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12,433 views
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

Best answer
55 votes
55 votes

Circuit behaves as shift register and $\mod6$ counter

$$\begin{array}{c|c}\hline \textbf{Clock Cycle}  &  \textbf{Output} \\\hline \text{1} & \text{100}\\ \text{2} & \text{110}\\ \text{3} & \text{111}\\ \text{4} & \text{011}\\ \text{5} & \text{001}\\ \text{6} & \text{000}\\\hline \end{array}$$

This is Johnson counter which is an application of Shift Register. And Johnson counter is mod 2N counter.

edited by
24 votes
24 votes

According to me, I think that A, C and D are the answer. Pleas, find the attached screen shot.

22 votes
22 votes
1st approach : Start from 000 and use conventional method (i.e. Truth table) to find modulus.

2nd approach : see, given circuit is nothing but Jonson Counter . Number of States in Jonson counter will be 2n where n is number of flip flops. here n is 3 so number of states are 6 . mod 6 counter..
edited by
9 votes
9 votes

A. Shift Register 
C. Mod-6 Counter


Given circuit diagram is for a Mod-6 Johnson Counter.

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Answer:

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