Can anyone tell me if an inverter actually uses additional energy to
convert DC to AC? Is it a percentage in direct proportion to the
conversion?
thank you
Bob
Can anyone tell me if an inverter actually uses additional energy to
convert DC to AC? Is it a percentage in direct proportion to the
conversion?
thank you
Bob
Yes, it uses additional energy that must escape as heat. There is some minimum loss, just to run the converter, unloaded, some proportional losses, and some losses proportional to the square of the load. If the designers concentrated on efficiency, all these losses may amount to 5% to 10% of the full load output (90to 95 % full load efficiency. All units are zero percent efficient at no load.
John gave you some good answers. Let me add one thing -- if you model the input power going to an inverter as something like:
Pin = A*Pout+B (watts)
....where A is perhaps 1.10 (~10% losses or ~90% inverter efficiency), the constant loss term (B) tends to be proportional to the _maximum_ power the inverter was designed for. It's not uncommon to see something like a 5,000 watt maximum inverter that'll sit there and dissipate 25 watts under no load, because the designers figured normally you'd probably be running 2-4kW and, hey, what's 25W vs. a few kilowatts? On the other hand, something like a little 300W inverter would hopefully dissipate, say, no more than 3W under no load.
Moral of the story (all other things equal): Don't buy an inverter _too_ much bigger than you need!
Have something to add? Share your thoughts — no account required.
Ask the community — no account required