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Chapter 3 Introduction to Quantum Theory of Solids W.K. Chen Electrophysics, NCTU 1 Outline Allowed and forbidden energy bands Electrical conduction in solid Extension to three dimension Density of state Statistical mechanics W.K. Chen Electrophysics, NCTU 2 3.1 Allowed and forbidden energy bands In one-electron atom Only discrete values of energy are allowed The probability function shows that the electron is not localized at a given radius We can extrapolate these singleatoms results to a crystal and quantitatively derive the concepts of allowed and forbidden energy bands − mo e 4 En = (4πε o ) 2 2h 2 n 2 W.K. Chen Electrophysics, NCTU 3 3.1.1 Formation of energy bands Fig. (a) shows the radial probability density function for the lowest electron energy state of the single, non-interacting hydrogen atom Fig. (b) shows the radial probability curves for two atoms that are in close proximity to each other Due to Pauli exclusion principle, no two electrons may occupy the same quantum state, the interaction of two electrons give each discrete quantized energy level splitting into two discrete energy levels W.K. Chen Electrophysics, NCTU 4 By pushing the numbers of hydrogen type atoms together with a regular periodic arrangement, the initial quantized energy level will split into a band of discrete energy levels The Pauli exclusion principle states that the joining of atoms to form a system (crystal) does not alter the total number of quantum states The discrete energy must split into a band of energies in order that each electron can occupy a distinct quantum state For a system with 1019 one-electron atoms with a width of allowed energy band of 1 eV, if the discrete energy states are equidistant, then the energy levels are separated by 10-19 eV, which is extremely small W.K. Chen Electrophysics, NCTU 5 Example 3.1 Small change of kinetic energy Consider an electron traveling at a velocity of 107 cm/s. Assume the velocity increases by a value of 1 cm/s. Find the increase in kinetic energy 1 1 4.56 ×10 −21 2 −31 5 2 − 21 E = mυ = (9.11× 10 )(10 ) = 4.56 ×10 J = = 2.85 ×10 − 2 eV −19 2 2 1.6 ×10 1 1 ΔE = mυ 22 − mυ12 2 2 Let υ 2 = υ1 + Δυ ⇒ υ 22 = (υ1 + Δυ ) 2 = υ12 + 2υ1Δυ + (Δυ ) 2 1 m(2υ1Δυ ) = mυ1Δυ 2 ΔE ≈ mυ1Δυ = (9.11× 10 −31 )(105 )(0.01) = 9.11×10 −28 J ⇒ ΔE ≈ 9.11×10 − 28 ΔE = = 5.7 ×10 −9 eV −19 1.6 ×10 W.K. Chen Electrophysics, NCTU 6 Discrete energy levels for a hydrogen atom − mo e 4 En = (4πε o ) 2 2h 2 n 2 E4 E3 E2 E1 = −13.6eV E1 = −3.4eV 22 E E3 = 21 = −1.51eV 3 E E4 = 21 = −0.85eV 4 ΔE = 0.66 eV ΔE = 1.89 eV E2 = W.K. Chen E1 7 Electrophysics, NCTU For a atom contains electrons up through the n=3 energy level, when we bought these atoms close together from initially far apart ΔE = 10.2 eV The n=3 energy shell will begin to interact initially, followed by n=2 energy shell interaction and finally the innermost n=1 shell interaction If ro is the equilibrium interatomic distance, then at this distance we have Bands of allowed energies (electrons may occupy) Bands of forbidden energies allowed band forbidden band W.K. Chen Electrophysics, NCTU 8 For silicon atoms, having 14 electrons, Isolated silicon atom For silicon crystal 3s states corresponds to n=3 and l=0 ( containing 2 quantum states) (n=3, l=0, s=-1/2, +1/2) 3p states corresponds to n=3 and l=1 ( containing 6 quantum states) n=3, l=+1, m=1, s=-1/2, +1/2 m=0, s=-1/2, +1/2 m=-1, s=-1/2, +1/2 W.K. Chen 2 electrons occupy n=1 energy level 8 electrons occupy n=2 energy level 4 remaining valence electrons occupying n=3 energy level are involved in chemical reactions Electrophysics, NCTU silicon crystal 9 As the interatomic distance decreases, the 3s and 3p states and gradually merge together to form broad band At the equilibrium interatomic distance, the broad band has again split with 4 quantum states per atom in the lower band and 4 quantum states per atom in the upper band The upper band is assigned as conduction band the lower band is call valence band Since each Si atom has only 4 valence electrons At 0 K, electrons are in the lowest wnergy states all states in the valence band will be full, and all states in the conduction band will be empty W.K. Chen Electrophysics, NCTU 10 3.1.2 The Kronig-Penney Model Fig. (a) shows potential function of a single, noninteracting, one-electron atom Fig. (b) shows potential function for the case when several atoms are in close proximity arranged, one-dimensional array Fig. (c) is the net potential function in Fig. (b) One-dimensional square periodic potential function is used in K.P model W.K. Chen 11 Electrophysics, NCTU Kronig-Penny Model ⎧0 V ( x) = ⎨ ⎩Vo 0< x<a −b < x < 0 Block Theorem The theorem states that all one-electron wave functions involving periodically varying potential energy functions must be of the form ψ ( x ) = u ( x )e u ( x) : periodic function with period(a + b) jkx e jkx :Q traveling wave The general time-independent wave function Ψ ( x, t ) = ψ ( x)φ (t ) = u ( x)e jkx ⋅ e − jωt W.K. Chen ω= Electrophysics, NCTU E h ( E = hω ) 12 Time-independent Schrodinger wave equation ∂ 2ψ ( x) 2m + 2 ( E − V ( x))ψ ( x) = 0 h ∂x 2 ψ ( x) = u ( x)e jkx ∂ 2ψ ( x) 2mE Region I : + 2 ψ ( x) = 0 h ∂x 2 (0 ≤ x ≤ a ) ∂ 2ψ ( x) 2m( E − Vo ) Region II : + ψ ( x) = 0 h2 ∂x 2 (−b ≤ x ≤ 0) W.K. Chen 13 Electrophysics, NCTU Region I: du ( x) d 2u1 ( x) + 2 jk 1 − (k 2 − α 2 )u1 ( x) = 0 2 dx dx α2 = 2mE h2 u1(x): the amplitude of the wave function in region I if Vo > E Region II : d 2u 2 ( x ) du ( x) + 2 jk 2 − (k 2 − β 2 )u2 ( x) = 0 2 dx dx β= u2(x): the amplitude of the wave function in region II u1 ( x) = Ae j (α − k ) x + Be − j (α + k ) x for 0 < x < a u2 ( x) = Ce j ( β − k ) x + Be − j ( β + k ) x for − b < x < 0 W.K. Chen Electrophysics, NCTU 2m( E − Vo ) = jγ h2 if E >Vo β2 =α2 − 2mVo h2 14 u1 ( x) = Ae j (α − k ) x + Be − j (α + k ) x for 0 < x < a u2 ( x) = Ce j ( β − k ) x + Be − j ( β + k ) x for − b < x < 0 4 unknowns, 4 BCs ψ ( x) = u ( x)e jkx The wave function ψ(x) and its derivative ∂ψ/∂x ∂ψ ( x) ∂u ( x) must be continuous ⇒u(x) and its derivative ∂u/∂x must be continuous ∂x − = ∂x e jkx + jku ( x)e jkx + ψ ( x − ) = ψ ( x + ) ⇒ u ( x − )e jkx = u ( x + )e jkx ⇒ u ( x − ) = u ( x − ) ∂ψ ( x) ∂ψ ( x) = ∂x ⇒ x ∂x − x+ − − + + ∂u ( x) ∂u ( x) e jkx + jku ( x − )e jkx = e jkx + jku ( x + )e jkx ∂x x − ∂x x + ⇒ ∂u ( x) ∂u ( x) = ∂x x − ∂x x + W.K. Chen Electrophysics, NCTU u1 ( x) = Ae j (α − k ) x + Be − j (α + k ) x for 0 < x < a u2 ( x) = Ce j ( β − k ) x + Be − j ( β + k ) x for − b < x < 0 B.C.1 : u2 (0 − ) = u1 (0 + ) ⇒ B.C.2 : ∂u2 ∂x B.C.3 : = 0 − ∂u2 ∂x W.K. Chen ∂u1 ∂x = x=a ∂u2 ∂x A+ B = C + D ⇒ (α − k ) A − (α + k ) B − ( β − k )C + ( β + k ) D = 0 0 + u1 (a) = u2 (−b) B.C.4 : 15 Ae j (α − k ) a + Be − j (α + k ) a = Ce − j ( β − k )b + Be + j ( β + k )b (α − k )e j (α − k ) a + (α + k )e − j (α + k ) a x =−b = ( β − k )e − j ( β − k ) b − ( β + k )e + j ( β + k ) b Electrophysics, NCTU 16 From the above 4 eq, derived from 4 BCs, we obtain the parameter k in relation of the total energy E (through the parameter α)and the potential function Vo (through the parameter β) Kronig-Penny Model − (α 2 + β 2 ) (sin α a )(sin β b) + (cos α a )(cos β b) = cos k (a + b) 2αβ ⎧0 V ( x) = ⎨ ⎩Vo 0< x<a −b < x < 0 W.K. Chen 17 Electrophysics, NCTU 3.1.3 The k-space diagram − (α 2 + β 2 ) (sin α a)(sin β b) + (cos α a )(cos β b) = cos k (a + b) 2αβ α= 2mE h2 β = jγ = j 2m(Vo − E ) , γ >0 h2 Case 1: Free particle: bVo is finite value, and b=0 Vo=0 V(x) 2mE β =α = h2 W.K. Chen E Electrophysics, NCTU 18 0 1 − (α + α ) (sin α a)(sin α b) + (cos α a )(cos α b) = cos k (a + b) 2α 2 2 2 cos αa = cos ka (free particle) ⇒ α =k 2mE 2m( 12 mυ 2 ) mυ p 2mE ⇒ =k = = = = k α= h2 h2 h2 h h ⎧ p = hk ⎪ ⇒⎨ p 2 h 2k 2 = ⎪E = 2m 2m ⎩ p: particle momentum k: wave number For the case of free particle, parabolic relation is valid between energy E and momentum p. p: momentum of wave motion W.K. Chen Electrophysics, NCTU 19 Case 2: bound electrons in crystal: bVo is finite value, but b→0 Periodic δ function potential barrier Vo→∞ ⇒ E<<Vo V(x) Periodic δ function potential barrier W.K. Chen Electrophysics, NCTU 20 − (α 2 + β 2 ) (sin α a )(sin β b) + (cos α a )(cos β b) = cos k (a + b) 2αβ α= 2mE h2 Q E << Vo ⇒ β = jγ = j 2m(Vo − E ) , γ >0 h2 decaying wave at barrier region (γ 2 − α 2 ) (sin α a)(sinh γ b) + (cos α a)(cosh γ b) = cos k (a + b) 2αγ 1 1 b→0 ⇒sinhγb≈ γb & coshγb→1 Q sin θ = [e jθ − e − jθ ] cos θ = [e jθ + e − jθ ] 2j 2 mVoba sin α a + (cos α a) = cos ka h2 αa sin α a E - k equation P' + (cos α a) = cos k a αa energy term momentum term P' = mVoba h2 α= 2mE h2 The above eq does not lend itself to an analytical solution, but must be solved using numerical or graphical techniques to obtain the relation between k, E and Vo. W.K. Chen 21 Electrophysics, NCTU Q sin θ = e x + e− x cosh x = 2 1 1 jθ [e − e − jθ ] cos θ = [e jθ + e − jθ ] 2j 2 e x − e− x sinh x = 2 e x + e− x cosh x = 2 x e x − e− x sinh x = 2 W.K. Chen Electrophysics, NCTU 22 f (αa) = P' sin αa + (cos αa) αa − 1 ≤ f (αa ) = cos ka ≤ +1 a is fixed, k varies The parameter α is related to the total energy E of the particle α= 2mE h2 W.K. Chen Electrophysics, NCTU 23 W.K. Chen Electrophysics, NCTU 24 E-k diagram The right figure shows the energy E as a function of the wave number k The plot shows the concept of allowed energy bands for particle propagating in the crystal lattice Since the energy E is discontinuities, we also have the concept of forbidden energy for particles in the crystal W.K. Chen 25 Electrophysics, NCTU Example 3.2 bandwidth Coefficient P’ =10, potential width a=5Å f (αa) = P' sin αa + (cos αa) = cos ka αa Solution The lowest allowed band occurs as ka changes from 0 to π For ka=0 P' sin αa + (cos αa) = +1 αa Using numerical technique (trial and error), we find α= Elower W.K. Chen αa = 2.628 2mEupper 2mE ⇒ = α a a = 2.628 h2 h2 (2.628) 2 h 2 (2.628) 2 (1.054 ×10−34 ) 2 = = = 1.68 ×10−19 J = 1.053 eV 2 −31 −10 2 2ma 2(9.11×10 )(5 ×10 ) Electrophysics, NCTU 26 For ka=π, as seen form the figure it happens αa=π αa = Eupper = 2mEupper h2 π 2h 2 2ma2 = a =π π 2 (1.054 ×10−34 ) 2 2(9.11×10−31 )(5 ×10−10 ) 2 = 2.407 ×10−19 J = 1.50 eV ΔE = Eupper − Elower = 1.50 − 1.053 = 0.447 eV W.K. Chen P' Electrophysics, NCTU sin αa + (cos αa ) = cos ka αa energy term 27 cos ka = cos(ka + 2nπ ) = cos(ka − 2nπ ) momentum term Displacement of portions of the curve in left figure by 2π still satisfy the E-k equation W.K. Chen Electrophysics, NCTU 28 The entire E versus k plot is contained within π/a<k<π/a For free electron, the particle momentum and the wave number k are related by momentum p = hk (free electron) Given the similarity between the free electron solution and the results of the single crystal, The parameter ħk in a single crystal is referred to as the crystal momentum, which is not the actual momentum of the electron in the crystal , but is a constant of motion that includes the crystal interaction crystal momentum : hk W.K. Chen Electrophysics, NCTU 29 3.2 Electrical conduction in solids For covalent bonding of silicon crystal, each silicon atom is surrounded by 8 valence electrons. 4 from itself, 4 from the 4 nearest Si neighbor For N silicon atoms in the crystal, there are 4 N energy states in lower valence band and 4N energy states in higher conduction band. Each energy state allow only one electron to reside At T=0K, all valence electrons are lying in their lowest energy states, valence band, i.e. the valence band is fully occupied and the conduction band is completely empty W.K. Chen Electrophysics, NCTU 30 Covalent bond breaking At T>0K, a few valence band (VB) electrons may gain enough thermal energy to break the covalent band and jump into conduction band (CB) VB: covalent bonding position CB: elsewhere Since the pure semiconductor is neutrally charged, as the negatively charged electron breaks away from its covalent bonding position, a positively charged “empty state” is created in the original covalent bonding position in the VB CB VB W.K. Chen Electrophysics, NCTU 31 At T=0K, the energy states in the VB in E-k diagram are completely full, and the states in CB are empty (Fig. (a)) At T>0K, a few valence band (VB) electrons may gain enough thermal energy to break the covalent band and jump into CB (Fig (b)) W.K. Chen Electrophysics, NCTU 32 3.2.3 Drift current l l N: total number of flow charge n: volume density of flow charge A: cross-sectional area υ: average drift velocity l: traveling length of carrier per Δt A N nAl nA(υΔt ) # = nAυ ( ) = = Δt Δt Δt sec Φ # ) Flux density φ = = nυ ( A sec ⋅ cm 2 Flux Φ = Drift current density J = qφ = qnυ d ( n J = q ∑υi A ) cm 2 Ampere υi: the velocity of the ith charged carrier i =1 W.K. Chen Electrophysics, NCTU 33 E-k diagram under external bias Under no external force, the electron distribution in CB is an even function of k, since the net momentum is zero under thermal equilibrium. Under applied force, electrons in CB can gain energy and a net momentum Work - Eenergy Theorem dE = Fdx = Fυdt W.K. Chen Electrophysics, NCTU 34 3.2.3 Electron effective mass The movement of an electron in a lattice is different from that of an electron in free space r Fext Ftotal = Fext + Fint = ma r Fext Fint: due to the interaction between the moving electron and other charged particles, such as ions, protons and else electrons in the lattice Fext = m*a Free space r Fext = ma crystal r Fext = m*a m*: effective mass Which takes into account of the particle mass and the effects of internal forces W.K. Chen Electrophysics, NCTU 35 E-k diagram and particle mass in free space (hk ) (parabolic curve) 1 p2 2 E = moυ = = 2 2mo 2mo 2 ( p = hk ) ∂E 2h 2 k hp hmoυ ⇒ = = = mo ∂k 2mo mo 1 ∂E = velocity : υ (free electron) ⇒ h ∂k h2 ∂ 2 E ∂ h 2k = ( )= mo ∂k 2 ∂k 2mo 1 1 ∂2E = ⇒ mass : mo h 2 ∂k 2 W.K. Chen The first derivative of E with k is related to the velocity of free particle The second derivative of E versus k is inversely proportional to the mass of free particle Electrophysics, NCTU 36 Electron effective mass & E-k diagram in crystal Conduction band The energy near the bottom of CB may be approximated by a parabola, just as that of free particle E − Ec = C1k 2 , C1 > 0 1 1 ∂2E = mass : mo h 2 ∂k 2 (free electron) W.K. Chen 1 1 ∂2E effective mass : * = 2 mn h ∂k 2 (Block electron) 37 Electrophysics, NCTU ∂E ∂2E = 2C1k (C1 > 0), = 2C1 ∂k ∂k 2 1 1 ∂ 2 E 2C1 ⇒ * = 2 = 2 h mn h ∂k 2 Thus acceleration of Block electron with external force is Fext = mn a * If we apply an electric field, the acceleration is a= Fext − eE = * mn* mn a Fext E: electric field E: total energy E Effective mass parameter is used to relate quantum mechanics and classical mechanics The larger the bowing of E-k diagram, the small the effective mass is W.K. Chen Electrophysics, NCTU 38 3.2.4 Concept of hole The empty state is created when a covalent bond is broken The movement of a valence electron into the empty state is equivalent to the movement of a positively charged empty state itself The above positive charge carrier is called “hole” The crystal now has a second equally important charge carrier, hole, that can give rise to a current W.K. Chen 39 Electrophysics, NCTU The drift current in VB Under external biased voltage, it is the electrons that moving in the VB, not the positively charged ions The drift current due to electrons in VB J = ( − e) ∑υ n Q J = q ∑υi , q = −e for electron i i ( filled ) J = ( − e) ∑ υ i + ( + e) i ( full ) i =1 ∑υ q = +e for + ion i i ( empty ) = + filled emptyfilled W.K. Chen Electrophysics, NCTU 40 (−e) ∑υi = ? (full valence band) i ( full ) The velocity of individual electron is given υe (E ) = 1 ∂E h ∂k The valence band is symmetric in k and every state in VB is occupied so that every electron with a velocity lυ l, there is a corresponding electron with a velocity -lυ l Since the valence band is full, the distribution of electrons with respect to k has no chance to be changed by an external applied. The net drift current density generated from a completely full band electrons, then, is zero ( − e) ∑ υ i = 0 i ( full ) W.K. Chen 41 Electrophysics, NCTU The net drift current density due to electrons in VB 0 J = ( − e) ∑ υ i + ( + e) i ( full ) ∑υ i i ( empty ) = ( + e) ∑υ i i ( empty ) The net drift current due to electrons in VB is entirely equivalent to placing a positively charged particle in the empty state (assigned as hole) and assuming all other states in the band are neutrally charged The associated hole velocity is exactly the velocity of the electron who occupy the empty state neutral W.K. Chen hole neutral Electrophysics, NCTU 42 Effective mass & E-k diagram in crystal Valence band The energy near the bottom of VB may be approximated by a parabola, just as that of free particle Eυ − E = C2 k 2 ∂E ∂2E = −2C2 k (C2 > 0), = −2C2 ∂k ∂k 2 1 1 ∂ 2 E − 2C 2 ⇒ *= 2 = <0 2 2 me h ∂k h An electron moving near the top of valence band behaves as if it has a negative mass W.K. Chen 43 Electrophysics, NCTU Thus acceleration of Block electron near the top of VB with external force is Fext = me*a = (−e) E We may rewrite me: negative quantity F F a = ext* = ext* > 0 me me = − eE + eE = * me* mp a Fext E filled emptyfilled m*p = − me* 1 1 1 ∂2E =− * =− 2 m*p me h ∂k 2 1 ∂E υ h ( E ) = −υ e ( E ) = − h ∂k W.K. Chen Electrophysics, NCTU E a Fext neutral hole neutral 44 Now we can model VB as having particles with a positive electronic charge and a positive effective mass, so it will move in the same direction as the direction of applied field The new particle is the hole The density of hole is the same as the density of empty states in VB W.K. Chen Electrophysics, NCTU 45 3.2.5 Metals, insulators and semiconductors Insulator: The allowed bands are either completely empty or completely full Since empty band and full band contribute no current at all, the resistivity of an insulator is very large The bandgap of insulator is usually on the order of 3.5 to 6 eV or larger W.K. Chen Electrophysics, NCTU 46 Semiconductor: The allowed bands are at conditions either almost empty or almost full Both the electrons in CB and holes in VB can contribute the current The bandgap of semiconductor is on the order of 1 eV The resistivity of semiconductor can be controlled and varied over many orders of magnitude W.K. Chen Electrophysics, NCTU 47 Metal: The band diagram for a metal may be in one of two forms partially full band overlapped conduction and valence bands In the case of partially full band, many electrons are available for conduction, so that the material can exhibit a large electrical conductivity For overlapped conduction and valence bands , there are large numbers of electrons as well as large numbers of holes can move, so this material can also exhibit a very high electrical conductivity W.K. Chen Electrophysics, NCTU 48 3.3 Extension to three dimensions The basic concept of allowed and forbidden energy bands comes from the electrons moving in periodic potential arrangement in crystal lattice For three dimensional crystal, the distance between atoms varies as the direction through the crystal changes W.K. Chen Electrophysics, NCTU 49 The E-k diagram is symmetric in k so that no new information is obtained by displaying the negative axis In following plot of E-k diagrams, E-k diagrams at two directions are plotted. [100] portion of the diagram with +k to the right and [111] portion of the diagram with +k to the left are plotted W.K. Chen Electrophysics, NCTU 50 Direct bandgap the minimum conduction band is at k=0 so that the transitions for electron between CB & VB can take place with no change in crystal momentum Indirect bandgap the minimum conduction band occurs not at k=0 so that the transitions for electron between CB & VB includes an interaction with the crystal W.K. Chen Electrophysics, NCTU 51 Effective mass concept in 3-dim crystal W.K. Chen Electrophysics, NCTU 52 3.4.1 Density of states for free electron To determine the density of allowed quantum states as a function of energy, we need to consider an appropriate mathematical model Consider a free electron confined to a three-dimensional cubic with length a with infinite potential well a V ( x, y , z ) = 0 for 0 < x < a for 0 < y < a for 0 < z < a a V ( x, y, z ) = ∞ elsewhere a Using the separation of variables technique in 3 dimensional Schrodinger’s wave equation, we can have r k 2mE = k 2 = k x2 + k y2 + k z2 2 h W.K. Chen Electrophysics, NCTU r kx r kz r ky 53 Two-dimensional plot of allowed quantum states in k space k x = nx π a , k y = ny π a , k z = nz π a nx, ny, nz: positive integers Positive and negative values of kx,ky or kz have the same energy and represent the same quantum state The volume Vk of a single quantum state in k-space is π Vk = ( ) 3 a W.K. Chen Electrophysics, NCTU 54 The density of quantum state in k-space for whole cubic box is 1 4πk 2 dk gT (k )dk = 2( ) Vk 8 2: due to two spin states allowed for each quantum state 1/8: due to the positive values of kx,ky, and kz gT (k )dk = k 2 dk π2 ⋅ a3 W.K. Chen 55 Electrophysics, NCTU Density of quantum states versus energy g k (k )dk = k 2 dk π2 ⋅a 4π (2m) 3 / 2 ⇒ g ( E )dE = h3 E dE 1 2mE h 1 1 1 1 m 2m ⋅ ⋅ dk = dE = dE 2 E h h 2E Qk2 = 2mE , h2 3 k= The number of energy states for whole cubic box between E and E+dE, i.e., between k and k+dk is given by g k (k )dk = k 2 dk π 2 ⋅ a 3 = g ( E )dE 2mE 1 1 m 4πa 3 3 g ( E )dE = 2 ⋅ 2 ⋅ dE ⋅ a = 3 ⋅ (2m) 3 / 2 ⋅ E dE h π h 2E h W.K. Chen Electrophysics, NCTU 56 The density of quantum states per unit energy per unit volume of the crystal g ( E )dE = gT ( E )dE gT ( E )dE = V a3 E 4π (2m) 3 / 2 g ( E )dE = h3 E dE g(E) The density of states is proportional to the energy E. As the energy of free particle becomes large, the number of available quantum states increase The density of states is dependent on the mass of the free particle W.K. Chen Electrophysics, NCTU 57 Example 3.3 Density of states Calculate the total density of states for free electron per unit volume with energies between 0 and 1 eV 4π (2m) 3 / 2 1eV N =∫ g ( E )dE = ∫0 E dE 0 h3 4π (2m) 3 / 2 2 3 / 2 N= ⋅ ⋅E h3 3 4π [2 ⋅ (9.11×10 −31 )]3 / 2 2 N= ⋅ ⋅ (1.6 ×10 −19 ) 3 / 2 = 4.5 ×10 27 m −3 −34 3 (6.625 ×10 ) 3 1eV N = 4.5 × 10 21 states/cm −3 The density of states is typically a large number ranging from 1018 to 1022 cm-3 The density of quantum states in semiconductor is usually less than the density of atoms in semiconductor crystal W.K. Chen Electrophysics, NCTU 58 3.4.2 Density of states for semiconductor The parabolic relationship between energy and momentum of a free electron is p 2 h 2 k2 = E= 2m 2m The E-k curve near k=0 at conduction band can be approximated as a parabola, so we have p 2 h 2 k2 E − Ec = = 2m 2m The density of states in CB is 4π (2mn* ) 3 / 2 g c ( E )dE = h3 W.K. Chen E − Ec dE Electrophysics, NCTU 59 Similarly, the E-k parabolic relationship for a free hole in VB h 2 k2 p2 = Eυ − E = 2m*p 2m*p The density of states for holes (electrons) in VB is gυ ( E )dE = W.K. Chen 4π (2m*p ) 3 / 2 h3 Eυ − E dE Electrophysics, NCTU 60 3.5 Statistical mechanics In dealing with large numbers of particles, we are interested only in the statistical behavior of the group as a whole rather than in the behavior of each individual particle There are three distribution laws determining the distribution of particles among available energy states W.K. Chen Particles are distinguishable No limit to the number of particles allowed in each energy state The exemplary case is low-pressure gas molecules in a container Bose-Einstein distribution 61 Maxwell-Boltzmann distribution Electrophysics, NCTU Particles are indistinguishable No limit to the number of particles allowed in each energy state The exemplary case is photon Fermi-Dirac distribution Particles are indistinguishable Only one particle is permitted in each energy state The exemplary case is electrons in crystal W.K. Chen Electrophysics, NCTU 62 3.5.2 The Fermi-Dirac Probability Function For ith energy level with gi quantum states Only one particle is allowed in each state First particle: gi ways of choosing to place the particle Second particle: (gi -1) ways of choosing to place the particle Ni particle: [gi –(Ni-1)] ways of choosing to place the particle Then the total number of ways of arranging Ni particles in the ith energy level is ( g i )( g i − 1) L ( g i − ( N i − 1)) = W.K. Chen gi ! ( g i − N i )! Electrophysics, NCTU 63 Since the particles are indistinguishable, the interchange of any two electrons does not produce a new arrangement The actual number of independent ways to distribute the Ni particles in the ith level is Wi = W.K. Chen gi ! N i !( g i − N i )! Electrophysics, NCTU 64 N ( E )dE = f E (E) = g ( E )dE 1 ⎛ E − Ef 1 + exp⎜⎜ ⎝ kT ⎞ ⎟⎟ ⎠ ( N ( E ) < g ( E )) N(E)dE (number density): Number of particles per unit volume per unit energy (E to E+dE) g(E)dE: number of quantum states per unit volume per unit energy W.K. Chen Electrophysics, NCTU 65 Meaning of Fermi energy At T=0K, The probability of a quantum state being occupied is unity for E<Ef The probability of a quantum state being occupied is zero for E>Ef f E (E) = W.K. Chen 1 ⎛ E − Ef 1 + exp⎜⎜ ⎝ kT ⎞ ⎟⎟ ⎠ f E (E) = 0 for E > E f f E (E) = 1 for E < E f Electrophysics, NCTU 66 3.5.3 Features of Fermi-Dirac Function For T>0K The probability of a state being occupied at E=Ef is always ½ f E (E) = 1 1 = 1 + exp(0 ) 2 W.K. Chen Electrophysics, NCTU 67 Example 3.6 3kT above the Fermi-energy fE (E) = 1 ⎛ E − Ef 1 + exp⎜⎜ ⎝ kT ⎞ ⎟⎟ ⎠ = 1 1 = = 4.74% ⎛ 3kT ⎞ 1 + 20.09 1 + exp⎜ ⎟ ⎝ kT ⎠ At energy above Ef, the probability of a state being occupied by an electron become significantly less than unity W.K. Chen Electrophysics, NCTU 68 Example 3.7 99% probability Assume Fermi energy is 6.25 eV, Calculate the temperature at which there is 1 % probability that a state 0.30 below the Fermi energy level will not contain an electron 1 − f E (E) = 1 − 1 ⎛ E − Ef 1 + exp⎜⎜ ⎝ kT ⎞ ⎟⎟ ⎠ 1 ⎛ − 0.30 ⎞ 1 + exp⎜ ⎟ ⎝ kT ⎠ ⇒ kT = 0.06529eV ⇒ T = 756K 0.01 = 1 − k = 1.38 ×10 −23 J/K = 8.62 ×10 −5 eV/K W.K. Chen 69 Electrophysics, NCTU Maxwell-Boltzmann Approximation f E (E) = 1 ⎛ E − Ef 1 + exp⎜⎜ ⎝ kT ⎞ ⎟⎟ ⎠ When E-Ef>>kT, the exponential term in the denominator is much greater than the unity ⎡ E − Ef ⎤ f E ( E ) ≈ exp ⎢− ⎥ for E − E f > 3kT kT ⎣ ⎦ kT = 0.026 eV at T = 300K The Fermi-Dirac probability for E-Ef> 3kT can be approximated by Boltzmann distribution For E-Ef>3kT, the Boltzmann approximation is slightly larger than the FermiDirac curve with inaccuracy less than 5% W.K. Chen Electrophysics, NCTU 70 Example 3.8 3kT Calculate the energy at which the difference between Boltzmann approximation and the Fermi-Dirac function is 5% ⎛ E − Ef exp⎜⎜ − kT ⎝ ⎞ 1 ⎟⎟ − ⎠ 1 + exp⎛⎜ E − E f ⎜ kT ⎝ 1 ⎛ E − Ef ⎞ ⎟⎟ 1 + exp⎜⎜ kT ⎝ ⎠ ⎞ ⎟⎟ ⎠ = 0.05 ⎛ E − Ef exp⎜⎜ − kT ⎝ ⎞ ⎧ ⎛ E − Ef ⎟⎟ ⋅ ⎨1 + exp⎜⎜ ⎠ ⎩ ⎝ kT ⎛ E − Ef exp⎜⎜ − kT ⎝ ⎞ 1 ⎟⎟ = 0.05 ⇒ E − E f = kT ln( ) ≈ 3kT 0.05 ⎠ W.K. Chen ⎞⎫ ⎟⎟⎬ − 1 = 0.05 ⎠⎭ Electrophysics, NCTU 71