Date: Fri, 7 Feb 2003 09:53:37 -0500
Reply-To: David Beierl <dbeierl@ATTGLOBAL.NET>
Sender: Vanagon Mailing List <vanagon@gerry.vanagon.com>
From: David Beierl <dbeierl@ATTGLOBAL.NET>
Subject: Re: Inverter Question
In-Reply-To: <000201c2ce78$a5a1d8c0$e9a2f5d1@hewlett2ih5nie>
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At 02:14 AM 2/7/2003, Loren A. Busch wrote:
>You are going to be much better off figuring out a circuit that will
>charge your batteries without using the inverter. The efficiency of
>jumping 12v up to 100 ac then back down (with your charger) to the 7.5v
>dc again to charge your batteries is going to eat a lot of juice. Check
>around. Though I've never seen one (never looked) I'm willing to bet
>that some of the RC model crowd has come up with chargers in the range
>you need that are designed to run off a 12v auto system. Or find an
>electronics guru that can gen up a circuit you can put together
>yourself.
Loren, a typical DC-powered charger (I believe) is going to use an IC or
transistor device as a shunt-type regulator to get the 12vdc down to 7.5,
which means that in the nature of the beast it will be about 60%
efficient. An AC-powered device might not use a regulator at all, or if it
did would start with a lower voltage. It would drop about a half volt in
the rectifier and probably not more than two in the regulator, so including
transformer losses it could be about .8-.9 (inverter) x .95 (transformer) x
.75 (diode bridge and regulator) = 57-65%, and could be a lot better if it
didn't use a regulator.
So it's not really a clear call which is better, and the difference
wouldn't be great either way.
I *suspect* that the numbers might be more favorable to the DC device for
something like a laptop power supply that has to use a DC-DC converter to
get a higher voltage out than the battery supplies -- but I don't
know. Likewise, a 12vdc-powered TV or such might be very inefficient
running on 12v, or very efficient (and the same on AC) -- I doubt the mfrs
put much emphasis on it. You can get a clue from how close the
internal-battery voltage is, if the device uses internal batteries;
whatever the efficiency is of that circuit, you can probably assume that
the 12v is going to be regulated down to nominal battery voltage or near it
with the attendant losses. Could even be less, since the device will have
to function down to somewhere between 0.8 and 1.1 volts/cell if it's going
to run on dry cells. Sometimes you just have to measure...
david
--
David Beierl - Providence RI USA -- http://pws.prserv.net/synergy/Vanagon/
'84 Westy "Dutiful Passage"
'85 GL "Poor Relation"
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