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Solid State Theory Physics 545 Band Theory III Each atomic orbital leads to a band of allowed states in the solid Band of allowed states Gap: no allowed states Band of allowed states Gap: no allowed states Band of allowed states Independent Bloch states F1 0 Solution of the tight binding model is periodic in k. Apparently have an infinite number of k-states k states for each allowed energy state. In fact the different k-states k states all equivalent. Bloch states kR ψ (r + R ) ≡ eik.R ψ (r ) E(k) -2 α= 10 -4 γ=1 -6 -8 -10 -12 -14 -16 -18 -4 Let k = ḱ′ + G where k′ is in the first Brillouin zone and d G is i a reciprocal i l lattice l i vector. -2 −π/a π/a 0 2 4 k [111] direction ψ(r + R ) ≡ eik′.R eiG.R ψ(r ) But G.R = 2πn, n-integer. Definition of the reciprocal lattice. So eiG.R = 1 and ψ(r + R ) ≡ eik′.R ψ(r ) e ik.R ≡ e ik ′.R k′ is exactly equivalent to k. k The only independent values of k are those in the first Brillouin zone. Reduced Brillouin zone scheme The only y independent p values of k are those in the first Brillouin zone. Discard for |k| > π/a Results of tight binding calculation 2π/a -2π/a Displace into 1st B. B Z Z. Results of nearly free electron calculation Reduced Brillouin zone scheme Extended, reduced and periodic B ill i zone schemes Brillouin h P i di Zone Periodic Z Reduced R d d Zone Z Extended E d d Zone Z All allowed states correspond to k-vectors in the first Brillouin Zone. Can draw E(k) in 3 different ways The number of states in a band Independent k-states in the first Brillouin zone, i.e. ⏐kx⏐ < π/a etc. Finite crystal: only discrete k k-states states allowed kx = ± 2πn x , n x = 0,1,2,.... etc. L Monatomic simple cubic crystal, lattice constant a, and volume V. One allowed k state per volume (2π)3/V in k-space. Volume of first BZ is (2π/a)3 Total number of allowed k-states in a band is therefore ⎛ 2π ⎞ ⎜ ⎟ ⎝ a ⎠ 3 (2π) = V V a 3 =N Precisely N allowed k-states i.e. 2N electron states (Pauli) per band This result is true for any lattice: each primitive unit cell contributes exactly one k-state to each band. Metals and insulators In full band containing 2N electrons all states within the first B. Z. are occupied. The sum of all the k-vectors in the band = 0. A partially filled band can carry current, a filled band cannot Insulators have an even integer number of electrons per primitive unit cell. E With an even number of electrons per unit cell can still have metallic behaviour due to ban overlap. Overlap in energy need not occur in the same k direction EF 0 π k a Metal M t l due d to t overlapping bands E E E EF π k a 0 π k a 0 Empty Band EF Energy Gap π k a 0 Partially Filled Band Full Band Part Filled Band Part Filled Band Energy Gap Full Band INSULATOR or SEMICONDUCTOR METAL METAL or SEMI-METAL Germanium Bands in 3D In 3D the band structure is much more complicated than in 1D because crystals y do not have spherical symmetry. The form of E(k) is dependent upon the direction as well as the magnitude of k. Figure removed to reduce file size • Chemical bonds and electron bands. a) Number of electrons in any band is finite because the density of states is finite finite. 8π 2 m ( E ) ρ3 D ( E ) = h3 3/ 2 E 1/ 2 8π 2 N= h3 b) Bands are formed from molecular orbitals. Etop 3 / 2 1/ 2 m ( E ) E dE ∫ Ebottom Filling of Energy Bands ⇒ An important property of a full band is that it is UNABLE to carry a net current since for each state in the band we can identify a corresponding state with equal and OPPOSITE momentum that is filled by an electron. To drive a net current through the crystal it is necessary to induce an IMBALANCE in the filling of momentum states ⇒ For an energy band that is filled completely however this requires that we excite electrons ACROSS the forbidden gap. E • Situation in which the lowest energy band is filled completely with electrons • the h only l way in which h h a net current can flow l is to excite electrons across the energy gap ENERGY GAP • if th the energy gap is i large l h however excitation it ti cannot be achieved and so no net current is allowed to flow k −π/a π/a ⇒By the same arguments if the energy band is PARTIALLY filled then it should be very EASY to generate a net current flow in the crystal ⇒ In this situation the forbidden gap lies FAR away from the highest filled electron states and so it is easyy to use an electric field to ggenerate an imbalance in the fillingg of momentum states ⇒ A small applied voltage will therefore generate a LARGE current as we discussed previously for free electrons E E ENERGY GAP ENERGY GAP k −π/a π/a NO APPLIED ELECTRIC FIELD k −π/a SMALL ELECTRIC FIELD APPLIED π/a ⇒Electronic band theory presents us a natural scheme for CLASSIFYING different types of materials ⇒ METALS should be materials whose uppermost energy band is only PARTIALLY filled with electrons. ⇒ This explains why these materials are GOOD conductors of electricity ⇒ We expect that insulators on the other hand should be materials whose energy bands are either COMPLETELY full or empty so that an energy gap BLOCKS current flow in these materials E FORBIDDEN GAP FILLED STATES METAL FORBIDDEN GAP FILLED STATES INSULATOR Band structure of metals monovalent l metals l multivalent metals, semimetals ⇒What types yp of elements produce p partial p or complete p fillingg of energy gy bands? ⇒ The GROUP I elements should be good METALS since these elements h have only l ONE valence l electron, l t whereas h coordination di ti number b is i 6-12. 6 12 ⇒ If we have a crystal composed of N atoms there will therefore be N g energy gy band valence electrons which will HALF-FILL a single ⇒ The GROUP IV elements should be INSULATORS since these elements have FOUR valence electrons and so in an N-atom crystal there will 4N valence electrons that FILL two energy bands completely E FILLED STATES FILLED STATES FILLED STATES GROUP I GROUP IV FILLING OF ENERGY LEVELS BY THE VALENCE ELECTRONS OF GROUP I & IV ELEMENTS Semiconductors • In certain materials known as SEMICONDUCTORS however the energy gap that separates the highest filled band in the ground state from the lowest empty band is SMALL * Such materials are INSULATORS at zero temperature since their ground state is one in which the energy bands are either completely full or empty * Since the forbidden gap is small however electrons can be EXCITED across it at higher temperatures to PARTIALLY fill the next band ⇒ The material will no longer g be an insulator at this temperature p but will CONDUCT electricity E E FORBIDDEN GAP FILLED STATES FILLED STATES INSULATOR T=0 INSULATOR T>0 FILLED STATES FILLED STATES SEMICONDUCTOR T=0 SEMICONDUCTOR T>0 Some general COMMENTS on semiconductors •The The energy band that holds the valence electrons in the ground state is known as the VALENCE BAND. It is usually formed by Bonding Orbitals. •The lowest empty band is known as the CONDUCTION BAND. BAND It is usually formed by antibonding orbitals. ⇒ The energy gap that separates these bands is usually denoted as Eg * Room R temperature t t semiconductors i d t are generally ll materials t i l in i which hi h Eg is i a FEW eV V (≤ 3 eV) ⇒ This should be compared to a thermal energy of approximately 40 meV that is available to electrons at room temperature (300 K) E SEMICONDUCTOR Si Ge InSb InAs InP GaP GaAs GaSb AlSb Eg (eV) 0K 300 K 1.17 0.74 0.23 0.43 1.42 2.32 1 52 1.52 0.81 1.65 1.11 0.66 0.17 0.36 1.47 2.25 1 1.43 43 0.68 1.60 CONDUCTION BAND Eg VALENCE BAND SEMICONDUCTOR T=0 Concept of a hole • At higher temperatures electrons in semiconductors may be excited into the conduction band where they are able carry an electrical current * Each electron leaves behind an EMPTY state in the valence band and to account for current flow in semiconductors we must ALSO consider the role of these HOLE states * If the valence band is COMPLETELY filled, then the total crystal momentum of this hi bband d iis equall to ZERO since i for f any occupied i d k-state k we can identify id if an corresponding filled state with OPPOSITE momentum E ENERGY GAP 2 1 k −π/a π/a • The total crystal momentum in a filled energy band is exactly equal to zero • to illustrate this consider the total momentum d to occupation of states 1 & 2 due • state 1 corresponds to an electron with positive momentum while state 2 corresponds to one with ith equall andd opposite it momentum t • the net crystal momentum of electrons occupying states 1 & 2 is zero and this pairing can be repeated for all other states in the band • When the valence band is completely filled with electrons we can write ∑k i =0 (17) * if we excite AN electron from the state with wavenumber ke in the valence band into the conduction band equation (17) for the valence band may now be REWRITTEN as ∑k ki ≠ k e i = − ke (18) * The empty state in the valence band may therefore be viewed as a HOLE which has OPPOSITE momentum to the electron that was excited out of that state kh = − ke (19) • Since the hole corresponds to a missing electron its energy may be written as Eh ( k h ) = − Ee ( − k e ) = − Ee ( k e ) (20) * Equation 10.4 shows that electrons and holes have OPPOSITE energy scales since moving DOWNWARD in the valence band implies INCREASING hole energy • Moving downwards in the valence band corresponds t increasing to i i hole h l energy E • in the picture shown here hole 1 therefore has more energy gy than hole 2 ENERGY GAP • this is not too difficult to understand if we think of the total energy of the electrons left in the band HOLE 2 HOLE 1 k −π π/a π/a • since hole 1 corresponds to a missing electron from a lower energy electron state than hole 2 the total energy gy off electrons in the band is higher g ffor hole 1 than for hole 2 • While the energy scales are oppositely directed for electrons and holes we can show that the hole VELOCITY is the SAME as that of the state from which the electron is missing * To do this we simply replace kh by –ke and Eh(kh) by –Ee(ke) vh = =− 1 d Eh ( k h ) = dkh 1 d 1 d ( − Ee ( ke )) = E e ( k e ) = ve = dke = dke * Using the same approach we can also show that the effective mass of the hole is a NEGATIVE quantity 1 1 d 1 d = E ( k ) = ( − (− Ee ( ke ))) h h * 2 2 mh = kh dkh = (− ke ) dke =− 1 d 1 ( ( )) E k = − e e = 2ke dke me* • It is important to appreciate that we do NOT actually have positively-charged carriers in the h semiconductor i d but b that h the h holes h l simply i l behave b h AS IF they h hhad d positive i i charge h * The basic idea is that EACH of the OCCUPIED electron states in the valence band responds to externally-applied fields as we would expect for a negatively-charged carrier. rri r The Th NET response r p n off the th band b nd however h r LOOKS like lik the th response r p n off a single in l particle with a POSITIVE charge! When we discuss conduction band properties of semiconductors or insulators we refer to electrons but when we discuss the valence properties, electrons, properties we refer to holes. holes This is because in the valence band only the missing electrons or holes lead to current flow. E E ENERGY GAP ELECTRIC FIELD ENERGY GAP k −π/a π/a k −π/a π/a AN ELECTRIC FIELD APPLIED IN THE +x DIRECTION ACCELERATES ELECTRONS IN THE –x DIRECTION AND SO THE HOLE STATE APPEARS TO BE ACCELERATED IN THE +x DIRECTION General comments Thermal vibrations or energy can be used to create a hole by exciting an electron from the valence band to the conduction band In I an intrinsic i t i i (undoped) ( d d) semiconductor, i d t the th number of holes in the valence band equals the electrons in the conduction band Holes can move about the valence band and recombine with electrons in the conduction band (to disappear) tatistics of electrons and holes in semiconductors One can use Boltzmann statistics for electrons and holes if their energy is small in comparison with EF ⎡ E − EF ⎤ 1 =≈ exp ⎢ − f (E) = , if E − EF > 3.5k BT ⎥ exp(( (( E − E F ) / k BT ) + 1 ⎣ k BT ⎦ Electron excitation on semiconductors n = N c e ( E F − Ec ) / k B T ⎡ 2 πme*k BT ⎤ , Nc = 2 ⎢ ⎥ 2 ⎣ h ⎦ p = N v e ( Ev − E F ) / k BT 3/ 2 ⎡ 2πmh*k BT ⎤ , Nv = 2 ⎢ ⎥ 2 ⎦ ⎣ h 3/ 2 For pure semiconductor EF ≈E Eg/2 n=p ≈C T2/3exp(-Eg /2kT) C is a material constant n p ≈C n·p C T2/3exp(-E exp( Eg /kT)=ni ni is i concentration t ti off intrinsic i ti i carriers Impurities in semiconductors: Doping n-type yp Band Diagram p yp Band p-type Diagram Impurities create levels in the forbidden gap of semiconductors. Impurities in semiconductors: concentration of carriers Ionized Dopant Concentration: 1 n= 1 ( N C N d ) 2 exp ⎛⎜ − E d e ⎞⎟ 2 ⎝ 2 kT ⎠ 1 1 ⎛ Eae ⎞ 2 ( N N ) exp p= V a ⎜− ⎟ 2 ⎝ 2 kT ⎠ Nd and Na is the concentration of donors and acceptors respectively Ed and Ea are in eV Comments on electron and hole concentration In pure semiconductor concentration of holes and of electrons are equal. equal Concentration of holes and of electrons depends on Eg.=> => Insulator are semiconductor with very large Eg Mobility in semiconductors has exactly the same 2 2 conductivity meaning as in metals. Total can be pe p τ neτe σ = neμ e + peμ p = * + * p expressed as: me mp Concentration C t ti off electrons l t and d h l holes i in semiconductors can be tailored by introducing impurities (doping) In this case concentration of electrons and Interesting to know $Band gap increases with increasing the strength of the chemical bonds in the semiconductor. Example: p Diamond > Silicon > Germanium $In many semiconductor alloys the band gap changes almost li linearly l with i h composition. ii Example: 1) GaAs – AlAs 2) HgTe – CdTe Chemistry of doping Doping activity of impurities (dopants) depend on the charge state of the impurity relative to the charge in unperturbed lattice. Doping activity depends on site, site at which dopant is incorporated. Some dopants may be both donors and acceptors (amphoteric). Impurities I iti may be b have h charge h more than th one. Some dopants are natural defects that are inherent to the material. Chemistry of doping: examples. 1 Elementary semiconductors: C (diamond), 1. (diamond) Si, Si Ge Band gap (eV) @300 K conductivity C (diamond) 5.5 < 10-16 (Ω·cm)-1 Si 1.11 4×10-6 (Ω·cm)-1 Ge 0 67 0.67 22.2×10 2×10-22 (Ω·cm)-11 Si Si Si Si Si e– As 1. Activity of substitutionally incorporated dopants is defined by the difference in number of valence electrons. 2. Interstitial dopants (mostly metals) are 2 donors. B h+ Li e– 3. Some interstitial dopants (Au, Ni) may be amphoteric. amphoteric 4. Influence of natural defects is negligible. Chemistry of doping: examples. 2. III-V semiconductors. Structure: cubic (sphalerite) or hexagonal Band gaps (eV) @300 K AlN 6.2 AlAs 2 15 w 2.15 GaN 3.44 w GaP 2.27 s GaAs 1.42 s InP 1.34 s InAs 0.35 s InSb 0.23 s w Sphalerite Wurtzite 1. Most 1 M off the h d defects f are substitutional. b i i l 2. The substitution site is largely defined by ionic radii. Example: Be substitutes Ga, Te substitute As in GaAs. Homework: Determine the doping action of Be and Te in GaAs. Interesting to know: both structures are polar (have a specific directio Chemistry of doping: examples. 3. II-VI semiconductors. Structure: cubic (sphalerite) or hexagonal ZnSe 2.8 w ZnTe 2.4 w CdS 24 2.4 w CdTe 1.4 s 1. Most of the defects are substitutional. 2. The substitution site is largely defined by ionic radii. Example: In substitutes Cd, As substitute Te in CdTe. 3. Electrical properties of II-VI crystals are stronly affected by natural defects. Example of natural defects: 1) Cadmium vacancies in CdTe may work as a single or double charged acceptor. Therefore, undoped CdTe is always ptype. yp Anion vacancies donors all II-VI In2)semiconductors withare a very largeinband gap semiconductors. acceptor-like defects are not thermodynamically stable if the energy of vacancy formation, EV, is smaller than the recombination energy energy, Er. C d i band Conduction C d i band Conduction Donor level Valence band Donor level Recombination energy, Er Acceptor level Valence band