A. Derivation of the Impact-Ionization Integral

In the following the derivation of the impact-ionization integral from Section 5.1.2 will be given. First, the differntial equation

$$\ensuremath{\frac{\ensuremath{ \mathrm{d}}I_p}{\ensuremath{ \ma... ...uremath{\alpha _p}- \ensuremath{\alpha _n}\right) I_p = \ensuremath{\alpha _n}I$$ (A.1)

has to be solved. Using the simplified notation

$$y' + P y = Q,$$ (A.2)

with $P(x)=-(\ensuremath{\alpha _p}- \ensuremath{\alpha _n})$ and $Q(x)=\ensuremath{\alpha _n}I,$ one can derive the homogenous solution

$$y_h = C \exp \left( - \int_0^x P \ensuremath{ \mathrm{d}}x \right) ,$$ (A.3)

where C is the constant of integration. Applying the method of variation of constants, the ansatz of the particular solution $y_p$ is derived from the homogenous solution, using $C \rightarrow C(x).$ This ansatz function $y_p$ can be differentiated to

$$y_p' = C' \exp \left( - \int_0^x P \ensuremath{ \mathrm{d}}x \right) - P y_p .$$ (A.4)

Comparison of the coefficients between equations (A.2) and (A.4) gives

$$Q = C' \exp \left( - \int_0^x P \ensuremath{ \mathrm{d}}x \right)$$ (A.5)

and $C(x)$ evaluates to

$$C = \int_0^x Q \exp \left( \int_0^x P \ensuremath{ \mathrm{d}}x' \right) \ensuremath{ \mathrm{d}}x.$$ (A.6)

This leads to the particular solution

$$y_p = \exp \left( - \int_0^x P \ensuremath{ \mathrm{d}}x \right)... ...t_0^x P \ensuremath{ \mathrm{d}}x' \right) \ensuremath{ \mathrm{d}}x \right ]$$ (A.7)

and together with the homogenous solution $y_h$ to the solution of (A.2)

$$y = \exp \left( - \int_0^x P \ensuremath{ \mathrm{d}}x \right) \... ...P \ensuremath{ \mathrm{d}}x' \right) \ensuremath{ \mathrm{d}}x + C \right ] .$$ (A.8)

Using (A.8) our initial problem (A.1) solves together with the boundary conditions $I_p(0) = I_{p0}$ and $I_p(W) = I = M_p I_{p0}$ to

$$I_p = I \exp \left( \int_0^x \left( \ensuremath{\alpha _p}- \ensu... ...h{ \mathrm{d}}x' \right) \ensuremath{ \mathrm{d}}x + \frac{1}{M_p} \right ] .$$ (A.9)

To simplify the solution (A.9) the following relationship can be used. Considering

$$U = \int_0^x y \ensuremath{ \mathrm{d}}x , \qquad \ensuremath{\f... ...}}}{\ensuremath{ \mathrm{d}}U}} \exp \left( U \right) = \exp \left( U \right),$$ (A.10)

the following simplification can be performed:

$$\int y \exp \left( \int_0^x y \ensuremath{ \mathrm{d}}x' \right)... ...left( U \right) = \exp \left ( \int_0^x y \ensuremath{ \mathrm{d}}x' \right) .$$ (A.11)

Making the relation (A.11) applicable, (A.9) can be rewritten at the position $x=W$ as

$$1 = \exp \left( \int_0^W \left( \ensuremath{\alpha _p}- \ensurema... ...,\mathrm{d}}x' \right) + 1} + \frac{1}{M_p} \left. \vphantom{\int_0^W} \right ]$$ (A.12)

and simplified to

$$1 - \frac{1}{M_p} = \int_0^W \ensuremath{\alpha _p}\exp \left( - ... ...pha _n}\right) \ensuremath{ \mathrm{d}}x' \right) \ensuremath{ \mathrm{d}}x .$$ (A.13)