# Chapter18-Semiconductors¶

## Ex2-pg539¶

In [1]:
import math
##Example 18.2
##calculation of probability

##given values
T=300.;##temp in K
kT=.026;##temperture equivalent at room temp in eV
Eg=5.6;##forbidden gap in eV

##calculation
f=1./(1.+math.e**(Eg/(2.*kT)));

print'%s %.3e %s'%('probability of an e being thermally promoted to conduction band is',f,'');

probability of an e being thermally promoted to conduction band is 1.698e-47


## Ex3-pg540¶

In [5]:
import math
##Example 18.3
##calculation of fraction of e in CB

##given values
T=300.;##temp in K
kT=.026;##temperture equivalent at room temp in eV
Eg1=.72;##forbidden gap of germanium in eV
Eg2=1.1;##forbidden gap of silicon in eV
Eg3=5.6;##forbidden gap of diamond in eV

##calculation
f1=math.e**(-Eg1/(2.*kT));
print'%s %.6f %s'%('fraction of e in  conduction band of germanium is',f1,'');
f2=math.e**(-Eg2/(2.*kT));
print'%s %.3e %s'%('fraction of e in  conduction band of silicon is',f2,'');
f3=math.e**(-Eg3/(2*kT));
print'%s %.3e %s'%('fraction of e in  conduction band of diamond  is',f3,'');
print'abpove results shows that larger the band gap and the smaller electrons that can go under into the conduction band'

fraction of e in  conduction band of germanium is 0.000001
fraction of e in  conduction band of silicon is 6.501e-10
fraction of e in  conduction band of diamond  is 1.698e-47
abpove results shows that larger the band gap and the smaller electrons that can go under into the conduction band


## Ex4-pg540¶

In [17]:
import math
##Example 18.4
##calculation of fractionional change in no of e

##given values
T1=300.;##temp in K
T2=310.;##temp in K
Eg=1.1;##forbidden gap of silicon in eV
k=8.6*10**-5.;##boltzmann's constant in eV/K

##calculation
n1=(10**21.7)*(T1**(3/2.))*10**(-2500.*Eg/T1);##no of conduction e at T1
n2=(10**21.7)*(T2**(3/2.))*10**(-2500.*Eg/T2);##no of conduction e at T2
x=n2/n1;
print'%s %.1f %s'%('fractional change in no of e is',x,'');
print 'in book he just worte ans but he didnt calculated final ans but here is i calculated'

fractional change in no of e is 2.1
in book he just worte ans but he didnt calculated final ans but here is i calculated


## Ex5-pg541¶

In [11]:
##Example 18.5
##calculation of resistivity

##given values
e=1.6*10**-19;
ni=2.5*10**19;##intrinsic density of carriers per m**3
ue=.39;##mobility of e
uh=.19;##mobility of hole

##calculation
c=e*ni*(ue+uh);##conductivity
r=1/c;##resistivity
print'%s %.2f %s'%('resistivity in ohm m is',r,'');

resistivity in ohm m is 0.43


## Ex6-pg548¶

In [12]:
import math
##Example 18.6
##calculation of conductivity of intrinsic and doped semiconductors

##given values
h=4.52*10**24;##no of holes per m**3
e=1.25*10**14;##no of electrons per m**3
ue=.38;##e mobility
uh=.18;##hole mobility
q=1.6*10**-19;##charge of e in C
##calculation
ni=math.sqrt(h*e);##intrinsic concentration
ci=q*ni*(ue+uh);
print'%s %.2f %s'%('conductivity of semiconductor(in S/m) is',ci,'');
cp=q*h*uh;
print'%s %.2f %s'%('conductivity of doped semiconductor (in S/m) is',cp,'');

conductivity of semiconductor(in S/m) is 2.13
conductivity of doped semiconductor (in S/m) is 130176.00


## Ex7-pg548¶

In [18]:
import math
##Example 18.7
##calculation of hole concentration

##given values
ni=2.4*10**19.;##carrier concentration per m**3
N=4*10**28.;##concentration of ge atoms per m**3

##calculation
ND=N/10**6.;##donor cocntrtn
n=ND;##no of electrones

p=ni**2./n;
print'%s %.3e %s'%('concentartion of holes per m^3 is',p,'');

concentartion of holes per m^3 is 1.440e+16


## Ex8-pg558¶

In [14]:
import math
##Example 18.8
##calculation of Hall voltage

##given values
ND=10**21.;##donor density per m**3
B=.5;##magnetic field in T
J=500.;##current density in A/m**2
w=3*10**-3.;##width in m
e=1.6*10**-19.;##charge in C

##calculation

V=B*J*w/(ND*e);##in volts
print'%s %.2f %s'%('Hall voltage in mv is',V*10**3,'');

Hall voltage in mv is 4.69