Chapter 4: Semiconductor Physics Crystal structures of solids Energy band structures of solids Charge carriers in semiconductors Carrier transport PHYS5320 Chapter Four (II) 1 Equilibrium Distribution of Electrons and Holes The densities of electrons and holes are related to the density density-of-states of states function and the Fermi distribution function. nE gc E fFE pE g v E 1 f FE The total electron/hole concentration per unit volume is found by integrating the corresponding function over the entire conduction/valence band energy. We need to determine the Fermi energy in order to find the thermalequilibrium electron and hole concentrations. We first W fi t consider id an intrinsic i t i i semiconductor. i d t An A id ideall intrinsic i ti i semiconductor is pure semiconductor without impurities or lattice defects. At T = 0 K,, all the energy gy states in the valence band are filled with electrons and all the energy states in the conduction band are empty. The Fermi energy must be somewhere between Ec and Ev. At T > 0 K, the number of electrons in the conduction band is equal to that of holes in the valence band. band PHYS5320 Chapter Four (II) 2 Equilibrium Distribution of Electrons and Holes gc E g v E 4 4 * 3/ 2 2mn h3 * 3/ 2 2mp h3 E E c Ev E If mn* = mp*, the Fermi energy is at th midgap the id energy. If mn* mp*, the th Fermi level for an intrinsic semiconductor will slightly shift away from the midgap energy. PHYS5320 Chapter Four (II) 3 The n0 and p0 Equations n0 gc E fFE dE d Because the Fermi distribution rapidly approaches zero with increasing energy, therefore n0 gc E fFE dE Ec If (EcEF) >> kT, then (EEF) >> kT, so that E EF 1 fFE E EF expp kT 1 exp kT n0 Ec If we let 4 * 3/ 2 2mn h3 E EF E Ec exp dE kT E Ec kT PHYS5320 Chapter Four (II) 4 Ec EF 1 / 2 n0 expp expp d 0 h3 kT The integral is a gamma function. Then 4 3/ 2 * 2mn kT 0 So 1/ 2 1 exp d 2 3/ 2 * 2mn kT n0 2 h2 We define a parameter Nc as 3/ 2 * 2mn kT Nc 2 Then Ec EF exp kT h 2 Nc is called the effective density-of-states function in the conduction band. Ec EF n0 Nc exp kT PHYS5320 Chapter Four (II) 5 Nc is called the effective density-of-states function in the conduction band. If we were to assume that mn* = m0, then at T = 300 K, 2 9.109 10 1.38110 Nc 2 34 2 6.626 10 F hholes: For l 31 23 300 3/ 2 2.5 1019 cm3 p0 g v E 1 fFE dE E EF exp 1 1 kT 1 f FE 1 E EF E EF EF E 1 exp 1 exp 1 exp kT kT kT If (EFEv) >> kT, kT then (EFE) >> kT. kT EF E 1 f FE exp kT PHYS5320 Chapter Four (II) 6 p0 Ev 4 h3 ' If we let Then p0 EF E Ev E exp dE kT * 3/ 2 2mp 4 3/ 2 * 2mpkT h3 Ev E kT EF Ev 0 1/ 2 exp ' exp ' d' kT EF Ev exp kT then 3/ 2 * 2mpkT p0 2 If we let h2 3/ 2 * 2mpkT EF Ev Nv 2 p0 N v exp 2 kT h Nv is i called ll d the h effective ff i density-of-states d i f f function i in i the h valence l band. b d PHYS5320 Chapter Four (II) 7 The Intrinsic Carrier Concentration We use ni and pi to denote electron and hole concentrations in an intrinsic semiconductor. Since ni = pi, we usually use ni to denote either the intrinsic electron or hole concentration. The Fermi level for an intrinsic semiconductor i called is ll d the th intrinsic i t i i Fermi F i energy, EFi. Ec EFi n0 ni Nc expp kT EFi Ev p0 pi ni N vexp kT Ec EFi EFi Ev 2 ni Nc N v exp exp kT kT ni2 Ec Ev Eg Nc N v exp Nc N v exp kT kT PHYS5320 Chapter Four (II) 8 At T = 300 K Nc, Nv, and ni are constant for a given semiconductor material at a given temperature. PHYS5320 Chapter Four (II) 9 3/ 2 * 2mn kT Nc 2 h 2 3/ 2 * 2mpkT Nv 2 ni2 h2 Eg Nc N v exp kT PHYS5320 Chapter Four (II) 10 The Intrinsic Fermi Level Position We can calculate the Fermi level pposition since the electron and hole concentrations are equal for an intrinsic semiconductor: Ec EFi EFi Ev n0 Nc exp p0 N v exp kT kT Nv Ec E Fi EFi Ev kT ln N c Nv EFi Ec Ev kT ln N 2 2 c 1 1 mp* EFi Emidgap kT ln m* 4 n 3 mp* EFi Emidgap kT ln m* 4 n 3 If mp* = mn*, the intrinsic Fermi level will be in the center of the bandgap. If mp* > mn*, the intrinsic Fermi level will be slightly above the center. If mp* < mn*, the th intrinsic i t i i Fermi F i level l l will be slightly below the center of the bandgap. bandgap PHYS5320 Chapter Four (II) 11 Donor Atoms and Energy Levels Real ppower of semiconductors is realized byy addingg controlled amounts of specific dopant, or impurity atoms. The doped semiconductor is called an extrinsic material. Doping is the primary reason that we can fabricate various semiconductor i d t devices. d i Add a ggroup p V element,, such as phosphorus, p p , to silicon as a substitutional impurity. The group V element has five valence electrons. Four of these will contribute to the covalent bonding with the silicon atoms, leaving the fifth more loosely bound to the phosphorus atom atom. PHYS5320 Chapter Four (II) 12 Donor Atoms and Energy Levels At very low temperatures temperatures, the extra electron is bound to the phosphorus atom. However, it should be clear that the energy required to elevate the extra electron into the conduction band is considerably smaller than that for the electrons involved in the covalent bonding. bonding The electron elevated into the conduction band can move through the crystal to generate a current, while the positively charged phosphorous atoms are fixed in the crystal. l This hi type off atom is i called ll d a donor impurity atom. The donor atoms add electrons to the conduction band without creating holes in the valence band. The resulting material is referred to as an n-type semiconductor. i d t PHYS5320 Chapter Four (II) 13 Acceptor Atoms and Energy Levels Consider adding a group III element, such as boron, which has three valence electrons. One covalent bonding position is empty. If an electron were to occupy this “empty” position, its energy would have to be greater than that of th valence the l electrons, l t since i the th nett charge h off the th B atom t would ld become b negative. However, the electron occupying this “empty” position does not have sufficient energy gy to be in the conduction band,, so its energy gy is far smaller than the conduction band energy. The “empty” position associated with the B atom can be occupied and other valence electron positions become vacated These other vacated electron positions can be thought of as holes. vacated. holes PHYS5320 Chapter Four (II) 14 Acceptor Atoms and Energy Levels The hole can move through the crystal to generate a current, while the negatively charged h d boron b atoms are fixed fi d in i the h crystal. l The group III atom accepts an electron from the valence band and so is referred to as an acceptor impurity atom. The acceptor atom can generate holes in the valence band without ith t generating ti electrons l t in i the th conduction band. This type of semiconductor material is referred to as a p-type p yp semiconductor. An extrinsic semiconductor will have either a preponderance of electrons (n-type) or a preponderance of holes (p-type). PHYS5320 Chapter Four (II) 15 Ionization Energy We can calculate the approximate distance of the donor electron from the donor impurity ion, and also the approximate energy required to elevate the donor electron into the conduction band. This energy is referred to as the ionization energy. We will use the Bohr model of the atom for these calculations. The permittivity Th i i i off the h semiconductor i d material i l instead i d off the h permittivity i i i off free space will be used, and the effective mass of the electron will be used. From the F h Coulomb C l b force f off attraction i being b i equall to the h centripetal i l force of the orbiting electron: e 2 * 2 mv 2 4rn rn Assume that the angular momentum is quantized, PHYS5320 Chapter Four (II) m vrn n * 16 n v * m rn The Bohr radius is n = 1,2,3, n 2 2 4 rn m *e 2 4 0 2 0.053nm a0 2 m0 e m0 n r m* a0 rn Then 2 If we consider n = 1 state, and if we consider silicon, for which r = 11.7, m*/m / 0 = 0.26, then h we hhave r1 45 a0 r1 2.39nm For silicon,, a = 0.543 nm. PHYS5320 Chapter Four (II) 17 e2 v 4n * 4 1 m e * 2 The kinetic energy is T mv 2 2 2 2n 4 2 * 4 e m e The p potential energy gy is V 4rn n 2 4 2 m *e 4 The total energy is E T V 2 2 2n 4 The lowest energy state of the hydrogen atom is E = 13.6 eV. For silicon,, the lowest energy is E = 25.8 meV (Eg = 1.12 eV at T = 300 K). PHYS5320 Chapter Four (II) 18 Quantum control of single atoms: t the th groundd state t t off the donor is within 0.1 nm and non-interacting, g, but the excited states expand to several nanometers and strongly interacting interacting. Solid-state implementation of coherent control of impurity wavefunctions Nature 2010, 465, 1057. PHYS5320 Chapter Four (II) 19 Group III III--V Semiconductors The donor and acceptor impurities in III III-V V compound semiconductors is more complicated than those in Si. When we talk about donors or acceptors in III-V semiconductors, we need to know for which atoms (III or V) impurity atoms are substituted. b i d For example, l for f Si atoms in i GaAs G A semiconductor, i d if Si atoms replace Ga atoms, Si impurities will act as donors. But if Si atoms replace As atoms, they will act as acceptors. For GaAs PHYS5320 Chapter Four (II) 20 Equilibrium Distribution of Electrons and Holes in Extrinsic Semiconductors Adding donor or acceptor impurity atoms to a semiconductor will change the distribution of electrons and holes in the material. Since the Fermi energy is related l t d to t the th distribution di t ib ti function, f ti the th Fermi energy will change as dopant atoms are added. In general, general when EF > Emidgap, the density of electrons is larger than that of holes, and the semiconductor is n-type. PHYS5320 Chapter Four (II) 21 Equilibrium Distribution of Electrons and Holes in the Extrinsic Semiconductor In general, when EF < Emidgap, the density of electrons is smaller than that of holes, holes and the semiconductor is p-type. Ec EF n0 Nc exp kT EF Ev p0 N v exp kT The above are ggeneral equations q for n0 and p0 in terms of the Fermi energy. The values of n0 and p0 will change with th Fermi the F i energy, EF. PHYS5320 Chapter Four (II) 22 Example: consider silicon at T = 300 K so that Nc = 2.8 1019 cm3 and Nv = 1.04 1019 cm3. If we assume the Fermi energy is 0.25 eV below the conduction band, calculate the thermal equilibrium concentrations of electrons and holes. The bandgap energy of silicon is 1.12 eV. Solution: Ec EF 0.25eV EF Ev 0.87eV 19 0 . 25 1 . 602 10 n0 2.8 1019 exp 1.8 1015cm3 1.38110 23 300 19 0 . 87 1 . 602 10 p0 1.04 1019 exp 2.6 104 cm3 1.38110 23 300 Comment: electron and hole concentrations change g by y orders of magnitude g from the intrinsic carrier concentrations (at 300 K, ni = 1.5 1010 cm3) as the Fermi energy changes by a few tenths of an eV. In an n-type semiconductor, i d n0 > p0, electrons l are referred f d to as majority carriers and holes as minority carriers. In a p-type semiconductor, p0 > n0, holes are referred to as majority carriers and electrons as minority carriers. PHYS5320 Chapter Four (II) 23 We can derive another form of the equations for the thermalequilibrium concentrations of electrons and holes: Ec EF Ec EFi EF EFi n0 Nc exp Nc exp k kT k kT EF EFi n0 ni exp kT EF Ev EFi Ev EFi EF p0 N v exp N v exp kT kT EFi EF EF EFi p0 ni exp ni exp kT kT PHYS5320 Chapter Four (II) 24 The n0p0 Product Ec EF EF Ev n0 p0 Nc N v exp exp kT kT Eg 2 n0 p0 Nc N v exp ni kT The product of n0 and p0 is always a constant for a given semiconductor material at a given temperature. It is one of the fundamental principles of semiconductors in thermal equilibrium equilibrium. It is important to keep in mind that the above equation is derived using the Boltzmann approximation. approximation We may think of the intrinsic concentration ni simply as a parameter of the semiconductor material. material PHYS5320 Chapter Four (II) 25 The Fermi Fermi--Dirac Integral n0 gc E fFE dE Ec n0 4 h If we define * 3/ 2 2mn Ec 3 E Ec F1 / 2 F 0 3/ 2 * 2mn kT h 2 F and kT n0 4 E Ec dE E EF 1 exp kT EF E c kT 1 / 2d 0 1 exp F 1 / 2d 1 exp F PHYS5320 Chapter Four (II) Fermi-Dirac integral 26 For holes: p0 g v E 1 fFE dE Ev p0 4 * 3 / 2 Ev 2mp 3 h If we define Ev E ' kT We need W d tto use th the F Fermii Dirac integral when EF is above Ec or below Ev. p0 4 Ev E dE EF E 1 exp kT and 'F 3/ 2 * 2mpkT h PHYS5320 Chapter Four (II) 2 0 Ev E F kT ' 1 / 2 d' 1 exp p' 'F 27 Degenerate and Nondegenerate Semiconductors When discussing donors and acceptors, the concentration of dopant atoms is assumed to be small compared to the density of host atoms. Impurities introduce discrete, non-interacting donor and acceptor energy states in an ntype andd p-type semiconductor, i d respectively. i l These Th types off semiconductors i d are referred to as nondegenerate semiconductors. As the A h iimpurity i concentration i is i increased, i d the h distance di between b i impurity i atoms decreases and the electrons from impurity atoms will begin to interact. When this occurs,, the discrete donor or acceptor p energy gy level will combine into a band of energies. As the impurity concentration is further increased, the band of donor or acceptor states widens and overlaps with the conduction b d bottom band b tt or the th valence l band b d top. t When Wh the th concentration t ti off electrons l t in i the conduction band exceeds the density of states Nc, the Fermi energy lies within the conduction band. This type yp of semiconductors is called degenerate g n-type semiconductors. When the concentration of holes exceeds the density of states Nv, the Fermi energy lies in the valence band. This type of semiconductors is called degenerate p p-type type semiconductors. semiconductors PHYS5320 Chapter Four (II) 28 Degenerate and Nondegenerate Semiconductors Nondegenerate Degenerate PHYS5320 Chapter Four (II) 29 Statistics of Donors and Acceptors The probability of the donor energy level being occupied is 1 1 Ed EF 1 exp kT 2 Each donor level can be empty, contain one electron of either spin, or two electrons of opposite spins. However, the C l b repulsion Coulomb l i off two t localized l li d electrons l t raises i the th energy of the doubly occupied level so high that double occupation is essentially prohibited. This is the reason for the factor of ½ appearing in the probability function. Nd nd 1 Ed EF 1 exp kT 2 nd Nd Nd Na pa 1 EF Ea 1 exp kT g pa Na Na nd is the density of electrons occupying the donor level. Nd is the concentration of donor atoms. Nd+ is the concentration of ionized donors. donors g is a degeneracy factor. The ground state g is normally taken as 4 for the acceptor level in Si and GaAs because of the detailed band structure. PHYS5320 Chapter Four (II) 30 Complete Ionization and Freeze Freeze--Out If ((Ed EF) >> kT,, we then have Nd Ed EF 2Nd exp nd 1 Ed EF kT 1 exp kT 2 Because Ec EF n0 Nc exp kT We can determine the percentage off electrons l in the donor state compared with the total number of electrons: Ed EF 2Nd exp nd kT nd n0 2N exp Ed EF N exp Ec EF d c kT kT nd 1 nd n0 1 Nc exp Ec Ed 2Nd kT PHYS5320 Chapter Four (II) 31 For phosphorus-doped silicon at T = 300 K, Nc = 2.8 1019 cm3, Nd = 1016 cm3, and the ionization energy is 0.045 eV: 1 nd 0.004 0.4% 19 nd n0 1 2.8 10 exp 0.045 16 0.0259 2 10 At room temperature, the donor states are almost completely ionized, which is also true for the acceptor states at room temperature. C Complete l t ionization i i ti PHYS5320 Chapter Four (II) 32 At T = 0 K, all electrons are in their lowest energy state. For an n-type semiconductor, each donor state must contain an electron, therefore, nd = Nd. expEd EF / kT 0 EF Ed T 0 The Fermi level is above the donor level for an n-type semiconductor at T =0K K. Similarly, Similarly the Fermi level will be below the acceptor level for a ptype semiconductor at T = 0 K. Freeze-out (T = 0 K) PHYS5320 Chapter Four (II) 33
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