reduce led current draw

[Telomostro]

Someone has yanked my alarm LED from my motorcycle.
I called the alarm manufacturer and they told me that alarm system is discontinued but that I could go to Radio Shack and buy any red LED at 12V with less than 5 mA draw. The smallest drawing LED I could find with that voltage was a 15 mA and so I picked that one up and installed it on my motorcycle: Great bright red light, more than I really needed but I figured it's an alarm LED, the brighter the better but sure enough the first time I did not use my motorcycle for 3 days in a row, I found the battery almost dead.
What should I do to reduce the current drawing to <5 mA...? I am willing to loose a lil brightness, someone told me 1K resistor in series, someone told me a few diodes in series.... I am confused....

 

[Comcokid]

Not knowing which LED you got from Radio Shack, it may require a series resistor, which is why it is so bright.

Probably the best thing to do is to use a series resistor. A 1K Ohm will limit the LED current to less than 10 ma(or even less if the LED really is 12V rated). If you find the LED is too dim then, drop the resistance to 680 or 430 Ohms. If still too bright, increase the resistor to 1200 to 2000 Ohms. The higher the resistance, the less the current, and the less draw from the battery when your motorcycle is not in use.

 

[Telomostro]

The LED I got from RS was a 12V rated one and I think it has a built in resistor but, if I understand correctly, you say that adding another resistor in series (i will experiment with the different values you sugested) will indeed drop the current that it draws from the battery... right...? Doesn't the resistor draw some current on it's own ? and if I end up being happy with the amount of light emited with the 1K, how many mA will this LED end up drawing...? and if I choose the 430 ohms how many mA will it draw then..?
thanx a lot !
these are the specs of the LED

Catalog #: 276-084

Typical MCD is 50. Typical wavelength is 635nm. Lens size is 6mm. Red lens color. Viewing angle is 45°. 15mA (max). Typical Voltage is 12, with a maximum voltage of 12V. Comes as package of 1. Mfg hole diameter is 3/16&quot;.

 

[Telomostro]

http://www.radioshack.com/product.asp?catalog_name=CTLG&product_id=276-084

that is the link where the LED specs are listed in more detail

 

[zeusfaber]

All red LEDs drop a pretty constant 2.5V when working irrespective of the current drain. (It varies a bit depending on the diode type, but not by more than half a volt). The series resistor has to drop the rest of the voltage, and thereby controls the current.

Your 12V, 15mA pack must be dropping 12 - 2.5 = 9.5V across the internal resistor, giving an internal resistance of 9.5 / 0.015 = 633ish ohms.

To get this down to 5 mA, you need a total resistance of 9.5 / 0.005 = 1900 ohms.

I'd suggest you try an additional series resistor of
1900 - 633 = 1.2k ish ohms.

A.

 

[kurmugn]

If it is assumed that the battery is 12-volts, that the forward voltage drop across the led is 1.75-volts, and that you want 5ma or less of current flowing through the led, then a 2.1 Kohm, 1/4 watt resistor in series with the led will work.

 

[Telomostro]

yes... I ended up putting a amp meter in series as I was trying different resistors and a 3k3 value brought it down to <5 mA. It was sorta difficult to read the exact value since the alarm makes the LED intermittent, so I only had a fraction of a second to read the meter and everytime the LED turned on the meter gave me a different reading so I had to avg it out. The brightness obviously went down, but it is still visible even during the day.

It's incredible though, since I made this mod (about 1 week)my motorcycle starts w/o a problem and all this difference just for dropping the current of an LED from 15 mA to 5 mA....????? on a motorvehicle battery...????