The MOSFET common-source amplifier shown here has 8m= 4 mA/V and ro= 20kQ2. Use the method of short-circuit time constants to determine the lower band frequency, f, in Hz.

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**MOSFET Common-Source Amplifier Analysis**

The MOSFET common-source amplifier depicted in the schematic features the following parameters: \( g_m = 4 \, \text{mA/V} \) and \( r_o = 20 \, \text{k}\Omega \). The task is to use the method of short-circuit time constants to determine the lower band frequency, \( f_L \), in Hz.

**Circuit Components:**
- \( R_{sig} = 200 \, \text{k}\Omega \): Input signal resistance
- \( C_1 = 0.1 \, \mu\text{F} \): Coupling capacitor at the input
- \( R_G1 = 2 \, \text{M}\Omega \): Gate resistor 1
- \( R_G2 = 1 \, \text{M}\Omega \): Gate resistor 2 
- \( R_S = 2 \, \text{k}\Omega \): Source resistor, with \( C_S = 5 \, \mu\text{F} \) bypass capacitor
- \( R_D = 10 \, \text{k}\Omega \): Drain resistor
- \( C_2 = 1 \, \mu\text{F} \): Coupling capacitor at the output
- \( R_L = 10 \, \text{k}\Omega \): Load resistor

**Explanation of Diagram:**
- The input signal is applied to the gate of the MOSFET \( Q1 \).
- \( C_1 \) serves as the input coupling capacitor, blocking DC while allowing AC signals through.
- \( R_G1 \) and \( R_G2 \) form a voltage divider biasing network for the MOSFET gate.
- \( R_S \) is the source resistor which is bypassed by \( C_S \) to ensure a stable DC bias without affecting the AC gain.
- The drain current develops a voltage across \( R_D \), providing the amplified output.
- \( C_2 \) is the output coupling capacitor that passes AC signals to the load while blocking DC.

**Objective:**
Calculate the lower band frequency \( f_L \) using short-circuit time constants, which involves analyzing how the reactive components (capacitors) affect the frequency response of the circuit.
Transcribed Image Text:**MOSFET Common-Source Amplifier Analysis** The MOSFET common-source amplifier depicted in the schematic features the following parameters: \( g_m = 4 \, \text{mA/V} \) and \( r_o = 20 \, \text{k}\Omega \). The task is to use the method of short-circuit time constants to determine the lower band frequency, \( f_L \), in Hz. **Circuit Components:** - \( R_{sig} = 200 \, \text{k}\Omega \): Input signal resistance - \( C_1 = 0.1 \, \mu\text{F} \): Coupling capacitor at the input - \( R_G1 = 2 \, \text{M}\Omega \): Gate resistor 1 - \( R_G2 = 1 \, \text{M}\Omega \): Gate resistor 2 - \( R_S = 2 \, \text{k}\Omega \): Source resistor, with \( C_S = 5 \, \mu\text{F} \) bypass capacitor - \( R_D = 10 \, \text{k}\Omega \): Drain resistor - \( C_2 = 1 \, \mu\text{F} \): Coupling capacitor at the output - \( R_L = 10 \, \text{k}\Omega \): Load resistor **Explanation of Diagram:** - The input signal is applied to the gate of the MOSFET \( Q1 \). - \( C_1 \) serves as the input coupling capacitor, blocking DC while allowing AC signals through. - \( R_G1 \) and \( R_G2 \) form a voltage divider biasing network for the MOSFET gate. - \( R_S \) is the source resistor which is bypassed by \( C_S \) to ensure a stable DC bias without affecting the AC gain. - The drain current develops a voltage across \( R_D \), providing the amplified output. - \( C_2 \) is the output coupling capacitor that passes AC signals to the load while blocking DC. **Objective:** Calculate the lower band frequency \( f_L \) using short-circuit time constants, which involves analyzing how the reactive components (capacitors) affect the frequency response of the circuit.
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