Answer to this reader question
This depends on the energy content in the smoothing capacitor(s). As a rough estimate one might say that energy from the capacitors can be used until their voltage has dropped down to half the peak of the rated input voltage. This would mean that 3/4 of the energy content can be used. The energy contained in a capacitance is calculated as C/2*u^2. Now take this enery and divide it by the DC output power you use, and you obtain an approximation of the ride-through time.
If you want a high ride-through time use a power supply that is made for operating on a wide range of voltages without any manual modifications (range switches). Today many of them are designed to work properly on all voltages from 90V to 264V (50Hz to 60Hz). Now this means that even at 90V the unit must have a reasonable ride-through time, at least something around 1 semi-wave (10ms at 50Hz and about 8ms at 60Hz). Now when operated at 230V the same device should have at least (230V/90V)^2=6.5 times the ride-through capability it has at 90V.
And this refers to a total outage, not just a sag. Actually a unit designed for a voltage range from 90V to 264V will have an unlimited ride-through capability for sags up to 60%.
Unfortunately if your system has a voltage rating as low as 115V you do not have such an easy and straightforward option. Then you must go back to the first step and try to find out how much smoothing capacitance there is in your unit.
Even more unfortunately, the smoothing capacitance will by default be much lower if the power unit has an active front end (electronic power factor correction - PFC). Above rules of thumb cannot be applied then because it is uncertain how the (low) smoothing capacitance is used and handled by the electronics.Log in to post comments