Click to expand...The heating is only dependent on the power/amplitude, not on the wavelength. One watt of IR will heat up a black object as much as one watt of visible light. Then there's the issue of waste heating of the LED itself (rather than heat projected in the beam), that has to do how efficient your LED is, and how hard you're driving it. Greater efficiency means less waste heat. I believe IR are usually the most efficient LEDs, followed by blue.
The heating is only dependent on the power/amplitude, not on the wavelength. One watt of IR will heat up a black object as much as one watt of visible light. Then there's the issue of waste heating of the LED itself (rather than heat projected in the beam), that has to do how efficient your LED is, and how hard you're driving it. Greater efficiency means less waste heat. I believe IR are usually the most efficient LEDs, followed by blue.However, driving an LED harder than recommended specs causes efficiency to go down, and temperature to go way up -- potentially damaging the LED, especially if it isn't heatsinked. Also, if an LED heats up, its resistance starts to go down, and it will draw more and more current until it burns out -- this is called thermal runaway. One thing to keep in mind is that with LEDs, small changes in voltage lead to big changes in current. Supplying an LED with a certain voltage is not sufficient -- instead, test out the current going to an LED with a multimeter, and wire in a resistor to reduce to the current as needed (do a google search for LED resistor caculators to find out which value is needed). Also, I'm not sure what sort of LED this is, but I'd also recommend attaching this to some sort of copper/aluminum heatsink if possible.
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