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Multiplexers in Digital Logic
  • Difficulty Level : Medium
  • Last Updated : 23 Dec, 2020

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 2n 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, 
2n : 1 MUX requires (2^n – 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 

……………… 

KN-1 / A = KN = 1 (till we obtain 1 count of MUX). 

And then add all the numbers of MUXes = K1 + K2 + K3 + …. + KN
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 Ath place so that they can be place in that manner. 

 

This article is contributed by Sumouli Choudhury.
 

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