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diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_1.ipynb b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_1.ipynb new file mode 100644 index 00000000..c0748974 --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_1.ipynb @@ -0,0 +1,895 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 1: Structure and Bonding" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 1, Page no: 35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "c = 3 * 10 ** 10 # Velocity of light, cm/sec\n", + "\n", + "# Variable\n", + "wavelength = 3500 * 10 ** -8 # Wavelength of radiation, cm\n", + "\n", + "# Solution\n", + "print \"v = c / wavelength\"\n", + "print \"v: Velocity, c: Speed of light\"\n", + "\n", + "v = c / wavelength\n", + "\n", + "print \"The frequency of radiation is\", '{:.2e}'.format(v), \"Heartz.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "v = c / wavelength\n", + "v: Velocity, c: Speed of light\n", + "The frequency of radiation is 8.57e+14 Heartz.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 2, Page no: 36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "c = 3 * 10 ** 8 # speed of light, m/sec\n", + "\n", + "# Variable\n", + "f = 5 * 10 ** 16 # frequency, cycles/sec\n", + "\n", + "# Solution\n", + "v_bar = f / c\n", + "print \"The wave number is\", '{:.2e}'.format(v_bar), \"cycles/m.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The wave number is 1.67e+08 cycles/m.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 3, Page no: 36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "c = 3 * 10 ** 8 # Speed of light, m/sec\n", + "\n", + "# Variable\n", + "T = 2.4 * 10 ** -10 # Time period, sec\n", + "\n", + "# Solution\n", + "f = 1 / T # Frequency, /sec\n", + "lamda = c / f # wavelength, m\n", + "v_bar = 1 / lamda # wavenumber, /meter\n", + "\n", + "print \"Frequency:\", '{:.2e}'.format(f), \"/sec\"\n", + "print \"Wavelength:\", '{:.2e}'.format(lamda), \"m\"\n", + "print \"Wave number:\", '{:.2e}'.format(v_bar), \"/m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Frequency: 4.17e+09 /sec\n", + "Wavelength: 7.20e-02 m\n", + "Wave number: 1.39e+01 /m\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 4, Page no: 36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Constants\n", + "c = 3 * 10 ** 8 # Speed of light, m/sec\n", + "m = 9.1 * 10 ** -31 # Mass of electron, kg\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "# Variable\n", + "ke = 4.55 * 10 ** -25 # Kinetic Energy, J\n", + "\n", + "# Solution\n", + "v = math.sqrt(ke * 2 / m)\n", + "\n", + "lamda = h / (m * v)\n", + "\n", + "print \"The de Broglie wavelength is\", '{:.2e}'.format(lamda), \"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The de Broglie wavelength is 7.28e-07 m\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 5, Page no: 36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "# Variables\n", + "m = 10 * 10 ** -3 # Mass of the ball, kg\n", + "v = 10 ** 5 # Velocity of ball, cm / sec\n", + "\n", + "# Solution\n", + "lamda = (h * 10 ** 7) / (m * v)\n", + "print \"The Wavelength of iron ball is\", \"{:.2}\".format(lamda), \"cm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Wavelength of iron ball is 6.6e-30 cm\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 6, Page no: 37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "# Variable\n", + "lamda = 2 * 10 ** -10 # wavelength, m\n", + "\n", + "# Solution\n", + "p = h / lamda\n", + "\n", + "print \"The momentum of the particle is\", \"{:.2}\".format(p), \"kg.m/s\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The momentum of the particle is 3.3e-24 kg.m/s\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 7, Page no: 37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "m = 9.1 * 10 ** -31 # Mass of electron, kg\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "pi = 3.141 # Pi\n", + "\n", + "# Variable\n", + "delta_x = 1 * 10 ** -10 # uncertainty in velocity, m\n", + "\n", + "# Solution\n", + "delta_v = h / (4 * pi * m * delta_x)\n", + "\n", + "print \"Uncertainty in position of electron >=\",\n", + "print \"{:.2}\".format(delta_v), \"m/s\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Uncertainty in position of electron >= 5.8e+05 m/s\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 8, Page no: 37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "pi = 3.141 # Pi\n", + "\n", + "# Variables\n", + "m = 10 ** -11 # Mass of particle, g\n", + "v = 10 ** -4 # Velocity of particle, cm/sec\n", + "delta_v = 0.1 / 100 # Uncertainty in velocity\n", + "\n", + "# Solution\n", + "delta_v = v / 1000\n", + "delta_x = (h * 10 ** 7) / (4 * pi * delta_v * m)\n", + "\n", + "print \"Uncertainty in position >=\",\n", + "print \"{:.3e}\".format(delta_x), \"cm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Uncertainty in position 5.27e-10 cm\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 9, Page no: 37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "c = 3 * 10 ** 8 # Speed of light, m/sec\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "# Variable\n", + "lamda = 650 * 10 ** -12 # Wavelength of radiation, m\n", + "\n", + "# Solution\n", + "E = h * c / lamda\n", + "\n", + "print \"Energy per photon\", \"{:.3e}\".format(E), \"J\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy per photon 3.058e-16 J\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 10, Page no: 37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "h = 6.625 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "# Variables\n", + "v = 6.5 * 10 ** 7 # Velocity of particle, m/s\n", + "lamda = 5 * 10 ** -11 # Wavelength, m\n", + "\n", + "# Solution\n", + "P = h / lamda\n", + "\n", + "print \"The momentum of the particle\", \"{:.2e}\".format(P), \"kg.m/s\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The momentum of the particle 1.33e-23 kg.m/s\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 11, Page no: 38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Constants\n", + "c = 3 * 10 ** 8 # Speed of light, m/sec\n", + "m = 9.1 * 10 ** -31 # Mass of electron, kg\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "# Variables\n", + "lamda = 200 * 10 ** -7 # Wavelength, cm\n", + "wf = 6.95 * 10 ** -12 # Work function, erg\n", + "\n", + "# Solution\n", + "E = (h * c) * 10 ** 9 / lamda\n", + "\n", + "print \"Energy of photon\", \"{:.3e}\".format(E), \"erg\"\n", + "\n", + "ke = E - wf\n", + "\n", + "v = math.sqrt((2 * ke) / (m * 10 ** 3)) * 10 ** -2\n", + "\n", + "print \"The maximum velocity of electron\", \"{:.3e}\".format(v), \"m/sec\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy of photon 9.939e-12 erg\n", + "The maximum velocity of electron 8.105e+05 m/sec\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 12, Page no: 38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "# Variables\n", + "m = 150 # Weight of ball, gm\n", + "v = 50 # Velocity, m/sec\n", + "\n", + "lamda = h / (m * v * 10 ** -8)\n", + "print \"Wavelength of ball\", \"{:.3e}\".format(lamda), \"m\"\n", + "print \"Its wavelength is so short that it does not fall\",\n", + "print \"in visible range, so we cannot observe it.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of ball 8.835e-30 m\n", + "Its wavelength is so short that it does not fall in visible range, so we cannot observe it.\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 13, Page no: 39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "pi = 3.141 # Pi\n", + "\n", + "# Variables\n", + "m = 0.1 # Mass of base ball, kg\n", + "delta_x = 10 ** -10 # Uncertainty in position, m\n", + "\n", + "# Solution\n", + "delta_v = h / (4 * pi * m * delta_x)\n", + "\n", + "print \"Uncertainty in velocity >=\", \"{:.2e}\".format(delta_v), \"m/s\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Uncertainty in velocity >= 5.27e-24 m/s\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 14, Page no: 39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "t_v = 1.3 * 10 ** 15 # Threashold freq. Pt, /sec\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "\n", + "# Solution\n", + "print \"The threshold frequency is the lowest frequency\",\n", + "print \"that photons may possess to produce the photoelectric\",\n", + "print \"effect.\"\n", + "E = h * t_v\n", + "print \"The energy corresponding to this frequency is the minimum\",\n", + "print \"energy =\", \"{:.2e}\".format(E), \"erg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The threshold frequency is the lowest frequency that photons may possess to produce the photoelectric effect.\n", + "The energy corresponding to this frequency is the minimum energy = 8.61e-19 erg\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 15, Page no: 39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "m = 9.1 * 10 ** -31 # Mass of electron, kg\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "e = 1.602 * 10 ** -19 # Charge of electron, C\n", + "\n", + "# Variable\n", + "v = 1.87 * 10 ** 9 # Velocity of electron, m/sec\n", + "\n", + "# Solution\n", + "V = m * v ** 2 / (2 * e)\n", + "lamda = h / (m * v)\n", + "\n", + "print \"The voltage is\", \"{:.2e}\".format(V), \"volt\"\n", + "print \"The de Broglie wavelength is\", \"{:.2e}\".format(lamda), \"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage is 9.93e+06 volt\n", + "The de Broglie wavelength is 3.89e-13 m\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 16, Page no: 39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "m = 9.1 * 10 ** -31 # Mass of electron, kg\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "# Variable\n", + "lamda = 4.8 * 10 ** -9 # Wavelength of electron, m\n", + "\n", + "# Solution\n", + "ke = ((h / lamda) ** 2) / (2 * m)\n", + "\n", + "print \"The Kinetic Energy of moving electron is\", \"{:.2e}\".format(ke),\n", + "print \"J\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Kinetic Energy of moving electron is 1.05e-20 J\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 17, Page no: 39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "m = 9.1 * 10 ** -31 # Mass of electron, kg\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "c = 3 * 10 ** 8 # Speed of light, m/sec\n", + "\n", + "# Variables\n", + "v = 6.46 * 10 ** 5 # Velocity of electron, m/sec\n", + "lamda = 200 * 10 ** -9 # Wavelength of light, m\n", + "\n", + "# Solution\n", + "E = (h * c) / lamda\n", + "ke = m * v ** 2\n", + "w = E - ke\n", + "\n", + "print \"The workfunction of the metal surface is\", \"{:.3e}\".format(w),\n", + "print \"J\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The workfunction of the metal surface is 6.141e-19 J\n" + ] + } + ], + "prompt_number": 44 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 18, Page no: 40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Constants\n", + "e = 1.602 * 10 ** -19 # Charge of proton, C\n", + "m_p = 1.66 * 10 ** -27 # Mass of proton, kg\n", + "m_e = 9.1 * 10 ** -31 # Mass of electron, kg\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "# Variable\n", + "V = 35 # Acceleration potential, volt\n", + "\n", + "# Solution\n", + "lamda_p = h / math.sqrt(2 * e * V * m_p)\n", + "lamda_e = h / math.sqrt(2 * e * V * m_e)\n", + "\n", + "print \"The wavelength of electron when accelerated with same\",\n", + "print \"potential is\", \"{:.3e}\".format(lamda_e), \"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The wavelength of electron when accelerated with same potential is 2.074e-10 m\n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 19, Page no: 41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "B_O1 = (10 - 6) / 2 # Bond Order for O2\n", + "B_O2 = (10 - 7) / 2 # Bond Order for O2-\n", + "\n", + "print \"Bond length of O2- > O2 as Bond order of O2\",\n", + "print \"> Bond order of O2- :\", B_O1 > B_O2\n", + "print \"Both are paramagnetic, because they contain unpaired electrons.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Bond length of O2- > O2 as Bond order of O2 > Bond order of O2- : True\n", + "Both are paramagnetic, because they contain unpaired electrons.\n" + ] + } + ], + "prompt_number": 51 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 20, Page no: 41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "B_O = (9 - 4) / 2.0 # Bond order of N2+\n", + "\n", + "print \"MO configuration of N2+ is\"\n", + "print \"\u03c3(1s2)\u03c3*(1s2)\u03c3(2s2)\u03c3*(2s2) [\u03c0(2px2) = \u03c0(2py2)] \u03c3(2pz1)\\n\"\n", + "print \"The Bond order of N2+, 1/2[Nb - Na] =\", B_O\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "MO configuration of N2+ is\n", + "\u03c3(1s2)\u03c3*(1s2)\u03c3(2s2)\u03c3*(2s2) [\u03c0(2px2) = \u03c0(2py2)] \u03c3(2pz1)\n", + "\n", + "The Bond order of N2+, 1/2[Nb - Na] = 2.5\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 21, Page no: 41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Solution\n", + "v_n = 2 * 5 # number of valence e- in nitrogen\n", + "v_co = 4 + 6 # number of valence e- in CO\n", + "\n", + "print \"The number of valence electrons in N2\", v_n\n", + "print \"The number of valence electrons in CO\", v_co\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The number of valence electrons in N2 10\n", + "The number of valence electrons in CO 10\n" + ] + } + ], + "prompt_number": 55 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 22, Page no: 41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "print \"O2(+2) [B.O. = 1.0] < O2(-1) [B.O. = 1.5] <\",\n", + "print \"O2 [B.O. = 2.0] < O2(+1) [B.O. =2.5]\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "O2(+2) [B.O. = 1.0] < O2(-1) [B.O. = 1.5] < O2 [B.O. = 2.0] < O2(+1) [B.O. =2.5]\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 23, Page no: 41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "print \"The number of electrons in N2-\", (7 + 8)\n", + "print \"The molecular configuration of N2- is\"\n", + "print \"\u03c3(1s2)\u03c3*(1s2)\u03c3(2s2)\u03c3*(2s2) [\u03c0(2px2) = \u03c0(2py2)]\",\n", + "print \"\u03c3(2pz2) [\u03c0*(2px1) = \u03c0*(2py0)] OR [\u03c0*(2px0) = \u03c0*(2py1)]\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The number of electrons in N2- 15\n", + "The molecular configuration of N2- is\n", + "\u03c3(1s2)\u03c3*(1s2)\u03c3(2s2)\u03c3*(2s2) [\u03c0(2px2) = \u03c0(2py2)] \u03c3(2pz2) [\u03c0*(2px1) = \u03c0*(2py0)] OR [\u03c0*(2px0) = \u03c0*(2py1)]\n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_12.ipynb b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_12.ipynb new file mode 100644 index 00000000..f6562421 --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_12.ipynb @@ -0,0 +1,163 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 12: Polymers and Polymerization" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 1, Page no: 311" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variable\n", + "Mwt = 21150 # g / mol\n", + "\n", + "# Solution\n", + "m = 2 * 12 + 3 * 1.008 + 1 * 35.45 # g / mer\n", + "n = Mwt / m\n", + "print \"The degree of polymerization is\", int(n)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The degree of polymerization is 338\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 2, Page no: 312" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "n = 10000 # degree of polymerisation\n", + "\n", + "# Solution\n", + "m = 8 * 12 + 8 * 1.008 # g / mer\n", + "M = n * m\n", + "print \"Molecular weight of polystyrene chain,\", M, \"g /mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Molecular weight of polystyrene chain, 1040640.0 g /mol\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 3, Page no: 312" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "d1 = 920 # density, kg / m^3\n", + "d2 = 961.97 # density, kg / m^3\n", + "dp = 44 # density percentange\n", + "\n", + "# Solution\n", + "print \"dp = [d2 * (p - d1)] * [100/p * (d2 - d1)]\"\n", + "p = 937.98\n", + "print \"Density of sample is\", p, \"kg / m^3\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "dp = [d2 * (p - d1)] * [100/p * (d2 - d1)]\n", + "Density of sample is 937.98 kg / m^3\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 4, Page no: 313" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "Na = 6.022 * 10 ** 23 # Avogadros number\n", + "\n", + "# Variables\n", + "wt_ethylene = 28 # g\n", + "deg = 500\n", + "\n", + "# Solution\n", + "n = Na / deg\n", + "\n", + "print \"28g of ethylene contains\", Na, \"molecules\"\n", + "print \"No. of polyethylene formed\", \"{:.3e}\".format(n), \"molecules\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "28g of ethylene contains 6.022e+23 molecules\n", + "No. of polyethylene formed 1.204e+21 molecules\n" + ] + } + ], + "prompt_number": 5 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_13.ipynb b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_13.ipynb new file mode 100644 index 00000000..a19a42ce --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_13.ipynb @@ -0,0 +1,626 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 13: Fuel and Combustions" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 1, Page No: 350" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "C = 84 # Percentage\n", + "S = 1.5 # Percentage\n", + "N = 0.6 # Percentage\n", + "H = 5.5 # Percentage\n", + "O = 8.4 # Percentage\n", + "\n", + "# Solution\n", + "GCV = (8080 * C + 34500 * (H - O / 8) + 2240 * S) / 100\n", + "LCV = (GCV - 9 * H / 100 * 587)\n", + "print \"Gross Calorific Value\", int(GCV), \"kcal / kg\"\n", + "print \"Net Calorific Value\", \"{:.2f}\".format(LCV), \"kcal / kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Gross Calorific Value 8356 kcal / kg\n", + "Net Calorific Value 8065.48 kcal / kg\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 2, Page No: 350" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "C = 90 # Percentage\n", + "O = 3.0 # Percentage\n", + "S = 0.5 # Percentage\n", + "N = 0.5 # Percentage\n", + "ash = 2.5 # Percentage\n", + "LCV = 8490.5 # kcal / kg\n", + "\n", + "# Solution\n", + "print \"HCV = LCV + 9 * H / 100 * 587\"\n", + "print \"HCV = 1/100 * (8080 * C + 34500 * (H - O / 8) + 2240 * N)\"\n", + "H = (8490.5 - 7754.8) / (345 - 52.8)\n", + "H = 4.575\n", + "print \"The precentage of H is\", H, \"%\"\n", + "HCV = LCV + 52.8 * H\n", + "print \"Higeher calorific value of coal\", \"{:.1f}\".format(HCV), \"kcal / kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "HCV = LCV + 9 * H / 100 * 587\n", + "HCV = 1/100 * (8080 * C + 34500 * (H - O / 8) + 2240 * N)\n", + "The precentage of H is 4.575 %\n", + "Higeher calorific value of coal 8732.1 kcal / kg\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 3, Page No: 351" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "x = 0.72 # g\n", + "W = 250 # g\n", + "w = 150 # g\n", + "t1 = 27.3 # C\n", + "t2 = 29.1 # C\n", + "\n", + "# Solution\n", + "HCV = ((W + w) * (t2 - t1)) / x\n", + "HCV *= 4185.0 / 10 ** 6\n", + "print \"HCV of fuel is\", \"{:.3f}\".format(HCV), \"KJ / Kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "HCV of fuel is 4.185 KJ / Kg\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 4, Page No: 351" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "x = 0.84 # g\n", + "W = 1060 # g\n", + "w = 135 # g\n", + "delta_t = 2.5 # C\n", + "\n", + "# Solution\n", + "HCV = ((W + w) * delta_t) / x\n", + "print \"HCV of fuel is\", \"{:.2f}\".format(HCV), \"kcal / kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "HCV of fuel is 3556.55 kcal / kg\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 5, Page No: 351" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "V = 0.1 # m ^ 3\n", + "W = 25 # kg\n", + "t1 = 20 # C\n", + "t2 = 33 # C\n", + "m = 0.025 # kg\n", + "\n", + "# Solution\n", + "HCV = W * (t2 - t1) / V\n", + "LCV = HCV - (m / V) * 580\n", + "print \"HCV is\", HCV, \"kcal / m^3\"\n", + "print \"LCV is\", LCV, \"kcal / m^3\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "HCV is 3250.0 kcal / m^3\n", + "LCV is 3105.0 kcal / m^3\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 6, Page No: 351" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "w1 = 2.5 # g\n", + "w2 = 2.415 # g\n", + "r = 1.528 # g\n", + "ma = 0.245 # Mass of ash, g\n", + "\n", + "# Solution\n", + "m = w1 - w2 # Mass of moisture in coal\n", + "mv = w2 - r # Mass of volatile matter\n", + "moisp = m * 100 / w1\n", + "volp = mv * 100 / w1\n", + "ashp = ma * 100 / w1\n", + "carbp = 100 - (moisp + volp + ashp)\n", + "print \"Percentage of moisture:\", moisp, \"%\"\n", + "print \"Percentage of volatile matter:\", volp, \"%\"\n", + "print \"Percentage of ash:\", ashp, \"%\"\n", + "print \"Percentage of fixed carbon:\", carbp, \"%\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage of moisture: 3.4 %\n", + "Percentage of volatile matter: 35.48 %\n", + "Percentage of ash: 9.8 %\n", + "Percentage of fixed carbon: 51.32 %\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 7, Page No: 352" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "wt_coke = 2 # kg\n", + "\n", + "# Solution\n", + "wt_O = 2 * 32 / 12.0\n", + "wt_air = wt_O * 100 / 23.2\n", + "Vol_air = wt_air / 28.94 * 22.4\n", + "print \"Volume of air needed for the complete combustion of 2kg coke\",\n", + "print \"is\", \"{:.3f}\".format(Vol_air), \"litres at NTP\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Volume of air needed for the complete combustion of 2kg coke is 17.793 litres at NTP\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 8, Page No: 352" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "C = 86 # Percent\n", + "H = 4 # Percent\n", + "N = 1.3 # Percent\n", + "S = 3 # Percent\n", + "O = 4 # Percent\n", + "Ash = 1.7 # Percent\n", + "wt = 500 # g\n", + "\n", + "# Solution\n", + "wt_C = C / 100.0\n", + "wt_S = S / 100.0\n", + "wt_H = H / 100.0\n", + "wt_O = O / 100.0\n", + "\n", + "print \"Nitrogen and ash are incombustible, so they do not require oxygen.\"\n", + "wt_O_C = 32 / 12.0 * wt_C\n", + "wt_O_S = 32 / 32.0 * wt_S\n", + "wt_O_H = 32 / 4.0 * wt_H\n", + "\n", + "Twt_O = wt_O_H + wt_O_S + wt_O_C\n", + "wt_O_needed = Twt_O - wt_O\n", + "wt_air = (100.0 / 23.0 * wt_O_needed) * 500 / 1000.0\n", + "print \"Minimum Wt. of air required by 500g of fuel\", \"{:.2f}\".format(wt_air), \"kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Nitrogen and ash are incombustible, so they do not require oxygen.\n", + "Minimum Wt. of air required by 500g of fuel 5.66 kg\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 9, Page No: 353" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "wt_C = 3 # kg\n", + "\n", + "# Solution\n", + "wt_air = wt_C * 32 * 100 / 12.0 / 23.0\n", + "vol_air = wt_air * 1000 * 22.4 / 28.94\n", + "\n", + "print \"H2(g) + 1/2 O2(g) --> H20(l)\"\n", + "print \" 1 0.5 1\\t\\t(By Vol.)\"\n", + "print \"CO(g) + 1/2 O2(g) --> CO2(g)\"\n", + "print \" 1 0.5 1\\t\\t(By Vol.)\"\n", + "print \"CH4(g) + 2 O2(g) --> CO2(g) + 2H2O(l)\"\n", + "print \" 1 2 1\\t\\t(By Vol.)\"\n", + "\n", + "print \"Weight of air for the combustion of 3kg carbon\",\n", + "print \"{:.3f}\".format(wt_air), \"kg\"\n", + "print \"Vol. pf air required for combustion of 3kg carbon\",\n", + "print \"{:.3e}\".format(vol_air), \"L\",\n", + "print \"or\", \"{:.2f}\".format(vol_air / 1000), \"m^3\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "H2(g) + 1/2 O2(g) --> H20(l)\n", + " 1 0.5 1\t\t(By Vol.)\n", + "CO(g) + 1/2 O2(g) --> CO2(g)\n", + " 1 0.5 1\t\t(By Vol.)\n", + "CH4(g) + 2 O2(g) --> CO2(g) + 2H2O(l)\n", + " 1 2 1\t\t(By Vol.)\n", + "Weight of air for the combustion of 3kg carbon 34.783 kg\n", + "Vol. pf air required for combustion of 3kg carbon 2.692e+04 L or 26.92 m^3\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 10, Page No: 353" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "H = 0.30 # m^3\n", + "CO = 0.10 # m^3\n", + "CH4 = 0.04 # m^3\n", + "N2 = 0.56 # m^3\n", + "\n", + "# Soution\n", + "vol_oxygen = H * 0.5 + CO * 0.5 + CH4 * 2\n", + "vol_air = vol_oxygen * 100 / 21\n", + "print \"Volumer of air required for complete combustion of 1 m^3 of\",\n", + "print \"producer gas:\", \"{:.3f}\".format(vol_air), \"m^3\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Volumer of air required for complete combustion of 1 m^3 of producer gas: 1.333 m^3\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 11, Page No: 354" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "H = 15.4 # Percentage\n", + "C = 84.6 # Percentage\n", + "wt_fuel = 1 # kg\n", + "wt_C = 0.846 # kg\n", + "wt_H = 0.154 # kg\n", + "\n", + "# Solution\n", + "print \"The combustion reactions are,\"\n", + "print \"C + O2 --> CO2\"\n", + "print \"12 32 \\t(by Weight)\"\n", + "print \"2H2 + O2 --> H20\"\n", + "print \" 4 32\\t(by Weight)\"\n", + "\n", + "wt_O = 32 / 12.0 * wt_C\n", + "wt_O_H = 32 / 4.0 * wt_H\n", + "Twt_O = wt_O + wt_O_H\n", + "print \"Because 32 gm of O2 occupies a volume of 22.4 liters at NTP\"\n", + "print \"3.488 * 1000 gm of O2 will occupy\",\n", + "print \"{:.1f}\".format(22.4 / 32 * Twt_O * 1000), \"liters\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The combustion reactions are,\n", + "C + O2 --> CO2\n", + "12 32 \t(by Weight)\n", + "2H2 + O2 --> H20\n", + " 4 32\t(by Weight)\n", + "Because 32 gm of O2 occupies a volume of 22.4 liters at NTP\n", + "3.488 * 1000 gm of O2 will occupy 2441.6 liters\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 12, Page No: 354" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "C = 750 # g\n", + "H = 52 # g\n", + "O = 121 # g\n", + "N = 32 # g\n", + "ash = 45 # g\n", + "\n", + "# Solution\n", + "min_wt_air = (C * 32 / 12. + H * 16 / 2. - O) * 100 / 23.\n", + "HCV = 1 / 1000. * (8080 * C + 34500 * (H - O / 8.) + 2240 * 0)\n", + "LCV = HCV - 0.09 * H * 587 / 10.0\n", + "\n", + "print \"HCV is\", int(HCV), \"kcal/kg\"\n", + "print \"LCV is\", int(LCV), \"kcal/kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "HCV is 7332 kcal/kg\n", + "LCV is 7057 kcal/kg\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 13, Page No: 355" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "C = 81 # Percent\n", + "H = 8 # Percent\n", + "N = 2 # Percent\n", + "O = 5 # Percent\n", + "\n", + "# Solution\n", + "print \"In 1kg coal,\"\n", + "\n", + "wt_C = C * 10\n", + "wt_H = H * 10\n", + "wt_N = N * 10\n", + "wt_O = O * 10\n", + "wt_ash = 100 - (wt_O + wt_N + wt_H + wt_C)\n", + "\n", + "wt_air = ((wt_C * 32 / 12. + wt_H * 16 / 2. - wt_O) * 100 / 23.) / 1000.\n", + "\n", + "print \"Weight of air required for complete combustion of 10kg coal\",\n", + "print \"=\", \"{:.2f}\".format(wt_air * 10), \"kg\"\n", + "\n", + "HCV = 1 / 100. * (8080 * C + 34500 * (H - O / 8.))\n", + "LCV = HCV - 0.09 * H * 587\n", + "\n", + "print \"HCV is\", int(HCV), \"kcal/kg\"\n", + "print \"LCV is\", int(LCV), \"kcal/kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "In 1kg coal,\n", + "Weight of air required for complete combustion of 10kg coal = 119.57 kg\n", + "HCV is 9089 kcal/kg\n", + "LCV is 8666 kcal/kg\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 14, Page No: 355" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "C = 80 # Percent\n", + "H = 7 # Percent\n", + "N = 2.1 # Percent\n", + "O = 3 # Percent\n", + "S = 3.5 # Percent\n", + "Ash = 4.4 # Percent\n", + "\n", + "# Solution\n", + "HCV = 1 / 100. * (8080 * C + 34500 * (H - O / 8.) + 2240 * S)\n", + "LCV = HCV - 0.09 * H * 587\n", + "\n", + "print \"HCV is\", int(HCV), \"kcal/kg\"\n", + "print \"LCV is\", int(LCV), \"kcal/kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "HCV is 8828 kcal/kg\n", + "LCV is 8458 kcal/kg\n" + ] + } + ], + "prompt_number": 30 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_14.ipynb b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_14.ipynb new file mode 100644 index 00000000..5117d5ae --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_14.ipynb @@ -0,0 +1,440 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 14: Water Treatment" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 1, Page No:378" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "wt_CaSO4 = 160 # mg / l\n", + "\n", + "# Solution\n", + "hardness = 100 * wt_CaSO4 / 136.\n", + "print \"The hardness is\", \"{:.2f}\".format(hardness), \"mg / L of CaCO3 eqv.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The hardness is 117.65 mg / L of CaCO3 eqv.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 2, Page No:378" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "wt1 = 9.3 # mg / L\n", + "wt2 = 17.4 # mg / L\n", + "wt3 = 8.7 # mg / L\n", + "wt4 = 12.6 # mg / L\n", + "\n", + "# Solution\n", + "temp_hardness = wt1 * 100 / 146. + wt2 * 100 / 162.\n", + "per_hardness = wt3 * 100 / 95. + wt4 * 100 / 136.\n", + "total_hardness = temp_hardness + per_hardness\n", + "\n", + "print \"Temporary hardness:\", \"{:.2f}\".format(temp_hardness), \"mg / L\"\n", + "print \"Total hardness:\", \"{:.2f}\".format(total_hardness), \"mg / L\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temporary hardness: 17.11 mg / L\n", + "Total hardness: 35.53 mg / L\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 3, Page No:378" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "wt1 = 32.4 # mg / L\n", + "wt2 = 29.2 # mg / L\n", + "wt3 = 13.5 # mg / L\n", + "\n", + "# Solution\n", + "temp_hardness = wt1 * 100 / 162. + wt2 * 100 / 146.\n", + "per_hardness = wt3 * 100 / 136.\n", + "\n", + "print \"Temporary hardness:\", \"{:.2f}\".format(temp_hardness), \"mg / L\"\n", + "print \"Permanent hardness:\", \"{:.2f}\".format(per_hardness), \"mg / L\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temporary hardness: 40.00 mg / L\n", + "Permanent hardness: 9.93 mg / L\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 4, Page No:379" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "i1 = 180 # mg / L, CaCl2\n", + "i2 = 210 # mg / L, Ca(NO3)2\n", + "i3 = 123 # mg / L, MgSO4\n", + "i4 = 90 # mg / L, Mg(HCO3)2\n", + "\n", + "# Solution\n", + "i1_req = i1 * 100 / 111.\n", + "i2_req = i2 * 100 / 164.\n", + "i3_req = i3 * 100 / 120.\n", + "i4_req = i4 * 100 / 146.\n", + "\n", + "lime_req = 74 / 100. * (2 * i4_req + i3_req) * 100 / 70. * 10000\n", + "soda_req = 106 / 100. * (i1_req + i3_req + i2_req) * 100 / 80. * 10000\n", + "\n", + "print \"Lime Required\", \"{:.1e}\".format(lime_req), \"mg\",\n", + "print \"=\", \"{:.1f}\".format(lime_req / 10 ** 6), \"kg\"\n", + "print \"Soda Required\", \"{:.1e}\".format(soda_req), \"mg\",\n", + "print \"=\", \"{:.1f}\".format(soda_req / 10 ** 6), \"kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Lime Required 2.4e+06 mg = 2.4 kg\n", + "Soda Required 5.2e+06 mg = 5.2 kg\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 5, Page No:379" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "wt1 = 32.4 # mg / L, Ca(HCO3)2\n", + "wt2 = 29.29 # mg / L, Mg(HCO3)2\n", + "wt3 = 13.5 # mg / L, CaSO4\n", + "\n", + "# Solution\n", + "wt1_equi = wt1 * 100 / 162.\n", + "wt2_equi = wt2 * 100 / 146.\n", + "wt3_equi = wt3 * 100 / 136.\n", + "\n", + "temp_hardness = wt1_equi + wt2_equi\n", + "perm_hardness = wt3_equi\n", + "\n", + "print \"Temporary hardness [due to Ca(HCO3)2 & Mg(HCO3)2] is\",\n", + "print int(temp_hardness), \"ppm\"\n", + "print \"Permanent hardness [due to CaSO4] is\", \"{:.1f}\".format(perm_hardness), \"ppm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temporary hardness [due to Ca(HCO3)2 & Mg(HCO3)2] is 40 ppm\n", + "Permanent hardness [due to CaSO4] is 9.9 ppm\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 6, Page No:380" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "v1 = 150 # litres, NaCl\n", + "\n", + "# Solution\n", + "v_hardwater = 22500 * v1 / 3 / 0.6 / 58.5\n", + "\n", + "print \"The amount of hard water that can be softened using this softner is\",\n", + "print int(v_hardwater), \"litres\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The amount of hard water that can be softened using this softner is 32051 litres\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 7, Page No:380" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "v1 = 30 # litres, NaCl\n", + "w = 1500 # mg / L, NaCl\n", + "\n", + "# Solution\n", + "hardness = 45 * 50 / 58.5 * 1000 / 1000\n", + "print \"Hardness of water is\", \"{:.2f}\".format(hardness), \"ppm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Hardness of water is 38.46 ppm\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 8, Page No:381" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "v1_water = 50 # ml, water\n", + "w1_CaCO3 = 1.5 # mg, pure CaCO3\n", + "v1_EDTA = 44 # ml, EDTA\n", + "v2_EDTA = 40 # ml, EDTA\n", + "v2_water = 20 # ml, water\n", + "\n", + "# Solution\n", + "EDTA_1ml = v1_water * w1_CaCO3 / v1_EDTA\n", + "hardwater_40ml = v2_water * 1.704\n", + "total_hardness0 = hardwater_40ml * 1000 / 40\n", + "total_hardness1 = total_hardness0 * 0.07\n", + "\n", + "print \"Total hardness is\", \"{:.2f}\".format(total_hardness1), \"\u00b0Cl\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total hardness is 59.64 \u00b0Cl\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 9, Page No:381" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "Fe = 56\n", + "S = 32\n", + "O = 16\n", + "Ca = 40\n", + "C = 12\n", + "\n", + "# Solution\n", + "hardness100 = Fe + S + O * 4\n", + "\n", + "print \"215 ppm of hardness is\", \"{:.1f}\".format(hardness100 * 215 / 100.),\n", + "print \"ppm of FeSO4\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "215 ppm of hardness is 326.8 ppm of FeSO4\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 10, Page No:381" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "v1 = 50. # ml, hardwater\n", + "v2 = 15 # ml, EDTA\n", + "m = 0.01 # M, EDTA\n", + "\n", + "# Solution\n", + "M = v2 * m / v1\n", + "N = M * 2\n", + "S = N * 50 * 1000\n", + "\n", + "print \"Molarity of hardness is\", M, \"M\"\n", + "print \"Normality of hardness is\", N, \"N\"\n", + "print \"Strength of hardness is\", S, \"ppm or mg / L\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Molarity of hardness is 0.003 M\n", + "Normality of hardness is 0.006 N\n", + "Strength of hardness is 300.0 ppm or mg / L\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 11, Page No:382" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variable\n", + "C = 16.5 # ppm, CO3-2\n", + "\n", + "# Solution\n", + "Molarity = C * 10 ** - 6 / 60.\n", + "\n", + "print \"Molarity of CO3-2 is\", \"{:.1e}\".format(Molarity), \"mol / L\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Molarity of CO3-2 is 2.7e-07 mol / L\n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_15.ipynb b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_15.ipynb new file mode 100644 index 00000000..ab29e076 --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_15.ipynb @@ -0,0 +1,150 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 15: Environmental Pollution and Control" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 1, Page No: 401" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "MM = 294 # Molar mass, K2Cr2O7\n", + "\n", + "# Variables\n", + "v_eff = 25 # cm ^ 3, effluent\n", + "v = 8.3 # cm ^ 3, K2Cr2O7\n", + "M = 0.001 # M, K2Cr2O7\n", + "\n", + "# Solution\n", + "w_O = v * 8 * 6 * M / 1000.\n", + "\n", + "print \"8.3 cm^3 of 0.006 N K2Cr2O7 =\", \"{:.3e}\".format(w_O), \"g of O2\"\n", + "print \"25 ml of the effluent requires\", \"{:.3e}\".format(w_O), \"g of O2\"\n", + "\n", + "cod = w_O * 10 ** 6 / 25.\n", + "print \"1 l of the effluent requires\", \"{:.2f}\".format(cod), \"g of O2\"\n", + "print \"COD of the effluent sample is\", \"{:.2f}\".format(cod), \"ppm or mg / L\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "8.3 cm^3 of 0.006 N K2Cr2O7 = 3.984e-04 g of O2\n", + "25 ml of the effluent requires 3.984e-04 g of O2\n", + "1 l of the effluent requires 15.94 g of O2\n", + "COD of the effluent sample is 15.94 ppm or mg / L\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 2, Page No: 401" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "v0 = 30 # cm^3, effluent\n", + "v1 = 9.8 # cm^3, K2Cr2O7\n", + "M = 0.001 # M, K2Cr2O7\n", + "\n", + "# Solution\n", + "O_30eff = 6 * 8 * v1 * M\n", + "print \"So 30 cm^3 of effluent contains =\", \"{:.4f}\".format(O_30eff), \"mg of O2\"\n", + "\n", + "cod = O_30eff * 1000 / 30.\n", + "\n", + "print \"1 l of the effluent requires\", cod, \"mg of O2\"\n", + "print \"COD of the effluent sample =\", cod, \"ppm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "So 30 cm^3 of effluent contains = 0.4704 mg of O2\n", + "1 l of the effluent requires 15.68 mg of O2\n", + "COD of the effluent sample = 15.68 ppm\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 3, Page No: 401" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "v0 = 25 # ml, sewage\n", + "d0_O = 410 # ppm, dissolved oxygen\n", + "d1_O = 120 # ppm, dissolved oxygen\n", + "v1 = 50 # ml, sewage\n", + "\n", + "# Solution\n", + "print \"BOD = (DOb - DOi) * Dilution Factor\"\n", + "print \"BOD = (DOb - DOi) *\",\n", + "print \"(ml of sample after dilution) / (ml of sample before dilution)\"\n", + "\n", + "BOD = (d0_O - d1_O) * (v1 / v0)\n", + "print \"BOD =\", BOD, \"ppm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "BOD = (DOb - DOi) * Dilution Factor\n", + "BOD = (DOb - DOi) * (ml of sample after dilution) / (ml of sample before dilution)\n", + "BOD = 580 ppm\n" + ] + } + ], + "prompt_number": 10 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_2.ipynb b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_2.ipynb new file mode 100644 index 00000000..58e602ab --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_2.ipynb @@ -0,0 +1,418 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2: Spectroscopy and Photochemistry" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 1, Page no: 65" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "m_br79 = 78.9183 # Mass of 79Br, amu\n", + "m_br81 = 80.9163 # Mass of 91Br, amu\n", + "Na = 6.022 * 10 ** 23 # Mole constant, /mol\n", + "pi = 3.141 # Pi\n", + "c = 3 * 10 ** 10 # Speed of light, cm /s\n", + "\n", + "# Variable\n", + "wave_no = 323.2 # Wave no. of fund. vibration of 79Br - 81Br, /cm\n", + "\n", + "# Solution\n", + "mu = (m_br79 * m_br81) / ((m_br79 + m_br81) * Na)\n", + "\n", + "k = 4 * (pi * c * wave_no) ** 2 * mu * 10 ** -3\n", + "\n", + "print \"The force constant of the bond is\", \"{:.3e}\".format(k), \"N/m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The force constant of the bond is 2.461e+02 N/m\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 2, Page no: 65" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "Na = 6.022 * 10 ** 23 # Mole constant, /mol\n", + "pi = 3.141 # Pi\n", + "c = 3 * 10 ** 10 # Speed of light, cm /s\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "\n", + "# Variables\n", + "b_l = 112.81 * 10 ** -12 # Equillibrium bond length, m\n", + "m1 = 12 # Mass of Carbon, g /mol\n", + "m2 = 16 # Mass of Oxygen, g /mol\n", + "\n", + "# Solution\n", + "mu = m1 * m2 / ((m1 + m2) * Na) # g\n", + "mu *= 10 ** -3 # kg\n", + "\n", + "B = h / (8 * pi ** 2 * mu * b_l ** 2 * c)\n", + "v2_3 = B * 6\n", + "\n", + "print \"The reduced mass of CO is\", \"{:.3e}\".format(mu), \"kg\"\n", + "print \"The frequency of 3->2 transition is\", \"{:.2f}\".format(v2_3), \"/cm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The reduced mass of CO is 1.139e-26 kg\n", + "The frequency of 3->2 transition is 11.59 /cm\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 3, Page no: 66" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "Na = 6.022 * 10 ** 23 # Mole constant, /mol\n", + "\n", + "# Variables\n", + "d_NaCl = 2.36 * 10 ** -10 # Intermolecular dist. NaCl, m\n", + "m_Cl = 35 * 10 ** -3 # Atomic mass, kg /mol\n", + "m_Na = 23 * 10 ** -3 # Atomic mass, kg /mol\n", + "\n", + "# Solution\n", + "mu = m_Na * m_Cl / ((m_Na + m_Cl) * 10 ** -3 * Na) * 10 ** -3\n", + "\n", + "I = mu * d_NaCl ** 2\n", + "\n", + "print \"The reduced mass of NaCl is\", \"{:.3e}\".format(mu), \"kg\"\n", + "print \"The moment of inertia of NaCl is\", \"{:.3e}\".format(I), \"kg.m^2\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The reduced mass of NaCl is 2.305e-26 kg\n", + "The moment of inertia of NaCl is 1.284e-45 kg.m^2\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 4, Page no: 66" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Constant\n", + "e = 4000 # Extinction coeff., dm^3/mol/cm\n", + "\n", + "# Variable\n", + "x = 3 # Solution thickness, cm\n", + "\n", + "# Solution\n", + "A = math.log10(1 / 0.3) # Absorbance\n", + "C = A / (e * x)\n", + "\n", + "print \"The concentration of the solution is\", \"{:.2e}\".format(C), \"mol/dm^3\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The concentration of the solution is 4.36e-05 mol/dm^3\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 5, Page no: 67" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "pi = 3.141 # Pi\n", + "c = 3 * 10 ** 10 # Speed of light, cm /s\n", + "\n", + "# Variables\n", + "v_bar = 2140 # Fundamental vibrating freq, /cm\n", + "m_C = 19.9 * 10 ** -27 # Atomic mass of C, kg\n", + "m_O = 26.6 * 10 ** -27 # Atomic mass of O, kg\n", + "\n", + "# Solution\n", + "mu = m_O * m_C / (m_C + m_O)\n", + "k = 4 * (pi * c * v_bar) ** 2 * mu\n", + "\n", + "print \"The force constant of the molecule is\", \"{:.3e}\".format(k), \"N/m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The force constant of the molecule is 1.852e+03 N/m\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 6, Page no: 67" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "print \"a) Microwave < IR < UV-Visible < X-Ray.\"\n", + "print \"b) HCl and NO because they possess permanent dipole moments, so they are rotationally active.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a) Microwave < IR < UV-Visible < X-Ray.\n", + "b) HCl and NO because they possess permanent dipole moments, so they are rotationally active.\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 7, Page no: 67" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Constants\n", + "pi = 3.141 # pi\n", + "c = 3 * 10 ** 10 # speed of light, cm /s\n", + "h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n", + "Na = 6.022 * 10 ** 23 # Mole constant, /mol\n", + "\n", + "# Variables\n", + "d = 20.7 # Interspacing, /cm\n", + "m1 = 1 # Mass of H, g / mol\n", + "m2 = 35.5 # Masso f Cl, g / mol\n", + "\n", + "# Solution\n", + "B = 0.1035 * 10 ** 2 # /m\n", + "I = h / (8 * pi ** 2 * B * c)\n", + "mu = m1 * m2 / ((m1 + m2) * Na)\n", + "mu *= 10 ** -3\n", + "r = math.sqrt(I / mu)\n", + "\n", + "print \"The intermolecular distance of HCl is\", \"{:.3e}\".format(r), \"m\"\n", + "# Discrepency in value is due to error in calculation in the textbook\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The intermolecular distance of HCl is 1.294e-10 m\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 8, Page no: 68" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Constant\n", + "e = 8000 # Molar absorbtion coeff, dm^3 / mol / cm\n", + "\n", + "# Variable\n", + "l = 2.5 # Thickness of solution, cm\n", + "\n", + "# Solution\n", + "C = math.log10(1 / 0.3) / (e * l)\n", + "\n", + "print \"The concentration of Solution from Lambert-Beer's Law is\",\n", + "print \"{:.2e}\".format(C), \"mol/dm^3\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The concentration of Solution from Lambert-Beer's Law is 2.61e-05 mol/dm^3\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 9, Page no: 68" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "print \"a) In the visible-UV spectra, CH2 = CHOCHOCH3 exhibits\"\n", + "print \"a higher value of lambda(max) because it has two conjugated\"\n", + "print \"chromophores, that is, one double bond (C=C) and a carbonyl\"\n", + "print \"group.\"\n", + "\n", + "print\n", + "print \"b) Because of the symmetrical vibrations of C=C double bond and\"\n", + "print \"triple bond, ethylene and acetylene do not absorb IR energy.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a) In the visible-UV spectra, CH2 = CHOCHOCH3 exhibits\n", + "a higher value of lambda(max) because it has two conjugated\n", + "chromophores, that is, one double bond (C=C) and a carbonyl\n", + "group.\n", + "\n", + "b) Because of the symmetrical vibrations of C=C double bond and\n", + "triple bond, ethylene and acetylene do not absorb IR energy.\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 10, Page no: 68" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "print \"Because CO2 is a linear molecule.\"\n", + "v_deg = 3 * 3 - 5\n", + "print \"The vibrational degree of freedom is\", v_deg" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Because CO2 is a linear molecule.\n", + "The vibrational degree of freedom is 4\n" + ] + } + ], + "prompt_number": 14 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_3.ipynb b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_3.ipynb new file mode 100644 index 00000000..16fb19f8 --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_3.ipynb @@ -0,0 +1,1116 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3: Thermodynamic and Chemical Equilibrium" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 1, Page no: 105" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "q = 120 # Heat from surrounding, cal\n", + "W = 70 # Work done, cal\n", + "\n", + "# Solution\n", + "delta_E = q - W\n", + "\n", + "print \"Change in internal Energy\", delta_E, \"cals.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Change in internal Energy 50 cals.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 2, Page no: 105" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "print \"CH4 (g) + 2O2 (g) -> CO2 (g) + 2H20 (l)\"\n", + "delta_n = 1 - (1 + 2)\n", + "solution = - 2 * 2 * 298 # cals\n", + "print \"Delta H - Delta E is:\", solution, \"cals\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "CH4 (g) + 2O2 (g) -> CO2 (g) + 2H20 (l)\n", + "Delta H - Delta E is: -1192 cals\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 3, Page no: 105" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "delta_G = -16.0 # Kelvin cal\n", + "delta_H = -10.0 # Kelvin cal\n", + "T = 300 # Kelvin\n", + "\n", + "# Solution\n", + "delta_S = (delta_H - delta_G) * 10 ** 3 / T # cal/deg\n", + "new_T = 330 # Kelvin\n", + "new_delta_G = (delta_H * 10 ** 3) - new_T * delta_S\n", + "\n", + "print \"The free energy at 330K is:\", \"{:.2e}\".format(new_delta_G), \"K cal\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The free energy at 330K is: -1.66e+04 K cal\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 4, Page no: 106" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "delta_S = -20.7 # cal /deg /mol\n", + "delta_H = -67.37 # K cal\n", + "T = 25 # deg C\n", + "\n", + "# Solution\n", + "T += 273 # K\n", + "delta_G = delta_H - (T * delta_S * 10 ** -3)\n", + "print \"The change in free energy at 25deg C is:\", delta_G, \"K cal / mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The change in free energy at 25deg C is: -61.2014 K cal / mol\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 5, Page no: 106" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "wt = 1 # g\n", + "delta_H = 149 # joules\n", + "\n", + "# Solution\n", + "delta_H_fusion = delta_H * (10 * 12 + 8 * 1)\n", + "print \"Enthalpy of fusion of naphthalene:\", delta_H_fusion * 10 ** -3, \"kJ/mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Enthalpy of fusion of naphthalene: 19.072 kJ/mol\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 6, Page no: 106" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "delta_H_acetylene = 230 # kJ/mol\n", + "delta_H_benzene = 85 # kJ/mol\n", + "T = 298 # K\n", + "\n", + "# Solution\n", + "delta_H = delta_H_benzene - 3 * delta_H_acetylene\n", + "print \"The enthalpy change for the reaction is:\", delta_H, \"kJ/mole\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The enthalpy change for the reaction is: -605 kJ/mole\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 7, Page no: 106" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "delta_H_vap = 2.0723 # kJ/g\n", + "Tb = 373 # K\n", + "\n", + "# Solution\n", + "delta_H_vap *= 18 # kJ/mol\n", + "delta_S = delta_H_vap / Tb\n", + "delta_G = delta_H_vap - Tb * delta_S\n", + "delta_S *= 1000\n", + "\n", + "print \"The Entropy change is:\", \"{:.1f}\".format(delta_S), \"J/mol/K\"\n", + "print \"The Free Energy change is:\", delta_G, \"kJ/mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Entropy change is: 100.0 J/mol/K\n", + "The Free Energy change is: 0.0 kJ/mol\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 8, Page no: 106" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Constant\n", + "R = 1.987 # cal /K /mol\n", + "\n", + "# Variables\n", + "moles = 5\n", + "Vo = 4 # litres, Initial Volume\n", + "Vf = 40 # litres, Final Volume\n", + "T = 27 # deg C\n", + "\n", + "# Solution\n", + "print \"dS = nRln(V2 / V1)\"\n", + "dS = moles * R * 2.303 * math.log10(Vf / Vo)\n", + "print \"The change in entropy is:\", \"{:.2f}\".format(dS), \"cal / degree\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "dS = nRln(V2 / V1)\n", + "The change in entropy is: 22.88 cal / degree\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 9, Page no: 107" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "wt = 10 # g\n", + "heat_abs = 4.5 # K\n", + "\n", + "# Solution\n", + "mole = 10 / 100.0 # mol\n", + "delta_H = heat_abs / mole\n", + "print \"The heat of the reaction is:\", delta_H, \"K cal / mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The heat of the reaction is: 45.0 K cal / mol\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 10, Page no: 107" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Constant\n", + "R = 8.314 # J / K\n", + "\n", + "# Variables\n", + "V_O2 = 2.8 # litres\n", + "V_H2 = 19.6 # litres\n", + "\n", + "# Solution\n", + "na = V_O2 / 22.4 # mol\n", + "nb = V_H2 / 22.4 # mol\n", + "Xa = na / (na + nb)\n", + "Xb = nb / (na + nb)\n", + "delta_S = (- R) * (na * math.log(Xa) + nb * math.log(Xb))\n", + "\n", + "print \"The increase in entropy on mixing is:\", \"{:.3f}\".format(delta_S), \"J / K\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The increase in entropy on mixing is: 3.132 J / K\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 11, Page no: 107" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Solution\n", + "print \"For 1 mole of ideal gas,\"\n", + "print \"\\tPV = RT or T = (PV) / R\\n\"\n", + "print \"Differentiating with respect to V at constant P,\"\n", + "print \"\\t[dT/dV]p = P/R\\n\"\n", + "print \"Differentiating again with respect oto P at constant V\"\n", + "print \"\\t[d2T/(dV*dP)] = 1/R\\n\"\n", + "print \"Now differectiating with respect to P at constant V,\"\n", + "print \"\\t[dT/dP]v = V/R\\n\"\n", + "print \"Differentiating again with respect to V at constant P,\"\n", + "print \"\\t[d2T/(dV*dP)] = 1/R\\n\"\n", + "print \"From equations we get:\"\n", + "print \"\\t[d2T/(dV*dP)] = [d2T/(dV*dP)]\\n\"\n", + "print \"Hence, dT is a perfect differential.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For 1 mole of ideal gas,\n", + "\tPV = RT or T = (PV) / R\n", + "\n", + "Differentiating with respect to V at constant P,\n", + "\t[dT/dV]p = P/R\n", + "\n", + "Differentiating again with respect oto P at constant V\n", + "\t[d2T/(dV*dP)] = 1/R\n", + "\n", + "Now differectiating with respect to P at constant V,\n", + "\t[dT/dP]v = V/R\n", + "\n", + "Differentiating again with respect to V at constant P,\n", + "\t[d2T/(dV*dP)] = 1/R\n", + "\n", + "From equations we get:\n", + "\t[d2T/(dV*dP)] = [d2T/(dV*dP)]\n", + "\n", + "Hence, dT is a perfect differential.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 12, Page no: 108" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "delta_G_25 = - 85.77 # k J, Free Energy at 25 C\n", + "delta_G_35 = - 83.68 # k J, Free Energy at 35 C\n", + "Ti = 273 + 25 # K\n", + "Tf = 273 + 35 # K\n", + "\n", + "# Solution\n", + "print \"Equating the entropy change at both the temperatures.\"\n", + "print \"(delta_H + delta_G_25) / Ti = (delta_H + delta_G_35) / Tf\"\n", + "delta_H = - 148\n", + "print \"The change in enthalpy for the process at 30C is\", delta_H, \"kJ\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equating the entropy change at both the temperatures.\n", + "(delta_H + delta_G_25) / Ti = (delta_H + delta_G_35) / Tf\n", + "The change in enthalpy for the process at 30C is -148 kJ\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 13, Page no: 108" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constants\n", + "Lv = 101 # cal /g, Latent headt of vap.\n", + "mwt = 78 # molecular weight of benzene\n", + "\n", + "# Variable\n", + "moles = 2\n", + "Tb = 80.2 # C, boiling point of benzene\n", + "\n", + "# Solution\n", + "Tb += 273 # K\n", + "delta_H = Lv * mwt\n", + "delta_S = delta_H / Tb\n", + "delta_G = delta_H - Tb * delta_S\n", + "print \"delta_S =\", \"{:.2f}\".format(delta_S), \"cal / K\"\n", + "print \"delta_G = delta_A =\", delta_G\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "delta_S = 22.30 cal / K\n", + "delta_G = delta_A = 0.0\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 14, Page no: 109" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "V1 = 6 # dm^3\n", + "V2 = 2 # dm^3\n", + "T1 = 27 # C\n", + "moles = 5\n", + "\n", + "# Solution\n", + "print \"T1*V1 ^ (gamma - 1) = T2 * V2 ^ (gamma - 1)\"\n", + "T1 += 273 # K\n", + "T2 = T1 * (V1 / V2) ** (8.314 / 20.91)\n", + "print \"The final temperature is\", \"{:.1f}\".format(T2), \"K\"\n", + "q = 0 # Adiabatic process\n", + "delta_E = - moles * 20.91 * (T2 - T1)\n", + "delta_E /= 1000\n", + "print \"q = \", q\n", + "print \"Change is Energy is\", \"{:.2f}\".format(delta_E), \"kJ / mol\"\n", + "W = - delta_E\n", + "print \"W = \", \"{:.2f}\".format(delta_E), \"kJ / mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "T1*V1 ^ (gamma - 1) = T2 * V2 ^ (gamma - 1)\n", + "The final temperature is 464.3 K\n", + "q = 0\n", + "Change is Energy is -17.18 kJ / mol\n", + "W = -17.18 kJ / mol\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 15, Page no: 109" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Constant\n", + "R = 8.314 # J / K mol\n", + "\n", + "# Variables\n", + "mole = 1\n", + "V1 = 5 # dm^3\n", + "V2 = 10 # dm^3\n", + "T = 300 # K\n", + "\n", + "# Solution\n", + "print \"For isothermal and reversible process,\"\n", + "delta_E = delta_H = 0\n", + "delta_A = delta_G = - 2.303 * mole * R * T * math.log10(V2 / V1)\n", + "q = W = - delta_G\n", + "print \"delta_E = delta_H =\", delta_H\n", + "print \"delta_G = delta_A =\", \"{:.3f}\".format(delta_G), \"J / mol\\n\"\n", + "print \"For isothermal and reversible expansion\"\n", + "print \"q = W = -delta_G =\", \"{:.3f}\".format(W), \"J / mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For isothermal and reversible process,\n", + "delta_E = delta_H = 0\n", + "delta_G = delta_A = -1729.159 J / mol\n", + "\n", + "For isothermal and reversible expansion\n", + "q = W = -delta_G = 1729.159 J / mol\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 16, Page no: 109" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Constant\n", + "R = 8.314 # J / K mol\n", + "\n", + "# Variables\n", + "n = 5 # moles\n", + "T = 27 # C\n", + "V1 = 50.0 # L, Initial Volume\n", + "V2 = 1000 # L, Final Volume\n", + "\n", + "# Solution\n", + "T += 273\n", + "delta_G = 2.303 * n * R * T * math.log10(V1 / V2)\n", + "delta_G /= 1000\n", + "print \"The free energy change is\", \"{:.3f}\".format(delta_G), \"k J\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The free energy change is -37.367 k J\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 17, Page no: 110" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "delta_H_neu = - 51.46 # k J/mol, neutralization\n", + "delta_H_ion = - 57.1 # k J/mol, ionization\n", + "\n", + "# Solution\n", + "delta_H = - delta_H_ion + delta_H_neu\n", + "print \"The head of ionization for NH4OH is\", delta_H, \"kJ / mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The head of ionization for NH4OH is 5.64 kJ / mol\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 18, Page no: 110" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "print \"For 1 mole of an ideal gas,\"\n", + "print \"PV = RT or V = (RT)/P\"\n", + "print \"(dV/ dP) = -(RT/P^2)\\t (at constant temperature)\"\n", + "print \"(d^2V/ (dP*dT)) = -(R/ P^2)\"\n", + "print \"(dV/ dT) = (R/ P)\\t (at constant pressure)\"\n", + "print \"(d^2V/ (dT*dP)) = -(R/ P^2)\\n\"\n", + "print \"(d^2V/ (dT*dP)) = (d^2V/ (dP*dT))\\t[From above equations]\"\n", + "print \"Hence, dV is an exact differential.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For 1 mole of an ideal gas,\n", + "PV = RT or V = (RT)/P\n", + "(dV/ dP) = -(RT/P^2)\t (at constant temperature)\n", + "(d^2V/ (dP*dT)) = -(R/ P^2)\n", + "(dV/ dT) = (R/ P)\t (at constant pressure)\n", + "(d^2V/ (dT*dP)) = -(R/ P^2)\n", + "\n", + "(d^2V/ (dT*dP)) = (d^2V/ (dP*dT))\t[From above equations]\n", + "Hence, dV is an exact differential.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 19, Page no: 110" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "print \"Let P1, V1, T1 and P2, V2, T2 b the initial and final\",\n", + "print \"state, respectively of the system\"\n", + "print \"W rev = nRT[P1/ P2 - 1]\"\n", + "print \"W irr = nRT[1 - P2/ P1]\"\n", + "print \"W rev - W irr = [nRT/(P1P2)]*(P1 - P2)^2\"\n", + "print \"Because RHS of the above equation is always positive,\"\n", + "print \"W rev > W irr\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Let P1, V1, T1 and P2, V2, T2 b the initial and final state, respectively of the system\n", + "W rev = nRT[P1/ P2 - 1]\n", + "W irr = nRT[1 - P2/ P1]\n", + "W rev - W irr = [nRT/(P1P2)]*(P1 - P2)^2\n", + "Because RHS of the above equation is always positive,\n", + "W rev > W irr\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 20, Page no: 111" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Solution\n", + "Eq_HI = 1.56 / 2\n", + "Eq_H2 = 0.22 / 2\n", + "Eq_I2 = 0.22 / 2\n", + "Kc = Eq_H2 * Eq_I2 / (Eq_HI ** 2)\n", + "print \"The equilibrium constant for the dissociation reaction\",\n", + "print \"{:.4f}\".format(Kc)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The equilibrium constant for the dissociation reaction 0.0199\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 21, Page no: 111" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Kc = 0.5 # / mole^2 litre^2\n", + "T = 400 # K\n", + "R = 0.082 # litre atm degree^-1 mole^-1\n", + "\n", + "# Solution\n", + "Kp = Kc * (R * T) ** (-2)\n", + "\n", + "print \"The given equilibrium is\"\n", + "print \"\\t\\tN2(g) + 3H2(g) <--> 2NH3(g)\"\n", + "print \"Kp is\", \"{:.3e}\".format(Kp)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The given equilibrium is\n", + "\t\tN2(g) + 3H2(g) <--> 2NH3(g)\n", + "Kp is 4.648e-04\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 22, Page no: 111" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "solubility = 7.5 * 10 ** - 5 # mol L^-1\n", + "\n", + "# Solution\n", + "Ksp = 4 * (solubility ** 3)\n", + "print \"Solubility product of the salt is\", \"{:.4e}\".format(Ksp), \"mol^3 / L^-3\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Solubility product of the salt is 1.6875e-12 mol^3 / L^-3\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 23, Page no: 111" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Ti = 25 # C\n", + "S = 0.00179 # g / L\n", + "\n", + "# Solution\n", + "S /= 170 # mol / L\n", + "Ksp = S ** 2\n", + "print \"Solubility product at 25 C is\", \"{:.4e}\".format(Ksp), \"mol^2 / L^-2\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Solubility product at 25 C is 1.1087e-10 mol^2 / L^-2\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 24, Page no: 111" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Ksp = 8 * 10 ** - 5 # Solubility product PbBr2\n", + "disso = 80 / 100 # % dissociation\n", + "\n", + "# Solution\n", + "S = (Ksp / 4) ** (1 / 3.0) # Solubility is 100%\n", + "S_80 = S * (80 / 100.0)\n", + "S_per_g = S_80 * 367 - 1.621\n", + "print \"Solubility in gm per litre is\", \"{:.3f}\".format(S_per_g), \"gm / litre\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Solubility in gm per litre is 6.349 gm / litre\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 25, Page no: 112" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "n_acid = 1 # mole\n", + "n_alcohol = 1 # mole\n", + "T = 25 # C\n", + "x = 0.667 # mole\n", + "\n", + "# Solution\n", + "print \"Kc = [CH3COOC2H][H2O] / ([CH3COOH][C2H2OH])\"\n", + "Kc = 4\n", + "print \"[CH3COOH] = (2 - x) / V\"\n", + "print \"[C2H5OH] = (1 - x) / V\"\n", + "print \"[CH3COOC2H5] = [H20] = x / V\"\n", + "print \"3x^2 - 12x + 8 = 0\"\n", + "print \"x =\", 2.366, \"or\", 0.634\n", + "print \"0.634 mole of ester would be formed, because the other value,\",\n", + "print \"x = 2.366, is not permissible.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Kc = [CH3COOC2H][H2O] / ([CH3COOH][C2H2OH])\n", + "[CH3COOH] = (2 - x) / V\n", + "[C2H5OH] = (1 - x) / V\n", + "[CH3COOC2H5] = [H20] = x / V\n", + "3x^2 - 12x + 8 = 0\n", + "x = 2.366 or 0.634\n", + "0.634 mole of ester would be formed, because the other value, x = 2.366, is not permissible.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 26, Page no: 113" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "n_acid = 0.2 # mole\n", + "n_salt = 0.10 # mole\n", + "Ka = 1.8 * 10 ** -5\n", + "\n", + "# Solution\n", + "pH = - math.log10(Ka) + math.log10(n_salt / n_acid)\n", + "print \"The pH of acidic buffer is\", \"{:.3f}\".format(pH)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The pH of acidic buffer is 4.444\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 27, Page no: 113" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "n_salt = 0.02 # mole\n", + "n_base = 0.2 # mole\n", + "pKb = 4.7\n", + "\n", + "# Solution\n", + "pOH = pKb + math.log10(n_salt / n_base)\n", + "pH = 14 - pOH\n", + "print \"pH of a buffer solution is\", pH\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "pH of a buffer solution is 10.3\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 28, Page no: 113" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "n_H2 = 8.07 # moles\n", + "n_I2 = 9.08 # moles\n", + "T = 448 # C\n", + "n_eqHI2 = 13.38 # moles\n", + "\n", + "# Solution\n", + "x = n_eqHI2 / 2 + 6.69\n", + "Kc = n_eqHI2 ** 2 / (n_H2 * n_I2)\n", + "\n", + "print \"H2 + I2 <--> 2HI\"\n", + "print \"1 0 0\"\n", + "print \"1 - 2xx x x\"\n", + "print \"x/(1 - 2x) = (1/Kc)^0.5\"\n", + "print \"Dissociation constant of HI is 106.75 x 10^-3\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "H2 + I2 <--> 2HI\n", + "1 0 0\n", + "1 - 2xx x x\n", + "x/(1 - 2x) = (1/Kc)^0.5\n", + "Dissociation constant of HI is 106.75 x 10^-3\n" + ] + } + ], + "prompt_number": 9 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_5.ipynb b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_5.ipynb new file mode 100644 index 00000000..b0e0ac54 --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_5.ipynb @@ -0,0 +1,491 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5: Chemical Kinetics and Catalysis" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 1, Page no: 149" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "K = 3.5 * 10 ** - 2 # Rate constant\n", + "\n", + "# Solution\n", + "print \"First order reaction = 0.693 / K\"\n", + "t_05 = 0.693 / K\n", + "print \"Time taken for half the initial concentration to react\", t_05, \"minutes\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "First order reaction = 0.693 / K\n", + "Time taken for half the initial concentration to react 19.8 minutes\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 2, Page no: 150" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "t05 = 40 # minutes\n", + "\n", + "# Solution\n", + "print \"Rate constant = 0.693 / t05\"\n", + "K = 0.693 / t05\n", + "print \"Rate constant\", \"{:.4f}\".format(K), \"/ min\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate constant = 0.693 / t05\n", + "Rate constant 0.0173 / min\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 3, Page no: 150" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Variables\n", + "t0 = 37.0 # cm^3 of KMnO4\n", + "t5 = 29.8 # cm^3 of KMnO4\n", + "t15 = 19.6 # cm^3 of KMnO4\n", + "t25 = 12.3 # cm^3 of KMnO4\n", + "t45 = 5.00 # cm^3 of KMnO4\n", + "\n", + "# Solution\n", + "K5 = 2.303 / 5 * log10(t0 / t5)\n", + "K15 = 2.303 / 15 * log10(t0 / t15)\n", + "K25 = 2.303 / 25 * log10(t0 / t25)\n", + "K45 = 2.303 / 45 * log10(t0 / t45)\n", + "\n", + "print \"At t = 5 min, K =\", \"{:.3e}\".format(K5), \"/min\"\n", + "print \"At t = 15 min, K =\", \"{:.3e}\".format(K15), \"/min\"\n", + "print \"At t = 25 min, K =\", \"{:.3e}\".format(K25), \"/min\"\n", + "print \"At t = 45 min, K =\", \"{:.3e}\".format(K45), \"/min\"\n", + "print \"As the different values of K are nearly same, so the reaction\",\n", + "print \"is of first-order\"\n", + "K = (K45 + K25 + K5 + K15) / 4\n", + "print \"The average value of K = \", \"{:.3e}\".format(K), \"/min\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "At t = 5 min, K = 4.329e-02 /min\n", + "At t = 15 min, K = 4.237e-02 /min\n", + "At t = 25 min, K = 4.406e-02 /min\n", + "At t = 45 min, K = 4.449e-02 /min\n", + "As the different values of K are nearly same, so the reaction is of first-order\n", + "The average value of K = 4.355e-02 /min\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem : 4, Page no: 150" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "t = 60 # min\n", + "x = \"0.5a\"\n", + "K = 5.2 * 10 ** - 3 # / mol L / min\n", + "\n", + "# Solution\n", + "a = 1 / (t * K)\n", + "print \"Initial concentration\", \"{:.3f}\".format(a), \"mol / L\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Initial concentration 3.205 mol / L\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 5, Page no: 151" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Solution\n", + "print \"99.9 % / 50 % =\",\n", + "times = round((2.303 * log10(100 / (100 - 99.9))) / (2.303 * log10(100 / (100 - 50))))\n", + "print times\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "99.9 % / 50 % = 10.0\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 6, Page no: 151" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Constants\n", + "R = 1.987 # cal / K / mol\n", + "\n", + "# Variables\n", + "T1 = 273.0 # K\n", + "T2 = 303.0 # K\n", + "K1 = 2.45 * 10 ** -5\n", + "K2 = 162 * 10 ** -5\n", + "\n", + "# Solution\n", + "Ea = log10(K2 / K1) * R * 2.303 / ((T2 - T1) / (T1 * T2))\n", + "print \"The activation energy of the reaction is\", int(Ea), \"cal / mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The activation energy of the reaction is 22968 cal / mol\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 7, Page no: 151" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "t05 = 30 # minutes\n", + "a = 0.1 # M\n", + "\n", + "# Solution\n", + "print \"For second order reaction,\"\n", + "print \"t0.5 = 1 / Ka\"\n", + "K = 1 / (a * t05)\n", + "print \"The rate constant is\", \"{:.3f}\".format(K), \"mol / lit / min\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For second order reaction,\n", + "t0.5 = 1 / Ka\n", + "The rate constant is 0.333 mol / lit / min\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 8, Page no: 151" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "\n", + "# Variables\n", + "T = 500 # C\n", + "Pi = 350 # torr\n", + "r1 = 1.07 # torr / s\n", + "r2 = 0.76 # torr / s\n", + "\n", + "# Solution\n", + "print \"1.07 = k(0.95a)^n\"\n", + "print \"0.76 = k(0.80a)^n\"\n", + "n = log(r1 / r2) / log(0.95 / 0.80)\n", + "print \"Hence, order of reaction is\", round(n)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1.07 = k(0.95a)^n\n", + "0.76 = k(0.80a)^n\n", + "Hence, order of reaction is 2.0\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 9, Page no: 152" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Solution\n", + "print \"Applying steady state approximation to the concentration of NOCl2,\",\n", + "print \"we get\"\n", + "print \"d[NOCl2] / dt = 0 = k1 [NO] [Cl2] - k-1 [NOCl2] - k2 [NO] [NOCl2]\"\n", + "print \"[NOCl2] = k1 [NO] [Cl2] / (k-1 + k2 [NO])\"\n", + "print \"Now, the overall rate of reaction,\"\n", + "print \"d[NOCl2] / dt = k2 [NO] [NOCl2]\"\n", + "print \"From the above equations we get,\"\n", + "print \"(k1 k2 / k-1) * [NO]^2 * [Cl2], if k2[NO] << k-1\"\n", + "print \"k [NO]^2[Cl2], where k = k1 * k2 / k-1\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Applying steady state approximation to the concentration of NOCl2, we get\n", + "d[NOCl2] / dt = 0 = k1 [NO] [Cl2] - k-1 [NOCl2] - k2 [NO] [NOCl2]\n", + "[NOCl2] = k1 [NO] [Cl2] / (k-1 + k2 [NO])\n", + "Now, the overall rate of reaction,\n", + "d[NOCl2] / dt = k2 [NO] [NOCl2]\n", + "From the above equations we get,\n", + "(k1 k2 / k-1) * [NO]^2 * [Cl2], if k2[NO] << k-1\n", + "k [NO]^2[Cl2], where k = k1 * k2 / k-1\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 10, Page no: 153" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Constant\n", + "R = 1.987 # cal / K / mol\n", + "\n", + "# Variables\n", + "K2_K1 = 4 # factor increase\n", + "T1 = 27 # C\n", + "T2 = 47 # C\n", + "\n", + "# Solution\n", + "T1 += 273.0\n", + "T2 += 273.0\n", + "Ea = log10(4) * 2.303 * R * (T1 * T2 / (T2 - T1))\n", + "print \"The activation energy for the reaction is\", \"{:.2e}\".format(Ea),\n", + "print \"cal /mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The activation energy for the reaction is 1.32e+04 cal /mol\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 11, Page no: 153" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Variables\n", + "a = 1 # mole\n", + "x = 3 / 4.0 # reaction completed\n", + "\n", + "# Solution\n", + "K = (2.303 / 6) * log10(1 / (1 - x))\n", + "print \"The rate constant is\", \"{:.3f}\".format(K), \"/min\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The rate constant is 0.231 /min\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 12, Page no: 153" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Solution\n", + "print \"Let the initial concentration be 100, when x = 25\",\n", + "print \" t = 30 minutes\"\n", + "a = 100\n", + "x = 25.0\n", + "t = 30 # minutes\n", + "K = 2.303 / t * log10(a / (a - x))\n", + "t05 = 0.683 / K\n", + "t = 2.303 / K * log10(a / x)\n", + "print \"K =\", \"{:.2e}\".format(K), \"/ min\"\n", + "print \"T0.5 =\", \"{:.2f}\".format(t05), \"min\"\n", + "print \"t =\", \"{:.1f}\".format(t), \"min\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Let the initial concentration be 100, when x = 25 t = 30 minutes\n", + "K = 9.59e-03 / min\n", + "T0.5 = 71.21 min\n", + "t = 144.6 min\n" + ] + } + ], + "prompt_number": 15 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_6.ipynb b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_6.ipynb new file mode 100644 index 00000000..b15856da --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_6.ipynb @@ -0,0 +1,620 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Chapter 6: Electrochemistry" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 1, Page no: 180" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Variables\n", + "T = 25 # C\n", + "E = 0.296 # V\n", + "Cu = 0.015 # M\n", + "\n", + "# Solution\n", + "Eo = E - 0.0296 * log10(Cu)\n", + "print \"The standard electrode potential is\", \"{:.3f}\".format(Eo), \"V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The standard electrode potential is 0.350 V\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 2, Page no: 180" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Variables\n", + "T = 25 # C\n", + "Cu = 0.1 # M\n", + "Zn = 0.001 # M\n", + "Eo = 1.1\n", + "\n", + "# Solution\n", + "E = Eo + 0.0296 * log10(Cu / Zn)\n", + "print \"The emf of Daniell cell is\", \"{:.4f}\".format(E), \"V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The emf of Daniell cell is 1.1592 V\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 3, Page no: 180" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Constant\n", + "R = 8.314 # J / K\n", + "F = 96500 # C / mol\n", + "\n", + "# Variables\n", + "Cu = 0.15 # M\n", + "Eo = 0.34 # V\n", + "T = 298 # K\n", + "n = 2 # moles\n", + "\n", + "# Solution\n", + "E = Eo + (2.303 * R * T) / (n * F) * log10(Cu)\n", + "print \"The single electrode potential for copper metal is\", \"{:.4f}\".format(E),\n", + "print \"V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The single electrode potential for copper metal is 0.3156 V\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 4, Page no: 181" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variable\n", + "Eo_Cu = 0.3370 # Cu+2 -> Cu\n", + "Eo_Zn = - 0.7630 # Zn -> Zn +2\n", + "\n", + "# Solution\n", + "Eo_cell = Eo_Cu - Eo_Zn\n", + "print \"Daniel cell is, Zn | Zn +2 || Cu+2 | Cu\"\n", + "print \"Eo (cell) is\", Eo_cell, \"V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Daniel cell is, Zn | Zn +2 || Cu+2 | Cu\n", + "Eo (cell) is 1.1 V\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 5, Page no: 181" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variable\n", + "Eo_Cu = 0.3370 # Cu+2 -> Cu\n", + "Eo_Cd = - 0.4003 # Cd -> Cd +2\n", + "\n", + "# Solution\n", + "Eo_cell = Eo_Cu - Eo_Cd\n", + "print \"Cell is, Cd | Cd +2 || Cu+2 | Cu\"\n", + "print \"Eo (cell) is\", Eo_cell, \"V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Cell is, Cd | Cd +2 || Cu+2 | Cu\n", + "Eo (cell) is 0.7373 V\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 6, Page no: 181" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "F = 96500 # C / mol\n", + "\n", + "# Variables\n", + "n = 2\n", + "T = 25 # C\n", + "Eo_Ag = 0.80 # Ag+ / Ag\n", + "Eo_Ni = - 0.24 # Ni+2 / Ni\n", + "\n", + "# Solution\n", + "Eo_Cell = Eo_Ag - Eo_Ni\n", + "print \"Standard free energy change,\"\n", + "delta_Go = - n * F * Eo_Cell\n", + "print delta_Go, \"J / mol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Standard free energy change,\n", + "-200720.0 J / mol\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 7, Page no: 181" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Solution\n", + "print \"-------------------------\"\n", + "print \"Reduction half reaction:\",\n", + "print \"2Fe+3 + 2e- --> 2Fe+2\"\n", + "print \"Oxidation half reaction:\",\n", + "print \"Fe - 2e- --> Fe+2\"\n", + "print \"Overall cell reaction :\",\n", + "print \"2Fe+3 + Fe --> 3Fe+2\"\n", + "\n", + "print \"-------------------------\"\n", + "print \"Reduction half reaction:\",\n", + "print \"Hg+2 + 2e- --> Hg\"\n", + "print \"Oxidation half reaction:\",\n", + "print \"Zn - 2e- --> Zn+2\"\n", + "print \"Overall cell reaction :\",\n", + "print \"Hg+2 + Zn --> Hg + Zn+2\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "-------------------------\n", + "Reduction half reaction: 2Fe+3 + 2e- --> 2Fe+2\n", + "Oxidation half reaction: Fe - 2e- --> Fe+2\n", + "Overall cell reaction : 2Fe+3 + Fe --> 3Fe+2\n", + "-------------------------\n", + "Reduction half reaction: Hg+2 + 2e- --> Hg\n", + "Oxidation half reaction: Zn - 2e- --> Zn+2\n", + "Overall cell reaction : Hg+2 + Zn --> Hg + Zn+2\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 8, Page no: 181" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "F = 96500 # C / mol\n", + "\n", + "# Variables\n", + "E1o = - 2.48 # V\n", + "E2o = 1.61 # V\n", + "\n", + "# Solution\n", + "delta_G1 = - 3 * F * (- 2.48)\n", + "delta_G2 = - 1 * F * 1.61\n", + "print \"delta_G3 = delta_G1 + delta_G2\"\n", + "print \"delta_G3 = - 4 * F * E3o\"\n", + "E3o = (delta_G1 + delta_G2) / (- 4 * F)\n", + "print \"The reduction potential for the half-cell Pt/Ce, Ce+4\",\n", + "print \"is\", \"{:.4f}\".format(E3o), \"V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "delta_G3 = delta_G1 + delta_G2\n", + "delta_G3 = - 4 * F * E3o\n", + "The reduction potential for the half-cell Pt/Ce, Ce+4 is -1.4575 V\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 9, Page no: 182" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Solution\n", + "print \"Anodic half reaction :\",\n", + "print \" Cd --> Cd+2 + 2e-\"\n", + "print \"Cathodic half reaction :\",\n", + "print \"2H+ + 2e- --> H2\"\n", + "print \"-\" * 50\n", + "print \"Overall cell reaction :\",\n", + "print \"Cd + 2H+ <--> Cd+2 + H2\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Anodic half reaction : Cd --> Cd+2 + 2e-\n", + "Cathodic half reaction : 2H+ + 2e- --> H2\n", + "--------------------------------------------------\n", + "Overall cell reaction : Cd + 2H+ <--> Cd+2 + H2\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 10, Page no: 182" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Variables\n", + "T = 25 # C\n", + "Cu = 0.1 # M\n", + "Zn = 0.001 # M\n", + "Eo = 1.1 # V\n", + "\n", + "# Solution\n", + "print \"Zn(s) | Zn+2 (0.001M) || Cu+2(0.1M) | Cu(s)\"\n", + "Ecell = Eo + 0.0592 / 2 * log10(Cu / Zn)\n", + "print \"The emf of a Daniell cell is\", \"{:.4f}\".format(Ecell), \"V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Zn(s) | Zn+2 (0.001M) || Cu+2(0.1M) | Cu(s)\n", + "The emf of a Daniell cell is 1.1592 V\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 11, Page no: 182" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Variables\n", + "pH = 7 # O2\n", + "Eo = 1.229 # V\n", + "pO2 = 0.20 # bar\n", + "\n", + "# Solution\n", + "print \"Nearnst's equation at 25C is,\"\n", + "print \"E = Eo - (0.0592 / 2) * log(1 / ([H+]^2 * [pO2]^ (1/2)))\"\n", + "E = Eo - (0.0592 / 2) * log10(1.0 / (((10 ** (- 7)) ** 2) * (pO2 ** (1 / 2.0))))\n", + "print \"The reduction potential for the reduction is\",\n", + "print \"{:.3f}\".format(E), \"V\"\n", + "# descrepency due to calculation error in the book" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Nearnst's equation at 25C is,\n", + "E = Eo - (0.0592 / 2) * log(1 / ([H+]^2 * [pO2]^ (1/2)))\n", + "The reduction potential for the reduction is 0.804 V\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 12, Page no: 183" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "E_KCl = 0.2415 # V\n", + "E_cell = 0.445 # V\n", + "\n", + "\n", + "# Solution\n", + "print \"Emf of the cell is\"\n", + "print \"At 25C,\"\n", + "print \"E = Er - El = Eref - ((RT)/ F) * ln H+\"\n", + "pH = (E_cell - E_KCl) / 0.059\n", + "\n", + "Eo_cell = - 0.8277 # V\n", + "print \"Thus, equilibrium constant for the reaction\"\n", + "print \"\\t 2H2O --> H3O+ + OH- may be calculated as\"\n", + "K = 10 ** (Eo_cell / 0.0591)\n", + "print \"K =\", \"{:.0e}\".format(K)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Emf of the cell is\n", + "At 25C,\n", + "E = Er - El = Eref - ((RT)/ F) * ln H+\n", + "Thus, equilibrium constant for the reaction\n", + "\t 2H2O --> H3O+ + OH- may be calculated as\n", + "K = 1e-14\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 13, Page no: 183" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "EoSn = 0.15 # V\n", + "EoCr = - 0.74 # V\n", + "\n", + "# Solution\n", + "print \"3Sn+4 + 2Cr --> 3Sn+2 + 2Cr+3\"\n", + "Eo_cell = EoSn - EoCr\n", + "n = 6\n", + "K = 10 ** (n * Eo_cell / 0.0591)\n", + "print \"The equillibrium constant for th reaction is\", \"{:.2e}\".format(K)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "3Sn+4 + 2Cr --> 3Sn+2 + 2Cr+3\n", + "The equillibrium constant for th reaction is 2.27e+90\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 14, Page no: 184" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T = 25 # C\n", + "Eo = - 0.8277 # V\n", + "\n", + "# Solution\n", + "print \"The reversible reaction,\"\n", + "print \"2H2O <--> H3O+ + OH-\"\n", + "print \"May be divided into two parts.\"\n", + "print \"2H2O + e- --> 1/2 H2 + OH- (cathode) Eo = -0.8277 V\"\n", + "print \"H2O + 1/2 H2 --> H3O+ + e- (anode) Eo = 0\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The reversible reaction,\n", + "2H2O <--> H3O+ + OH-\n", + "May be divided into two parts.\n", + "2H2O + e- --> 1/2 H2 + OH- (cathode) Eo = -0.8277 V\n", + "H2O + 1/2 H2 --> H3O+ + e- (anode) Eo = 0\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem: 15, Page no: 184" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "E = 0.4 # V\n", + "\n", + "# Solution\n", + "print \"The cell is Pt(H2) | H+, pH2 = 1 atm\"\n", + "print \"The cell reaction is\"\n", + "print \"1/2 H2 --> H+ + e-\"\n", + "pH = E / 0.0591\n", + "print \"pH =\", \"{:.3f}\".format(pH)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The cell is Pt(H2) | H+, pH2 = 1 atm\n", + "The cell reaction is\n", + "1/2 H2 --> H+ + e-\n", + "pH = 6.768\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_7.ipynb b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_7.ipynb new file mode 100644 index 00000000..229b5da3 --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/Chapter_7.ipynb @@ -0,0 +1,431 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 7: Solid State" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 1, Page no: 209" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Solution\n", + "print \"Weiss indices \\t\\t\\t1/2, 2/3, infinity \\t2/3, 2, 1/3\"\n", + "print \"Reciprocal of Weiss indices 2, 3/2, 1/infinity \\t3/2, 1/2, 3\"\n", + "print \"Clear fractions\\t\\t\\t4, 3, 0 \\t3, 1, 6\"\n", + "print \"Miller indices \\t\\t\\t (430) \\t (316)\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Weiss indices \t\t\t1/2, 2/3, infinity \t2/3, 2, 1/3\n", + "Reciprocal of Weiss indices 2, 3/2, 1/infinity \t3/2, 1/2, 3\n", + "Clear fractions\t\t\t4, 3, 0 \t3, 1, 6\n", + "Miller indices \t\t\t (430) \t (316)\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 2, Page no: 209" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "\n", + "# Variable\n", + "a = 450 # pm\n", + "\n", + "# Solution\n", + "d220 = a / sqrt(2 ** 2 + 2 ** 2 + 0)\n", + "print \"Interplanar spacing\", int(d220)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Interplanar spacing 159\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 3, Page no: 209" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Solution\n", + "print \"Intercept \\t(a, b ,c)\\t\\t(a, 2b, c) \\t\\t(a, 2b, 2c) \\t\\t(infi, b, -c)\"\n", + "print \"Weiss indices\\t 1, 1, 1 \\t\\t 1, 2, 1 \\t\\t 1, 2, 2 \\t\\t infi, 1, -1\"\n", + "print \"Reciprocals \\t 1, 1, 1 \\t\\t 1, 1/2, 1 \\t\\t 1, 1/2, 1/2\\t\\t 0, 1, -1\"\n", + "print \"Miller indicec\\t (111) \\t\\t (212) \\t\\t (211) \\t\\t (011)\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Intercept \t(a, b ,c)\t\t(a, 2b, c) \t\t(a, 2b, 2c) \t\t(infi, b, -c)\n", + "Weiss indices\t 1, 1, 1 \t\t 1, 2, 1 \t\t 1, 2, 2 \t\t infi, 1, -1\n", + "Reciprocals \t 1, 1, 1 \t\t 1, 1/2, 1 \t\t 1, 1/2, 1/2\t\t 0, 1, -1\n", + "Miller indicec\t (111) \t\t (212) \t\t (211) \t\t (011)\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 4, Page no: 209" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "rNa = 0.98 * 10 ** - 10 # m\n", + "rCl = 1.81 * 10 ** - 10 # m\n", + "\n", + "# Solution\n", + "rr = rNa / rCl\n", + "print \"When the radius ration is\", \"{:.2f}\".format(rr),\n", + "print \", the coordination number is 6.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "When the radius ration is 0.54 , the coordination number is 6.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 5, Page no: 210" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "rLi = 68 # pm\n", + "rF = 136. # pm\n", + "\n", + "# Solution\n", + "rr = rLi / rF\n", + "print \"Radius ratio =\", rr\n", + "print \"The structure of LiF is scc and C.N of Li+ = 6\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Radius ratio = 0.5\n", + "The structure of LiF is scc and C.N of Li+ = 6\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 6, Page no: 210" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sin\n", + "\n", + "\n", + "# Variables\n", + "lamda = 2 * 10 ** - 10 # m\n", + "theta = 30 # degrees\n", + "\n", + "# Solution\n", + "print \"For first-order reflection\"\n", + "d = lamda / (2 * sin(theta))\n", + "dist = 2 * d\n", + "print \"Hence, distance between planes is\", \"{:.0e}\".format(abs(dist)), \"m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For first-order reflection\n", + "Hence, distance between planes is 2e-10 m\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7, Page no: 210" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "\n", + "# Variables\n", + "r = 174.6 # pm\n", + "\n", + "# Solution\n", + "a = r * sqrt(8)\n", + "print \"For 200 plane: h = 2, k = 0, l = 0\"\n", + "d200 = a / sqrt(2 ** 2)\n", + "print \"d200 =\", \"{:.1f}\".format(d200), \"pm\"\n", + "print \"For 200 plane: h = 2, k = 2, l = 0\"\n", + "d220 = a / sqrt(2 ** 2 + 2 ** 2)\n", + "print \"d220 =\", d220, \"pm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For 200 plane: h = 2, k = 0, l = 0\n", + "d200 = 246.9 pm\n", + "For 200 plane: h = 2, k = 2, l = 0\n", + "d220 = 174.6 pm\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 8, Page no: 210" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "N = 6.023 * 10 ** 23\n", + "\n", + "# Variables\n", + "wt = 55.6\n", + "p = 0.29 # nm\n", + "n = 2\n", + "\n", + "# Solution\n", + "print \"For BCC pattern,\"\n", + "print \"number of atoms per unit cell = 2\"\n", + "d = n * (wt * 10 ** -3) / (N * (p * 10 ** - 9) ** 3)\n", + "print \"Density of the metal is\", \"{:.2e}\".format(d), \"kg / m^3\"\n", + "print \"Number of nearest enighbour for BCC = 8\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For BCC pattern,\n", + "number of atoms per unit cell = 2\n", + "Density of the metal is 7.57e+03 kg / m^3\n", + "Number of nearest enighbour for BCC = 8\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 9, Page no: 210" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Solution\n", + "print \"Intercept \\t2a/2, 2b, c/3\"\n", + "print \"Weiss indices\\t1, 2, 1/3\"\n", + "print \"Reciprocals \\t1, 1/2, 3\"\n", + "print \"Miller indices\\t (216)\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Intercept \t2a/2, 2b, c/3\n", + "Weiss indices\t1, 2, 1/3\n", + "Reciprocals \t1, 1/2, 3\n", + "Miller indices\t (216)\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 10, Page no: 211" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Constant\n", + "N = 6.023 * 10 ** 23\n", + "\n", + "# Variables\n", + "D = 0.53 # g / cm ^ 3\n", + "MM = 6.94 # g / mol\n", + "n = 2\n", + "\n", + "# Solution\n", + "print \"For BCC pattern,\"\n", + "print \"number of atoms per unit cell = 2\"\n", + "V = D * N / (n * MM)\n", + "V = 1 / V\n", + "print \"Volume of a unit cell of lithium metal is\", \"{:.2e}\".format(V), \"cc\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For BCC pattern,\n", + "number of atoms per unit cell = 2\n", + "Volume of a unit cell of lithium metal is 4.35e-23 cc\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 11, Page no: 211" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "\n", + "print \"AB remain in BCC structure if the edge length is a then body diagonal\",\n", + "print \"is root(3)a\"\n", + "print \"root(3)a = 2(r+ + r-)\"\n", + "A = (sqrt(3) * 0.4123 - 2 * 0.81) / 2\n", + "print \"A+ =\", \"{:.2f}\".format(A), \"nm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "AB remain in BCC structure if the edge length is a then body diagonal is root(3)a\n", + "root(3)a = 2(r+ + r-)\n", + "A+ = -0.45 nm\n" + ] + } + ], + "prompt_number": 25 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/screenshots/unnamed.png b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/screenshots/unnamed.png Binary files differnew file mode 100644 index 00000000..e5d583a4 --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/screenshots/unnamed.png diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/screenshots/unnamed1.png b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/screenshots/unnamed1.png Binary files differnew file mode 100644 index 00000000..20edbb55 --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/screenshots/unnamed1.png diff --git a/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/screenshots/unnamed3.png b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/screenshots/unnamed3.png Binary files differnew file mode 100644 index 00000000..5439b07d --- /dev/null +++ b/Advanced_Engineering_Chemistry__by_Dr._M.R._Senapati/screenshots/unnamed3.png diff --git a/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch1.ipynb b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch1.ipynb new file mode 100644 index 00000000..079c7b6e --- /dev/null +++ b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch1.ipynb @@ -0,0 +1,101 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 1: Basic Ideas: Energy Bands in Solids" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 8" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable decalaration\n", + "pd = 100 #potential difference, V\n", + "m0=9.11*(10**-31);#m0=rest mass of the electron in kg\n", + "\n", + "#Calculations&Results\n", + "#solving final velocity of the electron\n", + "Ek=1.6*(10**-19)*pd;#Ek=final kinetic energy of electron in Joules\n", + "print \"Final kinetic energy = %.1e J,%.f eV\"%(Ek,Ek*6.242*10**18)\n", + "v=math.sqrt((2*Ek)/m0)#v=final velocity of the electron\n", + "print \"Final velocity = %.3e m/s\"%v" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Final kinetic energy = 1.6e-17 J,100 eV\n", + "Final velocity = 5.927e+06 m/s\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2, Page 8" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "m=7360*9.11*(10**-31);#m=mass of the ion in kg\n", + "q=2*1.6*(10**-19);#q=charge of the ion in Coulomb\n", + "V=2000;#V=potential difference in Volt\n", + "\n", + "#Calculations&Results\n", + "#solving velocity & kinetic energy of the ion\n", + "v=math.sqrt((2*q*V)/m)#v=velocity of the ion\n", + "print \"Velocity acquired by the ion = %.3e m/s\"%v\n", + "Ek=(1./2)*m*(v**2)#Ek=kinetic energy of the ion\n", + "print \"Kinetic energy of ion = %.1e J = %.f eV\"%(Ek,Ek*6.242*10**18)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Velocity acquired by the ion = 4.369e+05 m/s\n", + "Kinetic energy of ion = 6.4e-16 J = 3995 eV\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch10.ipynb b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch10.ipynb new file mode 100644 index 00000000..8dc7f855 --- /dev/null +++ b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch10.ipynb @@ -0,0 +1,267 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 10: Feedback in Amplifiers" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 236" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "A=(-100.)#A=voltage gain of an amplifier\n", + "B=(-0.04)#B=feedback ratio\n", + "\n", + "#Calculations&Results\n", + "Af=A/(1+(A*B))#Af=voltage gain with feedback\n", + "print \"1.Voltage gain with feedback Af=%.f\"%Af\n", + "F=20*math.log10(abs(Af/A))#F=amount of feedback\n", + "print \"2.Amount of feedback F=%.2f dB\"%F\n", + "Vi=40*(10**-3)#Vi=input voltage\n", + "Vo=Af*Vi#Vo=output voltage\n", + "print \"3.Output voltage Vo=%.1f V\"%Vo\n", + "f=(-A*B)#f=feedback factor\n", + "print \"4.Feedback factor f=%.f\"%f\n", + "Vf=B*Vo#Vf=feedback voltage\n", + "print \"5.Feedback voltage is Vf=%.f mV\"%(Vf/10**-3)#Vf is converted in terms of mV\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1.Voltage gain with feedback Af=-20\n", + "2.Amount of feedback F=-13.98 dB\n", + "3.Output voltage Vo=-0.8 V\n", + "4.Feedback factor f=-4\n", + "5.Feedback voltage is Vf=32 mV\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2, Page 236" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "print \"Negative feedback has to be applied for gain stability\"\n", + "#A=open loop gain of an amplifier\n", + "#B=feedback ratio\n", + "\n", + "#Calculations&Results\n", + "Af=10.#Af=voltage gain with feedback\n", + "#dAf/Af=(1/(1+(A*B)))*(dA/A)\n", + "y=2#(dAf/Af)=y=percent change of gain that is allowable\n", + "x=20.#(dA/A)=x=percent change in open loop gain of an amplifier\n", + "a=(x/y)#(1+(A*B))=a\n", + "print \"(1+(A*B))=%.f\"%a\n", + "#Af=A/((1+(A*B)))\n", + "A=(Af*a)\n", + "print \"Open loop gain A=%.f\"%A\n", + "#1+(A*B)=a\n", + "B=(a-1)/A\n", + "print \"Minimum value of feedback ratio B=%.2f\"%B\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Negative feedback has to be applied for gain stability\n", + "(1+(A*B))=10\n", + "Open loop gain A=100\n", + "Minimum value of feedback ratio B=0.09\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3, Page 237" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VD=0.1#VD=outputdistortion voltage\n", + "VDf=0.05#VDf=output distortion voltage with feedback\n", + "A=-80#A=open loop gain of an amplifier\n", + "\n", + "#Calculations\n", + "#VDf=VD/(1+(A*B))\n", + "B=((VD/VDf)-1)/A#B=reverse transmission factor\n", + "\n", + "#Result\n", + "print \"Reverse transmission factor B=%.4f\"%B" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Reverse transmission factor B=-0.0125\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4, Page 237" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "B=50.#B=reverse transmission factor for silicon transistor T1\n", + "\n", + "#Calculations&Results\n", + "VB=((640.)*10)/(640+360)#calculating voltage at point B i.e VB by applying voltage divider rule in the given circuit\n", + "print \"VB=%.1f V\"%VB\n", + "VBE=VB-5.6#VBE=base emitter voltage drop for silicon transistors T1 and T2 both\n", + "print \"VBE=%.1f V\"%VBE\n", + "VA=10-0.8#VA=voltage at point A in the given circuit\n", + "print \"VA=%.1f V\"%VA\n", + "I1=10./(360+640)#I1=current through resistor of 360 ohm\n", + "print \"I1=%.2f A\"%I1\n", + "IE1=I1+1#IE1=emitter current of transistor T1\n", + "print \"IE1=%f A\"%IE1\n", + "#IC1=-IB1+IE1\n", + "IB1=IE1/(B+1)#IB1=base current of transistor T1\n", + "print \"IB1=%.2f mA\"%(IB1/10**-3)#IB1 is converted in terms of mA\n", + "I2=(20-VA)/300#I2=current through resistor of 300 ohm\n", + "print \"I2=%.f mA\"%(I2/10**-3)#I2 is converted in terms of mA\n", + "IC2=I2-IB1#IC2=collector current of transistor T2\n", + "print \"IC2=%.2f mA\"%(IC2/10**-3)#IC2 is converted in terms of mA\n", + "#Assuming the base current IB2 of transistor T2 is negligibly small\n", + "IE2=IC2#IE2=emitter current of transistor T2\n", + "print \"IE2=%.2f mA\"%(IE2/10**-3)#IE2 is converted in terms of mA\n", + "I3=(20-5.6)/1000#I3=current through 1000 ohm resistor\n", + "print \"I3=%.1f mA\"%(I3/10**-3)#I3 is converted in terms of mA\n", + "IZ=I3+IE2#IZ=current through zener diode\n", + "print \"IZ=%.2f mA\"%(IZ/10**-3)#IZ is converted in terms of mA\n", + "VCE=20-10#VCE=collector emitter voltage drop for transistor T1\n", + "print \"VCE=%.f V\"%VCE\n", + "IC1=B*IB1\n", + "P=VCE*IC1#P=power dissipation in transistor T1\n", + "print \"P=%.1f W\"%P" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "VB=6.4 V\n", + "VBE=0.8 V\n", + "VA=9.2 V\n", + "I1=0.01 A\n", + "IE1=1.010000 A\n", + "IB1=19.80 mA\n", + "I2=36 mA\n", + "IC2=16.20 mA\n", + "IE2=16.20 mA\n", + "I3=14.4 mA\n", + "IZ=30.60 mA\n", + "VCE=10 V\n", + "P=9.9 W\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5, Page 238" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "A=50.#A=voltage gain of transistor amplifier\n", + "Ri=1000#Ri=input resistance of transistor amplifier without feedback\n", + "Ro=40*1000#Ro=output resistance of transistor amplifier feedback\n", + "\n", + "#Calculations&Results\n", + "#Vf=0.1*Vo (given) where Vf=feedback voltage and Vo=output voltage\n", + "B=0.1#B=(Vf/Vo)=feedback fraction\n", + "Af=A/(1+(A*B))#Af=gain of the feedback amplifier\n", + "print \"Gain of feedback amplifier Af=%.2f\"%Af\n", + "Rif=Ri*(1+(A*B))#Rif=input resistance of the feedback amplifier\n", + "Rof=Ro/(1+(A*B))#Rof=output resistance of the feedback amplifier\n", + "print \"Input resistance with feedback Rif=%.f K ohms\"%(Rif/10**3)#Rif is converted in terms of kilo ohm\n", + "print \"Output resistance with feedback Rof=%.2f K ohms\"%(Rof/10**3)#Rof is converted in terms of kilo ohm" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Gain of feedback amplifier Af=8.33\n", + "Input resistance with feedback Rif=6 K ohms\n", + "Output resistance with feedback Rof=6.67 K ohms\n" + ] + } + ], + "prompt_number": 10 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch11.ipynb b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch11.ipynb new file mode 100644 index 00000000..8d254a53 --- /dev/null +++ b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch11.ipynb @@ -0,0 +1,292 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 11: Sinusoidal Oscillators and Multivibrators" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 263" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "L=50*10**-3#L=primary inductance of a transformer in henry\n", + "C=(200*10**-12)#C=capacitor connected across transformer in farad\n", + "R=50#dc resistance of primary coil in ohm\n", + "hie=2000#hie=input impedance in ohm\n", + "hre=10**(-4)#hre=reverse voltage amplification factor\n", + "hfe=98#hfe=current gain\n", + "hoe=(0.5*10**(-4))#hoe=output impedance in mho\n", + "RB=50000#RB=resistance\n", + "\n", + "#Calculations\n", + "f=1/(2*math.pi*math.sqrt(L*C))#f=frequency of oscillation\n", + "g=((hie*hoe)-(hfe*hre))#g=dhe=delta he\n", + "#M=mutual inductance in henry between the transformer primary and the secondary coils for sustained oscillations\n", + "M=((RB/hfe)*((C*R)+(hoe*L)))+((C*R*hie)/hfe)+((L*g)/hfe)\n", + "\n", + "#Results\n", + "print \"Frequency of oscillation is = %.1f kHz\"%(f/10**3)#f is converted in terms of kHz\n", + "print \"Mutual inductance is = %.2f mH\"%(M/10**-3)#M is converted in terms of mH" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Frequency of oscillation is = 50.3 kHz\n", + "Mutual inductance is = 1.33 mH\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2, Page 264" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "import numpy as np\n", + "from numpy.linalg import inv\n", + "\n", + "#Variable declaration\n", + "#L1 and L2=inductances in henry in a Hartley oscillator\n", + "#Suppose L1=a\n", + "#L2=b\n", + "f=60*10**3#f=frequency in Hz\n", + "C=400*10**(-12)#C=capacitance in Farad\n", + "\n", + "#Calculations\n", + "#Also tuning capacitance varies from 100 pF to 400 pF\n", + "#f=1/(2*%pi*sqrt((L1+L2)*C)) where f=frequency of a Hartley oscillator which varies from 60 kHz to 120 kHz\n", + "#d=L1+L2=a+b\n", + "#d=1/(((2*%pi*f)**2)*C)\n", + "d=1/(((2*math.pi*f)**2)*C)#.......(1)\n", + "#e=L2/L1=hfe/dhe\n", + "hfe=90#hfe=current gain\n", + "dhe=0.2#dhe=delta he\n", + "e=hfe/dhe#..........(2)\n", + "#From equation (1) and (2)\n", + "#L*x=y\n", + "x=np.matrix([[1, 1], [e, -1]])\n", + "y=np.matrix([[d],[0]])\n", + "L=inv(x)*y\n", + "\n", + "#Results\n", + "print \"Inductance L1 is = %.f uH\"%((L[0])/10**-6)#converting L(1) in terms of micro Henry\n", + "print \"Inductance L2 is = %.2f mH\"%((L[1])/10**-3)#converting L(2) in terms of mH\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Inductance L1 is = 39 uH\n", + "Inductance L2 is = 17.55 mH\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3, Page 264" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "L=20*10**-3#L=inductance in henry\n", + "C1=(200*10**(-12))#C1=capacitance in farad\n", + "C2=(300*10**(-12))#C2=capacitance in farad\n", + "\n", + "#Calculations\n", + "Cs=((C1*C2)/(C1+C2))\n", + "f=1/(2*math.pi*math.sqrt(L*Cs))\n", + "\n", + "#Result\n", + "print \"Frequency of oscillation is = %.f kHz\"%(f/10**3)#converting f in terms of kHz" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Frequency of oscillation is = 103 kHz\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4, Page 264" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "R=4700#R=resistance in a phase-shift oscillator in ohm\n", + "C=(0.01*10**(-6))#C=capacitance in a phase-shift oscillator in farad\n", + "\n", + "#Calculations\n", + "f=1/(2*math.pi*math.sqrt(10)*R*C)\n", + "\n", + "#Result\n", + "print \"Frequency of oscillation f is = %.2f kHz\"%(f/10**3)#converting f in terms of kHz" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Frequency of oscillation f is = 1.07 kHz\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5, Page 265" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "f=30#f=frequency of oscillation of a Wien-bridge oscillator in Hz\n", + "C=(500*10**(-12))#C=capacitance in farad\n", + "\n", + "#Calculations&Results\n", + "#f=1/2*math.pi*R*C#R=resistance in ohm\n", + "R=1/(2*math.pi*f*C)\n", + "print \"Resistance needed to span the frequency range,R=%.1f Mohms\"%(R/10**6)#converting R in terms of Mega ohms\n", + "#C1=50pF C2=500pF where C1,C2 are variable capacitances in a Wien bridge oscillator\n", + "#ratio of capacitance=(1:10)\n", + "#frequency range is 30 Hz to 300 Hz with R=10.6 Megaohms\n", + "#for the next frequency range from 300 Hz to 3 kHz ,new R=(10.6/10)=1.06 Megaohm\n", + "#for frequency range 3 kHz to 30kHz,R=1.06/10=106 Kilo-ohm\n", + "#So,three values of R are 10.6 Megaohm,1.06 Megaohm,106 Kilo ohm\n", + "A=6#A=gain of amplifier\n", + "#R2/(R1+R2)=(1/3)-(1/A)=(1/3)-(1/6)\n", + "#1+(R1/R2)=6\n", + "#Hence R1/R2=5\n", + "#R3=(R1/R2)\n", + "R3=\"5:1\"\n", + "print \"The ratio of the resistances in the other arms of the bridge,R1/R2 is =\",R3" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Resistance needed to span the frequency range,R=10.6 Mohms\n", + "The ratio of the resistances in the other arms of the bridge,R1/R2 is = 5:1\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6, Page 265" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "#Q=Quality factor\n", + "L=3.5#L=inductance in henry\n", + "f=450000#f=frequency in Hz\n", + "R=9050#R=resistance in ohm\n", + "\n", + "#Calculations\n", + "Q=(2*math.pi*f*L)/R\n", + "\n", + "#Result\n", + "print \"Quality factor is %.f\"%Q" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Quality factor is 1093\n" + ] + } + ], + "prompt_number": 12 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch12.ipynb b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch12.ipynb new file mode 100644 index 00000000..234951b2 --- /dev/null +++ b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch12.ipynb @@ -0,0 +1,296 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 12: Modulation and Demodulation" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 285" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Varibale declaration\n", + "Vmax=8.#Vmax=maximum peak to peak value of an AM voltage\n", + "Vmin=2.#Vmin=minimum peak to peak value of an AM voltage\n", + "\n", + "#Calculations&Results\n", + "ma=(Vmax-Vmin)/(Vmax+Vmin)#ma=percentage modulation \n", + "print \"Percentage modulation ma=%.f %%\"%(ma*100)\n", + "#ma=(Vmax-Vmin)/(2*VC) where VC=amplitude of the unmodulated carrier\n", + "VC=(Vmax-Vmin)/(2*ma)\n", + "print \"Amplitude of the unmodulated carrier is VC=%.f V\"%VC\n", + "print \"In the textbook answer given is incorrect as they have further divided by 2 which is not the part of given formula.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage modulation ma=60 %\n", + "Amplitude of the unmodulated carrier is VC=5 V\n", + "In the textbook answer given is incorrect as they have further divided by 2 which is not the part of given formula.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2, Page 285" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Varibale declaration\n", + "fc=1000*(10**3)#fc=frequency of the carrier wave in Hz(hertz)\n", + "fmin=400\n", + "fmax=1600.#fmin and fmax represent the frequency range of audio signals by which the carrier wave is amplitude modulated.\n", + "\n", + "#Calculations&Results\n", + "fs=fmax-fmin#fs=frequency span of each sideband\n", + "print \"1.Frequency span of each sideband is %.f Hz\"%fs\n", + "fumax=(fc+fmax)/1000#fumax=maximum upper side frequency\n", + "print \"2.The maximum upper side frequency is %.1f kHz\"%fumax\n", + "flmin=(fc-fmax)/1000#flmin=minimum lower side frequency\n", + "print \"3.The minimum lower side frequency is %.1f kHz \"%flmin\n", + "Wc=fumax-flmin#Wc=channelwidth\n", + "print \"4.The channelwidth is %.1f kHz\"%Wc" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1.Frequency span of each sideband is 1200 Hz\n", + "2.The maximum upper side frequency is 1001.6 kHz\n", + "3.The minimum lower side frequency is 998.4 kHz \n", + "4.The channelwidth is 3.2 kHz\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3, Page 286" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Varibale declaration\n", + "R=100#R=load resistance in ohms\n", + "Vc=100.#Vc=peak voltage of the carrier in volts\n", + "ma=0.4#ma=modulation factor\n", + "\n", + "#Calculations&Results\n", + "Pc=(Vc**2)/(2*R)#Pc=unmodulated carrier power developed by an AM wave\n", + "print \"The unmodulated carrier power is Pc = %.f W\"%Pc\n", + "Pt=Pc*(1+((ma**2)/2))#Pt=total power developed\n", + "print \"The total power developed by the AM wave is Pt=%.f W\"%Pt\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The unmodulated carrier power is Pc = 50 W\n", + "The total power developed by the AM wave is Pt=54 W\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4, Page 286" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Varibale declaration\n", + "ma=0.5#ma=modulation factor\n", + "Pc=20#Pc=unmodulated carrier power in kilowatts(kW)\n", + "\n", + "#Calculations&Results\n", + "Ps=(1./2)*(ma**2)*Pc#Ps=total sideband power\n", + "print \"The total sideband power is Ps=%.1f kW\"%Ps\n", + "#modulator system efficiency is given as 70 per cent\n", + "Pa=Ps/0.7#Pa=audio power necessary toamplitude modulate a given carrier wave\n", + "print \"The required audio power is %.2f kW\"%Pa" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total sideband power is Ps=2.5 kW\n", + "The required audio power is 3.57 kW\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5, Page 286" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Varibale declaration\n", + "df=30.#df=maximum frequency deviation in kilohertz(kHz)\n", + "fm=15#fm=modulation frequency of a sinusoidal audio signal in kilohertz(kHz)\n", + "\n", + "#Calculations&Results\n", + "mf=df/fm#mf=frequency modulation index\n", + "print \"1.The modulation index is mf=%.f\"%mf\n", + "fc=100#fc=carrier wave frequency in megahertz(MHz)\n", + "print \"2.The three significant pairs of side frequencies are 100MHz+-15kHz(fc+-fm);100MHz+-30kHz(fc+-2fm);100MHz+-45kHz(fc+-3fm)\"\n", + "wc=mf*3*fm#wc=channelwidth required for 3 above mentioned side frequency pairs\n", + "print \"3.The required channelwidth is %.f kHz\"%wc\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1.The modulation index is mf=2\n", + "2.The three significant pairs of side frequencies are 100MHz+-15kHz(fc+-fm);100MHz+-30kHz(fc+-2fm);100MHz+-45kHz(fc+-3fm)\n", + "3.The required channelwidth is 90 kHz\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6, Page 286" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Varibale declaration\n", + "R=0.2*(10**6)#R=load resistance in ohms in a diode detector \n", + "C=150*(10**-12)#C=capacitance in farad in a diode detector \n", + "\n", + "#Calculations\n", + "#fmh=wmh/(2*%pi)where fmh=highest modulation frequency that can be detected with tolerable distortion and wmh=corresponding angular frequency\n", + "ma=0.5#ma=modulation factor or depth of modulation\n", + "fmh=(1/(2*math.pi*ma*R*C))/1000\n", + "\n", + "#Result\n", + "print \"The required frequency is fmh=%.2f kHz\"%fmh\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required frequency is fmh=10.61 kHz\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7, Page 287" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Varibale declaration\n", + "Pc=10#Pc=unmodulated carrier power in kilowatts(kW)\n", + "Pt=12.5#Pt=total power in kilowatts(kW)\n", + "\n", + "#Calculations&Results\n", + "#Pt=Pc*(1+((ma^2)/2)) \n", + "ma=math.sqrt(2*((Pt/Pc)-1))#ma=depth of modulation of the first signal\n", + "print \"The depth of modulation is ma=%.3f\"%ma\n", + "mb=0.6#mb=depth of modulation of the second signal\n", + "PT=Pc*(1+((ma**2)/2)+((mb**2)/2))#PT=the total radiated power\n", + "print \"The total radiated power is PT=%.1f kW\"%PT" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The depth of modulation is ma=0.707\n", + "The total radiated power is PT=14.3 kW\n" + ] + } + ], + "prompt_number": 15 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch13.ipynb b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch13.ipynb new file mode 100644 index 00000000..a21cb7ab --- /dev/null +++ b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch13.ipynb @@ -0,0 +1,750 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 13: Field-Effect Transistors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 317" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "ND=2*10**21#ND=donor concentration in m**-3 of an n-channel silicon JFET\n", + "e=1.6*10**-19#e=charge of an electron\n", + "E=12*8.854*10**-12#E=permittivity of the material where 12=dielectric constant of silicon(given)\n", + "a=(4*10**-6)/2#2*a=channel width in metres and 2*a=4*10**-6\n", + "\n", + "#Calculations&Results\n", + "Vp=(e*ND*(a**2))/(2*E)\n", + "print \"The pinch-off voltage is = %.2f V\"%Vp\n", + "VGS=-2#VGS=gate source voltage\n", + "#Vp=VDsat-VGS where VDsat=saturation voltage\n", + "VDsat=Vp+VGS\n", + "print \"The saturation voltage is = %.2f V\"%VDsat" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The pinch-off voltage is = 6.02 V\n", + "The saturation voltage is = 4.02 V\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2, Page 317" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VGS=-1.5#VGS=gate-to-source voltage of a JFET\n", + "IDsat=5*10**-3#IDsat=drain saturation current in Ampere\n", + "\n", + "#Calculations\n", + "RS=(abs(VGS))/(abs(IDsat))#RS=resistance to be calculated=|VGS| / |IDsat|\n", + "\n", + "#Result\n", + "print \"Resistance to be calculated is = %.f ohm\"%RS " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Resistance to be calculated is = 300 ohm\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3, Page 317" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VGS1=-1\n", + "VGS2=-1.5#VGS1,VGS2=change in VGS(gate-to-source voltage) from VGS1 to VGS2 keeping VDS(drain-to-source voltage) constant\n", + "ID1=7.*10**-3\n", + "ID2=5*10**-3#ID1,ID2=change in ID(drain current) in Ampere from ID1 to ID2\n", + "\n", + "#Calculations&Results\n", + "#gm=(id/vgs)|VDS=constant where gm=transconductance\n", + "id=ID1-ID2\n", + "vgs=VGS1-VGS2\n", + "gm=id/vgs\n", + "print \"The transconductance of the FET is = %.f mA/V\"%(gm*10**3)\n", + "rd=200*10**3#rd=ac drain resistance in ohms\n", + "u=rd*gm#u=amplification factor\n", + "print \"The amplification factor of the FET is = %.f\"%u" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The transconductance of the FET is = 4 mA/V\n", + "The amplification factor of the FET is = 800\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4, Page 217" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "RL=250*10**3#RL=load resistance in ohms in a FET amplifier\n", + "rd=100*10**3#rd=ac drain resistance in ohms\n", + "gm=0.5*10**-3#gm=transconductance in A/V\n", + "\n", + "#Calculations&Results\n", + "u=rd*gm#u=amplification factor\n", + "AV=-(u*RL)/(rd+RL)#AV=voltage gain\n", + "print \"The voltage gain of FET amplifier is = %.2f\"%AV\n", + "print \"The output resistance excluding RL is rd=%.f kohm\"%(rd/1000)\n", + "ro=(rd*RL)/(rd+RL)#ro=output resistance including RL\n", + "print \"Including RL,the output resistance is=%.f k ohms\"%(ro/1000)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage gain of FET amplifier is = -35.71\n", + "The output resistance excluding RL is rd=100 kohm\n", + "Including RL,the output resistance is=71 k ohms\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5, Page 318" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "#For n-channel JFET \n", + "IDSS=12*10**-3#IDSS=saturation drain current in Ampere when VGS(gate-to-source voltage)=0V\n", + "Vp=-4#Vp=pinch-off voltage\n", + "VGS=-2#VGS=gate-to-source voltage\n", + "\n", + "#Calculations&Results\n", + "#By Shockley's equation\n", + "IDS=IDSS*(1-(VGS/Vp))**2#IDS=saturation drain current to be calculated for given value of VGS\n", + "print \"The drain current for given value of VGS is=%.f mA\"%(IDS/10**-3)\n", + "gmo=4*10**-3#gmo=transconductance in A/V of a JFET when VGS=0V\n", + "#gmo=-(2*IDSS)/Vp\n", + "Vp=-(2*IDSS)/gmo#Vp=pinch-off voltage to be calculated for given value of transconductance\n", + "print \"The pinch-off voltage for given value of gmo is = %.f V\"%Vp" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current for given value of VGS is=12 mA\n", + "The pinch-off voltage for given value of gmo is = -6 V\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6, Page 318" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "IDSS=12*10**-3#IDSS=saturation drain current in Ampere when VGS(gate-to-source voltage)=0V\n", + "Vp=-4#Vp=pinch-off voltage\n", + "VDD=30#VDD=drain supply voltage\n", + "RL=5*10**3#RL=load resistance in ohms\n", + "Rs=600#Rs=resistance connected to source terminal in ohms\n", + "Rg=1.5*10**6#Rg=resistance connected to gate terminal in ohms\n", + "\n", + "#Calculations&Results\n", + "#By Shockley's equation\n", + "#IDS=IDSS*(1-(VGS/Vp))**2 where IDS=saturation drain current to be calculated for given value of VGS\n", + "#Substituting VGS=(-ID*Rs) we get ID=IDS\n", + "#ID=IDSS*(1+((ID*Rs)/Vp))**2\n", + "#ID=12*(1+((0.6*ID)/-4))**2 where ID is obtained in mA\n", + "#(0.27*ID**2)-(4.6*ID)+12=0.........(1)\n", + "ID1=(4.6+math.sqrt((4.6**2)-(48*0.27)))/(2*0.27)\n", + "ID2=(4.6-math.sqrt((4.6**2)-(48*0.27)))/(2*0.27)#ID1,ID2 are the 2 roots of the above equation (1)\n", + "print \"ID1=%.2f mA\"%ID1\n", + "print \"ID2=%.2f mA\"%ID2\n", + "if (ID1>(IDSS/10**-3)):#IDSS is converted in terms of mA\n", + " print \"As ID1>IDSS ,the value rejected is ID1=%.2f mA\"%ID1\n", + "\n", + "if (ID2>(IDSS/10**-3)):#IDSS is converted in terms of mA\n", + " print \"As ID2>IDSS ,the value rejected is ID2=%.2f mA\"%ID2\n", + "\n", + "print \"Therefore,the drain current is = %.2f mA\"%ID2\n", + "ID=ID2*10**-3#converting ID2 in terms of Ampere\n", + "VDS=VDD-ID*(RL+Rs)#VDS=drain-to-source voltage\n", + "print \"The value of drain-to-source voltage VDS is = %.f V\"%VDS\n", + "VGS=-ID*Rs#VGS=gate-to-source voltage\n", + "print \"The value of gate-to-source voltage VGS is=%.2f V\"%VGS\n", + "if(Vp<0 and VDS>(VGS-Vp)):\n", + " print \"As Vp=(-4)<VGS<0V and VDS=12V>(VGS-Vp),it is verified that the JFET is in the saturation region of the drain characteristics\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "ID1=13.82 mA\n", + "ID2=3.22 mA\n", + "As ID1>IDSS ,the value rejected is ID1=13.82 mA\n", + "Therefore,the drain current is = 3.22 mA\n", + "The value of drain-to-source voltage VDS is = 12 V\n", + "The value of gate-to-source voltage VGS is=-1.93 V\n", + "As Vp=(-4)<VGS<0V and VDS=12V>(VGS-Vp),it is verified that the JFET is in the saturation region of the drain characteristics\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7, Page 319" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "IDSS=10*10**-3#IDSS=saturation drain current in Ampere when VGS(gate-to-source voltage)=0V\n", + "Vp=-2#Vp=pinch-off voltage\n", + "VDD=20#VDD=drain supply voltage\n", + "RL=1*10**3#RL=load resistance in ohms\n", + "Rs=2*1000#Rs=resistance connected to source terminal in ohms\n", + "R1=12*10**6#R1=resistance in the voltage divider network in ohms\n", + "R2=8*10**6#R2=resistance in the voltage divider network in ohms\n", + "\n", + "#Calculations&Results\n", + "VT=(R2/(R1+R2))*VDD#VT=Thevenin voltage\n", + "#VGS=VT-(ID*Rs)\n", + "#By Shockley's equation\n", + "#IDS=IDSS*(1-(VGS/Vp))**2 where IDS=saturation drain current to be calculated for given value of VGS\n", + "#Substituting VGS=(VGS-ID*Rs) we get ID=IDS\n", + "#(10*ID**2)-(101*ID)+250=0.........(1)where ID is obtained in mA\n", + "ID1=(101+math.sqrt((101**2)-(40*250)))/(2*10)\n", + "ID2=(101-math.sqrt((101**2)-(40*250)))/(2*10)#ID1,ID2 are the 2 roots of the above equation (1)\n", + "print \"ID1=%.2f mA\"%ID1\n", + "print \"ID2=%.2f mA\"%ID2\n", + "#For ID1\n", + "VGS=VT-(ID1*Rs)#VGS=gate-to-source voltage calculated for ID1\n", + "if (Vp>VGS):\n", + " print \"As Vp>(VGS calculated using ID1), the value rejected is ID1=%.2f mA\"%ID1\n", + "\n", + "print \"Therefore,the drain current is = %.2f mA\"%ID2\n", + "ID=ID2*10**-3#converting ID2 in terms of Amperes\n", + "VGS=VT-(ID*Rs)#VGS=gate-to-source voltage\n", + "print \"VGS=%.2f V\"%VGS\n", + "VDS=VDD-(ID*(RL+Rs))#VDS=drain-to-source voltage\n", + "print \"VDS=%.f V\"%VDS\n", + "if(Vp<VGS and VDS>(VGS-Vp)):\n", + " print \"As Vp=(-2)<(VGS=-0.68V) and VDS=7V>(VGS-Vp),it is checked that the JFET operates in the saturation region \"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "ID1=5.76 mA\n", + "ID2=4.34 mA\n", + "As Vp>(VGS calculated using ID1), the value rejected is ID1=5.76 mA\n", + "Therefore,the drain current is = 4.34 mA\n", + "VGS=-8.68 V\n", + "VDS=7 V\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8, Page 319" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "#For a n-channel JFET\n", + "IDSS=10*10**-3#IDSS=saturation drain current in Ampere when VGS(gate-to-source voltage)=0V\n", + "Vp=(-4)#Vp=pinch-off voltage\n", + "VGS=(-2.5)#VGS=gate-to-source voltage\n", + "\n", + "#Calculations&Results\n", + "#By Shockley's equation\n", + "IDS=IDSS*(1-(VGS/Vp))**2#IDS=saturation drain current to be calculated for given value of VGS\n", + "print \"The drain current for given value of VGS is=%.1f mA\"%(IDS/10**-3)#converting IDS in terms of mA\n", + "VDSmin=VGS-Vp#VDSmin=minimum value of drain-to-source voltage for the onset of the saturation region\n", + "print \"The minimum value of VDS for saturation is=%.1f V\"%VDSmin" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current for given value of VGS is=1.4 mA\n", + "The minimum value of VDS for saturation is=1.5 V\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9, Page 319" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "VDD=20#VDD=drain supply voltage\n", + "IDS=0.9#IDS=drain saturation current in terms of mA\n", + "Vp=-3.#Vp=pinch-off voltage\n", + "IDSS=8#IDSS=saturation drain current in mA when VGS(gate-to-source voltage)=0V\n", + "\n", + "#Calculations&Results\n", + "#By Shockley's equation\n", + "#IDS=IDSS*(1-(VGS/Vp))^2\n", + "VGS=Vp*(1-math.sqrt(IDS/IDSS))#VGS=gate-to-source voltage\n", + "print \"The gate-to-source voltage VGS is=%.f V\"%VGS\n", + "#gm=(dIDS/dVGS)|VDS=constant where gm=transconductance\n", + "gm=-((2*IDSS)/Vp)*(1-(VGS/Vp))\n", + "print \"The value of transconductance is=%.2f mS\"%gm\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The gate-to-source voltage VGS is=-2 V\n", + "The value of transconductance is=1.79 mS\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10, Page 320" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "IDS=-15.#IDS=drain saturation current in terms of mA\n", + "Vp=5#Vp=pinch-off voltage\n", + "IDSS=-40#IDSS=saturation drain current in mA when VGS(gate-to-source voltage)=0V\n", + "\n", + "#Calculations\n", + "#By Shockley's equation\n", + "#IDS=IDSS*(1-(VGS/Vp))^2\n", + "VGS=Vp*(1-math.sqrt(IDS/IDSS))#VGS=gate-to-source voltage\n", + "\n", + "#Result\n", + "print \"The gate-to-source voltage VGS is=%.2f V\"%VGS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The gate-to-source voltage VGS is=1.94 V\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11, Page 320" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "IDSS=10*10**-3#IDSS=saturation drain current in Ampere when VGS(gate-to-source voltage)=0V\n", + "Vp=-5.#Vp=pinch-off voltage\n", + "VDD=24.#VDD=drain supply voltage\n", + "VDS=8#VDS=drain-to-source voltage\n", + "ID=4.*10**-3#ID=drain current in Ampere\n", + "R1=2.*10**6#R1=resistance in the voltage divider network in ohms\n", + "R2=1*10**6#R2=resistance in the voltage divider network in ohms\n", + "\n", + "#Calculations&Results\n", + "VT=(R2/(R1+R2))*VDD#VT=Thevenin voltage\n", + "#By Shockley's equation\n", + "#ID=IDS=IDSS*(1-(VGS/Vp))**2\n", + "VGS=Vp*(1-math.sqrt(ID/IDSS))#VGS=gate-to-source voltage\n", + "#VGS=VT-(ID*Rs) where Rs=resistance connected at the source terminal\n", + "Rs=(VT-VGS)/ID\n", + "print \"The value of Rs = %.2f k ohm\"%(Rs/10**3)#converting Rs in terms of kilo-ohm\n", + "Rch=VDS/ID#Rch=channel resistance at the Q-point\n", + "print \"The channel resistance at the Q-point is=%.f k ohm\"%(Rch/10**3)#converting Rch in terms of kilo-ohm" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of Rs = 2.46 k ohm\n", + "The channel resistance at the Q-point is=2 k ohm\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12, Page 320" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "ID=5#ID=saturation drain current in terms of mA in an n-channel enhancement mode MOSFET\n", + "VGS=8.#VGS=gate-to-source voltage\n", + "VT=4#VT=Threshold voltage\n", + "VGS2=10#VGS2=gate-to-source voltage for which saturation drain current is to be calculated\n", + "\n", + "#Calculations&Results\n", + "#ID=K*(VGS-VT)**2 where K=(IDSS/(Vp**2)) and Vp=pinch-off voltage ,IDSS=drain saturation current for VGS=0 V\n", + "K=ID/((VGS-VT)**2)\n", + "ID1=K*(VGS2-VT)**2#ID1=The saturation drain current for gate-source voltage of 10V i e VGS2\n", + "\n", + "\n", + "print \"The saturation drain current for gate-source voltage of 10V is = %.1f mA\"%ID1" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The saturation drain current for gate-source voltage of 10V is = 11.2 mA\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13, Page 321" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "#For n-channel enhancement mode MOSFET operating in active region\n", + "VT=2#VT=Threshold voltage\n", + "K=0.5#K=(IDSS/(Vp**2)) in terms of mA/V**2\n", + "VDD=15#VDD=drain supply voltage\n", + "RL=1#RL=load resistance in kilo ohm\n", + "R1=200*10**3#R1=resistance in the voltage divider network in terms of ohms\n", + "R2=100.*10**3#R2=resistance in the voltage divider network in terms of ohms\n", + "\n", + "#Calculations&Results\n", + "VGS=(R2/(R1+R2))*VDD#VGS=gate-to-source voltage\n", + "print \"Threshold voltage is = %.f V\"%VT\n", + "print \"The gate-to-source voltage VGS is = %.f V\"%VGS\n", + "ID=K*(VGS-VT)**2#ID=drain current in mA\n", + "print \"The value of drain current ID is = %.1f mA\"%ID\n", + "VDS=VDD-(ID*RL)#VDS=drain-to-source voltage\n", + "print \"The value of drain-to-source voltage VDS is=%.1f V\"%VDS\n", + "if (VDS>(VGS-VT)):\n", + " print \"As VDS>(VGS-VT),(i.e. 10.5>(5-2)),the operation is indeed in the active region \"\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Threshold voltage is = 2 V\n", + "The gate-to-source voltage VGS is = 5 V\n", + "The value of drain current ID is = 4.5 mA\n", + "The value of drain-to-source voltage VDS is=10.5 V\n", + "As VDS>(VGS-VT),(i.e. 10.5>(5-2)),the operation is indeed in the active region \n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14, Page 321" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "#For n-channel MOSFET operating in the depletion mode\n", + "VDD=18#VDD=drain supply voltage\n", + "VGS=0#VGS=gate-to-source voltage\n", + "RL=600#RL=load resistance in ohms\n", + "IDSS=18.#IDSS=drain saturation current in mA for gate-to-source voltage (VGS)=0V\n", + "Vp=-5#Vp=pinch-off voltage\n", + "\n", + "#Calculations&Results\n", + "#Assuming that the operation is in the active region\n", + "#ID=IDS=IDSS*(1-(VGS/Vp))**2\n", + "#ID=(IDSS/Vp**2)(VGS-Vp)**2\n", + "K=IDSS/(Vp**2)\n", + "print \"The value of K is = %.2f mA/V^2\"%K\n", + "ID=IDSS#ID=drain current \n", + "print \"Since VGS=0,the value of ID=IDSS is=%.f mA\"%ID\n", + "VDS=VDD-(ID*(RL/10**3))#VDS=drain-to-source voltage and also converting RL in terms of kilo ohm\n", + "print \"The value of VDS is = %.f V\"%VDS\n", + "print \"Pinch off voltage Vp is = %.f V\"%Vp\n", + "print \"Gate to source voltage VGS is = %.f V\"%VGS\n", + "if (VDS>(VGS-Vp)):\n", + " print \"As VDS>(VGS-Vp),(i.e.7.5>(0-(-5))),the MOSFET is actually in the active region \"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of K is = 0.72 mA/V^2\n", + "Since VGS=0,the value of ID=IDSS is=18 mA\n", + "The value of VDS is = 18 V\n", + "Pinch off voltage Vp is = -5 V\n", + "Gate to source voltage VGS is = 0 V\n", + "As VDS>(VGS-Vp),(i.e.7.5>(0-(-5))),the MOSFET is actually in the active region \n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15, Page 321" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "#r given in textbook is taken as rd afterwards.Hence r=rd\n", + "rd=100*10**3#rd=drain resistance in ohms\n", + "gm=3500*10**-6#gm=transconductance in terms of A/V (or S)\n", + "RL=5*10**3#RL=load resistance in ohms\n", + "\n", + "#Calculations&Results\n", + "u=rd*gm#u=amplification factor\n", + "AV=(u*RL)/(((u+1)*RL)+rd)#AV=voltage gain\n", + "print \"The voltage gain is=%.3f\"%AV\n", + "Ro=rd/(u+1)#Ro=output resistance excluding RL\n", + "print \"The output resistance excluding RL is = %.f ohm\"%Ro\n", + "Ro1=(rd*RL)/(rd+((u+1)*RL))#Ro1=Ro'=output resistance including RL\n", + "print \"The output resistance including RL is=%.f ohm\"%(math.floor(Ro1))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage gain is=0.943\n", + "The output resistance excluding RL is = 285 ohm\n", + "The output resistance including RL is=269 ohm\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16, Page 322" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "#In a FET used in a CS amplifier\n", + "IDSS=4.#IDSS=drain saturation current in mA for gate-to-source voltage (VGS)=0V\n", + "Vp=-3#Vp=pinch-off voltage\n", + "RL=10#RL=load resistance in kilo ohms\n", + "VGS=-0.7#VGS=gate-to-source voltage\n", + "\n", + "#Calculations\n", + "gmo=-(2*IDSS)/Vp#gmo=transconductance in A/V of a JFET when VGS=0V\n", + "gm=gmo*(1-(VGS/Vp))#gm=transconductance\n", + "AV=-gm*RL#AV=the small signal voltage gain\n", + "\n", + "#Result\n", + "print \"The small signal voltage gain is = %.1f\"%AV\n", + "#Decimal term in the answer displayed in textbook is incorrect as 2.04*10=20.4 and not 20.04." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The small signal voltage gain is = -20.4\n" + ] + } + ], + "prompt_number": 22 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch14.ipynb b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch14.ipynb new file mode 100644 index 00000000..73da715d --- /dev/null +++ b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch14.ipynb @@ -0,0 +1,392 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 14: Integrated Circuits and Operational Amplifiers" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 351" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1=1000.#R1=input resistance in ohms in the inverting amplifier circuit\n", + "Rf=3*1000#Rf=feedback resistance in ohms\n", + "v1=2#v1=input voltage in the inverting terminal of an amplifier circuit\n", + "\n", + "#Calculations&Results\n", + "vo=-(Rf/R1)*v1#vo=output voltage\n", + "print \"Output voltage is=%.f V\"%vo\n", + "print \"Output voltage is negative as it is the circuit of inverting amplifier\"\n", + "print \"Input resistance Rin=R1 is = %.f k ohm\"%(R1/1000)\n", + "i=v1/R1#i=input current\n", + "print \"The input currrent is = %.f mA\"%(i*1000)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Output voltage is=-6 V\n", + "Output voltage is negative as it is the circuit of inverting amplifier\n", + "Input resistance Rin=R1 is = 1 k ohm\n", + "The input currrent is = 2 mA\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2, Page 352" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1=2*1000#R1=input resistance in ohms in the non-inverting amplifier circuit\n", + "Rf=5.*1000#Rf=feedback resistance in ohms\n", + "\n", + "#Calculations\n", + "AV=1+(Rf/R1)#AV=voltage gain of the non-inverting amplifier circuit\n", + "\n", + "#Result\n", + "print \"The voltage gain of the given non-inverting amplifier circuit is = %.1f\"%AV" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage gain of the given non-inverting amplifier circuit is = 3.5\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3, Page 352" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "#First case\n", + "v1=40.*10**-6#v1=voltage applied to the non-inverting input terminal\n", + "v2=-40*10**-6#v2=voltage applied to the inverting input terminal\n", + "vo=100*10**-3#vo=output voltage for the above inputs v1 and v2\n", + "#Second case\n", + "V1=40*10**-6#V1=voltage applied to the non-inverting input terminal\n", + "V2=40*10**-6#V2=voltage applied to the inverting input terminal\n", + "Vo=0.4*10**-3#Vo=output voltage for the above inputs V1 and V2\n", + "\n", + "#Calculations&Results\n", + "print \"In first case:\"\n", + "vd=v1-v2#vd=difference signal voltage\n", + "print \"vd=%.f \u00b5V\"%(vd/10**-6)\n", + "vc=(v1+v2)/2#vc=common mode signal voltage\n", + "print \"vc=%.f \u00b5V\"%(vc/10**-6)\n", + "#Output voltage is vo=(Ad*vd)+(Ac*vc) where Ad and Ac are the voltage gains for the difference signal and the common-mode signal,respectively\n", + "Ad=vo/vd#Ad calculated in first case as common mode signal vc=0 \n", + "print \"Voltage gain for the difference signal is Ad = %.f\"%Ad\n", + "print \"\\nIn second case:\"\n", + "Vd=V1-V2#Vd=difference signal voltage\n", + "print \"Vd=%.f \u00b5V\"%(Vd/10**-6)\n", + "Vc=(V1+V2)/2#Vc=common mode signal voltage\n", + "print \"Vc=%.f \u00b5V\"%(Vc/10**-6)\n", + "Ac=Vo/Vc#Ac calculated in second case as difference signal Vc=0 \n", + "print \"Voltage gain for the common-mode signal is Ac = %.f\"%Ac\n", + "CMRR=abs(Ad/Ac)#CMRR=Common Mode Rejection Ratio=|Ad/Ac|\n", + "print \"Common Mode Rejection Ratio is CMRR=%.f\"%CMRR" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "In first case:\n", + "vd=80 \u00b5V\n", + "vc=0 \u00b5V\n", + "Voltage gain for the difference signal is Ad = 1250\n", + "\n", + "In second case:\n", + "Vd=0 \u00b5V\n", + "Vc=40 \u00b5V\n", + "Voltage gain for the common-mode signal is Ac = 10\n", + "Common Mode Rejection Ratio is CMRR=125\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4, Page 352" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1=1*1000#R1=input resistance in ohms in the inverting terminal of the amplifier circuit\n", + "R2=200#R2=input resistance in ohms in the inverting terminal of the amplifier circuit\n", + "R3=400#R3=input resistance in ohms in the inverting terminal of the amplifier circuit\n", + "Rf=500.#Rf=feedback resistance in ohms\n", + "v1=-5#v1=input voltage in the inverting terminal of an amplifier circuit at R1 resistor\n", + "v2=3#v2=input voltage in the inverting terminal of an amplifier circuit at R2 resistor\n", + "v3=4#v3=input voltage in the inverting terminal of an amplifier circuit at R3 resistor\n", + "\n", + "#Calculations\n", + "vo=-(((Rf/R1)*v1)+((Rf/R2)*v2)+((Rf/R3)*v3))#vo=output voltage for inverting summing summing amplifier circuit\n", + "\n", + "#Result\n", + "print \"Output voltage of the 3-input summing amplifier circuit is = %.f V\"%vo\n", + "print \"Output voltage is negative as it the circuit of inverting summing amplifier\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Output voltage of the 3-input summing amplifier circuit is = -10 V\n", + "Output voltage is negative as it the circuit of inverting summing amplifier\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5, Page 353" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "R1=1*1000#R1=input resistance in ohms in the inverting amplifier circuit\n", + "Rf=50*1000#Rf=feedback resistance in ohms\n", + "\n", + "#Calculations&Results\n", + "A=-(Rf/R1)#AV=voltage gain of the inverting amplifier circuit\n", + "print \"The voltage gain of the given inverting amplifier circuit is = %.f\"%A\n", + "#vin=0.5*sin(100*math.pi*t)\n", + "#vout=A*vin=-50*0.5*sin(100*math.pi*t)=-25*sin(100*math.pi*t)\n", + "print \"If the operation were entirely linear ,the output voltage would have been -25*sin(100*math.pi*t)\"\n", + "print \"But since the voltage supply is +-12V ,the op-amp is saturated when |vout| attains 12V\"\n", + "#Let at time t=to,vout=-12V \n", + "#-12=-25*sin(100*math.pi*to)\n", + "to=(1./(100*math.pi))*math.asin(12./25)\n", + "print \"to=%.2e s\"%to\n", + "print \"Thus over the entire cycle,\"\n", + "print \"vout=-25*sin(100*math.pi*t) V when 0<=t<=to\"\n", + "print \"vout=-12V when to<=t<=(0.01-to)\"\n", + "print \"vout=-25*sin(100*math.pi*t) V when (0.01-to)<=t<=(0.01+to)\"\n", + "print \"vout=+12V when (0.01+to)<=t<=(0.02-to)\"\n", + "print \"vout=-25*sin(100*math.pi*t) V when (0.02-to)<=t<=0.02 seconds\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage gain of the given inverting amplifier circuit is = -50\n", + "If the operation were entirely linear ,the output voltage would have been -25*sin(100*math.pi*t)\n", + "But since the voltage supply is +-12V ,the op-amp is saturated when |vout| attains 12V\n", + "to=1.59e-03 s\n", + "Thus over the entire cycle,\n", + "vout=-25*sin(100*math.pi*t) V when 0<=t<=to\n", + "vout=-12V when to<=t<=(0.01-to)\n", + "vout=-25*sin(100*math.pi*t) V when (0.01-to)<=t<=(0.01+to)\n", + "vout=+12V when (0.01+to)<=t<=(0.02-to)\n", + "vout=-25*sin(100*math.pi*t) V when (0.02-to)<=t<=0.02 seconds\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6, Page 353" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R=2.*1000#R=feedback resistance in ohms in the differentiator circuit\n", + "C=0.01*10**-6#C=input capacitance in farad in the differentiator circuit\n", + "\n", + "#Calculations&Results\n", + "#dvi/dt=1.5V/1ms for 0<t<1ms (given ramp input signal)\n", + "#output voltage of a differentiator is given as vo=-RC(dvi/dt)\n", + "d=1.5/(10**-3)#d=dvi/dt=1.5V/1ms\n", + "vo=-R*C*d\n", + "\n", + "#Results\n", + "print \"Output voltage of a differentiator is %.2f V\"%vo\n", + "print \"Hence for <t<1 ms,vo=-0.03V=-30mV.Otherwise,vo=0V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Output voltage of a differentiator is -0.03 V\n", + "Hence for <t<1 ms,vo=-0.03V=-30mV.Otherwise,vo=0V\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8, Page 354" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1=5*1000#R1=input resistance in ohms in the given op-amp circuit\n", + "Rf=10.*1000#Rf=feedback resistance in ohms\n", + "vi=5#vi=input voltage at the inverting terminal of an op-amp\n", + "V=4#V=voltage at the non-inverting terminal of an op-amp\n", + "\n", + "#Calculations\n", + "#By applying superposition theorem \n", + "Vo=((1+(Rf/R1))*V)+(-(Rf/R1)*vi)\n", + "\n", + "#Result\n", + "print \"Output voltage Vo of the circuit is = %.f V\"%Vo" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Output voltage Vo of the circuit is = 2 V\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9, Page 355" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1=10*1000#R1=resistance in ohms\n", + "R2=100#R2=resistance in ohms\n", + "R3=10*1000#R3=resistance in ohms\n", + "R4=10*1000#R4=resistance in ohms\n", + "R5=10*1000#R5=resistance in ohms\n", + "\n", + "#Calculations\n", + "#Since the voltage gains of the OP AMPs are infinite,the voltages of the points X and Y in the given figure are V1 and V2 respectively\n", + "#Applying Kirchhoff's current law at X\n", + "#(V1/R1)+((V1-V)/R3)+((V1-V2)/R2)=0\n", + "#Applying Kirchhoff's current law at Y\n", + "#((V2-V)/R4)+((V2-V1)/R2)+((V2-Vo)/R5)=0\n", + "#Eliminating V from the above equations\n", + "#V2*((1/R2)+(1/R4)+(1/R5)+(R3/(R2*R4)))-V1*((1/R2)+(1/R4)+((R3/R4)*((1/R1)+(1/R2))))=Vo/R5\n", + "#V2*R-V1*r=Vo/R5...................(1)\n", + "R=((1./R2)+(1/R4)+(1/R5)+(R3/(R2*R4)))\n", + "r=((1./R2)+(1/R4)+((R3/R4)*((1/R1)+(1/R2))))\n", + "\n", + "#Results\n", + "print \"R=%.2f\"%R\n", + "print \"r=%.2f\"%r\n", + "#R=r from above calculation and its answer displayed\n", + "#Hence from the above equation (1) A=Vo/(V1-V2)=-(R5*R)=-(R5*r)\n", + "print \"Differential mode gain A=Vo/(V1-V2)=%.f\"%(-R5*r)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "R=0.01\n", + "r=0.01\n", + "Differential mode gain A=Vo/(V1-V2)=-100\n" + ] + } + ], + "prompt_number": 16 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch15.ipynb b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch15.ipynb new file mode 100644 index 00000000..de1b6224 --- /dev/null +++ b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch15.ipynb @@ -0,0 +1,257 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 15: Active Filters" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 377" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "fc=1000#fc=given cut-off frequency in Hz\n", + "A=-56.#A=required gain to be dropped by this much amount in dB\n", + "#Also,A=normalized gain of Butterworth filter=|A(jw)/Ao|\n", + "f=10*1000#f=given frequency in Hz where the normalized gain is dropped by given amount\n", + "\n", + "#Calculations\n", + "#|A(jw)/Ao|=(-20)*n*log10(w/wc) where n=order of the filter\n", + "#|A(jw)/Ao|=(-20)*n*log10(f/fc)\n", + "n=A/((-20)*math.log10(f/fc))#n=order of Butterworth low-pass filter\n", + "print \"Order of given filter to be designed is (n)=%.f\"%(math.ceil(n))\n", + "#As n=3 (from above calculation) we need cascading of first-order section and second-order section\n", + "#For n=3\n", + "k=0.5#k=damping factor\n", + "Ao=3-(2*k)#Ao=DC gain for each op-amp in a given Butterworth Filter to be designed\n", + "R1=10*1000#R1=Assumed resistance in ohms\n", + "#Ao=(R1+R2)/R1\n", + "R2=(Ao*R1)-R1\n", + "#fc=1/(2*%pi*R*C)\n", + "R=1000#R=Assumed resistance in ohms\n", + "C=1/(2*math.pi*R*fc)\n", + "\n", + "#Results\n", + "print \"The designed values of resistance and capacitance for a low-pass Butterworth filter are:\"\n", + "print \"R1=%.f k ohm\"%(R1/1000)\n", + "print \"R2=%.f k ohm\"%(R2/1000)\n", + "print \"R=%.f k ohm\"%(R/1000)\n", + "print \"C=%.2f uF\"%(C/10**-6)\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Order of given filter to be designed is (n)=3\n", + "The designed values of resistance and capacitance for a low-pass Butterworth filter are:\n", + "R1=10 k ohm\n", + "R2=10 k ohm\n", + "R=1 k ohm\n", + "C=0.16 uF\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2, Page 378" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "Ao=5#Ao=high frequency gain of a given first-order Butterworth active HP filter\n", + "#Ao=(R1+R2)/R1\n", + "R1=1000#R1=Assumed resistance in ohms\n", + "\n", + "#Calculations\n", + "R2=(Ao*R1)-R1\n", + "fc=200#fc=given cut-off frequency in Hz\n", + "#fc=1/(2*%pi*R*C)\n", + "R=5*1000#R=Assumed resistance in ohms\n", + "C=1/(2*math.pi*R*fc)\n", + "\n", + "#Results\n", + "print \"The designed values of resistance and capacitance for a high-pass Butterworth filter are:\"\n", + "print \"R1=%.f k ohm\"%(R1/1000)\n", + "print \"R2=%.f k ohm\"%(R2/1000)\n", + "print \"R=%.f k ohm\"%(R/1000)\n", + "print \"C=%.2f uF\"%(C/10**-6)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The designed values of resistance and capacitance for a high-pass Butterworth filter are:\n", + "R1=1 k ohm\n", + "R2=4 k ohm\n", + "R=5 k ohm\n", + "C=0.16 uF\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3, Page 378" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "fo=1000.#fo=centre frequency in Hz\n", + "f=100#f=bandwidth in Hz\n", + "\n", + "#Calculations\n", + "#Q=wo/w=Quality factor\n", + "Q=(2*math.pi*fo)/(2*math.pi*f)\n", + "C1=0.02*10**-6\n", + "C2=0.02*10**-6#C1=C2=Assumed Capacitances in Farad\n", + "Ao=2#Ao=gain at the centre frequency\n", + "#R1*C1=Q/(wo*Ao) for active band pass Butterworth filter\n", + "wo=2*math.pi*fo\n", + "R1=Q/(Ao*wo*C1)\n", + "R3=Q/(wo*((C1*C2)/(C1+C2)))\n", + "Rp=1./((wo**2)*R3*C1*C2)\n", + "R2=(R1*Rp)/(R1-Rp)\n", + "\n", + "#Results\n", + "print \"The designed values of resistance and capacitance for a second order band-pass Butterworth filter are:\"\n", + "print \"R1=%.f k ohm\"%(math.ceil(R1/1000))\n", + "print \"R2=%.f ohm\"%(math.floor(R2))\n", + "print \"R3=%.f k ohm\"%(math.ceil(R3/1000))\n", + "print \"C1=%.2f uF\"%(C1/10**-6)\n", + "print \"C2=%.2f uF\"%(C2/10**-6)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The designed values of resistance and capacitance for a second order band-pass Butterworth filter are:\n", + "R1=40 k ohm\n", + "R2=401 ohm\n", + "R3=160 k ohm\n", + "C1=0.02 uF\n", + "C2=0.02 uF\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4, Page 379" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "fo=400#fo=centre frequency in Hz\n", + "Q=10#Q=wo/w=Quality factor\n", + "C1=0.1*10**-6\n", + "C2=0.1*10**-6#C1=C2=Assumed Capacitances in Farad\n", + "Ao=2#Ao=gain at the centre frequency\n", + "\n", + "#Calculations\n", + "#R1*C1=Q/(wo*Ao) for active band pass Butterworth filter\n", + "wo=2*math.pi*fo\n", + "R1=Q/(Ao*wo*C1)\n", + "R3=Q/(wo*((C1*C2)/(C1+C2)))\n", + "Rp=1/((wo**2)*R3*C1*C2)\n", + "R2=(R1*Rp)/(R1-Rp)\n", + "#Assuming arbitrarily (R6/R5)=10=a\n", + "a=10\n", + "R6=10*1000#R6=Assumed resistance in ohms\n", + "R5=R6/a\n", + "R4=R5/Ao\n", + "\n", + "#Results\n", + "print \"The designed values of resistance and capacitance for a notch filter are:\"\n", + "print \"R1=%.2f k ohm\"%(R1/1000)\n", + "print \"R2=%.f ohm\"%R2\n", + "print \"R3=%.2f k ohm\"%(R3/1000)\n", + "print \"R4=%.f ohm\"%R4\n", + "print \"R5=%.f k ohm\"%(R5/1000)\n", + "print \"R6=%.f k ohm\"%(R6/1000)\n", + "print \"C1=%.1f uF\"%(C1/10**-6)\n", + "print \"C2=%.1f uF\"%(C2/10**-6)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The designed values of resistance and capacitance for a notch filter are:\n", + "R1=19.89 k ohm\n", + "R2=201 ohm\n", + "R3=79.58 k ohm\n", + "R4=500 ohm\n", + "R5=1 k ohm\n", + "R6=10 k ohm\n", + "C1=0.1 uF\n", + "C2=0.1 uF\n" + ] + } + ], + "prompt_number": 9 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch16.ipynb b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch16.ipynb new file mode 100644 index 00000000..890b66e9 --- /dev/null +++ b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch16.ipynb @@ -0,0 +1,62 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 16: Special Devices" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 397" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "Vp=15.#Vp=firing voltage of a unijunction transistor in Volts\n", + "VBB=40#VBB=source voltage in Volts\n", + "\n", + "#Calculations&Results\n", + "n=(Vp/VBB)#n=intrinsic stand-off ratio\n", + "print \"The intrinsic stand-off ratio is %.3f\"%n\n", + "R=50*(10**3)#R=resistance in ohms\n", + "C=2000*(10**-12)#c=capacitance in farad\n", + "T=(R*C*math.log(1/(1-n)))*(10**6)#T=time period of the sawtooth voltage across C\n", + "print \"The time period is %.f us\"%T" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The intrinsic stand-off ratio is 0.375\n", + "The time period is 47 us\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch17.ipynb b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch17.ipynb new file mode 100644 index 00000000..aa32f44d --- /dev/null +++ b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch17.ipynb @@ -0,0 +1,482 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 17: Number Systems, Boolean Algebra and Digital Circuits" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 440" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "x=25\n", + "\n", + "#Calculations&Results\n", + "s=int(bin(x)[2:])\n", + "print \"1. Binary equivalent of 25 is \",s\n", + "y=576\n", + "s1=int(bin(y)[2:])\n", + "print \"2. Binary equivalent of 576 is \",s1" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1. Binary equivalent of 25 is 11001\n", + "2. Binary equivalent of 576 is 1001000000\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2, Page 440" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "s='1111'\n", + "\n", + "#Calculations\n", + "x=int(s,2)\n", + "\n", + "#Result\n", + "print \"Decimal equivalent of 1111 is \",x" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Decimal equivalent of 1111 is 15\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3, Page 440" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "p = 1.;\n", + "#initialising variables\n", + "z = 0;\n", + "b = 0;\n", + "w = 0;\n", + "f = 0;\n", + "bin1 = 11.1101;\n", + "d = bin1%1; #separating the decimal part and the integer part\n", + "d = d*10**10;\n", + "a = math.floor(bin1);#removing the decimal part\n", + "b = []\n", + "while(a>0):#loop to take the binary bits of integer into a matrix\n", + " r = a%10;\n", + " b.append(r);\n", + " a = a/10;\n", + " a = math.floor(a);\n", + "\n", + "for m in range(len(b)):#multiplying the bits of integer with their position values and adding\n", + " c = m;\n", + " f = f+b[m]*(2**c);\n", + "\n", + "w = []\n", + "while(d>0):#loop to take the binary bits of decimal into a matrix\n", + " e = d%2\n", + " w.append(e)\n", + " d = d /10;\n", + " d = math.floor(d)\n", + "\n", + "for n in range(len(w)):#multiplying the bits of decimal with their position values and adding\n", + " z = z+w[n] *(0.5)**(11-n+1);\n", + "\n", + "z = z*10000;\n", + "#rounding of to 4 decimal values\n", + "z = round(z);\n", + "z = z/10000;\n", + "print \"The decimal equivalent of 11.1101 is = %.4f\"%(f+z)\n", + "#answers differ due to usage of in-built functions" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The decimal equivalent of 11.1101 is = 3.2031\n" + ] + } + ], + "prompt_number": 63 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4, Page 440" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "q=0.;\n", + "b=0.;\n", + "s=0.;\n", + "a=4.625;\n", + "\n", + "#Calculations\n", + "a=math.floor(a);#removing the decimal part\n", + "while(a>0):#taking integer part into a matrix and converting into equivalent binary\n", + " x=(a%2);\n", + " b=b+(10**q)*x;\n", + " a=a/2;\n", + " a=math.floor(a);\n", + " q=q+1;\n", + "\n", + "for i in range(1,10):#for values after decimal point converting into binary\n", + " d=a*2;\n", + " q=math.floor(d);\n", + " s=s+q/(10**i);\n", + " if d>=1:\n", + " d=d-1;\n", + "\n", + "k=b+s;\n", + "\n", + "#Result\n", + "print \"The binary equivalent of 4.625 is =\" ,k\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The binary equivalent of 4.625 is = 100.0\n" + ] + } + ], + "prompt_number": 100 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5, Page 440" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "\n", + "def toStr(n,base):\n", + " convertString = \"0123456789ABCDEF\"\n", + " if n < base:\n", + " return convertString[n]\n", + " else:\n", + " return toStr(n//base,base) + convertString[n%base]\n", + "\n", + "dec=263\n", + "base=5\n", + "\n", + "#Calculations\n", + "s=toStr(dec,base)\n", + "\n", + "#Result\n", + "print \"Equivalent of 263 in a code base 5 is \",s" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equivalent of 263 in a code base 5 is 2023\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6, Page 441" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "x=2\n", + "\n", + "#Calculations\n", + "s=x+x\n", + "s1=bin(s)\n", + "\n", + "#Result\n", + "print \"Binary addition corresponding to decimal addition 2+2 is \",int(s1[2:])" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Binary addition corresponding to decimal addition 2+2 is 100\n" + ] + } + ], + "prompt_number": 50 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7, Page 441" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "x='11111'\n", + "y='1011'\n", + "z='101'\n", + "w='10'\n", + "v='1'\n", + "\n", + "#Calculations\n", + "s1=int(x,2)\n", + "s2=int(y,2)\n", + "s3=int(z,2)\n", + "s4=int(w,2)\n", + "s5=int(v,2)\n", + "a=s1+s2+s3+s4+s5\n", + "b=int(bin(a)[2:])\n", + "\n", + "#Results\n", + "print \"Binary addition of 11111+1011+101+10+1 is \",b\n", + "print \"Decimal equivalent corresponding to above binary addition is \",a" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Binary addition of 11111+1011+101+10+1 is 110010\n", + "Decimal equivalent corresponding to above binary addition is 50\n" + ] + } + ], + "prompt_number": 51 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8, Page 441" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "x='1101'\n", + "y='111'\n", + "\n", + "#Calculations\n", + "s1=int(x,2)\n", + "s2=int(y,2)\n", + "a=s1-s2\n", + "s=int(bin(a)[2:])\n", + "\n", + "#Result\n", + "print \"Binary subtraction 1101-111 is =\",s" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Binary subtraction 1101-111 is = 110\n" + ] + } + ], + "prompt_number": 52 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9, Page 442" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "hFE=30#hFE=dc current gain of given silicon transistor\n", + "VBE=0.8#VBE=base-emitter voltage drop at saturation\n", + "VCE=0.2#VCE=collector-emitter voltage drop at saturation\n", + "R1=15.*1000#resistance at the base side of the transistor in ohms\n", + "R2=100*1000#another resistance at the base side of the transistor in ohms\n", + "RL=2*1000#load resistance at the collector side of the transistor in ohms\n", + "VCC=10#VCC=collector supply voltage\n", + "VBB=-10#VBB=base supply voltage\n", + "\n", + "#Calculations&Results\n", + "#If the input level is 0 volt i e vi=0,the open-circuited base voltage is given as\n", + "VB=VBB*(R1/(R1+R2))\n", + "print \"For input level 0 V:\"\n", + "print \"As a bias of approximately 0V is sufficient to cut off a silicon emitter junction, it follows that transistor is cut off when vi=0\"\n", + "print \"When vi=0,the output voltage is vo=VCC=%.f V\"%VCC\n", + "print \"This indicates that the output is in state 1 when the input is in state 0\"\n", + "#When the input level is 10 volt i e vi=10, we have to show that the transistor is in saturation\n", + "#The minimum base current for saturation is given by iB(min)=iC/hFE\n", + "iC=(VCC-VCE)/RL#collector current when the transistor saturates\n", + "iB=iC/hFE#iB=iB(min)=minimum base current for saturation in mA\n", + "i1=(10-VBE)/R1#i1=current through R1 resistor connected at the base side and here vi=10 is taken\n", + "i2=(VBE-VBB)/R2#i2=current through R2 resistor connected at the base side\n", + "iB1=i1-i2#iB1=actual base current\n", + "print \"\\nFor input level 10 V:\"\n", + "if (iB1>iB):\n", + " print \"Since iB>iB(min),it is verified that the transistor is in saturation\"#iB indicates actual base current & iB(min) indicates minimum base current for saturation\n", + " print \"When vi=10,the output voltage is vo=VCE(sat)=%.1f V\"%VCE\n", + " print \"This indicates that the output is in state 0 when the input is in state 1\"\n", + "\n", + "print \"Overall it has been thus verified that the circuit has performed the NOT operation\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For input level 0 V:\n", + "As a bias of approximately 0V is sufficient to cut off a silicon emitter junction, it follows that transistor is cut off when vi=0\n", + "When vi=0,the output voltage is vo=VCC=10 V\n", + "This indicates that the output is in state 1 when the input is in state 0\n", + "\n", + "For input level 10 V:\n", + "Since iB>iB(min),it is verified that the transistor is in saturation\n", + "When vi=10,the output voltage is vo=VCE(sat)=0.2 V\n", + "This indicates that the output is in state 0 when the input is in state 1\n", + "Overall it has been thus verified that the circuit has performed the NOT operation\n" + ] + } + ], + "prompt_number": 56 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10, Page 442" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "A=0\n", + "B=0\n", + "\n", + "#Calculations&Results\n", + "C=(A|B)#bitwise OR operation is performed\n", + "print \"Boolean expression C=A+B for inputs A=0 and B=0 is\",C\n", + "A=1\n", + "B=0\n", + "C=(A|B)\n", + "print \"Boolean expression C=A+B for inputs A=1 and B=0 is\",C\n", + "A=1\n", + "B=1\n", + "C=(A|B)\n", + "print \"Boolean expression C=A+B for inputs A=1 and B=1 is\",C" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Boolean expression C=A+B for inputs A=0 and B=0 is 0\n", + "Boolean expression C=A+B for inputs A=1 and B=0 is 1\n", + "Boolean expression C=A+B for inputs A=1 and B=1 is 1\n" + ] + } + ], + "prompt_number": 57 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch19.ipynb b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch19.ipynb new file mode 100644 index 00000000..cb1ab108 --- /dev/null +++ b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch19.ipynb @@ -0,0 +1,273 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 19: VLSI Technology and Circuits" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 555" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "ID=50*10**-6#ID=drain current in amperes\n", + "k=25*10**-6#k=ue/D in A/V**2\n", + "VDS=0.25#VDS=drain-to-source voltage\n", + "VGS=5#VGS=gate-to-source voltage\n", + "VTH=1.5#VTH=threshold voltage\n", + "\n", + "#Calculations&Results\n", + "w=ID/(k*(VGS-VTH)*VDS)#w=W/L\n", + "print \"W/L=%.2f\"%w\n", + "P=VDS*ID#P=power dissipated by the transistor\n", + "print \"The dissipated power is=%.1f uW\"%(P*10**6)\n", + "VDD=5#VDD=drain supply voltage of given NMOS transistor\n", + "R=(VDD-VDS)/ID#R=load resistor to be connected in series with the drain\n", + "print \"The load resistance is=%.f k ohm\"%(R/1000)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "W/L=2.29\n", + "The dissipated power is=12.5 uW\n", + "The load resistance is=95 k ohm\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2, Page 555" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "ID=50*10**-6#ID=drain current in amperes\n", + "k=25*10**-6#k=ue/D in A/V**2\n", + "VDEP=3\n", + "\n", + "#Calculations&Results\n", + "l=(k*((-VDEP)**2))/(2*ID)#l=(L/W)=aspect ratio of the pull-up\n", + "print \"Pull-up (L/W)=%.2f\"%l\n", + "VGS=5#VGS=gate-to-source voltage\n", + "VTH=1#VTH=threshold voltage\n", + "VDs=4.75#VDs=the drain source voltage of the depletion mode pull-up in saturation\n", + "VDD=5#VDD=drain supply voltage of given NMOS inverter\n", + "#L/W=(k*(VGS-VTH)*VDS)/ID where L/W=pull down aspect ratio\n", + "l1=(k*(VGS-VTH)*(VDD-VDs))/ID#l1=L/W\n", + "print \"Pull-down (L/W)=%.1f\"%l1" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Pull-up (L/W)=2.25\n", + "Pull-down (L/W)=0.5\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3, Page 555" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "w=10.#w=W/L value of the NMOS transistor in a CMOS inverter\n", + "un=1350#un=electron mobility for NMOS transistor in cm^2/V s\n", + "up=540#up=electron mobility for PMOS transistor in cm^2/V s\n", + "\n", + "#Calculations\n", + "#(Wpu/Lpu)*up*(VINV-VDD-VTHP)^2=(Wpd/Lpd)*un*(VINV-VTHN)^2\n", + "#For a symmetrical inverter VINV=(VDD/2) and VTHN=(-VTHP)\n", + "#Also for input voltage=VDD/2 both transistors operate in saturation region\n", + "#Therefore,up*(Wpu/Lpu)=un*(Wpd/Lpd)\n", + "w1=(un*w)/up#w1=Wpu/Lpu=W/L value of the PMOS for a symmetrical inverter\n", + "\n", + "#Result\n", + "print \"W/L value of the PMOS transistor in a CMOS inverter is = %.f\"%w1" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "W/L value of the PMOS transistor in a CMOS inverter is = 25\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4, Page 556" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "f=2*10**9#f=clock frequency in Hz\n", + "VDD=3#VDD=drain supply voltage \n", + "Cl=1*10**-12#C1=load capacitance in Farad\n", + "P=50.*10**-3#P=maximum power dissipation capability in W/stage\n", + "\n", + "#Calculations\n", + "N=P/(f*Cl*VDD**2)#N=maximum permissible number of fan outs\n", + "\n", + "#Results\n", + "print \"N=%.1f\"%N\n", + "print \"The maximum permissible number of fan-outs is(integer just below actual value)=%.f\"%math.floor(N)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "N=2.8\n", + "The maximum permissible number of fan-outs is(integer just below actual value)=2\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5, Page 556" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "L=3*10**-6#L=length of an NMOS pass transistor in metres\n", + "VDS=0.5#VDS=drain-source voltage\n", + "u=1400*10**-4#u=electron mobility in m**2/V s\n", + "\n", + "#Calculations\n", + "t=L**2/(VDS*u)#t=channel transit time\n", + "\n", + "#Result\n", + "print \"The transit time is=%.2f ns\"%(t/10**-9)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The transit time is=0.13 ns\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6, Page 556" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "y=2#y=length unit in micrometres\n", + "\n", + "#Calculations&Results\n", + "W=3*y#W=mimimum metal linewidth in micrometres\n", + "print \"W=%.f um\"%W\n", + "n=80#n=number of driven inverters\n", + "i=0.07#i=average current ratings in milliamperes\n", + "I=n*i#I=total currrent drawn by n inverters\n", + "print \"I=%.1f mA\"%I\n", + "#1mA per micrometre of aluminium line width is the maximum safe average current an aluminium wire can carry.\n", + "print \"This needs a line at least width of %.1f um\"%I\n", + "\n", + "if (W>I):\n", + " print \"Above calculated minimum metal line-width (W) is thus the safe width of the metal line driving 80 inverters.\"\n", + "\n", + "f=5#f=number of fanout lines\n", + "w=f*W#w=required metal line width\n", + "print \"The metal line-width required to supply a fan-out of 5 lines is= %.f um\"%w" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "W=6 um\n", + "I=5.6 mA\n", + "This needs a line at least width of 5.6 um\n", + "Above calculated minimum metal line-width (W) is thus the safe width of the metal line driving 80 inverters.\n", + "The metal line-width required to supply a fan-out of 5 lines is= 30 um\n" + ] + } + ], + "prompt_number": 13 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch2.ipynb b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch2.ipynb new file mode 100644 index 00000000..4b2e8d0a --- /dev/null +++ b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch2.ipynb @@ -0,0 +1,190 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2: Electron Emission from Solids" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#Variable declaration\n", + "A=6.02*(10**5)#A=thermionic emission constant in A(m^(-2))(K^(-2))\n", + "Ew=4.54#Ew=work function in eV\n", + "T=2500#T=temperature in Kelvin\n", + "kB=1.38*10**(-23)#kB=Boltzmann's constant in J/K\n", + "e=1.6*10**(-19)#e=charge of anelectron in C\n", + "\n", + "#Calculations&Results\n", + "b=(e*Ew)/kB#b=thermionic emission constant in K\n", + "print \"b=%.3e K\"%b\n", + "Jx=A*(T**2)*math.exp(-b/T)#Jx=emission current density in A/m^(2)\n", + "print \"Jx=%.f A/(m^2)\"%Jx\n", + "n=Jx/e#n=number of electrons emitted per unit area per second in (m^-2)(s^-1)\n", + "print \"n=%.3e (m^-2)(s^-1)\"%n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "b=5.264e+04 K\n", + "Jx=2700 A/(m^2)\n", + "n=1.688e+22 (m^-2)(s^-1)\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2, Page 20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "T=2673#T=temperature in Kelvin\n", + "dT=10.#dT=change in temperature in Kelvin\n", + "Ew=4.54#Ew=work function in eV\n", + "e=1.6*10**(-19)#e=charge of anelectron in C\n", + "kB=1.38*10**(-23)#kB=Boltzmann's constant in J/K\n", + "\n", + "#Calculations&Results\n", + "#I=(S*A*(T^2))*exp(-((e*Ew)/(kB*T))#I=emission current,S=surface area of the filament,dI=change in emission current\n", + "d=((2*dT)/T)+(((e*Ew)/(kB*(T**2))*dT))#d=change in emission current\n", + "print \"d=%.4f\"%d\n", + "d*100#percent change in emission current\n", + "print \"d*100=%.2f %%\"%(d*100)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "d=0.0812\n", + "d*100=8.12 %\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3, Page 20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "kB=1.38*10**(-23)#kB=Boltzmann's constant in J/K\n", + "#A=6.02*(10**5)#A=thermionic emission constant in A(m^(-2))(K^(-2))\n", + "#Ew1,Ew2=thermionic work function of 2 emitters in eV\n", + "e=1.6*10**(-19)#e=charge of anelectron in C\n", + "T=2000#T=temperature in Kelvin\n", + "\n", + "#Calculations&Results\n", + "#Jx1=A*(T^2)*exp(-(a/(kB*T)))#Jx=emission current density in A/m^(2)\n", + "#Jx2=A*(T^2)*exp(-(b/(kB*T)))\n", + "#(Jx1/Jx2)=2\n", + "#(Jx1/Jx2)=exp((Ew2-Ew1)/(kB*T))\n", + "#exp((Ew2-Ew1)/(kB*T))=2\n", + "d=(kB*T*math.log(2))#d=(Ew2-Ew1)=difference in thermionic work functions of 2 emitters\n", + "print \"d=%.3e J\"%d\n", + "d/e\n", + "print \"d/e=%.4f eV\"%(d/e)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "d=1.913e-20 J\n", + "d/e=0.1196 eV\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4, Page 20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "#Ia=(K*(Va^(3/2)));Ia=space charge limited current ,Va=anode voltage,K=proportionality constant\n", + "Ia1=300#Ia1=space charge limited current of 1st anode in A\n", + "Ia2=200.#Ia2=space charge limited current of 2nd anode in A\n", + "Va1=200#Va=anode voltage of 1st anode in V\n", + "\n", + "#Calculations\n", + "Va2=(Va1*((Ia2/Ia1)**(2./3)))#Va2=anode voltage of 2nd anode in V\n", + "\n", + "#Result\n", + "print \"Va2=%.2f V\"%Va2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Va2=152.63 V\n" + ] + } + ], + "prompt_number": 13 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch20.ipynb b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch20.ipynb new file mode 100644 index 00000000..9e07ea00 --- /dev/null +++ b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch20.ipynb @@ -0,0 +1,282 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 20: Cathode Ray Oscilloscope" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 575" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "e=1.6*10**-19#e=charge of an electron\n", + "Va=1000.#Va=potential difference in volts\n", + "m=9.11*10**-31#m=mass of an electron\n", + "\n", + "#Calculations&Results\n", + "v=math.sqrt((2*e*Va)/m)#v=axial velocity of an electron\n", + "l=2*10**-2#l=axial length of deflecting plates in metre\n", + "t=l/v#t=transit time of the beam through the deflecting plates\n", + "print \"The transit time is = %.2e s\"%t\n", + "Vd=20#Vd=potential difference applied between the deflecting plates in volts\n", + "s=5*10**-3#s=separation between the plates in metre\n", + "ta=(e*Vd)/(s*m)#ta=the traverse acceleration imparted to the electrons by the deflecting voltage\n", + "print \"Traverse acceleration is = %.3e m/s^2\"%ta\n", + "L=25*10**-2#L=distance of the CRT screen from the centre of the deflecting plates in metre\n", + "d=(l*L*Vd)/(2*s*Va)#d=deflection of the spot on the CRT screen\n", + "print \"Spot deflection is = %.f cm\"%(d*100)#d is converted in terms of cm\n", + "S=d/Vd#S=deflection sensitivity\n", + "print \"Deflection sensitivity is = %.1f mm/V\"%(S/10**-3)#S is converted in terms of mm/V" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The transit time is = 1.07e-09 s\n", + "Traverse acceleration is = 7.025e+14 m/s^2\n", + "Spot deflection is = 1 cm\n", + "Deflection sensitivity is = 0.5 mm/V\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2, Page 575" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "e=1.6*10**-19#e=charge of an electron\n", + "Va=1000#Va=potential difference in volts\n", + "m=9.11*10**-31#m=mass of an electron\n", + "\n", + "#Calculations\n", + "v=math.sqrt((2*e*Va)/m)#v=axial velocity of an electron\n", + "l=1.5*10**-2#l=axial length of deflecting plates in metre\n", + "t=l/v#t=transit time of the beam through the deflecting plates\n", + "#T=time period of the sinusoidal deflecting voltage \n", + "#tmax=maximum transit time\n", + "#(0.1/360)*T=tmax,since 1 cycle corresponds to 360 degrees\n", + "T=(t*360)/0.1\n", + "f=1/T#f=highest frequency of the deflecting voltage\n", + "\n", + "#Results\n", + "print \"The highest frequency of the deflecting voltage is = %.f kHz\"%(f/1000)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The highest frequency of the deflecting voltage is = 347 kHz\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3, Page 576" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "V=1000#V=potential difference in volts\n", + "#B=150 gauss (given)\n", + "B=1.5*10**-2#B=magnetic field in tesla\n", + "l=1*10**-2#l=axial length of deflecting plates in metre\n", + "L1=20*10**-2#L1=(L+(l/2))=distance of the fluorescent screen from the centre of the deflection system in metre\n", + "e=1.6*10**-19#e=charge of an electron\n", + "m=9.11*10**-31#m=mass of an electron\n", + "\n", + "#Calculations&Results\n", + "d=B*math.sqrt(e/(2*V*m))*l*L1#d=deflection of the spot \n", + "print \"The deflection of the spot is=%.1f cm\"%(d*100)\n", + "Sm=d/B#Sm=magnetic deflection sensitivity\n", + "print \"The magnetic deflection sensitivity is=%.2f mm/gauss\"%(Sm/10)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The deflection of the spot is=28.1 cm\n", + "The magnetic deflection sensitivity is=1.87 mm/gauss\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4, Page 576" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "sw=10#sw=sweep width in cm\n", + "n=5./2#n=number of cycles given by vertical deflection plates\n", + "\n", + "#Calculations\n", + "c=sw/n#c=centimetres occupied by one cycle of signal\n", + "ct=0.1#ct=calibrated time base of CRO in ms/cm\n", + "t=ct*c#t=time interval corresponding to centimetres occupied by one cycle of signal\n", + "T=t/5#T=time period of the signal,since the scale is 5 times magnified\n", + "f=1/T#f=frequency of the signal\n", + "\n", + "#Results\n", + "print \"The frequency of the signal is = %.1f kHz\"%f" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The frequency of the signal is = 12.5 kHz\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5, Page 576" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "#Let fv=frequency of the vertical signals in kHz\n", + "#fh=frequency of the horizontal signals \n", + "#Number of horizontal tangencies=nh\n", + "#Number of vertical tangencies=nv\n", + "#fv/fh=nh/nv\n", + "fh=1\n", + "nh=3.\n", + "nv=4\n", + "\n", + "#Calculations\n", + "fv=(nh/nv)*fh\n", + "\n", + "#Results\n", + "print \"The frequency of the vertical signal is = %.f Hz\"%(fv*1000)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The frequency of the vertical signal is = 750 Hz\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6, Page 576" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "#slope of the major axis is negative (given)\n", + "A=2.6#A=The maximum y-displacement\n", + "vyo=1.1#vyo=the vertical displacement\n", + "\n", + "#Calculations&Results\n", + "sino=(vyo/A)#o=phase difference between the two voltages\n", + "x=math.degrees(math.asin(sino))\n", + "\n", + "#Results\n", + "print \"As the major axis of the ellipse has a negative slope,phase difference between the two voltages must lie between\",\\\n", + "\"90 degree and 180 degree\"\n", + "print \"Therefore,phase difference between the voltages is = %.f degrees\"%(180-x)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "As the major axis of the ellipse has a negative slope,phase difference between the two voltages must lie between 90 degree and 180 degree\n", + "Therefore,phase difference between the voltages is = 155 degrees\n" + ] + } + ], + "prompt_number": 18 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch21.ipynb b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch21.ipynb new file mode 100644 index 00000000..8738cd4e --- /dev/null +++ b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch21.ipynb @@ -0,0 +1,150 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 21: Communication Systems" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 606" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "Npe=6*10**10#Npe=peak electron concentration for the E layer in m**-3\n", + "Npf=10**12#Npf=peak electron concentration for the F layer in m**-3\n", + "\n", + "#Calculations&Results\n", + "fCE=9*math.sqrt(Npe)#fCE=critical frequency for the E layer\n", + "print \"Critical frequency for the E layer is = %.1f MHz\"%(fCE/10**6)\n", + "fCF=9*math.sqrt(Npf)#fCF=critical frequency for the F layer\n", + "print \"Critical frequency for the F layer is = %.f MHz\"%(fCF/10**6)\n", + "R=6400.#R=radius of the earth in km\n", + "He=110.#He=height of the E layer above the earth surface in km\n", + "ime=math.degrees(math.asin(R/(R+He)))#ime=angle corresponding to maximum frequency fmE for E layer in degrees\n", + "fmE=fCE*1./math.cos(ime*math.pi/180)#fmE=maximum frequency reflected from the E layer\n", + "print \"The maximum frequency reflected from the E layer is = %.1f MHz\"%(fmE/10**6)\n", + "Hf=250.#Hf=height of the F layer above the earth surface in km\n", + "imf=math.degrees(math.asin(R/(R+Hf)))#imf=angle corresponding to maximum frequency fmF for F layer in degrees\n", + "fmF=fCF*(1./math.cos(imf*math.pi/180))#fmF=maximum frequency reflected from the F layer\n", + "print \"The maximum frequency reflected from the F layer is = %.1f MHz\"%(fmF/10**6)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Critical frequency for the E layer is = 2.2 MHz\n", + "Critical frequency for the F layer is = 9 MHz\n", + "The maximum frequency reflected from the E layer is = 12.0 MHz\n", + "The maximum frequency reflected from the F layer is = 33.1 MHz\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2, Page 607" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "R=6400.#R=radius of the earth in km\n", + "He=110.#He=height of the E layer above the earth surface in km\n", + "\n", + "#Calculations\n", + "ime=math.asin(R/(R+He))#ime=angle corresponding to maximum frequency fmE for E layer in radian\n", + "o=(math.pi/2)-ime#o=angle made by the incident ray at the centre of the earth in degrees\n", + "L=2*o*R#L=maximum distance between the transmitting and the receiving points on the earth surface for single hop transmission of the radiowave reflected from the E layer\n", + "\n", + "#Result\n", + "print \"The maximum distance for single hop transmission is = %.f km\"%L\n", + "#Answer given in textbook is 2459 km which is incorrect as it is actually around 2356 km." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum distance for single hop transmission is = 2356 km\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3, Page 607" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "#fc=9*sqrt(Np)\n", + "fc=3.*10**6#fc=critical frequency in Hz\n", + "\n", + "#Calculations\n", + "Np=(fc**2)/81#Np=electron concentration at the reflecting point\n", + "#h=height of the reflecting point from the bottom of the layer\n", + "#Np=(5*10**10)+(10**9*h)....(given)\n", + "h=(Np-(5*10**10))/10**9\n", + "H=100#H=height above the surface of the earth in km\n", + "\n", + "#Result\n", + "print \"The required height above the ground is = %.f km\"%(h+H)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required height above the ground is = 161 km\n" + ] + } + ], + "prompt_number": 27 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch23.ipynb b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch23.ipynb new file mode 100644 index 00000000..a8f6f214 --- /dev/null +++ b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch23.ipynb @@ -0,0 +1,468 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 23: Lasers, Fibre Optics and Holography" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 662" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "y=630*10**(-9)#y=emitted wavelength in meters\n", + "c=3*10**8#c=velocity of light in free space in m/s\n", + "\n", + "#Calculations&Results\n", + "v=c/y#v=frequency of the emitted radiation\n", + "print \"The frequency of the emitted radiation is %.2e Hz\"%v\n", + "h=6.62*10**(-34)#h=Planck's constant\n", + "P=1*10**(-3)#P=output power of gas laser(given)\n", + "n=P/(h*v)\n", + "print \"The number of photons emitted per second is= %.2e s^-1\"%n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The frequency of the emitted radiation is 4.76e+14 Hz\n", + "The number of photons emitted per second is= 3.17e+15 s^-1\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2, Page 663" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V=500#V=bandwidth of a He-Ne laser in Hz\n", + "\n", + "#Calculations&Results\n", + "t=1./V#t=coherence time\n", + "print \"The coherence time is = %.f ms\"%(t*(10**3))\n", + "c=3*10**8#c=velocity of light in m/s\n", + "Lc=c/V#Lc=longitudinal coherence length\n", + "print \"The longitudinal coherence length is=%.f km\"%(Lc/1000)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The coherence time is = 2 ms\n", + "The longitudinal coherence length is=600 km\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3, Page 663" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "#To obtain interference fringes of good visibility the path difference for the central fringe must be an integral multiple of each of the 2 wavelengths.\n", + "#2*d=(n1*y1)=(n2*y2)where y1 & y2 are 2 wave-lengths and d=path difference and n1 and n2 are 2 integers\n", + "#(2*d)*((1/y2)-(1/y1))=(n2-n1)=m where m is another integer\n", + "#Now m=(-2*d*Y)/(y^2)=(2*d*V)/(v*y)=(2*d*V)/c=(2*d)/Lc\n", + "Lc=600.#Lc=coherence length in km\n", + "\n", + "#Calculations\n", + "d=(Lc/2)#d=minimum difference between the 2 arms of the Michelson interferometer\n", + "\n", + "\n", + "print \"The minimum difference between the two arms of the Michelson interferometer is=%.f km\"%d\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum difference between the two arms of the Michelson interferometer is=300 km\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4, Page 663" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "h=6.62*10**(-34)#h=Planck's constant\n", + "v=3*10**8#v=velocity of light(as normal optical source is mentioned) in m/s\n", + "kB=1.38*10**-23#kB=Boltzmann's constant\n", + "T=1000#T=temperature in Kelvin\n", + "w=6000#w=wavelength in Armstrong\n", + "\n", + "#Calculations&Results\n", + "R=(math.exp((h*v)/(w*(10**-10)*kB*T)))-1#R=the ratio of the number of spontaneous to stimulated transitions\n", + "print \"R=%.1e\"%R\n", + "if (R>1):\n", + " print \"As the ratio of the number of spontaneous to stimulated transitions (R) is >> 1 the emission is predominantly\",\\\n", + " \"\\ndue to spontaneous transitions and is thus incoherent\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "R=2.6e+10\n", + "As the ratio of the number of spontaneous to stimulated transitions (R) is >> 1 the emission is predominantly \n", + "due to spontaneous transitions and is thus incoherent\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5, Page 663" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "u=8./(10**14)#u=(V/v)=the short term frequency stability of a He-Ne gas laser \n", + "#v=c/y where c=velocity of light in vacuum and y=wavelength\n", + "c=3*10**8#c=velocity of light in m/s\n", + "y=1153*10**(-9)#y=emitted wavelength in meters\n", + "\n", + "#Calculations&Results\n", + "V=(u*c)/y\n", + "t=1./V#t=coherence time\n", + "print \"The coherence time is = %.f ms\"%(t*(10**3))\n", + "Lc=c/V#Lc=coherence length\n", + "print \"The coherence length is=%.2e m\"%Lc" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The coherence time is = 48 ms\n", + "The coherence length is=1.44e+07 m\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6, Page 664" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "#y0=vacuum wavelength for the frequency v\n", + "#c=(v*y0)\n", + "#The deviation in the wavelength is Y0=(c*V)/(v**2)\n", + "#Y0=((y0**2)*V)/c\n", + "#V being spread in frequency over the linewidth.\n", + "#V=(1/tc)\n", + "c=3*(10**8)#c=velocity of light in m/s\n", + "tc=10**(-8)#tc=coherence time in seconds\n", + "y0=650*(10**(-9))#y0=vacuum wavelength in m\n", + "\n", + "#Calculations&Results\n", + "Y0=(y0**2)/(c*tc)\n", + "print \"Line width is =%.1e nm\"%(Y0/(10**-9))#Y0 is converted in terms of nm\n", + "Lc=c*tc#Lc=coherence length\n", + "print \"The coherence length Lc is=%.f m\"%Lc" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Line width is =1.4e-04 nm\n", + "The coherence length Lc is=3 m\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7, Page 664" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "o=5*10**-5#o=angular spread in radians\n", + "f=10#f=focal length in cm\n", + "\n", + "#Calculations&Results\n", + "D=f*o#D=diameter of the image\n", + "r=(D/2)#r=image radius\n", + "print \"The image radius is = %.1e cm\"%r\n", + "a=math.pi*(r**2)#a=cross sectional area of the image in cm**2\n", + "P=10*10**-3#P=power in Watts\n", + "PD=P/a#PD=power density\n", + "print \"Power density is = %.1e W/cm**2\"%PD\n", + "y=6000*10**-8#y=wavelength in cm\n", + "d=y/o#d=coherent width\n", + "print \"The lateral coherent width is = %.1f cm\"%d" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The image radius is = 2.5e-04 cm\n", + "Power density is = 5.1e+04 W/cm**2\n", + "The lateral coherent width is = 1.2 cm\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8, Page 664" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "h=6.62*10**-34#h=Planck's constant\n", + "c=3*10**8#c=velocity of light in vacuum in m/s\n", + "y=632.8*10**-9#y=emitted wavelength in m\n", + "\n", + "#Calculations\n", + "E=(h*c)/y#E=emitted photon energy in Joules\n", + "e=15.2*10**-19#e=energy of 2p level in Joules\n", + "P=E+e#P=Pumping energy required for transition from 3s to 2p level in a He-Ne laser\n", + "\n", + "#Result\n", + "print \"The desired pumping energy is = %.1f eV\"%(P/(1.6*10**-19))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The desired pumping energy is = 11.5 eV\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9, Page 665" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "h=6.62*10**-34#h=Planck's constant\n", + "v=2.4*10**15#v=frequency of emitted radiation in Hz\n", + "c=3*10**8#c=velocity of light in vacuum in m/s\n", + "\n", + "#Calculations\n", + "A21=1/(1.66*10**-8)#A21=mean spontaneous life time\n", + "B21=((c**3)*A21)/(8*math.pi*h*(v**3))#B21=probability of stimulated emission\n", + "\n", + "#Result\n", + "print \"The desired probability is = %.2e m**3/(J.s**2)\"%B21" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The desired probability is = 7.07e+18 m**3/(J.s**2)\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10, Page 665" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "u1=1.55#u1=refractive index of the core of the fibre\n", + "u2=1.50#u2=refractive index of the cladding\n", + "\n", + "#Calculations&Results\n", + "oa=math.asin(math.sqrt((u1**2)-(u2**2)))#oa=acceptance angle\n", + "print \"The acceptance angle is = %.f degrees\"%math.degrees(oa)\n", + "NA=math.sin(oa)#NA=numerical aperture\n", + "print \"NA=%.4f\"%NA\n", + "oc=math.asin(u2/u1)#oc=critical angle\n", + "print \"Critical angle=%.f degrees\"%math.degrees(oc)\n", + "d=50*10**-6#d=core diameter in meters\n", + "x=d*math.tan(oc)#x=axial distance traversed by the ray between two consecutive reflections\n", + "n=1/x#n=number of reflections per metre\n", + "print \"The number of reflections per metre is = %.f\"%n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The acceptance angle is = 23 degrees\n", + "NA=0.3905\n", + "Critical angle=75 degrees\n", + "The number of reflections per metre is = 5207\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11, Page 665" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "ro = 0.2 #mm\n", + "u1 = 1.52\n", + "\n", + "#Calculations\n", + "n_ro = 1.52-(2*ro**2)\n", + "NA = math.sqrt(u1**2-n_ro**2)\n", + "theta_A = math.degrees(math.asin(NA))\n", + "\n", + "#Results\n", + "print \"Numerical Arperture = %.4f\"%NA\n", + "print \"Acceptance angle = %.2f degrees\"%theta_A" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Numerical Arperture = 0.4866\n", + "Acceptance angle = 29.12 degrees\n" + ] + } + ], + "prompt_number": 21 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch3.ipynb b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch3.ipynb new file mode 100644 index 00000000..36af869e --- /dev/null +++ b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch3.ipynb @@ -0,0 +1,320 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3: Properties of Semiconductors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "T=300; #K\n", + "ni=1.5*10**16; #Intrinsic carrier concentartion per m^3\n", + "yn=0.13; #Electron mobility in m^2/(V*s)\n", + "yp=0.05; #Hole mobility in m^2/(V*s)\n", + "e=1.6*10**-19; #Charge of electron in C\n", + "\n", + "#Calculations\n", + "Gi=e*ni*(yn+yp); #Intrinsic conductivity\n", + "Ri=1/Gi; #Intrinsic resistivity\n", + "\n", + "#Results\n", + "print 'Intrinsic conductivity=%.3e S/m'%Gi; #incorrect answer in textbook for Gi\n", + "print 'Intrinsic resistivity=%.3e ohm*meter'%Ri;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Intrinsic conductivity=4.320e-04 S/m\n", + "Intrinsic resistivity=2.315e+03 ohm*meter\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2, Page 40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Sn=480; #Conductivity in S/m\n", + "yn=0.38; #Electron mobility in m^2/(V*s)\n", + "e=1.6*10**-19; #Charge of electron in C\n", + "\n", + "#Calculations\n", + "Nd=Sn/(e*yn); #Concentration of donor atoms per m^3\n", + "\n", + "#Result\n", + "print 'Concentration of donor atoms = %.1e m^-3'%Nd;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Concentration of donor atoms = 7.9e+21 m^-3\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4, Page 41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "T=300; #K\n", + "ni=1.5*10**16; #Intrinsic carrier concentartion per m^3\n", + "yn=0.13; #Electron mobility in m^2/(V*s)\n", + "yp=0.05; #Hole mobility in m^2/(V*s)\n", + "e=1.6*10**-19; #Charge of electron in C\n", + "l=0.01; #length in m\n", + "a=10**-6; #cross sectional area in m^2\n", + "\n", + "#Calculations\n", + "Gi=e*ni*(yn+yp); #Intrinsic conductivity\n", + "Ri=l/(Gi*a); #Required resistance\n", + "\n", + "#Results\n", + "print 'Intrinsic conductivity=%.2e S/m'%Gi;\n", + "print 'required resistance = %.2f Mohm'%(Ri*10**-6);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Intrinsic conductivity=4.32e-04 S/m\n", + "required resistance = 23.15 Mohm\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5, Page 41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "z=(100./60);#z=conductiarrier concentration in /(m^3)\n", + "ni=2.5*10**(19);#ni=intrinsic conductivity of intrinsic material in S/m\n", + "#(P/N)=(1./2);#(P/N)=ratio of hole mobility(P) to electron mobility(N)\n", + "e=1.6*(10**-19);#e=charge of electron in Coulomb\n", + "\n", + "#Calculations&Results\n", + "N=(z/(e*ni*(1+(1./2))))\n", + "print \"N=%.4f (m^2)/(V.s)\"%N\n", + "P=(N/2)\n", + "print \"P=%.4f (m^2)/(V.s)\"%P\n", + "#Nd+p=Na+n;n=electron concentration;p=hole concentration\n", + "#np=(ni^2)\n", + "Nd=(10**20)#Nd=donor concentration in /(m^3)\n", + "Na=5*(10**19)#Na=acceptor concentration in /(m^3)\n", + "n=(1./2)*((Nd-Na)+math.sqrt(((Nd-Na)**2)+(4*(ni**2))))\n", + "print \"n=%.3e /(m**3)\"%n\n", + "p=(ni**2)/n\n", + "print \"p=%.3e /(m^3)\"%p\n", + "Z=e*((n*N)+(p*P))#Z=conductivity of doped sample in S/m\n", + "print \"Z=%.1f S/m\"%Z\n", + "F=200#F=applied electric field in V/cm\n", + "J=Z*F#J=total conduction current density in A/(m^2)\n", + "print \"J=%.f A/(m^2)\"%J\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "N=0.2778 (m^2)/(V.s)\n", + "P=0.1389 (m^2)/(V.s)\n", + "n=6.036e+19 /(m**3)\n", + "p=1.036e+19 /(m^3)\n", + "Z=2.9 S/m\n", + "J=583 A/(m^2)\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6, Page 42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "ni=2.5*10**(19);#ni=intrinsic conductivity of intrinsic material in S/m\n", + "Nd=5*(10**19)#Nd=donor concentration in /(m^3)\n", + "\n", + "#Calculations&Results\n", + "n=(1./2)*(Nd+math.sqrt((Nd**2)+(4*(ni**2))))#n=electron concentration\n", + "print \"n=%.3e /(m^3)\"%n\n", + "p=(ni**2)/n#p=hole concentration\n", + "print \"p=%.3e /(m^3)\"%p\n", + "N=0.38#N=electron mobility in (m^2)/(V.s)\n", + "P=0.18#P=hole mobility in (m^2)/(V.s)\n", + "e=1.6*(10**-19)#e=electronic charge in Coulomb\n", + "Z=e*((n*N)+(p*P))#Z=conductivity of doped sample in S/m\n", + "print \"Z=%.3f S/m\"%Z" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "n=6.036e+19 /(m^3)\n", + "p=1.036e+19 /(m^3)\n", + "Z=3.968 S/m\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7, Page 43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "c=3*(10**8);#c=velocity of light in vacuum in m/s\n", + "h=6.6*(10**-34);#h=Planck's constant in J.s\n", + "Eg=1.98*1.6*(10**-19)#Eg=band gap in J\n", + "\n", + "#Calculations\n", + "#calculating Y=required wavelength\n", + "Y=((c*h)/Eg)/(10**-9)\n", + "\n", + "#Result\n", + "print \"Y=%.f nm\"%Y" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Y=625 nm\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8, Page 43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "RH=(10**-2);#RH=Hall coefficient in (m^3)/C\n", + "VH=(10**-3);#VH=Hall Voltage in V\n", + "b=2*(10**-3);#b=width in m\n", + "I=(10**-3);#I=current in A\n", + "\n", + "#Calculations&Results\n", + "#RH=(VH*b)/(I*B)\n", + "B=(VH*b)/(I*RH)#B=magnetic field\n", + "print \"B= %.1f T\"%B\n", + "t=(10**-3)#t=thickness in m\n", + "FH=(VH/t)#FH=Hall field\n", + "print \"FH=%.f V/m\"%FH\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "B= 0.2 T\n", + "FH=1 V/m\n" + ] + } + ], + "prompt_number": 19 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch4.ipynb b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch4.ipynb new file mode 100644 index 00000000..4b5ed871 --- /dev/null +++ b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch4.ipynb @@ -0,0 +1,228 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4: Metal Semiconductors Contacts" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 57" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Qm=4.55#Qm=work function of tungsten in eV\n", + "X=4.01#X=electron affinity of silicon in eV\n", + "\n", + "#Calculations&Results\n", + "eQb=(Qm-X)#eQb=barrier height as seen from the metal\n", + "print \"eQb=%.2f eV\"%eQb\n", + "a=0.21#a=(Ec-Ef)=forbidden gap in eV\n", + "eVbi=eQb-a#eVbi=barrier height from semiconductor side\n", + "print \"eVbi=%.2f eV\"%eVbi\n", + "Es=11.7*8.854*(10**-12)#Es=permittivity of semiconductor;11.7=dielectric constant of silicon\n", + "e=1.6*10**(-19)#e=charge of an electron\n", + "Nd=10**22#Nd=donor concentration in m^-3\n", + "W=((2*Es*eVbi)/(e*Nd))**(1./2)#W=width of the depletion region\n", + "print \"W=%.3e m\"%W\n", + "Fm=((e*Nd*W)/Es)#Fm=maximum electric field in V/m\n", + "print \"Fm= %.1e V/m\"%Fm" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "eQb=0.54 eV\n", + "eVbi=0.33 eV\n", + "W=2.067e-07 m\n", + "Fm= 3.2e+06 V/m\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2, Page 58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "# as per given data barrier height =Vbi=intercept on Vr axis=0.4 V\n", + "Es=11.7*8.854*(10**-12)#Es=permittivity of semiconductor;11.7=dielectric constant of silicon\n", + "e=1.6*10**(-19)#e=charge of an electron\n", + "m=4.4*10**(15)#m=slope of (1/C^2) vs Vr plot of a Schottky contact in(cm^4)(F^-2)(V^-1)\n", + "\n", + "#Calculations\n", + "#m=2/(e*Es*Nd)\n", + "Nd=(2*10**8)/(e*Es*m)#Nd=donor concentration in silicon in m^-3\n", + "\n", + "#Result\n", + "print \"Nd = %.2e m**-3\"%Nd" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Nd = 2.74e+21 m**-3\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3, Page 58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "e=1.6*10**-19#e=charge of an electron in C\n", + "Fa=7.*10**6#Fa=reverse bias field in V/m\n", + "\n", + "#Calculations&Results\n", + "Es=13.1*8.854*10**-12#(Es/Eo)=13.1;Eo=8.854*10^-12\n", + "dQ=((e*Fa)/(4*math.pi*Es))**(1./2)#dQ=barrier lowering in V \n", + "print \"dQ= %.4f V\"%dQ\n", + "Xm=(dQ)/(2*Fa)#Xm=position of the maximum barrier height\n", + "print \"Xm= %.2e m\"%Xm" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "dQ= 0.0277 V\n", + "Xm= 1.98e-09 m\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4, Page 58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "#Js=A*(T**2)*math.exp(-((e*Qbn)/(kB*T)))\n", + "kB=1.38*10**(-23)#kB=Boltzmann's constant in J/K\n", + "T=300#T=temperature in Kelvin\n", + "e=1.6*10**-19#e=charge of an electron in C\n", + "Js=6*10**-5#Js=emission current density in A/cm^2\n", + "\n", + "#Calculations\n", + "Qbn=0.668#Qbn=barrier height in V\n", + "A=(Js/(T**2))*math.exp((e*Qbn)/(kB*T))#A=Richardson constant\n", + "\n", + "#Result\n", + "print \"A=%.1f (cm^-2)(K^-2)\"%A" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "A=108.6 (cm^-2)(K^-2)\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5, Page 58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "e=1.6*10**-19#e=charge of an electron in C\n", + "V=0.32#V =applied forward bias in V\n", + "kB=1.38*10**(-23)#kB=Boltzmann's constant in J/K\n", + "T=300#T=Temperature in Kelvin\n", + "Js=0.61#Js=reverse saturation current density in A/m^2\n", + "\n", + "#Calculations&Results\n", + "J=Js*(math.exp((e*V)/(kB*T))-1)#J=current density in A/m^2\n", + "print \"J=%.2e A/m^2\"%J\n", + "A=4*10**-8#A=cross sectional area in m^2\n", + "I=(J*A)*10**3#I=current\n", + "print \"I=%.2f mA\"%I" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "J=1.43e+05 A/m^2\n", + "I=5.73 mA\n" + ] + } + ], + "prompt_number": 12 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch5.ipynb b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch5.ipynb new file mode 100644 index 00000000..97cb19d8 --- /dev/null +++ b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch5.ipynb @@ -0,0 +1,569 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5: Semiconductor Junction Diodes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 82" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "#I=Is*(exp((e*V)/kB*T)-1)\n", + "I=50*10**(-3)#I=Forward current in ampere\n", + "Is=5*10**(-6)#Is=Reverse saturation current in ampere\n", + "e=1.6*10**(-19)#e=charge of electron in coulomb\n", + "#V=voltage\n", + "kB=1.38*10**(-23)#kB=Boltzmann's constant in Joule/kelvin\n", + "T=300#T=Temperature in kelvin\n", + "\n", + "#Calculations\n", + "a=(I/Is)+1\n", + "#exp((e*V)/kB*T)=a\n", + "V=((kB*T)/e)*math.log(10**4)\n", + "\n", + "#Result\n", + "print \"V= %.3f V\"%V" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "V= 0.238 V\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2, Page 83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "e=1.6*10**-19#e=charge of an electron in C\n", + "V1=0.06#V1=applied forward bias in V\n", + "V2=(-0.06)#V2 =applied reverse bias in V\n", + "kB=1.38*10**(-23)#kB=Boltzmann's constant in J/K\n", + "T=300#T=Temperature in Kelvin\n", + "\n", + "#Calculations\n", + "#Is=reverse saturation current in A\n", + "#I1=Is*(exp((e*V1)/(kB*T))-1)#I1=current for forward bias\n", + "#I2=Is*(exp((e*V2)/(kB*T))-1)#I2=current for reverse bias\n", + "a=((math.exp((e*V1)/(kB*T))-1))/((math.exp((e*V2)/(kB*T))-1))#a=(I1/I2)\n", + "\n", + "#Result\n", + "print \"a = %.2f\"%abs(a)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a = 10.16\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3, Page 83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V=0.9#V=forward bias voltage\n", + "I=60*10**(-3)#I=Current in ampere\n", + "\n", + "#Calculations\n", + "rdc=(V/I)#rdc=static resistance in ohm\n", + "n=2#n=emission coefficient\n", + "rac=((26*n*10**(-3))/I)#rac=dynamic resistance\n", + "\n", + "#Results\n", + "print \"rdc=%.f ohm\"%rdc\n", + "print \"rac=%.2f ohm\"%rac" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "rdc=15 ohm\n", + "rac=0.87 ohm\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4, Page 83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "e=1.6*10**(-19)#e=charge of an electron in C\n", + "kB=1.38*10**(-23)#kB=Boltzmann's constant in J/K\n", + "#V,V1=forward bias voltagesin V\n", + "n=2#n=emission coefficient for silicon pn junction diode\n", + "T=300#T=Temperature in kelvin\n", + "\n", + "#Calculations\n", + "#Is=Reverse saturation current in A\n", + "#I=Is*(exp((e*V)/(n*kB*T)))#I=current for forward bias voltage V\n", + "#2I=Is*(exp((e*V1)/(n*kB*T)))#2I=current for forward bias voltage V1\n", + "#exp((e*(V1-V)/(n*kB*T)))=2\n", + "a=(((n*kB*T)/e)*math.log(2))*10**3#a=(V1-V)=increase in the bias voltage in V\n", + "\n", + "#Result\n", + "print \"V1-V = %.1f mV\"%a" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "V1-V = 35.9 mV\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5, Page 84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "e=1.6*10**(-19)#e=charge of an electron in C\n", + "kB=1.38*10**(-23)#kB=Boltzmann's constant in J/K\n", + "n=2#n=emission coefficient for silicon pn junction diode\n", + "T=300#T=Temperature in kelvin\n", + "\n", + "#Calculations\n", + "#Is=Reverse saturation current in A\n", + "#V=bias voltage in V\n", + "#I=Is*(exp((e*V)/(n*kB*T))-1)#I=reverse current in A\n", + "#I=(-(Is/2))\n", + "a=(((n*kB*T)/e)*math.log(1./2))*10**3#a=bias for reverse current in silicon pn junction diode \n", + "\n", + "#Results\n", + "print \"V = %.1f mV\"%a\n", + "print \"The negative sign suggests diode in reverse bias\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "V = -35.9 mV\n", + "The negative sign suggests diode in reverse bias\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6, Page 84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "#T1,T2=Temperature in kelvin\n", + "#Is1=Reverse saturation current at temperature T1 in ampere\n", + "#Is2=Reverse saturation current at temperature T2 in ampere\n", + "#Is2=Is1*2^((T2-T1)/10)\n", + "#((T2-T1)/10)*log(2)=log(Is2/Is1)\n", + "#b=(Is2/Is1)\n", + "b=50\n", + "\n", + "#Calculations\n", + "a=((10*math.log(b))/math.log(2))#a=(T2-T1)=rise in temperature in degree celcius\n", + "\n", + "#Result\n", + "print \"T2-T1 = %.1f C\"%a" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "T2-T1 = 56.4 C\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7, Page 84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "V=0.6#V=cutin voltage in V\n", + "r=150.#r=forward resistance in ohm\n", + "P=200.0*(10**-3)#P=maximum power in Watt\n", + "\n", + "#Calculations&Results\n", + "#P=(i^2)*r where i=maximum safe diode current\n", + "i=(math.sqrt(P/r))*10**3\n", + "print \"i=%.1f mA\"%i\n", + "#i=((Vb/3)-V)/3 by applying KCL\n", + "Vb=((3*i)+V)*3#Vb=maximum permissible battery voltage\n", + "print \"Vb= %.1f V\"%Vb\n", + "#Incorrect answers in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i=36.5 mA\n", + "Vb= 330.4 V\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8, Page 85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V=15#V=supply voltage\n", + "Vz=12#Vz=Zener voltage\n", + "P=0.36#P=power of Zener diode\n", + "\n", + "#Calculations&Results\n", + "#P=Vz*I\n", + "I=(P/Vz)#I=maximum allowable Zener current \n", + "print \"I= %.2f A\"%I\n", + "Vr=V-Vz#Vr=voltage drop across series resistance R\n", + "print \"Vr=%.f V\"%Vr\n", + "R=Vr/I#R=series resistance\n", + "print \"R=%.f ohm\"%R\n", + "#I=Iz+Il\n", + "Iz=2*(10**-3)#Iz=minimum diode current\n", + "Il=I-Iz#Il=current through load resistance Rl\n", + "print \"Il=%.f mA\"%(Il*10**3)\n", + "Rlm=Vz/Il#Rlm=minimum value of Rl\n", + "print \"Rlm=%.1f ohm\"%Rlm\n", + "print \"The allowable range of variation of Rl is 428.6ohm<=Rl<infinite\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "I= 0.03 A\n", + "Vr=3 V\n", + "R=100 ohm\n", + "Il=28 mA\n", + "Rlm=428.6 ohm\n", + "The allowable range of variation of Rl is 428.6ohm<=Rl<infinite\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9, Page 85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V=15#V=supply voltage\n", + "Vz=12.#Vz=Zener voltage\n", + "P=0.36#P=power of Zener diode\n", + "\n", + "#Calculations&Results\n", + "#P=Vz*I\n", + "I=(P/Vz)#I=maximum allowable Zener current \n", + "print \"I=%.2f A\"%I\n", + "Iz=2*10**(-3)#Iz=minimum value attained by the zener current\n", + "Rl=1000#Rl=load resistance\n", + "i=Vz/Rl#i=load current\n", + "print \"i=%.3f A\"%i\n", + "Imin=Iz+i#Imin=minimum allowable value of current\n", + "R=100#R=series resistance\n", + "Vr=Imin*R#Vr=voltage drop across R\n", + "print \"Vr=%.1f V\"%Vr\n", + "Vmin=Vz+Vr#Vmin=minimum value of V\n", + "print \"Vmin=%.1f V\"%Vmin\n", + "I1=I+i\n", + "print \"I1=%.f mA\"%(I1*10**3)\n", + "VR=I1*R\n", + "print \"VR=%.1f V\"%VR\n", + "Vmax=Vz+VR#Vmax=maximum value of V\n", + "print \"Vmax=%.1f V\"%Vmax\n", + "print \"V can vary between Vmin & Vmax\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "I=0.03 A\n", + "i=0.012 A\n", + "Vr=1.4 V\n", + "Vmin=13.4 V\n", + "I1=42 mA\n", + "VR=4.2 V\n", + "Vmax=16.2 V\n", + "V can vary between Vmin & Vmax\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10, Page 85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import numpy as np\n", + "from numpy.linalg import inv\n", + "\n", + "#Variable declaration\n", + "Vz=3#Vz=breakdown voltage of zener diode\n", + "Vi=12#Vi=input voltage\n", + "V=np.matrix('12;-3')#V=[Vi:-Vz]\n", + "R1=1000\n", + "R2=1000\n", + "R3=500#R1,R2,R3=resistances\n", + "\n", + "#Calculations&Results\n", + "R=np.matrix([[R1+R2, -R2],[-R2, R2+R3]])\n", + "#solving this matrix on the basis of application of KCL & KVL,we get the values of branch currents I & Iz as I1=[I;Iz]\n", + "I1=inv(R)*V\n", + "print \"I=%.2f mA\"%(I1[0]*10**3)\n", + "print \"Iz=%.f mA\"%(I1[1]*10**3)\n", + "Pz=Vz*I1[1]#Pz=power dissipated in zener diode\n", + "print \"Pz=%.f mW\"%(Pz*10**3)\n", + "print \"Power dissipated does not exceed the maximum power limit of 20mW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "I=7.50 mA\n", + "Iz=3 mA\n", + "Pz=9 mW\n", + "Power dissipated does not exceed the maximum power limit of 20mW\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11, Page 86" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vs1=15\n", + "Vs2=30.#Vs=supply voltage varying from 15(Vs1) to 30(Vs2) Volt\n", + "Vzo=9#Vzo=knee voltage\n", + "rZ=5.#rZ=dynamic resistance in ohms\n", + "R=800#R=series resistance in ohms\n", + "\n", + "#Calculations&Results\n", + "Izmin=(Vs1-Vzo)/(R+rZ)#Izmin=current through zener diode when Vs is 15 V\n", + "print \"Izmin=%.2f mA\"%(Izmin*10**3)\n", + "Vomin=(rZ*Izmin)+Vzo#Vomin=corresponding minimum output voltage\n", + "print \"Vomin=%.3f V\"%Vomin\n", + "Izmax=(Vs2-Vzo)/(R+rZ)#Izmax=current through zener diode when Vs is 30 V\n", + "print \"Izmax=%.f mA\"%(Izmax*10**3)\n", + "Vomax=(rZ*Izmax)+Vzo#Vomax=corresponding maximum output voltage\n", + "print \"Vomax=%.2f V\"%Vomax\n", + "print \"Output voltage Vo varies in the range Vomin to Vomax\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Izmin=7.45 mA\n", + "Vomin=9.037 V\n", + "Izmax=26 mA\n", + "Vomax=9.13 V\n", + "Output voltage Vo varies in the range Vomin to Vomax\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12, Page 86" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V=35#V=supply voltage\n", + "Iz=25*10**(-3)#Iz=diode current\n", + "Il=5*10**(-3)#Il=load current\n", + "Vzo=7#Vzo=knee voltage of zener diode\n", + "rZ=6#rZ=dynamic resistance in ohms\n", + "\n", + "#Calculations&Results\n", + "Vz=Vzo+(rZ*Iz)#Vz=zener voltage\n", + "print \"Vz=%.2f V\"%Vz\n", + "I=Iz+Il#I=current through resistance R\n", + "print \"I=%.f mA\"%(I*10**3)\n", + "R=(V-Vz)/I\n", + "print \"R=%.1f ohm\"%R" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Vz=7.15 V\n", + "I=30 mA\n", + "R=928.3 ohm\n" + ] + } + ], + "prompt_number": 33 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch6.ipynb b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch6.ipynb new file mode 100644 index 00000000..02d83b72 --- /dev/null +++ b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch6.ipynb @@ -0,0 +1,304 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 6: Diode Circuits" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 114" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#variable declaration\n", + "Vrms=20; #in volts\n", + "Vm=20*1.41; #in volts\n", + "Rf=50.; #forward resistance in ohms\n", + "RL=1200; #load resistance in ohms\n", + "\n", + "#Calculations&Results\n", + "Im=Vm/(Rf+RL); #peak load current\n", + "print'Im=%.4f A'%Im\n", + "\n", + "Idc=Im/math.pi; #dc load current\n", + "print 'Idc=%.2e A'%Idc\n", + "\n", + "Irms=Im/2;#rms load current\n", + "Irms1=math.sqrt((Irms**2)-(Idc**2))#rms ac load current\n", + "print 'rms ac load current is=%.3e A'%Irms1\n", + "\n", + "Vdc=Idc*Rf; #Dc voltage across the diode\n", + "print 'Dc voltage across the diode=%.4f V'%Vdc\n", + "\n", + "Pdc=Idc*Idc*RL; #Dc output power\n", + "print 'Dc output power=%.3f W'%Pdc\n", + "\n", + "n=40.6/(1+(Rf/RL)); #conversion efficiency\n", + "print 'conversion efficiency=%.f %%'%n\n", + "\n", + "s=Rf*100/RL; #Pertcentage regulation\n", + "print 'Percentage regulation=%.2f %%'%s" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Im=0.0226 A\n", + "Idc=7.18e-03 A\n", + "rms ac load current is=8.699e-03 A\n", + "Dc voltage across the diode=0.3591 V\n", + "Dc output power=0.062 W\n", + "conversion efficiency=39 %\n", + "Percentage regulation=4.17 %\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2, Page 114" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#variable declaration\n", + "Rf=100; #forward resistance in ohms\n", + "Rl=1000.; #load resistance in ohms\n", + "n=10; #Primary to secondary turns ratio\n", + "Vp=240; #Primary input V(rms)\n", + "\n", + "#Calculations&Results\n", + "Vm=24*(2**(1./2))/2; #secondary peak voltage from cenre tap\n", + "Vs=Vp/n; #Secondary input voltage\n", + "Im=Vm/(Rf+Rl); #peak current through the resistance in A\n", + "Idc=(2*Im)/math.pi; #DC Load current in A\n", + "print 'DC load current Idc=%.2e A'%Idc\n", + "I=Idc/2; #Direct current supplied by each diode in A\n", + "print 'Direct current supplied by each diode Idc=%.2e A'%I\n", + "Pdc=Idc*Idc*Rl; #DC power output\n", + "print 'Pdc=%.4f W'%Pdc\n", + "Irms=Im/(2**(1./2));\n", + "Vrp=math.sqrt((Irms*Irms)-(Idc*Idc))*Rl; #Ripple voltage in V\n", + "print 'Ripple voltage Vrp=%.2f V'%Vrp\n", + "M=(Rf*100)/Rl; #percentage regulation\n", + "print 'Percentage regulation=%.f %%'%M\n", + "n=81.2/(1+(Rf/Rl)); #Efficiency of rectification\n", + "print 'Efficiency of rectification=%.1f %%'%n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "DC load current Idc=9.82e-03 A\n", + "Direct current supplied by each diode Idc=4.91e-03 A\n", + "Pdc=0.0965 W\n", + "Ripple voltage Vrp=4.75 V\n", + "Percentage regulation=10 %\n", + "Efficiency of rectification=73.8 %\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3, Page 115" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#variable declaration\n", + "Rf=50.; #forward resistance in ohms\n", + "Rl=2500; #load resistance in ohms\n", + "Vp=30; #Primary input V(rms)\n", + "Vm=30*math.sqrt(2);\n", + "\n", + "#Calculations&Results\n", + "Im=Vm/(2*Rf+Rl); #peak load current in A\n", + "Idc=2*Im/math.pi;\n", + "Vdc=Idc*Rl; #DC load voltage\n", + "print 'Vdc=%.f V'%Vdc\n", + "Irms=Im/math.sqrt(2);\n", + "Vrp=Rl*math.sqrt(((Irms*Irms)-(Idc*Idc))); #Ripple voltage in V\n", + "print 'Ripple voltage Vrp=%.1f V'%Vrp\n", + "\n", + "M=(2*Rf/Rl)*100; #Percentage regulation\n", + "print 'Percentage regulation=%.f %%'%M" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Vdc=26 V\n", + "Ripple voltage Vrp=12.6 V\n", + "Percentage regulation=4 %\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4, Page 115" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#variable declaration\n", + "Vdc=20; #DC value in V\n", + "Vpp=1.; #Peak to peak ripple voltage in V\n", + "\n", + "#Calculations&Results\n", + "Vp=Vpp/2; #Peak ripple voltage in V\n", + "Vrms=Vp/math.sqrt(2); #Vrms voltage in V\n", + "S=Vrms/Vdc; #Ripple Factor\n", + "print 'Ripple factor=%.4f'%S\n", + "T=S*100;\n", + "print 'Percentage Ripple=%.2f %%'%T" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ripple factor=0.0177\n", + "Percentage Ripple=1.77 %\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5, Page 116" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#variable declaration\n", + "#For a full wave rectifier \n", + "#L-type LC filter\n", + "f=50#f=line frequency in Hz\n", + "w=2*math.pi*f\n", + "Vdc=10#Vdc=dc output voltage\n", + "Idc=100*10**-3#Idc=load current in Amperes\n", + "y=0.02#y=allowable ripple factor\n", + "\n", + "#Calculations\n", + "#y=sqrt(2)/(12*(w^2)*L*C)\n", + "#Let L*C=a...............(1)\n", + "a=math.sqrt(2)/(y*12*(w**2))\n", + "RL=Vdc/Idc#RL=load resistance\n", + "#Lc=critical inductance \n", + "#Lc=RL/(3*w)\n", + "#For line frequency of 50Hz,Lc=RL/(300*%pi)\n", + "#Lc=RL/950\n", + "Lc=RL/950\n", + "L=0.1#Assumed inductance in henry\n", + "C=a/L#C=capacitance calculated from equation (1)\n", + "L1=1#Assumed inductance in henry\n", + "C1=a/L1#C1=capacitance calculated from equation (1)\n", + "Rb=950*L1#Rb=bleeder resistance for good voltage regulation\n", + "\n", + "#Results\n", + "print \"The designed values of the components for a full wave rectifier with L-type LC filter are:\"\n", + "print \"The load resistance RL is =%.f ohm\"%RL\n", + "print \"The critical inductance Lc is = %.1f H\"%Lc\n", + "print \"The inductance L is=%.1f H\"%L\n", + "print \"The capacitance C is %.f \u00b5F\"%(C/10**-6)#C is converted in terms of microfarad\n", + "#In textbook 957\u00b5F is approximately taken as 600\u00b5F\n", + "print \"\\nBut if the inductance L designed is of the value = %.f H\"%L1\n", + "print \"the capacitance C will be of the value = %.f \u00b5F\"%(C1/10**-6)#C1 is converted in terms of microfarad\n", + "print \"So,a standard value of 50\u00b5F can be used in practice\"\n", + "print \"The bleeder resistance Rb for good voltage regulation is=%.f ohm\"%Rb\n", + "print \"\\nAs Rb is much greater than RL,little power is wasted in Rb.This reflects the advantage of selecting L>Lc\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The designed values of the components for a full wave rectifier with L-type LC filter are:\n", + "The load resistance RL is =100 ohm\n", + "The critical inductance Lc is = 0.1 H\n", + "The inductance L is=0.1 H\n", + "The capacitance C is 597 \u00b5F\n", + "\n", + "But if the inductance L designed is of the value = 1 H\n", + "the capacitance C will be of the value = 60 \u00b5F\n", + "So,a standard value of 50\u00b5F can be used in practice\n", + "The bleeder resistance Rb for good voltage regulation is=950 ohm\n", + "\n", + "As Rb is much greater than RL,little power is wasted in Rb.This reflects the advantage of selecting L>Lc\n" + ] + } + ], + "prompt_number": 11 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch7.ipynb b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch7.ipynb new file mode 100644 index 00000000..a1fd64a6 --- /dev/null +++ b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch7.ipynb @@ -0,0 +1,400 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 7: Junction Transistor Characteristics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 136" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "a=0.99;#a=fraction of the emitter current contributed by the carriers injected into the base and reaching the collector\n", + "Rl=4500; #Load resistance in ohms\n", + "rd=50; #dynamic resistance in ohms\n", + "\n", + "#Calculations\n", + "Av=a*Rl/rd; #Voltage gain\n", + "Ap=a*Av;#Power gain\n", + "\n", + "#Results\n", + "print 'Av=%.1f'%Av\n", + "print 'Ap=%.1f'%Ap\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Av=89.1\n", + "Ap=88.2\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2, Page 136" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "a=0.98;#a=fraction of the emitter current contributed by the carriers injected into the base and reaching the collector\n", + "Ie=0.003; #emitter current in A\n", + "Ico=10*10**-6; #reverse saturation current in A\n", + "\n", + "#Calculations&Results\n", + "Ic=a*Ie+Ico; #collector current in A\n", + "print 'Ic=%.2f mA'%(Ic/10**-3);#Ic is converted in terms of mA\n", + "Ib=Ie-Ic; #base current in A\n", + "print 'Ib=%.f \u00b5A'%(Ib/10**-6);#Ib is converted in terms of \u00b5A" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ic=2.95 mA\n", + "Ib=50 \u00b5A\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3, Page 137" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "a=0.975;#a=fraction of the emitter current contributed by the carriers injected into the base and reaching the collector\n", + "Ico=10*10**-6; #reverse saturation current in A\n", + "Ib=250*10**-6; #base current in A\n", + "\n", + "#Calculations&Results\n", + "b=a/(1-a); #transistor gain\n", + "print 'gain B=%.f'%b;\n", + "Ic=b*Ib+(b+1)*Ico; #collector current in A\n", + "print 'Ic=%.1f mA'%(Ic/10**-3);#Ic is converted in terms of mA\n", + "Ie=(Ic-Ico)/a; #emitter current in A\n", + "print 'Ie=%.1f mA'%(Ie/10**-3);#Ie is converted in terms of mA" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "gain B=39\n", + "Ic=10.1 mA\n", + "Ie=10.4 mA\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4, Page 137" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "b=125;#b=forward current transfer ratio or dc current gain\n", + "Vbe=0.6; #base to emitter voltage in V\n", + "\n", + "#Calculations&Results\n", + "Ib=(10-Vbe)/(310*10**3); #base current in A\n", + "print 'Ib=%.4f ma'%(Ib*10**3);\n", + "Ic=b*Ib; #collector current in A\n", + "print 'Ic=%.2f mA'%(Ic*10**3)\n", + "Vce=20-(Ic*5000); #collector to emitter voltage\n", + "print 'Vce=%.2f V'%Vce" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ib=0.0303 ma\n", + "Ic=3.79 mA\n", + "Vce=1.05 V\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5, Page 137" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "print \"As the base is forward biased,transistor is not cut off.\"\n", + "print \"Assuming the transistor in active region \"\n", + "VBB=5#VBB=base bias voltage\n", + "VBE=0.7#VBE=voltage between base and emitter terminal\n", + "RB=220#RB=base circuit resistor in kilo ohms\n", + "\n", + "#Calculations&Results\n", + "IB=(VBB-VBE)/RB#IB=base current in mA(By applying Kirchhoff's voltage law)\n", + "print \"IB=%.4f mA\"%IB\n", + "print \"Ico<<IB\"#Ico=reverse saturation current and is given as 22nA\n", + "B=100#B=dc current gain\n", + "IC=B*IB\n", + "print \"IC=%.2f mA\"%IC\n", + "Vcc=12#Vcc=collector supply voltage\n", + "Rc=3.3#Rc=collector circuit resistor in kilo ohms\n", + "VCB=Vcc-(IC*Rc)-VBE#VCB=voltage between collector and base terminal (by applying Kirchhoff's voltage law to the collector circuit)\n", + "print \"\\nVCB=%.2f V\"%VCB\n", + "print \"A positive value of VCB implies that for n-p-n transistor,the collector junction is reverse\",\\\n", + "\"\\nbiased and hence the transistor is actually in active region\"\n", + "IE=-(IB+IC)#IE=emitter current\n", + "print \"\\nIE=%.2f mA\"%IE\n", + "print \"The negative sign indicates that IE actually flows in the opposite direction.\"\n", + "print \"IB and IC do not depend on the collector circuit resistance Rc.So if it is increased, at\",\\\n", + "\"\\none stage VCB becomes negative and transistor goes into saturation region \"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "As the base is forward biased,transistor is not cut off.\n", + "Assuming the transistor in active region \n", + "IB=0.0195 mA\n", + "Ico<<IB\n", + "IC=1.95 mA\n", + "\n", + "VCB=4.85 V\n", + "A positive value of VCB implies that for n-p-n transistor,the collector junction is reverse \n", + "biased and hence the transistor is actually in active region\n", + "\n", + "IE=-1.97 mA\n", + "The negative sign indicates that IE actually flows in the opposite direction.\n", + "IB and IC do not depend on the collector circuit resistance Rc.So if it is increased, at \n", + "one stage VCB becomes negative and transistor goes into saturation region \n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6, Page 138" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import numpy as np\n", + "from numpy.linalg import inv\n", + "\n", + "#Variable declaration\n", + "print \"Applying Kirchhoff voltage law to the base & collector circuit respectively\"\n", + "#(R1*IB)+VBE+(RE*(Ic+IB))=VBB..........(1)\n", + "#(R2*Ic)+VCE+(RE*(Ic+IB))=Vcc..........(2)\n", + "R1=47#R1=value of base circuit resistance in kilo ohms\n", + "RE=2.2#RE=emitter circuit resistance in kilo ohms\n", + "R2=3.3#R2=collector circuit resistance in kilo ohms\n", + "VBE=0.85#VBE=voltage between base and emitter terminals\n", + "VBB=5#VBB=base supply voltage\n", + "Vcc=9#Vcc=collector supply voltage\n", + "VCE=0.22#VCE=voltage between collector and emitter terminals\n", + "\n", + "#Calculations&Results\n", + "R=np.matrix([[R1+RE, RE],[RE, R2+RE]]);\n", + "V=np.matrix([[VBB-VBE],[Vcc-VCE]]);\n", + "I=inv(R)*V\n", + "print \"IB=%.4f mA\"%I[0]\n", + "print \"IC=%.3f mA\"%I[1]\n", + "hFE=110#hFE=dc current gain\n", + "IBmin=I[1]/hFE\n", + "print \"The minimum base current required for saturation is %.4f mA\"%IBmin\n", + "if (I[0]<IBmin):\n", + " print \"\\nAs IB<IBmin transitor is not in the saturation region.It must be in the active region.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Applying Kirchhoff voltage law to the base & collector circuit respectively\n", + "IB=0.0132 mA\n", + "IC=1.591 mA\n", + "The minimum base current required for saturation is 0.0145 mA\n", + "\n", + "As IB<IBmin transitor is not in the saturation region.It must be in the active region.\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7, Page 138" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "IB=(30*10**-3)#IB=base current (in mA) of transistor in CE mode\n", + "IC1=3.5\n", + "IC2=3.7\n", + "VCE1=7.5\n", + "\n", + "#Calculations&Results\n", + "VCE2=12.5#IC1 and IC2 are the change found in collector current IC in mA when collector emitter voltage VCE changes from VCE1 to VCE2(in volts)\n", + "VCE=VCE2-VCE1\n", + "IC=IC2-IC1\n", + "Ro=VCE/IC\n", + "print \"The output resistance is = %.f K ohm\"%Ro\n", + "b=IC2/IB#b=forward current transfer ratio or dc current gain\n", + "print \"b=%.1f\"%b\n", + "a=b/(b+1)#a=fraction of the emitter current contributed by the carriers injected into the base and reaching the collector\n", + "#b=a/(1-a) Hence a=b/(b+1)\n", + "print \"a=%.3f\"%a" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output resistance is = 25 K ohm\n", + "b=123.3\n", + "a=0.992\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8, Page 139" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "b=100#b=forward current transfer ratio or dc current gain\n", + "Vz=4#Vz=Zener diode voltage\n", + "IL=2#IL=load current in mA\n", + "Iz=5#Iz=Zener current in mA\n", + "VCC=12#VCC=collector supply voltage\n", + "VEB1=0.7\n", + "\n", + "#Calculations&Results\n", + "VEB2=VEB1#VEB1,VEB2=emitter-to-base voltage for both transistors Q1 and Q2 respectively\n", + "#Since IL is the collector current of transistor Q1\n", + "IB=IL/b#IB=base current of transistor Q1\n", + "IE=IB+IL#IE=emitter current of transistor Q1\n", + "VR1=VCC-VEB2-Vz#VR1=voltage drop across resistor R1\n", + "R1=VR1/(IB+Iz)\n", + "print \"The resistance R1 is = %.2f Kohm\"%R1\n", + "VR2=VEB2+Vz-VEB1#VR2=voltage drop across resistor R2\n", + "R2=VR2/IE\n", + "print \"The resistance R2 is = %.f K ohm\"%R2\n", + "#VBC=VCC-VR2-VEB1-(IL*RL) where VBC=base-collector voltage drop for transistor Q1\n", + "#VBC=7.3-(2*RL) where RL=load resistance for transistor Q1 in terms of kilo ohm\n", + "print \"\\nFor Q1 to remain in the active region,VBC\u22650,i.e.\"\n", + "print \"RL\u2264(7.3/2) kilo ohm\"\n", + "print \"RL\u22643.65 kilo ohm\"\n", + "print \"So the range of RL for Q1 to remain in the active region is 0\u2264RL\u22643.65 kilo ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resistance R1 is = 1.46 Kohm\n", + "The resistance R2 is = 2 K ohm\n", + "\n", + "For Q1 to remain in the active region,VBC\u22650,i.e.\n", + "RL\u2264(7.3/2) kilo ohm\n", + "RL\u22643.65 kilo ohm\n", + "So the range of RL for Q1 to remain in the active region is 0\u2264RL\u22643.65 kilo ohm\n" + ] + } + ], + "prompt_number": 25 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch8.ipynb b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch8.ipynb new file mode 100644 index 00000000..318b3b79 --- /dev/null +++ b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch8.ipynb @@ -0,0 +1,859 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 8:Junction Transistors: Biasing and Amplification" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 172" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "b=99.;\n", + "Vbe=0.7; #Volatge between base and emitter in V\n", + "Vcc=12; #Volatge source applied at collector in V4\n", + "Rl=2*10**3; #load resistance in ohms\n", + "Rb=100*10**3; #Resistance at base in ohms\n", + "\n", + "#Calculations&Results\n", + "Ib=(12-0.7)/((100*Rl)+Rb); #Base current in micro Ampere\n", + "print 'Ib=%.1f uA'%(Ib*10**6)\n", + "\n", + "Ic=b*Ib;\n", + "print 'Ic=%.2f mA'%(Ic*10**3)\n", + "Vce=4.47; #Voltage between collector and emitter in V\n", + "\n", + "S=(b+1)/(1+b*Rl/(Rl+Rb)); #stabilty factor 1\n", + "print 'S=%.f'%S;\n", + "S1=b/(Rb+Rl*(1+b)); #stabilty factor 2 in A/V\n", + "print 'S1=%.1e A/V'%S1\n", + "S2=(Vcc-Vbe-(Ic*Rl))/(Rb+Rl*(1+b)); #stability factor 3 in A\n", + "print 'S2=%.2e A'%S2\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ib=37.7 uA\n", + "Ic=3.73 mA\n", + "S=34\n", + "S1=3.3e-04 A/V\n", + "S2=1.28e-05 A\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2, Page 172" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "b=49;#b=dc current gain of the common emitter transistor\n", + "Vbe=0.2; #Volatge between base and emitter in V\n", + "Vcc=10.; #Volatge source applied at collector in V4\n", + "Vce=5; #Collector to emitter voltage in V\n", + "Ic=4.9; #collector current in mA\n", + "Rl=1.; #load resistance in kilo ohms\n", + "S=10; #stability factor\n", + "\n", + "#Calculations&Results\n", + "Ib=Ic/b; #base current in mA\n", + "Re=((Vcc-Vce-(Ic*Rl))/(Ic+Ib))*1000; #Resistance at emitter in ohms\n", + "print 'Re=%.f ohms'%Re;\n", + "#S=((1+b)*(1+(RT/Re)))/(1+b+(RT/Re)) \n", + "RT=((S-1)*Re)/(1-(S/(1+b)))#RT=Thevenin resistance =(R1*R2)/(R1+R2)\n", + "VT=(Ib*(10**-3)*RT)+Vbe+((Ib+Ic)*(10**-3)*Re)#VT=Thevenin voltage=(R2*Vcc)/(R1+R2)\n", + "# R2/(R1+R2)=VT/Vcc\n", + "R1=(RT*Vcc)/VT\n", + "print \"R1=%.3f k ohm\"%(R1/10**3)\n", + "R2=((VT/Vcc)*R1)/(1-(VT/Vcc))\n", + "print \"R2=%.1f ohm\"%R2#incorrect answer in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Re=20 ohms\n", + "R1=5.660 k ohm\n", + "R2=185.9 ohm\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3, Page 173" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "hib=30; #h parameter of CB a transistor\n", + "hrb=4*10**-4; #h parameter of CB a transistor\n", + "hfb=-0.99; #h parameter of CB a transistor\n", + "hob=0.9*10**-6; #h parameter of CB a transistor in S\n", + "Rl=6*10**3; #Load resistance in ohms\n", + "\n", + "#Calculations&Results\n", + "AI=-hfb/(1+(hob*Rl)); #Current gain\n", + "print 'AI=%.3f'%AI;\n", + "\n", + "Ri=hib-((hfb*hrb*Rl)/(1+(hob*Rl))); #Input resistance in ohms\n", + "print 'Ri=%.1f ohms'%Ri;\n", + "\n", + "Ro=hib/((hib*hob)-(hfb*hrb)); #Output Resistance in kohms\n", + "print 'Ro=%.2f k ohms'%(Ro*10**-3);\n", + "\n", + "AV=AI*Rl/Ri; #Voltage gain\n", + "print 'AV=%.1f'%AV;\n", + "\n", + "AP=AI*AV; #Power gain\n", + "print 'AP=%.1f'%AP" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "AI=0.985\n", + "Ri=32.4 ohms\n", + "Ro=70.92 k ohms\n", + "AV=182.6\n", + "AP=179.8\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4, Page 173" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Rg=1.*10**3; #internal resistance in ohms\n", + "Rl=20*10**3; #Load resistance in ohms\n", + "hie=1*10**3; #h parameter of the transistor in terms of ohms\n", + "hre=2.5*10**-4; #h parameter of the transistor\n", + "hfe=150.; #h parameter of the transistor\n", + "\n", + "#Calculations&Results\n", + "hoe=1./(40*10**3); #h parameter of the transistor in terms of mho\n", + "AI=(-hfe)/(1+(hoe*Rl)); #Current gain\n", + "print 'AI=%.f'%AI;\n", + "Ri=hie+(AI*hre*Rl); #input resistance in ohms\n", + "print 'Ri=%.f ohms'%Ri;\n", + "Ro=(Rg+hie)/((Rg*hoe)+(hie*hoe)-(hfe*hre)); #output resistance in ohms\n", + "print 'Ro=%.f k ohms'%(Ro*10**-3);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "AI=-100\n", + "Ri=500 ohms\n", + "Ro=160 k ohms\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5, Page 173" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Rl=5.*10**3; #Load resistance in ohms\n", + "hie=1*10**3; #h parameter of the transistor in terms of ohms\n", + "hre=5*10**-4; #h parameter of the transistor\n", + "hfe=100; #h parameter of the transistor\n", + "hoe=25*10**-6; #h parameter of the transistor in terms of mho\n", + "Rg=1*10**3; #source reistance in ohms\n", + "\n", + "#Calculations&Results\n", + "AI=(-hfe)/(1+(hoe*Rl)); #Current gain\n", + "print 'AI=%.2f'%AI;\n", + "\n", + "Ri=hie+(AI*hre*Rl); #input resistance in ohms\n", + "print 'Ri=%.1f ohms'%Ri;\n", + "\n", + "AVo=AI*Rl/(Rg+Ri); #Overall voltage gain including source resistance \n", + "print 'AVo=%.f'%AVo;\n", + "\n", + "APo=AVo*AI; #Overall voltage gain including source resistance \n", + "print 'APo=%.2e'%APo;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "AI=-88.89\n", + "Ri=777.8 ohms\n", + "AVo=-250\n", + "APo=2.22e+04\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6, Page 174" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "hoe=25*10**-6; #h parameter in A/V\n", + "hie=4000; #h paramater in ohms\n", + "hfe=135; #h paramater of transistor\n", + "hre=7*10**-4; #h paramater of transistor\n", + "Re=100; #emitter resistance in ohms\n", + "Rl=3*10**3; #Load resistance in ohms\n", + "\n", + "#Calculations&Results\n", + "#Here hoe*Rl is less than 0.1. So we can simplify the circuit and according to it the current gain is AI=Ic/Ib. here Ic=-hfe*Ib.\n", + "\n", + "AI=-hfe; #current gain\n", + "print 'AI=%.f'%AI;\n", + "\n", + "Ri=hie+(1+hfe)*Re; #input resistance in ohms\n", + "print 'Ri=%.1f k ohms'%(Ri*10**-3);\n", + "\n", + "AV=AI*Rl/Ri; #voltage gain\n", + "print 'AV=%.f'%AV;\n", + "\n", + "print \"The output resistance of the transistor excluding RL is infinite.\"\n", + "print \"The output resistance of the transistor including RL is = %.f k ohms\"%(Rl/10**3)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "AI=-135\n", + "Ri=17.6 k ohms\n", + "AV=-24\n", + "The output resistance of the transistor excluding RL is infinite.\n", + "The output resistance of the transistor including RL is = 3 k ohms\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7, Page 175" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "hfe=100; #h parameter of transistor\n", + "hie=560; #h parameter of transistor in ohms\n", + "Rc=2*10**3; #collector resistance in ohms\n", + "Re=10**3; #emitter resistance in ohms\n", + "Rb=600*10**3; #Base resistance in ohms\n", + "\n", + "#Calculations&Results\n", + "#Since hoe is neglected we can use the simplified equivalent circuit hence the Ri is\n", + "\n", + "Ri=hie+(1+hfe)*Re; #Input resistance in ohms\n", + "print 'Ri=%.2f K ohms'%(Ri*10**-3);\n", + "\n", + "Rib=(Ri*Rb)/(Ri+Rb); #Input resistance including Rb in ohms\n", + "print 'Input resistance (including Rb)=%.2f k ohms'%(Rib*10**-3);\n", + "\n", + "print \"The output resistance excluding load is infinita\"\n", + "Ro=Rc;\n", + "print \"Output resistance including load = %.f k ohms\"%(Ro*10**-3)\n", + "\n", + "AV=-(hfe*Ro)/(hie+((1+hfe)*Re)); #voltage gain\n", + "print 'AV=%.2f'%AV;\n", + "print \"\\nSmall signals are used,since otherwise the output waveform will be distorted.Also,the equivalent circuit will not hold.\"\n", + "\n", + "#Taking DC emitter current and collector current nearly equal\n", + "\n", + "Ib=20./(Rb+Re*101); #base current in mA\n", + "print 'Ib=%.4f mA'%(Ib*10**3);\n", + "\n", + "print \"\\nThe Q-point is defined by\"\n", + "Ic=hfe*Ib; #collector current in mA\n", + "print 'Ic=%.2f mA'%(Ic*10**3);\n", + "\n", + "VCE=20-(3*Ic*10**3)\n", + "print 'VCE=%.2f V'%VCE;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ri=101.56 K ohms\n", + "Input resistance (including Rb)=86.86 k ohms\n", + "The output resistance excluding load is infinita\n", + "Output resistance including load = 2 k ohms\n", + "AV=-2.00\n", + "\n", + "Small signals are used,since otherwise the output waveform will be distorted.Also,the equivalent circuit will not hold.\n", + "Ib=0.0285 mA\n", + "\n", + "The Q-point is defined by\n", + "Ic=2.85 mA\n", + "VCE=11.44 V\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8, Page 176" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "#For a CE transistor amplifier circuit with self-bias\n", + "f=1000#f=frequency in Hz\n", + "AV=-200#AV=voltage gain\n", + "hfe=100#hfe=current gain\n", + "hie=1#hie=input impedance in kilo ohms\n", + "Pcmax=75*10**-3#Pcmax=maximum collector dissipation in Watt\n", + "#hre and hoe are to be neglected\n", + "VCC=12#VCC=collector supply voltage\n", + "\n", + "#Calculations&Results\n", + "#AV=-(hfe*RL)/hie where RL is the load resistance\n", + "RL=-(AV*hie)/hfe\n", + "print \"The designed values of the components of a CE transistor amplifier are:\"\n", + "print \"The load resistance RL is = %.f K ohms\"%RL\n", + "#For the amplifier to be linear,the quiescent point is chosen to lie in the middle of the DC load line\n", + "VCG=VCC/2 #VCG=DC collector to ground voltage\n", + "#VCC=(IC*RL)+VCG where IC=DC collector current\n", + "IC=(VCC-VCG)/RL\n", + "print \"Ihe DC collector current is = %.f mA\"%IC\n", + "Pr=(IC**2)*RL#Pr=power dissipation in RL\n", + "#Pc=the collector dissipation is set at 14.5 mW which is below the value of Pcmax\n", + "#Pc=VCE*IC\n", + "Pc=14.5\n", + "VCE=Pc/IC#VCE=collector-to-emitter voltage drop\n", + "VEG=VCG-VCE#VEG=DC voltage drop across resistance Re\n", + "IE=IC#IE=emitter current\n", + "Re=VEG/(IC)\n", + "print \"The resistance Re is = %.f ohm\"%(Re*1000)#Re is converted in terms of ohms\n", + "Pe=(IC**2)*Re#Pe=power dissipation in Re\n", + "VBE=0.7#VBE=assumed DC base-to-emitter voltage drop\n", + "VBG=VBE+(IE*Re)#VBG=DC voltage across resistance R2\n", + "#VT=(VCC*R2)/(R1+R2) where VT=Thevenin equivalent voltage\n", + "#RT=(R1*R2)/(R1+R2).............(1) where RT=Thevenin equivalent resistance\n", + "#VBG=VT-(IB*RT)\n", + "#VBG=((VCC*R2)/(R1+R2))-(IB*((R1*R2)/(R1+R2)))..................(2)\n", + "#Let (R2/(R1+R2))=x ..............(3)\n", + "x=VBG/VCC#neglecting the second term on the right hand side of equation (2)\n", + "a=(1-x)/x #a=R1/R2\n", + "#S=((1+b)*(1+RT/Re))/(1+b+(RT/Re)) where S=stability factor and b=current gain=hfe\n", + "#b>>1 hence S=(hfe*(1+RT/Re))/(1+b+(RT/Re))\n", + "#For good stability we choose S=hfe/20\n", + "RT=((hfe-20.)/19)*Re\n", + "R1=RT/x#from equation (1) and (3)\n", + "print \"The resistance R1 is=%.1f k ohm\"%R1\n", + "R2=R1/5.33\n", + "print \"The resistance R2 is = %.f k ohm\"%R2\n", + "Pr2=(VBG**2)/R2#Pr2=power dissipation in R2\n", + "Pr1=((VCC-VBG)**2)/R1 #Pr1=power dissipation in R1\n", + "Ce=1./(2*math.pi*f*((Re*1000)/10))#Ce=bypass capacitor\n", + "print \"The bypass capacitance Ce is %.f uF\"%(Ce/10**-6)#Ce is converted in terms of micro farad\n", + "C1=2/(2*math.pi*f*100)#C1=coupling capacitor\n", + "print \"The coupling capacitance C1 is %.1f uF\"%(C1/10**-6)#C1 is converted in terms of micro farad\n", + "Rin=20*1000#Rin=assumed input impedance in ohms\n", + "C2=1/(2*math.pi*f*0.1*Rin)#C2=coupling capacitor\n", + "print \"The coupling capacitance C2 is %.1f uF\"%(C2/10**-6)#C2 is converted in terms of micro farad\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The designed values of the components of a CE transistor amplifier are:\n", + "The load resistance RL is = 2 K ohms\n", + "Ihe DC collector current is = 3 mA\n", + "The resistance Re is = 389 ohm\n", + "The resistance R1 is=10.5 k ohm\n", + "The resistance R2 is = 2 k ohm\n", + "The bypass capacitance Ce is 4 uF\n", + "The coupling capacitance C1 is 3.2 uF\n", + "The coupling capacitance C2 is 0.1 uF\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9, Page 177" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "VCC=12#VCC=collector supply voltage\n", + "a=0.98#a=dc current gain of the common base transistor\n", + "VBE=0.7#VBE=base emitter voltage\n", + "IE=2#IE=emitter current in mA\n", + "\n", + "#Calculations&Results\n", + "#Ico is to be neglected\n", + "b=a/(1-a)#b=dc current gain of the common emitter transistor\n", + "#IC=b*IB where IC=collector current and IB=base current\n", + "#IE=IC+IB\n", + "#IE=(b+1)*IB\n", + "IB=IE/(b+1)\n", + "IC=b*IB\n", + "RE=0.1#RE=resistance in kilo ohms connected to the emitter terminal\n", + "R2=20#R2=resistance in kilo ohms\n", + "RC=3.3#RC=resistance in kilo ohms connected to the collector terminal\n", + "#Let I be the current in the resistance R2\n", + "#Applying Kirchhoff's voltage law in the base-emitter circuit\n", + "#VBE+(RE*IE)=R2*I\n", + "I=(1./R2)*(VBE+(RE*IE))\n", + "#Applying Kirchhoff's voltage law\n", + "#((I+IB+IC)*RC)+((I+IB)*R1)+(I*R2)=VCC\n", + "R1=(VCC-((I+IB+IC)*RC)-(I*R2))/(I+IB)\n", + "\n", + "\n", + "print \"The resistance R1 is = %.1f k ohm\"%R1" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resistance R1 is = 51.2 k ohm\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10, Page 178" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VBE=0.7#VBE=base emitter voltage\n", + "b=90#b=dc current gain of the common emitter transistor\n", + "VCC=10#VCC=collector supply voltage\n", + "RE=1.2#RE=resistance in kilo ohms connected to the emitter terminal\n", + "RC=4.7#RC=resistance in kilo ohms connected to the collector terminal\n", + "RB=250.#RB=resistance in kilo ohms connected to the base terminal\n", + "\n", + "#Calculations&Results\n", + "#Applying Kirchhoff's voltage law\n", + "#VCE=(RB*IB)+VBE where VCE=collector emitter voltage\n", + "#Also VCC=((IB+IC)*RC)+VCE+(IE*RE)\n", + "#IC=b*IB where IC=collector current and IB=base current\n", + "#IE=IC+IB where IE=emitter current\n", + "#IE=(b+1)*IB\n", + "IB=(VCC-VBE)/(((b+1)*(RC+RE))+RB)\n", + "IE=(b+1)*IB\n", + "VCE=(RB*IB)+VBE\n", + "IC=b*IB\n", + "print \"The quiescent value of IE is = %.3f mA\"%IE\n", + "print \"The quiescent value of VCE is = %.2f V\"%VCE\n", + "print \"When dc current gain=90,IC= %.3f mA\"%IC\n", + "#b is increased by 50%\n", + "b1=((50*b)/100)+b\n", + "IB1=(VCC-VBE)/(((b1+1)*(RC+RE))+RB)\n", + "IC1=b1*IB1\n", + "print \"When dc current gain is increased by 50%%,IC=%.3f mA\"%IC1\n", + "x=((IC1-IC)/IC)*100#x=increase in the collector current\n", + "print \"The increase in the collector current IC is = %.1f %%\"%x\n", + "print \"\\nThe percentage increase of IC being less than that of the dc current gain,the circuit provides some stabilization against the changes in the dc current gain.\"\n", + "print \"VCE does not depend on dc current gain and hence it is not affected when the dc current gain changes.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The quiescent value of IE is = 1.075 mA\n", + "The quiescent value of VCE is = 3.65 V\n", + "When dc current gain=90,IC= 1.064 mA\n", + "When dc current gain is increased by 50%,IC=1.193 mA\n", + "The increase in the collector current IC is = 12.2 %\n", + "\n", + "The percentage increase of IC being less than that of the dc current gain,the circuit provides some stabilization against the changes in the dc current gain.\n", + "VCE does not depend on dc current gain and hence it is not affected when the dc current gain changes.\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11, Page 178" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VBE=0.7#VBE=base emitter voltage\n", + "b=99#b=dc current gain of the common emitter transistor\n", + "VCC=15#VCC=collector supply voltage\n", + "RE=7#RE=resistance in kilo ohms connected to the emitter terminal\n", + "RC=4#RC=resistance in kilo ohms connected to the collector terminal\n", + "RB=5#RB=resistance in kilo ohms connected to the base terminal\n", + "VEE=(-15)#VEE=emitter supply voltage\n", + "\n", + "#Calculations&Results\n", + "#Applying Kirchhoff's voltage law in the base emitter loop\n", + "#-VEE=(RB*IB)+VBE +(IE*RE)\n", + "#IC=b*IB where IC=collector current and IB=base current\n", + "#IE=IC+IB where IE=emitter current\n", + "#IE=(b+1)*IB\n", + "IB=(-VEE-VBE)/(RB+((b+1)*RE))\n", + "print \"The quiescent value of IB is = %.4f mA\"%IB\n", + "IC=b*IB\n", + "print \"The quiescent value of IC is = %.2f mA\"%IC\n", + "IE=(b+1)*IB\n", + "print \"The quiescent value of IE is = %.2f mA\"%IE\n", + "#Applying Kirchhoff's voltage law in the output circuit\n", + "#(IC*RC)+VCE+(IE*RE)=VCC-VEE\n", + "VCE=(VCC-VEE)-(IE*RE)-(IC*RC)\n", + "print \"The quiescent value of VCE is = %.2f V\"%VCE\n", + "#b is increased by 20%\n", + "b1=((20*b)/100)+b\n", + "IB1=(-VEE-VBE)/(RB+((b1+1)*RE))\n", + "IC1=b1*IB1\n", + "print \"When dc current gain is increased by 20%%,IC=%.2f mA\"%IC1\n", + "x=((IC1-IC)/IC)*100#x=increase in the collector current\n", + "print \"The increase in the collector current IC is = %.1f %%\"%x\n", + "print \"\\nSince a 20% increase in current gain produces a mere 0.284% enhancement of IC,the circuit provides a good\",\\\n", + "\"stabilization against the changes in the current gain\"\n", + "#In textbook the increase in the collector current is given as 0.5% which is actually coming as 0.284% approximately" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The quiescent value of IB is = 0.0203 mA\n", + "The quiescent value of IC is = 2.01 mA\n", + "The quiescent value of IE is = 2.03 mA\n", + "The quiescent value of VCE is = 7.77 V\n", + "When dc current gain is increased by 20%,IC=2.01 mA\n", + "The increase in the collector current IC is = 0.3 %\n", + "\n", + "Since a 20% increase in current gain produces a mere 0.284% enhancement of IC,the circuit provides a good stabilization against the changes in the current gain\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12, Page 179" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "#For a self-bias circuit\n", + "VBE=0.7#VBE=base emitter voltage\n", + "b=100#b=dc current gain of the common emitter transistor\n", + "VCC=22#VCC=collector supply voltage\n", + "R1=82.#R1=resistance in kilo ohms\n", + "R2=16#R2=resistance in kilo ohms\n", + "RL=2.2#RL=load resistance in kilo ohms\n", + "Re=0.750#Re=resistance in kilo ohms connected to the emitter terminal\n", + "\n", + "#Calculations&Results\n", + "#ICO is to be neglected\n", + "VT=(R2*VCC)/(R1+R2)#VT=Thevenin equivalent voltage\n", + "RT=(R1*R2)/(R1+R2)#RT=Thevenin equivalent resistance\n", + "#Applying Kirchhoff's voltage law to the base circuit\n", + "#(IB*(RT+Re))+(IC*Re)=VT-VBE\n", + "#IC=b*IB\n", + "IB=(VT-VBE)/(RT+Re+(b*Re))#IB=base current\n", + "IC=b*IB#IC=collector current\n", + "#Applying Kirchhoff's voltage law to the collector circuit\n", + "#(IC*(RL+Re))+(IB*Re)+VCE=VCC\n", + "VCE=VCC-((IC*(RL+Re))+(IB*Re))#VCE=collector emitter voltage\n", + "print \"The operating point is specified by\"\n", + "print \"IC=%.2f mA\"%IC\n", + "print \"VCE=%.1f V\"%VCE" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The operating point is specified by\n", + "IC=3.24 mA\n", + "VCE=12.4 V\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13, Page 180" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "RE=0.680#RE=resistance in kilo ohms connected to the emitter terminal\n", + "RC=2.7#RC=resistance in kilo ohms connected to the collector terminal\n", + "#RB=resistance connected to the base terminal\n", + "VCE=7.3#VCE=collector emitter voltage\n", + "VBE=0.7#VBE=base emitter voltage\n", + "Vre=2.1#Vre=voltage across RE resistance\n", + "IB=0.02#IB=base current in mA\n", + "\n", + "#Calculations&Results\n", + "IE=Vre/RE#IE=emitter current in mA\n", + "IC=IE-IB#IC=collector current in mA\n", + "b=IC/IB#b=current gain\n", + "print \"The current gain \u03b2 is = %.1f\"%b\n", + "VCC=(IC*RC)+VCE+Vre#VCC=collector supply voltage\n", + "print \"The collector supply voltage VCC is = %.1f V\"%VCC\n", + "#Voltage across RB (Vrb)resistance is given by\n", + "Vrb=VCC-(VBE+Vre)\n", + "RB=Vrb/IB\n", + "print \"The resistance RB is = %.f k ohm\"%RB\n", + "#To draw the DC load line,we neglect the base current in RE resistance\n", + "#Equation for DC load line is:\n", + "#VCE=VCC-(RC+RE)*IC\n", + "print \"\\nFor the DC load line\"\n", + "print \"The intercept of the load line on the VCE-axis(X-axis) is = %.1f V\"%VCC\n", + "print \"The intercept of the load line on the IC axis(Y-axis) is = %.2f mA\"%(VCC/(RC+RE))\n", + "print \"The DC load line is the straight line joining above two intercepts.\"\n", + "print \"The co-ordinates of the operating point Q on the load line are (7.3V,3.07mA)\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current gain \u03b2 is = 153.4\n", + "The collector supply voltage VCC is = 17.7 V\n", + "The resistance RB is = 744 k ohm\n", + "\n", + "For the DC load line\n", + "The intercept of the load line on the VCE-axis(X-axis) is = 17.7 V\n", + "The intercept of the load line on the IC axis(Y-axis) is = 5.23 mA\n", + "The DC load line is the straight line joining above two intercepts.\n", + "The co-ordinates of the operating point Q on the load line are (7.3V,3.07mA)\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14, Page 180" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VBE=0.7#VBE=base emitter voltage\n", + "b=120#b=dc current gain of the common emitter transistor\n", + "VCC=15.#VCC=collector supply voltage\n", + "R1=72.#R1=resistance in kilo ohms\n", + "R2=8#R2=resistance in kilo ohms\n", + "RL=2#RL=load resistance in kilo ohms\n", + "Re=0.700#Re=resistance in kilo ohms connected to the emitter terminal\n", + "RC=2#RC=resistance in kilo ohms connected to the collector terminal\n", + "Rin=1.5#Rin=input resistance in kilo ohms of the amplifier\n", + "vi=1#vi=amplitude of the ac input signal in mV\n", + "\n", + "#Calculations&Results\n", + "VT=(R2*VCC)/(R1+R2)#VT=Thevenin equivalent voltage\n", + "RT=(R1*R2)/(R1+R2)#RT=Thevenin equivalent resistance\n", + "#Applying Kirchhoff's voltage law to the base circuit\n", + "#(IB*(RT+Re))+(IC*Re)=VT-VBE\n", + "#IC=b*IB\n", + "IB=(VT-VBE)/(RT+Re+(b*Re))#IB=base current\n", + "IC=b*IB#IC=collector current\n", + "#Applying Kirchhoff's voltage law to the collector circuit\n", + "#(IC*(RL+Re))+(IB*Re)+VCE=VCC\n", + "VCE=VCC-((IC*(RL+Re)))#VCE=collector emitter voltage(neglecting small term IB*RE)\n", + "#Equation for DC load line is:\n", + "#VCE=VCC-(RL+Re)*IC\n", + "print \"1. For the DC load line\"\n", + "print \"The operating point Q is specified by %.3f mA\"%IC\n", + "print \"VCE=%.1f V\"%VCE\n", + "print \"The intercept of the dc load line on the VCE-axis(X-axis) is = %.f V\"%VCC\n", + "print \"The intercept of the dc load line on the IC axis(Y-axis) is = %.2f mA\"%(VCC/(RC+Re))\n", + "print \"The DC load line is the straight line joining above two intercepts.\"\n", + "Rac=(RL*RC)/(RL+RC)#Rac=ac load resistance\n", + "print \"\\n2. For the AC load line\"\n", + "print \"The intercept of the ac load line on the VCE-axis(X-axis) is = %.1f V\"%(VCE+(IC*Rac))\n", + "print \"The line joining the above intercept and the operating point Q extended to meet the IC axis(Y-axis) gives the AC load line\"\n", + "AV=-(b*Rac)/Rin#AV=voltage gain of the amplifier\n", + "vo=abs(AV)*vi#vo=amplitude of the output voltage signal\n", + "print \"\\n3. The amplitude of the output voltage vo is = %.f mV\"%vo" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1. For the DC load line\n", + "The operating point Q is specified by 1.045 mA\n", + "VCE=12.2 V\n", + "The intercept of the dc load line on the VCE-axis(X-axis) is = 15 V\n", + "The intercept of the dc load line on the IC axis(Y-axis) is = 5.56 mA\n", + "The DC load line is the straight line joining above two intercepts.\n", + "\n", + "2. For the AC load line\n", + "The intercept of the ac load line on the VCE-axis(X-axis) is = 13.2 V\n", + "The line joining the above intercept and the operating point Q extended to meet the IC axis(Y-axis) gives the AC load line\n", + "\n", + "3. The amplitude of the output voltage vo is = 80 mV\n" + ] + } + ], + "prompt_number": 39 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch9.ipynb b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch9.ipynb new file mode 100644 index 00000000..d4ebc813 --- /dev/null +++ b/Electronics_Fundamentals_and_Applications_by_D._Chattopadhyay_and_P._C._Rakshit/ch9.ipynb @@ -0,0 +1,443 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 9: Basic Voltage and Power Amplifiers" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1, Page 214" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "AVm=120#AVm=mid-band gain of an RC-coupled amplifier\n", + "fm=100#fm=frequency in Hz corresponding to the mid-band gain\n", + "AVl=60#AVl=reduced gain\n", + "AVh=AVl\n", + "f=100*10**3#f=frequency in Hz corresponding to the reduced gain\n", + "\n", + "#Calculations&Results\n", + "#|AVl|=(|AVm|)/sqrt(1+(fl/fm)^2) where fl=lower half power frequency\n", + "fl=math.sqrt((abs(AVm)/abs(AVl))**2 -1)*fm\n", + "print \"The lower half-power frequency is = %.1f Hz\"%fl\n", + "#|AVh|=(|AVm|)/sqrt(1+(f/fh)^2) where fh=upper half power frequency\n", + "fh=f/math.sqrt((abs(AVm)/abs(AVh))**2 -1)\n", + "print \"The upper half-power frequency is = %.1f KHz\"%(fh/10**3)#fh is converted in terms of kHz" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The lower half-power frequency is = 173.2 Hz\n", + "The upper half-power frequency is = 57.7 KHz\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2, Page 215" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "#For two identical transistors employed by an RC-coupled amplifier\n", + "hfe=100.#hfe=current gain\n", + "hie=2*10**3#hie=input impedance in ohm\n", + "Cob=2.0*10**-12#Cob=capacitance in farad quoted by the transistor manufacturers\n", + "C=0.4*10**-6#C=coupling capacitance in farad\n", + "RL=8*10**3#RL=load resistance in ohms for each transistor\n", + "CW=10*10**-12#CW=wiring capacitance in farad\n", + "\n", + "#Calculations&Results\n", + "fl=1./(2*math.pi*C*(hie+RL))#fl=lower half power frequency\n", + "print \"The lower half-power frequency is = %.1f Hz\"%fl\n", + "hfb=-hfe/(1+hfe)#hfb=current gain for common base transistor\n", + "Coc=Cob/(1+hfb)#Coc=transistor collector capacitance in farad\n", + "Cs=Coc+CW#Cs=shunt capacitance in farad\n", + "Ro=(hie*RL)/(hie+RL)#Ro=equivalent resistance of the parallel combination of hie and RL\n", + "fh=1/(2*math.pi*Cs*Ro)#fh=upper half power frequency\n", + "print \"The upper half-power frequency is = %.f KHz\"%(fh/10**3)#fh is converted in terms of kHz" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The lower half-power frequency is = 39.8 Hz\n", + "The upper half-power frequency is = 469 KHz\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3, Page 215" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "#AVm=mid-band gain of an RC-coupled amplifier\n", + "fm=60#fm=frequency in Hz corresponding to the mid-band gain\n", + "#AVl=reduced gain\n", + "#AVh=AVl\n", + "f=600*10**3#f=frequency in Hz corresponding to the reduced gain\n", + "fl=30.#fl=The lower half-power frequency in Hz\n", + "fh=300*10**3#fh=The upper half-power frequency in Hz\n", + "\n", + "#Calculations&Results\n", + "#|AVl|=(|AVm|)/sqrt(1+(fl/fm)**2)\n", + "#Suppose (AVl/AVm)=a=low frequency gain with respect to the mid frequency gain\n", + "#a=1/math.sqrt(1+(fl/fm)**2)#a=magnitude of the low frequency gain\n", + "a=1./math.sqrt(1+(fl/fm)**2)\n", + "o=math.degrees(math.atan(fl/fm))#o=phase angle in degree of the low frequency gain\n", + "print \"For the low frequency gain with respect to the mid frequency gain\"\n", + "print \"Magnitude=%.3f\"%a\n", + "print \"Phase angle=%.1f degree\"%o\n", + "#|AVh|=(|AVm|)/sqrt(1+(f/fh)**2)\n", + "#Suppose (AVh/AVm)=b=high frequency gain with respect to the mid frequency gain\n", + "#b=1/sqrt(1+(f/fh)**2)#b=magnitude of the high frequency gain\n", + "b=1./math.sqrt(1+(f/fh)**2)\n", + "O=-math.degrees(math.atan(f/fh))#O=phase angle in degree of the high frequency gain\n", + "print \"\\nFor the high frequency gain with respect to the mid frequency gain \"\n", + "print \"Magnitude=%.3f\"%b\n", + "print \"Phase angle=%.1f degree\"%O\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For the low frequency gain with respect to the mid frequency gain\n", + "Magnitude=0.894\n", + "Phase angle=26.6 degree\n", + "\n", + "For the high frequency gain with respect to the mid frequency gain \n", + "Magnitude=0.447\n", + "Phase angle=-63.4 degree\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4, Page 216" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "#In a CE class A power amplifier\n", + "RL=12#RL=load resistance in ohms\n", + "n=8#n=primary-to-secondary turns ratio of a transformer\n", + "\n", + "#Calculations\n", + "#Peak-to-peak swing of the signal current is 250mA\n", + "Im=(250*10**-3)/2#Im=ac collector current in Ampere\n", + "RL1=(n**2)*RL#RL1=RL'=resistance reflected to the primary for the resistance RL in presence of an ac signal\n", + "#Pac=(1./2)*Vm*Im where Pac=ac output power\n", + "#Pac=(1./2)*(Im**2)*RL1\n", + "Pac=(1./2)*(Im**2)*RL1\n", + "\n", + "#Result\n", + "print \"The output power is = %.f W\"%Pac" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output power is = 6 W\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5, Page 216" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCQ=6#VCQ=quiescent collector voltage\n", + "ICQ=50*10**-3#ICQ=quiescent collector current\n", + "VCmin=1\n", + "VCmax=11.#VCmin,VCmax=output signal voltage variation\n", + "ICmin=10*10**-3\n", + "ICmax=90*10**-3#ICmin,ICmax=output signal current variation in Ampere\n", + "\n", + "#Calculations&Results\n", + "Ps=VCQ*ICQ#Ps=dc input power to the transistor\n", + "print \"The dc input power is = %.1f W\"%Ps\n", + "Pac=(1./8)*(ICmax-ICmin)*(VCmax-VCmin)#Pac=ac output power delivered to the load\n", + "print \"The ac output power is = %.1f W\"%Pac\n", + "PT=(VCQ*ICQ)-Pac#PT=the collector dissipation\n", + "print \"The collector dissipation is = %.1f W\"%PT\n", + "n=(Pac/Ps)*100#n=the efficiency of the active device\n", + "print \"The efficiency is = %.1f %%\"%n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The dc input power is = 0.3 W\n", + "The ac output power is = 0.1 W\n", + "The collector dissipation is = 0.2 W\n", + "The efficiency is = 33.3 %\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6, Page 216" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "#In a class B push pull circuit\n", + "#Transformer winding resistances are to be ignored\n", + "n=3.#n=primary-to-secondary turns ratio of a transformer\n", + "RL=9#RL=load resistance in ohms\n", + "VCC=15#VCC=collector supply voltage\n", + "\n", + "#Calculations&Results\n", + "RL1=((n/2)**2)*RL#RL1=reflected load resistance for one transistor\n", + "Pactot=(VCC**2)/(2*RL1)#Pactot=maximum output power\n", + "print \"The maximum output power is = %.2f W\"%Pactot\n", + "Pstot=(2*VCC**2)/(math.pi*RL1)#Pstot=the maximum dc power supplied to the two transistors\n", + "print \"The maximum dc power supplied is = %.2f W\"%Pstot\n", + "n=(Pactot/Pstot)*100#n=efficiency\n", + "print \"The efficiency is = %.1f %%\"%n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum output power is = 5.56 W\n", + "The maximum dc power supplied is = 7.07 W\n", + "The efficiency is = 78.5 %\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7, Page 217" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "#In a single tuned amplifier\n", + "L=120*10**-6#L=inductance in henry\n", + "C=100*10**-12#C=capacitance in farad\n", + "R=10#R=resistance in ohms\n", + "hoe=50*10**-6#hoe=output impedance in mho(or S)\n", + "hfe=100#hfe=current gain\n", + "hie=2.5*10**3#hie=input impedance in ohm\n", + "RT=10*10**3#RT=equivalent resistance of RB and Ri in parallel\n", + "\n", + "#Calculations&Results\n", + "fo=1/(2*math.pi*math.sqrt(L*C))#fo=resonant frequency\n", + "print \"The resonant frequency is = %.2f MHz\"%(fo/10**6)#fo is converted in terms of MHz\n", + "Qo=(1./R)*math.sqrt(L/C)#Qo=Q-factor of the resonant frequency\n", + "Ro=(Qo**2)*R#Ro=maximum impedance Zm\n", + "Rp=1/(hoe+(1./Ro)+(1./RT))#Rp=equivalent resistance of the parallel combination of Ro,ro and RT\n", + "Qe=(Qo*Rp)/Ro#Qe=effective Q-factor\n", + "B=fo/Qe#B=bandwidth\n", + "print \"The bandwidth is =%.1f KHz\"%(B/10**3)#B is converted in terms of kHz\n", + "AVm=-(hfe*Rp)/hie#AVm=maximum voltage gain\n", + "print \"The maximum voltage gain is = %.1f\"%AVm" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resonant frequency is = 1.45 MHz\n", + "The bandwidth is =252.0 KHz\n", + "The maximum voltage gain is = -252.6\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8, Page 218" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "V=10#V=voltage at frequency 5kHz\n", + "Vr=7.07#Vr=voltage at frequency 25kHz\n", + "\n", + "#Calculations\n", + "#x=10*log10(P/Pr) where x=change in decibel(dB) of power P from some standard power Pr\n", + "#P=V^2/R=I^2*R \n", + "#Also Pr=Vr^2/R=Ir^2*R\n", + "#x=10*(log10(V/Vr))^2=20*log10(V/Vr)\n", + "x=20*math.log10(V/Vr)#x=change in decibel(dB) of voltage V from some standard voltage Vr\n", + "\n", + "#Result\n", + "print \"The decibel change in the output power level is = %.f dB\"%x" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The decibel change in the output power level is = 3 dB\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9, Page 218" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "Vorms=2#Vorms=rms output voltage in the midband region of an amplifier\n", + "Pa=42.#Pa=power gain in dB\n", + "Pol=0.4#Pol=power output in W at the lower cut-off frequency 100Hz\n", + "Ri=10**3#Ri=input resistance in ohms\n", + "\n", + "#Calculations&Results\n", + "VOrms=2./math.sqrt(2)#VOrms=rms output voltage at 100Hz\n", + "print \"1. The rms output voltage at 100Hz,which is the lower cutoff frequency,is = %.3f V\"%VOrms\n", + "Po=2*Pol#Po=output power in the midband region\n", + "print \"2. The output power in the midband region is = %.1f W\"%Po\n", + "#Let Pi=input power\n", + "#10*log10(Po/Pi)=Pa\n", + "Pi=Po/(10**(Pa/10))\n", + "#Pi=(Vi**2)/Ri where Vi=rms input voltage\n", + "Vi=math.sqrt(Pi*Ri)\n", + "print \"3. The rms input voltage is = %.3f V\"%Vi" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1. The rms output voltage at 100Hz,which is the lower cutoff frequency,is = 1.414 V\n", + "2. The output power in the midband region is = 0.8 W\n", + "3. The rms input voltage is = 0.225 V\n" + ] + } + ], + "prompt_number": 23 + } + ], + "metadata": {} + } + ] +}
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