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I'm looking at the schematic for the Sparkfun Redboard and am wondering what is the function of this MOSFET. From what I understand the gate is held at ground and since it's a PFET, once the source sees a voltage the MOSFET will turn on, allowing VUSB to flow into the regulator.

I guess I'm not understanding the point of this, if the MOSFET is on, there has to be a voltage at the source, but once it turns on, VUSB is connected to the regulator, but wouldn't there already be voltage there?

I'd think that the MOSFET is there for some type of protection or to switch between the barrel jack/USB power but I can't seem to reason it out.

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    \$\begingroup\$ Yes, there's a lot of inflation, and things are expensive, but surely you could spare a link to the schematic? \$\endgroup\$
    – bobflux
    Jul 2, 2022 at 23:00

2 Answers 2

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The MOSFET gate is grounded through the normally closed switch on the barrel jack.

  • When no barrel plug is inserted then the gate will be grounded and USBVCC can directly power the 5V bus.
  • When a barrel plug is inserted the gate is pulled high by the 10k resistor R4 and the FET Q1 turns off, this prevents the bus from being back driven by the 5V regulator IC3.

schematic sample Image sourced from schematic here: http://cdn.sparkfun.com/datasheets/Dev/Arduino/Boards/RedBoard-V22.pdf

After some action in the comments it was brought to my attention that the barrel jack is connected incorrectly in the schematic, the ground and fet gate should be swapped. I grabbed the design files for this board from sparkfun's github and found that the connector is a Wurth 694108106102 and they've mistakenly swapped pins in their symbol to land translation:

schematic symbol to land pattern according to sparkfun

compare this with wurth's datasheet on this part, wurth pins 2 and 3 are opposite:

Wurth datasheet, with highlights

However all is not lost! look at the layout and you will see that the physically correct pin on the barrel jack drops out to a via which makes its way to the MOSFET, and the ground is correctly bonded to the plane: redboard production image

As to the MOSFET itself, it's a Diodes inc ZXMP6A13F, P channel enchancement device. I do believe it's connected correctly - if (as suggested in the comments) drain and source were to be reversed then the onboard 5V regulator could backdrive the USBVCC through the body diode.

redboard github: https://github.com/sparkfun/RedBoard wurth datasheet: https://www.we-online.com/katalog/datasheet/6941xx106102.pdf

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    \$\begingroup\$ Something doesn't look quite right for the input power jack in that schematic. I assume the upper rectangular symbol is for the inner conductor of the plug, which provides external 7-15 VDC to the input of the LM1117 through D1. But the sleeve would be contacted by the wiper arm of the switch, which goes to R4 and the gate of Q1. I think those connections should be reversed. The symbol is discussed here: electronics.stackexchange.com/questions/2381/… \$\endgroup\$
    – PStechPaul
    Jul 2, 2022 at 23:38
  • \$\begingroup\$ @PStechPaul great point! I think I just skipped that detail and got the spirit of the schematic and not the letter. I’m interested to know now where this error lies, or if the board even works as intended (I feel it must!). I will look for a BOM \$\endgroup\$
    – Bryan
    Jul 2, 2022 at 23:49
  • \$\begingroup\$ @PStechPaul the connections from power jack to GND and Q1 definitely look reversed \$\endgroup\$
    – jsotola
    Jul 3, 2022 at 0:30
  • \$\begingroup\$ Looks to me like Q1 is upside-down. Swap the source and drain connections. Or it is the wrong polarity. I can argue it either way. Is there a part number for Q1 to confirm what FET type it is? \$\endgroup\$
    – AnalogKid
    Jul 3, 2022 at 0:46
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    \$\begingroup\$ The PMOS FET is intentionally operated in the reverse direction to prevent back-feeding USBVCC. Using a FET instead of just a Schottky diode reduces the voltage drop to a few millivolts rather than a few hundred. This arrangement is common in power control circuit. \$\endgroup\$ Jul 3, 2022 at 16:29
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As others have said in the comments, the MOSFET is definitely drawn upside-down. Its SOURCE should be connected to the USB +5V (the source is the "end" where the gate-lead enters the "circle").

This circuit works as follows:

  • If nothing is plugged into the power jack, the power jack internal switch grounds the MOSFET's gate. If the circuit's USB input is live, the MOSFET's source will be at +5V from the USB, while its now-grounded gate is at zero. This turns the MOSFET "on" and +5V-USB powers the +3.3V regulator as well as touching the 1117's +5V out (the 1117 is unpowered in this case and being fed backwards -- I assume that's ok but it's a little "uncouth").
  • If, OTOH, a +7 ~ +15V external supply is plugged into the jack, the 1117 is now powered and supplies +5V to the output as well as the +3.3V regulator. Simultaneously, R4 pulls the (now, no-longer-grounded) gate of the MOSFET to +5V which means NO current flows thru the MOSFET regardless of whether the USB is powered or not (i.e. the MOSFET's source is +5V if the USB is connected, or "open" if it's not). Either way, the USB is now isolated and the circuit is no longer being powered from the USB -- instead, all its power comes from the external supply.
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    \$\begingroup\$ The PMOS FET is intentionally operated in the reverse direction to prevent back-feeding USBVCC. Using a FET instead of just a Schottky diode reduces the voltage drop to a few millivolts rather than a few hundred. This arrangement is common in power control circuit. \$\endgroup\$ Jul 3, 2022 at 16:30
  • \$\begingroup\$ "which means NO current flows thru the MOSFET" - I believe this part of the analysis neglects the body diode, which would allow current to flow from drain to source. \$\endgroup\$
    – Bryan
    Jul 3, 2022 at 19:59

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