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I have tiny christmas tree that has about 20 small LEDs, which are powered by 2 AA-batteries. So I'm assuming the LEDs are 3V (because of 2 AAs). This tree was bought in a store, as is.

What I want to do, is hook up the little tree to an Arduino and control the lights with some sensors (lights, sound, etc). I was planning to just connect the battery terminals to the 3V Arduino output.

Is it safe to assume that the battery pack has a built-in resistor somewhere and I can just safely connect it to a 3V output of the Arduino, or would it be wise to have an extra resistor somewhere?

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  • \$\begingroup\$ Are you talking about using the Arduino output as a replacement for the batteries? \$\endgroup\$ – ThreePhaseEel Nov 20 '17 at 12:47
  • \$\begingroup\$ @ThreePhaseEel, yes that's correct. I was going to play with controlling the LEDs and hopefully be able to play with PWM (maybe something similar to this: electronics.stackexchange.com/questions/23199/…) \$\endgroup\$ – PaulRmn Nov 20 '17 at 16:03
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A better way of doing this would be to test it yourself. Measure the off-load voltage of your battery pack, then connect a resistor with a known value (can measure it first) and then measure the on-load voltage.

Using Ohm's law, you should be able to find if there is any built in resistor in the battery pack, and an approximate value. If you find that the value is very low (probably less that 1 Ohm), chances are it will just be the batteries internal resistance and the battery pack does not have a built in resistor.

I would never say it is safe to assume anything, always better to test first!

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Thanks, it looks like you're right. Although my volt meter isn't working now, the wires run straight from battery to the string of LEDs, so I guess they use the internal resistance of the battery. I'll try to use a few resistors in series and remove them one by one to see when it seems right. I know, I should measure, but unfortunately that's not working now.. \$\endgroup\$ – PaulRmn Nov 21 '17 at 19:40
  • \$\begingroup\$ You can get multimeters very cheap from hardware stores these days. For applications such as this, a cheap one will do the job fine. Always a good idea to have one \$\endgroup\$ – MCG Nov 21 '17 at 20:47

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