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Find the temperature dependences of the chemical potential and hole con- centration of p-semiconductors. Answers: Hv = {€A + ] kgT1n94N P= NAN, 9A e-EX/2 kg T T <T, 10 10 10' p (cm) 10 p-Si: B 1013 10 1.2 E 1.0 0.8 p-Si: B MeV) 0.6 0.4 0.2 0.0 10 1000 100 T (K) Figure 4: Temperature dependences of the hole concentration and chemical po- tential for p-silicon (NA = 1015 cm-3, A = 45.6 meV; the temperature dependence of the energy gap E,(T) = E,(0) - T2/(T + 8), Eg(0) = 1.17 eV, a = 4.73. 10-4 eV/K, 8 = 636 K is taken into account)
Semiconductors Si (silicon) discovered in 1824 by J. J. Berzelius (Stockholm, Sweden) Ge (germanium) discovered in 1886 by C. A. Winkler (Freiberg, Germany) 2022) AIN Semiconductors Eg (eV) GaN Sic A A B AB ZnS InN 3. 3.5 3.3 a (A) violet (0.4) -0.4 ZnSe AIP Zn Te AIAS Cas minimal energy bandgap (eV) Gar blue (0.45) 0.5 green (0.55) 0.6 red (-0,7) -0.7 0.8 0.9 CdTe AISb wavelength (um) GaAs AWOOO ū ir MP GaSb fiber-optics (1.3-1.55) - 1.5 T= 300 K InSb InAs 0 5.4 5.6 5.8 6.0 6.4 Hg8.6.2 HgTe 6.6 lattice constant Semiconductors as semiconductors • Energy gap P semiconductor - 10-32cm - 10 cm RT: p.Ge - 104 cm, pSi~ 107 cm, p GaAs - 1010 12cm Oo = f(T) 1833 Faraday (Ag,S, Cu,0) O Semiconductors are made to be semiconductors not by nature,
Shallow impurities electron Si Si J. M. Luttinger and W. Kohn, Phys. Rev. 97, 869 (1955) Effective mass approximation: Si 72 DAS Si Si A- 2 W = EW KO! e? → e/k, mm €в me" n EB = ag = me? Donors Si EM 2nK7 theory Ge Si Se S Sn 5.67 6.1 5.8 5.9 GaAs: Ep (meV) 6.1 Si 5.9 Ge GaAs InSb Ko 11.5 16 12.8 18 theory P As Sb ER - 10 meV 4g 50 Å 9.81 12.9 14.2 10.3 Ge: Ep (meV) Si: &p (me V) 31.3 45.5 53.7 42.7 hole Si SiO Si Si CB Acceptors mass anisotropy [W. Kohn and J. M. Luttinger, Phys. Rev. 98, 915 (1955)] 10 ED Ind Si Si theory B ΑΙ Ga In 11 11.6 Si Ge: € (me V) 9.73 Si: &A (meV) 31.56 Luttinger hamiltonian 10.5 10.8 45.668.5 72 155 VB impurity potential Heavily doped semiconductors &B Nag-1 1/02 - 108 cm -13 imp 15 40 As in Ge 30 101 ionization energy (meV) ionization energy (meV) 20 B in Si 10 [P. Debye and E.M. Conwell, Phys. Rev. 93,693 (1954)] (T.E. Rosenbaum et al. Phys. Rev. Lett. 45,1723 (1980) 109 TOK 102 Si: P 10" 10" 10" 101 impurity concentration (cm) 102 10 10 10 10 10 impurity concentration (cm) 10' o('cm') The dielectric metal transition can occur before an impurity level is pushed into CB insulator metal 10' Physics of non-crystalline solids: N. Mott, Conduction in Non-Crystalline Materials (Oxford: Clarendon Press, 1993) .B. 1. Shklovskii and A. L. Efros, Electronic Properties of Doped Semiconductors (Springer, 1984) I 6 10 donor concentration (10'cm') Deep impurities CB 1.47 Ge S Se Si Sn
Electron distribution on impurities CB 0 He ED fp = 2 (-)/kg) FI 44p)/kg7 +1 VB Fermi-Dirac distribution describes non-interacting particles Gibbs distribution number of particles p=-e Z Lewř=Eyky7 z = { (UN-E)/37 e for Tele-p)/kg +1 np = NSD spinn E degeneracy factor g = 2 (GaAs), 12 (Si), 8 (Ge) UN - E P 0 0 1/Z 1 -- 个 1 PA=NASA -ep-Me + ED 1/2 exp(-(H. - Ep/k37) el 4-5x)/k57 +1 1 -&p - He+Ep 1/Zexp(-(M. - Ep/kp7) 14 2 8 00 0 DOS 3/2 P(E)= 41 (2m, j2 (27th) JE Nvalley density of states mass GaAs 1 2 1/3 Si 6 ma = N2/3 Valny (m?m, y3 , Ge 4 312 min 3/ 2 = 12hh 3/2 ma Charge carrier concentration in semiconductors (2m.)2 1 p(E)= 41 JE f =- (21h) E-u expl +1 kuT Electron concentration
3/2 Charge carrier concentration in semiconductors (2m.)2 1 P(E)= 41 JE (21) E-1 expl kuT +1 Electron concentration E CB f =e-Helkp?e-E/kg7 0 He 3/2 n= N e Melkg7 N =21 m,k3T 21 ħ? VB Hole concentration 1 CB E- expl +1 КТ 3/2 0 p=0.0 mila , = 24.7" /kp7 =2 mik,T ħ VB E
Law of mass action T (K) 500 400 1000 700 300 250 1019 10' p-Ge On-Ge np = N N e-/kp7 -0. 1017 10' Instrinsic semiconductors (np)" (cm) 1015 n=p charge neutrality condition 1014 103 -E n = p = = NN 15, 2kg7 ✓ 102 0.0008 0.0016 0.0040 0.0024 0.0032 1/T (K) 11 10" n-semiconductors &= 14.2 meV (AS) N = 10cm 10" Ge n (cm) = n+n) = p+N charge neutrality condition 7315 10 7-323F & 12 1013 0 0.02 0.06 Thermal activation from impurities 0.01 1/T (K) INN H. = } Ep + kgTin 18N No n= ep/237 8C Saturation ED kpT = N In 4g N N H = kgT In n=1 N D E kpT, = 2.N.N. In Intrinsic regime N 4. = * E; +į kpT in N ŁE, ND n= = VN.N, e»5./2697 e 12
Experimental data: n(T) 10" b-oo- 10' > 2 10" 100 107 n (cm) 6 10" SOC 16 5.5 10" Ge Ge 10" 10$ 00-00-0-000 10' 10? 7.5 105 1.7 103 0.04 0.08 0.12 n (cm) 0-0-0-0-0-0- 10' 10' 9.4 10' 10' GaAs 10' No=1 10" 13 cm O 0-0 0-0-0-0 (e-uɔ) u 5 10' 10 No=3.2 10 cm N = 0.9 10cm 10? 10" 10'? 0 0.10 0 0.05 0.15 0.20 0.02 0.04 0.06 0.08 1/T (K) 0.10 1/T (K) 13
CB Compensation O OO ОФо - VB 10' 0000-0-0-0 104) GaAs 5 10 N = 3.2 109 cm N=0.9 10cm 109 10" 0 0.05 0.10 0.15 0.20 1/T (K) Resumé NB • Electron concentration: n= N exp(-4./kpl) Semiconductors Impurities Shallow impurities Electron distribution on impurities Heavily doped semiconductors Charge carrier concentration Law of mass action Intrinsic semiconductors Doped semiconductors: •n-semiconductors . p-semiconductors • compensation 3/2 Ne =21 mkgT 27h • Bohr energy and radius: me" EB = 2nK a = DPK me? Problems: "p-puslaidininkiai" (0.75pt)