Auto LED dimly lit when off

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?

Answer

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.

^{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.