MTLE-6120: Advanced Electronic Properties of Materials. Intrinsic and extrinsic semiconductors. Reading: Kasap:

Transcription

1 MTLE-6120: Advanced Electronic Properties of Materials 1 Intrinsic and extrinsic semiconductors Reading: Kasap:

2 Band structure and conduction 2 Metals: partially filled band(s) i.e. bands cross Fermi level Semiconductors / insulators: each band either filled or empty (T = 0) Drude formula applicable, mobility µ = eτ m Effective mass m = 2 [ k k E n ( k)] 1 tensorial in general Filled band does not conduct: eτ dk(m ) 1 = 0 for each band Metals conduct due to carriers near Fermi level σ = g(e F )e 2 v 2 F τ/3 Semiconductors: g(e F ) = 0 (will show shortly) no conduction at T = 0

3 Band structure of silicon (diamond-cubic semiconductor) 6 3 E in ev 0 6 E c E v 12 L Λ Γ Δ Χ Σ Γ HOMO = Valence Band Maximum (VBM) with energy E v and LUMO = Conduction Band Minimum (CBM) with energy E c HOMO-LUMO gap E g = E c E v 1.1 ev HOMO and LUMO at different k indirect band gap Diamond: similar band structure, much larger gap ( 5.5 ev) insulator Valence electrons/cell = 8 (even), configuration: 3s 2 3p 2 (two Si/cell)

4 Band structure of GaAs (zinc-blende semiconductor) 4 HOMO-LUMO gap E g = E c E v 1.4 ev HOMO and LUMO at same k (Γ) direct band gap Valence electrons/cell = 8 (even), configuration: Ga(4s 2 4p 1 ), As(4s 2 4p 3 )

5 Density of states: silicon 5 Can calculate numerically from band structure Parabolic band approximation valid for narrow energy range near gap

6 Density of states: parabolic-band semiconductor 6 4 g(e) [ev -1 nm -3 ] Valence band, m * = -0.3 Conduction band, m * = E [ev] Parabolic bands near each band edge, with different effective masses Overall DOS reduces with reduced effective mass magnitude Set E v = 0 conventionally (overall energy not well-defined) Conduction band edge E c = E g Where is the Fermi level?

7 Where is the Fermi level? 7 At T = 0, valence band fully occupied f(e v = 0) = 1 E F > 0 At T = 0, conduction band fully empty f(e c = E g ) = 0 E F < E g Therefore, at T = 0, 0 < E F < E g i.e. Fermi level is in the band gap Chemical potential µ E F as T 0 In semiconductor physics, typically refer to E F (T ) instead of µ(t ) Therefore, Fermi functions will be f(e, T ) = 1 exp E E F (T ) + 1

8 Where is the Fermi level at T > 0? 8 Given Fermi level E F and density of states g(e) Number of electrons in conduction band is N e = E g deg(e)f(e) Number of holes in valence band is N h = 0 deg(e)(1 f(e)) Total number of electrons cannot change with T N e = N h deg(e)(1 f(e)) = deg(e) exp E E F exp E E F + 1 = deg(e)e (E F E)/( ) e E F /( ) ε dεg( ε)e 0 } {{ } N v E g E g deg(e)f(e) deg(e) 1 exp E E F + 1 E g deg(e)e (E E F )( ) e (E F E g)/( ) dεg(e g + ε)e 0 } {{ } N c Assuming E F, E g E F, ε energy from band edge ε

9 Band edge effective density of states Given density of states as a function of energy away from band edge 9 N c/v 0 dεg c/v (ε)e ε ( 2m ) 3 In parabolic band approximation g(ε) = 2π 4π ε for both bands (but with different m ; for tensor m, above defines DOS m eff ) Therefore band-edge effective of density states: N c/v 0 3 2m c/v dε 4π εe 2π ε 3 2m c/v 2πm c/v = 4πΓ(3/2)( ) 3/2 = 2 2π 2π N c/v (m c/v )3/2 (steeper g(ε) parabola) N c/v T 3/2 (climb higher up the g(ε) parabola) 3

10 Fermi level for T > 0 10 Charge neutrality imposes N v e E F /( ) = N c e (E F E g)/( ) Solve for Fermi level position: E F (T ) = E g 2 + k BT ln N v 2 N c At T 0, E F is exactly at the middle of the band gap At finite T, E F moves away E g (still close to gap center) Which way does the Fermi level move with increasing T? For electrons in metals (and classical gases), µ with T For semiconductors, E F (T ) with T iff N c > N v (more DOS in positive m band; negative m pulls µ other way)

11 Electron and hole concentrations 11 Number density of electrons n N e = N c e (Eg E F )/( ) Number density of holes p N h = N v e E F /( ) Which one is larger? So far, they are equal: charge neutrality! Note product np = N c N v e Eg/(k BT ) n 2 i, independent of E F Neutral pure semiconductor, n = p = n i, intrinsic carrier density If E F, then n and p (more electrons than holes) If E F, then n and p (more holes than electrons) But np = n 2 i, constant in all these cases This is an equilibrium constant, eg. [H + ][OH ] = M 2 in water How do you change E F? Doping! (Also later, gating)

12 Intrinsic semiconductor thermodynamics and transport 12 Fermi level far from band edges Boltzmann statistics in both bands Velocity distribution: Maxwell-Boltzmann distribution (classical gases) Internal energy of electrons n (E g + 3/2) Internal energy of holes p ( 3/2) (holes are missing electrons!) Net internal energy n (E g + 3/2) p ( 3/2) Drude theory conductivity σ = neµ e + peµ h

13 Intrinsic semiconductors: typical values at T = 300 K 13 E g [ev] m e/m e m h /m e N c [cm -3 ] N v [cm -3 ] n i [cm -3 ] Ge , , Si , , GaAs Note that m eff for N c/v is an average of longitudinal / transverse values (m eff = m 1/3 L m2/3 T ; for values see Table 5.1 in Kasap) N c and N v increase with m eff n i drops exponentially with increasing E g

14 Diamond-cubic structure: sp 3 bonding 14 Valence s and three p orbitals four sp 3 hybrid orbitals Orbitals point towards vertices of regular tetrahedron Si, C, Ge: 4 valence electrons each Form covalent bonds with four neighbours (8 shared electrons/atom) Bonding orbitals valence band, anti-bonding orbitals conduction band Tetrahedral network: FCC lattice with two atoms per cell

15 Zinc-blende structure: sp 3 bonding 15 Valence s and three p orbitals four sp 3 hybrid orbitals Orbitals point towards vertices of regular tetrahedron Combine Ga,In (3 electrons) with As,Sb (5 electrons) Form covalent bonds with four neighbours (8 shared electrons/atom) Bonding orbitals valence band, anti-bonding orbitals conduction band Tetrahedral network: FCC lattice with two atoms per cell With Al and N, tend to form closely related Wurtzite structure (FCC to HCP cell)

16 Doping: acceptors and donors 16 Extra / impurity Group III atoms: one less electron per atom Extra / impurity Group V atom: one extra electron per atom Covalent bonding theory: atoms want 8 (filled-shell) of shared electrons Group III acceptor : pick up electron from solid hole in valence band Group V donor : give electron to solid electron in conduction band Simple picture of doping: Density N a of acceptor atoms: charge en a Density N d of donor atoms: charge +en d Charge neutrality en + ep en a + en d = 0 n p = N d N a Change in n and p due to shift in E F, but np = n 2 i Solve for n and p, then find E F = Eg 2 + k BT 2 ln nnv pn c = E F 0 + k BT 2 ln n p Even simpler picture: usually N d, N a n i either p n or n p

17 Doping: p-type and n-type 17 n-type semiconductor: Donor impurities dominate N d > 0 (N a = 0 or < N d ) Typically N d N a n i n p (since p = n 2 i /n) Therefore n N d N a, p n 2 i /(N d N a ) E F = E F 0 + k BT 2 ln n p = E F 0 + ln N d N a n i Current predominantly carried by electrons p-type semiconductor: Acceptor impurities dominate N a > 0 (N d = 0 or < N a ) Typically N a N d n i p n (since n = n 2 i /p) Therefore p N a N d, n n 2 i /(N a N d ) E F = E F 0 + k BT 2 ln n p = E F 0 ln Na N d n i Current predominantly carried by holes (shifted towards CBM) (shifted towards VBM)

18 Doping: a more complete picture 18 Simple picture: donor atom donates an electron, becomes positively charged Positively charged donor ion can bind electrons: like a hydrogen atom Binding energy of pseudo-hydrogenic atom E b = ɛ 2 m r m e Ryd 0.05 ev Donor level: E d = E c E b (electrons bound relative to CBM) Exact argument for acceptors and holes, with charges swapped Acceptor level: E a = E v + E b (holes bound relative to VBM) Levels in Si: note some impurities introduce multiple levels Physics of Semiconductor Devices, S. M. Sze

19 Dopant levels in Ge and GaAs 19 For GaAs, Group II or Group VI are shallow dopants For GaAs, Group IV can be donor and acceptor dopants: how? Physics of Semiconductor Devices, S. M. Sze

20 Donor charge density 20 For each donor atom, degenerate donor levels typically with g d = 2 (spin) Electron occupation: zero or one for the whole atom (repulsions) Probability of occupation zero 1 Probability of occupation one g d exp E F E d Normalized probability of ionized donor (occupation zero): P + d = g d exp E F E d Therefore number density of ionized donors: N + d = N d 1 + g d exp E F E d (which is N d as long as E F several below E d )

21 Acceptor charge density 21 For each acceptor atom, degenerate acceptor levels typically with g a = 4 (two for spin, two for dgenerate hole bands) Hole occupation: zero or one for the whole atom (repulsions) Probability of hole occupation zero 1 Probability of hole occupation one g a exp Ea E F (tricky: flip energy axis when thinking in terms of holes) Normalized probability of ionized donor (hole occupation zero): P a = g a exp Ea E F Therefore number density of ionized acceptors: N a = N a 1 + g a exp Ea E F (which is N a as long as E F several above E a )

22 Charge neutrality 22 0 = ρ(e F ) = en + ep + en d en a = e N c e Eg E F } {{ } n + N v e E F } {{ } p + N d 1 + g d e E F E d N a 1 + g a e Ea E F If N a N d n i, then p n (p-type) If N d N a n i, then n p (n-type) Previous simple analysis holds if: Net doping is stronger than ni (one of the two regimes above), and Doping is small enough that EF is far above E a and far below E d (remember E F moves up/down k BT ln N d/a n i )

23 ρ(e F ): intrinsic semiconductor N a = 0, N d = ρ/e [cm -3 ] E [ev] Electrons increase with increaisng E F (hence ρ decreases) Cross-over point from + to is neutral E F

24 ρ(e F ): moderate p doping N a = 10 14, N d = ρ/e [cm -3 ] E [ev] Acceptors pull down Fermi level At moderate doping level, far from mid-gap and acceptor levels

25 ρ(e F ): moderate n doping N a = 0, N d = ρ/e [cm -3 ] E [ev] Donors pull up Fermi level At moderate doping level, far from mid-gap and donor levels

26 ρ(e F ): high n doping N a = 0, N d = ρ/e [cm -3 ] E [ev] At high n doping level, approach / cross donor level Donors may be partially ionized (simple model no longer works)

27 ρ(e F ): high p doping N a = 10 19, N d = ρ/e [cm -3 ] E [ev] At high p doping level, approach / cross acceptor level Acceptors may be partially ionized (simple model no longer works)

28 Degenerate doping 28 High-enough n doping: Fermi level enters conduction band ( n+ ) High-enough p doping: Fermi level enters valence band ( p+ ) One of our approximations breaks down for n+ case with E F > E g, n N c e (Eg E F )/( ) but instead n 1 3π 2 ( 2m (E F E g ) Similarly, for p+ case with E F < 0: p 1 3π 2 ( 2m ( E F ) ) 3 ) 3 These are the Fermi theory expressions with k F = 2m ε F / (where ε F is Fermi energy relative to band edge)! Important: partial donor / acceptor ionization in this regime

29 Partial donor ionization 29 Consider a p-type material (N d = 0) with E F = E a (we cross this point as we increase p-doping before getting to p+) Around this E F, acceptors are partially ionized When exactly does this occur? Since E F far from conduction band, neglect n p Charge neutrality yields N v e E F = N v e Ea = N a 1 + g a = N a 1 + g a exp Ea E F E a log (1+ga)Nv N a Therefore, this happens for high N a when E a But also, for lower than E a : dopant freeze-out Importance of shallow donor/acceptor levels! (for donors, replace E a E g E d )

30 Carrier density in ionization regime 30 If T much smaller than ionization threshold, neutrality: N v e E F = N v e E F N a 1 + g a exp Ea E F N a e Ea E F g a E F = E a 2 + k BT log g an v 2 N a Very similar to intrinsic case, except N a /g a replaces N c Effective gap between valence band and acceptor level! In this regime, p 2 = p Na = (N v N a /g a ) exp Ea Similar behavior for frozen-out donors in ionization regime for n-type

31 Temperature dependence of E F EF [ev] N a/d = N a/d = N a/d = N a/d = /1000 1/300 1/100 1/T [K 1 ] 1/50 Doping dominates at low T, approach intrinsic level at high T At low T, Fermi level decided by donor/acceptor level Note using E a/d = 0.1 ev from band edges to exaggerate effect

32 Temperature dependence of carrier concentration n [cm 3 ] Intrinsic N a/d = N a/d = N a/d = N a/d = Extrinsic Ionization 1/1000 1/300 1/100 1/50 1/T [K 1 ] 32 Ionization regime at low T upto threhsold which increases with N d/a Constant concentration in extrinsic regime; threshold increases with N d/a At high T, dopants don t matter: intrinsic regime

33 Intrinsic mobility 33 Drude theory: mobility µ = eτ/m Typically semiconductor m m e, so expect higher µ than metals At room temperature, µ e i-si 1400 cm2 /(Vs) and µ Ag 60 cm 2 /(Vs) Effective mass alone does not explain it! Remember τ 1 e-ph g(e)t For metals, g(e) g(e F ) since most carriers near Fermi level For semiconductors, carriers within few of band edge where g(e) E (and much smaller than metals) Averaged over carriers, g(e) T Therefore, τ 1 e-ph T 3/2 and µ τ T 3/2

34 Impurity scattering 34 Doped semiconductor contains ionized donors / acceptors Charged impurities cause electron PE 1/r near them Electrons with KE PE not scattered significantly Electrons with KE PE scattered most strongly Effective cross-section r 2 c, where P E(r c ) KE Therefore r c T 1 and cross-section σ cs T 2 Scattering time τ I = (N a/d σ cs v) 1 Average velocity v T Therefore τ I N 1 a/d T 3/2

35 Extrinsic mobility 35 Intrinsic scattering τ 1 e-ph T 3/2 Dopant / impurity scattering τ I N 1 a/d T 3/2 Net scattering τ ( T 3/2 + N a/d T 3/2) 1 (Mathiessen rule) Therefore mobility µ ( T 3/2 + N a/d T 3/2) 1 At high T, e-ph scattering dominates (intrinsic regime) At high doping concentration (or low T ), impurity scattering dominates See figures 5.18 and 5.19 in Kasap Conductivity σ = neµ e + peµ h : similar dependence as n (exponentials dominate over polynomial) µ effect visible mainly in extrinsic regime where n is constant See figure 5.20 in Kasap

36 Recombination 36 6 E in ev 0 6 E c E v 12 L Λ Γ Δ Χ Σ Γ Excite electrons and holes in semiconductor to higher energy e-ph scattering brings electrons and holes to band edges If np > n 2 i, equilibrium: nothing further happens What if you make more electron-hole pairs (eg. using light) such that np n 2 i? Electrons and holes will recombine to restore equilibrium

37 Recombination: direct vs indirect 37 6 E in ev 0 6 E c E v 12 L Λ Γ Δ Χ Σ Γ Direct gap: electrons and holes at band edges at same k Indirect gap: band edge carriers at different k Which will recombine faster? Direct gap: momentum conservation, recombine and emit light (usually) Indirect gap: cannot directly recombine: momentum not conserved

38 Recombination mechanisms 38 Recombination rate proportional to np n 2 i Radiative / direct recombination (direct gap materials) Trap-assisted (Shockley-Read-Hall recombination) Trap level in gap captures electron (hole), becoming (+) charged Later captures hole (electron), becoming neutral Energy from recombination emitted to phonons Probability of each capture Boltzmann factor of trap depth from band edge Net rate sech 2 E t E g/2 2k B (trap level E T t) Strongest for mid-band-gap states! Auger recombination Energy and momentum of e-h pair go to excite another e or h Need e or h to excite, so rate n, p Dominates at very high carrier concentrations Recombination rate = α(np n 2 i ) Minority carrier lifetime = 1/(max(n, p)α)