In a bipolar junction transistor (BJT) configured as common-emitter (C-E), the collector current Ic is related to the base current Ib by the current gain beta (β). If β = 100 and the base current is Ib = 30 µA, what is the resulting collector current Ic?

Difficulty: Easy

Correct Answer: 3 mA

Explanation:

Introduction / Context:In the common-emitter (C-E) configuration of a bipolar junction transistor (BJT), small changes in base current control much larger changes in collector current. The proportionality factor is the small-signal or DC current gain beta (β), making this relationship a staple of analog electronics and bias design.

Given Data / Assumptions:

  • Configuration: common-emitter.
  • Current gain β = 100 (assume approximately constant in this range).
  • Base current Ib = 30 µA.
  • Transistor operating in active region (not saturated or cut off).

Concept / Approach:The core relationship for a BJT in active mode is Ic = β * Ib. With β specified and Ib known, Ic follows by direct multiplication. This linear approximation is widely used for first-pass bias calculations before more detailed modeling (e.g., Early effect) is applied.

Step-by-Step Solution:1) Write the relationship: Ic = β * Ib.2) Substitute β = 100 and Ib = 30 µA.3) Calculate: Ic = 100 * 30 µA = 3000 µA.4) Convert units: 3000 µA = 3 mA.

Verification / Alternative check:A quick unit sense-check: for β ≈ 100, collector current is roughly two orders of magnitude larger than base current. 30 µA * 100 = 3000 µA, which is 3 mA—consistent and reasonable for small-signal devices.

Why Other Options Are Wrong:30 µA and 0.3 µA: These ignore the gain β (underestimate Ic heavily).

3 MA: Megaampere is absurdly large for a small-signal transistor—unit error by a factor of 10^6.

300 µA: This corresponds to β = 10, not 100.

Common Pitfalls:Confusing β with α (common-base gain), mixing up microampere and milliampere units, or applying the formula outside the active region (near saturation where Ic no longer equals β * Ib). Always verify region of operation when finalizing a bias network.

Final Answer:3 mA

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