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.
Mahi said:
1 decade ago
Formula of @Manimegala is correct
Nandan Roy said:
1 decade ago
If N flip flops are connected, then input frequency is divided by 2^n.
So in this problem
20.48 * 10^6 / 2^12 = 5000Hz.
So in this problem
20.48 * 10^6 / 2^12 = 5000Hz.
Shivaji said:
1 decade ago
N=2^n;where n= number of flip-flops,N=total number of states of a counter
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.
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
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)
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.
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.
Zoro said:
1 decade ago
Input frequency/no.of sequence of count.
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.
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