# Minimum No. of JK FlipFlops needed to design a user defined counter

We want to design a synchronous counter that counts the sequence 0−1−0−2−0−3 and then repeats. The minimum number of J-K flip-flops required to implement this counter is :

My doubt: We can design this with 3 flipflops if we use combinational logic and the output is taken from that computational logic we can get the possible output.My doubt is ,can it can be called as counting(if we use the o/p of combinatioanl logic)? or we should only consider the output from the Flipflop for counting/output(I mean the state of the flipflops)?

Possible Solution: Let output of 3 flip flops be S1, S2 , S3. Take them through below circuit.

So if we take B C as final output we can generate 0->1->0->2->0->3.

So I think it is possible with 3 flip flops.

Reference:Vinayk:www.Gateoverflow.in

• You can realize it very easily. Consider a mod-6 counter (counting from 0 to 5) made from whatever you want (in your case 3 JKs). And then just add a combinatorial logic to it's output, translating 0, 2, 4 into 0, and 1,3,5 into 1,2,3 respectively. – Eugene Sh. Feb 25 '16 at 16:16