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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 O O X O X O O X O O X O X 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 X 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)