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DIGITAL ELECTRONICS AND

MICROPROCESSORS

Rekha.AFaculty

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Gate level minimization refers to the design task of finding an

optimal gate level implementation of the boolean functions

describing a digital circuit.

The complexity of the digital logic gates that implement the booleanfunction is directly related to the complexity of the algebraic

expression from which the function is implemented.

The map method provides a procedure for minimizing the complexity

The map method is also known as the Karnaugh graph map or K-

map.

Gate Level Minimization and K-Map

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K-Map is a diagram made up of squares, with each square

representing one minterm or maxterm of the function.

Two variable Mapy

x 0 1

0

1

m0xy

m1

xy

m2

xym3

xy

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Three Variable K-Map

yz

x 00 01 11 10

0

1

m0

xyz

m1

x'yz

m3

xyz

m2

xyz

m4xyz

m5xyz

m7xyz

m6xyz

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4 Variable K-Map

yz

wx 00 01 11 10

00

01

11

10

mo m1 m3 m2

m4 m5 m7 m6

m12 m13 m15 m14

m8 m9 m11 m10

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The Karnaugh map uses the followin rules for the simplification of expressions

by groupingtogetheradjacent cells containing ones

1. Groups may not include any cell containing a zero

K-MAP Rules of simplification:

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2. Groups may be horizontal or vertical, but not diagonal

K-MAP Rules of simplification:

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3. Groups must contain 1, 2, 4, 8, or in general 2n cells.

That is if n = 1, a group will contain two 1's since 21 = 2.

If n = 2, a group will contain four 1's since 22 = 4.

K-MAP Rules of simplification:

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4. Each group should be as large as possible.

K-MAP Rules of simplification:

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5. Groups may overlap.

K-MAP Rules of simplification:

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6. Groups may wrap around the table. The leftmost cell in a row may be

grouped with the rightmost cell and the top cell in a column may be

grouped with the bottom cell.

K-MAP Rules of simplification:

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7 . There should be as few groups as possible, as long as this

does not contradict any of the previous rules.

K-MAP Rules of simplification:

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Flow chart for simplification using K-Map

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Simplify the following Boolean expression

F(x,y,z)=(2,3,4,5)

yz

x 00 01 11 10

0

11 1

1 1

F= xy+xy

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Simplify the following boolean expression

F(x,y,z) = (3,4,6,7)

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Simplify the following Boolean function

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

wx 00 01 11 10

00

01

11

10

F= y+wz+xz

Problems on K-map

1 1 1

1 1 1

1 1 1

1 1

1 1 1

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Dont Care Conditions

Functions that have unspecified outputs for some input

combinations are called incompletely specified functions.

The unspecified minterms of a function is called a dont care

conditions.

These dont care conditions are marked with X

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Simplify the boolean function F(w, x, y, z)=(1,3,7,11,15) which has the

dont care conditions d(w, x, y, z)=(0,2,5)

yz

wx 00 01 11 10

00

01

11

10

F= yz + wx

X 1 X

0 X 1 0

0 0 1 0

0 0 1 0

x 1 1 x

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Simplify the following Boolean function and Implement using gates.

f = ( a, b, c, d) = (2, 8, 11, 15) + d(3, 12, 14)

cd

ab 00 01 11 10

00

01

11

10

f = (a+b+c) ( a+c+d) (a+c+d)

x 0

x 0 x

0 0

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Simplify the following Boolean function into (a)Sum of products

(b)Product of sum

F=(A,B,C,D)=(0,1,2,5,8,9,10)

CD

AB 00 01 11 10

00

01

11

10

f= BD+BC+ACD

1 1 0 1

0 1 0 0

0 0 0 0

1 1 0 1

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F = CD+AB+BD

Apply Demorgans Theorem to F

CD

AB 00 01 11 10

00

01

11

10

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

1 1 0 1

0 1 0 0

0 0 0 0

1 1 0 1

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Variable Entered Mapping(VEM)

Simplification using K-map becomes slightly tedious if the

number of variables is more than 5.

In VEM method, the map includes entries like not only the

1s, 0s and dont cares but also Boolean expression or

variable.

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Consider the following truth Table

A B C F

0 0 0 0

0 0 1 0

0 1 0 1

0 1 1 1

1 0 0 1

1 0 1 0

1 1 0 X

1 1 1 X

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F = ABC + ABC + ABC + ABC + ABC + ABC + ABC + ABC

F.F= F = ABCF + ABCF + ABCF + ABCF +ABCF + ABCF +

ABCF + ABCF

=AB(CF+CF)+AB(CF+CF)+AB(CF+CF)+AB(CF+CF)

F= AB(C.0+C.0) + AB(C.1+C.1) + AB(C.1+C.0) + AB(C.X+C.X)

= AB(0) + AB(C+C) + AB(C) + AB(X)

= AB(0) + AB(1) + AB(C) + AB(X)

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0 C

1 X

A

B

0 1

0

1

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VEM Reading Procedure

First imagine that all the 1 entries in the map are replaced by the

map-entered variables ORed with its complement.

Perform looping over the single entry MEV.

Once all single MEV entries have been covered , rewrite the map.

Replace the MEV and MEV with 0.

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