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I have a magnifying glass LED lamp that I lost the power adapter to. There are no power requirement markings anywhere on the device, nor do I have the box/manual.

Short of trying to find the device's specs online (easily doable), I'm wondering if it's possible to determine what power rating of an adapter it would use based off the type and number of batteries it takes? The battery compartment takes 3x AA batteries.

Would that simply mean I need an adapter of 4.5 volts? Or is the rating between the two power supplies not as simple of a correlation as that?

I understand that amperage is "pulled", but would I also just add the amperage ratings of AA's together?

And finally, what about polarity? Is it possible to determine what adapter would have the correct polarity (short of looking it up)?

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Batteries can only supply a given amount of current. You could use an adapter that supplies the same voltage and exceeds the current capacity of the batteries to properly power the device.

I would recommend using a current meter to simply measure what the device uses: Get a 3X AA battery holder, and insert your meter between the batteries and the lamp. Measure with fresh batteries and you will have your current usage. An adapter that exceeds the value by at least 10% should suffice.

If your device has a DC power jack that isn't labeled, you will have to investigate as to whether the tip or ring is ground. Obviously any adapter you purchase could be modified to match a given polarity; many have plugs that can be attached either way. (I personally solder my own using adapters that come with bare wires.)

Make sure you get a regulated power supply. An unregulated one can supply more than 4.5 volts if not powering the load it was intended for.

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    \$\begingroup\$ You can probably do this for something that uses moderately large cells. But for dead simple products that use coin cells, it's possible the internal resistance of the cell could be critical to proper operation - replace the coin cell with a supply that does not sag under load and there's a small but worth considering possibility that a bright LED or something that was protected only by the behavior of the original supply could be blown out. \$\endgroup\$
    – Chris Stratton
    Commented Feb 11, 2013 at 1:32

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