Effective mass: from Newton s law. Effective mass. I.2. Bandgap of semiconductors: the «Physicist s approach» - k.p method

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1 Lecture 4 1/10/011 Effectie mass I.. Bandgap of semiconductors: the «Physicist s approach» - k.p method I.3. Effectie mass approximation - Electrons - Holes I.4. train effect on band structure - Introduction: strain, lattice mismatch - Elasticity theory - Band structure E V E 0 Wae ector q 1 * 1 V g = de = = d q 1± E0 / V E0 4E0 m m m m E Effectie mass: from Newton s law The crystal field affects the electron properties different electron mass Newton s law: d F = m* dt Electrons are described by their Bloch wae function: with Effectie mass If one applies an external force on an electron, its energy increases oer a certain distance dx by Its acceleration is then gien by The wae packet elocity is gien by and the effectie mass is defined as and the acceleration writes

2 Effectie mass Inersely proportional to the curature of the k-space energy dispersion Effectie mass: positie for the CB and negatie for the VB «Hole» in the VB like an electron in the CB. Actually, a hole is a missing electron in the VB! Effectie mass Direct bandgap (GaAs): the CB is isotropic around k=0 (s state) Isotropic effectie mass : The mobility does not depend on the electron motion direction in the crystal Indirect bandgap (i): the CB is anisotropic around the CB energy minimum Energy surface E(k)=constant is no longer a sphere, but a series of ellipsoids oriented along the 3 directions Effectie mass Effectie mass What will be the effectie mass of an electron in AlGaAs alloy?

3 Effectie mass Effectie mass Anisotropic case Isotropic case Anisotropic case Effectie mass Longitudinal and transerse masses Valence band structure: Effectie mass The energy dispersion E(k) can be deeloped in Taylor series - in the neighborhood of k=k 0, where the first deriatie is zero, E(k) can be expressed as follows Γ lh k hh E(k) Direct bandgap C: k 0 Two branches are degenerate in k=0: - Heay holes (hh) - Light holes (lh) Indirect bandgap C:, O band stands for spin-orbit coupling (due to lack of inersion symetry in zinc blende structure) m l and m t are the longitudinal and transerse effectie masses, respectiely

4 GaN Optical Properties Effectie mass emiconductors Electron effectie mass Hole effectie mass (m * /m 0 ) (m * /m 0 ) i 0.9 (m l ) 0.19 (m t ) 0.53 (m hh ) 0.16 (m lh ) Ge 1.59 (m l ) 0.08 (m t ) 0.35 (m hh ) (m lh ) GaAs (m hh ) (m lh ) InAs (m 0.60 hh ) (m lh ) GaN (m hh ) 0.6 (m lh ) Dingle et al. PRB 4 (1971) Chichibu in introduction to nitride semiconductor blue lasers and LEDs, Taylor & Francis Eds (000) Band structure - summary - Upper leel state of the VB in Γ (k=0) - Lower leel state in the CB Γ (direct) X or L (indirect) - Effectie mass gien by the dispersion energy curature - Isotropic for the CB in Γ - Anisotropic for the VB in X or L (longitudinal and transerse effectie masses) Bandga ap (ev) AlN GaN Lattice mismatch in III-V semiconductors InN AlP GaP GaAs AlAs InAs InP Lattice parameter (Å) The lattice parameters of semiconductors are neer the same train induced d in the semiconductor layers Read more: Yu and Cardona p

5 Epitaxial growth D nanostructures: quantum wells GaAs InGaAs GaAs CB E g Epitaxy: crystal growth proceeds layer-by- y layer and the layer structure complies with the substrate lattice VB Quantum wells Epitaxial growth (heteroepitaxy) Epitaxial growth Basic Principle and Problems Thermal expansion coefficient (TEC) canning tunneling microscopy (TM) in cross-section different TECs lead to strain in the lattice

6 An example: InGaAs/GaAs quantum wells No dislocation dislocations Critical thickness for a D layer Burgers ector TEM Edge type dislocation Elastic deformation (Coherent growth) Plastic deformation crew type dislocation Critical thickness for a D layer Critical thickness for a D layer Minimizing the total energy density with respect to the dislocation line density yields the critical thickness B (b II ): Burgers ector (edge component) α: energy of the dislocation core λ: angle between the Burgers ector and the interface

7 train and Heteroepitaxy Bi-axial strain (compression) Lattice mismatch: Δa/a = (a c -a s )/a s train: ε // = (a s -a c )/a c Ex: AlN/GaN a a GaN = Å and a AlN =3.11 Å a GaN on AlN: Δa/a=47%andε =.47 % and = -.4 4% ε < 0 ε >0 compressie strain tensile strain a // a // = a = a c Growth on a substrate with lattice parameter a s a // a // = a s a a =(1+ε ε )a c ε // = (a s -a c )/a c Elasticity theory Elastic deformation: σ xx σ yx σ zx σ 1 σ 6 σ 5 ε xx ε yx ε zx ε 1 ½ε 6 ½ε 5 Oeriew Hooke s law: relation between the tensors of deformations ε ij, ij stress σ ij, and elastic constants C ijkl : σ ij =C ijkl ε ij For cubic crystals defined by the crystallographic axes [100],[010] and [001]: σ 1 C 11 C 1 C ε 1 σ C 1 C 11 C ε riew Oe σ xy σ yy σ zy = σ 6 σ σ 4 and ε xy ε yy ε zy = ½ε 6 ε ½ε 4 σ xz σ yz σ zz σ 5 σ 4 σ 3 ε xz ε yz ε zz ½ε 5 ½ε 4 ε 3 R = (i, j, k) with directions [100], [010], [001] One defines R = (i, j, k ) for which h k i is along the growth axis σ 3 C 1 C 1 C ε 3 σ 4 = C x ε 4 ε ij = a ik a jl ε kl Note: x, y, z (R) and 1,, 3 (R ) σ ij =a ik a jl σ kl (aij) transfer matrix from R to R σ C 44 0 ε 5 σ C 44 ε 6

8 During growth, the surface is free of strain and can freely moe along the growth axis (3). Thus, it leads to σ 13 = 0 σ 3 = 0 σ 33 = 0 In the layer plane, the deformations are identical: ε 11 = ε = ε // ε 1 = 0 riew Oer Lt Let us write ε 0 = ε 33 and ε // = (a c -a c0 )/a 0 c with a c = a 0 s Problem: 1 unknown alues, with 6 components of the deformations and 6 components of the strain tensor, but with the 6 conditions just mentioned, only 6 are left with 6 equations goerned by the Hooke s law. After changing the coordinate system and taking into account the crystal symmetry, it follows: Epitaxial growth system [h,k,l] = [001]: ε 13 =0 σ 11 = σ = ε // (C 11 +C 1 )(C 11 -C 1 )/C 11 riew Oer [h,k,l] = [111]: ε = - C 1 /C 11 ε // ε 13 = 0 σ 11 = σ = ε // 6C 44( (C 11+C 1) )/(C 11+C 1+4C 44) ) ε = - (C 11 +C 1 -C 44 )/(C 11 +C 1 +4C 44 ) ε //

9 train effect on band structure The crystal symmetry is changed and thereby the band structure The Hamiltonian can be written as the sum of two components: a purely hydrostatic term H H and a shear strain term H. train effect on band structure Σ H =(1C (1-C 1 /C 11 )ε // Hydrostatic Σ = - (1+C 1 /C 11 )ε // hear «Hydrostatic» term H = H H +H 1. Conduction band Due to symmetry reasons, only the hydrostatic component plays on the CB E C changes by δe C = a c Σ H with a c the CB potential of deformation CB VB Hydrostatic (compression) shear hh lh. Valence band The hydrostatic component modifies the VB edge E by δe V = a Σ H with a the VB potential of deformation train effect on band structure «hear» term The shear strain deeply affects the ordering of the alence band leels l (hh, lh, and «spin-orbit»). Gien b the potential of deformation due to the shear strain and Δ O the «spin-orbit» splitting, the energy shifts of hh and lh write δe δe hh lh = b Σ Δ = O b Σ ( Δ O + b Σ ) + 8( b Σ ) train effect on band structure ilicon GaAs In summary, we hae: E = E e E E hh lh V = E = E + E + a Σ V V g + a Σ + a Σ H H c H b Σ Δ O b Σ ( Δ O + b Σ ) + 8( b Σ ) train Effect in emiconductors: Theory and Deice Applications train Effect in emiconductors: Theory and Deice Applications Yongke un, cott Thompson,Toshikazu Nishida

10 i/ige MOFET train effect on band structure Hydrostatic strain Equialent strain along x, y, and z Origin of the transistor performance enhancement? MIT (U) 003 Bi-axial strain train along directions: strained semiconductor epilayers