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Help with This??

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Kingdom Man

New Member
Hello,
I am new here and am eager to learn. I need a device or circuit that can have a ground on one side and power on the other side like a light bulb. What I mean is that I need to feed a ground signal back on the a 12vdc power wire when there is no power supplied, but then when the voltage is supplied it will not short out due to the fact that the other end is shorted to ground. It works exactly like a light bulb, a bulb has continuity and so there for ground will travel through the bulb to the source of the power, but when power is supplied the bulb lights up and there is no problem. however I cant use a light bulb for this application, i need something that replicates this principal but without the light bulb. I have a line that is 12vdc when turned on, and I need to tee this Device or whatever will do the job, to the 12vdc line and when it is switched off it will produce a ground on the whole 12vdc signal wire. Hope this makes some sense, as I found it kind of hard to get my point across in this application.
 

Boncuk

New Member
Hi,

your explanation is indeed hard to follow.

I suggest you make a small schematic. It will explain things easier than 1,000 words.

Boncuk
 

JimB

Super Moderator
Most Helpful Member
Why dont you describe the problem, not what you think the solution should be.

Depending on individual interpretation, what you have described is impossible.

JimB
 

Kingdom Man

New Member
Why dont you describe the problem, not what you think the solution should be.

Depending on individual interpretation, what you have described is impossible.

JimB
Ok I will try this, I will also draw a schematic for you and go from there, thanks for all your input!
 

Kingdom Man

New Member
New-1.JPG

The resistor is just there for reference, I dont know what to use in its position, that is why I am asking you guys. I need to have pretty much full continuity to Ground when there is no 12v present on the 12v line. and then when the 12v is switched on it needs to somehow not short to ground. Hope this makes more sense?
 

tcmtech

Banned
Most Helpful Member
sounds to me that if this is being used in an automotive application its to keep the other bulbs from lighting up dim when another circuit is on that has a bad ground and thusly back feeds from another point due to the lighting circuit having a better grounding current path than the other circuit has.
Am I close?
 

JimB

Super Moderator
Most Helpful Member
Hope this makes more sense?
No it does not, TELL US THE PROBLEM, not what you think the solution should be.
You have just re-iterated your nonsensical solution.

JimB
 

Techie7

Member
I thought my IQ is not working properly when read it first.:eek:
Hello,
I am new here and am eager to learn. I need a device or circuit that can have a ground on one side and power on the other side like a light bulb. What I mean is that I need to feed a ground signal back on the a 12vdc power wire when there is no power supplied, but then when the voltage is supplied it will not short out due to the fact that the other end is shorted to ground. It works exactly like a light bulb, a bulb has continuity and so there for ground will travel through the bulb to the source of the power, but when power is supplied the bulb lights up and there is no problem. however I cant use a light bulb for this application, i need something that replicates this principal but without the light bulb. I have a line that is 12vdc when turned on, and I need to tee this Device or whatever will do the job, to the 12vdc line and when it is switched off it will produce a ground on the whole 12vdc signal wire. Hope this makes some sense, as I found it kind of hard to get my point across in this application.
 

Kingdom Man

New Member
The whole problem stems from converting from normal headlight bulbs (which have full continuity through the bulb) to a ballast type HID Headlight system. The HID Ballast does not have continuity through its circuitry, so therefore it sets Headlight warning indicator on the instrument panel. To get rid of this warning the ground needs to flow through the (bulb) back on the power side in order to keep the warning out. If you tee a bulb in where I have put the resistor that will do it, but I am looking for other possible solutions as a bulb isn't the best idea.
 

Mikebits

Well-Known Member
You could always disconnect the warning light? I doubt they serve any purpose after your mod.
 

smanches

New Member
Sounds like a diode placed in parallel with the HID lamp from GND to +12v, but I can't say for certain as I'm still confused after reading all the posts.
 
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tcmtech

Banned
Most Helpful Member
I would use a relay that has the coil power and the common contact tied together and fed from the light circuit that goes to the HID units.
I would then have a 10 ohm 1 watt resistor on the NC output of the relay. That way when the power is off the relay coil and the ten ohm resistor look like a lamp filament to the sensor system.
However when you turn on the lights the relay closes and disconnects the 10 ohm resistor.
 

mneary

New Member
I think I understand what you're looking for, a friend invented a similar system. When the system first turns on, the computer applies a voltage to each bulb (not long enough to light it), just to see if the bulb is there. The common power line is monitored to see how much total system power is drawn. If any one "bulb" draws no incremental current, the computer declares a fault to the user.

But you want to fool the computer into thinking that there's a good bulb in place, without actually consuming unused power all the time (or lighting up the engine compartment).

First, you need to determine how you can tell when this test is performed. Does your car have a common power line with a CAN receiver at each headlamp site? This could make it pretty tough. What year/model car is it?

Maybe the computer can be told that it has HID headlamps retrofitted?

- -More to come as you do research, and verify my guesses.

[edit] tcmtech has a good idea; the relay might have to be delayed a few more milliseconds depending on how long the sensor system takes to make its readings. Also might require less than 10 ohms to look like a headlamp.
 
Last edited:

Kingdom Man

New Member
I think I understand what you're looking for, a friend invented a similar system. When the system first turns on, the computer applies a voltage to each bulb (not long enough to light it), just to see if the bulb is there. The common power line is monitored to see how much total system power is drawn. If any one "bulb" draws no incremental current, the computer declares a fault to the user.

But you want to fool the computer into thinking that there's a good bulb in place, without actually consuming unused power all the time (or lighting up the engine compartment).

First, you need to determine how you can tell when this test is performed. Does your car have a common power line with a CAN receiver at each headlamp site? This could make it pretty tough. What year/model car is it?

Maybe the computer can be told that it has HID headlamps retrofitted?

- -More to come as you do research, and verify my guesses.

[edit] tcmtech has a good idea; the relay might have to be delayed a few more milliseconds depending on how long the sensor system takes to make its readings. Also might require less than 10 ohms to look like a headlamp.
Ya the headlamps are less than 1 ohm, they have pretty much full continuity.
But I like where your going with this, also the car is a 1990 bmw, and therefore because of the age it is not super complex but way ahead of its time.
 

mneary

New Member
Maybe you might not need a relay if a big capacitor will do the job.

You can put a dummy load on your + line in parallel with your HID ballast. I would try a 1-2 ohm resistor (5W wirewound if you can get) in series with a diode and a 3300 µF capacitor. Bleed the cap with 270 ohms 1W. This will draw about 10A for 3 milliseconds, each time power is applied, and then disappear. When the power is removed, it will be discharged after several seconds. Don't leave the diode out, no telling what your computer will do if it sees voltage feeding back in when it thinks the light should be shut off.

The 1 ohm resistor will receive 3 millisecond surges of 150 watts so it must be wirewound. The capacitor should be 105c 25V because it's in a hostile environment.
 
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