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BCD to 7 segment display logic minimisation

  1. Jan 10, 2017 #1
    1. The problem statement, all variables and given/known data

    A 7-segment display is used to show a decimal digit and it is driven from 4-bit input. Each bar is assumed to light up when a logical 1 is applied to it.
    1. Draw the truth table to drive segment (d) of the display.

    2. Using the truth table you have obtained, draw a Karnaugh map to find the minimised logic expression in the 1-st canonical form (SOP, i.e. series of AND terms ORed together).

    3. Convert the expression from 2 to a function which uses only NAND gates and draw the equivalent circuit diagram in Logic Circuit and generate its truth table to compare with 1.
    seven-segment_example.png
    2. Relevant equations

    De-Morgan's laws:$$(A + B)' = A' \cdot B'$$$$A' + B' = (A \cdot B)'$$
    3. The attempt at a solution

    See attached for truth table and k-map.

    Canonical Form:$$A + C'D + CD' + A'B'$$

    Applying De-Morgan's Theorem:$$A'BCC'DD'$$

    but since ##C \cdot C' = 0## and ##D \cdot D' = 0## then this all reduces to 0.

    I'm not sure whether it is my K-Map minimisation or application of De-Morgan's laws that is wrong?
     

    Attached Files:

  2. jcsd
  3. Jan 10, 2017 #2

    berkeman

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  4. Jan 10, 2017 #3
    Yes that is correct thanks berkeman.
     
  5. Jan 10, 2017 #4

    berkeman

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    And this part of the problem statement implies you should only show 0-9 and blank on any number 10 and up? Or are you supposed to display A-F too...
     
  6. Jan 10, 2017 #5
    In the example that was provided for segment c, the decimal values 10 to 15 are shown as 'don't care' terms so I have kept it consistent with the example.
     
  7. Jan 10, 2017 #6

    berkeman

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    I'm confused though. You should end up with 7 truth tables, one for each segment drive, right?
     
  8. Jan 10, 2017 #7
    Yes in total, although this homework is only to derive the logic circuit for segment d.
     
  9. Jan 10, 2017 #8

    berkeman

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    Ah, okay. I missed that if you said it in your OP. Let me check your work now...
     
  10. Jan 10, 2017 #9

    berkeman

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  11. Jan 10, 2017 #10
    I did think this originally but then I checked the example solution that was provided in the homework material and found this.
     

    Attached Files:

  12. Jan 10, 2017 #11

    berkeman

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    Ah, fair enough. The issue I see is the blue rectangle and green rectangle that you have around 3 terms each in your K-map. I don't think you can do that... :smile:

    http://www.ee.surrey.ac.uk/Projects/Labview/minimisation/graphics/g3.gif
    g3.gif
     
  13. Jan 10, 2017 #12
    Great thanks for pointing that out, still got the same issue though:

    Canonical:

    ##A + BC'D + B'C + B'D' + CD'##

    Applying De-Morgan's law:

    ##A'BBB'CC'C'DDD'##

    Now the ##BB'##, ##CC'## and ##DD'## terms all equal zero.
     

    Attached Files:

  14. Jan 12, 2017 #13

    berkeman

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    That's not what I'm getting for a corrected version of your K-map. Can you post the corrected K-map and show your groupings?
     
  15. Jan 13, 2017 #14
    I think I am missing the ##A'## from the third minterm.

    ##A + BC'D + A'B'C +B'D' + CD'##
     

    Attached Files:

  16. Jan 13, 2017 #15

    berkeman

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    Your 3rd minterm does include A', and I get the same minterms as you now. :smile:
     
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