# Converting battery powered LED pucks to a DC power source

I have some (white) LED pucks that each run on (3) AAA batteries in series, or 4.5v. These are the kind of pucks that you would put underneath a kitchen cabinet, push on/off. Each puck has (3) white surface mount LEDs wired in parallel. I don't have data sheets available, and so I don't know the forward voltage drop for the LEDs, and also don't know how the equation changes when there are (3) LEDs in parallel. So the best I can do is measure the battery current thru the puck when using the (3) AAA batteries. The measured current is 62ma. I have a 12vdc power supply in the project (which powers other 12v things), which I am guessing should first be run through an LM regulator to get the voltage closer. I did do a test using 5v directly from a USB cable (no current limiting resistor), and I fried one of the LEDs. So, I know that I need minimally a current limiting resistor. Can someone help me calculate the value of that resistor?

• A battery is a DC power source. Apr 9 at 3:59
• Chances are battery voltage with almost fresh 3S AAA is about 4.2 V@62 mA (measure!), and $V_f$ is around 3.2…3.7 V (again). I'd start with an additional 22 Ω for 5 V or 150 Ω for 12 V and proceed as suggested by sai. With three pucks lighted in lockstep, try putting them in series for 12 V. (Consider explicating in your question: What is an LM regulator?) Apr 9 at 7:15
• Thanks, I will try the solution starting with 22 Ω. The (3) LEDs in parallel are in a single puck (surface mounted). Then there are 4 pucks, also in parallel, for which I would need 18v if in series. Apr 9 at 13:56
• @greybeard - Sorry, don't know where the LM came from, somehow I thought there was an M in the prefix - but I meant a 5V regulator something like the L7805 Apr 9 at 14:14