Hello everybody!
Sorry for the delay, but better late than never I suppose.
On Wednesday we were discussing what happens when we place a phosphorous doped semiconductor and a boron doped semiconductor beside one another.
Since the phosphorous atoms in the lattice have an extra valence electron compared to their silicon neighbors, the atoms will want to share electrons with the boron atoms in the lattice of the other semiconductor. The boron atoms having one fewer valence electron in relation to the silicon in the lattice.
Before the semiconductors were connected, the chemical potential \(\mu\) in the n-doped side was at 0.8 eV, and on the p-doped side \(mu=0.2\) eV. We are interested in finding at which value \(\mu\) the system reaches equilibrium in after the semiconductors are connected and electrons flow from n-doped to p-doped.
We started by investigating the field inside the slabs. The n-doped side would be have a negative net charge near the junction and the p-doped side would have a positive net charge. Then, we looked for the field.
If we consider that the distance \(X_{d}\) (called the depletion length) from the junction is ionized in the process, and we assume it carries a uniform charge density \(\rho\), then the field can be determined by Gauss's Law.
The electric field for just one of the slabs is determined as follows
\begin{align*}
\oint\vec{E}\cdot d\vec{a}&=AQ_{enc} \\
2EA&=\rho[AX_{d}]\dfrac{1}{\epsilon_{0}} \\
E&=\dfrac{X_{d}\rho}{2\epsilon_{0}}
\end{align*}
And together, they create a maximum field in the center of junction:
\begin{equation*}
E=X_{d}\dfrac{\rho}{\epsilon_{0}}
\end{equation*}
Since the field is zero at \(x=\pm X_{d}\) and \(X_{d}\dfrac{\rho}{\epsilon_{0}}\) at \(x=0\), and is a linear function of distance:
\begin{equation*}
E=\left\{\begin{array}{r r}
-\dfrac{\rho}{\epsilon_{0}}x+\dfrac{X_{d}\rho}{\epsilon_{0}} & x>0 \\
\dfrac{\rho}{\epsilon_{0}}x+\dfrac{X_{d}\rho}{\epsilon_{0}} & x<0 \\
\end{array}\right.
\end{equation*}
I know we need to find the potential energy of this function next, and also a way to represent \(\rho\) that is workable (density of states maybe). But overall I'm at a little bit of a loss.
Ideas?
edit: fixed a mistake with a factor of two
Thanks Ben.
ReplyDeleteIsn't pho = N_d*1.6 x 10^-19 coulombs where N_d is the donor dopant density in cm^-3?
DeletePlease note!
ReplyDeleteI think I forgot a factor of two in there for the field maximum. \textit{Each} charged slab contributes a \(E=\dfrac{1}{\epsilon_{0}}X_{d}\rho\) term, so the field maximum should actually read
\begin{equation*}
E=\dfrac{2X_{d}\rho}{\epsilon_{0}}
\end{equation*}
As a result, for the field we should get:
\begin{equation*}
E(x)=\left\{\begin{array}{l r}
\dfrac{2\rho}{\epsilon_{0}}\left(-x+X_{d}\right) & 0\leq x < X_{d} \\
\dfrac{2\rho}{\epsilon_{0}}\left(x+X_{d}\right) & -X_{d} < x < 0 \\
0 & \text{else}
\end{array}\right.
\end{equation*}
If I made another mistake in my correction, let me know!
This is wrong! Ignore this correction comment!
DeleteHi Ben, I do not remember my E&M very well...but it seems that your math is solid. Can you describe what you mean by Potential Energy?
ReplyDeleteI know we have an equation for the Electric field so cant we just use the fact that U (potential enrgy) = -Work done? Therefore U = Integral(1/2*e0*E^2) dV. So we have E all we need to do is integrate both our piece wise functions with respect to dV? Not sure.
And I agree with Zach that pho(rho?)= N_d*Charge that way we get C/cm^3 which is what we want.
Let me know what you think.
or for 1 dimension with respect to dx?
DeleteI had the electric potential of an electron in mind, and it sounds like you have a good handle on it! So first for the electric potential (voltage) you'd use (for the 1 dimensional case)
Delete\begin{equation*}
V=-\int E dx
\end{equation*}
And to get the work done on, say, an electron you'd multiply that by \(\text{e}^{-}\).
It's easier to just get the area underneath the \(\vec{E}\) field triangle, though. It's the same operation as integrating, you just gotta watch the signs.
I agree with that value of rho, too.
By the way, you can write equations in here by bounding math-y stuff with dollar signs (\$). To get mathematical symbols in the equations you just write backslash and the thing's name (no spaces). \int gives the integral symbol, \sigma would make a sigma, and \frac{numerator}{denominator} makes a fraction. It's LaTeX typesetting if you want to look up any documentation.
Not backslash dollar sign though! I'm also not 100% sure the dollar signs work here... If not, do backslash open-paren and backslash close-paren.
Delete$\tau\sigma\epsilon t$