boolean algebra to nand

Discussion in 'Homework Help' started by Wizardi, Mar 18, 2007.

  1. Wizardi

    Thread Starter New Member

    Mar 18, 2007
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    I have a problem getting this following boolean equation into to using only nand gates. AB+DA+CA+DCB ... just can't figure it out with the basic rules of boolean algebra.
     
  2. thingmaker3

    Retired Moderator

    May 16, 2005
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    Have another look at DeMorgan's Theoram. (Theorem? Theorom?) Have another look at DeMorgan's stuff.

    And remember... nobody restricted anyone to the use of two-input gates...
     
  3. Wizardi

    Thread Starter New Member

    Mar 18, 2007
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    That was given to me in the assignment that I can you more than 2 inputs. But my problem is how to get complenment lines in order to form nand gates? please elaborate your answer
     
  4. thingmaker3

    Retired Moderator

    May 16, 2005
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    You must use NAND gates exclusively, yes? Your outputs get inverted, yes?

    If AB=X, then (AB)'=X'.

    That's where DeMorgan's Theorem comes in.

    The circuit can be made with three gates from a 7400, one from a 7410, and one from a 7420.

    If you are still having trouble, try making a truth table for the problem.
     
  5. nomurphy

    AAC Fanatic!

    Aug 8, 2005
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    See attached:
     
  6. Dave

    Retired Moderator

    Nov 17, 2003
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    Please note [' = NOT] and I am not simplifying the expression in the following:

    AB+DA+CA+DCB

    Double NOT the expression (because double NOT does not alter the expression):

    (AB+DA+CA+DCB)''

    Using DeMorgans (A+B)' = A'.B'

    Therefore:

    ((A.B)'.(D.A)'.(CA)'.(DCB)')'

    Which can be implemented using purely NAND gates.

    You may be advised to look at simplifying the expression, although that depends on what your assignment requires.

    Dave
     
  7. Wizardi

    Thread Starter New Member

    Mar 18, 2007
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    I see your point dave but the way you did it requires a not gate and nand gates...And my assignment was to use only nand gates.
     
  8. Dave

    Retired Moderator

    Nov 17, 2003
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    It doesn't, it requires:

    - 1x4-I/P NAND
    - 1x3-I/P NAND
    - 3x2-I/P NAND

    Incidently, you can also make a NOT gate out of a 2-I/P NAND gate by tying the two inputs together.

    Dave
     
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