Help with JK flipflop counter?

Discussion in 'Homework Help' started by dob, Oct 12, 2006.

  1. dob

    Thread Starter New Member

    Oct 12, 2006
    1
    0
    ............
     
  2. n9352527

    AAC Fanatic!

    Oct 14, 2005
    1,198
    4
    First, you don't need JK FF 0, as you have observed already Q0 is always 0 whatever the conditions are. Why put something that does nothing in a circuit? Just ground the Q0 and you are set.

    Second, notice that if you have a truth table of the counter and you cover the Q0 column you are actually looking at an ordinary 2-bit up and down counter's truth table. So basically, counting 0, 2, 4, 6, 0 is equal to 2*0, 2*1, 2*2, 2*3, 2*0. This is because multiplying by two in binary domain is equal to shifting left by one bit.

    So, just design a 2-bit up/down counter, shift it left by one bit (by renaming the index to index + 1) and then add a grounded Q0.
     
Loading...