Difficulty: Easy
Correct Answer: each flip-flop
Explanation:
Introduction / Context:Ripple counters clock each successive flip-flop with the output of the previous stage. This creates a chain of sequential delays. Understanding which delays dominate sets expectations for maximum count frequency and informs when a synchronous counter is necessary.
Given Data / Assumptions:
Concept / Approach:The worst-case time for the most significant stage to settle equals the sum of the per-stage clock-to-Q delays of all preceding flip-flops. Thus, the flip-flop chain's propagation delays limit speed. External gates may appear in decoding, but they are not in the internal ripple clock path for basic counters.
Step-by-Step Solution:
Identify the number of stages N.Let t_pd(ff) be the per-flip-flop delay.Worst-case settle time to MSB ≈ N * t_pd(ff).Maximum count frequency must ensure period > worst-case settle time to avoid decoding hazards.Verification / Alternative check:Measure a ripple counter's outputs on a scope: higher-order bits lag the low-order bits by multiples of t_pd(ff), plainly showing the cumulative flip-flop delays.
Why Other Options Are Wrong:
Common Pitfalls:Assuming synchronous timing or ignoring accumulated delay when decoding multiple bits immediately after a transition (can cause glitches).
Final Answer:each flip-flop
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