How does the voltage difference develop when MOSFET is biased with Current Source?
In a MOSFET switched as a diode there's a local feedback going on.

Imagine that the current source is 100 uA, so Id = 100 uA. Now what if the Vgs of that transistor was very high, much higher than Vt. What would happen?

There would be almost no voltage across Vds right?

Since Vds = Vgs in this circuit, the above cannot be true. Vds cannot be very small.

What will Vds be then ?

Well equal to Vgs (obviously) so the Vds must end up at a value which results in a Vgs which makes Ids = 100 uA flow.

Suppose this goes wrong for some reason and we end up with a Vgs which is a little too low resulting in the NMOS wanting to make 90 uA flow instead of 100 uA.

So 100 uA is coming from above (current source), 90 uA is pulled from below (the NMOS). Now what happens to the voltage at the drain of the NMOS ?

The voltage will rise because the current source insists on making 100 uA flow so it raises the voltage hoping to make the NMOS conduct more so that it will allow 100 uA to flow instead of only 90 uA.

This rising voltage means Vds increases so also Vgs increases. And AHA our Vgs was a bit low. That's the feedback in action, Vgs is automatically  increased if it is too low.

Same is true for a Vgs that is too large, then the NMOS wants to conduct more current, for example 110 uA. That would make Vgs lower which would make the NMOS pull less current so that the 100 uA is decreased to 100 uA.