Download Materials and Devices in Electrical Engineering - KIT

Solution to Examination 2001
Materials and Devices in Electrical Engineering
Statements:
yes
no
correct incorrect
Quantum Numbers
The smaller the principal quantum number (n = 1,2,3,4…), the lower the energy
level.
There may be overlap in energy of a state in one shell with states in an adjacent
shell (the energy of a 3d state is greater than that for a 4s).
O
X
O
O
X
O
O
X
O
X
O
O
X
O
O
X
O
Electropositive elements accept electrons to form negatively charged ions.
O
O
X
Si and Ge show identical valence electron configuration.
O
X
O
O
O
X
O
O
X
O
O
X
O
O
X
O
X
O
O
O
X
O
O
X
O
O
X
Electrons in Atoms
The energies of electrons are quantized.
Electrons exhibit both wavelike and particle-like characteristics.
Electron Configurations
The Pauli exclusion principle stipulates that each electron state can hold no more
than two electrons, which must have opposite spins.
Valence electrons are those that occupy the outermost filled shell.
O
The Periodic Table
Atomic Bonding
There is a sharing of valence electrons between adjacent atoms when bonding is
ionic.
Solids having interatomic bonds that are partially ionic and partially covalent are
not existing.
Metallic Crystal Structures
For FCC crystal structure each corner atom is shared among 4 unit cells.
The atomic packing factor (APF) for FCC is 0.84, which is the maximum packing
possible for spheres all having the same diameter.
Only two atoms are wholly contained within each BCC unit cell.
Crystal Systems
The three (x, y, and z) axes are mutually perpendicular for four crystal systems—
cubic, hexagonal, tetragonal and orthorhombic.
Crystallographic Directions and Planes
In all crystal systems, crystallographic planes are specified by three Miller indices
as (hkl).
Imperfections in Solids
The concentrations of vacancies are significantly lower than that of self-interstitials.
yes
no
correct incorrect
The only way to specify composition of an alloy is weight percent.
Interfacial defects are two-dimensional, including external surfaces, grain
boundaries, twin boundaries, stacking faults, and phase boundaries.
O
O
X
O
X
O
The total interfacial energy is higher in large-grained materials than in fine-grained ones.
O
O
X
Diffusion
In most metal alloys, vacancy diffusion occurs more rapidly than interstitial
diffusion, since the vacancies are more mobile.
For steady-state diffusion, the mass transfer rate does not change with time.
O
O
X
O
X
O
O
X
O
X
O
O
O
X
O
O
X
In a metallic conductor all electrons are mobile charge carriers.
O
The Fermi energy Ef is the energy corresponding to the highest filled state at 273K.
For the electron band structure of metals such as copper, there is an overlap of filled
and empty outer bands.
The electron band structure found in the semiconductors is the same as for
insulators except that the band gap is relatively wide.
Crystalline defects and thermal vibrations serve as electron-scattering centers in
metals.
Semiconductors
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O
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O
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O
O
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O
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O
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O
The electrical behavior of intrinsic semiconductors is based on impurities.
O
O
X
The energy level of a donor state is below the conduction band.
O
X
O
Excess holes, introduced by donor impurities, are the predominant charges carriers
in n-type semiconductors.
O
O
X
The p-n rectifying junction is used for amplification of electrical signals.
O
O
X
O
O
X
O
O
X
O
O
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O
X
O
O
X
O
Mechanical Properties of Metals
Strain represents the amount of deformation induced by a stress.
There are four types of stress-strain tests: tension, compression, torsion, and shear.
Polymer Structures
Organic molecules that have single and double covalent bonds are termed
unsaturated.
Polymers are entirely amorphous and can not exhibit crystallinity.
O
Electrical Conduction
Dielectrics
A dipole moment is a vector that is directed from the positive to the negative
charge.
Electronic polarization results from the relative displacement of electrically charged
ions in response to an electric field.
Ionic polarization is found in all dielectric materials.
All three types of polarization are possible in PbTiO3.
Each polarization mechanism ceases to function when the applied electric field
frequency exceeds its relaxation or resonace frequency.
yes
no
correct incorrect
Magnetic Properties
There are three different sources of magnetic moments for electrons.
O
The relative permeability µr of diamagnetic materials is negative.
O
O
X
O
X
The atomic dipoles of a diamagnetic material are aligned opposite to the field
direction.
O
X
O
In ferromagnetic materials H >> M is valid.
O
O
X
Ferromagnetic materials exhibit no permanent magnetic moment for T < TC. (TC:
Curie temperature)
O
O
X
Soft magnetic materials exhibit a higher coercive force Hc than hard magnetic
materials.
O
O
X
Calculations
1. Metals (14 points)
Using the data in Table 19.1,
a) Compute the resistance of an aluminum wire 3 mm in diameter and 2 m long. (3 points)
R=ρ
l
l
=
=
A σA
2m
=
2Ω
= 7,45 × 10 −3 Ω
268,47
3
3,8 × 10 7 (Ω ⋅ m) −1 × 3,14 × ( × 10 −3 m) 2
2
b) If the potential drop across the ends of the wire is 0,05 V, what is the current density?
(3 points)
J = σE = σ
0,05V
V
= 3,8 × 10 7 (Ω ⋅ m) −1 ×
= 9,5 × 10 5 A / m 2
2m
l
c) A metal wire 2 mm in diameter is required to carry a current of 10A with a minmum of
0,1 V drop per meter of wire. Which of the metals listed in Table 19.1 are possible
candidates? (3 points)
1m
10 7 A
=
≥ 0,1V Þ σ ≤ 3,185 × 10 7 (Ω − m) −1
2
3
,
14
σ
⋅
m
σ × 3,14 × ( × 10 −3 m) 2
2
∴ Iron, Platinum.
V = IR = 10 A ×
d) Calculate the number of free electrons per cubic meter for gold assuming that there are 1,5
free electrons per gold atom. The density and atomic weight of Au is 19,32 g/cm3 and
196,97, respectively. Then compute the electron mobility of Au. (5 points)
n = 1,5 ×
19,32 × 10 6 g / m 3
× 6,023 × 10 23 / mol = 8.86 × 10 28 m −3
196,97 g / mol
3σ = n e µ e Þ µ e =
4,3 × 10 7 (Ω ⋅ m) −1
σ
=
= 3,03 × 10 −3 m 2 / V ⋅ s
28
−3
−19
n e 8.86 × 10 m × 1,6 × 10 C
2. Semiconductors (23 points)
a) Compute the number of free electrons and holes that exist in intrinsic silicon at room
temperature (300 K). (5 points)
σ = e0⋅(n⋅µn + p⋅µp) = 4⋅10-4 S/m
µn = 0,14 m²/Vs
µp = 0,05 m²/Vs
e0 = 1,602⋅10-19 As
n=p
Þ n = p = σ / (e0⋅(µn + µp)) = 1,314 ⋅1016 m-3
If you have not solved this task go on using n = p = 2 ⋅1016 m-3!
b) Calculate the number of holes per atom for silicon. (density of silicon: dSi = 2,33 g/cm³)
(5 points)
periodic table of the elements: atomic weight of Si mMol(Si) = 28,1 g/mole
NA = 6,023⋅1023 atoms/mole
dSi / mMol(Si) ⋅ NA = 2,33 g/cm³ / 28,1 g/mole ⋅ 6,023⋅1023 atoms/mole = 5⋅1022
atoms/cm³
number of holes per atom = 1,314 ⋅1016 m-3 / 5⋅1022 atoms/cm³ = 2,63⋅10-13 / atom
c) Compute the number of free electrons and holes that exist in intrinsic silicon at 600 K.
(band gap energy of silicon: Eg = 1,11 eV) (8 points)
n= p∝e
n(600k )
e
−
Eg
2 k ⋅600 K
−
=
Eg
2 kT
n(300k )
e
−
Eg
Þ n(600 K ) = n(300 K ) ⋅
2 k ⋅300 K
n(600K ) = p(600 K ) = 1,314 ⋅ 1016 m −3 ⋅ e
e
e
−
−
−
Eg
2 k ⋅600 K
Eg
= n(300 K ) ⋅ e
Eg
Eg
æ
− çç
−
è 2 k ⋅600 K 2 k ⋅300 K
ö
÷÷
ø
= n(300 K ) ⋅ e
−
Eg æ 1
1 ö
−
ç
÷
2 k è 600 K 300 K ø
2 k ⋅300 K
1,11eV ⋅K æ 1
1 ö
⋅ç
−
÷
2⋅8, 62⋅10− 5 eV è 600 K 300 K ø
= 6,01 ⋅ 10 20 m −3
If you have not solved this task go on using n = p = 2 ⋅1020 m-3!
d) Compute the electrical conductivity of intrinsic silicon at 600 K, assuming that the charge
carrier mobilities are constant (i.e. values at room temperature). (5 points)
σ = e0⋅(n⋅µn + p⋅µp) = e0⋅n⋅(µn + µp) = 1,602⋅10-19 As ⋅ 6,01⋅1020 m-3 ⋅ 0,19 m²/Vs = 18,3 S/m
3. Dielectrics (20 points)
a) A parallel-plate capacitor with an area of 5 × 10-3 m2 and a plate separation of 3 mm has
the capacitance of 40 pF (4,0×10-11 F), computer the dielectric constant. (5 points)
A
C ⋅ l 4,0 × 10 −11 F × 3 × 10 −3 m
= 2,4 × 10 −11 F / m
C =ε Þε =
=
−3
l
A
5 × 10 m
−11
ε
2,4 × 10
= 2,71
∴ε r =
=
ε 0 8,85 × 10 −12
b) The polarization P of a dielectric material positioned with a parallel-plate capacitor is to
be 2,0 × 10-6 C/m2. What must be the dielectric constant if an eletric field of 5 × 104 V/m
is applied? (5 points)
3 P = ε 0 (ε r − 1) E Þ ε r =
P
2,0 × 10 −6 C / m 2
+1 =
+ 1 = 5,52
ε0E
8,85 × 10 −12 F / m × 5 × 10 4 V / m
c) A charge of 5,5 × 1011 C is to be stored on each plate of a parallel-plate capacitor having
an area of 200 mm2 and a plate separation of 4 mm. If a material having a dielectric
constant of 4,8 is positioned within the plates, what voltage is required? (5 points)
Q
Q Q
Q
5,5 ×10−11 C
=
=
3C = Þ V = =
= 25,89V
A
A
V
C
200×10−6 m2
−12
ε
ε 0ε r
8,85×10 F / m × 4,8 ×
l
l
4 ×10−3 m
If you have not solved this task, go on using a voltage of 30V!
d) Computer the dielectric displacement and the polarization for part c. (5 points)
V
V
25,89V
= ε 0ε r = 8,85 × 10 −12 F / m × 4,8 ×
= 2,75 × 10 −7 C / m 2
l
l
4 × 10 −3 m
V
8,85 × 10 −12 F / m × 25,89V
P = D − ε 0 E = D − ε 0 = 2,75 × 10 −7 C / m 2 −
= 2,18 × 10 −7 C / m 2
−3
l
4 × 10 m
D = εE = ε
4. Magnetic Properties (18 points)
A ferromagnetic material exhibits the following properties at room temperature (300 K):
• remanence flux density: 1,4 tesla
• saturation flux density: 1,7 tesla at 70000 A/m
• coercive force: 50000 A/m
100000
90000
80000
70000
60000
50000
40000
30000
20000
10000
0
-10000
-20000
-30000
-40000
-50000
-60000
-70000
-80000
-90000
-100000
B/T
a) Sketch the entire hysteresis curve B(H) in the range H = -100000 to +100000 A/m. Scale
and label both coordinate axes. (6 points)
H / (A/m)
b) A bar of this material is positioned within a coil of wire 0,1 m long and having 1000 turns.
Calculate the current to achieve the saturation magnetization in the material. (6 points)
saturation flux density: 1,7 tesla at 70000 A/m
H = N⋅I/L Þ I = H⋅L/N = 70000 A/m ⋅ 0,1 m / 1000 = 7 A
c) Schematically sketch the B-H-behavior at
T1: 300 K < T1 < TC
T2: T2 > TC
TC: Curie temperature (6 points)