Electronics and Communication Engineering - Digital Electronics - Discussion


A JK flip flop has tpd= 12 ns. The largest modulus of a ripple counter using these flip flops and operating at 10 MHz is

[A]. 16
[B]. 64
[C]. 128
[D]. 256

Answer: Option D


Number of flip-flops = = 8.333 say 8

Modulus = 28 = 256.

Deepak said: (Jun 27, 2015)  
How they have calculated no of flip flops? Can anybody explain?

Sujideepika said: (Jul 12, 2016)  
Can anyone explain what is tpd?

Apurnak said: (Jul 23, 2016)  
@Sujideepika - TPD is propagation delay.

Javed Rizvi said: (Aug 4, 2016)  
Let's have an example.

I bought some eggs for 30 rupees, and I know the cost of 1 egg which is 5 rupee.

Then you can calculate the number of eggs I bought.

By dividing i.e. 30/5 = 6 eggs.


In the same ways in the given question, propogation delay time for 1 JK flip-flop is given and the combinational resulting operating frequency is also given,

Then we can change the combinational frequency into time by the simple conversion I. E (time = 1 / frequency).

And after that, we can calculate the number of flip-flop used by dividing I. E (total time taken / propagation time delay of one flip-flop).

After that.

If n flip-flop is given then the total modulus can be solved by the formula (modulus^n).

Hope I can explain clearly.

Javed Rizvi said: (Aug 4, 2016)  
At the last two lines. The formula used for the modulus will be (2^n).

Akshay Damre said: (Sep 21, 2016)  
Exact calculation formula for the number of flip-flops used is,

F = 1/(N * tpd).

N-> No of flip flops.
F-> Frequency.
Tpd-> prop. Delay time.

Santosh said: (Dec 30, 2016)  
Nice explanation @Akshay.

Karun Gupat said: (Feb 22, 2017)  
Thanks to all of you for explaining the question.

Sridevi said: (May 7, 2020)  
Fmax-max operating freq.
tpd- total propagation delay.
n - no.of Flipflops.
Fmax= 1/ n*tpd.
n= 1/ (10MHz*12ns).
n~ 8.
For mod counter its is 2^n.
so 2^8 =256.

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