0 votes 0 votes Initially all flips-flops in the circuit are cleared to zero , what is the mod value of the counter ___________ I got – > 3 sorry for the poor image quality Digital Logic digital-logic digital-counter made-easy-test-series + – Magma asked Jan 14, 2019 • edited Mar 4, 2019 by ajaysoni1924 Magma 654 views answer comment Share Follow See all 13 Comments See all 13 13 Comments reply Show 10 previous comments Shaik Masthan commented Jan 14, 2019 reply Follow Share no mam, those are toggled but not either J or J'. i mean to say, Q$_{0N}$ = 0 if J=0 and toggled if J=1 But you are saying Q$_{0N}$ = 0 if J=0 and 1 if J=1 ----- which is wrong 1 votes 1 votes srestha commented Jan 15, 2019 reply Follow Share @Shaik Masthan I got this $000\rightarrow 100\rightarrow 010\rightarrow 101\rightarrow 011\rightarrow {\color{red} {100}}$ $Q_{0}$ $Q_{1}$. $Q_{2}$ $J_{0}(Q_{2})'$ $K_{0}$. $J_{1}(Q_{0})$ $K_{1}$. $J_{2}$ $K_{2}$ $Q_{0}$ $Q_{1}$. $Q_{2}$ 0. 0. 0 1. 0. 0 0. 1. 0 1. 0 1 0. 1. 1 1. 0 0 1. 1. 0. 1. 0. 0 0 1. 1 1. 0. 0 1. 1. 0. 1. 1. 1 0. 1. 1 1. 0. 0 1. 1. 0. 1. 1. 1 1. 0. 0 0 1 0 1. 0 1 0. 1 1 1. 0 0 when Q1=0, then $J_{2}$ $K_{2}$ will not work, right? , but 000 will come one time -ve edge triggering will not any effect on the output 0 votes 0 votes Shaik Masthan commented Jan 15, 2019 reply Follow Share check the last transition mam, 011 ---> Q$_3$=1 means, $\overline{Q_3}$= 0, then Q$_{0N}$=0... continue it 0 votes 0 votes Please log in or register to add a comment.