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| author | Thomas Stephen Lee | 2015-09-04 22:04:10 +0530 |
|---|---|---|
| committer | Thomas Stephen Lee | 2015-09-04 22:04:10 +0530 |
| commit | 64419e47f762802600b3a2b6d8c433a16ccd3d55 (patch) | |
| tree | 14ad7c37c9547cd516f141494f3fa375621edbaa /Advance_Semiconductor_Devices_by_S._Sharma | |
| parent | 10f6fb8cd1d840a3042651dfaa6fd5af4924b94a (diff) | |
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diff --git a/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo1.ipynb b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo1.ipynb new file mode 100755 index 00000000..4d1dc38d --- /dev/null +++ b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo1.ipynb @@ -0,0 +1,800 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1> Chapter No 1 : Semiconductor Materials and\n",
+ "Their Properties<h1>"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.3, Page No 37 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Finding Density of silicon crystal\n",
+ "\n",
+ "#initialisation of variables\n",
+ "a=5.3 #in \u00c5\n",
+ "a=a*10**-10 #in m\n",
+ "N_A=6.023*10**23 \n",
+ "At_Si=28 #atomic weight of Si\n",
+ "n = 4\n",
+ "m=At_Si/N_A #in gm\n",
+ "m= m*10**-3 #in kg\n",
+ "V=a**3 #in m^3\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Rho=((m*n)/V)/1000 #in kg/m^3\n",
+ "\n",
+ "# Note: There is calculation error to find the value of density. So the answer in the book is wrong.#RESULTS\n",
+ "#RESULTS\n",
+ "print('Density of silicon crystal in kg/m^3 is = %.2f kg/m^3' %Rho)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Density of silicon crystal in kg/m^3 is = 1.25 kg/m^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.4, Page No 37 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "# Finding Density of the crystal \n",
+ "\n",
+ "#initialisation of variables\n",
+ "n = 4\n",
+ "r = 1.278 #in \u00c5\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "a = (4*r)/(math.sqrt(2)) #in\u00c5\n",
+ "a = a * 10**-10 #in m\n",
+ "V = (a)**3 #in m^3\n",
+ "At_W = 63.5 #atomic weight\n",
+ "N_A = 6.023*10**23\n",
+ "m = At_W /N_A #in gm\n",
+ "m = m*10**-3 #in kg\n",
+ "Rho = ((m*n)/V)/1000 #in kg/m^3\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Density of the crystal in kg/m^3 is =%.2f kg/m^3' %Rho)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Density of the crystal in kg/m^3 is =8.93 kg/m^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.5 Page No 43 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Finding Wavaelength of X ray\n",
+ "\n",
+ "import math\n",
+ "from numpy import *\n",
+ "#initialisation of variables\n",
+ "d=2.82 #in \u00c5\n",
+ "d=d*10**-10 #in m\n",
+ "n=1\n",
+ "theta1=10 #n degree\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "lembda=2*d*math.sin(math.radians(theta1)) #in m\n",
+ "lembda=lembda*10**10 #in \u00c5\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Wavaelength of X ray is =%.2f \u00c5 degrees' %lembda)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavaelength of X ray is =0.98 \u00c5 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.6, Page No 43"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Finding The spacing of atomic layer\n",
+ "\n",
+ "import math\n",
+ "#initialisation of variables\n",
+ "lembda = 1.6 #in \u00c5\n",
+ "theta = 14.2 #in degree\n",
+ "n = 1\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "d = (n*lembda)/(2*math.sin(math.radians(theta))) #in \u00c5\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The spacing of atomic layer in crystal is =%.2f in \u00c5 degrees' %d)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The spacing of atomic layer in crystal is =3.26 in \u00c5 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.7 Page No 43"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Finding The interplanner spacing\n",
+ "\n",
+ "import math\n",
+ "#initialisation of variables\n",
+ "n = 1\n",
+ "theta1 = 30 #in degree\n",
+ "lembda = 1.78 #in \u00c5\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "d = (n*lembda)/(2*math.sin(math.radians(theta1))) #in \u00c5\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The interplanner spacing is =%.2f in \u00c5 degrees' %d)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The interplanner spacing is =1.78 in \u00c5 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.8 Page No 43"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Finding Interplaner spacing of the crystal\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "lembda = 0.58 #in \u00c5\n",
+ "n = 1;\n",
+ "theta1 = 6.45 # in degree\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "d = (n*lembda)/(2*math.sin(math.radians(theta1))) #in \u00c5 \n",
+ "print('Part (i) : At angle of 6.45\u00b0, Interplaner spacing of the crystal is =%.2f \u00c5' %d)\n",
+ "theta2 = 9.15 #in degree\n",
+ "d1 = (n*lembda)/(2*math.sin(math.radians(theta2))) #in \u00c5 \n",
+ "print('Part (ii) : At angle of 9.15\u00b0, Interplaner spacing of the crystal is =%.2f \u00c5' %d1)\n",
+ "theta3 = 13 #in degree\n",
+ "n2 = 1\n",
+ "d2 = (n2*lembda)/(2*math.sin(math.radians(theta3))) #in \u00c5\n",
+ "print('Part (iii) : At angle of 13\u00b0, Interplaner spacing of the crystal is =%.2f \u00c5' %d2)\n",
+ "n=2\n",
+ "d2 = (n*lembda)/(2*math.sin(math.radians(theta3))) #in \u00c5 \n",
+ "\n",
+ "#RESULTS\n",
+ "print('Part (iv) : : The interplaner spacing is : =%.4f \u00c5' %d2)\n",
+ "print('The interplaner spacing for some other set of reflecting is : =%.2f \u00c5' %d1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Part (i) : At angle of 6.45\u00b0, Interplaner spacing of the crystal is =2.58 \u00c5\n",
+ "Part (ii) : At angle of 9.15\u00b0, Interplaner spacing of the crystal is =1.82 \u00c5\n",
+ "Part (iii) : At angle of 13\u00b0, Interplaner spacing of the crystal is =1.29 \u00c5\n",
+ "Part (iv) : : The interplaner spacing is : =2.5783 \u00c5\n",
+ "The interplaner spacing for some other set of reflecting is : =1.82 \u00c5\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.9, Page No 44 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Finding The glancing angle for a cubic in degree\n",
+ "\n",
+ "import math\n",
+ "from sympy.mpmath import *\n",
+ "import cmath\n",
+ "\n",
+ "#initialisation of variables\n",
+ "a = 2.814 #in \u00c5\n",
+ "l = 0\n",
+ "h = l\n",
+ "k = 0\n",
+ "d=a #in \u00c5\n",
+ "n = 2\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "lembda=0.710 #in \u00c5\n",
+ "theta = math.degrees(math.asin(n*lembda/(2*d)))\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The glancing angle for a cubic in degree is : =%.2f \u00c5 degrees' %theta)\n",
+ "print('Anwser is in points i.e = 14 36\"40\"..')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The glancing angle for a cubic in degree is : =14.61 \u00c5 degrees\n",
+ "Anwser is in points i.e = 14 36\"40\"..\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.10 Page No 44"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Finding Wavelength of X ray\n",
+ "\n",
+ "import math \n",
+ "\n",
+ "#initialisation of variables\n",
+ "a = 3.65 #in \u00c5\n",
+ "a = 3.65*10**-10 #in m\n",
+ "h = 1\n",
+ "k = 0\n",
+ "l = 0\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "d= a/(math.sqrt(h**2+k**2+l**2)) #in m\n",
+ "n = 1\n",
+ "theta=60 #in degree\n",
+ "lembda = 2*d*math.sin(math.radians(theta)) #in m\n",
+ "lembda = lembda * 10**10 #in \u00c5\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Wavelength of X ray is : =%.2f \u00c5' %lembda)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of X ray is : =6.32 \u00c5\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.11 Page No 44"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Finding The glancing angle in degree\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "lembda = 1.54 #in \u00c5\n",
+ "density = 9.024 #in gm/cc\n",
+ "n = 1\n",
+ "MI = 100\n",
+ "At_W = 63.54 #atomic weight\n",
+ "N_A = 6.023*10**23 \n",
+ "m = At_W/N_A #in gm\n",
+ "a =(density*m)**(1.0/3) #in cm\n",
+ "h = 1\n",
+ "k = 0\n",
+ "l = 0\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "d= a/(math.sqrt(h**2+k**2+l**2))\n",
+ "theta =math.degrees((lembda*10**-8)/(2*d)) #in degree\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The glancing angle is = %.2f degrees' %theta)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The glancing angle is = 4.48 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.12 Page No 45 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Finding To get the 2nd order spectrum the position of the detector in degree\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "a = 3.615 #in \u00c5\n",
+ "theta=22 #in degree\n",
+ "n=1\n",
+ "h=1\n",
+ "k=h\n",
+ "l=k\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "d = a/(math.sqrt( ((h)**2) + ((k)**2) + ((l)**2) )) #in \u00c5\n",
+ "lembda = 2*d*math.sin(math.radians(theta)) #in \u00c5\n",
+ "print('The wavelength of X ray is = %.2f \u00c5 ' %lembda)\n",
+ "theta2 =math.degrees(math.asin(lembda/d)) #in degree\n",
+ " #in degree\n",
+ "\n",
+ "#RESULTS\n",
+ "print('To get the 2nd order spectrum the position of the detector is = %.2f degrees ' %theta2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength of X ray is = 1.56 \u00c5 \n",
+ "To get the 2nd order spectrum the position of the detector is = 48.52 degrees \n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.13, Page No 45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "n = 1\n",
+ "lembda = 1.54 # in \u00c5\n",
+ "theta = 21.7 # in degree\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "d = lembda/(2*math.sin(math.radians(theta))) # in \u00c5\n",
+ "h = 1\n",
+ "k = h\n",
+ "l = k\n",
+ "a = d*math.sqrt(h**2+k**2+l**2) # in \u00c5\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Lattice constant is = %.2f in \u00c5 ' %a)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Lattice constant is = 3.61 in \u00c5 \n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.14, Page No 47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "h = 2\n",
+ "k = 1\n",
+ "l = 1\n",
+ "a = 4.8 # in \u00c5\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "d_211 = a/(math.sqrt(h**2+k**2+l**2)) #in \u00c5\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The distance between planes is =%.2f \u00c5 ' %d_211)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The distance between planes is =1.96 \u00c5 \n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.15, Page No 47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "r = 1.28 # in \u00c5\n",
+ "n = 4\n",
+ "M = 63.5\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "a = (4*r)/(math.sqrt(2)) #in \u00c5\n",
+ "a = a * 10**-8 # in cm\n",
+ "N_A = 6.023*10**23 \n",
+ "Rho = (n*M)/( N_A*((a)**3) ) # in gm/cc\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Density is =%.2f gm/cc ' %Rho)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Density is =8.89 gm/cc \n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.16, Page No 47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "M = 55.85\n",
+ "a = 2.9 # in \u00c5\n",
+ "Rho = 7.87 #in gm/cc\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "a = a * 10**-8 # in cm\n",
+ "N_A = 6.023*10**23\n",
+ "n = (Rho*N_A*((a)**3))/M # atom per unit\n",
+ "\n",
+ "#RESULTS\n",
+ "print('A lattice having =%.2f atom per unit cell is a BCC structure ' %n)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "A lattice having =2.07 atom per unit cell is a BCC structure \n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.17, Page No 47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "M = 60.0 # in gm/mole\n",
+ "Rho = 6.23 # in gm/cc\n",
+ "n = 4\n",
+ "N_A = 6.023*10**23 \n",
+ "\n",
+ "#CALCULATIONS\n",
+ "a = ((n*M)/(N_A * Rho))**(1.0/3) # in cm\n",
+ "r = (a*math.sqrt(2))/n #radius of atom in cm\n",
+ "r = r * 10**8 # in \u00c5\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Radius of atom is =%.2f \u00c5 ' %r)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of atom is =1.41 \u00c5 \n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.18, Page No 47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "Rho = 5.96 # in gm/cc\n",
+ "M = 50.0\n",
+ "n = 2\n",
+ "N_A = 6.023*10**23 \n",
+ "\n",
+ "#CALCULATIONS\n",
+ "a =((n*M)/(Rho*N_A))**(1.0/3) # in cm\n",
+ "r = (a*math.sqrt(3))/4 # in cm\n",
+ "P_f = (2*(4.0/3)*math.pi*((r)**3))/((a)**3) # packing factor\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Packing factor is =%.2f ' %P_f)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Packing factor is =0.68 \n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.19, Page No 48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "M = 120.0\n",
+ "n = 2.0\n",
+ "N_A = 6.023*10**23\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "m1 = M/N_A #mass of 1 atom in gm\n",
+ "m2 = n*m1 #mass of unit cell in gm\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Number of unit cell in 20 gms of element is : =%.2f X 10^22 unit cells ' %((20/m2)/(10**22)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of unit cell in 20 gms of element is : =5.02 X 10^22 unit cells \n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.20, Page No 48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "Rho = 2.48 # in gm/c.c\n",
+ "n = 4.0 \n",
+ "M = 58\n",
+ "N_A = 6.023*10**23\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "a = ( (n*M)/(Rho*N_A) )**(1.0/3) # in cm\n",
+ "a = a * 10**8 # in \u00c5\n",
+ "r = (a*math.sqrt(2))/n # in \u00c5\n",
+ "r = 2*r # in \u00c5\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The center to center distance between ions is =%.2f \u00c5' %r)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The center to center distance between ions is =3.80 \u00c5\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo2.ipynb b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo2.ipynb new file mode 100755 index 00000000..9bc8cffb --- /dev/null +++ b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo2.ipynb @@ -0,0 +1,1248 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1> Chapter No 2 : Semiconductor Materials and\n",
+ "Their Properties<h1>"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.1, Page No 65 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "lambda1 = 11000.0 #in \u00c5\n",
+ "lambda1 = lambda1 * 10**-10 \n",
+ "h = 6.625*10**-34 \n",
+ "c = 3*10**8\n",
+ "q = 1.6*10**-19 #in C\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "E_g = h*c/lambda1 #in J\n",
+ "E_g= E_g/q #in eV\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The energy gap in Si is = %.2f eV' %E_g)\n",
+ "\n",
+ "#Note: The answer in the book is not correct \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy gap in Si is = 1.13 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2, Page No 65 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "E_g = 0.75 #in eV\n",
+ "q=1.6*10**-19 #in C\n",
+ "E_g = E_g*q #in J\n",
+ "h = 6.63*10**-34 #in J\n",
+ "c = 3*10**8 #in m/s\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "lembda = (h*c)/E_g #in m\n",
+ "lembda = lembda * 10**10 #in \u00c5\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The wavelength is =%.f \u00c5 ' %lembda)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is =16575 \u00c5 \n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.3 Page No 81 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from numpy import *\n",
+ "\n",
+ "#initialisation of variables\n",
+ "del_E = 0.3 #value of E_C-E_F in eV\n",
+ "T1 = 330.0 #in K\n",
+ "T = 300 #in K\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "del_E1 = del_E*(T1/T) #value of E_C-E_F in eV\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The position of fermi level is =%.2f eV' %del_E1)\n",
+ "print('Hence the Fermi level will be %.2f eV below the conduction band' %del_E1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The position of fermi level is =0.33 eV\n",
+ "Hence the Fermi level will be 0.33 eV below the conduction band\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.4, Page No 81"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "K = 8.63*10**-5\n",
+ "T = 300.0 #in K\n",
+ "N_C = 2.8*10**19 #in cm^-3\n",
+ "del_E = 0.25\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "f_F = math.exp( (-del_E)/(K*T) )\n",
+ "n_o = N_C*math.exp( (-del_E)/(K*T) ) #in cm^-3\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The probability is =%.6f ' %f_F)\n",
+ "print('The thermal equillibrium electron concentration is =%.2f X 10^15 cm^-3' %(n_o/(10**15)))\n",
+ "print('Approximation error ')\n",
+ "# The answer is same as in book only thy have rounded off to 1.8 * 10^15 which is same as above"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The probability is =0.000064 \n",
+ "The thermal equillibrium electron concentration is =1.79 X 10^15 cm^-3\n",
+ "Approximation error \n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.5 Page No 82"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "N_V = 1.04*10**19 #in cm^-3\n",
+ "T1 = 400.0 #in K\n",
+ "T2 = 300.0 #in K\n",
+ "del_E = 0.27 #value of E_F-E_V in eV\n",
+ "K = 0.0259\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "N_V= N_V*(T1/T2)**(3/2) #in cm^-3\n",
+ "KT = K*(T1/T2) #in eV\n",
+ "p_o = N_V*math.exp( (-del_E)/(KT) ) #in /cm^3\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The hole concentration is =%.2f x 10^15 per cm^3' %(p_o/(10**15)))\n",
+ "print('Approximationa error ')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The hole concentration is =5.58 x 10^15 per cm^3\n",
+ "Approximationa error \n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.6 Page No 86"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "N = 6.02*10**23\n",
+ "A = 63.5 #atomic weight\n",
+ "Rho = 1.7*10**-6 #in ohm cm\n",
+ "d = 8.96 # in gm/cc\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "n = (N/A)*d #in /cc\n",
+ "e = 1.6*10**-19 #in C\n",
+ "Miu_e = 1/(Rho*n*e) #in cm^2/volt-sec\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The mobility of electron is =%.2f cm^2/volt.sec' %Miu_e)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mobility of electron is =43.28 cm^2/volt.sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.7, Page No 87"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from sympy.mpmath import *\n",
+ "import cmath\n",
+ "\n",
+ "\n",
+ "#initialisation of variables\n",
+ "d = 8.96 #in gm/cc\n",
+ "At = 63.5 #atomic weight\n",
+ "N_A = 6.02*10**23 #in /gm mole\n",
+ "l = 0.1 #in m\n",
+ "e = 1.6*10**-19 #in C\n",
+ "A = 1.7*10**-6 #in m^2\n",
+ "R = 0.1 #in ohm\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "n = (N_A/At)*d #in /cc\n",
+ "n = n * 10**6 #in /m^3\n",
+ "Rho = (R*A)/l #in ohm.m\n",
+ "Sigma = 1/Rho #in mho/m\n",
+ "Miu_e = Sigma/(n*e) #in m^2/V-sec\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The electron mobility is =%.3f X 10^-3 m^2/V-sec ' %(Miu_e*(10**3)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron mobility is =0.043 X 10^-3 m^2/V-sec \n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.8 Page No 87"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#initialisation of variables\n",
+ "N_A = 6.025*10**23 #in /gm mole\n",
+ "d = 10.5 #in gm/cc\n",
+ "At = 108.0 #atomic weight of \n",
+ "\n",
+ "#CALCULATIONS\n",
+ "n = (N_A/At)*d #in /cm^3\n",
+ "n = n * 10**6 #in /m^3\n",
+ "r = 10**-3 #in m\n",
+ "A = math.pi * ((r)**2) #in m^2\n",
+ "q = 1.6*10**-19 \n",
+ "I = 2 #in A\n",
+ "V = I/(n*q*A) #in m/s\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The drift velocity of an electron is =%.f X 10^-4 m/s' %(V*(10**5)))\n",
+ "print('Approximationa error ')\n",
+ "# answer is same as in book only they have rounded of 0.000068 to 7 * 10^-4"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The drift velocity of an electron is =7 X 10^-4 m/s\n",
+ "Approximationa error \n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.9 Page No 88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "e= 1.6*10**-19 #in C\n",
+ "d= 1.03 #in mm\n",
+ "d= d*10**-3 #in m\n",
+ "R= 6.51 #in ohm\n",
+ "l= 300.0 #in m\n",
+ "n= 8.4*10**28 #per m^3\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "r= d/2 #in m (radius)\n",
+ "A= math.pi*r**2 #in m^2\n",
+ "rho= R*A/l #in ohm meter\n",
+ "sigma= 1/rho #in mho/m\n",
+ "miu_e= sigma/(n*e) #m^2/V-sec\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The coductivity of copper is =%.2f mho/m' %sigma)\n",
+ "print('The mobility of charge carriers is : =%.6f m^2/V-sec' %miu_e)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The coductivity of copper is =55306469.41 mho/m\n",
+ "The mobility of charge carriers is : =0.004115 m^2/V-sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.9.ii Page No 88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "Miu_e = 1500.0 #in cm^2/volt.sec\n",
+ "Miu_h = 500.0 #in cm^2/volt.sec\n",
+ "n_i = 1.6*10**10 #in /cm^3\n",
+ "e = 1.6*10**-19 #in C\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Sigma_i = n_i*(Miu_e+Miu_h)*e #in mho/cm\n",
+ "Sigma = Sigma_i #in mho/cm\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The conductivity of pure silicon is =%.7f mho/cm' %Sigma)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The conductivity of pure silicon is =0.0000051 mho/cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 2.10 Page No 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "Miu_e = 1500.0 #in cm^2/volt.sec\n",
+ "Miu_h = 500.0 #in cm^2/volt.sec\n",
+ "n_i = 1.6*10**10 #in /cm^3\n",
+ "e = 1.6*10**-19 #in C\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Sigma_i = n_i*(Miu_e+Miu_h)*e #in mho/cm\n",
+ "Sigma = Sigma_i #in mho/cm\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The conductivity of pure silicon is =%.2f X 10^-6 mho/cm' %(Sigma*(10**6)))\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The conductivity of pure silicon is =5.12 X 10^-6 mho/cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 2.11 Page No 90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "Miu_d = 500.0 #in cm^2/V.S\n",
+ "Rho = 10 #in ohm cm \n",
+ "\n",
+ "#CALCULATIONS\n",
+ "e = 1.6*10**-19 #in C\n",
+ "n_d = 1/(Rho*e*Miu_d) #in /cm^3... correction\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The number of donor is =%.2f X 10^15 atom per cm^3' %(n_d/(10**15)))\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The number of donor is =1.25 X 10^15 atom per cm^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 2.12 Page No 90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "d = 5.32 #in gm/cc\n",
+ "N_A = 6.02*10**23 #in atoms/gm.mole\n",
+ "At = 72.6 #atomic weight\n",
+ "Miu = 3800 #in cm^2/v.s\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "n_d = (N_A/At) * d #in /cm^3\n",
+ "n_d = n_d * 10**-8 #in /cc\n",
+ "e = 1.6*10**-19 #in C\n",
+ "Sigma = n_d * Miu * e #in mho/cm\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The conductivity of specimen is =%.2f mho/cm' %Sigma)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The conductivity of specimen is =0.27 mho/cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 2.13 Page No 90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "Rho = 0.3623*10**-3 #in ohm m\n",
+ "d = 4.42*10**28 #Ge density in atoms/m^3\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Sigma = 1/Rho #in mho/m\n",
+ "n_d = d*10**-6 #in atoms/m^3\n",
+ "e = 1.6*10**-19 #in C\n",
+ "Miu = Sigma/(n_d*e) #in m^2/V.sec\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The electron mobility is =%.2f m^2/V-sec' %Miu)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron mobility is =0.39 m^2/V-sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 2.14 Page No 91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "N_A = 6.025*10**26 #in /kg.Mole\n",
+ "At = 72.59 #atomic weight\n",
+ "d = 5.36*10**3 #in kg/m^3\n",
+ "R = 0.42 #in ohm m\n",
+ "B_i = 10**-6 #rate of boron impurity in %\n",
+ "e = 1.6*10**-19 #in C\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "n = (N_A/At)*d #number of Ge atoms\n",
+ "h = n/10**8 #holes per unit volume\n",
+ "Miu_h = 1/(R*h*e) #in m^2/V.sec\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The Mobility of holes is =%.2f m^2/V-sec' %Miu_h)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Mobility of holes is =0.03 m^2/V-sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.15 Page No 90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "n_i = 2*10**19 #in /m^3\n",
+ "Miu_e = 0.36 #in m^2/v.s\n",
+ "Miu_h = 0.17 #in m^2/v.s\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "A = 1*10**-4 #in m^2\n",
+ "V = 2 #in Volts\n",
+ "l = 0.3 #in mm\n",
+ "l = l * 10**-3 #in m\n",
+ "e = 1.6*10**-19 #in C\n",
+ "Sigma_i = n_i * e * (Miu_e+Miu_h) #in mho/m\n",
+ "I = (Sigma_i * V*A)/l #in amp \n",
+ "\n",
+ "#RESULTS\n",
+ "print('The current in amp is =%.2f ' %I)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current in amp is =1.13 \n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.16 Page No 92"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "d = 4.2*10**28 #in atoms/m^3\n",
+ "n_d = d/10**6 #in atoms/m^3\n",
+ "e = 1.6*10**-19 #in C\n",
+ "Miu_e = 0.36 #in m^2/V-sec\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Sigma_n = n_d *e *Miu_e #in mho/m\n",
+ "Rho_n = 1/Sigma_n #ohm m\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The resistivity in \u03a9m is =%.2f X 10^-3' %(Rho_n*(10**3)))\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistivity in \u03a9m is =0.41 X 10^-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.17 Page No 92"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "n_i = 1*10**19 #in /m^3\n",
+ "Miu_e = 0.36 #in m^2/volt.sec\n",
+ "Miu_h = 0.17 #in m^2/volt.sec\n",
+ "A = 2 #in cm^2\n",
+ "A = A * 10**-4 #in m^2\n",
+ "t = 0.1 #in mm\n",
+ "t = t*10**-3 #in m\n",
+ "V = 4 #in volts\n",
+ "e = 1.6*10**-19 #in C\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Sigma_i = n_i * e * (Miu_e + Miu_h) #mho/m\n",
+ "J = Sigma_i * (V/t) #in Amp/m^2\n",
+ "I = J*A #in Amp\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The current in Amp is =%.2f ' %I)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current in Amp is =6.78 \n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 2.18 Page No 92"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "Miu_h = 500.0 #in cm^2/V.s\n",
+ "Miu_e = 1500.0 #in cm^2/V.s\n",
+ "n_i = 1.6*10**10 #in /cm^3\n",
+ "e = 1.6*10**-19 #in C\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Sigma_i = n_i * e * (Miu_e+Miu_h) #in mho/cm\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The conductivity of pure silicon in mho/cm is =%.2f X 10^-6 ' %(Sigma_i*(10**6)))\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The conductivity of pure silicon in mho/cm is =5.12 X 10^-6 \n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.19 Page No 96"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "Si_density = 10.5 #in gm/cc\n",
+ "N_A = 6.025*10**23 \n",
+ "At = 108.0 #atomic weight\n",
+ "B = 0.8 #in Tesla\n",
+ "w = 0.50 #in cm\n",
+ "w = w * 10**-2 #in m\n",
+ "t = 0.10 #in mm\n",
+ "t = t * 10**-3 #in m\n",
+ "A = w*t #in m^2\n",
+ "q = 1.6*10**-19 #in C\n",
+ "I = 2 #in ampere\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "n = (N_A/At) * Si_density #in /cc\n",
+ "n = n * 10**6 #in /m^3\n",
+ "V_H = (B*I*t)/(n*q*A) #in volts\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The hall voltage produced is =%.3f X 10^-7 volts' %(V_H*(10**7)))\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The hall voltage produced is =0.341 X 10^-7 volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.20 Page No 96"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "Sigma = 5.8*10**7 #in mho/m\n",
+ "l = 1 #in m\n",
+ "d = 1 #in cm\n",
+ "d = d * 10**-2 #in m\n",
+ "W = 1 #in mm \n",
+ "W = W*10**-3 #in m\n",
+ "I = 1 #in Amp\n",
+ "B = 1 #in Tesla\n",
+ "V_H = 0.074*10**-6 #in Volts\n",
+ "A = 10**-2 * 10**-3 #in m^2\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "R_H = (V_H*A)/(B*I*d) #in m^3/c\n",
+ "Miu = Sigma * R_H #in m^2/volt.sec\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Hall coefficient is =%.1f X 10^-11 m^3/c' %(R_H*(10**11)))\n",
+ "print('The mobility of electron is = %.2f X 10^-3 m^2/volt.sec ' %(Miu*(10**3)))\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Hall coefficient is =7.4 X 10^-11 m^3/c\n",
+ "The mobility of electron is = 4.29 X 10^-3 m^2/volt.sec \n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.21 Page No 97"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "n_i = 1.4*10**18 #in /m^3\n",
+ "n_D = 1.4*10**24 #in /m^3\n",
+ "n = n_D #in /m^3\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "p = n_i**2/n #in /m^3\n",
+ "R_e = n/p #Ratio of electron\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Concentration of is =%.2f X 10^12 holes per m^3 ' %(p/(10**12)))\n",
+ "print('Ratio of electron to hole concentration is =%.f 10^12 ' %(R_e/(10**12)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Concentration of is =1.40 X 10^12 holes per m^3 \n",
+ "Ratio of electron to hole concentration is =1 10^12 \n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.22 Page No 97"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "B = 0.48 #in Wb/m^2\n",
+ "R_H = 3.6 * 10**-4 #in m^3/c\n",
+ "R = 9*10**-3 #in ohm-m\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Sigma = 1/R #in (ohm-m)^-1\n",
+ "Rho = 1/R_H #in coulomb/m^3\n",
+ "e = 1.6*10**-19 #in C\n",
+ "n = Rho/e #in /m^3\n",
+ "Miu = Sigma * R_H #in m^2/volt-s\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The mobility of electron is =%.2f m^2/volt-s' %Miu)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mobility of electron is =0.04 m^2/volt-s\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.23 Page No 104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "e = 1.6*10**-19 #in C\n",
+ "R_H = 0.0145 #in m^3/coulomb\n",
+ "Miu_e = 0.36 #m^2/v-s\n",
+ "E = 100 #V/m\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "n = 1/(e*R_H) #in /m^3\n",
+ "J= n*e*Miu_e*E #in A/m^2\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The current density is =%.2f A/m^2 ' %J)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current density is =2482.76 A/m^2 \n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.24 Page No 105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "e = 1.6*10**-19 #in C\n",
+ "Miu_e = 7.04*10**-3 #in m^2/volt-sec\n",
+ "n = 5.8*10**28 #number of electron/m^3\n",
+ "m = 9.1*10**-31\n",
+ "E_F = 5.5 # in eV\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Torque = (Miu_e/e)*m # in sec \n",
+ "Rho = 1/(n*e*Miu_e) #in ohm cm\n",
+ "V_F = math.sqrt( (2*E_F*e)/m ) #in m/s\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Relaxtion time is =%.2f X 10^-15 sec ' %(Torque*(10**15)))\n",
+ "print('Resistivity of conductor is =%.2f X 10^-18 in \u03a9m' %(Rho*(10**8)))\n",
+ "print('Velocity of electron with the fermi energy is =%.2f X 10^-6 in m/s' %(V_F/(10**6)))\n",
+ "\n",
+ "# Note: The calculation of Part (ii) is wrong also the unit of resistivity of conductor is wrong\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Relaxtion time is =40.04 X 10^-15 sec \n",
+ "Resistivity of conductor is =1.53 X 10^-18 in \u03a9m\n",
+ "Velocity of electron with the fermi energy is =1.39 X 10^-6 in m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.25 Page No 105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "\n",
+ "E= 5.95 #in eV\n",
+ "EF= 6.25 #in eV\n",
+ "delE= 0.01\n",
+ "\n",
+ "#CALCULATIONS\n",
+ " #delE= 1-1/(1+exp((E-EF)/KT))\n",
+ "K=1.38*10**-23 #Boltzman Constant in J/K\n",
+ "T = ((E-EF)/math.log(1/(1-delE) -1)*1.6*10**-19)/K #in K\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The temperature is =%.2f K' %T)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The temperature is =756.95 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.26 Page No 105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "\n",
+ "T1 = 400.0 #in K\n",
+ "T2 = 300.0 #in K\n",
+ "N_V = 1.04*10**19 #in cm^-3\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "N1 = N_V*((T1/T2)**(3/2)) #in cm^-3\n",
+ "KT = 0.0259*(T1/T2) #in eV\n",
+ "FermiLevel= 0.27 #in eV\n",
+ "P_O = N1*math.exp( (-FermiLevel)/KT ) #in cm^-3\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The thermal equillibrium hole concentration is =%.2f X 10^15 cm^-3' %(P_O/(10**15)))\n",
+ "print('Approximationa error ')\n",
+ "# Answer in the book is wrong"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The thermal equillibrium hole concentration is =5.58 X 10^15 cm^-3\n",
+ "Approximationa error \n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.27 Page No 107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "T1 = 550.0 #in K\n",
+ "T2 = 300.0 #in K\n",
+ "N1 = 1.04*10**19 \n",
+ "\n",
+ "#CALCULATIONS\n",
+ "N_V = N1*((T1/T2)**(3)) \n",
+ "N_C = 2.8*10**19\n",
+ "E_g = -1.12\n",
+ "KT = 0.0259*(T1/T2)\n",
+ "n_i = math.sqrt(N_C*N_V*math.exp(E_g/KT)) #in cm^-3\n",
+ "Nd= math.sqrt(n_i**2/((1.05-1/2.0)**2-(1/2.0)**2))\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of n_i is =%.2f X 10^14 cm^-3' %(n_i/(10**14)))\n",
+ "print('The value of N_d is =%.2f X 10^15 cm^-3' %(Nd/(10**15)))\n",
+ "#Formula n_o= Nd/2+sqrt((Nd/2)^2+n_i^2) and n_o = 1.05*N_d;\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of n_i is =3.20 X 10^14 cm^-3\n",
+ "The value of N_d is =1.40 X 10^15 cm^-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.28 Page No 106"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "n_o = 10**15 #in cm^-3\n",
+ "n_i = 10**10 #in cm^-3\n",
+ "p_o = 10**5 #in cm^-3\n",
+ "del_p = 10**13 #in cm^-3\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "del_n = del_p #in cm^-3\n",
+ "KT= 0.0259 #in eV\n",
+ "Fermi_level1= KT*math.log(n_o/n_i) #in eV\n",
+ "Fermi_level2= KT*math.log((n_o+del_n)/n_i) #in eV\n",
+ "Fermi_level3= KT*math.log((p_o+del_p)/n_i) #in eV\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Fermi level for thermal equillibrium is : =%.2f eV ' %Fermi_level1)\n",
+ "print('Quasi-Fermi level for electrons in non equillibrium is : =%.2f eV' %Fermi_level2)\n",
+ "print('Quasi-Fermi level for holes in non equillibrium is: =%.2f eV' %Fermi_level3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Fermi level for thermal equillibrium is : =0.30 eV \n",
+ "Quasi-Fermi level for electrons in non equillibrium is : =0.30 eV\n",
+ "Quasi-Fermi level for holes in non equillibrium is: =0.18 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 55
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo3.ipynb b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo3.ipynb new file mode 100755 index 00000000..bc717890 --- /dev/null +++ b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo3.ipynb @@ -0,0 +1,992 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1> Chapter No 3 : Excess Carriers In Semiconductors<h1>"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.1, Page No 145 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "n_o = 10**17 #in/cm^3\n",
+ "n_i = 1.5*10**10 #in /cm^3\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "p_o = ((n_i)**(2))/n_o #in holes/cm^3\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The hole concentration is =%.f in holes/cm^3' %p_o)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The hole concentration is =2250 in holes/cm^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3, Page No 146"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "n_i = 1.5 * 10**10 #in /cm^3 for silicon\n",
+ "N_d = 10**17 #in atoms/cm^3\n",
+ "n_o = 10**17 #electrons/cm^3\n",
+ "KT = 0.0259\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "# E_r - E_i = KT * log(n_o/n_i)\n",
+ "del_E = KT * math.log(n_o/n_i) #in eV\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The energy band for this type material is Ei =%.2f eV' %del_E)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy band for this type material is Ei =0.41 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.4 Page No 147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "K = 1.38*10**-23 #in J/K\n",
+ "T = 27 #in degree C\n",
+ "T = T + 273 #in K\n",
+ "e = 1.6*10**-19 \n",
+ "Miu = 0.17 #in m^2/v-s\n",
+ "Miu1 = 0.025 #in m^2/v-s\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "D_n = ((K*T)/e)*Miu #in m^2/s\n",
+ "D_p = ((K*T)/e)*Miu1 #in m^2/s\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The diffusion coefficient of electrons is =%.2f X 10^-4 in m^2/s ' %(D_n*(10**4)))\n",
+ "print('The diffusion coefficient of holes is =%.2f X 10^-4 in m^2/s ' %(D_p*(10**4)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The diffusion coefficient of electrons is =43.99 X 10^-4 in m^2/s \n",
+ "The diffusion coefficient of holes is =6.47 X 10^-4 in m^2/s \n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.5, Page No 147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "Miu_n = 0.15 #in m^2/v-s\n",
+ "K = 1.38*10**-23 \n",
+ "T = 300 #in K\n",
+ "e = 1.6*10**-19 #in C\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "D_n = Miu_n*((K*T)/e) #in m^2/s\n",
+ "Torque_n = 10**-7 #in s\n",
+ "L_n = math.sqrt(D_n*Torque_n) #in m\n",
+ "del_n = 10**20 #electrons/m^3\n",
+ "J_n = (e*D_n*del_n)/L_n #in A/m^2\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The diffusion length is =%.2f X 10^-4 in m' %(L_n*(10**5)))\n",
+ "print('The diffusion current density is =%.2f X 10^3 in A/m^2 ' %(J_n/(10**3)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The diffusion length is =1.97 X 10^-4 in m\n",
+ "The diffusion current density is =3.15 X 10^3 in A/m^2 \n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.6 Page No 147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "Sigma_n = 0.1 #in (ohm-cm)^-1\n",
+ "Miu_n = 1300\n",
+ "q = 1.6*10**-19 #in C\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "n_n = Sigma_n/(Miu_n*q) #in electrons/cm^3 \n",
+ "n_i = 1.5*10**10\n",
+ "p_n = ((n_i)**2)/n_n #in holes/cm^3\n",
+ "p_n = p_n * 10**6 #in holes/m^3\n",
+ "n_n=n_n*10**6\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Concentration of electrons is =%.2f X 10^20 per m^3 ' %(n_n/(10**20)))\n",
+ "print('Concentration of holes is =%.2f X 10^11 per m^3 ' %(p_n/(10**11)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Concentration of electrons is =4.81 X 10^20 per m^3 \n",
+ "Concentration of holes is =4.68 X 10^11 per m^3 \n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.7 Page No 147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "L = 100*10**-6 #in m\n",
+ "Miu_e = 0.13 #in m^2/V-s\n",
+ "Torque_h = 10**-6 #in s\n",
+ "Miu_h = 0.05 #in m^2/v-s\n",
+ "V = 12 #in V\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Torque_n = ((L)**2)/(Miu_e*V) #in s\n",
+ "P = (Torque_h/Torque_n)*(1+(Miu_h/Miu_e))\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Electron transit time is =%.2f X 10^-9 in sec' %(Torque_n*(10**9)))\n",
+ "print('Photoconductor gain is =%.2f ' %P)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electron transit time is =6.41 X 10^-9 in sec\n",
+ "Photoconductor gain is =216.00 \n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.8, Page No 147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "q = 1.6*10**-19 #in C\n",
+ "n_i = 2.5*10**13\n",
+ "Miu_n = 3800 #in cm^2/V-s\n",
+ "Miu_p = 1800 #in cm^2/V-s\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Sigma = n_i*(Miu_n + Miu_p)*q #in (ohm-cm)^-1\n",
+ "Rho = 1/Sigma #in ohm-cm\n",
+ "N_D =4.4*10**22/10**8 #in atoms/cm^3\n",
+ "Sigma_n = N_D * Miu_n*q #in (ohm-cm)^-1 \n",
+ "Rho1 = 1/Sigma_n #in ohm cm\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The resistivity is =%.2f in ohm-cm ' %Rho)\n",
+ "print('The resistivity drops is =%.2f in ohm cm ' %Rho1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistivity is =44.64 in ohm-cm \n",
+ "The resistivity drops is =3.74 in ohm cm \n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.9 Page No 148"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#initialisation of variables\n",
+ "n_i = 10**16 #in /m^3\n",
+ "N_D = 10**22 #in /m^3\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "n = N_D #in /m^3\n",
+ "p = ((n_i)**2)/n #in /m^3\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The concentration of is =%.2f X 10^21 electrons per m^3' %(n/(10**21)))\n",
+ "print('The concentration of is =%.2f X 10^9 holes per m^3 ' %(p/(10**9)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The concentration of is =10.00 X 10^21 electrons per m^3\n",
+ "The concentration of is =10.00 X 10^9 holes per m^3 \n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.10 Page No 148"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "Rho = 9.6*10**-2 #ohm-m\n",
+ "Sigma_n = 1/Rho #in (ohm-m)^-1\n",
+ "Miu_n = 1300 #in cm^2/V-s\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Miu_n = Miu_n * 10**-4 #in m^2/V-s\n",
+ "q = 1.6*10**-19 #in C\n",
+ "N_D = Sigma_n/(Miu_n*q) #in atoms/m^3\n",
+ "d = 5*10**22 #in atoms/cm^3\n",
+ "d = d * 10**6 #// in atoms/m^3\n",
+ "R_d = N_D/d #Ratio \n",
+ "\n",
+ "#RESULTS\n",
+ "print('Ratio of donor atom to silicon atoms per unit volume is =%.2f X 10^-8 ' %(R_d*(10**8)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Ratio of donor atom to silicon atoms per unit volume is =1.00 X 10^-8 \n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.11 Page No 148"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "n_i = 1.5*10**10 #in /cm^3\n",
+ "n_n = 2.25*10**15 #in /cm^3\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "p_n = ((n_i)**2)/n_n #in /cm^3\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The concentration of is =%.2f X 10^5 holes per cm^3 ' %(p_n/(10**5)))\n",
+ "print('Donor impurity per cm^3 is =%.2f X 10^15 ' %(n_n/(10**15)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The concentration of is =1.00 X 10^5 holes per cm^3 \n",
+ "Donor impurity per cm^3 is =2.25 X 10^15 \n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 3.12 Page No 149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "N_A = 2*10**16 #in /cm^3\n",
+ "N_D = 10**16 #in /cm^3\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "C = N_A-N_D #in /cm^3\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Carrier concentration in holes/cm^3 is =%.2f X 10^16 ' %(C/(10**16)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Carrier concentration in holes/cm^3 is =1.00 X 10^16 \n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 3.13 Page No 149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "del_n = 10**15 #in /cm^3\n",
+ "Torque_p = 10*10**-6 #in sec\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "R_G = del_n/Torque_p #in electron hole pairs/sec/cm^3\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The rate of generation of minority carrier is : =%.2f X 10^20 electron hole pairs/sec/cm^3 ' %(R_G/(10**20)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The rate of generation of minority carrier is : =1.00 X 10^20 electron hole pairs/sec/cm^3 \n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 3.14 Page No 149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "V = 1/20.0 #in cm/\u00b5sec\n",
+ "V=V*10**6 #in cm/sec\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "E = 10 #in V/cm\n",
+ "Miu = V/E #in cm^2/V-sec\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The mobility of minority charge carrier is =%.f cm^2/V-sec' %Miu)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mobility of minority charge carrier is =5000 cm^2/V-sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 3.15 Page No 149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "q = 1.6 * 10**-19 #in C\n",
+ "N_D = 4.5 * 10**15 #in /cm^3\n",
+ "del_p = 10**21 \n",
+ "e=10 #in cm\n",
+ "A = 1 #in mm^2\n",
+ "A = A * 10**-14 #cm^2\n",
+ "l = 10 #in cm\n",
+ "Torque_p = 1 #in microsec\n",
+ "\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Torque_p = Torque_p * 10**-6 #in sec\n",
+ "Torque_n = 1 #in microsec\n",
+ "Torque_n = Torque_n * 10**-6 #in sec\n",
+ "n_i = 1.5 * 10**10 #in /cm^3\n",
+ "D_n = 30 #cm^2/sec\n",
+ "D_p = 12 #in cm^2/sec\n",
+ "n_o = N_D #in /cm^3\n",
+ "p_o = ((n_i)**2)/n_o #in /cm^3\n",
+ "l_n = math.sqrt(D_n * Torque_n) #in cm\n",
+ "l_p = math.sqrt(D_p * Torque_p) #in cm\n",
+ "x=34.6*10*8-4 #in cm\n",
+ "e1 = 1.88 * 10**1 #in cm\n",
+ "dnBYdx = del_p * e1 #in cm^4\n",
+ "dpBYdx = del_p *e #in cm^4\n",
+ "\n",
+ "J_P = -(q) * D_p * dpBYdx #in A/cm^2\n",
+ "J_n = q * D_n * dnBYdx #in A/cm^2\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Hole concentration at thermal equilibrium per cm^3 is =%.1f X 10^5 ' %(p_o/(10**5)))\n",
+ "print('Diffusion length of holes in cm is =%.1f X 10^-3 ' %(l_p*(10**3)))\n",
+ "print('Concentration gradient of holes at distance in cm^4 is =%.f X 10^22 ' %(dpBYdx/(10**22)))\n",
+ "print('Concentration gradient of electrons in per cm^4 is =%.1f X 10^22 ' %(dnBYdx/(10**22)))\n",
+ "print('Current density of holes due to diffusion in A/cm^2 is =%.2f X 10^4 ' %(J_P/(10**4)))\n",
+ "print('Current density of electrons due to diffusion in A/cm^2 is =%.2f X 10^4 ' %(J_n/(10**4)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Hole concentration at thermal equilibrium per cm^3 is =0.5 X 10^5 \n",
+ "Diffusion length of holes in cm is =3.5 X 10^-3 \n",
+ "Concentration gradient of holes at distance in cm^4 is =1 X 10^22 \n",
+ "Concentration gradient of electrons in per cm^4 is =1.9 X 10^22 \n",
+ "Current density of holes due to diffusion in A/cm^2 is =-1.92 X 10^4 \n",
+ "Current density of electrons due to diffusion in A/cm^2 is =9.02 X 10^4 \n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 3.16 Page No 151"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "h = 6.626 * 10**-34 #in J-s\n",
+ "q= 1.6*10**-19 #in C\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "h= h/q #in eV\n",
+ "c = 3*10**8\n",
+ "lembda = 5490*10**-10 #in m\n",
+ "E = h*c/lembda #in eV\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The energy band gap is =%.2f eV ' %E)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy band gap is =2.26 eV \n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 3.17 Page No 151"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "D_n = 35 #in cm^2/sec\n",
+ "q = 1.6*10**-19 #in C\n",
+ "y2 = 6*10**16 #in /cm^3\n",
+ "y1 = 10**17 #in /cm^3\n",
+ "x2 = 2*10**-4\n",
+ "x1 = 0\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "dnBYdx = (y2-y1)/(x2-x1)\n",
+ "J_n = q*D_n*dnBYdx #in A/cm^2\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The current density is =%.f in A/cm^2' %J_n)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current density is =-1120 in A/cm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 3.18 Page No 151"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "q = 1.6*10**-19 #in C\n",
+ "n_n = 5*10**20 #in /m^3\n",
+ "n_n = n_n * 10**-6 #in /cm^3\n",
+ "Miu_n = 0.13 #in m^2/V-sec\n",
+ "\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Miu_n = Miu_n * 10**4 #in cm^2/V-sec\n",
+ "Sigma_n = q*n_n*Miu_n #in ohm-cm^-1\n",
+ "Rho = 1/Sigma_n\n",
+ "A = 100 # in \u00b5m^2\n",
+ "A = A * 10**-8 #in cm^2\n",
+ "l = 0.1 #in cm\n",
+ "R = Rho * (l/A) #in ohm\n",
+ "R=R*10**-6\n",
+ "#RESULTS\n",
+ "print('The resistance of the bar is =%.2f in M ohm' %R)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistance of the bar is =0.96 in M ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 3.19 Page No 152"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "w = 3.0 #in \u00b5m\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "D = w/9 #in \u00b5m\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Depletion width on P side is =%.2f in \u00b5m ' %D)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Depletion width on P side is =0.33 in \u00b5m \n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 3.20 Page No 152"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "n_i = 1.5*10**16 #in /m^3\n",
+ "n_n = 5*10**20 #in /m^3\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "p_n = ((n_i)**2)/n_n #in /m^3\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The minority carrier density per m^3 is =%.2f X 10^11' %(p_n/(10**11)))\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minority carrier density per m^3 is =4.50 X 10^11\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 3.21 Page No 152"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "y2 = 10**14 #in /cm^3\n",
+ "y1 = 0\n",
+ "x1=-0.5 #in \u00b5m\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "x1= x1*10**-4 #in cm\n",
+ "x2=0\n",
+ "dnBYdx = (y2-y1)/(x2-x1) #in /cm^4\n",
+ "q = 1.6*10**-19 #in C\n",
+ "D_n = 25 #in cm^2/sec\n",
+ "J_n = q*D_n*dnBYdx #in A/cm^2\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The collector current density is =%.f in A/cm^2' %J_n)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The collector current density is =8 in A/cm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 3.22 Page No 153"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "h = 6.64*10**-34 ##in J-s\n",
+ "q=1.6*10**-19 #in C\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "h= h/q #in eV\n",
+ "c = 3*10**8 #in m/s\n",
+ "lembda = 0.87*10**-6 #in m\n",
+ "E_g = (h*c)/lembda #in eV\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The band gap in eV is =%.2f ' %E_g)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The band gap in eV is =1.43 \n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 3.23 Page No 153"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "alpha = 5*10**4 #in cm^-1\n",
+ "l = 0.46*10**-4 #in cm\n",
+ "hv = 2 #in eV\n",
+ "I_o = 10**-2 #in W\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "I_t = I_o*math.exp(-alpha*l) #in W\n",
+ "A_p = I_o-I_t #absorbed power in W or J/s\n",
+ "c = 1.43\n",
+ "A_E = (hv-c)/hv*A_p # in J/s\n",
+ "e = 1.6*10**-19 #in C\n",
+ "P = A_p/(e*hv) #Perfect quantum efficiency in photon/s\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Total energy absorbed is =%.f X 10^-3 in J/s' %(A_p*(10**3)))\n",
+ "print('Rate of excess thermal energy is =%.2f X 10^-3 in J/s' %(A_E*(10**3)))\n",
+ "print('Perfect quantum efficiency is =%.2f X 10^16 in photon/s' %(P/(10**16)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total energy absorbed is =9 X 10^-3 in J/s\n",
+ "Rate of excess thermal energy is =2.56 X 10^-3 in J/s\n",
+ "Perfect quantum efficiency is =2.81 X 10^16 in photon/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 3.24 Page No 154"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "Mu_p = 500 #in cm^2/v-s\n",
+ "kT = 0.0259\n",
+ "Toh_p = 10**-10 #in sec\n",
+ "p_o = 10**17 #in cm^-3\n",
+ "q= 1.6*10**-19 # in C\n",
+ "A=0.5 #in square meter\n",
+ "del_p = 5 * 10**16 #in cm^-3\n",
+ "n_i= 1.5*10**10 #in cm^-3 \n",
+ "\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "D_p = kT * Mu_p #in cm/s\n",
+ "L_p = math.sqrt(D_p * Toh_p) # in cm\n",
+ "x = 10**-5 #in cm\n",
+ "p = p_o+del_p*e**(x/L_p) #in cm^-3\n",
+ "Eip= math.log(p/n_i)*kT #in eV\n",
+ "Ecp= 1.1/2-Eip #value of E_c-E_p in eV\n",
+ "Ip= q*A*D_p/L_p*del_p*e**(x/L_p) #in A\n",
+ "Qp= q*A*del_p*L_p #in C\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The hole current is : =%.2f X 10^3 in A ' %(Ip*(10**3)))\n",
+ "print('Approximation error')\n",
+ "print('The value of Qp is :=%.2f X 10^-7 in C' %(Qp*(10**7)))\n",
+ "\n",
+ "#Note: There is a calculation error to evalaute the value of hole current hence the value of hole current in the book is wrong\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The hole current is : =8.61 X 10^3 in A \n",
+ "Approximation error\n",
+ "The value of Qp is :=1.44 X 10^-7 in C\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo4.ipynb b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo4.ipynb new file mode 100755 index 00000000..e76421ad --- /dev/null +++ b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo4.ipynb @@ -0,0 +1,928 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1> Chapter No 4 : Junctions and Interfaces<h1>"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.2 Page No 184"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "N_D = 10**17 * 10**6 #in atoms/m^3\n",
+ "N_A = 0.5*10**16*10**6 #in atoms/m^3\n",
+ "Epsilon_r = 10\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Epsilon_o = 8.85*10**-12\n",
+ "Epsilon = Epsilon_r*Epsilon_o #in F/m\n",
+ "e = 1.602*10**-19 #in C\n",
+ "V = 0\n",
+ "V_B = 0.7 #in V\n",
+ "W1 = math.sqrt( ((2*Epsilon*V_B)/e)*(1/N_A+1/N_D) ) #in m\n",
+ "V_o = V_B #in V\n",
+ "V1 = -10 #in V\n",
+ "V_B1 = V_o-V1 #in V\n",
+ "W = math.sqrt( ((2*Epsilon*V_B1)/e)*(1/N_A+1/N_D) ) #in m\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The junction width in meter when no external voltage is applied is =%.2f X 10^-6' %(W1*(10**6)))\n",
+ "print('Junction width in meter with an external voltage of -10V is =%.2f X 10^-6 m' %(W*(10**6)))\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The junction width in meter when no external voltage is applied is =0.39 X 10^-6\n",
+ "Junction width in meter with an external voltage of -10V is =1.54 X 10^-6 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4 Page No 208"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "V = 5 #in V\n",
+ "V_Gamma = 0.6 #in V\n",
+ "r_F = 12 #in ohm\n",
+ "R = 1 #in k ohm\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "R = R * 10**3 #in ohm\n",
+ "I_F = (V-V_Gamma)/(R+r_F) #in A\n",
+ "V_F = V_Gamma + (I_F*r_F) #in V\n",
+ "I_F=I_F*10**3 \n",
+ "#RESULTS\n",
+ "print('The forward diode current is =%.2f mA ' %I_F)\n",
+ "print('The diode voltage is =%.2f V ' %V_F)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The forward diode current is =4.35 mA \n",
+ "The diode voltage is =0.65 V \n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.5 Page No 214"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "n = 4.4*10**22 #total number of Ge atoms/cm^3\n",
+ "n_a = 1*10**8 #number of impurity atoms\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "N_A = n/n_a #in atoms/cm^3\n",
+ "N_A = N_A * 10**6 #in atoms/m^3\n",
+ "n_i = 2.5*10**13 #in atoms/cm^3\n",
+ "n_i = n_i * 10**6 #in atoms/m^3\n",
+ "N_D = 10**3 * N_A #in atoms/m^3\n",
+ "V_T = 26*10**-3 #in A\n",
+ "V_J = V_T*math.log( (N_A*N_D)/((n_i)**2) ) #in V\n",
+ "print('The contact difference of potential is =%.2f V For a silicon P-N junction ' %V_J)\n",
+ "n = 5*10**22\n",
+ "N_A = n/n_a #in atoms/cm^3\n",
+ "N_A = N_A * 10**6 #in atoms/m^3\n",
+ "N_D = 10**3 * N_A #in atoms/m^3\n",
+ "n_i = 1.5*10**10 #in /cm^3\n",
+ "V_J = V_T*math.log(N_A*N_D/n_i**2) #in V\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "\n",
+ "print('The contact difference of potential is =%.2f V' %V_J)\n",
+ "\n",
+ "#Note: There is a calculation error to find the value of V_J in the book, so the answer in the book is wrong.\n",
+ "\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The contact difference of potential is =0.33 V For a silicon P-N junction \n",
+ "The contact difference of potential is =1.44 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.6 Page No 214"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "Rho_p = 2 #in ohm-cm\n",
+ "Rho_n = 1 #in ohm cm\n",
+ "q = 1.6*10**-19 #in C\n",
+ "n_i = 2.5*10**13 #atoms per cm^3\n",
+ "Miu_p = 1800\n",
+ "Miu_n = 3800\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "N_A = 1/(Rho_p*q*Miu_p) #in /cm^3\n",
+ "N_D = 1/(Rho_n*q*Miu_n) #in /cm^3\n",
+ "V_T = 26 #in mV\n",
+ "V_T= V_T*10**-3 #in V\n",
+ "V_J = V_T*math.log((N_A*N_D)/((n_i)**2)) #in V\n",
+ "print('The height of the potential energy barrier is =%.2f in V ' %V_J)\n",
+ "Miu_p = 500\n",
+ "N_A = 1/(Rho_p*q*Miu_p) #in /cm^3\n",
+ "Miu_n = 1300\n",
+ "N_D = 1/(Rho_n*q*Miu_n) #in /cm^3\n",
+ "n_i = 1.5*10**10;\n",
+ "V_J = V_T*math.log((N_A*N_D)/((n_i)**2)) #in V\n",
+ "\n",
+ "#RESULTS\n",
+ "print('For silicon P-N juction The height of the potential energy barrier is =%.2f V' %V_J)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The height of the potential energy barrier is =0.22 in V \n",
+ "For silicon P-N juction The height of the potential energy barrier is =0.67 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.7 page No 215"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "Eta = 1\n",
+ "V_T = 26 #in mV\n",
+ "V_T= V_T*10**-3 #in V\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V= math.log(1-0.9)*V_T #in V\n",
+ "print(\"The voltage in volts is : %.2f \" %V)\n",
+ "V1=0.05 #in V\n",
+ "V2= -0.05 #in V\n",
+ "ratio= (math.e**(V1/(Eta*V_T))-1)/(math.e**(V2/(Eta*V_T))-1)\n",
+ "print(\"The ratio of the current for a forward bias to reverse bias is : %.2f \" %ratio)\n",
+ "\n",
+ "# Part (iii)\n",
+ "Io= 10 #in \u00b5A\n",
+ "Io=Io*10**-3 #in mA\n",
+ "#For \n",
+ "V=0.1 #in V\n",
+ "I = Io * (math.e**(V/(Eta*V_T)) - 1) #in mA\n",
+ "print(\"For v=0.1 V , the value of I is : %.2f mA \" %I)\n",
+ "\n",
+ "#For \n",
+ "V=0.2 #in V\n",
+ "I = Io * (math.e**(V/(Eta*V_T)) - 1) #in mA\n",
+ "print(\"For v=0.2 V , the value of I is : %.2f mA \" %I)\n",
+ "\n",
+ "#For \n",
+ "V=0.3 #in V\n",
+ "I = Io * (math.e**(V/(Eta*V_T)) - 1) #in mA\n",
+ "\n",
+ "#RESULTS\n",
+ "\n",
+ "I=I*10**-3\n",
+ "print(\"For v=0.3 V , the value of I is : %.2f mA\" %I)\n",
+ "print(\"From three value of I, for small rise in forward voltage, the diode current increase rapidly \")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The voltage in volts is : -0.06 \n",
+ "The ratio of the current for a forward bias to reverse bias is : -6.84 \n",
+ "For v=0.1 V , the value of I is : 0.46 mA \n",
+ "For v=0.2 V , the value of I is : 21.90 mA \n",
+ "For v=0.3 V , the value of I is : 1.03 mA\n",
+ "From three value of I, for small rise in forward voltage, the diode current increase rapidly \n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.8 Page No 216"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "# Part (i)\n",
+ "T1= 25 #in \u00b0C\n",
+ "T2= 80.0 #in \u00b0C\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "# Formula Io2= Io1*2**((T2-T1)/10)\n",
+ "AntiFactor= 2**((T2-T1)/10)\n",
+ "print('Anticipated factor for Ge is : =%.f ' %AntiFactor)\n",
+ "\n",
+ "# Part (ii)\n",
+ "T1= 25.0 #in \u00b0C\n",
+ "T2= 150.0 #in \u00b0C\n",
+ "AntiFactor= 2**((T2-T1)/10)\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Anticipated factor for Ge is : =%.f ' %AntiFactor)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Anticipated factor for Ge is : =45 \n",
+ "Anticipated factor for Ge is : =5793 \n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.9 Page No 216"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "I=5.0 #in \u00b5A\n",
+ "V=10.0 #in V\n",
+ "T1= 0.11 #in \u00b0C^-1\n",
+ "T2= 0.07 #in \u00b0C^-1\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Io= T2*I/T1 #in \u00b5A\n",
+ "I_R= I-Io #in \u00b5A\n",
+ "R= V/I_R #in M\u03a9\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The leakage resistance is : =%.1f M\u03a9 ' %R)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The leakage resistance is : =5.5 M\u03a9 \n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.10 Page No 216"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#initialisation of variables\n",
+ "Eta = 1.0\n",
+ "T = 125.0 #in \u00b0C\n",
+ "T = T + 273 #in K\n",
+ "V_T = 8.62 * 10**-5 * 398 #in V\n",
+ "I_o = 30 #in \u00b5A\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "I_o= I_o*10**-6 #in A\n",
+ "v = 0.2 #in V\n",
+ "r_f = (Eta * V_T)/(I_o * math.e**(v/(Eta* V_T))) #in ohm\n",
+ "print('The dynamic resistance in the forward direction is =%.2f \u03a9 ' %r_f)\n",
+ "\n",
+ "r_r = (Eta * V_T)/(I_o * math.e**(-v/(Eta* V_T))) #in ohm\n",
+ "r_r=r_r*10**-3\n",
+ "print('The dynamic resistance in the forward direction is =%.2f K\u03a9 ' %r_r)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The dynamic resistance in the forward direction is =3.36 \u03a9 \n",
+ "The dynamic resistance in the forward direction is =389.08 K\u03a9 \n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.11 Page No 217"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "epsilon = 16/(36 * math.pi * 10**11) #in F/cm\n",
+ "A = 1 * 10**-2\n",
+ "W = 2 * 10**-4\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "C_T = (epsilon * A)/W #in F\n",
+ "C_T=C_T*10**12\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The barrier capacitance is =%.2f pF ' %C_T)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The barrier capacitance is =70.74 pF \n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.12 Page No 217"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "A = 1.0 #in mm^2\n",
+ "A = A * 10**-6 #in m^2\n",
+ "N_A = 3 * 10**20 #in atoms/m^3\n",
+ "q = 1.6 *10**-19 #in C\n",
+ "V_o = 0.2 #in V\n",
+ "epsilon_r=16\n",
+ "epsilon_o= 8.854*10**-12 #in F/m\n",
+ "\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "epsilon=epsilon_r*epsilon_o\n",
+ "# Part (a)\n",
+ "V=-10 #in V\n",
+ "W = math.sqrt(((V_o - V) * 2 * epsilon)/(q * N_A)) #m\n",
+ "C_T1 = (epsilon * A)/W #in F\n",
+ "W=W*10**6\n",
+ "print('The width of the depletion layer for an applied reverse voltage of 10V is =%.2f \u00b5m ' %W)\n",
+ "\n",
+ "# Part (b)\n",
+ "V=-0.1 #in V\n",
+ "W = math.sqrt(((V_o - V) * 2 * epsilon)/(q * N_A)) #m\n",
+ "C_T2 = (epsilon * A)/W #in F\n",
+ "W=W*10**6\n",
+ "print('The width of the depletion layer for an applied reverse voltage of 0.1V is =%.2f \u00b5m ' %W)\n",
+ "\n",
+ "# Part (c)\n",
+ "V=0.1 #in V\n",
+ "W = math.sqrt(((V_o - V) * 2 * epsilon)/(q * N_A)) # m\n",
+ "W=W*10**6\n",
+ "print('The width of the depletion layer for an applied reverse voltage of 0.1V is =%.2f \u00b5m ' %W)\n",
+ "\n",
+ "#Part (d)\n",
+ "C_T1=C_T1*10**12\n",
+ "C_T2=C_T2*10**12\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The space charge capacitance for an applied reverse voltage of 10V is =%.2f pF ' %C_T1)\n",
+ "print('The space charge capacitance for an applied reverse voltage of 0.1V is =%.2f pF' %C_T2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The width of the depletion layer for an applied reverse voltage of 10V is =7.76 \u00b5m \n",
+ "The width of the depletion layer for an applied reverse voltage of 0.1V is =1.33 \u00b5m \n",
+ "The width of the depletion layer for an applied reverse voltage of 0.1V is =0.77 \u00b5m \n",
+ "The space charge capacitance for an applied reverse voltage of 10V is =18.26 pF \n",
+ "The space charge capacitance for an applied reverse voltage of 0.1V is =106.46 pF\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 4.13 Page No 218"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "I_o = 1.8 * 10**-9 #A\n",
+ "v = 0.6 #in V\n",
+ "Eta = 2\n",
+ "V_T = 26 #in mV\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V_T=V_T*10**-3 #in V\n",
+ "I = I_o *(math.e**(v/(Eta * V_T))) #in A\n",
+ "I=I*10**3\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The current in the junction is =%.2f mA ' %I)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current in the junction is =0.18 mA \n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 4.14 Page No 218"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "I_o = 2.4 * 10**-14\n",
+ "I = 1.5 #in mA\n",
+ "I=I*10**-3 #in A\n",
+ "Eta = 1\n",
+ "V_T = 26 #in mV\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V_T= V_T*10**-3 #in V\n",
+ "v =math.log((I + I_o)/I_o) * V_T #in V\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The forward biasing voltage across the junction is =%.2f V' %v)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The forward biasing voltage across the junction is =0.65 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 4.15 Page No 218"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "I_o = 10 #in nA\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "I = I_o * (-1) #in nA\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The Diode current is = %.f nA ' %I)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Diode current is = -10 nA \n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 4.16 Page No 218"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "R = 4.5 #in ohm\n",
+ "I = 44.4 #in mA\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "I=I*10**-3 #in A\n",
+ "V = R * I #in V\n",
+ "Eta = 1\n",
+ "V_T = 26 #in mV\n",
+ "V_T=V_T*10**-3 #in V\n",
+ "I_o = I/((math.e**(V/(Eta * V_T))) -1) #in A\n",
+ "V = 0.1 #in V\n",
+ "r_f = (Eta * V_T)/(I_o * ((math.e**(V/(Eta * V_T)))-1)) #in ohm\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The diode dynamic resistance is =%.2f \u03a9' %r_f)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The diode dynamic resistance is =27.78 \u03a9\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 4.18 Page No 219"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "V = 0.25 #in V\n",
+ "I_o = 1.2 #in \u00b5A\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "I_o = I_o * 10**-6 #in A\n",
+ "V_T = 26 #in mV\n",
+ "V_T = V_T * 10**-3 #in V\n",
+ "Eta = 1\n",
+ "r = (Eta * V_T)/(I_o * (math.e**(V/(Eta * V_T)))) #in ohm\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The ac resistance of the diode is =%.2f \u03a9' %r)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ac resistance of the diode is =1.45 \u03a9\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 4.19 Page No 219"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "q = 1.6 * 10**-19 #in C\n",
+ "N_A = 3 * 10**20 #in /m^3\n",
+ "A = 1 #in \u00b5m^2\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "A = A * 10**-6 #in m^2\n",
+ "V = -10 #in V\n",
+ "V_J = 0.25 #in V\n",
+ "V_B = V_J - V #in V\n",
+ "epsilon_o = 8.854 #in pF/m\n",
+ "epsilon_o = epsilon_o * 10**-12 #in F/m\n",
+ "epsilon_r = 16\n",
+ "epsilon = epsilon_o * epsilon_r\n",
+ "W = math.sqrt((V_B * 2 * epsilon)/(q * N_A)) #in m \n",
+ "\n",
+ "#RESULTS\n",
+ "C_T = (epsilon * A)/W #in pF\n",
+ "W=W*10**6\n",
+ "C_T=C_T*10**12\n",
+ "print('The width of depletion layer is =%.2f \u00b5m ' %W)\n",
+ "print('The space charge capacitance is =%.2f pF ' %C_T)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The width of depletion layer is =7.78 \u00b5m \n",
+ "The space charge capacitance is =18.21 pF \n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 4.20 Page No 220"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "C_T = 100.0 #in pF\n",
+ "C_T=C_T*10**-12 #in F\n",
+ "epsilon_r = 12\n",
+ "epsilon_o = 8.854 * 10**-12 #in F/m\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "epsilon = epsilon_r * epsilon_o\n",
+ "Rho_p = 5 #in ohm-cm\n",
+ "Rho_p = Rho_p * 10**-2 #in ohm-m\n",
+ "V_j = 0.5 #in V\n",
+ "V = -4.5 #in V\n",
+ "Mu_p = 500 #in cm^2\n",
+ "Mu_p = Mu_p * 10**-4 #in m^2\n",
+ "Sigma_p = 1/Rho_p #in per ohm-m\n",
+ "qN_A = Sigma_p/ Mu_p\n",
+ "V_B = V_j - V\n",
+ "W = math.sqrt((V_B * 2 * epsilon)/qN_A)\n",
+ "A = (C_T * W)/ epsilon #in m\n",
+ "D = math.sqrt(A * (4/math.pi)) #in m\n",
+ "D = D * 10**3 #in mm\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The diameter is =%.2f mm ' %D)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The diameter is =1.40 mm \n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 4.21 Page No 221"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "q = 1.6 * 10**-19 #in C\n",
+ "Mu_p = 500 #in cm^2/V-sec\n",
+ "Rho_p = 3.5 #in ohm-cm\n",
+ "Mu_n = 1500 #in cm^2/V-sec\n",
+ "Rho_n = 10 #in ohm-cm\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "N_A = 1/(Rho_p * Mu_p * q) # in /cm^3\n",
+ "N_D = 1/(Rho_n * Mu_n * q) # in /cm^3\n",
+ "V_J = 0.56 # in V\n",
+ "n_i = 1.5 * 10**10 #in /cm^3\n",
+ "V_T = V_J/math.log((N_A * N_D)/(n_i)**2) #in V\n",
+ "T = V_T * 11600 #in K\n",
+ "T = T - 273 #in \u00b0C\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The Temperature of junction is =%.2f \u00b0C ' %T)\n",
+ "print('Approximation error')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Temperature of junction is =14.28 \u00b0C \n",
+ "Approximation error\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 4.22 Page No 221"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "R = 5 #in ohm\n",
+ "I = 50 #in mA\n",
+ "I=I*10**-3 #in A\n",
+ "V = R * I #in V\n",
+ "Eta = 1\n",
+ "V_T = 26 #in mV\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V_T=V_T*10**-3 #in V\n",
+ "I_o = I/((math.e**(V/(Eta * V_T))) - 1) #in A\n",
+ "v1 = 0.2 #in V\n",
+ "r = (Eta * V_T)/(I_o * (math.e**(v1/(Eta * V_T)))) #in ohm\n",
+ "\n",
+ "#RESULTS\n",
+ "I_o=I_o*10**6\n",
+ "print('Reverse saturation current is =%.2f \u00b5A ' %I_o)\n",
+ "print('Dynamic resistance of the diode is =%.2f \u03a9 ' %r)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reverse saturation current is =3.33 \u00b5A \n",
+ "Dynamic resistance of the diode is =3.56 \u03a9 \n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo6.ipynb b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo6.ipynb new file mode 100755 index 00000000..44239148 --- /dev/null +++ b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo6.ipynb @@ -0,0 +1,76 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1> Chpater No 6: Microwave Diodes <h1>"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.1, Page No 254"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "C1_min= 5.0 #in pF\n",
+ "C1_max= 50.0 #in pF\n",
+ "C2_min= 5.0 #in pF\n",
+ "C2_max= 50.0 #in pF\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "C1_min=C1_min*10**-12 #in F\n",
+ "C2_min=C2_min*10**-12 #in F\n",
+ "C1_max=C1_max*10**-12 #in F\n",
+ "C2_max=C2_max*10**-12 #in F\n",
+ "L = 10.0 #in mH\n",
+ "L = L * 10**-3 #in H\n",
+ "C_T_min = (C1_min*C2_min)/(C2_min+C2_min) #in F\n",
+ "f_o_max = 1/( 2*math.pi*(math.sqrt(L*C_T_min)) ) #in Hz \n",
+ "f_o_max = f_o_max * 10**-6 #in MHz\n",
+ "C_T_max = (C1_max*C2_max)/(C2_max+C2_max) #in F\n",
+ "f_o_min = 1/( 2*math.pi*(math.sqrt(L*C_T_max)) ) #in Hz \n",
+ "f_o_min = f_o_min * 10**-3 #in kHz\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The tuning range for circuit will be =%.2f kHz' %f_o_min)\n",
+ "print(' To %.3f MHz' %f_o_max)\n",
+ "\n",
+ "\n",
+ "#print(\"The tuning range for circuit will be : \"+string(round(f_o_min))+\" kHz to \"+string(round(f_o_max))+\" MHz\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The tuning range for circuit will be =318.31 kHz\n",
+ " To 1.007 MHz\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo7.ipynb b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo7.ipynb new file mode 100755 index 00000000..431c4340 --- /dev/null +++ b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo7.ipynb @@ -0,0 +1,154 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1> Chpater No 7: Optoelectonic Devices <h1>"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.1, Page No 283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#initialisation of variables\n",
+ "Vout = 5.0 #in V\n",
+ "V = 1.5 #ON state voltage drop across LED in V\n",
+ "I = 5.0 #in mA\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "I = I*10**-3 #in A\n",
+ "R = (Vout-V)/I #in ohm\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Resistance is =%.f \u03a9' %R)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance is =700 \u03a9\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.2, Page No 306"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "N_A = 7.5*10**24 #in atoms/m^3\n",
+ "N_D = 1.5*10**22 #in atoms/m^3\n",
+ "D_e = 25*10**-4 #in m^2/s\n",
+ "D_n = 1*10**-3 #in m^2/s\n",
+ "V_T = 26*10**-3 #in V\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Torque_eo = 500*10**-9 #in sec\n",
+ "Torque_ho = 100*10**-9 #in sec\n",
+ "e = 1.6*10**-19 #in C\n",
+ "n_i = 1.5*10**16 #in /m^3\n",
+ "I_lambda = 12.5 #in mA/cm^2\n",
+ "I_lambda= I_lambda*10**-3 #in A/cm^2\n",
+ "L_e = math.sqrt(D_e*Torque_eo) #in m\n",
+ "L_n = math.sqrt(D_n*Torque_ho) #in m\n",
+ "J_s = e*((n_i)**2)*( ((D_e)/(L_e*N_A)) + ((D_n)/(L_n*N_D)) ) #in A/m^2\n",
+ "J_s= J_s*10**-4 #in A/cm^2\n",
+ "V_OC = V_T*(math.log(1+(I_lambda/J_s))) #in V\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Open circuit voltage is =%.2f volts ' %V_OC)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Open circuit voltage is =0.52 volts \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.3, Page No 307"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "Phi_o = 1.0*10**21 #in m^-2s^-1\n",
+ "alpha =1.0*10**5 #in m^-1\n",
+ "e= 1.6*10**-19 #in C\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "G_L1 = alpha*Phi_o #in m^-3s^-1\n",
+ "W = 26.0 #in \u00b5m\n",
+ "W = W * 10**-6 #in m\n",
+ "G_L2 = alpha*Phi_o*(math.e)**((-alpha)*W) #in m^-3s^-1\n",
+ "#temp=(1-math.e)**(-(alpha)*W)\n",
+ "J_L = e*Phi_o*(1-math.e**(-(alpha)*W)) #in A/m^2\n",
+ "J_L = J_L * 10**3*10**-4 #in mA/cm^2\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Photo current density is =%.1f mA/cm^2' %J_L)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Photo current density is =14.8 mA/cm^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo8.ipynb b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo8.ipynb new file mode 100755 index 00000000..47137bb2 --- /dev/null +++ b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo8.ipynb @@ -0,0 +1,576 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1> Chapter No 8 : Metal Semiconductor Field Effect Transistor <h1>"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.1, Page No 323"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "I_DSS = 15.0 #in mA\n",
+ "V_GS_off = -5.0 #in V\n",
+ "V_GS = 0 #in V\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "I_D = I_DSS*((1-(V_GS/V_GS_off))**2) #in mA\n",
+ "V_GS = -1 #in V\n",
+ "I_D1 = I_DSS*((1.0-(V_GS/V_GS_off))**2) #in mA\n",
+ "V_GS = -4 #in V\n",
+ "I_D2 = I_DSS*((1-(V_GS/V_GS_off))**2) #in mA\n",
+ "\n",
+ "#RESULTS\n",
+ "print('When V_GS=0, the drain current is =%.f mA ' %I_D)\n",
+ "print('When V_GS=-1V, the drain current is =%.2f mA ' %I_D1)\n",
+ "print('The interplanner spacing is =%.2f \u00c5' %I_D2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "When V_GS=0, the drain current is =15 mA \n",
+ "When V_GS=-1V, the drain current is =9.60 mA \n",
+ "The interplanner spacing is =0.60 \u00c5\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.4, Page No 327"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "I_DSS = 20.0 #in mA\n",
+ "V_P = -8 #in V\n",
+ "g_mo = 5000.0 #in \u00b5s\n",
+ "V_GS = -4.0 #in V\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "\n",
+ "I_D = I_DSS*((1-(V_GS/V_P))**2) #in mA\n",
+ "g_m = g_mo*(1-(V_GS/V_P)) #in \u00b5s\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of drain current is =%.f mA ' %I_D)\n",
+ "print('The transconductance is =%.f \u00b5s' %g_m)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of drain current is =5 mA \n",
+ "The transconductance is =2500 \u00b5s\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.5 Page No 327"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "\n",
+ "\n",
+ "#initialisation of variables\n",
+ "I_D1 = 10.0 #in mA\n",
+ "V_GS = -12.0 #in V\n",
+ "V_GSth = -3.0 #in V\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "K = I_D1/( (V_GS-V_GSth)**2 ) #in mA/V\n",
+ "V_GS= -6 #in V\n",
+ "I_D = K*((V_GS-V_GSth)**2) #in mA\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of I_D is =%.2f mA ' %I_D)\n",
+ "print('Approximation error')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D is =1.11 mA \n",
+ "Approximation error\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.7, Page No 327"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "V_GS = -2.0 #in V\n",
+ "V_P = -5.0 #in V\n",
+ "V_DS = V_GS-V_P #in V\n",
+ "I_DSS = 8.0 #in mA\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "I_D = I_DSS*((1-(V_GS/V_P))**2) #in mA\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The minimum value of V_DS is =%.f V ' %V_DS)\n",
+ "print('The drain current is =%.2f mA' %I_D)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum value of V_DS is =3 V \n",
+ "The drain current is =2.88 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.8 Page No 328"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "V_P = -2 #in V\n",
+ "I_DSS = 1.65 #in mA\n",
+ "I_D = 0.8 #in mA\n",
+ "V_DD = 24.0 #in V\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V_GS = V_P*(1- math.sqrt(I_D/I_DSS) ) #in V\n",
+ "g_mo = -(2*I_DSS)/V_P #in mS\n",
+ "g_m = g_mo*(1-(V_GS/V_P)) #in mS\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of V_GS is =%.2f V ' %V_GS)\n",
+ "print('The value of g_m is =%.2f mS' %g_m)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V_GS is =-0.61 V \n",
+ "The value of g_m is =1.15 mS\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.9 Page No 328"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "V_P = 5.0 #in V\n",
+ "I_DSS = -40.0 #in mA\n",
+ "I_D = -15.0 #in mA\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V_GS = V_P*(math.sqrt(I_D/I_DSS)-1 ) #in V\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The gate source voltage is =%.2f V' %(V_GS*-1))\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The gate source voltage is =1.94 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.10, Page No 328"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "I_D1 = 1.9 #in mA\n",
+ "I_D2 = 1.0 #in mA\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "del_I_D = I_D1-I_D2 #in mA\n",
+ "V_GS2 = -3.3 #in V\n",
+ "V_GS1 = -3.0 #in V\n",
+ "del_V_GS = V_GS1-V_GS2 #in V\n",
+ "g_m = del_I_D/del_V_GS #in mA/V\n",
+ "g_m = g_m * 10**3 #in \u00b5 mhos\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of transconductance is =%.f \u00b5 mhos ' %g_m)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of transconductance is =3000 \u00b5 mhos \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.11 Page No 329"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#initialisation of variables\n",
+ "V_DS1 = 14.0 #in V\n",
+ "V_DS2 = 5.0 #in V\n",
+ "del_V_DS = V_DS1-V_DS2 # in V\n",
+ "I_D1 = 3.3 #in mA\n",
+ "I_D2 = 3 # in mA\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "del_I_D = I_D1-I_D2 #in mA\n",
+ "r_d = del_V_DS/del_I_D #in k ohms\n",
+ "print('The drain resistance is =%.f k ohms' %r_d)\n",
+ "V_GS1 = 0.4 #in V\n",
+ "V_GS2 = 0.1 #in V\n",
+ "del_V_GS = V_GS1-V_GS2 #in V\n",
+ "I_D1 = 3.3 #in mA\n",
+ "I_D2 = 0.71 #in mA\n",
+ "del_I_D = I_D1-I_D2 # in mA\n",
+ "g_m = del_I_D/del_V_GS # in mA/V\n",
+ "g_m = g_m * 10**3 # in \u00b5mhos\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The transconductance is =%.f \u00b5mhos ' %g_m)\n",
+ "Miu =r_d*10**3*g_m*10**-6\n",
+ "print('Amplification factor is =%.f ' %Miu)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The drain resistance is =30 k ohms\n",
+ "The transconductance is =8633 \u00b5mhos \n",
+ "Amplification factor is =259 \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.12 Page No 334"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "q = 1.6*10**-19 #in C\n",
+ "N_D = 10**15*10**6 #electrons/m^3\n",
+ "a = 3*10**-4 #in cm\n",
+ "a=a*10**-2 #in m\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Epsilon_o = (36 * math.pi * 10**9)**-1\n",
+ "Epsilon = 12*Epsilon_o\n",
+ "V_P = (q*N_D*((a)**2))/(2*Epsilon) #in V\n",
+ "print('Pinch off voltage is =%.2f V ' %V_P)\n",
+ "V_GS = 1.0 #in V\n",
+ "V_P = 2.0 #in V\n",
+ "b = a*( 1-math.sqrt(V_GS/V_P) ) #in m\n",
+ "b = b * 10**6 #in \u00b5m\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The channel half width is =%.2f \u00b5m' %b)\n",
+ "# Note: In the book, the unit of channel half width is wrong."
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Pinch off voltage is =6.79 V \n",
+ "The channel half width is =0.88 \u00b5m\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.13 Page No 335"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "I_DSS = 8.0 #in mA\n",
+ "V_P = -4.0 #in V\n",
+ "a = 3*10**-4 #in cm\n",
+ "N_D = 10**15 #in electrons/cm^3\n",
+ "I_D = 3.0 #in mA\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V_GS = V_P*( 1-math.sqrt(I_D/I_DSS) ) #in V\n",
+ "V_DS_sat = V_GS-V_P #in V\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of V_GS is =%.2f V' %V_GS)\n",
+ "print('The value of V_DS_sat is =%.2f V ' %V_DS_sat)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V_GS is =-1.55 V\n",
+ "The value of V_DS_sat is =2.45 V \n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 8.14 Page No 335"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "V_P = -4 #in V\n",
+ "I_DSS = 9.0 #in mA\n",
+ "V_GS = -2.0 #in V\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "I_D = I_DSS*(( 1-(V_GS/V_P) )**2) #in mA\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The drain current is =%.2f mA ' %I_D)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The drain current is =2.25 mA \n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 8.15 Page No 335"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "I_DSS = 12.0 #in mA\n",
+ "V_P = -6.0 #in V\n",
+ "V_GS = -1.0 #in V\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "g_mo = (-2*I_DSS)/V_P #in mA/V\n",
+ "g_m = g_mo*(1-(V_GS/V_P)) #in mS\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of transconductance is =%.2f mS' %g_m)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of transconductance is =3.33 mS\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 8.16 Page No 336"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "I_DSS = 10.0 #in mA\n",
+ "V_P = -5.0 #in V\n",
+ "V_GS = -2.5 #in V\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "g_m = ((-2*I_DSS)/V_P)*(1-(V_GS/V_P)) #in mS .... correction\n",
+ "I_D = I_DSS * ((1-(V_GS/V_P))**2) #in mA\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The transconductance is =%.f mS ' %g_m)\n",
+ "print('The drain current is =%.2f mA ' %I_D)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The transconductance is =2 mS \n",
+ "The drain current is =2.50 mA \n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo9.ipynb b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo9.ipynb new file mode 100755 index 00000000..d2b16e2d --- /dev/null +++ b/Advance_Semiconductor_Devices_by_S._Sharma/ChapterNo9.ipynb @@ -0,0 +1,1684 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1> Chapter No 9 : MOS Transistors<h1>"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.1, Page No 361"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "V_TN = 0.7 #in V\n",
+ "W = 45*10**-4 #in cm\n",
+ "L = 4.0 #in \u00b5m\n",
+ "L = L * 10**-4 #in cm\n",
+ "t_ox = 450 #in \u00c5\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "t_ox = t_ox*10**-8 #in cm\n",
+ "V_GS = 1.4 #in V\n",
+ "Miu_n = 700 #in cm^2/V-s\n",
+ "Epsilon_ox = (8.85*10**-14)*(3.9) #in F/cm\n",
+ "k_n = (W*Miu_n*Epsilon_ox)/(2*L*t_ox) #A/V^2\n",
+ "I_D = k_n*((V_GS-V_TN)**2) #in A\n",
+ "k_n=k_n*10**3\n",
+ "I_D=I_D*10**3\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of k_n is : =%.2f mA/V^2' %k_n)\n",
+ "print('The current is =%.2f mA' %I_D)\n",
+ "print('Approximation error')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of k_n is : =0.30 mA/V^2\n",
+ "The current is =0.15 mA\n",
+ "Approximation error\n"
+ ]
+ }
+ ],
+ "prompt_number": 87
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.2, Page No 376"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "\n",
+ "I_Don = 6.0 #in mA\n",
+ "I_Don= I_Don*10**-3 #in A\n",
+ "V_GSon = 8 #in V\n",
+ "V_GSth = 3 #in V\n",
+ "V_DD = 12.0 #in V\n",
+ "R_D= 2*10**3 #in \u03a9\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "k= I_Don/(V_GSon-V_GSth)**2 #in A/V^2\n",
+ "A= R_D**2 #assumed\n",
+ "B= 2*R_D*V_GSth-2*R_D*V_DD-1/k #assumed\n",
+ "C= (V_DD-V_GSth)**2 #assumed\n",
+ "\n",
+ "I_DQ= 0.0027940 #in A\n",
+ "V_DSQ= V_DD-I_DQ*R_D #in V\n",
+ "I_DQ=I_DQ*10**3\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of I_DQ is : =%.2f mA' %I_DQ)\n",
+ "print('The value of V_DSQ is : =%.2f volts ' %V_DSQ)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_DQ is : =2.79 mA\n",
+ "The value of V_DSQ is : =6.41 volts \n"
+ ]
+ }
+ ],
+ "prompt_number": 88
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.3 Page No 378"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "V_GS = 6.0 #in V\n",
+ "I_D = 4.0 #in mA\n",
+ "V_GSth = 2.0 #in V\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V_DS = V_GS #in V\n",
+ "V_DD = 2*V_DS #in V\n",
+ "R_D = (V_DD-V_DS)/I_D #in k ohm\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of V_DD is =%.2f V ' %V_DD)\n",
+ "print('The value of R_D is =%.2f k ohm' %R_D)\n",
+ "print('The very high value for the gate to drain resistance is : 10 M\u03a9 ' )\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V_DD is =12.00 V \n",
+ "The value of R_D is =1.50 k ohm\n",
+ "The very high value for the gate to drain resistance is : 10 M\u03a9 \n"
+ ]
+ }
+ ],
+ "prompt_number": 89
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.4, Page No 379"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "I_Don = 3*10**-3 \n",
+ "V_GSon = 10.0 #in V\n",
+ "V_GSth= 5.0 #in V\n",
+ "R2= 18*10**6 #in \u03a9\n",
+ "R1= 22*10**6 #in \u03a9\n",
+ "R_S=820 #in \u03a9\n",
+ "R_D=3*10**3 #in \u03a9\n",
+ "V_DD= 40.0 #in V\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V_G= V_DD*R2/(R1+R2) #in V\n",
+ "k= I_Don/(V_GSon-V_GSth)**2 #in A/V**2\n",
+ "A= R_S**2 #assumed\n",
+ "B= 2*R_S*V_GSth-2*R_S*V_G-1/k #assumed\n",
+ "C= (V_G-V_GSth)**2 #assumed\n",
+ "I_D= 0.0067246 #in A\n",
+ "I_DQ= I_D #in A\n",
+ "V_GSQ= V_G-I_D*R_S #in V\n",
+ "V_DSQ= V_DD-I_DQ*(R_D+R_S) #in V\n",
+ "I_DQ=I_DQ*10**3\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of I_DQ is =%.2f mA' %I_DQ)\n",
+ "print('The value of V_GSQ is =%.2f volts ' %V_GSQ)\n",
+ "print('The value of V_DSQ is =%.2f volts ' %V_DSQ)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_DQ is =6.72 mA\n",
+ "The value of V_GSQ is =12.49 volts \n",
+ "The value of V_DSQ is =14.31 volts \n"
+ ]
+ }
+ ],
+ "prompt_number": 90
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.5 Page No 380"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "I_D= '0.3*(V_GS-V_P)^2' #given expression\n",
+ "V_DD= 30.0 #in V\n",
+ "V_P= 4 #in V\n",
+ "R_GS = 1.2*10**6 #in \u03a9\n",
+ "R_G = 1.2*10**6 #in \u03a9\n",
+ "I_D= 1.2 #in mA\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Req= R_GS/(R_GS+R_G) #in \u03a9\n",
+ "R_D= 15.0 #in \u03a9\n",
+ "A= (R_D*Req)**2 #assumed\n",
+ "B= -(2*R_D*Req*(V_DD*Req-V_P)+1/0.3) #assumed\n",
+ "C= (V_DD*Req-V_P)**2 #assumed\n",
+ "I_DSQ= I_D #in mA\n",
+ "V_GS= (V_DD-I_D*R_D) #in V\n",
+ "V_DS= Req*V_GS #in V\n",
+ "\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of I_DSQ is : =%.2f mA' %I_DSQ)\n",
+ "print('The value of V_GS is : =%.2f volts ' %V_GS)\n",
+ "print('The value of V_DS is : =%.2f volts ' %V_DS)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_DSQ is : =1.20 mA\n",
+ "The value of V_GS is : =12.00 volts \n",
+ "The value of V_DS is : =6.00 volts \n"
+ ]
+ }
+ ],
+ "prompt_number": 91
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.6 Page No 380"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "k = 0.1 #in mA/V^2\n",
+ "V_T = 1.0 # in V\n",
+ "R1 = 33.0 #in k ohm\n",
+ "R2 = 21.0 #in k ohm\n",
+ "V_DD = 6.0 #in V\n",
+ "R_D = 18 #in k ohm\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V_G = (R2/(R2+R1))*V_DD #in V\n",
+ "V_S = 0 #in V\n",
+ "V_GS = V_G-V_S #in V\n",
+ "I_D = k*((V_GS-V_T)**2) #in mA\n",
+ "V_DS = V_DD - (I_D*R_D) #in V\n",
+ "V_DSsat = V_GS-V_T #in V\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of I_D is =%.2f mA ' %I_D)\n",
+ "print('The value of V_DS(sat) is =%.2f V ' %V_DS)\n",
+ "print('The value of V_DS(sat) is =%.2f V ' %V_DSsat)\n",
+ "\n",
+ "if V_DS > V_DSsat :\n",
+ " print('MOSFET is in saturation region ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D is =0.18 mA \n",
+ "The value of V_DS(sat) is =2.80 V \n",
+ "The value of V_DS(sat) is =1.33 V \n",
+ "MOSFET is in saturation region \n"
+ ]
+ }
+ ],
+ "prompt_number": 92
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.8, Page No 382"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "R2 = 18.0 #in k ohm\n",
+ "R1 = 33.0 #in k ohm\n",
+ "V_DD = 6.0 #in V\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V_G = (R2/(R1+R2))*V_DD #in V\n",
+ "V_S = V_DD # in V\n",
+ "V_SG = V_S-V_G #in V\n",
+ "k = 0.1\n",
+ "V_T = -1 #in V\n",
+ "I_D = k*((V_SG+V_T)**2) #in mA\n",
+ "R_D = 3.0 # in k ohm\n",
+ "V_SD = V_DD - (I_D*R_D) #in V\n",
+ "V_SDsat = V_SG+V_T #in V\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of V_SG is =%.2f V ' %V_SG)\n",
+ "print('The value of I_D is =%.2f mA ' %I_D)\n",
+ "print('The value of V_SD is =%.2f V ' %V_SD)\n",
+ "print('The value of V_SD(sat) is =%.2f V ' %V_SDsat)\n",
+ "\n",
+ "if V_SD > V_SDsat :\n",
+ " print('The p MOSFET is indeed biased in the saturation region')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V_SG is =3.88 V \n",
+ "The value of I_D is =0.83 mA \n",
+ "The value of V_SD is =3.51 V \n",
+ "The value of V_SD(sat) is =2.88 V \n",
+ "The p MOSFET is indeed biased in the saturation region\n"
+ ]
+ }
+ ],
+ "prompt_number": 93
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9 Page No 384"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#initialisation of variables\n",
+ "V_G= 1.5 # in V\n",
+ "V_P= -3.0 # in V\n",
+ "R_S= 750.0 # in \u03a9\n",
+ "R_D= 1800 #in \u03a9\n",
+ "I_DSS= 6*10**-3 # in A\n",
+ "V_DD= 18 #in V\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "A= R_S**2\n",
+ "B=(2*R_S*(V_P-V_G)-V_P**2/I_DSS)\n",
+ "C=(V_P-V_G)**2\n",
+ "I_D= 0.0031170 #in A\n",
+ "I_DQ= I_D #in A\n",
+ "V_DS= V_DD-I_D*(R_D+R_S) # in V\n",
+ "V_DSQ= V_DS # in V\n",
+ "I_DQ=I_DQ*10**3\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of I_DQ is : =%.2f mA ' %I_DQ)\n",
+ "print('The value of V_DSQ is : =%.2f volts ' %V_DSQ)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_DQ is : =3.12 mA \n",
+ "The value of V_DSQ is : =10.05 volts \n"
+ ]
+ }
+ ],
+ "prompt_number": 94
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.10 Page No 384"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "V_GS = 4.0 #in V\n",
+ "V_P = 2.0 #in V\n",
+ "R2 = 10.0 #in k ohm\n",
+ "R1 = 30.0 #in k ohm\n",
+ "R_D= 2.5 #in kohm\n",
+ "I_D= 15 #in mA\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "I_D= I_D*10**-3 #in A\n",
+ "V_DD = 25 #in V\n",
+ "V_G = (V_DD/R_D)*V_DD/(R1+R2) #in V\n",
+ "R_S = (V_G-V_GS)/I_D # in ohm\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of R_S is =%.f ohm ' %R_S)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R_S is =150 ohm \n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.11 Page No 385"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "k= 0.1 #in mA/V^2\n",
+ "V_T= 1 # in V\n",
+ "R2= 87*10**3 #in \u03a9\n",
+ "R1= 110*10**3 # in \u03a9\n",
+ "R_S=2 #in k\u03a9\n",
+ "R_D=2 #in k\u03a9\n",
+ "V_DD= 6 #in V\n",
+ "V_SS= 6 #in V\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V_G= (V_DD+V_SS)*R2/(R1+R2) # in V\n",
+ "A= R_S**2 # assumed\n",
+ "B= -(2*R_S*(V_G+V_SS-V_T)+1/k) # assumed\n",
+ "C= (V_G+V_SS-V_T)**2 #assumed\n",
+ "I_D= 2.6001587 # in mA\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of I_D is : =%.2f mA' %I_D)\n",
+ "V_DS=V_DD+V_SS-I_D*R_D-I_D*R_S #in V\n",
+ "print('The value of V_DS is : =%.2f volts' %V_DS)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D is : =2.60 mA\n",
+ "The value of V_DS is : =1.60 volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 96
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 9.12 Page No 386"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "\n",
+ "k = 0.16 #in mA/V^2\n",
+ "V_T = 2.0 #in V\n",
+ "I_D = 0.5 #in mA\n",
+ "V_DD = 6.0 #in V\n",
+ "V_SS = -6 #in V\n",
+ "R_D = 10.0 #in k ohm\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V_GS = V_T + (math.sqrt(I_D/k)) #in V\n",
+ "R_S = 2 #in k ohm\n",
+ "V_S = (I_D*R_S) - V_DD # in V\n",
+ "V_G = V_GS+V_S # in V\n",
+ "I = 0.1*I_D #in mA\n",
+ "R2 = (V_G+V_DD)/I #in k ohm\n",
+ "print('The value of R2 is =%.2f k ohm' %R2)\n",
+ "R1 = (V_DD - V_G)/I #in k ohm\n",
+ "print('The value of R1 is =%.2f k ohm' %R1)\n",
+ "V_DS = (V_DD-V_SS) - (I_D*(R_S+R_D)) #in V\n",
+ "print('The value of V_DS is =%.2f V' %V_DS)\n",
+ "V_DSsat = V_GS-V_T # in V\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of V_DS(sat) is =%.2f V' %V_DSsat)\n",
+ "if V_DS > V_DSsat :\n",
+ " print('The MOSFET is in saturation region')\n",
+ "\n",
+ " # Note: The value of R1 is in k ohm but in the book it is wrong."
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R2 is =95.36 k ohm\n",
+ "The value of R1 is =144.64 k ohm\n",
+ "The value of V_DS is =6.00 V\n",
+ "The value of V_DS(sat) is =1.77 V\n",
+ "The MOSFET is in saturation region\n"
+ ]
+ }
+ ],
+ "prompt_number": 97
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 9.13 Page No 387"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "V_DD = 6.0 #in V\n",
+ "V_D = 3.0 #in V\n",
+ "R_D = 10.0 #in k ohm\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "I_DQ = (V_DD-V_D)/R_D #in mA\n",
+ "print('The value of I_DQ is =%.2f mA ' %I_DQ)\n",
+ "V_T = 0.8 #in V\n",
+ "k = 0.12 #in mA/V^2\n",
+ "V_GS = math.sqrt(I_DQ/k) + V_T #in V\n",
+ "V_S = -V_GS #in V\n",
+ "V_DS = V_D-V_S #in V\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of V_DS is =%.2f V' %V_DS)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_DQ is =0.30 mA \n",
+ "The value of V_DS is =5.38 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 98
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 9.14 Page No 388"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "I_D = 0.3 # in mA\n",
+ "k = 0.12 #in mA/V^2\n",
+ "V_T = 1.0 #in V\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V_GS = V_T + (math.sqrt(I_D/k)) #in V\n",
+ "V_S = -V_GS #in V\n",
+ "V_DD = 6.0 #in V\n",
+ "V_D = 3.0 # in V\n",
+ "I_DQ = 0.3 #in mA\n",
+ "R_D = (V_DD-V_D)/I_DQ #in k ohm\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of R_D is =%.2f k ohm' %R_D)\n",
+ "V_DS = V_D - V_S #in V\n",
+ "print('The value of V_DS is =%.2f V' %V_DS)\n",
+ "V_DSsat = V_GS - V_T #in V\n",
+ "print('The value of V_DS(sat) is =%.2f V' %V_DSsat)\n",
+ "if V_DS > V_DSsat :\n",
+ " print('The MOSFET is in saturation region')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R_D is =10.00 k ohm\n",
+ "The value of V_DS is =5.58 V\n",
+ "The value of V_DS(sat) is =1.58 V\n",
+ "The MOSFET is in saturation region\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 9.15 Page No 389"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "k= 0.05 #in mA/V^2\n",
+ "V_T= 1 #in V\n",
+ "V_DD= 6 # in V\n",
+ "R_S= 9.1 #in k\u03a9\n",
+ "I_D= 0.3631765 # in mA\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "A= R_S**2 # assumed\n",
+ "B=-(2*V_DD*R_S+1/k) # assumed\n",
+ "C= V_DD**2 #assumed\n",
+ "V_GS= V_DD-I_D*R_S #in V\n",
+ "V_DS= V_GS # in V\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of I_D is : =%.2f mA' %I_D)\n",
+ "print('The value of V_GS is : =%.2f volts' %V_GS)\n",
+ "print('The value of V_DS is : =%.2f volts' %V_DS)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D is : =0.36 mA\n",
+ "The value of V_GS is : =2.70 volts\n",
+ "The value of V_DS is : =2.70 volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 100
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 9.16 Page No 390"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "k1= 0.01 #in mA/V^2\n",
+ "k2= 0.05 #in mA/V^2\n",
+ "V_DD= 5 # in V\n",
+ "V_T1=1 # in V\n",
+ "V_T2=1 # in V\n",
+ "Vi= 5 #in V\n",
+ "Vo= 0.3485163 # in V\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "A=(k1+k2)\n",
+ "B=-(2*k1*(V_DD-V_T1)+2*k2*(Vi-V_T2))\n",
+ "C=k1*(V_DD-V_T1)**2\n",
+ "V_GS2= Vi # in V\n",
+ "V_DS2= Vo # in V\n",
+ "V_GS1= V_DD-Vo # in V\n",
+ "I_D1= k1*(V_GS1-V_T1)**2 #in mA\n",
+ "I_D2= I_D1 # in mA\n",
+ "print('Part (i) For Vi = 5 V')\n",
+ "print('The output voltage is : =%.2f volts' %Vo)\n",
+ "print('The value of I_D1 is : =%.2f mA' %I_D1)\n",
+ "print('The value of I_D2 is : =%.2f mA' %I_D2)\n",
+ "# Analysis for Vi= 1.5V\n",
+ "Vi= 1.5 #in V\n",
+ "Vo= V_DD-V_T1-math.sqrt(k2/k1)*(Vi-V_T2) # in V\n",
+ "I_D2= k2*(Vi-V_T2)**2 #in mA\n",
+ "I_D1= I_D2 #in mA\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Part (ii) For Vi = 1.5 V')\n",
+ "print('The output voltage is : =%.2f volts' %Vo)\n",
+ "print('The value of I_D1 is : =%.2f mA' %I_D1)\n",
+ "print('The value of I_D2 is : =%.2f mA' %I_D2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Part (i) For Vi = 5 V\n",
+ "The output voltage is : =0.35 volts\n",
+ "The value of I_D1 is : =0.13 mA\n",
+ "The value of I_D2 is : =0.13 mA\n",
+ "Part (ii) For Vi = 1.5 V\n",
+ "The output voltage is : =2.88 volts\n",
+ "The value of I_D1 is : =0.01 mA\n",
+ "The value of I_D2 is : =0.01 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 101
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 9.17 Page No 392"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "k = 0.12 #in mA/V^2\n",
+ "V_T = -2.5 #in V\n",
+ "V_GS = 0\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "I_D = k*((V_GS-V_T)**2) #in mA\n",
+ "print('The value of I_D is =%.2f mA' %I_D)\n",
+ "V_DD = 6 #in V\n",
+ "R_S = 4.7 #in k ohm \n",
+ "V_DS = V_DD -(I_D*R_S) #in V\n",
+ "print('The value of V_DS is =%.2f V' %V_DS)\n",
+ "V_S = 0 # in V \n",
+ "V_DSsat = V_S - V_T # in V\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of V_DS(sat) is =%.2f V' %V_DSsat)\n",
+ "if V_DS < V_DSsat :\n",
+ " print('The device is in the non saturation region')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D is =0.75 mA\n",
+ "The value of V_DS is =2.47 V\n",
+ "The value of V_DS(sat) is =2.50 V\n",
+ "The device is in the non saturation region\n"
+ ]
+ }
+ ],
+ "prompt_number": 102
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 9.18 Page No 394"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "k4 = 0.125 #in mA/V^2\n",
+ "k3 = k4 #in mA/V^2\n",
+ "k2 = k4 #in mA/V^2\n",
+ "k1 = 0.25 #in mA/V^2\n",
+ "V_T1 = 0.8 # in V\n",
+ "V_T2 = V_T1 # in V\n",
+ "V_T3 = V_T1 #in V\n",
+ "V_T4 = V_T1 #in V\n",
+ "V_SS = -5 #in V\n",
+ "V_DD = 5 #in V\n",
+ "R_D = 10 #in k ohm\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V_GS3 = ((math.sqrt(k4/k3) * (-V_SS - V_T4))+V_T3)/(1+math.sqrt(k4/k3)) #in V\n",
+ "I_Q = k2*((V_GS3-V_T2)**2) #in mA\n",
+ "I_D1 = I_Q #in mA\n",
+ "V_GS1 = V_T1 + (math.sqrt(I_D1/k1)) #in V\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of V_GS1 is =%.2f V ' %V_GS1)\n",
+ "V_DS2 = (-V_SS-V_GS1) #in V\n",
+ "print('The value of V_DS2 is =%.2f V ' %V_DS2)\n",
+ "V_DS1 = V_DD - (I_Q*R_D) - (V_SS + V_DS2) #in V\n",
+ "print('The value of V_DS1 is =%.2f V' %V_DS1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V_GS1 is =2.00 V \n",
+ "The value of V_DS2 is =3.00 V \n",
+ "The value of V_DS1 is =3.39 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 103
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 9.19 Page No 395"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "R2 = 20.0 # in k ohm\n",
+ "R1 = 30.0 #in k ohm\n",
+ "R_D = 20 # in k ohm\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "R_D=R_D*10**3 #in ohm\n",
+ "V_DD = 5 #in V\n",
+ "V_G = (R2/(R1+R2))*V_DD #in V\n",
+ "V_S = 0 #in V\n",
+ "V_GS = V_G #in V\n",
+ "k = 100*10**-6 #in A/V^2\n",
+ "V_T = 1 #in V\n",
+ "I_DQ = k*((V_GS-V_T)**2) #in A\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "V_DSQ = V_DD - (I_DQ*R_D) #in V \n",
+ "print('The value of V_DSQ is =%.f V ' %V_DSQ)\n",
+ "I_DQ=I_DQ*10**6\n",
+ "print('The value of I_DQ is =%.f \u00b5A' %I_DQ)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V_DSQ is =3 V \n",
+ "The value of I_DQ is =100 \u00b5A\n"
+ ]
+ }
+ ],
+ "prompt_number": 104
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 9.20 Page No 395"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "V_P= -8 # in V\n",
+ "R_S= 2.4 #in k\u03a9\n",
+ "I_DSS= 8 #in mA\n",
+ "V_DD= 20 #in V\n",
+ "R_D= 6.2 #in k\u03a9\n",
+ "I_D= 1.7668305 #in mA\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "A= R_S**2\n",
+ "B=(2*R_S*(V_P)-V_P**2/I_DSS)\n",
+ "C=(V_P)**2\n",
+ "I_DQ= I_D #in mA\n",
+ "V_GSQ= -I_D*R_S;\n",
+ "V_D= V_DD-I_D*R_D #in V\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of I_DQ is : =%.2f mA' %I_DQ)\n",
+ "print('The value of V_GSQ : =%.2f volts' %V_GSQ)\n",
+ "print('The value of V_D is : =%.2f volts' %V_D)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_DQ is : =1.77 mA\n",
+ "The value of V_GSQ : =-4.24 volts\n",
+ "The value of V_D is : =9.05 volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 105
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 9.21 Page No 396"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "k= 75*10**-3 #in mA/V^2\n",
+ "Vth= -0.8 #in V\n",
+ "R2 = 100 #in k ohm\n",
+ "R1 = 100 #in k ohm\n",
+ "R_S= 6 # in k\u03a9\n",
+ "R_D= 3 # in k\u03a9\n",
+ "V_SS = 10 # in V\n",
+ "V_S = 2.0602803\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V_G = (R2/(R1+R2))*V_SS #in V\n",
+ "V_GS= V_G-V_S #in V\n",
+ "I_D= k*(V_GS-Vth)**2 # in V\n",
+ "V_DS= -V_SS+I_D*(R_D+R_S) # in V\n",
+ "V_D= I_D*R_D #in V\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of I_D is : =%.2f mA ' %I_D)\n",
+ "print('The value of V_DS is =%.2f volts ' %V_DS)\n",
+ "print('The value of V_D is : =%.2f volts' %V_D)\n",
+ "print('The value of V_S is : =%.2f volts' %V_S)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D is : =0.12 mA \n",
+ "The value of V_DS is =-8.93 volts \n",
+ "The value of V_D is : =0.36 volts\n",
+ "The value of V_S is : =2.06 volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 106
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 9.22 Page No 396"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "V_T = 1.0 # in V\n",
+ "k = 160*10**-6 # in A/V^2\n",
+ "I_DQ = 160*10**-6 # in A\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V_GS = V_T + math.sqrt(I_DQ/k) #in V\n",
+ "V_DD = 5 # in V\n",
+ "V_DSQ = 3 # in V\n",
+ "R_D = (V_DD - V_DSQ)/(I_DQ) # in ohm\n",
+ "R_D = R_D * 10**-3 # in k ohm\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of R_D is =%.2f k ohm' %R_D)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R_D is =12.50 k ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 107
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 9.23 Page No 397"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "V_DD= 12 # in V\n",
+ "V_T= 2 # in V\n",
+ "kn= 0.5 # in mA/V^2\n",
+ "R1 = 2.2 # in M ohm\n",
+ "R2 = 1.8 # in M ohm\n",
+ "R_S= 1.5 # in k\u03a9\n",
+ "R_D= 3.9 # in k\u03a9\n",
+ "I_D= 1.2237155 # in mA\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V_G = (R2/(R1+R2))*V_DD # in V\n",
+ "V_GS= V_G-I_D*R_S # V\n",
+ "I_D= I_D-kn*(V_GS-V_T)**2 # in mA\n",
+ "I_DQ= I_D #in mA\n",
+ "V_DSQ= V_DD-I_D*(R_D+R_S) # in V\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of I_DQ is : =%.2f mA ' %I_DQ)\n",
+ "print('The value of V_DSQ is : =%.2f volts' %V_DSQ)\n",
+ "V_GS= V_G-I_D*R_S # V\n",
+ "V_DSsat= V_GS-V_T # in V\n",
+ "print('The value of V_DS %.2f is greater than the value of ' %V_DSQ)\n",
+ "print('V_DSsat %.2f So the MOSFET is in saturation region ' %V_DSsat)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_DQ is : =-0.00 mA \n",
+ "The value of V_DSQ is : =12.00 volts\n",
+ "The value of V_DS 12.00 is greater than the value of \n",
+ "V_DSsat 3.40 So the MOSFET is in saturation region \n"
+ ]
+ }
+ ],
+ "prompt_number": 108
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.24 Page No 397"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "kn= 0.5 # in mA/V^2\n",
+ "V_T= 1 # in V\n",
+ "R2 = 40 # in k ohm\n",
+ "R1 = 60 # in k ohm\n",
+ "R_S= 1 # in k ohm\n",
+ "R_D= 2 # in k ohm\n",
+ "V_DD = 5 # in V\n",
+ "V_SS = -5 # in V\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V_R2 = (R2/(R2+R1))*(V_DD-V_SS) # in V\n",
+ "V_G = V_R2 - V_DD # in V\n",
+ "V_S= I_D*R_S+V_SS # in V\n",
+ "V_GS= V_G-V_S # in V\n",
+ "I_D= 1.3542487 # in mA\n",
+ "I_DQ= I_D #in mA\n",
+ "V_S= I_D*R_S+V_SS # in V\n",
+ "V_GS= V_G-V_S # in V\n",
+ "V_DSQ= V_DD-V_SS-I_D*(R_D+R_S) # in V\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of I_DQ is :=%.2f mA ' %I_DQ)\n",
+ "print('The value of V_GS is : =%.2f volts ' %V_GS)\n",
+ "print('The value of V_DSQ is : =%.2f volts' %V_DSQ)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_DQ is :=1.35 mA \n",
+ "The value of V_GS is : =-1.35 volts \n",
+ "The value of V_DSQ is : =5.94 volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 109
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 9.25 Page No 398"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "R_S1 = 100*10**-3 # in k ohm\n",
+ "R_S2 = 100*10**-3 # k ohm\n",
+ "R_S = R_S1+R_S2 #in k ohm\n",
+ "R_D= 1.8 # in k ohm\n",
+ "I_DSS= 12 # in mA\n",
+ "Vp= -3.5 # in V\n",
+ "V_DD= 22 # in V\n",
+ "rd= 25 # in k ohm\n",
+ "R_L= 47 # in k ohm\n",
+ "I_D= 5.573521 # in mA\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V_GS= -I_D*R_S # in V\n",
+ "I_D= I_D-I_DSS*(1-V_GS/Vp)**2 # in mA\n",
+ "print('The value of I_D is : =%.2f mA' %I_D)\n",
+ "V_GS= -I_D*R_S # in V\n",
+ "print('The value of V_GS is : =%.2f volts ' %V_GS)\n",
+ "V_DS= V_DD-I_D*(R_D+R_S) # in V\n",
+ "print('The value of V_DS is : =%.2f volts ' %V_DS)\n",
+ "gmo= -2*I_DSS/Vp # in mS\n",
+ "gm= gmo*(1-V_GS/Vp) # in mS\n",
+ "miu= gm*rd\n",
+ "Av= -miu*R_D*R_L/(R_D+R_L)/(rd+R_D*R_L/(R_D+R_L)+(1+miu)*R_S1)\n",
+ "print('The value of Av is : =%.2f ' %Av)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D is : =-0.000000 mA\n",
+ "The value of V_GS is : =0.00 volts \n",
+ "The value of V_DS is : =22.00 volts \n",
+ "The value of Av is : =-6.76 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 9.26 Page No 399"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "V_T = 1 # in V\n",
+ "k = 0.5 # in mA/V^2\n",
+ "R2 = 40 # in k ohm\n",
+ "R1 = 60 # in k ohm\n",
+ "R_S= 1 # in k ohm\n",
+ "R_D= 2 # in k ohm\n",
+ "V_DD = 5 # in V\n",
+ "V_G = (R2/(R2+R1))*V_DD # in V\n",
+ "V_GS= V_G-I_D*R_S # in V\n",
+ "I_D= 0.2679492 # in mA\n",
+ "V_GS= V_G-I_D*R_S # in V\n",
+ "V_DS= V_DD-I_D*(R_D+R_S) # in V\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of I_D in mA is : =%.2f ' %I_D)\n",
+ "print('The value of V_GS in volts is :=%.2f ' %V_GS)\n",
+ "print('The value of V_DS in volts is : =%.2f ' %V_DS)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D in mA is : =0.27 \n",
+ "The value of V_GS in volts is :=-0.27 \n",
+ "The value of V_DS in volts is : =4.20 \n"
+ ]
+ }
+ ],
+ "prompt_number": 111
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 9.27 Page No 400"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "R_D = 7.5 # in k ohm\n",
+ "V_T = -0.8 # in V\n",
+ "k = 0.2 # in mA/V^2\n",
+ "R2 = 50 # in ohm\n",
+ "R1 = 50 # in ohm\n",
+ "V_DD = 5 # in V\n",
+ "V_S = 5 # in V\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V_G = (R2/(R2+R1))*V_DD # in V\n",
+ "V_GS = V_G - V_S # in V\n",
+ "I_D = k*((V_GS-V_T)**2) # in mA\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Drain current in is =%.2f mA' %I_D)\n",
+ "V_SD = V_DD - (I_D*R_D) # in V\n",
+ "print('Source to drain voltage is =%.2f V' %V_SD)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Drain current in is =3.53 mA\n",
+ "Source to drain voltage is =-21.46 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 112
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.28 Page No 400"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "I_Don = 5*10**-3 # in A\n",
+ "V_GSon = 6 # in V\n",
+ "V_GSth = 3 # in V\n",
+ "k = I_Don/(V_GSon-V_GSth)**2 # in A/V^2 \n",
+ "R2 = 6.8 #in M ohm\n",
+ "R1 = 10 #in M ohm\n",
+ "R_S= 750 # in ohm\n",
+ "R_D= 2.2*10**3 # in ohm\n",
+ "V_DD = 24 # in V\n",
+ "R_S = 750 # in ohm\n",
+ "V_G= R2*V_DD/(R1+R2) # in V\n",
+ "V_GS= V_G-I_D*R_S # in V\n",
+ "I_D= I_D-k*(V_GS-V_GSth)**2\n",
+ "I_D= 0.0049660 # in A\n",
+ "I_DQ= I_D # in A\n",
+ "V_GS= V_G-I_D*R_S # in V\n",
+ "V_GSQ= V_GS # in V\n",
+ "V_DSQ= V_DD-I_DQ*(R_D+R_S) # in V\n",
+ "I_D=I_D*10**3\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of I_D is : =%.2f mA ' %I_D)\n",
+ "print('The value of V_GSQ is : =%.2f volts ' %V_GSQ)\n",
+ "print('The value of V_DSQ is : =%.2f volts ' %V_DSQ)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D is : =4.97 mA \n",
+ "The value of V_GSQ is : =5.99 volts \n",
+ "The value of V_DSQ is : =9.35 volts \n"
+ ]
+ }
+ ],
+ "prompt_number": 113
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 9.29 Page No 401"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "I_Don = 4*10**-3 # in A\n",
+ "V_GSon = 6 # in V\n",
+ "V_GSth = 3 # in V\n",
+ "V_DS= 6 # in V\n",
+ "I_D= I_Don # in A\n",
+ "k = I_Don/((V_GSon-V_GSth)**2) # in A/V^2\n",
+ "V_GS= I_D-k*(V_GS-V_GSth)**2\n",
+ "V_GS= 6 # in V\n",
+ "V_DD= 2*V_DS # in V\n",
+ "R_D= (V_DD-V_GS)/I_D # in ohm\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of V_GS is : =%.f volts ' %V_GS)\n",
+ "print('The value of V_DD is : =%.f volts ' %V_DD)\n",
+ "print('The value of R_D is : =%.f k\u03a9' %R_D)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V_GS is : =6 volts \n",
+ "The value of V_DD is : =12 volts \n",
+ "The value of R_D is : =1500 k\u03a9\n"
+ ]
+ }
+ ],
+ "prompt_number": 114
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 9.31 Page No 402"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "V_DD = 5 # in V\n",
+ "V_T= 1 # in V\n",
+ "k= 1 # in mA/V^2\n",
+ "R1 = 1 # in M ohm\n",
+ "R2 = 1 # in M ohm\n",
+ "R_S= 2 # in k ohm\n",
+ "R_D= 2 # in k ohm\n",
+ "I1 = V_DD/(R1+R2) # in A\n",
+ "print('The value of I1 is : =%.2f \u00b5A' %I1)\n",
+ "V_A = (R2/(R2+R1))*V_DD # in V\n",
+ "print('The value of V_A and V_G is : =%.2f volts ' %V_A)\n",
+ "I_D= 0.4243061\n",
+ "V_C= I_D*R_S # in V\n",
+ "V_GS= V_A-V_C # in V\n",
+ "print('The value of I_D is : =%.2f mA' %I_D)\n",
+ "V_B= V_DD-I_D*R_D # in V\n",
+ "V_C= I_D*R_S # in V\n",
+ "V_DS= V_B-V_C # in V\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of V_B is : =%.2f volts ' %V_B)\n",
+ "print('The value of V_C is : =%.2f volts ' %V_C)\n",
+ "print('The value of V_DS is : =%.2f volts' %V_DS)\n",
+ "# Note: In the book, the calculated values are wrong, this is why the answer in the book is wrong.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I1 is : =2.00 \u00b5A\n",
+ "The value of V_A and V_G is : =0.00 volts \n",
+ "The value of I_D is : =0.42 mA\n",
+ "The value of V_B is : =4.15 volts \n",
+ "The value of V_C is : =0.85 volts \n",
+ "The value of V_DS is : =3.30 volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 115
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 9.32 Page No 403"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "I_DSS = 12 # in mA\n",
+ "I_DSS= I_DSS*10**-3 # in A\n",
+ "V_P = -3 # in V\n",
+ "r_d = 45 # in k ohm\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "r_d= r_d*10**3 # in ohm\n",
+ "g_m = I_DSS/abs(V_P) # in S\n",
+ "#Part (i)\n",
+ "R1 = 91 # in M ohm\n",
+ "R1=R1*10**6 #in ohm\n",
+ "R2 = 10 # in M ohm\n",
+ "R2= R2*10**6 # in ohm\n",
+ "Ri= R1*R2/(R1+R2) # in ohm\n",
+ "Ri=Ri*10**6\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of Ri is : =%.f Mohm' %(Ri/(10**12)))\n",
+ "#Part (ii)\n",
+ "R_S = 1.1 #in k ohm\n",
+ "R_S = R_S * 10**3 # in ohm\n",
+ "R_o= (R_S*1/g_m)/(R_S+1/g_m) # in ohm\n",
+ "print('The value of R_C is : =%.2f ohm' %R_o)\n",
+ "\n",
+ "# Part (iii)\n",
+ "R_desh_o= R_o*r_d/(R_o+r_d) # in ohm\n",
+ "print('The value of R_desh_o is : =%.2f ohm ' %R_desh_o)\n",
+ "# Part (iv)\n",
+ "Av= g_m*(R_S*r_d/(R_S+r_d))/(1+g_m*(R_S*r_d/(R_S+r_d)))\n",
+ "print('The value of Av is : =%.2f ' %Av)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Ri is : =9 Mohm\n",
+ "The value of R_C is : =203.70 ohm\n",
+ "The value of R_desh_o is : =202.79 ohm \n",
+ "The value of Av is : =0.81 \n"
+ ]
+ }
+ ],
+ "prompt_number": 116
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 9.34 Page No 404"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "V_S2 = -2 # in V\n",
+ "V_GS2 = -V_S2 # in V\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "I_DS2 = (V_GS2-1)**2 # in mA\n",
+ "I = 2 # in mA\n",
+ "I_DS1 = I-I_DS2 # in mA\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The current flow through M1 MOSFET is =%.f mA' %I_DS1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current flow through M1 MOSFET is =1 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 117
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 9.35 Page No 404"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "V_DD= 10 # in V\n",
+ "I_D= 0.4*10**3 # in A\n",
+ "W= 100 # in \u00b5m\n",
+ "L= 10 # in \u00b5m\n",
+ "uACox= 20 # in A/V**2\n",
+ "Vt= 2 # in V\n",
+ "R= 0.015\n",
+ "V_GS= V_DD-I_D*R # in V\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of R is : =%.f k\u03a9 ' %(R*10**3))\n",
+ "V_D= V_DD-I_D*R # in V\n",
+ "print('The value of V_D is : =%.f volts ' %V_D)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R is : =15 k\u03a9 \n",
+ "The value of V_D is : =4 volts \n"
+ ]
+ }
+ ],
+ "prompt_number": 118
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 9.36 Page No 405"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "V_GSth= 2 # in V\n",
+ "k= 2*10**-4 # in A/V^2\n",
+ "V_DD= 12 # in V\n",
+ "R_D= 5*10**3 # in ohm\n",
+ "I_D= 0.0014597\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V_DS= V_DD-I_D*R_D # in V\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of V_DS is : =%.1f volts ' %V_DS)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V_DS is : =4.7 volts \n"
+ ]
+ }
+ ],
+ "prompt_number": 119
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Advance_Semiconductor_Devices_by_S._Sharma/README.txt b/Advance_Semiconductor_Devices_by_S._Sharma/README.txt new file mode 100755 index 00000000..eec99ae9 --- /dev/null +++ b/Advance_Semiconductor_Devices_by_S._Sharma/README.txt @@ -0,0 +1,10 @@ +Contributed By: Santosh Pawar +Course: mca +College/Institute/Organization: ASM institute of computer studies +Department/Designation: MCA +Book Title: Advance Semiconductor Devices +Author: S. Sharma +Publisher: S. K. Kataria & Sons, New Delhi +Year of publication: 2012 +Isbn: 9789350142158 +Edition: 1
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