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Originally Posted By asmog
Many Thanks to everyone for replying. Sounds like I can do this without any problems. It was just that it had been mentioned that as the voltage/ amp would be different between the leds and standard bulbs it might upset the engine management computer. Sounds like this isn’t a problem.
Andrew


Andrew, you won't have any issues as it's not a sophisticated harness or CANbus, just be aware that if you swop the indicator bulbs to LED you may have to swop the flasher unit too, but head, side, tail, and brakelights will all be ok


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The flasher unit on my 2005 Roadster is incorporated into the PCB behind the dash. Too risky/difficult to change. It is easier to put a ballast resistor in parallel to each bulb. That way there is no drop in current so the fasher unit thinks there is are tungsten bulbs present.


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Originally Posted By Felix42
It is easier to put a ballast resistor in parallel to each bulb. That way there is no drop in current so the fasher unit thinks there is are tungsten bulbs present.

I would suggest a resistor in series to keep the same current flow.


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Originally Posted By BobtheTrain
Originally Posted By Felix42
It is easier to put a ballast resistor in parallel to each bulb. That way there is no drop in current so the fasher unit thinks there is are tungsten bulbs present.

I would suggest a resistor in series to keep the same current flow.


That wouldn't work Bob. The resistor (to emulate current draw of filament bulb) needs to shunt the LED.

You should know all about shunting wink


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In parallel 1/R = 1/R1 + 1/R2. If the LED has a lot less resistance than the tungsten, you will draw much more current this way.

I'm a Signal Engineer, not a Shunter!


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These are the instructions for the ballast resistors that I have used successfully:

"The dummy load resistor is designed to be used on vehicles that have electronic components that are not LED compatible. The use of an optional load resistor simulates the current draw used by a standard direction indicator bulb and enables the flasher unit to operate correctly.

For use when sporadic warning messages appear in your vehicle due to LED’s replacing incandescent globes for 12 volt use

Fitting Instructions

Join or splice one end of the resistor wire to the earth wire of the lamp

Join or Splice the other end of the resistor wire to the power or active wire of the lamp which is causing warning messages in your vehicle."


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Personally I'm not a fan of LED indicators on a Mog, I think the dwell (time it's on) is too short, with a very sharp on/off, compared to other modern's on the road you are very low and sometimes if you blink or briefly glance away you can miss an LED flash (try it) especially in bright sun

With a filament bulb the light decay is much slower so can be seen for a longer time

I'm happy with LED sidelights and brakes as they do improve visibility to others, but I don't swop my indicators


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I have no doubt you are right but I can't get my head round the logic.


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Originally Posted By BobtheTrain
In parallel 1/R = 1/R1 + 1/R2. If the LED has a lot less resistance than the tungsten, you will draw much more current this way.

I'm a Signal Engineer, not a Shunter!


Correct formula but not LED characteristic. You need to consider this Bob. For the common electro-mechanical indicator relay to flash normally it needs to see the current draw of two 21 watt bulbs = 42 watts - they often have this wattage marked on them.

Current draw of one bulb is nominally 21/12 = 1.75 amps. Ohms law gives resistance of bulb as 12/1.75 = 6.86 ohms. This is the resistance (÷2 as two indicator bulbs) the circuit needs to see to give correct current draw for indicator relay to flash at correct speed. We can ignore any extra low wattage repeaters.

LED bulbs present a high resistance, leaving a current draw so low as to be almost insignificant in this case. By placing a nominal 7 ohm resistor across the LED (shunting or in parallel with it) most of the current flows through this resistor (it needs to have a suitable power rating typically 50 watts) so the indicator relay is happy as it see's no difference to the presence of filament bulb and the LED is happy as the full 12 volts is across it. Any canbus monitoring is happy as well although not relevant to most classic Morgans.

BTW if you placed the LED directly in the circuit (with or without an extra resistor in series??) it's high resistance would provide a current draw so low the indicator relay would flash very quickly. Similar to the situation when one indicator bulb blows.

See here for further explanation.

Note most 12 volt LED's do have an internal series resistor to present correct voltage and limit current to LED itself but this is not relevant to my argument.

Phew!!! wink

Last edited by Richard Wood; 08/03/19 10:04 PM.

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Originally Posted By Richard Wood
Originally Posted By BobtheTrain
In parallel 1/R = 1/R1 + 1/R2. If the LED has a lot less resistance than the tungsten, you will draw much more current this way.

I'm a Signal Engineer, not a Shunter!


Correct formula but not LED characteristic. You need to consider this Bob. For the common electro-mechanical indicator relay to flash normally it needs to see the current draw of two 21 watt bulbs = 42 watts - they often have this wattage marked on them.

Thank you, Richard. What I didn't realise was that LEDs have a high resistance. I assumed that it was very low resistance in one direction and high in the other. Obviously, with this high resistance, you need a bypass resistor, not a series one. Cheers.


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