I try to resolve a problem, how to implement a xor gate with nand's gate.
A xor B = A'B + AB'
So, this becomes :
A xor B = A'B + AB' + AA' + BB' = A(A' + B') + B(A' + B') = (A + B)(A' + B') =
(with De Morgan apllied on the second term) = (A + B)(AB)' = ….. ?
In this point i am blocked.
If someone can help me, please.
Thank you.
Best Answer
You've got a good start. Just re-distribute the second term over the first:
(A + B)(AB)' = A(AB)' + B(AB)'
And then apply De Morgan to the whole thing:
A(AB)' + B(AB)' = ((A(AB)')'(B(AB)')')'
The Boolean expression gets to be a little hard to read, but it translates to the following circuit:
simulate this circuit – Schematic created using CircuitLab
Note that the exact same network works if all of the NAND gates are changed to NOR gates, except that you get an XNOR gate — the output is high if the inputs are equal.