Help with LED power

Hi all, 1st post here and I'm hoping I can get steered in the right direction....I'm doing a low budget DIY project and I want to make sure I don't go the wrong direction.

I have purchased a CREEXPE2-750-1 5000K single LED board for a project I'm doing, I arrived at this particular LED purely because the one I cannibalized from a flashlight seemed to be around this type. I'm planning on powering the LED with 2 18650 Li ion batteries in parallel. From what I have read I should be able to do this no problem at all if I purchase a "Buck puck" but I would prefer to use a PCB strictly because I'm still in prototype phase and I'm going to need some flexibility, I have also seen that I could use a resistor in line to regulate power but this apparently falls into a "not recommended" area.

I'm not an electronics guy but I've been trying to educate myself online as best I can in what is considered safe and the proper way to achieve this but I'm overwhelmed here. I'm hoping that you guys can help me sort thru all of the info and point me towards the correct path to success.

Thanks in advance for any and all help and if I didn't provide enough info please be gentle and let me know what I missed.

John

Simple resistors obey Ohm's Law: the current through it is proportional to the voltage you apply across it, and the power dissipated is current × voltage.

LEDs are semiconductors, so that rule doesn't apply. Their current draw is very non-linear, so they are very sensitive to the voltage you apply. They are best driven with a constant-current driver, which is a circuit that regulates its output voltage to maintain a constant current. The driver you choose needs to be matched to both recommended current for the LED(s) you are driving and the input voltage.

You could approximate a constant-current source with a series resistor, which makes the LED/resistor combination a bit more "Ohmic", but it's a trade-off between efficiency and voltage-insensitivity. The larger the resistor, the greater proportion of the voltage is across it, and the more power is wasted as heat. For a 2.2V 10mA red indicator LED in an automotive application with Vs of 10-14V, that may be acceptable: the resistor would be dropping about 10V at 10mA = 100mW and giving a ~17% efficiency overall, excluding the LED's own inefficiency.

For higher-power LEDs, the inefficiency drives higher waste heat, which gets unacceptable quite quickly.

The Cree you've chosen has a typical Vf of around 3.0-3.1V, so you'd need a driver that will convert the full range of the varying voltage of your nominal 3.7V battery pack to that sort of level whilst outputting up to 1000mA if you want to run it at full power.

Drivers can be "buck" (output voltage lower than input), "boost", (output higher than input) or both in one. Your proposed "buck puck" is the former, which makes sense given the numbers above. You just need to make sure it can still output ~3.1V at 1000mA at the lower end of your batteries' range when nearly depleted.