RS HID/LED fog lights

[zei20t]

the CH/LH looks good with the LEDs in the fog location

[dArK5HaD0w]

do u have "cornering foglights" activated?
if so, does it still do the fade in/out?

[Martin]

I saw that option and was tempted to turn it on - will have to experiment with it

[Mysticality]

do u have "cornering foglights" activated?
if so, does it still do the fade in/out?

It would. It's a voltage thing.
Same as DRL dim is a voltage thing, also.

The DRL's have 2 input circuits, both 12v, but the dimmer one would have resistors inside the DRL units to bring the power down and make the LED's dimmer

I saw that option and was tempted to turn it on - will have to experiment with it

Do it!!!

[zei20t]

yes they fade in and out. however, as there isn't as much load, it fades to a point then goes out.

a big resistor will fix the 'lamp warning' and also correct the fade function.

[Martin]

Ok - I was going to install my LED fog lights today but I had not thought about the bulb warnings
What value and wattage resistor do I need?

[zei20t]

if I knew id tell you

I haven't got around to doing it myself yet mate.

I would imagine a 5w ceramic style resister from jaycar should cover it. its one of the large ones with a big heatsink around it.

[Martin]

Ok, Sounds like I need to:

• Put a multimeter across the fog bulb to determine resistance
• Run the LED fog light on the bench and measure current (can then calculate wattage)

[zei20t]

there is a writeup on a german or Czech forum about the values he used. but I think it was just 5w. anything bigger and it will produce a lot of heat that youll need to get rid of.

[k_rider]

A better idea is to measure the current drawn by your original fog light. The resistance of the filament in an incandescent bulb increases as it heats up. Measuring the resistance of a cold bulb will throw your calculations way out!

I don't know exactly how the bulb checking is achieved (I'm happy to be educated ), but assuming that the system expects pretty much the full current consumption from the designed for bulb, then the actual wattage of the bulb will give you a ball park power rating for the resistor you need. But to get a more accurate power rating and the resistance value needed, is fairly simple by applying a couple of fundamental electrical laws - Ohm's and Kirchhoff's.

Once you've measured the current drawn by your original fog light and the replacement LED version, start by subtracting the value of the LED current from the value of the bulb current. The result tells you how much current needs to be drawn by your additional resistor. From this you can calculate the resistor value and power rating.

To illustrate: I'd guess the foglight would be about 50W (let's say 48W to keep the maths simple), that means it will draw 4A at 12V. Let's say the LED unit uses 4 x 3W LED emitters, a total of 12W, means it draws 1A at 12V. Subtracting that from your original bulb current gives a result of 3A extra you need to draw to keep the bulb monitor happy (this is basically Kirchhoff's Current Law in action).

This is easily done by connecting a resistor across the LED unit. The power rating of the resistor is calculated by multiplying the voltage across it by the current through it, that is, 12 x 3 = 36W. So, the resistor needs a power rating of at least 36W.

To work out the resistance value required, use Ohm's law. The value will be the voltage across the resistor divided by the current through it, so the resistor value will be 12 / 3 = 4 Ohms.

Note that the above numbers are purely for illustration - I have no idea off the top of my head, what the power rating of the vRS foglights are (and I'm too lazy to go and have a look at mine right now ) But as you can see, the values are pretty easy to calculate once you have your two current readings.

You may be able to go with a higher resistance (which will lower the power dissipation), it all depends on what the bulb monitoring circuit wants to see.

If you wanted to try a 5W resistor for sh*ts and giggles, then its resistance must be no lower than 30 Ohms (otherwise it'll just burn out), but this will only draw about 400mA, which may not be enough to stop a bulb failure warning.

Try RS Components or Element14 for high power resistors, it's unlikely that Jaycar carry anything higher than 10W.

As zei20t points out, they will generate a lot of heat (even a 5W resistor at rated dissipation will burn you), so be mindful of where you fit them!

[zei20t]

This is easily done by connecting a resistor across the LED unit.

I think you mean in series

anyway, good information, and yeah some basic maths will calculate what we need.

AFAIK the halogen fog lights are 50w. the LED fogs are as you say, 4 x 3w + around 3w for electronics to duty cycle the LEDs and the other circuit for 30% brightness.

[Mysticality]

AFAIK the halogen fog lights are 50w

I remember they were different - like 28W or something weird.

EDIT: Sorry, HB4 51W.

[k_rider]

Nah, I do mean in parallel!

Connecting the resistor in series with the LED fog light will reduce the current running through the fog light, which will dim it and also 'fool' the bulb monitoring circuit even less than if the resistor wasn't there at all...

The LED fog light unit/assembly will be designed to connect across 12V, so will already have the appropriate current limit circuit - which could be as simple as a series resistor, or may even be a transistor based constant current regulator, built in.

If it was simply a single LED itself (ie not in a 'bulb' or similar fitting) we're talking about, then yes, a series resistor is required to limit the current that will flow through the LED. Another technique often used is to connect a number of LED emitters in series, as this reduces the required power rating of the current limit circuit.

However, in this case, the aim is simply to draw enough current through the fog light circuit to keep the bulb monitor happy. You need to connect the resistor in parallel with (ie 'across') the fog light to do that.

The real question is, how much current actually needs to be drawn to achieve that "happy state"? It seems logical to start with the value that the original light was drawing and go from there. You could experiment by using a progressively higher resistance value (which will reduce the power rating required) until the point is found where the bulb failure warning is triggered.

Watch it, my inner nerd has been stirred!!

[zei20t]

Im not saying you are wrong, but if you run it in parallel it will half the average resistance. thus the CECM will see less load. I think with a resistor in series the LED will work fine. Yes, its lowering the voltage but the sensitivity of the LED, and the duty cycle circuit, im sure it will cope with the drop in voltage, as the LEDs don't run on 12v anyway.

Ill pull my finger out and actually do this one day and we will know for sure. But Im pretty sure a 5w ceramic resistor (cnat remember for sure) was used in Europe and it solved the issue. Mk5 golfs can also use these LEDs and they get them working fine also. Possibly with stopping the cold diagnostics on the fog circuit on their CECM

[k_rider]

As I pointed out originally, I don't know enough about how VAG have designed their bulb monitoring system, so I made the assumption it worked by measuring current in the circuit it was monitoring (the way I'd do it if I was designing it). But I take your point, quite often things can be made to work in a way that don't at first appear right on paper!

Anyway, I'd certainly be interested to hear how you get on with sorting out your "bulb out" warning. Not that I intend fitting the Fabia style fog light/DRL myself, but I am interested in fitting LEDs elsewhere (eg stop/tail, reverse, indicators).