Chapter 02 : Transport Phenomena in Semiconductor

Example 2.1, Page No 22

In [1]:
import math
#initialisation of variables
n=10.0**20
q=1.6*10**-19
mn=800     #cm^3
delta=1   #V/cm

#Calculations
J=n*q*mn*delta


#Results
print("The electron current density is= %.2f X 10^4 atom/cm^2 " %(J/(10**4)))

The electron current density is= 1.28 X 10^4 atom/cm^2

Example 2.2a, Page No 27

In [2]:
import math
#initialisation of variables
ni=10.0**10
Nd=10**12

#Calculations
n=(Nd+(math.sqrt(Nd+4*ni**2)))


#Results
print("The free electron  is= %.3f X 10^12 cm^3 " %(n/(10**12)))

The free electron  is= 1.020 X 10^12 cm^3

Example 2.2b, Page No 27

In [3]:
import math
#initialisation of variables
ni=10.0**10
Nd=10**18

#Calculations
n=(Nd+(math.sqrt(Nd+4*ni**2)))


#Results
print("The free electron  is= %.2f X 10^18 cm^3 " %(n/(10**18)))

The free electron  is= 1.00 X 10^18 cm^3

Example 2.3a, Page No 29

In [4]:
import math
#initialisation of variables
Av=6.02*(10**23)      #Avogadro No.
m=72.6                #Molar mass of germanium in gm/moles
d=5.32                #density in gm/cm^3

#Calculations
conc = (Av/m)*d       #Concentration of atom in germanium

#Results
print("The concentration of germanium atom is= %.2f X 10^22 atom/cm^3 " %(conc/(10**22)))

The concentration of germanium atom is= 4.41 X 10^22 atom/cm^3

Example 2.3b, Page No 29

In [5]:
import math

#initialisation of variables
Av=6.02*(10**23) #Avogadro No.
m=72.6        #Molar mass of germanium in gm/moles
d=5.32        #density in gm/cm^3
ni=2.5*(10**13)        #in cm^-3
n=ni
p=ni                #n=magnitude of free electrons, p=magnitude of holes, ni=magnitude of intrinsic concentration

#Calculations
q=1.6*(10**-19)          #Charge of an Electron
yn=3800.0         #in cm^2/V-s
yp=1800.0        #in cm^2/V-s

#Required Formula
A=ni*q*(yn+yp)          #Conductivity
print("Conductivity is = %.2f ohm-cm^-1 " %A)
R =1.0/A          #Resistivity
print("Resistivity is = %.2f ohm-cm " %R)

Conductivity is = 0.02 ohm-cm^-1 
Resistivity is = 44.64 ohm-cm

Example 2.3c Page No 29

In [6]:
import math

#initialisation of variables
print('We know that n=p=ni where n is conc of free electron p is conc of holes and ni is conc of intrinsic carriers')
#Resistivity if 1 donor atom per 10^8 germanium atoms
Nd=4.41*(10**14)           #in atoms/cm^3
ni=2.5*(10**13)         #in cm^3
yn=3800.0              #in cm^2/V-s

#Calculations
q=1.6*(10**-19)
n=Nd
p=(ni**2)/Nd

print("The concentration of holes is= %.2f holes/cm^3 " %p)
if n>p:
    A=n*q*yn        #Conductivity
    print("The conductivity is = %.2f ohm-cm^-1 " %A)
 
R=1.0/A           #Resistivity

#Results
print("The resistivity is = %.2f ohm-cm " %R)

We know that n=p=ni where n is conc of free electron p is conc of holes and ni is conc of intrinsic carriers
The concentration of holes is= 1417233560090.70 holes/cm^3 
The conductivity is = 0.27 ohm-cm^-1 
The resistivity is = 3.73 ohm-cm

Example 2.3d, Page No 29

In [7]:
import math
#initialisation of variables

print('We know that n=p=ni where n is conc of free electron p is conc of holes and ni is conc of intrinsic carriers')
#Ratio of Conductivities
Nd=4.41*(10**14)       #in atoms/cm^3
ni=2.5*(10**13)      #in cm^3
yn=3800.0           #in cm^2/V-s
q=1.6*(10**-19)

#Calculations
n=Nd
A=n*q*yn      #Conductivity

#If germanium atom were monovalent metal , ratio of conductivity to that of n-type semiconductor

n=4.41*(10**22)         #in electrons/cm^3


#Results
print('If germanium atom were monovalent metal')
A1=n*q*yn
print("The coductivity of metal is= %.2f ohm=cm^-1 x 10^7 " %(A1/10**7))
F=A1/A
print("The factor by which the coductivity of metal is higher than that of n type semiconductor is %.2f x 10^8 " %(F/10**8))

We know that n=p=ni where n is conc of free electron p is conc of holes and ni is conc of intrinsic carriers
If germanium atom were monovalent metal
The coductivity of metal is= 2.68 ohm=cm^-1 x 10^7 
The factor by which the coductivity of metal is higher than that of n type semiconductor is 1.00 x 10^8

Example 2.4, Page No 35

In [8]:
import math

#initialisation of variables
g=5*10**21    #Generation rate
tp=2*10**-6        #hole lifetime

#Calculations
p=g*tp

#Required Formula
print("Hole density is = %.2f cm^3 10^16 " %(p/10**16))

Hole density is = 1.00 cm^3 10^16