Maybe this will help clarify some of the issues that came up in today's class. This focuses on the nature of a metal-oxide-semiconductor(p-doped) sandwich, where-in the chemical potentials of the metal and doped semiconductor are perfectly aligned. In the left-hand picture the voltage is zero. If you apply a positive voltage, that leads to depletion in the p-doped semiconductor (center picture). All the equations pertain to the center illustration and to the use a a depletion ansatz for that case. It is interested to see how the applied voltage, V, influences the depletion width, $z_0$, and the band bending, $\Delta \Phi$.
Looking at these relationships, which seem pretty understandable, i am thinking we should have started with this simpler zero chemical potential difference case. For extra-credit, who can figure out/ calculate how a higher chemical ptential in the metal would effect the equations for the depletion width and $\Delta \Phi$.
PS. The vertical scale is meant to be related through the sequence of 3 pictures. Notice how the chemical potential on the far right side of the semiconductor in each picture is the same. What does that mean?
ReplyDeleteAlso, how/why does a positive voltage correspond to the chemical potential of the metal going down??
Applying a positive voltage across the junction causes the metal side to have a more positive charge and the semiconductor side to have a more negative charge, since this is the charge distribution that corresponds to a positive voltage. This charge imbalance comes from a migration of electrons from the metal to the semiconductor (or, equally valid, a migration of holes from the semiconductor to the metal). For a metal, the chemical potential or Fermi surface is the highest occupied orbital. If there are fewer electrons, fewer orbitals will be occupied. The chemical potential is therefore lower.
ReplyDeleteSo Zack and I were discussing the relationship between applied voltage and the
ReplyDeletecorresponding shift in the bands for a MOS capacitor that starts off with
$\mu_{metal} = \mu_{semi}$. He worked out a formula for this, making the
depletion approximation (ie, the assumption that the charge that builds up in
the semiconductor forms a square well of sorts--we're ignoring the exponential
increase in the n-carriers near the semiconductor oxide boundary). The result,
as posted earlier, is:
$\Delta \phi = \frac{\epsilon_{oxide}^2V^2}{2\epsilon_{semi}eN_aW_{oxide}^2}$.
Here, $\epsilon_{oxide}$ and $\epsilon_{semi}$ are the electrical permitivities
for the oxide and semiconductor, which are 3.9$\epsilon_0$ and 12$\epsilon_0$
respectively. $N_a$ is the number of acceptors; we assumed $10^{17}$cm$^{-3}$.
$W_{oxide}$ is the thickness of the oxide--we assumed 20nm. V is the applied
voltage and $\Delta \phi$ is the amount of band bending that occurs.
We can look at this formula another way, introducing the variable $V_c$:
$\Delta \phi = \frac{V}{V_c}V$, where $V_c =
\frac{2\epsilon_{semi}eN_aW_{oxide}^2}{\epsilon_{oxide}}$. When we plug in the
numbers (and please, check our results think them unreasonable), we find $V_c =
0.54V$.
So, $V_c$ is an energy scale we can use to gauge how much $\Delta \phi$ depends
on the applied voltage $V$. However, remember that this is only valid for
small applied voltages, when the depletion approximation remains valid.