Consider a long semiconductor bar is doped uniformly with donor atoms so that the concentration given by $n = N_D$ and is independent of position. Radiation falls upon the end of the bar at $x=0$, this light generates electron-hole pairs at $x=0$. light keeps on falling.


Because the material is n-doped (many electrons) the light does not significantly change the electron concentration. However, there are initially very few holes in the material, so the illumination does significantly change the number of holes. Holes in a n-type semiconductor are referred to as minority carriers.

Carrier transport in semiconductors takes place by drift and diffusion. The hole drift current can be ignored (We shall make the reasonable assumption that the injected minority concentration is very small compared with the doping level.

The statement that the minority concentration is much smaller than the majority concentration is called the low-level injection condition. Since the drift current is proportional to the concentration and we shall neglect the hole drift current but not the electron drift current and shall assume that $i_p$ is due entirely to diffusion. This assumption can be justified (see e.g. Electronic Principles, Paul E. Gray & Campbell L. Searle, John Wiley & Sons 1969, or Millman's Electronic Devices). The diffusion current density is proportional to the gradient in minority carrier concentration (in this case the holes) and diffusion coefficient,

$$j_p = -qD_p\frac{\partial p}{\partial x}$$ by Fick's law.

I wish to determine the time it takes for this system to reach steady state.

Steady state is the state at which the parameters (e.g current density and carrier concentration) at a particular position $x$ do not change with time. The continuity equation related to carrier current and generation and recombination rate is

$$\frac{\partial p}{\partial t} = -\frac{1}{q}\frac{\partial j_p}{\partial x} + G,$$

where $\tau_p$ is the mean life time, from the definition of mean life time and assuming that $\tau_p$ is independent of the magnitude of the hole concentration, $p_0$ is the value of $p$ in thermal-equilibrium value, $ g = p_0/\tau_p$ is the generation rate, $p/\tau_p$ is the recombination rate, and $G$ is the sum of generation rate and recombination rate.

Substituting the first equation and the value of $G$ into the second gives

$$\frac{\partial p}{\partial t} = D_p\frac{\partial^2 p}{\partial x^2} + \frac{p_0 - p}{\tau_p}.$$

In the steady state $p$ doesn't vary with time but vary w.r.t position and the concentration at $x=0$ will remain constant all the time hence we can put $$\frac{\partial p}{\partial t}=0;$$

hence when steady state is achieved we will have

$$\frac{\mathrm d^2 p}{\mathrm d x^2} = \frac{p - p_0}{D\tau_p}.$$

How much time will it take for the minority carrier concentration to reach this steady state value?


2 Answers 2


As I understand your question, the problem at hand is to solve the PDE for the minority carrier concentration:

$\dfrac{\partial p}{\partial t} = D_p\dfrac{\partial^2 p}{\partial x^2} + \dfrac{p_0 - p}{\tau_p}$

on the half-line $[0, \infty)$, subject to the initial condition:

$p(x, 0) = p_0$

And the boundary conditions:

$p(0, t) = p_0 + N_m$

Where $N_{m}$ is the excess minority carrier concentration due to the photoelectric effect (reference), and

$\lim_{x \to \infty} p(x,t) = p_0.$

To solve this, considering splitting the problem into the homogeneous PDE:

$\dfrac{\partial p}{\partial t} - D_p\dfrac{\partial^2 p}{\partial x^2} + \dfrac{p}{\tau_p} = 0 $

And the inhomogeneous PDE:

$\dfrac{\partial p}{\partial t} - D_p\dfrac{\partial^2 p}{\partial x^2} + \dfrac{p}{\tau_p} = \dfrac{p_0}{\tau_p}$.

Since the system is linear, if we have have a general solution to the homogeneous equation and a particular solution to the inhomogeneous equation, then if their sum satisfies the boundary conditons it will be a solution to the original problem.

It's easy to see that a particular solution to the inhomogeneous equation is $p(x,t)_p = p_0$.

Approaching the homogeneous PDE next, we can use the Laplace transform to find a general solution that satisfies the following BC and IC:

$p(0, t) = N_m$

$p(x, 0) = 0$, and

$\lim_{x \to \infty} p(x,t) = 0$.

Taking the Laplace transform:


In the time variable of the PDE gives us the ODE:

$s\hat{p}(x) - \mathcal{L}(p(x, 0)) = D_p\dfrac{d^2\hat{p}(x)}{dx^2} - \dfrac{\hat{p}(x)}{\tau_p} \implies $

$\dfrac{d^2\hat{p}(x)}{dx^2} = \hat{p}(x)\left(\dfrac{s}{D_p} + \dfrac{1}{D_p \tau_p}\right) = $

$\dfrac{d^2\hat{p}(x)}{dx^2} = C_n\hat{p}(x)$, where

$ C_n = \left(\dfrac{s}{D_p} + \dfrac{1}{D_p \tau_p}\right)$.

The general solution to this equation can be found by standard ODE methods and has the form:

$\hat{p}(x) = C_1e^{\sqrt{C_n}x} + C_2e^{-\sqrt{C_n}x}$.

Since we only desire solutions that decay as $x \to \infty$, the first solution is nonphysical and $C_1 = 0$.

The initial condition $p(0) = N_m\implies \hat{p}(0) = \dfrac{N_m}{s} \implies C_2 = \dfrac{N_m}{s}$,

so the solution to the ODE in the S domain is:

$\hat{p}(x) = \dfrac{N_m}{s}e^{-\sqrt{C_n}x}.$

Approaching the right hand side first, with some help from Wolfram Alpha, we find that the inverse Laplace transform of the function $P(s) = \frac{A}{s}e^{-\sqrt{s + \alpha}{x}}$ is:

$\frac{1}{2}Ae^{-\sqrt{\alpha}x}\left(\mathrm{erf}\left(\frac{2\sqrt{\alpha}t - x}{2\sqrt{t}}\right)+ e^{2\sqrt{\alpha}x}\mathrm{erfc}\left(\frac{2\sqrt{\alpha}t + x}{2\sqrt{t}}\right) + 1\right)$ for $x > 0$,

where $\mathrm{erfc}(z) = 1 - \mathrm{erf}(z)$ is the complementary error function.

Our function $\hat{p}(x)$ is of the form $\frac{A}{\beta}\frac{1}{{\frac{s}{\beta}}}e^{-\sqrt{\frac{s}{\beta} + \alpha}{x}}$, so applying the scaling theorem:

$\mathcal{L^{-1}}(P(\frac{s}{\beta})) = \beta p(\beta t)$ we finally get:

$\mathcal{L^{-1}}(\hat{p}(x)) = p(x,t)_h = \frac{1}{2}N_me^{-\sqrt{\alpha}x}\left(\mathrm{erf}\left(\frac{2\sqrt{\alpha}t\beta - x}{2\sqrt{t\beta}}\right)+ e^{2\sqrt{\alpha}x}\mathrm{erfc}\left(\frac{2\sqrt{\alpha}t\beta + x}{2\sqrt{t\beta}}\right) + 1\right)$,

where $\alpha = \frac{1}{D_p\tau_p}$, $\beta = D_p$, and the subscript "h" in $p(x,t)_h$ indicates that this is the homogeneous solution.

Summing the particular and homogeneous solution we get:

$p(x,t)_h + p(x,t)_p = p(x,t) = \frac{1}{2}N_me^{-\sqrt{\alpha}x}\left(\mathrm{erf}\left(\frac{2\sqrt{\alpha}t\beta - x}{2\sqrt{t\beta}}\right)+ e^{2\sqrt{\alpha}x}\mathrm{erfc}\left(\frac{2\sqrt{\alpha}t\beta + x}{2\sqrt{t\beta}}\right) + 1\right) + p_0$,

which satisfies the conditions of the original problem. It should be straightforward to take the limit of this expression as $t \to \infty$.

  • $\begingroup$ i don't think this is the correct solution it is not just about solving differential equations but your mathematics skills are great. at the very beginning fick's doesn't hold good so $∂p/∂t=D_p\ ∂^2p/∂x^2+ \frac{p_0−p}{τ_p}$ is not appropriate as you can see at t=0,x=0 $ \partial p/ \partial x$ is infinite and $\partial p/ \partial t$ will also be infinite which is obviously impossible.we need some physics here. $\endgroup$
    – user31782
    Dec 3, 2013 at 9:22

Actually if you integrate in space, you end up with a first order relaxation differential equation for the total number of carriers. So I would expect the equilibrium to be reached within $3\tau_p$.


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