1 votes 1 votes A three stage Johnson counter ring in figure is clocked at a constant frequency of fc from starting state of Q0Q1Q2 = 101. The frequency of output Q0Q1Q2 will be (a) fc / 2 (b) fc /6 (c) fc/ 3 (d) fc /8 Digital Logic digital-logic isro-ee + – sh!va asked Mar 10, 2017 • edited Mar 7, 2019 by Naveen Kumar 3 sh!va 2.3k views answer comment Share Follow See 1 comment See all 1 1 comment reply nikkey123 commented Aug 31, 2017 reply Follow Share please someone verify the answer. 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes It only Count two stages 101 and 010 hence fc/2 Neeraj Chandrakar answered Jul 23, 2017 Neeraj Chandrakar comment Share Follow See 1 comment See all 1 1 comment reply muthu kumar commented Nov 26, 2018 reply Follow Share Irrespective of states in will goes for no of flipflops. so, i think f/2n = f/6 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes Johnson counter is mod 2N counter where N is number of FF's. The output frequency will be fc/2N = fc/6 (option B) Rahul Jain25 answered Mar 10, 2017 Rahul Jain25 comment Share Follow See all 2 Comments See all 2 2 Comments reply Vishal Goel commented May 19, 2017 reply Follow Share I think your answer is valid when the initial value is 000. Here, the initial value is 101. And thus the counter is a mod-2 counter 101->010. So, the answer should be fc/2. Someone, please verify. 2 votes 2 votes Gyanu commented Jul 10, 2019 reply Follow Share @Vishal Goel sir, official key answer given is f/6. 0 votes 0 votes Please log in or register to add a comment.