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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.
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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

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Think of it as a state machine. Naively (and rightfully), it has 6 states, one for each input. 6 states require 3 bits to encode. 3 bits = 3 flipflops.

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  • \$\begingroup\$ So can we use the output of a combinational logic or is it possible to derive directly with the FF outputs for counting? \$\endgroup\$ – Mahesh Kumar Feb 25 '16 at 16:13
  • \$\begingroup\$ 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. \$\endgroup\$ – Eugene Sh. Feb 25 '16 at 16:16
  • \$\begingroup\$ So here we are using the output of the combinational logic right?.Can it be done in a counter?.I didn't found any reference for the output of a counter(counting) being fetched from combinational logic!.So I want to confirm that my approach is right. \$\endgroup\$ – Mahesh Kumar Feb 25 '16 at 16:20
  • \$\begingroup\$ That's an OK approach. \$\endgroup\$ – Eugene Sh. Feb 25 '16 at 16:23
  • \$\begingroup\$ Any standard reference for this approach?.Because, at last the proof matters.I didn't get one. \$\endgroup\$ – Mahesh Kumar Feb 25 '16 at 16:27

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