Not an electrical expert so please bear with me.
I bought a no-name brand LED lamp that I would like to use with my 9V DC adapter as power source. The lamp uses 3 AA batteries in series. Batteries supplied with the lamp are HDX brand and I found their specification via the seller's site. The only technical data I have for the lamp is that it's 500 lumens.
I opened the case and saw that the circuit leads from batteries/power supply to a PCB, and then from PCB via separate LED+ and LED- contacts to the light itself. There is also a push button on the PCB to turn the lamp on or off. So what I want to do is to take out only the LED light, without the PCB and use it with my 9V power supply.
From my understanding, if I want to change the power supply from 3.5V (AA batteries in series) to 9V (power brick), I need to add a resistor. To determine the resistor's capacity I need to know what the current through the LED is. I was able to measure only the voltage, as the whole circuit is already soldered. I'm reading 3.5V on the battery bank/PCB and 3.0V on the LED. From this I understand that I have a 0.5V voltage drop on the LED. Is this correct?
Next, I have two ways of approaching it:
- Max. current through the battery bank should be 1000mA, as per the spec document. In this case, R = (9 - 0.5) / 1 = 8.5 Ohm. This looks oddly low to me and this is another thing I'm not certain of.
- If I assume that 500 lumens is approx. 5W, and following rule of thumb from this thread that 5W will give me approx. 1500mA, I arrive at a similar value of 5.7 Ohm. Again, I'm really not sure if this makes sense.
How would you approach this problem? At the moment, I only plan to turn the light on occasionally, but I'd like to think about adding a heat sink and have it run continuously in the future.