The discussion focuses on why multicycle paths (MCP) do not require two-stage synchronizers despite potentially violating timing at intermediate clock edges. Participants clarify that multicycle constraints are typically applied within the same clock domain where clock tree balance and known uncertainties prevent the true metastability seen in cross-clock domain transitions.
The consensus is that while an intermediate edge might see a transition, the register is only intended to capture the stable value at the specified destination edge, and any transient instability is naturally resolved by the time the next valid sampling edge occurs.
- Multicycle paths are generally used within a single clock domain where clock skew and jitter are controlled.
- Metastability is primarily a concern for asynchronous clock domain crossings rather than synchronous multicycle paths.
- Intermediate clock edges in a multicycle path do not cause issues because the final sampling edge is guaranteed to meet setup and hold requirements.
- Correct implementation of multicycle paths often involves a clock enable signal to ensure data is only captured when stable.
- The probability of a metastable state persisting across multiple clock cycles is mathematically negligible at standard operating frequencies.
Hi All,
This thought occurred to me regarding multicycle path.
Assume I set multicycle of 2. Then this implies that data delay from launch edge(0) will respect the timing at latch edge(2). while edge(1) will not be respected and could well be violated. So how come metastability isn't an issue and no two stage synchronisers are needed. I know latch edge(2) will sample without violation but the events at edge(1) aren't absorbed just like any asynchronous case.
Regards
Nori
I see your point valid. I don't have the answer. One might say the latch edge is either enabled or not but this doesn't explain it since enable is not mandatory for multicycle.
I know multicycle works without problem if applied correctly so it could be this issue of metastability is questionable and a register will do its job as long as its sampling edge is not violated. Or it could be that clock enable is indeed needed for multicycle path.
Another thought is that the case of multicycle violation is different from case of clock crossing. With clock crossing any output sampled could go metastable but with multicycle the violated edge is not relevant as the next edge comes and samples input correctly overriding previous edge violation.
You have a good explication in http://www-classes.usc.edu/engr/ee-s/552/coursematerials/ee552-G1.pdf
Metastability occurs when signals cross clock domains as uncertainty is not known. Metastability exists (it is a probability) even if domains were to be on-chip whether in ASIC of FPGA(though here clock paths are already laid). Setting higher values of multicycle would not resolve metastability as uncertainty is still unknown between domains. Multicycle is relevant only in the same clock domain!!
To resolve metastability synchronizer is required. One must consider clocking domains being crossed i.e., slow-to-fast, fast-to-slow and clock frequency in the same range. Sometimes a string of flops may not be right and handshake would be required and it is a design issue. In the same domain metastability would not occur since clock tree is balanced and uncertainty is known.
