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 658 views answer comment Share Follow See all 13 Comments See all 13 13 Comments reply Shobhit Joshi commented Jan 14, 2019 reply Follow Share $101\rightarrow 010\rightarrow 101?$ So, $mod(2)$ 0 votes 0 votes Magma commented Jan 14, 2019 reply Follow Share Ooh no I also include 000--> 101----> 010 -_- 0 votes 0 votes Shaik Masthan commented Jan 14, 2019 reply Follow Share i am not getting how it is 101 state from 000 0 votes 0 votes Shobhit Joshi commented Jan 14, 2019 reply Follow Share i just did not see the clock to the $3^{rd}$ flip flop is from $q_2$, this needs correction 0 votes 0 votes Shaik Masthan commented Jan 14, 2019 reply Follow Share i am getting 000 → 100 → 010 → 101 → 011 → 000 total 5 states in loop note :- negative clock edge means, when transition takes from 1 to 0, then FF active 1 votes 1 votes Magma commented Jan 14, 2019 reply Follow Share mod 5 is correct brother But i didn't get you properly .. how it's 5 :/ 0 votes 0 votes Shobhit Joshi commented Jan 14, 2019 reply Follow Share @Shaik Masthan missed the transition part thanks for explanation 0 votes 0 votes Shaik Masthan commented Jan 14, 2019 reply Follow Share Q$_{0N}$ = $\left\{\begin{matrix} 0\;\;\;\;\;\;\;\;\;\; when\; \overline{Q_2} = 0 \\ Toggle\; when\; \overline{Q_2} = 1 \end{matrix}\right.$ Q$_{1N}$ = $\left\{\begin{matrix} 0\;\;\;\;\;\;\;\;\;\; when\; Q_0 = 0 \\ Toggle\; when\; {Q_0} = 1 \end{matrix}\right.$ Q$_{2N}$ = $\left\{\begin{matrix} Toggle\; when\; \overline{Q_1}\text{ is transistion from 1 to 0} \end{matrix}\right.$ 1 votes 1 votes Magma commented Jan 14, 2019 reply Follow Share Ooops I always did bad mistakes dunno why ~ _~ thanks @Shaik Masthan 0 votes 0 votes srestha commented Jan 14, 2019 reply Follow Share @Shaik $Q_{2}'=J_{0}$ and $Q_{0}=J_{1}$ these will change right? and what about $Q_{1}'$ 0 votes 0 votes 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.