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| author | kinitrupti | 2017-05-12 18:53:46 +0530 |
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| committer | kinitrupti | 2017-05-12 18:53:46 +0530 |
| commit | 6279fa19ac6e2a4087df2e6fe985430ecc2c2d5d (patch) | |
| tree | 22789c9dbe468dae6697dcd12d8e97de4bcf94a2 /Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap | |
| parent | d36fc3b8f88cc3108ffff6151e376b619b9abb01 (diff) | |
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diff --git a/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/README.txt b/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/README.txt new file mode 100755 index 00000000..5de12de7 --- /dev/null +++ b/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/README.txt @@ -0,0 +1,10 @@ +Contributed By: Jignesh Bhadani +Course: bca +College/Institute/Organization: Freelancer +Department/Designation: BCA +Book Title: Principles of Electrical Engineering Materials +Author: S. O. Kasap +Publisher: Mgh +Year of publication: 2000 +Isbn: 0072356448 +Edition: 1
\ No newline at end of file diff --git a/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch1.ipynb b/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch1.ipynb new file mode 100755 index 00000000..f890eade --- /dev/null +++ b/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch1.ipynb @@ -0,0 +1,507 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1d645bcf155c10fc3ccf4c5aa79440417efe4cfdb744fd0ce2e78c231ca774d3" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 1 : Elementary Materials Science Concepts" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.1 Page No : 4" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "A = 8.0*10**-77 # in J m**6\n", + "B = 1.12*10**-133 # in J m**12\n", + "\n", + "# lennard-Jones 6-12 potential Energy (PE)curve is E(r)=-A*r**-6+B*r**-12\n", + "# For bonding to occur PE should be minimum, hence differentiating the PE equation and setting it to Zero at r=ro we get\n", + "\n", + "# Calculations and Results\n", + "ro = (2.0*B/A)**(1.0/6)\n", + "print(\"Bond length in meters is {0:.5g}\".format(ro))\n", + "E_bond = -A*ro**-6+(B*ro**-12) # in J\n", + "E_bond = abs(E_bond/(1.6*10**-19))\n", + "print(\"Bond Energy for solid argon in ev is {0:.4f}\".format(E_bond))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Bond length in meters is 3.7543e-10\n", + "Bond Energy for solid argon in ev is 0.0893\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.2 Page No : 5" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "R = 8.314 # in J/mol/K\n", + "T = 27.0 # in degree celcius\n", + "T += 273.0 # in Kelvin\n", + "M_at = 14.0 # in g/mol\n", + "# From Kinetic Theory\n", + "\n", + "# Calculations and Results\n", + "V_rms = math.sqrt((3.0*R*T)/(2.0*M_at*10**-3))\n", + "print(\"rms velocity of Nitrogen molecule in atmosphere at 300K in m/s is {0:.4f}\".format(V_rms))\n", + "V_rmsx = V_rms/math.sqrt(3.0)\n", + "print(\"rms velocity in one direction in m/s is {0:.4f}\".format(V_rmsx))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "rms velocity of Nitrogen molecule in atmosphere at 300K in m/s is 516.9485\n", + "rms velocity in one direction in m/s is 298.4603\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.3 Page No : 10" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "R = 8.314 # in J/mol/K\n", + "M_at = 63.6 # in g/mol\n", + "# Acc. to Dulong -Petit rule Cm=3R for NA atoms\n", + "\n", + "# Calculations and Results\n", + "C_gram = 3.0*R/M_at\n", + "print(\"Heat Capacity of copper per unit gram in J/g/K is {0:.4f}\".format(C_gram))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat Capacity of copper per unit gram in J/g/K is 0.3922\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.4 Page No : 12" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "# Given\n", + "k = 1.38*10**-23 # in J/K\n", + "m = 9.1*10**-31 # in Kg\n", + "T = 300.0 # in Kelvin\n", + "\n", + "# Calculations and Results\n", + "v_av = math.sqrt(8.0*k*T/(math.pi*m))\n", + "print(\"Mean speed for a gas of non interacting electrons in Km is {0:.4f}\".format(v_av*10**-3))\n", + "v = math.sqrt(2.0*k*T/m)\n", + "print(\"Most probable speed for a gas of non interacting electrons in Km is {0:.4f}\".format(v*10**-3))\n", + "v_rms = math.sqrt(3.0*k*T/m)\n", + "print(\"rms velocity for a gas of non interacting electrons in Km is {0:.4f}\".format(v_rms*10**-3))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mean speed for a gas of non interacting electrons in Km is 107.6340\n", + "Most probable speed for a gas of non interacting electrons in Km is 95.3882\n", + "rms velocity for a gas of non interacting electrons in Km is 116.8262\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.5 Page No : 13" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "# Given\n", + "L = 100*10**-6 # in Henry\n", + "C = 100*10**-12 # in Farad\n", + "T = 300.0 # in Kelvin\n", + "R = 200*10**3 # in ohms\n", + "k = 1.38*10**-23 # in J/K\n", + "\n", + "# Calculations and Results\n", + "fo = 1.0/(2.0*math.pi*math.sqrt(L*C)) # resonant frequency\n", + "Q = 2.0*math.pi*fo*C*R # quality factor\n", + "B = fo/Q # Bandwidth of tuned RLC\n", + "# Acc. to Johnson resistor noise equation\n", + "\n", + "Vrms = math.sqrt(4*k*T*R*B) # in volts\n", + "Vrms /= 10**-6 # in micro volts\n", + "print(\"Minimum rms radio signal that can be detected in micro volts is {0:.4f}\".format(Vrms))\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum rms radio signal that can be detected in micro volts is 5.1338\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.7 Page No : 18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Given\n", + "n = 4.0\n", + "M_at = 63.55*10**-3 # Kg/mol\n", + "NA = 6.022*10**23 # mol**-1\n", + "R = 0.128 # in nm\n", + "c = 8.0 # no.of corners of unit cells\n", + "f = 6.0 # no.of faces of unit cells\n", + "\n", + "# Calculations and Results\n", + "# a\n", + "N = c*(1.0/8)+f*(1.0/2)\n", + "print(\"No. of atoms per unit cells is {0:.4f}\".format(N))\n", + "# b\n", + "# Lattice parameter\n", + "a = R*2*2**(1.0/2)\n", + "print(\"Lattice Parameter in nm is {0:.4f}\".format(a))\n", + "a *= 10**-9 # in m\n", + "# c\n", + "# APF=(No.of atoms in unit cell)*(Vol. of atom)/(Vol. of unit cell)\n", + "APF = 4**2*math.pi/(3*(2*math.sqrt(2.0))**3)\n", + "print(\"Atomic Packing Factor is {0:.4f}\".format(APF))\n", + "# d\n", + "p = n*M_at/(a**3*NA) # density\n", + "print(\"density of Copper in Kg/m3 is {0:.4f}\".format(p))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "No. of atoms per unit cells is 4.0000\n", + "Lattice Parameter in nm is 0.3620\n", + "Atomic Packing Factor is 0.7405\n", + "density of Copper in Kg/m3 is 8895.4906\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.8 Page No : 23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# lcm function\n", + "def lcm(*args):\n", + " return cal_lcm(args[0][0], cal_lcm(args[0][1], args[0][2]))\n", + "\n", + "\n", + "def cal_lcm(x, y):\n", + " if x > y:\n", + " greater = x\n", + " else:\n", + " greater = y\n", + " while True:\n", + " if(greater % x == 0) and (greater % y == 0):\n", + " result = greater\n", + " break\n", + " greater += 1\n", + " return result\n", + "\n", + "\n", + "# Given\n", + "a = 1.0/float(\"inf\")\n", + "b = -1.0/1\n", + "c = 2.0/1\n", + "p = (1.0, 1.0, 1.0)\n", + "# 1/%inf = 0 ; (0/1 -1/1 2/1) hence lcm is taken for [1 1 1]\n", + "\n", + "# Calculations and Results\n", + "LCM = lcm(p)\n", + "h = a*float(LCM)\n", + "k = b*float(LCM)\n", + "l = c*float(LCM)\n", + "print(\"miller indices = {0:.4f} {1:.4f} {2:.4f}\".format(h, k, l))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "miller indices = 0.0000 -1.0000 2.0000\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.9 Page No : 28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Given\n", + "k = 1.38*10**-23 # J/K\n", + "T = 300.0 # kelvin\n", + "Ev = 0.75 # eV/atom\n", + "Ev = Ev*1.6*10**-19 # in J\n", + "T1 = 660.0 # degree celcius\n", + "T1 += 273.0 # in kelvin\n", + "# at room temperature\n", + "# let nv/N=nv_N for convenience\n", + "\n", + "# Calculations and Results\n", + "nv_N = math.exp(-Ev/(k*T))\n", + "print(\"Fractional concentration of vacancies in the aluminium crystal at room temperature is {0:.4g}\".format(nv_N))\n", + "# at melting temperature\n", + "# let nv/N=nv_N for convenience\n", + "nv_N = math.exp(-Ev/(k*T1))\n", + "print(\"Fractional concentration of vacancies in the aluminium crystal at melting temperature is {0:.4g}\".format(nv_N))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fractional concentration of vacancies in the aluminium crystal at room temperature is 2.581e-13\n", + "Fractional concentration of vacancies in the aluminium crystal at melting temperature is 8.961e-05\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.10 Page No : 33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Given\n", + "NA = 6.023*10**23 # mol**-1\n", + "d = 2.33 # density of Si in g/cm3\n", + "Mat = 28.09 # g/mol\n", + "Ev = 2.4 # ev/atom\n", + "Ev = 2.4*1.6*10**-19 # J/atom\n", + "k = 1.38*10**-23 # J/K\n", + "T = 300.0 # kelvin\n", + "T1 = 1000.0 # degree celsius\n", + "T1 += 273.0 # in kelvin\n", + "\n", + "# Calculations and Results\n", + "N = (NA*d)/Mat\n", + "# at room temperature\n", + "nv = N*math.exp(-(Ev/(k*T)))\n", + "print(\"concentration of vacancies in a Si crystal at room temperature in cm**-3 is {0:.4g}\".format(nv))\n", + "# at 1000 degree celsius\n", + "nv = N*math.exp(-(Ev/(k*T1)))\n", + "print(\"concentration of vacancies in a Si crystal at 1000 degree celsius in cm**-3 is {0:.4g}\".format(nv))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "concentration of vacancies in a Si crystal at room temperature in cm**-3 is 2.608e-18\n", + "concentration of vacancies in a Si crystal at 1000 degree celsius in cm**-3 is 1.605e+13\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.11 Page No : 34" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "# from fig 7.1\n", + "# at 210 degree celsius\n", + "print(\"At 210 degree celsius\")\n", + "C_L = 50.0 # CL=50% Sn\n", + "C_alpha = 18.0 # C_alpha=18% Sn\n", + "Co = 40.0 # solidification of alloy\n", + "\n", + "# Calculations and Results\n", + "# lever rule\n", + "W_alpha = (C_L-Co)/(C_L-C_alpha)\n", + "print(\"weight fraction of alpha in the alloy is {0:.4f}\".format(W_alpha*100))\n", + "W_L = 1-W_alpha\n", + "print(\"weight fraction of liquid phase in the alloy is {0:.4f}\".format(W_L*100))\n", + "\n", + "# Given\n", + "# at 183.5 degree celsius\n", + "print(\"At 183.5 degree celsius\")\n", + "C_L = 61.9 # CL=50% Sn\n", + "C_alpha = 19.2 # C_alpha=18% Sn\n", + "Co = 40.0 # solidification of alloy\n", + "\n", + "# Calculation and Results\n", + "# lever rule\n", + "W_alpha = (C_L-Co)/(C_L-C_alpha)\n", + "print(\"weight fraction of alpha in the alloy is {0:.4f}\".format(W_alpha*100))\n", + "W_L = 1-W_alpha\n", + "print(\"weight fraction of liquid phase in the alloy is {0:.4f}\".format(W_L*100))\n", + "\n", + "# Given\n", + "# at 182.5 degree celsius\n", + "print(\"At 182.5 degree celsius\")\n", + "C_beta = 97.5 # CL=50% Sn\n", + "C_alpha = 19.2 # C_alpha=18% Sn\n", + "Co = 40.0 # solidification of alloy\n", + "\n", + "# Calculations and Results\n", + "# lever rule\n", + "W_alpha = (C_beta-Co)/(C_beta-C_alpha)\n", + "print(\"weight fraction of alpha in the alloy is {0:.4f}\".format(W_alpha*100))\n", + "W_beta = 1-W_alpha\n", + "print(\"weight fraction of beta phase in the alloy is {0:.4f}\".format(W_beta*100))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "At 210 degree celsius\n", + "weight fraction of alpha in the alloy is 31.2500\n", + "weight fraction of liquid phase in the alloy is 68.7500\n", + "At 183.5 degree celsius\n", + "weight fraction of alpha in the alloy is 51.2881\n", + "weight fraction of liquid phase in the alloy is 48.7119\n", + "At 182.5 degree celsius\n", + "weight fraction of alpha in the alloy is 73.4355\n", + "weight fraction of beta phase in the alloy is 26.5645\n" + ] + } + ], + "prompt_number": 12 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch2.ipynb b/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch2.ipynb new file mode 100755 index 00000000..0bb2ce2c --- /dev/null +++ b/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch2.ipynb @@ -0,0 +1,628 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:bfb56d32f2fccdba55a3a9ca27883b3fb3f7cd5c9e884d2c3ad898549e047a85" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2 : Electrical and thermal conduction in solids" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.2 Page No : 43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "sigma = 5.9*10**5 # in ohm**-1*cm**-1\n", + "e = 1.6*10**-19 # Coulombs\n", + "d = 8.93 # g/cm**3\n", + "Mat = 63.5 # g/mol\n", + "NA = 6.02*10**23 # mol**-1\n", + "\n", + "# Calculations and Results\n", + "n = d*NA/Mat\n", + "u_d = sigma/(e*n) # electron drift mobility\n", + "print(\"Drift mobility of electrons in copper at room temperature in cm2/V/s is {0:.4f}\".format(u_d))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Drift mobility of electrons in copper at room temperature in cm2/V/s is 43.5570\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.3 Page No : 49" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "u_d = 3.2*10**-3 # in m**2/V/s\n", + "u = 1.2*10**6 # m/s\n", + "v_dx = 0.1*u\n", + "\n", + "# Calculations and Results\n", + "# drift velocity of conduction electrons is v_dx=u_d*E\n", + "E = v_dx/u_d\n", + "print(\"Applied electric field in V/m is {0:.4g}\".format(E))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Applied electric field in V/m is 3.75e+07\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.4 Page No : 50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "T_summer = 20.0 # in degree celsius\n", + "T_summer += 273.0 # in kelvin\n", + "T_winter = -30.0 # in degree celsius\n", + "T_winter += 273.0 # in kelvin\n", + "\n", + "# Calculations and Results\n", + "# we have R is proportional to A*T\n", + "# Hence\n", + "R = (T_summer-T_winter)/T_summer\n", + "R *= 100\n", + "print(\" Percentage change in the resistance of a pure metal wire from Saskatchewans summer \"\n", + " \"too winter in % is {0:.4f}\".format(R))\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Percentage change in the resistance of a pure metal wire from Saskatchewans summer too winter in % is 17.0648\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.5 Page No : 52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "d = 8.96*10**3 # in Kg/m3\n", + "NA = 6.02*10**23 # mol**-1\n", + "Mat = 63.56*10**-3 # Kg/mol\n", + "k = 1.38*10**-23 # J/K\n", + "T = 300.0 # kelvin\n", + "e = 1.6*10**-19 # in coulombs\n", + "m_e = 9.1*10**-31 # in Kg\n", + "u = 1.25*10**6 # m/s\n", + "f = 4*10**12 # frequency in s**-1\n", + "\n", + "# Calculations and Results\n", + "Ns = d*NA/Mat # atomic concentration in m**-3\n", + "M = Mat/NA\n", + "w = 2*math.pi*f # angular frequency of the vibration\n", + "# by virtue of Equipartition of energy theorem\n", + "a = math.sqrt((2*k*T)/(M*w**2))\n", + "S = math.pi*a**2 # cross sectional area\n", + "t = 1.0/(S*u*Ns) # mean free time\n", + "u_d = e*t/m_e # drift velocity\n", + "u_d *= 10**4 # change in units\n", + "Ns /= 10**6 # in cm**-3\n", + "sigma = e*Ns*u_d # conductivity\n", + "print(\"drift velocity of electrons in m2/V/s is {0:.4f}\".format(u_d))\n", + "print(\"conductivity of copper in ohm**-1/cm is {0:.4g}\".format(sigma))\n", + "# slight change in the answer is due to the computation method, otherwise answer is matching with textbook\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "drift velocity of electrons in m2/V/s is 42.4947\n", + "conductivity of copper in ohm**-1/cm is 5.77e+05\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.7 Page No : 53" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "n = 1.2\n", + "To = 293.0 # in kelvin\n", + "\n", + "# Calculations and Results\n", + "alpha_o = n/To\n", + "print(\"Theoretical value of TCR at 293K is {0:.4f} which is in \"\n", + " \"well agreement with experimental value\".format(alpha_o))\n", + "alpha_o = 0.00393 # experimental value\n", + "n = alpha_o*To\n", + "print(\"Theoretical value of n at 293K (n = {0:.4f}) is in well agreement with experimental value\".format(n))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Theoretical value of TCR at 293K is 0.0041 which is in well agreement with experimental value\n", + "Theoretical value of n at 293K (n = 1.1515) is in well agreement with experimental value\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.9 Page No : 58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Given\n", + "P = 40.0 # in Watt\n", + "V = 120.0 # in Volts\n", + "D = 33*10**-6 # in meter\n", + "L = 0.381 # in meter\n", + "To = 293.0 # in kelvin\n", + "P_radiated = 40.0 # in watt\n", + "epsilon = 0.35\n", + "sigma_s = 5.6*10**-8 # in W/m2/K4\n", + "\n", + "# Calculations and Results\n", + "I = P/V\n", + "A = math.pi*D**2/4.0\n", + "R = V/I # resistance of the filament\n", + "p_t = R*A/L # resistivity of tungsten\n", + "p_o = 5.51*10**-8 # resistivity at room temperature in ohm*m\n", + "# p_t=p_o*(T/To)**1.2\n", + "T = To*(p_t/p_o)**(1.0/1.2)\n", + "print(\"Temperature of the bulb when it is operated at the rated voltage in Kelvin is {0:.4f}\".format(T))\n", + "A = L*math.pi*D\n", + "# Stefans Law\n", + "T = (P_radiated/(epsilon*sigma_s*A))**(1.0/4.0)\n", + "print(\"Temperature of the filament in kelvin is {0:.4f}\".format(T))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature of the bulb when it is operated at the rated voltage in Kelvin is 2746.7599\n", + "Temperature of the filament in kelvin is 2681.0430\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.10 Page No : 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "M_Au = 197.0\n", + "w = 0.1\n", + "M_Cu = 63.55\n", + "p_exp = 108.0 # n*ohm*m\n", + "\n", + "# Calculations and Results\n", + "X = M_Au*w/((1-w)*M_Cu+(w*M_Au))\n", + "C = 450.0 # n*ohm*m\n", + "p_Au = 22.8 # resistivity in n*ohm*m\n", + "p = p_Au+C*X*(1-X) # Nordheim rule\n", + "x = ((p-p_exp)/p)*100\n", + "print(\"resistivity of the alloy in n*ohm*m is {0:.4f}\".format(p))\n", + "print(\"The difference in the value from experimental value in % is {0:.4f}\".format(x))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistivity of the alloy in n*ohm*m is 108.5513\n", + "The difference in the value from experimental value in % is 0.5079\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.11 Page No : 68" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "u = 1.58*10**6 # in m/s\n", + "N = 8.5*10**28 # m**-3\n", + "e = 1.6*10**-19 # in coulombs\n", + "me = 9.1*10**-31 # in Kg\n", + "N_I = 0.01*N\n", + "\n", + "# Calculations and Results\n", + "l_I = N_I**(-1/3)\n", + "t_I = l_I/u\n", + "p = me/(e**2*N*t_I)\n", + "print(\" worst case resistivity in ohm*m {0:.4g}\".format(p))\n", + "# slight change in answer due to computational method\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " worst case resistivity in ohm*m 5.616e+11\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.13 Page No : 74" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "Xd = 0.15\n", + "p_c = 1*10**-7 # ohm*m\n", + "\n", + "# Calculations and Results\n", + "p_eff = p_c*((1+0.5*Xd)/(1-Xd))\n", + "print(\"Effective resistivity in ohm m is {0:.4g}\".format(p_eff))\n", + "# slight change in the answer due to printing the answer\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Effective resistivity in ohm m is 1.265e-07\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.14 Page No : 75" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "Xd = 0.15\n", + "p_c = 4*10**-8 # ohm*m\n", + "\n", + "# Calculations and Results\n", + "p_eff = p_c*((1+0.5*Xd)/(1-Xd))\n", + "print(\"Effective resistivity in ohm m is {0:.4g}\".format(p_eff))\n", + "# change in the answer due to coding\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Effective resistivity in ohm m is 5.059e-08\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.16 Page No : 76" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "# at f=10MHz\n", + "a = 10**-3 # in m\n", + "f = 10*10**6 # in Hz\n", + "w = 2*math.pi*f\n", + "sigma_dc = 5.9*10**7 # in m**-1\n", + "u = 1.257*10**-6 # in Wb/A/m\n", + "\n", + "# Calculations and Results\n", + "delta = 1.0/math.sqrt(0.5*w*sigma_dc*u)\n", + "# let r=r_ac/r_dc=a/(2*delta)\n", + "r = a/(2*delta)\n", + "print(\"Change in dc resistance of a copper wire at 10MHz is {0:.4f}\".format(r))\n", + "# part(b)\n", + "f = 1*10**9 # in Hz\n", + "w = 2*math.pi*f\n", + "delta = 1.0/math.sqrt(0.5*w*sigma_dc*u)\n", + "# let r=r_ac/r_dc=a/(2*delta)\n", + "r = a/(2*delta)\n", + "print(\"Change in dc resistance of a copper wire at 1GHz is {0:.4f}\".format(r))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Change in dc resistance of a copper wire at 10MHz is 24.1345\n", + "Change in dc resistance of a copper wire at 1GHz is 241.3452\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.18 Page No : 79" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "sigma = 5.9*10**7 # ohm**-1*m**-2\n", + "RH = -0.55*10**-10 # m**3/A/s\n", + "\n", + "# Calculations and Results\n", + "u_d = -RH*sigma\n", + "print(\"drift mobility of electrons in copper in m2/V/s is {0:.4g}\".format(u_d))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "drift mobility of electrons in copper in m2/V/s is 0.003245\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.19 Page No : 80" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "no = 8.5*10**28 # in m3\n", + "e = 1.6*10**-19 # in coulombs\n", + "u_d = 3.2*10**-3 # m2/V/s\n", + "sigma = 5.9*10**7 # in ohm**-1*m**-1\n", + "\n", + "# Calculations and Results\n", + "n = sigma/(e*u_d)\n", + "print(\"concentration of conduction electrons in copper in m**-3 is {0:.4g}\".format(n))\n", + "A = n/no\n", + "print(\"Average number of electrons contributed per atom is {0:.4f}\".format(A))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "concentration of conduction electrons in copper in m**-3 is 1.152e+29\n", + "Average number of electrons contributed per atom is 1.3557\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.20 Page No : 85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "sigma = 1*10**7 # ohm**-1*m**-1\n", + "T = 300.0 # kelvin\n", + "C_WFL = 2.44*10**-8 # W*ohm/K2\n", + "X_d = 0.15\n", + "\n", + "# Calculations and Results\n", + "K_c = sigma*T*C_WFL\n", + "K_eff = K_c*((1-X_d)/(1+0.5*X_d))\n", + "print(\"Thermal Conductivity at room temperature in W/m/K is {0:.4f}\".format(K_eff))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thermal Conductivity at room temperature in W/m/K is 57.8791\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.21 Page No : 89" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "# Chapter2\n", + "# Ex_21\n", + "# Given\n", + "sigma = 50*10**-9 # in ohm\n", + "T = 300.0 # kelvin\n", + "C_WFL = 2.45*10**-8 # in W*ohm/K2\n", + "L = 30*10**-3 # in m\n", + "d = 20*10**-3 # in m\n", + "Q = 10.0 # in W\n", + "\n", + "# Calculations and Results\n", + "# Wiedemann-Franz Lorenz Law\n", + "k = sigma**-1*T*C_WFL # thermal conductivity\n", + "A = math.pi*(d**2)/4.0\n", + "theta = L/(k*A) # thermal resistance\n", + "delta_T = theta*Q\n", + "print(\"Temperature drop across the disk in degree celsius is {0:.4f}\".format(delta_T))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature drop across the disk in degree celsius is 6.4961\n" + ] + } + ], + "prompt_number": 15 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch3.ipynb b/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch3.ipynb new file mode 100755 index 00000000..283128fd --- /dev/null +++ b/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch3.ipynb @@ -0,0 +1,731 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a1c57dcea36d3741ad4c663237ed74b9446d043dd515ecab430a8c5aaad770ab" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3 : Elementary Quantum Physics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.1 Page No : 101" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "lambda_val = 450*10**-9 # in nm\n", + "h = 6.6*10**-34 # in J s\n", + "e = 1.6*10**-19 # in coulombs\n", + "c = 3*10**8 # in m/s\n", + "\n", + "# Calculations and Results\n", + "E_ph = h*c/lambda_val # in J\n", + "E_ph /= e # in eV\n", + "print(\" Energy of blue photon in eV is {0:.4f}\".format(E_ph))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Energy of blue photon in eV is 2.7500\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.2 Page No : 103" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "lambda_o = 522*10**-9 # in nm\n", + "lambda_val = 250*10**-9 # in nm\n", + "h = 6.6*10**-34 # in J s\n", + "c = 3*10**8 # in m/s\n", + "e = 1.6*10**-19 # in coulombs\n", + "I = 20*10**-3 # in W/cm2\n", + "I = 20*10**-3*10**4 # in J/s/m2\n", + "\n", + "# Calculations and Results\n", + "# part(a)\n", + "phi = h*c/(lambda_o*e) # in eV\n", + "print(\"Work function of sodium in eV is {0:.4f}\".format(phi))\n", + "KE = h*c/(lambda_val*e)-phi\n", + "print(\"Kinetic energy of photo emitted electrons in eV is {0:.4f}\".format(KE))\n", + "J = (e*I*lambda_val)/(h*c)\n", + "print(\"Photoelectric current density in A/m2 is {0:.4f}\".format(J))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Work function of sodium in eV is 2.3707\n", + "Kinetic energy of photo emitted electrons in eV is 2.5793\n", + "Photoelectric current density in A/m2 is 40.4040\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.4 Page No : 104" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "theta = 15.2 # in degree\n", + "d = 0.234 # in nm\n", + "V = 100.0 # in V\n", + "\n", + "# Calculations and Results\n", + "lambda_val = 2*d*math.sin(math.radians(theta)) # Braggs condition\n", + "print(\"Wavelength of electron in nm is {0:.4f}\".format(lambda_val))\n", + "lambda_val = 1.226/math.sqrt(V) # debroglie wavelength in nm\n", + "print(\"de Broglie Wavelength of electron in nm is {0:.4f}\".format(lambda_val))\n", + "print(\"de Broglie Wavelength is in excellent agreement with that determined from Braggs condition\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of electron in nm is 0.1227\n", + "de Broglie Wavelength of electron in nm is 0.1226\n", + "de Broglie Wavelength is in excellent agreement with that determined from Braggs condition\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.5 Page No : 105" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "h = 6.6*10**-34 # in J s\n", + "c = 3*10**8 # in m/s\n", + "n = 1.0\n", + "m = 9.1*10**-31 # in Kg\n", + "a = 0.1*10**-9 # in m\n", + "e = 1.6*10**-19 # in coulombs\n", + "\n", + "# Calculations and Results\n", + "E1 = (h**2*n**2)/(8*m*a**2)\n", + "E1 /= e # in eV\n", + "print(\"Ground Energy of the electron in J is {0:.4f}\".format(E1))\n", + "# part(b)\n", + "n = 3.0\n", + "E3 = E1*n**2\n", + "print(\"Energy required to put the electrons in third energy level in eV is {0:.4f}\".format(E3))\n", + "E = E3-E1\n", + "print(\"Energy required to take the electron from E1 to E3 in eV is {0:.4f}\".format(E))\n", + "lambda_val = h*c/(E*e)\n", + "print(\"wavelength of the required photon in nm is {0:.4g}\".format(lambda_val))\n", + "print(\"which is an X-ray photon\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ground Energy of the electron in J is 37.3970\n", + "Energy required to put the electrons in third energy level in eV is 336.5728\n", + "Energy required to take the electron from E1 to E3 in eV is 299.1758\n", + "wavelength of the required photon in nm is 4.136e-09\n", + "which is an X-ray photon\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.6 Page No : 110" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "h = 6.6*10**-34 # in J s\n", + "c = 3*10**8 # in m/s\n", + "n = 1.0\n", + "m = 0.1 # in Kg\n", + "a = 1.0 # in m\n", + "\n", + "# Calculations and Results\n", + "E1 = (h**2*n**2)/(8*m*a**2)\n", + "v = math.sqrt(2*E1/m)\n", + "print(\"Minimum speed of the object in m/s {0:.4g}\".format(v))\n", + "# calculation of quantum number n\n", + "v = 1.0 # in m/s\n", + "E_n = m*v**2/2.0\n", + "n = math.sqrt((8*m*a**2*E_n)/h**2)\n", + "print(\"Quantum number if the object is moving with a minimum speed of 1m/s is {0:.4g}\".format(n))\n", + "delta_E = (h**2/(8*m*a**2))*(2*n+1) # delta_E=E_n+1-En\n", + "print(\"Separation of energy levels of the object moving with speed of 1 m/s in Joules is {0:.4g}\".format(delta_E))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum speed of the object in m/s 3.3e-33\n", + "Quantum number if the object is moving with a minimum speed of 1m/s is 3.03e+32\n", + "Separation of energy levels of the object moving with speed of 1 m/s in Joules is 3.3e-34\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.8 Page No : 114" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "h_bar = 1.054*10**-34 # in J s\n", + "delta_x = 0.1*10**-9 # in m\n", + "m_e = 9.1*10**-31 # in Kg\n", + "\n", + "# Calculations and Results\n", + "delta_Px = h_bar/delta_x\n", + "print(\"uncertainty in momentum in Kg m/s is {0:.4g}\".format(delta_Px))\n", + "delta_v = delta_Px/m_e\n", + "KE = delta_Px**2/(2*m_e)\n", + "print(\"Uncertainty in Kinetic Energy in J is {0:.4g}\".format(KE))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "uncertainty in momentum in Kg m/s is 1.054e-24\n", + "Uncertainty in Kinetic Energy in J is 6.104e-19\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.9 Page No : 115" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "h_bar = 1.054*10**-34 # in J s\n", + "delta_x = 1.0 # in m\n", + "m = 0.1 # in Kg\n", + "\n", + "# Calculations and Results\n", + "delta_Px = h_bar/delta_x\n", + "delta_v = delta_Px/m\n", + "print(\"minimum uncertainty in the velocity in m/s is {0:.4g}\".format(delta_v))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "minimum uncertainty in the velocity in m/s is 1.054e-33\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.10 Page No : 121" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "h_bar = 1.054*10**-34 # in J s\n", + "m = 9.1*10**-31 # in Kg\n", + "e = 1.6*10**-19 # in coulombs\n", + "Vo = 10.0 # in ev\n", + "Vo *= e # in J\n", + "E = 7.0 # in eV\n", + "E *= e # in J\n", + "a = 5*10**-9 # in m\n", + "\n", + "# Calculations and Results\n", + "alpha = math.sqrt(2*m*(Vo-E)/h_bar**2)\n", + "To = 16*E*(Vo-E)/Vo**2\n", + "T = To*math.exp(-2*alpha*a)\n", + "print(\"Transmission coefficient of conduction electrons in copper is {0:.4g}\".format(T))\n", + "a = 1*10**-9 # in m\n", + "T = To*math.exp(-2*alpha*a)\n", + "print(\"Transmission coefficient if the oxide barrier is 1 nm is {0:.4g}\".format(T))\n", + "# slight change in the answer due to approximations in alpha value\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Transmission coefficient of conduction electrons in copper is 1.033e-38\n", + "Transmission coefficient if the oxide barrier is 1 nm is 6.666e-08\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.11 Page No : 122" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "h_bar = 1.054*10**-34 # in J s\n", + "m = 100.0 # in Kg\n", + "g = 10.0 # in m/s2\n", + "h = 10.0 # in m\n", + "h1 = 15.0 # in m\n", + "a = 10.0 # in m\n", + "E = m*g*h # total energy of carriage\n", + "Vo = m*g*h1 # PE required to reach the peak\n", + "\n", + "# Calculations and Results\n", + "alpha = math.sqrt(2*m*(Vo-E)/h_bar**2)\n", + "To = 16*E*(Vo-E)/Vo**2\n", + "T = To*math.exp(-2*alpha*a)\n", + "print(\"Transmission probability is {0:.4g}\".format(T))\n", + "h_bar = 10*10**3 # Js\n", + "alpha = math.sqrt(2*m*(Vo-E)/h_bar**2)\n", + "D = Vo**2/(4*E*(Vo-E))\n", + "T = (1+(math.sinh(alpha*a))**2)**-1\n", + "print(\"transmission probability in a universe where h_bar is 10KJs is {0:.4g}\".format(T))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Transmission probability is 0\n", + "transmission probability in a universe where h_bar is 10KJs is 0.42\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.12 Page No : 124" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "x = 9\n", + "\n", + "# Calculation and Results\n", + "for n1 in range(1, x+1):\n", + " for n2 in range(1, x+1):\n", + " for n3 in range(1, x+1):\n", + " y = n1**2+n2**2+n3**2 # let y=N**2=n1**2+n2**2+n3**2\n", + " if y == 41:\n", + " print('{0}\\t{1}\\t{2}'.format(n1, n2, n3))\n", + "print(\"Thus there are nine possible states\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1\t2\t6\n", + "1\t6\t2\n", + "2\t1\t6\n", + "2\t6\t1\n", + "3\t4\t4\n", + "4\t3\t4\n", + "4\t4\t3\n", + "6\t1\t2\n", + "6\t2\t1\n", + "Thus there are nine possible states\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.13 Page No : 126" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "h = 6.6*10**-34 # in J s\n", + "c = 3*10**8 # in m/s\n", + "m = 9.1*10**-31 # in Kg\n", + "e = 1.6*10**-19 # in coulombs\n", + "v = 2.1*10**6 # in m/s\n", + "E = m*v**2/2.0 # in J\n", + "E /= e # in eV\n", + "E1 = -13.6 # in eV\n", + "\n", + "# Calculations and Results\n", + "# change in the energy is E=En-E1\n", + "n = math.sqrt(-13.6/(E+E1))\n", + "print(\" the electron gets excited to {0:.4f} level\".format(n))\n", + "n = 3.0\n", + "E3 = -13.6/n**2\n", + "delta_E31 = E3-E1 # in eV\n", + "delta_E31 *= e # in J\n", + "lambda_31 = h*c/delta_E31\n", + "print(\"wavelength of emitted radiation from n=3 to n=1 in nm is {0:.4f}\".format(lambda_31*10**9))\n", + "# Another probability is transition fromm n=3 to n=2\n", + "n = 2.0\n", + "E2 = -13.6/n**2\n", + "delta_E32 = E3-E2 # in eV\n", + "delta_E32 *= e # in J\n", + "lambda_32 = h*c/delta_E32\n", + "print(\"wavelength of emitted radiation from n=3 to n=2 in nm is {0:.4f}\".format(lambda_32*10**9))\n", + "# Another probability is transition fromm n=2 to n=1\n", + "E2 = -13.6/n**2\n", + "delta_E21 = E2-E1 # in eV\n", + "delta_E21 *= e # in J\n", + "lambda_21 = h*c/delta_E21\n", + "print(\"wavelength of emitted radiation from n=2 to n=1 in nm is {0:.4f}\".format(lambda_21*10**9))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the electron gets excited to 3.5835 level\n", + "wavelength of emitted radiation from n=3 to n=1 in nm is 102.3667\n", + "wavelength of emitted radiation from n=3 to n=2 in nm is 655.1471\n", + "wavelength of emitted radiation from n=2 to n=1 in nm is 121.3235\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.14 Page No : 127" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "Z = 2.0\n", + "n = 1.0\n", + "\n", + "# Calculations and Results\n", + "E1 = -Z**2*13.6/n**2\n", + "E1 = abs(E1)\n", + "print(\"Energy required to ionize He+ further in eV is {0:.4f}\".format(E1))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy required to ionize He+ further in eV is 54.4000\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.15 Page No : 130" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "Z = 1.0\n", + "n1 = 2.0\n", + "n2 = 3.0\n", + "R_inf = 1.0974*10**7 # in m**-1\n", + "\n", + "# Calculations and Results\n", + "# Let x=1/lambda\n", + "x = R_inf*Z**2*((1/n1**2)-(1/n2**2))\n", + "lambda_val = 1.0/x\n", + "print(\"Wavelength of first spectral line in Angstroms is {0:.4f}\".format(lambda_val*10**10))\n", + "n1 = 2.0\n", + "n2 = 4.0\n", + "x = R_inf*Z**2*((1/n1**2)-(1/n2**2))\n", + "lambda_val = 1.0/x\n", + "print(\"Wavelength of second spectral line in Angstroms is {0:.4f}\".format(lambda_val*10**10))\n", + "print(\"These spectral lines correspond to H_alpha and H_beta lines of Hydrogen\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of first spectral line in Angstroms is 6560.9623\n", + "Wavelength of second spectral line in Angstroms is 4859.9721\n", + "These spectral lines correspond to H_alpha and H_beta lines of Hydrogen\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.16 Page No : 133" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "h = 6.6*10**-34 # in J s\n", + "e = 1.6*10**-19 # in coulombs\n", + "E1 = 13.6 # in eV\n", + "E1 *= e # in J\n", + "Z = 1.0\n", + "n1 = 109.0\n", + "n2 = 110.0\n", + "ao = 52.918*10**-12 # in m\n", + "\n", + "# Calculations and Results\n", + "v = Z**2*E1*((1.0/n1**2)-(1.0/n2**2))/h\n", + "print(\"Frequency of radiation in MHz is {0:.4f}\".format(v*10**-6))\n", + "print(\"The frequency of radiation in the transition from n1=109 to n2=110 is same as that \"\n", + " \"of the detected frequency .Hence, the radiation comes from excited hydrogen atoms in\"\n", + " \" the give transition\")\n", + "x = 2*n2**2*ao\n", + "print(\"The size of the atom in micro meter is {0:.4f}\".format(x*10**6))\n", + "# slight difference in the answer is due to approximations\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Frequency of radiation in MHz is 5022.5088\n", + "The frequency of radiation in the transition from n1=109 to n2=110 is same as that of the detected frequency .Hence, the radiation comes from excited hydrogen atoms in the give transition\n", + "The size of the atom in micro meter is 1.2806\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.20 Page No : 139" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "P_out = 2.5*10**-3 # in Watt\n", + "I = 5*10**-3 # in Amp\n", + "V = 2000.0 # in volts\n", + "\n", + "# Calculations and Results\n", + "P_in = V*I\n", + "E = (P_out/P_in)*100\n", + "print(\"Efficiency of the laser in % is {0:.4f}\".format(E))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Efficiency of the laser in % is 0.0250\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.21 Page No : 142" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "lambda_o = 632.8*10**-9 # in m\n", + "c = 3*10**8 # in m/s\n", + "T = 127.0 # in degree celsius\n", + "T += 273.0 # in Kelvin\n", + "m_A = 20.2*10**-3 # in Kg/mol\n", + "NA = 6.023*10**23 # mol**-1\n", + "k = 1.38*10**-23 # in J/K\n", + "\n", + "# Calculations and Results\n", + "m = m_A/NA # in Kg\n", + "vx = math.sqrt(k*T/m)\n", + "vo = c/lambda_o\n", + "delta_v = 2*vo*vx/c\n", + "print(\"delta_v in GHz is {0:.4g}\".format(delta_v))\n", + "delta_lambda = delta_v*(-lambda_o/vo)\n", + "print(\"delta_lambda in meters is {0:.4g}\".format(abs(delta_lambda)))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "delta_v in GHz is 1.282e+09\n", + "delta_lambda in meters is 1.711e-12\n" + ] + } + ], + "prompt_number": 19 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch4.ipynb b/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch4.ipynb new file mode 100755 index 00000000..0bd1d8ee --- /dev/null +++ b/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch4.ipynb @@ -0,0 +1,379 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:964094f6f27ad1a01c6c3e380eaf82de54433e63e5c47ff1d4d1fcbe0e988294" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4 : Bonding the Band Theory of Solids and Statistics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.6 Page No : 154" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "E_FO = 7.0 # in eV\n", + "e = 1.6*10**-19 # in coulombs\n", + "E_FO *= e # in Joules\n", + "me = 9.1*10**-31 # in Kg\n", + "\n", + "# Calculations and Results\n", + "v_f = math.sqrt(2*E_FO/me)\n", + "print(\"Speed of the conduction electrons in m/s is {0:.4g}\".format(v_f))\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Speed of the conduction electrons in m/s is 1.569e+06\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.7 Page No : 159" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "e = 1.6 * 10 ** -19 # in coulombs\n", + "Eg = 1.1 # in eV\n", + "Eg *= e # in Joules\n", + "h = 6.6 * 10 ** -34 # in Js\n", + "c = 3 * 10 ** 8 # in m/s\n", + "\n", + "# Calculations and Results\n", + "lambda_val = h * c / Eg\n", + "print(\"Wavelength of light that can be absorbed by an Si photo-detector\"\n", + " \" at Eg=1.1 eV in micro meter is {0:.4g}\".format(lambda_val * 10 ** 6))\n", + "print(\"Hence the light of wavelength 1.31 micro meter and 1.55 micro meter will not \"\n", + " \"be absorbed by Si and thus cannot be detected by detector\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of light that can be absorbed by an Si photo-detector at Eg=1.1 eV in micro meter is 1.125\n", + "Hence the light of wavelength 1.31 micro meter and 1.55 micro meter will not be absorbed by Si and thus cannot be detected by detector\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.8 Page No : 162" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "e = 1.6*10**-19 # in coulombs\n", + "h = 6.626*10**-34 # in Js\n", + "me = 9.1*10**-31 # in Kg\n", + "# let x=k*T\n", + "x = 0.026 # in eV\n", + "E = 5.0 # in ev\n", + "E *= e # in Joules\n", + "\n", + "# Calculations and Results\n", + "g_E = (8*math.pi*math.sqrt(2))*(me/h**2)**(3./2)*math.sqrt(E) # in J**-1*m**-3\n", + "# conversion of units\n", + "g_E = g_E*10**-6*e # in eV**-1 cm**-3\n", + "print(\"density of states at the center of the band in cm**-3*J**-1 is {0:.4g}\".format(g_E))\n", + "# part(b)\n", + "n_E = g_E*x # in cm**-3\n", + "print(\"No.of states per unit volume within kT about the center in cm**-3 is {0:.4g}\".format(n_E))\n", + "# part(c)\n", + "# Given\n", + "E = 0.026 # in eV\n", + "E *= e # in joules\n", + "\n", + "# Calculations and Results\n", + "g_E = (8*math.pi*math.sqrt(2))*(me/h**2)**(3./2)*math.sqrt(E) # in J**-1*m**-3\n", + "# conversion of units\n", + "g_E = g_E*10**-6*e # in eV**-1 cm**-3\n", + "print(\"density of states at at kT above the band in cm**-3*J**-1 is {0:.4g}\".format(g_E))\n", + "# part(d)\n", + "n_E = g_E*x # in cm**-3\n", + "print(\" No.of states per unit volume within kT about the center in cm**-3 is {0:.4g}\".format(n_E))\n", + "# solved using the values taken from the solution of textbook\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "density of states at the center of the band in cm**-3*J**-1 is 1.518e+22\n", + "No.of states per unit volume within kT about the center in cm**-3 is 3.946e+20\n", + "density of states at at kT above the band in cm**-3*J**-1 is 1.095e+21\n", + " No.of states per unit volume within kT about the center in cm**-3 is 2.846e+19\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.9 Page No : 165" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Given\n", + "e = 1.6*10**-19 # in coulombs\n", + "h = 6.626*10**-34 # in Js\n", + "me = 9.1*10**-31 # in Kg\n", + "d = 10.5 # in g/cm\n", + "Mat = 107.9 # g/mol\n", + "NA = 6.023*10**23 # mol**-1\n", + "E_ctr = 5.0 # in ev\n", + "E_ctr *= e # in Joules\n", + "\n", + "# Calculations and Results\n", + "S_band = 2*(16*math.pi*math.sqrt(2)/3)*(me/h**2)**(3./2)*E_ctr**(3./2) # in states m**-3\n", + "# conversion of units\n", + "S_band *= 10**-6 # in states cm**-3\n", + "print(\"No. of states in the band in states cm**-3 is {0:.4g}\".format(S_band))\n", + "n_Ag = d*NA/Mat\n", + "print(\"No.of atoms per unit volume in silver in atoms per cm3 is {0:.4g}\".format(n_Ag))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "No. of states in the band in states cm**-3 is 1.012e+23\n", + "No.of atoms per unit volume in silver in atoms per cm3 is 5.861e+22\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.10 Page No : 169" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Given\n", + "e = 1.6*10**-19 # in coulombs\n", + "h = 6.626*10**-34 # in Js\n", + "me = 9.1*10**-31 # in Kg\n", + "d = 8.96 # in g/cm\n", + "Mat = 63.5 # g/ mol\n", + "NA = 6.023*10**23 # mol**-1\n", + "n = d*NA/Mat # in cm**-3\n", + "n *= 10**6 # in m**-3\n", + "\n", + "# Calculations and Results\n", + "E_FO = (h**2/(8*me))*(3*n/math.pi)**(2./3) # in J\n", + "E_FO /= e # in eV\n", + "print(\"Fermi energy at 0 Kelvin in eV is {0:.4f}\".format(E_FO))\n", + "E_FO = (h**2/(8*me))*(3*n/math.pi)**(2./3) # in J\n", + "v_e = math.sqrt(6*E_FO/(5*me))\n", + "print(\"Average speed of conduction electrons in m/s is {0:.4g}\".format(v_e))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fermi energy at 0 Kelvin in eV is 7.0653\n", + "Average speed of conduction electrons in m/s is 1.221e+06\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.11 Page No : 174" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "e = 1.6*10**-19 # in coulombs\n", + "me = 9.1*10**-31 # in Kg\n", + "u_d = 43*10**-4 # in cm2/V/s\n", + "v_e = 1.22*10**6 # in m/s\n", + "\n", + "# Calculations and Results\n", + "T = u_d*me/e\n", + "l_e = v_e*T\n", + "print(\"Mean free path of electrons in meters is {0:.4g}\".format(l_e))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mean free path of electrons in meters is 2.984e-08\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.13 Page No : 178" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "# Given\n", + "e = 1.6*10**-19 # in coulombs\n", + "T = 373.0 # in kelvin\n", + "To = 273.0 # in kelvin\n", + "k = 1.38*10**-23 # in m2 kg /k/s2\n", + "# from table 4.3\n", + "E_FAO = 11.6 # in eV\n", + "E_FAO *= e # in J\n", + "x_A = 2.78\n", + "E_FBO = 7.01 # in eV\n", + "E_FBO *= e # in J\n", + "x_B = -1.79\n", + "\n", + "# Calculations and Results\n", + "# Mott jones Equation\n", + "V_AB = (-math.pi**2*k**2/(6*e))*((x_A/E_FAO)-(x_B/E_FBO))*(T**2-To**2)\n", + "print(\"EMF in micro volts available from Al and Cu thermocouple with the given respective\"\n", + " \" temperatures at the junctions is {0:.4f}\".format(V_AB*10**6))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "EMF in micro volts available from Al and Cu thermocouple with the given respective temperatures at the junctions is -391.2988\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.14 Page No : 182" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "# Given\n", + "phi = 2.6 # in eV\n", + "e = 1.6*10**-19 # in coulombs\n", + "phi *= e # in Joules\n", + "Be = 3*10**4 # schottky coefficient in A/m2/K2\n", + "T = 1600.0 # in degree celsius\n", + "T += 273.0 # in Kelvin\n", + "k = 1.38*10**-23 # m2 kg s-2 K-1\n", + "d = 2*10**-3 # in m\n", + "l = 4*10**-2 # in in m\n", + "\n", + "# Calculations and Results\n", + "# Richardson-Dushman Equation\n", + "J = Be*T**2*math.exp(-phi/(k*T))\n", + "A = math.pi*d*l\n", + "I = J*A\n", + "print(\"Saturation current in Amperes if the tube is operated at 1873 kelvin is {0:.4g}\".format(I))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Saturation current in Amperes if the tube is operated at 1873 kelvin is 2.708\n" + ] + } + ], + "prompt_number": 8 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch5.ipynb b/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch5.ipynb new file mode 100755 index 00000000..0abd7b0a --- /dev/null +++ b/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch5.ipynb @@ -0,0 +1,524 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:0a2e92e1bbe3dfa1c8302057620bde3bdeb0efb1fb47bd9ddc100369dbe575f1" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5 : Semiconductors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.1 Page No : 193" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "e = 1.6*10**-19 # in coulombs\n", + "h = 6.6*10**-34 # in J s\n", + "m = 9.1*10**-31 # in Kg\n", + "me = 1.08*m\n", + "mh = 0.56*m\n", + "T = 300.0 # in Kelvin\n", + "Eg = 1.10 # in eV\n", + "ue = 1350.0 # in cm2/V/s\n", + "uh = 450.0 # in cm2/V/s\n", + "k = 1.38*10**-23 # m2 kg s-2 K-1\n", + "\n", + "# Calculations and Results\n", + "Nc = 2*((2*math.pi*me*k*T)/h**2)**(3.0/2.0) # in m**-3\n", + "Nc *= 10**-6 # in cm**-3\n", + "Nv = 2*((2*math.pi*mh*k*T)/h**2)**(3.0/2.0) # in m**-3\n", + "Nv *= 10**-6 # in cm**-3\n", + "ni = math.sqrt(Nc*Nv)*math.exp(-Eg*e/(2*k*T))\n", + "print(\"Intrinsic concentration of Si in cm**-3 is {0:.4g}\".format(ni))\n", + "sigma = e*ni*(ue+uh)\n", + "p = 1.0/sigma\n", + "print(\"Intrinsic resistivity of Si in ohm cm is {0:.4g}\".format(p))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Intrinsic concentration of Si in cm**-3 is 1.02e+10\n", + "Intrinsic resistivity of Si in ohm cm is 3.404e+05\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.2 Page No : 194" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "# Given\n", + "T = 300.0 # in kelvin\n", + "k = 1.38*10**-23 # in m2 kg s-2 K-1\n", + "me = 9.1*10**-31 # in Kg\n", + "m = 0.26*me\n", + "\n", + "# Calculations and Results\n", + "Ve = math.sqrt(3*k*T/m)\n", + "print(\"Mean speed of electrons in conduction band in m/s is {0:.4g}\".format(Ve))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mean speed of electrons in conduction band in m/s is 2.291e+05\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.3 Page No : 195" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "e = 1.6*10**-19 # in coulombs\n", + "ue = 1350.0 # in cm2/V/s\n", + "uh = 450.0 # in cm2/V/s\n", + "ni = 1.45*10**10 # in cm**-3\n", + "L = 1.0 # in cm\n", + "A = 1.0 # in cm2\n", + "N_Si = 5*10**22 # in cm**-3\n", + "\n", + "# Calculations and Results\n", + "sigma = e*ni*(ue+uh)\n", + "R = L/(sigma*A)\n", + "print(\"Resistance of a pure Silicon crystal in ohms is {0:.4g}\".format(R))\n", + "Nd = N_Si/10**9\n", + "n = Nd # at room temperature\n", + "p = ni**2/Nd\n", + "sigma = e*n*ue\n", + "R = L/(sigma*A)\n", + "print(\"Resistance in ohms of Silicon crystal when doped with Arsenic with 1 in 10**9 is {0:.4f}\".format(R))\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Resistance of a pure Silicon crystal in ohms is 2.395e+05\n", + "Resistance in ohms of Silicon crystal when doped with Arsenic with 1 in 10**9 is 92.5926\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.4 Page No : 197" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "Na = 10**17 # acceptor atoms /cm3\n", + "Nd = 10**16 # donor atoms /cm3\n", + "p = Na-Nd # in cm**-3\n", + "ni = 1.45*10**10 # in cm**-3\n", + "\n", + "# Calculations and Results\n", + "n = ni**2/p\n", + "print(\"Electron concentration in cm**-3 is {0:.4f}\".format(n))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Electron concentration in cm**-3 is 2336.1111\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.5 Page No : 202" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "Na = 2*10**17 # acceptor atoms /cm3\n", + "Nd = 10**16 # acceptor atoms /cm3\n", + "ni = 1.45*10**10 # in cm**-3\n", + "K = 0.0259 # in eV\n", + "\n", + "# Calculations and Results\n", + "# math.since Nd>>ni\n", + "n = Nd\n", + "# let EFn-EFi=E\n", + "E = K*math.log(Nd/ni)\n", + "print(\"Position of the fermi energy w.r.t fermi energy in intrinsic Si in eV is {0:.4f}\".format(E))\n", + "# for intrinsic Si\n", + "# ni=Nc*exp(-(Ec-E_Fi)/(k*T))\n", + "# for doped Si\n", + "# Nd=Nc*exp(-(Ec-E_Fn)/(k*T))\n", + "# let x=Nd/ni\n", + "# let K=k*T\n", + "p = Na-Nd\n", + "# let E=EFp-EFi\n", + "# let n=p/ni\n", + "E = -K*math.log(p/ni)\n", + "print(\"Position of the fermi energy w.r.t fermi energy in n-type case in eV is {0:.4f}\".format(E))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Position of the fermi energy w.r.t fermi energy in intrinsic Si in eV is 0.3482\n", + "Position of the fermi energy w.r.t fermi energy in n-type case in eV is -0.4245\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.7 Page No : 208" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "Nd = 10**15 # in cm**-3\n", + "Nc = 2.8*10**19 # in cm**-3\n", + "Ti = 556.0 # in Kelvin\n", + "k = 8.62*10**-5 # in eV/K\n", + "delta_E = 0.045 # in eV\n", + "T = 300.0 # in kelvin\n", + "\n", + "# Calculations and Results\n", + "# part(a)\n", + "print(\"From fig 5.16 the estimated temperature above which the si sample behaves as if intrinsic is 556 Kelvin\")\n", + "# part(b)\n", + "Ts = delta_E/(k*math.log(Nc/(2*Nd)))\n", + "Nc_Ts = Nc*(Ts/T)**(3.0/2.0)\n", + "print(\"Lowest temperature in kelvin is {0:.4f}\".format(Ts))\n", + "# the improved temperature\n", + "Ts = delta_E/(k*math.log(Nc_Ts/(2*Nd)))\n", + "print(\"Extrinsic range of Si is {0:.4f} K to 556 K\".format(Ts))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "From fig 5.16 the estimated temperature above which the si sample behaves as if intrinsic is 556 Kelvin\n", + "Lowest temperature in kelvin is 54.6823\n", + "Extrinsic range of Si is 74.6473 K to 556 K\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.9 Page No : 214" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "e = 1.6*10**-19 # in coulombs\n", + "Nd = 10**17 # in cm**-3\n", + "Na = 9*10**16 # in cm**-3\n", + "\n", + "# Calculations and Results\n", + "# part(a)\n", + "ue1 = 800.0 # at 300 kelvin ue in cm2/V/s\n", + "sigma1 = e*Nd*ue1\n", + "ue2 = 420.0 # at 400 kelvin ue in cm2/V/s\n", + "sigma2 = e*Nd*ue2\n", + "print(\"when Si sample is doped with 10**17 arsenic atoms/cm3, the conductivity of \"\n", + " \"the sample at 300K and 400K in ohm**-1*cm**-1 is {0:.4f}, {1:.4f} respectively\".format(sigma1, sigma2))\n", + "# part(b)\n", + "ue1 = 600.0 # at 300 kelvin ue in cm2/V/s\n", + "sigma1 = e*(Nd-Na)*ue1\n", + "ue2 = 400.0 # at 400 kelvin ue in cm2/V/s\n", + "sigma2 = e*(Nd-Na)*ue2\n", + "print(\"when n-type Si is further doped with 9*10**16 boron atoms /cm3, the conductivity of\"\n", + " \" the sample at 300K and 400K in ohm**-1*cm**-1 is {0:.4f}, {1:.4f} respectively\".format(sigma1, sigma2))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "when Si sample is doped with 10**17 arsenic atoms/cm3, the conductivity of the sample at 300K and 400K in ohm**-1*cm**-1 is 12.8000, 6.7200 respectively\n", + "when n-type Si is further doped with 9*10**16 boron atoms /cm3, the conductivity of the sample at 300K and 400K in ohm**-1*cm**-1 is 0.9600, 0.6400 respectively\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.11 Page No : 218" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "# part(a)\n", + "h = 6.63*10**-34 # in Js\n", + "c = 3*10**8 # in m/s\n", + "e = 1.6*10**-19 # in coulombs\n", + "ue = 0.034 # in m2/V/s\n", + "uh = 0.0018 # in m2/V/s\n", + "t = 1*10**-3 # in seconds\n", + "L = 1*10**-3 # in m\n", + "D = 0.1*10**-3 # in m\n", + "W = 1*10**-3 # in m\n", + "I = 1.0 # mW/cm**2\n", + "I = I*10**-3*10**4 # conversion of units to W/m**2\n", + "n = 1.0 # quantum efficiency\n", + "lambda_val = 450*10**-9 # in m\n", + "V = 50.0 # in volts\n", + "\n", + "# Calculations and Results\n", + "# part(a)\n", + "A = L*W # in m3\n", + "EHP_ph = (A*n*I*lambda_val)/(h*c)\n", + "print(\"No.of EHP/s generated per second is {0:.4g}\".format(EHP_ph))\n", + "# part(b)\n", + "delta_sigma = e*n*I*lambda_val*t*(ue+uh)/(h*c*D)\n", + "print(\"Photo conductivity of the sample in ohm**-1 m**-1 is {0:.4g}\".format(delta_sigma))\n", + "# part(c)\n", + "A = 0.1*10**-6 # m2\n", + "E = V/W\n", + "delta_J = E*delta_sigma\n", + "delta_I = A*delta_J\n", + "print(\"Photo-current produced in mA is {0:.4g}\".format(delta_I*10**3))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "No.of EHP/s generated per second is 2.262e+13\n", + "Photo conductivity of the sample in ohm**-1 m**-1 is 1.296\n", + "Photo-current produced in mA is 6.48\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.13 Page No : 223" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "e = 1.6*10**-19 # in coulombs\n", + "T = 300.0 # in kelvin\n", + "ue = 1300.0 # in cm2/V/s\n", + "\n", + "# Calculations and Results\n", + "# V=k*T/e\n", + "V = 0.0259 # thermal voltage in Volts\n", + "# D=ue*k*T/e\n", + "D = ue*V\n", + "print(\"Diffusion coefficient of electrons at room temperature in cm2/s is {0:.4f}\".format(D))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diffusion coefficient of electrons at room temperature in cm2/s is 33.6700\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.17 Page No : 229" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "Eg = 1.42 # in eV\n", + "# letE=hc/lambda=hf\n", + "E = 1.96 # in eV\n", + "P_L = 50.0 # in mW\n", + "kT = 0.0259 # in eV\n", + "\n", + "# Calculations and Results\n", + "delta_E = E-(Eg+(3.0/2.0)*kT)\n", + "P_H = (P_L/E)*delta_E\n", + "print(\"Amount of power dissipated as heat in mW is {0:.4f}\".format(P_H))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Amount of power dissipated as heat in mW is 12.7844\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.18 Page No : 234" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Given\n", + "phi_m = 4.28 # in eV\n", + "e = 1.6*10**-19 # in coulombs\n", + "X = 4.01 # in eV\n", + "kT = 0.026 # in eV\n", + "Vf = 0.1 # in V\n", + "T = 300.0 # in kelvin\n", + "Be = 30.0 # A/K2/cm2\n", + "A = 0.01 # cm2\n", + "\n", + "# Calculations and Results\n", + "# part(a)\n", + "phi_B = phi_m-X\n", + "print(\"Theoretical barrier height in eV is {0:.4f}\".format(phi_B))\n", + "# part(b)\n", + "phi_B = 0.5 # in eV\n", + "Io = A*Be*T**2*math.exp(-phi_B/kT)\n", + "print(\"Saturation current in micro amperes is {0:.4f}\".format(Io*10**6))\n", + "# let/E=e*Vf #in eV\n", + "E = 0.1 # in eV\n", + "If = Io*(math.exp((E/kT))-1)\n", + "print(\"Forward current in milli amperes is {0:.4f}\".format(If*10**3))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Theoretical barrier height in eV is 0.2700\n", + "Saturation current in micro amperes is 120.1011\n", + "Forward current in milli amperes is 5.5021\n" + ] + } + ], + "prompt_number": 11 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch6.ipynb b/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch6.ipynb new file mode 100755 index 00000000..4b62c1bc --- /dev/null +++ b/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch6.ipynb @@ -0,0 +1,595 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:ba6cb48fc1189e2b468babe0bc479e8b749b88611384eb78034e3edb673abb92" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 6 : Semiconductor Devices" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.1 Page No : 248" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "# let K=kT/e\n", + "K = 0.0259 # in V\n", + "Nd = 10**17 # in cm**-3\n", + "Na = 10**16 # in cm**-3\n", + "ni_Si = 1.45*10**10 # in cm**-3\n", + "ni_Ge = 2.40*10**13 # in cm**-3\n", + "ni_GaAs = 1.79*10**6 # in cm**-3\n", + "\n", + "# Calculations and Results\n", + "# Vo=(k*T/e)*math.log(Nd*Na/ni**2)\n", + "Vo_Si = K*math.log(Nd*Na/ni_Si**2)\n", + "print(\"Built in potential for Si in Volts is {0:.4f}\".format(Vo_Si))\n", + "Vo_Ge = K*math.log(Nd*Na/ni_Ge**2)\n", + "print(\"Built in potential for Ge in Volts is {0:.4f}\".format(Vo_Ge))\n", + "Vo_GaAs = K*math.log(Nd*Na/ni_GaAs**2)\n", + "print(\"Built in potential for GaAs in Volts is {0:.4f}\".format(Vo_GaAs))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Built in potential for Si in Volts is 0.7560\n", + "Built in potential for Ge in Volts is 0.3721\n", + "Built in potential for GaAs in Volts is 1.2222\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.2 Page No : 253" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "# let K=kT/e\n", + "K = 0.0259 # in V\n", + "Na = 10**18 # in cm**-3\n", + "Nd = 10**16 # in cm**-3\n", + "e = 1.6*10**-19 # in coulombs\n", + "Eo = 8.85*10**-12 # in m-3 kg-1 s4 A2\n", + "Er = 11.9\n", + "E = Eo*Er\n", + "ni = 1.45*10**10 # in cm**-3\n", + "\n", + "# Calculations and Results\n", + "# Vo=(k*T/e)*math.log(Nd*Na/ni**2)\n", + "Vo = K*math.log(Nd*Na/ni**2)\n", + "print(\"{0:.4f}\".format(Vo))\n", + "Nd *= 10**6 # in m**-3\n", + "Wo = math.sqrt(2*E*Vo/(e*Nd))\n", + "print(\"Depletion width in micro meters is {0:.4f}\".format(Wo*10**6))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "0.8157\n", + "Depletion width in micro meters is 0.3277\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.3 Page No : 258" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "# part(a)\n", + "# let K=k*T/e\n", + "K = 0.0259 # in V\n", + "Te = 5*10**-9 # in s\n", + "Th = 417*10**-9 # in s\n", + "ue = 120.0 # in cm2/V/s\n", + "uh = 440.0 # in cm2/V/s\n", + "Na = 5*10**18 # in cm**-3\n", + "Nd = 10**16 # in cm**-3\n", + "T1 = 300.0 # in kelvin\n", + "T2 = 373.0 # in kelvin\n", + "Tg = 10**-6 # in seconds\n", + "Vr = 5.0 # in volts\n", + "ni_300 = 1.45*10**10 # in cm**-3 at 300K\n", + "ni_373 = 1.2*10**12 # in cm**-3 at 373K\n", + "A = 0.01 # in cm2\n", + "e = 1.6*10**-19 # in coulombs\n", + "epsilon_o = 8.85*10**-12 # in F/m\n", + "epsilon_r = 11.9\n", + "V = 0.6 # in v\n", + "\n", + "# Calculations and Results\n", + "# De=k*T*ue/e\n", + "De = K*ue\n", + "Dh = K*uh\n", + "Le = math.sqrt(De*Te)\n", + "Lh = math.sqrt(Dh*Th)\n", + "print(\"Diffusion length of electrons in cm is {0:.4g}\".format(Le))\n", + "print(\"Diffusion length of holes in cm is {0:.4g}\".format(Lh))\n", + "# part(b)\n", + "# Vo=(k*T/e)*math.log(Nd*Na/ni**2)\n", + "Vo = K*math.log(Nd*Na/ni_300**2)\n", + "print(\"Built-in potential in volts is {0:.4f}\".format(Vo))\n", + "# part(C)\n", + "Iso_300 = A*e*ni_300**2*Dh/(Lh*Nd)\n", + "# I=Iso*exp(eV/kT)\n", + "I = Iso_300*math.exp(V/K)\n", + "print(\"Current when there is a forward bias of 0.6 V at 300K in Amperes is {0:.4g}\".format(I))\n", + "# part(d)\n", + "Iso_373 = Iso_300*(ni_373/ni_300)**2\n", + "I = Iso_373*math.exp((V/K)*(T1/T2))\n", + "print(\"Current when there is a forward bias of 0.6 V at 373K in Amperes is {0:.4f}\".format(I))\n", + "# part(e)\n", + "Nd *= 10**6 # in m**-3\n", + "epsilon = epsilon_o*epsilon_r\n", + "W = math.sqrt(2*epsilon*(Vo+Vr)/(e*Nd))\n", + "W *= 10**2 # in cm\n", + "ni = 1.45*10**10 # in cm**-3\n", + "I_gen = e*A*W*ni/Tg\n", + "print(\"Thermal generation current in Amperes is {0:.4g}\".format(I_gen))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diffusion length of electrons in cm is 0.0001247\n", + "Diffusion length of holes in cm is 0.00218\n", + "Built-in potential in volts is 0.8574\n", + "Current when there is a forward bias of 0.6 V at 300K in Amperes is 0.002023\n", + "Current when there is a forward bias of 0.6 V at 373K in Amperes is 0.1488\n", + "Thermal generation current in Amperes is 2.037e-09\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5 Page No : 261" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "A = 10**-6 # in m2\n", + "Vo = 0.856 # in V\n", + "I = 5*10**-3 # in Amperes\n", + "Iso = 0.176*10**-12 # in Amperes\n", + "e = 1.6*10**-19 # in coulombs\n", + "Eo = 8.85*10**-12 # in m-3 kg-1 s4 A2\n", + "Er = 11.9\n", + "Th = 417*10**-9 # in seconds\n", + "Nd = 10**22 # in m**-3\n", + "# let K=kT/e\n", + "K = 0.0259 # in V\n", + "\n", + "# Calculations and Results\n", + "# Vo=(k*T/e)*math.log(I/Iso)\n", + "V = K*math.log(I/Iso)\n", + "I = 5.0 # in mA\n", + "rd = 25.0/I\n", + "print(\"Incremental diode resistance in ohms is {0:.4f}\".format(rd))\n", + "E = Eo*Er\n", + "C_dep = A*math.sqrt((e*E*Nd)/(2*(Vo-V)))\n", + "print(\"Depletion capacitance of the diode in Farads {0:.4g}\".format(C_dep))\n", + "C_diff = Th*I/25\n", + "print(\"Incremental diffusion coefficient in Farads is {0:.4g}\".format(C_diff))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Incremental diode resistance in ohms is 5.0000\n", + "Depletion capacitance of the diode in Farads 6.019e-10\n", + "Incremental diffusion coefficient in Farads is 8.34e-08\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6 Page No : 264" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "e = 1.6*10**-19 # in coulombs\n", + "Nd = 10**16 # in cm**-3\n", + "Ebr = 4*10**5 # in V/cm\n", + "epsilono = 8.85*10**-12*10**-2 # in F/cm\n", + "epsilonr = 11.9\n", + "\n", + "# Calculations and Results\n", + "epsilon = epsilono*epsilonr\n", + "Vbr = epsilon*Ebr**2/(2*e*Nd)\n", + "print(\"Reverse break down voltage of the Si diode in Volts is {0:.4f}\".format(Vbr))\n", + "# part(b)\n", + "Nd = 10**17 # in cm**-3\n", + "Ebr = 6*10**5 # in V/cm\n", + "Vbr = epsilon*Ebr**2/(2*e*Nd)\n", + "print(\"Reverse break down voltage in Volts when phosphorous doping is increased \"\n", + " \"to 10**17 cm**-3 is {0:.4f}\".format(Vbr))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Reverse break down voltage of the Si diode in Volts is 52.6575\n", + "Reverse break down voltage in Volts when phosphorous doping is increased to 10**17 cm**-3 is 11.8479\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.7 Page No : 266" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "# part(a)\n", + "Th = 250*10**-9 # in seconds\n", + "A = 0.02*10**-2 # in cm2\n", + "Av = 10.0 # voltage gain\n", + "ni = 1.45*10**10 # in cm**-3\n", + "Nd = 2*10**16 # in cm**-3\n", + "W_B = 2*10**-4 # in cm\n", + "uh = 410.0 # in cm2/V/s\n", + "I_E = 2.5*10**-3 # in Amperes\n", + "# let K=kT/e\n", + "K = 0.0259 # in V\n", + "\n", + "# Calculations and Results\n", + "# Dh=(kT/e)*uh\n", + "Dh = K*uh\n", + "Tt = W_B**2/(2*Dh)\n", + "e = 1.6*10**-19 # in coulombs\n", + "alpha = 1-(Tt/Th)\n", + "print(\"CB current transfer ratio is {0:.4f}\".format(alpha))\n", + "# funcprot(0)\n", + "beta = alpha/(1-alpha)\n", + "print(\"current gain is {0:.4f}\".format(beta))\n", + "# part(c)\n", + "I_EO = e*A*Dh*ni**2/(Nd*W_B)\n", + "# V_EB=(k*T/e)*math.log(I_E/I_EO)\n", + "V_EB = K*math.log(I_E/I_EO)\n", + "print(\"V_EB in volts is {0:.4f}\".format(V_EB))\n", + "# re=(k*T/e)/IE=25/IE(mA)\n", + "I_E = 2.5 # in mA\n", + "re = 25/I_E\n", + "print(\"small signal input resistance in ohms is {0:.4f}\".format(re))\n", + "# part(d)\n", + "R_C = Av*re\n", + "print(\"R_C in ohms is {0:.4f}\".format(R_C))\n", + "# part(e)\n", + "I_E = 2.5*10**-3 # in Amperes\n", + "I_B = I_E*(1-alpha)\n", + "print(\"base current in micro amperes is {0:.4f}\".format(I_B*10**6))\n", + "# part(f)\n", + "f = 1.0/Tt\n", + "print(\"upper frequency range limit in MHz is {0:.4f}\".format(f*10**-6))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "CB current transfer ratio is 0.9925\n", + "current gain is 131.7375\n", + "V_EB in volts is 0.6647\n", + "small signal input resistance in ohms is 10.0000\n", + "R_C in ohms is 100.0000\n", + "base current in micro amperes is 18.8342\n", + "upper frequency range limit in MHz is 530.9500\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.8 Page No : 271" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "# part(c)\n", + "Nd = 2*10**16 # in cm**-3\n", + "Na = 10**19 # in cm**-3\n", + "W_B = 2*10**-4 # in cm\n", + "W_E = 2*10**-4 # in cm\n", + "ue = 110.0 # in cm2/V/s\n", + "uh = 410.0 # in cm2/V/s\n", + "Th = 250*10**-9 # in seconds\n", + "# let K=kT/e\n", + "K = 0.0259 # in V\n", + "\n", + "# Calculations and Results\n", + "# Dh=(kT/e)*uh\n", + "Dh = K*uh\n", + "Tt = W_B**2/(2*Dh)\n", + "gamma = 1.0/(1+((Nd*W_B*ue)/(Na*W_E*uh)))\n", + "print(\"Injection frequency is {0:.4f}\".format(gamma))\n", + "alpha = gamma*(1-(Tt/Th))\n", + "print(\"Modified alpha is {0:.4f}\".format(alpha))\n", + "beta = alpha/(1-alpha)\n", + "print(\"modified current gain is {0:.4f}\".format(beta))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Injection frequency is 0.9995\n", + "Modified alpha is 0.9919\n", + "modified current gain is 122.9784\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.9 Page No : 276" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "# rms output voltage\n", + "Ic = 2.5 # in mA\n", + "Rc = 1000.0 # in ohms\n", + "beta = 100.0\n", + "vs = 1.0 # in mV\n", + "Rs = 50.0 # in ohms\n", + "\n", + "# Calculations and Results\n", + "r_be = beta*25/Ic # Ic in mA\n", + "gm = Ic/25.0 # Ic in mA\n", + "# Av=v_ce/v_be=gm*Rc\n", + "Av = gm*Rc\n", + "v_be = vs*r_be/(r_be+Rs) # in mV\n", + "v_ce = Av*v_be\n", + "print(\"rms output voltage in mV is {0:.4f}\".format(v_ce))\n", + "v_be *= 10**-3 # in volts\n", + "Ap = beta*Av\n", + "P_in = v_be**2/r_be\n", + "print(\"Input power in watts is {0:.4f}\".format(P_in*10**9))\n", + "P_out = P_in*Ap\n", + "print(\"output power in watts is {0:.4f}\".format(P_out*10**6))\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "rms output voltage in mV is 95.2381\n", + "Input power in watts is 0.9070\n", + "output power in watts is 9.0703\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.10 Page No : 278" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "V_GS = -1.5 # in Volts\n", + "V_GS_off = -5.0 # in Volts\n", + "I_DSS = 10*10**-3 # in A\n", + "R_D = 2000.0 # in ohms\n", + "\n", + "# Calculations and Results\n", + "I_DS = I_DSS*(1-(V_GS/V_GS_off))**2 # in A\n", + "gm = -2*math.sqrt(I_DSS*I_DS)/V_GS_off\n", + "Av = -gm*R_D\n", + "print(\"voltage amplification for small signal is {0:.4f}\".format(Av))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "voltage amplification for small signal is -5.6000\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.11 Page No : 283" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "Z = 50*10**-6 # in m\n", + "L = 10*10**-6 # in m\n", + "t_ox = 450*10**-10 # in m\n", + "V_GS = 8.0 # in V\n", + "V_th = 4.0 # in V\n", + "V_DS = 20.0 # in V\n", + "lambda_val = 0.01\n", + "ue = 750*10**-4 # in m2/V/s\n", + "epsilon_r = 3.9\n", + "epsilon_o = 8.85*10**-12 # F/m2\n", + "\n", + "# Calculations and Results\n", + "epsilon = epsilon_r*epsilon_o\n", + "K = (Z*ue*epsilon)/(2*L*t_ox)\n", + "I_DS = K*(V_GS-V_th)**2*(1+lambda_val*V_DS)\n", + "print(\"drain current in mA is {0:.4g}\".format(I_DS*10**3))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "drain current in mA is 2.761\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.13 Page No : 288" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "e = 1.6*10**-19 # in coulombs\n", + "I = 10**-3 # in A\n", + "Th = 10**-6 # in s\n", + "\n", + "# Calculations and Results\n", + "B = 1/Th # in Hz\n", + "i_sn = math.sqrt(2*e*I*B)\n", + "print(\"shot noise current in amperes is {0:.4g}\".format(i_sn))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "shot noise current in amperes is 1.789e-08\n" + ] + } + ], + "prompt_number": 11 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch7.ipynb b/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch7.ipynb new file mode 100755 index 00000000..99322bfb --- /dev/null +++ b/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch7.ipynb @@ -0,0 +1,576 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:bf05b593414ed786160c113f8f3bb8234f72964f8f7ff07fb2368e2be036655e" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 7 : Dielectric Materials and Insulation" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.1 Page No : 301" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "NA = 6.023*10**23 # in mol**-1\n", + "d = 1.8 # g/cm3\n", + "Mat = 39.95 # in mol**-1\n", + "epsilon_o = 8.85*10**-12 # F/m2\n", + "alpha_e = 1.7*10**-40 # F*m2\n", + "\n", + "# Calculations and Results\n", + "N = NA*d/Mat # in cm**-3\n", + "N *= 10**6 # in m**-3\n", + "epsilon_r = 1+(N*alpha_e/epsilon_o)\n", + "print(\"Dielectric constant of solid Ar is {0:.4f}\".format(epsilon_r))\n", + "# using clausius-mossotti equation\n", + "epsilon_r = (1+(2*N*alpha_e/(3*epsilon_o)))/(1-(N*alpha_e/(3*epsilon_o)))\n", + "print(\"using clausius-mossotti equation, Dielectric constant of solid Ar is {0:.4f}\".format(epsilon_r))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Dielectric constant of solid Ar is 1.5213\n", + "using clausius-mossotti equation, Dielectric constant of solid Ar is 1.6309\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.2 Page No : 307" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "N = 5*10**28 # in m**-3\n", + "e = 1.6*10**-19 # in coulombs\n", + "Z = 4.0\n", + "me = 9.1*10**-31 # in Kg\n", + "epsilon_o = 8.85*10**-12 # F/m2\n", + "epsilon_r = 11.9\n", + "\n", + "# Calculations and Results\n", + "# part(a)\n", + "alpha_e = (3*epsilon_o/N)*((epsilon_r-1)/(epsilon_r+2))\n", + "print(\"Electronic polarizability in F/m2 {0:.4g}\".format(alpha_e))\n", + "# part(b)\n", + "# let x=E_loc/E\n", + "x = (epsilon_r+2)/3.0\n", + "print(\"Local field is a factor of {0:.4f} greater than applied field\".format(x))\n", + "# part(c)\n", + "wo = math.sqrt(Z*e**2/(me*alpha_e))\n", + "fo = wo/(2*math.pi)\n", + "print(\"resonant frequency in Hz is {0:.4g}\".format(fo))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Electronic polarizability in F/m2 4.164e-40\n", + "Local field is a factor of 4.6333 greater than applied field\n", + "resonant frequency in Hz is 2.616e+15\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.3 Page No : 311" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "# let epsilon=E\n", + "Eo = 8.85*10**-12 # in F/m\n", + "Ni = 1.43*10**28 # in m**-3\n", + "alpha_e_Cs = 3.35*10**-40 # F m2\n", + "alpha_e_Cl = 3.40*10**-40 # F m2\n", + "alpha_i = 6*10**-40 # F m2\n", + "\n", + "# Calculations and Results\n", + "# (Er-1)/(Er+2)=(1/(3*E0))*(Ni*alpha_e(Cs+)+Ni*alpha_e(Cl-)+Ni*alpha_i)\n", + "# let x=(1/(3*E0))*(Ni*alpha_e(Cs+)+Ni*alpha_e(Cl-)+Ni*alpha_i)\n", + "# after few mathematical steps we get\n", + "# Er=(2*x+1)/(1-x)\n", + "x = (1.0/(3*Eo))*(Ni*alpha_e_Cs+Ni*alpha_e_Cl+Ni*alpha_i)\n", + "Er = (2*x+1)/(1-x)\n", + "print(\"Dielectric constant at low frequency is\")\n", + "# similarly\n", + "# let y=(1/(3*E0))*(Ni*alpha_e(Cs+)+Ni*alpha_e(Cl-))\n", + "# after few mathematical steps we get\n", + "# Erop=(2*x+1)/(1-x)\n", + "y = (1.0/(3*Eo))*(Ni*alpha_e_Cs+Ni*alpha_e_Cl)\n", + "Erop = (2*y+1)/(1-y)\n", + "print(\"Dielectric constant at optical frequency is {0:.4f}\".format(Erop))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Dielectric constant at low frequency is\n", + "Dielectric constant at optical frequency is 2.7137\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.6 Page No : 315" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Given\n", + "# power dissipated at a given voltage per unit capacitance depends only on w*math.tan(delta)\n", + "# at f=60 #in Hz.\n", + "f = 60.0 # in Hz.\n", + "w = 2*math.pi*f\n", + "# let x=math.tan(delta)\n", + "x_PC = 9*10**-4 # Ploy-carbonate\n", + "x_SR = 2.25*10**-2 # Silicone rubber\n", + "x_E = 4.7*10**-2 # Epoxy with mineral filler\n", + "\n", + "# Calculations and Results\n", + "p_PC = w*x_PC\n", + "p_SR = w*x_SR\n", + "p_E = w*x_E\n", + "a = min(p_PC, p_SR, p_E)\n", + "print(\"The minimum w*math.radians(math.tan(delta) is {0:.4f} which corresponds to polycarbonate\".format(a))\n", + "print(\"Hence the lowest power dissipation per unit capacitance at a given voltage \"\n", + " \"corresponds to polycarbonate at 60Hz\")\n", + "\n", + "# Given\n", + "# at f=1 # in MHz.\n", + "f = 10**6 # in Hz.\n", + "w = 2*math.pi*f\n", + "# let x=math.tan(delta)\n", + "x_PC = 1*10**-2 # Ploy-carbonate\n", + "x_SR = 4*10**-3 # Silicone rubber\n", + "x_E = 3*10**-2 # Epoxy with mineral filler\n", + "\n", + "# Calculations and Results\n", + "p_PC = w*x_PC\n", + "p_SR = w*x_SR\n", + "p_E = w*x_E\n", + "a = min(p_PC, p_SR, p_E)\n", + "print(\"The minimum w*math.radians(math.tan(delta) is {0:.4f} which corresponds to Silicone rubber\".format(a))\n", + "print(\"Hence, the lowest power dissipation per unit capacitance at a given \"\n", + " \"voltage corresponds to Silicone rubber at 1MHz\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum w*math.radians(math.tan(delta) is 0.3393 which corresponds to polycarbonate\n", + "Hence the lowest power dissipation per unit capacitance at a given voltage corresponds to polycarbonate at 60Hz\n", + "The minimum w*math.radians(math.tan(delta) is 25132.7412 which corresponds to Silicone rubber\n", + "Hence, the lowest power dissipation per unit capacitance at a given voltage corresponds to Silicone rubber at 1MHz\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.7 Page No : 320" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "# at 60 Hz\n", + "f = 60.0 # Hz\n", + "E = 100*10**3*10**2 # in V/m\n", + "# values taken from table 7.3\n", + "epsilon_o = 8.85*10**-12 # in F/m\n", + "epsilon_r_HLPE = 2.3\n", + "epsilon_r_Alumina = 8.5\n", + "# let x=math.tan(delta)\n", + "x_HLPE = 3*10**-4\n", + "x_Alumina = 1*10**-3\n", + "\n", + "# Calculations and Results\n", + "W_vol_HLPE = 2*math.pi*f*E**2*epsilon_o*epsilon_r_HLPE*x_HLPE # in W/m3\n", + "W_vol_HLPE /= 10**3 # in mW/cm3\n", + "print(\"Heat dissipated per unit volume of HLPE at 60 Hz in mW/cm3 is {0:.4f}\".format(W_vol_HLPE))\n", + "W_vol_Alumina = 2*math.pi*f*E**2*epsilon_o*epsilon_r_Alumina*x_Alumina\n", + "W_vol_Alumina /= 10**3 # in mW/cm3\n", + "print(\"Heat dissipated per unit volume of Alumina at 60 Hz in mW/cm3 is {0:.4f}\".format(W_vol_Alumina))\n", + "\n", + "# Given\n", + "# at 1 MHz\n", + "f = 10**6 # Hz\n", + "x_HLPE = 4*10**-4\n", + "x_Alumina = 1*10**-3\n", + "\n", + "# Calculations and Results\n", + "W_vol_HLPE = 2*math.pi*f*E**2*epsilon_o*epsilon_r_HLPE*x_HLPE # in W/m3\n", + "W_vol_HLPE /= 10**6 # in W/cm3\n", + "print(\"Heat dissipated per unit volume of HLPE at 1 MHz in mW/cm3 is {0:.4f}\".format(W_vol_HLPE))\n", + "W_vol_Alumina = 2*math.pi*f*E**2*epsilon_o*epsilon_r_Alumina*x_Alumina\n", + "W_vol_Alumina /= 10**6 # in W/cm3\n", + "print(\"Heat dissipated per unit volume of Alumina at 1 MHz in mW/cm3 is {0:.4f}\".format(W_vol_Alumina))\n", + "print(\"The heats at 60Hz are small comparing to heats at 1MHz\")\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat dissipated per unit volume of HLPE at 60 Hz in mW/cm3 is 0.2302\n", + "Heat dissipated per unit volume of Alumina at 60 Hz in mW/cm3 is 2.8359\n", + "Heat dissipated per unit volume of HLPE at 1 MHz in mW/cm3 is 5.1158\n", + "Heat dissipated per unit volume of Alumina at 1 MHz in mW/cm3 is 47.2653\n", + "The heats at 60Hz are small comparing to heats at 1MHz\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.10 Page No : 324" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "# part(C)\n", + "d = 0.5 # cm\n", + "a = d/2.0 # in cm\n", + "t = 0.5 # in cm\n", + "Ebr_X = 217.0 # in kV/cm from table 7.5\n", + "Ebr_S = 158.0 # in kV/cm from table 7.5\n", + "\n", + "# Calculations and Results\n", + "b = a+t\n", + "Vbr_X = Ebr_X*a*math.log(b/a)\n", + "print(\"breakdown voltage of XLPE in kV is {0:.4f}\".format(Vbr_X))\n", + "Vbr_S = Ebr_S*a*math.log(b/a)\n", + "print(\"breakdown voltage of Silicone rubber in kV is {0:.4f}\".format(Vbr_S))\n", + "# part(d)\n", + "# Given\n", + "# letE=epsiolon\n", + "Er_X = 2.3 # for XLPE\n", + "Er_S = 3.7 # for Silicone rubber\n", + "# Eair_br=Ebr\n", + "Eair_br_X = 100.0 # in kV/cm\n", + "Eair_br_S = 100.0 # in kV/cm\n", + "\n", + "# Calculations and Results\n", + "# Vair_br=Eair_br*a*math.log(b/a)/Er\n", + "Vair_br_X = Eair_br_X*a*math.log(b/a)/Er_X\n", + "print(\"Voltage for partial discharge in a microvoid for XLPE in kV is {0:.4f}\".format(Vair_br_X))\n", + "Vair_br_S = Eair_br_S*a*math.log(b/a)/Er_S\n", + "print(\"Voltage for partial discharge in a microvoid for Silicone rubber in kV is {0:.4f}\".format(Vair_br_S))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "breakdown voltage of XLPE in kV is 59.5997\n", + "breakdown voltage of Silicone rubber in kV is 43.3952\n", + "Voltage for partial discharge in a microvoid for XLPE in kV is 11.9414\n", + "Voltage for partial discharge in a microvoid for Silicone rubber in kV is 7.4231\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.11 Page No : 327" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "# letE=epsiolon\n", + "Er_100c = 2.69\n", + "Er_25c = 2.60\n", + "f = 1*10**3 # in Hz\n", + "w = 2*math.pi*f\n", + "C_25c = 560*10**-12 # in Farads\n", + "# Gp=w*C*math.tan(delta)\n", + "# let x=math.tan(delta)=0.002\n", + "x = 0.002\n", + "\n", + "# Calculations and Results\n", + "Gp = w*C_25c*x\n", + "print(\"Equivalent parallel conductance at 25 degree celsius in ohm**-1 is {0:.4g}\".format(Gp))\n", + "\n", + "# Given\n", + "# at 100 c\n", + "x = 0.01\n", + "\n", + "# Calculations and Results\n", + "C_100c = C_25c*Er_100c/Er_25c\n", + "Gp = w*C_100c*x\n", + "print(\"Equivalent parallel conductance at 100 degree celsius in ohm**-1 is {0:.4g}\".format(Gp))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equivalent parallel conductance at 25 degree celsius in ohm**-1 is 7.037e-09\n", + "Equivalent parallel conductance at 100 degree celsius in ohm**-1 is 3.64e-08\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.12 Page No : 331" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "Eo = 8.85*10**-12 # F/m2\n", + "Er = 1000.0\n", + "D = 3*10**-3 # in m\n", + "V = 5000.0 # in V\n", + "d = 200*10**-12 # in m/V\n", + "L = 10*10**-3 # in mm\n", + "\n", + "# Calculations and Results\n", + "A = math.pi*(D/2.0)**2\n", + "F = Eo*Er*A*V/(d*L)\n", + "print(\"Force required to spark the gap in Newton is {0:.4f}\".format(F))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Force required to spark the gap in Newton is 156.3924\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.13 Page No : 333" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "fs = 1.0 # in MHz\n", + "k = 0.1\n", + "\n", + "# Calculations and Results\n", + "fa = fs/(math.sqrt(1-k**2))\n", + "print(\"fa value in MHz for given fs is {0:.4f}\".format(fa))\n", + "print(\"thus fa-fs is only {0:.4f} kHz (which means they are very close)\".format((fa-fs)*10**3))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "fa value in MHz for given fs is 1.0050\n", + "thus fa-fs is only 5.0378 kHz (which means they are very close)\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.14 Page No : 334" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "Co = 5.0 # in pF\n", + "fa = 1.0025 # in MHz\n", + "fs = 1.0 # in MHz\n", + "R = 20.0 # in ohms\n", + "\n", + "# Calculations and Results\n", + "C = Co*((fa/fs)**2-1)\n", + "print(\"Capacitance value in the equivalent circuit of the crystal in pF is {0:.4f}\".format(C))\n", + "L = 1/(C*(2*math.pi*fs)**2)\n", + "print(\"Inductance value in the equivalent circuit of the crystal in Henry is {0:.4f}\".format(L))\n", + "\n", + "# Given\n", + "fs *= 10**6 # in Hz\n", + "C *= 10**-12 # in F\n", + "\n", + "# Calculations and Results\n", + "Q = 1.0/(2*math.pi*fs*R*C)\n", + "print(\"Quality factor of the crystal is {0:.4g}\".format(Q))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Capacitance value in the equivalent circuit of the crystal in pF is 0.0250\n", + "Inductance value in the equivalent circuit of the crystal in Henry is 1.0119\n", + "Quality factor of the crystal is 3.179e+05\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.15 Page No : 340" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "P = 380*10**-6 # in C/m2/K\n", + "c = 380.0 # in J/Kg/K\n", + "# let epsilon=E\n", + "Eo = 8.85*10**-12 # in F/m\n", + "Er = 290.0\n", + "rho = 7000.0 # in Kg/m3\n", + "delta_V = 0.001 # in V\n", + "delta_t = 0.2 # in seconds\n", + "\n", + "# Calculations and Results\n", + "I = (P/(rho*c*Eo*Er))**-1*delta_V/delta_t\n", + "print(\"Minimum radiation intensity that can be measured in W/m2 is {0:.4f}\".format(I))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum radiation intensity that can be measured in W/m2 is 0.0898\n" + ] + } + ], + "prompt_number": 11 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch8.ipynb b/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch8.ipynb new file mode 100755 index 00000000..07126dd3 --- /dev/null +++ b/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap/ch8.ipynb @@ -0,0 +1,147 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:041d5244aa5d603722207511a0c716199565294cf0a34c3074cf07858de21983" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 8 : Magnetic properties and conductivity" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.3 Page No : 354" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "Mat = 55.85*10**-3 # in Kg/mol\n", + "NA = 6.022*10**23 # in mol**-1\n", + "p = 7.86*10**3 # in kg/m3\n", + "Msat = 1.75*10**6 # in A/m\n", + "# funcprot(0)\n", + "beta = 9.27*10**-24 # in J/tesla\n", + "\n", + "# Calculations and Results\n", + "n_at = p*NA/Mat\n", + "x = Msat/(n_at*beta)\n", + "print(\"In the solid each Fe atom contributes only {0:.4f} bohr magneton\".format(x))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "In the solid each Fe atom contributes only 2.2275 bohr magneton\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.5 Page No : 358" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "u_o = 4*math.pi*10**-7 # in H/m\n", + "u_ri = 2*10**3\n", + "N = 200.0 # no. of turns\n", + "d = 0.005 # in m\n", + "D = 2.5*10**-2 # in m\n", + "\n", + "# Calculations and Results\n", + "A = math.pi*(d**2)/4\n", + "l = math.pi*D\n", + "L = u_ri*u_o*N**2*A/l\n", + "print(\"Approximate inductance of the coil in Henry is {0:.4f}\".format(L))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Approximate inductance of the coil in Henry is 0.0251\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.7 Page No : 359" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Given\n", + "N = 500.0 # no.of turns\n", + "B = 5.0 # in Tesla\n", + "l = 1.0 # in m\n", + "r = 10**-3 # in m\n", + "uo = 4*math.pi*10**-7 # in H/m\n", + "d = 10*10**-2 # in m\n", + "\n", + "# Calculations and Results\n", + "I = (B*l)/(uo*N)\n", + "print(\"current in Amperes is {0:.4f}\".format(I))\n", + "E_vol = B**2/(2*uo)\n", + "v = math.pi*l*d**2/4.0\n", + "E = E_vol*v\n", + "print(\"Energy stored in the solenoid in joules is {0:.4f}\".format(E))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "current in Amperes is 7957.7472\n", + "Energy stored in the solenoid in joules is 78125.0000\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
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