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Semiconductors At zero temperature semiconductors are insulators with completely filled bands. Conduction Band (empty at T=0) Eg Valence Band (full at T=0) At higher temperatures they conduct at due to the thermal excitation of electrons across a relatively small band gap. In a semiconductor the highest energy filled band is called the valence band and the lowest energy band called the conduction band. We will consider states near the top of the valence band to be holes (particles of charge +e) with free electron like dynamics but effective mass m*h We will consider states near the bottom of the conduction band to be electrons with free electron like dynamics but effective mass m*e 1 Direct Gap Semiconductors Direct gap semiconductors: the top of the valence band and the bottom of the conduction band occur at the same k-vector. Conduction band Energy Gap Valence band k Gallium Arsenide GaAs 2 Indirect Gap Semiconductors Indirect gap semiconductors: the top of the valence band and the bottom of the conduction band occur at different k-vector. Germanium (Ge) Indirect bandgap 0.8eV Direct bandgap, @ k =0, is 0.66eV 3 Direct Optical absorption Direct gap semiconductor: sharp onset of absorption when the photon energy is equal to the bandgap Optical absorption in the direct gap semiconductor InSb at 4K Conduction Band Energy k hn =Eg Valence Band Bandgap of InSb at 4K is 0.23eV Photon creates an “electron-hole pair” 4 Indirect Optical absorption A transition across an indirect band gap requires a photon to be absorbed and a phonon to be absorbed or emitted. Conduction Band Energy hnphoton ~Eg kphonon k Valence Band 5 Ge: Indirect Optical absorption Indirect gap semiconductor: no sharp onset of absorption 0.80 0.88 0.73 0.66 For T = 300 K Eg (indirect gap) = 0.66 eV and EG1 (direct gap) = 0.8 eV For T = 77K Eg (indirect gap) = 0.73 eV and EG1 (direct gap) = 0.88 eV Note: numerical values for E(k) for Germanium in figure in Kittel, reproduced earlier. 6 Number of electrons in conduction band Silicon @ 300K n ~ 2x1016 m-3 Note Units are cm-3 Electron density increases ~ exponentially with temperature 7 Density of States Assume: bottom of conduction band and the top of valence band parabolic i.e. conduction band E = Ec + 2k2/2me* valence band E = Ev - 2k2/2mh* 1 2me Conduction De (E) = 2 2 2 band valence band D(E) Conduction Band Valence Band 0 Eg Ec = Eg 3/2 1 (E E g )2 3/2 * 1 1 2mh 2 Dh (E) = 2 2 ( E) 2 * Conduction Band Ev = 0 Valence Band Note. I do not set Ev = 0 until later in notes document E 8 Chemical potential or Fermi level At a temperature T the probability that an electron state is occupied is give by the Fermi-Dirac function -1 E- fe (E) = exp + 1 kB T The chemical potential, , is the energy for which f = ½ . Fermi energy: all energy states are occupied below EF at T = 0. In discussing semiconductors is often referred to as the Fermi level ! Will find that is within the energy gap. In the conduction band Therefore E- exp 1 kB T (E - ) fe (E) exp kB T 9 Electrons density in conduction band 3/2 Density of states 1 2me De (E) = 2 2 2 Distribution function (E - ) fe (E) exp kB T * 1 (E E g )2 Total number density of occupied states n D ( E ) f (E)dE e e Eg 3 me * k B T 2 n 2 exp ( Eg ) / k B T ) 2 2 n Nc exp ( Eg ) / k B T ) where “Fermi level” E g k BT ln n / N c m k T N c 2 2 Silicon Eg=1.15eV m*e = 0.2me & n~1016 m-3 @ 300K. Gives * e B 2 3 2 E g 0.5eV Fermi level near middle of gap 10 Number of holes and electrons Exactly same argument for holes in the valence band gives ( E ) f h (E) 1 f e ( E ) exp kBT Distribution function m k T Total number p 2 density of 2 occupied states * h B 2 3 2 exp ( ) / k B T Nv exp ( ) / k B T n Nc exp ( Eg ) / k B T ) 3 3 k T np 4 B 2 (m*e m*h ) 2 exp( E g / k BT) Independen t 2 This last result is particularly important of True for both intrinsic and extrinsic semiconductors 11 Intrinsic semiconductors n = p In pure “intrinsic” semiconductors the electrons and holes arise only from excitation across the energy gap. Therefore n = p 3 k BT * * 32 np 4 (me mh ) exp( E g / k BT ) N v N c exp( E g / k BT ) 2 2 3 k BT 2 * * 3 4 1/ 2 ni n p 2 ( m m ) exp( E / 2 k T ) ( N N ) exp( Eg / 2k BT ) e h g B v c 2 2 Chemical potential? m mh* 1 3 Eg k BT n * 2 4 me Eg 2 * e 3 2 0.026 exp ( Ec ) / k B T ) m * h 3 2 exp / k B T Silicon Eg=1.15eV m*e = 0.2me & m*e = 0.8me @ 300K. Gives eV Fermi level near middle of gap 12 Hydrogenic Donors & Acceptors An electron added to an intrinsic semiconductor at T=0 would go into the lowest empty state i.e. at the bottom of the conduction band. When one adds a donor atom at T=0 the extra electron is bound to positive charge on the donor atom. -e +ve ion Conduction Band DE The electron bound to the positive Ion is in an energy state ED = Eg- DE where DE is the binding energy. An electron which moves on to an acceptor atom has energy EA +e -ve ion Ec = Eg ED EA Valence Band Ev = 0 13 Magnitude of binding energy Similar to a hydrogen atom. Ground state wavefunction is 3 1 1 (r ) exp[ r / a0 ] a0 2 +ve ion -e The Bohr radius, a0 = 4ro2/mee2 determines the spatial extent of the wavefunction. Hydrogen atom (r = 1 ) a0 = 0.53 Å. Binding energy of an electron in the ground state of a hydrogen atom is 4 EB e me 13.6 eV 2 2(40 r ) Typical Semiconductor me* ~0.15 me and r ~15. 0.53 r a0 * me / me 13.6 m*e EB 2 eV r me ~ 50 Å ~10 meV. 14 Number of Electrons in the conduction band Consider a semiconductor with ND donor atoms per m3 ND0 and ND+ – number density of neutral and ionised donors Conduction Band DE Ec = Eg ED At T = 0 all electrons in the lowest available energy states. No electrons are excited from donor states into conduction band n = 0 ; ND0 = ND Valence Band Ev = 0 15 T ~ Room temperature kBT > ~ (EC -ED) & number of available states in conduction band >> ND . Therefore almost all the donors will be ionised. n = ND+ ~ ND Relevant regime for all electronics. Note that the density of electrons is ~ independent of temperature. The chemical potential is well below EC and the expression obtained for n in an extrinsic semiconductor can be used to give N E g k B T n C n where Silicon Nc ~ 2.6x1024 so for n ~ 1022 m * k BT N C 2 2 2 3 2 Eg 6k B T Eg 150meV T >> Room temperature In this limit number of electrons excited across the bandgap becomes larger than number of donors. Behaves like an intrinsic semiconductor. 16 n-type semiconductors n exp Eg / 2k B T ) n Cons tan t ND ln(n) n exp (Eg ED ) / k B T ) Number of electrons in conduction band ln(T) Eg ED ln() “Fermi level” Eg / 2 ln(T) 17 p-type semiconductors p exp E g / 2k B T ) p Cons tan t NA ln(p) p exp EA / kBT) Number of holes in valence band ln(T) Eg / 2 ln() “Fermi level” EA ln(T) 18 Compensated semiconductor Compensated semiconductor: both donors and acceptors present. Conduction Band ND donors per m3 and NA acceptors per m3 For ND > NA have an n-type semiconductor with n ~ ND - NA for T ~ 300K Ec = Eg ED NA electrons fall into acceptor states For NA > ND have an p-type semiconductor with p ~ NA - ND for T ~ 300K EA Valence Band Ev = 0 19 Impurity Bands Have considered the impurities as isolated atoms. Reasonable as doping level normally ~ one donor per 106 semiconductor atoms. At very high donor concentrations, one has substantial overlap between the donor or acceptor wavefunctions. f(r) aB + b + Above a critical doping level one has an impurity energy band with a finite conductivity. Electron density at which this “metal insulator transition” occurs? aB ~ 50 Å & lattice constant, a ~ 2.5 Å. Need b ~ aB = 20a . i.e. one donor per 203 = 8000 semiconductor atoms 20 Mobility of semiconductors E Both electrons and holes carry current in the same direction in a semiconductors. -e Conductivity: s nee + peh +e e: electron mobility, h: holes mobility In considering scattering of electrons and holes it is important to consider mobility as the numbers of carriers varies with temperature. Conductivity /me* se = ne2tp /me* Mean Free Path Le tpv Mobility e = se/ne = etp so e = e Le /vme* ~ Le /v The electron and hole distributions are non-degenerate and <E> ~ kBT <v> ~ < 2E/me* > ~ T1/2. T > ~ TD number of phonons increases as T. L ~ T-1 . ~ T-3/2 T << TD ionised impurity scattering dominates. Similar to Rutherford scattering & scattering cross section ~ E-2 ~ T-2. So L ~ T2 and ~ T3/2 21 Hall Effect in Semiconductors n-type semiconductors (n>>p) RH ~ -1/ne p-type semiconductors (p>>n) RH ~ +1/pe. The Hall effect is used to obtain the carrier densities in semiconductors. In an electric field electrons and holes drift in opposite directions. j Consider case of n = p and e = h. F Have no Hall field. B -e B F +e The free carrier Hall coefficient is generally (Hook and Hall p153) RH = ( p h2 - n e2 )/e(n e + p h )2 22 Semiconductor devices: Inhomogeneous semiconductors All solid-state electronic and opto-electronic devices are based on doped semiconductors. In many devices the doping and hence the carrier concentrations are non-homogeneous. In the following section we will consider the p-n junction which is an important part of many semiconductor devices and which illustrated a number of key effects 23 The p-n semiconductor junction: p-type / n-type semiconductor interface We will consider the p-n interface to be abrupt. This is a good approximation. n-type ND donor atoms per m3 p-type NA acceptor atoms per m3 Consider temperatures ~300K Almost all donor and acceptor atoms are ionised. p-n interface at x=0. impurity atoms m-3 ND NA p-type n-type x=0 xa ND (x) = ND (x>0) = 0 (x<0) NA (x) = NA (x<0) (x>0) = 0 (x>0) (x<0) 24 p-type semiconductor Electron and hole transfer n-type semiconductor Electrons EC EC EV EV Holes Consider bringing into contact p-type and n-type semiconductors. n-type semiconductor: Chemical potential, below bottom of conduction band p-type semiconductor: Chemical potential, above top of valence band. Electrons diffuse from n-type into p-type filling empty valence states. 25 EC Electrons Band Bending EC EV Holes eDf0 EV p-type semiconductor n-type semiconductor Electrons diffuse from n-type into p-type filling empty valence band states. The p-type becomes negatively charged with respect to the n-type material. Electron energy levels in the p-type rise with respect to the n-type material. A large electric field is produced close to the interface. Dynamic equilibrium results with the chemical potential (Fermi level) constant throughout the device. Note: Absence of electrons and hole close to interface -- depletion region 26 Electrostatic voltage drop, Df0 EC Electrons In equilibrium constant. Electrostatic voltage difference, Df0, between nand p- regions. EC EV Holes eDf0 EV p-type semiconductor n-type semiconductor Nc Nc E k T n c B N N N A D D For x >> 0 Ec k BT n for x << 0 Nv Nv Ev k BT n Ev k BT n N A ND NA Since Eg = Ec – Ev ND N A E g eD 0 k BT n Nc Nv 27 Depletion region Depletion region Assume the electric field in the region of the junction removes all the free carriers creating a depletion region for –dp<x < dn. n,p Electron and Hole Density r The ionised impurities are fixed in the lattice. So charge density is r +eND per m3 for 0 <x <dn r –eNA per m-3 for -dp < x < 0. Net charge density E -dp 0 Depletion region dn xa The total charge in the depletion region must be zero as the number of electrons removed from the right equals the number of holes removed from the left i.e. NDdn = NAdp. 28 Electric field E(x) Can calculate the electrostatic potential, f(x) from the Poisson’s equation 2f ( x) r ( x) / 0 r 2 x Charge density: r(x) = eND for 0 < x < dn r(x) = -eNA for –dp < x < 0 Boundary condition: E = 0 for x > dn and x < –dp Depletion region So integration gives N A e E ( x d p ) for d p x 0 x 0 r r Net charge density E N D e E ( x d n ) for 0 x d n x 0 r Electric Field (negative) 0 xa 29 Electrostatic potential, f(x) Integration of E gives the potential (x). Since 0 for x < –dp and D0 for x < dn. eN A ( x) (x d p )2 2 0 r ( x) D 0 (x) is continuous at x = 0 so eN D (x dn )2 2 0 r for 0 x dn e( N D d n N A d p ) 2 D 0 So since NDdn = NAdp dp x 0 for 2 0 r N A D 0 dn eN ( N N ) D D A 2 2 0 r 1 2 2 N D 0 dp 0 r D eN ( N N ) D A A 1 2 Resulting depletion width is ~100nm to 1m. Self consistent 30 Electron and Hole Density n,p r Net charge density E Electric Field (negative) Electrostatic Potential Energy of Conduction band edge f(x) EC 0 Depletion layer xa 31 p-n junction with a forward bias Electric Current Net flow of electrons EC EC eV EV e(Df0V Net flow of holes p-type biased positive EV n-type biased negative Forward bias: p-type region biased positive with respect to n-type region. The voltage is dropped across depletion region since the free carrier density is low and therefore the resistivity is high. Total potential across the depletion layer is D = D0 – V 32 Generation Current EC EC EV eDf0 EV p-type semiconductor n-type semiconductor Electron-hole pairs created in the depletion region move apart in the strong electric field. A generation current, Jgen, in the negative x-direction results. Magnitude of the generation current density is Jgen = A exp(-Eg/2kBT) where A is a constant. 33 Recombination Current EC eDf0 Photons Out EC EV eDf0 EV p-type semiconductor n-type semiconductor Electrons with energies greater than eΔ0 can move into the p-type material where they recombine with holes. A recombination current, Jrec, in the positive x-direction results Jrec = B exp(-e Δ0/2kBT) where B is ~ constant. For direct band gap semiconductors recombination leads to photon emission 34 Current-Voltage Characteristic At equilibrium, without a bias voltage Jgen + Jrec = 0 With external positive voltage V the Jgen is ~ unchanged, but Jrec becomes - (eD 0 - eV) J rec (V ) = B exp kBT eV 1) Total net current density is J J rec (V ) J gen (0) J gen (0)(exp kBT J Reverse bias negative p n positive Check Forward bias positive p n negative V -Jgen 35 Applications of p-n junctions • p-n junction diodes: Excellent diodes, which can be used for rectification of AC signals. • Light emitting diodes (LEDs) and lasers: In forward bias one has an enhanced recombination current. For direct band gap semiconductors light is emitted. • Solar cells: If photons with hn>Eg are absorbed in the depletion region of a p-n junction one has an enhanced generation current. The energy of the photons can be converted to electrical power in solar cells based on this mechanism. • p-n-p junction transistors: Transistors based on the properties of p-n junctions can also be produced. See Hook and Hall p184-8. 36