If you inputs are A, B (going to the first AND) C (additional input to the
OR) and D (additional input to the last AND) then your diagram is:

Out = ( A * B + C ) * D

Using a 4 gate NAND chip, you can get

Out = /( A * B + C ) * D

In other words, the exact same thing, but with the output inverted or you
can get

Out = ( A * B + /C ) * D


The tricks are: 
- An AND gate with both inputs and the output inverted works just like an
OR.
- A NAND gate with both inputs tied together acts like an inverter.

NAND Input
Gate a  b  Output
---- -- -- ------
 1   A  B  2a
 2   1o C  3a
 3   2o D  4a & 4b
 4   3o 3o Output = ( A * B + /C ) * D

NAND Input
Gate a  b  Output
---- -- -- ------
 1   A  B  2a
 2   1o 4o 3a
 3   2o D  Output = /( A * B + C ) * D
 4   C  C  2b


Another option (if you can find the chip) is a 7456 Dual AND-OR-INVERT gate.
http://focus.ti.com/lit/ds/sdls113/sdls113.pdf

It does Out = /(a * b + c * d) and has two of them. Actually, I don't think
that can do what you need... Hummm... Well, I hope that helps.

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> -----Original Message-----
> From: piclist-bounces@mit.edu 
> [mailto:piclist-bounces@mit.edu] On Behalf Of Denny Esterline
> Sent: 2004 Oct 14, Thu 13:12
> To: piclist
> Subject: [EE] AND-OR-AND - simplest implementation?
> 
> I need a simple logic function, a two input AND feeding a two 
> input OR feeding another two input AND [see picture]. Since I 
> don't need any other logic in the project I hate waste the 
> space and use two chips, is anyone else's Boolean algebra 
> less rusty that mine? Is there a one part implementation 
> (other than a PAL or CPLD)?
> 
> Thanks
> -Denny
> 
> 
> 


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