ELCT201: DIGITAL LOGIC DESIGN Dr. Eng. Haitham Omran, haitham.omran@guc.edu.eg

Dr. Eng. Wassim Alexan, wassim.joseph@guc.edu.eg

Lecture 4

Following the slides of Dr. Ahmed H. Madian

هــ 1441محرم

Winter 2019

COURSE OUTLINE

1. Introduction

2. Gate-Level Minimization

3. Combinational Logic

4. Synchronous Sequential Logic

5. Registers and Counters

6. Memories and Programmable Logic

2

LECTURE OUTLINE

• Combinational Logic Circuits • Steps of Combinational Logic Design

• The XOR Function

• Half Adder

• Full Adder

• Binary Adder/Subtractor

• Binary Multiplier

• Magnitude Comparator

• Parity Generator/Checker

3

COMBINATIONAL LOGIC

• Combinational logic circuits can have any number of inputs and outputs

• The logic states of the inputs at any given instant determine the state of the output

4

• Sequential circuits, which we will discuss later in this course, will feature circuits in which the outputs are not determined solely by the inputs at the same time

HOW TO DESIGN A COMBINATIONAL LOGIC CIRCUIT?

1. From the specifications of the circuit, determine the required number of inputs and outputs and assign a letter (symbol) to each

2. Derive the truth table that defines the required relationship between the inputs and outputs

3. Obtain the simplified Boolean functions for each output as a function of the input variables (using a K-map)

4. Sketch the logic diagram

5

6

DESIGN PROBLEM

Design a digital system whose output is defined as logically

low if the 4-bit input binary number is a multiple of 3;

Otherwise, the output will be logically high. The output is

defined if and only if the input binary number is greater

than 2

7

INPUT/OUTPUT RELATIONSHIP AND TRUTH TABLE

• Design a digital system

whose output is defined as

logically low if the 4-bit

input binary number is a

multiple of 3; otherwise, the

output will be logically high

• The output is defined if and

only if the input binary

number is greater than 2

8

BOOLEAN FUNCTION SIMPLIFICATION USING A K-MAP

9

BOOLEAN FUNCTION SIMPLIFICATION USING A K-MAP

SOP POS

𝑌𝑆𝑂𝑃 = 𝐵′𝐷′ + 𝐴′𝐶′ + 𝐴′𝐵𝐷 + 𝐵𝐶′𝐷 + 𝐴𝐵′𝐶 + 𝐴𝐶𝐷′

𝑌𝑃𝑂𝑆 = (𝐴 + 𝐵)(𝐵 + 𝐶 + 𝐷′)(𝐴 + 𝐶′ + 𝐷)(𝐴′ + 𝐵′ + 𝐶 + 𝐷)(𝐴′ + 𝐵′ + 𝐶′ + 𝐷′)

10

SKETCHING THE LOGIC DIAGRAM

THE XOR FUNCTION

The XOR symbol is denoted as ⊕

Its Boolean operation is 𝑥 ⊕ 𝑦 = 𝑥𝑦′ + 𝑥′𝑦

The XNOR symbol is denoted as ⊙

Its Boolean operation is 𝑥 ʘ 𝑦 = 𝑥𝑦 + 𝑥′𝑦′

The identities of the XOR operation are

given by:

𝑥 ⊕ 0 = 𝑥 𝑥 ⊕ 1 = 𝑥′

𝑥 ⊕ 𝑥 = 0 𝑥 ⊕ 𝑥′ = 1

Commutative and associative:

A ⊕ 𝐵 = 𝐵 ⊕ 𝐴

(A ⊕ 𝐵) ⊕ 𝐶 = 𝐴 ⊕ (𝐵 ⊕ 𝐶) = 𝐴 ⊕ 𝐵 ⊕ 𝐶

11

Z Y X

0

1

1

0

0

1

0

1

0

0

1

1

Z Y X

1

0

0

1

0

1

0

1

0

0

1

1

XOR

XNOR

𝑥⊙𝑦 = 𝑥 ⊕ 𝑦

THE XOR IMPLEMENTATION

12

• The output analysis for the first

circuit is very easy!

• The output at each of the NAND

gates for the second circuit is as

follows:

• At L1: 𝑥𝑦 = 𝑥′ + 𝑦′

• At L2:

𝑥(𝑥′ + 𝑦′) = 𝑥𝑥′ + 𝑥𝑦′ = 𝑥𝑦′ = 𝑥′ + 𝑦

• At L3:

𝑦(𝑥′ + 𝑦′) = 𝑥′𝑦 + 𝑦𝑦′ = 𝑥′𝑦 = 𝑥 + 𝑦′

• At L4:

L1

L2

L3

L4

(𝑥′+𝑦)(𝑥 + 𝑦′) = (𝑥′ + 𝑦) + (𝑥 + 𝑦′) = 𝑥𝑦′ + 𝑥′𝑦 = 𝑥 ⊕ 𝑦

13

ARITHMETIC CIRCUITS

• We will continue with the design of digital logic circuits

• One of the famous digital logic circuits is the calculator

• How to design it?

14

ARITHMETIC CIRCUITS

• An arithmetic circuit is a combinational circuit that performs arithmetic operations such as: • Addition

• Subtraction

• Multiplication

• Division

• A combinational circuit that performs the addition of two bits is called a Half Adder

outputs one digit

outputs two digits!

carry sum

So, we need two binary outputs to

represent the addition block

(carry & sum)

0 + 0 = 0

0 + 1 = 1

1 + 0 = 1

1 + 1 = 10

15

HALF ADDER

It is required to design a combinational circuit that adds two binary numbers and produces the output in the form of two bits, sum and carry

Solution:

1. We need to determine the inputs and output of the system and give letters for all of them: Our system has two inputs (x, y) and two outputs (S, C)

Half Adder

x

y C

S

16

HALF ADDER

2. Derive the truth table according to the given relation between inputs and outputs

Outputs Inputs

S C y x

0

1

1

0

0

0

0

1

0

1

0

1

0

0

1

1

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HALF ADDER

3. Obtain the simplified Boolean functions for each output as a function of the input variables using a K-map

𝐶 = 𝑥𝑦 𝑆 = 𝑥𝑦′ + 𝑥′𝑦

= 𝑥 ⊕ 𝑦

18

HALF ADDER

4. Sketch the logic diagram

19

FULL ADDER

It is required to design a combinational circuit that adds three binary numbers and produces the output in the form of two bits, sum and carry

Solution:

1. We need to determine the inputs and outputs of the system and give letters for all of them: Our system has three inputs (x, y, z) and two outputs (S, C)

Full Adder y

S

C z

x

20

FULL ADDER

2. Derive the truth table according to the given relation between the inputs and outputs

Decimal

Equivalent

Outputs Inputs

S C z y x

0

1

1

2

1

2

2

3

0

1

1

0

1

0

0

1

0

0

0

1

0

1

1

1

0

1

0

1

0

1

0

1

0

0

1

1

0

0

1

1

0

0

0

0

1

1

1

1

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FULL ADDER

3. Obtain the simplified Boolean functions for each output as a function of the input variables using a K-map

𝑆 = 𝑥′𝑦′𝑧 + 𝑥′𝑦𝑧′ + 𝑥𝑦′𝑧′ + 𝑥𝑦𝑧

= 𝑧 𝑥′𝑦′ + 𝑥𝑦 + 𝑧′ 𝑥′𝑦 + 𝑥𝑦′

= 𝑧(𝑥′𝑦 + 𝑥𝑦′) + 𝑧′(𝑥′𝑦 + 𝑥𝑦′)

= 𝑧 𝑥 ⊕ 𝑦 + 𝑧′ 𝑥 ⊕ 𝑦

= 𝑧 ⊕ 𝑥 ⊕ 𝑦

= 𝑥 ⊕ 𝑦 ⊕ 𝑧

Remember that:

𝑥⊙𝑦 = 𝑥 ⊕ 𝑦, 𝑥 ⊕ 𝑦 = 𝑥𝑦′ + 𝑥′𝑦, 𝑥⊙𝑦 = 𝑥𝑦 + 𝑥′𝑦′

22

FULL ADDER

3. Obtain the simplified Boolean functions for each output as a function of the input variables using a K-map

𝐶 = 𝑥𝑦 + 𝑥𝑧 + 𝑦𝑧

23

FULL ADDER

4. Sketch the logic diagram

𝑆 = 𝑥′𝑦′𝑧 + 𝑥′𝑦𝑧′ + 𝑥𝑦′𝑧′ + 𝑥𝑦𝑧 𝐶 = 𝑥𝑦 + 𝑥𝑧 + 𝑦𝑧

24

FULL ADDER

• The logic circuit for the full adder could also be sketched using two half adders and a single OR gate

𝐶 = 𝑥 ⊕ 𝑦 𝑧 + 𝑥𝑦

= 𝑥𝑦′ + 𝑥′𝑦 𝑧 + 𝑥𝑦

= 𝑥𝑦′𝑧 + 𝑥′𝑦𝑧 + 𝑥𝑦

𝑆 = 𝑥 ⊕ 𝑦 ⊕ 𝑧

Half adder Half adder

Compare the obtained Boolean expression

for 𝐶 here and the one obtained in slide 22

25

4-BIT BINARY RIPPLE CARRY ADDER

• Connecting n full adders in cascade allows us to add n-bit binary numbers together

• Example:

Connecting 4 full adders in cascade allows us to add 1011 to 0011.

𝐴 = 𝐴3𝐴2𝐴1𝐴0 = 1 0 1 1 𝐵 = 𝐵3𝐵2𝐵1𝐵0

= 0 0 1 1

26

4-BIT BINARY RIPPLE CARRY ADDER 𝑨𝟎 = 𝟏

𝑪𝟎 = 𝟎

𝑺𝟎 = 𝟎 𝑪𝟒 = 𝟎

𝟏 𝟏 𝟎

𝑨𝟏 = 𝟏 𝑨𝟐 = 𝟎 𝑨𝟑 = 𝟏 𝑩𝟎 = 𝟏 𝑩𝟏 = 𝟏 𝑩𝟐 = 𝟎 𝑩𝟑 = 𝟎

𝑺𝟏 = 𝟏 𝑺𝟐 = 𝟏 𝑺𝟑 = 𝟏

This adder is extremely

slow, as each stage must

wait for the previous one to

get the carry from it!

27

BINARY SUBTRACTOR

• The subtraction of binary numbers can be easily done using complements

• The subtraction 𝐴 − 𝐵 is done by taking the 2’s complement of 𝐵 and adding it to 𝐴

• The 2’s complement can be obtained by taking the 1’s complement and adding 1 to the least significant bit (LSB)

• The 1’s complement can be implemented easily with an inverter gate

• We can add 1 to the sum by making the initial input carry of

the parallel adder equal to 1

28

BINARY ADDER/SUBTRACTOR

• Subtractor

• Adder/Subtractor

FA

FA

𝑆

𝐶𝑜𝑢𝑡

𝐴

𝐶𝑖𝑛 = 1

𝐵

𝐵 𝐴

𝑆

𝐶𝑜𝑢𝑡

This is equivalent

to 𝐴 plus the 2’s

complement of 𝐵

Remember that 𝐵 ⊕ 0 = 𝐵

and 𝐵 ⊕ 1 = 𝐵′

If 𝐶𝑖𝑛 = 0, circuit acts as an Adder

If 𝐶𝑖𝑛 = 1, circuit acts as a Subtractor

𝐶𝑖𝑛

29

BINARY MULTIPLIER

2 bits × 2 bits = max 4 bits

(11)2× (11)2= (1001)2

(3)10× (3)10= (9)10

𝑀3 𝑀2 𝑀1 𝑀0 𝑀3 𝑀2 𝑀1 𝑀0

𝑆 𝐶 𝐶 𝑆

30

MAGNITUDE COMPARATOR

It is required to design a circuit to compare between two inputs 𝐴 = {𝐴1, 𝐴2} and 𝐵 = {𝐵1, 𝐵2}. Both inputs consist of two binary bits and the circuit has three outputs: Greater than, Less than or Equal

Solution:

1. We need to determine the inputs and outputs of the system and give letters for all of them: Our system has four inputs (𝐴0, 𝐴1, 𝐵0, 𝐵1) and three outputs (𝐺, 𝐿, 𝐸)

Magnitude

Comparator

𝐴0

𝐴1

𝐵1

𝐵0

𝐺

𝐿

𝐸

31

MAGNITUDE COMPARATOR

2. Derive the truth table that defines the required relationship between the inputs and outputs

32

MAGNITUDE COMPARATOR

3. Obtain the simplified Boolean functions for each output as a function of the input variables using a K-map

𝐺 𝐴1, 𝐴0, 𝐵1, 𝐵0 = 𝐴1𝐴0𝐵0′ + 𝐴0𝐵1

′ 𝐵0′ + 𝐴1𝐵1

′

33

MAGNITUDE COMPARATOR

3. Obtain the simplified Boolean functions for each output as a function of the input variables using a K-map

𝐸 𝐴1, 𝐴0, 𝐵1, 𝐵0 = 𝐴1′ 𝐴0

′ 𝐵1′ 𝐵0

′ + 𝐴1′ 𝐴0𝐵1

′ 𝐵0 +𝐴1𝐴0𝐵1𝐵0 + 𝐴1𝐴0

′ 𝐵1𝐵0′

34

MAGNITUDE COMPARATOR

3. Obtain the simplified Boolean functions for each output as a function of the input variables using a K-map

𝐿 𝐴1, 𝐴0, 𝐵1, 𝐵0 = 𝐴1′ 𝐴0

′ 𝐵0 + 𝐴0′ 𝐵1𝐵0 + 𝐴1

′ 𝐵1

35

MAGNITUDE COMPARATOR

4. Sketch the logic diagram for the output 𝐺

𝐺 𝐴1, 𝐴0, 𝐵1, 𝐵0 = 𝐴1𝐴0𝐵0′ + 𝐴0𝐵1

′ 𝐵0′ + 𝐴1𝐵1

′

𝐴1

𝐴0

𝐵0′

𝐴0

𝐵1′

𝐵0′

𝐴1

𝐵1′

𝐺

36

4. Sketch the logic diagram for the output 𝐸

𝐴1′

𝐴0′

𝐵1′

𝐵0′

𝐸 𝐴1, 𝐴0, 𝐵1, 𝐵0 = 𝐴1′ 𝐴0

′ 𝐵1′ 𝐵0

′ + 𝐴1′ 𝐴0𝐵1

′ 𝐵0 +𝐴1𝐴0𝐵1𝐵0 + 𝐴1𝐴0

′ 𝐵1𝐵0′

𝐸

𝐴1′

𝐴0

𝐵1′

𝐵0 𝐴1

𝐴0

𝐵1

𝐵0 𝐴1

𝐴0′

𝐵1

𝐵0′

MAGNITUDE COMPARATOR

37

MAGNITUDE COMPARATOR

4. Sketch the logic diagram for the output 𝐿

𝐿 𝐴1, 𝐴0, 𝐵1, 𝐵0 = 𝐴1′ 𝐴0

′ 𝐵0 + 𝐴0′ 𝐵1𝐵0 + 𝐴1

′ 𝐵1

𝐴1′

𝐴0′

𝐵0

𝐴0′

𝐵1

𝐵0

𝐴1′

𝐵1

𝐿

38

PARITY BITS

• XOR functions are very useful in systems requiring error detection and correction (e.g. communication systems)

• A parity bit is used to detect errors that occur during the transmission of binary information

• A parity bit is an extra bit included with a binary message to make the number of 1s either even or odd

• The message, including the parity bit, is transmitted and then checked for errors at the receiver

39

PARITY BITS

• An error is detected if the checked parity does not correspond with the one transmitted

• The circuit that generates the parity bit in the transmitter is called a parity generator

• The circuit that checks the parity bit in the receiver is called a parity checker

40

PARITY BITS: AN EXAMPLE OF EVEN PARITY

• Consider a three-bit message to be transmitted together with an even-parity bit

• For even parity, the bit P must be generated to make the total number of 1s even

• P is an odd function, thus we can express it using the XOR operator as

𝑃 = 𝑥 ⊕ 𝑦 ⊕ 𝑧

41

• The three-bits in the message, as well as the parity bit, are transmitted to the destination

• At the destination, a parity checker circuit is used to check for possible errors

• Since the information was transmitted with even parity, the four bits received must have an even number of 1s

• An error occurs during the transmission if the four bits received have an odd number of 1s (this means that a bit has changed in value)

• The output of the parity checker circuit will be equal to 1 if an error occurs

PARITY BITS: AN EXAMPLE OF EVEN PARITY

42

• The output of this circuit is equal to 1 whenever there is an odd number of 1s in the inputs

PARITY BITS: AN EXAMPLE OF EVEN PARITY

Does having a value of 𝐶 = 0 insures

that there are no errors?

43

• A parity generator/checker circuit can be implemented in a single circuit

• This is done if the input P is connected to logic 0 and the output is marked with P

• This is possible because

𝑧 ⊕ 0 = 𝑧

PARITY GENERATOR AND CHECKER CIRCUITS