Wednesday, Jun 07, 2006 at 23:08
Careful about applying input and output wattages and voltages in the same sentences. How long do you think the fuse on the input (which won't be more than 20 or 30A) is going to let it draw the suggested 125A for? The 1500W peak is referring to an output rating. Not the input. If it was even getting close to that for an instant the current draw would blow the cigarette lighter fuse rather than allow the inverter to "konk out".
I = 1500W / 240v = 6.25 Amps output.
To get that in a process that isn't 100% efficient - and may not be even 80% efficient will require maybe as much as 9 or 10Amps continuous input.
P = IV = 12.5 * 10 = 125W
It's probably not a true sine wave inverter, so a multimeter on AC volts isn't necessarily going to report the same voltage you expect.
A 240v AC output from the wall socket at
home is made up of a sine wave that is peaking at 339v. Because it is a sine wave the multimeter reads the RMS value of 240v.
RMS stands for Root-Mean-Square, RMS is a standard method for determining
the effective value of a varying or alternating voltage. The RMS value
is the DC voltage that would produce the same amount of heat as the
voltage you are measuring, if you were to apply both voltage across
identical resistors.
The value produced by a square wave that is 50% duty cycle and peaks at 340v if measured as RMS will be about half that at 170v.
You are probably chasing a problem that doesn't exist, but is rather a function of the fact that it's a cheap square wave inverter.
If it drives what you want it to drive, don't worry about it.
Dave
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