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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/ > > > > > > > > > __________________________________ > Do you Yahoo!? > Yahoo! Mail Address AutoComplete - You start. We > finish. > http://promotions.yahoo.com/new_mail >

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