# Creating a circuit with only NOR gates

#### anonkun123

Joined Mar 31, 2020
10
Start from scratch.

Do you know DeMorgan's Theorems?

View attachment 203057
Yes, I do. !B + !D = !(BD).
In this case, the diagram shows !(BD)(B+C), right?

#### MrChips

Joined Oct 2, 2009
21,134
Yes, I do. !B + !D = !(BD).
In this case, the diagram shows !(BD)(B+C), right?
No, no, no!
We are so blind that the eye cannot see.

Replace the AND gate in the circuit diagram with its DeMorgan equivalent.

#### anonkun123

Joined Mar 31, 2020
10
No, no, no!
We are so blind that the eye cannot see.

View attachment 203058

Replace the AND gate in the circuit diagram with its DeMorgan equivalent.
(!B+!D)(B+C) = !BC+!DB+!DC = (!B+!C)+(!B+!D)+(!C+!D) = !B+!C+!D

#### MrChips

Joined Oct 2, 2009
21,134
(!B+!D)(B+C) = !BC+!DB+!DC = (!B+!C)+(!B+!D)+(!C+!D) = !B+!C+!D
Oh, for heaven's sake!

What is the DeMorgan equivalent of X AND Y?

#### djsfantasi

Joined Apr 11, 2010
6,430
(!B+!D)(B+C) = !BC+!DB+!DC = (!B+!C)+(!B+!D)+(!C+!D) = !B+!C+!D
Try to answer the question asked and not guess what it means or the next step.

#### anonkun123

Joined Mar 31, 2020
10
Oh, for heaven's sake!

What is the DeMorgan equivalent of X AND Y?
!X OR !Y

#### MrChips

Joined Oct 2, 2009
21,134

#### dl324

Joined Mar 30, 2015
10,758

#### crutschow

Joined Mar 14, 2008
24,978
When using only one type of gate, I like to think of DeMorgan's rule as being that a NOR gate for positive logic becomes a NAND gate for negative logic.
Similarly, a NAND gate for positive logic becomes a NOR for negative logic.
So you look where you need NAND and NOR gates, and then invert the logic as needed.
Once you have that, then you can then simplify by eliminating unnecessary inverters.

If you start with your original logic configuration and go from there, with the above in mind, you should be able to see the simplest way to do it all with NOR gates.

#### djsfantasi

Joined Apr 11, 2010
6,430