In semiconductor, Fermi level lies : (a) Close to Valence Band (VB) in p-type Which is correct ?
(b) Close to Conduction Band (CB) in p-type
(c) Close to Conduction Band in n-type
(d) Close to Valence Band in n-type
(a), (c)
The Fermi level marks where the probability of occupation is one half, so it moves towards whichever band is more heavily populated.
In n-type material the donors contribute electrons to the conduction band, so \(n\gg p\) and the Fermi level rises towards the conduction band — statement (c). Quantitatively,
\(E_{F}=E_{C}-kT\ln\left(\dfrac{N_{C}}{N_{D}}\right)\)
so the heavier the donor doping ND, the closer EF creeps to EC.
In p-type material the acceptors create holes in the valence band, so \(p\gg n\) and the Fermi level falls towards the valence band — statement (a):
\(E_{F}=E_{V}+kT\ln\left(\dfrac{N_{V}}{N_{A}}\right)\)
So (a) and (c) — option 3. Statements (b) and (d) are these two exactly reversed.
| Material | Majority carrier | EF position |
|---|---|---|
| Intrinsic | n = p | Near mid-gap |
| n-type | Electrons | Near EC |
| p-type | Holes | Near EV |
The memory aid that never fails : the Fermi level moves towards the carriers. Electrons live at the top of the gap, holes at the bottom, so n-type pulls EF up and p-type pulls it down.
Two consequences worth knowing.
Degeneracy. Dope heavily enough — beyond roughly \(10^{19}\ \text{cm}^{-3}\) — and EF moves right into the band itself. The material is then degenerate and behaves much like a metal, which is exactly the condition exploited by the tunnel diode and by the heavily doped contact regions of every transistor.
The built-in potential of a junction. When p and n material are joined, the two Fermi levels must line up at equilibrium — a single EF throughout, since no net current flows. The bands must therefore bend, and the amount of bending is the built-in potential:
\(V_{bi}=\dfrac{kT}{q}\ln\left(\dfrac{N_{A}N_{D}}{n_{i}^{2}}\right)\)
That single requirement — a flat Fermi level at equilibrium — is the starting point of every pn-junction, MOS and heterojunction band diagram.
Temperature pushes EF back towards mid-gap: as the intrinsic carrier concentration rises, the doping matters proportionally less, and at high enough temperature an extrinsic semiconductor behaves intrinsically — which is what ultimately limits a device's operating temperature.
Hence, the correct statements are (a) and (c).
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