need some info about the logic circuits

Discussion in 'Homework Help' started by shaider530, Feb 9, 2008.

  1. shaider530

    Thread Starter New Member

    Jul 7, 2007
    hi! Please help me, my prof. gave us a simplified boolean expression. (F= AC + BCbarD). And he wants to convert it into a logic circuits using NAND gates only.
    What would be the easier way to solve that problem.
    Or better yet what is the answer? (At least the simplified boolean expression using NAND gates.)

  2. scubasteve_911

    Senior Member

    Dec 27, 2007
    I'm kind of rusty at doing boolean reduction, I usually stick to K-maps, but I gave it a shot.

    F= AC + BC'D
    = (AC + BC'D)"
    = ((AC)'(BC'D)')'

    AC are inputs to a 2-input NAND1
    C is a shorted input to a 2-input NAND2 (make an inverter)
    C'BD are inputs to a 3-input NAND3

    I think I'm right, but I could be wrong

  3. Dave

    Retired Moderator

    Nov 17, 2003
    The Boolean reduction is correct. You need one more 2-input NAND gate whose inputs are from the outputs of NAND1 and NAND3.

    The trick to these questions is as follows:

    1. Where relevant take out an common factors (this didn't need to be done in this question).

    2. If relevant use the double NOT for the whole function, i.e. F => {NOT gate} => NOT(F) => {NOT gate} => NOT(NOT(F)) = F. Why do you need to do this? Consider DeMorgans theorems (part 3).

    3. Apply DeMorgans theorems, in particular (A + B)' = A'B'. DeMorgans may need to be applied several times to remove the OR operands.

    4. Remember you can create a NOT gate using a NAND gate by tying the inputs together (as Steve states above).

  4. scubasteve_911

    Senior Member

    Dec 27, 2007
    Oh yeah, my bad? :p