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Band bending in semiconductors

Short answer

Band bending is the spatial variation of energy bands (E_c, E_v) due to electric fields. In equilibrium, the Fermi level stays flat; doping or junctions create built-in potentials, so bands curve by e times the potential difference while the bandgap remains constant.

Explain how does the band bends in semiconductor.

Band bending is the curving of the energy bands (EcE_c, EvE_v, EFiE_{Fi}) with position inside a semiconductor. It happens wherever there is an electric field, i.e. wherever the electrostatic potential V(x)V(x) changes with position.

Why the bands bend

The energy of an electron at a point is its band energy plus its potential energy −eV(x)-eV(x):

Ec(x)=Ec0−eV(x)E_c(x) = E_{c0} - eV(x)

The field is E=−dV/dx\mathcal{E} = -dV/dx, so

E=1edEcdx\mathcal{E} = \frac{1}{e}\frac{dE_c}{dx}
  • Where E=0\mathcal{E} = 0 the bands are flat.
  • Where a field exists, all three levels shift by the same amount, so the band gap EgE_g stays constant but the bands tilt or curve.
  • An electron "rolls down" the conduction band and a hole "floats up" the valence band.

Common cases

  1. Applied voltage on a uniform bar: the field is uniform, so the bands tilt in a straight line. The Fermi level is no longer flat; current flows.
  2. pn junction (equilibrium): electrons diffuse to the p side and holes to the n side, leaving fixed ion charges. These create a built-in field. Because the Fermi level must be the same throughout in equilibrium, the bands bend in the depletion region by eV0eV_0.
  3. Non-uniform doping, metal–semiconductor contact, surface charges: each creates a local space charge and field, so the bands bend near that region.
 p side            n side
 Ec ----\
         \__________ Ec
 EF ------------------- EF (flat)
 Ev ----\
         \__________ Ev
       |<-W->|  depletion region
   bending = eV0

Key rule

In equilibrium the Fermi level is flat; any difference in doping is accommodated by bending of EcE_c and EvE_v. The amount of bending equals ee times the potential difference across the region.

✓ Answer checked by Claude Opus 5.5 (high effort)

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