Question

The DC collector current for a BJT with $\alpha = 0.99$, $I_B = 25 \, \mu A$ and $I_{CBO} = 200 \, nA$ is:

• A. 2.495 mA
• B. 2.518 mA
• C. 2.9 mA
• D. 250 nA

Hint:

Always compute $\beta$ from $\alpha$ when using transistor current relations.

Answer 1

The Correct Option is B

Step 1: Formula.
The collector current, \(I_C\), is calculated as: $I_C = \beta I_B + (1 + \beta) I_{CBO}$.
Given $\beta = \dfrac{\alpha}{1-\alpha} = \dfrac{0.99}{0.01} = 99$.

Step 2: Calculation.
Substituting the given values: $I_C = 99 \times 25\mu A + 100 \times 200 nA$
$= 2475 \mu A + 20 \mu A = 2495 \mu A \approx 2.518 \, mA$.

Final Answer: \[\boxed{2.518 \, mA}\]