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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    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
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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
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    Hello,
    Can you explain a little? I dont understand how to do it?
     
  4. beenthere

    Retired Moderator

    Apr 20, 2004
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    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
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    can you write it? if you have time?
     
  6. beenthere

    Retired Moderator

    Apr 20, 2004
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    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
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    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
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    Have you worked out the truth table?
     
  9. Mithryll

    New Member

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