Electronic – Boolean expression to logic circuit

digital-logic

I am having trouble trying to convert the boolean expression A and B or Not C to an all nor gate circuit i have tried a couple of times but cannot seem to get the circuit right using ALL NOR GATES only. Any help would be appreciated. I think I need about 6 gates in total but i keep getting lost and confused

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this is what i submitted i will go back and change the curve on them so they appear to look more like NOR but the circuit is correct to the truth table

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Hoefully this image looks a bit clearer. Note this is how the software performs meaning i can only select and insert imgages i cannot really change the curve so much of the gate itself only drag it out ever so slightly

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that is the expression i came up with and a hand sketch of the circuit instructor just went on to say my simulation were not NOR gates

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Best Answer

A two-level implementation with NOR gates requires that the function is simplified into product-of-sums (POS) form.

Your function \$F=AB+C' \$ is not in that form. Convert it into POS form, invert it twice, and use DeMorgan's Law to obtain a function that is realizable using only NOR gates.

Edit

The first part is to convert the your function into POS form. This can be done elegantly by using the AND-distributive law $$F=(A+C')(B+C') $$ Inverting this once gives $$F'=[(A+C')(B+C')]'=(A+C')'+(B+C')' $$

All you have to do is invert the expression one last time, and you can realise the function with only NOR gates.

Edit 2

From the comments

i have already gotten the function down to the last step in the second last image i have been told that much is true, I think my actual software schematic may have been unreadable as when i test it against the truth table it all checks out

I think you have worked on this long enough. I will show you my solution.

Inverting \$F' \$ from the first edit leaves us with the final expression. $$F''=F=[(A+C')'+(B+C')']' $$

So the circuit can be realized with 4 NOR gates like this. enter image description here

Note that we here use that a NOR gate with duplicate inputs acts as an inverter. However, this is necessary as OP only can use NOR gates.

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