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This question already has an answer here:

I'm looking to use the P-channel MOSFET approach to reverse polarity protection as described here and here (fig 5). Basically this picture:

enter image description here

I've seen that circuit topology used, for example, here, but I'm not convinced that schematic is accurate (maybe you can comment). The way I see it, I can either connect the Drain-pin to the Input Voltage, or the Source-pin to the Input Voltage. Does it matter which way I connect the P-channel MOSFET, or is the protection symmetric about the drain-source channel and all that matters is that the D-S channel is in series with the input voltage?

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marked as duplicate by Matt Young, Chetan Bhargava, Joe Hass, Daniel Grillo, PeterJ Mar 28 '14 at 21:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    \$\begingroup\$ I've never used or studied this circuit, but I imagine it won't provide much protection if the body diode is forward biased under the "reverse battery" condition. So the battery has to be on the left and the load on the right in your diagram. \$\endgroup\$ – The Photon Mar 28 '14 at 16:01
  • \$\begingroup\$ @ThePhoton so would you agree that the schematic I referenced (for example) is not accurate? \$\endgroup\$ – vicatcu Mar 28 '14 at 17:07
  • \$\begingroup\$ Bear in mind that the reverse current is stopped only by the inner diode of the mosfet ic. It's limitation might be very low. I also have doubt about the use of the mosfet in the opposite direction, It may work but I wouldn't use it with significant amount of power. Idealy, one should use a P-mosfet without protection diode in the right direction. But I don't think it exists anymore. \$\endgroup\$ – Fredled Jan 24 at 22:19
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It should work. Here's another diagram of this approach.

enter image description here (fig.5 from Maxim App Note 636)

When the battery has a proper polarity (as shown in the diagram):

  • The body diode of the MOSFET is forward biased.
  • VGS = -Vbatt+Vdiode < 0V, and the P-ch MOSFET is turned on.
    It is necessary for the body diode to be forward biased with battery in proper polarity. When the FET turns on, Voltage drop is reduced to Vdrop=I*RDSon which is significantly lower than the voltage drop of the diode.

When the battery has reversed polarity:

  • The body diode of the MOSFET is reverse biased.
  • VDG = Vbat > 0V, and the P-ch MOSFET is turned off (FETs are symmetric devices, in this condition drain and source are swapped).

    • There is no reverse current through the load.

P.S.

The O.P. references an Arduino shield schematic as an example.

enter image description here

  1. It's poorly drawn. It's not difficult to come to a conclusion that the guy who drew it dislikes his clients (or everybody).
  2. It's incorrect (I think).
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Source pin is the unknown polarity input voltage and ensure that it doesn't get too close to the breakdown rating of the mosfet's gate-source voltage. Probably about 15 volts is maximum but check the data sheet.

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You put the anode of the body diode on the Vin direction so you get the negative Vgs.

> Does it matter which way I connect the P-channel MOSFET<

Yes, when you apply a positive voltage on the D side, this voltage will also appear (minus diode drop) on the S side. With the G grounded, you will have Vgs = -Vin and the device will conduct. As a P-MOSFET conducts with negative Vgs.

You could get it work by reversing D and S but then the body diode will conduct if you place a negative power source at the input. Which might get you in trouble.

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  • \$\begingroup\$ Can you be more expressive in your answer with regard to which particular aspects of my question you are addressing? What is "It" you are referring to. \$\endgroup\$ – vicatcu Mar 28 '14 at 17:01
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    \$\begingroup\$ Sorry, I'm drunk. \$\endgroup\$ – Dejvid_no1 Mar 28 '14 at 17:15
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    \$\begingroup\$ @Dejvid_no1 Where was captcha when it was needed? \$\endgroup\$ – Nick Alexeev Mar 28 '14 at 20:48

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