retagged by
840 views

1 Answer

Best answer
2 votes
2 votes

Initially  $\rightarrow 000$

  • After $1$st clock  $\rightarrow 111$ . How ?

The first FF will always toggle on every input .Now, we goto $3$rd FF and take feedback from $Q_{2}' = 1$. This $1$ reaches $2$nd FF and $Q_{0}$= clock input = 1 makes the $2$nd FF toggle giving $Q_{1} = 1$ = input for $3$rd FF with clock being high makes it toggle too.

This gives the next state $111$.

  • After $2$nd clock  $\rightarrow 011$
  • After $3$rd clock  $\rightarrow 110$
  • After $4$th clock  $\rightarrow 010$ . How ?

The first FF will always toggle on every input .Now, we goto $3$rd FF and take feedback from $Q_{2}' = 1$. This $1$ reaches $2$nd FF and $Q_{0}$ = clock input = 0 makes the $2$nd FF not toggle giving $Q_{1} = 1$ = input for $3$rd FF with clock being $0$ makes it not toggle too.

  • After $5$th clock  $\rightarrow 100$
  • After $6$th clock  $\rightarrow 000$
selected by

Related questions

0 votes
0 votes
0 answers
1
rahul sharma 5 asked Dec 23, 2016
366 views
Is it possible to use asynchronous /ripple counter to design random sequence?
0 votes
0 votes
2 answers
3
Na462 asked Oct 10, 2018
453 views
In the given synchronous Counter circuit, initially all Outputs are Reset. It is required to replace FF2 with AB Flip Flop. The FF2 inputs would be:- A. A = Q1 and B = Q1...
0 votes
0 votes
0 answers
4
Mk Utkarsh asked Jan 15, 2019
1,152 views
MOD-8 synchronous down counterMOD-8 asynchronous up counterMOD-10 asynchronous up counterMOD-8 asynchronous down counterPlease explain why it is down counter?