## Q. 5.2

Semiconductor Doping
One step in manufacturing transistors, which function as electronic switches in integrated circuits, involves diffusing impurity atoms into a semiconductor material such as silicon. Suppose a silicon wafer 0.1 cm thick, which originally contains one phosphorus atom for every 10 million Si atoms, is treated so that there are 400 phosphorus atoms for every 10 million Si atoms at the surface (Figure 5-10). Calculate the concentration gradient (a) in atomic percent/cm and (b) in $\frac{atoms}{cm^3 \cdot cm}$. The lattice parameter of silicon is 5.4307 Å.

## Verified Solution

a. First, let’s calculate the initial and surface compositions in atomic percent:

$c_{i}=\frac{1 \ P \ atom}{10^7 \ atoms} × 100=0.00001 \ at$% P

$c_{s}=\frac{400 \ P \ atoms}{10^7 \ atoms} × 100=0.004 \ at$% P

$\frac{\Delta c}{\Delta x}=\frac{0.00001 \ – \ 0.004 \ at\% \ P}{0.1 \ cm}=-0.0399 \frac{at\% \ P}{cm}$

b. To find the gradient in terms of $\frac{atoms}{cm^3 \cdot cm}$, we must find the volume of the unit cell. The crystal structure of Si is diamond cubic (DC). The lattice parameter is 5.4307 × $10^{-8}$ cm. Thus,

$V_{cell}=(5.4307 × 10^{-8} \ cm)^3=1.6 × 10^{-22} \frac{cm^3}{cell}$

The volume occupied by $10^7$ Si atoms, which are arranged in a DC structure with 8 atoms/cell, is

$V=[\frac{10^7 \ atoms}{8 \ \frac{atoms}{cell}}][1.6 × 10^{-22}(\frac{cm^3}{cell})]=2 × 10^{-16} \ cm^3$

The compositions in atoms/cm³ are

$c_{i}=\frac{1 \ P \ atom}{2 × 10^{-16} \ cm^3} =0.005 × 10^{18} \ \frac{P \ atoms}{cm^3}$
$c_{s}=\frac{400 \ P \ atoms}{2 × 10^{-16} \ cm^3} =2 × 10^{18} \ \frac{P \ atoms}{cm^3}$

$\frac{\Delta c}{\Delta x}=\frac{0.005 × 10^{18} \ – \ 2 × 10^{18} \ \frac{P \ atoms}{cm^3} }{0.1 \ cm}$
$=-1.995 × 10^{19} \ \frac{P \ atoms}{cm^3 \cdot cm}$