# Circuit Using only NAND and NOT Gates

I have the following expressions for some logic which will each power a motor:

1. A OR (NOT B) OR D
2. NOT(A AND C) or (NOT C AND D)

I have a limitation on the number of integrated circuits I can use, with each IC being able to hold 6 of any 1 gate type on it.

I have tried below to turn both of the expressions below into circuits using only two types of gate, NOT and NAND gates, I know that I can create a NOT gate using a NAND gate with both of its inputs being into the NAND but I don't believe it makes too much of a difference for my requirements.

Circuit 1

simulate this circuit – Schematic created using CircuitLab

Circuit 2:

simulate this circuit

If possible, I want to be able to combine the circuits together, having 4 inputs A B C D with each of the outputs to the motor being separate I'm looking to keep the number of IC's as little as possible - 2 MAX.

Formatted the question since the first post made hardly any sense, Sorry!

• please show some effort to solve the problem – jsotola Mar 3 '18 at 18:44
• web search karnaugh map – jsotola Mar 3 '18 at 18:46
• Sorry if i didnt make it clear, i already have solutions for both logic ciruits as follows 1. NOT(NOT A AND B AND NOT D) 2. NOT(A AND NOT D)AND NOT C. i just dont know how to turn these into only NAND/NOT circuits – Harry Grace Mar 3 '18 at 18:50
• There's no need for both NAND and NOT. You can create a NOT from a NAND (A NAND 1 => NOT A). – Jerry Coffin Mar 3 '18 at 18:54
• Welcome to the site, Harry. Please edit your question to add detail to it, don't put that into comments. Please can you put a schematic for your original expressions, the schematic editor here is a breeze to use. The better the quality of your question, the better the quality of the answers it will attract. Again, a warm welcome to the site. – TonyM Mar 3 '18 at 18:54

You really cannot simplify $$\A + \bar B + D\$$.

But $$\ \overline {A C} + \bar C D\$$ can be simplified. Take deMorgan's on first term.

$$\ \bar A + \bar C + \bar C D = \bar A + \bar C ( 1 + D ) = \bar A + \bar C = \overline {A C}\$$

No possibility of combining the two circuits.

So using NANDs as NOTs, you have 6 NANDs total.