GC ignition key light wiring

[ScoobySam8]

Bought myself a jdm ignition key light for my GC, and now I need a little help wiring it up. I've figured out how it works by touching the wires on my car battery, but now I just don't know which wire to connect to

There's 3 wires: 1 connected to constant 12V (no problem), 1 connected to constant ground (no problem), and the last I am having trouble with. The last wire needs ground when the door is openned, then needs the ground to 'drop-out', or have a negative power draw, to light up then dim out after 10 secs after the door is closed. So...my question is, what can I connect this last wire to!? I've spend the last 30 mins in my car with a multi-meter trying to find a wire with 0V (when door is open) and -12V (when door is closed). And no, hooking up to the dome light doesn't seem to work (or you could explain to me how it can work )

Thanks for reading everyone, and I hope you guys can help me out.

* Registered users of the site do not see these ads.

[RE3Rotor]

I have electrical diagrams for 97-98 if it's of help to you.

[showoval]

So you need a ground when the door is open and then need it to disappear when the door is closed. Our cars use a negitive door trigger, that should give you the ground your looking for when the door is open and go away when the door is closed.

[rbimdxe]

Originally Posted by showoval

So you need a ground when the door is open and then need it to disappear when the door is closed. Our cars use a negitive door trigger, that should give you the ground your looking for when the door is open and go away when the door is closed.

But the issue with that, is as soon as the door closes, the light will go out.

You need to use a capacitor in the system, what value, I can't say. It depends ont he current draw of the light.

Possibly even using a relay setup can make it work. Similar to this

Obviously, the wiring diagram will need modification, but the principle behind the capacitor is what you're after.

I don't know if such a wire exists *already* in the GC's, but Instead of trying to find something that might not exist, if you're handy with soldering and a bit of electrical schematics, you can design your own circuit.

It's too late here in Oz, and I'm tied up with a funeral all weekend, but early next week I'll try and organize a schematic if you haven;t devised one by then.

Good luck with it.

PS; I hope I understood your request correctly

[jnorth85]

Originally Posted by rbimdxe

But the issue with that, is as soon as the door closes, the light will go out.

You need to use a capacitor in the system, what value, I can't say. It depends ont he current draw of the light.

Possibly even using a relay setup can make it work. Similar to this

Obviously, the wiring diagram will need modification, but the principle behind the capacitor is what you're after.

I don't know if such a wire exists *already* in the GC's, but Instead of trying to find something that might not exist, if you're handy with soldering and a bit of electrical schematics, you can design your own circuit.

It's too late here in Oz, and I'm tied up with a funeral all weekend, but early next week I'll try and organize a schematic if you haven;t devised one by then.

Good luck with it.

PS; I hope I understood your request correctly

This is correct in the newer vehicles that have this key ring light equipped stock they have a BCM that controls this and other interior lighting functions through a control module and micro chip.

[ScoobySam8]

Originally Posted by showoval

So you need a ground when the door is open and then need it to disappear when the door is closed. Our cars use a negitive door trigger, that should give you the ground your looking for when the door is open and go away when the door is closed.

I'll test the door trigger this weekend and see what kinda readings I get.

In terms of figuring out a whole circuit, the light already comes with a circuit board with capacitors and all that jazz. As mentioned in my 1st post, I've tested the light (which has 3 wires), and thats the only way it works.

[rbimdxe]

^^ My mistake.

As showoval said; the door trigger will be what your after.

It will show 0v(Ground) when the door is open, but will be an open circuit(no 'finish'. There is no such thing as a -12v line. You can only have a -ve voltage in two circumstances. Your multimeter is connected the 'wrong' way. Or, you're measuring voltage drop across a component.).

[ScoobySam8]

That'll probably be my answer then

An interesting thing I noticed, and also wanted to test again this weekend, was that the dome light reads 12V on either side of the bulb when door is closed, then one side drops to 0V (light goes on) allowing voltage to flow when door opens

I'll see how it goes this weekend. Thanks for all the help guys, much appreciated

[rbimdxe]

^^ What you're seeing is the effect of the voltage drop across the light bulb.

When the door is open, the circuit is open, with no ground. hence +12v either side of the diode(light bulb). Once the door is open, and the circuit grounded, one side reads +12v, but the other reads 0v, due to the bulb 'using' that electricity.

Good luck with it.

[ScoobySam8]

Originally Posted by showoval

So you need a ground when the door is open and then need it to disappear when the door is closed. Our cars use a negitive door trigger, that should give you the ground your looking for when the door is open and go away when the door is closed.

This was exactly what I needed, wired up and looks awesome
Thanks again for your help guys