From our discussion today it sounds like our steady-state continuity equation is $\frac{\partial J_{p}}{\partial x}=\frac{e(p(x)-p^{o})}{\tau_{p}}\large$.
Then 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 $A=0$.
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.
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.