Unit-3.pptx

ELE101: Fundamentals of Electrical and

Electronics Engineering

Module 3 Contents

 _{Basic logic circuit concepts}

 Basic Gates and Universal Gates

 _{Representation of numerical data in binary form}  Binary to decimal, Octal, Hexadecimal

 _{Boolean algebra}

 _{Combinational logic circuits- Half adder, full adder}  Synthesis of logic circuits

 _{Minimization of logic circuits}  Sequential logic circuits

 _{Computer organization}  Memory types

Digital Electronics



Digital Electronics represents information (0,

1) with only two discrete values.



Ideally

Gates

 The most basic digital devices are called gates.

 Gates got their name from their function of

allowing or blocking (gating) the flow of digital information.

 A gate has one or more inputs and produces an

output depending on the input(s).

 A gate is called a combinational circuit.

 Three most important gates are: AND, OR, NOT

 Other gates are NAND, NOR, XOR and XNOR

gates

Digital Logic

Universal Gates: NAND Gate as an

Inverter Gate

Equivalent to Inverter

X Z 

X X

NAND Gate as an AND Gate

X Y Z 0 0 0 0 1 0 1 0 0 1 1 1 X

Y

NAND Gate Inverter

Equivalent to AND Gate

Y X Y

X Z  

NAND Gate as an OR Gate

X Y Z 0 0 0 0 1 1 1 0 1 1 1 1

Equivalent to OR Gate

X Y NAND Gate Inverters Y X Y X Y X

Z      X

NAND Implementation

Example:

Design a NAND Logic Circuit that is equivalent to the circuit shown below.

C A C

NAND Implementation

NAND Implementation

NAND Implementation

NAND Implementation

Proof of Equivalence

C _{B C}

Number Systems

 R is the radix or base of the number system

 Must be a positive number

 R digits in the number system: [0 .. R-1]

 Important number systems for digital systems:

 _{Base 2 (binary): [0, 1]}

 Base 8 (octal): [0 .. 7]

 _{Base 16 (hexadecimal): [0 .. 9, A, B, C, D, E,}

Conversion of Decimal

Integer

 Use repeated division to convert to any

base

 N = 57 (decimal)

 _{Convert to binary (R = 2) and octal (R}

= 8)

57 / 2 = 28: rem = 1 = a0 28 / 2 = 14: rem = 0 = a1 14 / 2 = 7: rem = 0 = a2 7 / 2 = 3: rem = 1 = a3 3 / 2 = 1: rem = 1 = a4 1 / 2 = 0: rem = 1 = a5

57 = 111001

57 / 8 = 7: rem = 1 = a0 7 / 8 = 0: rem = 7 = a1

57₁₀ = 71₈

 User power series expansion to

Conversion of Decimal

Fraction

 Use repeated multiplication to convert to

any base

 N = 0.625 (decimal)

 _{Convert to binary (R = 2) and octal (R =}

8)

0.625 * 2 = 1.250: a-1 = 1 0.250 * 2 = 0.500: a-2 = 0 0.500 * 2 = 1.000: a-3 = 1

0.625₁₀ = 0.101₂

0.625 * 8 = 5.000: a-1 = 5

0.625₁₀ = 0.5₈

 Use power series expansion to

 In some cases, conversion results in a

repeating fraction

 Convert 0.7₁₀ to binary

0.7 * 2 = 1.4: a-1 = 1 0.4 * 2 = 0.8: a-2 = 0 0.8 * 2 = 1.6: a-3 = 1 0.6 * 2 = 1.2: a-4 = 1 0.2 * 2 = 0.4: a-5 = 0 0.4 * 2 = 0.8: a-6 = 0

Number System Conversion

 Conversion of a mixed decimal number is

implemented as follows:

 _{Convert the integer part of the number}

using repeated division.

 _{Convert the fractional part of the decimal}

number using repeated multiplication.

 _{Combine the integer and fractional}

 Conversion between binary and octal can be

carried out by inspection.

 _{Each octal digit corresponds to 3 bits}

 101 110 010 . 011 001

2 = 5 6 2 . 3 18  010 011 100 . 101 001

2 = 2 3 4 . 5 18  7 4 5 . 3 2

8 = 111 100 101 . 011 0102  3 0 6 . 0 5

 Conversion between binary and hexadecimal

can be carried out by inspection.

 Each hexadecimal digit corresponds to 4

bits

 1001 1010 0110 . 1011 0101

2 = 9 A 6 . B 516  1100 1011 1000 . 1110 0111

2 = C B 8 . E 716  E 9 4 . D 2

16 = 1110 1001 0100 . 1101 00102  1 C 7 . 8 F

Boolean Expressions

 Literals – variables and their complements

 Examples

 F = A.B'.C + A'.B.C' + A.B.C + A'.B'.C'

 F = (A+B+C').(A'+B'+C).(A+B+C)

Boolean Expressions

Example:

Evaluate the following Boolean expression, for all combination of inputs, using a Truth

table.

Boolean Expressions

 Two Boolean expressions are equivalent if

they have the same value for each

combination of the variables in the Boolean expression.

 F₁ = (A + B)'

 F₂ = A'.B'

 How do you prove that two Boolean

expressions are equivalent?

 Truth table

Boolean Expressions

Example:

Using a Truth table, prove that the following two Boolean expressions are equivalent.

Boolean Algebra: Basic Laws and

Theorems

Commutative Law A + B = B + A A.B = B.A

Associative Law A + (B + C) = (A + B) + C A . (B . C) = (A . B) . C

Distributive Law A.(B + C) = AB + AC A + (B . C) = (A + B) . (A + C)

Null Elements A + 1 = 1 A . 0 = 0

Identity A + 0 = A A . 1 = A

A + A = A A . A = A

Complement A + A' = 1 A . A' = 0

Involution A'' = A

Absorption (Covering) A + AB = A A . (A + B) = A Simplification A + A'B = A + B A . (A' + B) = A . B DeMorgan's Rule (A + B)' = A'.B' (A . B)' = A' + B' Logic Adjacency (Combining) AB + AB' = A (A + B) . (A + B') = A

Complement

A + A' = 1

F = ABC'D + ABCD F = ABD.(C' + C)

A + AB = A

F = A'BC + A' F = A'

G = XYZ + XY'Z + X'Y'Z' +

XZ

G = XYZ + XZ + X'Y'Z' G = XZ + X'Y'Z'

H = D + DE + DEF H = D

A.(A + B) = A

F = A'.(A' + BC) F = A'

Boolean Algebra

Example:

Using Boolean Algebra, simplify the following Boolean expression.

Boolean Algebra

Example:

Using Boolean Algebra, simplify the following Boolean expression.

DeMorgan's Laws

 Can be stated as follows:

 The complement of the product (AND) is

the sum (OR) of the complements.

 (X.Y)' = X' + Y'

 _{The complement of the sum (OR) is the}

product (AND) of the complements.

 (X + Y)' = X' . Y‘

.

Proving DeMorgan's Law

Student Exercise:

Draw the AND-OR circuits for the following

Sum-of-Products (SOP) expressions:

1. F

= A'B + AC' + B'C

Student Exercise:

Draw the OR-AND circuits for the following

Product-of-Sums (POS) expressions:

1. F

= (A+B').(A'+C).(B+C')

2. F

= (A+B+D).(B'+C+D').(A'+B+C).

Single-bit Adder Circuits : The Full

Adder

The Full Adder

Cin

Cin

Cin

Cin

S Cout

S = X xor Y xor Cin

The Full Adder

Y Ci n

S

The Full Adder

Half Adder _{Half Adder}

Cin

Cin

Cin