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-{


- "metadata": {


-  "name": "",


-  "signature": "sha256:04a6842d7b11c3362cf74422ffc19db29b3f2ea1cd75fe882bcb992815e56505"


- },


- "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": {}


-  }


- ]


-}
\ No newline at end of file