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Tried replacing all my 2003 Z06's external lights with LED asemblies (except headlights of course). There are a couple of internet sources which sell multi LED modules assembled as direct replacements for all the 3057, 4057, 194's, etc. The first thing I learned is that the circuit resistence changes when the filament bulbs are replaced by the LED assemblies. The solution, according to the vendors, is to put either a 3 or 6 ohm resistor across the + and - wire to each of the car's turn signal lights. They provided nontechnical diagrams for those who have blinker bulbs alone, i.e., Euro rear light assemblies, and those who have multi-filament bulbs that double as brake lights, i.e., standard U.S. C5 rear light assemblies. I did the bulb / LED assembly swaps, installed the resistors where I thought they belonged and gave it a few days to stabilize. It didn't. Flasher switch blinked the turn signal LED's at least twice as fast as the filament bulbs (resistors were supposed to take care of that), then, to use a technical term, electrical weirdness manifested itself:
1) cruise control worked intermittently
2) daylight driving lights would wink out when the brake pedal was pressed
3) during the daylight, all inside dash lights (like the clock LED, radio LED, ambient temp LED, etc.) would dim when the brake pedal was pressed
4) at night with the headlights on, the brake lights would come on and stay on, not changing when the brake pedal was pressed

I went back to the stock filament bulbs, yanked the resistor assemblies from the turn signal lights, and everything returned to normal.

I had heard that the C5's BCM was real sensitive to any non-stock after-market electrical thingy's put online. Seems to be the case. Luckily, no permanent damage (yet).

Any EE's or people experienced with the C5's lighting system have any comments or suggestions?
:usa:
 

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Watts = V*I. So a 30W lamp (which, if I recall, is about what a turnsignal bulb dissipates) draws 2.5A at 12V. Knowing how much current the bulb draws lets you calculate it's effective resistance (really, in an incandescent bulb it's "impedance", but you can use these terms interchangably for this math). Resistance = V/I tells us the bulb has a resistance of about 5 ohms.

So to work correctly, your LED replacement bulbs need to have a series resistance that's fairly close to this 5 Ohm value.

Without knowing the specifics of the LED replacements you used, it's not possible to say exactly what needs to be changed to fix the problem. LEDs have very low series resistances and they require a current-limiting series resistor in order to work. That resistance is usually pretty high (100s of Ohms) to keep the current low because LEDs don't survive long at high currents. But because the LED assembly resistance is high, it looks like a burned out bulb to the Flasher, which is why the LEDs blink too fast. In this situation, putting a resistor in parallel is the right thing to do.

You didn't say what exact value you used. 3 Ohms is probably too low, though. Assuming a 100 Ohms or so resistance for the LED assembly, a 3 Ohm resistor is going to result in an effective resistance of the assembly that's less than 3 Ohms. Even assuming 3 Ohms, that'll result in an increase current draw of about 60% over an incandescent bulb.

This is certainly what bummed out the BCM.

If you have an Ohm meter, you could measure the forward resistance of the LED lamp assembly (reverse resistance will be very large since LEDs are diodes, they only conduct in one direction). Knowing this information would allow me to specify an exact solution. If you don't have an Ohm meter, assuming you didn't use 6 Ohm resistors, my recommendation would be that you re-do your hookup with this larger value component.

Be aware that the resistor needs to be a high wattage ceramic type, not the small barrel style resistors you see on Printed Circuit Boards. Power = I*I*R. If you're putting 2 amps across a 6 Ohm resistor (which is about what it'll be, given V=IR dictates that 12V and 6 Ohms equals 2 Amps), that resistor is going to dissipate 2*2*6 = 24 Watts, and it'll get *hot*. You need to be sure you've chosen a component rated for this kind of power or you're either going to just burn out the resistor or possibly set fire to your car.
 

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:jawdrop: :bang:
 

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"............or possibly set fire to your car."

I was following along just fine till I got there. :jawdrop:
 
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