Charge Carriers in Semiconductors
Charge carriers = “carriers” = electrons and holes
An undoped (“intrinsic”) semiconductor has very few electrons in the conduction band,
because a typical band gap of 1 eV is large compared to the thermal energy kBT 25 meV
(at room temperature). That makes the probability extremely low for having an electron
excited across the band gap. To make this argument quantitative, one has to determine the
occupancy at the conduction band minimum, which is determined by the Fermi-Dirac
distribution f(E). Since f(E) contains the Fermi level EF (see Lecture 14), one needs to
know where in the band gap EF is located. The position of EF is determined by charge
neutrality. The density of electrons in the conduction band has to match the density of
holes in the valence band. That puts EF close to mid-gap for intrinsic semiconductors.
The figure illustrates how to determine the Fermi level position from charge neutrality.
electrons
holes
To calculate the density of electrons and holes (dn/dE and dp/dE in the figure), multiply
the density of states D(E) by the Fermi-Dirac function f(E) and [1f(E)], respectively. If
the density of states in the conduction band DC(E) is smaller than that in valence band,
the Fermi level has to move slightly upwards to compensate the decrease in DC(E) by an
increase in f(E), as shown in Panel b on the right. The difference is small, since f(E)
varies very rapidly with E near the band edges.
For intrinsic silicon (Eg  1.1 eV), with EF mid-gap, the probability of having an electron
excited into the conduction band minimum (CBM) at room temperature is roughly:
exp[(CBMEF)/kBT ] = exp[½Eg /kBT ] = e0.55eV / 0.025 eV  e22  3  1010
(Approximate the Fermi-Dirac function by exp[(EEF)/kBT ] for (EEF) » kBT . )
Holes in Semiconducturs = Antiparticles
Creating an electron-hole pair with a photon (which has negligible momentum):
E
ke
Ee
CBM
Eg
Ephoton
EF=0
VBM
Eh
kh
G/2
0
uncompensated
momentum
+G/2
k
Hole = (N1) Electrons + N Counter-Charges (Ions)
Use conservation laws to determine the quantum numbers of a hole:
Quantum
Number:
Hole
h
Charge q
+e
Electron
e
e
(E-VBM) +(E-CBM)
>0
>0
Momentum ħk
ke
+ke
Energy E
Conservation Law
qh + qe = qphoton = 0
Eh + Ee + Eg = Ephoton
kh + ke = kphoton  0
Holes are the equivalent of antiparticles in quantum field theory. In quantum
electrodynamics, for example, positrons are sometimes characterized as missing electrons
with negative energy that are moving backwards in time. We would simply say that they
are holes. Likewise, the vacuum of quantum electrodynamics is actually a semiconductor.
The Dirac equation for electrons has solutions for both positive energy (the conduction
band) and negative energy (the valence band, called the Dirac sea in particle physics).
The band gap of the vacuum is 2mec2  1 MeV, which corresponds to the minimum
energy required to create an electron-positron pair by a   photon.
Doping
To achieve a significant electron density in the conduction band, one has to bring the
Fermi level closer to the conduction band minimum to allow thermal excitations of an
electron into the conduction band. That is achieved by adding a donor, i.e. an atom that
has an occupied level close to the conduction band minimum. The Fermi level has to lie
above the donor level (at low temperature), since the level is occupied. Likewise, on uses
an acceptor atom to push the Fermi level close to the valence band and create holes. An
acceptor has an empty level close to the valence band maximum. Donors are typically
one row to the right of the semiconductor in the Periodic Table (e.g. P, As, Sb for Si),
while acceptors are one row to the left (B for Si).
E
EF
E
Donor
EF
D(E)
Acceptor
D(E)
Calculating Donor / Acceptor Levels
The H atom is used to model the extra electron orbiting a donor atom (or the hole orbiting
an acceptor atom). The Rydberg constant Ry  e4 me and the Bohr radius a0  [e2 me]1
(Lect. 20, p. 3) need to be adapted to a solid by substituting e2  e2 /ε and me  meff ∙ me .
With typical values of 10 and meff  0.1, the donor binding energy Ed is reduced from
1Ry = 13.6 eV by a factor of meff /2  103 . That makes it comparable to the thermal
energy kBT 25 meV and allows thermal excitation of electrons from the donor level into
the conduction band (or from the valence band into the acceptor level).
Ed = Ry  meff / 2
ad = a0   /meff
Ed
T[K]
E [meV]
to n=  (dotted line).
are extrapolated
Fermi Level EF and Carrier Density n versus Temperature and Doping
As the temperature increases, EF moves towards the middle of the gap.
Low T
High T
Intrinsic
Low T
Saturation
Freeze-out
Temperature T
High T
Room
Temp.
Mobility versus Temperature for Various Materials
The mobility  is defined
as the electron velocity
divided by the electric field
(Lect. 25). High mobility is
important
efficient
for
fast
electronics.
and
To
optimize the mobility, one
has to avoid scattering of
the
electrons.
Phonon
scattering can be reduced
by freezing out phonons at
low temperature. To avoid
scattering by defects and
impurities one needs to
perfect crystals. However,
dopants also act as defects.
Si
They can be eliminated by
modulation doping, where
the dopants are spatially
separated from the carriers
(Lect. 33, p. 1,2). Silicon
Cu
has a mobility of about
103 cm2/Vs. Electrons tend
to have higher mobility
than holes, because the
effective
mass
of
the
conduction band minimum
is often smaller than that of
the valence band maximum (see the preceding page).
Pentacene