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I bought an LED bulb, W5W, to replace the position lights, number plate lights and maybe the interior reading lamps. However they stay dimly lit when they're off.

I am thinking that the current is already flowing through the wires when the lights are off and the key is out.

So the circuit is closed, but with an incandescent lamp, not enough current exists to heat the coil, but with LEDs there is enough of it to light it.

Some people advise to put a resistor before the bulb, to fix this.

Will these LEDs drain the battery if they stay partially lit (i dont mind them staying like this)? If I add a resistor, won't that one drain the battery too?

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    \$\begingroup\$ So, you are worried that partially lit LEDs will drain your battery but you are OK with incandescent lamps drawing the current without lighting up? Check your wiring, something is definitely wrong with it. \$\endgroup\$ – Maple Aug 26 '18 at 17:53
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    \$\begingroup\$ This seems odd, and more a question on automotive systems than original electronic design. Try measuring the voltage across the bulb, when the vehicle is off. And make sure it is really off, not some overly clever minute-long "walk to your door without stumbling in the dark" courtesy delay. \$\endgroup\$ – Chris Stratton Aug 26 '18 at 17:53
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    \$\begingroup\$ It won't. It can't if your car is wired correctly. It's not the lamp (how should the lamp even know?!) that shuts off or on – it's a switch somewhere else. So if there's current flowing through the LED, there was also current flowing through the incandescent lamp. \$\endgroup\$ – Marcus Müller Aug 26 '18 at 18:20
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    \$\begingroup\$ In all probability this is current flowing from the lights ECU to be able to test if lamps are blown. A small resistor may be required on each of the lamp circuits. \$\endgroup\$ – Jack Creasey Aug 26 '18 at 18:24
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    \$\begingroup\$ @eonootz You are refusing to understand what is said to you. Let's try again: if you LEDs draw current when power is off then your incandescent lights also draw current when power is off. Most likely more than LEDs. \$\endgroup\$ – Maple Aug 26 '18 at 19:19
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Cars often pass a little bit of current through "off" bulbs in order to determine if the (previously incandescent) bulb was working normally.

Depending on the replacement LED bulb circuitry, this can either cause intermittent flashing (when the LED driver circuitry builds up enough charge to light up), or constantly light an LED dimly (if it's a simple resistor in series with the LED). Yours sounds like the latter.

Olin is right with the suggested workaround - adding a bleed resistor in parallel with your LED bulb. The parallel resistor can provide an alternate path for the small "off" state current, meaning the voltage across the LEDs never gets high enough to cause them to light up.

schematic

simulate this circuit – Schematic created using CircuitLab

Please note that all values here are placeholders. You'll need to do some experimentation to find a bleed resistor value that works. 10k is probably a good starting point.

In "on" mode, the parallel resistor will just waste a tiny bit of power - it make the bulb visibly dimmer.

The replacement LED bulb will almost certainly be wasting less power than the original incandescent when "off". You could confirm that by checking the voltage across a working incandescent and your new LED bulb, and calculating the current through the bulb check resistor with V=I*R. Lower voltage across the bulb means higher voltage across the bulb check resistor, which means higher current wasted.

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    \$\begingroup\$ I understand that car designers are not rocket scientists, but constantly sending "testing" current while ignition is OFF sounds ridiculous to the point of unbelievable. A reference to such testing system of any car make would be appreciated. Typical outage detection circuits work when lights are ON by sending current through flasher, so frequency changes when bulb is burnt. Besides, 0.1mA in your schematics is probably not enough to dimly lit 45 mA LED. \$\endgroup\$ – Maple Aug 27 '18 at 8:37
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First, it doesn't sound right that there is a little current thru the lights when they are supposed to be off. You really should check into that.

However, if you are truly OK with that current, then put a resistor across the LED. Find what the current is, then size the resistor to produce about 1 V with that current. That won't be enough to light the LED, but should draw so little extra current when the LED is lit to be irrelevant.

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  • \$\begingroup\$ Note that so called "load equalizer" 6 Ohm resistors often installed in parallel to LEDs. Basically returning current consumption to incandescent level. Not irrelevant when lights are on and certainly huge draw when power is off. \$\endgroup\$ – Maple Aug 27 '18 at 8:44
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It is not unusual to have leakage currents through relay coils or transistors that are supposed to be fully OFF. The driving source may leak a few milli-amps of current, not enough to turn on the relay or a series indicator lamp.

12 volt relay coils are about 400 to 1,000 ohms of DC resistance, enough to turn off a conventional lamp but not an LED.

To fix the problem add a small value resistor across the LED such that the LED voltage drops to <= 2 volts down to 1 volt. That is below the 'ON' threshold of most LED's.

The leak is already there so a resistor to bypass the LED is not going to increase the leakage current. Remember there was a bulb there at one time, not impeding the flow of leakage current at all.

Measure the leak current if you can. A 1 K resistor creates a 1 volt drop across it per each mA of current flowing, so a 1 K 1/4 W resistor is a good starting point. It may have to be as low as 200 ohms if several mA of leakage current is present, but you said "dimly lit", which implies only 1 or 2 mA of leakage current.

NOTE: 'OFF' is not always an ideal OFF, as this case proves. I had to use diodes often to block leakage currents from electronic modules used to control overhead lights, and sometimes used to trigger a third-party alarm. Often it is a transistor that is not fully OFF simply because of a cheap design. It worked at the factory with standard light bulbs, not expecting LED replacements.

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    \$\begingroup\$ I am totally confused. Would you mind posting a circuit or a link to a circuit illustrating how "relay coils" would pass current trough the exterior and interior lights (not indicator lights) when the ignition and corresponding light switches are off? \$\endgroup\$ – Maple Aug 26 '18 at 23:10
  • \$\begingroup\$ FWIW, my friend was repairing his Hummer recently and was astonished to find that AC fan is wired directly into the battery via PWM controlled MOSFET. Which means even with ignition off there is some substantial leakage current. But I don't think the lights mentioned by OP have anything like that. \$\endgroup\$ – Maple Aug 26 '18 at 23:26
  • \$\begingroup\$ @Maple. I have nothing to post. It comes from years of doing car alarms and stereos. 'OFF' is not always an ideal OFF, as this case proves. I had to use diodes often to block leakage currents from electronic modules used to control overhead lights, and sometimes used to trigger a third-party alarm. Often it is a transistor that is not fully OFF simply because of a cheap design. It worked at the factory with standard light bulbs, not expecting LED replacements. \$\endgroup\$ – Sparky256 Aug 27 '18 at 0:38

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