{ "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": {} } ] }