3-8 decoder with 2 2-4 decoder

Discussion in 'Homework Help' started by Leite33, Dec 6, 2015.

  1. Leite33

    Thread Starter Member

    Nov 28, 2015
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    0
    Dear friends.
    I looked a lot at google but i cant find a solution. How i can make one 3-8 decoder with (2) 2-4 decoders with out use enable input and without inverse outputs.
     
  2. bertus

    Administrator

    Apr 5, 2008
    15,646
    2,344
    Hello,

    You will NEED the select or enable inputs for making a 3->8 using 2 X 2->4 decoders.

    Bertus
     
  3. Leite33

    Thread Starter Member

    Nov 28, 2015
    57
    0
    But my exercise says "Design a circuit that will produce a digit of even parity for a number of 3 binary digit (xyz) using:
    a) decoder 2-4 (without using enable input and inverse outputs) and gates OR
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    In other words, the problem is NOT asking you to create a 3-to-8 decoder at all!

    How would you make an even parity generator for a three-bit input using XOR or XNOR gates?

    How would you make an XOR of XNOR gate using 2-4 decoders and OR gates?
     
  5. Leite33

    Thread Starter Member

    Nov 28, 2015
    57
    0
    So you say to me that exercise is wrong?
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    There is nothing wrong with the exercise as you presented it in Post #3:

    But this is NOT the same problem that you described in your original post:

    These are two essentially unrelated problems. The only thing they happen to share is that each involves 2-4 decoders.

    Which problem are you trying to solve?
     
  7. Leite33

    Thread Starter Member

    Nov 28, 2015
    57
    0
    Sorry the first one.
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    Which brings us back to my previous question, but let's rephrase it a bit.

    Produce a circuit that produces an even-parity output for a three-bit input using nothing but XOR and XNOR gates.

    Do that much first, and then we can worry about using 2-4 decoders and OR gates.
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    Actually, you don't. You can use a 2->4 decoder (without enable) to produce an AND gate and to produce a NOT gate. That means that it is a universal gate and that you can implement ANY Boolean function using just a bunch of 2-> decoders. Now, it may not be pretty, but it can be done.
     
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