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Just barreling along Needs to Get Out More!
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Just barreling along Needs to Get Out More!
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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
Jon M
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Has a lot to Say!
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Has a lot to Say!
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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.
John
Silver 2005 S1 Roadster V6 - Henrietta
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Talk Morgan Sage
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Talk Morgan Sage
Joined: Apr 2014
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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.
Best Regards Lang may yer lum reek
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Joined: Feb 2016
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Needs to Get Out More!
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Needs to Get Out More!
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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 
Richard
2018 Roadster 3.7 1966 Land Rover S2a 88 2024 Royal Enfield Guerrilla 450 1945 Guzzi Airone
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Joined: Apr 2014
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Talk Morgan Sage
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Talk Morgan Sage
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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!
Best Regards Lang may yer lum reek
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Joined: Dec 2013
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Has a lot to Say!
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Has a lot to Say!
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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."
John
Silver 2005 S1 Roadster V6 - Henrietta
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Just barreling along Needs to Get Out More!
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Just barreling along Needs to Get Out More!
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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
Jon M
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Talk Morgan Sage
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Talk Morgan Sage
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I have no doubt you are right but I can't get my head round the logic.
Best Regards Lang may yer lum reek
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Needs to Get Out More!
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Needs to Get Out More!
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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! 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!!! 
Last edited by Richard Wood; 08/03/19 10:04 PM.
Richard
2018 Roadster 3.7 1966 Land Rover S2a 88 2024 Royal Enfield Guerrilla 450 1945 Guzzi Airone
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Joined: Apr 2014
Posts: 6,957 Likes: 116
Talk Morgan Sage
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Talk Morgan Sage
Joined: Apr 2014
Posts: 6,957 Likes: 116 |
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.
Best Regards Lang may yer lum reek
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