Difficulty: Medium · Puzzle 6 of 6 · Topic: Digital Logic Puzzles
A synchronizer MTBF is modeled as MTBF proportional to exp(Tresolve/tau). If adding one extra sync flop increases settling time by 2 ns and tau=50 ps, by what factor does MTBF improve?
Compute exp(deltaT/tau) using ratio 2 ns / 50 ps.
1) MTBF ratio = exp((Tnew - Told)/tau) = exp(2 ns / 50 ps).
2) Convert units: 2 ns = 2000 ps.
3) Exponent = 2000/50 = 40.
4) Ratio = e^40 approx 2.35 x 10^17.
5) So one extra stage can massively improve reliability when timing budget allows.
6) Caveat: the extra stage only buys the resolution time left after clk-to-q and setup of the next flop, so in a real design the gain is e^((T_clk - t_cq - t_setup)/tau) per stage, and a 2 ns gain assumes a slow clock (at 1 GHz a stage buys well under 1 ns).Answer: MTBF improves by e^40, approximately 2.35e17 times.
Checked by running a model of the rule in a Python model. A computational check: exp(2 ns / 50 ps) = exp(40) = 2.35e17.
Expecting a fixed additive gain per extra stage (like +1 year); the exponent adds, so MTBF multiplies. The opposite mistake is quoting e^40 for a fast clock without subtracting clk-to-q and setup.
Why does this argument not guarantee safe transfer of one-cycle pulses? Model answer: MTBF only measures the chance that a sampled value resolves too late. A pulse shorter than the destination period can fall between two destination edges and never be sampled. Use a toggle or handshake (see the clock-domain pulse puzzle).
The ratio of the two MTBFs is e to the power of the extra resolution time over tau. The extra stage adds 2 nanoseconds, which is 2000 picoseconds, and tau is 50 picoseconds, so the exponent is 40. The improvement is e to the 40, about 2.35 times ten to the 17. The trap is expecting an additive gain, but the exponent adds so MTBF multiplies. Also, in a real design the gain is the clock period minus clock-to-q and setup, so a 2 nanosecond gain assumes a slow clock.