XNOR gate using NAND

As we know, the logical equation of a 2-input XNOR gate is given as below:
                      Y = A (xnor) B = (A' B '   +    A B)
Let us take an approach where we consider and A' as different variables for now (optimizations related to this, if any, will consider later). Thus, the logic equation, now, becomes:
                       Y = (CD    +    A B)           -----   (i)
     where
                      C = A'     and      D = B'
De-Morgan's law states that
                                m + n = (m'n')'

Taking this into account,
                     Y = ((CD)'(AB)')' = ((A' B')'  (A B)')'
Thus, Y is equal to ((A' nand B') nand (A nand B)). No further optimizations to the logic seem possible to this logic. Figure 1 below shows the implementation of XOR gate using 2-input NAND gates.
A 2 -input XOR gate implementation using NAND, XOR gate using NAND
Figure 1: 2-input XNOR gate implementation using NAND gates

2 comments:

  1. "Thus, Y is equal to ((A' nand B) nand (A nand B')). "

    Shouldn't it be ((A' nand B') nand (A nand B))?

    ReplyDelete

Thanks for your valuable inputs/feedbacks. :-)