Lecture 4 Nand, Nor Gates,Circuit Minimization

and Karnaugh Maps

Prof. Sin-Min Lee

Department of Computer Science

Boolean Algebra to Logic Gates

• Logic circuits are built from components called logic gates.

• The logic gates correspond to Boolean operations +, *, ’.

• Binary operations have two inputs, unary has one

OR+

AND*

NOT’’

Logic Circuits ≡ Boolean Expressions• All logic circuits are equivalent to Boolean expressions and any boolean expression can be

rendered as a logic circuit.• AND-OR logic circuits are equivalent to sum-of-products form.• Consider the following circuits:

A

CB

ABC ABC

A*B

y=ABC+AB*C+A*B

Y

A

B

C

Y

Y=AB*+B*C

AB*C

NAND and NOR Gates

• NAND and NOR gates can greatly simplify circuit diagrams. As we will see, can you use these gates wherever you could use AND, OR, and NOT.

NAND

NOR

A B AB

0 0 1

0 1 1

1 0 1

1 1 0

A B AB

0 0 1

0 1 0

1 0 0

1 1 0

XOR and XNOR Gates• XOR is used to choose between two mutually exclusive

inputs. Unlike OR, XOR is true only when one input or the other is true, not both.

XOR

XNOR

A B AB

0 0 0

0 1 1

1 0 1

1 1 0

A B A B

0 0 1

0 1 0

1 0 0

1 1 1

NAND and NOR as Universal Logic Gates

• Any logic circuit can be built using only NAND gates, or only NOR gates. They are the only logic gate needed.

• Here are the NAND equivalents:

NAND and NOR as Universal Logic Gates (cont)

• Here are the NOR equivalents:

• NAND and NOR can be used to reduce the number of required gates in a circuit.

Example Problem• A hall light is controlled by two light switches, one

at each end. Find (a) a truth function, (b) a Boolean expression, and (c) a logic network that allows the light to be switched on or off by either switch.

x y f(x,y)

0 0 0

0 1 1

1 0 1

1 1 0

(What kind of gate has this truth table?

Let Let xx and and yy be the switches: be the switches:

Example (cont)

• One possible equation is the complete sum-of-products form:

f(X,Y) = XY* + X*Y• Use The Most Complex Machine xLogicCircuit

Module to implement the

equation.

x y f(x,y)

0 0 0

0 1 1

1 0 1

1 1 0

How to use NAND gates to build an OR gate?

Truth TableA B C D Q

0 0 1 1 0

0 1 1 0 1

1 0 0 1 1

1 1 0 0 1Hint 1 : Use 3 NAND gates

Hint 2 : Use 2 NAND gates to build 2 NOT gates

Hint 3 : Put the 3rd NAND gate after the 2 “NOT” gates

A

B

C

DQ

How to use NAND gates to build a NOR gate?

Truth TableA B C D E Q

0 0 1 1 0 1

0 1 1 0 1 0

1 0 0 1 1 0

1 1 0 0 1 0

A

B

C

DQ

E

Hint 4 : Put the “NOT” gate after “OR” gate

Hint 3 : Use a NOR gate to build a NOT gate

Hint 2 : Use 3 NAND gates to build an OR gate

Hint 1 : Use 4 NAND gates

How to use NOR gate to build a NOT gate?Truth Table

A B C Q

0 0 0 1

1 1 1 0Hint!

Link inputs B & C together (to a same source).

A Q

B

C

When A = 0, B = C = A = 0When A = 1, B = C = A = 1

How to use NOR gates to build an OR gate?Truth Table

Hint 1 : Use 2 NOR gates

AQ

B

C

Hint 2 : From a NOR gate, build a NOT gate Hint 3 : Put this “NOT” gate after a NOR gate

D

E

A B C D E Q

0 0 1 1 1 0

0 1 0 0 0 1

1 0 0 0 0 1

1 1 0 0 0 1

NOR NOT

How to use NOR gates to build an AND gate?Truth Table

Hint 1 : Use 3 NOR gatesHint 2 : From 2 NOR gates, build 2 NOT gates Hint 3 : Each “NOT” gate is an input to the 3rd NOR gate

A B C D Q

0 0 1 1 0

0 1 1 0 0

1 0 0 1 0

1 1 0 0 1

A

B

C

D

Q

How to use NOR gates to build a NAND gate?Truth Table

Hint 2 : Use 3 NOR gates to build a NAND gate (previous lesson)

Hint 3 : Use the 4th NOR gate to build a NOT gate

Hint 4 : Insert “NOT” gate after “NAND” gate

Hint 5 : NOT-NAND = AND

Hint 1 : Use 4 NOR gates

A

B

C

DQ

E A B C D E Q

0 0 1 1 0 1

0 1 1 0 0 1

1 0 0 1 0 1

1 1 0 0 1 0

How to use NAND gates to build a NOT gate?Truth Table

Hint! Link inputs B & C together (to a same source).

When A = 0, B = C = A = 0When A = 1, B = C = A = 1

A Q

C

B A B C Q

0 0 0 1

1 1 1 0

How to use NAND gates to build an AND gate?Truth Table

AQ

B

A B C Q

0 0 1 0

0 1 1 0

1 0 1 0

1 1 0 1

C

Hint 1 : Use 2 NAND gatesHint 2 : From a NAND gate, build a NOT gate

Hint 3 : Put this “NOT” gate after a NAND gate

NAND NOT

Hint 4 : NOT-NAND = AND

How to use NAND gates to build an OR gate?Truth Table

A B C D Q

0 0 1 1 0

0 1 1 0 1

1 0 0 1 1

1 1 0 0 1Hint 1 : Use 3 NAND gates

Hint 2 : Use 2 NAND gates to build 2 NOT gates

Hint 3 : Put the 3rd NAND gate after the 2 “NOT” gates

A

B

C

DQ

How to use NAND gates to build a NOR gate?Truth Table

A B C D E Q

0 0 1 1 0 1

0 1 1 0 1 0

1 0 0 1 1 0

1 1 0 0 1 0Hint 1 : Use 4 NAND gates

Hint 2 : Use 3 NAND gates to build an OR gate

Hint 3 : Use a NOR gate to build a NOT gate

A

B

C

DQ

E

Hint 4 : Put the “NOT” gate after “OR” gate

Since functions can be represented by a sum of product form of minterms, any function can be shown in the map by placing each a 1 in each square which represents a minterm in the function.

EXAMPLE: Draw the map for f = X + Y

1. Determine which minterms are needed by using truth table.

X Y f = X + Y minterm0 0 0 m0

0 1 1 m1

1 0 1 m2

1 1 1 m3

2. Draw the map and place a 1 in each square for required minterms.

1 1

1

1Y

X

0

X 1

Y

f = X + Y = m1 + m2 + m3 = X´ Y + X Y´ + X Y

Use the 2-Variable Karnaugh Mapfor Minimization

Example: Given f (X,Y) = (0, 2)

Find: Simplified sum of products

XY

X

Y

m0

m2 m3

m1

2-Variable Karnaugh Map

1. Place the 1’s corresponding to the minterms on the map.

XY

X

Y

1

1 0

0

f (X,Y) = (0, 2)

2-Variable Karnaugh Map

2. Now group the 1’s into columns or rows; in this casewe can group them in the first column.

XY

X

Y

1

1 0

0

f (X,Y) = (0, 2)

3. This column is Y´ so the simplified function f (X,Y) = Y´

2-Variable Karnaugh Map

Example 2: Given f (X,Y) = (0, 2, 3)

Find: Simplified sum of products

XY

X

Y

m0

m2 m3

m1

2-Variable Karnaugh Map

Example 3: Given f (X,Y) = (0, 1)

Find: Simplified sum of products

XY

X

Y

m0

m2 m3

m1

Three Variable Map

m0 m1 m3 m2

m4 m5 m7 m6

X´ Y´ Z´ X´ Y´ Z X´ Y Z X´ YZ´

X Y´ Z´ XY´Z X Y Z X Y Z´1X

0

0 0 0 1 1 1 10

Y

YZX

Z

There are 2N = 23 = 8 squares.

The minterms are arranged so that only one variable changes from 0 to 1 or from 1 to 0 as you move from square to square in the vertical or horizontal direction.

For any two adjacent squares, only one literal changes from complemented to non-complemented (normal).From this property the left and right ends of the map are “adjacent.”

Three Variable Map

Using the distributive and complement properties, any two adjacent minterms can be combined and simplified to a single term with one less literal.

X´Y´Z´ X´ Y´ Z X´ Y Z X´ Y Z´

X Y´Z´ X Y´ Z X Y Z X YZ´1X

0

0 0 0 1 1 1 10

Y

YZX

Z

Example: combine adjacent squares for X´ Y´ Z and XY´Z

X´Y´ Z + XY´Z = Y´ Z ( X´ + X) = Y´ Z · 1 = Y´ Z

Example: Combine adjacent squares for X Y´Z´ and X YZ´

XY´Z´ + X YZ´ = X Z´ ( Y´ + Y) = X Z´ · 1 = X Z´

Combining terms on the Karnaugh Map simplifies Boolean functions.

X´Y´Z´ X´ Y´ Z X´ Y Z X´ Y Z´

X Y´Z´ X Y´ Z X Y Z X YZ´1X

0

0 0 0 1 1 1 10

YYZ

X

Z

Example: Simplify f = X´Y´ Z´ + X Y Z + X´Y´ Z + X´ Y Z

1X

0

0 0 0 1 1 1 10

Y

YZX

f = X´Y´ + Y Z

Z

1

1

1 1

Even if the function is not in its simplest form, we can stilluse the map to simplify it further.

Example: Simplify f = X´ Y´ + Y´ Z + X´ Z + X Y Z

1X

0

0 0 0 1 1 1 10

Y

YZX

Z f = Z + X´ Y´ (by further grouping of minterms)

1 1

1

1

1

3-Variable Karnaugh Map

Example: Given f (X,Y,Z) = (0, 2, 3, 4, 7)

Find: Simplified sum of products

XYZ

X

Y

m0

m4m5 m7

m6

m2m3m1

Z

3-Variable Karnaugh Map

Example: Given f (X,Y,Z) = (0, 2, 3, 4, 7)

Simplified sum of products: f = YZ + YZ + XY

XYZ

X

Y

1

1 0 1 0

110

Z

Truth Table

f (X,Y,Z) = (0, 2, 3, 4, 7)

Simplified sum of products: f = Y´Z´ + YZ + X´Y

0 0 0 10 0 1 0 0 1 0 10 1 1 11 0 0 11 0 1 01 1 0 01 1 1 1

The truth table for f:

X Y Z f

f

Y

Z

Y

Z

X

f (X,Y,Z) = (0, 2, 3, 4, 7) = Y´Z´ + YZ + X´Y

Y

3-Variable Karnaugh Map

Example 2: Given f (X,Y,Z) = (2, 3, 4, 5)

Find: Simplified sum of products

XYZ

X

Y

m0

m4m5 m7

m6

m2m3m1

Z

3-Variable Karnaugh Map

Example 2: Given f (X,Y,Z) = (1, 2, 5, 6, 7)

Find: Simplified sum of products

XYZ

X

Y

m0

m4m5 m7

m6

m2m3m1

Z

3-Variable Karnaugh Map

Example 3: Given f (X,Y,Z) = X´Z + X´Y + XY´Z + YZ

Find: “Sum of minterms” expression

XYZ

X

Y

m0

m4m5 m7

m6

m2m3m1

Z

X Y f0 0 0 0 1 11 0 11 1 0

Exclusive OR XOR

f = X´Y + XY´ = X Y

Exclusive OR

f (X,Y) = (1, 2) = X Y

XY

X

Y

0

1 0

1

XOR Truth Table

f (X,Y,Z) = X Y Z = (1, 2, 4, 7)

0 0 0 00 0 1 1 0 1 0 10 1 1 01 0 0 11 0 1 01 1 0 01 1 1 1

The truth table for f:

X Y Z f

Exclusive OR

Y

X

Zf (X,Y,Z) = X Y Z

Four Variable Map

m0 m1 m3 m2

m4 m5 m7 m6

m12 m13 m15 m14 m8 m9 m11 m10

W x´y´z

01

W

00

0 0 0 1 1 1 10YZ

WX

Z

11

10

Y

w x y´z

X

N = 4 variables 2N = 24 = 16 square (minterms)

Row and column are numbered using a reflected-code sequence. The minterm number can be obtained by concatenation of the row and column number .

Example: Row 4 = 10, Column 2 = 01 giving 1001 = 9 decimal for W X´ Y´ Z.

Four Variable Map

wx´y´z 1 1wx´y´z´

1 1

01

00

0 0 0 1 1 1 10 YZ

11

10

wx´y´z´ w´x´y z´

W

X

Z

Notice that top and bottom edges and right and left edges are “adjacent.”

Y

1 square = a term with 4 literals2 square = a term with 3 literal 4 square = a term with 2 literals 8 square = a term with 1 literal 16 square = a function equal to 1

Z

WX

Y

01

00

0 0 0 1 1 1 10YZ

11

10

X

W

Z

WX

Y

01

00

0 0 0 1 1 1 10YZ

11

10

X

Simplify: f (W, X, Y, Z) = W X´ Z + W X Z + W´ Y Z

f = W Z + Y Z f = Z ( W + Y)

W

Simplify: f (W, X, Y, Z) = ( 0, 1, 4, 5, 6, 8, 9, 12, 13, 14)

f = Y´ + WZ´ + XZ´

Z

W

Y

01

00

0 0 0 1 1 1 10YZ

11

10

X

8 square 4 square 4 square

WX

Simplify: f = W´ X´ Y´ + X´Y Z´ + W´ X Y Z´ + W X´ Y´

f = X´ Z´ + X´ Y´ + W´ Y Z´

Z

WX

Y

01

00

0 0 0 1 1 1 10YZ

11

10

X

Simplification of kmap

• Generate all PIs

• Find EPIs

• If EPIs can cover all minterms, then it is answer. Otherwise choose some non-essential PIs (which has less cost) such that all minterms are cover.

Step 1

• Generate PIs– Blue circle are PIs

– They are the largest circle you can drawn on kmap

1 1

1 1

1 1

1

1 1

00 01

00

01

11

10

x1x2

x3x4

0

1

3

2

4

5

7

6

12

13

15

14

8

9

11

10

11 10

Step 2

• Find EPIs– Red circle are EPIs

• Minterm 5, 14, 11 can only be cover by these 3 red circle 1 1

1 1

1 1

1

1 1

00 01

00

01

11

10

x1x2

x3x4

0

1

3

2

4

5

7

6

12

13

15

14

8

9

11

10

11 10

Step 3

• EPIs cannot cover minterm 7– Choose between

green/blue circle to cover minterm 7

– Green is chosen as it is larger

• Less cost

1 1

1 1

1 1

1

1 1

00 01

00

01

11

10

x1x2

x3x4

0

1

3

2

4

5

7

6

12

13

15

14

8

9

11

10

11 10

Final

• Final result is obtained– x3x4’ + x2’x3 + x1’x3 + x2x3’x4

1 1

1 1

1 1

1

1 1

00 01

00

01

11

10

x1x2

x3x4

0

1

3

2

4

5

7

6

12

13

15

14

8

9

11

10

11 10

On the K Map a 1 was placed in the squares which representedminterms for a function. The squares not used represent thecomplement of the function. Mark these squares with 0, combine them and read the complement as a sum of productsfunction, f´. Using De Morgan’s Law on f´ will give f, the originalFunction. It will be in a product of sums form.

Example: Determine the Product of sums formfor f ( W, X, Y, Z) = (0, 1, 2, 5, 8, 9, 10)

f´ = W X + YZ + X Z´ from the mapf = (W´ + X´) ( Y´ + Z´) ( X´ + Z) by De Morgan’s

Complement and Product of Sums

Also, can determine the product of sums form directly fromThe map by recognizing that the 0’s on the map representmaxterms. Remember that a 1 on the edge of the map Represents a complemented literal for maxterms.

Thus, for f ( W, X, Y, Z) = (0, 1, 2, 5, 8, 9, 10)…….

(combine 0’s on k-map)…

f = ( Y´ + Z´) (W´ + X´) ( X´ + Z)

Complement and Product of Sums

01

00

0 0 0 1 1 1 10YZ

Z

11

10

Y

0 1 0 0

0 0 0 0

1 1 0 1

W

WX

f = ( Y´ + Z´ ) ( W´ + X´ ) ( X´ + Z)

1 1 0 1