Problem Sheet 1 From material in Lectures 2 to 5

Transcription

1 lectrons in Solids ) Problem Sheet From material in Lectures to 5 ) [Standard derivation] Consider the free electron model for a - dimensional solid between x = and x = L. For this model the time independent Schrödinger equation has the form: d dx ψ = ψ a) Write down the boundary conditions. b) Show that it has a general solution of the form: ψx) = Aexpik x x) and find values of wavevector k x. c) Find a real solution for A. d) Find the dispersion relationship. e) First consider only positive values of k x. Find the density of states D). f) How does this differ from D) for the -dimensional particlein-a-box / infinite well model? g) Find the value for D) for the situation when both positive and negative values of k x are allowed. ) Consider a nanoscale solid whose smallest sub-unit has length d. Treat the solid as a -dimensional infinite potential well containing noninteracting electrons. The nanoscale solid can increase its length according to L = pd, where p is an integer number of sub-units. Let each sub-unit contain 5 electrons. Find the variation of the smallest electronic transition Δ with p for both p even and p odd, and show that at large p they both tend to: Δ 5π pd 3) For ) above, find the general equation for the Fermi level for p even and p odd. If d =.5 nm, calculate the limiting value of in ev) at large p. 4) In Lecture 3 we showed that the finite potential well lead to intermediate solutions of the form: ψ I x) = C expβx) ψ II x) = Acosαx) + Bsinαx) α = β = V ) ψ III x) = D exp βx) in the three different regions. By considering the continuity of ψ and ψ at x = -a and +a, find the full even and odd solutions.

2 lectrons in Solids ) Problem Sheet From material in Lectures 5 to 8 ) lectrons in solids obey Fermi-Dirac statistics. a) Define the Fermi energy. b) State the Fermi-Dirac distribution function, and sketch how it varies with temperature. c) Define the chemical potential µ. d) What is the limit in which = µ? e) By considering the value of the Fermi temperature T F, is it possible to approximate by µ at 3K? Sketch the room temperature Fermi-Dirac distribution for a typical metal according to the free electron model. ) [Standard derivation] Consider the 3-dimensional free electron model 3D FM) for a cube which lies between x = and x = L, y = and y = L and z = and z = L, and has total volume V = L 3. For this model the time independent Schrödinger equation has the form: ψ = ψ a) Write down the boundary conditions in terms of x, y and z. b) Show that it has a general solution of the form: ψr) = C expik.r) and find the dispersion relationship. c) Find the values of the wavevector components k x, k y and k z. d) Find a real solution for C. 3) Consider the 3-dimensional free electron model from question ). a) Find the volume occupied per state in k-space. b) By considering the number of states in a spherical shell in k- space, find dn/dk. c) Hence find the density of states D). d) From D) and the answer to )a), find the Fermi energy. e) From the value of find the Fermi wavevector k F. f) By rewriting the answer to part )a) in terms of k instead of, find a second method to calculate k F. g) By considering a sphere in k-space occupied by N carriers, find a third method to calculate k F. h) From 3)g) or h), find, and hence D). 4) How does the 3-dimensional free electron model explain the electrical conductivity σ Ωm) of metals? Consider both the charge carrier density and the charge carrier mobility.

3 lectrons in Solids ) Problem Sheet 3 From material in Lectures 8 to ) For the 3-dimensional free electron model 3D FM) it is possible to calculate a fundamental thermal property of solids, the specific heat. a) Give a simple definition of the specific heat C V. Write down equations for it in units of J K -, J K - mol - and J K - kg -. b) What is C V J K - ) for a classical ideal gas of N particles? c) Consider the 3D FM for a cube of volume V containing N electrons. Find the thermal energy ΔU of the electrons within kt B of the surface of the Fermi sphere. d) Using the values of D) and, show that for the 3D FM: D ) = 3N e) Hence, find the 3D FM specific heat C V J K - ). How does this and the answer to )b) compare to the measured C V for metals at low temperatures? ) What method can be used to measure the Fermi surfaces of real metals? Do these Fermi surfaces differ from that predicted by the 3D FM? What are the successes and failures of the FM? 3) A -dimensional D) periodic potential has the form: Ux) = Ux + na) a) State Bloch s theorem for this potential. b) How is the Bloch wavefunction ψx) related to itself in the nearest-neighbour unit cell ie. when n = )? c) Show that the wavefunction is equally distributed in both unit cells. d) What are the allowed wavevector k x values for these states. e) How does the meaning of k x differ from that in the D FM. 4) The Kronig-Penney model assumes a -dimensional periodic potential made up of repeating Dirac delta-functions: where in the regions between the δ-functions the potential U =.

4 4) continued a) In unit cell I, a trial solution ψιx) has the form of an incident and reflected traveling wave: ψ I x) = A n exp iqx) + B n expiqx) Rewrite this in the form of Csinqx) and Dcosqx), and find a value for q. b) Using Bloch s theorem, write down a solution for the wavefunction ψιιx) in unit cell II. c) Using continuity at x =, find an equation relating ψιx) to ψιιx). d) Show that this leads to: C D = cosqa) exp ik a) x sinqa) e) Find the first derivative with x of ψιx) and ψιιx). f) Using the Dirac delta-function cusp) condition at x = : ψ II ) ψ I) = U ψ x x I ) find an equation relating dψιx)/dx to dψιιx)/dx. g) Shows that this leads to: C D = expik a) sinqa) U x q expik x a) cosqa) h) Using d) and g), find the Kronig-Penney equation: cosk x a) = cosqa) + mau sinqa) qa 5) Consider the Kronig-Penney model in question 4). a) Sketch the variation of the right hand side of the Kronig- Penney equation: cosqa) + mau sinqa) qa with qa for both U = and U >. b) Sketch the Kronig-Penney dispersion curve and compare it to that of thed FM. At what k x values do discontinuities occur in energy? c) How many states are there per energy band? d) What is the variation of the chemical potential with number of electrons per unit cell?

5 lectrons in Solids ) Problem Sheet 4 From material in Lectures to 7 ) Consider the Kronig-Penny band theory model. a) Sketch the Kronig-Penney dispersion curve showing the st to 4 th energy bands and the st to 4 th Brillouin zones. b) All Bloch wavefunction wavevector values k outside the st Brillouin zone BZ) can be translated into it using the equation: k = k + π a n where n is an integer and k lies inside the st BZ. Find the values for n when k lies inside the nd, 3 rd and 4 th BZ for both positive and negative k. c) Sketch the extended, periodic and reduced zone schemes. ) Sketch the band structure of a semiconductor in real space and in k- space, the latter using the reduced zone scheme representation, showing the energies of the conduction band minima C and valence band maxima V, the energy gap G and the chemical potential µ. 3) The Kronig-Penney dispersion relationship has the form: cosk x a) = cosqa) + mau sinqa) qa where symbols have their usual meaning. Show that when U and k x, and therefore, are small the effective mass m* is given by: m* = m mau + 3 4) The conduction and valence band dispersion curves C k) and V k) of a semiconductor at the edge of the energy gap can be closely approximated by: C k) = C + k nc nv V k) = V + k where symbols have there usual meaning. a) By considering the density of states DOS) for the 3D free electron model, write down equations for the DOS in the semiconductor at the conduction and valence band edges.

6 b) What are the equations which govern the total number of electrons N and holes P in the conduction and valence bands of the semiconductor? c) Show that for -µ >> k B T, the Fermi-Dirac distribution function for the occupied states in the conduction band and the unoccupied states in the valence band are given respectively by: d) Using the integral: f ) = exp µ) / k B f ) = exp µ ) / k B x exp x)dx = π and appropriate substitutions, show that the electron and hole carrier densities are given by: n = N C exp C µ) / k B p = N V exp µ V ) / k B and find values of N C and N V. 5) Using sketches of the band diagrams in real space, explain the difference between an intrinsic, a p-type extrinsic and a n-type extrinsic semiconductor.

7 lectrons in Solids ) Problem Sheet SOLUTIONS Covering material in Lectures to 5 ) a) D Born von Karman boundary conditions:- ψx) = ψx + L) b) ψx) = Aexpik x x) d ψ dx = ψ k xψ = ψ = k x ψx) = ψx + L) ψ) = ψ + L) A = Aexpik x L) k x L = πn k x = πn L Allow both positive & negative values of k x n = ±, ±, ±3, ±4, c) If both positive & negative values of k x : = k x k = k x k =

8 d) L ψx).ψ x) dx = L A expikx).a exp ikx) dx L = A dx = A x [ ] L = A L = A = L e) Considering only +k x values: k = k x n = n x Therefore: Also: k = Therefore: With spin: k = n π L n = L π k dn dk = L π dk d = dn d = dn dk dk d = L 4π D) = L π

9 f) For the infinite potential well: k = n π L Therefore: dn dk = L π Also: k = Therefore DOS is x result for e): D) = L π g) Considering both +k x and k x values. There are now two k values per n. For an increase of n to n+dn, the length of D k-space occupied increases from k to k+dk. Therefore: dn = L π dk dn dk = L π This leads to the same result as for f): D) = L π ) nergy of level n: = k k = π L n n = π n L Number of electrons is 5p. For p =,, 3, 4, 5, 6, 7 highest filled level is n=3, 5, 8,, 3, 5, 8. For even p, n = 5/)p. For odd p, n=+5p- )/=5/)p+/. Lowest empty level is n+. Let L = pd.

10 For p even: For p odd: n = π n = π p 5 / L p ) d n + = π n +) L ) p d = π 5 / ) p + 5p + ) Δ = n + n = π 5p + 5π p d pd n = π n = π 5 / )p +/ L p ) d n + = π n +) L = π 5 / )p +/ p +) d ) = π p d 5 / )p +/ ) + 5 / )p +/ ) + ) Δ = n + n = π 5p + 5π p d pd 3) For p even: = n + Δ = π p p d 5 / ) + π 5p + 4mp d ) = π p d 5 / ) p + 5 / )p +/ ) For p odd: = n + Δ = π p d 5 / )p +/ ) + π 5p + 4mp d ) = π p d 5 / )p +/ ) + 5 / )p +) = π p d 5 / ) p + 5p + 5 / 4)

11 At large p: = π p d 5 / ) p ) = 5π 8md = 9.33eV 4) ψ I x) = C expβx) ψ II x) = Acosαx) + Bsinαx) ψ III x) = D exp βx) α = β = V ) Continuity of ψ and ψ at x = -a & +a: Acosαa) Bsinαa) = C exp βa) αa sinαa) + αb cosαa) = βc exp βa) Acosαa) + Bsinαa) = D exp βa) αa sinαa) + αb cosαa) = βd exp βa) Combine: B sinαa) = C D) exp βa) αa sinαa) = βc + D) exp βa) A cosαa) = C + D) exp βa) αb cosαa) = βc D) exp βa) This leads to:- ither: B = C = D Or: α cot αa = β And either: A = C = D and or: α tanαa = β Both equations cannot be true as leads to: tan αa = making α imaginary & β negative => not possible from definitions Also do not want trivial solution: A = B = C = D =

12 Therefore we have two possible solution sets: ither: B = C = D α tanαa = β Or: A = C = D α cot αa = β Leads to a set of even solutions: ψ I x) = C expβx) ψ II x) = Acosαx) ψ III x) = C exp βx) α tanαa = β And odd solutions: ψ I x) = C expβx) ψ II x) = Bsinαx) ψ III x) = C exp βx) α cot αa = β To solve the dispersion relationship then requires a graphical solution. Note: I would not require you to do the graphical solution under exam conditions. Note: The most important part of the above solution is to understand how the exponential tails in regions I and III leads to an extended wavefunction and tunnelling across the short distances between nano-scale solids, and how this tunnelling distance relates to mobility.

13 ) lectrons in Solids ) Problem Sheet SOLUTIONS Covering material in Lectures 5 to 8 a) The Fermi energy Fermi level) is the energy to which the system is filled in the ground state the lowest energy state, which is at K). It lies between the highest occupied and lowest unoccupied state at K). When number of electrons N is large, energy difference between these states tends to zero, then: N = where D) is the density of states. " D)d b) The Fermi-Dirac distribution function defined for a free electron system at thermal equilibrium) is: f ) = exp " µ) / k B + where T is temperature and µ is the chemical potential. It gives the probability f) that the state at energy is occupied. d) At K the ground state): e) The Fermi temperature is: FM Fermi energies lie in the range of about to ev, so T F is in the range of about 4 to 5 K. This is x to x greater than room temperature. Hence, for the electrons in this system, at 3K the Fermi-Dirac distribution is in the limit of T. Hence, at temperatures around 3K, µ. It is a very good approximation. For a typical metal with about ev: - )/k B T T/T F =. lim T " µ = T F = k B f). c) The chemical potential is chosen for a particular problem such that for that D) and T the total number of the particles in the system comes out correctly that is, equal to N: N = = " " f )D)d D)d exp $ µ) / k B +

14 ) a) 3D Born von Karman boundary conditions b) Trial solution: TIS: "x,y,z) = "x + L,y,z) "x,y,z) = "x,y + L,z) "x,y,z) = "x,y,z + L) r = x,y,z) k = k x,k y,k z ) "r) = "x,y,z) = C expik.r) = C expik x x + k y y + k z z)) = C expik x x) expik y y) expik z z) = CX x)yy)zz) X x) = expik x x) Yy) = expik y y) Zz) = expik z z) " $ = $ $ = " $ % d $ dx + d $ dy + d $ * = " $ & dz ) % C YZ d X dx + XZ d Y dy + XY d Z * = " CXYZ & dz ) % X " X + Y" Y + Z" * = " & Z ) X "= "k x expik x x) X " / X = "k x "k x " k y " k z ) = " = k c) For n x, n y and n z =, ±, ±, ±3, ±4, d) "x,y,z) = "x + L,y,z) ",,) = " + L,,) C = C expik x L) k x L = n x k x = n x L k y = n y L k z = n z L L L L $ $ $ "x,y,z)." x,y,z) dxdydz = L L L $ $ $ C dxdydz = C L 3 = C = V "r) = V expik.r)

15 3) a) In the 3D FM the volume in k-space per state is: "k x = L "k y = L "k z = L $ "k x "k y "k z = & ) % L b) Number of states dn in a spherical shell of radius k and thickness dk is: 4"k dk dn = k x k y k z ) = L 3 $ & ) 4"k dk % " dn dk = L 3 $ & ) 4"k % " c) From 3D FM dispersion relationship: As wavevector increases from k to k+dk, energy increases from to +d: dk = = k k = k = dk d = = k d Definition of density of states: D) " dn d 3 Initially ignoring the effect of spin: With spin twice as many states per unit energy range) and V = L 3, this leads to the 3D FM DOS: d) From )a): Leads to: dn d = dn dk dk d dn dk = L 3 & 4"k = L 3 & 4" $ " $ " % $ dn dk dk d = L 3 & 4" $ " % $ = L 3 & " & $ " $ D) = V & " $ N = N = ) " = V & " $ D)d V & " $ * = 3" N& % $ V & % $ & & e) From first part of 3)c): [Note: k F is the radius of the Fermi sphere in k-space it is a magnitude] k = k = k F = 3 -, / +. / 3 3" N& $ V / 3 = L3 & 4" $ d = V & 3" $ = 3" N& $ V / 3

16 f) From part )a): N = " F D)d = Due to the interrelationship between and k and and k F can also write: N = By considering the answer to part 3)a) and taking into account spin x) gives: g) The Fermi sphere in k-space containing N electrons has volume: Volume of k-space per state is: " k F dn dk dk k dn N = " F dk dk = L 3 $ k & ) 4 " F k dk % = 8 L 3 $ * k 3 - & ), / % + 3. $ N k F = & 3 ) % V / 3 k F V sphere = 4 3 "k 3 F " = V 3 k 3 F $ "k x "k y "k z = & ) % L 3 dn d d Therefore: h) To find : Rearrange: V sphere = NV electron 4 3 "k 3 F = N 3 " & = 4N " 3 $ L V N& k F = % 3" $ V / 3 = k = k F = 3" N& $ V N = V & 3" $ As N varies with at exactly the same rate as n with, and = when n = N: dn = V & " $ dn = V & " $ D) = V & " $ / 3 d d Given electrons per state, the volume of k-space per electron is: V electron = " & $ L 3

17 4) lectrical conductivity is: where n m -3 ) is the electron density and µ n m /Vs) is the electron mobility. [Note: do not confuse n and µ here with the number of states and the chemical potential in ), ) & 3) unfortunately they use the same symbols.] Carrier density In the 3D FM: " = e nµ n n = N V It assumes that all valence electrons are involved in conduction, and that their density is given by the number of valence electrons per unit volume. The valence electron density will be the number of electrons per unit cell this is the number of valence electrons per atom multiplied by the number of atoms per unit cell) divided by the volume of the unit cell. In the 3D FM, the value of n leads directly to the values of and k F. Carrier mobility The electron mobility can be directly related to the charge carrier drift velocity v d by: v d = "µ n field where field is the applied electric field. The velocity is limited by scattering from impurities, lattice defects and quantized thermal lattice vibrations, and can be related to the acceleration a due to the applied field by: v d = a" = e field m " where is the scattering, or relaxation, time. In the 3D FM, the velocity of the electrons v can be directly related to their wavevector value k by: v = m k Hence, the drift velocity can be related to a shift in the position of the Fermi sphere in 3D k-space by k. v d = m "k The Fermi sphere shifts in the opposite direction to the applied field.

18 ) lectrons in Solids ) Problem Sheet 3 SOLUTIONS From material in Lectures 8 to a) The specific heat C V is the change in the internal energy U for a given change in temperature T at constant volume V. For an arbitrary system eg. of N particles) this is: C V = "U & $ "T V in units of J K -. For a mole of material: C V = "U& N A $ "T V where N A is Avagadros number and C V is in units of J K - mol -. For a mass m of material: C V = "U& m $ "T V and c V is in units of J K - kg -. b) For a classical ideal gas of N particles: U = 3 Nk B T C V = 3 Nk B c) Width of region of thermally excited electrons at surface of Fermi sphere: " k B T Density of states dn/d at surface of sphere is D). Therefore, number of thermally excited electrons at surface is: n kt = D )" = D )k B T As each of these electrons acquires thermal energy k BT, internal energy is therefore: ) "U = n kt.k B T = D ) k B T d) Density of states: D) = V & " $ Fermi level: = 3" N& $ V N = V 3" & % F $ = V & & 3 $ " $ V & & $ " $ / 3 = 3 N Therefore: D ) = V & & $ " $ D ) = 3 N e) Therefore: ) "U = D ) k B T = 3 N C V = 3 N k B k B For 3D FM total internal thermal energy is the energy of these thermally excited electrons at surface of sphere. Thus U is U. Therefore: C V = "U & $ "T V = 3 N k B For the measured C V of metals at low temperatures, the FM can successfully explain both the magnitude and T variation. The classical approach cannot explain the T variation & gives a value which is orders of magnitude too large.

19 ) Fermi surfaces of real metals can be measured using the de Haas van Alphen effect. This measures oscillations, at low temperatures, of the magnetic susceptibility with magnetic field strength. The magnetic field results in quantized electron orbitals - a series of concentric cylinders in k-space known as Landau levels. If one of these intersects the belly of the Fermi sphere, it will produce a strong response. As the B-field increases, the cylinders contract, resulting in the oscillations as they pass through the sphere belly. Can be used to measure the area of the Fermi surface perpendicular to B. By changing direction of field can map out Fermi surface in 3D Real Fermi surfaces for some metals alkali metals) do indeed have an approximately spherical Fermi surface, which is a key prediction of the 3D FM. This indicates that the states are only weakly perturbed by the periodic lattice potentials. However, other Fermi surfaces differ radically, involving more than one non-spherical surface as well as having enclosed cavities in which holes can also orbit. The FM successfully predicts the filling of the states in k-space and the existence of Fermi surfaces in metals. It also successfully explains the general electrical and thermal properties of metals. However, it cannot explain the detailed behaviour of many metals. Crucially, it also cannot explain the existence of semiconductors and semiconductors it predicts that everything is a metal. b) c) "x + a) = u k x + a) expik x x + a)) = "x) expik x a) " x + a)."x + a) = " x) exp$ik x a)."x) expik x a) = " x)."x) % " x + a)."x + a) = " x)."x) d) D BvK: "x + L) = "x) For N unit cells: L = Na "x + Na) = "x) expik x Na) "x + L) = "x) expik x L) = "x) expik x L) = k x = L n x n x = ±,±,±3,... 3) a) Potential is in D, so Bloch s Theorem D Periodic potential: Ux) = Ux + na) Bloch s Theorem: The eigenstates of the one electron Hamiltonian for a periodic potential can be chosen to have the form of a plane wave multiplied by a function with the periodicity of the lattice: "x) = u k x) expik x x) where: u k x) = u k x + na) for all n. assuming k x can have both positive and negative values) e) For a D Bloch wavefunction k x is the quantum number of that state. The wavefunction is a wavepacket constructed from different k values. k x is the wavevector value for the plane wave part of the wavefunction. It cannot be directly related to momentum p and velocity v. For a D FM state the value of k x is also the quantum number of that state. As the wavefunction is a plane wave, k x is also the actual wavevector value of that state. k x can be directly related to momentum p and velocity v.

20 4) a) " I x) = A n expiqx) + B n expiqx) = A n + B n ) sinqx) + ia n + B n ) cosqx) = C sinqx) + D cosqx) In regions between functions, Ux) = so: " $ I x) = $ I x) x ) = $ I x) " "q $ I x) = q q = b) Bloch: "x + na) = expik x na)"x) So: " II x + a) = expik x a)" I x) ) = expik x a) C sinqx) + D cosqx) e) Derivatives: " I x) = q C cosqx) $ D sinqx) "x " II x) "x ) = q expik x a) C cosqx $ a)) $ D sinqx $ a)) ) f) Cusp condition for delta function: " II ) $ " ) I = U "x "x I ) In region I: " I x = ) = qc cos) $ qd sin) = qc "x In region II: " II x = ) = expik x a) qc cosq $ a)) $ qd sinq $ a)) "x = q expik x a) C cosqa) + D sinqa) Also: ) ) " I x = ) = C sin) + D cos) = D Leads to: " II x) = expik x a) C sinqx a)) + D cosqx a)) ) c) Continuity at x = C sinqx) + D cosqx) = expik x a) C sinqx " a)) + D cosqx " a)) ) ) D = expik x a) "C sinqa) + D cosqa) d) = expik x a) " C & % sinqa) + cosqa) $ D exp"ik x a) " cosqa) = " C D sinqa) C D = cosqa) " exp"ik a) x sinqa) Therefore: ) " qc = U q expik x a) C cosqa) + D sinqa) D g) " q expik x a) C % $ cosqa) + sinqa) q C D & D = U q expik x a) C D cosqa) q C D = U q expik x a) sinqa) U + q expik C D = x a) sinqa) q expik x a) cosqa) + q = U q + expik x a) sinqa) expik x a) cosqa)

21 h) C D = "exp"ik " U + expik xa) + cosqa) q x a) sinqa) = sinqa) " expik x a) cosqa) 5)a) 4 "exp"ik x a) + cosqa) ) " expik x a) cosqa) ) 3 U = U > = " U q sinqa) + expik x a) sin qa) "exp"ik x a) + cosqa) + cosqa) " expik x a) cos qa) = "exp"ik x a) + cosqa) " expik x a) cos qa) RHS of KP quation = " U q sinqa) + expik x a) sin qa) - "exp"ik x a) + cosqa) " expik x a) cos qa) + sin qa)) = " U sinqa) q - -6/a -4/a -/a /a 4/a 6/a qa "exp"ik x a) " expik x a) = " U q sinqa) " cosqa) "cosk x a) = " U q sinqa) " cosqa) cosk x a) = cosqa) + mau sinqa) qa b) nergy KP FM -4/a -3/a -/a -/a /a /a 3/a 4/a k Discontinuities occur at k x values of ±/a

22 c) From part 3)d) L = Na k x = " L n x = " Na n x n x = Na " k x Range of k x in st band: " a $ k x $ a Therefore range of n x in st band: " N k x N Therefore band has N possible k x states With spin band therefore has N possible states Inspections shows each band also contains the same number of k x values, therefore the same number of N states d) 7N 6N nergy 5N FM KP -4/a -3/a -/a -/a /a /a 3/a 4/a k N unit cells If electron per unit cell, have N electrons => half fill first band If electrons per unit cell, have N electrons => fill first band If 3 electrons per unit cell, have 3N electrons => half fill second band tc. As the number of electrons per unit cell increases, the chemical potential µ rises in the band diagram. If odd number => µ in band => metal If even number => µ in band gap => semiconductor or insulator 4N 3N N N

23 lectrons in Solids ) Problem Sheet 4 SOLUTIONS From material in Lectures to 7 c) xtended zone scheme: ) a) 4 th BZ 3 rd BZ nd BZ st BZ nd BZ 3 rd BZ 4 th BZ nergy 4 th Band nergy 3 rd Band -6/a -4/a -/a /a 4/a 6/a k Band -4/a -3/a -/a -/a /a /a 3/a 4/a k nd Band st Periodic zone scheme: b) k " = k + a n " k = k " $ a n " nergy By inspection of the above diagram:- If k is in the nd BZ in the positive direction: n = +. If k is in the 3 rd BZ in the positive direction: n = +. If k is in the 4 th BZ in the positive direction: n = +. If k is in the nd BZ in the negative direction: n = -. If k is in the 3 rd BZ in the negative direction: n = -. If k is in the 4 th BZ in the negative direction: n = -. -4/a -3/a -/a -/a /a /a 3/a 4/a k

24 Reduced zone scheme: 3) ) nergy -/a /a k cosk x a) = cosqa) + mau sinqa) qa U = mau cosk x a = cosqa + U sinqa qa cosk x a " k xa) cosqa " qa) sinqa " qa qa)3 3 k x a) k xa) = q = qa) + U $ qa)3 & qa ) qa % 3 = qa) + U $ & qa) ) = U qa) U + 3 % 6 6 ) k xa = U a U a 6 6 = U + 3 U + 3 = 3U a U + 3 $ d m* = & ) % dk ) U + ) k x ) + d dk = 3 m U + 3) $ m* = m U & % 3 + ) $ m* = m mau & +) % 3 k x 3 U + 3) a )

25 4) a) 3D FM DOS 3D FM k) C k) = C + k " nc D) = V & " $ = k By comparison DOS at conduction band edge: By comparison DOS at conduction band edge: V k) = V + k " nv b) The equations which govern the total number of electrons N and holes P in the conduction and valence bands of the semiconductor are: C k) " C ) = " C D C ) = V % nc * $ & * ) ) = k nc " C V k) " V ) = k = " k nv V " V k)) = V " D V ) = V % nv * $ & * ) f ) = N = P = $ $ nv ) = k exp " µ) / k B + C V f )D C )d " f ))D V )d nv V " c) d) N = $ C = V & * ) nc, % ), + x = " C k B T " f ) = " exp " µ) / k B + = k B Tx + C d = k B Tdx = exp " µ) / k B + exp " µ) / k B + " exp " µ) / k B + = exp " µ) / k B exp " µ) / k B + = exp " µ) / k B exp" " µ) / k B & % exp " µ) / k B +$ exp" " µ) / k B = expµ " ) / k B + " µ >> k B T f ) = D C ) exp" " µ) / k B d $ C " C ) exp" " µ) / k B d Lim x - C ) = Lim x) = - $ " C ) exp" " µ) / k B d C = $ k B Tx) exp"k B Tx + C " µ) / k B k B Tdx = k B exp" C " µ) / k B $ x exp"x)dx = k B exp" C " µ) / k B N = V & * ) nc, 4% ), exp " µ) / k B + exp " µ) / k B = exp" " µ) / k B " f ) = expµ " ) / k B + expµ " ) / k B = exp"µ " ) / k B % k B exp" C " µ) / k B

26 P = V = V & % ) nv + $ + * x = V " k B T D V ) exp"µ " ) / k B d V V " ) exp"µ " ) / k B d 5) An intrinsic semiconductor is a pure material which contains no dopants. If n i and p i are the intrinsic carrier densities: n = p = n i = p i = "k B Tx + V d = "k B Tdx Lim x, V ) = Lim x, ) = V - Lim x k B T >> ). /, V V " ) exp"µ " ) / k B d = " k B Tx) exp"µ + k B Tx " V ) / k B k B Tdx / = " k B exp"µ " V ) / k B x exp"x)dx = k B exp"µ " V ) / k B P = V & % ) nv + 4$ + * / $ k B exp"µ " V ) / k B A p-type extrinsic semiconductor is a semiconductor which contains impurity dopant atoms known as acceptors. These acceptors have a valency less than the semiconductor, and can trap electrons, thus creating holes in the valence band. They have an energy close to that of the valence band edge so are fully occupied at room temperature. They have density N A. For a p-type semiconductor: N A " p n << p m " " n = m nc n = N V n = $ " n k B T & ) 4 % n = N C exp* C * µ) / k B $ N C = m " nk B T & ) % m " " p = m nv exp* C * µ) / k B p = P V p = $ " p k B T 4 & % ) p = N V exp*µ * V ) / k B $ N V = m " pk B T & % ) exp*µ * V ) / k B

27 A n-type extrinsic semiconductor is a semiconductor which contains impurity dopant atoms known as donors. These donors have a valency greater than the semiconductor, and can donate electrons to the conduction band. They have an energy close to that of the conduction band edge so are fully ionized at room temperature. They have density N D. For a n-type semiconductor: N D " n n >> p