从 NOR 门构建 AND 门

Question

One of the alternate to getting the same output as an AND Gate, is to put together NOR gates. The operator for NOT gates is (A+B)' , However there is a problem, I can't get the same output an AND gate would.

This image is from https://en.wikipedia.org/wiki/NOR_logic

So for example if we take:

A = 1 B = 0 , I would expect 0 as a result, but that does not seem to be the case here since,

= (1 NOR 0) NOR (1 NOR 0)
= (0) NOR (0)
= 1

What am I doing wrong?

Answer 1

There is indeed an error in your expression.

The key here is to see that the first two ports each take a pair of values that are duplicates. So we have

(A NOR A)
(B NOR B) 

This evaluates a NOT operation, so we get:

(NOT A)
(NOT B)

These two are then the operands of the final NOR, giving the desired result.

For your example with A = 1 B = 0 :

(1 NOR 1) == 0
(0 NOR 0) == 1

And

0 NOR 1 == 0

All possibilities

A	B	A NOR A	B NOR B	(A NOR A) NOR (B NOR B)
0	0	1	1	0
0	1	1	0	0
1	0	0	1	0
1	1	0	0	1