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+1 vote

in Digital Logic by Loyal (6.4k points) | 243 views

T CLK  >= T propagation_delay

Propagation delay = T FF + T combinational

TCLK >= 15 + 5  = 20 ns

Frequency CLK = 1/20

                          = 50 Mhz

In case of Syn counters all the flipflops will get the clocks at the same time, so we need not wait for the previous flipflop to produce the result which means we can run everything in parallel==> one time of flipflop delay.

Similarly the combinational circuits doesn't need to wait for any thing ==> one time of combinations ckt delay

===>15+5= 20ns

Freq= 1/time=> 50Mhz
When op of one flip flop goes as an input to another flip flop then why don't we consider propogation delay inclusive of all the N flip flops .
All the flipflops will operate at the same time. N-1th flipflops previous state o/p will be the input as current input to  nth. So there won't be any delay. All will run in parallel
I have just one doubt , say I have 3 FF ,Now output of 1st flip flop is connected to input of 2nd FF and output of 2nd FF is connected to input of 3rd FF .

Now until we get op of 2nd FF , how can we get OP of 3rd FF , so we should take into consideration the delay of all flip flops .

1 Answer

0 votes
Its synchronous counter so ,

Propagation delap = T(f-f)+ T(combinational)

T(pdsyn) = 15 + 5 =20 ns

F= 1/T(pdsyn) = 50 HZ
by Junior (977 points)
So answer will be option A as we require min clock of 20ns . but clock time more than 20ns is allow but less is not allowed
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