Turn Signal Relay (Electronic Flasher) ??
#1
Turn Signal Relay (Electronic Flasher) ??
Im planning on swapping my turn signals bulbs out and putting in LED's. But in order to do so i must replace the turn signal relay (electronic flasher box) with one made for LED's, so they do no hyperflash. Does anybody know where this relay (electronic Flasher) is located?? I looked under the hood at the fuses near the battery but those werent it. I took a quick glance at the fuses inside the car but i dont think those were them either. Also i was looking at 2 different model aftermarket relays (electronic flashers) for leds. Theres ones that are ISO and ones that are JSO does anyone know which one it is?? Thanks
http://frickinbrite.com/catalog/prod...roducts_id=137
http://frickinbrite.com/catalog/prod...roducts_id=298
http://frickinbrite.com/catalog/prod...roducts_id=137
http://frickinbrite.com/catalog/prod...roducts_id=298
#2
Thats not true
You can get a 20 watt 20 ohm resister and wire it in series with the LED, it's not hard. Its how I did it on my motorcycle. The resisters are big white ones, that look like they're ceramic, like .5"x.5"x2"
Also, I THINK that our flashers will not flash faster... I THINK it just SOUNDS Like it is, but the proper flash rate will be retained (since there's really no flasher, that clicking is generated just to hear the click, the flashing is done with solid state relays in the BCM)
You can get a 20 watt 20 ohm resister and wire it in series with the LED, it's not hard. Its how I did it on my motorcycle. The resisters are big white ones, that look like they're ceramic, like .5"x.5"x2"
Also, I THINK that our flashers will not flash faster... I THINK it just SOUNDS Like it is, but the proper flash rate will be retained (since there's really no flasher, that clicking is generated just to hear the click, the flashing is done with solid state relays in the BCM)
#5
Originally Posted by flyt89
BCM?? also ceamic flashers get red hot
#6
#7
Originally Posted by KPierson
A 20 ohm resistor by itself will draw 720mA, which equals 10.3 watts at 14.4vdc. At a 50% buffer and a 20 watt resistor would be the right choice. If you go with a 1 watt resistor it will most likely live a short life, and become a fire hazard.
Example: You need 100 mA going through the LED's. Assume you get 14.4V going to the turn signals, and you drop about 3V across the LED's. That means 9V will drop across the resistor. Using R=14.4/.1 = 144 Ohm. That will ensure you get 100 mA through the LED's, then for the power rating, you can do V^2/R=I^2*R=V*I = 1.44 Watt resistor.
So if these LED's are like all the little 5mm ones you see in a lot of common projects, they are generally 20mA forward current. High flux LED's are around 70mA forward current. And luxeons can be between 1-1.5A forward current.
Personally I just think a 20W resistor is overkill for most available LED's out there.
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#8
Originally Posted by MidnightG35X
Not many LED's can take 720 mA of current going through them. It would be helpful to know which LED's are being used. You have to know the voltage drop across the LED's along with the amount of current you want to put through them. Knowing that, you can figure out what resistance you need in the resistor. And then finally you can figure out what type of power handling you need.
Example: You need 100 mA going through the LED's. Assume you get 14.4V going to the turn signals, and you drop about 3V across the LED's. That means 9V will drop across the resistor. Using R=14.4/.1 = 144 Ohm. That will ensure you get 100 mA through the LED's, then for the power rating, you can do V^2/R=I^2*R=V*I = 1.44 Watt resistor.
So if these LED's are like all the little 5mm ones you see in a lot of common projects, they are generally 20mA forward current. High flux LED's are around 70mA forward current. And luxeons can be between 1-1.5A forward current.
Personally I just think a 20W resistor is overkill for most available LED's out there.
Example: You need 100 mA going through the LED's. Assume you get 14.4V going to the turn signals, and you drop about 3V across the LED's. That means 9V will drop across the resistor. Using R=14.4/.1 = 144 Ohm. That will ensure you get 100 mA through the LED's, then for the power rating, you can do V^2/R=I^2*R=V*I = 1.44 Watt resistor.
So if these LED's are like all the little 5mm ones you see in a lot of common projects, they are generally 20mA forward current. High flux LED's are around 70mA forward current. And luxeons can be between 1-1.5A forward current.
Personally I just think a 20W resistor is overkill for most available LED's out there.
The 20 ohm resistor, in this case, is a load resistor, NOT a current limiting resistor. The point of the resistor is to load the electromechanical flasher to the point that it 'thinks' that a bulb is in the circuit. The current added by the load resistor will represent the current lost by removing the bulb. You will still need your current limiting resistor to protect the bulb. If you use ONLY the current limiting reisistor (say a 470ohm, 1/2 watt) your turn signals will flash quickly because it will think a bulb is burnt out. When you put the 20 ohm load resistor in the circuit (in parallel with the LED AND the current limiting resistor) the flasher will think the bulb is still there, and will flash at the normal rate.
But, what do I know? I've never, in my life, replaced a bulb with an LED in a car!
Last edited by KPierson; 05-21-2007 at 04:57 PM.
#9
Originally Posted by KPierson
You're looking at this completely wrong.
The 20 ohm resistor, in this case, is a load resistor, NOT a current limiting resistor. The point of the resistor is to load the electronic flasher to the point that it 'thinks' that a bulb is in the circuit. The current added by the load resistor will represent the current lost by removing the bulb. You will still need your current limiting resistor to protect the bulb. If you use ONLY the current limiting reisistor (say a 470ohm, 1/2 watt) your turn signals will flash quickly because it will think a bulb is burnt out. When you put the 20 ohm load resistor in the circuit (in parallel with the LED AND the current limiting resistor) the flasher will think the bulb is still there, and will flash at the normal rate.
The 20 ohm resistor, in this case, is a load resistor, NOT a current limiting resistor. The point of the resistor is to load the electronic flasher to the point that it 'thinks' that a bulb is in the circuit. The current added by the load resistor will represent the current lost by removing the bulb. You will still need your current limiting resistor to protect the bulb. If you use ONLY the current limiting reisistor (say a 470ohm, 1/2 watt) your turn signals will flash quickly because it will think a bulb is burnt out. When you put the 20 ohm load resistor in the circuit (in parallel with the LED AND the current limiting resistor) the flasher will think the bulb is still there, and will flash at the normal rate.
I digress, I was looking at this as a current limiting resistor. Side question: What sort of current is needed that the BCM will think there is still a bulb in there? Just seems so inefficient to burn power for no reason
But, what do I know? I've never, in my life, replaced a bulb with an LED in a car!
#10
Originally Posted by MidnightG35X
I digress, I was looking at this as a current limiting resistor. Side question: What sort of current is needed that the BCM will think there is still a bulb in there? Just seems so inefficient to burn power for no reason
I sense sarcasm in your typing
I sense sarcasm in your typing
Actually, I really have never done this. I've installed LEDs, but never replaced bulbs with LEDs. I've read about it, and I understand the concept of how/why it all works.
The best way to find out how much current is needed it to measure the actual resistance of the bulb you are replacing. You can the calculate it all out, if you want to account for the current of the LEDs, or you can just use close to the same size resistor. It is a waste of energy, but its the same amount of energy your old bulbs were using anyway, so you arn't using any additional energy. You could also probably reduce power consumption by 25% without fear of 'fast blinking'. So, in the end, you will have 'better' looking lights and be saving energy (in a really weird round about way).
I do believe, though, in our G35s that the alternator spins constantly and is clutchless, and the regulator is switched (only turns on when battery is charging). If this is in fact the case, then you won't be saving or losing any energy at all, as the energy source in question here is gasoline.
#11
Originally Posted by KPierson
Actually, I really have never done this. I've installed LEDs, but never replaced bulbs with LEDs. I've read about it, and I understand the concept of how/why it all works.
The best way to find out how much current is needed it to measure the actual resistance of the bulb you are replacing. You can the calculate it all out, if you want to account for the current of the LEDs, or you can just use close to the same size resistor. It is a waste of energy, but its the same amount of energy your old bulbs were using anyway, so you arn't using any additional energy. You could also probably reduce power consumption by 25% without fear of 'fast blinking'. So, in the end, you will have 'better' looking lights and be saving energy (in a really weird round about way).
I hope I don't have to put a resistor in there just to burn power off, but time will tell.
I do believe, though, in our G35s that the alternator spins constantly and is clutchless, and the regulator is switched (only turns on when battery is charging). If this is in fact the case, then you won't be saving or losing any energy at all, as the energy source in question here is gasoline.
BTW... sorry for the thread-jack
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