5.4 Current flow through a non degenerated semiconductor

We investigate the charge flow in an electrical field for a non degenerated semiconductor. Combining the equations (5.18) and (5.21) we get

 \begin{equation*} \vec{j} = - \frac{q}{4\pi^3}\int_{V_k} \tau(\vec{k}) \vec{v}(\vec{k}) \frac{\partial f_0}{\partial E} \left\langle \left(q\vec{E} - \vec{\nabla}_r \mu - (E - \mu) \vec{\nabla}_r\ln(T) \right),\vec{v}\right\rangle d^3k \qquad . \end{equation*}(5.24)

We will neglect gradients in the temperature, so \((E - \mu) \vec{\nabla}_r\ln(T)\) vanishes. Since the semiconductor is not degenerated we can simplify the Fermi statistics by the Boltzmann statistics, i.e.

 \begin{equation*} \frac{\partial f_0}{\partial E}= - \frac{f_0}{k T} \qquad . \end{equation*}(5.25)

We take the band energies of the free electron gas

 \begin{equation*} E = E_0 + \frac{m^*}{2}\left|\vec{v}(\vec{k}) \right|^2 \; \mbox{, so} \quad v_i = \frac{\hbar}{m^*}k_i \quad . \end{equation*}(5.26)

In addition we assume \(\tau\) to be independent of \(\vec{k}\). Summing up all approximations we find for the particle current

 \begin{equation*} \vec{j}= - \frac{q \tau \hbar^2}{4 \pi^3 k T(m^*)^2} \left[e\vec{E}-\vec{\nabla}\mu \right] \tilde{M} \qquad , \end{equation*}(5.27)

and \(\tilde{M}\) is a matrix with the components

 \begin{equation*} \tilde{M}_{ij}=\int_{V_k} k_i k_j f_0(E(k))d^3k = \int_k dk k^4 f_0(E(k)) \int_{S_k}d\Omega\frac{k_i k_j}{k^2} \qquad . \end{equation*}(5.28)

Using \(k^2 = k_x^2 + k_y^2 + k_z^2\) and integrating over the surface of a sphere we get

 \begin{equation*} \int_{S_k}d\Omega\frac{k_i k_j}{k^2} = \frac{4 \pi}{3} \delta_{ij} \qquad . \end{equation*}(5.29)

Using

 \begin{equation*} dE = \frac{\hbar^2}{m^*} k dk \qquad , \end{equation*}(5.30)

the current density is written as

 \begin{equation*} \vec{j}= - \frac{q \tau}{3\pi^2kTm^*}\left[q\vec{E}-\vec{\nabla}\mu \right] \left(\frac{2m^*}{\hbar^2} \right) \int_{E_0}^{\infty} dE (E-E_0)^{\frac{3}{2}} \exp\left(- \frac{E-\mu}{k T} \right) \qquad . \end{equation*}(5.31)

After partial integration we get

 \begin{equation*} \vec{j}= - \frac{q \tau}{3\pi^2kTm^*}\left[q\vec{E}-\vec{\nabla}\mu \right] \left(\frac{2m^*}{\hbar^2} \right) \frac{-3kT}{2} \int_{E_0}^{\infty} dE (E-E_0)^{\frac{1}{2}} \exp\left(- \frac{E-\mu}{k T} \right) \qquad . \end{equation*}(5.32)

Taking into account the density of state of free electrons

 \begin{equation*} D(E) = \frac{1}{2\pi^2} \left(\frac{2m^*}{\hbar^2} \right)^{\frac{3}{2}} (E-E_0)^{\frac{1}{2}} \qquad , \end{equation*}(5.33)

we finally get

 \begin{equation*} \vec{j}= \frac{q \tau}{m^*}\left[q\vec{E}-\vec{\nabla}\mu \right] \int_{E_0}^{\infty} dE D(E) f_0(E) = \frac{q \tau n_e}{m^*}\left[q\vec{E}-\vec{\nabla}\mu \right] \label{j_final}\qquad . \end{equation*}(5.34)


With frame Back Forward as PDF

© J. Carstensen (Stat. Meth.)