Chapter - Boolean Logic Design

AND

A

0011

B

0101

X

0001

X A B=

OR

A

0011

B

0101

X

0111

X A B+=

NOT

A

01

X

10

X A=

EOR

A

0011

B

0101

X

0110

X A B=

NAND

A

0011

B

0101

X

1110

X A B=

NOR

A

0011

B

0101

X

1000

X A B+=

AB

AB

AB

AB

AB

AX X X

X X X

Boolean Axioms A A+ A= A A A=

Idempotent

A B+ C+ A B C+ += A B C A B C =

Associative

A B+ B A+= A B B A=

Commutative

A B C + A B+ A C+ = A B C+ A B A C +=

Distributive

A 0+ A= A 1+ 1=

Identity

A 0 0= A 1 A=

A A+ 1= A A=

Complement

A A 0= 1 0=

A B+ A B= A B A B+=

DeMorgan’s

Example Equation

X A B C+ A B C+ +=

Write the ladder logic

Some logic form examples

NAND

X A B=

X A B+=

NOR

X A B+=

X A B=

A

B

X

A B X

EOR

X A B=

X A B A B+=A B X

A B C D E C+ + F C+ =

Boolean Example

Example Description

Process Description:A heating oven with two bays can heat one ingot in each bay. When the heater is on it provides enough heat for two ingots. But, if only one ingot is present the oven may become too hot, so a fan is used to cool the oven when it passes a set temperature.

Control Description:If the temperature is too high and there is an ingot in only one bay then turn on fan.

Define Inputs and Outputs:B1 = bay 1 ingot presentB2 = bay 2 ingot presentF = fanT = temperature overheat sensor

The equation

F T B1 B2 =

Simplified

Ladder Logic

Simplified Further

Ladder Logic

Circuits can be reverse engineered to Boolean equations

A

B

C

B

C

A

X

And then to Ladder Logic

Simplify the following and write ladder logic

A B+ A B+

Simplify the following and write the ladder logic

ABCD ABCD ABCD ABCD+ + +

A B C D+ + B C+ A B C D+ ++

Simplify the following and write the ladder logic

Example Case

Problem: Design a controller using an equation that will turn on an output to a cylinder ‘C’ when only one of two optical sensors, ‘A’ or ‘B’, is on. The system will be disabled if an input switch ‘D’ is on.

Hint: start with a truth table