Electronic – Reverse polarity protection – PMOS vs Schottkey diode

The bigger problem you're likely to run into is operation under forward bias conditions. Schottky diodes still have a voltage drop under forward bias, say 0.25V.

That means at 100mA, you're dissipating 25mW of power. Better than a standard silicon diode, but not great especially for a battery constrained device.

A better way to get reverse bias protection is to use a P-Channel MOSFET. MOSFET's act more like a resistor when saturated, and it's possible to get MOSFETs with low on resistances.

Let's assume we have a 1 ohm on resistance. At 100mA, that's a 0.1 V drop across the MOSFET and 10mW dissipation. 1 ohm on resistance is kind of lousy for a MOSFET, you can get some which have significantly less on resistance. I'm not entirely sure about the leakage current through MOSFET's, but I seem to remember it being quite small.

To hook up the mosfet:

Connect the drain to the positive battery terminal, connect the gate to the negative terminal, and connect your load to the source. For added protection you can add a zener diode and a resistor across the source/gate.

A more complete explanation can be found here.

This is a very handy reverse polarity protection scheme.

A P channel MOSFET turns on when \$V_{gs}\$ is negative, probably starting around -3V to -5V. When power is switched on, \$V_{gs}=0V\$. This is where the parasitic diode drawn across the MOSFET in the datasheet comes into play (drawn for explanation purposes only, do not put a diode between the drain and source). It allows current to flow, drain to source, and drops about 1V.

^{simulate this circuit – Schematic created using CircuitLab}

Assuming a supply voltage of 10V:

$$V_{gs}=V_g-V_s=0V-9V=-9V$$

The MOSFET is driven into saturation. Since the MOSFET's on resistance, \$R_{ds(on)}\$, is around \$25m\Omega\$, the parasitc diode gets shorted.

If the polarity is reversed, the parasitic diode never conducts, thus never allowing the MOSFET to turn on.

You need to be mindful of the maximum \$V_{gs}\$ in the datasheet. The MOSFET may be able to switch 100V, but the maximum \$V_{gs}\$ may only be 20V. In that case, you must put some kind of voltage divider in place to protect the gate of the MOSFET:

^{simulate this circuit}