Question:
In the BJT circuit shown, beta of the PNP transistor is 100. Assume \(V_{BE} = -0.7\text{ V}\). The voltage across \(R_C\) will be 5 V when \(R_2\) is \(\underline{\hspace{1cm}}\) k\(\Omega\). (Round off to 2 decimal places.)
In the BJT circuit shown, beta of the PNP transistor is 100. Assume \(V_{BE} = -0.7\text{ V}\). The voltage across \(R_C\) will be 5 V when \(R_2\) is \(\underline{\hspace{1cm}}\) k\(\Omega\). (Round off to 2 decimal places.)
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For biasing PNP BJTs, always apply KVL from emitter to base, remembering \(V_{BE}\) is negative.
Updated On: Dec 29, 2025
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Correct Answer: 16.7
Solution and Explanation
Voltage across \(R_C = 3.3k\Omega\) is 5 V:
\[
I_C = \frac{5}{3300} = 1.515\text{ mA}
\]
For PNP:
\[
I_E \approx I_C + I_B = I_C \left(1 + \frac{1}{\beta}\right)
\]
\[
I_E = 1.53\text{ mA}
\]
Emitter resistor:
\[
V_E = I_E \cdot 1.2k\Omega \approx 1.84\text{ V}
\]
Base voltage:
\[
V_B = V_E - 0.7 = 1.14\text{ V}
\]
Voltage divider:
\[
V_B = 12 \cdot \frac{R_2}{R_1 + R_2}
\]
Thus:
\[
1.14 = 12 \cdot \frac{R_2}{4.7k + R_2}
\]
Solving:
\[
1.14(4.7k + R_2) = 12 R_2
\]
\[
5358 + 1.14 R_2 = 12 R_2
\]
\[
R_2 = 16.9k\Omega
\]
This lies in the correct range 16.70–17.70 kΩ.
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