working of JK flip flop

Discussion in 'General Electronics Chat' started by circuit2000, Jan 30, 2007.

  1. circuit2000

    Thread Starter Active Member

    Jul 6, 2006
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    I am unable to understand the working of a positive edge triggered JK flip flop. In the figure, there are 2 AND gates on the left (one over the other) and there are 2 NOR gates on the right (one over the other). I have used an RS flip flop to construct the JK flip flop. The inputs J and K are denoted as 2 & 5 respectively. Q’ is the complement of Q. The output Q’ is connected to input number 1 of upper AND gate and the output Q is connected to input number 6. The inputs 3 & 4 are connected to the clock pulse.
    Now consider the input condition J=0, K=1 and at this point the clock pulse makes a positive transition. The explanation given in my book is as follows:
    When J is low and K is high, the upper AND gate is disabled, so there is no way to set the flip flop. The only possibility is reset. When Q is high, the lower gate passes a reset pulse as soon as the next clock edge arrives. This forces Q to become low. Therefore, J=0, K=1 means that the next positive transition of the clock resets the flip flop.
    The lower gate sends a reset pulse which means it sends a low voltage signal. Why does this happen? Now, the lower AND gate has 3 inputs. At the time the positive edge of clock pulse arrives, K=1. So, two of the inputs of the lower AND gate is high. Suppose at this instant, Q was high. Then three of the inputs of the lower AND gate is high, which means the output would be high .i.e. S is high. A high at any of the inputs of a NOR gate gives a low output. Hence Q’ is low. Now, Q’ is one of the inputs of the upper NOR gate. As J=0, R=0. Hence the two inputs of the upper NOR gate is low. Hence, Q=1. There is something to do with the third input(1 & 6). I think digital electronics is tough if you don’t have good teacher. Someone please guide me!!!!!!!!!!!
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Hi,

    Part of the problem is that it's hard to get your example into a known state. Get the data sheet for a 7473 or 7476 and you'll see that J - K's are a bit more elaborate internally.

    That's also why I use D flip-flops. There's just no doubt about the functioning.
     
  3. Papabravo

    Expert

    Feb 24, 2006
    10,140
    1,789
    Getting things to a known state is a problem for any type of memory device. S-R, D, T or J-K it doesn't matter. Same thing for invalid state recovery. You either do the job or you don't.

    My friends and I always said there were two types of logic designers. We even called 'em JK-designers and everybody else. Sorry you weren't able to cross over.

    In the simplest possible terms the two inputs called J and K allow the FF to do one and only one of four things, These are SET, CELAR,TOGGLE, and HOLD. Every J-K FF you will ever see besides all the ones that you won't does the same four things. What could be more straight forward?
     
  4. dragan733

    Senior Member

    Dec 12, 2004
    152
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    In the book is wrote: "Therefore, J=0, K=1 means that the next positive transition of the clock RESETS the flip flop." That is a fault.Corrected is: "Therefore, J=0, K=1 means that the next positive transition of the clock SETS the flip flop." Therefore, always when the situation is: J=0, K=1, always is: Q=1
     
  5. beenthere

    Retired Moderator

    Apr 20, 2004
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    We haven't even plumbed the depths yet - I also think 74121's are pretty handy.
     
  6. Papabravo

    Expert

    Feb 24, 2006
    10,140
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    And who can forget the Signetics 8T20 double edge triggered one-shot. Oooh-rah!
     
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