# Implement a full adder for two 2 bit binary numbers by using (4:1) multiplexer

Discussion in 'Homework Help' started by Nec, Nov 21, 2008.

1. ### Nec Thread Starter New Member

Nov 21, 2008
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0
Hi all,

How can implement a full adder for two bit binary numbers by using (4:1) multiplexer?

2. ### beenthere Retired Moderator

Apr 20, 2004
15,815
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Make a truth table for the various combinations of the bits. Arrange the multiplexer to select an output based on the truth table.

3. ### Mithryll New Member

Nov 22, 2008
6
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Hello,
Can you explain a little? I dont understand how to do it?

4. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
Look at the data sheet for a multiplexer like a 74153. See what it takes to select the input that gets connected to the output. Should be pretty obvious from there.

5. ### Mithryll New Member

Nov 22, 2008
6
0
can you write it? if you have time?

6. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
Write the data sheet? It's already been done - www.[B]datasheet[/B]catalog.com/datasheets_pdf/7/4/1/5/74153.shtml - 15k.

If you are referring to the truth table for a two bit ADD, surely you can work that out?

7. ### Mithryll New Member

Nov 22, 2008
6
0
not so detailed we saw.. question just wants the implement a full adder for two 2-bit binary numbers by using 4:1 multiplexer modules. and it says don't use additional gates.

8. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
Have you worked out the truth table?

9. ### Mithryll New Member

Nov 22, 2008
6
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yes i can draw a simple truth table but still i didn't understand the solution of this.

10. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
Use each pair of bits to address a multiplexer. insure the inputs are arranged such that the output agrees with the truth table.

11. ### Mithryll New Member

Nov 22, 2008
6
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is this a true truth table?

A1 A0 B1 B0 | S1 S2 C1 C2
--------------------------
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
0 | 0 | 0 | 1 | 1 | 0 | 0 | 0
0 | 0 | 1 | 0 | 0 | 1 | 0 | 0

12. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
It is a fraction of one. Make the full truth table with the carries included, and you will be able to tell how many multiplexers will be needed.

Yes, I'm being a bit dense, but it is your assignment. You really need to do some thinking on your own instead of leaning on us.

13. ### Mithryll New Member

Nov 22, 2008
6
0
thanks, i'm trying on it.