Digital Electronics - Flip-Flops - Discussion
Discussion Forum : Flip-Flops - General Questions (Q.No. 1)
1.
Determine the output frequency for a frequency division circuit that contains 12 flip-flops with an input clock frequency of 20.48 MHz.
Discussion:
35 comments Page 2 of 4.
Utkarsh Arora said:
10 years ago
A counter or a flip flop in series combination acts as a frequency divider and divides the frequency in the power of 2^n, where n represents the number of flip flops used.
Therefore, (20.48*10^6)/2^12.
Therefore, (20.48*10^6)/2^12.
Sowmiya said:
1 decade ago
What are the two stable states of flipflops?
John Brown said:
1 decade ago
What then is the relation between propagation delay and clock frequency of flip-flop?
Khitab Oza said:
1 decade ago
Output frequency = input frequency/n^12.
Where n=2 for divide-by-2 divider normally it takes n=2.
Where n=2 for divide-by-2 divider normally it takes n=2.
Deepika said:
1 decade ago
As the input given, we have to find the output frequency. The formula is:
Output frequency = input frequency/2^n.
Where n is no.of flip-flops.
Output frequency = input frequency/2^n.
Where n is no.of flip-flops.
Devaraj said:
1 decade ago
1FF are used divide the clock frequency by 2.
12ff divide the frequency by 2^(12).
Given i/p frequency = 20.48MHZ.
Then o/p frequency = (20.48*10^6)/(2^12).
= 5 khz.
12ff divide the frequency by 2^(12).
Given i/p frequency = 20.48MHZ.
Then o/p frequency = (20.48*10^6)/(2^12).
= 5 khz.
Zoro said:
1 decade ago
Input frequency/no.of sequence of count.
DSingh said:
1 decade ago
The output frequency after each of the flip flop divided by 2.
So after 12 stages the clock frequency becomes
= 20.48*10^6/2^12=~5 KHz.
Hence option B is correct.
So after 12 stages the clock frequency becomes
= 20.48*10^6/2^12=~5 KHz.
Hence option B is correct.
Sai said:
1 decade ago
They said 12 flip-flops that mean we require 12 <= 2^n. So we are taking n as 4. Then we know that the output frequency = input frequency/2^n.
So 20.48/2^4=5.
So 20.48/2^4=5.
Pushp thakur said:
1 decade ago
12 ff contain 4096 bits and each ff has two states(2^12)
and we've given input frequency i.e.
f=20.48
so the output frequency=input frequency*(1/2^no.of ff)
and the output frequency=20.48*(1/4096)
i.e.o/p frequency=0.005hz(5khz)
and we've given input frequency i.e.
f=20.48
so the output frequency=input frequency*(1/2^no.of ff)
and the output frequency=20.48*(1/4096)
i.e.o/p frequency=0.005hz(5khz)
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