UGC NET CSE | January 2017 | Part 2 | Question: 8

Question

A binary $3$-bit down counter uses $J$-$K$ flip-flops, $FF_{i}$ with inputs $J_{i}$, $K_{i}$ and outputs $Q_{i}$, $i$ = $0, 1, 2$ respectively. The minimized expression for the input from following is :

1. $J_{0} = K_{0} = 0$
2. $J_{0} = K_{0} = 1$
3. $J_{1} = K_{1} = Q_{0}$
4. $J_{1} = K_{1} = \overline{Q}_{0}$
5. $J_{2} = K_{2} =Q_{1} Q_{0}$
6. $J_{2} = K_{2} =\overline{Q}_{1} \overline{Q}_{0}$

• A. I, III, V
• B. I, IV, VI
• C. II, III, V
• D. II, IV, VI

Comments

option 4

Answer 1

ans is 4  

refer http://digitalsystemtutorial.blogspot.in/2008/12/synchronous-3-bit-count-down-counter.html

Answer 2

is option 4 answer?