**Numerical solution of transport equations for semiconductor devices**

The numerical method applied in this study employs Newton's algorithm to obtain a steadystate solution to the set of transport Eqs. (9)–(11) using an initial distribution of the electrostatic potential in thermodynamic equilibrium that is obtained from conditions of electrical neutral‐ ity. Poisson's equation under equilibrium conditions gives the first solution function in the iterative procedure that leads to the solution of the set of transport equations under nonequi‐ librium conditions.

First, we have to solve the Poison's equation in steady-state conditions. If the charge density in the semiconductor structure does not depend on the gradient of the potential, the Poisson's Eq. (11) can be represented in the following form:

$$
\varepsilon \varepsilon\_0 \nabla^2 \psi + \varepsilon\_0 \nabla \psi \nabla \varepsilon + q \,\rho = 0 \tag{A1}
$$

where

$$
\rho = p - n + N\_D^\* - N\_A^- \tag{A2}
$$

is the difference between the density of positive and negative charge carriers (the hole concentrations *p* plus the concentration of ionized donors *ND* <sup>+</sup> minus the electron concentra‐ tions *n* and the concentration of ionized acceptors *NA* − ).

To find the numerical solution of the nonlinear Poisson Eq. (A1) by using a diffusion-equation differential scheme, we replaced it by an equivalent diffusion equation as it was postulated in reference [31]:

$$\frac{\partial \Psi}{\partial t} = \varepsilon \varepsilon\_0 \nabla^2 \Psi + \varepsilon\_0 \nabla \Psi \nabla \varepsilon + q \,\rho\tag{A3}$$

where *t* is the pseudotime.

The initial values of the electrostatic potential *ψ*<sup>0</sup> at every point of the semiconductor structure can be numerically calculated assuming the electrical neutrality conditions. It is expressed by an equation of electrical neutrality:

$$n(\psi') - n(\psi) + N\_D^+(\psi') - N\_A^-(\psi') = 0 \tag{A4}$$

The concentration of ionized acceptors and donors are expressed as

$$N\_{\mathcal{A}}^{-} = \frac{N\_{\mathcal{A}}}{1 + a \exp[\left(E\_{\mathcal{A}} - E\_{\mathcal{H}}\right) / k\_{\mathcal{B}}T]} \tag{A5}$$

HgCdTe Mid- and Long-Wave Barrier Infrared Detectors for Higher Operating Temperature Condition http://dx.doi.org/10.5772/63943 87

$$N\_D^\* = \frac{N\_D}{1 + b \exp[\left(E\_{Fl} - E\_D\right)/k\_B T]} \tag{A6}$$

where *EA* and *ED* are the ionization energies of acceptors and donors, respectively. Then, an iterative approach should be used to solve the set of transport Eqs. (9)–(11). Let us bring an iterative algorithm to solve Poisson's equation which was presented in reference [31]. Eq. (A1) may be expressed in the following form:

$$L(\wp \prime) = 0 \tag{A7}$$

The iterative form of Eq. (A7) can be expressed as follows:

$$L(\boldsymbol{\psi}^{n+1}) = L(\boldsymbol{\psi}^{n}) + \frac{\partial L(\boldsymbol{\psi}^{n})}{\partial \boldsymbol{\psi}^{n}} \delta \boldsymbol{\psi}^{n+1} \tag{A8}$$

where *Ψn+1* is a vector of *n* + 1 iterative correction of vector *Ψ*. Similarly, Eq. (A3) can be expressed as

$$\frac{\partial \boldsymbol{\psi}}{\partial t} = L(\boldsymbol{\psi}) \tag{A9}$$

and the iterative form of Eq. (A9) can be expressed as

$$
\delta\psi\nu^{\ast\ast 1} = \Delta t L(\psi\nu^{\ast}) + \frac{1}{2}\Delta t \frac{\partial L(\psi\nu^{\ast})}{\partial \varphi\nu^{\ast}} \delta\psi\nu^{\ast\ast 1} \tag{A10}
$$

where Δ*t* is the pseudotime step.

potential in thermodynamic equilibrium that is obtained from conditions of electrical neutral‐ ity. Poisson's equation under equilibrium conditions gives the first solution function in the iterative procedure that leads to the solution of the set of transport equations under nonequi‐

First, we have to solve the Poison's equation in steady-state conditions. If the charge density in the semiconductor structure does not depend on the gradient of the potential, the Poisson's

*D A*

− ).

is the difference between the density of positive and negative charge carriers (the hole

To find the numerical solution of the nonlinear Poisson Eq. (A1) by using a diffusion-equation differential scheme, we replaced it by an equivalent diffusion equation as it was postulated in

0 0 *q*

The initial values of the electrostatic potential *ψ*<sup>0</sup> at every point of the semiconductor structure can be numerically calculated assuming the electrical neutrality conditions. It is expressed by

() () () () 0 *D A pnN N*

 y

1 exp[( ) / ] *A*

*a E E kT*

*A Fi B*

 y

2

¶ = Ñ + ÑÑ+

ee y e y e r

Ñ + ÑÑ+ = *q* 0 (A1)

*pnN N* + - = -+ - (A2)

¶ (A3)

+ - -+ - = (A4)


<sup>+</sup> minus the electron concentra‐

librium conditions.

86 Modeling and Simulation in Engineering Sciences

where

reference [31]:

where *t* is the pseudotime.

an equation of electrical neutrality:

Eq. (11) can be represented in the following form:

2 0 0

r

concentrations *p* plus the concentration of ionized donors *ND*

*t* y

yy

The concentration of ionized acceptors and donors are expressed as

*<sup>N</sup> <sup>N</sup>*

*A*

tions *n* and the concentration of ionized acceptors *NA*

ee y e y e r

> An iteration method allows the calculation of corrections to electrical potential, quasi-Fermi levels, and temperature [32]:

$$
\boldsymbol{\Psi}' = \boldsymbol{\Psi}'^0 + \boldsymbol{\delta}\boldsymbol{\Psi}' \boldsymbol{\nu} \\
\boldsymbol{\Phi}\_{\boldsymbol{\eta}} = \boldsymbol{\Phi}\boldsymbol{\rho}^0 + \boldsymbol{\delta}\boldsymbol{\Phi}\_{\boldsymbol{\eta}'} \boldsymbol{\Phi}\_{\boldsymbol{\rho}} = \boldsymbol{\Phi}\boldsymbol{\rho}^0 + \boldsymbol{\delta}\boldsymbol{\Phi}\_{\boldsymbol{\rho}'} \boldsymbol{\Gamma} \\
= \boldsymbol{T}^{\boldsymbol{\ast}0} + \boldsymbol{\delta}\boldsymbol{\Gamma} \tag{A11}
$$

and consequently to other physical parameters:

$$
\delta m = n^0 + \frac{\partial n}{\partial \nu} \delta \nu + \frac{\partial n}{\partial \Phi\_n} \delta \Phi\_n + \frac{\partial n}{\partial T} \delta T \tag{A12}
$$

$$p = p^0 + \frac{\partial p}{\partial \nu} \delta \nu + \frac{\partial p}{\partial \Phi\_p} \delta \Phi\_p + \frac{\partial p}{\partial T} \delta T \tag{A13}$$

$$G - R = G^0 - R^0 + \delta(G - R) \tag{A14}$$

where

$$\begin{split} \delta \delta (G - R) &= \frac{\partial (G - R)}{\partial n} \bigg| \frac{\partial n}{\partial \nu} \delta \nu + \frac{\partial n}{\partial \mathcal{O}\_n} \delta \mathcal{O}\_n + \frac{\partial n}{\partial T} \delta T \Bigg) \\ &+ \frac{\partial (G - R)}{\partial p} \bigg( \frac{\partial p}{\partial \nu} \delta \nu + \frac{\partial p}{\partial \mathcal{O}\_p} \delta \mathcal{O}\_n + \frac{\partial p}{\partial T} \delta T \Bigg) \end{split} \tag{A15}$$

Finally, knowing the electric potential *Ψ*, the electron affinity *X*, and the bandgap energy *Eg*, we can determine the energy values for the edge of the conduction and valence bands:

$$E\_c = -X - q\varphi \tag{A16}$$

$$E\_\chi = -X - q\varphi - E\_g \tag{A17}$$
