Chapter 8, Problem 8.105PAE

(a)

Interpretation Introduction

Interpretation:The mass of dopant required to generate one metric ton n-type silicon semiconductor have to determine.

Concept introduction:In n-type silicon semiconductor, the dopant must contain 5 valence electrons, so that after bonding with silicon (4 electron system) there remains an extra electron. The dopant frequently used for the commercially available silicon semiconductor is phosphorous (P), as the phosphorous impurity in silicon doesn’t increase the weight of the semiconductor much. The bonding pattern and roaming of extra electron of phosphorous generates n-type (negative charged or electron mediated) semiconductor. However, on the amount of dopant the silicon semiconductor may be subdivided into two category: light and heavy semiconductor.

In light silicon semiconductor 1 impurity atom is present per 1,000,000,000 or ppb (parts per billion) silicon atoms. On the other hand for generation of heavy silicon semiconductor 1 atom of impurity needed per 1,000 atom of silicon.

Answer to Problem 8.105PAE

Solution:The mass of phosphorous required for the doping of silicon for making light and heavy semiconductor are $1.107\times {10}^{-3}g$ and $1.107\times {10}^3$ gmrespectively.

Explanation of Solution

1 metric ton of silicon = ${10}^6$ gm of silicon.

On conversion of ${10}^6$ gm of silicon to moles of silicon, $\frac{{10}^6}{28}=3.571\times {10}^4$ moles (as molecular weight of silicon is 28).

1 mole of silicon is equivalent to $6.023\times {10}^{23}$ number of atoms, thus $3.571\times {10}^4$ moles is equivalent to $\left(6.023\times {10}^{23}\times 3.571\times {10}^4\right)=2.151\times {10}^{28}$ atoms of silicon.

For making light semiconductor 1 atom of dopant is required per ${10}^9$ atoms of silicon.

Henceforth, number of dopant atoms required is $\frac{2.151\times {10}^{28}}{{10}^9}=2.151\times {10}^{19}$.

The number of moles of phosphorous required as dopant is $\frac{2.151\times {10}^{19}}{6.023\times {10}^{23}}=3.571\times {10}^{-5}$.

The mass of phosphorous dopant required $\left(3.571\times {10}^{-5}\times 31\right)=1.107\times {10}^{-3}$ gm (As the atomic mass of phosphorous is 31g).

Henceforth to make the phosphorous incorporated light semiconductor per metric ton of silicon required $1.107\times {10}^{-3}$ gm of phosphorous.

To prepare heavy silicon semiconductor number of dopant atoms required is $\frac{2.151\times {10}^{28}}{{10}^3}=2.151\times {10}^{25}$.

The number of moles of phosphorous required as dopant is $\frac{2.151\times {10}^{25}}{6.023\times {10}^{23}}=35.713$.

The mass of phosphorous dopant required $\left(35.713\times 31\right)=1.107\times {10}^3$ gm.

Henceforth to make the phosphorous incorporated heavy semiconductor per metric ton of silicon required $1.107\times {10}^3$ gm of phosphorous.

(b)

Interpretation Introduction

Interpretation:The mole fraction of the dopant required to generate commercially available silicon n-type semiconductor.

Concept introduction:The mole number of the impurity or dopant present per unit total mole number of the dopant and substrate in a semiconductor is called the mole fraction of the dopant. It can be expressed as-

$\begin{array}{l}\text{The mole fraction of the dopant in semicondutor =}\\ \frac{\text{The mole number of dopant present}}{\text{The mole number of dopant + mole number of substance}}\end{array}$

Answer to Problem 8.105PAE

Solution:The mole fraction of dopant phosphorous present in light and heavy 1 metric ton silicon semiconductor are $1\times {10}^{-9}$ and $9.992\times {10}^{-4}$ respectively.

Explanation of Solution

For light silicon semiconductor doped by phosphorous the mole number of dopant and substrate are $3.571\times {10}^{-5}$ and $3.571\times {10}^4$ respectively.

On plugging the values in the equation,

$\text{The mole fraction of the dopant in semicondutor = }\frac{3.571\times {10}^{-5}}{\text{(3}\text{.571}\times {\text{10}}^{-5}\text{) + (3}\text{.571}\times {\text{10}}^4\text{)}}$

So, $\text{The mole fraction of the dopant = }\frac{3.571\times {10}^{-5}}{\text{3}\text{.571}\times {\text{10}}^4}=1\times {10}^{-9}$

Thus in the light semiconductor the mole fraction of the dopant is $1\times {10}^{-9}$.

On the other side for heavy semiconductor doped by phosphorous the mole number of dopant and substrate are 35.713 and $3.571\times {10}^4$ respectively.

On plugging the values in the equation,

$\text{The mole fraction of the dopant in semicondutor = }\frac{35.713}{\text{35}\text{.713 + (3}\text{.571}\times {\text{10}}^4\text{)}}$

So, $\text{The mole fraction of the dopant = }\frac{35.713}{\text{3}\text{.574}\times {\text{10}}^4}=9.992\times {10}^{-4}$

Thus in the light semiconductor the mole fraction of the dopant is $9.992\times {10}^{-4}$.