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.
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?
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.
PARASHUUUUU said:
7 years ago
12 flip-flop =2^12 =4096.
Input clock frequency of 20.48 MHz.
F=I/P CLOCK FREQUENCY/2^N.
F=(20.48X10^6)/2^12 =5KHZ.
Input clock frequency of 20.48 MHz.
F=I/P CLOCK FREQUENCY/2^N.
F=(20.48X10^6)/2^12 =5KHZ.
(6)
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.
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.
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.
Pratyusha said:
1 decade ago
12 flip flops = 2^12 = 4096.
=> 20.48*10^6=20480000.
20480000/4096 = 5000 i.e., 5 kHz.
=> 20.48*10^6=20480000.
20480000/4096 = 5000 i.e., 5 kHz.
John Brown said:
1 decade ago
What then is the relation between propagation delay and clock frequency of flip-flop?
Shivaji said:
1 decade ago
N=2^n;where n= number of flip-flops,N=total number of states of a counter
Ramakrishna said:
2 years ago
Output frequency = Input Freq./2^N.
Here, N is a number of flip-flops.
Here, N is a number of flip-flops.
(7)
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