answersLogoWhite

0

You'l need 5 4 to 1 muxes for making a 16 to 1 mux if your inputs are say W(0)-W(15) i.e 16 inputs ..... you start of with giving 4 inputs each to the 4 to 1 muxes the select lines for all 4 4 to 1 muxes will be common now each of the four 4 to 1 muxes is giving you one o/p so ..... take each of those 4 outputs and give them to the fifth 4 to 1 mux and voila you have a final o/p corresponding to 16 inputs !!! THIS IS HOW IT WILL LOOK LIKE inputs outputs mux 1 : w(0)w(1)w(2)w(3) m(1) mux2 : w(4)w(5)w(6)w(7) m(2) mux3 : w(8)w(9)w(10)w(11) m(3) mux4 : w(12)w(13)w(14)w(15) m(4) taking the above 4 outputs and giving them 2 mux5 mux5 : m(1)m(2)m(3)m(4) m(5) m(5) is the final output corresponding to 16 inputs W(0)-W(15)

User Avatar

Wiki User

15y ago

Still curious? Ask our experts.

Chat with our AI personalities

JudyJudy
Simplicity is my specialty.
Chat with Judy
DevinDevin
I've poured enough drinks to know that people don't always want advice—they just want to talk.
Chat with Devin
BlakeBlake
As your older brother, I've been where you are—maybe not exactly, but close enough.
Chat with Blake
More answers

You can design an 8-to-1 multiplexer using two 4-to-1 multiplexers, and a 2-1 multiplexor.

The 8 inputs would be connected to the two 4-1's using two of the selector inputs and the outputs of the two 4-1's would be connected to the 2-1 using the third selector input.

If the 4-1's have tri-state ouputs, you can eliminate the 2-1, and use the third selector input, and its complement, to drive the two 4-1's. You will need an inverter in this case. You just need to be careful that the 4-1's do not drive the output at the same time - this could result in large current spikes on GND and VCC, and you don't want that - open collector outputs, as opposed to totem pole outputs, are a wonderful solution to this problem - it all depends on required propagation delay time.

User Avatar

Wiki User

15y ago
User Avatar

A multiplexer will have 2n inputs, n selection lines and 1 output.

An 8 input multiplexer accepts 8 inputs i. e. 23. We also know that an 8:1 multiplexer needs 3 selection lines.

A 4 input multiplexer accepts 4 inputs i. e. 22. We also know that a 4:1 multiplexer needs 2 selection lines.

To realize an 8:1 multiplexer, two 4:1 multiplexers are required. They provide 8 inputs (4+4).

Join the two selection lines of each MUX. Now we require 8 combinations from selection lines. i. e.

000, 001, 010, 011

-------------

100, 101, 110,111.

We know that 00, 01, 10 11 are common. Only the first bit differs (0 or 1). Hence, apply the third selection line as it is (i. e. 1) to upper 4:1 MUX and apply it complimented (i. e. 0) to lower MUX. Now it acts as 8:1 MUX.

User Avatar

Wiki User

12y ago
User Avatar

it can be done using two 8x1 mux accepting 16 inputs ,output of each 8x1 mux goes to 2x1 mux with A,B,C as input selector to 8x1 mux (with A as MSB) and D as selector to 2x1 mux.

User Avatar

Wiki User

11y ago
User Avatar

5

User Avatar

Wiki User

12y ago
User Avatar

Add your answer:

Earn +20 pts
Q: How do you construct 16x1 multiplexer circuit using 8x1 multiplexer using diagram?
Write your answer...
Submit
Still have questions?
magnify glass
imp