113 votes 113 votes We want to design a synchronous counter that counts the sequence $0-1-0-2-0-3$ and then repeats. The minimum number of $\text{J-K}$ flip-flops required to implement this counter is _____________. Digital Logic gatecse-2016-set1 digital-logic digital-counter flip-flop normal numerical-answers + – Sandeep Singh asked Feb 12, 2016 • retagged Aug 4, 2017 by Arjun Sandeep Singh 52.2k views answer comment Share Follow See all 15 Comments See all 15 15 Comments reply Show 12 previous comments mrinmoyh commented Jul 3, 2019 reply Follow Share @Shaik Masthan If the sequence was 0 - 1 - 0 - 2 - 3 .... then we would require 3 FFs right???? 0 votes 0 votes JashanArora commented Jan 23, 2020 reply Follow Share Please update this. The answer is 3 or 4 as per the official key. 3 votes 3 votes sonic commented Aug 17, 2021 reply Follow Share anwer is 4 , the sequence has 3 distinct numbers 1,2,3 to store that 2 bit is required and the zeros are 3 distinct zeros as 0-1,0-2,0-3 they are not the same , therefore for 3 distinct zeros you need 2 more bits to store them . final answer is 2+2 = 4 bits means 4 FF . as 1 ff stores 1 but of info 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes As the question just speaks about synchronous counters, I thought of using a Johnson Counter in which with n flip-flops we could count mod(2n). Considering that, the question is counting 6 bits therefore we require 3 flip-flops. aditya_cracks2021 answered Oct 17, 2020 aditya_cracks2021 comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes 0 1 0 2 0 3 It means 00 01 00 10 00 11 but to differentiate 000 001 100 110 000 011 Still need one more bit 0000 0001 0100 0110 1000 1011 0----->1----->4---->6---->8---->11---->0 The counter is working like this. So 4 flip flops shashankrustagi answered Nov 18, 2020 shashankrustagi comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes …………………………………... Nitesh_Yadav answered Jan 17, 2022 Nitesh_Yadav comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Total States: The counter needs to represent 6 distinct states: $0, 1, 0, 2, 0, 3$ the repeat.. Required Flip-Flops: Each flip-flop can represent $2$ states (0 or 1). To represent 6 states, we need $2^n >= 6$, where n is the number of flip-flops. The smallest value of n that satisfies this condition is $n = 4$. The answer could be 3 also as he condition $2^n >= 6$ is satisfied for $n=3$ as well. Therefore, $4$ or $3$ J-K flip-flops are required to implement this counter. rajveer43 answered Jan 12 rajveer43 comment Share Follow See all 0 reply Please log in or register to add a comment.