I have a GreenLux E27 7W light bulb (AC 100-240V, 60 Hz) and tried to get a minimal light output from it by connecting ~2.40V 2 series connected AAA batteries. The bulb consumed ~60 mV (voltage dropped to ~1.80V) in several seconds and did nothing.

I was also told at the store that it is impossible to power an AC light bulb with DC batteries, even if I would not bother to series connect dozens of 1.5-3V batteries and may not need the full 7W output.

So am curious why so, if power (in Watts) = V*I, so is directly proportional to voltage.

  • 1
    \$\begingroup\$ Incandescent lightbulb : yes. Your DC supply must be at its AC rated voltage for full brightnesss though. Economy bulbs with electronic controllers or ballasts? Who knows? It all depends on the electronics. Some LED bulbs can be dismantled and the AC supply thrown away, to run on 12V DC. \$\endgroup\$
    – user16324
    Nov 11, 2020 at 21:51
  • \$\begingroup\$ The bulb consumed ~60 mV (voltage dropped to ~1.80V) That is not how electric power works. What you mean: the battery voltage has dropped 60 mV. If the lightbulb was designed to work on 2.4 V, why would the manufacturer list "(AC 100-240V, 60 Hz)" on the bulb? Being able to use it on 2.4 V DC would be a nice feature. The circuit in the lightbulb need a high voltage of at least 100 V AC. It has to do with how the circuit is deigned. Go watch BigClive on Youtube to learn how LED bulbs work: youtube.com/watch?v=RESvg1LvgC0 \$\endgroup\$ Nov 11, 2020 at 21:53
  • \$\begingroup\$ Yes, it is a LED light bulb. I connected manually: the 2 voltage terminals of the battery to the metal side and bottom of the bulb, fixed by tape. The light bulb seems not an economy class. I was curious if I could get at least some mW (or 7 photons) out of powering with DC batteries. \$\endgroup\$ Nov 11, 2020 at 22:00
  • \$\begingroup\$ "Being able to use it on 2.4 V DC would be a nice feature." - if such a circuit will be designed, let me know. \$\endgroup\$ Nov 11, 2020 at 22:09
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    \$\begingroup\$ "AC 100-240V" means it has a universal supply inside, can handle AC power anywhere in the world. But you gotta give it at least 100V (AC or DC) to have any hope of it lighting up correctly. THis also means that you're not going to be able to control the brightness by controlling the applied voltage. It will (in theory) light up the same regardless the input voltage (so long as it's in the 100-240V range). Your solution is to get a different light bulb.... \$\endgroup\$
    – Kyle B
    Nov 11, 2020 at 22:16

1 Answer 1


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Figure 1. A typical arrangement for a cheap capacitive dropper power-supply LED bulb. Image source: Dimmers for LEDs.

You have two likely problems:

  1. The lamp will contain multiple series-connect LEDs which each have a forward voltage of several volts. Your batteries haven't enough voltage to get them to turn on.
  2. Many of these lamps use a capacitor (C1 in Figure 1) to drop the AC voltage without generating a load of heat. (R1 is a very high-value resistor just to discharge the capacitor when the power is turned off.) As the capacitor symbol suggests, it is a pair of plates with a gap between them so, while it conducts AC to an extent which increases with frequency, it blocks DC completely.

The bulb consumed ~60 mV (voltage dropped to ~1.80V) in several seconds and did nothing.

No, the battery lost that voltage due to it's internal resistance. There was only 1.8 V left to be applied across your lamp. That suggests that either something in the lamp is drawing some significant current or that the lamp is drawing a small current but that your batteries are flat. Without knowing the innards of that lamp it is not possible to be more specific.

  • \$\begingroup\$ I have the impression some bulbs even skip the bridge rectifier in favour of antiparallel LEDs. \$\endgroup\$
    – Janka
    Nov 11, 2020 at 22:50
  • \$\begingroup\$ @Janka Hm, I have a feeling that a rectifying bridge will be far cheaper than having twice the number of LEDs \$\endgroup\$
    – pipe
    Nov 11, 2020 at 23:06
  • \$\begingroup\$ It's the same number of LEDs. You have to adjust C. \$\endgroup\$
    – Janka
    Nov 12, 2020 at 0:13

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