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In a common-emitter BJT configuration with DC current gain β_dc = 100, determine the collector current when the base current is 10 μA. Assume forward-active operation.

Difficulty: Easy

Correct Answer: 1 mA

Explanation:


Introduction / Context:
Small-signal and bias calculations in bipolar junction transistors (BJTs) often start from the simple relationship between collector current and base current: I_C = β I_B in the forward-active region. This question asks for a direct substitution, reinforcing basic DC analysis in common-emitter configurations.


Given Data / Assumptions:

  • β_dc (also denoted h_FE) = 100.
  • Base current I_B = 10 μA.
  • Transistor operates in forward-active region (not saturated or cut off).


Concept / Approach:
In forward-active region, collector current is proportional to base current by β_dc. The emitter current is approximately I_E ≈ (1 + β) I_B, but here only I_C is requested, so we use I_C = β_dc * I_B directly.


Step-by-Step Solution:

I_C = β_dc * I_B = 100 * 10 μA.Compute: 100 * 10 μA = 1000 μA = 1 mA.Thus, the collector current is 1 mA under the stated conditions.


Verification / Alternative check:

If V_CE is sufficiently large to avoid saturation (e.g., V_CE > 0.2–0.3 V typical), the forward-active assumption holds and the relation is valid.


Why Other Options Are Wrong:

10 μA: This equals I_B, which would imply β = 1 (not the given 100).100 μA or 0.5 mA: Do not satisfy the β = 100 relationship.10 mA: Would require I_B = 100 μA or β = 1000 instead.


Common Pitfalls:

Confusing β with α; α ≈ β / (β + 1) is used in common-base analysis, not needed here.


Final Answer:

1 mA
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