Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 247 of 494
As the temperature of an N-type semiconductor is raised to very high levels (e.g., above \( 600\text{ K} \)), its Fermi level \( E_F \):
A
Moves deeper into the conduction band
B
Remains permanently pinned to the donor level
C
Drops instantly to zero eV
D
Shifts downward toward the middle of the forbidden bandgap (intrinsic behavior)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: Shifts downward toward the middle of the forbidden bandgap (intrinsic behavior)
Concept:

At elevated temperatures, intrinsic thermal generation across the bandgap exceeds the fixed donor concentration (\( n_i \gg N_D \)), causing the semiconductor to behave intrinsically and shifting \( E_F \) back toward the middle of the bandgap.

Formula:

$$E_F(T \to \text{high}) \longrightarrow E_i = \frac{E_c + E_v}{2} \quad (\text{as } n_i(T) \gg N_D)$$

Solution:

  • At moderate temperatures, \( n \approx N_D \) (extrinsic range) and \( E_F \) sits near the conduction band.


  • At high temperatures, thermal electron-hole pair generation dominates (\( n_i(T) \gg N_D \)), so \( n \approx p \approx n_i \).


  • As the material transitions to intrinsic behavior, \( E_F \) shifts downward toward the middle of the forbidden gap.


Why other options are incorrect:

  • Option A: \( E_F \) moves into the conduction band under heavy low-temperature doping (degeneracy), not at high temperatures.
  • Option B: The Fermi level is dynamic and changes position with temperature.
  • Option C: \( E_F \) shifts continuously; it does not drop discontinuously to zero.

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