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LED Indicators?


stu1986

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I have worked it out to be a 1x 25w resistor for each side :)

 

because the indicators on each side (e.g. all left and / all right side) are linked if you change the front and rear indicators then all you need is one 25w reistor on the indicator wire on each side, easiest place in the boot

Edited by stu1986
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The reason I ask is that it is not about the wattage rating of the resistor but the resistance value itself that is important to make it work. The wattage rating is important so that the resistor can handle the power flowing through it which is a factor of the current flowing through it and the voltage dropped across it. This is all dependent on the resistor value and the voltage drop across the LED.

 

You could put a 500W resistor in there and it would work providing the resistance value is correct.

 

The purpose of increasing the resistance is to trick the circuit into believing that the right amount of current is flowing through it. With two bulbs wired in parallel there is more current flowing than if there was just one. When one bulb blows, less current flows and is detected by the circuit causing the on off cycle to be quicker. This is what alerts you to a failed bulb. LEDs draw much less current than bulbs which is why the circuit is flashing as though a bulb is missing. So you need to make the circuit draw more current and this is achieved by putting a resistor in parallel with each LED such that the resistor draws almost as much current as a normal bulb with the LED drawing the rest. In practice it probably is nor possible to get it just right.

 

So the original bulb is 21W. Using James Watts power law of electrickery, W = I x V (Ivy Watts is how I remember it). The current flow I = 12 / 21 = 1.75 amps.

 

The equivalent resistance to use would be (using ohms law V = I x R) R = 12 / 1.75 = 6.9 ohms. The power rating of the resistor would need to be 21W or higher as the same current is flowing as that of the bulb and the voltage across it is also the same.

 

So you need a 6.9 ohm 21 Watt power resistor. Maplins only do power resistors up to 10W - oops! Not to worry. If you have two 10W resistors of half the resistance in series then the power in each is halved so 10W would do.

 

So you need two 3.9R 10 Watt power resistors from maplins. 3.9 is the closest. Solder one end to each other and then the free ends across the wires of the LED. Repeat for each LED.

 

I would also use some heat shrink tubing over the resistors and all connections to properly insulate.

 

If you use lower resistance you will draw more current and risk blowing the circuitry but that's not to say it may not work. You will also draw more current than the wiring might be able to handle which could potentially cause a fire - but again unlikely as the cables are normally quite thick and as long as you are not blowing fuses it should be just fine. The bulbs are never an exact Wattage anyway so there will be a tolerance.

 

Hope that helps.

 

You may also be able to buy a replacement relay that switches at the correct rate for the lower current draw meaning no soldering etc needed.

 

The disadvantage of using resistors is the if the LED fails, the circuit will not flash faster and you will not be alerted to the failed indicator.

Edited by scubapics
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It would appear that there is no flasher relay in the Z. The 350z at least, The indicators are controlled by the BCM (body control module). So resistors are the only way but you run the risk of not knowing about a failed LED bulb and mixing it up with the law or worst having an accident. The lifespan of an LED bulb should be way longer than a normal bulb and may never fail for the lifetime of the vehicle though. But don't take my word on it - check for yourself.

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