Hi,
I'm trying to build a battery reverse polarity protection circuit for my 3.3V battery. I don't know what is the purpose of zener diode and how to select the right zener diode for my appliation and also how to calculate the value of the resistor. I'm attaching a reference schematic below.

Thanks in advance.

 

The zener protects the gate from seeing too high of voltage. For 3 or 5 volts, it's fine to leave it out.

Just be sure to use a logic-level P-channel mosfet.

 

The zener protects the gate from seeing too high of voltage. For 3 or 5 volts, it's fine to leave it out.

Just be sure to use a logic-level P-channel mosfet.

Thanks

 

Without the zener, you can leave out the resistor and connect the gate directly the the negative rail. That's how I do it on my micro boards.

 

Two problems here:
Besides not needing the Zener, with source follower configuration the likely large gate-to source voltage will reduce your 3.3 volts greatly and you red to swap the drain and source otherwise the MOSFET will not work the way you want.

Some MOSFETS of that type may not work with such a low voltage.

 

That MOSFET has too high maximum threshold voltage (-4V) to reliably work with a 3.3V battery (below):
You need a logic-level type that has a maximum Vgs(th) of ≤-2V.

 

....with source follower configuration the likely large gate-to source voltage will reduce your 3.3 volts greatly and you red to swap the drain and source otherwise the MOSFET will not work the way you want.

No, the circuit is correct (the version I use all the time is below). When power is applied, the mosfet body diode conducts from drain to source, which turns the mosfet on. The only voltage drop is the I × R drop across the mosfet.

This circuit was popularized by musicians – the effects pedals used by guitarists are almost all powered by center-negative coaxial jacks which are often destroyed when a random wall wart was used to power them. Adding a reverse protection diode dropped the voltage too much to use the normally supplied wall wart.

 

When power is applied, the mosfet body diode conducts from drain to source, which turns the mosfet on.

Yes, the body diode conducts until the load voltage goes above the MOSFET Vgs(th) at which point the MOSFET turns on and carries the load current.
The circuit works because MOSFETs conduct equally well in either direction when biased on.

 

You are both right, it will work if you can find a MOSFET that can supply enough current with a diode drop less gate drive.

 

if you can find a MOSFET that can supply enough current with a diode drop less gate drive.

Sorry, I don't understand what a diode-drop has to do with a MOSFET(?).

 

The body diode internal to the MOSFET supplies the source voltage.

 

The body diode internal to the MOSFET supplies the source voltage.

Only until the MOSFET starts to turn on.

 

True, that is load dependent.
Have we beaten this to death yet? :- )