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 3 of 4.
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.
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.
John Brown said:
1 decade ago
What then is the relation between propagation delay and clock frequency of flip-flop?
Sowmiya said:
1 decade ago
What are the two stable states of flipflops?
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.
Pankaj said:
10 years ago
What is RS Flip flop? Explain with electronic circuit and truth table?
Vyshnavi said:
10 years ago
Which is best? Can you give me an explanation.
Shreyan said:
9 years ago
Output freq = Input freq/ 2^n.
Where 'n' is number of flipflops = 20.48 * 10^6/ 4096 = 10^6/200 = 5000 = 5kHz.
Where 'n' is number of flipflops = 20.48 * 10^6/ 4096 = 10^6/200 = 5000 = 5kHz.
Lalit Arora said:
9 years ago
Each flip flop reduces frequency by factor of 2. Hence 12 flip flops so f (out) = (1/ (2^12) )f (input).
f(out) = (20.48 * 10^6) / (2^12) = 5000Hz = 5KHz.
f(out) = (20.48 * 10^6) / (2^12) = 5000Hz = 5KHz.
Mounika said:
9 years ago
What is the main difference for the relation between propagation delay and clock frequency of flip-flop?
Can anyone tell me?
Can anyone tell me?
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