The use of a MOSFET for reverse voltage protection is very straight forward.
Some of your references are correct but of low relevance and are tending to make the problem look more complex than it is. The key requirements (which you have essentially already identified) are
MOSFET must have enough Vds_max rating for maximum voltage applied
MOSFET Ids_max rating more than ample
Rdson as low as sensibly possible.
Vgs_max not exceeded in final circuit.
Power dissipation as installed able to sensibly handle operating power of I_operating^2 x Rdson_actual
Power dissipation as installed able to handle turn on and off higher dissipation regions.
Gate driven to cutoff "rapidly enough" in real world circuit.
(Worst case - apply Vin correctly and then reverse Vin instantaneously. Is cutoff quick enough?)
In practice this is easily achieved in most cases.
Vin has little effect on operating dissipation.
Rdson needs to be rated for worst case liable to be experienced in practice. About 2 x headlined Rdson is usually safe OR examine data sheets carefully. Use worst case ratings - DO NOT use typical ratings.
Turn on may be slow if desired but note that dissipation needs to be allowed for.
Turn off under reverse polarity must be rapid to allow for sudden application of protection.
What is Iin max ?
You don't say what I_in_max is and this makes quite a difference in practice.
You cited:
"If the drain-to-source voltage is zero, the drain current also becomes zero regardless of gate–to-source voltage. This region is at the left side of the VGS– VGS(th)= VDS boundary line (VGS – VGS(th) > VDS > 0).
and
Even if the drain current is very large, in this region the power dissipation is maintained by minimizing VDS(on)."
Note that these are relatively independent thoughts by the writer. The first is essentially irrelevant to this application.
The second simply says that a low Rdson FET is a good idea.
You said:
Does this configuration fall under the VDS = 0 classification? That seems like a somewhat dangerous assumption to make in a noisy environment (this will be operating in the vicinity of various types of motors), as any voltage offsets between input supply ground and local ground could cause current to flow. Even with that possibility, I'm not sure I need to spec for my maximum load current on the drain current ID. It would then follow that I don't need to dissipate very much power either. I suppose I could mitigate the problem by Zener clamping VGS closer to VGS(th) to reduce drain current/voltage?
Too much thinking :-).
When Vin is OK get FET turned on asap.
Now Vds is as low as it is going to get and is set by Ids^2 x Rdson
Ids = your circuit current.
At 25C ambient Rds will start at value cited at 25C in spec sheet and will rise if/as FET heats. In most cases FET will not heat vastly.
eg 1 20 milliOhm FET at 1 amp gives 20 mW heating. Temperature rise is very low in any sensible pkg with minimal heatsinking. At 10A the dissipation = 10^2 x 0.020 = 2 Watts. This will need a DPAk or TO220 or SOT89 or better pkg and sensible heatsinking. Die temperature may be in 50-100C range and Rdson will increase over nominal 25C value. Worst case you may get say 40 milliOhm and 4 Watts. That is still easy enough to design for.
Added: Using the 6A max you subsequently provided.
PFet = I^2.R. R = P/i^2.
For 1 Watt disspation max you want Rdson = P/i^2 = 1/36 ~= 25 milliohm.
Very easily achieved.
At 10 milliohm P = I^2.R = 36 x 0.01 = 0.36W.
At 360 mW a TO220 will be warm but not hot with no heatsink but good airflow. A trace of flag heatsink will keep it happy.
The following are all under $1.40/1 & in stock at Digikey.
LFPACK 60V 90A 6.4 milliohm !!!!!!!!!!!
TO252 70V 90A 8 milliohm
TO220 60V 50A 8.1 milliohm
You said:
I suppose I could mitigate the problem by Zener clamping VGS closer to VGS(th) to reduce drain current/voltage?
No!
Best saved for last :-).
This is the exact opposite of what is required.
Your protector needs to have minimal impact on the controlled circuit.
The above has mjaximum impact and increases dissipation in protector over what can be achieved by using a sensibly low Rdson FET and turning it on hard.
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.
Best Answer
No.
In this circuit the MAX6326 is powered directly from the external source. Absolute maximum input voltage rating is -0.3V to +6V, so reverse polarity will kill it. And you require the circuit to accept 12V, which will also kill it. There is no simple solution to this problem.
Reverse battery protection is also iffy. If the battery is connected in reverse with external power off then the FET's internal body diode will stop the reverse voltage from getting to the output. However if external power is then applied the FET will turn on for at least 100ms until the MAX6326 comes out of reset. During this time a very large current will flow from the external power supply into the battery. The components will probably survive this momentary surge current, but the high stress could reduce reliability.
Given the problems with this circuit you will have to look for other solutions that don't involve a MAX6326. The simplest is to just use a Schottky diode in place of Q1. If you choose a high current Schottky the voltage drop could be less than 0.3V. If this is too much then just add another cell to the battery to make 4.5V, then a standard silicon diode will do the job.