Can this NAND circuit be more simplified?

Discussion in 'Homework Help' started by Tera-Scale, Jan 7, 2011.

  1. Tera-Scale

    Thread Starter Active Member

    Jan 1, 2011
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    I've constructed a 4-1 multiplexor with and not and or gates, then i converted it to nand gates. I removed the extra not gates to simplify and was left with the circuit attached. Do you guys think it is simplified enough?

    thanks

    Brandon
     
  2. Tera-Scale

    Thread Starter Active Member

    Jan 1, 2011
    164
    5
    here is the 4-1 mux circuit.
     
  3. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    You're crazy if you think I'm going to open a 'zipped' file on this computer from a source I don't know.

    Use your account here and upload the picture as a .png file. Then post it here using the BB[​IMG]
     
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  4. Tera-Scale

    Thread Starter Active Member

    Jan 1, 2011
    164
    5
    thanks for attracting my attention. i managed to uploaded as a png. I previously zipped the file because it was a jpg and it exceeded the size limit.
    I would be really grateful if you consider my post.

    thanks

    Brandon
     
  5. Wendy

    Moderator

    Mar 24, 2008
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    Why NAND gates? Part of the assignment? Basically if the requirement is NAND gates there isn't much you can do.
     
  6. Tera-Scale

    Thread Starter Active Member

    Jan 1, 2011
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    yeah, it's part of the assignment... I constructed the mux with normal gates as well but then he asks me to convert to nand gates. The thing is that in the last part he asks me about my conclusions from the two circuits :s

    I found something interesting for a 4-1mux which uses tree architecture (Attached). Do you know if I can obtain it from karnaugh maps or where i can read about it?
     
  7. Georacer

    Moderator

    Nov 25, 2009
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    I don't think Karnaugh maps will point to your second image. Not even to the first actually. The first circuit is a total overkill. The second one is as elegant as it gets, I think.

    A comparison with a OR and AND gate implementation would be to count the number of gates used, or make a price comparison based on your closest retailer. Don't forget that in a typical 74XX IC the OR, AND and NAND gates come 4-in-an-IC whereas there are 6 NOTs in an IC.

    A pure Karnaugh map approach would be to build a 6-input map (ABCDS0S1) and solve it, but that's too much and I doubt it will give anything as elegant as the second circuit.
     
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