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function design using 74lvc1g57 or 99

Other Parts Discussed in Thread: SN74AHC00

I have a very simple boolean eq to implement:

I have A, B & Z coming from three palces.

Y=AB(NAND)*Z(INV)

OR

Y=[ A(INV)+B(INV)]Z(INV)

I could not fit it in either '57 or '99

any one can fit  the logic  inside one of these?

thnx

-robin

 

  • Hi!

    No, not really. But why do you want to use one of these very special logic gates? You could use a simple quadruple 2-input positive-nand gate. E.g. the SN74AHC00 in VQFN package. This is small too.

    Best regards,

    Edwin Krasser

  • Ed:

    thnx for the time.

    Of course one can always do brute force way.

    I have a board that is already built.. Done. finito.

    Now comes the need to modify the logic.

    Too bad these days they do not have the good ol' programmable logic in say tiny 8-pin with loads of combinatorial & sequential componets. Then you could reconfigure any way you want.

    I still think some smarty logician can propose a wacky way to fit it in.

    Neve know.

    -robin'

     

  • Hi,

    well, it's not very difficult to find a solution, but I did not know what you want to do with it. If I understand correctly, the function to implement is /(AB)*/Z, and if you do not need this function as high speed implementation, there is a solution:

    implement the nand by using A=input C,5 and B=input B,3, tie input A,2 to 0 and D,6 to 1 => you get: AB xor 1 what's the same as /(AB)

    implement the following */Z by using the /OE input: Z=input /OE, and a pull down resistor at the output Y,7: Z=1 => /(AB)*/Z = 0 because the output is turned off and the pull down sets the output to 0, Z=0 => the output is enabled, Y=/(AB)

    Well, it's a solution for the logical function.

    Best regards and good luck,
    Edwin Krasser

  • Ed:

    yeah! there you go: sounds like a good way out.

    We can add a pull up/down chip resistor. did not think about the  output pull up/down & I left it at "z" state being unalterable state of output.

    thnx a lot.

    Will do.