{ "metadata": { "celltoolbar": "Raw Cell Format", "name": "", "signature": "sha256:be03421cc765abd4c9572b7c61bb823243fbea415c12e649bb60ed73fc4375e6" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 2: Semiconductor Physics" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.21.1,Page number 2-47" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#given data\n", "\n", "ro=1.72*10**-8 #resistivity of Cu\n", "s=1/ro #conductivity of Cu\n", "n=10.41*10**28 #no of electron per unit volume\n", "e=1.6*10**-19 #charge on electron\n", "\n", "u=s/(n*e)\n", "\n", "print\"mobility of electron in Cu =\",\"{0:.3e}\".format(u),\"m^2/volt-sec\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "mobility of electron in Cu = 3.491e-03 m^2/volt-sec\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.21.2,Page number 2-47" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#given data\n", "\n", "m=63.5 #atomic weight\n", "u=43.3 #mobility of electron\n", "e=1.6*10**-19 #charge on electron\n", "N=6.02*10**23 #Avogadro's number\n", "d=8.96 #density\n", "\n", "Ad=N*d/m #Atomic density\n", "\n", "n=1*Ad\n", "\n", "ro=1/(n*e*u)\n", "\n", "print\"Resistivity of Cu =\",\"{0:.3e}\".format(ro),\"ohm-cm\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Resistivity of Cu = 1.699e-06 ohm-cm\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.21.3,Page number 2-47" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#given data\n", "\n", "e=1.6*10**-19 #charge on electron\n", "ne=2.5*10**19 #density of carriers\n", "nh=ne #for intrinsic semiconductor\n", "ue=0.39 #mobility of electron\n", "uh=0.19 #mobility of hole\n", "\n", "s=ne*e*ue+nh*e*uh #conductivity of Ge\n", "\n", "ro=1.0/s #resistivity of Ge\n", "\n", "print\"Resistivity of Ge =\",round(ro,4),\"ohm-m\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Resistivity of Ge = 0.431 ohm-m\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.21.5,Page number 2-48" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#given data\n", "\n", "Eg=1.2 #energy gap\n", "T1=600 #temperature\n", "T2=300 #temperature\n", "\n", "#since ue>>uh for intrinsic semiconductor\n", "\n", "#s=ni*e*ue\n", "\n", "K=8.62*10**-5 #Boltzman constant\n", "\n", "s=1l\n", "\n", "s1=s*exp((-Eg)/(2*K*T1))\n", "\n", "s2=s*exp((-Eg)/(2*K*T2))\n", "\n", "m=(s1/s2)\n", "\n", "print'Ratio between conductivity =',\"{0:.3e}\".format(m)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Ratio between conductivity = 1.092e+05\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.21.6,Page number 2-49" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#given data\n", "\n", "c=5*10**28 #concentration of Si atoms\n", "e=1.6*10**-19 #charge on electron\n", "u=0.048 #mobility of hole\n", "s=4.4*10**-4 #conductivity of Si\n", "\n", "#since millionth Si atom is replaced by an indium atom\n", "\n", "n=c*10**-6\n", "\n", "sp=u*e*n #conductivity of resultant\n", "\n", "print\"conductivity =\",(sp),\"mho/m\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "conductivity = 384.0 mho/m\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.21.7,Page number 2-49" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#given data\n", "\n", "m=28.1 #atomic weight of Si\n", "e=1.6*10**-19 #charge on electron\n", "N=6.02*10**26 #Avogadro's number\n", "d=2.4*10**3 #density of Si\n", "p=0.25 #resistivity\n", "\n", "#no. of Si atom/m**3\n", "\n", "Ad=N*d/m #Atomic density\n", "\n", "#impurity level is 0.01 ppm i.e. 1 atom in every 10**8 atoms of Si\n", "\n", "n=Ad/10**8 #no of impurity atoms\n", "\n", "#since each impurity produce 1 hole\n", "\n", "nh=n\n", "\n", "print\"1) hole concentration =\",round(n,4),\"holes/m^3\"\n", "\n", "up=1/(e*p*nh)\n", "\n", "print\"2) mobility =\",round(up,4),\"m^2/volt.sec\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "1) hole concentration = 5.14163701068e+20 holes/m^3\n", "2) mobility = 0.0486 m^2/volt.sec\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.22.1,Page number 2-50" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#given data\n", "\n", "t=27 #temp in degree \n", "T=t+273 #temp in kelvin\n", "K=8.62*10**-5 #Boltzman constant in eV\n", "Eg=1.12 #Energy band gap\n", "\n", "#For intrensic semiconductor (Ec-Ev)=Eg/2\n", "\n", "#let (Ec-Ev)=m\n", "\n", "m=Eg/2\n", "\n", "a=(m/(K*T))\n", "\n", "#probability f(Ec)=1/(1+exp((Ec-Ev)/(K*T))\n", "\n", "p=1/(1+exp(a))\n", "\n", "\n", "print\"probability of an electron being thermally excited to conduction band=\",\"{0:.3e}\".format(p)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "probability of an electron being thermally excited to conduction band= 3.938e-10\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.22.2,Page number 2-50" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#given data\n", "\n", "T=300 #temp in kelvin\n", "K=8.62*10**-5 #Boltzman constant in eV\n", "m=0.012 #energy level(Ef-E)\n", "\n", "a=(m/(K*T))\n", "\n", "#probability f(Ec)=1/(1+exp((Ec-Ev)/(K*T))\n", "\n", "p=1.0/(1+exp(a))\n", "\n", "p1=1-p\n", "\n", "print\"probability of an energy level not being occupied by an electron=\",round(p1,4)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "probability of an energy level not being occupied by an electron= 0.614\n" ] } ], "prompt_number": 19 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.22.3,Page number 2-51" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#given data\n", "\n", "t=20 #temp in degree \n", "T=t+273 #temp in kelvin\n", "K=8.62*10**-5 #Boltzman constant in eV\n", "Eg=1.12 #Energy band gap\n", "\n", "#For intrensic semiconductor (Ec-Ev)=Eg/2\n", "\n", "#let (Ec-Ev)=m\n", "\n", "m=Eg/2\n", "\n", "a=(m/(K*T))\n", "\n", "#probability f(Ec)=1/(1+exp((Ec-Ev)/(K*T))\n", "\n", "p=1.0/(1+exp(a))\n", "\n", "\n", "print\"probability of an electron being thermally excited to conduction band=\",\"{0:.3e}\".format(p)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "probability of an electron being thermally excited to conduction band= 2.348e-10\n" ] } ], "prompt_number": 21 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.22.4,Page number 2-51" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#given data\n", "\n", "T=300 #temp in kelvin\n", "K=8.62*10**-5 #Boltzman constant in eV\n", "Eg=2.1 #Energy band gap\n", "\n", "#probability f(Ec)=1/(1+exp((Ec-Ev)/(K*T))\n", "\n", "m=K*T\n", "\n", "#for f(E)=0.99\n", "\n", "p1=0.99\n", "\n", "b=1.0-(1.0/p1)\n", "\n", "a=math.log(b) #a=(E-2.1)/m\n", "\n", "E=2.1+m*a\n", "\n", "print\"1) Energy for which probability is 0.99=\",(E),\"eV\"\n", "\n", "#for f(E)=0.01\n", "\n", "p2=0.01\n", "\n", "b2=1-1.0/p2\n", "\n", "a1=math.log(b2) #a=(E-2.1)/m\n", "\n", "E1=2.1+m*a1\n", "\n", "print\"2)Energy for which probability is 0.01=\",(E1),\"eV\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "ename": "ValueError", "evalue": "math domain error", "output_type": "pyerr", "traceback": [ "\u001b[1;31m---------------------------------------------------------------------------\u001b[0m\n\u001b[1;31mValueError\u001b[0m Traceback (most recent call last)", "\u001b[1;32m\u001b[0m in \u001b[0;36m\u001b[1;34m()\u001b[0m\n\u001b[0;32m 17\u001b[0m \u001b[0mb\u001b[0m\u001b[1;33m=\u001b[0m\u001b[1;36m1.0\u001b[0m\u001b[1;33m-\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;36m1.0\u001b[0m\u001b[1;33m/\u001b[0m\u001b[0mp1\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m 18\u001b[0m \u001b[1;33m\u001b[0m\u001b[0m\n\u001b[1;32m---> 19\u001b[1;33m \u001b[0ma\u001b[0m\u001b[1;33m=\u001b[0m\u001b[0mmath\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0mlog\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mb\u001b[0m\u001b[1;33m)\u001b[0m \u001b[1;31m#a=(E-2.1)/m\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0m\u001b[0;32m 20\u001b[0m \u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m 21\u001b[0m \u001b[0mE\u001b[0m\u001b[1;33m=\u001b[0m\u001b[1;36m2.1\u001b[0m\u001b[1;33m+\u001b[0m\u001b[0mm\u001b[0m\u001b[1;33m*\u001b[0m\u001b[0ma\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n", "\u001b[1;31mValueError\u001b[0m: math domain error" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.23.1,Page number 2-52" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#given data\n", "\n", "ni=2.4*10**19 #density of intrensic semiconductor\n", "n=4.4*10**28 #no atom in Ge crystal\n", "Nd=n/10**6 #density\n", "Na=Nd\n", "e=1.6*10**-19 #charge on electron\n", "T=300 #temerature at N.T.P.\n", "K=1.38*10**-23 #Boltzman constant\n", "\n", "Vo=(K*T/e)*log(Na*Nd/(ni**2))\n", "\n", "print\"Potential barrier for Ge =\",round(Vo,4),\"Volts\"\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Potential barrier for Ge = 0.3888 Volts\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.23.2,Page number 2-52" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#given data\n", "\n", "B=0.6 #magnetic field\n", "d=5*10**-3 #distancebetween surface\n", "J=500 #current density\n", "Nd=10**21 #density\n", "e=1.6*10**-19 #charge on electron\n", "\n", "Vh=(B*J*d)/(Nd*e) #due to Hall effect\n", "\n", "print\"Hall voltage =\",\"{0:.3e}\".format(Vh),\"Volts\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Hall voltage = 9.375e-03 Volts\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.23.3,Page number 2-53" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#given data\n", "\n", "Rh=6*10**-7 #Hall coefficient\n", "B=1.5 #magnetic field\n", "I=200 #current in strip\n", "W=1*10**-3 #thickness of strip\n", "\n", "Vh=Rh*(B*I)/W #due to Hall effect\n", "\n", "print\"Hall voltage =\",(Vh),\"Volt\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Hall voltage = 0.18 Volt\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.23.4,Page number 2-53" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#given data\n", "\n", "Rh=2.25*10**-5 #Hall coefficient\n", "u=0.025 #mobility of hole\n", "\n", "r=Rh/u\n", "\n", "print\"Resistivity of P type silicon =\",\"{0:.3e}\".format(r),\"ohm-m\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Resistivity of P type silicon = 9.000e-04 ohm-m\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.23.5,Page number 2-53" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#given data\n", "\n", "B=0.55 #magnetic field\n", "d=4.5*10**-3 #distancebetween surface\n", "J=500 #current density\n", "n=10**20 #density\n", "e=1.6*10**-19 #charge on electron\n", "Rh=1/(n*e) #Hall coefficient\n", "\n", "Vh=Rh*B*J*d #Hall voltage\n", "\n", "print\"1) Hall voltage =\",round(Vh,4),\"Volts\"\n", "\n", "print\"2) Hall coefficient =\",(Rh),\"m^3/C\"\n", "\n", "u=0.17 #mobility of electrom\n", "\n", "m=math.atan(u*B)\n", "\n", "a=m*180/math.pi #conversion randian into degree\n", "\n", "print\"3) Hall angle =\",round(a,4),\"degree\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "1) Hall voltage = 0.0773 Volts\n", "2) Hall coefficient = 0.0625 m^3/C\n", "3) Hall angle = 5.3416 degree\n" ] } ], "prompt_number": 14 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.23.6,Page number 2-54" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#given data\n", "\n", "Rh=3.66*10**-4 #Hall coefficient\n", "r=8.93*10**-3 #resistivity \n", "e=1.6*10**-19 #charge on electron\n", "\n", "#Hall coefficient Rh=1/(n*e)\n", "\n", "n=1/(Rh*e) #density\n", "\n", "print\"1) density(n) =\",round(n,4),\"/m^3\"\n", "\n", "u=Rh/r #mobility of electron\n", "\n", "print\"2) mobility (u) =\",round(u,4),\"m^2/v-s\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "1) density(n) = 1.70765027322e+22 /m^3\n", "2) mobility (u) = 0.041 m^2/v-s\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.23.7,Page number 2-55" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#given data\n", "\n", "B=0.2 #magnetic field\n", "e=1.6*10**-19 #charge on electron\n", "ue=0.39 #mobility of electron\n", "l=0.01 #length\n", "A=0.001*0.001 #cross section area of bar\n", "V=1*10**-3 #Applied voltage\n", "d=0.001 #sample of width \n", "\n", "r=1/(ue*e) #resistivity\n", "R=r*l/A #resistance of Ge bar\n", "\n", "#using ohm's law\n", "\n", "I=V/R\n", "Rh=r*ue #hall coefficient\n", "\n", "#using formulae for hall effect\n", "\n", "J=I/A #current density\n", "Vh=Rh*B*J*d\n", "\n", "print\"Hall voltage =\",(Vh)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Hall voltage = 7.8e-06\n" ] } ], "prompt_number": 17 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.24.1,Page number 2-55" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#given data\n", "\n", "x1=0.4 #difference between fermi level and conduction band(Ec-Ef)\n", "T=300 #temp in kelvin\n", "K=8.62*10**-5 #Boltzman constant in eV\n", "\n", "#ne=N*e**(-(Ec-Ef)/(K*T))\n", "#ne is no of electron in conduction band\n", "#since concentration of donor electron is doubled\n", "\n", "a=2 #ratio of no of electron\n", "\n", "#let x2 be the difference between new fermi level and conduction band(Ec-Ef')\n", "\n", "x2=-math.log(a)*(K*T)+x1 #arranging equation ne=N*e**(-(Ec-Ef)/(K*T))\n", "\n", "print\"Fermi level will be shifted towards conduction band by\",round(x2,4),\"eV\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Fermi level will be shifted towards conduction band by 0.3821 eV\n" ] } ], "prompt_number": 19 }, { "cell_type": "code", "collapsed": false, "input": [], "language": "python", "metadata": {}, "outputs": [] } ], "metadata": {} } ] }