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A Nanowire with charge neutrality in band gap

  1. Mar 24, 2017 #1
    I'm working on a finite element simulation of the electrostatic potential [itex]V(r)[/itex] in and around semiconductor nanowires, based on solving Poisson's equation. While the details of the problem will follow shortly, the crux of where I run into trouble is that for nanowires it is important to include the effects of surface states. This is usually done in the form of a constant surface state density, but I want to include that the surface represents a capacitor with a variable charge as a function of the Fermi level.

    To do so, I will include the so called charge neutrality level (CNL) [itex]\Phi_{NL}[/itex], following the paper Resolving ambiguities in nanowire field-effect transistor characterization by Heedt et al. (which should be open access). According to this, one can take the surface states and the charge neutrality level into account by assuming a finite density of surface states [itex]D_s[/itex] and a surface charge density given by
    \sigma_s(r) = D_s\left[\Delta\Phi_{NL} - eV(r) - \chi_{InAs}\right]
    where [itex]\Delta\Phi_{NL} = \Phi_{NL} - E_c[/itex] is the location of the charge neutrality level w.r.t. the conduction band edge [itex]E_c[/itex] and [itex]\chi_{SE}[/itex] is the electron affinity of the semiconductor. Note that this expression is what sets the model apart from many others.

    Now, the above gives us the surface density. What remains is to define the the space charge density. This can be done with the Kane model in the context of [itex]k \cdot p[/itex] perturbation theory, where one writes the electron concentration in terms of its density of states integrated with the fermi-dirac distribution
    n(r) = \int_{E_c}^{\infty} \frac{(2m_e^*)^{3/2}}{2 \pi^2 \hbar^3}\sqrt{E-E_c} \frac{1}{\exp(\frac{E-V(r)}{k_B T})}dE
    and similarly for the hole concentration
    p(r) = \int_{-\infty}^{E_c-E_g} \frac{(2m_p^*)^{3/2}}{2 \pi^2 \hbar^3}\sqrt{E-(E_c-E_g)} \left(1-\frac{1}{\exp(\frac{E-V(r)}{k_B T})}\right)dE
    leading to a space charge density of
    \rho(r) = e\left[p(r) - n(r)\right]
    In the above [itex]m_{\{e,p\}}[/itex] are the effective electron and hole masses, and [itex]E_g[/itex] is the band gap.

    Okay, great! At this stage we actually have all that is needed to set up a finite element simulation by drawing the hexagonal wire, defining the space charge and the surface charge density, surrounding the wire with something like air and filling in appropriate parameters.

    But here we are getting to my problem. What the above model is supposed to do (if I understand it correctly) is find [itex]V(r)[/itex] so that it can match [itex]\rho(r)[/itex] and [itex]\sigma(r)[/itex], essentially bending the bands. This works for [itex]\chi_{NL} > E_c[/itex], where the CNL lies in the conduction band and results in an electron accumulation layer, and also for [itex]\chi_{NL} < E_c - E_g[/itex], where it lies in the valence band. But how should the above model work for [itex]E_c - E_g <\Delta\chi_{NL} < E_c[/itex]? In this region the space charge density expression is simply zero, as the integrals are only evaluated outside of the bandgap. So the model can't exactly modify [itex]V(r)[/itex] to bend the bands into the region.

    If I understand it correctly the model thus needs a modification that allows the bands to go into the gap around the surface, and I am not sure how to do this. What I've come to realise is that what one essentially needs is that [itex]E_c \rightarrow E_c(V(r))[/itex]; this would make the limits of the integrals move with the potential, as they should. But how does the conduction band edge respond to the potential? That seems rather tricky..

    [1]: http://pubs.rsc.org/en/Content/ArticleLanding/2015/NR/C5NR03608A#!divAbstract
  2. jcsd
  3. Mar 29, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
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