[Digital Systems]NAND-Gate Implementation

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The discussion focuses on implementing the output function Z=A'C+BC' using only NAND gates. Participants emphasize that K-Maps are unnecessary for this simple expression, as it can be constructed directly with NAND gates. The solution involves applying DeMorgan's theorem to derive Z' and using inverters to complete the function. It is noted that all basic logic gates can be created using NAND gates, allowing for a modular approach to circuit design. The conversation also touches on the concept of redundant gates and the cancellation of inverters in the final implementation.
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Homework Statement


For the output function Z=A'C+BC'

Where A,B,C are inputs and A',B',C' are the complements of the inputs respectively

Implement the design using only NAND GATES


Homework Equations


K-Map, maybe?


The Attempt at a Solution


I haven't taken digital system design course in years, so I don't even know where to start this problem. Can somebody start me off with some hints then I'll attempt the problem?
 
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this one is pretty simple. You don't need to worry about K-Maps because K-Maps are for simplifying messy boolean expressions. The expression you have, Z=A'C+BC' is a very simple one.

In practice, constructing a digital circuit is cheapest by implementing it using only NAND gates, so I guess the idea is to show you how it's done.

Basically each component, NOT, OR, AND, can each be made up of an assortment of NAND gates. All you do is put little blocks of these NAND gate configurations together the same as you would with regular NOT, OR, and AND gates.

Here's a good quick tutorial on it:
www.colchsfc.ac.uk/electronics/pdf_files/Doing it All With NAND Gates and NOR Gates.pdfHope this helps.
 
verd said:
this one is pretty simple. You don't need to worry about K-Maps because K-Maps are for simplifying messy boolean expressions. The expression you have, Z=A'C+BC' is a very simple one.

In practice, constructing a digital circuit is cheapest by implementing it using only NAND gates, so I guess the idea is to show you how it's done.

Basically each component, NOT, OR, AND, can each be made up of an assortment of NAND gates. All you do is put little blocks of these NAND gate configurations together the same as you would with regular NOT, OR, and AND gates.

Here's a good quick tutorial on it:
www.colchsfc.ac.uk/electronics/pdf_files/Doing it All With NAND Gates and NOR Gates.pdf


Hope this helps.

(Hits head) Of course, Demorgan's!

Ok so basically

Z = A'C+BC'

Setting A'C = X, BC' = Y

Z = X + Y
Z' = (X'Y')'
Z' = [(A'C)'(BC')']'

Since DeMorgan's theorem gives Z', for Z, just make an inverter at the end to complement the function. Is this a valid solution?
 
You can use NAND gates to make up all of the other gates. The way I look at it is like blocks, I just put them together and cancel redundant gates when everything is assembled.

I made the following to illustrate what I'm trying to say:
http://img36.picoodle.com/img/img36/9/9/3/f_logicdiagram_f42f88a.jpg


Hope this helps
 
why u remove implements in last step?
 
awais bashir said:
why u remove implements in last step?

two inverters cancel out each other.
in other words, if you have a's complement's complement, that's just a itself!
 
what are registers& and flip floop..and difference b/w them.?
 
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