**ECE 1406, Homework #6, Due May 13, 2003**

1. 4.15. (see equation 4.13 for part b.) Note also that Irvin's curves do not assume a constant mobility.

Assume that the diffusion length [(DT)^1/2] is 0.4 microns. From this information, you can find the background concentration (which is necessary for using Irvin's curves). Note that for the contribution of the different parts of the diffusion, you will have to integrate numerically.

You can do the numerical integration using any method you want. I generally use MATLAB, so I am including an example file to use for finding the integral of a gaussian. (I used this in the solution to homework #5) To use this file in MATLAB, first edit it with the correct parameters for the gaussian, and then use the quad8 function as follows:

D=quad8('gausint', 0, 2e-4)

This integrates the function in gausint from 0 to microns (assuming all your units are in cm!).

The m-file for matlab is given below, % indicates a comment line.

function N=gausint(x);

% gausint is the function that contains a gaussian to be integrated.

% No and DT are the surface concentration and the square of the

% diffusion length, and are adjusted for

% each new gaussian to be integrated.

No=1.0e20;

DT=0.16e-8;

%

N=No*exp(-x.^2/(4*DT));

2. Compare the resistivity found using the mobility expression in problem 4.24
with that in figure 4.8. Compare at 10^{20}, 10^{19},
and 10^{18} cm^{-3}, for both
n and p-type doping. Note that the resistivity is the reciprocal of the conductivity.
Also note that the conductivity, s=qnm_{n}
or qpm_{p}. (This equation
has greek letters. If they did not display correctly, s corresponds to sigma,
the conductivity, and m is the mobility, mu.)

3. 4.24. The easiest way to do this is to use matlab or some other mathmatical package with integration capability or a calculator which can do numerical integration.

4. 5.6. Hint: Use Irvin's curves to find the surface concentration and then
work backward using N_{0} and x_{j}
to find the required implant dose. (You don't need to find the energy.) Also
note that only half of the implanted dose goes into the silicon.