Please convert the below content to rendered equations without any changes and errors. The MOSFET stage shown is a **common-drain (source follower)** amplifier using **2N7002 MOSFET**.
We will find:

Question

1. Small-signal **input resistance (R_{in})**
2. **Output resistance (R_{out})**
3. **Approximate voltage gain (A_v)**

Given:

* (g_m = 0.32,S)
* (R_1 = 750,\Omega)
* (R_2 = 400k\Omega)
* (R_3 = 400k\Omega)
* Drain connected to (15V) (AC ground)
* Coupling capacitors (C_1) and (C_2) are **large**, so treated as **short circuits for AC analysis**.

---

# 1️⃣ Small-Signal Input Resistance (R_{in})

For a MOSFET:

* Gate current ≈ **0**
* So input resistance is determined by the **bias resistors**

[
R_{in} = R_2 \parallel R_3
]

[
R_{in} = \frac{400k 	imes 400k}{400k + 400k}
]

[
R_{in} = 200k\Omega
]

✅ **Input Resistance**

[
\boxed{R_{in} = 200,k\Omega}
]

---

# 2️⃣ Small-Signal Output Resistance (R_{out})

For a **source follower**:

[
R_{out} \approx \left(\frac{1}{g_m}ight) \parallel R_1
]

First compute:

[
\frac{1}{g_m} = \frac{1}{0.32}
]

[
\frac{1}{g_m} = 3.125\Omega
]

Now parallel with (R_1):

[
R_{out} = 3.125 \parallel 750
]

Since (750 \gg 3.125),

[
R_{out} \approx

VIPSolutions · Accepted Answer

# Common-Drain (Source Follower) MOSFET Amplifier Analysis

The MOSFET is a **common-drain ( follower)** amplifier using a **2N7002 MOSFET**.

We will find:

1. Small-signal **input resistance** $(R_{in})$
2. **Output resistance** $(R_{out})$
3. **Approximate voltage gain** $(A_v)$

## Given

- $g_m = .32\,\mathrm{S}$
- $R_1 = 750\,\Omega$
- $R_2 = 400\,\text{k}\Omega$
- $R_3 = 400\,\text{k}\Omega$
- Drain connected to $15\,\mathrm{V}$ (AC ground)
- Coupling capacitors $C_1$ and $C_2$ are **large** (AC short circuits)

---

## 1️⃣ Small-Signal Input Resistance $(R_{in})$

For a MOSFET:

- Gate current is approximately zero.
- Input resistance is determined by the bias resistors.

$$
R_{in} = R_2 \parallel R_3
$$

$$
R_{in} = \frac{400k \times 400k}{400k + 400k}
$$

$$
R_{in} = 200k\Omega
$$

**Input Resistance:**

$$
\boxed{R_{in} = 200\,k\Omega}
$$

---

## 2️⃣ Small-Signal Output Resistance $(R_{out})$

For a **source follower**:

$$
R_{out} \approx \left(\frac{1}{g_m}\right) \parallel R_1
$$

Compute:

$$
\frac{1}{g_m} = \frac{1}{.32}
$$

$$
\frac{1}{g_m} = 3.125\,\Omega
$$

Parallel with $R_1$:

$$
R_{out} = 3.125 \parallel 750
$$

Since $750 \gg 3.125$:

$$
R_{out} \approx 3.1\,\Omega
$$

**Output Resistance:**

$$
\boxed{R_{out} \approx 3.1\,\Omega}
$$

---

## 3️⃣ Voltage Gain $(A_v)$

For a **source follower**:

$$
A_v = \frac{g_m R_1}{1 + g_m R_1}
$$

Calculate:

$$
g_m R_1 = .32 \times 750
$$

$$
g_m R_1 = 240
$$

So,

$$
A_v = \frac{240}{241}
$$

$$
A_v \approx .996
$$

**Voltage Gain:**

$$
\boxed{A_v \approx .996\ (\approx 1)}
$$

A voltage gain near 1 is expected for a source follower.

---

## ✅ Final Answers

| Quantity          | Value            |
| ----------------- | ---------------- |
| Input Resistance  | $200\,k\Omega$       |
| Output Resistance | $\approx 3.1\,\Omega$      |
| Voltage Gain      | $\approx .996$ ($\approx 1$) |

---

## ✅ Quick Intuition

This circuit acts as a **buffer amplifier**:

- **Very high input resistance**
- **Very low output resistance**
- **Gain ≈ 1**

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