In the circuit shown MOSFET has \({\mu _n}{C_{ox}}\frac{W}{L} = 2

SOLUTION

Concept:

The current in a MOSFET operating in saturation is given by:

\({I_{DS}} = \frac{{{K_N}}}{2}{\left( {{V_{GS}} - {V_t}} \right)^2} \)

For MOSFET to remains in the saturation region:

V_DS ≥ V_GS - V_T

Application:

V_GS = 2V  {since I_G ≃ ; No drop in 1 MΩ}

\({I_{DS}} = \frac{{{K_N}}}{2}{\left( {{V_{GS}} - {V_t}} \right)^2} \)

\(I_{DS}= \frac{2}{2}{\left( {2 - 1} \right)^2}\)

I_DS = 1 mA

V_DS = V_DD - I_DS R_D

V_DS = 12 - R_D

For MOSFET to remain in saturation Region.

12 - R_D ≥ V_GS - V_t

12 - R_D ≥ 2 – 1

12 - R_D ≥ 1

R_D – 12 ≤ -1

R_D ≤ 11 kΩ 

For R_D ≤ 11 kΩ; MOSFET is in saturation Region

For R_D > 11 kΩ; MOSFET is in Triode Region.

Option (2) and (4) are correct.