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diff --git a/Electronic_Devices_and_Circuits/Chapter1.ipynb b/Electronic_Devices_and_Circuits/Chapter1.ipynb
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-{

- "metadata": {

-  "name": ""

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter 01 : Basic semiconductor and pn junction theory"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 1.1, Page No 15"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "#initialisation of variables\n",

-      "Nd=3*10**14\n",

-      "Na=0.5*10**14                       #all in atom/cm**3\n",

-      "ni=1.5*10**10\n",

-      "\n",

-      "#Calculations\n",

-      "print(\"resultant densities of free electrons and hole\")\n",

-      "ne=(-(Na-Nd)+(math.sqrt(((Na-Nd)**2)+4*ni**2)))/2\n",

-      "print(\"Electron densities = %.1f x 10^14 electron/cm**3\" %(ne/(10**14)))     #electron densities in electron/cm**3\n",

-      "Nd>Na\n",

-      "n=Nd-Na\n",

-      "print(n)\n",

-      "p=(ni**2)/n\n",

-      "\n",

-      "#Results\n",

-      "\n",

-      "print(\"densities of hole is =%.1f X 10^6 dhole/cm3\" %(p/(10**6)))\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "resultant densities of free electrons and hole\n",

-        "Electron densities = 2.5 x 10^14 electron/cm**3\n",

-        "2.5e+14\n",

-        "densities of hole is =0.9 X 10^6 dhole/cm3\n"

-       ]

-      }

-     ],

-     "prompt_number": 1

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 1.2, Page No 18"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "#initialisation of variables\n",

-      "l=1*10**-3\n",

-      "E=10\n",

-      "\n",

-      "#Calculations\n",

-      "un=1500*10**-4\n",

-      "up=500*10-4\n",

-      "Vn=-(un*E)/l\n",

-      "\n",

-      "#Results\n",

-      "print(\"drift current is =%.2dm/s\\n\" %Vn)\n",

-      "print(\"drift current of hole\")\n",

-      "Vp=(up*E)/l\n",

-      "print(\"drift current is =%.f dm/s\\n\" %(Vp/10**5))"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "drift current is =-1500m/s\n",

-        "\n",

-        "drift current of hole\n",

-        "drift current is =500 dm/s\n",

-        "\n"

-       ]

-      }

-     ],

-     "prompt_number": 2

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 1.3 Page No 19"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "\n",

-      "#initialisation of variables\n",

-      "l=1*10**-3\n",

-      "a=0.1*10**-4\n",

-      "ni=1.5*10**10\n",

-      "p=1.5*10**10\n",

-      "un=1500\n",

-      "up=500           #in cm3/V.s\n",

-      "q=1.6*10**-19\n",

-      "\n",

-      "#Calculations\n",

-      "m=q*((ni*un)+(p*up))*10**6\n",

-      "print( \"mobility is =%.1fmicro/ohmcm\" %m)\n",

-      "R=l/(m*a)\n",

-      "print(\" resistance is =%.1fMohm\" %R)\n",

-      "\n",

-      "#for doped material\n",

-      "n=8*10**13\n",

-      "p=(ni**2)/n\n",

-      "m=q*((n*un)+(p*up))\n",

-      "\n",

-      "#Results\n",

-      "print(\"mobility is =%3.4f/ohmcm\" %m)\n",

-      "R=l/(m*a)\n",

-      "print(\" resistance is %.2f Kohm\" %(R/1000))"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "mobility is =4.8micro/ohmcm\n",

-        " resistance is =20.8Mohm\n",

-        "mobility is =0.0192/ohmcm\n",

-        " resistance is 5.21 Kohm\n"

-       ]

-      }

-     ],

-     "prompt_number": 3

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 1.4, Page No 25"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "#initialisation of variables\n",

-      "T1=25.0\n",

-      "T2=35.0\n",

-      "T3=45.0\n",

-      "I0=30.0       # nA\n",

-      "print(\"I0(35)=I0*2**(T2-T1)/10\")\n",

-      "#on solving\n",

-      "I035=I0*2**((T2-T1)/10)\n",

-      "print(\"Current at 35c is =%.2f nA\\n\" %I035)\n",

-      "print(\"I0(45)=I0*2**(T3-T1)/10\")\n",

-      "#on solving\n",

-      "I045=30*2**2\n",

-      "print(\"current at 45c is =%.2f nA\\n\" %I045)\n",

-      "I_CS=100.0 \n",

-      "V_CC=200.0 \n",

-      "t_on=40*10**-6"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "I0(35)=I0*2**(T2-T1)/10\n",

-        "Current at 35c is =60.00 nA\n",

-        "\n",

-        "I0(45)=I0*2**(T3-T1)/10\n",

-        "current at 45c is =120.00 nA\n",

-        "\n"

-       ]

-      }

-     ],

-     "prompt_number": 4

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 1.5, Page No 28"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "#initialisation of variables\n",

-      "I0=30\n",

-      "Vd=0.7\n",

-      "n=2\n",

-      "\n",

-      "#Calculations\n",

-      "Vt=26.0*10**-3\n",

-      "k=Vd/(n*Vt)\n",

-      "Id=I0*((2.7**k)-1)*10**-6     #Junction current\n",

-      "print(\" a) Forward bais current is =%.2f mA\\n\" %Id)\n",

-      "Vd=-10                #reverse bais\n",

-      "k=Vd/(n*Vt)\n",

-      "Id=I0*((2.7**k)-1)\n",

-      "\n",

-      "#Results\n",

-      "print(\" b) Forward bais current is =%.2f nA\" %Id)"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " a) Forward bais current is =19.23 mA\n",

-        "\n",

-        " b) Forward bais current is =-30.00 nA\n"

-       ]

-      }

-     ],

-     "prompt_number": 5

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 1.6, Page No 29"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "#initialisation of variables\n",

-      "Id=.1*10**-3\n",

-      "n=2\n",

-      "vt=26*10**-3\n",

-      "I0=30*10**-9\n",

-      "\n",

-      "#Calculations\n",

-      "Vd=(n*Vt)*math.log(Id/I0)*10**3\n",

-      "print(\" a) Forward bais current is =%.2f mV\\n\" %Vd)\n",

-      "Id=10*10**-3\n",

-      "Vd=(n*Vt)*math.log(Id/I0)*10**3\n",

-      "\n",

-      "#Results\n",

-      "print(\"b) forward bais current is %dmV\\n\" %Vd)"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " a) Forward bais current is =421.81 mV\n",

-        "\n",

-        "b) forward bais current is 661mV\n",

-        "\n"

-       ]

-      }

-     ],

-     "prompt_number": 6

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file