Adding LED alternator indicator. Question about wiring

[will e]

Going to add an indicator for the alternator charging. I have a volt meter but, let's face it, I don't keep on eye on it while driving. So an indicator to draw attention to the problem is a plus.

I poked around the internet and the wiring is pretty straight forward.

led lite wiring.jpg

Although I do not understand why the additional diode is added, doesn't the LED already prevent possible feedback?

But on to my other question. I want to be able to turn the indicator off when I am hanging out (playing the stereo or whatever) and the accessory on.

Does anyone see a problem with this circuit? The LED is fed from a different ignition on source.

led lite 3.jpg

[tgreese]

I have not made this mod, but I think generally you are correct. Possible the extra diode is there to protect the LED from any alternating current back-feeding from the alternator. LEDs do not like to be reverse biased. However, in this case I would expect the added diode to be between the alternator and the LED.

I would also expect the resistor to be in series, not parallel, with the LED. Normally you use the resistor to current-limit the LED. V=IR. If you are using a naked LED, its spec sheet will tell you what the voltage drop is and what the maximum current is. You can put enough diodes and LEDS in series so the drop adds up to your supply voltage, thereby limiting the current. The inherent drop across the diode or LED follows V=IR like a resistor. Maybe this is what they have in mind with this circuit. Look up the drop across the two diodes, put the R in parallel, and find the current through each branch. Compare this to the maximum current allowed for your LED.

Which alternator? Alternator lights all work on the same principle.

The alternator is self-generating. That is, it uses part of the current it makes to power its field, which is turned into more current as it spins. When not turning, the ignition switch is a source, and the alternator is a sink. When the alternator self-generates (operates normally), the alternator becomes a source and the light turns off.

I'd point out, the alternator continues to be a source until the engine stops. That resistor will feed current into the ignition switch when the engine is running. You may not be able to turn the ignition off with the resistor in there.

[tgreese]

I would expect the accessory terminal on the switch would not power the ignition. Try it - turn the key to accessory and measure if you have power at the coil. If you connect your light to the ignition and not accessory, it should not light on accessory. I believe that's how it's supposed to work.

[tgreese]

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.

[tgreese]

Postscript - I'd also point out that you need to size the resistors big enough to handle the heat they generate.

Recall P=IV. 20 ma at 12V is 0.02*12 = 0.24 or 1/4 watt from R1. 14*0.052 = 0.72 or 3/4 watt from R2. A good rule of thumb is at least 2x oversized from expected dissipation for resistors, so 1/2W for R1 and 3W for R2. 3x would be better; round up to available sizes for 1W and 5W resp.