Analytical evaluation of the charge carrier density of organic materials with a Gaussian density of states revisited

Abstract

An analytical solution for the calculation of the charge carrier density of organic materials with a Gaussian distribution for the density of states is presented and builds upon the ideas presented by Mehmetoğlu (J Comput Electron 13:960–964, 2014) and Paasch et al. (J Appl Phys 107:104501-1–104501-4, 2010). The integral of interest is called the Gauss–Fermi integral and can be viewed as a particular type of integral in a family of the more general Fermi–Dirac-type integrals. The form of the Gauss–Fermi integral will be defined as

$$\begin{aligned} G\left( \alpha ,\beta ,\xi \right) =\mathop {\displaystyle \int }\limits _{-\infty }^{\infty }\frac{ e^{-\alpha \left( x-\beta \right) ^{2}}}{1+e^{x-\xi }}\hbox {d}x\text {,} \end{aligned}$$

where $G\left( \alpha ,\beta ,\xi \right)$ is a dimensionless function. This article illustrates a technique developed by Selvaggi et al. [3] to derive a mathematical formula for a complete range of parameters $\alpha$, $\beta$, and $\xi$ valid $\forall$ $\alpha$ $\varepsilon$ ${\mathbb {R}} \ge 0$, $\forall$ $\beta$ $\varepsilon$ ${\mathbb {R}}$, and $\forall$ $\xi$ $\varepsilon$ ${\mathbb {R}}$.

A Appendix

A derivation of the Gauss–Fermi function begins with (8) rewritten below.

$$\begin{aligned} G\left( \alpha ,\beta ,\xi \right) =\mathop {\displaystyle \int }\limits _{0}^{\infty }\frac{ e^{-\alpha \left( x-\beta \right) ^{2}}}{1+e^{x-\xi }}\hbox {d}x+\mathop {\displaystyle \int }\limits _{0}^{ \infty }\frac{e^{-\alpha \left( x+\beta \right) ^{2}}}{1+e^{-(x+\xi )}}\hbox {d}x. \end{aligned}$$

(14)

Rewrite (14) as follows

$$\begin{aligned} G\left( \alpha ,\beta ,\xi \right)= & {} \mathop {\displaystyle \int }\limits _{0}^{\xi }\frac{ e^{-\alpha \left( x-\beta \right) ^{2}}}{1+e^{-(\xi -x)}}\hbox {d}x \nonumber \\&\,+\,\mathop {\displaystyle \int }\limits _{\xi }^{\infty }\frac{e^{-\alpha \left( x-\beta \right) ^{2}}e^{-\left( x-\xi \right) }}{1+e^{-\left( x-\xi \right) }}\hbox {d}x \nonumber \\&\,+\,\mathop {\displaystyle \int }\limits _{0}^{\infty }\frac{e^{-\alpha \left( x+\beta \right) ^{2}}}{ 1+e^{-(x+\xi )}}\hbox {d}x. \end{aligned}$$

(15)

The denominators in each of the integrals in Eqs. (15) can be expanded in a binomial expansion. Similar methods can be found in Çopuro ğlu et al. [18, 19]. Each expansion results in a convergent integral within its limits of integration. This expansion leads to the following result.

$$\begin{aligned} G\left( \alpha ,\beta ,\xi \right)= & {} \sum _{p=1}^{\infty }(-1)^{p-1}\mathop {\displaystyle \int }\limits _{0}^{\xi }e^{-\alpha \left( x-\beta \right) ^{2}}e^{-(p-1)(\xi -x)}\hbox {d}x \nonumber \\&\,+\,\sum _{p=1}^{\infty }(-1)^{p-1}\mathop {\displaystyle \int }\limits _{\xi }^{\infty }e^{-\alpha \left( x-\beta \right) ^{2}}e^{-p(x-\xi )}\hbox {d}x \nonumber \\&\,+\,\sum _{p=0}^{\infty }(-1)^{p}\mathop {\displaystyle \int }\limits _{\xi }^{\infty }e^{-\alpha \left( x+\beta \right) ^{2}}e^{-p(x+\xi )}\hbox {d}x, \nonumber \\= & {} \mathop {\displaystyle \int }\limits _{0}^{\xi }e^{-\alpha \left( x-\beta \right) ^{2}}\hbox {d}x+\sum _{p=1}^{\infty }(-1)^{p} \nonumber \\&\,\times \left[ \mathop {\displaystyle \int }\limits _{0}^{\xi }e^{-\alpha \left( x-\beta \right) ^{2}}e^{-p(\xi -x)}\hbox {d}x\right. \nonumber \\&\left. \,-\, \mathop {\displaystyle \int }\limits _{\xi }^{\infty }e^{-\alpha \left( x-\beta \right) ^{2}}e^{-p(x-\xi )}\hbox {d}x\right] \nonumber \\&\,+\,\sum _{p=0}^{\infty }\left( -1\right) ^{p}\mathop {\displaystyle \int }\limits _{0}^{\infty }e^{-\alpha (x+\beta )^{2}-p(x+\xi )}\hbox {d}x. \end{aligned}$$

(16)

Equation (16) can now be transformed into a more recognizable form. Define the following:

$$\begin{aligned}&g_{0}\left( \alpha ,\beta ,\xi \right) =\mathop {\displaystyle \int }\limits _{0}^{\xi }e^{-\alpha \left( x-\beta \right) ^{2}}\hbox {d}x, \end{aligned}$$

(17)

$$\begin{aligned}&g_{1}\left( \alpha ,\beta ,\xi \right) =\mathop {\displaystyle \int }\limits _{0}^{\xi }e^{-\alpha \left( x-\beta \right) ^{2}}e^{-p(\xi -x)}\hbox {d}x, \end{aligned}$$

(18)

$$\begin{aligned}&g_{2}\left( \alpha ,\beta ,\xi \right) =\mathop {\displaystyle \int }\limits _{\xi }^{\infty }e^{-\alpha \left( x-\beta \right) ^{2}}e^{-p(x-\xi )}\hbox {d}x, \end{aligned}$$

(19)

$$\begin{aligned}&g_{3}\left( \alpha ,\beta ,\xi \right) =\mathop {\displaystyle \int }\limits _{0}^{\infty }e^{-\alpha (x+\beta )^{2}-p(x+\xi )}\hbox {d}x. \end{aligned}$$

(20)

In (17), let $y^{2}=\alpha \left( x-\beta \right) ^{2}$. This results in the following expression.

$$\begin{aligned} g_{0}\left( \alpha ,\beta ,\xi \right)= & {} \frac{1}{\sqrt{\alpha }} \mathop {\displaystyle \int }\limits _{-\beta \sqrt{\alpha }}^{\left( \xi -\beta \right) \sqrt{\alpha }}e^{-y^{2}}\hbox {d}y, \nonumber \\= & {} \frac{1}{\sqrt{\alpha }}\left( \mathop {\displaystyle \int }\limits _{0}^{\beta \sqrt{\alpha } }e^{-y^{2}}\hbox {d}y+\mathop {\displaystyle \int }\limits _{0}^{\left( \xi -\beta \right) \sqrt{\alpha } }e^{-y^{2}}\hbox {d}y\right) , \nonumber \\= & {} \frac{1}{2}\sqrt{\frac{\pi }{\alpha }}\left\{ \hbox {Erf}\left( \beta \sqrt{ \alpha }\right) \right. + \nonumber \\&\left. \hbox {Erf}\sqrt{\alpha }\left( \xi -\beta \right) \right\} , \end{aligned}$$

(21)

where $\hbox {Erf}(\cdot )$ is the error function. In (18), let $y=x-\beta$. This results in the following expression.

$$\begin{aligned} g_{1}\left( \alpha ,\beta ,\xi \right) =e^{-p\left( \xi -\beta \right) + \frac{p^{2}}{4\alpha }}\mathop {\displaystyle \int }\limits _{-\beta }^{\xi -\beta }e^{-\alpha \left( y-\frac{p}{2\alpha }\right) ^{2}}\hbox {d}y. \end{aligned}$$

(22)

In (22), let $v^{2}=\alpha \left( y-\frac{p}{2\alpha }\right) ^{2}$. This results in the following expression.

$$\begin{aligned} g_{1}\left( \alpha ,\beta ,\xi \right)= & {} \frac{e^{\frac{p^{2}}{4\alpha } -p\left( \xi -\beta \right) }}{\sqrt{\alpha }}\mathop {\displaystyle \int }\limits _{-\sqrt{\alpha } \left( \beta +\frac{p}{4\alpha }\right) }^{\sqrt{\alpha }\left( \xi -\beta - \frac{p}{2\alpha }\right) }e^{-v^{2}}\hbox {d}v, \nonumber \\= & {} \frac{e^{\frac{p^{2}}{4\alpha }-p\left( \xi -\beta \right) }}{\sqrt{ \alpha }}\left( \mathop {\displaystyle \int }\limits _{0}^{\sqrt{\alpha }\left( \beta +\frac{p}{ 4\alpha }\right) }e^{-v^{2}}\hbox {d}v\right. + \nonumber \\&\left. \mathop {\displaystyle \int }\limits _{0}^{\sqrt{\alpha }\left( \xi -\beta -\frac{p}{2\alpha }\right) }e^{-v^{2}}\hbox {d}v\right) , \nonumber \\= & {} \frac{1}{2}\sqrt{\frac{\pi }{\alpha }}e^{\frac{p^{2}}{4\alpha }-p\left( \xi -\beta \right) }\left\{ \hbox {Erf}\left( \frac{p}{2\sqrt{\alpha }}+\beta \sqrt{ \alpha }\right) \right. + \nonumber \\&\left. \hbox {Erf}\left( -\frac{p}{2\sqrt{\alpha }}+\sqrt{\alpha }\left( \xi -\beta \right) \right) \right\} . \end{aligned}$$

(23)

The identical steps used in deriving $g_{1}\left( \alpha ,\beta ,\xi \right)$ are used in deriving $g_{2}\left( \alpha ,\beta ,\xi \right)$. The expression is given as

$$\begin{aligned} g_{2}\left( \alpha ,\beta ,\xi \right)= & {} \frac{e^{\frac{p^{2}}{4\alpha } -p\left( \beta -\xi \right) }}{\sqrt{\alpha }}\mathop {\displaystyle \int }\limits _{\sqrt{\alpha } \left( \xi -\beta +\frac{p}{2\alpha }\right) }^{\infty }e^{-v^{2}}\hbox {d}v, \nonumber \\= & {} \frac{1}{2}\sqrt{\frac{\pi }{\alpha }}e^{\frac{p^{2}}{4\alpha }+p\left( \xi -\beta \right) } \nonumber \\&\,\times {\text {Erfc}}\left( \frac{p}{2\sqrt{\alpha }}+\sqrt{\alpha }\left( \xi -\beta \right) \right) \text {.} \end{aligned}$$

(24)

Likewise, $g_{3}\left( \alpha ,\beta ,\xi \right)$ can be written as follows:

$$\begin{aligned} g_{3}\left( \alpha ,\beta ,\xi \right)= & {} \mathop {\displaystyle \int }\limits _{0}^{\infty }e^{-\alpha (x+\beta )^{2}-p(x+\xi )}\hbox {d}x \nonumber \\= & {} e^{-p\left( \xi -\beta \right) }\mathop {\displaystyle \int }\limits _{0}^{\infty }e^{-\alpha (x+\beta )^{2}-p(x+\beta )}\hbox {d}x. \end{aligned}$$

(25)

In (25), let $u=x+\beta$. This leads to

$$\begin{aligned} g_{3}\left( \alpha ,\beta ,\xi \right) =e^{\frac{p^{2}}{4\alpha }-p\left( \xi -\beta \right) }\mathop {\displaystyle \int }\limits _{\beta }^{\infty }e^{-\alpha \left( u+\frac{ p}{2\alpha }\right) ^{2}}\hbox {d}u. \end{aligned}$$

(26)

In (26), let $v^{2}=$ $\alpha \left( u+\frac{p}{2\alpha }\right) ^{2}$. This transformation leads to the following:

$$\begin{aligned} g_{3}\left( \alpha ,\beta ,\xi \right)= & {} \frac{1}{2}\sqrt{\frac{\pi }{ \alpha }}e^{\frac{p^{2}}{4\alpha }-p\left( \xi -\beta \right) } \nonumber \\&{\text {Erfc}}\left[ \left( \frac{p}{2\sqrt{\alpha }}+\beta \sqrt{\alpha }\right) \right] . \end{aligned}$$

(27)

The Gauss–Fermi function is now given by

$$\begin{aligned} G\left( \alpha ,\beta ,\xi \right)= & {} g_{0}\left( \alpha ,\beta ,\xi \right) +\sum _{p=1}^{\infty }(-1)^{p}\left\{ g_{1}\left( \alpha ,\beta ,\xi \right) \right. \nonumber \\&\left. -\,g_{2}\left( \alpha ,\beta ,\xi \right) \right\} +\sum _{p=0}^{\infty }(-1)^{p}g_{3}\left( \alpha ,\beta ,\xi \right) , \nonumber \\= & {} \frac{1}{2}\sqrt{\frac{\pi }{\alpha }}{\text {Erfc}}\left( \sqrt{\alpha }\left( \beta -\xi \right) \right) \nonumber \\&\,+ \frac{1}{2}\sqrt{\frac{\pi }{\alpha }}e^{-\alpha \left( \xi -\beta \right) ^{2}}\sum _{p=1}^{\infty }\left( -1\right) ^{p} \nonumber \\&\,\times \left\{ e^{\left( \frac{p}{2\sqrt{\alpha }}-\sqrt{\alpha }\left( \xi -\beta \right) \right) ^{2}} \right. \nonumber \\&\,\times {\text {Erfc}}\left( \frac{p}{2\sqrt{\alpha }}-\sqrt{\alpha }\left( \xi -\beta \right) \right) \nonumber \\&\, -\,e^{\left( \frac{p}{2\sqrt{\alpha }}+\sqrt{\alpha }\left( \xi -\beta \right) \right) ^{2}} \nonumber \\&\left. \, \times {\text {Erfc}}\left( \frac{p}{2\sqrt{\alpha }}+\sqrt{\alpha }\left( \xi -\beta \right) \right) \right\} . \end{aligned}$$

(28)

This proves the result given by (10) and is valid $\forall$ $\alpha$ $\varepsilon$ ${\mathbb {R}} \ge 0$, $\forall$ $\beta$ $\varepsilon$ ${\mathbb {R}}$, and $\forall$ $\xi$ $\varepsilon$ ${\mathbb {R}} \ge 0$. The Gauss–Fermi function valid $\forall$ $\alpha$ $\varepsilon$ ${\mathbb {R}} \ge 0$, $\forall$ $\beta$ $\varepsilon$ ${\mathbb {R}}$, and $\forall$ $\xi$ $\varepsilon$ ${\mathbb {R}} \le 0$ follows a similar proof.