{
"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<ipython-input-4-0fb7e85ec399>\u001b[0m in \u001b[0;36m<module>\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": {}
}
]
}