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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter - 3 : Junction Diode and Junction Capacitance"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example -  3.1 : Page No - 90"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "#Given data\n",
      "Co= 20 # in pF\n",
      "Vr= 5 # in V\n",
      "V_T= 26 # in mV\n",
      "V_T= V_T*10**-3 # in V\n",
      "C_T= Co/(1+(Vr/V_T)) # in pF\n",
      "print \" The transition capacitance of diode = %0.2f pF\" %C_T"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The transition capacitance of diode = 0.10 pF\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example -  3.2 : Page No - 91"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given data\n",
      "toh= 10**-6 # in sec\n",
      "I=10 # in mA\n",
      "I=I*10**-3 # in A\n",
      "n=1 \n",
      "V_T= 26 # in mV\n",
      "V_T= V_T*10**-3 # in V\n",
      "C_D= toh*I/(n*V_T) # in F\n",
      "print \" The diffusion capacitance in p-n junction diode = %0.f nF\" %(C_D*10**9)\n",
      "\n",
      "# Note: There are two mistake in the book. First one is this that they put the wrong value of I to evaluating\n",
      "#  the value of C_D because the value of I is given 10mA (i.e. 10*10**-3= 10**-2 amp) but they put 10**-3 at place \n",
      "#          of 10**-2 and second one is calculation error. So the answer in the book is wrong."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The diffusion capacitance in p-n junction diode = 385 nF\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example -  3.3 : Page No - 91"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import exp\n",
      "#Given data\n",
      "T=300 # in K\n",
      "V_T= T/11600 # in V\n",
      "v= 0.3 # forward bias voltage in volt\n",
      "I= 10 # leakage current in micro amp\n",
      "I=I*10**-6 # in amp\n",
      "id= I*(exp(v/V_T)) # in amp\n",
      "print \" The diode current = %0.2f amp\" %id"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The diode current = 1.09 amp\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example -  3.4 : Page No - 92"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import log\n",
      "#Given data\n",
      "Vd_1= 0.3 # in V\n",
      "V_T= 25 # in mV\n",
      "V_T= V_T*10**-3 # in V\n",
      "# when Id_1= 1 mA\n",
      "Id_1= 1 # in mA\n",
      "Id_1=Id_1*10**-3 # in A\n",
      "# Formula Id_1= Io*[%e**(Vd/(n*V_T))-1]= Io*[e**(Vd/(n*V_T))]\n",
      "# Id_1= Io*[e**(Vd_1/(n*V_T))]            (i)\n",
      "\n",
      "# when Id_2= 200 mA\n",
      "Id_2= 200 # in mA\n",
      "Id_2=Id_2*10**-3 # in A\n",
      "Vd_2= 0.45 # in V\n",
      "# Id_2= Io*[e**(Vd_2/(n*V_T))]            (ii)\n",
      "# Dividing (ii) by (i), we have\n",
      "n= (Vd_2-Vd_1)/(log(Id_2/Id_1)*V_T) \n",
      "print \" The value of the constant for the diode = %0.2f \" %n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The value of the constant for the diode = 1.13 \n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example -  3.5 : Page No - 92"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given data\n",
      "T=300 # in K\n",
      "V_T= T/11600 # in V\n",
      "n=1 # assuming value\n",
      "Jd=10**5 # in A/m**2\n",
      "Jo=250 # in mA/m**2\n",
      "Jo= Jo*10**-3 # in A/m**2\n",
      "#Formula Id= Io*(%e**(Vd/V_T)-1) and after dividing both the sides by area of the junction, we have\n",
      "# Jd= Jo*(%e**(Vd/V_T)) # approx by neglecting 1 \n",
      "Vd= V_T*log(Jd/Jo) # in volt\n",
      "print \" Voltage to be applied across a p-n junction = %0.2f volt\" %Vd"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Voltage to be applied across a p-n junction = 0.33 volt\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example -  3.6 : Page No - 92"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import log\n",
      "#Given data\n",
      "J=10**4 # in A/m**2\n",
      "Jo=200 # in mA/m**2\n",
      "Jo= Jo*10**-3 # in A/m**2\n",
      "T=300 # in K\n",
      "V_T= T/11600 # in V\n",
      "e=1.6*10**-19 # electrone charge\n",
      "k= 1.38*10**-23 \n",
      "n=1 # assuming value\n",
      "#Formula I= Io*(%e**(e*V/(n*k*T))-1) and after dividing both the sides by area of the junction, we have\n",
      "# J= Jo*(%e**(e*V/(n*k*T))) # approx by neglecting 1 \n",
      "V= n*k*log(J/Jo)/e \n",
      "print \" Voltage to be applied across the junction = %0.2e volts\" %V"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Voltage to be applied across the junction = 9.33e-04 volts\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example -  3.7 : Page No - 93"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given data\n",
      "n=2 \n",
      "V_T=26 # in mV\n",
      "Io= 30 # in mA\n",
      "# (i) when\n",
      "I_D= 0.1 # in mA\n",
      "V_D= n*V_T*log(I_D/Io) # in mV\n",
      "print \" (i) When I_D is 0.1 mA, The junction forward-bias voltage = %0.f mV\" %V_D\n",
      "# (ii) when\n",
      "I_D= 10 # in mA\n",
      "V_D= n*V_T*log(I_D/Io) # in mV\n",
      "print \" (ii) When I_D is 10 mA, The junction forward-bias voltage = %0.f mV\" %V_D\n",
      "\n",
      "# Note: There is calculation error in the book so answer in the book is wrong."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " (i) When I_D is 0.1 mA, The junction forward-bias voltage = -297 mV\n",
        " (ii) When I_D is 10 mA, The junction forward-bias voltage = -57 mV\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example -  3.8 : Page No - 93"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given data\n",
      "I_by_Io= -0.9 \n",
      "V_T=26 # in mV\n",
      "V_T=V_T*10**-3 #in V\n",
      "n=1 \n",
      "# From Diode equation I= Io*[e**(e*V/(n*V_T))-1]\n",
      "V= n*V_T*log(1+I_by_Io) # in volt\n",
      "print \"The value of voltage = %0.1f mV \" %(V*10**3)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The value of voltage = -59.9 mV \n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example -  3.9 : Page No - 94"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given data\n",
      "nita= 2 \n",
      "T1= 25 # in \u00b0C\n",
      "T2= 150 # in \u00b0C\n",
      "k= 8.62*10**-5 \n",
      "V_T150= k*(T2+273) # in V\n",
      "V_T25= k*(T1+273) # in V\n",
      "V= 0.4 # in V\n",
      "# Io150= Io25*2**(T2-T1) \n",
      "Io150byIo25= 2**((T2-T1)/10) \n",
      "I150byI25= Io150byIo25 *(exp(V/(nita*V_T150))-1)/(exp(V/(nita*V_T25))-1) \n",
      "print \" The value of factor = %0.f\" %I150byI25"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The value of factor = 578\n"
       ]
      }
     ],
     "prompt_number": 19
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example -  3.10 : Page No - 94"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given data\n",
      "I_F= 100 # in mA\n",
      "I_F=I_F*10**-3 # in A\n",
      "V_F= 0.75 # in V\n",
      "R_F= V_F/I_F # in ohm\n",
      "print \" Forward resistance = %0.1f ohm \" %R_F\n",
      "# At\n",
      "V_R= 50 # in V\n",
      "I_R= 100 # in nA\n",
      "I_R= I_R*10**-9 # in A\n",
      "R_R= V_R/I_R # in ohm\n",
      "print \" Reverse resistance = %0.f Mohm \" %(R_R*10**-6)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Forward resistance = 7.5 ohm \n",
        " Reverse resistance = 500 Mohm \n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example -  3.11 : Page No - 95"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given data\n",
      "I_F= 70 # in mA\n",
      "V_F= 26 # in mV\n",
      "delta_I_F= 60 # in mA\n",
      "delta_I_F=delta_I_F*10**-3 # in A\n",
      "delta_V_F= 0.025 # in V\n",
      "r_d= delta_V_F/delta_I_F # in ohm\n",
      "print \" Dynamic resistance = %0.2f ohm\" %r_d\n",
      "# and the stimated value of the dynamic resistance is\n",
      "r_d= V_F/I_F # in ohm\n",
      "print \" The estimated value of the Dynamic resistance = %0.2f ohm\" %r_d"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Dynamic resistance = 0.42 ohm\n",
        " The estimated value of the Dynamic resistance = 0.37 ohm\n"
       ]
      }
     ],
     "prompt_number": 21
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example -  3.12 : Page No - 95"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given data\n",
      "Io= 1 # in micro amp\n",
      "Io=Io*10**-6 # in amp\n",
      "V_F= 0.52 # in V\n",
      "V_R= -0.52 # in V\n",
      "nita= 1 \n",
      "T=300 # in K\n",
      "V_T= T/11600 # in volt\n",
      "V_T=round(V_T*10**3) # in mV\n",
      "\n",
      "# (i)\n",
      "r_F= nita*V_T*10**-3/(Io*exp(V_F/(nita*V_T*10**-3))) \n",
      "print \"(i) : Dynamic resistance in the forward biased condition = %0.2e ohm\" %r_F\n",
      "\n",
      "# (ii)\n",
      "r_r= nita*V_T*10**-3/(Io*exp(V_R/(nita*V_T*10**-3))) \n",
      "print \"(ii) : Dynamic resistance in the reverse biased condition = %0.2e ohm\" %r_r"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(i) : Dynamic resistance in the forward biased condition = 5.36e-05 ohm\n",
        "(ii) : Dynamic resistance in the reverse biased condition = 1.26e+13 ohm\n"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example -  3.13 : Page No - 96"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given data\n",
      "V_F= 0.2 # in V\n",
      "T=300 # in K\n",
      "V_T= T/11600 # in volt\n",
      "Io= 1 # in micro amp\n",
      "Io=Io*10**-6 # in amp\n",
      "Id= Io*(exp(V_F/V_T)-1)\n",
      "I_F=Id \n",
      "r_dc= V_F/I_F # in ohm\n",
      "print \" Dynamic resistance = %0.1f ohm\" %r_dc\n",
      "r_ac= .026/I_F # in ohm\n",
      "print \" Static resistance = %0.1f ohm\" %r_ac"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Dynamic resistance = 87.6 ohm\n",
        " Static resistance = 11.4 ohm\n"
       ]
      }
     ],
     "prompt_number": 25
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example -  3.14 : Page No - 96"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given data\n",
      "# Part (i)\n",
      "I_D=2 # in mA\n",
      "I_D=I_D*10**-3 # in amp\n",
      "V_D= 0.5   # in volt\n",
      "R_DC= V_D/I_D # in ohm\n",
      "print \"(i) : DC resistance levels for the diode   = %0.f ohm\"  %R_DC\n",
      "\n",
      "# Part (ii)\n",
      "I_D=20 # in mA\n",
      "I_D=I_D*10**-3 # in amp\n",
      "V_D= 0.8   # in volt\n",
      "R_DC= V_D/I_D # in ohm\n",
      "print \"(ii) : DC resistance levels for the diode  = %0.f ohm\" %R_DC\n",
      "\n",
      "# Part (iii)\n",
      "I_D=-1 # in micro amp\n",
      "I_D=I_D*10**-6 # in amp\n",
      "V_D= -10   # in volt\n",
      "R_DC= V_D/I_D # in ohm\n",
      "print \"(iii) : DC resistance levels for the diode = %0.f Mohm\" %(R_DC*10**-6)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(i) : DC resistance levels for the diode   = 250 ohm\n",
        "(ii) : DC resistance levels for the diode  = 40 ohm\n",
        "(iii) : DC resistance levels for the diode = 10 Mohm\n"
       ]
      }
     ],
     "prompt_number": 27
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example -  3.15 : Page No - 97"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given data\n",
      "T1= 25 # in \u00b0C\n",
      "T2= 100 # in \u00b0C\n",
      "deltaT= T2-T1 # in \u00b0C\n",
      "deltaV_F= -1.8*10**-3 # in mV/\u00b0C\n",
      "I_F= 26 # in mA\n",
      "V_F1= 0.7 # in V (at T1)\n",
      "V_F2= V_F1+(deltaT*deltaV_F) # in V (at T2)\n",
      "# At 25\u00b0C\n",
      "T= 25+273 # in K\n",
      "rd= 26/I_F*T/298 # in \u03a9\n",
      "print \" Junction dynamic resistance at 25\u00b0C  = %0.f \u03a9 \" %rd\n",
      "# At 100\u00b0C\n",
      "T= 100+273 # in K\n",
      "rd= 26/I_F*T/298 # in \u03a9\n",
      "print \" Junction dynamic resistance at 100\u00b0C = %0.2f \u03a9 \" %rd"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Junction dynamic resistance at 25\u00b0C  = 1 \u03a9 \n",
        " Junction dynamic resistance at 100\u00b0C = 1.25 \u03a9 \n"
       ]
      }
     ],
     "prompt_number": 28
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example -  3.16 : Page No - 97"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given data\n",
      "I= 2 # in mA\n",
      "I=I*10**-3 # in A\n",
      "V_T= 25 # in mV\n",
      "V_T= V_T*10**-3 # in V\n",
      "nita= 1 \n",
      "r_F= nita*V_T/I # in \u03a9\n",
      "print \" The dynamic resistance of a diode = %0.1f \u03a9\" %r_F "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The dynamic resistance of a diode = 12.5 \u03a9\n"
       ]
      }
     ],
     "prompt_number": 30
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example -  3.17 : Page No - 98"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given data\n",
      "I= 30 # in \u00b5A\n",
      "I=I*10**-6 # in A\n",
      "T=125+273 # in K\n",
      "r_F= T/(11600*I*exp(-0.32/T)*11600) # in \u03a9\n",
      "print \" The dynamic resistance = %0.3f m\u03a9\" %(r_F*10**3)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The dynamic resistance = 98.672 m\u03a9\n"
       ]
      }
     ],
     "prompt_number": 32
    }
   ],
   "metadata": {}
  }
 ]
}