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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 02 : Transport Phenomena in Semiconductor"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.1, Page No 22"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"n=10.0**20\n",
"q=1.6*10**-19\n",
"mn=800 #cm^3\n",
"delta=1 #V/cm\n",
"\n",
"#Calculations\n",
"J=n*q*mn*delta\n",
"\n",
"\n",
"#Results\n",
"print(\"The electron current density is= %.2f X 10^4 atom/cm^2 \" %(J/(10**4)))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The electron current density is= 1.28 X 10^4 atom/cm^2 \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.2a, Page No 27"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"ni=10.0**10\n",
"Nd=10**12\n",
"\n",
"#Calculations\n",
"n=(Nd+(math.sqrt(Nd+4*ni**2)))\n",
"\n",
"\n",
"#Results\n",
"print(\"The free electron is= %.3f X 10^12 cm^3 \" %(n/(10**12)))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The free electron is= 1.020 X 10^12 cm^3 \n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.2b, Page No 27"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"ni=10.0**10\n",
"Nd=10**18\n",
"\n",
"#Calculations\n",
"n=(Nd+(math.sqrt(Nd+4*ni**2)))\n",
"\n",
"\n",
"#Results\n",
"print(\"The free electron is= %.2f X 10^18 cm^3 \" %(n/(10**18)))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The free electron is= 1.00 X 10^18 cm^3 \n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.3a, Page No 29"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"Av=6.02*(10**23) #Avogadro No.\n",
"m=72.6 #Molar mass of germanium in gm/moles\n",
"d=5.32 #density in gm/cm^3\n",
"\n",
"#Calculations\n",
"conc = (Av/m)*d #Concentration of atom in germanium\n",
"\n",
"#Results\n",
"print(\"The concentration of germanium atom is= %.2f X 10^22 atom/cm^3 \" %(conc/(10**22)))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The concentration of germanium atom is= 4.41 X 10^22 atom/cm^3 \n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.3b, Page No 29"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"Av=6.02*(10**23) #Avogadro No.\n",
"m=72.6 #Molar mass of germanium in gm/moles\n",
"d=5.32 #density in gm/cm^3\n",
"ni=2.5*(10**13) #in cm^-3\n",
"n=ni\n",
"p=ni #n=magnitude of free electrons, p=magnitude of holes, ni=magnitude of intrinsic concentration\n",
"\n",
"#Calculations\n",
"q=1.6*(10**-19) #Charge of an Electron\n",
"yn=3800.0 #in cm^2/V-s\n",
"yp=1800.0 #in cm^2/V-s\n",
"\n",
"#Required Formula\n",
"A=ni*q*(yn+yp) #Conductivity\n",
"print(\"Conductivity is = %.2f ohm-cm^-1 \" %A)\n",
"R =1.0/A #Resistivity\n",
"print(\"Resistivity is = %.2f ohm-cm \" %R)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Conductivity is = 0.02 ohm-cm^-1 \n",
"Resistivity is = 44.64 ohm-cm \n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.3c Page No 29"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"print('We know that n=p=ni where n is conc of free electron p is conc of holes and ni is conc of intrinsic carriers')\n",
"#Resistivity if 1 donor atom per 10^8 germanium atoms\n",
"Nd=4.41*(10**14) #in atoms/cm^3\n",
"ni=2.5*(10**13) #in cm^3\n",
"yn=3800.0 #in cm^2/V-s\n",
"\n",
"#Calculations\n",
"q=1.6*(10**-19)\n",
"n=Nd\n",
"p=(ni**2)/Nd\n",
"\n",
"print(\"The concentration of holes is= %.2f holes/cm^3 \" %p)\n",
"if n>p:\n",
" A=n*q*yn #Conductivity\n",
" print(\"The conductivity is = %.2f ohm-cm^-1 \" %A)\n",
" \n",
"R=1.0/A #Resistivity\n",
"\n",
"#Results\n",
"print(\"The resistivity is = %.2f ohm-cm \" %R)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"We know that n=p=ni where n is conc of free electron p is conc of holes and ni is conc of intrinsic carriers\n",
"The concentration of holes is= 1417233560090.70 holes/cm^3 \n",
"The conductivity is = 0.27 ohm-cm^-1 \n",
"The resistivity is = 3.73 ohm-cm \n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.3d, Page No 29"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"\n",
"print('We know that n=p=ni where n is conc of free electron p is conc of holes and ni is conc of intrinsic carriers')\n",
"#Ratio of Conductivities\n",
"Nd=4.41*(10**14) #in atoms/cm^3\n",
"ni=2.5*(10**13) #in cm^3\n",
"yn=3800.0 #in cm^2/V-s\n",
"q=1.6*(10**-19)\n",
"\n",
"#Calculations\n",
"n=Nd\n",
"A=n*q*yn #Conductivity\n",
"\n",
"#If germanium atom were monovalent metal , ratio of conductivity to that of n-type semiconductor\n",
"\n",
"n=4.41*(10**22) #in electrons/cm^3\n",
"\n",
"\n",
"#Results\n",
"print('If germanium atom were monovalent metal')\n",
"A1=n*q*yn\n",
"print(\"The coductivity of metal is= %.2f ohm=cm^-1 x 10^7 \" %(A1/10**7))\n",
"F=A1/A\n",
"print(\"The factor by which the coductivity of metal is higher than that of n type semiconductor is %.2f x 10^8 \" %(F/10**8))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"We know that n=p=ni where n is conc of free electron p is conc of holes and ni is conc of intrinsic carriers\n",
"If germanium atom were monovalent metal\n",
"The coductivity of metal is= 2.68 ohm=cm^-1 x 10^7 \n",
"The factor by which the coductivity of metal is higher than that of n type semiconductor is 1.00 x 10^8 \n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.4, Page No 35"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"g=5*10**21 #Generation rate\n",
"tp=2*10**-6 #hole lifetime\n",
"\n",
"#Calculations\n",
"p=g*tp\n",
"\n",
"#Required Formula\n",
"print(\"Hole density is = %.2f cm^3 10^16 \" %(p/10**16))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Hole density is = 1.00 cm^3 10^16 \n"
]
}
],
"prompt_number": 8
}
],
"metadata": {}
}
]
}
|
