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.