Ripple counter timing — recycle delay through 4 cascaded flip-flops A 4-bit ripple counter uses flip-flops with 15 ns clock-to-Q propagation delay each. How long does it take to recycle from 1111 back to 0000 (worst-case total ripple time)?

Introduction / Context:
Ripple (asynchronous) counters are built by daisy-chaining flip-flops so that each stage is clocked by the previous stage’s output. This simplicity comes with a cost: output bits do not change simultaneously. Instead, transitions propagate sequentially from the least significant stage to the most significant stage, producing a cumulative timing delay that matters in high-speed applications.

Given Data / Assumptions:

• Counter width: 4 bits (four cascaded flip-flops).
• Clock-to-Q propagation delay per flip-flop: 15 ns.
• We are concerned with the recycle transition from 1111 to 0000 (the longest ripple event).

Concept / Approach:

In a ripple counter, each flip-flop toggles after the previous stage’s output transition has propagated. Thus, the total time for the most significant bit to settle equals the sum of the individual propagation delays through all stages involved. The worst case is typically the all-ones to all-zeros transition, requiring a toggle at each stage in sequence.

Step-by-Step Solution:

Verification / Alternative check:

Timing diagram inspection shows staggered transitions: Q0 at t0+15 ns, Q1 at t0+30 ns, Q2 at t0+45 ns, Q3 at t0+60 ns. Only after the MSB settles has the state truly become 0000.

Why Other Options Are Wrong:

15 ns: counts only one stage, not the ripple through all four.

30 ns or 45 ns: account for 2 or 3 stages, still short of the full chain.

Common Pitfalls:

Assuming simultaneous bit changes as in synchronous counters; ignoring that ripple delay limits maximum input clock frequency and can cause decoding glitches if outputs are read during the transition window.

Final Answer:

60 ns