1 1 vote No of JK / T FFs needed to design synchronous counter for sequence 1,4,2,3,1,4,2,3,1,4... Digital Logic digital-logic digital-counter + – jatin khachane 1 1.9k views answer comment Share Follow Print See all 6 Comments 6 6 Comments reply Show 3 previous comments Hemanth_13 commented Nov 9, 2018 reply Follow flag For ring counter we need 4 FF Johnson we need 2FF 0 0 replyShare jatin khachane 1 commented Nov 9, 2018 reply Follow flag Can we get any such seq of n distinct states with n FFs Example ,,0,7,9,2,0,7,9,2,...for this is it possible with 2 FFs ?? 0 0 replyShare Hemanth_13 commented Nov 9, 2018 reply Follow flag I think its mod 4 counter but we don't see what are the values,we just assign the mod part to timing circuit. to generate the clock pulses. 0 0 replyShare Please log in or register to add a comment.
0 0 votes I think $2$ ffs should be enough. This is how I think to implement it. $01->00->10->11$. I will complement the bits & add $1$ in output the moment both bits are found to be $0$. I believe in this manner, one can implement any counter of $n$ states in $log_2n$ flip flops. Still, I'm ready for counter arguments. See this as well. 2019_Aspirant answered Nov 9, 2018 2019_Aspirant comment Share Follow See all 4 Comments 4 4 Comments reply Pavan Karthik commented Nov 11, 2018 reply Follow flag I THINK THE LARGEST DIGIT IS 4 TO REPRESENT 4 (100) WE NEE 3 FLIPFLOPS SO 3 FF ARE NEEDED 0 0 replyShare 2019_Aspirant commented Nov 11, 2018 reply Follow flag Did you read the explanation? 0 0 replyShare Kshitij Hansda commented Nov 11, 2018 reply Follow flag why complement?? 0 0 replyShare 2019_Aspirant commented Nov 11, 2018 reply Follow flag to show $00$ which is output as $100$. 0 0 replyShare Please log in or register to add a comment.
