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 1 of 4.
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)
Sujeet Kumar Layek said:
1 decade ago
We know that a single Flip Flop contains 2 states. For 1 FF frequency is 1/2, for 2 FF frequency is 1/4.
i.e. 1/2^n, where n is the no. of FF.
so for 12 FF, frequency is 1/2^12 = 1/4096.
So, o/p frequency = (i/p frequency)*(1/4096)...
We also called it as a divider ckt.
i.e. 1/2^n, where n is the no. of FF.
so for 12 FF, frequency is 1/2^12 = 1/4096.
So, o/p frequency = (i/p frequency)*(1/4096)...
We also called it as a divider ckt.
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.
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.
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.
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.
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.
Anand said:
1 decade ago
Maximum number stored by 12-Flip Flops is 4096.
Freq division =(20.48X10^6)/4096
=5000
Output frequency of the 12 FF's freq divisor is =5KHz
Freq division =(20.48X10^6)/4096
=5000
Output frequency of the 12 FF's freq divisor is =5KHz
Vbnbh said:
9 years ago
Mod no = Fin/Fout -->(1)
Mod no. = 2^N -->(2) Where N is the no. of flip flop
Hence Mod no = 2^(12)= Fout = 20.48 / [2^(12)].
Mod no. = 2^N -->(2) Where N is the no. of flip flop
Hence Mod no = 2^(12)= Fout = 20.48 / [2^(12)].
(2)
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