(a) How energy bands arise:
First, consider two atoms far apart. The wave functions of their valence (outermost-orbit)
electrons are too far apart to interact. As they are brought closer together, the electrons
from each atom start to “feel” the Coulombic force and potential from the other atom, and
the wavefunctions start to overlap and interact. The two energy levels from the two
formerly independent atoms (systems) now have become two slightly different levels
(“bonding” with a slightly lower energy level and “anti-bonding” with a slightly higher one).
Repeat this for N times or Avogadro’s number of atoms. The split energy levels are very
close to each other and form “energy bands” where the N energy levels are effectively
continuous, and different energy bands are separated by band gaps where no energy levels
are allowed. For N atoms, there are 2N states in each band.
(b) Sketch the E-k structure:
Notice that in semiconductors, the bandgap Eg is relatively small (e.g. 1.1 eV for silicon), and
it is larger in insulators (e.g. about 8.9 eV for typical SiO2 used in semiconductor devices).
(c) Difference between metals, semiconductors, and insulators:
In metals, the bandgap is zero Eg=0, and the conduction band is always partially full,
allowing for significant electric current conduction.
In semiconductors, at temperature T = 0 Kelvins, the valence band is completely filled while
the conduction band is completely empty. Both bands do not conduct. However, the
bandgap Eg is relatively small. So, at temperatures T > 0 Kelvins, some (a tiny fraction of)
electrons in the VB can acquire enough energy and move up into the CB. Now that the VB is
slightly empty and the CB slightly filled, both bands can conduct.
In insulators, the bandgap Eg is large, and the VB is always completely filled while the CB is
always completely empty, making conduction impossible.
(d) When an energy band is full, there are no open states for an electron to move into, so it
cannot conduct electricity under externally applied electric field (or voltage difference).
(e) A non-conduction electron comes from the scenario described in
A conduction electron comes from the opposite case; when the energy band has empty
states for the electron to move into, it can conduct electricity.
(f) A hole is a vacancy of electron when an electron in the VB gets promoted into the CB. Now
that there is an empty state in the VB, all other electrons can move into it, giving rise to
conduction in presence of an externally applied field. We can show that mathematically, it is
equivalent to treat the conduction by all the electrons in the VB as a positive charge of
equal magnitude (elementary charge) conducting current in the opposite direction. In a
nutshell, an electron vacancy in the VB responds to an externally applied field as a positive
charge, and we call it a hole.
(g) The general form of the wavefunction for electrons in a periodic potential of a crystalline
solid is expressed as a Bloch function, and in 1D it is:
Ψ() = ()
The function () is periodic: () = ( + ), where is an integer, and is the lattice
constant. The complex exponential represents a travelling (plane) wave with a wave
vector depending on the (crystal’s) energy state.
(h) The general form of the Schrödinger equation for a crystal is:
− 2
() = ( + )
The periodic potential energy can be represented by a Fourier series as:
() = ∑exp ( 2
=−∞
Combining everything together, we have the general form of SE in crystals (1D):
− 2
2
2Ψ()
Ψ() = Ψ()
(i) The effective mass ∗ of an electron in a crystalline solid represents how it accelerates
under an externally applied electric field.
It is calculated using the second-order derivative (or “curvature”) of the E-k relationship as:
∗ = 2
( 2 2
)
Under an external field (notice it’s not the energy ), the electron feels force = −,
and the acceleration is:
∗ = −

∗
Typically in a semiconductor, ∗ is different than the free-space electron mass
0 = 9.109 × 10−31kg.
2. (5.2) Intrinsic electron and hole concentrations versus temperature for silicon
Equations used are:
2 )
3 2
Effective electron masses are: ∗ = 1.080 (CB electrons) and
∗ = 0.810 (VB holes).
Special notes: We should use the effective masses given in Appendix C for density of state
calculations. This solution is using these values.
The values given in Question 5.2 need to be adjusted to account for the full band structure
of silicon, which we will learn about if we take more semiconductor physics classes.
From now on, please use ∗ = . and
∗ = . for density of states.
If you have used the values given in Question 5.2, you will get full credits as long as
everything else is the reasonable result when those numbers are used. But it is for this time
only.
Popular mistakes: Please make sure the units are consistent. The concentrations should be
in cm−3 and the temperatures need to be in Kelvins to calculate and (but we may use
Celsius in the graphs).
Energy units: 1Joule = 1kgm2s−2 and 1eV = 1.602 × 10−19Joules
Concentration units: 1m−3 = 10−6cm−3
The plot is on the next page.
At room temperature = 27 = 300K, the intrinsic carrier concentrations for electrons
are holes are = = 1.33 × 1010cm−3.
3. (5.3) Number of atoms in silicon
Method 1: Using the concepts related to unit cells for crystals, we will need to apply the
knowledge learned from the course and look up material properties from the Internet.
Si lattice constant (conventional unit cell) = 5.43 = 5.43 × 10−8cm
Si has diamond crystal structure, so a conventional unit cell has 8 atoms.
Therefore, the number of Si atoms per unit volume is:
= 8
3 = 5.0 × 1022cm−3
Method 2: Using properties related to mass, we can verify the first result as the following:
Si atomic mass: 28.1 amu (14 protons + 14 neutrons)
“Atomic mass constant” 1amu = 1.66 × 10−24g
Si mass density: 2.33g/cm3 (mass of 1cm3 of Si is 2.33g)
Therefore, by counting the mass of Si per 1cm3, we have:
= massper1cm3
massperatom =
4. (5.4) Intrinsic carrier concentration and Fermi level
(a) () as a function of bandgap :
Functions used are:
2 )
When the bandgap increases, the concentration exponentially decreases. In this question,
we use ∗ =
)
For this graph, we set the reference point at the valence band top, so = 0. Solve the
above equation for the Fermi level and get:
= − ln (
) = ln (
(c) Use formula for electron concentration:
= exp (− −
)
For this graph, we set the reference point at the conduction band bottom, so = 0. Solve
the above equation for the Fermi level and get:
= + ln (
) = −ln (
Since > , < 0.
Critical results that can be used for checking for correctness are:
() = () = ln (
) = −ln (
)
0.1eV 3.6 × 1018cm−3 0.05eV −0.05eV 1.1eV 1.4 × 1010cm−3 0.55eV −0.55eV 6.0eV 1.0 × 10−31cm−3 3.0eV −3.0eV
(d) When we have equal effective masses for CB electrons and VB holes ∗ =
∗ , we have
= . Besides, since at equilibrium, for intrinsic semiconductors, = = = , the
following results are true:
− = −
2
This means the Fermi level is located at the middle of the bandgap.
5. (5.6) Effects of dopants
Each boron atom has three valence electrons, that is one less than a silicon atom has. When
introduced into the silicon crystal (and gets activated), it provides one less electron to the
conduction band, or effectively, it provides one hole to the valence band. Thus, boron is an
acceptor dopant.
Each phosphorous atom has five valence electrons, so it works in the opposite way. Thus, it
is a donor dopant.
(In general, the dopant atoms also need to have similar atoms sizes as silicon, and the
donor/acceptor levels need to be close to the conduction/valance bands.)
6. (5.7) Extrinsic semiconductor carrier concentrations
= 1×1016 cm3⁄ , = 5×1014 cm3⁄ , use = 1×1010cm-3.
Method 1: Using approximation: Since , ≈
≈ = 1×1016 cm3⁄
= 2
= −
= 2
7. (5.8) Conduction band energy versus wavenumber
First, we solve for the range for the wavenumber :
= √ 2∗
2
With the given energy upper limit for plotting = 0.5eV and the effective mass of
conduction band electrons ∗ = ∗ = 0.260, we find that the range for the
wavenumber is (be careful with the units):
−1.85×109m-1 ≤ ≤ 1.85×109m-1
The - curve should be a parabola as follows:
8. (5.9) Velocity versus wavenumber
With a parabolic band:
= 1
∗
Note that the electron velocity can be negative (in 1D) or in the negative direction (with
velocity vectors in higher dimensions), depending on the value of the derivative
.
0 ≤ ≤ 2×109m-1
The velocity range is:
0 ≤ ≤ 8.9×107cm/s
The velocity-wavenumber relationship is linear (a straight line) as in the following graph:
9. (5.10) Comparing effective masses
= 1
The first-order derivative
is the “slope” of the band structure that is the - curve. The
“steeper” it goes, the higher the instantaneous velocity is, which means the faster the
electron moves in the crystal.
∗ = 2
( 2 2
The second-order derivative 2
2 is the “curvature” of the band structure. The more it
“bends”, the smaller the effective mass is, which means the easier an electron in that band
can accelerate under an externally applied electric field.
The largest velocity will be from the “X” location near “Γ”, and the smallest velocity
(actually zero) is found at the “X” location at “L”.
The effective masses for the three parabolic bands are put in order of lowest to highest as:
Γ ∗ <