A 12 MHz clock frequency is applied to a cascaded counter containing a modulus-5 counter, a modulus-8 counter and a modulus-10 counter. The lowest output frequency possible is ________
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Cascaded counters divide the clock by the product of their moduli. Here the total division factor is 5×8×10 = 400, so 12 MHz / 400 = 30 kHz.
Formula / Key Relation
Mtotal = 5 × 8 × 10 = 400
fout = 12 MHz / 400
= 30 kHz
Detailed Explanation
Concept & Reasoning
For cascaded divide-by-N stages, the output of one stage clocks the next, so frequency division factors multiply. The lowest available frequency is at the final stage. This is a direct modulus-product calculation.
How to Solve It in the Exam
Multiply all moduli first, then divide the input frequency.
Important Exam Point
Use the final-stage output for the lowest frequency.
Related Revision Path
This question sits in the revision path Digital Electronics → Sequential Logic → Cascaded Counters .
Quick Trick
Multiply all moduli first, then divide the input frequency.
Common Mistake
Adding the moduli instead of multiplying their division ratios.
Exam Tip
Use the final-stage output for the lowest frequency.
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