# Digital-GrandTest1-Q12

+2 votes

A certain J-K FF has a propagation delay of 12ns. The largest MOD of the counter such that, the counter can be designed from these FFs which will operate up to 10 MHz?

asked Jun 19 in Digital
reshown Jun 20

## 3 Answers

+2 votes

Best answer

$\\ Time \ period \ of\ clock\ pulse=1/10MHz=10^{-7}sec\\ \\Propagation\ delay\ of \ each\ flip\ flop=12*10^{-9} sec\\ \\No. \ of\ FF\leq t_{clk}/t_{pd}=10^{-7}/12*10^{-9}=8.33=8\\ \\MOD\leqslant 2^{N}=2^{8}=256$

Answer : 256

answered Jun 20 by (10,800 points)
selected Jun 21 by gbmentor
0 votes
256 is correct one
answered Jun 20 by (17,270 points)
+1 vote
Answer will vary depending on the type of counter. (Ripple, Ring and Johnson)
answered Jun 20 by (280 points)
edited Jun 20
Can you please provide a method/ formula as well as the answer to respective counters?
Answer: