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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 04 - JUNCTIONS"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E02 - Pg 165"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Exa 4.2 - 165\n",
      "import math\n",
      "# Given data\n",
      "C1= 5.*10.**-12.;# in F\n",
      "C2= 5.*10.**-12.;# in F\n",
      "L= 10.*10.**-3.;# in H\n",
      "C_Tmin= C1*C2/(C1+C2);# in F\n",
      "f_omax= 1./(2.*math.pi*math.sqrt(L*C_Tmin));# in Hz\n",
      "C1= 50.*10.**-12.;# in F\n",
      "C2= 50.*10.**-12.;# in F\n",
      "C_Tmax= C1*C2/(C1+C2);# in F\n",
      "f_omin= 1./(2.*math.pi*math.sqrt(L*C_Tmax));# in Hz\n",
      "f_omax= f_omax*10.**-6.;# in MHz\n",
      "f_omin= f_omin*10.**-3.;# in kHz\n",
      "print '%s %.f' %(\"The maximum value of resonant frequency in MHz is : \",f_omax)\n",
      "print '%s %.f' %(\"The minimum value of resonant frequency in kHz is : \",f_omin)\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The maximum value of resonant frequency in MHz is :  1\n",
        "The minimum value of resonant frequency in kHz is :  318\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E03 - Pg 188"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Exa 4.3 - 188\n",
      "import math\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",
      "# The contact potential for Ge semiconductor,\n",
      "V_J = V_T * math.log((N_A * N_D)/(n_i)**2.);# in V\n",
      "print '%s %.3f' %(\"The contact potential for Ge semiconductor in V is\",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",
      "# The contact potential for Si P-N junction,\n",
      "V_J = V_T * math.log((N_A * N_D)/(n_i)**2.);# in V\n",
      "print '%s %.3f' %(\"The contact potential for Si P-N junction in V is\",V_J);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The contact potential for Ge semiconductor in V is 0.329\n",
        "The contact potential for Si P-N junction in V is 0.721\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E04 - Pg 188"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Exa 4.4 - 188\n",
      "import math \n",
      "# 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",
      "# The height of the energy barrier for Ge,\n",
      "V_J = V_T * math.log((N_A * N_D)/(n_i)**2);# in V\n",
      "print '%s %.2f' %(\"For Ge the height of the energy barrier in V is\",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",
      "# The height of the energy barrier for Si p-n junction,\n",
      "V_J = V_T * math.log((N_A * N_D)/(n_i)**2.);# in V\n",
      "print '%s %.3f' %(\"For Si p-n junction  the height of the energy barrier in V is\",V_J);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "For Ge the height of the energy barrier in V is 0.22\n",
        "For Si p-n junction  the height of the energy barrier in V is 0.666\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E05 - Pg 189"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Exa 4.5 - 189\n",
      "import math\n",
      "# Given data\n",
      "Eta = 1.;\n",
      "V_T = 26.;# in mV\n",
      "V_T= V_T*10.**-3.;# in V\n",
      "#From equation of the diode current,  I = I_o * (%e**(V/(Eta*V_T)) - 1) and I = -(0.9) * I_o\n",
      "V= math.log(1-0.9)*V_T;#voltage in V\n",
      "print '%s %.3f' %(\"The voltage in volts is : \",V)\n",
      "# Part (ii)\n",
      "V1=0.05;# in V\n",
      "V2= -0.05;# in V\n",
      "# The ratio of the current for a forward bias to reverse bias \n",
      "ratio= (math.e**(V1/(Eta*V_T))-1.)/(math.e**(V2/(Eta*V_T))-1.)\n",
      "print '%s %.2f' %(\"The ratio of the current for a forward bias to reverse bias is : \",ratio)\n",
      "# Part (iii)\n",
      "Io= 10.;# in uA\n",
      "Io=Io*10.**-3.;# in mA\n",
      "#For \n",
      "V=0.1;# in V\n",
      "# Diode current\n",
      "I = Io * (math.e**(V/(Eta*V_T)) - 1.);# in mA\n",
      "print '%s %.3f' %(\"For V=0.1 V , the value of I in mA is : \",I)\n",
      "#For \n",
      "V=0.2;# in V\n",
      "# Diode current\n",
      "I = Io * (math.e**(V/(Eta*V_T)) - 1);# in mA\n",
      "print '%s %.1f' %(\"For V=0.2 V , the value of I in mA is : \",I)\n",
      "#For \n",
      "V=0.3;# in V\n",
      "# Diode current\n",
      "I = Io * (math.e**(V/(Eta*V_T)) - 1.);# in mA\n",
      "print '%s %.2f' %(\"For V=0.3 V , the value of I in A is : \",I*10**-3)\n",
      "print '%s' %(\"From three value of I, for small rise in forward voltage, the diode current increase rapidly\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The voltage in volts is :  -0.060\n",
        "The ratio of the current for a forward bias to reverse bias is :  -6.84\n",
        "For V=0.1 V , the value of I in mA is :  0.458\n",
        "For V=0.2 V , the value of I in mA is :  21.9\n",
        "For V=0.3 V , the value of I in A is :  1.03\n",
        "From three value of I, for small rise in forward voltage, the diode current increase rapidly\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E06 - Pg 189"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Exa 4.6 - 189\n",
      "import math\n",
      "# Given data\n",
      "# Part (i)\n",
      "T1= 25.;# in degreeC\n",
      "T2= 80.;# in degreeC\n",
      "# Formula Io2= Io1*2**((T2-T1)/10)\n",
      "AntiFactor= 2.**((T2-T1)/10.);\n",
      "print '%s %.f' %(\"Anticipated factor for Ge is : \",round(AntiFactor))\n",
      "# Part (ii)\n",
      "T1= 25.;# in degreeC\n",
      "T2= 150.;# in degreeC\n",
      "#AntiFactor= 2.**((T2-T1)/10.);\n",
      "AntiFactor=2.**12.5\n",
      "print '%s %.f' %(\"Anticipated factor for Si is : \",round(AntiFactor))\n",
      "#answer in textboo is wrong due to rounding error"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Anticipated factor for Ge is :  45\n",
        "Anticipated factor for Si is :  5793\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E07 - Pg 190"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Exa 4.7 - 190\n",
      "import math\n",
      "# Given data\n",
      "I=5.;# in uA\n",
      "V=10.;# in V\n",
      "T1= 0.11;# in degreeC**-1\n",
      "T2= 0.07;# in degreeC**-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 uA\n",
      "I_R= I-Io;# in uA\n",
      "R= V/I_R;# in Mohm\n",
      "print '%s %.1f' %(\"The leakage resistance in Mohm is : \",R)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The leakage resistance in Mohm is :  5.5\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E08 - Pg 190"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Exa 4.8 - 190\n",
      "import math\n",
      "# Given data\n",
      "Eta = 1.;\n",
      "T = 125.;# in degreeC\n",
      "T = T + 273.;# in K\n",
      "V_T = 8.62 * 10.**-5. * 398.;# in V\n",
      "I_o = 30.;# in uA\n",
      "I_o= I_o*10.**-6.;# in A\n",
      "v = 0.2;# in V\n",
      "# The dynamic resistance in the forward direction \n",
      "r_f = (Eta * V_T)/(I_o * math.e**(v/(Eta* V_T)));# in ohm\n",
      "print '%s %.2f' %(\"The dynamic resistance in the forward direction in ohm is \",r_f);\n",
      "# The dynamic resistance in the reverse direction \n",
      "r_r = (Eta * V_T)/(I_o * math.e**(-v/(Eta* V_T)));# in ohm\n",
      "r_r= r_r*10.**-3.;# in k ohm\n",
      "print '%s %.2f' %(\"The dynamic resistance in the reverse direction in kohm is\",r_r);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The dynamic resistance in the forward direction in ohm is  3.36\n",
        "The dynamic resistance in the reverse direction in kohm is 389.08\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E09 - Pg 191"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Exa 4.9 - 191\n",
      "import math\n",
      "# Given data\n",
      "epsilon = 16./(36. * math.pi * 10.**11.);# in F/cm\n",
      "A = 1. * 10.**-2.;\n",
      "W = 2. * 10.**-4.;\n",
      "# The barrier capacitance \n",
      "C_T = (epsilon * A)/W;# in F\n",
      "C_T= C_T*10.**12.;# in pF\n",
      "print '%s %.2f' %(\"The barrier capacitance in pF is\",C_T);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The barrier capacitance in pF is 70.74\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E10 - Pg 191"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Exa 4.10 - 191\n",
      "import math\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 = math.sqrt(((V_o - V) * 2. * epsilon)/(q * N_A));# m\n",
      "W= W*10.**6.;# in um\n",
      "print '%s %.2f' %(\"The width of the depletion layer  for an applied reverse voltage of 10V in um is \",W);\n",
      "W= W*10.**-6.;# in m\n",
      "C_T1 = (epsilon * A)/W;# in F\n",
      "C_T1= C_T1*10.**12.;# in pF\n",
      "# Part (b)\n",
      "V=-0.1;# in V\n",
      "W = math.sqrt(((V_o - V) * 2. * epsilon)/(q * N_A));# m\n",
      "W= W*10.**6.;# in um\n",
      "print '%s %.2f' %(\"The width of the depletion layer  for an applied reverse voltage of 0.1V in um is \",W);\n",
      "W= W*10.**-6.;# in m\n",
      "C_T2 = (epsilon * A)/W;# in F\n",
      "C_T2= C_T2*10.**12.;# in pF\n",
      "# Part (c)\n",
      "V=0.1;# in V\n",
      "W = math.sqrt(((V_o - V) * 2. * epsilon)/(q * N_A));# m\n",
      "W= W*10.**6.;# in um\n",
      "print '%s %.3f' %(\"The width of the depletion layer  for an applied for a forward bias of 0.1V in um is \",W);\n",
      "# Part (d)\n",
      "print '%s %.2f' %(\"The space charge capacitance for an applied reverse voltage of 10V  in pF is\",C_T1);\n",
      "print '%s %.2f' %(\"The space charge capacitance for an applied reverse voltage of 0.1V  in pF is\",C_T2);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The width of the depletion layer  for an applied reverse voltage of 10V in um is  7.76\n",
        "The width of the depletion layer  for an applied reverse voltage of 0.1V in um is  1.33\n",
        "The width of the depletion layer  for an applied for a forward bias of 0.1V in um is  0.768\n",
        "The space charge capacitance for an applied reverse voltage of 10V  in pF is 18.26\n",
        "The space charge capacitance for an applied reverse voltage of 0.1V  in pF is 106.46\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E11 - Pg 192"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Exa 4.11 - 192\n",
      "import math\n",
      "# 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",
      "# The current in the junction\n",
      "I = I_o *(math.e**(v/(Eta * V_T)));# in A\n",
      "I= I*10.**3.;# in mA\n",
      "print '%s %.3f' %(\"The current in the junction in mA is\",I);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The current in the junction in mA is 0.185\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E12 - Pg 192"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Exa 4.12 - 192\n",
      "# Given data\n",
      "import math\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",
      "# The forward biasing voltage across the junction\n",
      "v =math.log((I + I_o)/I_o) * V_T;# in V\n",
      "print '%s %.4f' %(\"The forward biasing voltage across the junction in V is\",v);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The forward biasing voltage across the junction in V is 0.6463\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E13 - Pg 192"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Exa 4.13 - 192\n",
      "# Given data\n",
      "I_o = 10.;# in nA\n",
      "# I = I_o * ((e**(v/(Eta * V_T))) - 1) as diode is reverse biased by large voltage\n",
      "# e**(v/(Eta * V_T)<< 1, so neglecting it\n",
      "I = I_o * (-1.);# in nA\n",
      "print '%s %.f' %(\"The Diode current in nA is \",I);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Diode current in nA is  -10\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E14 - Pg 192"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Exa 4.17 - 193\n",
      "import math\n",
      "# 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",
      "# The junction potential\n",
      "V_J = V_T * math.log((N_A * N_D)/(n_i)**2.);# in V\n",
      "print '%s %.3f' %(\"The junction potential in V is\",V_J)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The junction potential in V is 0.329\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E15 - Pg "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Exa 4.15 - 192\n",
      "%matplotlib inline\n",
      "import math\n",
      "import numpy as np\n",
      "import matplotlib.pyplot as plt\n",
      "# Given data\n",
      "## in V\n",
      "# V_S = i*R_L + V_D\n",
      "V_S = 10## in V (i * R_L = 0)\n",
      "print '%s %.f' %(\"when diode is OFF, the voltage in volts is : \",V_S)#\n",
      "R_L = 250.## in ohm\n",
      "I = V_S/R_L## in A\n",
      "print '%s %.f' %(\"when diode is ON, the current in mA is\",I*10**3)#\n",
      "V_D2=np.linspace(10,0,num=100)## in V\n",
      "j=0;\n",
      "I2 = np.zeros(100)\n",
      "for x in V_D2:\n",
      "\tI2[j] = (V_S- x)/R_L*1000.;\n",
      "\tj+=1\n",
      "plt.plot(V_D2,I2)\n",
      "plt.xlabel(\"V_D in volts\")#\n",
      "plt.ylabel(\"Current in mA\")\n",
      "plt.title(\"DC load line\")#\n",
      "plt.show()\n",
      "print '%s' %(\"DC load line shown in figure\")"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "when diode is OFF, the voltage in volts is :  10\n",
        "when diode is ON, the current in mA is 40\n"
       ]
      },
      {
       "metadata": {},
       "output_type": "display_data",
       "png": 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r6YskfRcAn4uIu+veF+BuYFjSXcAXgRdL+nOSG6k/CvxdE38vs5b5fgJmZiXm\n00FmZiXm00E2kCQdDDSe4vmviHhrHvWYFZVPB5mZlZhPB5mZlZgbATOzEnMjYGZWYm4EzMxKzI2A\nmVmJ/X860Ze3zrWA4AAAAABJRU5ErkJggg==\n",
       "text": [
        "<matplotlib.figure.Figure at 0x95c11f0>"
       ]
      },
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "DC load line shown in figure\n"
       ]
      }
     ],
     "prompt_number": 31
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E16 - Pg"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Exa 4.16 - 193\n",
      "# Given data\n",
      "import math\n",
      "V = 0.25;# in V\n",
      "I_o = 1.2;# in uA\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",
      "# The ac resistance of the diode \n",
      "r = (Eta * V_T)/(I_o * (math.e**(V/(Eta * V_T))));# in ohm\n",
      "print '%s %.3f' %(\"The ac resistance of the diode in ohm is\",r);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The ac resistance of the diode in ohm is 1.445\n"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E17 - Pg 193"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Exa 4.17 - 193\n",
      "import math\n",
      "# 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",
      "# The junction potential\n",
      "V_J = V_T * math.log((N_A * N_D)/(n_i)**2.);# in V\n",
      "print '%s %.3f' %(\"The junction potential in V is\",V_J)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The junction potential in V is 0.329\n"
       ]
      }
     ],
     "prompt_number": 19
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E18 - Pg 194"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Exa 4.18 - 194\n",
      "import math\n",
      "# 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 degreeC\n",
      "T = T + 273.;# in K\n",
      "q = 1.6 * 10.**-19.;# in C\n",
      "V_T = (K*T)/q;# in V\n",
      "# The forward dynamic resistance,\n",
      "r_f = (Eta * V_T)/(I_o * (math.e**(v/(Eta * V_T))));# in ohm\n",
      "print '%s %.3f' %(\"The forward dynamic resistance in ohm is\",r_f);\n",
      "# The Reverse dynamic resistance\n",
      "r_f1 = (Eta * V_T)/(I_o * (math.e**(-(v)/(Eta * V_T))));# in ohm\n",
      "r_f1= r_f1*10.**-3.;# in k ohm\n",
      "print '%s %.2f' %(\"The Reverse dynamic resistance in kohm is\",r_f1);\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The forward dynamic resistance in ohm is 3.391\n",
        "The Reverse dynamic resistance in kohm is 386.64\n"
       ]
      }
     ],
     "prompt_number": 24
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E19 - Pg 194"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Exa 4.19 - 194\n",
      "import math\n",
      "# Given data\n",
      "q = 1.6 * 10.**-19.;# in C\n",
      "N_A = 3 * 10.**20.;# in /m**3\n",
      "A = 1.;# in um**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",
      "# The width of depletion layer,\n",
      "W = math.sqrt((V_B * 2. * epsilon)/(q * N_A));# in m \n",
      "W=W*10.**6.;# in um\n",
      "print '%s %.2f' %(\"The width of depletion layer in um is\",W);\n",
      "W=W*10.**-6.;# in m\n",
      "# The space charge capacitance,\n",
      "C_T = (epsilon * A)/W;# in pF\n",
      "C_T=C_T*10.**12.;# in pF\n",
      "print '%s %.4f' %(\"The space charge capacitance in pF is\",C_T);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The width of depletion layer in um is 7.78\n",
        "The space charge capacitance in pF is 18.2127\n"
       ]
      }
     ],
     "prompt_number": 25
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E20 - Pg 194"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Exa 4.20 - 194\n",
      "import math\n",
      "# 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",
      "C_T= C_T*10.**12.;# in pF\n",
      "print '%s %.3f' %(\"The barrier capacitance in pF is\",C_T);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The barrier capacitance in pF is 70.832\n"
       ]
      }
     ],
     "prompt_number": 26
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E21 - Pg 195"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Exa 4.21 - 195\n",
      "import math\n",
      "# 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 = math.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 = math.sqrt(A * (4./math.pi));# in m\n",
      "D = D * 10.**3.;# in mm\n",
      "print '%s %.2f' %(\"The diameter in mm is\",D);\n",
      " \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The diameter in mm is 1.40\n"
       ]
      }
     ],
     "prompt_number": 27
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E22 - Pg 195"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Exa 4.22 - 195\n",
      "import math\n",
      "# 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/math.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 - 273;# in degreeC\n",
      "print '%s %.3f' %(\"The Temperature of junction in degree C is\",T);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Temperature of junction in degree C is 14.276\n"
       ]
      }
     ],
     "prompt_number": 28
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E23 - Pg 196"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Exa 4.23 - 196\n",
      "import math\n",
      "# 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 = math.log(IbyIo) * V_T;# in V\n",
      "print '%s %.5f' %(\"The reverse bias voltage in volts is\",V);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The reverse bias voltage in volts is -0.05987\n"
       ]
      }
     ],
     "prompt_number": 29
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E24 - Pg 196"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Exa 4.24 - 196\n",
      "import math\n",
      "# 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",
      "# The reverse saturation current \n",
      "I_o = I/((math.e**(V/(Eta * V_T))) - 1.);# in A\n",
      "I_o= I_o*10.**6.;# in uA\n",
      "print '%s %.2f' %(\"Reverse saturation current in uA is\",I_o);\n",
      "I_o= I_o*10.**-6.;# in A\n",
      "v1 = 0.2;# in V\n",
      "# The dynamic resistance of the diode,\n",
      "r = (Eta * V_T)/(I_o * (math.e**(v1/(Eta * V_T))));# in ohm\n",
      "print '%s %.3f' %(\"Dynamic resistance of the diode in ohm is\",r);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Reverse saturation current in uA is 3.33\n",
        "Dynamic resistance of the diode in ohm is 3.558\n"
       ]
      }
     ],
     "prompt_number": 30
    }
   ],
   "metadata": {}
  }
 ]
}