Homework 4 and today's class are completely linked. I got the sense that today's class was difficult and confusing; but I think we can simplify and clarify the issues by breaking them down into parts.
From a mathematical perspective, we obtained a differential equation and solved it with boundary conditions at $x=x_0$ and $\infty$. However, the boundary condition we used at $x=x_0$ was very non-obvious and important. In fact, the entire essence of p-n junction phenomenology, including diode behavior, LED's and perhaps solar cells, hinges on understanding the origin and implications of this b.c..
There are two parts which are described in more detail in the 1st comment below:
1) the first part involved getting a solve-able differential equation for the hole current in the region $x \ge x_0$, which is discussed in the 1st comment below.
2) the second part involved establishing the boundary conditions for that solution.
Boundary conditions are usually pretty boring, but in this case the boundary condition at $x = x_0$ is not at all boring and in fact involves a bold step that takes us outside the realm of equilibrium physics in a way that most of us have not experienced before. Let's focus on that first, and let's separate explaining what we did from justifying it.
What we did was to take the chemical potential that was valid on the far left, and use it at $x = +x_0$. Pretty weird, huh? The consequence of that assumption was that we obtained an enhanced value for p(x) at $x_0$ which leads to an enhanced p(x) for $x \ge x_0$ and an enhanced and unopposed $J_p (x)$ for $x \ge x_0$. That enhancement is exponential and it leads to a hole diffusion current at $x = x_0$ of $J_p (x) \approx (e n_{i}^2/N_d) [e^{eV/kT} - 1] (D/\tau_p)^{=1/2}$. There should be a similar electron diffusion current on the other side ($x \le -x_0$. These represent charge current flow in the same direction; holes to the right, electrons to the left and may provide a model basis fro understanding current-voltage relation for a p-n junction.
One implication of the enhanced minority carrier concentration (e.g., enhanced p(x) for x gt x_0) is that the ordinary relationships between n, p, $\mu$ and KT are not applicable. I believe you might also describe the regions around $x_0$ as "hot", though that may be more confusing than helpful.
Please post discussion of this below. There are a number of aspects to focus on including:
Added HW problem and discussion point.
How would you justify using $p(x_0) = (n_{i}^2/N_d) e^{eV/kT}$?
(or, equivalently, $p(x_0) = (N_a) e^{-e(\Delta \Phi-V)/kT}$? )
(did i get that right?)
There are two parts which are described in more detail in the 1st comment below:
1) the first part involved getting a solve-able differential equation for the hole current in the region $x \ge x_0$, which is discussed in the 1st comment below.
2) the second part involved establishing the boundary conditions for that solution.
Boundary conditions are usually pretty boring, but in this case the boundary condition at $x = x_0$ is not at all boring and in fact involves a bold step that takes us outside the realm of equilibrium physics in a way that most of us have not experienced before. Let's focus on that first, and let's separate explaining what we did from justifying it.
What we did was to take the chemical potential that was valid on the far left, and use it at $x = +x_0$. Pretty weird, huh? The consequence of that assumption was that we obtained an enhanced value for p(x) at $x_0$ which leads to an enhanced p(x) for $x \ge x_0$ and an enhanced and unopposed $J_p (x)$ for $x \ge x_0$. That enhancement is exponential and it leads to a hole diffusion current at $x = x_0$ of $J_p (x) \approx (e n_{i}^2/N_d) [e^{eV/kT} - 1] (D/\tau_p)^{=1/2}$. There should be a similar electron diffusion current on the other side ($x \le -x_0$. These represent charge current flow in the same direction; holes to the right, electrons to the left and may provide a model basis fro understanding current-voltage relation for a p-n junction.
One implication of the enhanced minority carrier concentration (e.g., enhanced p(x) for x gt x_0) is that the ordinary relationships between n, p, $\mu$ and KT are not applicable. I believe you might also describe the regions around $x_0$ as "hot", though that may be more confusing than helpful.
Please post discussion of this below. There are a number of aspects to focus on including:
- the math,
- the assumptions,
- the results, including the magnitude of the current, its dependence on voltage,
- and the picture that emerges from all this including where recombination occurs, etc.
Added HW problem and discussion point.
How would you justify using $p(x_0) = (n_{i}^2/N_d) e^{eV/kT}$?
(or, equivalently, $p(x_0) = (N_a) e^{-e(\Delta \Phi-V)/kT}$? )
(did i get that right?)
From our discussion today it sounds like our steady-state continuity equation is $\frac{\partial J_{p}}{\partial x}=\frac{e(p(x)-p^{o})}{\tau_{p}}\large$.
ReplyDeleteThen we substitute our equation for $J_p$ for $x>x_o$
\[J_p(x>x_o)=-eD\frac{\partial p(x)}{\partial x}\]into the continuity equation with the intention of solving for the hole concentration for $x>x_o$.
Noting that $\frac{\partial J_{p}}{\partial x}=-eD\frac{\partial^2 p(x)}{\partial x^2}\large$ the continuity equation becomes
\[-eD\frac{\partial^2 p(x)}{\partial x^2}=\frac{e(p(x)-p^{o})}{\tau_{p}}\]Canceling out the e's and bringing $\tau_p$ to the other side yields\[-D\tau_p\frac{\partial^2 p(x)}{\partial x^2}=p(x)-p^o.\]It is easiest to visualize how to solve this differential equation by rewriting it as
\[(\frac{\partial^2 }{\partial x^2}+\frac{1}{D\tau_p})p(x)=\frac{p^o}{D\tau_p}\]The complementary solution is
\[p_c(x)=Ae^\frac{x}{\sqrt{D\tau_p}}+Be^\frac{-x}{\sqrt{D\tau_p}}\]and the particular solution is
\[p_p(x)=p^o.\]As $x\rightarrow \infty\large$ we require that $p(x)\large$ is finite.
Thus $A=0$.
Therefore $p(x)=p^o+Be^\frac{-x}{\sqrt{D\tau_p}}.\large$
Our boundary condition is that at $x=x_o$, $p=p(x_o)$ so
\[p(x_o)=p^o+Be^\frac{-x_o}{\sqrt{D\tau_p}}\]\[p(x_o)-p^o=Be^\frac{-x_o}{\sqrt{D\tau_p}}\]\[B=(p(x_o)-p^o)e^\frac{x_o}{\sqrt{D\tau_p}}\]Therefore,\[p(x)=p^o+(p(x_o)-p^o)e^\frac{-(x-x_o)}{\sqrt{D\tau_p}}\]It is important to note that the units of $D\tau_p$ is $length^2$, so the term in the exponential is unitless.
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ReplyDeleteNice post. Everyone is encouraged to comment and/or ask questions about the above math, the interpretation, and the the value we assume for the very important p(x_0) term. (Where does it come from? Why is it important? What does it imply about enhancement of p(x) and J(x)?) What is J(x) as x approaches x_0 (from the right) and why is that important?
ReplyDeleteAlso, what is the significance of the length scale associated with $(D \tau_p$ ?
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ReplyDeletemake that: ($D \tau_p )^{1/2}$.
ReplyDeleteWhat are the units of that anyway?
What does it look like when you rewrite it in terms of $\mu$ ? (Einstein rel'n)
The units of $(D\tau_p)^{1/2}$ are length; recall, $J=-eD\frac{dp}{dx}$. Since $p$ has units of 1/length^3, and dx is units of length, and e is charge, and J is charge/time/length^2, it must be that D has units of length^2/time. $\tau_p$ is a time, and so $D\tau_p$ has units of length^2. Thus, $\sqrt{D\tau_p}$ has units of length.
ReplyDeleteRecall from the Einstein relation that $D=\mu kT/e$. Moreover, $\mu=e\tau_{\mbox{collision}}/m^*$, where $\tau_{\mbox{collision}}$ is the average time between collisions (the time for an electron or hole to scatter off a lattice imperfection or a phonon) and $m^*$ is the reduced mass. Moreover, kT is the thermal energy of the particle, which will have a thermal velocity $v$ such that $kT = 1/2m^*v^2$. Combining these, we have that $\sqrt{D\tau_p} = \sqrt{1/2v^2\tau_p\tau_{\mbox{collision}}}$. We can think of $v\tau_p$ as the recombination mean free path, and $v\tau_{\mbox{collision}}$ as the collision mean free path.
Note that the geometric mean of $\{a_1,a_2,...,a_k\}$ is $(a_1a_2...a_k)^{1/k}$. So, we see that $\sqrt{D\tau_p}$ is proportional to the geometric mean of the recombination mean free path and the collision mean free path.
This result seems very interesting to me, because it shows that the diffusion length, $\sqrt{D\tau_p}$, is directly related to how far the charge carrier can go before scattering (collision mean free path) and before recombining with the opposite type carrier (collision mean free path).
Correction/Question:
ReplyDeleteI don't think $kT=1/2m^*v^2$, but rather there is some other multiplicative factor (instead of 1/2). Can anyone clarify this for me?
There's a 3/2 in front of the kT--you get a kT/2 for each degree of freedom.
ReplyDeleteKelsey nailed it. Geno your post is great and was really easy to follow (well laid out), clearing stuff up along the way. Thanks! Conceptually I am doing okay but my math gets really muddled and messy.
ReplyDelete