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Semiconductor Lattice Structures Diamond Lattices Chapter 3: semiconductor science and light emitting diodes ¾ The diamond-crystal lattice characterized by four covalently bonded atoms. ¾ The lattice constant, denoted by ao, is 0.356, 0.543 and 0.565 nm for diamond, silicon, and germanium, respectively. ¾ Nearest neighbors are spaced ( 3ao / 4) units apart. ¾ Of the 18 atoms shown in the figure, only 8 belong to the volume ao3. Because the 8 corner atoms are each shared by 8 cubes, they contribute a total of 1 atom; the 6 face atoms are each shared by 2 cubes and thus contribute 3 atoms, and there are 4 atoms inside the cube. The atomic density is therefore 8/ao3, which corresponds to 17.7, 5.00, and 4.43 X 1022 cm-3, respectively. (After W. Shockley: Electrons and Holes in Semiconductors, Van Nostrand, Princeton, N.J., 1950.) Semiconductor Lattice Structures How Many Silicon Atoms per cm-3? Diamond and Zincblende Lattices Diamond lattice can be though of as an FCC structures with an extra atoms placed at a/4+b/4+c/4 from each of the FCC atoms • Number of atoms in a unit cell: • 4 atoms completely inside cell • Each of the 8 atoms on corners are shared among cells Æ count as 1 atom inside cell • Each of the 6 atoms on the faces are shared among 2 cells Æ count as 3 atoms inside cell ⇒ Total number inside the cell = 4 + 1 + 3 = 8 • Cell volume: (.543 nm)3 = 1.6 x 10-22 cm3 Diamond lattice Si, Ge Zincblende lattice GaAs, InP, ZnSe The Zincblende lattice consist of a face centered cubic Bravais point lattice which contains two different atoms per lattice point. The distance between the two atoms equals one quarter of the body diagonal of the cube. • Density of silicon atoms = (8 atoms) / (cell volume) = 5 x 1022 atoms/cm3 Semiconductor Materials for Optoelectronic Devices The Si Crystal • Each Si atom has 4 nea rest neighbors • lattice constant = 5.431Å “diamond cubic” lattice 400~450 450~470 470~557 557~567 567~572 572~585 585~605 605~630 630~700 Pure Blue Blue Pure Green Green Yellow Green Yellow Amber Orange Red GaSb In0.7Ga0.3As0.66P0.34 In0.14Ga0.86As InP GaAs P 0.55 0.45 GaP(N) GaAs InGaN SiC(Al) In0.57Ga0.43As0.95P0.05 Semiconductor Materials for Optoelectronic Devices Semiconductor Optical Sources GaAs1-yPy x = 0.43 In1-xGaxAs1-yPy AlxGa1-xAs In0.49AlxGa0.51-xP 0.6 Red Green Yellow Orange 0.5 Blue Violet 0.4 0.7 0.8 λ 0.9 I n f r ar e d (um) 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 Quantization Concept Periodic Table of the Elements Group** Period plank constant 1 IA 1A 1 1 H 1.008 3 2 Semiconductor Materials IV Compounds SiC, SiGe III-V Binary Compounds AlP, AlAs, AlSb, GaN, GaP, GaAs, GaSb, InP, InAs, InSb III-V Ternary Compounds AlGaAs, InGaAs, AlGaP III-V Quternary Compounds AlGaAsP, InGaAsP II-VI Binary Compounds ZnS, ZnSe, ZnTe, CdS, CdSe, CdTe II-VI Ternary Compounds HgCdTe 7 6 7 16 17 VIA VIIA 6A 7A 8 9 2 He 4.003 10 B C N O F Ne 12.01 14.01 16.00 19.00 20.18 12 Na Mg 24.31 20 3 IIIB 3B 21 4 IVB 4B 22 5 VB 5B 23 K Ca Sc Ti V 39.10 40.08 44.96 47.88 50.94 38 39 40 41 6 VIB 6B 24 7 VIIB 7B 25 8 9 10 ------- VIII ------------- 8 ------26 27 28 11 IB 1B 29 12 IIB 2B 30 34 35 36 Se Br Kr 78.96 79.90 83.80 42 43 44 45 46 47 48 Ru Rh Pd Ag Cd 101.1 106.4 107.9 95.94 74 (98) 75 76 102.9 77 78 Cs Ba La* Hf Ta W Re Os Ir Pt 132.9 137.3 180.9 183.9 186.2 190.2 190.2 195.1 108 200.5 50 51 52 53 54 In Sn Sb Te I Xe 114.8 118.7 121.8 127.6 126.9 131.3 81 82 83 84 85 86 Tl Pb Bi Po At Rn 204.4 207.2 209.0 (210) (210) (222) 109 110 111 112 114 116 118 Ra Ac~ Rf Db Sg Bh Hs Mt --- --- --- --- --- --- (226) (260) (262) (265) (266) () () () () () () (263) 107 80 Au Hg 197.0 49 74.92 Fr (257) 106 79 112.4 72.59 (223) (227) 105 33 Ga Ge As 69.72 Nb Mo Tc 104 32 Zn 65.39 92.91 89 31 Cu Zr 88 Ar 39.95 63.55 91.22 73 18 Cl 35.45 Ni Y 72 17 S 32.07 58.69 88.91 178.5 16 P Co 55.85 Sr 57 15 30.97 58.47 54.94 87.62 138.9 14 Si 28.09 Cr Mn Fe Rb 56 13 Al 26.98 52.00 85.47 87 Valence electrons 5 15 VA 5A 10.81 55 6 4 14 IVA 4A Be 37 5 13 IIIA 3A 9.012 19 4 2 IIA 2A Li 22.99 Core electrons 8A 6.941 11 3 18 VIIIA Semiconductor Materials Atomic Bonding Covalent Bonding Bonding forces in Solids a. b. c. d. e. Ionic bonding (such as NaCl) Metallic bonding (all metals) Covalent bonding (typical Si) Van der Waals bonding (water…) Mixed bonding (GaAs, ZnSe…, ionic & covalent) Quantization Concept Quantization Concept The Shell Model plank constant L shell with two sub shells Nucleus 1s Core electrons K L 2s 2p 1s22s22p2 or [He]2s22p2 Valence electrons ¾ The shell model of the atom in which the electrons are confined to live within certain shells and in sub shells within shells. Energy Band Formation (I) Energy Band Formation (I) Band theory of solids Two atoms brought together to form molecule ¾“splitting” of energy levels for outer electron shells = Allowed energy levels of an electron acted on by the Coulomb potential of an atomic nucleus. Energy Band Formation (II) Pauli Exclusion Principle Energy Band Formation (III) Conceptual development of the energy band model. where ‘no’ states exist As atoms are brought closer towards one another and begin to bond together, their energy levels must split into bands of discrete levels so closely spaced in energy, they can be considered a continuum of allowed energy. Electron energy Electron energy N isolated Si-atoms Only 2 electrons, of spin ± 1/2, can occupy the same energy state at the same point in space. ¾ Strongly bonded materials: small interatomic distances. ¾ Thus, the strongly bonded materials can have larger energy bandgaps than do weakly bonded materials. The electrical properties of a crystalline material correspond to specific allowed and forbidden energies associated with an atomic separation related to the lattice constant of the crystal. p s n=3 6N p-states total 2N s-states total (4N electrons total) p Crystalline Si N -atoms Electron energy Energy Bandgap Splitting of energy states into allowed bands separated by a forbidden energy gap as the atomic spacing decreases. 4N allowed-states (Conduction Band) No states 4N allowed-states (Valance Band) 4N empty states s 2N+2N filled states isolated Si atoms Decreasing atom spacing Si lattice spacing Electron energy → ← Mostly empty Eg Mostly filled Etop Ec Ev Ebottom Energy Band Formation (IV) Energy Band ¾ Broadening of allowed energy levels into allowed energy bands separated by forbidden-energy gaps as more atoms influence each electron in a solid. Energy band diagrams. N electrons filling half of the 2N allowed states, as can occur in a metal. One-dimensional representation Two-dimensional diagram in which energy is plotted versus distance. Metals, Semiconductors, and Insulators Metals, Semiconductors, and Insulators Typical band structures of Metal Electron Energy, E Typical band structures of Semiconductor Covalent bond Free electron Vacuum level A completely empty band separated by an energy gap Eg from a band whose 2N states are completely filled by 2N electrons, representative of an insulator. 3s Band Si ion core (+4e) Electron energy, E Ec+χ E =0 2 p Band 3p 3s 2p 2s 2 s Band ConductionBand(CB) Empty of electrons at 0 K. Ec Overlapping energy bands Band gap = Eg Ev Electrons Valence Band (VB) Full of electrons at 0 K. 1s ATOM 0 1s SOLID ¾ In a metal the various energy bands overlap to give a single band of energies that is only partially full of electrons. ¾ There are states with energies up to the vacuum level where the electron is free. A simplified two dimensional view of a region of the Si crystal showing covalent bonds. The energy band diagram of electrons in the Si crystal at absolute zero of temperature. Metals, Semiconductors, and Insulators Metals, Semiconductors, and Insulators Typical band structures at 0 K. Carrier Flow for Metal Carrier Flow for Metals.mov Carrier Flow for Semiconductor Carrier Flow for Semiconductors.mov Insulator Metals, Semiconductors, and Insulators Semiconductors Electrons within an infinite potential energy well of spatial width L, its energy is quantized. Conductors En = Many ceramics Superconductors Alumina Diamond Inorganic Glasses Polypropylene PVDF Soda silica glass Borosilicate Pure SnO2 PET SiO2 10-18 10-15 Degenerately doped Si Alloys Intrinsic Si 10-9 10-6 10-3 kn = nπ L ∞ n = 1,2,3... h k n : electron momentum E3 infinite square potential ∞ ∞ Conductivity (Ωm)-1 106 109 1012 -a/2 0 ϕ2 x E2 ∞ n=3 x n=2 -a/2 V=0 V(x) 103 ∞ E1 n=1 Te Graphite NiCrAg 100 ∞ ϕ m Amorphous Intrinsic GaAs As2Se3 10-12 ( h k n2 ) 2me Energy increases parabolically with the wavevector kn. Metals Mica Metal Energy Band Diagram Range of conductivities exhibited by various materials. Insulators Semiconductor +a/2 x Energy state 0 +a/2 -a/2 Wavefunction 0 +a/2 Probability density ¾ This description can be used to the behavior of electron in a Metal within which their average potential energy is V(x) ≈ 0 inside, and very large outside. 3.1 Energy Band Diagram 3.1 Energy Band Diagram within the Crystal! E-k diagram, Bloch function. E-k diagram, Bloch function. PE(r) Potential Energy of the electron around an isolated atom r When N atoms are arranged to form the crystal then there is an overlap of individual electron PE functions. V(x) a a 0 a 2a Surface Moving through Lattice.mov ¾ The electron potential energy [PE, V(x)], inside the crystal is periodic with the same periodicity as that of the crystal, a. ¾ Far away outside the crystal, by choice, V = 0 (the electron is free and PE = 0). 3.1 Band Diagram m = 1,2,3... E-k diagram There are many Bloch wavefunction solutions to the one-dimensional crystal each identified with a particular k value, say kn which act as a kind of quantum number. Each ψk (x) solution corresponds to a particular kn and represents a state with an energy Ek. The energy is plotted as a function of the wave number, k, along the main crystallographic directions in the crystal. The Energy Band Diagram E k ¾ 3.1 Energy Band Diagram E-k diagram of a direct bandgap semiconductor Si Ge GaAs CB Conduction Band (CB) eE g Valence Band (VB) Periodic Potential Periodic Wave function Surface Crystal Schrödinger equation Bloch Wavefunction x=L 3a V ( x ) = V ( x + ma ) Ψk ( x ) = U k ( x ) e i k x PE of the electron, V(x), inside the crystal is periodic with a period a. x x=0 d 2 Ψ 2m e + 2 [ E − V ( x )] ⋅ Ψ = 0 dx 2 h h+ Empty ψ E c E v k E hυ E Occupied ψ k ec v hυ h+ VB k – π /a π/a ¾ The E-k curve consists of many discrete points with each point corresponding to a possible state, wavefunction Ψk (x), that is allowed to exist in the crystal. ¾ The points are so close that we normally draw the E-k relationship as a continuous curve. In the energy range Ev to Ec there are no points [no Ψk (x) solutions]. The bottom axis describe different directions of the crystal. 3.1 Energy Band Diagram 3.1 Energy Band E-k diagram A simplified energy band diagram with the highest almost-filled band and the lowest almost-empty band. E E E CB Direct Bandgap Ec Eg Indirect Bandgap, Eg Photon CB Ev kcb VB k –k VB kvb –k GaAs Er Ev k vacuum level CB Ec Ec Phonon Ev VB –k χ : electron affinity k Si with a recombination center Si conduction band edge In GaAs the minimum of the CB is directly above the maximum of the VB. direct bandgap semiconductor. Recombination of an electron In Si, the minimum of the CB is displaced from the maximum of and a hole in Si involves a recombination center. the VB. indirect bandgap semiconductor valence band edge 3. 1 Electrons and Holes Electrons and Holes Electrons: Electrons in the conduction band that are free to move throughout the crystal. Generation of Electrons and Holes Holes: Missing electrons normally found in the valence band (or empty states in the valence band that would normally be filled). Electron energy, E Ec+χ CB hυ > Eg Ec Free e– hυ Eg Ev Hole h+ hole e– VB 0 A photon with an energy greater then Eg can excitation an electron from the VB to the CB. Each line between Si-Si atoms is a valence electron in a bond. When a photon breaks a Si-Si bond, a free electron and a hole in the Si-Si bond is created. These “particles” carry electricity. Thus, we call these “carriers” 3.1 Effective Mass (I) 3.1 Carrier Movement Within the Crystal An electron moving in respond to an applied electric field. E E Density of States Effective Masses at 300 K within a Vacuum F = − q E = m0 dv dt within a Semiconductor crystal F = − q E = m n∗ dv dt ¾ It allow us to conceive of electron and holes as quasi-classical particles and to employ classical particle relationships in semiconductor crystals or in most device analysis. 3.1 Effective Mass (II) Electrons are not free but interact with periodic potential of the lattice. Wave-particle motion is not as same as in free space. Ge and GaAs have “lighter electrons” than Si which results in faster devices 3.1 Energy Band Diagram The energy is plotted as a function of the wave number, k, along the main crystallographic directions in the crystal. Si Ge GaAs Moving through Lattice.mov The bottom axis describe different directions of the crystal. Curvature of the band determine m*. m* is positive in CB min., negative in VB max. 3.1 Mass Approximation Covalent Bonding The motion of electrons in a crystal can be visualized and described in a quasi-classical manner. In most instances ¾ The electron can be thought of as a particle. ¾ The electronic motion can be modeled using Newtonian mechanics. The effect of crystalline forces and quantum mechanical properties are incorporated into the effective mass factor. ¾ m* > 0 : near the bottoms of all bands ¾ m* < 0 : near the tops of all bands Carriers in a crystal with energies near the top or bottom of an energy band typically exhibit a constant (energy-independent) effective mass. ` d 2E 2 = constant : near band edge dk Covalent Bonding Band Occupation at Low Temperature Band Occupation at High Temperature Band Occupation at High Temperature Band Occupation at High Temperature Band Occupation at High Temperature Band Occupation at High Temperature Impurity Doping The need for more control over carrier concentration ¾ Without “help” the total number of “carriers” (electrons and holes) is limited to 2ni. ¾ For most materials, this is not that much, and leads to very high resistance and few useful applications. ¾ We need to add carriers by modifying the crystal. ¾ This process is known as “doping the crystal”. Regarding Doping, ... Concept of a Donor “Adding extra” Electrons Concept of a Donor “Adding extra” Electrons Concept of a Donor “Adding extra” Electrons Concept of a Donor “Adding extra” Electrons Band diagram equivalent view Concept of a Donor “Adding extra” Electrons Concept of a Donor “Adding extra” Electrons V(x), PE (x) V(x) n-type Impurity Doping of Si x Electron Energy Energy Band Diagram in an Applied Field PE (x) = – eV Electron Energy E CB e– As+ Ec ~0.05 eV E d As+ As+ Ec EF As+ As+ E c − eV E F − eV Ev Ev x Distance into crystal E v − eV A 6 As atom sites every 10 Si atoms The four valence electrons of As allow it to bond just like Si but the 5th electron is left orbiting the As site. The energy required to release to free fifth- electron into the CB is very small. Energy band diagram for an n-type Si doped with 1 ppm As. There are donor energy levels just below Ec around As+ sites. n-Type Semiconductor B ¾ Energy band diagram of an n-type semiconductor connected to a voltage supply of V volts. ¾ The whole energy diagram tilts because the electron now has an electrostatic potential energy as well. ¾ Current flowing V Concept of a Acceptor “Adding extra” Holes Hole Movement All regions of material are neutrally charged One less bond means the acceptor is electrically satisfied. One less bond means the neighboring Silicon is left with an empty state. Hole Movement Another valence electron can fill the empty state located next to the Acceptor leaving behind a positively charged “hole”. Empty state is located next to the Acceptor Hole Movement The positively charged “hole” can move throughout the crystal. (Really it is the valance electrons jumping from atom to atom that creates the hole motion) Hole Movement Hole Movement The positively charged “hole” can move throughout the crystal. The positively charged “hole” can move throughout the crystal. (Really it is the valance electrons jumping from atom to atom that creates the hole motion) (Really it is the valance electrons jumping from atom to atom that creates the hole motion) Hole Movement Concept of a Acceptor “Adding extra” Holes Band diagram equivalent view Region around the “acceptor” has one extra electron and thus is negatively charged. Region around the “hole” has one less electron and thus is positively charged. The positively charged “hole” can move throughout the crystal. (Really it is the valance electrons jumping from atom to atom that creates the hole motion) Concept of a Acceptor “Adding extra” Holes p-type Impurity Doping of Si Intrinsic, n-Type, p-Type Semiconductors Energy band diagrams Electron energy B atom sites every 106 Si atoms Ec x Distance into crystal h+ CB Ec EFi B– Ea Ev Boron doped Si crystal. B has only three valence electrons. When it substitute for a Si atom one of its bond has an electron missing and therefore a hole. B– B– h+ B– B– ~0.05 eV Ec Ev EFp Ev VB VB Energy band diagram for a p-type Si crystal doped with 1 ppm B. There are acceptor energy levels just above Ev around B- site. These acceptor levels accept electrons from the VB and therefore create holes in the VB. Impurity Doping Ev Ec EFn n-type semiconductors Intrinsic semiconductors ¾ In all cases, p-type semiconductors np=ni2 ¾ Note that donor and acceptor energy levels are not shown. Impurity Doping Valence Band Valence Band Impurity Doping Position of energy levels within the bandgap of Si for Energy Band Energy band diagrams. common dopants. ¾ Energy-band diagram for a semiconductor showing the lower edge of the conduction band Ec, a donor level Ed within the forbidden band gap, and Fermi level Ef, an acceptor level Ea, and the top edge of the valence band Ev. 3.2B Semiconductor Statistics Density of States Concept 3.2B Semiconductor Statistics Density of States Concept Quantum Mechanics tells us that the number of available states in a cm3 per unit of energy, the density of states, is given by: gc ( E ) dE gv ( E ) dE General energy dependence of gc (E) and gv (E) near the band edges. The number of conduction band states/cm3 lying in the energy range between E and E + dE (if E ≥ Ec). The number of valence band states/cm3 lying in the energy range between E and E + dE (if E ≤ Ev). Density of States in Conduction Band Density of States in Valence Band 3.2B Fermi- Dirac function How do electrons and holes populate the bands? Probability of Occupation (Fermi Function) Concept Probability of Occupation (Fermi Function) Concept ¾ Now that we know the number of available states at each energy, how do the electrons occupy these states? then ¾ We need to know how the electrons are “distributed in energy”. “Fer ¾ Again, Quantum Mechanics tells us that the electrons follow the mi-distribution function”. f (E) = 1 ( E−E f ) / kT 1+ e Ef ≡ Fermi energy (average energy in the crystal) k ≡ Boltzmann constant (k=8.617×10-5eV/K) T ≡Temperature in Kelvin (K) “The Fermi function f (E) is a probability distribution function that tells one the ratio of filled to total allowed states at a given energy E” f(E) is the probability that a state at energy E is occupied. 1-f(E) is the probability that a state at energy E is unoccupied. ¾ Fermi function applies only under equilibrium conditions, however, is universal in the sense that it applies with all materials-insulators, semiconductors, and metals. 3.2B Semiconductor Statistics Fermi Function Fermi-Dirac Distribution • Probability that an available state at energy E is occupied: f ( E) = 1 1 + e( E − EF ) / kT • EF is called the Fermi energy or the Fermi level Ef There is only one Fermi level in a system at equilibrium. If E >> EF : If E << EF : If E = EF : 3.2B Semiconductor Statistics Probability of Occupation (Fermi function) Concept Maxwell Boltzmann Distribution Function Boltzmann Approximation TYU If EF − E > 3kT , f ( E ) ≅ 1 − e( E − EF ) / kT • Assume the Fermi level is 0.30eV below the conduction band energy (a) determine the pro bability of a state being occupied by an electr on at E=Ec+KT at room temperature (300K). If E − EF > 3kT , f ( E) ≅ e − ( E − EF ) / kT Probability that a state is empty (occupied by a hole): ( E − EF ) / kT − ( EF − E ) / kT 1 − f ( E) ≅ e =e TYU • Determine the probability that an allowed ene rgy state is empty of electron if the state is be low the fermi level by (i) kT (ii) 3KT (iii) 6 KT How do electrons and holes populate the bands? Example 2.2 The probability that a state is filled at the conduction band edge (Ec) is precisely equal to the probability that a state is empty at the valence band edge (Ev). Where is the Fermi energy locate? Solution The Fermi function, f(E), specifies the probability of electron occupying states at a given energy E. The probability that a state is empty (not filled) at a given energy E is equal to 1- f(E). f ( EC ) = 1 − f ( EV ) f (EC ) = EC − EF kT How do electrons and holes populate the bands? Probability of Occupation Concept The density of electrons (or holes) occupying the states in energy between E and E + dE is: gc ( E ) f ( E ) dE Electrons/cm3 in the conduction band between E and E + dE (if E ≥ Ec). gv ( E ) f ( E ) dE Holes/cm3 in the conduction band between E and E + dE (if E ≤ Ev). 0 Otherwise 1 1 + e ( EC − E F ) / kT = EV − E F kT 1 − f ( EV ) = 1 − EF = 1 1 = 1 + e ( EV − E F ) / kT 1 + e ( E F − EV ) / kT EC + EV 2 How do electrons and holes populate the bands? Probability of Occupation Concept How do electrons and holes populate the bands? Developing the Mathematical Model for Electrons and Holes concentrations units of n and p are [ #/cm3] Typical band structures of Semiconductor E The Density of Electrons is: g (E) ∝ (E–Ec)1/2 E Ec+χ E Probability the state is filled [1–f(E)] CB For electrons Area = ∫ n E ( E ) dE = n Ec number of states per unit energy per unit volume EF Ec probability of occupancy of a state EF Ev Ev For holes nE(E) number of electrons per unit energy per unit volume The area under nE(E) vs. E is the electron concentration. pE(E) Area = p Number of states per cm-3 in energy range dE The Density of Hole is: VB Probability the state is empty 0 fE) nE(E) or pE(E) Fermi-Dirac probability function g(E) X f(E) Energy density of electrons in the CB g(E) Energy band diagram Density of states Number of states per cm-3 in energy range dE Electron Concentration (no) TYU Calculate the thermal equilibrium electron concen tration in Si at T=300K for the case when the F ermi level is 0.25eV below the conduction band . EC EF EV 0.25eV Hole Concentration (no) TYU • Calculate thermal equilibrium hole concentrati on in Si at T=300k for the case when the Fermi level is 0.20eV above the valance band energy Ev. EC EF EV 0.20eV Degenerate and Nondegenerate Semiconductors Developing the Mathematical Model for Electrons and Holes Useful approximations to the Fermi-Dirac integral: Nondegenerate Case n = NCe ( E f − EC ) kT p = NV e Semiconductor Statistics The intrinsic carrier concentration no = N C e (E f − EC ) kT po = NV e (EV − E f ) kT When n = ni, Ef = Ei (the intrinsic energy), then ni = N C e ( E i − EC ) kT or N C = ni e ( EC − E i ) kT and ( EV − Ei ) kT N = n e ( E i − EV ) kT or ni = NV e V i (EV − E f ) kT Other useful relationships: n⋅p product: ni = N C e ( E i − EC ) kT and ni = N V e ( EV − Ei ) kT ni = N C NV e − ( EC − EV ) kT = N C NV e 2 ni = N C NV e − E g 2 kT − E g kT TYU Determine the intrinsic carrier concentration in GaAs (a) at T=200k and (b) T=400K Law of mass action Law of mass Action Since (Ei − E f ) kT (E − E ) kT no = ni e f i and po = pi e nopo = ni Example An intrinsic Silicon wafer has 1x1010 cm-3 holes. When 1x1018 cm-3 donors are added, what is the new hole concentration? 2 It is one of the fundamental principles of semiconductors in thermal equilibrium nopo = ni 2 TYU if N D 〉〉 N A n ≅ ND and and N D 〉〉 ni p≅ ni2 ND TYU : The concentration of majority carrier electron is no=1 x 1015 cm-3 at 300K. D etermine the concentration of phosphorus th at are to be added and determine the concentr ation minority carriers holes. • Find the hole concentration at 300K, if the electron concentration is no=1 x 1015 cm-3, which carrier is majority carrier and which carrier is minority carrier? Energy band diagram showing negative charges Partial Ionization, Intrinsic Energy and Parameter Rel ationships. 3.5 Carrier concentration-effects of doping Charge Neutrality: 3.5 Developing the Mathematical Model for Electrons and Holes Charge Neutrality: Total Ionization case ¾ If excess charge existed within the semiconductor, random motion of charge would imply net (AC) current flow. ¨ Not possible! ¾ Thus, all charges within the semiconductor must cancel. [( p + N ) = (N + n )] q ⋅ [( p − N ) + (N − n )] + d Mobile - charge ¼ Immobile - charge ¼ − A ¾ NA¯ = Concentration of “ionized” acceptors = ~ NA ¾ ND+ = Concentration of “ionized” Donors ( p − N )+ (N − A − a Immobile + charge ¼ + d Mobile + charge ¼ Energy band diagram showing positive charges =0 + d = ~ ND ) −n =0 Electron concentration versus temperature for n-type Semiconductor. The intrinsic carrier concent ration as a function of temperature. Carrier Concentration vs. Temper ature position of Fermi Energy level no = N c e ( ) [ − Ec − E f ] kT Ec − EF = kT ln( Nc / no) Nd >> ni Ec − EF = kT ln( Nc / Nd ) Note: If we have a compensated semiconductor , then the Nd term in the above equation is simply replaced by Nd-Na. position of Fermi Energy level po = N V e (EV − E f ) kT position of Fermi level as a function of carrier concentration EF − Ev = kT ln( Nv / po) Na >> ni EF − Ev = kT ln( Nv / Na ) Note: If we have a compensated semiconductor , then the Na term in the above equation is simply replaced by Na-Nd. Where is Ei ? Extrinsic Material: TYU • Determine the position of the Fermi level with res pect to the valence band energy in p-type GaAs at T=300K. The doping concentration are Na=5 x 1 016 cm-3 and Na=4 x 1015 cm-3. Note: The Fermi-level is pictured here for 2 separate cases: acceptor and donor doped. position of Fermi Energy level Extrinsic Material: no = ni e (E f − E fi ) kT po = ni e (E fi − E f ) kT Solving for (Ef - Efi) p n E f − E fi = kT ln = − kT ln n ni i for N D 〉〉 N A and N D 〉〉 ni N E f − E fi = kT ln D ni for N A 〉〉 N D and N A 〉〉 ni N E f − E fi = −kT ln A ni TYU 3.8 • Calculate the position of the Fermi level in ntype Si at T=300K with respect to the intrinsi c Fermi energy level. The doping concentrati on are Nd=2 x 1017 cm-3 and Na=3 x 1016 cm-3 . Mobile Charge Carriers in Smiconductor devices • Three primary types of carrier action occur inside a semiconductor: – Drift: charged particle motion under the influence of an electric field. – Diffusion: particle motion due to concentration gradient or temperature gradient. EC EF – Recombination-generation (R-G) EFi EV Carrier Dynamics Direction of motion Carrier Drift Describe the mechanism of the carrier drift and drift current due to an applied electric field. ¾ Holes move in the direction of the electric field. (⊕F\) Carrier Motion Electron Drift Electron Diffusion Hole Drift Hole Diffusion ¾ Electrons move in the opposite direction of the electric field. (\F⊕) ¾ Motion is highly non-directional on a local scale, but has a net direction on a macroscopic scale. ¾ Average net motion is described by the drift velocity, vd [cm/sec]. ¾ Net motion of charged particles gives rise to a current. Instantaneous velocity is extremely fast Drift Drift of Carriers Drift Schematic path of an electron in a semiconductor. E Electric Field Drift of electron in a solid E The ball rolling down the smooth hill speeds up continuously, but the ball rolling down the stairs moves with a constant average velocity. Random thermal motion. Combined motion due to random thermal motion and an applied electric field. µ [cm2/Vsec] : mobility Drift Drift Conduction process in an n-type semiconductor Random thermal motion. Combined motion due to random thermal motion and an applied electric field. Thermal equilibrium Under a biasing condition Drift Drift Jdrf = vd J p drf = q ⋅ p ⋅ vd At Low Electric Field Values, Jp Drift = e⋅ p⋅µp ⋅ E and Jn Drift = e ⋅ n ⋅ µn ⋅ E Given current density J ( I = J x Area ) flowing in a semiconductor block with face area A under the influence of electric field E, ρ is volume density, the component of J due to drift of carriers is: J drf = ρvd J n drf = −e ⋅ n ⋅ vd and J p drf = e ⋅ p ⋅ vd Hole Drift Current Density Electron Drift Current Density Electron and Hole Mobilities µ has the dimensions of v/ : cm/s cm2 = V/cm V ⋅ s Electron and hole mobilities of selected intrinsic semiconductors (T=300K) µ n (cm /V·s) µ p (cm2/V·s) 2 Si Ge GaAs InAs 1350 3900 8500 30000 480 1900 400 500 ¾ µ [cm2/V·sec] is the “mobility” of the semiconductor and measures the ease with which carriers can move through the crystal. ¾ The drift velocity increases with increasing applied electric field.: Jdrf = J p Drift + Jn Drift = q ( µ p p + µ n n) ⋅ E EX 4.1 • Consider a GaAs sample at 300K with dopin g concentration of Na=0 and Nd=1016 cm-3. Assume electron and hole mobitities given in table 4.1. Calculate the drift current density if the applied electric filed is E=10V/cm. Mobility ¾ µ [cm2/Vsec] is the “mobility” of the semiconductor and measures the ease with which carriers can move through the crystal. µn ~ 1360 cm2/Vsec µp ~ 460 cm2/Vsec µn ~ 8000 cm2/Vsec µp ~ 400 cm2/Vsec µn, p = qτ m n* , p for Silicon @ 300K for Silicon @ 300K for GaAs @ 300K for GaAs @ 300K [ cm 2 V sec] ¾ <τ > is the average time between “particle” collisions in the semiconductor. ¾ Collisions can occur with lattice atoms, charged dopant atoms, or with other carriers. Saturation velocity Drift velocity vs. Electric field in Si. Saturation velocity Drift velocity vs. Electric field Designing devices to work at the peak results in faster operation 1/2mvth2=3/2kT=3/2(0.0259) =0.03885eV µn, p = qτ m n* , p [ cm 2 V sec] ¾ Ohm’s law is valid only in the low-field region where drift velocity is independent of the applied electric field strength. ¾ Saturation velocity is approximately equal to the thermal velocity (107 cm/s). Negative differential mobility Drift Electron distributions under various conditions of electric Drift velocity vs. Electric field in Si and GaAs. fields for a two-valley semiconductor. Note that for n-type GaAs, there is a region of negative differential mobility. µn, p = qτ m n* , p [ cm Negative differential mobility Velocity-Field characteristic of a Two-valley semiconductor. 2 V se m*n=0.55mo m*n=0.067mo TYU • Silicon at T=300K is doped with impurity concentration of Na=5 X 1016 cm-3 and Nd=2 x 1016 cm-3. (a) what are the electron and hole mobilities? (b) Determine the resistivity and conductivity of the material. Figure 3.24. Mean Free Path l = vthτ mp • Average distance traveled between collisions EX 4.2 Using figure 4.3 determine electron and hole nobilities. EX 4.2 Using figure 4.3 determine electron and hole mobilities in (a) Si for Nd=1017 cm-3, Na=5 x 1016 cm-3 and (b) GaAs for Na=Nd=1017cm-3 Ex 4.2 Mobility versus temperature Mobility versus temperature Effect of Temperature on Mobility Effect of Temperature on Mobility ¾ Electron mobility in silicon versus temperature for various donor concentrations. ¾ Since the slowing moving carrier is likely to be scattered more strongly by an interaction with charged ion. ¾ Impurity scattering events cause a decrease in mobility with decreasing temperature. ¾ Insert shows the theoretical temperature dependence of electron mobility. ¾ A carrier moving through the lattice encounters atoms which are out of their normal lattice positions due to the thermal vibrations. ¾ The frequency of such scattering increases as temperature increases. ¾ At low temperature, thermal motion of the carriers is slower, and ionized impurity scattering becomes dominant. At low temp. lattice scattering is less important. As doping concentration increase, impurity scattering increase, then mobility decrease. Temperature dependence of mobility with both lattice and impurity scattering. Effect of Doping concentration on Mobility Resistivity and Conductivity Ohms’ Law 300 K J =σ ⋅E = E [A cm ] 2 Ohms Law σ [1 ohm ⋅ cm ] Conductivity ρ ρ [ohm ⋅ cm ] ¾ Electron and hole mobilities in Silicon as functions of the total dopant concentration. Resistivity semiconductor conductivity and resistivity Adding the Electron and Hole Drift Currents (at low electric fields) Jdrf = J p Drift + Jn Drift = e( µ p p + µ n n ) ⋅ E Drift Current σ = e( µ p p + µ n n ) ρ= 1 σ [ = 1 e( µ n n + µ p p ) Conductivity ] Resistivity ¾ But since µn and µp change very little and n and p change several orders of magnitude: σ ≅ e µn n σ ≅ eµ p p for n-type with n>>p for p-type with p>>n µn, p = qτ m n* , p 2 V sec] Diffusion Diffusion Particles diffuse from regions of higher concentration to regions of lower concentration region, due to random thermal motion. [ cm Nature attempts to reduce concentration gradients to zero. Example: a bad odor in a room, a drop of ink in a cup of water. ¾ In semiconductors, this “flow of carriers” from one region of higher concentration to lower concentration results in a “Diffusion Current”. Jp Diffuse Diffusion Jn Diffusion Diffuse Visualization of electron and hole diffusion on a macroscopic scale. Diffusion Current J N,diff = eDN dn dx J P,diff = −eDP dp dx x x D is the diffusion constant, or diffusivity. Diffusion current density Total Current Fick’s law ¾ Diffusion as the flux, F, (of particles in our case) is proportional to the gradient in concentration. F = − D∇η η : Concentration D : Diffusion Coefficient ¾ For electrons and holes, the diffusion current density ( Flux of particles times ± q ) Jp Diffusion = − q ⋅ D p∇p Jn Diffusion = q ⋅ Dn∇n The opposite sign for electrons and holes J = JN + JP ε + qD JN = JN,drift + JN,diff = qnµn N ε – qD JP = JP,drift + JP,diff = qpµp P dn dx dp dx Total Current TYU Total Current = Drift Current + Diffusion Current Jp = Jp Drift + Jp Diffusion = q ⋅ µ p p E − q ⋅ D p∇p Jn = Jn Drift + Jn Diffusion = q ⋅ µ n n E + q ⋅ Dn∇n • Consider a sample of Si at T=300K. Assume that electron concentration varies linearly with distance, as shown in figure.The diffusion current density is found to be Jn=0.19 A/ cm2. If the electron diffusio n coefficient is Dn=25cm2/sec, determine the electr on concentration at x=0. J = J p + Jn J N,diff = eDN dn dx Jp=0.270 A/cm2 Dp=12 cm2/sec Find the hole concentration at x=50um J P,diff = −eDP dp dx Graded impurity distribution Energy band diagram of a semiconductor in thermal equilibrium with a nonuniform donor impurity concentration Generation and Recombination Carrier Generation Generation Mechanism Band-to-Band Generation Thermal Energy or Light Band-to-band generation Gno=Gpo ¾ Band-to-Band or “direct” (directly across the band) generation. ¾ Does not have to be a “direct bandgap” material. ¾ Mechanism that results in ni. ¾ Basis for light absorption devices such as semiconductor photodetectors, solar cells, etc… Recombination Mechanism Band-to-Band Recombination n = ∆n + n0 Photon (single particle of light) or Rno=Rpo multiple phonons (single quantum of lattice vibration - equivalent to saying thermal energy) ¾ Band to Band or “direct” (directly across the band) recombination. ¾ Does not have to be a “direct bandgap” material, but is typically very slow in “indirect bandgap” materials. ¾ Basis for light emission devices such as semiconductor Lasers, LEDs, etc… In thermal equilibrium: Gno=Gpo=Rno=Rpo Excess minority carrier lifetime Carrier Lifetime ∆p (t ) = ∆p0 e − t τ po ∆n(t ) = ∆n0 e − t τ no Schematic diagram of photoconductivity decay measurement. Light Pulses Rs VA Semiconductor Excess carrier Recombination and Generation + I RL VL _ Oscilloscope and p = ∆p + p0 In Non-equilibrium, n·p does not equal ni2 Low-Level-Injection implies ∆p << n0 , n ≈ n0 in a n-type material ∆n << p0 , p ≈ p0 in a p-type material Example low level injection case Nd=1014/cm3 doped Si at 300K subject to a perturbation where ∆p =∆n =109/cm3. Ä n0 ≅ Nd =1014/cm3 and p0 ≅ ni2/Nd = 106/cm3 Ä n = n0 + ∆n ≅ n0 and ∆p ≅ = 109/cm3 << n0 ≅ 1014/cm3 Although the majority carrier concentration remains essentially unperturbed under low-level injection, the minority carrier concentration can, and routinely does, increase by many orders of magnitude. Material Response to “Non-Equilibrium” Relaxation Concept ¾ Consider a case when the hole concentration in an n-type sample is not in equilibrium, i.e., n·p ≠ ni2 R' n = ∂p ∆p (t ) =− τ po ∂t τpo is the minority carrier lifetime ¾ The minority carrier lifetime is the average time a minority carrier can survive in a large ensemble of majority carriers. ¾ If ∆p is negative ¼ Generation or an increase in carriers with time. ¾ If ∆p is positive ¼ Recombination or a decrease in carriers with time. ¾ Either way the system “tries to reach equilibrium” ¾ The rate of relaxation depends on how far away from equilibrium we are. Material Response to “Non-Equilibrium” Relaxation Concept Generation and Recombination process Recombination-Generation center recombination ¾ Likewise when the electron concentration in an p-type sample is not in equilibrium, i.e., n·p does NOT equal ni2 ∂n ∂t Thermal =− R −G ∆n τn τn is the minority carrier lifetime Indirect recombinationgeneration processes at thermal equilibrium. Recombination Mechanism Generation and Recombination process Recombination-Generation (R-G) Center Recombination Energy loss can result in a Photon but is more often multiple phonons ¾ Also known as Shockley-Read-Hall (SRH) recombination. ¾ Two steps: 1 1st carrier is “trapped” (localized) at an defect/impurity (unintentional/intentional ). 2 2nd carrier (opposite type) is attracted and annihilates the 1st carrier. ¾ Useful for creating “fast switching” devices by quickly “killing off” EHP’s. Generation Mechanism Recombination-Generation (R-G) Center Generation Effects of recombination centers on solar cell performance b Thermal Energy ¾ Two steps: 1 A bonding electron is “trapped” (localized) at an unintentional defect/impurity generating a hole in the valence band. 2 This trapped electron is then promoted to the conduction band resulting in a new EHP. ¾ Almost always detrimental to electronic devices. AVOID IF POSSIBLE! The light-generated minority carrier can return to the ground state through recombination center before being collected by the junction: i) through path (a) ii) through path (c) Without recombination centers paths (b) and (d) are dominated Light d EC a c EV Auger Recombination 3.3 p-n Junction Diode Auger Recombination Auger – “pronounced O-jay” p-Type Material n-Type Material Requires 3 particles. p-n Junction ¾ Two steps: 1 1st carrier and 2nd carrier of same type collide instantly annihilating the electron hole pair (1st and 3rd carrier). The energy lost in the annihilation process is given to the 2nd carrier. 2 2nd carrier gives off a series of phonons until it’s energy returns to equilibrium energy (E~=Ec) This process is known as thermalization. p-n Junction principles p-n Junction p-Type Material n-Type Material p-n Junction p-Type Material n-Type Material ¾ A p-n junction diode is made by forming a p-type region of material directly next to a n-type region. p-n Junction Diode But when the device has no external applied forces, no current can flow. Thus, the Fermi-level must be flat! We can then fill in the junction region of the band diagram as: EC EC EF Ei Ei EF EV EV p-Type Material n-Type Material p-n Junction Diode p-n Junction Diode Built-in-potential But when the device has no external applied forces, no current can flow. Thus, the Fermi-level must be flat! p-Type Material We can then fill in the junction region of the band diagram as: n-Type Material EC - qVbi Ei EF EC EC Ei EF EF Ei EV EF Ei EV EV EV Electrostatic Potential p-Type Material V= 1 ( EC − E ref ) q n-Type Material VBI Built-in-potential p-n Junction Diode Built-in-potential Electrostatic Potential Electric Field dV dx x x Charge Density Electric Field E=− Electric Field E=− 1 ( EC − E ref ) q VBI Built-in-potential x p-n Junction Diode Built-in-potential V= EC dV dx Charge Density ρ = KS ⋅ε0 x qND + - dE dx + - qNA + x Built-In Potential Vbi TYU 5.1 qVbi = q(ΦS p−side + ΦS n −side ) = ( Ei − EF ) p−side + ( EF − Ei ) n −side • Calculate the built-in-potential barrier in a Si pm junction at T=300K for (a) Na=5 x 1017c m-3, Nd=1016cm-3 (b)Na=1015cm-3 For non-degenerately doped material: p ( Ei − EF ) p−side = kT ln ni N = kT ln A ni n ( EF − Ei )n−side = kT ln ni N = kT ln D ni p p-n Junction Diode n As + Bh+ e– Built-in-potential M Metallurgical Junction Electric Field dV E=− dx Neutral p-region M E (x ) Neutral n-region E0 -Wp 0 -Wn x –Eo x V (x ) M Wp lo g (n ), lo g (p ) Vo Space charge region Wn p po Charge Density ρ = KS ⋅ε0 + - + - qNA P E (x ) ni dE dx eV o pno n po Hole o Potential Energy PE (x) x x = 0 ρ net qND x nno Charge Density x M + x eN d Electron Potential Energy PE (x) – Wp x Wn -e N a –eV o p-n Junction Principles Movement of Electrons and Holes when Forming the Junction p-n Junction Depletion Region Approximation: Step Junction Solution Poisson’s Equation Charge Density (NOT Resistivity) Electric Field ∇⋅E = − ρ = q( p − n + N D − N A ) ρ ( x) = −qNA ρ ( x) = qND ρ dE ρ − qNA / D =− = K S ⋅ ε 0 dx KS ⋅ε0 KS ⋅ε0 in 1-dimension Relative Permittivity of Semiconductor (εr) Permittivity of free space Number of negative charges per unit area in the p region is equal to the number of positive charges per unit area in the n-region Electric potential V(x) 0r φ(x) Depletion Region Approximation: Step Junction Solution Space charge width(depletion layer width) Depletion Region Approximation: Step Junction Solution n-region space charge width p-region space charge width TYU5.2 • A silicon pn junction at T=300k with zero ap plied bias has doping concentration of Nd= 5 x 1016cm-3 and Na=5 x 1015 cm-3. Determine xn, xp, W, and |Ex(max)|, Vbi 0.718V pn junction reverse applied bias Depletion Region Approximation: Step Junction Solution pn junction reverse/forward applied bias Depletion Region Schematic representation of depletion layer width and energy band diagrams of a p-n junction under various biasing conditions. Thermal-equilibrium Forward-bias condition Diode under Forward Bias.mov Diode under no Bias.mov Reverse-bias condition Diode under Reverse Bias.mov pn junction forward bias applied bias Depletion Region Approximation: Step Junction Solution Thus, only the boundary conditions change resulting in direct replacement of Vbi with (Vbi-VA) with VA ≠ 0. pn junction reverse/forward applied Depletion Region Approximation: Step Junction Solution with VA ≠ 0 Consider a p+n junction (heavily doped p-side, lightly doped n side) Movement of Electrons and Holes when Forming the Junction Forward bias condition Movement of Electrons and Holes when Forming the Junction Reverse bias condition Space charge width and Electric field 2εs (Vbi + VR ) Nd 1 xp = e Na N N + a d 2εs (Vbi + VR ) N a xn = e Nd 1/ 2 1 Na + Nd 1/ 2 2εs (Vbi + VR ) N A + Nd W = xn + xp = e NaNd 1/ 2 p-n EX 5.3 Junction I-V Characteristics In Equilibrium (no bias) Total current balances due to the sum of the individual components no net current! • A Si pn junction at 300K is reverse bias at V R=8V, the doping concentration are Na=5 x 1015cm-3 and Nd= 5 x 1016 cm-3. Determin e xn, xp and W, repeat for VR=12V. Electron Diffusion Current Electron Drift Current Hole Drift Current Hole Diffusion Current Diode under no Bias.mov p-n p-n Junction I-V Characteristics Forward Bias (VA > 0) In Equilibrium (no bias) IN Total current balances due to the sum of the individual components p-Type Material q VBI EC ++ EV + + ++ + + + + + + + + + + + + + EC EF Ei EV p vs. E Jn = Jn Drift + Jn Diffusion = q ⋅ µ n nE + q ⋅ Dn∇n = 0 Jp = Jp + Jp Drift Diffusion = q ⋅ µ p pE + q ⋅ D p ∇p = 0 no net current! Electron Diffusion Current Electron Drift surmount potential barrier Current n vs. E n-Type Material E EF i Junction I-V Characteristics Lowering of potential hill by VA VA Hole Diffusion Current Current flow is dominated by majority carriers flowing across the junction and becoming minority carriers IP Current flow is proportional to e(Va/Vref) due to the exponential decay of carriers into the majority carrier bands Hole Drift Current I = IN + IP I Diode under Forward Bias m p-n p-n Junction I-V Characteristics Reverse Bias (VA < 0) Electron Drift Current Current flow is constant due to thermally generated carriers swept out by E fields in the depletion region Increase of potential hill by VA Junction I-V Characteristics Where does the Reverse Bias Current come from? ¾ Generation near the depletion region edges “replenishes” the current source. Electron Diffusion Current negligible due to large energy barrier Hole Diffusion Current negligible due to large energy barrier Current flow is dominated by minority carriers flowing across the junction and becoming majority carriers Hole Drift Current Diode under Reverse Bias.m p-n µ P-N Junction Diodes ¸ Junction I-V Characteristics Putting it all together Current Flowing through a Diode I-V Characteristics Quantitative Analysis (Math, math and more math) -I0 for Ideal diode Vref = kT/q p-n Quantitative Junction I-V Characteristics Diode Equation p-n Diode Solution Assumptions: qV I = I 0 exp ηkT 1) 2) 3) 4) 5) − 1 Steady state conditions Non- degenerate doping One- dimensional analysis Low- level injection No light (GL = 0) Current equations: J p = J p ( x )+ J n ( x ) η : Diode Ideality Factor Continuity Equations Steady state : n(x) is time invariant. Transient state : n(x) is time dependent. ∆x J p ( x − ∆x ) J p ( x) Area, A cm 2 x ∂n ∂F 1 ∂J =− = ∂t ∂ x q ∂x x + ∆x Continuity Equation F: Particle Flux J: Current Density dn J n = qµ n nE − qD n dx dp J p = qµ p pE − qD p dx Ways Carrier Concentrations can be Altered Continuity Equations Ways Carrier Concentrations can be Altered Continuity Equations ∂n ∂t ∂p ∂t ∂n ∂n = ∂t ∂t ∂p ∂p = ∂t ∂t + Drift + Drift ∂n ∂t ∂p ∂t + ∂n ∂n + All other processes ∂t Thermal R −G ∂t such as light ... + ∂p ∂p + ∂t Thermal R −G ∂t Diffusion Diffusion + Drift + Drift ∂n ∂t ∂p ∂t 1 ∂J Nx ∂J Ny ∂J Nz + + ∂y ∂z q ∂x = Diffusion Diffusion 1 = ∇ ⋅ JN q ∂J Py ∂J Pz 1 ∂J = − 1 ∇ ⋅ JP = − Px + + ∂y ∂z q ∂x q ∂n 1 ∂n ∂n = ∇ ⋅ JN + + All other processes q ∂t ∂t Thermal R −G ∂t such as light ... ∂p ∂p ∂p 1 = − ∇ ⋅ JP + + All other processes ∂t ∂t Thermal R −G ∂t such q as light ... All other processes such as light ... ¾ There must be spatial and time continuity in the carrier concentrations. Continuity Equations: Special Case known as “Minority Carrier Diffusion Equation” Simplifying Assumptions: Continuity Equations: Special Case known as “Minority Carrier Diffusion Equation” ¾ Because of (3) no electric field E = 0 2) We will only consider minority carriers. 5) Low-level injection conditions apply. 6) SRH recombination-generation is the main recombination-generation mechanism. Minority Carrier Diffusion Equation + Jn Diffusion 0 Diffusion = q ⋅ µ n nE + q ⋅ D n ∇n = q ⋅ D n ∇n ∂ 2 ( n0 + ∆n) ∂ 2n ∂ 2 ( ∆n ) 1 1 ∂J N ∇ ⋅ JN = = Dn 2 = Dn = Dn 2 q q ∂x ∂x ∂x ∂x 2 ¾ Because of (5) - low level injection 7) The only “other” mechanism is photogeneration. Continuity Equations Drift = Jn 3) Electric field is approximately zero in regions subject to analysis. 4) The minority carrier concentrations IN EQUILIBRIUM are not a function of position. 0 Jn = Jn 1) One dimensional case. We will use “x”. ∂n ∂t =− Reombination − Generation ∆n τn ¾ Finally ∂n ∂ ( n0 + ∆n) ∂ ( ∆n) = = ∂t ∂t ∂t ¾ Because of (7) - Photogeneration ∂n ∂t All other processes such as light ... = GL Continuity Equations: Special Case known as “Minority Carrier Diffusion Equation” Further simplifications (as needed): Continuity Equations ∂n ∂n = ∂t ∂t 0 + Drift Continuity Equations: Special Case known as “Minority Carrier Diffusion Equation” ∂n ∂t + Diffusion ∂n ∂t + Reombination − Generation ∂n ∂t ¾ Steady State … All other processes such as light ... ∂ ( ∆n p ) ∂t →0 ∂ ( ∆ pn ) →0 ∂t and ¾ No minority carrier diffusion gradient … DN ∂ ( ∆n p ) ∂t = DN ∂ 2 ( ∆n p ) ∂x 2 − ( ∆n p ) τn + GL ∂ 2 ( ∆n p ) ∂x 2 →0 ∂ ( ∆ pn ) ∂ ( ∆ pn ) ( ∆ pn ) = DP − + GL ∂t ∂x 2 τp 2 ∆n τn =0 and ¾ Consider a semi-infinite p-type silicon sample with NA=1015 cm-3 constantly illuminated by light absorbed in a very thin region of the material creating a steady state excess of 1013 cm-3 minority carriers (x=0). Semi-infinite sample Light absorbed in a thin skin. ∂t No excess carrier = DN DN ∂x 2 ∂ 2 ( ∆n p ) ∂x 2 = − ( ∆n p ) τn + GL 0generation ( ∆n p ) τn =0 Steady-state carrier injection from one side. Semiconductor x ∂ 2 ( ∆n p ) ∆p τp Solutions to the “Minority Carrier Diffusion Equation” ¾ What is the minority carrier distribution in the region x> 0 ? 0 − GL → 0 Solutions to the “Minority Carrier Diffusion Equation” ∂ ( ∆n p ) ∂ 2 ( ∆pn ) →0 ∂x 2 ¾ No light … Minority Carrier Diffusion Equations Steady state DP ¾ No SRH recombination-generation … − Light and Sample with thickness W Direct generation and recombination of electron-hole pairs: at thermal equilibrium under illumination. Solutions to the “Minority Carrier Diffusion Equation” Continue General Solution 0 ( − x LN ) ∆n p ( x ) = A ⋅ e + B ⋅ e ( + x LN ) where LN ≡ DN ⋅ τ N Surface recombination at x = 0. The minority carrier distribution near the surface i s affected by the surface recombination velocity. Solutions to the “Minority Carrier Diffusion Equation” ¾ Consider a p-type silicon sample with NA=1015 cm-3 and minority carrier lifetime τ =10 µsec constantly illuminated by light absorbed uniformly throughout the material creating an excess 1013 cm-3 minority carriers per second. The light has been on for a very long time. At time t=0, the light is shut off. ¾ What is the minority carrier distribution in for t < 0 ? ¾ LN is the “Diffusion length” the average distance a minority carrier can move before recombining with a majority carrier. ¾ Boundary Condition … ∆n p ( x = 0) = 1013 cm −3 = A + B ∆n p ( x = ∞ ) = 0 = A( 0) + Be (+ ∞ LN ) → B=0 ∆n p ( x ) = 1013 e (− x LN )cm − 3 Semiconductor Light absorbed uniformly Semiconductor x ∂ ( ∆n p ) ∂t 0 = DN Light Uniform 0 distribution ∂ 2 ( ∆n p ) ∂x 2 − ( ∆n p ) τn + GL ∆n p (all x , t < 0) = G L ⋅ τ n = 10 7 cm −3 Solutions to the “Minority Carrier Diffusion Equation” Quantitative Continue p-n Diode Solution Application of the Minority Carrier Diffusion Equation ¾ In the previous example: What is the minority carrier distribution in for t > 0 ? Quisineutral Region Semiconductor Light absorbed uniformly x Quisineutral Region Light minority carrier diffusion eq. ∂ ( ∆n p ) ∂t = DN ∂ 2 ( ∆n p ) ∂x 2 0 − ( ∆n p ) τn + GL 0 0 0 minority carrier diffusion eq. Since electric fields exist in the depletion region, the minority carrier diffusion equation does not apply here. 0 0 ∆n p ( t ) = [∆n p ( t = 0)] ⋅ e ( − t τ n ) ∆n p ( t ) = 10 7 e ( − t 1e − 5 ) Quasi - Fermi Levels Equilibrium n0 = ni e Non-Equilibrium ( E f − E i ) kT n = ni e ( FN − E i ) kT ( E i − E f ) kT p = ni e ( E i − FP ) kT p0 = ni e ¾ The Fermi level is meaningful only when the system is in thermal equilibrium. ¾ The non-equilibrium carrier concentration can be expressed by defining QuasiFermi levels Fn and Fp . Equilibrium Non-Equilibrium Quantitative p-n Diode Solution (At the depletion regions edge) Quisineutral Region Quisineutral Region quasi-Fermi levels formalism np = n i e ( F N − F P 2 ) kT Quantitative p-n Diode Solution Quisineutral Region Quantitative Quisineutral Region Quisineutral Region Quisineutral Region x”=0 dn J n = q µ n nE + D n dx 0 = qD n = qD n ( d n 0 + ∆n p ) dx dp J p = q µ p pE + D p dx 0 ? = qD p d∆n p = qD p dx d ( p 0 + ∆p n dx d∆p n dx ) p-n Diode Solution x’=0 Approach: ¾ Solve minority carrier diffusion equation in quasineutral regions. ¾ Determine minority carrier currents from continuity equation. ¾ Evaluate currents at the depletion region edges. ¾ Add these together and multiply by area to determine the total current through the device. ¾ Use translated axes, x t x’ and -x t x’’ in our solution. Quantitative p-n Diode Solution Quisineutral Region Quisineutral Region x”=0 x’=0 Holes on the n-side Quantitative Quisineutral Region p-n Diode Solution Quisineutral Region x”=0 x’=0 Holes on the n-side Quantitative p-n Diode Solution Quisineutral Region Quantitative p-n Diode Solution Quisineutral Region Thus, evaluating the current components at the depletion region edges, we have… x”=0 J = Jn (x”=0) +Jp (x’=0) = Jn (x’=0) +Jn (x”=0) = Jn (x’=0) +Jp (x’=0) x’=0 Similarly for electrons on the p-side… Ideal Diode Equation Shockley Equation Note: Vref from our previous qualitative analysis equation is the thermal voltage, kT/q Quantitative p-n Diode Solution Quisineutral Region Quisineutral Region x”=0 x’=0 Total on current is constant throughout the device. Thus, we can characterize the current flow components as… J Example A silicon pn junction at T=305K has the following parameters: NA = 5x1016 cm-3, ND = 1x1016 cm-3 , Dn = 25 cm2/sec, Dp = 10 cm2/sec, τn0 = 5x10-7 sec, and τp0 = 1x10-7 sec, ni305K = 1.5x1010 cm-3,. The cross-sectional area is A=10-3 cm2, and the forward-bias voltage is Va = 0.625 V. Calculate the (a) minority electron diffusion current at the space charge region. (b) minority hole diffusion current at the space charge edge. (c) total current in the pn junction diode. Solution Minority electron diffusion current density -xp xn Ln = Dnτ n Example Example A silicon pn junction at T=305K has the following parameters: NA = 5x1016 cm-3, ND = 1x1016 cm-3 , Dn = 25 cm2/sec, Dp = 10 cm2/sec, τn0 = 5x10-7 sec, and τp0 = 1x10-7 sec, ni305K = 1.5x1010 cm-3,. The cross-sectional area is A=10-3 cm2, and the forward-bias voltage is Va = 0.625 V. Calculate the (a) minority electron diffusion current at the space charge region. (b) minority hole diffusion current at the space charge edge. (c) total current in the pn junction diode. Solution Ln = Minority electron diffusion current density 2 Jn = Dnτ n Given figure is a dimensioned plot of the steady state carrier concentrations inside a pn junction diode maintained at room temperature. (a) Is the diode forward or reverse biased? Explain how you arrived at your answer. (b) Do low-level injection conditions prevail in the quasineutral regions of diode? Explain how you arrived at your answer. (c) Determine the applied voltage, VA. n or p (log scale) pp 25 (1.5 × 1010 ) 0.625 2 exp − 1 = 0.154 mA / cm 5 × 10 − 7 5 × 1016 0.0259 2 = (1.6 × 10 −19 ) 1015 1010 np 2 qD p nn 0 qVa D p ni 2 qVa exp exp −1 = q −1 L p kT τ p 0 N D kT pn 108 Minority hole diffusion current density Jp = nn 1017 2 qDn n p 0 qVa Dn ni qVa exp exp −1 = q −1 Ln kT τ n 0 N A kT The diode is forward biased. There is pile-up or minority carrier excess (∆np>0 and ∆pn>0 ) at the edges of the depletion region. 103 -xp 2.1x10-2cm 105 x xn 1.6x10-2cm 10 (1.5 × 1010 ) 0.625 2 ) exp − 1 = 1.09 mA / cm 1 × 10 − 7 1 × 1016 0.0259 2 = (1.6 × 10 − 19 Total current density= 1.24 mA/cm2 Total current = AxJ=10-3x1.24 =1.24 µA pn-junction diode structure used in the discussion of currents. The sketch shows the dimensions and the bias convention. The cross-sectional area A is assumed to be uniform. Hole current (solid line) and recombining electron current (dashed line) in the quasi-neutr al n-region of the long-base diode of Figure 5.5. The sum of the two currents J (dot-dash l ine) is constant. Hole density in the quasi-neutral n-region of an ideal short-base diode under forward bias of Va volts. The current components in the quasi-neutral regions of a long-base diode under moderate forward bias: J(1) injected minority-carrier current, J(2) majority-carrier current recombining with J(1), J(3) majority-carrier current injected across the junction. J(4) space-charge-region recombination current. Quantitative p-n Diode Solution J p-region SCL n-region J = J elec + J h ole T otal current M ajority carrier diffusion and drift current J h ole J elec (a) Transient increase of excess stored holes in a long-base ideal diode for a constant current drive applied at time zero with the diode initially unbiased. Note the constant gradient at x = xn as time increases from (1) through (5), which indicates a constant injected hole current. (Circuit shown in inset.) (b) Diode voltage VD versus time. M inority carrier diffusion current –W p Wn x ¾ The total current anywhere in the device is constant. ¾ Just outside the depletion region it is due to the diffusion of minority carriers. Current-Voltage Characteristics of a Typical Silicon Quantitative p-n Junction p-n Diode Solution Examples Diode in a circuit Quantitative Summary p-n Diode Solution Current flow in a pn junction diode ∇⋅E = ρ KS ⋅ε0 Poisson’s Equation Built-in-Potential (a) under equilibrium, both diffusion currents are cancelled by opposing drift currents. (b) under reverse bias, only a small number of carriers are available to diffuse across the junction (once within the junction they drift to the other side). With increasing reverse bias the reverse current increases due to tunneling and carrier multiplication. W = x p + xn = 2 K sε 0 ( N A + N D ) (Vbi − (V A )) q N AND Width of Depletion Region qVa J = J 0 exp −1 η kT (c) under forward bias, the drift current is slightly reduced but the diffusion current is greatly increased. (d) current-voltage characteristic qDn n p 0 qD p pn 0 Diode + J0 = L p Equation Ln EX 5.7 • A silicon pn junction dide at T=300K is forw ard biased. The reverse saturation current is IS=4x 10-14A. Determine the required dide v oltage to induce a diode current of ID=4.25m A.
