Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 209 of 494
In semiconductor physics, which pair correctly matches the minority charge carriers in N-type and P-type semiconductors, respectively?
A
Holes in N-type; Free electrons in P-type
B
Free electrons in N-type; Holes in P-type
C
Positive ions in N-type; Negative ions in P-type
D
Protons in N-type; Neutrons in P-type
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: Holes in N-type; Free electrons in P-type
Concept:

Doping introduces donor or acceptor states that create a large majority carrier population. Thermally generated carriers of the opposite polarity make up the minority carrier population.

Formula:

$$\text{N-type: } n_n \gg p_n \implies p_n = \frac{n_i^2}{N_D} \text{ (Minority: Holes)}$$

$$\text{P-type: } p_p \gg n_p \implies n_p = \frac{n_i^2}{N_A} \text{ (Minority: Electrons)}$$

Solution:

  • In N-type material, pentavalent dopants make electrons the majority carriers, leaving thermally generated holes as the minority carriers.


  • In P-type material, trivalent dopants make holes the majority carriers, leaving thermally generated electrons as the minority carriers.


Why other options are incorrect:

  • Option B: This lists the majority carriers for each material.
  • Option C: Fixed donor and acceptor ions are immobile lattice sites, not mobile carriers.
  • Option D: Protons and neutrons are bound in atomic nuclei.

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