Chapter 18 : Electrical Properties

Example 18.1 Page No 682

In [4]:
sigma=10**-6   # (Ohm-m)**-1  Electrical Conductivity
e=1.6*10**-19  #Coulomb Charge on electron
m_e=0.85       # m**2/V-s  Mobility of electron
m_h=0.04       # m**2/V-s  Mobility of holes


print"Intrinsic Carrier Concentration is",round(ni,-11),"m**-3"
Intrinsic Carrier Concentration is 7e+12 m**-3

Example 18.2 Page No 689

In [8]:
e=1.6*10**-19    #Coulomb Charge on electron
ni=4*10**19      #For Si at 423 K   (m**-3)
m_e=0.06        #m**2/V-s  Mobility of electron
m_h=0.022       #m**2/V-s  Mobility of holes


print"Electrical Conductivity is ",round(sigma,2),"(ohm-m)**-1"
Electrical Conductivity is  0.52 (ohm-m)**-1

Example 18.3 Page No 690

In [9]:
n=10**23 #m**-3  Carrier Concentration
e=1.6*10**-19  #Coulomb  Charge on electron
m_e=0.07  #m**2/V-s  Mobility of electron


m_e2=0.04  #m**2/V-s  Mobility of electron

print"Conductivity at n=10**23 is ",sigma,"(Ohm-m)**-1"
print"Conductivity at T=373 K becomes ",sigma2,"(Ohm-m)**-1"
Conductivity at n=10**23 is  1120.0 (Ohm-m)**-1
Conductivity at T=373 K becomes  640.0 (Ohm-m)**-1

Design Example 18.1, Page No: 691

In [8]:
c=50           #ohm**-1, room temprature conductivity
Na1=10**22      #m**-3, assumed impurity content value
mu1=0.04        #m**2/Vs, assumed electrical mobility
e=1.6*10**-19  #Electronic charge
NA=6.023*10**23     #Avagadro no

C=Na1*e*mu1    #Conductivity
Na2=10**21      #m**-3, 
mu2=0.045       #m**2/Vs,
rho=2.33       # g/cm**3
Asi=28.09      # g/mole

print"The concentration of acceptor impurities is",round(Ca,7)
print"Thus a Silicon material having conductivity 50 ohm**-1  \nmust contain",round(Ca,7),"% boron,aluminium,Gallium or indium ."
The concentration of acceptor impurities is 1.6e-05
Thus a Silicon material having conductivity 50 ohm**-1  
must contain 1.6e-05 % boron,aluminium,Gallium or indium 

Example 18.4 Page No 693

In [13]:
sigma=3.8*10**7  #(Ohm-m)**-1  Electrical Conductivity
m_e=0.0012       #m**2/V-s  Mobility of electron
Rh=-m_e/sigma    #Hall coefficient
Ix=25            #Ampere(A) Current
d=15*10**-3      #m Thickness
Bz=0.6           #Tesla  Magnetic field


print"Hall coefficient is ",round(Rh,13),"V-m/A-Tesla"
print"Hall Voltage is ",round(Vh,10),"V"
Hall coefficient is  -3.16e-11 V-m/A-Tesla
Hall Voltage is  -3.16e-08 V

Example 18.5 Page No 707

In [15]:
A=6.45*10**-4    #m**2, area
d=2*10**-3       #m. Plate separation
V=10             #V Potential
Er=6             #Dielectric constant
Eo=8.85*10**-12  #F/m  Constant dielectric constant

print"The Capacitance is",round(C,13),"F"
print"The magnitude of charge stored is ",round(Q,12),"C"
print"The Dielectric displacement is is",round(D,9),"C/m**2"
print"The Polarization is",round(P,9),"C/m**2"
The Capacitance is 1.71e-11 F
The magnitude of charge stored is  1.71e-10 C
The Dielectric displacement is is 2.65e-07 C/m**2
The Polarization is 2.21e-07 C/m**2
In [ ]: