Originally posted by INRETECH:
Ohms law can be a wonderfull tool, but its just a start - high current on semiconductors can lead to different results - always have a current meter handy
<font size="2" face="Verdana, Arial">Well said! Ohm's law does _not_ apply very well to diodes....well, it really does, but you have to treat R as a variable which changes with V and with temperature.
Zinje's post above points out to one of the confusing aspects of powering LEDs. 'Just what is the proper voltage for them anyway?'
The datasheet will show a range of Vf, which for Luxeons goes from something like 2.8V to 3.99V. This _DOES NOT_ mean that you could power any particular Luxeon at any voltage in that range.
Each Luxeon will require a particular Vf to supply normal operating current. The Vf of each Luxeon will fall _somewhere_ in the datasheet range. Some Luxeons only need 3.2V to reach full power, others need 3.6V. If you connect a 3.2V Luxeon to a 3.8V supply, then you will toast it.
Finally, as Inretech noted, people are not driving the Luxeons with 4.5V. They are using 3 alkaline cells with a nominal voltage of 1.5V each, for a total _nominal_ voltage of 4.5V. But when you put alkaline cells under load, their voltage drops and the actual voltage delivered to the Luxeon is considerably lower, in the 3.5 to 4V range (depending upon the cells and the Luxeon).
With these same 'direct drive' flashlights, using 3 NiMH cells, with a nominal total voltage of 3.75V, you actually get more current flowing through the LED, which quickly toasts.
As StuU noted, the voltage dropped in the resistor is wasted in heat production. However the other side of this coin is that the more voltage dropped in heat production, the less sensitive the system is to changes in voltage.
Good luck!
Jon