To solve the given questions, we will go step by step using the given data. Below are the problem details and corresponding solutions.

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(a) State the conditions to verify the assumptions regarding the Saturation Mode of a MOSFET.

For the MOSFET to be in the Saturation Mode, the following conditions must be satisfied:

1. $V_{GS} > V_T$: The gate-to-source voltage must exceed the threshold voltage $V_T$.
2. $V_{DS} \geq V_{GS} - V_T$: The drain-to-source voltage must be greater than or equal to $V_{GS} - V_T$.

Given:

• $V_T = 1 \, \text{V}$.

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(b) Analyze the circuit and calculate $V_{out}$ for $R_D = 0.1 \, k\Omega$. Validate the assumption whether the MOSFET is in saturation mode.

Step 1: Analyze the circuit.

For Input = Logic HIGH,

• $V_{in} = 5 \, \text{V}$.
• The MOSFET gate voltage, $V_{GS} = V_{in} = 5 \, \text{V}$.
• The drain voltage is $V_{out}$, and $V_{DS} = V_{out} - 0 \, \text{V} = V_{out}$.

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Step 2: Check the MOSFET mode.

1. Check $V_{GS} > V_T$:

   • $V_{GS} = 5 \, \text{V}, V_T = 1 \, \text{V}$.
   • $5 > 1$: This condition is satisfied.

2. Check $V_{DS} \geq V_{GS} - V_T$:

   • $V_{DS} = V_{out}, V_{GS} - V_T = 5 - 1 = 4 \, \text{V}$.
   • Assume saturation mode: $V_{out} \geq 4 \, \text{V}$.
   • We will validate this later.

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Step 3: Apply MOSFET equations to calculate $V_{out}$.

In saturation mode,
$I_{D} = \frac{k}{2} (V_{GS} - V_T)^2$

Substitute given values:

• $k = 4 \, \text{mA/V}^2, V_{GS} = 5 \, \text{V}, V_T = 1 \, \text{V}$.

$I_{D} = \frac{4}{2} (5 - 1)^2 = 2 \cdot 16 = 32 \, \text{mA}.$

Now, calculate $V_{out}$:
$V_{out} = V_{DD} - I_{D} R_{D},$
where $V_{DD} = 10 \, \text{V}, R_{D} = 0.1 \, \text{k}\Omega = 100 \, \Omega$.

$V_{out} = 10 - (32 \times 0.1) = 10 - 3.2 = 6.8 \, \text{V}.$

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Step 4: Validate saturation assumption.

For saturation mode, $V_{out} \geq 4 \, \text{V}$.
Here, $V_{out} = 6.8 \, \text{V}$, so the MOSFET is indeed in saturation mode.

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(c) Analyze the modified circuit for $R_D = 1 \, k\Omega$ and calculate $V_{out}$.

Step 1: Recalculate $I_{D}$ using the same formula.

$I_{D} = \frac{4}{2} (5 - 1)^2 = 32 \, \text{mA}.$

Step 2: Recalculate $V_{out}$:

$V_{out} = V_{DD} - I_{D} R_{D},$
where $R_{D} = 1 \, \text{k}\Omega = 1000 \, \Omega$.

[ V_{out} = 10 - (32 \times 1) = 10 - 32 = -22 , \text{V}. \