Determine the most probable failures, if any, in indicated measurements. circuit, based on the Vcc +12 V 20v R 10 kN Rc 1.0 kN 0.71v 00mV BDc = 180 RE 100 N R2 1.0 kN

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**Title: Analyzing Circuit Failures Based on Measurements**

**Objective:**  
Determine the most probable failures, if any, in the circuit based on the specified measurements.

**Circuit Description:**

- **Power Supply:**
  - \( V_{CC} = +12 \, \text{V} \)

- **Components:**
  - **\( R_1 = 10 \, \text{k}\Omega \)** resistor connected from \( V_{CC} \) to the base of the transistor.
  - **\( R_2 = 1.0 \, \text{k}\Omega \)** resistor connected in parallel with the transistor’s base to ground.
  - **\( R_C = 1.0 \, \text{k}\Omega \)** resistor connected from \( V_{CC} \) to the collector of the transistor.
  - **\( R_E = 100 \, \Omega \)** resistor connected from the emitter of the transistor to ground.
  - **Transistor with \( \beta_{DC} = 180 \)**

- **Measured Voltages:**
  - Base-Emitter Voltage (\( V_{BE} \)): 0.71 V
  - Collector-Emitter Voltage (\( V_{CE} \)): 12.0 V
  - Voltage across \( R_E \): 100 mV

**Analysis:**

The given circuit is a common emitter amplifier. The measured values can help diagnose any issues:

1. **\( V_{BE} = 0.71 \, \text{V} \):** This value is typical for a silicon transistor, indicating that the base-emitter junction is properly biased.

2. **\( V_{CE} = 12.0 \, \text{V} \):** This value suggests that the transistor is not conducting (in cutoff mode), as the entire supply voltage appears across the collector-emitter junction with no voltage drop across \( R_C \).

3. **Voltage across \( R_E = 100 \, \text{mV} \):** This indicates little to no current flowing through \( R_E \), consistent with the transistor being in cutoff.

**Probable Failures:**

- **Open Transistor:** The transistor may be open, causing it not to conduct at all, which explains the full \( V_{CC} \) reading across \( V_{CE} \).

- **Base Resistor
Transcribed Image Text:**Title: Analyzing Circuit Failures Based on Measurements** **Objective:** Determine the most probable failures, if any, in the circuit based on the specified measurements. **Circuit Description:** - **Power Supply:** - \( V_{CC} = +12 \, \text{V} \) - **Components:** - **\( R_1 = 10 \, \text{k}\Omega \)** resistor connected from \( V_{CC} \) to the base of the transistor. - **\( R_2 = 1.0 \, \text{k}\Omega \)** resistor connected in parallel with the transistor’s base to ground. - **\( R_C = 1.0 \, \text{k}\Omega \)** resistor connected from \( V_{CC} \) to the collector of the transistor. - **\( R_E = 100 \, \Omega \)** resistor connected from the emitter of the transistor to ground. - **Transistor with \( \beta_{DC} = 180 \)** - **Measured Voltages:** - Base-Emitter Voltage (\( V_{BE} \)): 0.71 V - Collector-Emitter Voltage (\( V_{CE} \)): 12.0 V - Voltage across \( R_E \): 100 mV **Analysis:** The given circuit is a common emitter amplifier. The measured values can help diagnose any issues: 1. **\( V_{BE} = 0.71 \, \text{V} \):** This value is typical for a silicon transistor, indicating that the base-emitter junction is properly biased. 2. **\( V_{CE} = 12.0 \, \text{V} \):** This value suggests that the transistor is not conducting (in cutoff mode), as the entire supply voltage appears across the collector-emitter junction with no voltage drop across \( R_C \). 3. **Voltage across \( R_E = 100 \, \text{mV} \):** This indicates little to no current flowing through \( R_E \), consistent with the transistor being in cutoff. **Probable Failures:** - **Open Transistor:** The transistor may be open, causing it not to conduct at all, which explains the full \( V_{CC} \) reading across \( V_{CE} \). - **Base Resistor
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