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switching bench power supplies overloads my epever inverter

You've run into the difference between watts and volt-amps.

Inverters are often rated in VA, while loads are often rated in watts. In a DC circuit they are the same, but not in AC.

The power factor describes the relationship for a given load. In your case, that Velleman LABPS3020SM power supply has an abysmal power factor rating of 0.61.

The equation for power factor is PF = W/VA. So if you have the power factor and watts, then VA = W/PF. You power supply is, therefore, drawing 1.6x the power it uses. Note that this doesn't mean you're losing that power - as a reactive load, it draws on one portion of the cycle, and then feeds back on the other portion, so the total consumed energy is still the same as watts, however the inverter sees a 1.6x load during portions of its cycle, even though the power supply isn't actually consuming that much energy.

So at 22.2v and 16.7A output, your power supply is providing 370w. With a conversion efficiency of perhaps 85% it's consuming 436w, so generating about 66w in heat, and 370w in output energy. Given the conversion efficiency and the power factor, the maximum VA that power supply can use is right about 1,200VA, which is a common power design limit for electronics.

If it's actually consuming 436w, and has a power factor of 0.61, then its apparent power is 436/0.61 --> 715VA

Given that you're running a least your lights, and possibly other loads, off that 1,500VA, and most of them are likely similar in terms of power factor. Cheap LED lights are often just barely above 0.5 power factor - so they consume nearly twice the apparent power (ie, VA) than their rated wattage.

There are some things you can do to match the loads to your inverter, but honestly it's best to simply buy the inverter that is rated for the continuous VA that your loads need, or buy loads that have a 0.95+ power factor so the watts equals the volt-amps.

This is why a lot of people ignore VA vs W and simply handwave the issue by suggesting you always buy an inverter 2x larger than your loads. It's not a bad way of doing things, but it's also not the most efficient or effective way to design a system.
wow thanks. This makes sense. Thanks alot =)
 
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