Date: Wed, 20 Oct 2004 11:06:10 -0700
Reply-To: Reinhard Vehring <rvehring@YAHOO.COM>
Sender: Vanagon Mailing List <vanagon@gerry.vanagon.com>
From: Reinhard Vehring <rvehring@YAHOO.COM>
Subject: Re: Inverter usage (was "Dual Battery Installation Instructions
on Ebay")
In-Reply-To: <20041020162726.82434.qmail@web14602.mail.yahoo.com>
Content-Type: text/plain; charset=us-ascii
Jeffrey,
not an easy question, because efficiency depends on
the type of inverter and the system setup.
Generally the inverter efficiency is around 90 % at
full load, maybe 95% for more expensive models and
larger inverters. The efficiency drops for smaller
loads. To find out how much, you need an efficiency
graph for your particular model. They are ususally not
easily avaliable for the cheap models (for a reason).
A way to estimate how well the inverter works at low
loads is to look at the "no load current draw" For a
500 W inverter it maybe around 500 mA.(So don't leave
it switched on if you don't need it) Performance at no
load and low load can be quite good, but you'll have
to pay for it is my guess.
I've seen an efficiency graph for a 400 W inverter
that had 75 % at 50 W load and 60 % at 25 W load.
Now overall efficiency is an entirely different
matter: In the case of a microwave load this is quite
interesting. There are basically two types of
inverters around: True sine wave inverters ($$$) and
"modified" sine wave inverters. The latter control the
RMS output by changing the width of the rectangular
pulses they produce. Most loads run fine on this type
of waveform but it affects efficiency. Microwave
cooking power depends on peak voltage not RMS voltage,
so you only get the full cooking power if the inverter
produces reasonably high peak voltage. It only does
that if you feed it at about 14 to 14.5 V. Unless you
charge the battery by running the engine when you
cook, this is not going to happen. In fact drawing a
large current draw from a small battery will drop the
voltage to maybe around 11V or less. At that DC
voltage the peak output voltage of the inverter drops
and the efficiency of the microwave is lousy.
Basically you need to run it longer to get your stuff
hot. Now the overall efficiency of your system maybe
only 40 to 50 %. To get it up you'd need a larger
battery that tolerates the current draw. This is why I
don't like the cheap small battery - cheap inverter
solution.
BTW: The gas burner is the way to go: Just one energy
conversion:fuel - heat. Microwave in a car needs 4
conversions: fuel - mechanical (alternator) -
electrical DC - electrical AC - heat.
Hope this helps, sorry about the rambling
Reinhard
--- Jeffrey Earl <jefferrata@yahoo.com> wrote:
>
> Reinhard,
>
> This brings up a question I've had since recently
> installing 400W inverter on my aux. battery: is the
> power consumption largely dependent on the total
> amperage draw of the devices connected thru the
> inverter? Or does the inverter itself draw enough to
> soon drain the aux. battery?
>
> Appreciate any advice you can offer.
>
> Jeffrey Earl
> 1983 diesel Westfalia "Vanasazi"
> http://www.vanthology.com/
>
>
>
>
>
>
>
>
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