reverse-polarity
I am working on reverse polarity protection circuit. I found out that most of the people are using the circuit as is mentioned below.
I have few concerns regarding the direction of MOSFET
Did you see (and understand) Spehro Pefhany's comment? I take the liberty of turning it into an answer because that's also what I'd suggest.
Add a diode in front of your circuit that shorts the voltage if the battery is reversed (=V2). In that case the (resettable) fuse will break the current flow. If the battery is not reversed (=V1) there is almost no voltage drop (only a small one across the fuse because it won't have 0 resistance):
simulate this circuit – Schematic created using CircuitLab
The diode must pass enough current at reversed polarity to deactivate the fuse. The fuse, of course, must be able to sustain enough current for supplying the circuit in normal operation but must blow/deactivate if the battery is shorted via the diode.
- Main question: I have got no clue whatsoever what the capacitor is doing here. What is it for and how do I choose its value?
The capacitor is to ensure the circuit works well when there is a rapid change in input voltage polarity. It will decharge the gate-source capacitance very fast, turning off the mosfet as fast as possible.
From the picture above, you can see that without the capacitor C1 the mosfet keeps conducting because Vgs=V(load, gate) > Vth for a short time. Check also I(R2) which becomes about minus 150 mA.
Regarding the capacitor:
The capacitor will cause an inrush current when an input voltage is applied. Care should be taken whether the mosfet and zener diode can handle this peak current.
If (the value of) the capacitor is too large, this inrush current may be present for too long, still burning the mosfet and/or zener diode eventually.
If it is too small, it may not decharge the gate capacitance fast enough, as C is proportional to this current ( i = C du / dt ).
- How to choose the value of R? Do I only take the max current that the Zener diode can handle?
The value of R can choosen based on the input voltage, zener voltage and zener (test) current that is used to define the deviation of the zener voltage. (Datasheets typically state the min and max zener voltage for a given (test) current. $$ R = \frac{ V_{in}-V{zener} }{ Iz } $$ For low voltage zeners, this test current typically is 5 mA. You could choose to lower the zener current to reduce power dissipation, at the expense of reaching the rated zener voltage.
Best Answer
The P-FET is connected in reverse (drain towards +) so that when the input is backwards, not only is the FET gate OFF, but the body diode is reverse-biased. So that case is covered: reverse current is blocked with the backwards battery.
When the battery connected normally, the gate-drain voltage Vgs turns on the FET and it conducts with a very low Rds(on).
As it happens, the body diode never really comes into play although it doesn't actually hurt anything. This is because when the FET Vgs voltage is ON, the current flows from drain to source (yes, this works - enhancement FETs conduct in either direction). Because Rds(on) is low, not enough voltage develops to forward-bias the body diode above its threshold, so it doesn't conduct.
So the 'backwards' connected FET behaves as a nearly ideal diode, with almost no forward voltage drop. Neat, huh?
Now, what's that Zener for? Because this is a MOSFET, there's a maximum gate-source (Vgs) voltage it can sustain, which is usually +/- 10V for most power FETs. The Zener limits Vgs to a safe level (in this case, 10V) to prevent damaging the device. If the input voltage were only ~10V this would not be an issue, and the gate could be directly connected to GND.
Simulate it here.