Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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Question
(Note: Insert the image of the transistor circuit here)
#### Step-by-Step Solution
1. **Solve for \( I_B \) (Base Current):**
Using the formula for base current,
\[
I_B = \frac{V_{CC} - V_{BE}}{R_B + (\beta_{DC} \cdot R_E)}
\]
Assuming \( V_{BE} \approx 0.7V \) (typical for silicon transistors),
\[
I_B = \frac{12V - 0.7V}{390k\Omega} \approx \frac{11.3V}{390k\Omega} \approx 29 \mu A
\]
2. **Solve for \( I_C \) (Collector Current):**
Using the current gain relationship,
\[
I_C = \beta_{DC} \cdot I_B
\]
\[
I_C = 150 \cdot 29 \mu A \approx 4.35 mA
\]
3. **Solve for \( V_{CE} \) (Collector-Emitter Voltage):**
Using the formula for \( V_{CE} \),
\[
V_{CE} = V_{CC} - (I_C \cdot R_C)
\]
\[
V_{CE} = 12V - (4.35mA \cdot 1.5k\Omega) \approx 12V - 6.525V \approx 5.475V
\]
####](https://content.bartleby.com/qna-images/question/22d8487a-aee7-41d5-95a0-1e0950421a9b/ae3c48fe-2825-4c97-ae1a-d1664726752a/3oflas_processed.jpeg)
Transcribed Image Text:### Transistor Biasing Circuit Analysis
#### Problem Statement
In the figure below, solve for \( I_B \), \( I_C \), and \( V_{CE} \). Also, construct a DC load line showing the values of \( I_{C(sat)} \), \( V_{CE(off)} \), \( I_{CQ} \), and \( V_{CEQ} \).
#### Circuit Diagram
The given circuit diagram is as follows:
- The supply voltage \( V_{CC} \) is 12 V.
- The base resistor \( R_B \) is 390 kΩ.
- The collector resistor \( R_C \) is 1.5 kΩ.
- The current gain \( \beta_{DC} \) is 150.
(Note: Insert the image of the transistor circuit here)
#### Step-by-Step Solution
1. **Solve for \( I_B \) (Base Current):**
Using the formula for base current,
\[
I_B = \frac{V_{CC} - V_{BE}}{R_B + (\beta_{DC} \cdot R_E)}
\]
Assuming \( V_{BE} \approx 0.7V \) (typical for silicon transistors),
\[
I_B = \frac{12V - 0.7V}{390k\Omega} \approx \frac{11.3V}{390k\Omega} \approx 29 \mu A
\]
2. **Solve for \( I_C \) (Collector Current):**
Using the current gain relationship,
\[
I_C = \beta_{DC} \cdot I_B
\]
\[
I_C = 150 \cdot 29 \mu A \approx 4.35 mA
\]
3. **Solve for \( V_{CE} \) (Collector-Emitter Voltage):**
Using the formula for \( V_{CE} \),
\[
V_{CE} = V_{CC} - (I_C \cdot R_C)
\]
\[
V_{CE} = 12V - (4.35mA \cdot 1.5k\Omega) \approx 12V - 6.525V \approx 5.475V
\]
####
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