[1604]

Hello All,

Wrapping up the Ford 5.0 HO/T18 swap into my 1980 CJ-7. I'm using a Ford alternator that came with the Mustang. When I wired the alternator up I didn't connect the excitation line because at the time I thought it's only purpose was to turn the Battery Lamp on when a failure occurred. I've now learned that the excitation lead is also used to energize to field before the alternator produces an output. Looking at the schematic it shows a 500 Ohm resistor connected in parallel to the Battery Lamp wired in series to Hot in Run or Start. I'd like to use a single resistor value to replicate the equivalent resistance of the Lamp and 500 Ohm resistor. I've read elsewhere regarding generic alternators that they require about 125mA of excitation current. This suggests a resistance of ~100 Ohms. Too much current will damage the alternator and an insufficient amount will not allow the field to energize. What would be a good value?
Thanks,
Lee

 

[517]

That will probably be just fine. I have to wonder if the resistor is there in case the bulb burns out, there will still be excitation current. I don't know about Fords, but on a GM the resistance is there not to regulate current, but to prevent the alternator from feeding power back into the ignition system after you turn off the key. You can replace a GM resistance wire with a diode with the stripe (cathode) towards the alternator (effectively the ground before it starts to charge.).

Personally, I like the idiot light. If you want to use one, find one that has an insulated case and two leads, as you don't want the case to be ground.

 

[1604]

Thanks for the advice. I'll give 100 Ohms a shot and see what happens.
Lee