Hi

http://www-s.ti.com/sc/psheets/schs045a/schs045a.pdf

On page 3-139 (bottom of the page) you will find the possible solutions and
on page 3-140 the function truth table of the same.
You will require 2 IC's, both programmed as NAND gates.

The first 8 bits are connected to the first CD4048B. The EXP input of 
the first CD4048B must be low.

The other 8 bits are connected to the second CD4048B.
When you connect the output of the first CD4048B to the EXP input of the 
second CD4048B you will get a 16 input NAND.

girish

> I am attempting to connect a pair of CD4048 8-input
> expandable gates together to form a 16-input nand gate.  I did
> not do a very thorough job of writing down some of the notes
> which, as I recall weren't all that good to begin with so I have
> a question or two.
>
>       Pin 15 is labeled as Expand.  Is this an input or an
> output?  What do you do with it on a CD4048 that is being used as
> an 8-input device by itself?
>
>       When cascading multiple gates, do you connect the J
> output pin to the Expand pin of the following gate?
>
>       The rest, I think I can figure out.  Hu rah for wire-wrap!
>
http://www.chipdir.org/giicm/4048.txt

It's an 'or' or a 'nor' gate. 

Use OR & invert your main output. Connect 8-16 into any input for a 15 input

gate. To make 16, you'd need an extra 'or' gate. That's how it seems to me.


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        Declan Moriarty

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