Electronic – Do DC/DC converters provide reverse polarity protection

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}

Given the additional information that it is an LT8610, there looks like something that can be done.

The EN/UV input has a 1.0V +/-6% threshold and it has almost no (+/-20nA) bias current. That's with the input rising, it's 40mV less typically with the input falling.

You already have a voltage divider on the input. All you need is a resistor from Vout to EN/UV.

^{simulate this circuit – Schematic created using CircuitLab}

When the UV input is RISING, \$V_{OUT} = 0\$ and thus it will trigger at \$V_{IN}\$ =\$ 1.0V \cdot \$\$(R1 + R2 || R3) \over (R2 ||R3)\$

It is easy\$^{TM}\$ to show that the hysteresis: is \$0.04V\cdot\$ \$R1\over R1||R2||R3\$ + \$V_X \cdot( \$ \$ R1 \over R3\$), where \$V_X \$is the output voltage.