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.
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
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.
Shivaji said:
1 decade ago
N=2^n;where n= number of flip-flops,N=total number of states of a counter
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.
Mahi said:
1 decade ago
Formula of @Manimegala is correct
Csk said:
1 decade ago
20.48 Mhz= 20480 Khz;
12 flip flops indicating 2 power 12 counter i.e 2 power 12 = 4096;
20480/4096=5 khz.
12 flip flops indicating 2 power 12 counter i.e 2 power 12 = 4096;
20480/4096=5 khz.
Arunalex said:
1 decade ago
N=number of Flip Flop, Input Frequency in MHz so 10^6
Nisha said:
1 decade ago
@manimegala
Yes correct.
Yes correct.
Manimegala said:
1 decade ago
Output frequency = Input Freq./2^N.
Here N is no of filp-flops.
Here N is no of filp-flops.
Divyanshu said:
1 decade ago
@Kowsalya
20.48 MHz = 20.48*10^6 Hz
20.48 MHz = 20.48*10^6 Hz
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