Full Band Energy Gap Part Filled Band E

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

Solution of the tight binding model -2 α= 10 is periodic in k. Apparently have -4 γ = 1 an infinite number of k -states for -6 E(k) each allowed energy state. -8 In fact the different k-states all -10 -12 equivalent. -14

-16 ikRk.R Bloch states ψ (r + R) ≡ e ψ (r) -18 -4 -2−π/a 0π/a 2 4 Let k = k′́ + G where k′ is in the first Brillouin zone k [111] direction andGid G is a reci procal ll latti ce vector. ψ(r + R) ≡ eik′.ReiG.Rψ(r) But G.R = 2πn, n-integer. Definition of the reciprocal lattice. So eiG.R = 1 and ψ(r + R) ≡ eik′.Rψ(r) eik.R ≡ eik′.R k′ is exactly equivalent to k. The only independent values of k are those in the first Brillouin zone. Reduced Brillouin zone scheme The onlyyp independent 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 BZB. Z.

Results of nearly free electron calculation Reduced Brillouin zone scheme Extended, reduced and periodic Br illoui n zone sch emes

PidiZPeriodic Zone R Rdeduced dZ Zone E xtend ddZed Zone 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. 2πn FiniteFinite ccrystal:rystal: oonlynly ddiscreteiscrete k-states aallowedllowed k = ± x , n = 0,1,2,.... etc. x L x 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 3 ⎛ 2π ⎞ (2π) V ⎜ ⎟ = = N ⎝ a ⎠ V a 3

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 E of electrons per primitive unit cell.

With an even number of electrons per unit cell can still have metallic behaviour due to ban overlap.

EF Overlap in energy need not occur in the same k direction

0 π k a MtldMetal due t o overlapping bands E E

0 π k 0 π k a a 0 π k a Empty Band Partially Energy Gap Filled Band Part Filled Band EF Full Band Part Filled Band Energy Gap

Full Band

INSULATOR METAL METAL or SEMICONDUCTOR or SEMI-METAL Bands in 3D Germanium

In 3D the band structure is much more complicated than Figure removed to in 1D because crystals do not reduce file size have spherical symmetry.

The form of E(k) is dependent upon the direction as well as the magnitude of k. • Chemical bonds and electron bands. a) Number of electrons in any band is finite because the density of states is finite.

E 3/ 2 1/ 2 top 8π 2m(E) E 8π 2 3/ 2 1/ 2 ρ (E) = N = m(E) E dE 3D 3 h3 ∫ h Ebottom b) Bands are formed from molecular orbitals. 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.

• Situation in which the lowest energy band is filled completely with electrons

ENERGY GAP • the only way in whhhich a net current can fl ow is to excite electrons across the energy gap

• if the energy gap is large however excitati on 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 easyyg to use an electric field to generate an imbalance in the filling 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 k

−π/a π/a −π/a π/a

NO APPLIED ELECTRIC FIELD SMALL ELECTRIC FIELD APPLIED ⇒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

FORBIDDEN GAP FORBIDDEN GAP

FILLED STATES FILLED STATES

METAL INSULATOR Band structure of metals monovalllent metals multivalent metals, semimetals ⇒What types of elements p roduce p artial or comp lete filling of energy bands? ⇒ The GROUP I elements should be good METALS since these elements have only ONE valence elltectron, whereas coordi ditination numb er is 6-12. ⇒ If we have a crystal composed of N atoms there will therefore be N valence electrons which will HALF-FILL a singggyle energy band ⇒ 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

FILLED STATES

FILLED STATES FILLED STATES FILLING OF ENERGY LEVELS BY THE VALENCE ELECTRONS OF GROUP I & IV GROUP I GROUP 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 longgper be an insulator at this temperature but will CONDUCT electricity

E E

FORBIDDEN GAP

FILLED STATES FILLED STATES FILLED STATES FILLED STATES

INSULATOR INSULATOR SEMICONDUCTOR SEMICONDUCTOR T = 0 T > 0 T = 0 T > 0 Some general COMMENTS on semiconductors •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. It is usually formed by antibonding orbitals.

⇒ The energy gap that separates these bands is usually denoted as Eg * Room temperat ure semi cond uct ors are generall y mat eri al s in whi ch Eg is a FEW eV (≤ 3 eV) ⇒ This should be compared to a thermal energy of approximately 40 meV that is available to electrons at room temperature (300 K)

SEMICONDUCTOR Eg (eV)

0 K 300 K CONDUCTION BAND Si 1.17 1.11 Ge 0.74 0.66 Eg InSb 0.23 0.17 InAs 0.43 0.36 InP 1.42 1.47 VALENCE BAND GaP 2.32 2.25 GaAs 1521.52 1431.43 GaSb 0.81 0.68 AlSb 1.65 1.60

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 ban d is equa l to ZERO since for any occupi ided k-state we can ident ify an corresponding filled state with OPPOSITE momentum E • The total crystal momentum in a filled energy band is exactly equal to zero • to illustrate this consider the total momentum due to occupation of states 1 & 2 ENERGY GAP • state 1 corresponds to an electron with positive 2 1 momentum while state 2 corresponds to one with equal and opposite momentum • the net crystal momentum of electrons occupying states 1 & 2 is zero and this pairing can be repeated k for all other states in the band

−π/a π/a • When the valence band is completely filled with electrons we can write

k = 0 ∑ i (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 (18) ∑ ki = − ke ki ≠ ke

* 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 (20) Eh (kh ) = − Ee (−ke ) = − Ee (ke )

* 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 E to iiincreasing hlhole energy

• in the picture shown here hole 1 therefore has more energy than hole 2 ENERGY GAP • this is not too difficult to understand if we think of HOLE 2 the total energy of the electrons left in the band

HOLE 1 • since hole 1 corresponds to a missing electron from k a lower energy electron state than hole 2 the total

−π/a π/a energy of electrons in the band is higher for 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) 1 d vh = Eh (kh ) = dkh 1 d 1 d = − (−Ee (ke )) = Ee (ke ) = 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 * = 2 Eh (kh ) = 2 (− (−Ee (ke ))) mh = kh dkh = (−ke ) dke 1 d 1 = − 2 (Ee (ke )) = − * = ke dke me • It is important to appreciate that we do NOT actually have positively-charged carriers in the sem icond uctor but that the hhloles simply l bbhehave AS IF they hdhad pos iiitive ch arge * 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 . The NET resp on se o f the band however LOOKS like the resp on se of a singl in le 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, we refer to holes. This is because in the valence band only the missing electrons or holes lead to current flow.

ELECTRIC E E FIELD

ENERGY GAP ENERGY GAP

k k

−π/a π/a −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 Inanitiintrins ic (d(undope d)semidticonductor, the 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

1 ⎡ E − EF ⎤ f (E) = =≈ exp⎢− ⎥ , if E − EF > 3.5kBT exp((E − EF ) / kBT ) +1 ⎣ kBT ⎦ Electron excitation on semiconductors

3/ 2 ⎡ * ⎤ (EF −Ec ) / kBT 2πmekBT n = Nc e , Nc = 2 ⎢ 2 ⎥ ⎣ h ⎦

3/ 2 ⎡ * ⎤ (Ev −EF ) / kBT 2πmhkBT p = Nv e , Nv = 2 ⎢ 2 ⎥ ⎣ h ⎦

FForor pupurere sesemiconductormiconductor EF ≈Eg/2 n=p ≈C T2/3exp(-E /2kT) g C is a material constant npn·p ≈CTC T2/3exp(-E /kT)=n g i ni ittifitiiis concentration of intrinsic carriers Impurities in semiconductors: Doping n-type Band Diagram

p-type Band Diagram

Impurities create levels in the forbidden gap of semiconductors. Impurities in semiconductors: concentration of carriers

Ionized Dopant Concentration: 1 1 1 E e 1 2 ⎛ Eae ⎞ n= 2 ⎛ d ⎞ p= (N N ) exp⎜ − ⎟ (N C N d ) exp⎜ − ⎟ V a 2 ⎝ 2kT ⎠ 2 ⎝ 2kT ⎠

Nd and Na is the concentration of donors and acceptors respectively

Ed and E a areineVare in eV Comments on electron and hole concentration

In pure semiconductor concentration of holes and of electrons are equal. Concentration of holes and of electrons depends on

Eg.=>

Insulator are semiconductor with very large Eg Mobility in semiconductors has exactly the same meaning as in metals. Total2 conductivity2 can be neτe peτp expressed as:σ = neμe + peμp = * + * me mp

CttiConcentration of eltlectrons and hlholes 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: Diamond > Silicon > Germanium

$In many semiconductor alloys the band gap changes almost linearl y wi th composi ti on. 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, at which dopant is incorporated. Some dopants may be both donors and acceptors (amphoteric). IitiImpurities may be have charge more than one. Some dopants are natural defects that are inherent to the material. Chemistry of doping: examples. 1. Elementary semiconductors: C (diamond) , Si , Ge

Band gap (eV) conductivity Si Si @300 K Si -16 -1 C (diamond) 5.5 < 10 (Ω·cm) Si Si Si 1.11 4×10-6 (Ω·cm)-1 -2 -1 Ge 0670.67 222.2×10 (Ω·cm) e– As 1. Activity of substitutionally incorporated dopants is defined by the difference in number of valence electrons. 2. Interstitial dopants (mostly metals) are donors. B h+

Li 3. Some interstitial dopants (Au, Ni) may be amphoteric . e– 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 w AlAs 2152.15 w GaN 3.44 w GaP 2.27 s GaAs 1.42 s InP 1.34 s InAs 0.35 s InSb 0.23 s Sphalerite Wurtzite 1. MfhdfMost of the defects are sub biilstitutional. 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 22.8.8 w 1. Most of the defects are substitutional. 2. The substitution site is largely defined by ionic radii. ZnTe 2.4 w Example: In substitutes Cd, As substitute Te in CdTe. CdS 242.4 w 3. Electrical properties of II-VI crystals are stronly CdTe 1.4 s 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 p- type. In2) semiconductors Anion vacancies with are a very donors large in band all II-VI gap semiconductors.acceptor-like defects are not thermodynamically stable if the energy of vacancy formation, EV, is smaller than the recombination energy, E . CdConducti on banr d CdConducti on band Donor level Donor level Recombination

energy, E r Acceptor level Valence band Valence band