Diodes do have a resistance and voltage-drop. Just not perfectly linear. Forward-voltage (Vf) to amps flowed is in the XM-L datasheet and you can derive a resistance from that.
Originally Posted by unterhausen
Regulators actually don't change voltage in order to maintain current. Linear regulators like the LM317T uses a voltage-divider and shunt-to-ground in order to provide a fixed current output. It takes whatever voltage is fed to it and splits it into two legs (Y). For example, with a 6v input:
LEG1 = 3.35v output to LED in order to flow 3000ma
LEG2 = 6-3.35v = 2.65v @ 3000ma is shunt to ground
The power-source in this case (batteries, transformers, etc.) will be outputting 6v @ 3000ma or 18w total. This is then split by the regulator to send 3.35v @3000ma to the LED (10w) and 2.65v@3000ma to ground (8w). Based upon the total consumption, 18w, and 10w actually used, you can consider this regulator to be 55% efficient.
More modern regulators, such as the AMC 7135 PIC, sends full-voltage across to the driven LED. Current is regulated with PWM, like a fuel-injector. There is no shunt-to-ground to dissipate the un-needed voltage that would've resulted in too much current flowing. A 13v battery can actually run a 3.35v Vf LED @3000ma by driving it at 25% duty-cycle. When battery-voltage drops near 3.35v, the regulator will use 100% duty-cycle (direct-drive) and connect the power-source directly to LED. In this case, you can consider this regulator to be 100% efficient.
If the 7135 is configured in a buck-boost configuration, it can even raise the supply voltage if it's less than the Vf of the LED. Common department-store flashlights (Husky from Home Depot), that uses 2x 1.5 AA or D batteries in series for 3v supply uses a boost circuit to send a higher 3.35v to 4.5v to the LED, like a P7.