Sorry if there is an obvious answer to this - I'm pretty new at Arduino and electronics. I've searched a bunch, but I think I'm too vague - and don't know what terms to look for.

I have a pretty simple project using a 3.3v Arduino Pro Mini (https://www.adafruit.com/products/2377) and a 12v button/LED (https://www.adafruit.com/products/1185).

I'd like to give the LED the full 12v so the light is a bright as possible. The system will be battery operated.

The question is: Do I need a 12v power supply or can I use a 3.7v power supply and upgrade it somehow? I know I can power the LED with 3.7v but then it's really dim. Is there a way to "upgrade" the 3.7v to 12v? I would assume this would draw more power and drain the battery faster, but I think that would be ok.

I'm hoping to get away with a 3.7v power supply because I want to use a lithium ion polymer battery (https://www.adafruit.com/products/328) - since it's small and rechargeable.

Seems like if I go with 12v source then I need a much bigger battery pack, and I'm hoping to keep the project small. Not sure if it would help to know, but I only need the device to run for 4+ hours between charges.

I'm guessing people will say to go with 12v since the Pro Mini can downgrade the 12v supply so there'd be less parts I have to deal with (I can already make this work). But, again, I was hoping to get away with a small rechargeable power supply. Or, are there good, small 12v batteries - I can't seem to find any on Adafruit or SparkFun.

Any recommendations? Thanks!

  • The product description states "The LED has a built in resistor so you can run it up to 12V"., But, of couse, at 3.7 you won't get enough bright. I'd probably feed the Arduino with a 12 vdc power source and divert some of the power for the button. Maybe you need to find a lower voltage button, or test the one you have with the 3.7 v supply,
    – fabrosell
    Feb 9, 2016 at 14:37

2 Answers 2


Since your project is to be battery powered, you want to make your system as efficient as is reasonable.

So I am going to take an approach that ends up improving the efficiency of your project.

The button you describe in your link states :

This button has a domed red plastic cover with a black retaining ring around it. There's an optional LED included

Further, the LED included has a "built in resistor", so that the LED can operate from 12 volts.

Since you would have to install the "optional LED", why not purchase a Red LED that operates from lower voltage. You will still need a resistor but you won't need to have a voltage converter.

Example : Their LED probably has a 2 volt LED (approximate) that operates at perhaps 20 mA). This would require a 500 ohm resistor (which they incorporate into the LED). The power loss (wasted) in the resistor is 0.2 watts (200 mW).

Improved Example : With LED of about 2 volts operating from 3.3 volt power source. This would require a 68 ohm resistor. The power loss in the resistor would be 0.027 watts (27 mW). A substantial improvement in efficiency.

The manufacturer of the button does not give specifications for the LED (the one with the resistor included). You should connect their LED to a 12 volt power supply and measure the current drawn. It is possible their LED draws more current than I have shown in the examples. This would require a re-calculation of the resistor values.

But in any case, the improvement in efficiency will still exist.

  • Thanks! Would a 2v LED be way dimmer than the 12v one? I don't know a lot of about electronics (clearly), but I guess I assumed the higher the volts the brighter it would be? For my project a nice bright LED is desirable.
    – brian h
    Feb 9, 2016 at 19:25
  • I reread your comment... Realized I missed your point. Yeah, if the LED is only 2v then your solution would be perfect! I'll try to find the specs on the LED... That being said, am I still going to be limited on my light output based on 3.3v? I think I'll have to do some tests... Thanks for your response!
    – brian h
    Feb 9, 2016 at 19:40
  • I'm thinking this is what I'll need: adafruit.com/products/754 Thanks again for the suggestion!
    – brian h
    Feb 9, 2016 at 20:34
  • Would a 2v LED be way dimmer than the 12v one? - LEDs have a standard voltage drop (around 2V depending on the colour). When you get a "12V LED" it just has a bigger resistor, as Marla said. See Wikipedia for a bit more detail.
    – Nick Gammon
    Feb 9, 2016 at 23:12
  • The ones you linked from Adafruit have a slightly higher voltage drop: "3.0V Typical Forward Voltage, at 20mA current". You still need the current-limiting resistor though.
    – Nick Gammon
    Feb 9, 2016 at 23:14

Basically you have two choices. There is a third, but you don't wan to take that route.

The third route is to directly power the system with 12V and rely on the Arduino's regulator to drop it to the right voltage for the Arduino. That is very very inefficient and wasteful, and your battery won't last very long - plus the regulator will get hot. So don't even consider that option.

So the two sensible options - start with 3.7V and step it up to 12V using a boost regulator, or start at 12V and step it down using a buck regulator.

Which of those will be the easiest and most efficient? Well, there's not much to choose between them.

With either of those solutions you are converting voltage and current. With a buck regulator you are converting voltage into current, and with a boost regulator you are converting current into voltage.

If you start at 12V and step it down yes you need more cells, but each cell can be smaller since each one will have to supply less current. With 3.7V and stepping up you will have just one cell, but it will need to be bigger since it will have to supply more current.

From a charging perspective it is easier to charge a single cell since you don't have to worry about cell balancing.

From a price and availability point of view, buck (step-down) regulators are cheaper and easier to get hold of on places like eBay than boost (step-up) regulators as pre-built circuits.

So really the choice is yours - 12V (well, 11.4V it would really be - 3S LiIon / Li-Poly) and buck regulator, or 3.7V and boost regulator. Both will work, both will be about as efficient as each other.

Incidentally there are boost regulators specifically designed for driving LEDs from battery sources - you might like to look around to see what is available in that way.

  • Thank you for your reply! I think I will dig into your suggestions a bit - since I think the brightness of my LED may require more power. For now, however, I think Marla's answer above my be better for my project.
    – brian h
    Feb 9, 2016 at 19:42

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