# 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
17,595
5,470
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?

4. ### WBahn Moderator

Mar 31, 2012
23,398
7,106
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?

Mar 31, 2012
23,398
7,106