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33 votes
33 votes

For the initial state of $000$, the function performed by the arrangement of the $\text{J-K}$ flip-flops in figure is:

  1. Shift Register
  2. $\text{Mod- 3}$ Counter
  3. $\text{Mod- 6}$ Counter
  4. $\text{Mod- 2}$ Counter
  5. None of the above
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9 Answers

2 votes
2 votes

Short method:

Johnson’s counter is Mod 2N counter and ring is Mod N counter.

Detailed  Explanation:

If you look closely at the question, you will notice that the input of J and K are complements of each other and if you recall, if the input values are the complement of each other, JK FF behaves as D FF

Here in the diagram we can see how D FF is formed from JK FF.

So this means the JK in our question will behave as D FF. You can also replace the JK FF with D FF if you want for easier understanding.

So now we can equate the values of J with Q,

Hence we can get the state diagram. And we will have 6 states since it was mentioned that it will start from 000 thereby eliminating the possibility of MOD 2 counter.

Also the shift register option is correct.

0 votes
0 votes
There are two counters present for the given circuit.

1. Mod 6 counter (States 000,001,011,111,110,100)

2. Mod 2 counter (As the remaining states 010 and 101 are also doing mod 2 counting.)

And the question is also saying multiple choices can be correct.

So answer should be C and D.
0 votes
0 votes
It is johnson counter of three bit. For Johnson counter I there is n bit then it will have 2n state =6 (here) so this is shift reg with mod 6
Answer:

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