Chapter 5, Problem 5.56P

Interpretation Introduction

Interpretation:

The surface concentration of $P$ and $B$ needs to be calculated.

Concept introduction:

The Fick's second law is used to determine the amount of phosphorus and boron at any depth $x$ under the surface.

The expression for the Fick's second law is as follows:

  $\frac{\partial c}{\partial t}=\frac{\partial }{\partial x}\left(D\frac{\partial c}{\partial x}\right)$

After derivation:

  $\frac{c_s-c_x}{c_s-c_0}=\text{erf}\left(\frac{x}{2\sqrt{Dt}}\right)\mathrm{..........}\left(1\right)$

Here, $c_s$ is the concentration of diffusion at the surface, $c_0$ is the concentration of diffusion at x, $c_0$ is the uniform concentration of diffused atoms in the material, $c_x$ is the concentration of atoms diffusing at x and D is diffusion coefficient.

Answer to Problem 5.56P

Surface concentration for $n$ type after $\text{1}\text{h}$, is $1.1\times {10}^{18}{\text{ atoms/cm}}^3$.

Surface concentration for $p$ type after $\text{1}\text{h}$ is $1.1\times {10}^{18}{\text{ atoms/cm}}^3$

Explanation of Solution

For the n type region:

Here, the constant concentration of diffusing atoms at the surface is $c_s$.

The uniform concentration of diffusing atoms in the material $c_0=0$ (concentration is Zero because at the starting of process there will be no diffusion of atom through the barrier due to the lack of threshold energy).

The concentration of diffusing atoms at location $x=1\times {10}^{-5}cm{\text{ is 10}}^{18}{\text{ atoms/cm}}^3$.

Time $t$ when concentration is $c_x\text{ is 3600 }s$.

The diffusion coefficient $D\text{ is 6}\text{.5}\times {\text{10}}^{-13}c{m}^2/s$.

Substitute the above values in Fick's equation $\left(1\right)$ as given below.

  $\begin{array}{c}\frac{c_s-c_x}{c_s-c_0}=\text{erf}\left(\frac{x}{2\sqrt{Dt}}\right)\\ =\text{erf}\left(\frac{1\times {10}^{-5}}{2\sqrt{\left(6.5\times {10}^{-13}\right)\left(3600\right)}}\right)\\ =\text{erf}\left(0.103\right)\end{array}$

Find the value of $\left(\text{erf}\frac{x}{2\sqrt{Dt}}\right)$ when the value of $\left(\frac{x}{2\sqrt{Dt}}\right)\text{ is 0}\text{.103}$, using table $\left(1\right)$.

Error function values for Fick's second law
Argument of the error   $\frac{x}{2\sqrt{Dt}}$	Values of the error function   $\text{erf}\frac{x}{2\sqrt{Dt}}$
$0$   $$	$0$
$0.10$	$0.1125$
$0.20$	$0.2227$
$0.30$	$0.3286$
$0.40$	$0.4284$
$0.50$	$0.5205$
$0.60$	$0.6039$
$0.70$	$0.6778$
$0.80$	$0.7421$
$0.90$	$0.7969$
$1.00$	$0.8427$
$1.50$	$0.9961$
$2.00$	$0.9953$

  $\begin{array}{c}\text{erf}\left(0.103\right)=0.1125\\ \frac{c_s-c_x}{c_s-c_0}=0.1125\\ \frac{c_s-{10}^{18}}{c_s-0}=0.1125\\ c_s-{10}^{18}=0.1125c_s\\ c_s=\frac{{10}^{18}}{0.8875}\\ c_s=1.1\times {10}^{18}{\text{ atoms/cm}}^3\end{array}$

Hence, Surface concentration for $n$ type after $1hr$, will be $1.1\times {10}^{18}{\text{ atoms/cm}}^3$.

For the $p$ type region

Here, the constant concentration of diffusing atoms at the surface is $c_s$.

The uniform concentration of diffusing atoms in the material $c_0=0$ (concentration is Zero because at the starting of process there will be no diffusion of atom through the barrier due to the lack of threshold energy).

The concentration of diffusing atoms at location $x=1\times {10}^{-5}{\text{cm is 10}}^{18}{\text{ atoms/cm}}^3$.

Time $t$ when concentration is $c_x\text{ is 3600 }s$.

The diffusion coefficient $D\text{ is 6}\text{.1}\times {\text{10}}^{-13}{\text{cm}}^{\text{2}}\text{/s}$.

Substitute the above value in equation $\left(1\right)$ as given below.

  $\begin{array}{c}\frac{c_s-c_x}{c_s-c_0}=\text{erf}\left(\frac{x}{2\sqrt{Dt}}\right)\\ =\text{erf}\left(\frac{1\times {10}^{-5}}{2\sqrt{\left(6.1\times {10}^{-13}\right)\left(3600\right)}}\right)\\ =\text{erf}\left(0.1\right)\end{array}$

Find the value of $\left(\text{erf}\frac{x}{2\sqrt{Dt}}\right)$ when the value of $\left(\frac{x}{2\sqrt{Dt}}\right)\text{ is 0}\text{.1}$, using table $\left(1\right)$.

  $\begin{array}{c}\text{erf}\left(0.103\right)=0.1125\\ \frac{c_s-c_x}{c_s-c_0}=0.1125\\ \frac{c_s-{10}^{18}}{c_s-0}=0.1125\\ c_s-{10}^{18}=0.1125c_s\\ c_s=\frac{{10}^{18}}{0.8875}\\ c_s=1.1\times {10}^{18}{\text{ atoms/cm}}^3\end{array}$

Hence, surface concentration for $p$ type after $\text{1}\text{hr}$ will be $1.1\times {10}^{18}{\text{ atoms/cm}}^3$

The value of surface concentration for both $p$ type and $n$ type are the same, this is because of the little difference in their diffusion coefficient.

Conclusion

Thus, the surface concentration for $n$ type after $1hr$, will be $1.1\times {10}^{18}{\text{ atoms/cm}}^3$.

Surface concentration for $p$ type after $1hr$ will be $1.1\times {10}^{18}{\text{ atoms/cm}}^3$