Alright, I'm pitching someone a really nice slow-pitch, right down the center of the strike zone here.
How the heck does an alternator charge indicator work? I'm finishing up a swap, and had a light in the car before (generator) and have a wire coming from the alternator now. I'm having a hard time grasping why the light comes on when you AREN'T getting charge. Seems like with a wire from the alternator (I'm assuming this is the hot side), the light would be on where you WERE getting charge, and would flicker/go out when there was a problem.
What am I misunderstanding?
Simple explaination. When the alternator is working properly it has the same voltage as the battery therefore there is no difference in voltage across the lamp. No voltage difference equals no lit bulb. If the alternator fails to put out as much voltage as the battery the light glows.
Okay, so I need battery voltage to one side of the bulb and alternator voltage to the other side, correct? What's the deal with the resistor shown above?
JohnInKansas wrote:
Okay, so I need battery voltage to one side of the bulb and alternator voltage to the other side, correct? What's the deal with the resistor shown above?
It's a controlled leak. Basically it backfeeds the alternator when you start the car up, getting the alternator to start charging quicker.
Once the alternator is up and running, everything goes to equal voltage, so nothing moves in that circuit.
If you're discharging (light is on), even more juice is being wasted through that resistor.
In reply to foxtrapper:
So... its a good thing if the alternator is working (jumpstarts the charging process) and a bad thing if it isn't working (continually tries to jumpstart a disfunctional alternator).
So as the alternator begins pushing current back toward the bulb, the voltage drop across the resistor decreases and reduces the current to the alternator. Will the circuit that passes through the bulb do the same thing, without the variability, or would I just have to count on the alternator to start making charge without the jumpstart?
Thanks for the help so far, this is making more and more sense.
Simple explaination part two: When you turn the ignition on the current flows from the battery to the alternator exciter in order to allow the alternator to generate power. If the lamp is blown the current doesn't have a path to the exciter therefore the alternater won't work. The resistor is rated higher than the lamp focing the current to flow through the lamp rather than the resistor. If you take out the resistor the current will take the path of least resistance and bypass the lamp.
So basically that bulb runs on the difference in the positive outputs of the alternator and battery?
Well I learned something new, I thought it was hooked up to some kind of voltage-sensing device and just came on when it dropped below a good charging voltage.
I was talking more about removing the resistor and associated wiring.
The bulb can do the job of the resistor, but it greatly limits the amount of current that can backfeed or jump start the alternator.
When you're doing something like installing a GM alternator onto an old Ford tractor, and you want to eliminate the bulb and or resistor, it means it can take several minutes for the alternator to wake up and start charging. Especially if you don't rev the tractor engine up, and keep it up for a while. That's how many 1-wire GM alternators work. And it takes them a little while to start charging as a result.
An old fashioned generator starts charging quicker (residual magnatism), and at a lower rpm compared to the fancier alternator. That's why a generator can often times be a better choice for the likes of a tractor.
Sounds like I ought to have a resistor.
Is that something I should try to scavenge off the donor harness ('91 Subaru Legacy)? Or is there an agreed-upon resistance that I could add in-line between the ignition switch and the alternator?
Dunno the specs on a Subaru alternator. If you've got a parts car there to scavenge, I would use its parts.
(scurries off to research Subaru resistors)
Thanks for the input, everybody. 
