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Building a logical gate function with NAND gates only

  1. Feb 12, 2012 #1

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 12, 2012 #2

    I like Serena

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    No exaggeration. :smile:
    Building a circuit from NAND gates does require a lot of gates.


    But shouldn't each NAND gate have exactly 2 inputs?
    Like this:
    120px-NAND_ANSI_Labelled.svg.png



    And you appear to have drawn a thin rectangle before each NAND gate.
    What does that represent?

    Did you perhaps intend something like:
    120px-NOT_from_NAND.svg.png
    Since this is the way to construct a NOT gate from a NAND gate.
     
    Last edited: Feb 12, 2012
  4. Feb 12, 2012 #3

    I like Serena

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    Let's zoom in on the second sets of inputs for a minute:

    attachment.php?attachmentid=43808&stc=1&d=1329049299.jpg

    The first NAND gate has the output ##\overline{AD}##.

    And the second NAND gate has the output ##\overline{\overline{AD} C} = AD + \overline{C}##.


    Is that what you intended?
    Or did you really want ##A\overline{C}D=AD\overline{C}##?
     

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  5. Feb 12, 2012 #4

    Ouabache

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    3 input NAND Gates are allowed and often used. Unless the question specified using 2-input gates, but it doesn't appear so.
     
  6. Feb 12, 2012 #5

    Femme_physics

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    Yes, Quabache is right we're allowed to use 3 inputs, though 1 input probably not...

    You're right-- I'm not sure where I've seen it, but I can see that you're right (as per usual), that's not the sign for a NAND gate.

    I intended to try and get the answer for the question but if there's a + C then I made a mistake it appears. What I really wanted is what you wrote at the end of that post. Is it all about just trying to build it via experience or is it about building a truth table and doing it methodically and schematically in a straightforward fashion?
     
  7. Feb 12, 2012 #6

    I like Serena

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    Whatever works for you.
    Systematic and methodical is good. :)


    On the wiki page for NAND logic, they give a couple of building blocks to create the various logic operations:
    http://en.wikipedia.org/wiki/NAND_logic
    With it you can build up your schematic.



    Myself I like to work backward as follows.

    You know you want to end up with ##AD\overline{C}##.
    Last thing in a NAND gate is the NOT.
    So before the NOT, you want to have ##\overline{AD\overline{C}}=\overline{AD} + C##.
    But you can't make an OR directly from an NAND gate, so first you would use a NAND as a NOT gate.

    Working backward you would have the 3 inputs ##A##, ##D##, and ##\overline{C}## for a first NAND gate, followed by the NOT construction.

    You're left with only constructing ##\overline{C}## and I think you already know how. ;)
     
    Last edited: Feb 12, 2012
  8. Feb 12, 2012 #7

    Femme_physics

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    But that's the question. Is there another method, or is this all about trial and error?

    Yes, thanks, I've used a similar page trying to do this :) I'll give it another go
     
  9. Feb 12, 2012 #8

    I like Serena

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    I have the impression you were following a proper method.
    Verifying your result showed a little mistake however. ;)
    I'm sure you'll get it.
     
  10. Feb 12, 2012 #9

    LCKurtz

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    Look at what happens if you factor the ##A\overline B C## out of the last two terms before you draw the circuit.
     
  11. Feb 13, 2012 #10

    NascentOxygen

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    There is a lot of redundancy here. The expression can be simplified before you start to implement it using gates. LCKurtz gave you a hint, which raises the question in my mind: are you certain that you have correctly reproduced the expression that you are realizing with gates?
     
  12. Feb 14, 2012 #11

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  13. Feb 14, 2012 #12

    NascentOxygen

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    Method looks right.
     
  14. Feb 14, 2012 #13

    I like Serena

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    Very good! You got it!
    And very creative to put a double bar over it and break it up! :smile:
    I also like how clearly you explained what you did and how you showed a nice simple solution.

    This time around there appear to be no beetles scurrying away. ;)
     
    Last edited: Feb 14, 2012
  15. Feb 14, 2012 #14

    Femme_physics

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    LOL

    Thanks for the help ILS, everyon. Glad I got it right.
     
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