Alternator charge warning lamp - replace with LED

superheat6k

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I wish to replace a signal lamp bulb 1.2W 12V incandescent with a super bright LED. The bulb provides the initial excitation current to the associated alternator.

The LED I have chosen draws 20mA at 12v so 0.24 W.

It is safe to use up to 24vdc, although pre-excitation the voltage would be at most ~ 13v, and after excitation no current will flow because the voltage each side of the excitation lamp will equalise. So I do not need to consider the excited voltage.

So if I wire in parallel I work out I need a resistor to dissipate a further ~1 W. so at 12v this 0.08 amps. From Ohm's Law this is 150 ohms

So a 150 ohm resistor will this satisfy the alternator excitation when using the LED instead of the incandescent and a 2w rated resistor will have plenty of spare capacity.

The total power in the alternator circuit for excitation will remain close to 1.2W

Can someone please check my maths and Ohms Law please.

Should I also wire in a Zener Diode to ensure the warning LED turns off once the alternator is excited. I was thinking a 4v Zener. But do I need to account for power loss in the Zener. I intended to try it without the Zener and see if the warning light works without initially.

Thanks.
 
Some engines do have an excitation circuit via a resistor in addition to the warning light

They are usually either 50 or 33 ohms, 5 watt
You don't need a zenner


( Re the maths. There is no need to work out the amps then the resistance because W = V²/R or R=V²/W )
 
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I used an LED designed for the job - less than a quid - it has resistor already in the base ... all I had to do was make sure I connected it correct polarity.
An led will need a series resistor for use on 12 volts but what superheat is asking about is a parallel circuit to provide sufficient initial excitation current or are you saying you obtained an led combined with both series and parallel resistors.
A parallel resistor in the 30-50 ohms range will need to be a 5 watt one
.
 
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An led will need a series resistor for use on 12 volts but what superheat is asking about is a parallel circuit to provide sufficient initial excitation current or are you saying you obtained an led combined with both series and parallel resistors.
A parallel resistor in the 30-50 ohms range will need to be a 5 watt one
.

All I know is I went to Auto Shop - told them I wanted an LED to do excite and show ignition live when engine stopped. Guy put hand in draw and handed me the LED .. said match the polarity ...

I went back to boat ... took out the old lamp .. wired it in ... and it works ..

It excites and once alternator is outputting - light goes out ... couldn't be simpler ...

What's inside the small body of the LED he gave - ?? - Its been on the boat for nigh on 20yrs now .. the red lamp part is just under 10mm diameter ..
 
I have LED's in the excitation alternators connections.

They consist of a LED with a current limiting resistor with a resistor to provide the current needed to excavate the alternator so it will start charging

2 calculations are needed 24V - the forward voltage of the LED divided by the forward current of the LED 20 mA

The parallel resistor will be (24 V) ² divided by the wattage of the existing Lamp

The extra 20 mA flowing through the LED will give a little extra excitation current but should not give any issues
 
........................................... I intended to try it without the Zener and see if the warning light works without initially.

Thanks.
Notwithstanding the good advice given above, should for some reason you find you do need a zener then something like a BZX55C 6V2 500mW would be a good place to start.
 
Thanks for the sound advice, especially the reference to the specific zener, and now I will explain it is for a motorcycle instrument display head I have just acquired, and where the original signal bulbs are barely visible in daylight and through a covering fly screen.IMG_4939.jpeg
 
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