How does this power mosfet work in a constant current load?

[onemilimeter]

Imagine that you MOSFETs, for a drain current (Id) of 10A at a given drain voltage,  one requires a Vgs of 3v and the other a Vgs of 5v.
Plug the first MOSFET into the circuit and adjust the pot for 10A, the Op-Amp output will be Vgs +(0.1 ohm * Id) or 4v.
Change to the second MOSFET without changing the pot setting.
Initially the drain current will be lower leading to a reduced voltage across the 0.1 ohm resistor, this will cause the input voltage to the Op-Amp to rise (ie difference between the voltage set by the pot and the voltage across the 0.1ohm resistor).
Thus the output of the Op-Amp will rise until equilibrium is reached.
For the Op-Amp output to rise by 2v (from 4 to 6v) the voltage across the 0.1 ohm resistor has to change by 2v/gain or 20uV for an Op-Amp gain of 100,000.
Through the action of negative feedback the Op-Amp has reduced the effect of a change in Vgs of 2v to a change in Id of 20uV/0.1ohm = 200uA.

Hi...

Great explanation! I love reading it... Thanks a lot!