Leds Wired Led
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having resistor trouble wiring leds?
I am working on a motorcycle project and keep having trouble. I am wiring 5 leds in my peg brackets, wired in series. Now I am no soldering professional, but I got one set done and it worked, and then after a week of not touching it, it stopped working and one led is out. I want to wire them individually. I have checked many resistor calculators and most say to use a 3watt 33ohm resistor per led. I want to shrink tube the resistors when finished.
It id getting more expensive the more leds die on me and having to change them out after they are mounted. The leds are red, have a 2.2 forward voltage, 300mA.
Should I use a resistor higher than 3 watts? Will I get the same brightness and less heat?
These are the leds I have- I saw a vid on youtube of a guy who did a similar project with the same leds.
http://cgi.ebay.com/5-PCs-1W-8mm-140-StrawHat-RED-LED-210-000mcd-300mA-/370403890416?pt=LH_DefaultDomain_0&hash=item563dcad4f0
The charger likely puts out about 14.5volts. If you had all 5 in series with no resistor, that is about 11V total. Pretty likely to burn out one of the LED's, sooner rather than later.
I would leave the 5 in series. If you put an 11.7Ω or greater resistor in series, with
E = I*R, 14.5 - 11V = 0.3 * R ==> R ≥ 11.67Ω
Power = I²R = 0.3² * 11.67Ω = 1.05W
I would measure the voltage of the bike with the engine running first to verify voltage. If it is a little lower, you can put resistor a little lower. OK to use a larger resistor, but it will reduce brightness slightly. Also, can string resistors together to get value you want, you can rarely find odd values like this. Measure the resistance of the resistor(s) before you solder them in, then verify voltage drop across resistor when you are done to assure the LED's are not getting too hot. Use E = I*R = 0.300A * 11.67Ω = 3.5V. If the drop is larger than this, too much current, need a larger resistor. If you leave the 5 together, when the bike is off, relatively large drop in brightness.
With single diode, E = I*R, 14.5 - 2.2V = 0.3 * R ==> R ≥ 41Ω
Power = I²R = 0.3² * 41Ω = 3.7W
That is a pretty good sized resistor.
With single diode, relatively little drop in brightness when engine off.
I would put 2 or 3 in series. For 2:
14.5 - 4.4V = 0.3 * R ==> R ≥ 33.67Ω
Power = I²R = 0.3² * 33.67Ω = 3.02W
14.5 - 6.6V = 0.3 * R ==> R ≥ 26.33Ω
Power = I²R = 0.3² * 26.33Ω = 2.37W
Good luck!
PS, power used by resistor is determined by its resistance and the circuit it is in. The power rating of resistor is how much heat it can dissipate without smoking. All resistor of the same resistance work pretty much the same until the one with a low power rating overheats too badly and melts. I say pretty much because they do change resistance a tiny bit as they heat up.
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