Chapter3 : Excess Carriers in Semiconductor

Example 3.2 Page No 111

In [1]:
#Example 3.2
#What is Minimum required energy 

#given data
l=6000                 #in Angstrum
h=6.6*10**(-34)             #Planks constant
c=3*10**8                   #speed of light in m/s
e=1.602*10**(-19)            #Constant

#calculation
phi=c*h/(e*l*10**(-10))

#result
print"Minimum required energy is",round(phi,2),"eV "

Minimum required energy is 2.06 eV

Example 3.3 Page No 112

In [6]:
#Exa 3.3
#calculate Work function of the cathode material

#given data
Emax=2.5                   #maximum energy of emitted electrons in eV 
l=2537.0                #in Angstrum

#Calculation
EeV=12400.0/l           #in eV
phi=EeV-Emax               #in eV

#result
print "Work function of the cathode material is ",round(phi,2),"eV"

Work function of the cathode material is  2.39 eV

Example 3.4 Page No 115

In [1]:
#Example 3.4
#Find (i)The fraction of each photon energy unit which is converted into heat",f
#(ii)Energy converted into heat in ,((2-1.43)/2)*0.009,"J/s"
#(iii)Number of photons per second given off from recombination events ",0.009/(e*2)

#given data
t=0.46*10**-4                 #in centi meters
hf1=2                          #in ev
hf2=1.43
Pin=10                        #in mW
alpha=50000                   # in per cm
e=1.6*10**-19                 #constant
Io=0.01                   #in mW

import math

#Calculation
It=Io*math.exp(-alpha*t)           #in mW
Iabs=Io-It
f=(hf1-hf2)/hf1
E=f*Iabs
N=Iabs/(e*hf1)

#result
print"(i)Thus power absorbed is ",round(Iabs,3),"J/s"
print"(ii)Energy converted into heat is",round(E,4),"J/s"
print"(iii)Number of photons per second given off from recombination events ",round(N,-14)
#In book there is calculation mistake in Number of photons.

(i)Thus power absorbed is  0.009 J/s
(ii)Energy converted into heat is 0.0026 J/s
(iii)Number of photons per second given off from recombination events  2.81e+16

Example 3.5 Page No 123

In [29]:
#Example 3.5
#What is Photoconductor gain 
#Electron transit time.

#given data
L=100                     #in uM
A=10&-7                   #in cm**2
th=10**-6                 #in sec
V=12                      #in Volts
ue=0.13                   #in m**2/V-s
uh=0.05                   #in m**2/V-s

#Calculation
E=V/(L*10**-6)            #in V/m
tn=(L*10**-6)/(ue*E)
Gain=(1+uh/ue)*(th/tn)

#result
print"Electron transit time in sec is ",round(tn,10),"s"
print"Photoconductor gain is ",Gain

Electron transit time in sec is  6.4e-09 s
Photoconductor gain is  216.0

Example 3.6 Page No128

In [30]:
#Example3.6
#Calculate Current flowing through diode .

#given datex
import math
Io=0.15                    #in uA
V=0.12                     #in mVolt
Vt=26                      #in mVolt

#calculation
I=Io*10**-6*(math.exp(V/(Vt*10**-3))-1)          #in A

#result
print"Current flowing through diode is ",round(I*10**6,2),"micra A"

Current flowing through diode is  15.0 micra A

Example 3.7 Page No 128

In [31]:
#Exa 3.7
#Determine the Forward voltage 

#given data
import math
Io=2.5              #in uA
I=10                #in mA
Vt=26               #in mVolt
n=2                 #for silicon

#Calculation
V=n*Vt*10**-3*math.log((I*10**-3)/(Io*10**-6))

#Result
print "Forward voltage  is ",round(V,2),"V"

Forward voltage  is  0.43 V

Example 3.8 Page No 128

In [33]:
#Example 3.8
#What is Reverse saturation current density 

#given data
ND=10**21                #in m**-3
NA=10**22                #in m**-3
De=3.4*10**-3            #in m**2-s**-1
Dh=1.2*10**-3            #in m**2-s**-1
Le=7.1*10**-4            #in meters
Lh=3.5*10**-4            #in meters
ni=1.6*10**16            #in m**-3
e=1.602*10**-19          #constant

#calculation
IoA=e*ni**2*(Dh/(Lh*ND)+De/(Le*NA))

#Result
print"Reverse saturation current density is ",round(IoA*10**6,2),"uA "

Reverse saturation current density is  0.16 uA