Resistor & LED Power Dissipation Calculations in DEPTH

WBahn said:

Generally you want to work with percentages (though hp1729 is choosing a value based on drive capability).

If you have an LED rated for a maximum current of, say, 20 mA, then it is common practice not to exceed about 85% of this in operation. The LED will then last virtually forever. A lot of people will run it at 50%. A change in current by a factor of two in an LED is often hard to notice unless they are right next to each other.

Also, one thing I didn't see mentioned is that the thing that makes running LEDs directly in parallel a bad thing is a phenomenon known as thermal runaway. Yes, when you hook them all up there will be some differences in the forward current at the applied forward voltage, even if all of the LEDs came from the same wafer. So one LED will be carrying a bit more current than any of the others. As the LED junction warms up, it's forward voltage drops at the same current or, saying it in a way that is more applicable to this situation, the current will increase at the same forward voltage drop. Since the LED with the most current is dissipating the most power, it will also be the one that heats up the most quickly and so its current will increase the quickest. But the total current into all of them is being held pretty must constant (due to the single resistor) so the current in the others will decrease. This crowding out effect results in one LED hogging most of the current. If you only have two or three LEDs in parallel, it is quite possible that having one hog all the current will not lead to its rapid failure. But if you have ten or twenty in series, then it will likely be so overloaded that it burns out very quickly. That starts a chain reaction in which another LED will become the weak link and burnout even quicker. And so on and so on.

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