# NAND circuit to NOR

Discussion in 'Homework Help' started by nanduko, Oct 31, 2009.

1. ### nanduko Thread Starter New Member

Oct 31, 2009
11
0
Hello everyone !

So i have to turn this circuit that is made of nand gates into the equivalent of this but this time only with NOR gates. This is the circuit :

First i tryed to turn the circuit into some boolen algebra statement but nothing happned, actually it became harder =/ i am strungling trying to solve this loads hours ....help me please...

2. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Take a look at DeMorgan's Theorem.

hgmjr

3. ### nanduko Thread Starter New Member

Oct 31, 2009
11
0
Well ok De morgan's theorem, but i need an expression 1st for the circuit, there is where i am strungling , i am getting confused all the time....

This is stupid but anyway... my conclusion is: (((((pp)'p)')'(rr)')')'

this is not right, right?

4. ### hgmjr Moderator

Jan 28, 2005
9,030
214
First observation is that there is no q in your boolean expression.

Have you noticed that four of the nand gates are configured as simple inverters?

hgmjr

5. ### nanduko Thread Starter New Member

Oct 31, 2009
11
0
indeed theres no q, it was meant to be like this (((((pp)'q)')'(rr)')')'

but still not sure if this is .

6. ### hgmjr Moderator

Jan 28, 2005
9,030
214
HINT: If you feed the same signal into both inputs of a NAND gate, all that is doing is inverting the input.

hgmjr

7. ### nanduko Thread Starter New Member

Oct 31, 2009
11
0
yeah i know that, thats why so many '''''' xD

ill try again and post

8. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Using your boolean notation, I get simply:

(p'q)r'

because (pp)' = p'

hgmjr

9. ### nanduko Thread Starter New Member

Oct 31, 2009
11
0
i tryed again in a bit different way,

isnt it ( (p' NAND q)' NAND (r') )' ?

10. ### hgmjr Moderator

Jan 28, 2005
9,030
214
That looks correct although it would be clearer if some simplification were performed.

hgmjr

11. ### nanduko Thread Starter New Member

Oct 31, 2009
11
0
hmmmm i tried to simplify it a bit, not sure again if its right

( (p' NAND q)' NAND (r') )' =

((p' q) nand (r') )' =

(((p' q) (r') )' )' =

((p+q') + r)' =

(p'q) r'

i believe theres a mistake somewhere

12. ### hgmjr Moderator

Jan 28, 2005
9,030
214
If you take a look at post #8 in this thread, you will see that I got the same simplified answer you have in post #11.

I guess your next step is to implement this boolean expression using NOR gates. Right?

hgmjr

13. ### nanduko Thread Starter New Member

Oct 31, 2009
11
0
yeah as long as this result is ok for the given circuit, ah btw forgot to mention that i hd already proved that p'q = p nor (p nor q) , the problem is the r' now, should i call for example p'q=d and say r'd = r nor ( r nor d) ?

14. ### hgmjr Moderator

Jan 28, 2005
9,030
214
My thought was that if you needed an inversion, you would feed the same signal into both inputs of a NOR gate just as was done in the NAND gate version of the circuit.

hgmjr

15. ### nanduko Thread Starter New Member

Oct 31, 2009
11
0

indeed i could feed a nor gate with 2x r so the result is r' , but still theres a multiply , how can do that as a nor gate . i mean --> (p'q) multiply r'

16. ### hgmjr Moderator

Jan 28, 2005
9,030
214
You do realize that there is not an actual multiply operation involved. That is simply the indication of an AND function.

hgmjr

17. ### hgmjr Moderator

Jan 28, 2005
9,030
214
See if the information at this website helps you out.

hgmjr

18. ### nanduko Thread Starter New Member

Oct 31, 2009
11
0
ehm lol yeah, i think i finally figured out, ill post the scematic as soon as possible

19. ### nanduko Thread Starter New Member

Oct 31, 2009
11
0
nor version of this (p'q) r' is :

hopefully

EDIT: i think that the second NOR (where r entrance is ) is not needed tho.

Last edited: Oct 31, 2009
20. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Your diagram has a couple of errors. Take another stab.

hgmjr