A pure semiconductor is a material with poor electrical conductivity. However the conductivity can be increased by order of magnitude by the addition of a small concentration of dopant atoms or by applying an electric field. Semiconductors form the basis of computation, communication, and display technologies. They are an important component of consumer electronics and transporation technologies. There is a strong increase in the use of semiconductors in solar cells and light-emitting diodes. The most commonly produced lasers are semiconductor lasers.
Kittel chapter 8: Semiconductor Crystals or R. Gross und A. Marx: Halbleiter
For the exam you should
be able to look at a bandstructure diagram for a semiconductor and be able to identify the conduction band, the valence band, the energy gap, whether the semiconductor is direct or in direct, and to be able to determine the effective masses of electrons and holes.
be able to define donors, acceptors, intrinsic semiconductors, extrinsic semiconductors, holes, and Bloch oscillations.
know the formulas n = Ncexp((μ-Ec)/kBT) and p = Nvexp((Ev-μ)/kBT). From these formulas you should be able to calculate the density of electrons in the conduction band, the density of holes in the valence band, the intrinsic carrier concentration ni, and the chemical potential for an intrinsic semiconductor or extrinsic semiconductor.
be able to describe how a light emitting diode, a solar cell, and a laser diode works.
know how thermal transport is coupled to electrical transport in semiconductors. Be able to describe the Seebeck effect and the Peltier effect.
Effective density of states in conduction band (300 K) Nc
2.78 × 1025 m-3
1.04 × 1025 m-3
4.45 × 1023 m-3
Effective density of states in valence band (300 K) Nv
9.84 × 1024 m-3
6.0 × 1024 m-3
7.72 × 1024 m-3
Effective mass electrons m*/m0
ml* = 0.98 mt* = 0.19
ml* = 1.64 mt* = 0.082
m* = 0.067
Effective mass holes m*/m0
mlh* = 0.16 mhh* = 0.49
mlh* = 0.044 mhh* = 0.28
mlh* = 0.082 mhh* = 0.45
5.0 × 1028
4.42 × 1028
4.42 × 1028
For a doped semiconductor, the density of electrons in the conduction band is,
the density of holes in the valence band is,
the density of ionized donors is,
and the density of ionized acceptors is,
The factor of 4 is valid in the formula for the acceptors if the semiconductor has a light hole and a heavy hole band as Si and Ge do.
The four quantities n, p, Nd, and Na can only be determined if the chemical potential, μ, is known. Typically, μ must first be determined from the charge neutrality condition,
n + Na- = p + Nd+.
The chemical potential can be found by solving the charge neutrality condition numerically. One way to do this is to program the formulas for n, p, Nd+, and Na- in a spreadsheet. Then choose a temperature and calculate n, p, Nd+, Na- for every value of the chemical potential between Ev and Ec. For one of these μ values, the charge neutrality condition will be satisfied.
When n + Na- and , p + Nd+ are ploted as a function of μ, the chemical potential is where the two lines cross. A plot like the one below can be generated by pressing the button below the plot.
The Fermi energy can be calculated as a function of temperature by determining where n + Na- = p + Nd+ for every temperature. A plot like the one below can be generated by pressing the button below the plot.
Normally it is not necessary to determine μ numerically and the following approximation is sufficient.
n-typeNd > Na:
n = Nd - Na p = ni²/n
μ = Ec - kBTln(Nc/(Nd - Na))
p-typeNa > Nd:
p = Na - Nd n = ni²/p
μ = Ev + kBTln(Nv/(Na - Nd))