Question 6 **Transcription for Educational Website:**---**BJT Amplifier Analysis****Problem 6:**For the BJT amplifier given below, with a \( \beta_{DC} = 200 \), perform DC analysis to calculate the emitter DC current. The tasks are:a. Determine the voltage gain for this unloaded amplifier.b. If a load resistance of 3300 Ohms is placed on the amplifier output, what is the voltage gain?**Circuit Diagram:**- **Power Supply:** - \( V_{CC} = +10 \text{ V} \)- **Components:** - \( R_1 = 33 \text{ k}\Omega \) - \( R_2 = 6.8 \text{ k}\Omega \) - \( R_C = 3.3 \text{ k}\Omega \) - \( R_{E1} = 330 \text{ }\Omega \) - \( R_{E2} = 120 \text{ }\Omega \)- **Capacitors:** - \( C_1 = 1 \mu \text{F} \) - \( C_2 = 10 \mu \text{F} \) - \( C_3 = 1 \mu \text{F} \)- **Inputs and Outputs:** - \( V_{in} \) at node before \( C_1 \) - \( V_{out} \) after \( C_3 \)**Description:**The circuit is a common-emitter amplifier with a voltage divider biasing network. The supply voltage \( V_{CC} \) is 10 V. The network consists of \( R_1 \) and \( R_2 \) used for biasing, and \( R_C \), \( R_{E1} \), and \( R_{E2} \) are used for amplifier configuration.In the provided circuit, capacitors \( C_1 \), \( C_2 \), and \( C_3 \) are coupling and bypass capacitors to ensure AC signal passage while blocking DC components. The schematic layout has \( V_{in} \) at the input side with \( C_1 \), and \( V_{out} \) on the output side with \( C_3 \).The DC analysis will provide the emitter current necessary for calculating the voltage

Answered: Image /qna-images/answer/aa1dc463-4090-4b84-bcd4-b17b5eef419a.jpg

6- For the BJT amplifier given below, Bpc = 200, use DC analysis to calculate the emitter DC current, then: a- Determine the voltage gain for this unloaded amplifier. If a load resistance of 3300 Ohm is placed on the amplifier output, what is the voltage gain. Vcc +10 V RC 3.3 kn C3 - 33 kN out IµF Vin I µF REI 330 N 10 ΚΩ RE2 330 N C2 10 µF

6- For the BJT amplifier given below, Bpc = 200, use DC analysis to calculate the emitter DC current, then: a- Determine the voltage gain for this unloaded amplifier. If a load resistance of 3300 Ohm is placed on the amplifier output, what is the voltage gain. Vcc +10 V RC 3.3 kn C3 - 33 kN out IµF Vin I µF REI 330 N 10 ΚΩ RE2 330 N C2 10 µF

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Question

Question 6

**Transcription for Educational Website:**

---

**BJT Amplifier Analysis**

**Problem 6:**

For the BJT amplifier given below, with a \( \beta_{DC} = 200 \), perform DC analysis to calculate the emitter DC current. The tasks are:

a. Determine the voltage gain for this unloaded amplifier.

b. If a load resistance of 3300 Ohms is placed on the amplifier output, what is the voltage gain?

**Circuit Diagram:**

- **Power Supply:**
- \( V_{CC} = +10 \text{ V} \)

- **Components:**
- \( R_1 = 33 \text{ k}\Omega \)
- \( R_2 = 6.8 \text{ k}\Omega \)
- \( R_C = 3.3 \text{ k}\Omega \)
- \( R_{E1} = 330 \text{ }\Omega \)
- \( R_{E2} = 120 \text{ }\Omega \)

- **Capacitors:**
- \( C_1 = 1 \mu \text{F} \)
- \( C_2 = 10 \mu \text{F} \)
- \( C_3 = 1 \mu \text{F} \)

- **Inputs and Outputs:**
- \( V_{in} \) at node before \( C_1 \)
- \( V_{out} \) after \( C_3 \)

**Description:**

The circuit is a common-emitter amplifier with a voltage divider biasing network. The supply voltage \( V_{CC} \) is 10 V. The network consists of \( R_1 \) and \( R_2 \) used for biasing, and \( R_C \), \( R_{E1} \), and \( R_{E2} \) are used for amplifier configuration.

In the provided circuit, capacitors \( C_1 \), \( C_2 \), and \( C_3 \) are coupling and bypass capacitors to ensure AC signal passage while blocking DC components. The schematic layout has \( V_{in} \) at the input side with \( C_1 \), and \( V_{out} \) on the output side with \( C_3 \).

The DC analysis will provide the emitter current necessary for calculating the voltage

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