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{
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"name": "",
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},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"\n",
"Chapter6:SEMICONDUCTOR JUNCTIONS WITH METALS AND INSULATORS"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex6.1:pg-226"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"e = 1.6*10**-19\n",
"n = 10**22\n",
"rho = 2.7*10**(-6)\n",
"print\"using following terms J = Current density ; s(sigma) = 1/rho = conductivity ; F = Electric field \"\n",
"print\"Using relations J = s*F = n*e*v = n*e*u*F ; we get\"\n",
"mu_ = 1.0/(n*e*rho)\n",
"print\"The mobility of electrons in aluminium is ,mu_ =\",\"{:.2e}\".format(mu_),\"cm**2(Vs)**-1\"\n",
"#The answer given in the book is 240.4 cm**2/Vs which is wrong\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"using following terms J = Current density ; s(sigma) = 1/rho = conductivity ; F = Electric field \n",
"Using relations J = s*F = n*e*v = n*e*u*F ; we get\n",
"The mobility of electrons in aluminium is ,mu_ = 2.31e+02 cm**2(Vs)**-1\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"\n",
"Ex6.2:pg-232"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"e = 1.6*10**-19\n",
"apsilen = 11.9*8.85*10**-12\n",
"A= 7.85*10**-9\n",
"S= 3*10**24\n",
"Nd = (2/(S*e*apsilen*(A**2)))\n",
"print\"The doping density in silicon is ,Nd =\",\"{:.2e}\".format(Nd),\"m**-3\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The doping density in silicon is ,Nd = 6.42e+20 m**-3\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex6.3:pg-236"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"Nd = 10**16\n",
"Nc = 2.8*10**19\n",
"kBT=0.026\n",
"Vf=0.3\n",
"e = 1.6*10**-19\n",
"A= 10**-3\n",
"print\" for W-n type Si schottky barrier \"\n",
"T = 300\n",
"phi_b = 0.67\n",
"print\"schottky barrier heights(in volts) =\",\"{:.2e}\".format(phi_b),\"eV\"\n",
"R = 110\n",
"Is = A*R*(T**2)*(exp(-(phi_b)/(kBT)))\n",
"print\"The reverse saturation current is ,Is =\",\"{:.2e}\".format(Is),\"A\"\n",
"print\"using relation I= Is*(exp((e*V)/(nkBT))-1) and neglecting 1\"\n",
"I = Is*(exp((Vf)/(kBT)))\n",
"print\"I=\",\"{:.1e}\".format(I),\"A\"\n",
"print\" for Si p+ -n junction diode \"\n",
"Na = 10**19\n",
"Db = 10.5\n",
"Tb = 10**-6\n",
"Lb = sqrt(Db*Tb)\n",
"print\"The electron carrier diffusion length is,Lb =\",\"{:.2e}\".format(Lb),\"cm\"\n",
"pn = 2.2*10**4\n",
"Io = A*e*pn*(Db/Lb)\n",
"print\"The saturation current current is Io =\",\"{:.1e}\".format(Io),\"A\"\n",
"I1 = Io*(exp((Vf)/(kBT)))\n",
"print\"The diode current for HBT is ,I =\",\"{:.1e}\".format(I1),\"A\"\n",
"print\"Since diode current for HBT is almost 6 orders of magnitude smaller than the value in the Schottky diode \"\n",
"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\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" for W-n type Si schottky barrier \n",
"schottky barrier heights(in volts) = 6.70e-01 eV\n",
"The reverse saturation current is ,Is = 6.37e-08 A\n",
"using relation I= Is*(exp((e*V)/(nkBT))-1) and neglecting 1\n",
"I= 6.5e-03 A\n",
" for Si p+ -n junction diode \n",
"The electron carrier diffusion length is,Lb = 3.24e-03 cm\n",
"The saturation current current is Io = 1.1e-14 A\n",
"The diode current for HBT is ,I = 1.2e-09 A\n",
"Since diode current for HBT is almost 6 orders of magnitude smaller than the value in the Schottky diode \n",
"hence for the p-n diode to have the same current that the schottky dode has at .3 V , the voltage required is .71V\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"\n",
"Ex6.4:pg-237"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"kBT=0.026\n",
"mo = 9.1*10**-31\n",
"m=0.08*mo\n",
"T = 300\n",
"phi_b1 = 0.7\n",
"phi_b2 = 0.6\n",
"R = 120*(m/mo)\n",
"print\"The effective richardson constant is ,R* =\",round(R,2),\" A cm**-2 k**-2\"\n",
"Js1 = R*(T**2)*(exp(-(phi_b1)/(kBT)))\n",
"print\"The saturation current density is ,Js(phi_b=0.7) =\",\"{:.1e}\".format(Js1),\"A/cm**2\"\n",
"Js2 = R*(T**2)*(exp(-(phi_b2)/(kBT)))\n",
"print\"The saturation current density is ,Js(phi_b=0.6) =\",\"{:.2e}\".format(Js2),\"A/cm**2\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The effective richardson constant is ,R* = 9.6 A cm**-2 k**-2\n",
"The saturation current density is ,Js(phi_b=0.7) = 1.8e-06 A/cm**2\n",
"The saturation current density is ,Js(phi_b=0.6) = 8.21e-05 A/cm**2\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex6.5:pg-239"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"apsilen = 11.9*8.85*10**-12\n",
"Nd = 10**16\n",
"Nc = 2.8*10**19\n",
"kBT = 0.026\n",
"I=10*10**-3\n",
"e = 1.6*10**-19\n",
"A= 10**-3\n",
"print\" for W-n type Si schottky barrier \"\n",
"T = 300\n",
"phi_b = 0.67\n",
"R = 110\n",
"Is = A*R*(T**2)*(exp(-(phi_b)/(kBT)))\n",
"V = kBT*(log(I/Is))\n",
"E = kBT*log(Nc/Nd)\n",
"print\"The fermi level positionin the neutral semiconductor(Efs) with respect to the conduction band is,Ec-Efs= E = \",\"{:.2e}\".format(E),\"eV\"\n",
"Vbi= phi_b-(E)\n",
"print\"The built in voltage is ,Vbi=\",\"{:.2e}\".format(Vbi),\"V\"\n",
"Cd = A*sqrt((e*Nd*apsilen)/(2*(Vbi-V)))\n",
"print\"The diode capacitance is ,Cd =\",\"{:.2e}\".format(Cd),\"F\"\n",
"R = kBT/I\n",
"print\"The resistance is ,R =\",\"{:.2e}\".format(R),\"ohm\"\n",
"RC = R*Cd\n",
"print\"The RC time constant is ,RC(schottky) =\",\"{:.2e}\".format(RC),\"s\"\n",
"print\" for Si p+ -n junction diode \"\n",
"Tb = 10**-6\n",
"print\"In the p-n diode the junction capacitance and the small signal resistance will be same as those in the schottky diode\"\n",
"Cdiff = ((I*Tb)/(kBT))\n",
"print\"The diffusion capacitance is ,Cdiff = (I*Tb)/(kBT) = \",\"{:.2e}\".format(Cdiff),\"F\"\n",
"RC1 = R*Cdiff\n",
"print\"The RC time constant is ,RC(p-n) = \",\"{:.2e}\".format(RC1),\"s\"\n",
"print\"From the above RC time constant value it can be concluded that p-n diode is almost 1000 times slower\"\n",
"# Note: due to approximation, the value of diode capicitance and diffusion capacitance are differ from that of the textbook"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" for W-n type Si schottky barrier \n",
"The fermi level positionin the neutral semiconductor(Efs) with respect to the conduction band is,Ec-Efs= E = 2.06e-01 eV\n",
"The built in voltage is ,Vbi="
]
},
{
"output_type": "stream",
"stream": "stdout",
"text": [
" 4.64e-01 V\n",
"The diode capacitance is ,Cd = 7.43e-10 F\n",
"The resistance is ,R = 2.60e+00 ohm\n",
"The RC time constant is ,RC(schottky) = 1.93e-09 s\n",
" for Si p+ -n junction diode \n",
"In the p-n diode the junction capacitance and the small signal resistance will be same as those in the schottky diode\n",
"The diffusion capacitance is ,Cdiff = (I*Tb)/(kBT) = 3.85e-07 F\n",
"The RC time constant is ,RC(p-n) = 1.00e-06 s\n",
"From the above RC time constant value it can be concluded that p-n diode is almost 1000 times slower\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"\n",
"Ex6.6:pg-242"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"apsilen = 11.9*8.85*10**-14\n",
"phi_b = 0.66\n",
"mo = 9.1*10**-31\n",
"m=0.34*mo\n",
"e = 1.6*10**-19\n",
"h = 1.05*10**-34\n",
"n1 = 10**18\n",
"n2 = 10**20\n",
"print\"Assume that the built in potential Vbi is same as barrier potential becouse of highly doped semiconductor\"\n",
"W1 = (sqrt((2*apsilen*phi_b)/(e*n1)))/10**-8\n",
"print\"The depletion width is ,W(n=10**18) =\",\"{:.2e}\".format(W1),\" Angstrom\"\n",
"W2 = (sqrt((2*apsilen*phi_b)/(e*n2)))/10**-8\n",
"print\"The depletion width is ,W(n=10**20) =\",\"{:.2e}\".format(W2),\" Angstrom\"\n",
"F1 = phi_b/(W1*10**-8)\n",
"print\"The average field in depletion region for(n=10**18), F1 =\",\"{:.2e}\".format(F1),\"V/cm\"\n",
"F2 = phi_b/(W2*10**-8)\n",
"print\"The average field in depletion region for(n=10**18), F2 =\",\"{:.2e}\".format(F2),\"V/cm\"\n",
"F1 = F1/10**-2\n",
"F2 = F2/10**-2\n",
"T = exp(-(4.0*(2.0*m)**.5*(e*phi_b)**(3.0/2.0))/(3.0*e*F1*h))\n",
"print\"The tunneling current for(n=10**18),T =\",\"{:.2e}\".format(T),\"V/cm\"\n",
"T1 = exp(-(4.0*(2.0*m)**.5*(e*phi_b)**(3.0/2.0))/(3.0*e*F2*h))\n",
"print\"The tunneling current for(n=10**20), T1 =\",\"{:.2e}\".format(T1),\"V/cm\"\n",
"# in the textbook author has used approximate value for depletion width and hence it affect the value of all other answer\n",
"# NOTE: In the textbook author has used approximate answer for tunneling current"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Assume that the built in potential Vbi is same as barrier potential becouse of highly doped semiconductor\n",
"The depletion width is ,W(n=10**18) = 2.95e+02 Angstrom\n",
"The depletion width is ,W(n=10**20) = 2.95e+01 Angstrom\n",
"The average field in depletion region for(n=10**18), F1 = 2.24e+05 V/cm\n",
"The average field in depletion region for(n=10**18), F2 = 2.24e+06 V/cm\n",
"The tunneling current for(n=10**18),T = 2.79e-42 V/cm\n",
"The tunneling current for(n=10**20), T1 = 6.99e-05 V/cm\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex6.7:pg-248"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"n = 10**18\n",
"W = 25*10**-4\n",
"R = 100*10**3\n",
"e = 1.6*10**-19\n",
"D= 5000*10**-8\n",
"mu_=100.0\n",
"Ro = 1.0/(n*e*mu_*D)\n",
"print\"The sheet resistance of the film is ,Ro =\",\"{:.2e}\".format(Ro),\" ohm/square\"\n",
"L = (R*W)/Ro\n",
"print\"The length of the desired resistor is ,L =\",\"{:.2e}\".format(L),\" cm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The sheet resistance of the film is ,Ro = 1.25e+03 ohm/square\n",
"The length of the desired resistor is ,L = 2.00e-01 cm\n"
]
}
],
"prompt_number": 1
}
],
"metadata": {}
}
]
}
|
