How do you construct 16x1 multiplexer circuit using 8x1 multiplexer using diagram?

Answer 1

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)

Answer 2

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.

Answer 3

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

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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.

Answer 4

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.

Answer 5

5