Date:         Tue, 12 Jul 2011 08:10:49 -0700
Reply-To:     Shawn Wright <vwdiesels@GMAIL.COM>
Sender:       Vanagon Mailing List <vanagon@gerry.vanagon.com>
From:         Shawn Wright <vwdiesels@GMAIL.COM>
Subject:      Re: Fwd: Greek Single Cab + alternator help...
Comments: To: David Beierl <dbeierl@attglobal.net>
In-Reply-To:  <4e1bb0c7.907fe50a.6655.fffff5c1@mx.google.com>
Content-Type: text/plain; charset=ISO-8859-1

David,

Thanks, I isolated each of 120R and they checked out fine (I cut one end of
each, then soldered back), so I will try the current draw test. Now I am
suspecting a wiring fault between the panel and the alternator. Will have
another go at it tonight.

Thanks for the help!

On Mon, Jul 11, 2011 at 7:26 PM, David Beierl <dbeierl@attglobal.net> wrote:

> At 09:27 PM 7/11/2011, Shawn Wright wrote:
>
>> I have two big 120ohm resistors, and a 470ohm in series with the LED. All
>> three check out ok, yet the resistance between terminal 8 & 11 is 540ohms,
>> not the 160 it should be. Or is that also an old spec in Bentley?
>>
>
> From the diagram for the '88, 97.126 tracks 52-4...it shows two resistors
> in parallel with the LED/resistor pair.  If each of those resistors is 120R,
> the effective resistance of that pair is 60R.
>
> The diagram shows a diode in series with the two parallel resistors and
> LED/resistor.  That will make ohmmeter readings a matter very much dependent
> on the individual characteristics of the particular ohmmeter in use.
>
> Put +12 on the panel and see how much current you can draw from pin 11.
>  Roughly:
>
> You're starting with 12-12.5 V.  Deduct .6 V for the series diode, that
> leaves 11.5-12.
>
> I = E / R.  Taking into account just the two 120R resistors, I = 11,5/60 =
> .192 A or 192 mA.  Add 15 mA or so for the LED/resistor pair to get 200+ mA.
>
> If you can only pull half that then one of the 120R is open.  If only ~15
> mA then both are open.
>
> Easiest way may be to unplug the fridge relay and alternator D+, then plug
> in the panel and ground either of those leads through your milliammeter.
>
> Power dissipated would be P = E^2 / R = 12^2 / 120 = 144/120 = 1.2 W at
> each of the 120R resistors, so I'd take your measurement promptly and
> disconnect.
>
> To track which resistor is open, put a pea bulb of some sort instead of
> your milliammeter, then use the voltmeter to track voltages on the
> resistors.  The one(s) with the same voltage at both ends is the open
> one(s).
>
> Yours,
> David
>



--
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Shawn Wright