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.
simulate this circuit – Schematic created using CircuitLab
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perhaps the diode in the MOSFET block diagram is coming into play? – Grady Player Mar 21 '13 at 03:55
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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.
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 parasitic 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:
JYelton
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Matt Young
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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. – IamDp Mar 21 '13 at 04:47
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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. – Matt Young Mar 21 '13 at 04:53
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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. – IamDp Mar 21 '13 at 05:12
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Awesssome ! I never observed that point. Thank you so much Matt. I completely got it now. – IamDp Mar 21 '13 at 05:23
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You have the values of the resistors in your voltage divider swapped. – Dave Tweed Mar 21 '13 at 11:23
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You write, "The MOSFET is driven into saturation". But ["saturation" means something else for MOSFETs](http://en.wikipedia.org/wiki/MOSFET#Modes_of_operation) than it does for BJTs, and I don't think is what's desired, or what happens here. – Phil Frost Aug 03 '13 at 12:36
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I'm sorry for being negative, but the explanation seems wrong. Consider the most simple (and the most common) PMOS: a symmetric PMOS used to implement digital logic. The Source and Drain terminals of this PMOS are identical. It means that if such PMOS was used in your example, then its source would have been defined by the most positive voltage. This would mean that this PMOS is initially biased to conduct, regardless of all the parasitic diodes. Now, either I'm wrong, or this explanation is wrong, or this explanation is valid for particular MOSFETs. Matt Young, which one of the above? – Vasiliy Aug 03 '13 at 13:21
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And, as @PhilFrost has already indicated, the MOSFET in your example will not (in general) operate in saturation mode. – Vasiliy Aug 03 '13 at 13:22
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A PMOS only conducts when gate < source, right? And when polarity is reversed gate>source (and also gate>=drain). – avl_sweden Jul 01 '15 at 18:35
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