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In a bipolar junction transistor (BJT), which terminal current is the largest under normal operation, and how are the three currents related?

Difficulty: Easy

Correct Answer: emitter current

Explanation:


Introduction / Context:
Kirchhoff’s current law (KCL) applied to the transistor junctions provides a direct relation among emitter (IE), base (IB), and collector (IC) currents. Recognizing that IE is the sum of the other two is vital for quick bias calculations and understanding of current gain definitions α and β.


Given Data / Assumptions:

  • Steady-state operating point in any BJT configuration.
  • KCL at the transistor: IE = IB + IC.
  • Typically, IB ≪ IC (for a transistor with reasonable β), hence IE is marginally larger than IC.


Concept / Approach:
Since the emitter injects carriers that split into the collector (mostly) and the base (small recombination), IE is necessarily the largest terminal current. With β = IC/IB high (e.g., 100), IB is small; IC is slightly less than IE by the amount of IB. Therefore ordering is IE > IC ≫ IB.


Step-by-Step Solution:

Write KCL: IE = IC + IB.Given IB is small, IE ≈ IC + a small increment, making IE the largest.Hence, the correct selection is “emitter current”.


Verification / Alternative check:

For β = 100 and IC = 10 mA, IB = 0.1 mA, IE = 10.1 mA → IE largest, confirming the relation.


Why Other Options Are Wrong:

Collector current is slightly less than IE by IB.Base current is the smallest in a well-designed BJT bias.“Either … or …” contradicts the fixed KCL relation.


Common Pitfalls:

Forgetting KCL or misinterpreting β and α, leading to incorrect ordering of currents.


Final Answer:

emitter current
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