Even number counter using jk flip-flops

Discussion in 'Homework Help' started by lloret09, Sep 30, 2013.

  1. lloret09

    Thread Starter New Member

    Sep 30, 2013
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    0
    hi im new here.:)
    i have this problem in my science class we are going to put this

    design a 4 bit even counter design using jk flip-flop
    (0,2,4,6,8,0 so on..)

    the state are
    0-0000
    2-0010
    4-0100
    6-0110
    8-1000

    next state
    2-0010
    4-0100
    6-0110
    8-1000
    0-0000

    this is what i got from my k-map
    Q0
    Jo= 0 Ko=1 LSB
    Q1
    J1=Q0(bar) K1=1
    Q2
    J2=Q2 K2=Q2
    Q3
    J3=Q2 K3=1 MSB

    i tried putting this on my breadboard but the output isn't right.
    0110
    0111
    it keeps repeating on that sequence
    I did something wrong but i don't know what it is i'm just 12. :(
    can you tell me?:confused:
     
  2. MrChips

    Moderator

    Oct 2, 2009
    12,449
    3,365
    Learn to think outside the box.
    Build a 3-bit counter. No need to include the LSB since it is always 0.
     
  3. WBahn

    Moderator

    Mar 31, 2012
    17,775
    4,804
    Actually, I'm impressed that you are doing this kind of stuff at your age. What grade are you in? 6th?

    If you were asked to build a 3-bit counter that counted 0-1-2-3-4-0..., could you do that?

    Getting back to the design you came up with. If you were given the circuit as you describe (the J and K logic for each JKFF), could you analyze it to determine what the sequence would be that would result?

    You definitely got a mismatch between your logic and your breadboard. If you have J0=0 and K0=1, that should result in Q0 always being 0. So focus on that on your breadboard, first. Make sure that any preset/reset inputs are tied to logic levels that ensure that they are not asserted -- do NOT leave them floating! Then ensure that you are putting the 0 and 1 into the right inputs of the LSB flipflop.

    Look into that and see what you find.

    If you can, please post your actual schematic (a scanned, handdrawn schematic is fine).
     
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  4. lloret09

    Thread Starter New Member

    Sep 30, 2013
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    [​IMG]

    here is my schematic i did
     
  5. WBahn

    Moderator

    Mar 31, 2012
    17,775
    4,804
    Let's think about that schematic a bit.

    Notice that there is no connection between Q0/Q1 and Q2/Q3. That means that the state that Q2 and Q3 transistion to are not dependent in any way on the value of either Q0 or Q1. Is that consistent with your next-state table?

    What will the inputs to the J and K inputs of Q2 and Q3 be if Q2 is LO? What will happen from that point on?

    Do these JKFFs have asynchronous reset and preset inputs? If so, how are they connected?
     
  6. lloret09

    Thread Starter New Member

    Sep 30, 2013
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    0
    [​IMG]

    i really don't quite understand..:confused:

    i was able to make 0,2,4,6,0 using a 3 bit counter...
    [​IMG]
    is it possible to make it to 8 using just 3 bit counters?

    im really :confused:
     
    Last edited: Oct 1, 2013
  7. MrChips

    Moderator

    Oct 2, 2009
    12,449
    3,365
    Counting 0,2,4,6,0,... is the same as 0,1,2,3,0,...
    For this you need a 2-bit counter plus one static output of 0.

    Counting 0,2,4,6,8,0... is the same as 0,1,2,3,4,0,...
    There are four unique states. Hence you need a 3-bit counter plus one static output of 0.

    Focus on a 3-bit counter cycling 0,1,2,3,4,0,...
     
    lloret09 likes this.
  8. lloret09

    Thread Starter New Member

    Sep 30, 2013
    7
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    tnx..
    i will give it try..
    :)
     
  9. lloret09

    Thread Starter New Member

    Sep 30, 2013
    7
    0
    im really sorry but can you check my kmap?
    im having trouble making this work on a breadboard i think ive done something wrong
    but i dont know what.
    Previous State
    qa-qb-qc

    =
    0--0--0
    0--0--1
    0--1--0
    0--1--1
    1--0--0

    Next State
    qa qb qc

    .
    0--0--1
    0--1--0
    0--1--1
    1--0--0
    0--0--0
    .
    ja ka

    .
    0--0
    0--x
    0--x
    0--x
    x--1
    jb kb

    .
    0--x
    1--x
    x--0
    x--1
    0--x
    jc kc

    .
    1--x
    x--1
    1--x
    x--1
    x--0
    KMAP
    JA--(C)
    (qAqB)0 0
    00----0-1
    01----0-x
    11----x-x
    10----x-x

    JA=qB KA=1

    JB--(qC)
    (qAqB)0 1
    00----0-1
    01----x-x
    11----x-x
    10----0-x

    JB=qC

    KB--(C)-
    (qAqB)0-1
    00----x--x
    01----0--1
    11----x--x
    10----x--x
    KB=qC

    JC--(C)
    (qAqB)0-1
    00----1-x
    01----1-x
    11----x-x
    10----x-x
    JC=1


    KC--(qC)
    (qAqB)0 1
    00----x-1
    01----x-x
    11----x-x
    10----0-x
    Kc=qA(bar)
     
    Last edited: Oct 1, 2013
  10. MrChips

    Moderator

    Oct 2, 2009
    12,449
    3,365
  11. lloret09

    Thread Starter New Member

    Sep 30, 2013
    7
    0

    im very sorry but i can't get my head to get the values right
    comes with being a kid i guess..

    can you check if this will work for a 0,1,2,3,4,0 counter
    im using a 74LS73AP IC if that helps

    [​IMG]

    im still confused im a slow learner :confused:
     
    Last edited: Oct 1, 2013
  12. lloret09

    Thread Starter New Member

    Sep 30, 2013
    7
    0
    i think this is it.. i did something wrong on my k-map and it seems my breadboard
    keeps producing a positive charge on everything on my breadboard that is confusing confused:

    here is the work I've done though
    [​IMG]
     
  13. WBahn

    Moderator

    Mar 31, 2012
    17,775
    4,804
    Note that the equation next to your OR gate doesn't match the way it is wired.

    So let's see what behavior your schematic should produce.

    Q3 Q2 Q1 Q0 J3 K3 J2 K2 J1 K1 J0 K0 Q3' Q2' Q1' Q0'
    0 0 0 0 0 1 0 0 1 1 0 1 0 0 1 0
    0 0 1 0 0 1 1 1 1 1 0 1 0 1 0 0
    0 1 0 0


    Do you see how I am building up this table?

    See if you can finish it.
     
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