The field current bypasses the light, so you shouldn't need the power resistor. When the alternator fails, the isolation diodes can't keep the field at a voltage higher than the battery - so at that point the current runs from the battery runs through the light to the field windings, yes.
But when that happens, it doesn't matter, because the alternator is dead anyway.
Hi, most alternators in cars I have worked on, seem to have a 12 Volt 3 Watt indicator lamp on the dash, to assist in exciting the armature of the alternator.
That equates to around 250 mA, which yields a 48 Ohm resistor.
Get a 47 Ohm, 5 or 10 watt resistor to replace the lamp and put the led in parrallel with it and have a series R to limit the current through the led to 20 mA or less.
The alternator indicator lamp may be just a 194 wedge bulb. This type, and many others, already have LED replacements available to fit the existing socket.
I looked up the operation of a typical alternator and the circuitry does not need a bulb with any specific resistance. If voltage drops out from the diode trio that powers the regulator, the bulb just turns on.
Diver300 said:It is also a good idea to have the warning light start to work at a low voltage. Putting a bridge rectifier and a voltage divider (made from the two 5 Ω resistors) is a bad idea and you won't be able to tell if the alternator is behaving badly. If a diode fails, you can get the warning light glow because it is being fed with just a few volts. You want to know about that, and having the LED need 6 V to light would hide that.
I looked up the operation of a typical alternator and the circuitry does not need a bulb with any specific resistance. If voltage drops out from the diode trio that powers the regulator, the bulb just turns on.
RODALCO said:Of course, if all else fails and during experimenting a standard voltmeter seems the best and easiest option to monitor the accu voltage.
And how much voltage / current does a filament lamp require to light up sufficiently to be noticed?
I looked up the operation of a typical alternator and the circuitry does not need a bulb with any specific resistance. If voltage drops out from the diode trio that powers the regulator, the bulb just turns on.
Obviously there may be some fine tuning to do. It's been suggested that the bridge is not required; but if it's not there, how can the potential be normalised, bearing in mind that it MUST be since the LED is a unidirectional device, and current may flow in either direction?
I haven't mentioned it before, but my requirement is to provide an optoisolated input to an MCU. In this scenario, by the way,
Carlos, your circuit will enable the light to be plugged in either way, no fooling with a voltmeter necessary. Your original circuit was the right circuit. I shouldn't have suggested otherwise.
duffy said:I was hoping you were going to build this marvel of the 21st century, the alternator light that doesn't burn out! The dashboard light that gives you piece of mind that a burned-out filament bulb will not cause you grief as you roll up to the fifties-era diner for the car-club meet. You were going to make this simple device, market it because there isn't one, put a stupid-high price tag on it, sell a couple thousand to hard core motorheads!
Now it sounds more like you're making a Nightrider dashboard mod or something. Ok, points for style, but...
Wow. I happen to be building something like this for my car today and I found your thread. Brilliant!Your idea of implementing a bridge rectifier would add an additional layer of reliability to my design. Thank you. I haven't checked your math on the resistors, yet. I found this website which may be of some use. Current limiting Resistor calculator for leds Are your calculations based on a voltage of 14.2?
Axle Roads said:I wonder what would happen if we used a bi-color LED. Say red/green. Could we make it so the green LED flashes when charging and red LED flashes when discharging?
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