1

I have a project with an Arduino Nano 33 BLE Sense. My project is working great, but I need to transition it to battery power, and I'm having a bit of confusion about exactly what's required.

I have a small 3.7V Li-ion battery I'd like to use (similar to this one {0}) if possible.

I have read what seems to be some conflicting information, so I'd like to see which of the following are true:

  • I can power the Arduino directly from the 3.3v pin? I'm suspicious of this advice since on the datasheet the 3.3v pin is marked output only. There's a note on the product page that says:

Note: for ultra low-power functionality, you should cut the 3V3 jumper on the back of the board, and use an external battery at 3V3.

Is this something I should do? It seems difficult to reverse, I'm worried I might break my project.

  • I can power the Arduino directly from the VIN pin? This again seems to contradict the datasheet, which says the VIN pin can only handle 4.5 - 21V. 3.7V is less than this.

  • I could use a Step-up DC-DC transformer - like this {1}? I'd rather not wait for shipping from China.

Furthermore, I'd like to be able to charge the battery, so if I use a charging board like this one {2}, would that just work?

I don't want to connect the battery anywhere it might not be welcome for the risk of frying anything. Any help would be appreciated.


Link references (in case ebay listings go away):

0: Lithium Ion Battery - 110mAh

enter image description here

1: Mini 3V 3.3V 3.7V 5V 9V to 12V Step Up Boost Converter Module Voltage Regulator

enter image description here

2: 4.2V 5V 1A Micro USB Li-ion Battery Charger TP4056 Protection Chip Lithium-ion

enter image description here

  • the pads near the 3.3 V pin where you can cut the tiny trace between them are easy to bridge with solder. – Juraj Mar 7 at 14:24
  • Thanks, that's good to know. So if I cut the bridge, enabling 3.3V mode, the USB cable will stop working, but I can power directly from that pin. It seems to suggest there's no voltage conversion going on, so I could damage the board if I supply 3.7+? – fredley Mar 7 at 14:29
  • the power from USB will stop, not the data – Juraj Mar 7 at 15:15
  • Did you find a solution to this problem? – shi Apr 29 at 18:11
1

The VIN (and the USB) goes into a 3.3V switching regulator. This has a minimum input voltage of 4.5V.

You isolate the output of that regulator from the rest of the circuit by cutting that one link (which is easy to re-solder). You are then free to feed 3.3V from your own power source directly into the board.

Yes, if you provide more than 3.3V you are liable to damage the board (note that a Li-Poly can be around 4.2V when fully charged).

You need to take your battery power and feed it through a nice efficient switching voltage regulator that has a low enough "dropout" voltage to allow good use from the battery. Note that many devices "rated" at 3.3V will also work much lower. For example:

  • Nina-B3 series module will run at down to 1.7V.
  • The APDS-9960 will operate down to 2.4V.
  • The HTS221 will run at down to 1.7V.
  • MP34DT06J can go as low as 1.6V.
  • LSM9DS1 can run from 1.9V.
  • LPS22 is 1.7V minimum

So it could be good to reduce your power to 2.5V instead of 3.3V to give more life from the battery. You should check each of the components in the both the device schematic and the rest of your circuit (by reading the datasheets) to find the highest "minimum" voltage they will work on.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.