In a p-type semiconductor where the Fermi level is initially 0.3 eV above the valence band, what happens to the Fermi level when acceptor concentration is increased?

Difficulty: Medium

Correct Answer: 0.28 eV above valence band

Explanation:


Introduction / Context:
The Fermi level indicates the chemical potential for electrons. In doped semiconductors, its position shifts with dopant concentration and type, affecting carrier densities and device behavior.


Given Data / Assumptions:

  • Material: p-type semiconductor.
  • Initial EF position: 0.3 eV above Ev (valence band edge).
  • Action: Increase acceptor concentration.


Concept / Approach:
Increasing acceptor doping raises hole concentration, pulling EF closer to the valence band. Thus EF moves downward toward Ev (i.e., decreases the separation EF − Ev).


Step-by-Step Solution:

Step 1: Recognize trend: stronger p-type doping → EF approaches Ev.Step 2: From 0.3 eV above Ev, a higher acceptor density reduces this offset.Step 3: Among the choices, 0.28 eV above Ev reflects a modest shift closer to Ev, consistent with increased acceptor density.


Verification / Alternative check:

Qualitative band diagram analysis confirms EF movement toward Ev with higher p-type doping; quantitative shift depends on temperature and exact dopant concentration.


Why Other Options Are Wrong:

0.5 eV above Ev: Moves away from Ev, opposite of expected behavior.0.1 eV above Ev: Could occur for a much larger increase but is not the best modest-step representation.Below Ev: EF cannot lie inside the valence band in equilibrium for a semiconductor (without degeneracy effects not implied here).Toward midgap regardless of doping: Incorrect trend for p-type doping.


Common Pitfalls:

Confusing the directions EF moves for n-type vs p-type; assuming EF crosses into bands under standard nondegenerate conditions.


Final Answer:

0.28 eV above valence band

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