# boolean algebra to nand

Discussion in 'Homework Help' started by Wizardi, Mar 18, 2007.

1. ### Wizardi Thread Starter New Member

Mar 18, 2007
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I have a problem getting this following boolean equation into to using only nand gates. AB+DA+CA+DCB ... just can't figure it out with the basic rules of boolean algebra.

2. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
Have another look at DeMorgan's Theoram. (Theorem? Theorom?) Have another look at DeMorgan's stuff.

And remember... nobody restricted anyone to the use of two-input gates...

3. ### Wizardi Thread Starter New Member

Mar 18, 2007
5
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That was given to me in the assignment that I can you more than 2 inputs. But my problem is how to get complenment lines in order to form nand gates? please elaborate your answer

4. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
You must use NAND gates exclusively, yes? Your outputs get inverted, yes?

If AB=X, then (AB)'=X'.

That's where DeMorgan's Theorem comes in.

The circuit can be made with three gates from a 7400, one from a 7410, and one from a 7420.

If you are still having trouble, try making a truth table for the problem.

5. ### nomurphy AAC Fanatic!

Aug 8, 2005
567
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See attached:

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6. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
Please note [' = NOT] and I am not simplifying the expression in the following:

AB+DA+CA+DCB

Double NOT the expression (because double NOT does not alter the expression):

(AB+DA+CA+DCB)''

Using DeMorgans (A+B)' = A'.B'

Therefore:

((A.B)'.(D.A)'.(CA)'.(DCB)')'

Which can be implemented using purely NAND gates.

You may be advised to look at simplifying the expression, although that depends on what your assignment requires.

Dave

7. ### Wizardi Thread Starter New Member

Mar 18, 2007
5
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I see your point dave but the way you did it requires a not gate and nand gates...And my assignment was to use only nand gates.

8. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
It doesn't, it requires:

- 1x4-I/P NAND
- 1x3-I/P NAND
- 3x2-I/P NAND

Incidently, you can also make a NOT gate out of a 2-I/P NAND gate by tying the two inputs together.

Dave