[summary: plan on responding to over-current in 1.16 microsecond]
The thermal time constant of a cubic meter of silicon is 11,600 seconds (the inverse of the thermal_diffusion), computed by multiplying the specific-heat of a cubic meter with the thermal-resistance of a cubic meter.
The thermal time constant of 0.1 cubic meter (4" on a side) is 100X faster, at 11600/100 or 116 seconds.
The thermal time constant of 0.01m (1centimeter or 0.4 inches) is another factor of 100X faster, at 11600/10000 = 1.6 seconds.
This size --- 1cm --- is much larger than the junction depth of a MOSFET. Which encapsulated the channel where the heat is generated, and will vary depending on the MOSFET physical design.
The active part of the FET will be about 100 microns, so the heat can be easily dumped into a COPPER TAB and pulled out of the package.
We now have a size we can compute with ------ 100 microns, or 0.1 milliMeters of depth.
What is the thermal time constant of 100 microns?
That ---- 100 microns ---- is 10*10 thinner than our previous numbers at 1cm, thus the thermal time constant is
11600 / (100 * 100 * 100 * 100) = 11600 / 100,000,000
== 11600 sec no (this is for 1 cubic meter)
== 116 sec no ( this is for 0.1 cubic meter)
== 1.16 sec no (this is for 1cm cube)
== 0.0116 sec no (this is for 1mm cube)
== 0.000116 sec YES (this, 116 microsecond, is for 100 micron cube)
Thus your protection circuit needs to be AT LEAST THIS FAST.
However, some FETs use just the top 10 microns or so for the FET action, and you need to protect that 10micron region from overheating, by protecting in 1.16 microseconds.
Thus, in summary, you need to aim for 1.16 microsecond response time.