I'm trying to integrate an ESP32-PICO-KIT/V4.1. into a small project.

The documentation for the PICO power supply says this:

There are three mutually exclusive ways to provide power to the board:

1) Micro USB port, default power supply

2) 5V / GND header pins

3) 3V3 / GND header pins


The power supply must be provided using one and only one of the options above, otherwise the board and/or the power supply source can be damaged.

I have chosen to power the board with a 3.3V supply coming from a linear regulator. The PICO board has two pins for 3.3V, and I have connected each one to the 3.3V Vout leg of the regulator. I've also connected a 0.1uF capacitor next to each 3.3V pin.

My question is about the 5V pin on the PICO. It will not be used. Should I leave it unconnected, tie it with a large pull down to ground, or do something else? It's not an input pin so I'm not sure what to do.

Thank you!

  • 3
    \$\begingroup\$ In this sort of situation, the unused power inputs are usually left open. \$\endgroup\$ Commented May 27, 2020 at 9:27

1 Answer 1


You don't need to do anything with the 5V pin, just leave it open, but you might want to consider what would happen if someone plugged a USB cable into the connector. From the official schematic:

enter image description here

There is an SS14 1A Schottky diode between the USB and the AMS1117 input, so the 3.3V cannot backfeed the USB, but the USB could backfeed whatever is connected to that 3.3V supply.

It's not clear from the AMS1117 datasheet but perhaps the 3.3V backfeeds the EXT_5V node and causes the LED to illuminate (albeit more dimly than if a 5V supply or USB was connected) (that would be my guess).

  • \$\begingroup\$ I'm powering this with the 3.3V output of a LD1117 linear regulator. If someone were to plug in a USB cable while the first supply was active then the 3.3V pin (and everything downstream) would see 3.3V from both the AMS1117 and the LD1117. Is this bad or something I should try to prevent? If so, is there something I could add to the circuit that would automatically sense if both become active and choose one or the other source? Thank you! \$\endgroup\$
    – par
    Commented May 28, 2020 at 1:06
  • 1
    \$\begingroup\$ I don't think it would hurt anything, with those particular regulators. \$\endgroup\$ Commented May 28, 2020 at 19:33
  • 1
    \$\begingroup\$ @par You should not give any voltage back over USB. This is forbidden by the USB specification and some host hardware can get into latch-up condition when this happens, which causes the chips to shortcircuit if the computer is started. This goes for both the USB data lines as VCC. This has hard bricked macbooks in the past. The USB data lines are less likely to be a problem, but can still give the same problems \$\endgroup\$
    – Ferrybig
    Commented Jan 1, 2022 at 11:55

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