# Chapter 2 - Semiconductor Physics¶

## Example E1 - Pg 87¶

In :
#calculate Conductivity of pure silicon crystal
print '%s' %("Conductivity of pure silicon crystal is given by")
print '%s' %("   sigma = ni*e*(ue + uh)")
uh=480. #mobility in cm**2/Volt-sec
ue=1350. #mobility of electrons in cm**2/Volt-sec
e=1.6*10.**(-19.) #electron charge
ni=1.072*10.**10. #density of electron hole
sigma_i=ni*e*(uh+ue) #conductivity of silicon
print '%s %.2f' %("\n Conductivity of pure silicon crystal = %.8f ohm**(-1)/cm \n",sigma_i)
print '%s' %("Resistivity of silicon is given by")
rho=1./sigma_i #resistivity of silicon
print '%s %.2f' %("\n Resistivity of pure silicon crystal = %.0f ohm-cm \n",rho)

Conductivity of pure silicon crystal is given by
sigma = ni*e*(ue + uh)

Conductivity of pure silicon crystal = %.8f ohm**(-1)/cm
0.00
Resistivity of silicon is given by

Resistivity of pure silicon crystal = %.0f ohm-cm
318591.47


## Example E2 - Pg 87¶

In :
#Ex2_2 Pg-87
#calculate Conductivity of pure silicon crystal, Resistivity of pure phosphorous
print '%s' %("sigma = u*e*n")
u=1200. #mobility
e=1.6*10.**(-19.) #electron charge
n=10.**13. #phosphorous concentration
sigma=u*e*n #conductivity
print '%s %.2f' %("\n Conductivity of pure silicon crystal = %.5f ohm**(-1)/cm \n",sigma)
rho=1./sigma #resistivity
print '%s %.2f' %("\n Resistivity of pure phosphorous = %.0f ohm-cm \n",rho)

sigma = u*e*n

Conductivity of pure silicon crystal = %.5f ohm**(-1)/cm
0.00

Resistivity of pure phosphorous = %.0f ohm-cm
520.83


## Example E4 - Pg 88¶

In :
#Ex2_4 Pg-88
#calculate the ratio of n_p/n_i
print '%s' %("(n_i)**2 = n*p = n_p*N_a")
ni=2.5*10.**19. #density of electron hole
Na=1.1*10.**20. #acceptor density
np=(ni**2.)/Na
N=np/ni
print '%s %.2f' %("\n The ratio of n_p/n_i = %.4f",N)

(n_i)**2 = n*p = n_p*N_a

The ratio of n_p/n_i = %.4f 0.23


## Example E5 - Pg 88¶

In :
#Ex2_5 Pg-88
#calculate number density of donor atoms
print '%s' %("sigma_n = u_n*e*N_d")
e=1.6*1e-19 #electron charge
sigma=5. #conductivity in mho/cm
un=3850. #mobility of electrons
Nd=sigma/(e*un) #concentration
print '%s %.2f' %("Number density of donor atoms = %.1f*1e16 per cm**3",Nd*1e-16)
#the answer in the book is wrong the correct answer is 0.8*1e16 per cm**3

sigma_n = u_n*e*N_d
Number density of donor atoms = %.1f*1e16 per cm**3 0.81


## Example E6 - Pg 88¶

In :
#Ex2_6 Pg-88
#calculate Eg
print '%s' %("We know forbidden energy gap between conduction and valence bands for a semiconductor is nearly 1 eV.\nFor Ge and Si,energy gap is 0.785 eV and 1.21 eV respectively at 0K.\nEnergy gap decreases with increase in temperature which is represented by the expression.\nObviously, Si and Ge will remain semiconductors at 1000K ambient temperature.\n")
print '%s' %("Eg(T) = 1.21 - 3.6*10**(-4)*T (For Si)")
T=1000. #temperature
Eg=1.21 - 3.6*10**(-4)*T
print '%s %.2f %s' %("\nEg(1000)=",1.21 - 3.6*10**(-4)*T,"eV\n")

We know forbidden energy gap between conduction and valence bands for a semiconductor is nearly 1 eV.
For Ge and Si,energy gap is 0.785 eV and 1.21 eV respectively at 0K.
Energy gap decreases with increase in temperature which is represented by the expression.
Obviously, Si and Ge will remain semiconductors at 1000K ambient temperature.

Eg(T) = 1.21 - 3.6*10**(-4)*T (For Si)

Eg(1000)= 0.85 eV



## Example E7 - Pg 88¶

In :
#Ex2_7 Pg-88
#calculate Avg relaxation time of electrons
print '%s' %("Relaxation time in terms of mobility is given by")
print '%s' %("t=m*u/e")
print '%s' %("\n\nTaking effective mass of electron an holes in consideration,\nrelaxation time is given by\n")
print '%s' %("t=m_star*u/e")
print '%s' %("(a)For electrons,m_star = 0.259*m_0")
m0=9.1*10.**(-31.)
ue=0.135 #mobility of electrons
e=1.6*10.**(-19.) #electron charge
t_e=(0.259*m0*ue)/e
print '%s %.2f %s' %("\nAverage relaxation time of eletrons =",t_e*1e13,"*1e-13 secs\n")
uh=0.048 #mobility of holes
print '%s' %("(b)For holes in the valance band,m=0.537*m_0")
t_h=(0.537*m0*uh)/e
print '%s %.2f %s' %("\nAverage relaxation time of eletrons =",t_h*1e13,"*1e-13 secs\n")

Relaxation time in terms of mobility is given by
t=m*u/e

Taking effective mass of electron an holes in consideration,
relaxation time is given by

t=m_star*u/e
(a)For electrons,m_star = 0.259*m_0

Average relaxation time of eletrons = 1.99 *1e-13 secs

(b)For holes in the valance band,m=0.537*m_0

Average relaxation time of eletrons = 1.47 *1e-13 secs



## Example E8 - Pg 89¶

In :
#Ex2_8 Pg-89
#calculation of Total conduction current density
import math
print '%s' %("Conductivity of an intrinsic material is given by ")
print '%s' %("sigma = e*ni*un*(1+up/un)")
sigma_i=100./60.
ni=2.5*10.**19. #concentration of intrinsic carrier in m**3
up_un=0.5
e=1.6*10.**(-19.) #electron charge
un=(sigma_i/(e*ni*(1+(up_un)))) #concentration of electrons
up=un/2. #holes concentratin
print '%s' %("Let n be eletron concentration and p be hole con-centration for doped sample. Sincethe sample is electrically neutral,we have")
print '%s' %("Nd + p = Na + n")
print '%s' %("Where Nd is donor concentration and Na is acceptor concentration,assumed to be fully ionized. From mass action law,we have np =ni**2*S0")
print '%s' %("Nd + (ni**2)/n = Na + n")
print '%s' %("or n**2 + (Na-Nd)*n-ni**2 = 0")
print '%s' %("or n=0.5*([Nd-N])(+/-) sqrt(Nd-Na)**2 + 4*ni**2)")
print '%s' %("n is positive and hence we can drop negative sign before the radical")
Nd=10.**20. #number density of donor atoms /m**3
Na=5.*10.**19. #number of acceptor atoms in /m**3
n=0.5*((Nd-Na)+ math.sqrt((Nd-Na)**2 + 4*ni**2)) #electron concentration for doped sample
p=ni**2./n #hole concentration for doped sample
conduct_doped=e*(n*un+p*up) #conductivity of doped sample(value in textbook is wrong)
print '%s %.2f' %("\n Conductivity of doped sample = %.2f S/m \n",conduct_doped)

print '%s' %("We have assumed that carrier mobilities in the doped signal sample are the same as those in the intrinsic material.This assumption is justified by the low doping level, and Coulomb scaterring Applied by the ionized impurities is weak at 300K.")
print '%s' %("Applied electric field")

F_cm=2. #applied electric field in V/cm
F_m=2.*100. #applied electric field in V/m
J=conduct_doped*F_m #total conduction current density
J_cm=J/1000. #cm**2 to m**2s

print '%s %.2f' %("Total conduction current density = %.0f A/m**2",J)
print '%s %.2f' %("\n                               = %.2f A/cm**2",J_cm)

Conductivity of an intrinsic material is given by
sigma = e*ni*un*(1+up/un)
Let n be eletron concentration and p be hole con-centration for doped sample. Sincethe sample is electrically neutral,we have
Nd + p = Na + n
Where Nd is donor concentration and Na is acceptor concentration,assumed to be fully ionized. From mass action law,we have np =ni**2*S0
Nd + (ni**2)/n = Na + n
or n**2 + (Na-Nd)*n-ni**2 = 0
or n=0.5*([Nd-N])(+/-) sqrt(Nd-Na)**2 + 4*ni**2)
n is positive and hence we can drop negative sign before the radical

Conductivity of doped sample = %.2f S/m
2.91
We have assumed that carrier mobilities in the doped signal sample are the same as those in the intrinsic material.This assumption is justified by the low doping level, and Coulomb scaterring Applied by the ionized impurities is weak at 300K.
Applied electric field
Total conduction current density = %.0f A/m**2 582.52

= %.2f A/cm**2 0.58


## Example E9 - Pg 91¶

In :
#Ex2_9 Pg-91
print '%s' %("For P-type,Nd=0. By charge neutrality and mass action law,")
print '%s' %("      p + Nd = p = +na = (ni)**2/n")
print '%s' %("or    n**2 + Na*n - (ni)**2 = 0")
print '%s' %("Solving the quadratic equation for n and discharging the extraneous negative root,one obtains")
print '%s' %("      n = 0.5*(-Na + sqrt(Na**2 + 4*ni**2))")
print '%s' %("Knowing n,one obtains from mass action law p = ni**2/n")
print '%s' %("For N-type doping,Na=0. By analogous procedure,")
print '%s' %("      p = 0.5*(-Nd +sqrt(Nd**2 + 4*Ni**2));  n=ni**2/p")

For P-type,Nd=0. By charge neutrality and mass action law,
p + Nd = p = +na = (ni)**2/n
or    n**2 + Na*n - (ni)**2 = 0
Solving the quadratic equation for n and discharging the extraneous negative root,one obtains
n = 0.5*(-Na + sqrt(Na**2 + 4*ni**2))
Knowing n,one obtains from mass action law p = ni**2/n
For N-type doping,Na=0. By analogous procedure,
p = 0.5*(-Nd +sqrt(Nd**2 + 4*Ni**2));  n=ni**2/p


## Example E10 - Pg 91¶

In :
#Ex2_10 Pg-91
e=1.6*10.**(-19.) #electron charge
un=700. #mobility of silicon
n=10**17. #concentration of phosphorous atoms
sigma=e*un*n #conductivity
print '%s %.2f' %("Conductivity = %.1f (ohm-cm)**-1",sigma)
res=sigma**(-1) #resistivity
print '%s %.2f' %("\n Resistivity = %.4f ohm-cm",res)
Rh=-(e*n)**(-1) #hall coefficient
print '%s %.2f' %("\n Hall coefficient = %.1f cm**3/C",Rh)
Ix=10**-3.
Bz=10**(-5.)
x=10**(-2.)
Vh=(Ix*Bz*Rh)/x
print '%s %.2f' %("\n Hall Volage = %.1f uV",Vh*10.**6.)

Conductivity = %.1f (ohm-cm)**-1 11.20

Resistivity = %.4f ohm-cm 0.09

Hall coefficient = %.1f cm**3/C -62.50

Hall Volage = %.1f uV -62.50


## Example E11 - Pg 91¶

In :
#Ex2_11 Pg-91
print '%s' %("(a) Hall coefficient is")
print '%s' %("Rh = Eh/(Jx*B)")
Vh=21.4*10.**(-3.) #hall voltage
Eh=Vh/b #electric field
t=0.052*10.**(-3.) #thickness
I=200*10.**(-6.) #current
Jx=I/(b*t) #current density
B=0.5 #magnetic field
Rh=Eh/(Jx*B)
print '%s %.2f' %("\n Hall coefficient = %.6f m**3/C \n\n",Rh)

print '%s' %("(b) Electrons per unit volume,")
e=1.6*10.**(-19.) #electron charge
n=1/(Rh*e) #electrons per unit volume
print '%s %.2f' %("\n\n Electrons per unit volume = %.0f electrons/m**3 \n\n",n)
V=195.*10.**(-3.) #voltage
I=200.*10.**(-6.) #current
R=V/I #resistance
print '%s' %("Since R=(l/(A*sigma) = (l/(A*e*n*R))")
l=2.65*10.**(-2.) #length
w=1.7*10.**(-2.) #width
t=0.052*10.**(-3.) #thicknes
A=t*w #area
sigma=l/(A*e*n*R) #conductivity
print '%s %.2f' %("\n\n Conductivity = %.3f m**3/Vs",sigma)

(a) Hall coefficient is
Rh = Eh/(Jx*B)

Hall coefficient = %.6f m**3/C

0.01
(b) Electrons per unit volume,

Electrons per unit volume = %.0f electrons/m**3

561646297627606056960.00
Since R=(l/(A*sigma) = (l/(A*e*n*R))

Conductivity = %.3f m**3/Vs 0.34