# Digital Circuits

#### The_Stewie

Joined Dec 6, 2011
3
Hi all,

It's 20 to midnight, so obviously I'm desperate!

I'm very new to digital circuits and I just don't get how to change this circuit from OR and AND gates to NAND and/or NOR gates, without complicating the circuit further:

F=SL+SI+SC+CIL

Any help with this would be much appreciated, at this desperate hour.

If it helps at all, the original expression was:

F=S'LCI+SL'C'I+SL'CI'+SL'CI+SLC'I'+SLC'I+SLCI'+SLCI

I used k-map to reduce it, pretty confident I did THAT bit right, just can't convert it ... confused.com

Cheers guys
Stew

#### RiJoRI

Joined Aug 15, 2007
536
This is probably too late, but:

I check the K-map results with a truth table. If the reduced result is the same as the un-reduced result, I've done it right.

"SL+SI+SC+CIL" can be further simplified as S*(L+I+C)+CIL.

I'd then draw the positive logic circuit; after which I'd apply DeMorgan's theorems:
A+B = /(/A*/B) and A*B = /(/A+/B)

Let us know how it all turned out!

--Rich

#### The_Stewie

Joined Dec 6, 2011
3
Thank You so much

I didn't even think to simplify it further, I got my K-Map result and thought that was it - job done! I suppose that's midnight problem solving for you

Still not sure why using NAND gates makes the circuit simpler though, I'm gonna need more components to make the circuit work???

#### justtrying

Joined Mar 9, 2011
429
glad you figured it out. About NAND gates - it's not so much about the amount of gates you use as it is about the fact that you use the same type of gate. And since most chips come with multiple gates anyway, if you can turn a circuit that uses say inverter and an AND gate into a circuit that uses only NAND gates you will need one chip instead of two. That's the logic behind the logic...