Electronic – How to feed the right current to a superbright LED with a 3.7V LiPo battery


I'm trying to do something that seemed simple at first but is proving to be complicated (shocker, right?) I want to power a super bright LED (Luxeon K2, Cool White, 350mA. It's discontinued, but the datasheet is still the first hit on Google) using a 3.7V LiPo battery from SparkFun. The instructions I have for the LED tell me to regulate the voltage with an LM317 voltage regulator.

Problem being, the LED has a forward voltage of somewhere between 2.8 and 3.4V, and the LM317 appears to eat up too much voltage to make that possible. I don't want to wire 2 batteries in series because my enclosure is pretty small, and there's no space for another battery.

So I found this charging/step-up breakout board that can bump me up to 5V. I'm extremely new to electronics and this is my first project where exact voltage/current/resistance are an issue. My question is threefold:

  1. If I'm using the breakout board to regulate my voltage, do I even need the LM317? Or can I just use resistors to get the current to where I want it?

  2. How can I know what the current will be when it comes out of the breakout board? The datasheet for the voltage regulator on the board seems to say I can program the output current to anything between 15 and 500mA, but I don't understand how to do it.

  3. Am I right about needing the step-up? Or could I potentially just run this with the battery and the LM317?

I'm doing this project in order to learn, so I would prefer answers that show me how to figure this out for myself as opposed to just doing it for me.

Best Answer

I would simply find a voltage regulator or even better, a current regulator that has a low drop-out voltage. In that way, you don't need the sparkfun board and you can control your LED without having to boost it and then regulate it back down. That seems like a lot of effort and wasted energy going through two stages instead of one since your battery can handle the voltage requirements of the LED.