NIELIT 2017 OCT Scientific Assistant A (IT) - Section D: 10

Question

A $4$ bit ripple counter and a $4$ bit synchronous counter are made using flip-flops having a propagation delay of $10$ ns each. If the worst case delay in the ripple counter and the synchronous counter be $R$ and $S$ respectively, then

• A. $R = 10$ ns, $S = 40$ ns
• B. $R = 40$ ns, $S = 10$ ns
• C. $R = 10$ ns, $S = 30$ ns
• D. $R = 30$ ns, $S = 10$ ns

Comments

https://gateoverflow.in/344429/Nielit-2017-oct-scientific-assistant-a-cs-section-d-10

Answer 1

option (b) is correct.

for rippel counter:- $T_{delay}$=nx$T_{ff}$+$T_{cc}$

for synchronous counter:- $T_{delay}$=$T_{ff}$+$T_{cc}$

Answer 2

For Ripple Counter

for $2^n$  state no of Flipflop required is n. As this is async counter so Delay = $N*Propagation Delay$ so $4*10 ms = 40 ms$

For Sync Counter

Delay = $1*Propagation Delay$ as Sync counter has Clock  same clock

Hence $Delay = 10ms$

Correct Ans is: $B$

Answer 3

B is the answer.
Time delay is constant since it’s synchronous. Ripple will be 10 + 10 + 10 + 10 = 40

Answer 4

ANS : B

R=40R=40 ns, S=10S=10 ns