A bipolar junction transistor (BJT) carries a collector current of 2 mA. If the current gain (beta, β) is 135, what is the base current I_B in microamperes?

Introduction / Context:
For a bipolar junction transistor (BJT) operating in the forward-active region, the small base current controls a much larger collector current. The DC current gain, beta (β), defines this relationship and is a staple calculation in analog electronics and bias network design. Correctly converting among milliampere and microampere units and using the β relation is essential to size bias resistors and estimate input drive requirements.

Given Data / Assumptions:

• Collector current I_C = 2 mA.
• Current gain β = 135 (I_C / I_B).
• Assume forward-active operation and negligible leakage.

Concept / Approach:
The relationship is I_C = β * I_B, or equivalently I_B = I_C / β. Because units are mixed (mA vs µA), convert carefully: 1 mA = 1000 µA. After computing I_B, compare with the listed options.

Step-by-Step Solution:
Write the formula: I_B = I_C / β.Substitute: I_B = 2 mA / 135.Convert units: 2 mA = 2000 µA → I_B = 2000 µA / 135.Compute: 2000 / 135 ≈ 14.81 µA → round to 14.8 µA.

Verification / Alternative check:
Reverse-multiply to confirm: 14.8 µA * 135 ≈ 1998 µA ≈ 1.998 mA, which rounds to the given 2 mA. The small rounding error is expected from decimal truncation and is within typical component tolerance bands.

Why Other Options Are Wrong:
10.3 µA, 7.8 µA, and 7.75 µA correspond to incorrect β usage or arithmetic. “None of the above” is false because 14.8 µA matches the computed value.

Common Pitfalls:
Forgetting to convert mA to µA; using α (common-base gain) instead of β; assuming β is constant across operating points—it varies with current and temperature, so designs should include margin.

Final Answer:
14.8 µA