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(a) Write the truth tables of (i) AND gate and (ii) NOT gate.

(b) Show how an OR gate may be obtained with the combination of NAND gates.

 

 
 
 
 
 

Answers (1)

(a) AND gate

A logic gate which performs the Boolean function Y = A.B

A B AB
0 0 0
0 1 0
1 0 0
1 1 1

(ii) NOT gate

A logic which performs the Boolean function Y = \overline{A}

A Y = \overline{A}
0 1
1 0

(b) NAND gate

A logic gate which performs the Boolean function Y = \overline{A.B}

NAND gate is a universal gate

A B Y = \overline{A.B}
0 0 1
0 1 1
1 0 1
1 1 0

By using NAND gate we can make any gate.

For OR logic

OR logic is C = A + B

NAND logic C = \overline{AB}

By applying Demorgan's thereom

\overline{AB}= \overline{A}+\overline{B}

\overline{\overline{A}.\overline{B}}= \overline{\overline{A}}+\overline{\overline{B}}=A+B              OR gate

By using 3 NAND gate we can make an OR gate.

Posted by

Safeer PP

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