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I have a project where Arduino Nano 33 BLE (3.3V operating voltage) is powered by a LiPo battery. The battery is charged using Adafruit's MicroLipo Charger it is then connected to Adafruit's MiniBoost to boost the voltage to 5V and that powers the Arduino through Vin as well as a logic level converter.

Ideally I'd like to use the device while it's charging. The MicroLipo charger provides a USB C breakout board providing a 5V pin. Is it a bad idea to connect 5V output from the boost converter and from USB C to th same power rail as shown in the frtizing sketch? My gut says try it, my brain says this is a dumb idea.

Is there a way I could use something as a switch which takes in 5V from both the cable and the battery only uses the battery when cable voltage is 0V (not connected)?

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Ivan Novikov
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  • I would definitely not do that. You try to short the USB 5 V with the 3.7 V of the LiPo battery. If VBat on the charger is not protected, 5V will produce a loading current that might destroy the battery. I also have to tell you, the resistor for the LED on your schematics has no meaning. Both legs of the resistor and one leg of the LED are on the same rail of breadboard. For a power selector the following link might help https://arduino.stackexchange.com/questions/39805/can-you-charge-and-use-a-lipo-battery-at-the-same-time – Peter Paul Kiefer Mar 20 '23 at 14:14
  • For example Seeed studio Xiao BLE has the same MCU and also battery "pads" + charging circuit And of course then you can't use constructs like while (!Serial); in setup – KIIV Mar 21 '23 at 12:19
  • "Ideally I'd like to use the device while it's charging", this is not an Arduino question, it is an electronic question. In order to use the device while charging, you need a "load sharing" circuit, otherwise your battery might not get fully charged. Read this Microchip App Note. The load sharing circuitry disconnect the battery from the load (i.e. your device) while charging and directly supply the 5V from USB to the load, and when the USB/5V is not present, the battery will supply the power to the load. – hcheung Mar 21 '23 at 13:59

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