Answer:
Step-by-step explanation:
To determine the drain current (ID) for each of the given values of VGS and VDS, we can use the following equation:
ID = kn/2 * (VGS - VT)^2 * (1 + λVDS)
where kn is the process transconductance parameter, VT is the threshold voltage, λ is the channel-length modulation parameter, and VDS is the drain-source voltage. In this case, kn = 0.18 mA/V^2, WL = 8, and VT = 0.4 V.
(a) For VGS = 0.8 V and VDS = 0.2 V:
ID = kn/2 * (VGS - VT)^2 * (1 + λVDS)
= 0.18 mA/V^2 / 2 * (0.8 V - 0.4 V)^2 * (1 + 0.01 V^-1 * 0.2 V)
= 0.036 mA
Therefore, the drain current is 0.036 mA for VGS = 0.8 V and VDS = 0.2 V.
(b) For VGS = 0.8 V and VDS = 1.2 V:
ID = kn/2 * (VGS - VT)^2 * (1 + λVDS)
= 0.18 mA/V^2 / 2 * (0.8 V - 0.4 V)^2 * (1 + 0.01 V^-1 * 1.2 V)
= 0.0846 mA
Therefore, the drain current is 0.0846 mA for VGS = 0.8 V and VDS = 1.2 V.
(c) For VGS = 0.8 V and VDS = 2.5 V:
ID = kn/2 * (VGS - VT)^2 * (1 + λVDS)
= 0.18 mA/V^2 / 2 * (0.8 V - 0.4 V)^2 * (1 + 0.01 V^-1 * 2.5 V)
= 0.1733 mA
Therefore, the drain current is 0.1733 mA for VGS = 0.8 V and VDS = 2.5 V.
(d) For VGS = 1.2 V and VDS = 2.5 V:
ID = kn/2 * (VGS - VT)^2 * (1 + λVDS)
= 0.18 mA/V^2 / 2 * (1.2 V - 0.4 V)^2 * (1 + 0.01 V^-1 * 2.5 V)
= 1.1916 mA
Therefore, the drain current is 1.1916 mA for VGS = 1.2 V and VDS = 2.5 V.