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:

  1. 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.
  2. 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.

  • \$\begingroup\$ The number of lumens per watt is a function of the LED color and the quality or efficiency of the LED, itself. You don't mention the color. And even if you did, there's a range of efficiencies that are possible. You don't know what the circuit board does and we don't have a picture of it, either. The battery is 1.5 V, so it's 4.5 V total, not 3.5 V, for three of them. So I think you need to perform some testing for us, before we can offer any useful advice. You will need to characterize the LED. It would be best if you could do this at 3 different currents. Do you have a "bench" supply? \$\endgroup\$
    – jonk
    Oct 25, 2020 at 15:09
  • \$\begingroup\$ "Max. current through the battery bank should be 1000mA, as per the spec document" where does it say that? \$\endgroup\$
    – Finbarr
    Oct 25, 2020 at 15:40
  • \$\begingroup\$ You measured 3.5V on the battery when it is powering the LED because the 4.5V Chinese battery is almost dead. You measured 3V on the LED which is its forward voltage drop when it is powered from the almost dead battery. \$\endgroup\$
    – Audioguru
    Oct 25, 2020 at 17:54
  • \$\begingroup\$ Thank you all for the replies. jonk, having the difficulties to carry this out myself and my limited knowledge is what prompted me to ask the question. I guess 3.5V I measured comes from the batteries not being at full capacity. I can get to a benchtop supply in a couple of days and do the tests. Finbarr, that's my understanding from the second table on page 2 of 6. Audioguru, noted. The funny thing is, even at 3.5V this thing shines really bright! \$\endgroup\$
    – mk1138
    Oct 25, 2020 at 22:32

2 Answers 2


3V on the LED is what you'd expect from a white LED.

Forget about the 9V supply. Instead use a USB "charger" ie a 5V 1A power supply. This should be much more useful as you'll be able to plug the light on a powerbank, on a cellphone charger in the car, etc. Also the resistor power dissipation will be manageable.

Cut a USB cable and connect to the lamp's PCB on the battery terminals. 3 AAs is close enough to 5V so it should work.

Otherwise you can connect the LED directly to the 5V with a resistor. Assuming the "500 lumens" are "cheap lumens" let's go for 3W. So that's 1 Amp at 3 volts on the LED. With a 5V supply, that leaves 1.9 volts on the resistor and 0.1 volts on the cheapo USB cable, total 5V. A 1.9 ohms resistor will do the job. It will dissipate close to 2 watts, so get a 5 watts model. It will get pretty hot. With a 9V power supply you'd need a 6 ohms resistor, and it would dissipate 6 watts, that would be a problem.

  • 1
    \$\begingroup\$ This is super useful, thanks. I'd never thought of using USB cable as a power supply. That's actually what I use to power my Arduino and 5V is what Arduino delivers on Vin/Gnd connections. I'm just bit vary of bringing 5V directly to the PCB. It is a no-name lamp after all. I'll try to go with the resistor and direct connection to the lamp. Another thing I was considering is simply getting a 4.5V adapter with a DC jack and run the whole thing that way. "Assuming the '500 lumens' are 'cheap lumens' let's go for 3W. So that's 1 Amp at 3 volts on the LED." - this makes total sense! \$\endgroup\$
    – mk1138
    Oct 25, 2020 at 22:28
  • \$\begingroup\$ Thanks;) Yes you can get a 4.5V adapter although it'll probably work on 5V. You also get to keep the dimming/disco flashing modes if the light has any... \$\endgroup\$
    – bobflux
    Oct 25, 2020 at 22:59

Me, I would wire an ammeter in series with the LED. Then measure the amps.

Then, I'd get a constant-current driver that runs that exact current (at that general range of voltage).

But then, I'm presuming I have ordinary LEDs which are not ohmic (i.e. do not obey Ohm's Law or random websites in fancy fonts). If I had super-magic ohmic LEDs that mysteriously appeared in random no-name products off the boat, then yeah, I'd size a resistor lol.

  • \$\begingroup\$ Will try, however my question arose from my difficulties to measure the current. \$\endgroup\$
    – mk1138
    Oct 25, 2020 at 22:29
  • \$\begingroup\$ @mk1138 since 1000ma is the likely maximum battery current, that is well within the range of many DVM's onboard ammeters. So you need to insert the DVM in series with the lines from PCB to LED. If that isn't feasible, you'll need to insert an ammeter shunt (resistor of very low but precisely known value). \$\endgroup\$ Oct 25, 2020 at 23:00

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