Let's solve this problem step by step.

Given dataLet's solve this problem step by step.

Given

• $V_{CC} = 24 \, V$
• $R_C = 10 \, k\Omega$
• $R_e = 270 \, \Omega$
• $\beta = 45$
• $V_{CE} = 5 \, V$
• The transistor is silicon, so $V_{BE} \approx 0.7 \, V$.

Step 1: Determine $I_C$

From the relation between $V_{CC}$, $V_{CE}$, $I_C$, and $R_C$, we have: $V_{CC} = V_{CE} + I_C (R_C + R_e)$ Substitute the values: $24 = 5 + I_C (10,000 + 270)$ $I_C \cdot 10,270 = 19$ $I_C = \frac{19}{10,270} \approx 1.85 \, mA$

Step 2: Calculate $I_B$

Using the relation $I_C = \beta \cdot I_B$: $I_B = \frac{I_C}{\beta} = \frac{1.85 \, mA}{45} \approx 0.0411 \, mA$

Step 3: Calculate $V_{B}$

Since $V_{B} = V_{BE} + I_E R_e$ and $I_E \approx I_C$ (as $I_C \approx I_E$ for large $\beta$): $V_B = 0.7 + (1.85 \, mA) \cdot 270 \, \Omega$ $V_B = 0.7 + 0.5 \approx 1.2 \, V$

Step 4: Find $R$ using voltage division

To find $R$, we use the voltage divider rule with the base current: $V_{CC} = V_B + I_B \cdot R$ $24 = 1.2 + 0.0411 \, mA \cdot R$ $0.0411 \, mA \cdot R = 22.8$ $R = \frac{22.8}{0.0411 \, mA} \approx 107 \, k\Omega$

Thus, the calculated value of $R$ is approximately 107 kΩ, which matches the given answer.

Would you like more details or have any questions about this solution?

Here are some related questions to deepen your understanding:

1. How would the solution change if $\beta$ had a different value?
2. Why is the $V_{BE}$ drop assumed to be $0.7 \, V$ for silicon transistors?
3. What would happen to $I_C$ if $R_e$ was increased?
4. How does the transistor’s $\beta$ affect the overall stability of this circuit?
5. What role does the voltage divider play in setting the base voltage $V_B$?

Tip: When analyzing transistor circuits, always verify the assumed operating mode (active, cutoff, saturation) to ensure the calculations align with the expected behavior.