Consider the NMOS circuit shown

The drain to source voltage is VDS = (Vt + 2) volts, then drain current iD is:

\(\left( Assume~{{K}_{n}}={{\mu }_{n}}C_{ox}~\left( \frac{W}{L} \right) \right)\)

1
iD = 2 Kn
2
iD = Kn
3
\({{i}_{D}}=\frac{{{K}_{n}}}{2}\)
4
\({{i}_{D}}=\frac{{{K}_{n}}}{4}\)

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