Consider an ideal long channel nMOSFET (enhancement-mode) with gate length 10 $\mu$m and width 100 $\mu$m. The product of electron mobility ($\mu_n$) and oxide capacitance per unit area ($C_{OX}$) is $\mu_n C_{OX} = 1\ \text{mA/V}^2$. The threshold voltage of the transistor is 1 V. For a gate-to-source voltage $V_{GS} = [2 - \sin(2t)]$ V and drain-to-source voltage $V_{DS} = 1$ V (substrate connected to source), the maximum value of the drain-to-source current is ________________.
Show Hint
- 40 mA
- 20 mA
- 15 mA
- 5 mA
The Correct Option is C
Solution and Explanation
Step 1: Determine maximum $V_{GS$.}
\[ V_{GS}(t) = 2 - \sin(2t). \] Since $\sin(2t)$ ranges from $-1$ to $+1$: \[ V_{GS,\max} = 2 - (-1) = 3\ \text{V}. \]
Step 2: Check region of operation.
Threshold voltage $V_T = 1$ V.
At maximum $V_{GS} = 3$ V, \[ V_{GS} - V_T = 2\ \text{V}. \] Given $V_{DS} = 1$ V<$(V_{GS} - V_T) = 2$ V, the MOSFET operates in the \textit{linear (triode) region}.
Step 3: Drain current in linear region.
\[ I_D = k_n \left[ (V_{GS} - V_T) V_{DS} - \frac{V_{DS}^2}{2} \right]. \] Substitute maximum values: \[ I_{D,\max} = 10\ \text{mA/V}^2 \left[ (2)(1) - \frac{1}{2} \right] = 10\ \text{mA/V}^2 \left[ 2 - 0.5 \right] = 10 \times 1.5. \] \[ I_{D,\max} = 15\ \text{mA}. \]
Thus, the maximum drain current is 15 mA.
Final Answer: 15 mA
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