I recently bypassed my vehicles electrical system for my headlights and now my vehicle comes up with headlight out warnings. I have been told if I installed resisters in the original wiring to simulate the load of the light bulbs then my problem will go away.
Based on ohm calculations, I've figured to simulate 55watt on a 12 volt circuit I'd need a 2.6ohm resistor. Optionally I can change the computer to think they are different lights and only need 35watts, which would require a 4.1ohm resistor.
I see resistors all over the net, radioshack, etc, but they are tiny little things that I do not think are sufficient for this application.
Two things I would appreciate help on:
Are my calculations correct? And is there a specific type of resistor i need to simulate 12v 35watt or just any 4.1ohm resistor?
Thanks for the source for those resistors! I wonder, just how hot do those get? The only place to mount them is on the inside plastic of the headlight unless I breach the shell of the headlight to run wires outside (which kinda defeats my oem look purpose and potentially allows moisture in).
The vehicle checks the lights whenever they are switched on, so if they are switched on before i power the vehicle it throws the warning up until the vehicle is off, and also if i turn them on while driving, it then springs the warning on me right then. If a bulb burns out while driving, it pops the warning too I believe, so it seems to be constantly monitoring.
My other option is to disable the bulb out sensor, but nobody happens to know where it is in my car (2002 Audi A4).
Compact fluorescents are a good deal more lumen-efficient than incandescents. Unfortunately, they're not small and are somewhat fragile.
High-output LEDs are better than small incandescents, by a factor of
2:1 or 3:1. They still dissipate quite a bit of heat.
And, in truth, it doesn't really matter at all how lumen-efficient the bulb is, if you're planning to have it in an enclosed space. The light emitted by it will strike the inside of the enclosed space, be absorbed, and will turn into heat.
If you're running 3 amps at 12 volts into a lamp, you're going to have
36 watts of heat to get rid of, one way or another. Through one path or another, every bit of power you pump into the lamp ends up as heat, somewhere.
I'm curious - what is/was your original goal. You mentioned having "bypassed" your original lights - just what did you do, and what are the advantages of having done so?
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Ahh, good point on the light striking surface and turning into heat.. I hadn't thought of that.. pretty futile maneauver except perhaps to spread the heat a little i guess.
Well, I installed HID (xenon) lights in my non-xenon equipped vehicle. The projectors are designed for xenon's though, just it didn't come w/ that option. I bought an OEM kit, and everything is there, except the stock wiring doesn't support the high amp load the startup of the HID's draw which can be upwards of 100watts I understand. So I used the existing wiring to a relay, to turn on/off my wiring harness which brings power direct from the battery thru fuses to the HID lights.
Now since I bypassed the cars internal wiring, the car see's no light bulbs in there, so it complains.
It may only be a case of satisfying the monitoring circuit's detection current rather than simulating the high-power consumption of a real main-beam lamp.
I'd suggest you substitute whatever you have in the nature of small panel lamps, dome-light bulbs to see if any of them will do the trick.
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Maybe the headlight warning system doesn't need to 'see' the full headlight load, maybe 15 ohms is enough to fool it, personally I'd try something higher than that, reducing the value until you get a reliable no fault indication. Bear in mind you need to dissipate the heat generated and you need to increase the wattage of the resistor as you lower the resistance, a higher ohm value will produce less heat and be physically smaller in a lower wattage package.
I am curious as to why you want to by-pass the car electrical system which monitors lamp current. I assume that by-passing still means you have the original lamps in operation.
If this is the case it seems rather self defeating to bypass the monitor circuitry while having to install dummy load resistors which will consume (waste)the same amount of energy as the original lamp load in addition to the original lamps themselves. Without a very good reason, it would appear to me to be futile to double the normal lamp current drawn from your battery.
This assumes that the load sensors in the ciruitry will "sense" that headlights are installed with 15ohm loads which could very well be true as it comes from xenondepot. With a 15 ohm resistor and the vehicle voltage around 14V, Ohms law tells us this resistor needs to be rated at over 13W.
Watts = Volts^2 / Ohms 13.066 = 14^2 / 15
You can order 15 Ohm / 25 Watt rated resistors from Mouser.com for $2.18 US. Part # is 71-HL25-06Z-15. You will need 2 of these, 1 for each headlight.
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If you go this route, make sure you order the correct resistance. If you put that part # into Mouser's part search, their first few results will be the incorrect ones as they show the 150 Ohm and 15k Ohm resistors first.
If you want to experiment a little, you can try going to any electronic store around you like a radioshack and picking up a few 1.5k resistors either 1/4 or 1/2 watt will do as you will be only passing .13 watts through them at 14V. These come in packs of 5 or so from Radioshack for 99 cents. It is quite possible that your sensors will see even this small of a load as a working light. If this doesnt work then your only out a buck and I would suggest going with the 15 Ohm resistors.
Excellent info! thank you! Last night I plugged a 2 watt lightbulb into the headlight circuit and still had the warning, then i plugged a
5 watt bulb in, and warning went away! that tells me I dont have to generate a full 55watts, which is a really good thing, that would be a lot of heat!
Given that information, a 33 Ohm 10 Watt resistor will allow you to pass 5.9W through the resistor and be well within the rated power dissipation for the resistor. If you can find a 47 Ohm 5 Watt resistor, you *might* be able to save a few cents but doing this your only going to have a 4.1W load. Just pointing out your options.
Here is a link to a 40 cent 33 Ohm 10 Watt 5% resistor.
On Sun, 19 Feb 2006 13:52:18 +1100, Franc Zabkar put finger to keyboard and composed:
Another approach is to leave the original resistor within the monitoring device and add a second resistor to the heavy gauge harness. The downside is that the current would then be shared between two paths. To avoid disturbing the module, you may be able to determine the value of the internal resistor by metering the BAT and LAMP OUTPUT pins.
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I ordered for testing 2 x 15ohm 25watt 1% resistors... they were about $5 each, which I'm ok to spend just to learn..
I wired them up tonight to my drivers side stock wire headlight, and they are definitly receiving power, they get very hot within about 30 seconds.. to hot to touch. Turn the lights on and the computer still spits out the bulb out warning.
So, in series with the resistor I put the 2watt lightbulb, still got an error. In series I then put the 5watt light bulb, and error goes away.
I'm not super savvy with power stuff, but could perhaps anyone tell me, is the resistor drawing a different type of current/power/something than the lightbulb? Clearly the resistor at 25watt is drawing more power than the 5watt lightbulb.
You did not say if you installed the 15 ohm resistors in series or in parallel.
When you finally got it working you had the resistor (2 15 ohm resistors) at either 7.5 ohms or 30 ohms in series with either one or 2 bulbs.
To get an answer, please tell us what the final circuit was. How many bulbs were in series with the two resistors, and were those resistors in series or in parallel?
This is 2 completely separate circuits I'm speaking about, one for the left headlight, and one for the right headlight. So, effectively if I get one working the same setup on the other will work as well (cross fingers). So, each side will only have 1 resistor. The resistor is a
15ohm 25watt resistor.
When I did my tests last night, the first test which still made my car beep and cry about the light out was just a single resistor, and that is it. Here is my best text drawing.
--------------- | 15ohm |
- ---------------
My 2nd test, in parallel behind the resistor I put the 2 watt lightbulb, and still got the error from the car. Drawing:
------------------------ | | 15ohm 2watt bulb | |
- ------------------------
My 3rd test, same as the 2nd test except with a 5 watt bulb. This made the error go away.
------------------------ | | 15ohm 5watt bulb | |
- ------------------------
Thing is, in my earlier tests before I had resistors to play with, this setup also made the error go away:
------------------------ | 5watt bulb |
- ------------------------
The resistor in parallel doesn't seem to do anything. =o Although, it gets very hot so it's definitly sucking power.
OK! My guess is that the sensing circuit is looking for a very low resistance. Once it finds it, it stops looking. The bulb, when cold, presents a very low resistance, far lower than 15 ohms. It appears that satisfies the circuit.
It very well may be that the circuit needs to stay "satisfied" by a higher resistance. You could test with a switch between the 5 watt bulb and the 15 ohm resistor. One the circuit is "happy" switch the bulb out of the circuit to see if the circuit stays "happy". Of course, the 15 ohm resistor is not the equivalent resistance of the bulb (once the bulb is hot) When hot and dissipating 5 watts at ~13.8 volts, the bulb is drawing ~362 mA. That makes the resistance equal to about
38 ohms.
If you want to go with a resistor load instead of the 5 watt bulb, you may be able to use something conceptually like this:
The 30 ohm resistor can be the two 15 ohm 25 watt resistors you already have, but it will dissipate only about 7 watts. That keeps the heat down a bit. The 1 ohm resistor is 20 watts or more. There needs to be some time delay before the relay energizes, so RY1 should be energized by an adjustable time delay circuit. Also, 1 ohm is a guess.
The nice thing about the circuit is that the heavy power dissipation in the 1 ohm resistor occurs only for a brief time, until the relay is energized.
And it could be that you don't need the 30 ohm resistor (2 15 ohm resistors in series) at all - the new style headlamp may draw enough to keep the sensing circuit "happy", once the initial test is satisfied. You could test by placing the 5 watt lamp in the circuit, turning things on, and then switching the 5 watt lamp out. If the sensing circuit does not object, than all you need is the time delay relay (delay on) and a suitable load resistor (the one ohm resistor I guessed at)
With some experimention with the length of the time delay, and the value of the 1 ohm resistor, and possible elimination of the 30 ohms resistance, you may find an ideal method to make the car's computer happy with your new headlights. Ideal would be R1 completely out of the circuit. If it has to be there, the higher the resistance the better, as long as it yields 100% correct operation. (The higher the resistance, the lower the heat produced.)
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