Kinda think this is what you want -
altLEDiagram (600 x 303).jpg
The usual indicator light is a bulb like an 1816. That's 1W at 14V. P=IV, so 1/14 = 0.072 or 72 milliamps. Your typical red LED has about a 2V drop and operates at about 20 ma. This may not be enough to excite the alternator. We know the 1816 can, so if we make a circuit that lights an LED, delivers 72 ma and blocks back-feeding, it should work.
Subtract the LED drop (2V) from the applied voltage with the current needed to light the LED (20 ma) to get 12V. Apply Ohm's law V=IR to calculate the resistance R1 of 600 ohms. The 2V drop across the LED at 20 ma is an equivalent resistance of 100 ohms (apply Ohm's law again). Thus R1 plus the LED is 700 ohms at 20 ma and 14V.
To get the bypass resistor value R2, you need to do a little algebra. Parallel resistances can be found from the formula (R1 * R2) / (R1 + R2). Thus we have 0.072 * (700 * R2) / (700 + R2) = 14. Some algebra and we have R2 = 270 ohms.
Lacking algebra, you could observe that we need 72 - 20 = 52 ma through R2 at 14V; apply Ohm's law again to get 270 ohms.
Putting R1 and R2 in parallel is 14 / ((700 * 270) / (700 + 270)) = 0.072 amps.
R1 plus R2 plus the LED may be enough to stop back feeding, but I added a blocking diode to be certain. This will add a voltage drop of 0.6V to the circuit, but it's not significant.
This is basically the same as the circuit you show originally, with the extra diode dropping the voltage for the LED. I'm not sure that would be enough voltage drop to keep the LED alive; maybe. Your circuit also assumes the added resistor passes enough current to excite the alternator but not so much as to back feed. Also maybe. I think my circuit is more likely to work. You can tweek the values based on the components you select. Let us know how it goes.
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