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 "cells": [
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   "metadata": {},
   "source": [
    "# Chapter3 : Excess Carriers in Semiconductor"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 3.2 Page No 111"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Minimum required energy is 2.06 eV \n"
     ]
    }
   ],
   "source": [
    "#Example 3.2\n",
    "#What is Minimum required energy \n",
    "\n",
    "#given data\n",
    "l=6000                 #in Angstrum\n",
    "h=6.6*10**(-34)             #Planks constant\n",
    "c=3*10**8                   #speed of light in m/s\n",
    "e=1.602*10**(-19)            #Constant\n",
    "\n",
    "#calculation\n",
    "phi=c*h/(e*l*10**(-10))\n",
    "\n",
    "#result\n",
    "print\"Minimum required energy is\",round(phi,2),\"eV \"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 3.3 Page No 112"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Work function of the cathode material is  2.39 eV\n"
     ]
    }
   ],
   "source": [
    "#Exa 3.3\n",
    "#calculate Work function of the cathode material\n",
    "\n",
    "#given data\n",
    "Emax=2.5                   #maximum energy of emitted electrons in eV \n",
    "l=2537.0                #in Angstrum\n",
    "\n",
    "#Calculation\n",
    "EeV=12400.0/l           #in eV\n",
    "phi=EeV-Emax               #in eV\n",
    "\n",
    "#result\n",
    "print \"Work function of the cathode material is \",round(phi,2),\"eV\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 3.4 Page No 115"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "(i)Thus power absorbed is  0.009 J/s\n",
      "(ii)Energy converted into heat is 0.0026 J/s\n",
      "(iii)Number of photons per second given off from recombination events  2.81e+16\n"
     ]
    }
   ],
   "source": [
    "#Example 3.4\n",
    "#Find (i)The fraction of each photon energy unit which is converted into heat\",f\n",
    "#(ii)Energy converted into heat in ,((2-1.43)/2)*0.009,\"J/s\"\n",
    "#(iii)Number of photons per second given off from recombination events \",0.009/(e*2)\n",
    "\n",
    "#given data\n",
    "t=0.46*10**-4                 #in centi meters\n",
    "hf1=2                          #in ev\n",
    "hf2=1.43\n",
    "Pin=10                        #in mW\n",
    "alpha=50000                   # in per cm\n",
    "e=1.6*10**-19                 #constant\n",
    "Io=0.01                   #in mW\n",
    "\n",
    "import math\n",
    "\n",
    "#Calculation\n",
    "It=Io*math.exp(-alpha*t)           #in mW\n",
    "Iabs=Io-It\n",
    "f=(hf1-hf2)/hf1\n",
    "E=f*Iabs\n",
    "N=Iabs/(e*hf1)\n",
    "\n",
    "#result\n",
    "print\"(i)Thus power absorbed is \",round(Iabs,3),\"J/s\"\n",
    "print\"(ii)Energy converted into heat is\",round(E,4),\"J/s\"\n",
    "print\"(iii)Number of photons per second given off from recombination events \",round(N,-14)\n",
    "#In book there is calculation mistake in Number of photons."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 3.5 Page No 123"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 29,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Electron transit time in sec is  6.4e-09 s\n",
      "Photoconductor gain is  216.0\n"
     ]
    }
   ],
   "source": [
    "#Example 3.5\n",
    "#What is Photoconductor gain \n",
    "#Electron transit time.\n",
    "\n",
    "#given data\n",
    "L=100                     #in uM\n",
    "A=10&-7                   #in cm**2\n",
    "th=10**-6                 #in sec\n",
    "V=12                      #in Volts\n",
    "ue=0.13                   #in m**2/V-s\n",
    "uh=0.05                   #in m**2/V-s\n",
    "\n",
    "#Calculation\n",
    "E=V/(L*10**-6)            #in V/m\n",
    "tn=(L*10**-6)/(ue*E)\n",
    "Gain=(1+uh/ue)*(th/tn)\n",
    "\n",
    "#result\n",
    "print\"Electron transit time in sec is \",round(tn,10),\"s\"\n",
    "print\"Photoconductor gain is \",Gain"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 3.6 Page No128"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 30,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Current flowing through diode is  15.0 micra A\n"
     ]
    }
   ],
   "source": [
    "#Example3.6\n",
    "#Calculate Current flowing through diode .\n",
    "\n",
    "#given datex\n",
    "import math\n",
    "Io=0.15                    #in uA\n",
    "V=0.12                     #in mVolt\n",
    "Vt=26                      #in mVolt\n",
    "\n",
    "#calculation\n",
    "I=Io*10**-6*(math.exp(V/(Vt*10**-3))-1)          #in A\n",
    "\n",
    "#result\n",
    "print\"Current flowing through diode is \",round(I*10**6,2),\"micra A\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 3.7 Page No 128"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 31,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Forward voltage  is  0.43 V\n"
     ]
    }
   ],
   "source": [
    "#Exa 3.7\n",
    "#Determine the Forward voltage \n",
    "\n",
    "#given data\n",
    "import math\n",
    "Io=2.5              #in uA\n",
    "I=10                #in mA\n",
    "Vt=26               #in mVolt\n",
    "n=2                 #for silicon\n",
    "\n",
    "#Calculation\n",
    "V=n*Vt*10**-3*math.log((I*10**-3)/(Io*10**-6))\n",
    "\n",
    "#Result\n",
    "print \"Forward voltage  is \",round(V,2),\"V\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 3.8 Page No 128"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 33,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Reverse saturation current density is  0.16 uA \n"
     ]
    }
   ],
   "source": [
    "#Example 3.8\n",
    "#What is Reverse saturation current density \n",
    "\n",
    "#given data\n",
    "ND=10**21                #in m**-3\n",
    "NA=10**22                #in m**-3\n",
    "De=3.4*10**-3            #in m**2-s**-1\n",
    "Dh=1.2*10**-3            #in m**2-s**-1\n",
    "Le=7.1*10**-4            #in meters\n",
    "Lh=3.5*10**-4            #in meters\n",
    "ni=1.6*10**16            #in m**-3\n",
    "e=1.602*10**-19          #constant\n",
    "\n",
    "#calculation\n",
    "IoA=e*ni**2*(Dh/(Lh*ND)+De/(Le*NA))\n",
    "\n",
    "#Result\n",
    "print\"Reverse saturation current density is \",round(IoA*10**6,2),\"uA \""
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": []
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