# 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
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0
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.)

Thanks..., 2. ### scubasteve_911 Senior Member

Dec 27, 2007
1,202
1
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

Steve

3. ### Dave Retired Moderator

Nov 17, 2003
6,960
170
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).

Dave

4. ### scubasteve_911 Senior Member

Dec 27, 2007
1,202
1
Oh yeah, my bad? 