nMOSFET Problem Overview

This problem involves analyzing the behavior of an ideal long channel nMOSFET (enhancement-mode) to determine its maximum drain-to-source current. We are provided with the transistor's physical dimensions, material properties, threshold voltage, and time-varying gate-to-source voltage, along with a constant drain-to-source voltage.

Given Parameters and Values

Let's list the key parameters provided in the question:

• Transistor type: Ideal long channel nMOSFET (enhancement-mode)
• Gate length ($L$): $10 \, \mu\text{m}$
• Width ($W$): $100 \, \mu\text{m}$
• Product of electron mobility ($\mu_n$) and oxide capacitance per unit area ($C_{OX}$): $\mu_n C_{OX} = 1 \, \text{mA/V}^2$
• Threshold voltage ($V_T$): $1 \, \text{V}$
• Gate-to-source voltage ($V_{GS}$): $[2 - \sin(2t)] \, \text{V}$
• Drain-to-source voltage ($V_{DS}$): $1 \, \text{V}$
• Substrate is connected to the source.

Transistor Operation Analysis

MOSFET Transconductance Parameter

First, we calculate the transconductance parameter, often denoted as $K$ or $\beta$, which is crucial for current calculations in an nMOSFET. It is given by:

$$K = \mu_n C_{OX} \frac{W}{L}$$

Substitute the given values:

$$K = (1 \, \text{mA/V}^2) \left( \frac{100 \, \mu\text{m}}{10 \, \mu\text{m}} \right)$$

$$K = (1 \, \text{mA/V}^2) \times 10$$

$$K = 10 \, \text{mA/V}^2$$

Gate-to-Source Voltage Range

The gate-to-source voltage ($V_{GS}$) is a time-varying signal given by $V_{GS} = [2 - \sin(2t)] \, \text{V}$. To find the maximum drain-to-source current, we need to consider the range of $V_{GS}$.

The sine function, $\sin(2t)$, oscillates between $-1$ and $1$.

• Maximum $V_{GS}$ value: This occurs when $\sin(2t)$ is at its minimum, which is $-1$. $$V_{GS,max} = 2 - (-1) = 3 \, \text{V}$$
• Minimum $V_{GS}$ value: This occurs when $\sin(2t)$ is at its maximum, which is $1$. $$V_{GS,min} = 2 - 1 = 1 \, \text{V}$$

So, the gate-to-source voltage ranges from $1 \, \text{V}$ to $3 \, \text{V}$. We are looking for the maximum drain-to-source current ($I_{DS}$). Generally, for an enhancement-mode nMOSFET, $I_{DS}$ increases with $V_{GS}$ when the transistor is ON. Therefore, the maximum $I_{DS}$ will occur at the maximum $V_{GS}$, which is $3 \, \text{V}$.

Determining Operating Region

An enhancement-mode nMOSFET operates in different regions based on the terminal voltages. For it to conduct current, $V_{GS}$ must be greater than $V_T$. In our case, $V_{GS,min} = 1 \, \text{V}$ which is equal to $V_T = 1 \, \text{V}$. This means the transistor can be in cut-off (no current), triode (linear), or saturation regions.

We are interested in the condition for maximum $I_{DS}$, which occurs at $V_{GS} = V_{GS,max} = 3 \, \text{V}$. Let's determine the operating region for $V_{GS} = 3 \, \text{V}$ and $V_{DS} = 1 \, \text{V}$.

First, calculate the overdrive voltage ($V_{OV}$), which is $V_{GS} - V_T$:

$$V_{OV} = V_{GS} - V_T = 3 \, \text{V} - 1 \, \text{V} = 2 \, \text{V}$$

Now, compare $V_{DS}$ with $V_{GS} - V_T$ to identify the operating region:

• If $V_{DS} < V_{GS} - V_T$, the MOSFET is in the Triode (Linear) Region.
• If $V_{DS} \ge V_{GS} - V_T$, the MOSFET is in the Saturation Region.

Given $V_{DS} = 1 \, \text{V}$ and $V_{GS} - V_T = 2 \, \text{V}$.

Since $1 \, \text{V} < 2 \, \text{V}$ ($V_{DS} < V_{GS} - V_T$), the nMOSFET is operating in the Triode (Linear) Region at the point where $I_{DS}$ is maximum.

Drain Current Calculation

Maximum Drain-to-Source Current

For an ideal long channel nMOSFET operating in the Triode (Linear) Region, the drain-to-source current ($I_{DS}$) is given by the formula:

$$I_{DS} = K \left[ (V_{GS} - V_T)V_{DS} - \frac{1}{2}V_{DS}^2 \right]$$

Now, substitute the values for maximum $I_{DS}$ (i.e., with $V_{GS} = 3 \, \text{V}$):

• $K = 10 \, \text{mA/V}^2$
• $V_{GS} = 3 \, \text{V}$
• $V_T = 1 \, \text{V}$
• $V_{DS} = 1 \, \text{V}$

$$I_{DS,max} = (10 \, \text{mA/V}^2) \left[ (3 \, \text{V} - 1 \, \text{V})(1 \, \text{V}) - \frac{1}{2}(1 \, \text{V})^2 \right]$$

$$I_{DS,max} = (10 \, \text{mA/V}^2) \left[ (2 \, \text{V})(1 \, \text{V}) - \frac{1}{2}(1 \, \text{V}^2) \right]$$

$$I_{DS,max} = (10 \, \text{mA/V}^2) \left[ 2 \, \text{V}^2 - 0.5 \, \text{V}^2 \right]$$

$$I_{DS,max} = (10 \, \text{mA/V}^2) \left[ 1.5 \, \text{V}^2 \right]$$

$$I_{DS,max} = 15 \, \text{mA}$$

Final Answer

The maximum value of the drain-to-source current is $15 \, \text{mA}$.