Boolean function to Boolean function with NAND gate only

Discussion in 'Homework Help' started by Tipao, Feb 9, 2013.

  1. Tipao

    Thread Starter New Member

    Feb 9, 2013
    3
    0
    I have to create the circuit for this function: A+(BC)'+(CD)'=Z, using NAND gates.
    After asking some friends about how to do this, and searching in the forums, I been using this method:

    So I got this:

    Z= A + (BC)' + (CD)'

    Z= [A+(BC)'+(CD)']''

    Using De Morgan's...

    Z=[A'.(BC)''.(CD)'']'

    Then finally...

    Z=[A'.(BC).(CD)]'

    And I don't know what to do from now on, since (B.C) are AND, not NAND, what am I missing?
     
  2. MrChips

    Moderator

    Oct 2, 2009
    12,431
    3,360
    Put the NAND back into (BC) by double NOT gates.

    (BC) = ((BC)')'
     
  3. Tipao

    Thread Starter New Member

    Feb 9, 2013
    3
    0
    Thanks for the fast and accurate answer.
    I made a diagram of the final circuit, is it ok?

    [​IMG]
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,720
    4,788
    You final circuit doesn't use just NAND gates. You have inverters in there as well. You need to implement those with NAND gates, too. Also, you have a 3-input NAND. Be sure that that is acceptable since problems like this frequently only allow you to use 2-input NAND gates.
     
  5. Tipao

    Thread Starter New Member

    Feb 9, 2013
    3
    0
    I tried to correct it, the function remains the same this way?

    [​IMG]
     
  6. WBahn

    Moderator

    Mar 31, 2012
    17,720
    4,788
    Nope. Look at your expression carefully. Does it match your circuit?
     
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