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[mrwilsn]

It does both things -- they're equivalent statements. If you apply 12V to an LED, it will draw massive current, until it burns out. The series resistor, by limiting the current that can flow, drops the voltage for the LED. That's what resistors do.

Sorry, what I meant is that if you put the resistor in series with the LED the amount of current flowing through the LED and the resistor will be equivalent.  If the LED turn signals are designed for a 12V supply then putting the resistor in series is going to drop the voltage the LED sees and depending on the cutoff and the size of the resistor the LED won't work.

Uh, no. The parallel resistor has no effect on the current through the LED/series resistor combo. It just increases the current through the flasher, so the flasher will feel "normal" current and will operate normally.

But that's not what I said.  If you put a resistor in parallel with the LED the current is split between the resistor and the LED.  How much current goes to the LED vs. how much goes through the resistor depends on the value of the resistor and the resistance of the LED.  But yes, the total current draw for the combo stays the same.

[pacificcricket]

It does both things -- they're equivalent statements. If you apply 12V to an LED, it will draw massive current, until it burns out. The series resistor, by limiting the current that can flow, drops the voltage for the LED. That's what resistors do.

Sorry, what I meant is that if you put the resistor in series with the LED the amount of current flowing through the LED and the resistor will be equivalent.  If the LED turn signals are designed for a 12V supply then putting the resistor in series is going to drop the voltage the LED sees and depending on the cutoff and the size of the resistor the LED won't work.

Uh, no. The parallel resistor has no effect on the current through the LED/series resistor combo. It just increases the current through the flasher, so the flasher will feel "normal" current and will operate normally.

But that's not what I said.  If you put a resistor in parallel with the LED the current is split between the resistor and the LED.  How much current goes to the LED vs. how much goes through the resistor depends on the value of the resistor and the resistance of the LED.  But yes, the total current draw for the combo stays the same.

+1

[mrwilsn]

ok,

it's not polarity, just for giggles i reversed everything and still nothing.

i know the lights work, i can test them with 9v battery and they illuminate, it also tells me the polarity was correct the first time.

the LED's i am using are THESE. they do not mention anything about requiring additional resistors but who knows.

if i take 1 of the led's out of the loop (front), and use the stock bulb, everything works, just flashes quickly. so, once again wiring is correct, i just need to solves the resistance and location questions.

what resistor should i use, and where is best to tie in?

thanks,

laramie 

Measure the resistance of the stock bulb.  If you use a resistor of that value in parallel (front only or back only) then everything should work but flash fast just like with the stock bulb up front but LED in back.  Using a higher value will slow things down but the higher you go the less energy saving benefits from using the LED.

[Skidz]

I also thought about a parallel resistor per circuit, but you need to keep in mind that the resistor has to dissipate the energy in heat. So if stock bulbs are 10W each, you need to dissipate 20Watts resulting in a resistor of 7,18 ohms, but because they aren't really available you'd have to use a 6,8 ohms resistor or a 7,5 ohms resistor. At Element14 they cost around $8 a piece, and with a duty cycle of 50% you don't need to cool them I think but YMMV

[Francois]

a normal light use 5w a led use 1w sometime less
the best is to change the signal box for one witch is working with 2w
link this
http://fr.eachbuyer.com/4-x-motorcycle-amber-12-led-turn-signal-light-bulb-lamp-flasher-p180106.html?currency=EUR&from=pla&gclid=CNqT6OCj2s8CFUEaGwod3acJjg
you are going to use less courant

[Fred]

But that's not what I said.  If you put a resistor in parallel with the LED the current is split between the resistor and the LED.  How much current goes to the LED vs. how much goes through the resistor depends on the value of the resistor and the resistance of the LED.  But yes, the total current draw for the combo stays the same.

That really isn't correct. Add a resistor in parallel and (assuming your constant voltage power supply isn't struggling) the LED will draw the same amount of current and power as it did before. The new resistor will draw some too. Your battery will go flat a bit quicker.

[wijnand71]

yeah, I've been strugling with this also. But ended up with the easy install of Weiser's turn signal indicators. Looks great and they are BRIGHT! Plug and play, with resistors and polarity protect diode soldered on the backside.

[laramie LC4]

ya, i saw those plug and play ones but the whole point of doing this was to lose that ridiculous rear end and huge blinkers.

so, i just went and tested resistance on the stock bulb. it came back @ 1.5 ohms. does this mean is just need to find 4 resistors of that value or higher, then put them on the hot wire leading into the light?

correct?

since i started this, i have received about 5 different answers telling me 5 diff things to do to make it work and so far none have worked. never would have thought this was such a challenge.

thanks,

laramie 

[pacificcricket]

so, i just went and tested resistance on the stock bulb. it came back @ 1.5 ohms. does this mean is just need to find 4 resistors of that value or higher, then put them on the hot wire leading into the light?

No, I believe that means your new load needs to come back at roughly 1.5Ohm total. That's why I suggested you also measure your new blinkers for resistance. If they come with higher resistance, you will need to add a resistor in parallel, to reduce the overall resistance of the load. If they come back with less resistance, you will add a resistor in series to increase the resistance.

[Doug S]

Measuring the cold resistance of a light bulb doesn't tell you its operating current. The resistance of a light bulb increases quite a bit as it heats up enough to create light. A fixed 1.5-ohm resistor, at 13.3V, would draw 13.3V/1.5 ohms = 8.9 amps of current. That represents 13.3V * 8.9A = 118 watts of power, which is far more power than even your headlight draws. In reality, once it's up to operating temp, a 12W bulb will draw just about an amp from the vehicle's ~13.3V operating voltage, with its hot resistance somewhere in the 13-ohm range.

Try this: Put one of the old right-hand signals back on, say the front. Verify that it works when you turn on your right blinker (it may operate fast because it's only running one light, but it should operate). Put your voltmeter on the leads going to the right rear signal, and see which one goes positive when the signal blinks. Attach the new signal with that polarity. It should operate with the front one. Mark the wires with tape so you can get it re-attached correctly with 100% confidence. Now do the same procedure for the front: Put the old signal back on the rear, verify proper operation, check which front lead goes positive with the rear signal, attach the new one, verify it works, mark the wires. Now attach both new signals; you should, at this point, have 100% confidence they're correctly polarized.

If at this point they don't work, it's because the blinker requires more load to operate correctly...LED turn signals don't draw enough current to make some blinkers work at all, other blinkers will operate but too fast. You can then either change the blinker to a more modern one that blinks properly even with a light load, or add resistance (probably something in the 15-ohm, 15W range would work well) in parallel with one of the left-hand signals, and one of the right-hand blinkers. I'd suggest upgrading the blinker instead of adding parallel resistance.

[pacificcricket]

Good point about the incandescent bulbs having variable resistance! Can't we just use the Ohm's law to calculate the resistance then based on Wattage ?

Say the bulb is rated for 5W.

A = 5W / 12V = ~0.42A

R = 5W / (0.42A * 0.42A) = ~28 Ohm

[laramie LC4]

ok, im a dumb ass. should have listened earlier to others when they said to just replace the flasher. took all of 5 mins to find and remove. a quick trip to the auto parts store and $12 spent. now everything works perfect.

thanks for the help guys.

laramie 

[Doug S]

Good point about the incandescent bulbs having variable resistance! Can't we just use the Ohm's law to calculate the resistance then based on Wattage ?

Say the bulb is rated for 5W.

A = 5W / 12V = ~0.42A

R = 5W / (0.42A * 0.42A) = ~28 Ohm

Yes. I did that in my post. 5W is low for a turn signal; 12W is much more typical.

[acacia1731]

should have listened earlier to others when they said to just replace the flasher. took all of 5 mins to find and remove. a quick trip to the auto parts store and $12 spent. now everything works perfect.

Thanks for the report laramie.  Can you post which flasher you picked up (pics or part number)?

I didn't have any luck with my new flasher.  Not sure if it was my lights, or a bad flasher, or the wrong flasher.  I tried replacing the rear turn signals with LED versions and put LED bulbs into the original front housings.  Neither the old or new flasher gave me any light though (just appeared dead).  I finally settled on the original flasher and incandescent front bulbs, combined with the rear LED signals, which works but flashes about 3X faster than normal.

[laramie LC4]

i just dug until i saw one that said "LED" and had the same plug pattern as the OEM flasher.

i am going to do a full write up but for now at least this is out there.

OEM flasher is on the left, the new LED flasher on the right.

laters,

laramie