Question

In the transistor circuit given in the figure, the emitter-base junction has a voltage drop of 0.7 V. A collector-emitter voltage of 14 V reverse biases the collector. Assuming the collector current to be the same as the emitter current, the value of \(R_B\) is ........... k\(\Omega\).
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Hint:

In transistor bias circuits, base current is much smaller than collector current. The assumption \(I_C \approx I_E\) simplifies calculations significantly.

Answer 1

Correct Answer: 860

Step 1: Write down given data.
\[ V_{CC} = 20\,V, V_{CE} = 14\,V, V_{BE} = 0.7\,V, \beta = 100 \] \[ R_C = 2\,k\Omega, R_E = 1\,k\Omega \]

Step 2: Apply KVL to the collector-emitter loop.
\[ V_{CC} = I_C R_C + V_{CE} + I_E R_E \] Since \(I_E \approx I_C\): \[ 20 = I_C(2 + 1) \times 10^3 + 14 \] \[ I_C = \frac{20 - 14}{3000} = 2\,mA \]

Step 3: Calculate base current.
\[ I_B = \frac{I_C}{\beta} = \frac{2 \times 10^{-3}}{100} = 20\,\mu A \]

Step 4: Apply KVL to base circuit.
\[ V_{CC} = I_B R_B + V_{BE} + I_E R_E \] \[ 20 = 20 \times 10^{-6} R_B + 0.7 + 2 \times 10^{-3} \times 10^3 \] \[ 20 = 20 \times 10^{-6} R_B + 0.7 + 2 \] \[ R_B = \frac{17.3}{20 \times 10^{-6}} = 865 \times 10^3 = 865\,k\Omega \] Hence, approximately \(R_B = 865\,k\Omega\).

Step 5: Conclusion.
The required base resistor is \(R_B = 865\,k\Omega\).