Chapter 5, Problem 5.85DP

Interpretation Introduction

Interpretation:

Considering the depth of material, the concentration profile should be determined. Also, the graph for the profile P for pre-deposition process should be plotted.

Concept introduction:

Constant source diffusion is the relationship of diffusion source with depth, diffusion coefficient and time. The relationship is given by using the below equation,

  $\xi =\frac{x}{2\sqrt{Dt}}$

Where,

$\xi$ is the constant source diffusion.

D is the diffusion coefficient.

t is the time.

x is the depth.

Constant source diffusion is of two types:

1. Pre deposition step: It is the first step for constant source diffusion in which a small quantity of impurity is doped using the constant source.
2. Drive in deposition: It is the second step for constant source diffusion in which the material is maintaining the higher value of temperature for a certain period of time.

Equation used for calculating the diffusion is,

Fick's Law of diffusion: This law states that molar flux is directly proportional to concentration gradient. The law is stated as:

  $\text{J}\text{=}-\text{D}\frac{{\text{dC}}_{\text{A}}}{\text{dx}}$

Where,

J is the molar flux defined as the number of atoms passing per unit area per unit time.

D is the diffusion coefficient in ${\text{cm}}^{\text{2}}\text{/sec}$.

  $\frac{{\text{dC}}_{\text{A}}}{\text{dx}}$ is the concentration gradient in $\frac{\text{atoms}}{{\text{cm}}^{\text{3}}\text{.cm}}$.

Factors affection diffusion are as follows:

1. Temperature
2. Diffusion coefficient

The following equation is stated as:

  $\text{D}={\text{D}}_{\text{0}}\exp \left(\frac{\text{-Q}}{\text{RT}}\right)$

Where,

Q is the activation energy in calorie/ mole.

R is universal gas constant in $\frac{\text{cal}}{\text{mole}\text{K}}$.

T is the absolute temperature in kelvin.

  ${\text{D}}_{\text{0}}$ is constant for given diffusion system and its value is defined as $\frac{\text{1}}{\text{T}}=0$ or $\text{T}\text{=}\infty$.

Explanation of Solution

Given Information:

The given equation for basis is:

  $c\left(x,t\right)=C_s\left[1-erf\left(\frac{x}{4Dt}\right)\right]$

Where,

c is the concentration considering pre deposition function.

D is the diffusion coefficient.

X is the depth.

T is the time.

erf is the error function.

The given boundary conditions are:

Depth = $0.25\text{μm}$.

Concentration of P at depth of $0.25\text{μm}$ is ${10}^{13}\text{atoms}\text{/}{\text{cm}}^{\text{3}}$.

The temperature at which material is conducted is ${1000}^{°}\text{C}$.

Value of diffusion coefficient is $2.5\times {10}^{-4}{\text{cm}}^2/\sec$.

Total time for carrying out the process is $8\text{hr}$.

Equation considered for calculation is,

  $c\left(x,t\right)=C_s\left[1-erf\left(\frac{x}{4Dt}\right)\right]$

On integrating the above equation within limits,

  $\begin{array}{l}c\left(x,t\right)=C_s\left[1-\frac{2}{\sqrt{\pi }}{\displaystyle \underset{0}{\overset{\frac{x}{2\sqrt{Dt}}}{\int }}{e}^{-y^2}}\text{dy}\right]\\ c\left(x,t\right)=C_s\times erfc\left(\frac{x}{2\sqrt{Dt}}\right)\end{array}$

Thus, the equation considered for calculation is,

  $c\left(x,t\right)=C_s\times erfc\left(\frac{x}{2\sqrt{Dt}}\right)$

Assuming the various values of depth required for calculation. Assumed values of depth are,

  $\begin{array}{l}x=0.25\text{μm}=2.5\times {10}^{-5}\text{cm}\\ x=0.50\text{μm}=5\times {10}^{-5}\text{cm}\\ x=0.75\text{μm}=7.5\times {10}^{-5}\text{cm}\\ x=1\text{μm}=1\times {10}^{-4}\text{cm}\\ x=1.25\text{μm}=0.000125\text{cm}\\ x=1.5\text{μm}=0.00015\text{cm}\\ x=1.75\text{μm}=0.000175\text{cm}\\ x=2\text{μm}=0.0002\text{cm}\\ x=2.25\text{μm}=0.000225\text{cm}\\ x=2.5\text{μm}=0.00025\text{cm}\end{array}$

Calculation of concentration with respect to time and depth. Considering depth as

  $x=0.25\text{μm}=2.5\times {10}^{-5}\text{cm}$

Value of diffusion coefficient (D) is $2.5\times {10}^{-4}{\text{cm}}^2/\sec$.

Total time for carrying out the process is $8\text{hr}$ = $8\text{hr}\times 60\frac{\text{sec}}{\text{hr}}=480\sec$.

Substituting the given value in the above relation,

  $c\left(x,t\right)=C_s\times erfc\left(\frac{x}{2\sqrt{Dt}}\right)$

  $\begin{array}{l}c\left(x,t\right)=C_s\times erfc\left(\frac{x}{2\sqrt{Dt}}\right)\\ c\left(x,t\right)={10}^{21}\times erfc\left(\frac{2.5\times {10}^{-5}}{2\sqrt{2.5\times {10}^{-14}\times 480}}\right)\\ c\left(x,t\right)={10}^{21}\times erfc\left(3.60\right)\end{array}$

Using table for calculation of error function,

  $\begin{array}{l}erfc\left(3.60\right)=0.99\\ c\left(x,t\right)={10}^{21}\times 0.99\\ c\left(x,t\right)=9.9\times {10}^{20}\text{atoms}{\text{/cm}}^{\text{3}}\end{array}$

Thus, concentration at a depth of $0.25\text{μm}$ is $c\left(x,t\right)=9.9\times {10}^{20}\text{atoms}{\text{/cm}}^{\text{3}}$.

Calculation of concentration with respect to time and depth. Considering depth as $x=0.50\text{μm}=5\times {10}^{-5}\text{cm}$

Value of diffusion coefficient (D) is $2.5\times {10}^{-4}{\text{cm}}^2/\sec$.

Total time for carrying out the process is $8\text{hr}$ = $8\text{hr}\times 60\frac{\text{sec}}{\text{hr}}=480\sec$.

Substituting the given value in the above relation,

  $c\left(x,t\right)=C_s\times erfc\left(\frac{x}{2\sqrt{Dt}}\right)$

  $\begin{array}{l}c\left(x,t\right)=C_s\times erfc\left(\frac{x}{2\sqrt{Dt}}\right)\\ c\left(x,t\right)={10}^{21}\times erfc\left(\frac{5\times {10}^{-5}}{2\sqrt{2.5\times {10}^{-14}\times 480}}\right)\\ c\left(x,t\right)={10}^{21}\times erfc\left(7.216\right)\end{array}$

When the value of error function exceeds the range of limit 4. The formula used for calculation of error function is,

  $\begin{array}{l}erf\left(x\right)=1-\frac{e^{-x^2}}{\sqrt{\pi }x}\\ x=7.216\\ erf\left(7.216\right)=1-\frac{e^{-{7.216}^2}}{\sqrt{\pi }\times 7.216}\\ erf\left(7.216\right)=1-\frac{2.43\times {10}^{-23}}{12.790}\\ erf\left(7.216\right)=1\\ erf\left(7.216\right)=0.8427\\ c\left(x,t\right)={10}^{21}\times 0.8427\\ c\left(x,t\right)=8.427\times {10}^{21}\text{atoms}\text{/}{\text{cm}}^{\text{3}}\end{array}$

Thus, concentration at a depth of $0.50\text{μm}$ is $c\left(x,t\right)=8.427\times {10}^{21}\text{atoms}{\text{/cm}}^{\text{3}}$.

Calculation of concentration with respect to time and depth. Considering depth as $x=0.75\text{μm}=7.5\times {10}^{-5}\text{cm}$

Value of diffusion coefficient (D) is $2.5\times {10}^{-4}{\text{cm}}^2/\sec$.

Total time for carrying out the process is $8\text{hr}$ = $8\text{hr}\times 60\frac{\text{sec}}{\text{hr}}=480\sec$.

Substituting the given value in the above relation,

  $c\left(x,t\right)=C_s\times erfc\left(\frac{x}{2\sqrt{Dt}}\right)$

  $\begin{array}{l}c\left(x,t\right)=C_s\times erfc\left(\frac{x}{2\sqrt{Dt}}\right)\\ c\left(x,t\right)={10}^{21}\times erfc\left(\frac{7.5\times {10}^{-5}}{2\sqrt{2.5\times {10}^{-14}\times 480}}\right)\\ c\left(x,t\right)={10}^{21}\times erfc\left(10.825\right)\end{array}$

When the value of error function exceeds the range of limit 4. The formula used for calculation of error function is,

  $\begin{array}{l}erf\left(x\right)=1-\frac{e^{-x^2}}{\sqrt{\pi }x}\\ x=10.825\\ erf\left(10.825\right)=1-\frac{e^{-{10.825}^2}}{\sqrt{\pi }\times 10.825}\\ erf\left(10.825\right)=1-\frac{1.2855\times {10}^{-51}}{19.186}\\ erf\left(10.825\right)=1\\ erf\left(10.825\right)=0.8468\\ c\left(x,t\right)={10}^{21}\times 0.8468\\ c\left(x,t\right)=8.468\times {10}^{20}\text{atoms}\text{/}{\text{cm}}^{\text{3}}\end{array}$

Thus, concentration at a depth of $0.75\text{μm}$ is $c\left(x,t\right)=8.468\times {10}^{20}\text{atoms}{\text{/cm}}^{\text{3}}$.

Calculation of concentration with respect to time and depth. Considering depth as $x=1\text{μm}=1\times {10}^{-4}\text{cm}$

Value of diffusion coefficient (D) is $2.5\times {10}^{-4}{\text{cm}}^2/\sec$.

Total time for carrying out the process is $8\text{hr}$ = $8\text{hr}\times 60\frac{\text{sec}}{\text{hr}}=480\sec$.

Substituting the given value in the above relation,

  $c\left(x,t\right)=C_s\times erfc\left(\frac{x}{2\sqrt{Dt}}\right)$

  $\begin{array}{l}c\left(x,t\right)=C_s\times erfc\left(\frac{x}{2\sqrt{Dt}}\right)\\ c\left(x,t\right)={10}^{21}\times erfc\left(\frac{1\times {10}^{-4}}{2\sqrt{2.5\times {10}^{-14}\times 480}}\right)\\ c\left(x,t\right)={10}^{21}\times erfc\left(14.433\right)\end{array}$

When the value of error function exceeds the range of limit 4. The formula used for calculation of error function is,

  $\begin{array}{l}erf\left(x\right)=1-\frac{e^{-x^2}}{\sqrt{\pi }x}\\ x=14.433\\ erf\left(14.433\right)=1-\frac{e^{-{14.433}^2}}{\sqrt{\pi }\times 14.433}\end{array}$

  $\begin{array}{l}erf\left(14.433\right)=0.9999\\ erf\left(14.433\right)=0.838\\ c\left(x,t\right)={10}^{21}\times 0.838\\ c\left(x,t\right)=8.38\times {10}^{20}\text{atoms}\text{/}{\text{cm}}^{\text{3}}\end{array}$

Thus, concentration at a depth of $1\text{μm}$ is $c\left(x,t\right)=8.38\times {10}^{20}\text{atoms}\text{/}{\text{cm}}^{\text{3}}$.

Calculation of concentration with respect to time and depth. Considering depth as $x=1.25\text{μm}=0.000125\text{cm}$.

Value of diffusion coefficient (D) is $2.5\times {10}^{-4}{\text{cm}}^2/\sec$.

Total time for carrying out the process is $8\text{hr}$ = $8\text{hr}\times 60\frac{\text{sec}}{\text{hr}}=480\sec$.

Substituting the given value in the above relation,

  $c\left(x,t\right)=C_s\times erfc\left(\frac{x}{2\sqrt{Dt}}\right)$

  $\begin{array}{l}c\left(x,t\right)=C_s\times erfc\left(\frac{x}{2\sqrt{Dt}}\right)\\ c\left(x,t\right)={10}^{21}\times erfc\left(\frac{0.000125}{2\sqrt{2.5\times {10}^{-14}\times 480}}\right)\\ c\left(x,t\right)={10}^{21}\times erfc\left(18.042\right)\end{array}$

When the value of error function exceeds the range of limit 4. The formula used for calculation of error function is,

  $\begin{array}{l}erf\left(x\right)=1-\frac{e^{-x^2}}{\sqrt{\pi }x}\\ x=18.042\\ erf\left(18.042\right)=1-\frac{e^{-{18.042}^2}}{\sqrt{\pi }\times 18.042}\end{array}$

  $\begin{array}{l}erf\left(18.042\right)=1\\ erf\left(18.042\right)=0.842\\ c\left(x,t\right)={10}^{21}\times 0.842\\ c\left(x,t\right)=8.42\times {10}^{20}\text{atoms}\text{/}{\text{cm}}^{\text{3}}\end{array}$

Thus, concentration at a depth of $1.25\text{μm}$ is $c\left(x,t\right)=8.42\times {10}^{20}\text{atoms}\text{/}{\text{cm}}^{\text{3}}$.

Calculation of concentration with respect to time and depth. Considering depth as $x=1.50\text{μm}=0.00015\text{cm}$.

Value of diffusion coefficient (D) is $2.5\times {10}^{-4}{\text{cm}}^2/\sec$.

Total time for carrying out the process is $8\text{hr}$ = $8\text{hr}\times 60\frac{\text{sec}}{\text{hr}}=480\sec$.

Substituting the given value in the above relation,

  $c\left(x,t\right)=C_s\times erfc\left(\frac{x}{2\sqrt{Dt}}\right)$

  $\begin{array}{l}c\left(x,t\right)=C_s\times erfc\left(\frac{x}{2\sqrt{Dt}}\right)\\ c\left(x,t\right)={10}^{21}\times erfc\left(\frac{0.00015}{2\sqrt{2.5\times {10}^{-14}\times 480}}\right)\\ c\left(x,t\right)={10}^{21}\times erfc\left(21.650\right)\end{array}$

When the value of error function exceeds the range of limit 4. The formula used for calculation of error function is,

  $\begin{array}{l}erf\left(x\right)=1-\frac{e^{-x^2}}{\sqrt{\pi }x}\\ x=21.650\\ erf\left(21.650\right)=1-\frac{e^{-{21.650}^2}}{\sqrt{\pi }\times 21.650}\end{array}$

  $\begin{array}{l}erf\left(21.650\right)=1\\ erf\left(21.650\right)=0.842\\ c\left(x,t\right)={10}^{21}\times 0.85\\ c\left(x,t\right)=8.5\times {10}^{20}\text{atoms}\text{/}{\text{cm}}^{\text{3}}\end{array}$

Thus, concentration at a depth of $1.50\text{μm}$ is $c\left(x,t\right)=8.5\times {10}^{20}\text{atoms}\text{/}{\text{cm}}^{\text{3}}$.

Calculation of concentration with respect to time and depth. Considering depth as $x=1.75\text{μm}=0.000175\text{cm}$.

Value of diffusion coefficient (D) is $2.5\times {10}^{-4}{\text{cm}}^2/\sec$.

Total time for carrying out the process is $8\text{hr}$ = $8\text{hr}\times 60\frac{\text{sec}}{\text{hr}}=480\sec$.

Substituting the given value in the above relation,

  $c\left(x,t\right)=C_s\times erfc\left(\frac{x}{2\sqrt{Dt}}\right)$

  $\begin{array}{l}c\left(x,t\right)=C_s\times erfc\left(\frac{x}{2\sqrt{Dt}}\right)\\ c\left(x,t\right)={10}^{21}\times erfc\left(\frac{0.000175}{2\sqrt{2.5\times {10}^{-14}\times 480}}\right)\\ c\left(x,t\right)={10}^{21}\times erfc\left(25.259\right)\end{array}$

When the value of error function exceeds the range of limit 4. The formula used for calculation of error function is,

  $\begin{array}{l}erf\left(x\right)=1-\frac{e^{-x^2}}{\sqrt{\pi }x}\\ x=25.259\\ erf\left(25.259\right)=1-\frac{e^{-{25.259}^2}}{\sqrt{\pi }\times 25.259}\end{array}$

  $\begin{array}{l}erf\left(25.259\right)=1\\ erf\left(25.259\right)=0.842\\ c\left(x,t\right)={10}^{21}\times 0.85\\ c\left(x,t\right)=8.5\times {10}^{20}\text{atoms}\text{/}{\text{cm}}^{\text{3}}\end{array}$

Thus, concentration at a depth of $1.75\text{μm}$ is $c\left(x,t\right)=8.5\times {10}^{20}\text{atoms}\text{/}{\text{cm}}^{\text{3}}$.

Calculation of concentration with respect to time and depth. Considering depth as $x=2\text{μm}=0.0002\text{cm}$.

Value of diffusion coefficient (D) is $2.5\times {10}^{-4}{\text{cm}}^2/\sec$.

Total time for carrying out the process is $8\text{hr}$ = $8\text{hr}\times 60\frac{\text{sec}}{\text{hr}}=480\sec$.

Substituting the given value in the above relation,

  $c\left(x,t\right)=C_s\times erfc\left(\frac{x}{2\sqrt{Dt}}\right)$

  $\begin{array}{l}c\left(x,t\right)=C_s\times erfc\left(\frac{x}{2\sqrt{Dt}}\right)\\ c\left(x,t\right)={10}^{21}\times erfc\left(\frac{0.0002}{2\sqrt{2.5\times {10}^{-14}\times 480}}\right)\\ c\left(x,t\right)={10}^{21}\times erfc\left(28.867\right)\end{array}$

When the value of error function exceeds the range of limit 4. The formula used for calculation of error function is,

  $\begin{array}{l}erf\left(x\right)=1-\frac{e^{-x^2}}{\sqrt{\pi }x}\\ x=28.867\\ erf\left(28.867\right)=1-\frac{e^{-{28.867}^2}}{\sqrt{\pi }\times 28.867}\end{array}$

  $\begin{array}{l}erf\left(28.867\right)=1\\ erf\left(28.867\right)=0.89\\ c\left(x,t\right)={10}^{21}\times 0.89\\ c\left(x,t\right)=8.9\times {10}^{20}\text{atoms}\text{/}{\text{cm}}^{\text{3}}\end{array}$

Thus, concentration at a depth of $\text{2μm}$ is $c\left(x,t\right)=8.9\times {10}^{20}\text{atoms}\text{/}{\text{cm}}^{\text{3}}$.

Calculation of concentration with respect to time and depth. Considering depth as $x=2.25\text{μm}=0.000225\text{cm}$.

Value of diffusion coefficient (D) is $2.5\times {10}^{-4}{\text{cm}}^2/\sec$.

Total time for carrying out the process is $8\text{hr}$ = $8\text{hr}\times 60\frac{\text{sec}}{\text{hr}}=480\sec$.

Substituting the given value in the above relation,

  $c\left(x,t\right)=C_s\times erfc\left(\frac{x}{2\sqrt{Dt}}\right)$

  $\begin{array}{l}c\left(x,t\right)=C_s\times erfc\left(\frac{x}{2\sqrt{Dt}}\right)\\ c\left(x,t\right)={10}^{21}\times erfc\left(\frac{0.000225}{2\sqrt{2.5\times {10}^{-14}\times 480}}\right)\\ c\left(x,t\right)={10}^{21}\times erfc\left(32.475\right)\end{array}$

When the value of error function exceeds the range of limit 4. The formula used for calculation of error function is,

  $\begin{array}{l}erf\left(x\right)=1-\frac{e^{-x^2}}{\sqrt{\pi }x}\\ x=32.475\\ erf\left(32.475\right)=1-\frac{e^{-{32.475}^2}}{\sqrt{\pi }\times 32.475}\\ erf\left(32.475\right)=1\end{array}$

  $\begin{array}{l}erf\left(32.475\right)=1\\ erf\left(32.475\right)=0.89\\ c\left(x,t\right)={10}^{21}\times 0.89\\ c\left(x,t\right)=8.9\times {10}^{20}\text{atoms}\text{/}{\text{cm}}^{\text{3}}\end{array}$

Thus, concentration at a depth of $\text{2}\text{.5}\text{μm}$ is $c\left(x,t\right)=8.9\times {10}^{20}\text{atoms}\text{/}{\text{cm}}^{\text{3}}$.

The table represents the data of concentration with variation in depth,

Depth	Concentration
$0.25\text{μm}$	$c\left(x,t\right)=9.9\times {10}^{20}\text{atoms}\text{/}{\text{cm}}^{\text{3}}$
$0.50\text{μm}$	$c\left(x,t\right)=8.427\times {10}^{21}\text{atoms}\text{/}{\text{cm}}^{\text{3}}$
$0.75\text{μm}$	$c\left(x,t\right)=8.468\times {10}^{20}\text{atoms}\text{/}{\text{cm}}^{\text{3}}$
$1\text{μm}$	$c\left(x,t\right)=8.38\times {10}^{20}\text{atoms}\text{/}{\text{cm}}^{\text{3}}$
$1.25\text{μm}$	$c\left(x,t\right)=8.42\times {10}^{20}\text{atoms}\text{/}{\text{cm}}^{\text{3}}$
$\text{1}\text{.5}\text{μm}$	$c\left(x,t\right)=8.5\times {10}^{20}\text{atoms}\text{/}{\text{cm}}^{\text{3}}$
$1.75\text{μm}$	$c\left(x,t\right)=8.5\times {10}^{20}\text{atoms}\text{/}{\text{cm}}^{\text{3}}$
$2\text{μm}$	$c\left(x,t\right)=8.9\times {10}^{20}\text{atoms}\text{/}{\text{cm}}^{\text{3}}$
$2.5\text{μm}$	$c\left(x,t\right)=8.9\times {10}^{20}\text{atoms}\text{/}{\text{cm}}^{\text{3}}$

Based on the given values drawing graph of concentration versus depth,

  

Thus, the required graph of concentration versus depth is shown above. As for pre deposition process concentration has dependency on time and depth, blue line is showing variation of concentration versus time and yellow line represents the variation of concentration with time.