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load resistors for LED turnsignal retrofits

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  • load resistors for LED turnsignal retrofits

    I measured the factory current draw of a standard 7440 bulb. The measurement at 12.5V was 1.8amps.

    I built an LED array that draws 600ma at the same voltage. I am still 1200ma short of the draw that the factory bulb does.

    Since I need to draw 1200ma MORE, can't I just use an appropriate resistor in parallel to draw this much more current?

    Every "LED load resistor" I've seen is 6ohms, which will put me well over on the current draw compared to factory.

    According to my math, 12.5V divided by 1.2amps should be achieved with about 10.4 ohms. Can't I just buy a 10 ohm resistor of sufficient wattage? The draw from the 10 ohm resistor, and my 600ma LED array should together put me at the factory bulb current draw levels- preventing "hyperflash".

    Has anyone tried this like this to prevent "excessive" draw?

    I was looking at these: https://www.mouser.com/ProductDetail...iNVzEC1WdQs%3D


  • #2
    For low current draw (like turn signals), a regular resistor may be ok. Another reason why it is in your favor is that turn signals only run intermittently. I did just that for an auxiliary turn signal many years ago; it still works.

    Or, you can get 3 of the 6ohm resistors. Put two in parallel, then 1 in series with them. For the parallel resistors, you get 1/ (1/6 + 1/6) = 3ohms. Then, put the 3rd one in series to get 3 + 6 = 9 ohms. Maybe it will be close enough? Or get 7, put pairs of 3 in parallel (making them effectively 2ohms), put them in series, making the sum 4ohms, and then put a lone resistor in series to get 4+6=10ohms
    Jul 2012 ROTM (3-way quad headlight) ; Sep 2015 ROTM (custom muli-lens 7" fogs)

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    • #3
      Originally posted by satrya View Post
      For low current draw (like turn signals), a regular resistor may be ok. Another reason why it is in your favor is that turn signals only run intermittently. I did just that for an auxiliary turn signal many years ago; it still works.

      Or, you can get 3 of the 6ohm resistors. Put two in parallel, then 1 in series with them. For the parallel resistors, you get 1/ (1/6 + 1/6) = 3ohms. Then, put the 3rd one in series to get 3 + 6 = 9 ohms. Maybe it will be close enough? Or get 7, put pairs of 3 in parallel (making them effectively 2ohms), put them in series, making the sum 4ohms, and then put a lone resistor in series to get 4+6=10ohms
      The resistor I provided a link for is 10 ohms already, and they aren't that much money. I don't see why it wouldn't be any different but I didn't know if anyone had tried this before.

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