8 replies to this topic

[ghf]

Have a 1990 mini being refurbed and are changing the ignition, oil, indicators etc bulbs to LED's. All work fine except for the ignition. The problem is the LED fails to go out, changed it back to a bulb and everything works OK. Appears as thouigh it needs to see a resistive element as opposed to an LED. With this in mind I used arelay. (Coil being resistive) but that did not work. Any suggestions would be appreciated.

Thanks


Geoff

[Sammy D]

Get a selection of resistors and put one inline as well?

[ghf]

Thanks but these LEDS are 12V's and the resistors are built in. The resistors are usually there to restrict the current so the LED doesn't burn out unless you know differently? If this doesn't work are there any other suggestions?

Thanks for the reply.

[dklawson]

You are correct. You need to add resistors. The resistors are not only necessary to make the light go out, they are necessary to allow the proper amount of current to flow from the ignition switch to the alternator during startup.

Start by placing a 40 Ohm, 10 Watt resistor in series with the wire from the ignition switch to the small terminal on the alternator. The 40 Ohm resistor will allow current to flow almost like the regular light bulb.

Take your LED (with a properly rated/sized voltage dropping resistor if you started with a discrete LED) and wire it in PARALLEL to the 40 Ohm resistor added above. That is to say, connect the high side of the LED to the ignition switch, connect the other side of the LED (after its small dropping resistor) to the small terminal on the alternator.

When you turn the key on (engine off) current will flow through BOTH the LED and the 40 Ohm resistor. This will allow the coils in the alternator to energize AND light the LED.
When you start the engine, the alternator will start making electricity and its small terminal will be brought up to about 14V. At that point the LED will go out just like the light bulb.

Remember that the 40 Ohm resistor needs to be rated for power applications... not a tiny 1/4 or 1/8 Watt part like you'd find on a circuit board.

[stevede]

You are correct. You need to add resistors. The resistors are not only necessary to make the light go out, they are necessary to allow the proper amount of current to flow from the ignition switch to the alternator during startup.

Start by placing a 40 Ohm, 10 Watt resistor in series with the wire from the ignition switch to the small terminal on the alternator. The 40 Ohm resistor will allow current to flow almost like the regular light bulb.

Take your LED (with a properly rated/sized voltage dropping resistor if you started with a discrete LED) and wire it in PARALLEL to the 40 Ohm resistor added above. That is to say, connect the high side of the LED to the ignition switch, connect the other side of the LED (after its small dropping resistor) to the small terminal on the alternator.

When you turn the key on (engine off) current will flow through BOTH the LED and the 40 Ohm resistor. This will allow the coils in the alternator to energize AND light the LED.
When you start the engine, the alternator will start making electricity and its small terminal will be brought up to about 14V. At that point the LED will go out just like the light bulb.

Remember that the 40 Ohm resistor needs to be rated for power applications... not a tiny 1/4 or 1/8 Watt part like you'd find on a circuit board.

I'm really interested in this as I have changed all mine to LED's but not got the engine running as yet. (Ethel did warn me the ignition lamp might not be straight forward - I solved the indicator tell tale with 2 x diodes)



As I have just finished most of the wiring in my engine bay and don't want to disturb it, is it feasible to put the 40 ohm, 10 watt resistor at the clock end of the cable and take the feed return for the led and join at the alternator side of the resistor. I think I am right in saying that the small wire connected to the alternator is exclusive to the warning light.

Thanks in advance

Steve

Edit: Spelling

Edited by stevede, 05 August 2010 - 10:19 PM.

[dklawson]

You can certainly locate both the 40 Ohm resistor and the LED dropping resistor at the speedo end of the wiring. The important thing to keep in mind is the function you are trying to achieve... 40 Ohms in series with the wire to the alternator and the indicator LED (and its dropping resistor) in parallel to the 40 Ohm resistor.

Your setup looks nice. However, it looks like you may need a bit more resistance on the supply to that high-beam indicator blue LED. It's WAY bright. That's great but you don't want that lamp blinding you when driving at night.

[stevede]

Doug

Just a quick update to say thanks for the advice.

I bought the resistor and wired as you advised and now that my engine is in and running, can confirm it works a treat.



I stuck the resistor to the back of the speedo with a hot glue gun.

Part if anyone needs it - Maplins, H47R (10W W/W 47R) for the fine sum of 58 pence!

Thanks again

Steve

ps the blue lamp isn't quite as bright in reality as it shows on the photo.

Edited by stevede, 30 August 2010 - 09:24 AM.

[krazy]

how was the signal indicator wired up? On my mini, 2 wires converges into 1 socket.

Edited by krazy, 23 November 2010 - 12:21 AM.

[stevede]

how was the signal indicator wired up? On my mini, 2 wires converges into 1 socket.

Mine were the same.

Get 2 x 1/2 watt diodes and put one on each end of the wire coming from the indicators. (band facing the indicators) Connect the 2 diodes together at the other end & connect to the LED anode via a resistor. (I used 470 ohm, but check suitable to your led) Connect the LED cathode to ground.

Should work a treat

Regards

Steve