Let's solve the problem **step-by-step** for the common-source amplifier with the given circuit. --- ## 1. **Small-Signal Model** The circuit shown is a standard **common-source MOSFET amplifier** with a resistive load. We'll use small-signal analysis for voltage gain, input resistance, and output resistance. Let the MOSFET parameters be: - \( g_m \): transconductance - \( r_{ds} \): output resistance of MOSFET --- ## 2. **Voltage Gain (\(A_v\))** ### **Small-Signal Equivalent** - The input signal \( v_{in} \) is coupled to the gate through \( C_1 \). - **Gate** draws negligible current (MOSFET input is open circuit for DC). - The output is taken at the drain. #### **Voltage Gain Derivation** The small-signal voltage gain \(A_v\) is: \[ A_v = \frac{v_{out}}{v_{in}} \] \[ A_v = -g_m \cdot (R_D \parallel r_{ds} \parallel R_L) \] - \( g_m \): MOSFET transconductance - \( R_D \): Drain resistor - \( r_{ds} \): MOSFET output resistance - \( R_L \): Load resistor (if connected across output) **The negative sign** indicates a 180° phase shift (common-source characteristic). --- ## 3. **Input Resistance (\(R_{in}\))** The input resistance is determined by the resistance seen at the gate: \[ R_{in} = R_G = R_1 \parallel R_2 \] - \( R_1, R_2 \): Gate biasing resistors If the signal source is coupled through a series resistance \(R\), include that as well: \[ R_{in,total} = R + (R_1 \parallel R_2) \] But looking into the gate of MOSFET itself, it is infinite for an ideal MOSFET. --- ## 4. **Output Resistance (\(R_{out}\))** Output resistance is the resistance seen **looking back into the drain** of the MOSFET **with the input grounded**. \[ R_{out} = R_D \parallel r_{ds} \] If \(R_L\) is connected, output resistance seen by the load is: \[ R_{out,total} = (R_D \parallel r_{ds}) \parallel R_L \] --- ## 5. **Summary Table** | Quantity | Expression | |------------------|------------------------------------------------------------------------------| | **Voltage Gain** | \(A_v = -g_m \cdot (R_D \parallel r_{ds} \parallel R_L)\) | | **Input Resistance** | \(R_{in} = R_1 \parallel R_2\) (or include source resistance if any) | | **Output Resistance** | \(R_{out} = R_D \parallel r_{ds}\) (without \(R_L\)), \((R_D \parallel r_{ds}) \parallel R_L\) (with \(R_L\)) | --- ## 6. **Explanation** - **Voltage Gain**: Caused by the transconductance \(g_m\) of the MOSFET and the total load resistance at the drain. - **Input Resistance**: High, determined by the biasing resistors, since the MOSFET gate draws negligible current. - **Output Resistance**: The parallel combination of the drain resistor and MOSFET output resistance. --- ## **Final Answers** ### **Voltage Gain:** \[ \boxed{A_v = -g_m \cdot (R_D \parallel r_{ds} \parallel R_L)} \] ### **Input Resistance:** \[ \boxed{R_{in} = R_1 \parallel R_2} \] ### **Output Resistance:** \[ \boxed{R_{out} = R_D \parallel r_{ds}} \] or with load, \[ \boxed{R_{out,total} = (R_D \parallel r_{ds}) \parallel R_L} \] If you need a step-by-step small-signal circuit diagram or a numerical example, let me know!

Let's analyze the common-source amplifier circuit step-by-step and derive the required expressions for voltage gain, input resistance, and output resistance. --- ### **Step 1: Understanding the circuit** From the given image, the key elements are: - **Gate**: connected to input voltage \( v_{in} \) through coupling capacitor \( C_1 \). - **Source**: may be grounded or connected via a resistor (not specified here, so assume grounded for simplicity). - **Drain**: connected to \( V_{DD} \) through a drain resistor \( R_D \). - **Load**: possibly connected at the output, but generally, the output is taken at the drain. - **MOSFET parameters**: transconductance \( g_m \), drain-source resistance \( r_{ds} \). --- ### **Step 2: Small-signal model** In the small-signal equivalent: - Gate: high impedance (input resistance \( R_{in} \)) - Drain: connected to \( R_D \) and possibly load \( R_L \) - Source: grounded (assuming common source configuration) --- ### **Step 3: Derive the voltage gain \(A_v\)** The small-signal drain current: \[ i_d = g_m v_{gs} \] Since the gate is driven by \( v_{in} \), and assuming the source is grounded: \[ v_{gs} = v_{in} \] The output voltage at the drain: \[ v_{out} = - i_d (R_{D} \parallel r_{ds}) = - g_m v_{in} (R_{D} \parallel r_{ds}) \] Including the load resistor \( R_L \), the effective load at drain: \[ R_{load} = R_D \parallel r_{ds} \parallel R_L \] So, the voltage gain: \[ \boxed{ A_v = \frac{v_{out}}{v_{in}} = - g_m R_{load} = - g_m (R_D \parallel r_{ds} \parallel R_L) } \] --- ### **Step 4: Input resistance \( R_{in} \)** The input resistance seen at the gate: - The gate of the MOSFET is effectively insulated; input resistance is very high. - The biasing network \( R_1 \) and \( R_2 \) set the bias point and are connected to ground. Thus: \[ \boxed{ R_{in} = R_1 \parallel R_2 } \] If the source has a resistor \( R_s \), then the input resistance looking into the gate remains dominated by the biasing resistors, as the gate draws negligible current. --- ### **Step 5: Output resistance \( R_{out} \)** Looking back into the drain: - The drain resistance is \( R_D \). - The transistor's output resistance is \( r_{ds} \). Therefore, the output resistance: \[ \boxed{ R_{out} = R_D \parallel r_{ds} } \] If there's an external load \( R_L \), the total output resistance seen at the output terminal is: \[ R_{out, total} = R_D \parallel r_{ds} \parallel R_L \] --- ### **### Final Summary:** | Quantity | Expression | |------------|------------| | **Voltage Gain** | \(\boxed{A_v = - g_m (R_D \parallel r_{ds} \parallel R_L)}\) | | **Input Resistance** | \(\boxed{R_{in} = R_1 \parallel R_2}\) | | **Output Resistance** | \(\boxed{R_{out} = R_D \parallel r_{ds}}\) (or include load \( R_L \) for total) | --- ### **Note:** - The negative sign indicates phase inversion typical of common-source amplifiers. - The transconductance \( g_m \) depends on the bias point: \( g_m = \sqrt{2 I_D \mu C_{ox} \frac{W}{L}} \) or is often provided. --- **If you need numerical values or further explanation, feel free to ask!**