So I just hooked up these 5mm LEDs into my license plate lights,
I had done this before in my 97 audi A6, and it was fine( no technology then)
I took a chance with the bulb out warning and i knew i'd get one.
Now, Im trying to calculate what size resistor I need to make the warning go away.

The LEDs I am running are with these stats:
[FONT=Verdana][SIZE=2][SIZE=4][FONT='Times New Roman'][/FONT][/SIZE][/SIZE][/FONT]
Looking at it, its running at 80mw, correct? So thats .08 watts,
From this link, it seems that the oem bulbs run at about 5watts?
https://www.northamericanmotoring.co...se+plate+watts

So, I need to make up for that 4.92 watts? Or is it the amperage I need to make up for?
With the LED I have connected, I have a 1/4W 510ohm +/- 5% resistor soldered to the positive leg of it, so that it can run with 12-14V running through it. This would be cnnected in series. But it seems that I need to hook up a resistor in parallel (meaning from the positive leg to the negative leg of the LED (before or after?? the first resistor in series)?

Thats question number 1. Question two is what size resistor will I need?
would a 5watt resistor of any ohms work? Or do I need to select the correct resistance--> R=V/I ...whats the amperage running through these lights? I don't have a multimeter, so if someone does could they check, if this is indeed what I need???

THANKS SO MUCH IN ADVANCE EVERYONE!!!
(pics once I get everything working of course) <<incentive to help me out =]

 

Okay,
Maybe I have answered my own question...

So right now using my LED, it is using
3.3V avg of the LED ... V=IR (12V-3.3V)=.03A * R,
says I should be using a 290ohm resistor in series to be able to run at 12V,
considering while the car is running its around 13-14v, I have a 500ohm 1/4w +/-5% resistor on there, good enough.

Now, using P=V^2/R in Watts, I assume that a filament bulb is at 5w, so, P =5w (as the LED is only .08w), 5w=8.7^2 / R to find the resistance of the 5W resistor I need, 75.69/5 = R, R= 15.138ohms...
So, attaching a 15+ohm, 5W resistor in parallel to the LED should "trick" the car to think there is a fillament bulb there running at the correct voltage/current?

 

Looks like you got your calculations right. Just remember that 5Watt resistor will become quite hot, so mount it somewhere where it cant do no harm.

 

It seems like thats what i've been reading... how "hot" would it become? There is definitely room inside of the boot "handle" where the light fixtures are, but that is a plastic piece =\

It seems you can also just hook it in parallel anywhere along the circuit? would this work if i just tapped both ends of the resistor into the positive line running to the lights?

...If so... where would this wire be located? i suck at realoem.com... would it be on there?

 

When I installed my strip of LEDs under the rear package shelf instead of the single filiment OEM bulb, I had to add a 1000ohm resister to the circuit to keep them from flickering. 1000ohms did even get warm. Just a sugestion.

YD

 

I believe the ohms dont have to do with the heat, its the wattage..
I have 1000ohms on my auto sport button mod, but its only 1/2 watt resistor and doesnt get hot,

My led now in the license plate is 1/4 watt but 510ohms, and it doesnt get hot..

i could be wrong though, not sure =\

 

LEDs use so little power that you can pretty much ignore their values when selecting a load resistor to fool the bulb-out detector. So just choose a resistor according to the bulb being replaced (5W in this case).

If you use a resistor that has exactly the same value as the bulb, then you can expect it to get just as hot as the bulb. You can use stand-off spacers/brackets to keep it away from plastic that may melt. To minimize heat and power, ideally you would use the highest ohm value you can get away with, that would still fool the bulb-out detector. 1000 ohms might be too high but it would be great if it works.

FWIW, superlumination.com uses 100 ohms as the load resistor they sell for replacing a 5W bulb with LED.

 

I went to radioshack and got 10 watt 10% tolerance 50 ohm wirewound resistors...
will this mean its going to get as hot as if there was a 10 watt bulb in there? =\

would this resistor work?
its white-block looking resistor, never used one this massive before...

 

No it all depneds on how much current is being drawn thru the resistor.

By using the formulas on this website we can calculate the wattage (heat) that the resistor will generate.
http://www.the12volt.com/ohm/ohmslaw.asp

So I=V/R or amps = voltage/ohms or I=14/50 or 0.28amps

Then W=V*I or watts = voltage * amps or W=14*0.28 or 3.9 watts.

or

W=V2/R or watts = voltage squared / ohms or W= (14.2*14.2) / 50 or W=201.2/50 or 4.2 watts

So that resistor will generate less than 5 watts of heat.

So for your purposes just figure that the resistor will get just as hot as a 5W bulb.

Keep it away form plastic that will melt and you should be okay.

You may want to try a 100 ohm 5 watt resistor just to see if it will work. That is what many of the LED replacement light manufactures use to fool the bulb out indicators.

The 100 ohm 5 watt will produce about 2.5 watts of heat or just more than half the heat and will also be smaller.

 

Thanks for that! that makes a lot of sense.
The only thing is, with producing 2.5 watts of heat, it will still "fool" the bulb out because it is a 5 watt resistor, correct? Or does that just mean that it is a resistor that can hold up to 5 watts of power? and yeah, these are pretty large, i'm going to have to come up with a clever way to hook them up...
I wish I could do it inside of the trunk under the panels or somewhere if i Knew which wire was feeding the license plate bulbs, that way i could have more room to work with also to dissapate heat, and not worry about melting anything.

 

Voltage, current, and resistance generate heat just as rubbing your hands together creates heat.

What is of concern here is to draw enough current to fool the bulb sensing circuit.

A 5W bulb in this case draws about 0.357 amps of current. Most bulb detection systems are designed to accept a range of currents as all bulbs are not exactly equal. A warmer (more yellow glow) will draw less current than a bright white bulb (normally). So what many designers do is to set the minimum current draw that is detected by the circuit to about 1/3 of the normal. This ensures that as the bulb wears and draws less current that it will not actually signal a blown bulb.

So by putting in a 100 ohm resistor it will draw about .15 amps which should (but may not be enough) current to make the circuit think there is a bulb there.

 

perfect. That is what I was after originally is how many amps the 5w bulbs draw. thank you. Since I have the 50ohm ones here, i think im going to try them and see how it works! thanks