I was about to design a small custom board using ATmega32U4, replicating the Arduino Micro. I know that T1 is used to block VUSB when VIN is connected.

But why T2? The gate is 5V, so it will be always OFF! Or does it act as a diode?

Schemtic https://www.arduino.cc/en/uploads/Main/arduino-micro-schematic.pdf


T1 and T2 implement a power supply ORing of the VUSB and VIN voltages. Read this for extended information.

The body diode of each MOSFET performs the basic ORing function. When the MOSFET is turned OFF it just behaves as a diode. But when the MOSFET is turned ON, the diode is shorted by the conduction channel, thus avoiding its voltage drop (and achieving a behaviour closer to an ideal diode).


  1. When VUSB is present and VIN is not connected (but pulled down to ground by R9), T1 turns ON and the +5V rail is connected to VUSB. Also, T2 turns OFF and the regulator input is isolated from VIN and the regulator simply doesn't operate (thus, avoiding a potential conflict because its output +5V is being forced to VUSB).

  2. When VUSB is not connected and VIN is present, T1 turns OFF and VUSB is isolated from the +5V rail. Also T2 turns ON and VIN is connected to the regulator input, which then regulates the voltage down into the +5V rail.

  3. When both VUSB and VIN are connected, the behaviour is the same as in case 2 above. Because VIN is supposed to be >7V and VUSB approx 5V, T1 will isolate VUSB and avoid a conflict with the regulator output. Also, a VIN >7V will allow T2 to turn on hard enough to bypass the body diode and avoid its voltage drop.

  • i got it. Could you please tell why "7V" in 3rd case? – Sanu - Open Maker Jul 6 '17 at 14:53
  • The input voltage VIN for the Arduino Micro is specified in the 7-12 V range. – Enric Blanco Jul 6 '17 at 14:56
  • So it could be any volt that can be handled by voltage regulator? – Sanu - Open Maker Jul 10 '17 at 12:30
  • 1
    The input capacitor probably can't withstand as much voltage as the regulator itself. I wouldn't operate the Micro outside its recommended voltage range. – Enric Blanco Jul 11 '17 at 14:30

It's known as a "zero drop-out diode". When the power is first applied the body diode of the MOSFET conducts and powers the system. That imposes a (roughly) 0.7V drop on the input voltage.

Once the power gets through the regulator and is regulated to 5V that is then fed back into the gate of the MOSFET, which turns it on (as long as -5V is above the source voltage less the diode drop, so -5.7V minimum [it's a P channel so thresholds etc are negative]). That then bypasses the internal body diode getting rid of the 0.7V drop.

If the input voltage is too low the 5V gate no longer turns the MOSFET on and the body diode blocks any reverse current.

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