Help with JK flipflop counter?

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

  1. dob

    Thread Starter New Member

    Oct 12, 2006
  2. n9352527

    AAC Fanatic!

    Oct 14, 2005
    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.