Transistor current gain calculation: A BJT has a DC current gain β = 250. If the base current is IB = 20 µA, what is the collector current IC (assume active region)?

Difficulty: Easy

Correct Answer: 5 mA

Explanation:


Introduction:
In the active region of a BJT, the collector current is approximately proportional to base current via the DC current gain β (also called hFE). This relationship is used to estimate required base drive and to size bias networks for desired collector currents.


Given Data / Assumptions:

  • Current gain β = 250.
  • Base current IB = 20 µA.
  • Transistor operating in active region (not saturated).


Concept / Approach:

The basic relation is IC = β * IB. Multiply the given IB by β, keeping careful track of microampere-to-ampere unit conversions, then express the result in milliamperes for clarity.


Step-by-Step Solution:

Write IC = β * IB.Substitute values: IC = 250 * 20 µA.Compute: 250 * 20 µA = 5,000 µA.Convert units: 5,000 µA = 5 mA.


Verification / Alternative check:

Reverse check: IB = IC / β = 5 mA / 250 = 0.02 mA = 20 µA, matching the given value. If the transistor were in saturation, IC would be lower than β * IB, but the problem specifies normal operation.


Why Other Options Are Wrong:

  • 500 µA: Corresponds to β = 25 or IB = 2 µA, not the given numbers.
  • 50 mA: Implies β = 2,500 for IB = 20 µA, unrealistically high.
  • 5 A: Six orders of magnitude too large; unit error.
  • 2.5 mA: Would require β = 125, not 250.


Common Pitfalls:

  • Miscalculating microampere to milliampere conversions.
  • Forgetting that β varies with operating conditions; use as a nominal design estimate unless specified.


Final Answer:

5 mA

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