+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 by gbmentor (73,870 points)
reshown Jun 20 by gbmentor

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 tskushagra-guptacse (11,660 points)
selected Jun 21 by gbmentor
0 votes
256 is correct one
answered Jun 20 by tsnikhilsharmagate2018 (24,690 points)
+1 vote
Answer will vary depending on the type of counter. (Ripple, Ring and Johnson)
answered Jun 20 by tsjeganbalajibs (390 points)
edited Jun 20 by tsjeganbalajibs
Can you please provide a method/ formula as well as the answer to respective counters?