# Detrapping rates arising from Fermi-Dirac statistics

Have you ever wondered how Fermi-Dirac statistics arises? I already have in the previous couple of posts, but the more important question is why I suddenly started to care about Fermi-Dirac statistics. It is still all related to [1]. Reviewers raised doubts about our assumption about uniform detrapping rates, asking us to provide experimental evidence for our claim. As far as I am able to read the experimental works, it doesn't seem possible to provide any direct evidence. Still as a theoretician I can provide theoretical justification which is indirectly supported by experimental works.

This is why in this post we'll talk about detrapping rates arising from a simple model explored in an earlier post.

## Derivation of rate distribution

From earlier the earlier post we already know that the occupation probability of trap levels follows Fermi-Dirac statistics:

\begin{equation} F\left(E\right)=\frac{1}{1+\exp\left(\frac{E-E_{F}}{k_{B}T}\right)}. \end{equation}

Then the probability to find certain trap level unoccupied is given by:

\begin{equation} V\left(E\right)=1-F(E)=\frac{1}{1+\exp\left(-\frac{E-E_{F}}{k_{B}T}\right)}. \end{equation}

This is not yet the quantity we are interested in. We are interested in the detrapping process, but for it to occur first the particle needs to be trapped. And the trapping from the conduction level \( E_C \) to an arbitrary trap level \( E \) occurs with probability:

\begin{equation} P\left(E_{C}\rightarrow E\right)=\frac{V\left(E\right)}{\int_{E_{1}}^{E_{2}}V\left(E\right)dE}. \end{equation}

The integral is required for normalization purposes only, while the weights are given by \( V(E) \). Calculating the integral:

\begin{equation} C^{-1}=\int_{E_{1}}^{E_{2}}V\left(E\right)dE=E_{2}-E_{1}+k_{B}T\ln\left[\frac{\exp\left(\frac{E_{1}}{k_{B}T}\right)+\exp\left(\frac{E_{F}}{k_{B}T}\right)}{\exp\left(\frac{E_{2}}{k_{B}T}\right)+\exp\left(\frac{E_{F}}{k_{B}T}\right)}\right]. \end{equation}

We have previously assumed that all moves between energy levels are determined by the Boltzmann factor. Specifically for detrapping process this law is known as Arrhenius law:

\begin{equation} \gamma=A\exp\left(\frac{E_{A}}{k_{B}T}\right). \end{equation}

So, now we know the probability for trapping event to specific trap level to occur. We also know how the depth of the trap level, \( E_{A} = E-E_{C} \), is related to the detrapping rate \( \gamma \). All we need to do is to transform the stochastic variable from \( E \) to \( \gamma \). From the conservation of probability density we have that:

\begin{equation} p\left(\gamma\right)=P\left(E_{C}\rightarrow E\right)\frac{dE}{d\gamma}=\frac{C k_{B}T}{\gamma+\gamma\exp\left(-\frac{E-E_{F}}{k_{B}T}\right)}=\frac{C k_{B}T}{A\exp\left(\frac{E_{C}-E_{F}}{k_{B}T}\right)+\gamma}. \end{equation}

Note that for \( \gamma \ll A\exp\left(\frac{E_{C}-E_{F}}{k_{B}T}\right) \), the distribution is uniform! Which is exactly what the model presented in [1] requires! Hooray, but does my derivation work? Check the interactive app below.

## Interactive app

In many regards this app is identical to the one from the previous post. The main difference is that this app also plots the observed detrapping rates \( \gamma \). These are calculated by observing the time particles spend in each state \( \tau_{i} \) (where the index corresponds to event number), and then taking the inverse of this time, i.e., \( \gamma_{i} = \tau_{i}^{-1} \). On average, the distribution of observed detrapping rates should roughly correspond to the detrapping rate distribution over the trap levels.

## References

- A. Kononovicius, B. Kaulakys. 1/f noise in semiconductors arising from the heterogeneous detrapping process of individual charge carriers. Journal of Statistical Mechanics 2024: 113201 (2024). doi: 10.1088/1742-5468/ad890b. arXiv:2306.07009 [math.PR].