Solving Metastability Using Double-Register Approach

Question

For solving metastability caused by different clock domains in Verilog, double-register method is used.

But as far as I know, the final output of metastability is undetermined. Output is independent of input.

So, my question is how to guarantee the correctness of output using double-register method?

Thanks.

Answer 1

You cannot be completely sure that you avoided metastability. As you mentioned, the output of a metastable flip-flop is unpredictable so you can potentially propagate a wrong value when you have metastability even with the 'two-register' approach.

This method however never intended to solve metastability but tries to reduce the probability that a metastable value enters your circuit. What is called here MTBF (Mean Time Between Failure). To reduce the MTBF you can even chain more that 2 registers.

Even if this does not solve the unpredictive-ness of a value, it is interesting to use these double registers because when a value is metastable, it will oscillate until it stabilize to 0 or 1.

This oscillation will make your circuit toggle and then use a lot of energy for nothing as each transition consumes energy. That for this reason that it is important to use double registers for clock-domain crossing.

To ensure that you data is valid though, you can use a request-acknowledge mechanism between the two clock-domains.

Quick example:

1. Set the data to the bus (input of a double registers)
2. Wait 1(or more) clock cycle to be sure the data is well established on the other side
3. Send a request signal (input of a double register)
4. Worst case : The request signal is metastable and stays at 0 once stabilized. Next clock cycle it will be at 1 as it would be already set to 1 for at least 1 clock cycle. Best case : next cycle the destination will accept the data
5. The data is stable, the request is stable and at 1 -> the data can be consumed. Send an acknowledgement to the source.
6. The acknowledgement arrives (on a double register in case of metastability). If metastable it can take a clock cycle more to arrive.
7. Request falls.
8. Another data can be sent via the bus

This protocol is called a 4-phase protocol. You can find a lot of documentation about it on the web as it is a classic protocol for asynchronous designs.

It is quite simple to understand and to implement. Keep in mind though that it will generate an overhead in area that can be quite important.

Hope it helps.