# How can we convert multi-input NOR gate diagram to a 2-input NOR gate diagram?

The question asks us to simplify the Boolean expression and implement it using 2-input NOR gates. I used a K-map and got the POS and drew the diagram without nor gates at first then replaced every OR with and OR-invert and the AND with invert-AND.Then I replaced the invert-AND with NOR. Lastly, I compensated the NOR bubble by complementing the output literal.

Problem: My diagram doesn't work since I used a 3-input NOR, I should use 2 input NOR.

Note:
W=A
X=B
Y=C
Z=C

My diagram:

Solution Manual:

• Do you know about Boolean algebra? Commented Jul 7, 2017 at 16:38
• Definitely, why?
– Joe
Commented Jul 7, 2017 at 16:40
• What is your question? Where is the original Boolean expression? Commented Jul 7, 2017 at 16:56
• @GlennW9IQ Assuming the solution manual version is correct, the minimized equation is: F = w' x y z'
– jonk
Commented Jul 7, 2017 at 17:55
• @GlennW9IQ Ah. I couldn't tell your purpose. (I also wouldn't have produced that solution, doing this by hand. Just as a note. But still would have probably used 7 gates.)
– jonk
Commented Jul 7, 2017 at 17:57

convert multi-input NOR gate diagram to a 2-input NOR gate diagram?

This is not a feature of any schematic drawing package that I am aware of. The answer is therefore probably "you can't". You have to delete the existing diagram and re-draw it using different gates.

This seems to be especially true since your diagram appears to be ASCII art intead of drawn with something that understands the underlying logic functions.

Just re-draw it.

My diagram doesn't work since I used a 3-input NOR, I should use 2 input NOR.

_A + _B + _C = (_A + _B) + (0 + _C)

or _A + _B + _C = (_A + _B) + (_C + _C)

Since your equation uses A, B, C and D with NOTs and the solution uses W, X, Y and Z with no NOTs, I'm going to go out on a limb and say, the solution does not match up with the question.

But to help you $X = \overline {A+B+C}$.

simulate this circuit – Schematic created using CircuitLab

Your solution is: $$Y = \overline {\overline{(A + \bar D)} + \overline{(\bar B + \bar D)} + \overline{(\bar A + \bar B + \bar C)}}$$ $$Y = (A + \bar D)(\bar B + \bar D)(\bar A + \bar B + \bar C)$$

Solution manual (W=A, X=B, Y=C & Z=D): $$F = \overline{\overline{\overline{\overline{(C + D)} + \overline{(A + B)}}} + \overline{\overline{( A + D)}}}$$ $$F = \overline{\overline{(C + D)} + \overline{(A + B)}} \ \overline{( A + D)}$$ $$F = (C + D) (A + B) \overline{( A + D)}$$

F is not even close to Y.

What was your initial Boolean expression?