Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 263 of 494
Why does the Zener breakdown voltage \( V_Z \) of a heavily doped diode DECREASE as its operating temperature increases?
A
Thermal energy narrows the forbidden band gap \( E_g \), increasing the probability of valence electrons tunneling into the conduction band
B
The depletion layer expands to infinite thickness
C
Carrier mobility increases toward infinity
D
All covalent bonds dissociate completely into plasma
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: Thermal energy narrows the forbidden band gap \( E_g \), increasing the probability of valence electrons tunneling into the conduction band
Concept:

Zener breakdown is governed by quantum mechanical band-to-band tunneling. Because semiconductor band gaps narrow slightly with increasing temperature, the tunneling barrier height decreases, allowing breakdown to occur at a lower reverse voltage.

Formula:

$$E_g(T) = E_g(0) - \frac{\alpha T^2}{T + \beta} \implies \text{Tunneling probability increases} \implies V_Z \text{ decreases}$$

Solution:

  • As temperature rises, the energy band gap \( E_g \) decreases.


  • This reduces the energy barrier that valence electrons must tunnel through to reach the conduction band.


  • Consequently, tunneling occurs at lower reverse electric fields, decreasing the Zener breakdown voltage \( V_Z \) (negative temperature coefficient).


Why other options are incorrect:

  • Option B: The depletion layer thickness is determined by doping and voltage, and does not expand to infinity.
  • Option C: Carrier mobility decreases with temperature due to increased lattice scattering.
  • Option D: Standard operating temperatures do not turn semiconductors into plasma.

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