Discussion

In a flipflop-based, 4-stage, ripple counter, each flipflop has a propagation delay of 25 ns. Find the clock frequency.

Answer: B

In a ripple counter each flipflop is clocked by the output of the one before it, so the delays add up. For four stages

\[ t_{total} = 4 \times 25\ \text{ns} = 100\ \text{ns} \]

The clock period must be at least the total delay, so the highest clock frequency is

\[ f = \frac{1}{100 \times 10^{-9}} = 10\ \text{MHz} \]

The value 100 MHz is the frequency for a single flipflop of 10 ns, and 1 MHz and 100 kHz are far below the limit. The answer is 10 MHz, option B.

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