Hi everyone,

I will be as clear and concise as possible. I am attempting to model how much dust accumulates on a given surface as a function of time. The surface (a wafer) is exposed to an environment (a chamber) for a minute, then is removed. This cycle is repeated (say n times). I counted the dust particles at time (cycle) n on the wafer Wn, and the dust in the chamber as Cn. Now, the chamber starts off very contaminated, say with C0 = 1000000 particles (C0 for n = 0). And the wafer can be considered clean (W0 = 0). But, not only is dust going on the wafer, but it is also getting removed from the chamber by vents. So, at each cycle, some fraction (l) of the chambers particles are removed, and some fraction (k) is deposited onto the wafer. I think I am correct in expressing it like this:
Wn = Wn-1 kCn (or maybe this should be kCn-1, not kCn?)
Cn = Cn-1(1 - k - l)

I smell something like e^(-cn) or something, based on a quick plot I did of data I made using a C program to simulate the steps summarized by the two equations above. But, depending on the parameters, sometimes the result looks linear. I would like a nice way to express Wn = f(W0, C0, k, l, n). If this doesn't exist then any tips would certainly be appreciated.

Thanks very much!
Eric

This is a differential equation. To get a closed-form answer we need to use continuous time and get rid of the notion of "cycles".

w = volume of particles on the wafer
c = density of particles in the chamber

The rate of accumulation on the wafer is proportional to the density of chamber particles.

dw/dt = kc
dc/dt = -akc-lc ; let l' = ak+l where a measures the ratio of units
dc/dt = -l'c
dc/c = -l' dt
ln c = -l't + C1
c = exp(-l't+C1)
and we want to choose C1 so that c(0) = the desired concentration

dw/dt = k(exp(-l't)+C1)
dw = k(exp(-l't)+C1)dt
w = -k/l' exp(-l't) + C1t + C2
and we want to choose C2 so that w(0) = 0 (it'll be k/l' )

so it is always exponential. However, if the ratio -k/l' is small compared to C1, or if l't is large making exp(-l't) small compared to C1, the function will behave very much like C1t + C2.