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I am currently working on an stm32-based design. I am trying to understand how to power the stm32 through USB. We came up with the following solution:

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The idea is that +3v3_power is the main rail of our product generated when it's powered through sector voltage. V_usb is the power coming from the USB cable (Product as device, not providing power supply). We are making an or-ing thanks to the PMOS. When +3v3_power and V_usb is not here, +3V3 = +3v3_power (pmos on). When +3v3_power is not here, and V_USB is present, PMOS is off, and V_USB is regulated by the Zener diode to supply the stm32. Does this work ?

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  • \$\begingroup\$ It is not ideal to supply +3V3 through a 2.2K resistor. Imagine your load need a tiny 1mA or maybe 2mA of current. That wont work. If you need 3.3V after the USB, use a LDO, if not already present? \$\endgroup\$ Commented Nov 24 at 9:45
  • \$\begingroup\$ There are a lot of very-inexpensive 5v to 3.3v linear regulators available as low as $0.0045 (e.g. TPAP2125K-3.3) while a BZX84C3x is around $0.009, both from LCSC. \$\endgroup\$ Commented Nov 24 at 9:45
  • \$\begingroup\$ Thank you both for your quick answers. I wanted to ask more advises, because the more I dig into this topic, the more I think it's no reliable. I thought about reducing the 2.2 kΩ resistor to supply more current to the STM32. But using the Zener as a “regulator” is not reliable: it would dissipate too much. Also, if both +3V3_POWER and VBUS are present, there is a risk of backfeeding through the PMOS body diode. Is my thinking correct ? So your solution with LDO definitely sounds better. \$\endgroup\$ Commented 2 days ago
  • \$\begingroup\$ There wont be any backfeeding. In the present schematic and if both sources are present. Then there will be 5V on the gate and less on the source so the mosfet will not conduct. Then you have two diodes feeding into +3V3. \$\endgroup\$ Commented 2 days ago

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No. It does not work.

You can't power a MCU from 5V through 2k2 series resistor.

Unless your MCU consumes max 0.7 mA, which is likely less than what your whole board with MCU needs.

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