Rob, our previous spat aside, maybe you could answer a question for me? How can you determine the proper resistance resistor to use to duplicate the correct flash rate of an incandescant turn signal in an LED turn signal? Is it simply a matter of measuring the resistance of the OEM turn signal, and then measuring the resistance of the LED turn signal and using resistors to close the gap? Is there something more to it? or am I just overthinking it?
LEDs have a much higher resistance than incandescants, so you need to lower the effective resistance of the circuit back to somwhere near its original level. High resistance means less power dissipation (one of the advantages of using leds) but the relay depends on the heating effects of the power dissipated through it to operate the bi-metal strip which controls the flash rate.
The resistor sits in parallel between the led and ground. The easiest way is to measure the resistance of the original bulb and use a resistor of that value - the resistance of the led is so much higher that the small error won't matter. If you want to work it out, read on.
With resistors in parallel the effective resistance is derived from 1/Re = 1/R1 + 1/R2 + .........1/Rx
eg, for two 6 ohm resistors in parallel, 1/Re = 1/6+1/6 which equals 2/6 or 1/3.
if 1/Re = 1/3, then Re = 3 ohms
So in this case the effective resistance of the circuit is half of the value of each individual resistor.
In a DC circuit power (in Watts) can be taken as Volts * Amps.
If you have an 18W incandescant indicator bulb on a 12V circuit, then current throughh that bulb can be calculated from I = W/V.
I (amps) = 18 (Watts) /12 (Volts) = 1.5A.
Resistance = V/I = 12/1.5 = 8 ohm.
Given that the led power dissipation is probably about 1 or 2 watts, if you work out the resistance of the led then use that in the equation for the effective resistance of the circuit you'll see that the resistance of the led makes very little difference to the answer, so you can just ignore it.
As the resistor will be carrying about the same current as the original bulb it should ideally have a power rating equal to that bulb, so an 8 ohm 18W resistor would be the correct replacement for an 18W bulb. That's a heavy duty resistor, and as it's flashing and only used for short periods you can get away with an 8 ohm 3W unit probably.
Without working it out, for a 21W incandescant a 7 ohm resistor is probably about right.
Going from what you've seen about effective resistance in parallel circuits, if you replace both front and rear indies, then rather than using two resistors you can use a single resistor to serve both leds, of half the value in ohms but double the power rating. In most flsher circuits the outputr from the relay feeds the flasher switch, so you could actually fit a single resistor to ground between the relay output and the switch that would satisfy the loading for all four indies. Size it for one pair, as normally only one side is on at a time.
This won't work if you have a hazard warning option that flashes all 4 together unless you don't mind the flash rate being wrong in hazard warning mode.
Read up on Ohm's law, and Kirchoff's Theorem for the derivation of effective resistance in combination parallel and series circuits.
You've also seen that installing the resistors causes the overall circuit to draw just as much power as it did with incandescants, so you've lost the potential power saving of fitting the leds. An electronic relay designed for leds lets you run them without resistors and saves all the above work.
Rob