One of the very popular digital design interview questions is how to make a gate using a 2:1 multiplexer, or a mux as we usually refer to. This tests your knowledge of the basic boolean logic or boolean algebra.

Key is to know basic manipulation rules of boolean algebra. There are several ways to deal with this question. The way I would start is from the equation of mux. Reduce it to the equation of a gate that you want.

Equation of a mux with inputs A & B, select S and output Out is following.

Out = S * A + (S)bar * B

We know that equation of inverter is :

Out = (In)bar

We can see that equation of mux has one ‘bar’ term, which is ‘(S)bar’. We just need to keep this bar term and we need to zero out the other terms ‘S * A’. We can achieve that by tying A to ‘0’. We can keep ‘(S)bar’ term by tying B to ‘1’.

Out = S * A + (S)bar * B

Tie A to ‘0’ and B to ‘1’.

Out = S * 0 + (S)bar * 1

Out = 0 + (S)bar

Out = (S)bar

Out inverts S.

Here is the figure for such a circuit.

Figure 1. Inv using 2:1 MUX

How can we make an AND gate now ?

-SS.