References S.M. Sze, Physics of semiconductor devices (Wiley) Chap 1,2

1. Metodi Sperimentali della Fisica Moderna P-N junctions Short review of semiconductor properties P-n junctions References  S.M. Sze, Physics of semiconductor devices (Wiley) Chap 1,2 Ashcroft-Mermin, Solid State Physics (Saunders) Chap 4-5, 28-29 Grosso- Parravicini, Solid State Physics Chap 13-14

2. Crystal lattice Bravais lattice = set of all points such that R = n1a1+ n2a2+ n3a3 a1, a2, a3 = primitive vectors n1 , n2 , n3 = integers Simple cubic (SC) Body centered cubic (BCC) Face centered cubic (FCC) a1 = (y+z)a/2 a1 = (-x+y+z)a/2 a2 = (x+z)a/2 a3 = (x+y)a/2 a2 = (x-y+z)a/2 a3 = (x+y-z)a/2

3. Crystal lattice a1; a2; a3 = primitive vectors are not unique Primitive unit cell = volume of space that can fill up all the available space with no overlapping (not unique)

4. Crystal lattice with basis The crystal can be described by different primitive vectors, adding also the atoms within the unit cell using a basis Lattice with two point basis Diamond lattice Primitive vectors Basis

5. Directions and planes [n1, n2, n3] = crystal direction [1 1 1] [1 1 0] [1 0 0] (n1, n2, n3) = crystal plane Miller indexes different surface atomic density

6. Reciprocal lattice Plane wave Bravais lattice R = n1a1+ n2a2+ n3a3 The set of K yelding plane waves with the periodicity of a Bravais lattice is the reciprocal lattice The vectors defining the reciprocal lattice are If k is a linear combination To be a k of the reciprocal space the ki must be integers

7. Reciprocal lattice Real lattice

8. L X K W Reciprocal lattice Brillouin zone: region of points closer to a lattice point than to any other lattice point in the reciprocal space. The main symmetry directions are labelled , K, X, W, L bcc fcc What is the physical meaning of k? Periodic potential U(r) Crystal momentum Free electron

9. Fermi level Consider N non-interacting electrons confined in a volume V = L3 Non interacting The ground state is obtained from the single e- levels in V and filling them up with the N electrons  = Energy of the level 1 e- Schroedinger equation Represent the electron confinement to V with a boundary condition  (x+L, y, z) =  (x, y, z)  (x, y +L, z) =  (x, y, z)  (x, y, z +L) =  (x, y, z) This leads to running waves, while choosing as boundary condition for  to vanish at the surface would give rise to standing waves…… STANDING WAVES

10. a) Neglect the boundary condition Normalization condition The solution of the Schroedinger equation is With energies The one-electron level k(r) is an eigenstate of the momentum operator with eigenvalue that is the electron momentum But k is also a wave vector of the plane wave corresponding to the wavelength

11. b) Consider the boundary condition The bc are satisfied by the plane wave only if ez = 1 only for z = 2m wavevectorcomponents the area per point is (2/L)2 The allowed wavevectors in the k space are those with components made by integer multiples of 2/L

12. Fermi wavevector How many allowed k are contained in a region  of K space large compared to 2/L? area per point (2/L)2 Volume occupied by one point The k space density of levels is

13. build the N-electron (non interacting) ground state by filling up the one-electron levels Each level can contain 2 electrons (one for each spin) For N large, we get a sphere volume the sphere radius is defined kF Number of allowed k within the sphere For N non interacting e- in a volume V The ground state is formed by occupying all single-particle levels Occupied for k < kF Unoccupied for k > kF FERMI WAVE VECTOR

15. Electronic structure of a solid We start by assuming the positions of atoms R The Hamiltonian describing the electronic structure is Electron Kinetic energy Electron-ion interaction Electron-electron interaction many-body problem Impossible to calculate solutions Hartree-Fock is not good for solids, it dumps the many body problem in the correction term

16. 1-D band theory: Tight-binding model Potential Semiconductors V atomic The lattice potential is constructed from a superposition of N free atoms potentials Va(r) arranged on a chain with lattice constant a (but it is not the true lattice potential) V between Atomic planes V along the ion line Atomic Schroedinger equation (r) is the eigenstate of the isolated atom

17. The Schroedinger equation for the crystal is The solution can be constructed using a linear combination of the type (r) are Wannier functions, similar to the solutions  of the atomic Schroedinger equations cn : coefficient to be found Insert the solution into the Schroedinger equation  matrix Eigenvector of two generic sites Only the on-site and first neighbor coefficients are retained thus defining the matrix elements , Recalling that the  must be of the type eikr

18. Reducing the matrix elements allows to obtain the expansion coefficients cn Hence the eigen energy corresponding to a solution of the Schroedinger equation is Dispersion relation for energy bands Energy band Wavevector along the Brillouin directions It is important how much the (r) overlaps between neighboring sites. The bandwidth depends on the overlap. The potential breaks the degeneracy

19. The potential is also affecting the bands at the high symmetry points of the brillouin zone giving rise to prohibited energies, i.e. energy gaps

20. Energy bands for electrons T dependence of energy gap Ge Si GaAs

21. Fermi level in metals The ground state of N electrons in a solid is made similarly to the free electron case The electron levels are identified by quantum numbers n and k, and the structure of the solid gives rise to the electronic bands When all levels are filled with electrons, in a metal the last band is only partially filled Fermi level is the energy below which the one-electron levels are occupied and above which are unoccupied

22. Semiconductors The semiconductors are characterized by a gap between the last occupied band, defined as VALENCE BAND and the first unoccupied band, defined as CONDUCTION BAND CB Hence the definition of Fermi level as "the energy below which the one-electron levels are occupied and above which are unoccupied" is valid for all energies in the gap. EF is not univocally determined One has to define a number telling the energy position of the electrons with the lowest binding energy VB At T  0 there is a finite probability that some electrons will be thermally excited across the gap, and there will be conduction of ELECTRONS and HOLES will see that GaAs: EG = 1.4 eV at T = 300 K, KBT  0.025 eV and N  7 1013 InSb: EG = 0.16 eV at T = 300 K, KBT  0.025 eV and N  4 1016

23. Semiconductors Thermally excited electrons will be mostly close to CBM and holes close to VBM so the band dispersion can be approximated by a parabolic relation 0.28m 0.044m The tensor M can be diagonalized so 0.49m 0.16m Ge Si Effective masses Constant energy surfaces for e- Constant energy surfaces for h

24. Chemical potential & Fermi-Dirac function Gibbs energy for a system containing N particles At constant T and P Chemical potential If the system loses 1 particle, the free energy changes System of identical fermions The average number of fermions in a single-particle state i Fermi–Dirac (F–D) distribution Average number of fermions with energy i = F–D distribution ni X the degeneracy gi (i.e. the number of states with energy i) For solids g() = DOS

25. Carrier concentration in Semiconductors To understand the equivalent of Fermi level in semiconductors, we have to evaluate the number of carriers in thermal equilibrium g() = DOS electrons holes The carrier density depends on the chemical potential  Suppose

26. Since only the carriers within KBT of the band edges contribute, the effective mass approximation is good and the DOS g() is NC, PV Effective density of states integrated over all energies (mC)3, (mV)3 effective masses Carrier density Therefore it is the chemical potential that sets the density of carriers n,p and also the energy position of the states within the gap. This is usually referred to Fermi level in semiconductors but it is NOT a Fermi level

27. Carrier density NC, PV effective DOS for carriers

28. Intrinsic semiconductors At finite temperature thermal excitation of e- leaves an equal number of holes in the VB so that n = p = ni intrinsic carrier density depends only on the gap energy Mass action law

29. Intrinsic semiconductors What is the chemical potential for the intrinsic case i?

30. Intrinsic semiconductors So for T  0, the chemical potential i, is in the middle of the energy gap Since (mV/mC)  1, the chemical potential i does not change more than ~ KBT Are valid

31. Intrinsic semiconductors Considering (mV/m)  (mC/m) 1, at RT Silicon: EG= 1.1 eV GaAs: EG= 1.4 eV InSb: EG= 0.17 eV But mC and mV are 0.01

32. Fermi level for intrinsic semiconductors band diagramdensity of statescarrierconcentrations

33. Donors and acceptors P: 5 valence e- Si: 4 valence e- B: 3 valence e- n-type: a Si atom is replaced by P atom with an extra e-. The P atom is called donor p-type: a Si atom is replaced by B atom with an extra hole. The B atom is called acceptor Donors and acceptors introduce extra energy levels, whose energy can be estimated using the H atom model

34. Donors and acceptors Why hydrogen model? • Neglect the ion core of the inserted P atom • The P is represented by a Si atom with 1 hole (e+) fixed in the site + 1 e- Isolated atom Atom inside crystal Charge field reduced by the macroscopic dielectric constant  (13 for Si) e- binding energy = 10.486 eV (P ionization potential) The e- moving in the lattice has energies of the type E=Ec(k) (k =crystal wavevector) The allowed energy levels has to be close to conduction band minimum to minimize energy Assume parabolic bands with effective masses The electron of the donor impurity is represented as a particle of charge –e and and mass m* moving in the presence of attractive charge e/ Impurity H atom r0 up to 10 nm

35. Extrinsic semiconductors The density of donors or acceptors is above 1012/cm3 The level energies are very small compared to the energy gap so it is very easy to excite an electron from a donor level or a hole from an acceptor level

36. Fermi level for extrinsic semiconductors Introduce impurities not all dopants are necessarily ionized: depends on the impurity energy level and T DONOR ND (cm-3) = donor concentration D = energy of donor level g = ground state degeneracy = 2 The donor level are localized hence cannot accommodate 2 electrons due to charge repulsion Probability of occupation of donor levels Probability of finding ionized donor levels Density of ionized donors

37. Fermi level for extrinsic semiconductors Introduce impurities not all dopants are necessarily ionized: depends on the impurity energy level and T ACCEPTOR NA (cm-3) = acceptor concentration A = energy of acceptor level g = ground state degeneracy = 4 each acceptor impurity level can accept one hole of either spin The impurity level is doubly degenerate at K =0 Probability of occupation of acceptor levels Probability of finding ionized acceptor levels Density of ionized acceptor

38. Fermi level for extrinsic semiconductors With impurity atoms introduced preserve charge neutrality Total negative charges (electrons and ionized acceptors) Total positive charges (holes and ionized donors) extrinsic intrinsic intrinsic extrinsic if I addbothdonors and acceptors Mass action law stillvalid Add DONORS only h density in VB e density in CB Ionizeddonordensity

39. Fermi level for extrinsic semiconductors DONORS Graphically solved to determine EF Plot for two different values of D

40. Fermi level for extrinsic semiconductors DONORS n-type semiconductor band diagramdensity of statescarrierconcentrations modificare le E della figura in epsilon

41. Fermi level for extrinsic semiconductors ACCEPTORS p-type semiconductor band diagramdensity of statescarrierconcentrations modificare le E della figura in epsilon

42. For a set of dopant concentrations it is possible to estimate the Fermi level

43. Carrier concentration temperature dependence Charge neutrality D = energy level of the donor impurity TDKB = (C-D) ionization impuritytemperature 1) T<< TD = D/KB freezing out region D < EF < C neglect p Carrier density in the conduction band

44. Carrier concentration temperature dependence 2) TD < T << EG/KB saturation region D < EF < C At RT almost all electron (donors) and holes (acceptors ) are excited = saturation condition Intrinsic electrons negligible Majority carriers Minority carriers p(T) Intrinsic Silicon N-type Silicon 3) TD < EG/KB < T intrinsic region

45. Mobility Why? We need to know what is happening to carriers for concentrations out of equilibrium - Carrier transport for a semiconductor in the presence of an electric field E - 2 parabolic bands (valence and conduction) and effective masses mv and mc mobility Current density • = scattering time n = carrier density • depends on carrier interaction within the lattice and on temperature For e- and holes

46. Diffusion coefficient Consider a concentration gradient in the solid Suppose the flux goes from regions of high concentration to regions of low concentration with a magnitude that is proportional to the concentration gradient Dn is the diffusion coefficient For a doped semiconductor with a carrier concentration gradient and applied electric field the total current density is Drift Diffusion

47. Mobility-diffusion coefficient relation The mobility p, and the diffusion coefficient D, are not independent Consider an n-typesemiconductor with nonuniform (n=n(x)) doping concentration and without an externalappliedfield Drift current balances the diffusion current the nonuniform doping generates a potential (x) that rigidly shifts the energy levels of the semiconductor internal electric field

48. The p-n junction p-type semiconductor n-type semiconductor All donors and acceptors ionized (saturation condition) ND,A (cm-3) = donor,acceptor concentration

49. The p-n junction Bring together the two regions p-type semiconductor n-type semiconductor • Some e- move from n-type to p and recombine with h • Some h move from p-type to n and recombine with e- • In the n region close to x=0 remains non neutralized donors (+) • In the p region close to x=0 remains non neutralized acceptors (-) A region depleted of majority carriers is formed at the interface (space charge region) A strong electric field is built up opposing further diffusion of majority carriers, reaching equilibrium

50. At equilibrium and no external field E applied , no current flows so The Fermi level must remain constant throughout the depletion layer and the sample since the system is at equilibrium This means that the band in the two regions bend to adjust across the p-n junction