$\text{An n-channel MOSFET is connected such that}{\mathrm{ V}}_G=V_D.\text{Assume}{\mathrm{ V}}_{TH}=1\text{V,}{\mathrm{ V}}_{DD}=5\text{V, and}{\mathrm{ \mu }}_n{C}_{\mathrm{ox}}\left(\frac{W}{L}\right)=2\text{mA}{V}^{-2}.$

$\text{The gate voltage}\left(V_G\right)\text{of the n-channel MOSFET (in Volt) is ----.}$ $\text{(rounded off to two decimal places)}$

$\text{<br>}$

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The correct answer is 2.56.

Based on the provided circuit diagram:
- The supply voltage is $V_{DD}=5\text{V}$.
- The drain resistor is $R_D=1\text{k}\Omega$.
- The gate and drain terminals of the n-channel MOSFET are connected together, which means $V_G=V_D$.
- The source terminal of the MOSFET is connected directly to ground, so $V_S=0\text{V}$.
- Consequently, the gate-to-source voltage is $V_{GS}=V_G-V_S=V_G$ and the drain-to-source voltage is $V_{DS}=V_D-V_S=V_G$.

For an n-channel MOSFET to conduct, we must have $V_{GS}>V_{TH}$, where $V_{TH}=1\text{V}$.
Since $V_{DS}=V_{GS}$, the condition for the saturation region is satisfied because:

$V_{DS}>V_{GS}-V_{TH}\Rightarrow V_{GS}>V_{GS}-1$

Thus, the MOSFET operates in the saturation region.

The drain current $I_D$ flowing through the drain resistor $R_D$ is given by Ohm's law:

$I_D=\frac{V_{DD}-V_D}{R_D}=\frac{5-V_G}{1\text{k}\Omega }=(5-V_G)\text{mA}$

The drain current in the saturation region is also given by the MOSFET current equation:

$I_D=\frac{1}{2}{\mu }_n{C}_{ox}\left(\frac{W}{L}\right){(V_{GS}-V_{TH})}^2$

Substituting the given parameters ${\mu }_n{C}_{ox}\left(\frac{W}{L}\right)=2{\text{mA V}}^{-2}$ and $V_{TH}=1\text{V}$:

$I_D=\frac{1}{2}·2·{(V_G-1)}^2\text{mA}={(V_G-1)}^2\text{mA}$

Equating the two expressions for the drain current in mA:

${(V_G-1)}^2=5-V_G$

Expanding the left-hand side gives:

$V_G^2-2V_G+1=5-V_G$

Rearranging the quadratic equation:

$V_G^2-V_G-4=0$

Using the quadratic formula to solve for $V_G$:

$V_G=\frac{-(-1)\pm \sqrt{{(-1)}^2-4\left(1\right)(-4)}}{2\left(1\right)}$

Simplifying the expression under the square root:

$V_G=\frac{1\pm \sqrt{1+16}}{2}=\frac{1\pm \sqrt{17}}{2}$

Since $V_G$ must be greater than the threshold voltage $V_{TH}=1\text{V}$ for the MOSFET to be turned on, we discard the negative root:

$V_G=\frac{1+\sqrt{17}}{2}$

Calculating the value:

$V_G\approx \frac{1+4.1231}{2}\approx 2.5615\text{V}$

Rounding off to two decimal places gives $V_G=2.56\text{V}$.