Help re inverters please.

[robp]

This sounds like a bit of a basic question but anyway here goes, if anyone can help?
If say 75 Watts at 230V is being supplied by an inverter, is the ratio straightforward at the 12V side? I.e does 0.33A at 230V become roughly 6.25A at 12V? All allowing for the inefficiencies of a cheap inverter of course. Or is there some other dark secret about switch mode inverters I don't know about? I couldn't find an ammeter man enough to tell me yesterday!

 

[davidej]

Yes -but the efficiency is lowish. I am no expert but would guess 75-80% so this means you would need to gross the amperage up by a factor of 1.25-1.33

 

[Billjratt]

Yup, thick wires on the low voltage side!
The inverter will take a little 'off-load' and add more for each watt drawn at the output, plus a percentage for inefficiency. If you shunt your ammeter with a bit of fusewire it will read higher, then you can measure a known load, and compare.

 

[aluijten]

Actually this can be a bit more complicated. There are two big families of inverters: The pure sinewave and the modified sinewave.
The first one are more expensive. However some appliances can increase their consumption on a modified sinewave inverter, or even not work at all. The Senseo Coffee machine is a well known example.

Generally speaking more expensive inverters also have better efficiency (even up to 95% for the well build ones) this also makes them cooler to run.
Both types of inverters can be efficient to a lower or higher extent. It basically depends on the quality of the electronics.

Cheers,

Arno

 

[robp]

Thanks for the replies. Seems then that a straight conversion using Ohm's law will do it, allowing for about 70% to 80% efficiency.
Thanks for the suggestion of a shunt. It's funny how one forgets what we learnt 50 odd years ago. I'll check it with a known 12V bulb. Doh!