[304065]

Doug,

So in normal circumstances, here's what happens. Current flows out of the battery to the ignition switch, through the warning light bulb on the back of the oil pressure gauge, back through the blue wire to the voltage regulator, then over to the alternator D+/61 terminal, where it flows into the rotor and to ground. Usually, the reason it lights is because the voltage coming from the battery (which gets fed back into the rotor by the DF wire or directly to the rotor in an internally regulated setup) is less than the voltage being put out by the battery.

So consider what's happening when the ignition switch is OFF. The +12v connection is open, one side of the bulb, the one with the red and black wire going to it, is shorted to ground through the ignition switch.

OK, so we have a ground on one side of the bulb. Now we need power to the other side. It has to come in through the blue wire. The blue wire goes to the regulator and over to the alternator, where it connects to one of the brushes. The other brush is connected to ground. Also connected to the same brush is the DF or "dynamo field" wire. The DF wire comes from the voltage regulator, and once the alternator is producing current, the DF wire regulates the field strength, regulating the output voltage from the stator. So the DF wire gets its power from the stator.

When the diodes are working correctly, current is only supposed to flow in one direction- from the stator to the B+ terminal and down to the battery. When the diodes are defective, current flows from the Battery past the diodes, into the stator, and also to the DF wire, where it flows to the rotor, where it flows into the D+/61 wire, then down to the ignition switch, providing the "hot" side of the circuit!

Now, what kind of alternator do you have, internal or external regulator? If you want to post a picture here we can help.

Good luck!