Arduino mega designI'm trying to understand the Arduino design.

LM358 and FDN306P is connected in between USB input and primary source input. In presence of USB and primary supply the Arduino mega board is drawing current from both of the sources. If it is used for power switching then why is the load sharing happening?

Here is the schematic. Is this mosfet used for allowing one of the power source at a time or for load sharing?

  • 1
    \$\begingroup\$ Please provide schematic. \$\endgroup\$
    – smajli
    Oct 15, 2018 at 7:46
  • \$\begingroup\$ like this if u ask qtn, no body can answer.. out of air we can't understand things. \$\endgroup\$
    – user19579
    Oct 15, 2018 at 8:49
  • \$\begingroup\$ Please find the Arduino mega design schematic. \$\endgroup\$ Oct 15, 2018 at 8:56
  • 2
    \$\begingroup\$ Please find the Arduino mega design schematic. – user201284 Nobody want to looking for a schematic around the web for answer your question. Here there's people to work and help you in spare time. If you want an answer you should provide everything necessary to enable us to help you. \$\endgroup\$
    – Antonio
    Oct 15, 2018 at 10:33
  • \$\begingroup\$ And he gets 1 upvote for providing figure in answer... meanwhile all i get for my questions is tumbleweed badge.... no offense though \$\endgroup\$
    – Deep
    Oct 15, 2018 at 12:33

1 Answer 1


Do you know that the power is shared between both? The point of this arrangement is to prevent just that. IC5B is used as a comparator, when Vin is above 6.6V (since it's divided by 2 and compared against the 3.3V supply) the comparator goes high, switching T2 off and isolating the supply from the USB. If there's no supply at Vin, there's no output from the regulators (top left on https://www.arduino.cc/en/uploads/Main/arduino-mega2560-schematic.pdf ) and T2 is on and draws from the USB. This does leave the regulators reverse biased, presumably they tolerate this.

  • \$\begingroup\$ Thanks for the response, I have a doubt on the gate voltage.As the mosfet is a P channel Enhance type, Gate requires a -ve supply to work. So how does it work in +ve supply? \$\endgroup\$ Oct 15, 2018 at 14:34
  • \$\begingroup\$ The gate will be negative with respect to the source when the comparator output is low. Note that the FET is effectively backwards to the way you'd normally use for switching, so that the body diode will conduct and bring the +5V line up even before the FET is brought to conduction, allowing the comparator to operate and then the FET bypasses the diode bringing the +5V line up to the full USB supply. \$\endgroup\$
    – Phil G
    Oct 15, 2018 at 15:09
  • \$\begingroup\$ I wasn't aware of the FET body diode @PhilG Thank you. Your answer was helpful. \$\endgroup\$ Oct 17, 2018 at 12:57

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