GATE EC 2022 Set 1 — Question 40

The nMOSFET parameters are given as:
Threshold voltage $V_T = 1 \text{ V}$
Electron mobility-oxide capacitance product $\mu_n C_{ox} = 1 \text{ mA/V}^2$
Gate length $L = 10 \text{ µm}$
Width $W = 100 \text{ µm}$
Gate-to-source voltage $V_{GS} = [2 - \sin(2t)] \text{ V}$
Drain-to-source voltage $V_{DS} = 1 \text{ V}$First, calculate the aspect ratio: $W/L = 100 \text{ µm} / 10 \text{ µm} = 10$.The range of $V_{GS}$ is determined by the range of $\sin(2t)$, which is $[-1, 1]$:
Minimum $V_{GS} = 2 - 1 = 1 \text{ V}$ (when $\sin(2t) = 1$)
Maximum $V_{GS} = 2 - (-1) = 3 \text{ V}$ (when $\sin(2t) = -1$)For the transistor to conduct, $V_{GS} \ge V_T$. Since $V_T = 1 \text{ V}$, the transistor conducts for $V_{GS} \in [1, 3] \text{ V}$.Next, determine the operating region based on $V_{DS}$ and $V_{GS} - V_T$.
$V_{GS} - V_T = (2 - \sin(2t)) - 1 = 1 - \sin(2t)$.1. Saturation Region: $V_{DS} \ge V_{GS} - V_T$
$1 \text{ V} \ge 1 - \sin(2t) \implies \sin(2t) \ge 0$.
Also, for saturation, $V_{GS} > V_T$, which means $2 - \sin(2t) > 1 \implies \sin(2t) < 1$.
So, the saturation region is for $0 \le \sin(2t) < 1$.
The drain current in saturation is $I_D = \frac{1}{2} \mu_n C_{ox} \frac{W}{L} (V_{GS} - V_T)^2$.
$I_D = \frac{1}{2} (1 \text{ mA/V}^2) (10) (1 - \sin(2t))^2 = 5 (1 - \sin(2t))^2 \text{ mA}$.
To maximize $I_D$ in this region, we need to maximize $(1 - \sin(2t))^2$. This occurs when $\sin(2t)$ is minimum in the range $[0, 1)$, i.e., $\sin(2t) = 0$.
Maximum $I_D$ in saturation = $5 (1 - 0)^2 = 5 \text{ mA}$.2. Linear (Triode) Region: $V_{DS} < V_{GS} - V_T$
$1 \text{ V} < 1 - \sin(2t) \implies \sin(2t) < 0$.
So, the linear region is for $-1 \le \sin(2t) < 0$.
The drain current in the linear region is $I_D = \mu_n C_{ox} \frac{W}{L} \left( (V_{GS} - V_T) V_{DS} - \frac{1}{2} V_{DS}^2 \right)$.
$I_D = (1 \text{ mA/V}^2) (10) \left( (1 - \sin(2t)) (1) - \frac{1}{2} (1)^2 \right)$.
$I_D = 10 \left( 1 - \sin(2t) - \frac{1}{2} \right) = 10 \left( \frac{1}{2} - \sin(2t) \right) \text{ mA}$.
To maximize $I_D$ in this region, we need to minimize $\sin(2t)$. The minimum value of $\sin(2t)$ is $-1$.
Maximum $I_D$ in linear region = $10 \left( \frac{1}{2} - (-1) \right) = 10 \left( \frac{3}{2} \right) = 15 \text{ mA}$.Comparing the maximum currents from both regions, the overall maximum drain-to-source current is $15 \text{ mA}$. The phrase "appropriately biased in the saturation region" refers to the typical operating point, but the question asks for the maximum current, which occurs when the device enters the linear region due to the varying $V_{GS}$.The final answer is $\mathbf{15 \text{ mA}}$.