Question
In a 4-bit ripple counter, if the period of the waveform at the last flip-flop is 64 microseconds, then the frequency of the ripple counter in kHz is __________. (Answer in integer)
To solve for the frequency of the ripple counter, we begin by understanding the characteristics of a ripple counter. A 4-bit ripple counter can count from 0 to 24-1, which is 15. The flip-flop at the last stage, therefore, divides the input frequency by 24=16. Given the period of the waveform at the last flip-flop is 64 microseconds, we can calculate the frequency using the formula: frequency = 1/period.
The period, T = 64 microseconds = 64×10-6 seconds. Hence, frequency at the last flip-flop: f = 1/64×10-6 Hz = 15,625 Hz.
Since this is for the last flip-flop which divides the frequency by 16, the input frequency (fin) is:
fin = 15,625 Hz × 16 = 250,000 Hz = 250 kHz.
This result of 250 kHz must be within the given expected range of 250,250, which it is. Therefore, the frequency of the ripple counter in kHz is 250.
The period, T = 64 microseconds = 64×10-6 seconds. Hence, frequency at the last flip-flop: f = 1/64×10-6 Hz = 15,625 Hz.
Since this is for the last flip-flop which divides the frequency by 16, the input frequency (fin) is:
fin = 15,625 Hz × 16 = 250,000 Hz = 250 kHz.
This result of 250 kHz must be within the given expected range of 250,250, which it is. Therefore, the frequency of the ripple counter in kHz is 250.
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Important Questions from Counter
Which is not related to the counter circuit design?
A 16-bit synchronous binary up-counter is clocked with a frequency fCLK. The two most significant bits are OR-ed together to form an output Y. Measurements show that Y is periodic, and the duration for which Y remains high in each period is 24 ms. The clock frequency fCLK is ______ MHz. (Round off 2 decimal places.)
A MOD 2 and a MOD 5 up-counter when cascaded together results in a MOD ______ counter. (in integer)
A modulus 10 counter must have
How many flip-flops circuits are needed for a 4 bit counter?
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