Question

How many two input NAND gates are required to perform the action of a two input OR gate?

Answer 1

3 `"NAND"` gates

Explanation:

We know the following:

`"NAND" = bar(A*B)`

`"OR"=A+B`

De Morgan's Law gives us:

`bar(A*B) = barA + barB`

This means that a `"NAND"` gate is equal to an `"OR"` gate with the inputs inverted.

You can invert the inputs by connecting both inputs of a `"NAND"` to the same input:

`bar(A*A)=barA` and `bar(B*B)=barB`

This is the final equation:

`A+B = bar((bar(A*A))*(bar(B*B)))`

The following schematic should make it clear:

The answer is 3 `"NAND"` gates.