Question

In an \( npn \) transistor circuit, the collector current is \( 10 \) mA. If \( 95\% \) of the electrons emitted reach the collector, then the base current is nearly:

• A. \( 5.3 \) mA
• B. \( 53 \) mA
• C. \( 35 \) mA
• D. \( 0.53 \) mA

Hint:

For an \( npn \) transistor, \( I_E = I_B + I_C \), and \( \eta \) is the percentage of electrons reaching the collector.

Answer 1

The Correct Option is D

Step 1: Use Transistor Current Relation For an \( npn \) transistor: \[ I_E = I_B + I_C \] The efficiency \( \eta \) is given by: \[ \eta = \frac{I_C}{I_E} \] Step 2: Compute Base Current Given \( \eta = 95\% \): \[ I_E = \frac{I_C}{0.95} = \frac{10}{0.95} = 10.53 \text{ mA} \] \[ I_B = I_E - I_C = 10.53 - 10 \] \[ I_B = 0.53 \text{ mA} \] Thus, the correct answer is \( 0.53 \) mA.

Answer 2

Given:
- Collector current, \( I_C = 10 \) mA = \( 10 \times 10^{-3} \) A
- Percentage of electrons reaching the collector = 95% = 0.95

Step 1: Understand the currents in an npn transistor:
- Electrons emitted from the emitter current (\( I_E \)) split into two parts:
- \( I_C \): Electrons collected by the collector
- \( I_B \): Electrons recombining in the base (base current)

Step 2: Since 95% electrons reach collector, 5% recombine in the base:
\[ I_C = 0.95 I_E \] \[ I_B = 0.05 I_E \]

Step 3: Express \( I_E \) in terms of \( I_C \):
\[ I_E = \frac{I_C}{0.95} = \frac{10 \times 10^{-3}}{0.95} \approx 10.526 \times 10^{-3} \, \text{A} \]

Step 4: Calculate the base current \( I_B \):
\[ I_B = 0.05 \times I_E = 0.05 \times 10.526 \times 10^{-3} = 0.5263 \times 10^{-3} \, \text{A} = 0.5263 \, \text{mA} \]

Step 5: Round to two decimal places:
\[ I_B \approx 0.53 \, \text{mA} \]

Therefore, the base current is nearly:
\[ \boxed{0.53 \, \text{mA}} \]