# How to calculate the number of required flip-flop stages needed for clock-domain crossing?

In a given scenario where I have two clock domains driven by a 200MHz and a 30 MHz external independent clocks, what would be the best way to calculate the number of flip-flop stages needed for proper clock-domain crossing?

At the moment, I'm assuming the calculation for the number of stages should go something like this

200/30 = 6.67


As such, I'd need 7 flip-flop stages to be able to meet the clock-domain crossing requirement. Is this a correct assumption?

The system has multiple control signals and a couple of data-buses going back and forth. The widest bus is 32-bits while the smallest bus is 24-bits.

• Sounds right to enable flow control plus margin for prop.delay and latency. But could be more. Commented Nov 10, 2021 at 16:47
• Frequency of clocks and no. of flops in synchronizer chain has no relationship like you calculated. 2 stage FF synchronizer is enough for low/medium speed designs.....but you should provide us more information on what is passed and in what direction. Is it multi bit/single bit signal etc.. Commented Nov 10, 2021 at 17:06
• @mitu-raj, thanks for the reply! Just added some more details to my inquire. Commented Nov 10, 2021 at 17:30
• ... to meet the clock-domain crossing requirement What clock-domain crossing requirement? Go 200->30, and you cannot of course capture all the data. Go 30->200, and only one flop is needed for the data, with 2 or 3 for control signals to meet metastability requirements. Commented Nov 10, 2021 at 18:41
• I think you only need 3 FF's, properly connected and driven, to guarantee metastable-free signal crossing from one cock domain to another, irrespective of what the two domain clock frequencies are. Commented Nov 11, 2021 at 0:30