An $n$-channel MOSFET is connected as shown in the figure below. Assume $V_{TH}=1$ V, $V_{DD}=5$ V, and $\mu C_{ox}\left(\dfrac{W}{L}\right)=2\text{ mA V}^{-2}$, and neglect channel length modulation effects. The gate voltage $V_G$ of the $n$-channel MOSFET (in Volt) is . (Rounded off to two decimal places)

Question

An NMOS transistor operating in the linear region has $I_{DS} = 5 \, \mu{A}$ at $V_{DS} = 0.1 \, {V}$. Keeping $V_{GS}$ constant, the $V_{DS}$ is increased to $1.5 \, {V}$. Given that: \[ \mu_n C_{ox} \frac{W}{L} = 50 \, \mu{A}/{V}^2, \] the transconductance at the new operating point (in $\mu{A}/{V}$, rounded off to two decimal places) is $\_\_\_\_$.

Answer 1

Step 1: Understanding the Concept:
A MOSFET with its gate tied to its own drain always ends up biased in saturation, because the drain to source voltage automatically stays above the value needed to keep the channel pinched off. This lets us use a single saturation current equation together with the resistor's voltage drop to pin down the gate voltage.

Step 2: Key Formula or Approach:
Let $x=V_G-1$ be the overdrive voltage, since $V_{TH}=1\text{ V}$. The saturation current is $I_D=1\cdot x^2$ mA, because half of $2\text{ mA/V}^2$ is $1\text{ mA/V}^2$, and the resistor equation gives $I_D=5-(x+1)=4-x$ mA.

Step 3: Detailed Explanation:
Setting the two expressions for $I_D$ equal,
\[ x^2 = 4-x \]
\[ x^2+x-4=0 \]
Solve using the quadratic formula:
\[ x=\frac{-1\pm\sqrt{1+16}}{2}=\frac{-1\pm\sqrt{17}}{2} \]
Taking the positive root since the overdrive voltage must be positive for the device to conduct,
\[ x=\frac{-1+4.123}{2}\approx1.5616\text{ V} \]
Then the gate voltage is
\[ V_G=x+1\approx1.5616+1=2.5616\text{ V} \]
which rounds to $2.56$ V, matching the value obtained by solving directly for $V_G$.

Step 4: Final Answer:
\[ \boxed{V_G\approx2.56\text{ V}} \]

Show Hint

Correct Answer: 2.56

Solution and Explanation

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