I am designing two synchronous counters one mod 21 and the other one mod 30. Is there a way to know which kind of flip flop (D, SR, JK, T) will use the smallest number of logic gates? I know we can build all of them and then check for the answer but that will be a lot of work. I am just wondering if there is a simple way to know that before design.

  • \$\begingroup\$ Thinnk that the nr of extern logic gates depends on the truth table of counter states, state(n) vs state(n+1), if there is a binary pattern then you will require minimal gates. \$\endgroup\$ Jan 17 '16 at 23:38
  • \$\begingroup\$ I assume you are asking about logic gates outside the flops, because flops themselves are made up of logic gates? \$\endgroup\$
    – mattgately
    Jan 19 '16 at 1:55

D, JK or T flip flops can all be configured such that they divide by 2. SR flip-flops, by themselves, can not be wired to toggle that way. So they are not applicable.

There are really only two types of counters you can build; a ripple counter, which divides by 2\$^N\$ using N flip-flops, and a ring counter, which divides by N using N flip-flops. The ripple counter's output can be short-circuited to divided by a number less than 2\$^N\$, for example a decade counter using four flip-flops. So we just need ripple counters, the ring counter will always use more flip-flops and can be discarded.

So to minimize the number of flip-flops, you need to break the counter value into prime factors, e.g. 21 breaks up into 3 x 7, and 30 breaks up into 2 x 3 x 5.

In the first case (21), since it doesn't divide by two you can't use a single flip-flop to divide the input by half. But you can cascade two ripple counters, one counting to 3 and the other counting to 7 (the output of the first fed into the input of the second). The total number of flip-flops needed would be 2 + 3 = 5.

In the second case (30), you can divide the input by two using a single flip-flop, and once again cascade two ripple counte3rs, one counting to 3 and the other counting to 5. The total number of flip-flops needed would be 1 + 2 + 3 = 6. (Or, you could use one ripple counter counting to 5, and another counting to 6, using the same number of flip-flops.)


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