How to design circuit count odd numbers with 7 segment decoder 1-3-5-7-9

Discussion in 'The Projects Forum' started by jasmine21, Dec 4, 2014.

  1. jasmine21

    Thread Starter New Member

    Dec 4, 2014
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    Hello my friends ! Please help...How to design circuit count odd numbers with 7 segment decoder (1-3-5-7-9 ) ?
    Thanks for everything..:)
     
  2. Papabravo

    Expert

    Feb 24, 2006
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    What have you done so far to solve this problem?
    How about writing the numbers in binary to see if there is a pattern.
     
  3. jasmine21

    Thread Starter New Member

    Dec 4, 2014
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    I made a circuit that counts from zero to nine..but I can't do that....and I want to truth table and karnaugh map of the odd counter...How is designed ?
    writing the numbers in binary can be.
     
  4. Papabravo

    Expert

    Feb 24, 2006
    10,144
    1,791
    I asked you to write out the numbers in the sequence, one underneath the other, for a reason. Do it and you will see a pattern that suggests an obvious solution. If you don't see the pattern immediately write the boolean equations for each column. It will come to you eventually.
     
  5. jasmine21

    Thread Starter New Member

    Dec 4, 2014
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    okey thanks for the everything :)
     
  6. Papabravo

    Expert

    Feb 24, 2006
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    So, were you able to solve the problem?
     
  7. jasmine21

    Thread Starter New Member

    Dec 4, 2014
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    No, I don't think so :(
     
  8. jasmine21

    Thread Starter New Member

    Dec 4, 2014
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    and I apologize to all for my terrible English.. :D :(
     
  9. Dodgydave

    Distinguished Member

    Jun 22, 2012
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    you can just detect when the 1 bit in binary is high, if its low its even...
     
  10. Papabravo

    Expert

    Feb 24, 2006
    10,144
    1,791
    Did you write out the numbers 1, 3, 5, 7, 9 in binary? If you did that can you show us what it looks like to you?
     
  11. jasmine21

    Thread Starter New Member

    Dec 4, 2014
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    I must see one bit in the seven segment display.
     
  12. Papabravo

    Expert

    Feb 24, 2006
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    I don't think so. You have 2 separate problems which you can solve independently.
    1. Make a 4-bit counter that goes {0001, 0011, 0111, 1001}+
    2. Make a decoder that maps a 4-bit input into segments
    The plus after the curly brackets means that the sequence repeats one or more times. There you got me to write the sequence of states for you. Does that help you to solve the problem?

    I don't know why you are completely unwilling to do as I have asked you to do. Is it a cultural thing that prevents you from doing it because you think you will lose face? If it is you need to work very hard to overcome that tendency, or success in engineering will be very hard to come by.

    In some cultures, the nail that sticks up gets hammered down.
    In western engineering environments, the nail that stays down gets ignored, or worse, that nail gets fired.
    You have to choose, get ignored or take a risk that you might get hammered down.
    The choice is the essence of freedom, and is ever so precious
     
    Last edited: Dec 4, 2014
  13. lightingman

    Senior Member

    Apr 19, 2007
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    22
    A counter 1 to 10, then a flip-flop on the clock input. Have the flip-flop reset at zero count and then use the divide by 2 from the flip-flop to latch the display number into the display driver.

    So, a binary counter (4029) into a 7 seg decoder (4511). The counter will count 0,1,2,3,4,5,6,7,8,9, but the flip-flop will only load the new value into the display (LE input) on every other count (1,3,5,7,9). The zero count can be detected by a 4 input NAND gate on the four binary outputs from the counter. When all inputs are low, you will get a high from that to reset the fli-flop (4013) through a monostable (4047) to give a pulse to the LE input on the driver.

    Daniel.
     
  14. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
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    Excuse me. Forget everything about this problem for a second. Forget 7 segment displays. Forget truth tables. Forget karnaugh maps. Clear your mind…

    Cleared?

    Now write in a column, the numbers 1,3,5,7,9 in binary.

    Take a cleansing breath. Look at the binary numbers. Uh uh, forget 7 segment decoders! Do you see a pattern? Look carefully. There it is! See it!
     
  15. Dodgydave

    Distinguished Member

    Jun 22, 2012
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    no, just the last bit is always 1
     
  16. lightingman

    Senior Member

    Apr 19, 2007
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    Yep. 1 is always high. Count 0,1,2,3 into inputs 2,4,8, and there is 1,3,5,9.
     
  17. MrChips

    Moderator

    Oct 2, 2009
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    Come on guys. This is probably homework. Allow the OP/TS to figure this out on her own.
     
  18. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
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    I don't think so, lightingman
     
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