|MC...what?||We start with these...||Make a few of these...|
|Carriers are alive?||Some hints...||Related Topics|
1. You have the problem, now what do you do with it?
First of all, READ THE PROBLEM CAREFULLY! Take note of the information given. In one way or another, the type of semiconductor material, the temperature, time, presence or absence of light, doping, etc., will be given in the wording of the problem. Look for keywords like, "uniform", "infinite", "room temperature", "for a long time", "suddenly", "in the sun", or "in the dark".
2. Determine which equation you need.
Once you understand the problem, you need to determine the type of material you’re working with, whether it's p-type or n-type, and the minority carrier you’re solving for. Remember, for a p-type material, electrons are the minority carrier, and for an n-type material, holes are the minority carrier. So if you're solving for electrons in a p-type material, you would use the following equation:
If you're solving for holes in an n-type material, you would use this equation:
3. Simplify the equation.
Next, simplify the equation using the information you obtained in step 1. Also keep in mind the assumptions on pages 122, 241, and 389 in your book, depending on the problem you are solving.
If the semiconductor has been a certain way for a long time, we can assume that it is in steady state. Remember that steady state means that nothing changes as a function of time, so time derivatives go to 0. So, in steady state:
Otherwise, if "suddenly at time t = ?" a condition changed, that term probably stays in the equation. If something like this occurs, you may have to solve for both instances, before the condition changed, and after because one represents the initial condition.
Diffusion Current or Change in Concentration
When you see the words "uniformly doped", we can assume that the semiconductor is doped equally throughout the region. In other words, the concentration does not vary as a function of position. This is a necessary condition to use the MCDEs. (Why?)
The spatial derivatives go to zero when there is no variation in the concentration of n or p with position:
This one is hard to decide sometimes. Look for any indication that there are at least two points where the values are not the same. Note that one or both of the partial derivatives (time or space) must be eliminated to be able to find a "nice" solution to the problem.
If the problem asks you to solve with respect to position, chances are it stays and the time derivative will be eliminated (steady state problem). If the problem asks for the change in concentration of the minority carrier with respect to time, the spatial derivative must go but, for full credit you must explain why. Look for clues in the problem statement.
Pay close attention to what is said about minority carrier lifetime or diffusion length and the width of the region of interest. This can give you clues as to whether or not the following is true:
We assume thermal R-G occurs when the minority carrier diffusion length is less than or equal to the width of the region we are analyzing. In the event that it is stated or assumed that the minority carrier diffusion length is much greater than the width of the region, as in the base region of an ideal BJT, we can assume that there isn't any thermal R-G taking place in that region. The reasoning behind that is we assume the carriers won't have time to recombine before they cross the region and are removed.
The presence of light causes photogeneration in the semiconductor, which means that carriers are being generated in the semiconductor. The clues to look for are phrases like "illuminated", "it has been outside", or "in the dark". If it is in the dark GL = 0. However, there are times when light is shining on the semiconductor, but GL = 0. Figure out what is happening to the light and whether it is penetrating the semiconductor.
4. Obtain a general solution.
Once you have simplified the equation by taking out some of the terms, match the equation you have to the general differential equation solutions found on your cheat sheet given in class and at the exam. Write down the general solution corresponding to the differential equation remaining after simplifying. If you have a spatial partial derivative, you should have two unknowns to solve for because it is a second order differential equation. If the time derivative remained, you only need to solve for one unknown because it is a first order differential equation.
5. Boundary Conditions
The hardest part of solving for the minority carrier concentration is determining the boundary conditions. Boundary conditions are used to solve for the unknowns and find the exact solution.
First you must identify the location(s) of the boundary(ies). If you have a time varying general solution, you need an initial condition, the value of the quantity at t = 0. If you have a spatial problem, you need to know the boundaries of the region of interest. If the boundary is "a long way away", it is typically at x = +-infinity. Semi-infinite semiconductors have boundaries at x = 0 and x = infinity. pn junctions have boundaries at +-infinity, -xp, and xn.
Next, determine the values at the boundary(ies). Sometimes these values are explicitly stated in the problem, but under most circumstances they must be inferred from the situation described. Here are some rules (assuming we are talking about holes, for electrons, substitute n for p where needed )
- If the minority carrier concentration "is at its equilibrium value" at a boundary, Dp = p - p0 = 0 because p = p0.
- If "all the minority carriers are removed" at a boundary, i.e. by an ohmic contact, p = p0 + Dp = 0, so Dp = -p0.
- If the boundary is at the edge of a pn junction depletion region, the law of the junction holds, so
- If a boundary is "a long way away" from things that are happening, but the region is "uniformly illuminated":
- If a boundary is in the dark and "a long way away" from what we call the "perturbation", usually the minority carrier concentration has returned to its equilibrium value (see 1. above).
6. Obtain Exact Solution.
Now you’re ready to find the exact solution. You do this by "plugging in" the values of x or t that you used to determine your boundary conditions, i.e. x = 0, x = ± infinity, t = 0, or t = ± infinity, into your general equation and setting it equal to the value you obtained as your boundary condition. Solve for your constants. WATCH YOUR ALGEBRA! This is what gets you into trouble most of the time.
7. Verifying Your Solution.
Finally! You've solved for the constants, you have the exact equation, but does it look right? The only way you can find out is to check it. "Plug in" the values you used to find your boundary conditions into your equation, the result should be the boundary condition you used in step 5 for that value. You can check the units of the result, as well.