Question 5.9: Silicon Device Fabrication Devices such as transistors are m......

We assume that we can use one of the solutions to Fick’s second law (i.e., Equation 5-7):
\frac{c_{s} \ – \ c_{x}}{c_{s} \ – \ c_{0}}= erf ( \frac{x}{2\sqrt{Dt}})       (5-7)
We will use concentrations in atoms/cm³, time in seconds, and D in \frac{cm^2}{s} . Notice that the left-hand side is dimensionless. Therefore, as long as we use concentrations in the same units for c_{s}, c_{x}, and c_{0}, it does not matter what those units are.
\frac{c_{s} \ – \ c_{x}}{c_{s} \ – \ c_{0}}=  \frac{10^{20} \ \frac{atoms}{cm^3} \ – \ 10^{18} \ \frac{atoms}{cm^3}}{10^{20} \ \frac{atoms}{cm^3} \ – \ 0 \ \frac{atoms}{cm^3}}=0.99
= erf \left[\frac{x}{2 \sqrt{(6.5 × 10^{-13} \ \frac{cm^2}{s})(3600 \ s)}} \right]
= erf \left( \frac{x}{9.67 × 10^{-5}} \right)
From the error function values in Table 5-3 (or from your calculator/computer), if erf(z) = 0.99, z = 1.82; therefore,
1.82 = \frac{x}{9.67 × 10^{-5}}
or
x = 1.76 × 10^{-4} \ cm
or
x = (1.76 × 10^{-4} \ cm)(\frac{10^4 \ μm}{cm})
x = 1.76 μm

Note that we have expressed the final answer in micrometers since this is the length scale that is appropriate for this application. The main assumptions we made are (1) the D value does not change while phosphorus gets incorporated in the silicon wafer and (2) the diffusion of P is only in one dimension (i.e., we ignore any lateral diffusion).