Rank: Newbie Groups: Member
Joined: 11/24/2009 Posts: 1 Points: 3 Location: Auburn
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Im trying to make my own LED taillight for my motorcycle and Im having a hard time figuring out what resistor to use and what type. I have 72 LED's run in parallel that have a forward voltage of 3.2 to 3.8 volts. Max continuous current is 30mA. Max peak forward current is 75mA. The motorcycle obviously runs on 12-14.5 volts. I wanted to try and do this as simple as possible. I have 2 power wires and a single ground wire. The running light is constant power when ever the key is on and the brake light power is controlled by two switches, the handle bar control brake switch, and the foot control brake switch. I just want a little guidance since Im obviously not a master at this. Thanks for any help in advance!! Brandon
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Rank: Newbie Groups: Member
Joined: 2/2/2010 Posts: 1 Points: 3 Location: Edmonton, AB
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Why r u using so many LED's?
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Rank: Newbie Groups: Member
Joined: 2/28/2010 Posts: 5 Points: 15
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@yamyvmax
This is approximately how to do it;
Lets say one uses a 220 ohm resister for 5.0Volts with a 20ma led. You want ten percent more current at 30ms. So lets say that makes 200Ohms for a 30ma led at 5 volts. Midrange lets say your voltage is 13VDC. We divide 13 by 5 and get about 2.6 times for the operating the voltage. This means we need to limit the current with a 200 ohm* 2.6 times or about 520 ohm resistor. So one 30ma led at 13. Volts uses a 520 Ohm resistor. It turns out that both 510 Ohms and 560 Ohms are standard resistor values so we will pick the 510 Ohms at lets say 10% tolerance. Ok…But now we have to calculate the wattage or power dissipation of the leds. P=E*I 13volts times 30ma = 390milliwatts each, so lets be on the safe side and round that to 400mw dissipation (almost .5watt) per led. The total power is highly dependent on the number of leds you have in parallel. We now multiply that 400mw by 72 leds to get 28800mwatts or 28.8 watts or about a 30Watt disipation resistor. So you would need a 510 Ohm 30watt 10% tolerance (probably) wire wound resistor for 72 30ma leds at 13VDC.
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I can recommend that you use at least two leds in series each, rather than putting them all in parallel. This will reduce the size of your ballast resistor in terms of wattage probably to a more manageable 15Watts. To reduce the size of the resistor further you may want to consider a dynamic power regulation. The number of leds groups will now be 36 of two leds in each. Note the amount of power limiting in the resistors with the leds in series goes down because the led group do their own current limiting.
A 30watt, or even a 15watt constant current draw is a bit much for a motorcycle though it may be about right for an automobile. You may want to consider finding the correct resistor value for three leds or even four leds in series in each group. This would probably cut the resistor wattage in half to about 5 watts or so. If you use led grouping you cannot use a single low value resistor without guaranteeing that a certain number of leds are always present. If you can’t, then break the single resistor into large resistance value small wattage resistors per group.
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These are all rough calulations but I think if you go through them a few times your self you will have in mind the ideas that you will need to design things in an sucessful experimental manner...
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