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I guess my question was not clear.
Let me split my questions.
Assume VDD=10V.
Requirement for Iout = 10mA.
What should be R value?How to calculate it?
MOSFET PN: IRFL210
Saturation is the mode in the MOSFET where it is fully operational conducting (non-linear mode or amplifying mode).
Based on that, the circuit you are presenting is a double MOSFET in mirror mode (Current in M1 is approx the same current that in M2 supposing that M1 and M2 are almost identical transistors)
R is calculated based on the desired current and the output Iout, Vdd and the drop voltage of M1 (that based on your PDF is around 4V, using an Iref = 10mA
Based on that
\[R=\frac{ { V}_{DD }- { V}_{GS }}{ {I }_{REF }}\]
In the datasheet, VGS min=2 & max=4V.
On What basis Vgs would vary?
on keeping the Vdd within +/-1% accuracy, R +/-1%, with varrying Vout. What is the variation expected in Vgs?
No, in the datasheet the \[{V }_{ GS}\] may vary from 2V to 4V. But that is the case when \[{V }_{ GS} = {V }_{ DS}\], check always for the test conditions.
By the way, I got the information using Fig 1 and Fig 3 from the datasheet.
As you can see, for a \[{V }_{ DS}=10V\], the current you are trying to use (\[{10 }^{-2 }A\]), won't produce any given \[{V }_{ GS}\], thus your transistor can disipate much more power than the case you are giving to it.
Based on the Fig 3, the \[{V }_{ GS}\] is tending to be 4V when \[{I }_{ D}\] is very small, so I use this value for \[{V }_{ GS}\]
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