The common-emitter discrete amplifier shown has the following values: RB1 = 400 kΩ, RB2 = 100 kΩ, RC = 4 kΩ, RE = 500 Ω, Rsig = 1 kΩ, RL = 10 kΩ and VCC = 15 V. Assume VBE = 0.7 V and β = 90. Also assume that the capacitors are shorts for ac. The bias circuit has been analyzed and IC = 1.65 mA, Rin = 1.34 kΩ and Rout = RC. If vsig = 10 mVpeak, determine the peak value of vout.

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The common-emitter discrete amplifier shown has the following values: RB1 = 400 kΩ, RB2 = 100 kΩ, RC = 4 kΩ, RE = 500 Ω, Rsig = 1 kΩ, RL = 10 kΩ and VCC = 15 V. Assume VBE = 0.7 V and β = 90. Also assume that the capacitors are shorts for ac.

The bias circuit has been analyzed and IC = 1.65 mA, Rin = 1.34 kΩ and Rout = RC.

If vsig = 10 mVpeak, determine the peak value of vout.

The image displays a circuit diagram of a common-emitter discrete amplifier. Below is a detailed description of the circuit and its components, followed by the problem statement and multiple-choice answers:

**Circuit Diagram:**
- The amplifier consists of a BJT transistor with its collector connected to \( V_{CC} \) through a resistor \( R_C \).
- The base is connected to a voltage signal source \( v_{sig} \) through resistors \( R_{B1} \) and \( R_{B2} \).
- An emitter resistor \( R_E \) is connected to the emitter terminal.
- The output voltage \( v_{out} \) is taken across the load resistor \( R_L \).
- Capacitors are present but considered short circuits for AC signals.

**Given Values:**
- \( R_{B1} = 400 \, k\Omega \), \( R_{B2} = 100 \, k\Omega \)
- \( R_C = 4 \, k\Omega \)
- \( R_E = 500 \, \Omega \)
- \( R_{sig} = 1 \, k\Omega \)
- \( R_L = 10 \, k\Omega \)
- \( V_{CC} = 15 \, V \)
- Assume \( V_{BE} = 0.7 \, V \) and \( \beta = 90 \)
- Assume capacitors are shorts for AC.

**Bias Circuit Analysis:**
- Collector current \( I_C = 1.65 \, mA \)
- Input resistance \( R_{in} = 1.34 \, k\Omega \)
- Output resistance \( R_{out} = R_C \)

**Problem Statement:**
Determine the peak value of \( v_{out} \) when \( v_{sig} = 10 \, mV_{peak} \).

**Multiple-Choice Answers:**
- \( v_{outpeak} = 1.08 \, V \)
- \( v_{outpeak} = 3.22 \, V \)
- \( v_{outpeak} = 2.24 \, V \)
- \( v_{outpeak} = 2.98 \, V \)

The correct peak value of \( v_{out} \) has been selected as \( v_{outpeak} = 1.08 \, V \).
Transcribed Image Text:The image displays a circuit diagram of a common-emitter discrete amplifier. Below is a detailed description of the circuit and its components, followed by the problem statement and multiple-choice answers: **Circuit Diagram:** - The amplifier consists of a BJT transistor with its collector connected to \( V_{CC} \) through a resistor \( R_C \). - The base is connected to a voltage signal source \( v_{sig} \) through resistors \( R_{B1} \) and \( R_{B2} \). - An emitter resistor \( R_E \) is connected to the emitter terminal. - The output voltage \( v_{out} \) is taken across the load resistor \( R_L \). - Capacitors are present but considered short circuits for AC signals. **Given Values:** - \( R_{B1} = 400 \, k\Omega \), \( R_{B2} = 100 \, k\Omega \) - \( R_C = 4 \, k\Omega \) - \( R_E = 500 \, \Omega \) - \( R_{sig} = 1 \, k\Omega \) - \( R_L = 10 \, k\Omega \) - \( V_{CC} = 15 \, V \) - Assume \( V_{BE} = 0.7 \, V \) and \( \beta = 90 \) - Assume capacitors are shorts for AC. **Bias Circuit Analysis:** - Collector current \( I_C = 1.65 \, mA \) - Input resistance \( R_{in} = 1.34 \, k\Omega \) - Output resistance \( R_{out} = R_C \) **Problem Statement:** Determine the peak value of \( v_{out} \) when \( v_{sig} = 10 \, mV_{peak} \). **Multiple-Choice Answers:** - \( v_{outpeak} = 1.08 \, V \) - \( v_{outpeak} = 3.22 \, V \) - \( v_{outpeak} = 2.24 \, V \) - \( v_{outpeak} = 2.98 \, V \) The correct peak value of \( v_{out} \) has been selected as \( v_{outpeak} = 1.08 \, V \).
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