Electrical Engineering (EE) Exam > Electrical Engineering (EE) Questions > A pulse train with a frequency of 1 MHz is co...
Start Learning for Free
A pulse train with a frequency of 1 MHz is counted using a mod-1024 ripple counter built with J-K flip-flops. For proper operation of the counter the maximum permissible propagation delay per flip-flop stage is
- a)100 ns
- b)50 ns
- c)20 ns
- d)10 ns
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
A pulse train with a frequency of 1 MHz is counted using a mod-1024 ri...
Concept:
If we pass the input signal to a single T-flip flop, we will get half of the frequency at the output.
Similarly, when we pass the input signal into an n-bit flip flop counter, the output frequency (fout) will be:
Calculation:
Similarly, when we pass the input signal into an n-bit flip flop counter, the output frequency (fout) will be:
Calculation:
Given is a mod-1024 ripple counter which means that it can count 1024 states.
To count 1024 number of flipflop required is:
1024 = 2n
n = 10 flip-flops
tpdff = Propagation delay of flip flops
tpdff = 1/f
tpdff = 10-6 sec.
Propagation delay per flipflop =
= 100 nsec
Free Test
FREE
| Start Free Test |
Community Answer
A pulse train with a frequency of 1 MHz is counted using a mod-1024 ri...
The maximum permissible propagation delay per flip-flop stage in a mod-1024 ripple counter built with J-K flip-flops can be calculated based on the maximum frequency of the input pulse train and the desired operation of the counter.
Given that the frequency of the input pulse train is 1 MHz, we can calculate the period of each pulse as follows:
Period = 1 / Frequency
= 1 / 1 MHz
= 1 μs
Since the counter is a mod-1024 ripple counter, it will count up to 1024 before resetting to zero. Therefore, the counter needs to complete 1024 clock cycles within the period of the input pulse train in order to properly operate.
To calculate the maximum permissible propagation delay per flip-flop stage, we need to determine the maximum time allowed for each stage of the counter to propagate the signal. This can be calculated by dividing the period of the input pulse train by the number of stages in the counter.
Propagation Delay per Flip-Flop Stage = Period / Number of Stages
In this case, since the counter is a mod-1024 ripple counter, the number of stages is 10 (log2(1024) = 10).
Propagation Delay per Flip-Flop Stage = 1 μs / 10
= 100 ns
Therefore, the maximum permissible propagation delay per flip-flop stage is 100 ns (option A).
In a ripple counter, the output of each flip-flop is connected to the clock input of the next flip-flop in a chain. As a result, the output of each flip-flop is affected by the propagation delay of the previous flip-flop. If the propagation delay per flip-flop stage exceeds the maximum permissible value, the output of the counter may not change at the correct time, leading to incorrect counting.
Hence, it is crucial to ensure that the propagation delay per flip-flop stage is within the specified limit to guarantee proper operation of the counter.
Given that the frequency of the input pulse train is 1 MHz, we can calculate the period of each pulse as follows:
Period = 1 / Frequency
= 1 / 1 MHz
= 1 μs
Since the counter is a mod-1024 ripple counter, it will count up to 1024 before resetting to zero. Therefore, the counter needs to complete 1024 clock cycles within the period of the input pulse train in order to properly operate.
To calculate the maximum permissible propagation delay per flip-flop stage, we need to determine the maximum time allowed for each stage of the counter to propagate the signal. This can be calculated by dividing the period of the input pulse train by the number of stages in the counter.
Propagation Delay per Flip-Flop Stage = Period / Number of Stages
In this case, since the counter is a mod-1024 ripple counter, the number of stages is 10 (log2(1024) = 10).
Propagation Delay per Flip-Flop Stage = 1 μs / 10
= 100 ns
Therefore, the maximum permissible propagation delay per flip-flop stage is 100 ns (option A).
In a ripple counter, the output of each flip-flop is connected to the clock input of the next flip-flop in a chain. As a result, the output of each flip-flop is affected by the propagation delay of the previous flip-flop. If the propagation delay per flip-flop stage exceeds the maximum permissible value, the output of the counter may not change at the correct time, leading to incorrect counting.
Hence, it is crucial to ensure that the propagation delay per flip-flop stage is within the specified limit to guarantee proper operation of the counter.
Attention Electrical Engineering (EE) Students!
To make sure you are not studying endlessly, EduRev has designed Electrical Engineering (EE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Electrical Engineering (EE).
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Similar Electrical Engineering (EE) Doubts
Top Courses for Electrical Engineering (EE)View all
A pulse train with a frequency of 1 MHz is counted using a mod-1024 ripple counter built with J-K flip-flops. For proper operation of the counter the maximum permissible propagation delay per flip-flop stage isa)100 nsb)50 nsc)20 nsd)10 nsCorrect answer is option 'A'. Can you explain this answer?
Question Description
A pulse train with a frequency of 1 MHz is counted using a mod-1024 ripple counter built with J-K flip-flops. For proper operation of the counter the maximum permissible propagation delay per flip-flop stage isa)100 nsb)50 nsc)20 nsd)10 nsCorrect answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A pulse train with a frequency of 1 MHz is counted using a mod-1024 ripple counter built with J-K flip-flops. For proper operation of the counter the maximum permissible propagation delay per flip-flop stage isa)100 nsb)50 nsc)20 nsd)10 nsCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A pulse train with a frequency of 1 MHz is counted using a mod-1024 ripple counter built with J-K flip-flops. For proper operation of the counter the maximum permissible propagation delay per flip-flop stage isa)100 nsb)50 nsc)20 nsd)10 nsCorrect answer is option 'A'. Can you explain this answer?.
A pulse train with a frequency of 1 MHz is counted using a mod-1024 ripple counter built with J-K flip-flops. For proper operation of the counter the maximum permissible propagation delay per flip-flop stage isa)100 nsb)50 nsc)20 nsd)10 nsCorrect answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A pulse train with a frequency of 1 MHz is counted using a mod-1024 ripple counter built with J-K flip-flops. For proper operation of the counter the maximum permissible propagation delay per flip-flop stage isa)100 nsb)50 nsc)20 nsd)10 nsCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A pulse train with a frequency of 1 MHz is counted using a mod-1024 ripple counter built with J-K flip-flops. For proper operation of the counter the maximum permissible propagation delay per flip-flop stage isa)100 nsb)50 nsc)20 nsd)10 nsCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A pulse train with a frequency of 1 MHz is counted using a mod-1024 ripple counter built with J-K flip-flops. For proper operation of the counter the maximum permissible propagation delay per flip-flop stage isa)100 nsb)50 nsc)20 nsd)10 nsCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.
Here you can find the meaning of A pulse train with a frequency of 1 MHz is counted using a mod-1024 ripple counter built with J-K flip-flops. For proper operation of the counter the maximum permissible propagation delay per flip-flop stage isa)100 nsb)50 nsc)20 nsd)10 nsCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A pulse train with a frequency of 1 MHz is counted using a mod-1024 ripple counter built with J-K flip-flops. For proper operation of the counter the maximum permissible propagation delay per flip-flop stage isa)100 nsb)50 nsc)20 nsd)10 nsCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A pulse train with a frequency of 1 MHz is counted using a mod-1024 ripple counter built with J-K flip-flops. For proper operation of the counter the maximum permissible propagation delay per flip-flop stage isa)100 nsb)50 nsc)20 nsd)10 nsCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A pulse train with a frequency of 1 MHz is counted using a mod-1024 ripple counter built with J-K flip-flops. For proper operation of the counter the maximum permissible propagation delay per flip-flop stage isa)100 nsb)50 nsc)20 nsd)10 nsCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A pulse train with a frequency of 1 MHz is counted using a mod-1024 ripple counter built with J-K flip-flops. For proper operation of the counter the maximum permissible propagation delay per flip-flop stage isa)100 nsb)50 nsc)20 nsd)10 nsCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Electrical Engineering (EE) tests.
|
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup