4x1 mux using NAND gates

In the post 2x1 mux using NAND gates, we discussed how we can use NAND gates to build a 2x1 multilexer. In this post, we will discuss how we can use NAND gates to build a 4x1 mux:

1. Using structural approach: As we know that a 4x1 mux can be structurally built from 2x1 muxes as shown in figure 1 below. Thus, in the same way, we can arrange the 2-input NAND gates to build 4x1 muxes as shown in figure 1.

Figure 1: 4x1 mux using NAND gates with structural approach

2. Building 4x1 mux directly from NAND gates: The logical equation of a 4x1 multiplexer is given as:

Y = (S1' S0' A + S1' S0 B + S1 S0' C + S1 S0 D)

where S1 and S0 are the selects of the multiplexer and A, B, C and D are the multiplexer inputs.

Now,  using De-morgan's law (m + n = (m'n')')

The above equation turns into,

Y = ((S1' S0' A)'  (S1' S0 B)' (S1 S0' C)' (S1 S0 D)')'

In other words,

Y = NAND (NAND(S1',S0',A),NAND(S1',S0,B),NAND(S1,S0',C),NAND(S1,S0,D)) 

Thus, we require four 3-input NAND gates and a 4-input NAND gate to implement a 4x1 mux. The implementation is shown in figure 2 below.

DESIGN PROBLEM : 4-bit increment by 2 circuit

Problem: Derive the logical expression for a 4-bit increment by 2 circuit and draw the architecture of it.

Solution: The task here is to design a circuit that increments its count by two. Since, it is a 4-bit circuit, the total number of possible states is 16. Each state transitions to the state which has a binary value two greater than it. Now, there are two possible scenarios based upon the initial state that the counter gets into:

1. It can count 0 -> 2 -> 4 -> 6 -> 8 -> 10 -> 12 -> 14 -> 0 (their binary equivalents)

2. It can count 1 -> 3 -> 5 -> 7 -> 9 -> 11 -> 13 -> 15 -> 1 (their binary equivalents)

The state transition table can be represented as shown below:

We can find the expression for outputs using K-maps as below.

Expression for D3(next): Let us first derive the expression for D3(next). The K-map can be represented as below:

The expression for D3(next) as derived from K-map is:

D3(next) = D3.D2' + D3.D1' + D3'.D2.D1

D3(next) = D3.(D2'+D1') + D3'.D2.D1.

D3(next) = D3.(D2.D1)'+D3'.(D2.D1)

D3(next) = D3 (exor) (D2.D1) 

Expression for D2(next): Given below is the K-map derived from state transition table for D2(next).

The expression for D2(next) as derived from K-map is:

D2(next) = D2'.D1 + D2.D1' = D2 (exor) D1

Expression for D1(next):  Given below is the K-map derived from state transition table for D1(next).

The expression for D1(next) is derived from K-map as:

D1(next) = D1'

Expression for D0(next): Given below is the K-map for D0(next).

The expression for D0(next) is:

D0(next) = D0

Combining all the expressions, the circuit is as given below:

Can you come up with a better solution for this problem? Let us know your views in comments.

This question was asked by one Himadri Roy Pramanik on our post your query page. You can also post your queries there. We will try to answer using our limited knowledge.