# Multiplexers in Digital Logic

It is a combinational circuit which have many data inputs and single output depending on control or select inputs. For N input lines, log n (base2) selection lines, or we can say that for 2^{n} input lines, n selection lines are required. Multiplexers are also known as **“Data n selector, parallel to serial convertor, many to one circuit, universal logic circuit”**. Multiplexers are mainly used to increase amount of the data that can be sent over the network within certain amount of time and bandwidth.

Now the implementation of 4:1 Multiplexer using truth table and gates.

Multiplexer can act as universal combinational circuit. All the standard logic gates can be implemented with multiplexers.

**a) Implementation of NOT gate using 2 : 1 Mux**

**NOT Gate :**

We can analyze it

Y = x’.1 + x.0 = x’

It is NOT Gate using 2:1 MUX.

The implementation of NOT gate is done using “n” selection lines. It cannot be implemented using “n-1” selection lines. Only NOT gate cannot be implemented using “n-1” selection lines.

**b) Implementation of AND gate using 2 : 1 Mux**

**AND GATE**

This implementation is done using “n-1” selection lines.

**c) Implementation of OR gate using 2 : 1 Mux using “n-1” selection lines.**

**OR GATE**

Implementation of NAND, NOR, XOR and XNOR gates requires two 2:1 Mux. First multiplexer will act as NOT gate which will provide complemented input to the second multiplexer.

**d) Implementation of NAND gate using 2 : 1 Mux**

**NAND GATE**

**e) Implementation of NOR gate using 2 : 1 Mux**

**NOR GATE**

**f) Implementation of EX-OR gate using 2 : 1 Mux**

**EX-OR GATE**

**g) Implementation of EX-NOR gate using 2 : 1 Mux**

**EX-NOR GATE**

**Implementation of Higher order MUX using lower order MUX**

**a) 4 : 1 MUX using 2 : 1 MUX**

**Three(3)** 2 : 1 MUX are required to implement 4 : 1 MUX.

Similarly,

While 8 : 1 MUX require **seven(7)** 2 : 1 MUX, 16 : 1 MUX require **fifteen(15)** 2 :1 MUX, 64 : 1 MUX requires **sixty three(63)** 2 : 1 MUX.

Hence, we can draw a conclusion,

2^{n} : 1 MUX requires **(2n- 1)** 2 : 1 MUX.

**b) 16 : 1 MUX using 4 : 1 MUX**

In general, to implement B : 1 MUX using A : 1 MUX , one formula is used to implement the same.

B / A = K1,

K1/ A = K2,

K2/ A = K3

………………

K_{N-1} / A = K_{N} = 1 (till we obtain 1 count of MUX).

And then add all the numbers of MUXes = K1 + K2 + K3 + …. + K_{N}.

For example : To implement 64 : 1 MUX using 4 : 1 MUX

Using the above formula, we can obtain the same.

64 / 4 = 16

16 / 4 = 4

4 / 4 = 1 (till we obtain 1 count of MUX)

Hence, total number of 4 : 1 MUX are required to implement 64 : 1 MUX = 16 + 4 + 1 = 21.

**An example to implement a boolean function if minimal and don’t care terms are given using MUX.**

f ( A, B, C) = Σ ( 1, 2, 3, 5, 6 ) with don’t care (7) using 4 : 1 MUX using as

**a) AB as select** : Expanding the minterms to its boolean form and will see its 0 or 1 value in Cth place so that they can be placed in that manner.

**b) AC as select** : Expanding the minterms to its boolean form and will see its 0 or 1 value in Bth place so that they can be place in that manner.

**c) BC as select** : Expanding the minterms to its boolean form and will see its 0 or 1 value in A^{th} place so that they can be place in that manner.

This article is contributed by** Sumouli Choudhury.**

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