Semiconductor carrier types: in which material are holes the minority carriers under normal doping and equilibrium conditions?

Difficulty: Easy

Correct Answer: n-type

Explanation:


Introduction / Context:
Knowing which carriers are majority and which are minority is crucial for understanding diode behavior, transistor action, and recombination processes. This question probes identification of minority carriers for a common doping type.


Given Data / Assumptions:

  • Thermal equilibrium conditions.
  • Standard n-type and p-type doping without high-level injection.
  • No illumination or extreme temperature effects changing carrier populations drastically.


Concept / Approach:
In n-type material, donor impurities add electrons, making electrons the majority carriers and holes the minority carriers. In p-type material, acceptor impurities create holes as majority carriers, making electrons the minority carriers. Intrinsic material has equal concentrations of electrons and holes.


Step-by-Step Solution:
Step 1: Identify n-type doping adds electrons as majority carriers.Step 2: Recognize that the other carrier type (holes) becomes the minority.Step 3: Conclude that holes are minority carriers in n-type semiconductors.


Verification / Alternative check:
Position of the Fermi level near the conduction band in n-type material reflects higher electron concentration and lower hole concentration, consistent with holes as minority carriers.


Why Other Options Are Wrong:

  • Extrinsic: Generic term for doped semiconductors; does not specify majority type.
  • Intrinsic: Electrons equal holes, so there is no minority in the sense of imbalance.
  • p-type: Holes are majority there, not minority.
  • None of the above: Incorrect because n-type is correct.


Common Pitfalls:
Confusing majority and minority roles between n-type and p-type, or overlooking that illumination or injection can temporarily change populations; here we assume normal equilibrium.


Final Answer:
n-type.

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion