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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 07 : Integrated Circuit Fabrication and Characteristic"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.1, Page No 215"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"\n",
"print('At distance equal to x=xi at which N = concentration n of doped silicon wafers , the net impurity density is zero. Thus xi is the distance at which junction is formed')\n",
"q = 1.6*(10**-19) #Charge of electron\n",
"yn=1300.0 #mobility of silicon\n",
"p = 0.5 #resistivity in ohm=cm\n",
"y=2.2\n",
"\n",
"#Calculations\n",
"t=2.0*3600 #in sec.\n",
"xi = 2.7*(10**-4) #Junction Depth in cm.\n",
"n = 1/(p*yn*q) #Concentration of doped silicon wafer\n",
"print(\"The concentration n = %.2f cm^-3 x 10^16\" %(n/10**16))\n",
"print('The junction is formed when N = n')\n",
"\n",
"#y = xi/(2*(D*t)^0.5)\n",
"D=((xi)**2/((2*y)**2*t)) #Diffusion Constant\n",
"\n",
"#Results\n",
"print(\"The value of Diffusion Constant for Boron = %.2f cm^2/sec X 10^-13\" %(D*10**13))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"At distance equal to x=xi at which N = concentration n of doped silicon wafers , the net impurity density is zero. Thus xi is the distance at which junction is formed\n",
"The concentration n = 0.96 cm^-3 x 10^16\n",
"The junction is formed when N = n\n",
"The value of Diffusion Constant for Boron = 5.23 cm^2/sec X 10^-13\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.2, Page No 215"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"d=5.2*10**-13 #from previous example\n",
"depth=1.7*10**-4\n",
"t=2*3600.0\n",
"c=2.5*10**17 # boron concentration cm^3\n",
"\n",
"#Calculations\n",
"y = depth/(2*(math.sqrt(d*t)))\n",
"q=(c*(math.sqrt(math.pi*4*10**-13*3420)))/(math.exp(-((depth**2)/(4*4*10**-13*3420))))\n",
"\n",
"\n",
"#Results\n",
"print(\"The value of Y is = %.2f \" %(y))\n",
"print(\"The value of Q is = %.2f cm2 X 10^15 \" %(q/10**15))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of Y is = 1.39 \n",
"The value of Q is = 3.22 cm2 X 10^15 \n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.3, Page No 222"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"y=100.0*10**-4 #mm\n",
"h=500.0 #cm^2/V-s\n",
"p=10.0**16 #boron of concentration\n",
"\n",
"\n",
"#Calculations\n",
"Rs=1.0/(1.6*10**-19*h*p*y)\n",
"\n",
"#Results\n",
"print(\"The value of Rs sheet resistance is = %.2f ohm/sqare\" %(Rs))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of Rs sheet resistance is = 125.00 ohm/sqare\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.4, Page No 223"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"Rs=100.0 #ohm/square\n",
"l=50.0 #mm\n",
"w=10 #mm\n",
"\n",
"\n",
"#Calculations\n",
"R=Rs*(l/w)\n",
"\n",
"#Results\n",
"print(\"The resistance of defused resistor is = %.2f ohm\" %(R))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The resistance of defused resistor is = 500.00 ohm\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.5, Page No 225"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"A=100*10**-8 #mm^2\n",
"q=1.6*10**-19\n",
"Nd=10**16 #donor concentration /cm^3\n",
"e=11.9*8.85*10**-14\n",
"Vj=0.82 #v\n",
"\n",
"\n",
"#Calculations\n",
"C=A*math.sqrt((q*Nd*e)/(2*Vj))\n",
"\n",
"#Results\n",
"print(\"The capacitance is = %.f fF\" %(C*10**15))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The capacitance is = 32 fF\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.6, Page No 225"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"A=100*10*10**-8 #mm^2\n",
"q=1.6*10**-19\n",
"e=11.9*8.85*10**-14\n",
"Vj=0.98 #v\n",
"Mn=1300.0\n",
"pn=0.01\n",
"\n",
"\n",
"\n",
"#Calculations\n",
"Nd=1/(q*Mn*pn) #donor concentration /cm^3\n",
"C=A*math.sqrt((q*Nd*e)/(2*Vj))\n",
"\n",
"#Results\n",
"print(\"The capacitance is = %.f pF\" %(C*10**12))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The capacitance is = 2 pF\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.7, Page No 226"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"e=3.9*8.85*10**-14\n",
"d=20*10**-8\n",
"\n",
"\n",
"#Calculations\n",
"C=(e/d)*(10**9/10**8)\n",
"\n",
"#Results\n",
"print(\"The capacitance per unit area is = %.2f fF/mM^2\" %(C*10**6))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The capacitance per unit area is = 17.26 fF/mM^2\n"
]
}
],
"prompt_number": 14
}
],
"metadata": {}
}
]
}
|
