Re: Capacitance
I think you are referring to the so-called hold-up time in a power supply, powered from a PFC or 400VDC bus.
400V sounds like the DC bus of a PFC.
I think the question should be asked like this: what is the value of a cap that will keep the power supply operating for x ms (usually 20ms= one AC cycle), at a nominal output of 2500W, if the voltage on the cap is to be no less than X volts, starting from 400V?
The answer is in the energy required. Assume constant power drain (normal with switching power supplies). The minimum voltage on the cap will be assumed 300V.
Then, the energy in the cap is E=0.5*C*U^2
The starting energy is then Estart=0.5*C*1600
The energy at the end of hold-up time is :
Eend=0.5*C*900
The difference should equal the power * time
Erequired=2500W*20ms=50 Joules
Then: Estart- Eend=50J= 0.5*C*(1600-900)=350*C=Erequired
C=50/350=0.143F
If the power supply can work at less than 300V, then use that value to calculate the energy at the end of hold-up time and the cap will be lower.
If the time must be longer than 20ms, the cap will be larger.
Generally, you should consider the start voltage to be less the 400V, namely the minimum that you can have on the bus under normal conditions, usually 385V. That will increase the cap dramatically. And the final value of the cap must be corrected with the tolerance, since you must assume the cap can be 20% lower and still hold up. Finally, add some margin to all that, say another 10-25%.