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I've got a wee challenge on my hands. My ex-girlfriend gave me this light as a b-day present 3D printed. I've been told that the LED is a 10mm Ultrabrite LED.

See the picture attached for the battery pack at the bottom. How would I go about changing it from running off batteries, to running off the mains? She said her uncle who made it says I'll probably need a '14v power limiter'. Haven't touched a soldering iron since high-school; willing to do so.

Cheers for any input - its much appreciated.

Pictures Of Lamp

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    \$\begingroup\$ I'd grab one one of the DC wall adapters scattered around and connect instead of the batteries. No soldering is necessary. \$\endgroup\$
    – Eugene Sh.
    May 18, 2021 at 13:17
  • \$\begingroup\$ that looks more like 24V to me. 24V wall plugs are fairly easy to find. perhaps it's actually some sort of automotive LED lamp in there \$\endgroup\$
    – Jasen Слава Україні
    May 18, 2021 at 13:40
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    \$\begingroup\$ Why is one 3.3V LED powered by a 24V battery?? \$\endgroup\$
    – Audioguru
    May 18, 2021 at 16:12
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    \$\begingroup\$ Is there only the one LED inside the lamp as seen in the third picture? Why are there so many wires exiting the base in the center? Do they connect to something else? \$\endgroup\$
    – Theodore
    May 18, 2021 at 19:37

1 Answer 1

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For one blue/white 10mm LED just use 3.3V or drop down with a Schottky diode to 3.0 for <half brightness roughly . Use a series 1 ohm to test current for 20mV +/- 50% (or more)

Where you get 3.3Vdc is up to you. E.g. 2V drop from 5Vdc / 20mA = 250 ohms. Higher V means bigger power resistor. [ohms & watts]

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  • \$\begingroup\$ Drop what? The OP is asking about running it from mains. \$\endgroup\$
    – Eugene Sh.
    May 18, 2021 at 13:31
  • \$\begingroup\$ Yes I assumed he knows the difference \$\endgroup\$
    – Tony Stewart EE75
    May 18, 2021 at 13:37
  • \$\begingroup\$ doesn't eplain how it's powered by two 12V batteries in series. \$\endgroup\$
    – Jasen Слава Україні
    May 18, 2021 at 13:40

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