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BJT Emitter Follower: Draw the small-signal model for the common-collector amplifier shown below. Derive an expression for the small-signal voltage gain (4=v/v). Based on your answer for A, can you explain why this circuit is often referred to as an emitter-follower? ▲Vcc RB V₁O-W -OV RL

BJT Emitter Follower: Draw the small-signal model for the common-collector amplifier shown below. Derive an expression for the small-signal voltage gain (4=v/v). Based on your answer for A, can you explain why this circuit is often referred to as an emitter-follower? ▲Vcc RB V₁O-W -OV RL

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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**BJT Emitter Follower:** **Instruction:** Draw the small-signal model for the common-collector amplifier shown below. Derive an expression for the small-signal voltage gain (\(A = \frac{v_o}{v_i}\)). Based on your answer for \(A\), can you explain why this circuit is often referred to as an emitter-follower? **Circuit Diagram:** The diagram depicts a BJT common-collector amplifier (also known as an emitter-follower). The components and connections are as follows: - **\(V_i\)**: This represents the input voltage. - **\(R_B\)**: This is the base resistor connected between the input voltage \(V_i\) and the base of the BJT. - **Transistor (BJT)**: The transistor has its base connected to \(R_B\), the collector connected to \(V_{CC}\), and the emitter connected to the output \(V_O\) and the load resistor \(R_L\). - **\(V_O\)**: This represents the output voltage taken from the emitter. - **\(V_{CC}\)**: The supply voltage connected to the collector of the transistor. - **\(R_L\)**: The load resistor connected between the emitter and ground. **Analysis:** To analyze the small-signal voltage gain, \(A = \frac{v_o}{v_i}\), perform the following steps: 1. Replace the BJT with its small-signal model. 2. Apply small-signal analysis techniques to derive the voltage gain. **Explanation:** The common-collector amplifier configuration is often called an emitter-follower because the output voltage at the emitter follows the input voltage applied at the base, with a voltage gain close to unity (but slightly less than 1). This behavior results from the fact that the emitter voltage \(V_O\) is approximately equal to the base voltage \(V_i\) minus the base-emitter voltage drop \(V_{BE} \approx 0.7V\).
Transcribed Image Text:**BJT Emitter Follower:** **Instruction:** Draw the small-signal model for the common-collector amplifier shown below. Derive an expression for the small-signal voltage gain (\(A = \frac{v_o}{v_i}\)). Based on your answer for \(A\), can you explain why this circuit is often referred to as an emitter-follower? **Circuit Diagram:** The diagram depicts a BJT common-collector amplifier (also known as an emitter-follower). The components and connections are as follows: - **\(V_i\)**: This represents the input voltage. - **\(R_B\)**: This is the base resistor connected between the input voltage \(V_i\) and the base of the BJT. - **Transistor (BJT)**: The transistor has its base connected to \(R_B\), the collector connected to \(V_{CC}\), and the emitter connected to the output \(V_O\) and the load resistor \(R_L\). - **\(V_O\)**: This represents the output voltage taken from the emitter. - **\(V_{CC}\)**: The supply voltage connected to the collector of the transistor. - **\(R_L\)**: The load resistor connected between the emitter and ground. **Analysis:** To analyze the small-signal voltage gain, \(A = \frac{v_o}{v_i}\), perform the following steps: 1. Replace the BJT with its small-signal model. 2. Apply small-signal analysis techniques to derive the voltage gain. **Explanation:** The common-collector amplifier configuration is often called an emitter-follower because the output voltage at the emitter follows the input voltage applied at the base, with a voltage gain close to unity (but slightly less than 1). This behavior results from the fact that the emitter voltage \(V_O\) is approximately equal to the base voltage \(V_i\) minus the base-emitter voltage drop \(V_{BE} \approx 0.7V\).
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