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I have an idea for a hack I'd like to build, but I need some help figuring out the right transformer. Let me know if there's a better place to post this sort of request.

I have some exterior LED path lights that work on 4 C-Cell batteries each. There are 6 of them, so having to replace the batteries is relatively expensive and a bit of a pain. Looking at the lights, I see that I could easily solder wires to the battery contacts and replace the batteries with a transformer. I could run a low voltage wire indoors to a transformer and never have to worry about batteries again. My problem is that I'm not sure what transformer to get.

I did some simple math but I have no electronics background, so I would appreciate getting guidance from folks that have more knowledge about this stuff.

Here's what I know:

Each light takes 4 C-cell batteries in line. That is they are stacked one on top of the other with positive terminal touching negative. I think that means I'd need to add the voltages together meaning 4 x 1.5V = 6V required.

I have no idea how much power each light draws (specs don't say anywhere) and the math I did doesn't seem to make sense. I know that alkaline batteries are normally rated for about 8000 mAh. The manufacturer of the light claims that the lights last about 50 hours before needing the batteries replaced. 8000 / 50 = 160 mAh draw. That seems really low to me. Is 160 mAh a reasonable amount for an LED light (supposedly 80 lumens output if that helps)?

So if I multiply that out 6 * 160 = 960 mAh required. I want to be able to add a few lights so let's double that (so just under 2A).

Does that mean I need a 6V transformer that can do 2 amps? That seems really low to me, so I'm hoping someone with electronics knowledge can provide some help figuring out what I really need.

Any guidance or nudges in the right direction would be greatly appreciated.

Thanks.

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  • \$\begingroup\$ mAh is a measure of charge. mA is a measure of current. \$\endgroup\$ – Hearth Nov 5 '18 at 1:00
  • \$\begingroup\$ And it's not low, in fact! LEDs are remarkably efficient, especially if what you're used to is incandescent lights, aka bright heaters. \$\endgroup\$ – Hearth Nov 5 '18 at 1:02
  • \$\begingroup\$ @felthry Thanks. So I'm going in the wrong direction then? how would I figure this out? \$\endgroup\$ – Ben Riga Nov 5 '18 at 1:03
  • \$\begingroup\$ You're not going in the wrong direction. 8000mAh/50h = 160mA. You just forgot that you're dividing by hours. Though you might have some difficulty using a transformer--these lights want DC, and transformers only work with AC! \$\endgroup\$ – Hearth Nov 5 '18 at 1:04
  • \$\begingroup\$ @Felthry hmm. I was able to find something like that on amazon for pretty cheap. Wouldn't something like this work: amazon.com/inShareplus-Mounted-100-240V-Transformer-Connector/… \$\endgroup\$ – Ben Riga Nov 5 '18 at 1:08
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160mA at 6V is about 1W that would be easily suffucuent for a fairly bright LED path light.

If you're not equipped to measure the current consumption of the lights maybe go for a 2A 6V supply that should be plenty if you estimate is in the right ballpark.

You could probably even use a 5V 2A "phone charger" as battery powered devices are usually designed to operate from as low as 1.1V per cell so 5V should be plenty.

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