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I read somewhere that this can be used for the reverse polarity protection in the circuit. But I am quite confused with its operation. Can somebody help me with this.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ perhaps the diode in the MOSFET block diagram is coming into play? \$\endgroup\$ – Grady Player Mar 21 '13 at 3:55
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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.

schematic

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:

schematic

simulate this circuit

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  • \$\begingroup\$ Hi Matt, I totally agree with your body-diode concept but when the reverse voltag is applied that is say 10V to Source and the drain is grounded. So now the Vgs is -10V which is well above the threshold and is good enough to keep the MOS in saturation region. So it is conducting in the opposite direction also now. Please correct me if i have said something which is incorrect. \$\endgroup\$ – Durgaprasad Mar 21 '13 at 4:47
  • \$\begingroup\$ Look back at the schematic. If the polarity is reversed, the diode will not conduct. No current flows, so the gate and source will be at 10V. \$V_{gs}=10V-10V=0V\$. The transistor does not turn on. \$\endgroup\$ – Matt Young Mar 21 '13 at 4:53
  • \$\begingroup\$ I got your point with the diode. But when the source is at 10V and gate is grounded and drain is also grounded. Then the Vgs is = -10V. Right? So I am thinking that the MOS should conduct (Usual configuration of PMOS), though the body diode is reverse biased. \$\endgroup\$ – Durgaprasad Mar 21 '13 at 5:12
  • \$\begingroup\$ When the polarity is reversed, ground becomes 10V. \$\endgroup\$ – Matt Young Mar 21 '13 at 5:18
  • \$\begingroup\$ Awesssome ! I never observed that point. Thank you so much Matt. I completely got it now. \$\endgroup\$ – Durgaprasad Mar 21 '13 at 5:23

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