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.
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.
(6)
Jasso said:
2 years ago
The Output Clock frequency = 20480000/4096 = 5000 i.e., 5 kHz.
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.
(5)
Yaswant said:
8 years ago
According to me, it's 20.48M/2^12.
(2)
Vbnbh said:
8 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)].
(1)
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?
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.
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.
Vyshnavi said:
9 years ago
Which is best? Can you give me an explanation.
Pankaj said:
9 years ago
What is RS Flip flop? Explain with electronic circuit and truth table?
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