Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 425 of 494
In an extrinsic N-type semiconductor, how does the Fermi energy level (\( E_F \)) shift relative to its intrinsic position?
A
It shifts upward closer to the bottom edge of the conduction band
B
It shifts downward closer to the top edge of the valence band
C
It remains at the exact midpoint of the forbidden energy gap
D
It drops below the bottom of the valence band
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: It shifts upward closer to the bottom edge of the conduction band
Concept:

Donor dopants add free electrons to the conduction band, increasing the electron occupation probability and shifting the Fermi level closer to the conduction band.

Formula:

$$E_F - E_i = k_B T \ln\left(\frac{N_D}{n_i}\right) > 0$$

Solution:

  • Doping with donor atoms increases the electron concentration (\( n \approx N_D \gg n_i \)).


  • To reflect the increased probability of finding electrons in the conduction band, the Fermi level (\( E_F \)) shifts upward toward the conduction band edge \( E_c \).


Why other options are incorrect:

  • Option B: A downward shift toward the valence band occurs in P-type semiconductors.
  • Option C: The Fermi level sits at the midpoint only in undoped intrinsic semiconductors.
  • Option D: Shifting below the valence band corresponds to degenerate P-type doping.

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