Asynchronous counter delay accumulation A 5-bit ripple (asynchronous) binary counter uses five flip-flops, each with propagation delay 12 ns. What is the total worst-case propagation delay tp(tot)?

Introduction / Context:
In ripple counters, the clock signal propagates through cascaded flip-flops, so output bits do not change simultaneously. The worst-case propagation delay equals the sum of the individual stage delays up to the most significant bit used, which limits the maximum counting frequency and affects decoding hazards.

Given Data / Assumptions:

• Number of stages (flip-flops): 5.
• Propagation delay per stage: 12 ns (worst case).
• Asynchronous (ripple) topology: each stage toggles on the previous stage’s output.

Concept / Approach:
Because the toggling ripples from the least significant to the most significant stage, the worst-case time for the final bit to settle is the sum of delays through all five flip-flops. Therefore, tp(tot) = N * t_pd for N cascaded stages in the ripple path.

Step-by-Step Solution:

Verification / Alternative check:
Compare to a 4-bit ripple counter: 4 * 12 ns = 48 ns. Extending to 5 bits adds another 12 ns, consistent with linear accumulation.

Why Other Options Are Wrong:

• 12 ms: Milliseconds are three orders of magnitude too large and mismatch per-stage specs.
• 24 ns / 48 ns: Correspond to 2 or 4 stages, not 5.

Common Pitfalls:
Confusing ripple with synchronous counters (which do not accumulate delays in the same way), or forgetting worst-case analysis when decoding multiple bits immediately after a transition.

Final Answer:
60 ns