Delft University of Technology
Exam Solid State Physics
ET8027
Technische Universiteit Delft
April 7, 2010
14:00 - 17:00
Preface
Please write down
1. your name (given and family names in this order)
2. your student number, and
3. the course code:
• ET8027 (MSc students)
This exam consists of assignments to be found on page 2 to 13.
Formula sheets and other provided information On page 14 until 19
some information you may need is provided. It is not implied however that one
needs this information.
Weight The relative weight of the assignments is indicated by stars.
Multiple choice questions Unless stated otherwise, to the multiple-choice
questions only one of the alternative answers is correct. Indicate the right
answer; just the symbol suffices, no further explanation is required.
Language You may hand in answers in either English or Dutch.
1
Exam Solid-State Physics ((ET8027) April 7, 2010
2
Question 1. (*)
Ion implantation dopes impurity atoms to a semiconductor material. After ion
implantation we need an annealing step.
Explain why and how we perform the annealing step.
end of question 1.
Solution: In the ion implantation, the crystal is damaged by the penetration
of dopant atoms because of collisions between the incident dopant atoms and
the host atoms. Most of the damage can be removed by thermal annealing the
semiconductor at an elevated temperature. By the annealing with which wafers
are heated by a furnace or light, the both host and dopant atoms in the damaged
region will be placed at proper positions of crystal lattice. (pp.16 and Prologue
xxiii)
Figure 1: The (100) crystal plane and [100] direction of the simple cubic lattice.
Exam Solid-State Physics ((ET8027) April 7, 2010
3
Question 2. (*)
Figure 1 shows the (100) crystal plane and [100] direction of the simple cubic
lattice.
(a) In a similar way, sketch the (362) plane and the [230] direction.
(b) The lattice constant of the simple cubic lattice is 5.63 Å. Calculate the distance
between the nearest (110) planes.
end of question 2.
Solution: See the attached figure 7April2010Q2.jpg.
Question 3. (*)
Consider an electron traveling at a velocity of 108 cm/sec.
(a) Calculate the De Broglie wavelength of the electron.
(b) If an x-ray has the same wavelength, how much energy should be associated
with the x-ray?
end of question 3.
Solution: See the attached figure 7April2010Q3.jpg.
Figure 2: A 1D step potential.
Exam Solid-State Physics ((ET8027) April 7, 2010
4
Question 4. (***)
For the 1D step potential function shown in Figure 2, assume that the total energy
of electrons E > V0 and that the electrons are incident from the +x direction and
are traveling in the −x direction.
(a) Write the wave equation in the region I (x < 0) and II (x > 0).
(b) Write the general solutions of the equation for each region.
(c) The traveling wave in the +x direction in the region I should not exist. Explain
why.
(d) Derive expressions for the reflection coefficient R. (R can be obtained by
R = (AA∗ )/(BB ∗ ), where A and B are the amplitude of the wave-function
for the reflected and the incident electrons, respectively.)
end of question 4.
Solution:
(a) and (b)
For the region I, x < 0, V=0,
∂ψ1 (x) 2mE
+ 2 ψ1 (x) = 0.
∂x2
h̄
(1)
The general solution is of the form
ψ1 (x) = A1 ejK1 x + B1 e−jK1 x
√
where
K1 =
2mE
h̄2
(2)
(3)
For the region II, x > 0, V=V0
∂ψ2 (x) 2m(E − V0 )
+
ψ2 (x) = 0.
∂x2
h̄2
(4)
General form of the solution is
ψ2 (x) = A2 ejK2 x + B2 e−jK2 x
√
where
K2 =
2m
(E − V0 )
h̄2
(5)
(6)
Term with B2 represents incident wave (←), and term with A2 represents the
reflected wave (→).
Exam Solid-State Physics ((ET8027) April 7, 2010
5
(c)
The term involving B1 represents the transmitted wave (←) in the region I.
If a particle is transmitted into region I, it will not be reflected because there is
no potential barrier so that
A1 = 0.
(7)
(d)
Then
ψ1 (x) = B1 e−jK1 x
(8)
ψ2 (x) = A2 ejK2 x + B2 e−jK2 x
(9)
Boundary conditions:
(1)
(2)
ψ1 (x = 0) = ψ2 (x = 0)
∂ψ1 (x)
∂x
=
x=0
∂ψ2 (x)
∂x
(10)
x=0
Applying the boundary conditions to the solutions, we find
B1 = A2 + B2
(11)
K2 A2 − K2 B2 = −K1 B1
(12)
Combining these two equations, we find
(
)
K 2 − K1
A2 =
B2
K 2 + K1
)
(
2K2
B2
B1 =
K2 + K 1
(13)
(14)
The reflection coefficient is
√ √
|A2 |2
A2 A∗2
|K2 − K1 |2
2E − E E − V0 − V0
√ √
→R=
R=
=
=
B2 B2∗
|B2 |2
|K1 + K2 |2
2E + E E − V0 − V0
(15)
Exam Solid-State Physics ((ET8027) April 7, 2010
6
Question 5. (*)
Consider a regular periodic arrangement of atoms in which each atom contains
electrons up to the n = 3 energy level. If the atoms are brought closer together,
the n = (A) shell will begin to interact initially. Since two electrons can not
have the same quantum number, the discrete energy must split into a band of
energies. The Kronig-Penny model models the energy band formation in a 1D
crystal by assuming (B) potential function. To solve the wave equation, it uses
(C). Solving the equation gives an E versus (D) plot which can explain the energy
band formation.
(A): 3, 2, 1
(B): a parabolic, a periodic, an infinite
(C): Bloch’s theorem, Paul’s exclusion principle, the wave-particle duality principle
(D): x, k, t
end of question 5.
Solution:
(A): 3
(B): periodic
(C): Bloch’s theorem
(D): k
Exam Solid-State Physics ((ET8027) April 7, 2010
7
Question 6.
Question 6.a. (*)
The distribution of electrons in the conduction band is given by
A: (density of quantum states) × (energy of a state)
B: (density of quantum states) × (probability a state is occupied)
C: (energy of quantum states) × (probability a state is occupied)
D: (energy of quantum states) × (chemical potential of a state)
end of question 6.a.
(Answer 6.a: B)
Solution: See: [1, pg. 104.]
Question 6.b. (*)
Consider an ideal intrinsic semiconductor in thermal equilibrium. No external
forces or fields are applied to this semiconductor.
Let n denote the concentration of electrons in the conduction band while p
denotes the number of holes in the valence band.
In an ideal intrinsic semiconductor at room temperature, we have
A: n = p.
B: n ≈ p but p 6= n.
C: n p.
D: n p.
end of question 6.b.
(Answer 6.b: A)
Solution: See: [1, pg. 104.]
Question 6.c. (*)
Consider an electron in the conduction band of homogeneously doped n-type silicon
~ act on this electron. There is no magnetic
(Si). Let a non-zero electric field E
~
field. Let ||E|| be the strength of the electric field.
~ > 30 kV/cm, as a result of a scattering process, the
For high electric fields, ||E||
electron
~
A: will reach a constant quasi-Fermi level that becomes independent of ||E||.
~
B: will reach a constant avarage mobility that becomes independent of ||E||.
~
C: will reach a constant potential energy that becomes independent of ||E||.
~
D: will reach a constant avarage drift velocity that becomes independent of ||E||.
end of question 6.c.
(Answer 6.c: D)
Solution: See: [1, pg. 168]
end of question 6.
Exam Solid-State Physics ((ET8027) April 7, 2010
8
Question 7. (**)
Consider a piece of silicon in thermal equilibrium. Let the Fermi energy level EF
be equal to the lower edge of the conduction band energy Ec .
1. If one would apply Boltzmann’s approximation, what value would one assign
to the electron concentration? (hint: see part 2 of this assignment!)
2. At part 1 of this assignment, a numerical answer cannot be achieved. Why
not?
3. The concentration of electrons, n, can be expressed as
2
n = √ Nc F1/2 (ηF ) ,
π
where
ηF =
EF − Ec
,
kB T
and where Nc is the effective density of states in the conduction band, F1/2
is the Fermi-Dirac integral function and kB is Boltzmann’s constant. These
expressions are also valid if the Boltzmann approximation is not valid.
Derive an approximate value for the concentration of electrons in the conduction band. Present your answer in the form of the numerical value of
n/Nc . Explain clearly how you arrive at your result.
end of question 7.
Solution: See: [1, pg. 107, and pg. 126]
1. If Boltzmann’s approximation applies, the electron concentration is given
by
[
]
Ec − EF
n = Nc exp −
.
kB T
Hence, according to this approximate expression, if EF = Ec , n would take
the value Nc , where Nc is the quantum concentration associated with the
conduction band.
2. The quantum concentration Nc depends on temperature. As the temperature is not specified for this assignment, one cannot evaluate the numerical
value of Nc .
3. According to the expressions given, when EF = Ec we find
2
n
= √ F1/2 (0) .
Nc
π
Exam Solid-State Physics ((ET8027) April 7, 2010
9
According to the graph provided by the formula sheet of this exam, F1/2 (0) ≈
0.5. Hence
n
1
≈ √ ≈ 0.6 .
Nc
π
Exam Solid-State Physics ((ET8027) April 7, 2010
10
Question 8.
Question 8.a. (*)
Consider an ideal intrinsic semiconductor in thermal equilibrium. No external
forces or fields are applied to this semiconductor.
At temperatures above 0 K, the electron concentration in the conduction band is
non-zero because
A: some electrons from dopant atoms will overcome the bandgap by gained thermal
energy.
B: some electrons from dopant atoms will overcome the ionization energy by gained
thermal energy.
C: some electrons from the conduction band will overcome the bandgap by gained
thermal energy.
D: some electrons from the valence band will overcome the bandgap by gained
thermal energy.
end of question 8.a.
(Answer 8.a: D)
Solution: See: [1, pg. 104.]
Question 8.b. (*)
Dopant atoms, when added to an intrinsic semiconductor,
A: introduce quantum states that are close to the edges of the forbidden band.
B: introduce quantum states that are near the center of the forbidden band.
C: increase the energy of electrons in the valence band.
D: increase the energy of electrons in the conduction band.
end of question 8.b.
(Answer 8.b: A)
Solution: See: [1, pg. 115.]
Question 8.c. (*)
The “ionization energy of an acceptor in a semiconductor” refers to the difference
between:
A: the energy of an electron bound to the acceptor and the top of the valence
band.
B: the energy of an electron bound to the acceptor and the bottom of the conduction band.
C: the energy of an electron bound to the acceptor and the bottom of the valence
band.
D: the energy of an electron bound to the acceptor and the top of the conduction
band.
end of question 8.c.
(Answer 8.c: A)
Solution: See: [1, pg. 117.]
end of question 8.
Exam Solid-State Physics ((ET8027) April 7, 2010
11
Question 9. (***)
Given are the following expressions for electron- (n) and hole (p)- concentration
in a semiconductor:
(
)
Ec − EF
n = Nc exp −
,
kB T
(
and
p = Nv exp
Ev − EF
kB T
)
.
1. From the expressions given above derive an expression for the intrinsic Fermi
level EFi , as a function of Nc , Nv , Ec , Ev , kB , and T .
2. Given the result for EFi as derived in the previous assignment, why, and
under what conditions, is the following expression,
∇EFi = −q ∇ψ ,
valid? In this expression, q is the elementary charge and ψ is the electrostatic
d
potential. (You may restrict your solution to one dimension, where ∇ = dx
.)
3. Assume that in a given 1-dimensional piece of extrinsic, n-doped semiconducting material, the density of electrons in the conductionband n is equal
to the donor concentration Nd . Let this doping concentration depend on the
position coordinate x (in one direction only):
n = Nd [x] .
Given that the material is in thermal equilibrium, derive an expression for
~ in the semiconductor, as a function of kB , T ,
the built-in electric field E
q, Nd and ∇Nd . You may restrict your solution to one dimension, where
d
∇ = dx
.
4. Assume one would like to realize a built-in electric field in the x-direction.
The x component of this field should depend linearly on the x coordinate:
Ex [x] = a x ,
where a is a real constant. Furthermore it is required that the value of the
electric field vanishes at the origin (x = 0):
Ex [0] = 0 .
Derive an expression for the required doping concentration Nd [x] as a function of position. Give an expression for this doping profile as a function of
temperature T , position x, basic physical constants, the parameter a specified above and the doping concentration Nd [0] at the origin.
end of question 9.
Exam Solid-State Physics ((ET8027) April 7, 2010
12
Solution: See: [1, pg. 175, eqn. (5.40)]:
1. The intrinsic Fermi level EFi is the value the Fermi level would take in an
intrinsic (i.e. pure, non-doped) semiconductor. In such a semiconductor,
n = p. Hence:
(
)
(
)
Ec − EFi
Ev − EFi
Nc exp −
= Nv exp
.
kB T
kB T
Solving for EFi gives
EFi
[
]
1
Nv
1
= (Ev + Ec ) + kB T ln
.
2
2
Nc
2. In the expression for EFi as derived in part1 of this assignment, Ec and Ev
are the total energies of an electron at the edges of the conduction band
and the valence band, respectively. At the edges of the energybands,
the kinetic energy is zero, and the total energy equals the total potential
energy. This potential energy is the sum of, firstly, microscopic energy
(c , v , respectively) due to interaction of electrons with the crystal, and,
secondly, the potential energy due to the interaction of the electrons with
macroscoscopic electrostatic field ψ:
Ec = c − qψ ,
and
Ev = v − qψ .
Under conditionm that the material is homogeneous, so that c , v , Nc ,
Nv and T do not depend on spatial position, the given expression, ∇EFi =
−q ∇ψ, directly follows by differentiation of the result derived under part 1
with respect to the spatial position.
3. Given
(
Ec − EF
Nd [x] = Nc exp −
kB T
[
it follows
Ec − EF = −kB T ln
or, using the same notation as above,
Nd [x]
Nc
[
)
,
]
,
Nd [x]
−qψ + c − EF = −kB T ln
Nc
]
.
In thermal equilibrium EF is spatially constant, so taking the derivative
with respect to spatial coordinates we find
~ = − kB T ∇Nd [x] ,
E
q Nd [x]
~ = −∇ψ is the electric field.
where E
Exam Solid-State Physics ((ET8027) April 7, 2010
13
4. In a 1-dimensional situation, the result derived in the previous part of this
assignment can be written as:
Ex [x] = −
kB T 1 dNd [x]
.
e Nd [x] dx
This is equivalent to:
Ex [x] = −
or
kB T d ln[Nd [x]]
,
e
dx
d ln[Nd [x]]
e
=−
Ex [x] .
dx
kB T
Integrate both sides (in thermal equilibrium T does not depend on position!) with respect to x:
∫ x
Nd [x]
e
ln[
]=−
Ex [ξ] dξ .
Nd [0]
kB T 0
Apply exp to both sides and multiply by Nd [0]:
∫ x
e
Nd [x] = Nd [0] exp[−
Ex [ξ] dξ] .
kB T 0
For Ex [x] = ax, where a is a constant, we find
∫ x
∫ x
1
Ex [ξ] dξ =
a ξ dξ = ax2 ,
2
0
0
so that
Nd [x] = Nd [0] exp[−
ae 2
x ] .
2 kB T
Formula sheet for Final Exam Solid-State Physics (ET8027) April 7, 2010
{
Quantum mechanics
n0 = ni exp
E = hν = h λc
h
λ = hp = mv
h
h̄ = 2π
√
k = 2mE
h̄
∆x × ∆p ≥ h̄
∆E × ∆t ≥ h̄
2
−h̄2 ∂ Ψ(x,t)
+ V (x)Ψ(x, t) = jh̄ ∂Ψ(x,t)
2m
∂x2
∂t
p0 = ni exp
∂ 2 ψ(x)
∂x2
+ 2m
(E − V (x))ψ(x) = 0
h̄2
Ψ(x, t) = ψ(x)φ(t) = ψ(x)e−j(E/h̄)t
vp = ωk
Quantum theory of solid states
ψ(x) = u(x)ejkx
(
)
∂ 2 u1 (x)
1 (x)
+ 2jk ∂u∂x
− k 2 − α2 u1 (x) = 0
∂x2
α2 = 2mE
h̄2
2m
0
(E
−
V0 ) = α2 − 2mV
= β2
h̄2
h̄2
(
)
2
∂ u2 (x)
2 (x)
0
+ 2jk ∂u∂x
− k 2 − α2 + 2mV
u2 (x) = 0
∂x2
h̄2
0 sin αa
P αa + cos αa = cos ka
P 0 = mVh̄02ba
p
1 ∂E
h̄ ∂k = m = v
1
1 ∂2E
m = h̄2 ∂k2
3 √
2
g(E) = 4π(2m)
E
h3
N (E)
n1
o
(E−EF )
g(E) = fF (E) =
1+exp
kB T
Semiconductor
(
Nc = 2
2πm∗
n kB T
h2
) 32
(
) 32
2πm∗
p kB T
Nv = 2
h2
n(E) = gc (E)fF (E)
( ∗)
m
EFi − Emidgap = 34 kB T ln mp∗
n
{
}
Fi )
ni = Nc exp −(EkcB−E
T
{
}
−(Ec −EF )
n0 = Nc exp
{ kB T }
v)
p0 = Nv exp −(EkFB−E
{ T
}
v)
n2i = Nc Nv exp −(EkcB−E
T
nd =
{
14
(EF −EFi )
kB T
}
−(EF −EFi )
kB T
Nd
Ed −EF
1+ 12 exp k
BT
Na
EF −Ea
1
1+ g exp k T
}
= Nd − Nd+
= Na − Na−
√(
)
Nd −Na 2
a
n0 = Nd −N
+
+ n2i
2
√( 2 )
Na −Nd 2
d
p0 = Na −N
+
+ n2i
2
( 2 )
c
Ec − EF = kB T ln N
(n0 )
EF − EF i = kB T ln nn0i
( )
v
EF − Ev = kB T ln N
(p0 )
EF i − EF = kB T ln np0i
pa =
B
n0 p0 = n2i
Mathematics
sin(x ± y) = sin x cos y ± cos x sin y
cos(x ± y) = cos x cos y ∓ sin x sin y
sin 2x = 2 sin x cos y
x
sin2 21 x = 1−cos
2
cos 2x = cos2 x − sin2 x
x
cos2 21 x = 1+cos
2
∂ 2 ψ(x)
∂x2
− c2 ψ(x) = 0
When −c2 > 0, ψ(x) = Aej|c|x + Be−j|c|x
When −c2 < 0, ψ(x) = Aecx + Be−cx
Carrier transport
J = enµn E + epµp E + eDn ∇n − eDp ∇p
µL ∝ T −3/2
+3/2
µI ∝ T NI
v
eτ
µp = Edp = mcp
∗
p
(
)
I
V
=
σ
A
L
qEy = qvx Bz
VH = E H W
1 Ix
vdx = Jepx = ep
Wd
x Bz
VH = Iepd
µh = epVIxxLW d
kB T
D
µ = e
Electric Field
~ = −∇φ = 1 ∇EFi ;
3D: E
e
1 dEFi
1D: Ex = − dφ
=
dx
e dx
Formula sheet for Final Exam Solid-State Physics (ET8027) April 7, 2010
Properties of Si
Bandgap energy = 1.12eV
Effective density of states in conduction band at 300K Nc = 2.8 × 1019 cm−3
Effective density of states in valence band at 300K Nv = 1.04 × 1019 cm−3
Commonly accepted intrinsic carrier concentration at 300K ni = 1.5 × 1010 cm−3
m∗
Effective mass for electrons mn0 = 1.08
Effective mass for holes
m∗
p
m0
= 0.56
Properties of GaAs
Bandgap energy = 1.42eV
Effective density of states in conduction band at 300K Nc = 4.7 × 1017 cm−3
Effective density of states in valence band at 300K Nv = 7.0 × 1018 cm−3
Commonly accepted intrinsic carrier concentration at 300K ni = 1.8 × 106 cm−3
m∗
Effective mass for electrons mn0 = 0.067
Effective mass for holes
m∗
p
m0
= 0.48
Properties of Ge
Bandgap energy = 0.66eV
Effective density of states in conduction band at 300K Nc = 1.04 × 1019 cm−3
Effective density of states in valence band at 300K Nv = 6.0 × 1018 cm−3
Commonly accepted intrinsic carrier concentration at 300K ni = 2.4 × 1013 cm−3
m∗
Effective mass for electrons mn0 = 0.55
Effective mass for holes
m∗
p
m0
= 0.37
15
Formula sheet for Final Exam Solid-State Physics (ET8027) April 7, 2010
Table 1: Orbits
l
Letter symbol
Spectroscopy
0
s
sharp
1
p
principal
2
d
diffuse
3
f
fundamental
Table 2: Quantum numbers
Principal quantum number
Azimutal quantum number
Magnetic quantum number
n
l
|ml |
= 1, 2, 3, ..
= n − 1, n − 2, n − 3, ...0
= l, l − 1, ...0
16
Formula sheet for Final Exam Solid-State Physics (ET8027) April 7, 2010
Figure 3: Mobility as a function of impurity concentration
Figure 4: Periodic table
17
Formula sheet for Final Exam Solid-State Physics (ET8027) April 7, 2010
Figure 5: The Fermi-Dirac Integral function
18
Formula sheet for Final Exam Solid-State Physics (ET8027) April 7, 2010
19
References
[1] Donald A. Neamen. Semiconductor Phsyics and Devices: Basic Principles. McGraw-Hill, New York, 3rd
edition, 2003.
This is the last page.