elct201: digital logic design - guc · • digital circuits are frequently constructed with nand or...

ELCT201: DIGITAL LOGIC DESIGN Prof. Dr. Eng. Tallal El-Shabrawy, [email protected]

Dr. Eng. Wassim Alexan, [email protected]

Lecture 3

هــ 1441محرم

Spring 2020

Following the slides of Dr. Ahmed H. Madian

mailto:[email protected]

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

4-VARIABLE MAP

• Notice the order of the minterms

• Remember that the cells in the top row

are adjacent to the cells in the bottom row

• Remember that cells in the most left

column are adjacent to cells in the most

right column

• Remember that cells in the four corners

are adjacent to each other

3

NOTES ON A 4-VARIABLE MAP

• The number of adjacent squares that may be combined must always represent a number that is a power of two, such as 1, 2, 4, 8 and 16

• As more adjacent squares are combined, we obtain a product term with fewer literals

• One square represents one minterm, giving a term with 4 literals

• Two adjacent squares represent a term with 3 literals

• Four adjacent squares represent a term with 2 literals

• Eight adjacent squares represent a term with 1 literal

• Sixteen adjacent squares encompass the entire map and produce a function that is always equal to logic 1

4

4-VARIABLE MAP: EXAMPLE I Simplify the Boolean expression:

𝐹 𝐴, 𝐵, 𝐶, 𝐷 = Σ(0,1,2,4,5,7,8,9,10,12,13)

5

4-VARIABLE MAP: EXAMPLE I Simplify the Boolean expression:

𝐹 𝐴, 𝐵, 𝐶, 𝐷 = Σ(0,1,2,4,5,7,8,9,10,12,13)

𝐹 𝐴, 𝐵, 𝐶, 𝐷 = 𝐶′ + 𝐵′𝐷′ + 𝐴′𝐵𝐷

6

4-VARIABLE MAP: EXAMPLE II Simplify the Boolean expression:

𝐹 𝑤, 𝑥, 𝑦, 𝑧 = Σ(0,1,2,4,5,6,8,9,12,13,14)

7

4-VARIABLE MAP: EXAMPLE II Simplify the Boolean expression:

𝐹 𝑤, 𝑥, 𝑦, 𝑧 = Σ(0,1,2,4,5,6,8,9,12,13,14)

𝐹 𝑤, 𝑥, 𝑦, 𝑧 = 𝑦′ +𝑤′𝑧′ + 𝑥𝑧′

8

We can simplify a function by using larger blocks

Do we really need all blocks?

Can we leave some out to further simplify an expression?

Any function needs to contain a special type of blocks

These are called Essential Prime Implicants

We need to define some new terms:

Implicant

Prime implicant

Essential prime implicant

9

CHOICE OF BLOCKS

Implicant (I)

Any product term in the SOP form

A block of 1s in a K-map

Prime implicant (PI)

Block of 1s that cannot be further increased

Product term that cannot be further reduced

Essential prime implicant (EPI)

A prime implicant on a K-map which covers at least one 1 which is not covered by any other prime implicant is called an Essential Prime Implicant

10

TERMINOLOGY

Is 𝐶’ an essential prime implicant?

1. Generate all Prime Implicants of the function

2. Include all Essential Prime Implicants

3. For the remaining minterms not included in the Essential Prime Implicants, select a set of other Prime Implicants to cover them, with minimal overlap in the set

4. The resulting simplified function is the logical OR of the product terms selected above

11

THE SYSTEMATIC PROCEDURE FOR SIMPLIFYING BOOLEAN FUNCTIONS

ILLUSTRATING THE TERMS: EXAMPLE I

12

The Prime Implicants are:

• Of which only two are Essential:

1 1

1 1 1

1 1

1

1

1

ILLUSTRATING THE TERMS: EXAMPLE I

13


• 𝐴′𝐷 𝑔𝑟𝑎𝑦 , • 𝐴𝐶 𝑟𝑜𝑠𝑒 , • 𝐵𝐶′𝐷′ 𝑝𝑖𝑛𝑘 , • 𝐶𝐷 𝑝𝑢𝑟𝑝𝑙𝑒 , • 𝐴𝐵𝐷′ 𝑔𝑟𝑒𝑒𝑛 , • 𝐴′𝐵𝐶′ (𝑦𝑒𝑙𝑙𝑜𝑤).

• Of which only two are Essential:

• 𝐴′𝐷 𝑔𝑟𝑎𝑦 𝑎𝑛𝑑 𝐴𝐶 𝑟𝑜𝑠𝑒

ILLUSTRATING THE TERMS: EXAMPLE II

14


• Of which only four are Essential:

1

1 1 1

1 1

1

1

ILLUSTRATING THE TERMS: EXAMPLE II

15


• 𝐵𝐷 𝑔𝑟𝑎𝑦 , • 𝐴′𝐵𝐶′ 𝑦𝑒𝑙𝑙𝑜𝑤 , • 𝐴𝐶′𝐷 𝑝𝑢𝑟𝑝𝑙𝑒 , • 𝐴𝐵𝐶 𝑔𝑟𝑒𝑒𝑛 , • 𝐴′𝐶𝐷 𝑟𝑜𝑠𝑒 .

• Of which only four are Essential:

𝐴′𝐵𝐶′ 𝑦𝑒𝑙𝑙𝑜𝑤 , 𝐴𝐶′𝐷 𝑝𝑢𝑟𝑝𝑙𝑒 , 𝐴𝐵𝐶 𝑔𝑟𝑒𝑒𝑛 𝑎𝑛𝑑 𝐴′𝐶𝐷 𝑟𝑜𝑠𝑒 .

• Use the SOP simplification on the zeros of the function in the K-map to get 𝐹’

• Find the complement of 𝐹’, i.e. 𝐹’ ′ = 𝐹

• Recall that the complement of a Boolean function can be obtained by (1) taking the dual and (2) complementing each literal

• Or by using DeMorgan’s theorem

16

PRODUCT OF SUMS SIMPLIFICATION USING K-MAPS

17

PRODUCT OF SUMS MINIMIZATION

How to generate a POS from a K-map?

• Use duality of Boolean algebra (DeMorgan’s law)

• Look at the 0s in map instead of the 1s

• Generate blocks around the 0s

This gives the inverse of 𝐹

Use duality to generate POS

Example:

𝐹 = (0,1,2,5,8,9,10)

0

0

0

0

0 0 0

1 0 0

1 1 1

1 1 1

00 01 11 10

00

01

11

10

AB

CD

18








Example:

𝐹 = (0,1,2,5,8,9,10)

0

0

0

0

0 0 0

0 0

00 01 11 10

00

01

11

10

AB

CD

19








Example:

𝐹 = (0,1,2,5,8,9,10)

𝐹′ = 𝐴𝐵 + 𝐶𝐷 + 𝐵𝐷′

𝐹 = (𝐴′ + 𝐵′)(𝐶′ + 𝐷′)(𝐵′ + 𝐷)

GATE IMPLEMENTATION

20

SOP using:

• The 1s in the k-map

POS using:

• The 0s in the k-map

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

21

EXAMPLE ON POS MINIMIZATION

Given the K-map below, produce the 𝐹′ from the zeros in the map and then obtain 𝐹 from it

𝐹′ = AB′+AC′+A′BCD ′

𝐹 = (AB′)(AC′)(A′BCD′ ) 𝐹 = (𝐴′ + 𝐵)(𝐴′ + 𝐶)(𝐴 + 𝐵′ + 𝐶′ + 𝐷)

22

DON’T CARE CONDITIONS

There may be a combination of input values which: Will never occur,

If they do occur, the output is of no concern

The function value for such combinations is called a don’t care

They are usually denoted with an 𝑋

Each 𝑋 may be arbitrarily assigned the value 0 or 1 in an implementation

Don’t cares can be used to further simplify a function

23

MINIMIZATION USING DON’T CARES

Treat don’t cares as if they are 1s to generate Prime Implicants

Delete Prime Implicants that cover only don’t care minterms

Treat the covering of the remaining don’t care minterms as optional in the selection process (i.e. they maybe, but need not be covered)

𝐹 𝑤, 𝑥, 𝑦, 𝑧 = (1,3,7,11,15) and 𝑑 𝑤, 𝑥, 𝑦, 𝑧 = (0,2,5)

What are the possible solutions?

24

MINIMIZATION EXAMPLE

24

𝐹 = 𝑦𝑧 + 𝑤′𝑥′ 𝐹 = 𝑦𝑧 + 𝑤′𝑧

Simplify the function whose K-map is shown at the right

25

EXAMPLE INVOLVING X

𝐹 = 𝐴′𝐶′𝐷 + 𝐴𝐵′ + 𝐶𝐷′ + 𝐴′𝐵𝐶′ or

𝐹 = 𝐴′𝐶′𝐷 + 𝐴𝐵′ + 𝐶𝐷′ + 𝐴′𝐵𝐷′

26

ANOTHER EXAMPLE

Simplify the function whose K-map is shown at the right

𝐹 = 𝐴′𝐶′ + 𝐴𝐵

or

𝐹 = 𝐴′𝐶′ + 𝐵𝐷′

• Digital circuits are frequently constructed with NAND or NOR gates

rather than AND & OR gates

• NAND and NOR gates are easier to fabricate with electronic components and are the basic gates used in all IC digital logic families

27

NAND AND NOR IMPLEMENTATIONS

A NAND gate has the smallest

propagation time delay among the

other gates, except for the inverter!

NAND AND NOR IMPLEMENTATIONS

• Each NAND or NOR gate requires only 4 transistors

28

Vdd Vdd

Gnd Gnd

NAND IMPLEMENTATION

• Each NAND gate requires only 4 transistors

• C = (AB)’

29

0 1

0 1

vdd

Gnd

A B

A

B

C

A B C

0 0 1

0 1 1

1 0 1

1 1 0

NAND IMPLEMENTATION


• C = (AB)’

30

0 1

0 1

vdd

Gnd

0 0

0

0

1

Gnd

vdd

vdd

A B C

0 0 1

0 1 1

1 0 1

1 1 0

NAND IMPLEMENTATION


• C = (AB)’

31

0 1

0 1

vdd

Gnd

0 1

0

1

1

Gnd

vdd

vdd

A B C

0 0 1

0 1 1

1 0 1

1 1 0

NAND IMPLEMENTATION


• C = (AB)’

32

0 1

0 1

vdd

Gnd

1 0

1

0

1

Gnd

vdd

vdd

A B C

0 0 1

0 1 1

1 0 1

1 1 0

NAND IMPLEMENTATION


• C = (AB)’

33

0 1

0 1

vdd

Gnd

1 1

1

1

0

Gnd

vdd

Gnd

A B C

0 0 1

0 1 1

1 0 1

1 1 0

NOR IMPLEMENTATION

• Each NOR gate requires only 4 transistors

• C = (A+B)’

34

0 1

0 1

vdd

Gnd

A B

C

A B C

0 0 1

0 1 0

1 0 0

1 1 0

A

B

AND NAND-BASED IMPLEMENTATION

• Each AND gate requires 6 transistors

• C = AB

35

0 1

0 1

vdd

Gnd

A B

A

B

C

A B C

0 0 0

0 1 0

1 0 0

1 1 1

Gnd

C’



• C = AB

36

0 1

0 1

vdd

Gnd

A B

A

B

C = 0

A B C

0 0 0

0 1 0

1 0 0

1 1 1

Gnd

C’=1

vdd



• C = AB

37

0 1

0 1

vdd

Gnd

A B

A

B

C = 1

A B C

0 0 0

0 1 0

1 0 0

1 1 1

Gnd

C’=0

Gnd

NOT, AND, and OR can be implemented with NAND!

38

LOGIC OPERATION WITH NAND GATE

The complement

operation is obtained

from a one-input NAND

gate that behaves

exactly like an inverter

Minimized expressions are AND-OR combinations

There are two illustrations for NAND gates

Key observation: two “bubbles” eliminate each other

Two bubbles equal a straight wire

How to generate a sum of minterms using NAND?

Use AND-invert for minterms

Use invert-OR for sum

39

CONVERSION TO NAND IMPLEMENTATION

• Sum of minterms

• Replace AND with AND-invert and

OR with invert-OR

Still the same circuit!

• Replace AND-invert and invert-OR

with NAND

𝐹 = (𝐴𝐵)′(𝐶𝐷)′ = 𝐴𝐵 + 𝐶𝐷

40


NAND

Representations

A

B

C

D

A

B

C

D

• Sum of minterms

• Replace AND with AND-invert and

OR with invert-OR

Still the same circuit!

• Replace AND-invert and invert-OR

with NAND

𝐹 = (𝐴𝐵)′(𝐶𝐷)′ = 𝐴𝐵 + 𝐶𝐷

41


NAND

Representations

Minimize and implement the function

𝐹(𝑥, 𝑦, 𝑧) = (1,2,3,4,5,7),

using only NAND gates

42

NAND EXAMPLE IMPLEMENTATION

𝐹 = 𝑥𝑦′ + 𝑥′𝑦 + 𝑧

NAND

Representations

Multilevel circuits conversion rules:

1. Convert all AND gates to NAND with AND-invert symbols

2. Convert all OR gates to NAND with invert-OR symbols

3. Check all bubbles in the diagram. For every bubble that is not compensated by another bubble, insert an inverter.

Example:

43

MULTILEVEL NAND CIRCUITS





Example:

44


NAND

Representations





Example:

45


NAND

Representations

𝑪𝑫

𝑩

𝑨

𝑩𝑪

𝑪𝑫

𝑩

𝑨

𝑩 𝑪





Example:

46


NAND

Representations

𝑩 + 𝑪𝑫

𝑨

𝑩𝑪

𝑩 + 𝑪𝑫

𝑨

𝑩 𝑪





Example:

47


NAND

Representations

𝑩+ 𝑪𝑫 𝑨

𝑩𝑪

(𝑩 + 𝑪𝑫)𝑨

𝑩 𝑪

48

MULTILEVEL NAND CIRCUITS: AN EXAMPLE

𝐹 = (𝐴𝐵′ + 𝐴′𝐵)(𝐶 + 𝐷′)

NAND

Representations

49


𝐹 = (𝐴𝐵′ + 𝐴′𝐵)(𝐶 + 𝐷′)

NAND

Representations

50


𝐹 = (𝐴𝐵′ + 𝐴′𝐵)(𝐶 + 𝐷′)

NAND

Representations

51


𝐹 = (𝐴𝐵′ + 𝐴′𝐵)(𝐶 + 𝐷′)

NAND

Representations

NOR can also replace NOT, AND & OR

There are two representations of the NOR gate:

52

LOGIC OPERATION WITH NOR GATE

The complement

operation is obtained

from a one-input NOR

gate that behaves

exactly like an inverter

Same rules as for NAND implementations

With NOR

53

CONVERTING TO NOR IMPLEMENTATIONS

𝐹 = (𝐴𝐵′ + 𝐴′𝐵)(𝐶 + 𝐷′)

𝐹 = (𝐴𝐵′ + 𝐴′𝐵)(𝐶 + 𝐷′)

NOR

Representations

elct201: digital logic design - guc · • digital circuits are frequently constructed with nand or...

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