Question

Show how AND, OR Not gate functions can be implemented using only:

NAND gate

Answer 1

1. NOT Gate using NAND

Connect both inputs of a NAND gate to the same input A.

Y = (A ⋅ A)′ = A′

2. AND Gate using NAND

First NAND the inputs A and B, then use another NAND gate as a NOT gate.

X = (A ⋅ B)′

Y = (X ⋅ X)′ = A ⋅ B

Therefore,

A ⋅ B = [(A ⋅ B)′(A ⋅ B)′]′

3. OR Gate using NAND

First obtain A′ and B' using NAND gates:

A′ = (A ⋅ A)′ 

B′ = (B ⋅ B)′

Then NAND these outputs:

Y = (A′ ⋅ B′)′

By De Morgan's theorem,

Y = A + B