So I'm looking at the schematics of the Arduino Micro and I have a few questions.

  1. On their website, Arduino claims that:

The Micro can be powered via the micro USB connection or with an external power supply. The power source is selected automatically.

I'm looking at their schematics, and more specifically here: Voltages area of the Micro

And can't really understand how T1 works, assuming only USB is connected. If there is no Vin, how is there any Vgs voltage, and how does the transistor T1 open/close?

  1. What is the function of the Diode D2?

Diode D2?

Thank you.

  • 1
    \$\begingroup\$ I suspect they are just relying on the intrinsic diode of T1. \$\endgroup\$
    – Joren Vaes
    Jun 17, 2018 at 5:51
  • \$\begingroup\$ @JorenVaes I thought so too, but that just means there is voltage on the drain and source. What about the gate? \$\endgroup\$
    – nettek
    Jun 17, 2018 at 5:54
  • 3
    \$\begingroup\$ There's a 10k resistor to ground on the gate of T1. Presumably, that pulls the gate to ground when there's no voltage on Vin. \$\endgroup\$
    – JRE
    Jun 17, 2018 at 6:02
  • \$\begingroup\$ @JRE Wow I did not notice that... now it makes sense. Thanks! \$\endgroup\$
    – nettek
    Jun 17, 2018 at 6:10
  • 1
    \$\begingroup\$ One of the disadvantages of chopping up a schematic with tags. It makes it less cluttered, but obscures some of the connections. \$\endgroup\$
    – JRE
    Jun 17, 2018 at 6:12

1 Answer 1


When 5 V is available and Vin is not available

  1. In the beginning only the body diode of T1 conducts.
  2. Hence, there will be (5V - diode drop) at the source of T1.
  3. that is only momentary,because, now we have a valid Source voltage. Gate is pulled low to ground via R9 (10kohm).
  4. Now, Since, Vgs is more negative, PMOS is ON. Once it is on, diode gets bypassed by the MOSFET. Hence, one will get almost 5V finally for the 3.3 V regulator .
  5. One can assume, the low on-resitance of PMOS will certainly create very low ohmic drop.

When 5 V is available and Vin is also available

  1. The body diode may conduct in the beginning if 5 V is plugged in first.
  2. The MOSFET too may turn on sooner and provide almost 5 V to 3.3 V regulator
  3. As soon as Vin is connected (assuming that Vin is greater than 5 V), the MOSFET turns off because VGS is no more a negative value.
  4. The current from USB will no longer be used because, the strong 5 V is available from 5 V regulator which is now powered from Vin.

When 5 V is not available and Vin is available

  1. T1 will not be in the picture
  2. T2 body diode can conduct and supply Vin to the U2 ( 5V regulator)
  3. 5 V regulator will feed the 3.3 V regulator as previous case

Function of D2

There is a reset switch. Any spike arising out of the reset switch due to debounce can be sometimes higher than 5 V supply itself. This may inherently cause damage to the RESET pin of the MCU. Hence, placing the diode will start to conduct when voltage higher than 5.4 V (example) is present at the output of the switch.

Consider it like a clamping circuit and clamping voltage is set to 5 V plus the forward voltage of the diode.

Normally, there will also be diodes internal to the MCU pin for protection, but placing the external diode is definitely a good practice. The internal one is for emergency use only, i would say.

  • \$\begingroup\$ Thanks. What about the D2 diode? \$\endgroup\$
    – nettek
    Jun 17, 2018 at 6:46
  • \$\begingroup\$ just a tip - it is good practice to wait for a few good answers or review of existing answer before accepting.. you may miss a different view on the problem because of that. \$\endgroup\$
    – User323693
    Jun 17, 2018 at 6:46
  • \$\begingroup\$ Please see update about D2 \$\endgroup\$
    – User323693
    Jun 17, 2018 at 6:50

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