1 0 1 1 0 1 11 1 1 0 1 0 0 11 1 1 1 1 0 0 0

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Alphanumeric Codes

An alphanumeric code represents all of the various characters and functions that are found in a standard typewriter (or computer) keyboard.yp ( p ) y

ASCII Code: The most widely used alphanumeric code, the American standard code for Information Interchange as is used in most micro computers and minicomputers and in many mainframes. The ASCII d i 7 bit d d it h 27 128 ibl dcode is a 7 bit code and so it has 27 =128 possible code

groups.

Character 7- Bit ASCII Hex A 1000001 41B 1000010 42C 1000011 43... …. …... …. …

Z 1011010 5A0 0110000 30

… … …

9 0111001 39

a, b, … blank, etc.

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Example : When writing a BASIC Programme,

instruction GO TO 25

G 01000111

O 01001111O 01001111

T 01010100

O 01001111

2 00110010

5 00110101

0 is added because the codes must be stored as bytes (8bits).

This adding of an extra bit is called padding with 0s.

Boolean Algebra

Algebra on Binary numbers

A variable x can take two values {0,1} 0

False

No

Low voltage

1

True

Yes

High voltage

Basic operations:

1 2AND: y = x . x

Y is 1 if and only if both x1 and x2 are 1, otherwise zero

x1 x2 y

0 0 0 0 1 0

1 0 0 1 1 1

Truth Table

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Basic operations:

1 2OR: y = x + x

Y is 1 if either x1 and x2 is 1. Or y= 0 if and only if both variables are zero

x x yx1 x2 y

0 0 0 0 1 1

1 0 1 1 1 1

NOT: y = xyx

1

1

0

0

What next……

• Digital Cont..

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Documents

(Course Code: EE222) Lecture 0 3: Digital Electronics ...home.iitj.ac.in/~sptiwari/DLD/Lecture0_3_DLD.pdf · i) Familiarization with logic gates and logic building. ii) Encoders and