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 52.1k 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.
3 votes 3 votes Both 3 and 4 were given the correct answer to this question in GATE this year bad_engineer answered Mar 29, 2016 bad_engineer comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes both 3 and 4 are correct in the official key icecool answered Mar 20, 2016 • reshown Mar 28, 2016 by abhilashpanicker29 icecool comment Share Follow See 1 comment See all 1 1 comment reply abhilashpanicker29 commented Mar 28, 2016 reply Follow Share Actually only one can be correct.. :P IISC people are clever.. they just increased the error allowed in the answer - 3.0 to 4.0 :) 2 votes 2 votes Please log in or register to add a comment.
2 votes 2 votes Can we think like this ? First we build a circuit for the sequence ; 0 - 1 - 4 - 6 - 8 - 11 To build this we need 4 FF, and after doing this, just connect $Q_1$ and $Q_0$ terminals to the decoder. dd answered Sep 28, 2016 dd comment Share Follow See all 2 Comments See all 2 2 Comments reply srestha commented Nov 20, 2016 reply Follow Share What is advantage of it? 0 votes 0 votes rishi71662data4 commented Dec 12, 2017 reply Follow Share Can we just ignore the two most significant bit and take only the least 2 significant bit. No need of decoder. Can we ? 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes 2no of flip flops needed >= mod of the circuit in this question mod of counter is 6 and let no of flip flops be X; 2x >= 6 answer = 3 Bharani Viswas answered Mar 6, 2016 Bharani Viswas comment Share Follow See 1 comment See all 1 1 comment reply Sona Barman commented Oct 28, 2017 i edited by Sona Barman Oct 28, 2017 reply Follow Share Answer is either 3 or 4 as per IISC.But i think this approach is right also. 0 votes 0 votes Please log in or register to add a comment.