4 bit sync counter

Discussion in 'The Projects Forum' started by hawk360, Jun 9, 2013.

  1. hawk360

    Thread Starter New Member

    Jun 9, 2013
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    I tried to make a 4 bit synchronous counter using J K which in multisim which goes through following stages:
    1011->1100->1110->1111->0000->1011.....

    For this i made the transition table n then with the help of k-map i found out the required i/p to flip flop. (for reference i'm attaching my K-maps.) but inspite going through all this i'wasn't able to get the required count. I'm even uploading my circuit.
    I wanna know why this ckt is not working and is there any way to get it to wrk.

    thank u!

    u can dwnd my ckt frm 4shared.com: here's the link-
    http://www.4shared.com/file/YnoZqsPT/count.html

    [​IMG]

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  2. tshuck

    Well-Known Member

    Oct 18, 2012
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    I don't know about you, but I can't read your K-maps on a 160 x 120 pixel image.

    Also, since I don't(and I assume others as well) have Multisim, how about posting a screenshot of the circuit?

    You can upload a zipped folder of all of your references here, so you don't have to resort to a off-site host, something I'm always wary of.

    Upload you images here, using the small paperclip button. Then, you can add the links from the uploaded images as the reference URL for the image button in your post.

    Also, your last three images don't show up...
     
    hawk360 likes this.
  3. hawk360

    Thread Starter New Member

    Jun 9, 2013
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    as per your suggestion i hv re uploaded my files.
     
  4. tshuck

    Well-Known Member

    Oct 18, 2012
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    Thank you for reposting those, however, I don't see your state table. That explains what it is the K-maps are doing and your reasoning in doing what you've done.

    Where are the variables A, B, C, and D defined? I could probably guess what they are, it isn't a good idea to let the grader guess when you want points...
     
  5. hawk360

    Thread Starter New Member

    Jun 9, 2013
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    here's my state table n my multisim ckt!
    srry for not mentioning abt A,B,C,D.
    they're next state variable names in order of MSB TO LSB.
     
  6. LDC3

    Active Member

    Apr 27, 2013
    920
    160
    When I analyze your circuit, after reset, I get it cycling between 2 states: 1010 and 1011 (unless I have the state table for a J-K flip flop wrong).
    And yes, there should be a way to fix your circuit.

    Edit: Sorry my mistake, I made an error in my calculations.
     
    Last edited: Jun 9, 2013
  7. tshuck

    Well-Known Member

    Oct 18, 2012
    3,531
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    You should look over your state table again, particularly with the transition from 1100 -> 1110...
     
  8. tshuck

    Well-Known Member

    Oct 18, 2012
    3,531
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    You also left out the 1 in your J0 K-map at 0000
     
  9. hawk360

    Thread Starter New Member

    Jun 9, 2013
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    its right! I'hv checked it twice now!
     
  10. hawk360

    Thread Starter New Member

    Jun 9, 2013
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    no I haven't...u can chk the pic!
     
  11. hawk360

    Thread Starter New Member

    Jun 9, 2013
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    this ckt goes thru following stages:
    1111->0101->0000

    N.B- you hv to take into a/c negative logic used in multisim i.e if the bulb doesn't glow it signifies 1 and vice versa!
     
  12. tshuck

    Well-Known Member

    Oct 18, 2012
    3,531
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    [​IMG]
     
  13. tshuck

    Well-Known Member

    Oct 18, 2012
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    In order to make a JK flip-flop transition from a 0 to a 1, when Q = 0, what are the required inputs?

    EDIT: Disregard, I was being careless..... sorry...
     
    Last edited: Jun 9, 2013
  14. hawk360

    Thread Starter New Member

    Jun 9, 2013
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    oops...i didn't noticed that...so J0 = A + C(bar)...let me edit this in my circut...
     
  15. hawk360

    Thread Starter New Member

    Jun 9, 2013
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    0->1 => J= 1 n K= X ( don't care)
     
  16. tshuck

    Well-Known Member

    Oct 18, 2012
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    Thanks, but the mistake was with me. I was looking at the wrong cell.:rolleyes:
     
  17. hawk360

    Thread Starter New Member

    Jun 9, 2013
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    still this ckt is not wrkin as required...:(
    its going through following stages now:
    1111->0100->0001->0110->0001->0110....
     
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  18. tshuck

    Well-Known Member

    Oct 18, 2012
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    Post your updated K-maps....
     
  19. hawk360

    Thread Starter New Member

    Jun 9, 2013
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    thers's not much change...jst J0 = A+ C(bar) ...bcoz of octet using 1st two rows...rest remain same...n i'hv shown it in the previous ckt attached!

    N.B- i hv attached the updated K map...take a look!
     
  20. tshuck

    Well-Known Member

    Oct 18, 2012
    3,531
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    You are being quite sloppy in your notation. What relationship is there to J_0 and A + C'? You could have written J_B = A + C', or J_0 = Q_1 + Q_3', but 0 has no relationship with the letters...

    As you have it, the order of outputs from MSB to LSB is CDAB, not exactly a meaningful sequence. While this may not be a big problem, it makes the person looking at your answer need to spend the extra time to understand what you have done.

    Now, allow me to rewrite your results with the ABCD notation, where A is the MSB:
    J_A = 1
    K_A = BD

    J_B = C
    K_B = D

    J_C = 1
    K_C = D

    J_D = A + C'
    K_D = 1

    Now, examine your circuit, is this what you have?

    Note, it might be easier to see if you orient your flip flops vertically...
     
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