Alan

Glen - when I guess at something - which isn't all that often - I make a note to say that I'm not sure. I'm not guessing at this one.

The bulb really serves 2 purposes... in an S4 it isn't really even needed as an indicator since the digi dash monitors the 61 circuit directly and gives you a nice text message instead... its still there.

A low enough resistor value will work also - but then you consume more power & have to dissipate it while the ignition is on with engine not running...

Bulbs are really cheap & mostly reliable solutions...

Alan

GlenL

Check Tass928's comment: The bulb alone didn't work. It's not the surge when the bulb comes on that bootstraps the alternator. Think it through: if the bulb had to "come on" to start the alternator's generation then once it was on the alternator would never start.

You are guessing and you are wrong.

Alan

Read what I said - I never suggested JUST the bulb... is that what all this crap is about??

I also did not talk about an initiation benefit - its at high load idle when the field is collapsing that the bulb helps most.

Alan

WallyP

"Forgot to mention....One of the things I did try was going through a bulb alone...it didn't work."

Neither the bulb alone nor the 68 Ohm resistor alone will reliably excite the alternator.

M. Requin

+1 based on my experience troubleshooting my 86.5. Alan, thanks for the comment about bulb resistance cold vs hot. Crap, I knew that! but somehow forgot... Explains a lot. But what a weird solution Porsche chose!

Alan

Actually this is the common implementation on most vehicles - hardly Porsche specific.

Alan

LT Texan

sorry to jump in, but what is the voltage drop across the bulb/resistor?

Is the alt exciter wire programmed to draw a certain current?

edit: I only ask because the bulb, when hot, is about 120ohms but only when there's 12volts across it

Alan

When running and generating normally there is no significant voltage dropped across the bulb. With ignition on and engine stopped there is probably about 10-11v across the bulb. The only time the exciter circuit provides much current is when the alternator isn't generating enough to match battery voltage either because its not yet rotating or because it has insufficient power generation to match the load its trying to drive.

The problematic case here is when its overloaded at idle (esp when hot) and its regulated output is drooping below nominal battery level - in this case the feed to the regulator to drive the rotor field coils droops too - this causes the ouput power to drop further. The low resistance of the cold bulb & the resistor in this case helps supplement the field current to help maintain a sufficient rotating field to maintain decent generation.

The rotor field current is intended to be supplied solely by the alternator after it initiates - but the exciter circuit is always connected while running - so current can be drawn as needed based on the voltage differential. The current is designed to vary substantially based on need, but ~no current flows here when running normally at non-idle rpms.

Alan

Tass 928

The bulb alone was an idea I culled from a VW forum....so I gave it a shot. Some said use a bulb, some said use a resistor, nobody said use both

I had no luck at Radio Shack today so when I get home tonight I'm going to dig through my stash of stripped out wiring. I'm hoping to find the original equipment resistor ( somewhere on the pod panel?) and then I'll wire it in series with an indicator-light bulb.

Thanks for all the responses, I think my problem is solved. (fingers crossed)

a4sfed928

Here is a clip from the training books.

A shunt resistor is installed from Mod 79 because the charge Indicator lamp was reduced in size to
1.2 W (formerly 3 W) The internal resistance of the lamp must be reduced in order to allow a
sufficiently large field current to flow. A 3-Watt lamp has an internal resistance of ,Approx 60 ohms
and a 1.2 Watt lamp one of approx 170 ohms.
Example: Current I at 3 W I = V/R = 14 Volt/R 60 ohms= 0.230 A = 230 mA
Current I at 1 2 W I = V/R = 14 Volt/R 170 ohms = 0.080 A = 80 mA
In order to reduce the high Internal resistance of the 1.2 Watt lamp a resistor is wired In parallel with
the lamp (68 ohms) This produces a total resistance of 48.6 ohms.
The current that flows is thus I = 14.0 V / 48.6 ohms = 290 mA
Without the shunt resistor the field current would only be 80 mA - much too little to reliably excite
the alternator.

LT Texan

Alan, Thanks very much for a great explanation.

Tass 928

I still had the stock instrument panel and I pulled what I believe to be the parts that I need. Can someone verify ? I soldered some leads on to a stock indicator bulb. I think you can read the numbers on the resistor (hopefully reduced picture is clear enough).

WallyP

Dave,

When I was doing endurance racing, we wanted an alternator light where the driver could see it. This let us know if the belt or the alternator failed, rather than finding out when the car quit on-track.

If you are going to have to use a bulb anyway, there is virtually no additional weight putting it where you can see it.

Alan

Well sure looks like 150 ohms - 10% tolerance. Not ideal - was this the stock resistor on the very early cars - sure looks similar in the mounting style.

Bulb holder is fine, assume you will locate somewhere dry & not in eng compartment.

Alan

Tass 928

Wally, I agree.
Once I get through this testing stage and the car is fully sorted I'm going to make a new dash that incorporates flush-mount gauges, indicator lights and the switches I need. I'm currently running a homebuilt switch & fuse box that has an indicator light for each circuit. It would be easy to add this charging indicator charging light to my new dash plan.