# Chapter6:SEMICONDUCTOR JUNCTIONS WITH METALS AND INSULATORS¶

## Ex6.1:pg-226¶

In :
e = 1.6*10**-19
n = 10**22
rho = 2.7*10**(-6)
print"using following terms   J = Current density ; s(sigma) = 1/rho = conductivity ; F = Electric field "
print"Using relations  J = s*F = n*e*v = n*e*u*F ; we get"
mu_ = 1.0/(n*e*rho)
print"The mobility of electrons in aluminium is ,mu_ =","{:.2e}".format(mu_),"cm**2(Vs)**-1"
#The answer given in the book is 240.4 cm**2/Vs which is wrong

using following terms   J = Current density ; s(sigma) = 1/rho = conductivity ; F = Electric field
Using relations  J = s*F = n*e*v = n*e*u*F ; we get
The mobility of electrons in aluminium is ,mu_ = 2.31e+02 cm**2(Vs)**-1


## Ex6.2:pg-232¶

In :
e = 1.6*10**-19
apsilen = 11.9*8.85*10**-12
A= 7.85*10**-9
S= 3*10**24
Nd = (2/(S*e*apsilen*(A**2)))
print"The doping density in silicon is ,Nd =","{:.2e}".format(Nd),"m**-3"

The doping density in silicon is ,Nd = 6.42e+20 m**-3


## Ex6.3:pg-236¶

In :
Nd = 10**16
Nc = 2.8*10**19
kBT=0.026
Vf=0.3
e = 1.6*10**-19
A= 10**-3
print"        for W-n type Si schottky barrier          "
T = 300
phi_b = 0.67
print"schottky barrier heights(in volts) =","{:.2e}".format(phi_b),"eV"
R = 110
Is = A*R*(T**2)*(exp(-(phi_b)/(kBT)))
print"The reverse saturation current is ,Is =","{:.2e}".format(Is),"A"
print"using relation I= Is*(exp((e*V)/(nkBT))-1) and neglecting 1"
I = Is*(exp((Vf)/(kBT)))
print"I=","{:.1e}".format(I),"A"
print"        for Si p+ -n junction diode          "
Na = 10**19
Db = 10.5
Tb = 10**-6
Lb = sqrt(Db*Tb)
print"The electron carrier diffusion length is,Lb =","{:.2e}".format(Lb),"cm"
pn = 2.2*10**4
Io = A*e*pn*(Db/Lb)
print"The saturation current current is Io =","{:.1e}".format(Io),"A"
I1 = Io*(exp((Vf)/(kBT)))
print"The diode current for HBT is ,I =","{:.1e}".format(I1),"A"
print"Since diode current for HBT is almost 6 orders of magnitude smaller than the value in the Schottky diode "
print"hence for the p-n diode to have the same current that the schottky dode has at .3 V , the voltage required is .71V"

        for W-n type Si schottky barrier
schottky barrier heights(in volts) = 6.70e-01 eV
The reverse saturation current is ,Is = 6.37e-08 A
using relation I= Is*(exp((e*V)/(nkBT))-1) and neglecting 1
I= 6.5e-03 A
for Si p+ -n junction diode
The electron carrier diffusion length is,Lb = 3.24e-03 cm
The saturation current current is Io = 1.1e-14 A
The diode current for HBT is ,I = 1.2e-09 A
Since diode current for HBT is almost 6 orders of magnitude smaller than the value in the Schottky diode
hence for the p-n diode to have the same current that the schottky dode has at .3 V , the voltage required is .71V


## Ex6.4:pg-237¶

In :
kBT=0.026
mo = 9.1*10**-31
m=0.08*mo
T = 300
phi_b1 = 0.7
phi_b2 = 0.6
R = 120*(m/mo)
print"The effective richardson constant is ,R* =",round(R,2)," A cm**-2 k**-2"
Js1 = R*(T**2)*(exp(-(phi_b1)/(kBT)))
print"The saturation current density is ,Js(phi_b=0.7) =","{:.1e}".format(Js1),"A/cm**2"
Js2 = R*(T**2)*(exp(-(phi_b2)/(kBT)))
print"The saturation current density is ,Js(phi_b=0.6) =","{:.2e}".format(Js2),"A/cm**2"

The effective richardson constant is ,R* = 9.6  A cm**-2 k**-2
The saturation current density is ,Js(phi_b=0.7) = 1.8e-06 A/cm**2
The saturation current density is ,Js(phi_b=0.6) = 8.21e-05 A/cm**2


## Ex6.5:pg-239¶

In :
apsilen = 11.9*8.85*10**-12
Nd = 10**16
Nc = 2.8*10**19
kBT = 0.026
I=10*10**-3
e = 1.6*10**-19
A= 10**-3
print"        for W-n type Si schottky barrier          "
T = 300
phi_b = 0.67
R = 110
Is = A*R*(T**2)*(exp(-(phi_b)/(kBT)))
V = kBT*(log(I/Is))
E = kBT*log(Nc/Nd)
print"The fermi level positionin the neutral semiconductor(Efs) with respect to the conduction band is,Ec-Efs= E = ","{:.2e}".format(E),"eV"
Vbi= phi_b-(E)
print"The built in voltage is ,Vbi=","{:.2e}".format(Vbi),"V"
Cd = A*sqrt((e*Nd*apsilen)/(2*(Vbi-V)))
print"The diode capacitance is ,Cd =","{:.2e}".format(Cd),"F"
R = kBT/I
print"The resistance is ,R =","{:.2e}".format(R),"ohm"
RC = R*Cd
print"The RC time constant is ,RC(schottky) =","{:.2e}".format(RC),"s"
print"        for Si p+ -n junction diode          "
Tb = 10**-6
print"In the p-n diode the junction capacitance and the small signal resistance will be same as those in the schottky diode"
Cdiff = ((I*Tb)/(kBT))
print"The diffusion capacitance is ,Cdiff = (I*Tb)/(kBT) = ","{:.2e}".format(Cdiff),"F"
RC1 = R*Cdiff
print"The RC time constant is ,RC(p-n) = ","{:.2e}".format(RC1),"s"
print"From the above RC time constant value it can be concluded that p-n diode is almost 1000 times slower"
# Note: due to approximation, the value of diode capicitance and diffusion capacitance are differ from that of the textbook

        for W-n type Si schottky barrier
The fermi level positionin the neutral semiconductor(Efs) with respect to the conduction band is,Ec-Efs= E =  2.06e-01 eV
The built in voltage is ,Vbi= 4.64e-01 V
The diode capacitance is ,Cd = 7.43e-10 F
The resistance is ,R = 2.60e+00 ohm
The RC time constant is ,RC(schottky) = 1.93e-09 s
for Si p+ -n junction diode
In the p-n diode the junction capacitance and the small signal resistance will be same as those in the schottky diode
The diffusion capacitance is ,Cdiff = (I*Tb)/(kBT) =  3.85e-07 F
The RC time constant is ,RC(p-n) =  1.00e-06 s
From the above RC time constant value it can be concluded that p-n diode is almost 1000 times slower


## Ex6.6:pg-242¶

In :
apsilen = 11.9*8.85*10**-14
phi_b = 0.66
mo = 9.1*10**-31
m=0.34*mo
e = 1.6*10**-19
h = 1.05*10**-34
n1 = 10**18
n2 = 10**20
print"Assume that the built in potential Vbi is same as barrier potential becouse of highly doped semiconductor"
W1 = (sqrt((2*apsilen*phi_b)/(e*n1)))/10**-8
print"The depletion width is ,W(n=10**18) =","{:.2e}".format(W1)," Angstrom"
W2 = (sqrt((2*apsilen*phi_b)/(e*n2)))/10**-8
print"The depletion width is ,W(n=10**20) =","{:.2e}".format(W2)," Angstrom"
F1 = phi_b/(W1*10**-8)
print"The average field in depletion region for(n=10**18), F1 =","{:.2e}".format(F1),"V/cm"
F2 = phi_b/(W2*10**-8)
print"The average field in depletion region for(n=10**18), F2 =","{:.2e}".format(F2),"V/cm"
F1 = F1/10**-2
F2 = F2/10**-2
T = exp(-(4.0*(2.0*m)**.5*(e*phi_b)**(3.0/2.0))/(3.0*e*F1*h))
print"The tunneling current for(n=10**18),T =","{:.2e}".format(T),"V/cm"
T1 = exp(-(4.0*(2.0*m)**.5*(e*phi_b)**(3.0/2.0))/(3.0*e*F2*h))
print"The tunneling current for(n=10**20), T1 =","{:.2e}".format(T1),"V/cm"
# in the textbook author has used approximate value for depletion width and hence it affect the value of all other answer
# NOTE: In the textbook author has used approximate answer for tunneling current

Assume that the built in potential Vbi is same as barrier potential becouse of highly doped semiconductor
The depletion width is ,W(n=10**18) = 2.95e+02  Angstrom
The depletion width is ,W(n=10**20) = 2.95e+01  Angstrom
The average field in depletion region for(n=10**18), F1 = 2.24e+05 V/cm
The average field in depletion region for(n=10**18), F2 = 2.24e+06 V/cm
The tunneling current for(n=10**18),T = 2.79e-42 V/cm
The tunneling current for(n=10**20), T1 = 6.99e-05 V/cm


## Ex6.7:pg-248¶

In :
n = 10**18
W = 25*10**-4
R = 100*10**3
e = 1.6*10**-19
D= 5000*10**-8
mu_=100.0
Ro = 1.0/(n*e*mu_*D)
print"The sheet resistance of the film is ,Ro =","{:.2e}".format(Ro)," ohm/square"
L = (R*W)/Ro
print"The length of the desired resistor is ,L =","{:.2e}".format(L)," cm"

The sheet resistance of the film is ,Ro = 1.25e+03  ohm/square
The length of the desired resistor is ,L = 2.00e-01  cm