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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter : 3 - Junction Properties"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example :  3.10.1 - Page No : 3-43"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "from math import log\n",
      "# Given data\n",
      "t = 4.4 * 10**22 # total number of Ge atoms/cm**3\n",
      "n = 1 * 10**8 # number of impurity atoms\n",
      "N_A = t/n # in atoms/cm**3\n",
      "N_A = N_A * 10**6 # in atoms/m**3\n",
      "N_D = N_A * 10**3 # in atoms/m**3\n",
      "n_i = 2.5 * 10**13 # in atoms/cm**3\n",
      "n_i = n_i * 10**6 # in atoms/m**3\n",
      "V_T = 26 #in mV\n",
      "V_T= V_T*10**-3 # in V\n",
      "V_J = V_T * log((N_A * N_D)/(n_i)**2) # in V\n",
      "print \"Part (a) : The contact potential = %0.3f V \" %V_J\n",
      "# Part (b)\n",
      "t = 5* 10**22 # total number of Si atoms/cm**3\n",
      "N_A = t/n # in atoms/cm**3\n",
      "N_A = N_A * 10**6 # in atoms/m**3\n",
      "N_D = N_A * 10**3 # in atoms/m**3\n",
      "n_i = 1.5 * 10**10 # in atoms/cm**3\n",
      "n_i = n_i * 10**6 # in atoms/m**3\n",
      "V_T = 26 #in mV\n",
      "V_T= V_T*10**-3 # in V\n",
      "V_J = V_T * log((N_A * N_D)/(n_i)**2) # in V\n",
      "print \"Part (b) : The contact potential  = %0.3f V \" %V_J"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Part (a) : The contact potential = 0.329 V \n",
        "Part (b) : The contact potential  = 0.721 V \n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example :  3.10.2 - Page No : 3-44"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "V_T = 26 # in mV\n",
      "V_T=V_T*10**-3 # in V\n",
      "n_i = 2.5 * 10**13 \n",
      "Sigma_p = 1 \n",
      "Sigma_n = 1 \n",
      "Mu_n = 3800 \n",
      "q = 1.6 * 10**-19 # in C\n",
      "Mu_p = 1800 \n",
      "N_A = Sigma_p/(2* q * Mu_p) # in /cm**3\n",
      "N_D = Sigma_n /(q * Mu_n) # in /cm**3\n",
      "V_J = V_T * log((N_A * N_D)/(n_i)**2) # in V\n",
      "print \"For Ge the height of the energy barrier = %0.2f V \" %V_J\n",
      "# For Si p-n juction\n",
      "n_i = 1.5 * 10**10 \n",
      "Mu_n = 1300 \n",
      "Mu_p = 500 \n",
      "N_A = Sigma_p/(2* q * Mu_p) # in /cm**3\n",
      "N_D = Sigma_n /(q * Mu_n) # in /cm**3\n",
      "V_J = V_T * log((N_A * N_D)/(n_i)**2) # in V\n",
      "print \"For Si p-n junction  the height of the energy barrier = %0.3f V \" %V_J"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "For Ge the height of the energy barrier = 0.22 V \n",
        "For Si p-n junction  the height of the energy barrier = 0.666 V \n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.10.3 - Page No : 3-45"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import exp\n",
      "# Given data\n",
      "Eta = 1 \n",
      "V_T = 26 # in mV\n",
      "V_T= V_T*10**-3 # in V\n",
      "# I = I_o * (exp(V/(Eta*V_T)) - 1) and I = -(0.9) * I_o \n",
      "V= log(1-0.9)*V_T # in V\n",
      "print \"The voltage = %0.2f volts \" %V\n",
      "# Part (ii)\n",
      "V1=0.05 # in V\n",
      "V2= -0.05 # in V\n",
      "ratio= (exp(V1/(Eta*V_T))-1)/(exp(V2/(Eta*V_T))-1)\n",
      "print \"The ratio of the current for a forward bias to reverse bias = %0.2f\" %ratio\n",
      "# Part (iii)\n",
      "Io= 10 # in \u00b5A\n",
      "Io=Io*10**-3 # in mA\n",
      "#For \n",
      "V=0.1 # in V\n",
      "I = Io * (exp(V/(Eta*V_T)) - 1) # in mA\n",
      "print \"For v=0.1 V , the value of I = %0.3f mA \" %I\n",
      "#For \n",
      "V=0.2 # in V\n",
      "I = Io * (exp(V/(Eta*V_T)) - 1) # in mA\n",
      "print \"For v=0.2 V , the value of I = %0.1f mA \" %I\n",
      "#For \n",
      "V=0.3 # in V\n",
      "I = Io * (exp(V/(Eta*V_T)) - 1) # in mA\n",
      "print \"For v=0.3 V , the value of I = %0.2f A \" %(I*10**-3)\n",
      "print \"From three value of I, for small rise in forward voltage, the diode current increase rapidly\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The voltage = -0.06 volts \n",
        "The ratio of the current for a forward bias to reverse bias = -6.84\n",
        "For v=0.1 V , the value of I = 0.458 mA \n",
        "For v=0.2 V , the value of I = 21.9 mA \n",
        "For v=0.3 V , the value of I = 1.03 A \n",
        "From three value of I, for small rise in forward voltage, the diode current increase rapidly\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.10.4 - Page No : 3-46"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "# Part (i)\n",
      "T1= 25 # in \u00b0C\n",
      "T2= 80 # in \u00b0C\n",
      "# Formula Io2= Io1*2**((T2-T1)/10)\n",
      "AntiFactor= 2**((T2-T1)/10) \n",
      "print \"Anticipated factor for Ge = %0.f \" %AntiFactor\n",
      "# Part (ii)\n",
      "T1= 25 # in \u00b0C\n",
      "T2= 150 # in \u00b0C\n",
      "AntiFactor= 2**((T2-T1)/10) \n",
      "print \"Anticipated factor for Si = %0.f \" %AntiFactor"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Anticipated factor for Ge = 45 \n",
        "Anticipated factor for Si = 5793 \n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.10.5 - Page No : 3-46"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "I=5 # in \u00b5A\n",
      "V=10 # in V\n",
      "T1= 0.11 # in \u00b0C**-1\n",
      "T2= 0.07 # in \u00b0C**-1\n",
      "# Io+I_R=I                            (i)\n",
      "# dI_by_dT= dIo_by_dT      (ii)\n",
      "# 1/Io*dIo_by_dT = T1 and 1/I*dI_by_dT = T2, So\n",
      "Io= T2*I/T1 # in \u00b5A\n",
      "I_R= I-Io # in \u00b5A\n",
      "R= V/I_R # in M\u03a9\n",
      "print \"The leakage resistance = %0.1f M\u03a9 \" %R"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The leakage resistance = 5.5 M\u03a9 \n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.10.6 - Page No : 3-47"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "Eta = 1 \n",
      "T = 125 # in \u00b0C\n",
      "T = T + 273 # in K\n",
      "V_T = 8.62 * 10**-5 * 398 # in V\n",
      "I_o = 30 # in \u00b5A\n",
      "I_o= I_o*10**-6 # in A\n",
      "v = 0.2 # in V\n",
      "r_f = (Eta * V_T)/(I_o * exp(v/(Eta* V_T))) # in ohm\n",
      "print \"The dynamic resistance in the forward direction = %0.2f ohm \" %r_f\n",
      "r_r = (Eta * V_T)/(I_o * exp(-v/(Eta* V_T))) # in ohm\n",
      "print \"The dynamic resistance in the reverse direction = %0.2f kohm \" %(r_r*10**-3)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The dynamic resistance in the forward direction = 3.36 ohm \n",
        "The dynamic resistance in the reverse direction = 389.08 kohm \n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example :  3.10.7 - Page No : 3-48"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from numpy import pi\n",
      "# Given data\n",
      "epsilon = 16/(36 * pi * 10**11) # in F/cm\n",
      "A = 1 * 10**-2 \n",
      "W = 2 * 10**-4 \n",
      "C_T = (epsilon * A)/W # in F\n",
      "print \"The barrier capacitance = %0.2f pF \" %(C_T*10**12)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The barrier capacitance = 70.74 pF \n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.10.8 - Page No : 3-48"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import sqrt\n",
      "# Given data\n",
      "A = 1 # in mm**2\n",
      "A = A * 10**-6 # in m**2\n",
      "N_A = 3 * 10**20 # in atoms/m**3\n",
      "q = 1.6 *10**-19 # in C\n",
      "V_o = 0.2 # in V\n",
      "epsilon_r=16 \n",
      "epsilon_o= 8.854*10**-12 # in F/m\n",
      "epsilon=epsilon_r*epsilon_o \n",
      "# Part (a)\n",
      "V=-10 # in V\n",
      "# V_o - V = 1/2*((q * N_A )/epsilon) * W**2\n",
      "W = sqrt(((V_o - V) * 2 * epsilon)/(q * N_A)) # m\n",
      "C_T1 = (epsilon * A)/W # in F\n",
      "print \"The width of the depletion layer  for an applied reverse voltage of 10V     = %0.2f \u00b5m \" %(W*10**6)\n",
      "# Part (b)\n",
      "V=-0.1 # in V\n",
      "W = sqrt(((V_o - V) * 2 * epsilon)/(q * N_A)) # m\n",
      "C_T2 = (epsilon * A)/W # in F\n",
      "print \"The width of the depletion layer  for an applied reverse voltage of 0.1V    = %0.2f \u00b5m \" %(W*10**6)\n",
      "# Part (c)\n",
      "V=0.1 # in V\n",
      "W = sqrt(((V_o - V) * 2 * epsilon)/(q * N_A)) # m\n",
      "print \"The width of the depletion layer  for an applied for a forward bias of 0.1V = %0.3f \u00b5m \" %(W*10**6)\n",
      "# Part (d)\n",
      "print \"The space charge capacitance for an applied reverse voltage of 10V          = %0.2f pF \" %(C_T1*10**12)\n",
      "print \"The space charge capacitance for an applied reverse voltage of 0.1V         = %0.2f pF \" %(C_T2*10**12)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The width of the depletion layer  for an applied reverse voltage of 10V     = 7.76 \u00b5m \n",
        "The width of the depletion layer  for an applied reverse voltage of 0.1V    = 1.33 \u00b5m \n",
        "The width of the depletion layer  for an applied for a forward bias of 0.1V = 0.768 \u00b5m \n",
        "The space charge capacitance for an applied reverse voltage of 10V          = 18.26 pF \n",
        "The space charge capacitance for an applied reverse voltage of 0.1V         = 106.46 pF \n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example :  3.10.9 - Page No : 3-49"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "I_o = 1.8 * 10**-9 # A\n",
      "v = 0.6 # in V\n",
      "Eta = 2 \n",
      "V_T = 26 # in mV\n",
      "V_T=V_T*10**-3 # in V\n",
      "I = I_o *(exp(v/(Eta * V_T))) # in A\n",
      "print \"The current in the junction = %0.3f mA \" %(I*10**3)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The current in the junction = 0.185 mA \n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example :  3.10.10 - Page No : 3-49"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "I_o = 2.4 * 10**-14 \n",
      "I = 1.5 # in mA\n",
      "I=I*10**-3 # in A\n",
      "Eta = 1\n",
      "V_T = 26 # in mV\n",
      "V_T= V_T*10**-3 # in V\n",
      "v =log((I + I_o)/I_o) * V_T # in V\n",
      "print \"The forward biasing voltage across the junction = %0.4f V \" %v"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The forward biasing voltage across the junction = 0.6463 V \n"
       ]
      }
     ],
     "prompt_number": 22
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example :  3.10.11 - Page No : 3-50"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "I_o = 10 # in nA\n",
      "# I = I_o * ((e**(v/(Eta * V_T))) - 1)\n",
      "# e**(v/(Eta * V_T)<< 1, so neglecting it\n",
      "I = I_o * (-1) # in nA\n",
      "print \"The Diode current = %0.f nA \" %I"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Diode current = -10 nA \n"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example :  3.10.12 - Page No : 3-50"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "R = 4.5 # in ohm\n",
      "I = 44.4 # in mA\n",
      "I=I*10**-3 # in A\n",
      "V = R * I # in V\n",
      "Eta = 1 \n",
      "V_T = 26 #in mV\n",
      "V_T=V_T*10**-3 # in V\n",
      "I_o = I/((exp(V/(Eta * V_T))) -1) # in A\n",
      "# At\n",
      "V = 0.1 # in V\n",
      "r_f = (Eta * V_T)/(I_o * ((exp(V/(Eta * V_T)))-1)) # in ohm\n",
      "print \"The diode dynamic resistance = %0.2f \u03a9 \" %r_f"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The diode dynamic resistance = 27.78 \u03a9 \n"
       ]
      }
     ],
     "prompt_number": 26
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example :  3.10.13 - Page No : 3-50"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "%matplotlib inline\n",
      "import matplotlib.pyplot as plt\n",
      "import numpy as np\n",
      "# Given data\n",
      "V_D = 10 # in V\n",
      "# V_S = i*R_L + V_D\n",
      "V_S = V_D # in V (i * R_L = 0)\n",
      "print \"when diode is OFF, the voltage = %0.f volts \" %V_S\n",
      "R_L = 250 # in ohm\n",
      "I = V_S/R_L # in A\n",
      "print \"when diode is ON, the current = %0.f mA \" %(I*10**3)\n",
      "V_D= np.arange(0,10,0.1) # in V\n",
      "I= (V_S-V_D)/R_L*1000 # in mA\n",
      "plt.plot(V_D,I)\n",
      "plt.xlabel('V_D in volts') \n",
      "plt.ylabel('Current in mA')\n",
      "plt.title('DC load line')\n",
      "plt.axis([0, 12, 0, 50])\n",
      "plt.show()\n",
      "print \"DC load line shown in figure\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "when diode is OFF, the voltage = 10 volts \n",
        "when diode is ON, the current = 40 mA \n"
       ]
      },
      {
       "metadata": {},
       "output_type": "display_data",
       "png": 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MjFgxEJGyabVaaLXaBu9f5z2Bbt264cKFC0b/xX7t2jU0bdoUjo6OuHPnDoYPH45Fixbh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\nCsYkQESkYP8HqYZH+eX0mvAAAAAASUVORK5CYII=\n",
       "text": [
        "<matplotlib.figure.Figure at 0x7fa3f4063650>"
       ]
      },
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "DC load line shown in figure\n"
       ]
      }
     ],
     "prompt_number": 56
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example :  3.10.14 - Page No : 3-51"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "V = 0.25 # in V\n",
      "I_o = 1.2 # in \u00b5A\n",
      "I_o = I_o * 10**-6 # in A\n",
      "V_T = 26 # in mV\n",
      "V_T = V_T * 10**-3 # in V\n",
      "Eta = 1 \n",
      "r = (Eta * V_T)/(I_o * (exp(V/(Eta * V_T)))) # in ohm\n",
      "print \"The ac resistance of the diode = %0.3f ohm \" %r"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The ac resistance of the diode = 1.445 ohm \n"
       ]
      }
     ],
     "prompt_number": 37
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example :  3.10.15 - Page No : 3-52"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "t = 4.4 * 10**22 # in total number of atoms/cm**3\n",
      "n = 1 * 10**8 # number of impurity\n",
      "N_A = t/n # in atoms/cm**3\n",
      "N_A = N_A * 10**6 # in atoms/m**3\n",
      "N_D = N_A * 10**3 # in atoms/m**3\n",
      "V_T = 26 # in mV\n",
      "V_T = V_T * 10**-3 # in V\n",
      "n_i = 2.5 * 10**19 # in /cm**3\n",
      "V_J = V_T * log((N_A * N_D)/(n_i)**2) # in V\n",
      "print \"The junction potential = %0.3f V \"%V_J"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The junction potential = 0.329 V \n"
       ]
      }
     ],
     "prompt_number": 38
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example :  3.10.16 - Page No : 3-53"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "Eta = 1 \n",
      "I_o = 30 # in MuA\n",
      "I_o = I_o * 10**-6 # in A\n",
      "v = 0.2 # in V\n",
      "K = 1.381 * 10**-23 # in J/degree K \n",
      "T = 125 # in \u00b0C\n",
      "T = T + 273 # in K\n",
      "q = 1.6 * 10**-19 # in C\n",
      "V_T = (K*T)/q # in V\n",
      "r_f = (Eta * V_T)/(I_o * (exp(v/(Eta * V_T)))) # in ohm\n",
      "print \"The forward dynamic resistance = %0.3f ohm \" %r_f\n",
      "r_f1 = (Eta * V_T)/(I_o * (exp(-(v)/(Eta * V_T)))) # in ohm\n",
      "print \"The Reverse dynamic resistance = %0.2f k\u03a9 \" %(r_f1*10**-3)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The forward dynamic resistance = 3.391 ohm \n",
        "The Reverse dynamic resistance = 386.64 k\u03a9 \n"
       ]
      }
     ],
     "prompt_number": 40
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example :  3.10.17 - Page No : 3-53"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "q = 1.6 * 10**-19 # in C\n",
      "N_A = 3 * 10**20 # in /m**3\n",
      "A = 1 # in \u00b5m**2\n",
      "A = A * 10**-6 # in m**2\n",
      "V = -10 # in V\n",
      "V_J = 0.25 # in V\n",
      "V_B = V_J - V # in V\n",
      "epsilon_o = 8.854 # in pF/m\n",
      "epsilon_o = epsilon_o * 10**-12 # in F/m\n",
      "epsilon_r = 16 \n",
      "epsilon = epsilon_o * epsilon_r \n",
      "W = sqrt((V_B * 2 * epsilon)/(q * N_A)) # in m \n",
      "print \"The width of depletion layer = %0.2f \u00b5m \" %(W*10**6)\n",
      "C_T = (epsilon * A)/W # in pF\n",
      "print \"The space charge capacitance = %0.4f pF \" %(C_T*10**12)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The width of depletion layer = 7.78 \u00b5m \n",
        "The space charge capacitance = 18.2127 pF \n"
       ]
      }
     ],
     "prompt_number": 41
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example :  3.10.18 - Page No : 3-54"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "W = 2 * 10**-4 # in cm\n",
      "W = W * 10**-2 # in m\n",
      "A = 1 # in mm**2\n",
      "A = A * 10**-6 # in m**2\n",
      "epsilon_r = 16 \n",
      "epsilon_o = 8.854 * 10**-12 # in F/m\n",
      "epsilon = epsilon_r * epsilon_o \n",
      "C_T = (epsilon * A)/W # in F\n",
      "print \"The barrier capacitance = %0.3f pF \" %(C_T*10**12)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The barrier capacitance = 70.832 pF \n"
       ]
      }
     ],
     "prompt_number": 42
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example :  3.10.19 - Page No : 3-54"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "C_T = 100 # in pF\n",
      "C_T=C_T*10**-12 # in F\n",
      "epsilon_r = 12 \n",
      "epsilon_o = 8.854 * 10**-12 # in F/m\n",
      "epsilon = epsilon_r * epsilon_o \n",
      "Rho_p = 5 # in ohm-cm\n",
      "Rho_p = Rho_p * 10**-2 # in ohm-m\n",
      "V_j = 0.5 # in V\n",
      "V = -4.5 # in V\n",
      "Mu_p = 500 # in cm**2\n",
      "Mu_p = Mu_p * 10**-4 # in m**2\n",
      "Sigma_p = 1/Rho_p # in per ohm-m\n",
      "qN_A = Sigma_p/ Mu_p \n",
      "V_B = V_j - V \n",
      "W = sqrt((V_B * 2 * epsilon)/qN_A) # in m\n",
      "#C_T = (epsilon * A)/W \n",
      "A = (C_T * W)/ epsilon # in m\n",
      "D = sqrt(A * (4/pi)) # in m\n",
      "D = D * 10**3 # in mm\n",
      "print \"The value of diameter = %0.3f mm \" %D\n",
      " "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The value of diameter = 1.398 mm \n"
       ]
      }
     ],
     "prompt_number": 44
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example :  3.10.20 - Page No : 3-55"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "q = 1.6 * 10**-19 # in C\n",
      "Mu_p = 500 # in cm**2/V-sec\n",
      "Rho_p = 3.5 # in ohm-cm\n",
      "Mu_n = 1500 # in cm**2/V-sec\n",
      "Rho_n = 10 # in ohm-cm\n",
      "N_A = 1/(Rho_p * Mu_p * q) # in /cm**3\n",
      "N_D = 1/(Rho_n * Mu_n * q) # in /cm**3\n",
      "V_J = 0.56 # in V\n",
      "n_i = 1.5 * 10**10 # in /cm**3\n",
      "V_T = V_J/log((N_A * N_D)/(n_i)**2) # in V\n",
      "# V_T = T/11600\n",
      "T = V_T * 11600 # in K\n",
      "T = T /19.78 # in \u00b0C   ( 1 degree K = 19.78 degree C)\n",
      "print \"The Temperature of junction = %0.3f \u00b0C \" %T"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Temperature of junction = 14.524 \u00b0C \n"
       ]
      }
     ],
     "prompt_number": 51
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example :  3.10.21 - Page No : 3-56"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "V_T = 26 # in mV\n",
      "V_T = V_T * 10**-3 # in V\n",
      "Eta = 1 \n",
      "# I = -90% for Io, so\n",
      "IbyIo= 0.1 \n",
      "# I  = I_o * ((e**(v/(Eta * V_T)))-1)\n",
      "V = log(IbyIo) * V_T # in V\n",
      "print \"The reverse bias voltage = %0.5f volts \" %V"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The reverse bias voltage = -0.05987 volts \n"
       ]
      }
     ],
     "prompt_number": 52
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example :  3.10.22 - Page No : 3-57"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "R = 5 # in ohm\n",
      "I = 50 # in mA\n",
      "I=I*10**-3 # in A\n",
      "V = R * I # in V\n",
      "Eta = 1 \n",
      "V_T = 26 # in mV\n",
      "V_T=V_T*10**-3 # in V\n",
      "I_o = I/((exp(V/(Eta * V_T))) - 1) # in A\n",
      "print \"Reverse saturation current = %0.2f \u00b5A \" %(I_o*10**6)\n",
      "v1 = 0.2 # in V\n",
      "r = (Eta * V_T)/(I_o * (exp(v1/(Eta * V_T)))) # in ohm\n",
      "print \"Dynamic resistance of the diode = %0.3f \u03a9 \" %r"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Reverse saturation current = 3.33 \u00b5A \n",
        "Dynamic resistance of the diode = 3.558 \u03a9 \n"
       ]
      }
     ],
     "prompt_number": 55
    }
   ],
   "metadata": {}
  }
 ]
}