Question

Arsenic diffusion in Si: Arsenic is diffused into a thick slice of silicon with no previous arsenic in it at 1100ºC. If the surface concentration of the arsenic is 5.0 × 1018 atoms/cm3 and its concentration at 1.2 µm below the silicon surface is 1.5 × 1016 atoms/cm3 , how long must be the diffusion time? (D = 3.0 × 10-14 cm2 /s for As diffusing in Si at 1100ºC.)

Answer 1

Answer:

Diffusion time is 7.42 h

Solution:

As per the question:

Temperature, T = $1100^{\circ}C$

Surface concentration of arsenic, $C_{S} = 5.0\times 10^{18}\ atoms/cm^{3}$

Surface concentration below Silicon surface, $C_{x} = 1.5\times 10^{16}\ atoms/cm^{3}$

D = $3.0\times 10^{- 14}\ cm^{2}/s$

x = $1.2\mu m = 1.2\times 10^{- 4}\ cm$

Initial concentration at t = 0, $C_{o} = 0$

Now, by using Flick's second eqn:

$\frac{C_{S} - C_{x}}{C_{x} - C_{o}} = erf(\frac{x}{\sqrt{Dt}})$

Thus by putting appropriate values:

$\frac{5.0\times 10^{18} - 1.5\times 10^{16}}{5.0\times 10^{18}} = erf(\frac{1.2\times 10^{- 4}}{2\sqrt{3.0\times 10^{- 14}t}})$

$0.997 = erf(\frac{364.4}{\sqrt{t}})$              (1)

Now,

$erf(z) = 0.997$

Now, from error function values tabulation:

For z = 2.0, erf(z) = 0.998

For z = 2.2, erf(z) = 0.995

Now,

With the help of linear interpolation method:

$\frac{z - 2}{2.2 - 2.0} = \frac{0.997 - 0.995}{0.998 - 0.995}$

z = 2.12

Now, using eqn (1) and above value:

$\frac{364.4}{\sqrt{t}} = 2.12$

$t = (\frac{364.4}{2.12})^{2}$ = 26700 s

t = $\frac{26700}{3600} = 7.42\ h$