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3-input AND - NAND equivalent?

  1. Feb 16, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm trying to convert the 3-input AND gate shown below using only NAND gates...but am having troubles. Is it possible to use only 2 NANDS for the conversion?


    http://www.doctronics.co.uk/images/4081_03.gif [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 16, 2012 #2

    LCKurtz

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    What about a 3 input NAND followed by a two input NAND used as an inverter?
     
  4. Feb 17, 2012 #3
    I'm trying to convert this expression: ga + za + sgz

    using just 2-input nand gates...more specifically the 7400 ic chip.

    I'm trying to use as little NAND gates as possible. I've got (ga + za) down to 5 NAND gates currently...I can only use 8 total NANDS for this.
     
  5. Feb 17, 2012 #4

    LCKurtz

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    I can do that expression with six 2-input NANDs. I don't see how to show you how without showing the solution. Is this a homework problem to hand in?
     
  6. Feb 18, 2012 #5
    Here is what I got....this is a practice problem to prepare for a midterm, but I would like to see how you used 6 NAND gates.

    http://img827.imageshack.us/img827/7766/schematic.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  7. Feb 18, 2012 #6

    LCKurtz

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    OK. Here's my circuit. One NAND is used as an inverter.

    forumcircuit.jpg
     
  8. Feb 20, 2012 #7
    thank you. I have one more expression for practice problems:

    ab~ce + b~cde + cde + abc + acd~e (The ~ symbol represent "not")

    I'm supposed to do this expression in 12 gates.

    I currently have the last 3 condensed to: c(de + ab + ad~e)

    The first two: b~ce(a + d)

    so the final condensed form looks like this: b~ce(a + d) + c(de + ab + ad~e)

    if you can try helping me convert this expression into a NAND diagram that would be great...thanks for your help again.
     
    Last edited: Feb 20, 2012
  9. Feb 20, 2012 #8

    NascentOxygen

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    You have written ab&ce

    Is this any different from a&b&c&e ?
     
  10. Feb 20, 2012 #9
    I changed the above post...it should be tilde symbols ~ for "not".
     
  11. Feb 20, 2012 #10

    LCKurtz

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    But in the expression ab~cd is it the b or the c that is the not? You can make the expressions much more readable with tex, like this: ##ab\bar cd##. Here's how you enter it:
    Code (Text):
    ##ab \bar cd##
     
  12. Feb 20, 2012 #11

    NascentOxygen

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    Consider de + ad¬e
    Take out the term d,
    d (e + a¬e)

    when e is TRUE, the bracketed expression = e
    when e is FALSE, the bracketed expression evaluates = a

    So,
    d (e + a¬e) = d (e + a) = de + da

    So c(de + ab + ad~e) = c (de + ab + ad) = c( d(e+a) + ab)

    In the absence of better advice, I would implement paired terms, e.g.,
    I would form E+A
    then AND it with D
    then form AB
    then OR these two terms (using NAND gates)
    then AND with C
    then ....

    This seems rather pedestrian, hopefully someone else has a better idea. :smile:
     
  13. Feb 20, 2012 #12

    NascentOxygen

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    duplicate
     
    Last edited: Feb 20, 2012
  14. Feb 20, 2012 #13
    original form: ##ab \bar ce## + ##b \bar cde## + ##cde## + ##abc## + ##acd \bar e##

    condensed form: ##b \bar c(ae + de)## + ##c(de + ab + ad \bar e)##

    partial circuit diagram using NANDS ##c(de + ab + ad \bar e)##:

    I made that portion with 7 NANDS...I still have to finish the ##b \bar c(ae + de)## portion of the diagram...still have 5 NANDS left.

    hjkh.jpg
     
  15. Feb 20, 2012 #14
    i figured it out thanks to oxygen's logic...I checked all 32 combinations for all the letters related to the problem...and all of the outputs came out correct.

    final condensed form: c(de + da + ab) + b(ae + de)

    heres my diagram:

    tit.jpg
     
    Last edited: Feb 21, 2012
  16. Feb 21, 2012 #15

    NascentOxygen

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    Good.

    http://s16.postimage.org/mb1ot390j/tit.jpg

    I think you unnecessarily duplicated D NAND E
    when you could have used the output twice?
     
    Last edited: Feb 21, 2012
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