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This is the electrical schematic of the circuit that provides 3.3V on the Arduino UNO board: enter image description here

My question is: What is the purpose of the circuit?

Here is what I understand: RN1A and RN1B divide Vin by 2 before the comparison (not sure why), but what is the point of the MOSFET, since a positive voltage from USBVCC can cause current flow from the diode pictured anyways, without requiring the MOSFET to be turned on? LP2985's datasheet mentions that it requires a positive voltage to operate, so it does not make sense to talk about what would happen if USBVCC is negative.

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    \$\begingroup\$ Note that Vin is 9V from the barrel jack. An important part of this schematic is missing. Vin powers a 5V regulator, output connected to +5V. This +5V powers all the on-board 5V circuits. \$\endgroup\$ Commented Sep 11 at 14:18

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Basically when VIN is present and if it is larger than 6.6V, it will make the output of U1A go to +5V and shut off T1.

Even if USBVCC is present and considered to also be 5V, no current will flow through the body-diode of T1 and +5V will supply U2.

If only USBVCC is present and VIN is not connected, U1A will be close to GND and current will first flow through the body-diode and the voltage will initially be a diode drop lower at +5V until it turns on T1 and USBVCC is equal to +5V.

If there is another regulator at +5V, it should be able to handle USBVCC at its output.

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  • \$\begingroup\$ I think the issue is that T1 is placed in the wrong direction on the schematic, S and D should be flipped, so that T1 is flipped horizontally. \$\endgroup\$ Commented Sep 11 at 12:55
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    \$\begingroup\$ Schematic is correct. If T1 S-D are swapped, 5V can feed back to your PC USB if the PC is powered OFF, you don't want this to happen. \$\endgroup\$ Commented Sep 11 at 14:22
  • \$\begingroup\$ @Mattman944 you are right. Also, following your second comment, the purpose of the circuit and Tyrassin's answer becomes clear when considering the +5v part of the schematic. In conclusion, if I understand correctly, the purpose of circuit is to supply the 3v3 regulator with either the barrel jack (or Vin pin) or through USB without mixing the voltages (which would have caused damage if we had just shorted them) when both are present. \$\endgroup\$ Commented Sep 12 at 7:22
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    \$\begingroup\$ @BramFran Yes it is to supply the 3.3V and to seperate them and also to make Vin the first priority. \$\endgroup\$ Commented Sep 12 at 7:26

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