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Hi everyone, I recently switched out my rear 1156 turn signal bulbs with LEDs (projector style, not the el cheapo dim ones).
I installed 25W 8ohm resistors to fix the hyper flashing and they have been working very well so far.

I searched around on this forum and saw that others have used 50W 6ohm resistors instead, and I'm wondering what the difference, if any, there is between the two resistors.
Does one use more power (unnecessarily) than the other? Produce more heat?
I just need to prevent hyperflash, so the one that wastes as little power as possible would be preferable.

The LED bulbs I installed supposedly uses 15W if it matters. Hopefully someone with electrical knowledge can chime in.

Thanks :)
 

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If they are truly 15 watts ea (thats hard to believe if they are LED unless they use 3 - 5 watt CREE LEDs in each them) 25 watt resistors are fine.

The important part is they need to be mounted to metal to dissipate the heat. I went a step further and used Arctic Silver (high end CPU / heatsink heat transfer grease used in PCs) between my resistors and the car body metal.
 

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Discussion Starter #3
If they are truly 15 watts ea (thats hard to believe if they are LED unless they use 3 - 5 watt CREE LEDs in each them) 25 watt resistors are fine.

The important part is they need to be mounted to metal to dissipate the heat. I went a step further and used Arctic Silver (high end CPU / heatsink heat transfer grease used in PCs) between my resistors and the car body metal.
If I were to switch my current resistors with 50w ones, would anything differ from an electrical point of view? Since they are in parallel with the light, I assume it will not draw 50w of power anyway, right?

Also where/how did you mount your resistors, especially with the thermal compound?
I couldn't find any holes to screw them into, so I just used some magnets to stick them on the body. Definitely not ideal...
 

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based on your first post, yes, there would be a difference. You would be going from 8 ohm to 6 ohm. This would be less resistance and your blink rate would increase in speed by 1/4 as compared between the 8 ohm and none at all. The watt ratings would not have any play on this installation since both are over the "quoted rated wattage" of your bulbs.

I used 50 watt, 6 ohm for my installation and I included 1 image at the end of my first post showing where I mounted them in the back. I also posted some videos showing the difference with/without so you can see what to expect with the 6 ohm vs nothing.

http://www.hyundai-forums.com/231-md-2011-elantra/161029-led-turn-signals-installed-bumper-reflector-mod.html
 

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based on your first post, yes, there would be a difference. You would be going from 8 ohm to 6 ohm. This would be less resistance and your blink rate would increase in speed by 1/4 as compared between the 8 ohm and none at all. The watt ratings would not have any play on this installation since both are over the "quoted rated wattage" of your bulbs.
No, the blink rate is controlled by a computer on the Elantra. The fast blinking is generated by the computer to tell the driver that a bulb is burned. So the resistance will not change the blinking rate, since it's not controlled by a thermistor.

The total watts rating must match the watts rating of the replaced bulb. So the total wattage of the LED bulb + the resistor should be around 27 watts for the system to work. There is no point in putting a 50W load in the system.
 

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Discussion Starter #6
Thanks for the insight guys.
Any idea if the front signals require separate resistors parallel with their bulbs, or if I can add one 50W resistor in the back to handle two front/back turn signal LED conversions?
 

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Thanks for the insight guys
More insight is below...

Any idea if the front signals require separate resistors parallel with their bulbs
No parallel... only in series.




Let me put it this way - the resistor it POWER resistor, what mean it consumes current to produce heat. Nothing more.
Then, rating of 50W means nothing, but how much power it can safely distribute without heatsinsk (I assume the resistor came with stock heatsink).

Now, to know what power you will produce you need to know resistance of the power resistor and voltage you will be passing through.

To know what resistor to use you need to compare the OEM set up.

Say you had 24W (for simple calculations) tail turn signal, 24W front and 12W side (if there was any). All together on one side you have 24+24+12=60W. That will need 60W/12V=5A.

Now, the resistance is 12V/5A=2.4 Ohm.

Your current setup say includes 12W and 12W (front/rear) and 6W side.
Total 30W, gives 2.5A, what gives 4.8 Ohm.

And here the BCM (or turn signal relay, which is RESISTANCE dependent) starts flashing faster, because it detects 4.8 Ohm, which is twice what should be.



What you need to do is to calculate what exactly power consumption is for each turn signal and choose the right resistor.

Power will be easy, if the LED is 15W, OEM was 27W (I think that is the correct number) you need at least 12W power resistor. Its resistance depends on the setup. Do not fall for the Wattage only.


Keep in mind that 50W resistor at 8 Ohm will behave differently at 12V and at 24V. See: 8 Ohm at 12V will pass 12/8=1.5A, but at 24V will pass 3A. Hence, its real power will be: 12Vx1.5A=18W, while at 24V is 24Vx3A=72W.

Do you see the difference now?

So 50W is not always 50W, but 8 Ohm stays 8 Ohm (unless you overheat it... which is another story - resistance increases with heat)

I hope that clears the air here...




Finally, one could fit electronic relay with adjustable resistance/flash rate, but those are not so common and currently most of the turn signal relays are hidden/buried in BCM or deep inside dash...
 

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No parallel... only in series.
This is so... wrong! You want to connect the resistor in parallel with the bulb. Put them in series, and you actually DECREASE the current going through the bulb instead of INCREASING the total load of the circuit...

igob8a said:
Any idea if the front signals require separate resistors parallel with their bulbs
As for the question about rear vs. front, you need a separate resistor front and back; because the car has two independent circuits for them
 

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This is so... wrong! You want to connect the resistor in parallel with the bulb. Put them in series, and you actually DECREASE the current going through the bulb instead of INCREASING the total load of the circuit...
You do not change the current going though the bulb, as there is no bulb anymore...

My mistake in this particular matter - you add LOAD resistor in parallel to add load, but NOT because you would limit power, as the LED is the limiting factor here.
In addition, the LED pulls less current than the resistor so adding resistor would not change much here.


Regardless: LOAD resistor should be added in parallel to the LED. All other info applies as how to find the minimal resistor.
 

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My mistake in this particular matter - you add LOAD resistor in parallel to add load...Regardless: LOAD resistor should be added in parallel to the LED.
You're in good company...I made the same faux pas. :wallbash: Brain fart, old age, take your pick.
 

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Wow... Bunch of rocket scientist here... well appreciated...

Where do I mount the resistor to in for the front, since there is no plug, but instead the twist cap.... I know i tapped the yellow wire on the main headlight hardness to run DRLs but I rather not poke and guess where the turn signal tap is....

Your help is appreciated.
 

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Wow... Bunch of rocket scientist here... well appreciated...

Where do I mount the resistor to in for the front, since there is no plug, but instead the twist cap.... I know i tapped the yellow wire on the main headlight hardness to run DRLs but I rather not poke and guess where the turn signal tap is....

Your help is appreciated.
Avante MD/Elantra Headlight Pin Array You can guinea pig off the main harness if you wish. Just make sure the resistor wires are long enough to go wherever you're going to mount them.
 

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OK, here is the MATH, People.

The Load Resistor is placed in parallel with the LED Bulb, NOT in series. That means from the Hot Signal Wire to Ground Wire. Try not to short out your Tail Lights or you will need Fuses. Lots of Fuses! And maybe a new Flasher...

The 1156 Incandescent Bulb is rated for 26.9 Watts and 12 Volts (What your car uses). This means it draws about 2.25 Amps (because P = I x E; see below; hence I = P / E). The LED Bulbs draws about 3 watts (or .25 Amps). The resistor is intended to fool the flasher into thinking there is an old incandescent bulb there. In theory the resistor should draw 2 Amps. Since the formula is R = E / I (below) or 12 Volts / 2, this resistance should technically be 6 ohms and will generate Power of P = E x I or 12 Volts x 2 Amps = 24 Watts. If you use a 25 Watt resistor, it will run hot and may burn out. Hence the 50 Watt recommendation.

But... Consider that 8 Ohms will still draw Current of (I = E / R = 12 / 8) 1.5 Amps, hence the Wattage would be 18 Watts. You should be good with a 25 Watt resistor. The 1.75 Amps total may be enough to satisfy your flasher circuit (no guarantees here). Some flashers will work fine with slightly lower loads. If you have TWO Bulbs being replaced. You likely need to add another resistor. Again, 8 Ohms might work but 6 Ohms gives you the rated current for the 1156 Incandescent Bulb.

I am looking to install a new flasher rated for 1 to 10 bulbs in my Beastly Old Motorhome which has 4 - 1157 bulbs on either side (Each rated at 27 Watts). That will mean 1 Amp through the 4 LEDs meaning I am 1.25 Amps short of the load expected by the flasher for that 1 Incandescent Bulb.

  • Since E = I x R then R = E / I; hence R = 12 / 1.25 = 9.6 Ohms. This is the load needed to make up the shortfall.
  • I will try 2 - 8 Ohm (10 Watt) Resistors in series giving me 16 Ohms (20 Watts) to draw .75 Amps (12 Watts load).
  • If that doesn't work, add another series set for 8 Ohms (40 Watt) combined drawing 1.5 Amps (24 Watts load).
The bad news about these resistor loads is that, in the unlikely event that an LED Bulb burns out, the circuit may think it is still OK and not warn you of the fault. Best check your lights every so often so you don't draw the attention of those guys who hand out tickets.

FYI;
The Magic Formulas are E = I x R (yielding R = E / I) and P = I x R
-Where 'V' is the Voltage; 'I' is the Current in Amps; 'R' is the Resistance in Ohms; 'P' is the Power in Watts.

Also, For Your Edifications: The 3 Watt LED Bulbs I am using throw 800 Lumens whereas the 27 Watt 1157 (Thomas Edison Special) only throws about 402. LED's are Twice as Bright and Consume One Ninth the Power. If only we didn't need those pesky resistors to steal away this LED magic.
 
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