# Boolean expression to circuit diagram

#### kishanpatel50

Joined May 20, 2009
3
Hi, I came across an exam question today which asked me to draw a circuit diagram using only 1 NOT gate, 1 XOR gate and 1 OR gate of the following expression

f = A'B'C' + A'B'C + AB'C + A'BC
Which i've simplified to

f=A'B' + B'C + A'C

Ive been trying for hours but i cant seem to find an answer.
I don't think its possible There was a hint that said that it may be appropriate to use P=A+B but the final answer must be in terms of A,B and C
The gates can have any number of inputs.

the problem could also be that i interpreted the question wrong and
f = A'B'C' + A'BC + ABC + AB'C which simplifies to
f = A'B'C' + AC + BC

Either way, i cant make the circuit.

#### PRS

Joined Aug 24, 2008
989
Dog gone it, kish, I used to know this stuff. Your question makes me feel like I have forgotten too much. But if you reasoned out your simplification right, could you not realize the first term, A'B' with the Or gate (A or B)? For the second and third terms C is required in either case but also you need this A' + B'. So you have the case C(A'B + A'B') The latter is a case of ANDing C to an XOR gate with A and B inputs. So, can you draw a logic gate diagram from that?

#### mik3

Joined Feb 4, 2008
4,843
Are you sure for the kind of gates you have to use?

#### kishanpatel50

Joined May 20, 2009
3
@Mik3 Yeah, im sure about the gates

@PRS A'B' is the same as (A or B)' ie all notted. If you meant that then fair enough.
This would lead to an equation (A+B)' + C(A xor B).
I understand how you to get there but doesnt allow me to only use the gates specified. with the above equation, i still need an AND gate and an extra NOR gate.

Last edited:

#### Ratch

Joined Mar 20, 2007
1,070
kishanpatel50,

the problem could also be that i interpreted the question wrong and
f = A'B'C' + A'BC + ABC + AB'C which simplifies to
f = A'B'C' + AC + BC

Either way, i cant make the circuit.
Let's continue on with what you have so far.

F = (A'B')(C)'+(A+B)C
F = (A'B')(C)'+(A'B')'C by DeMorgan's theorem
F = (A'B') XOR C

So, take a 2-input OR gate and attach inputs A and B. Feed the output of the OR gate to the NOT gate and the output of the NOT gate to one input of a XOR gate. Attach the C input to the other input of the XOR gate and the output of the XOR gate should be what you want.

Ratch

#### mik3

Joined Feb 4, 2008
4,843