Page 221 - Semiconductor For Micro- and Nanotechnology An Introduction For Engineers
P. 221

Transport Theory
                                                                             q j
                                                                            –
                                                                                 for elec-
                             We use (6.71) to calculate the electric current density i =
                                                                               n
                             trons from the form given in (6.11) and assume that the material is homo-
                             geneous, i.e., ∇µ =  0  . In this way we obtain three contributions
                                                               q 4
                                        2
                                               3
                                                         –
                                                          1
                                   i =  q K E +  q K B ×  ( M E) +  --------K EB(  )MB  (6.72)
                                          1       2                3
                                                               M
                             where the components of the tensors K   are given by [6.2]
                                                            i
                                             1         τ i     f ∂  0
                                               3 ∫
                                                                       3
                                      K =  – ---------  --------------------------------------------v ⊗  vd k  (6.73)
                                        i
                                                      2 2
                                            4π    1 +  q τ    ∂ E
                                                     ----------BMB
                                                      M
                                                V k
                             The rank two tensors K   are called the generalized transport coefficients.
                                                i
                             They reflect the fully anisotropic character of constitutive equations in a
                             crystal. To simplify the discussion, we now write (6.72) by defining a
                             generalized constitutive equation for the current density
                                                    i =  σ∗ E                     (6.74)
                             where the effective conductivity σ∗   is defined by
                                                               q  4
                                                       3
                                               2
                                         σ∗ =  q K I +  q K A +  --------K A      (6.75)
                                                  1      2  2      3  3
                                                               M
                             and where   is the identity matrix. The matrices A 2   and A 3   are given by
                                      I
                                                   0   – B m  B m
                                                         3  2  2  3
                                           A =   – B m   0   – B m               (6.76a)
                                            2      3  1         1  3
                                                 – B m  B m    0
                                                   2  1  1  2
                                                  2
                                                B m 1  B B m B B m 1
                                                            1
                                                               1
                                                 1
                                                        1
                                                          2
                                                                 3
                                         A =   B B m   B m    B B m              (6.76b)
                                                         2
                                          3
                                                2  1  2  2  2  2  3  2
                                                                2
                                               B B m B B m     B m
                                                3  1  3  3  2  3  3  3
                             Note that these equations are given in coordinates of the crystal direc-
                             tions.  Thus, for the effective masses we may chose the two different
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