A junction is formed between a metal on the left and an n-type semiconductor on
the right. Before forming the junction, the Fermi level EF of the metal lies below
that of the semiconductor. Then which of the following schematics are correct for
the bands and the I-V characteristics of the junction?
(A)
(B)
(C)
(D)
EF
EF
Conduction Band
Valence Band
Metal
n-type semiconductor
EF
Conduction Band
EF
Valence Band
Metal
n-type semiconductor
I
V
-V0
I
V
-V0
Official Answer Key:
A,C
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.57)**
**1. Metal-Semiconductor Junction Principles (Schottky Barrier)**:
- Junction formed between metal and n-type semiconductor.
- Before contact: Fermi level $E_F$, work functions $\Phi_m$ and $\Phi_s$.
- In equilibrium, Fermi levels align across the interface:
- Band bending occurs in the semiconductor to match electrostatic potential.
- A depletion region forms if $\Phi_m > \Phi_s$, giving a rectifying Schottky barrier.
- Alignment of Fermi levels and majority carrier depletion correspond to Statements (A, C).
**2. Official Verification**:
The verified official IIT Master Answer is **Options (A, C)**.
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