Added(A)/Deleted(D) following books

 A  Electronics_Devices_and_Circuits_by_G._S._N._Raju/README.txt
A  Electronics_Engineering_by_P._Raja/chapter_1_1.ipynb
A  Electronics_Engineering_by_P._Raja/chapter_2_1.ipynb
A  Electronics_Engineering_by_P._Raja/chapter_3_1.ipynb
A  Electronics_Engineering_by_P._Raja/chapter_4_1.ipynb
A  Electronics_Engineering_by_P._Raja/chapter_5_1.ipynb
A  Electronics_Engineering_by_P._Raja/chapter_6_1.ipynb
A  Electronics_Engineering_by_P._Raja/chapter_7_1.ipynb
A  Electronics_Engineering_by_P._Raja/chapter_8_1.ipynb
A  Electronics_Engineering_by_P._Raja/chapter_9_1.ipynb
A  Electronics_Engineering_by_P._Raja/screenshots/7_1.png
A  Electronics_Engineering_by_P._Raja/screenshots/snap-3_1.png
A  Electronics_Engineering_by_P._Raja/screenshots/snap-6_1.png
A  Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_1_1.ipynb
A  Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_2_1.ipynb
A  Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_3_1.ipynb
A  Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_4_1.ipynb
A  Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_5_1.ipynb
A  Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_6_1.ipynb
A  Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_7_1.ipynb
A  Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_8_1.ipynb
A  Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/1_1.png
A  Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/6_1.png
A  Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/7_1.png
A  Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/README.txt
A  Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/README.txt
A  Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/README.txt
A  Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap_/README.txt
A  Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_10_SILICON_CONTROLLED_RECTIFIER_1.ipynb
A  Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_1_CRYSTAL_STRUCTURES_1.ipynb
A  Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_6_ELECTRICAL_BREAKDOWN_IN_PN_JUNCTIONS_1.ipynb
A  Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_2_ENERGY_BAND_THEORY_OF_SOLIDS_1.ipynb
A  Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_3_CARRIER_TRANSPORT_IN_SEMICONDUCTOR_1.ipynb
A  Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_4__EXCESS_CARRIER_IN_SEMICONDUCTOR_1.ipynb
A  Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_5_PN_JUNCTION_DIODE_1.ipynb
A  Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_7_BIPOLAR_JUNCTION_TRANSISTORB_1.ipynb
A  Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_8_THE_FIELD_EFFECT_TRANSISTOR_1.ipynb
A  Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2015-11-05_at_11.31.11_pm_1.png
A  Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2015-11-05_at_11.32.51_pm_1.png
A  Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2015-11-05_at_11.33.45_pm_1.png

11 files changed, 4970 insertions, 0 deletions

diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_1_1.ipynb b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_1_1.ipynb
new file mode 100644
index 00000000..4377829c
--- /dev/null
+++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_1_1.ipynb
@@ -0,0 +1,707 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter : 1 - Semiconductor Materials And Crystal Properties"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 1.18.3 - Page No : 1-43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# Given data\n",
+      "At = 28;# Atomic weight\n",
+      "n = 4;\n",
+      "N_A = 6.023*10**23;\n",
+      "a = 5.3;# in \u00c5\n",
+      "a = a * 10**-10;# in m\n",
+      "m = At/N_A;# in gm\n",
+      "m = m*10**-3;# in kg\n",
+      "V = (a)**3;# in m**3\n",
+      "Rho = (m*n)/V;# in gm/m**3\n",
+      "Rho = Rho * 10**-3;# in kg/m**3\n",
+      "print \"The volume density = %0.3f kg/m**3 \" %Rho"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The volume density = 1.249 kg/m**3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 1.18.4 - Page No : 1-43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "# Given data\n",
+      "r = 1.278;# in \u00c5\n",
+      "a = (4*r)/(sqrt(2));# in \u00c5\n",
+      "a = a * 10**-10;# in m\n",
+      "V = (a)**3;# in m**3\n",
+      "At = 63.5;# atomic weight\n",
+      "N_A = 6.023*10**23;\n",
+      "m = At/N_A;# in gm\n",
+      "m = m * 10**-3;# in kg\n",
+      "n = 4;\n",
+      "Rho = (m*n)/V;# in gm/m**3\n",
+      "Rho = Rho * 10**-3;# in kg/m**3\n",
+      "print \"The density of copper crystal = %0.3f kg/m**3\" %Rho"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The density of copper crystal = 8.929 kg/m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 1.21.1 - Page No : 1-50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sin, pi\n",
+      "# Given data\n",
+      "d = 2.82;# in \u00c5\n",
+      "d = d * 10**-10;# in m\n",
+      "n = 1;\n",
+      "theta = 10;# in degree \n",
+      "#Formula 2*d*sin(theta) = n*lembda;\n",
+      "lembda = 2*d*sin(theta*pi/180);# in m\n",
+      "lembda = lembda * 10**10;# in \u00c5\n",
+      "print \"The wavelength of X-ray = %0.3f \u00c5\" %lembda"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength of X-ray = 0.979 \u00c5\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 1.21.2 - Page No : 1-51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "lembda = 1.6;# in \u00c5\n",
+      "lembda = lembda * 10**-10;# in m\n",
+      "theta = 14.2;# in degree\n",
+      "n = 1;\n",
+      "#Formula 2*d*sin(theta) = n*lembda;\n",
+      "d = (n*lembda)/(2*sin(theta*pi/180));# in m\n",
+      "d = d * 10**10;# in \u00c5\n",
+      "print \"The spacing of atomic layer in the crystal = %0.2f \u00c5\" %d\n",
+      "# Note : In the book the unit of the answer is wrong."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The spacing of atomic layer in the crystal = 3.26 \u00c5\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 1.21.3 - Page No : 1-51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "n = 1;\n",
+      "theta = 30;# in degree \n",
+      "lembda = 1.78;# in \u00c5\n",
+      "lembda = lembda * 10**-10;# in m\n",
+      "#Formula 2*d*sin(theta) = n*lembda;\n",
+      "d = (n*lembda)/(2*sin(theta*pi/180));# in m\n",
+      "d = d * 10**10;# in \u00c5\n",
+      "print \"The interplaner spacing = %0.2f \u00c5\" %d"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The interplaner spacing = 1.78 \u00c5\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 1.21.4 - Page No : 1-51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "lembda = 0.58;# in \u00c5\n",
+      "n = 1;\n",
+      "theta1 = 6.45;# in degree\n",
+      "d = (n*lembda)/(2*sin(theta1*pi/180));# in \u00c5 \n",
+      "print \"Part (i) : At angle of 6.45\u00b0, Interplaner spacing of the crystal = %0.3f \u00c5\" %d\n",
+      "theta2 = 9.15;# in degree\n",
+      "d1 = (n*lembda)/(2*sin(theta2*pi/180));# in \u00c5 \n",
+      "print \"Part(ii) : At angle of 9.15\u00b0, Interplaner spacing of the crystal = %0.3f \u00c5\" %d1\n",
+      "theta3 = 13;# in degree\n",
+      "n2 = 1;\n",
+      "d2 = (n2*lembda)/(2*sin(theta3*pi/180));# in \u00c5 \n",
+      "print \"Part(iii) : At angle of 13\u00b0, Interplaner spacing of the crystal = %0.3f \u00c5\" %d2\n",
+      "# For \n",
+      "n=2;\n",
+      "d2 = (n*lembda)/(2*sin(theta3*pi/180));# in \u00c5 \n",
+      "print \"Part (iv) : The interplaner spacing = %0.3f \u00c5\" %d2\n",
+      "print \"The interplaner spacing for some other set of reflecting = %0.3f \u00c5\" %d1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part (i) : At angle of 6.45\u00b0, Interplaner spacing of the crystal = 2.582 \u00c5\n",
+        "Part(ii) : At angle of 9.15\u00b0, Interplaner spacing of the crystal = 1.824 \u00c5\n",
+        "Part(iii) : At angle of 13\u00b0, Interplaner spacing of the crystal = 1.289 \u00c5\n",
+        "Part (iv) : The interplaner spacing = 2.578 \u00c5\n",
+        "The interplaner spacing for some other set of reflecting = 1.824 \u00c5\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 1.21.5 - Page No : 1-52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import asin\n",
+      "# Given data\n",
+      "a = 2.814;# in \u00c5\n",
+      "h = 1;\n",
+      "k = 0;\n",
+      "l = 0;\n",
+      "d = a/(sqrt( ((h)**2) + ((k)**2) + ((l)**2) ));# in \u00c5\n",
+      "n = 2;\n",
+      "lembda = 0.710;# in \u00c5\n",
+      "#Formula 2*d*sin(theta) = n*lembda;\n",
+      "theta = asin(n*lembda/(2*d) )*180/pi;# in degree\n",
+      "print \"The glacing angle = %0.2f degree\" %theta"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The glacing angle = 14.61 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 1.21.6 - Page No : 1-52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "a = 3.65;# in \u00c5\n",
+      "h = 1;\n",
+      "k = 0;\n",
+      "l = 0;\n",
+      "d = a/(sqrt( ((h)**2) + ((k)**2) + ((l)**2) ));# in \u00c5\n",
+      "n = 1;\n",
+      "theta = 60;# in degree\n",
+      "#Formula 2*d*sin(theta) = n*lembda;\n",
+      "lembda = 2*d*sin(theta*pi/180);# in \u00c5\n",
+      "print \"Wavelength of X ray = %0.4f \u00c5\" %lembda"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength of X ray = 6.3220 \u00c5\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 1.21.7 - Page No : 1-53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "lembda = 1.54;# in \u00c5\n",
+      "lembda= lembda*10**-8;# in cm\n",
+      "At = 63.54;# atomic weight\n",
+      "density = 9.024;# in gm/cc\n",
+      "n = 1;\n",
+      "N_A = 6.023*10**23;\n",
+      "m = At/N_A;# mass\n",
+      "a =(density*m)**(1/3);# in cm\n",
+      "h = 1;\n",
+      "k = 0;\n",
+      "l = 0;\n",
+      "d = a/(sqrt( ((h)**2) + ((k)**2) + ((l)**2) ));# in cm\n",
+      "n = 1;\n",
+      "#Formula 2*d*sin(theta) = n*lembda;\n",
+      "theta = asin( (lembda)/(2*d) )*180/pi;# in degree\n",
+      "print \"The glancing angle = %0.4f degree\" %theta\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The glancing angle = 4.4893 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 1.21.8 - Page No : 1-53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "a = 3.615;# in \u00c5\n",
+      "theta = 22;# in degree\n",
+      "n = 1;\n",
+      "h = 1;\n",
+      "k = 1;\n",
+      "l = 1;\n",
+      "d = a/(sqrt( ((h)**2) + ((k)**2) + ((l)**2) ));# in \u00c5\n",
+      "# 2*d*sin(theta) = n*lembda;\n",
+      "lembda = 2*d*sin(theta*pi/180);# in \u00c5\n",
+      "print \"The wavelength = %0.3f \u00c5\" %lembda\n",
+      "n = 2;\n",
+      "#Formula 2*d*sin(theta) = n*lembda;\n",
+      "theta = asin(n*lembda/(2*d) )*180/pi;# in degree\n",
+      "print \"The glacing angle for second order = %0.2f degree\" %theta\n",
+      "print \"To get the 2nd order spectrum the position of the detector should be %0.2f \u00b0\" %(2*theta)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength = 1.564 \u00c5\n",
+        "The glacing angle for second order = 48.52 degree\n",
+        "To get the 2nd order spectrum the position of the detector should be 97.04 \u00b0\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 1.21.9 - Page No : 1-54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "n = 1;\n",
+      "lembda = 1.54;# in \u00c5\n",
+      "theta = 21.7;# in degree\n",
+      "#Formula 2*d*sin(theta) = n*lembda;\n",
+      "d = (lembda*n)/(2*sin(theta*pi/180));# in \u00c5\n",
+      "h = 1;\n",
+      "k = 1;\n",
+      "l = 1;\n",
+      "# Formula d = a/(sqrt( ((h)**2) + ((k)**2) + ((l)**2) ));\n",
+      "a =  d*(sqrt( ((h)**2) + ((k)**2) + ((l)**2) ));# in \u00c5\n",
+      "print \"Lattice constant = %0.3f \u00c5\" %a"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Lattice constant = 3.607 \u00c5\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 1.22.1 - Page No : 1-56"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "a = 4.8;# in \u00c5\n",
+      "h = 2;\n",
+      "k = 1;\n",
+      "l = 1;\n",
+      "d = a/(sqrt( ((h)**2) + ((k)**2) + ((l)**2) ));# in \u00c5\n",
+      "print \"The distance between d_211 plains = %0.2f \u00c5\" %d"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The distance between d_211 plains = 1.96 \u00c5\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 1.22.2 - Page No : 1-56"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "r = 1.28;# in \u00c5\n",
+      "#Formula r = (a*sqrt(2))/4;\n",
+      "a = (4*r)/(sqrt(2));# in \u00c5\n",
+      "a = a * 10**-8;# in cm\n",
+      "n = 4;\n",
+      "M = 63.5;\n",
+      "N_A = 6.023*10**23;\n",
+      "Rho = (n*M)/(N_A*((a)**3));# in gm/cc\n",
+      "print \"The density = %0.2f gm/cc\" %Rho"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The density = 8.89 gm/cc\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 1.22.3 - Page No : 1-57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "a = 2.9;# in \u00c5\n",
+      "a = a * 10**-8;# in cm\n",
+      "Rho = 7.87;# in gm/cc\n",
+      "N_A = 6.023*10**23;\n",
+      "M = 55.85\n",
+      "# Rho = (n*M)/(N_A*((a)**3))\n",
+      "n = (Rho*N_A*((a)**3))/M;# in atoms per unit cell\n",
+      "print \"The number of atoms per unit cell = %0.f \" %n\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The number of atoms per unit cell = 2 \n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 1.22.4 - Page No : 1-57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "M =60;# in gm/mole\n",
+      "Rho = 6.23;# in gm/cc\n",
+      "n = 4;\n",
+      "N_A = 6.023*10**23;\n",
+      "# Rho = (n*M)/(N_A*((a)**3));\n",
+      "a = ( (n*M)/(N_A*Rho) )**(1/3);# in cm\n",
+      "r = (a*sqrt(2))/4;# in cm\n",
+      "r = r * 10**8;# in \u00c5\n",
+      "print \"The radius = %0.3f \u00c5\" %r"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The radius = 1.414 \u00c5\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 1.22.5 - Page No : 1-57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "Rho = 5.96;# in gm/cc\n",
+      "M = 50;\n",
+      "n = 2;\n",
+      "N_A = 6.023*10**23;\n",
+      "#Formula Rho = (n*M)/(N_A*((a)**3));\n",
+      "a = ( (n*M)/(N_A*Rho) )**(1/3);# in cm\n",
+      "r = (a*sqrt(3))/4;# in cm\n",
+      "P_F = (2*(4/3)*pi*((r)**3))/((a)**3);# Packing factor\n",
+      "print \"The Packing factor = %0.2f\" %P_F"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Packing factor = 0.68\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 1.22.6 - Page No : 1-58"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "d = 5.2;# in gm/cc\n",
+      "n = 2;\n",
+      "M = 120;\n",
+      "N_A = 6.023*10**23;\n",
+      "m = M/N_A;#mass of 1 atom in gm\n",
+      "m = n*m;#mass of unit cell in gm\n",
+      "g = 20;# in gm\n",
+      "m = g/m;# in unit cells\n",
+      "print \"The number of unit cell in its 20 gm = %0.3e\" %m"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The number of unit cell in its 20 gm = 5.019e+22\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 1.22.7 - Page No : 1-58"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "Rho = 2.48;# in gm/cc\n",
+      "K = 39;# molecular weight of K\n",
+      "F = 19; # molecular weight of F\n",
+      "M = K+F;# molecular weight of KF\n",
+      "n = 4;\n",
+      "N_A = 6.023*10**23;\n",
+      "#Formula Rho = (n*M)/(N_A*((a)**3));\n",
+      "a = ( (n*M)/(N_A*Rho) )**(1/3);# in cm\n",
+      "a = a * 10**8;# in \u00c5\n",
+      "r = (a*sqrt(2))/4;# in \u00c5\n",
+      "r = 2*r;# in \u00c5\n",
+      "print \"The distance between ions = %0.1f \u00c5\" %r\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The distance between ions = 3.8 \u00c5\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_2_1.ipynb b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_2_1.ipynb
new file mode 100644
index 00000000..64ef8b00
--- /dev/null
+++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_2_1.ipynb
@@ -0,0 +1,1136 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:146b2a737d67bfa0a93cceee1f0df7461bc034ad0becba42baf98a47d6983e5d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter : 2 - Energy Bands And Charge Carriers In Semiconductor"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.6.1 - Page No : 2-10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# Given data\n",
+      "h = 6.625 * 10**-34 # in J\n",
+      "c = 3 * 10**8 # in J\n",
+      "lambda_Gr = 17760 * 10**-10 # in m\n",
+      "lambda_Si = 11000 # in A\u00b0\n",
+      "lambda_Si = lambda_Si * 10**-10 # in m\n",
+      "E_g = (h*c)/lambda_Si # in J\n",
+      "E_g = E_g /(1.6*10**-19) # in eV\n",
+      "print \"The energy gap of Si = %0.3f eV \" %E_g"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The energy gap of Si = 1.129 eV \n"
+       ]
+      }
+     ],
+     "prompt_number": 53
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  2.6.2 - Page No : 2-10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "E_g = 0.75# in eV\n",
+      "E_g = 0.75 * 1.6 * 10**-19 # in J\n",
+      "h = 6.63 * 10**-34 # in J\n",
+      "c = 3 * 10**8 # in m/s \n",
+      "# hv = E_g\n",
+      "#E_g = (h*c)/lambda\n",
+      "Lambda=(h*c)/E_g # in m\n",
+      "Lambda=Lambda * 10**10 # in A\u00b0\n",
+      "print \"The wavelength at which germanium starts to absorb light = %0.f A\u00b0 \" %Lambda"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength at which germanium starts to absorb light = 16575 A\u00b0 \n"
+       ]
+      }
+     ],
+     "prompt_number": 54
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.15.1\n",
+      " - Page No : 2-30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "del_E = 0.3 # in eV\n",
+      "T1 = 300 # in K\n",
+      "T2 = 330 # in K\n",
+      "# del_E = K * T1 * log(N/N_c) where del_E= E_C-E_F\n",
+      "# del_E1 = K * T2 * log(N/N_c) where del_E1= E_C-E_F at T= 330 \u00b0K\n",
+      "del_E1 = del_E*(T2/T1) # in eV \n",
+      "print \"The Fermi level will be \",round(del_E1,2),\" eV below the conduction band\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Fermi level will be  0.33  eV below the conduction band\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.15.2\n",
+      " - Page No : 2-31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import exp\n",
+      "# Given data\n",
+      "N_c = 2.8 * 10**19 # in cm**-3\n",
+      "del_E = 0.25 # fermi energy in eV\n",
+      "KT = 0.0259 \n",
+      "f_F = exp(-(del_E)/KT) \n",
+      "print \"The probaility in the condition band is occupied by an electron = %0.2e\"  %f_F\n",
+      "n_o = N_c * exp(-(del_E)/KT) # in cm**-3\n",
+      "print \"The thermal equilibrium electron concentration = %0.1e cm**-3 \" %n_o"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The probaility in the condition band is occupied by an electron = 6.43e-05\n",
+        "The thermal equilibrium electron concentration = 1.8e+15 cm**-3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.15.3\n",
+      " - Page No : 2-33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "T1 = 300 # in K\n",
+      "T2 = 400 # in K\n",
+      "del_E = 0.27 # Fermi level in eV\n",
+      "KT = (0.0259) * (T2/T1) # in eV\n",
+      "N_v = 1.04 * 10**19 # in cm**-3\n",
+      "N_v = N_v * (T2/T1)**(3/2) # in cm**-3 \n",
+      "p_o = N_v * exp(-(del_E)/KT) # in per cm**3\n",
+      "print \"The thermal equilibrium hole concentration = %0.2e per cm**3 \" %p_o"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The thermal equilibrium hole concentration = 6.44e+15 per cm**3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  2.16.1 - Page No : 2-37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "At = 63.5;# atomic weight\n",
+      "Rho = 1.7*10**-6;# in ohm cm\n",
+      "d = 8.96;# in gm/cc\n",
+      "N_A = 6.02*10**23;# in /gm.mole\n",
+      "e = 1.6*10**-19;# in C\n",
+      "n = (N_A/At)*d;\n",
+      "Miu_e = 1/(Rho*n*e);# in cm**2/volt.sec\n",
+      "print \"The electron mobility = %0.3f cm**2/volt-sec\" %Miu_e"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The electron mobility = 43.281 cm**2/volt-sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  2.16.2 - Page No : 2-38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "l = 0.1;# in m\n",
+      "A = 1.7;# in mm**2\n",
+      "A = A * 10**-6;# in m**2\n",
+      "R = 0.1;# in ohm\n",
+      "At = 63.5;# atomic weight\n",
+      "N_A = 6.02*10**23;\n",
+      "d = 8.96;# in gm/cc\n",
+      "n = (N_A/At)*d;# in /cc\n",
+      "n = n * 10**6;# in /m**3\n",
+      "#Formula R = Rho*(l/A);\n",
+      "Rho = (R*A)/l;# in ohm m\n",
+      "Sigma = 1/Rho;# in mho/m\n",
+      "e = 1.6*10**-19;\n",
+      "# Formula Sigma = n*e*Miu_e\n",
+      "Miu_e = Sigma/(n*e);# in m**2/V.sec\n",
+      "print \"The mobility = %0.2e m**2/V-sec\" %Miu_e"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mobility = 4.33e-05 m**2/V-sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  2.16.3 - Page No : 2-39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# Given data\n",
+      "d = 10.5;# in gm/cc\n",
+      "At = 108;# atomic weight\n",
+      "N_A = 6.025*10**23;# in /gm mole\n",
+      "r = 10**-3;# in m\n",
+      "q = 1.6*10**-19;# in C\n",
+      "n = (N_A/At)*d;# in /cm**3\n",
+      "n = n * 10**6;# in /m**3\n",
+      "A = pi*((r)**2);# in m**2\n",
+      "I = 2;# in A\n",
+      "# I = q*n*A*v;\n",
+      "v = I/(n*q*A);# in m/s\n",
+      "print \"The drift velocity = %0.e m/s\" %v\n",
+      "\n",
+      "# Note: There is calculation error to find the value of drift velocity (i.e v), so the answer in the book is wrong."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The drift velocity = 7e-05 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  2.16.4 - Page No : 2-39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "d = 1.03;# in mm\n",
+      "d = d *10**-3;# in m\n",
+      "r = d/2;# in m\n",
+      "R = 6.51;# in ohm\n",
+      "l = 300;# in mm\n",
+      "e = 1.6*10**-19;\n",
+      "n = 8.4*10**28;# in /m**3\n",
+      "A = pi*r**2;\n",
+      "# R = Rho*(l/A);\n",
+      "Rho = (R*A)/l;#in ohm m\n",
+      "Sigma = 1/Rho;# in mho/m\n",
+      "print \"The conductivity of copper = %0.3e mho/m\" %Sigma\n",
+      "#Formula sigma = n*e*Miu_e\n",
+      "Miu_e = Sigma/(n*e);# in m**2/V.sec\n",
+      "print \"The mobility = %0.3e m**2/V-sec\" %Miu_e"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The conductivity of copper = 5.531e+07 mho/m\n",
+        "The mobility = 4.115e-03 m**2/V-sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 49
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  2.16.5 - Page No : 2-40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "At = 63.5;# atomic weight\n",
+      "d = 8.96;# in gm/cc\n",
+      "Miu_e = 43.28;# in cm**2/V.sec\n",
+      "N_A = 6.02*10**23;# in /gm mole\n",
+      "e = 1.6*10**-19;# in C\n",
+      "n = (N_A/At)*d;# in /cc\n",
+      "Rho = 1/(n*e*Miu_e);# in ohm-cm\n",
+      "print \"The resistivity = %0.1e ohm-cm\" %Rho"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistivity = 1.7e-06 ohm-cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 52
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  2.17.1 - Page No : 2-42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "Mu_e = 1500 # in cm**2/volt sec\n",
+      "Mu_h = 500 # in cm**2/volt sec\n",
+      "n_i = 1.6 * 10**10 # in per cm**3\n",
+      "e = 1.6 * 10**-19 # in C\n",
+      "Sigma = n_i * (Mu_e + Mu_h) * e # in mho/cm\n",
+      "print \"The conductivity of pure semiconductor = %0.2e mho/cm \" %Sigma"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The conductivity of pure semiconductor = 5.12e-06 mho/cm \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  2.17.2 - Page No : 2-42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "Rho = 10 # in \u03a9-cm\n",
+      "Mu_d = 500 # in cm**2/v.s.\n",
+      "e = 1.6*10**-19 \n",
+      "n_d = 1/(Rho * e * Mu_d) # in per cm**3\n",
+      "print \"The number of donor atom must be added to achieve = %0.2e per cm**3 \" %n_d"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The number of donor atom must be added to achieve = 1.25e+15 per cm**3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.17.3 - Page No : 2-43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given data\n",
+      "AvagadroNumber = 6.02 * 10**23 # in atoms/gm.mole\n",
+      "at_Ge = 72.6 # atom weight of Ge\n",
+      "e = 1.6 * 10**-19 # in C\n",
+      "D_Ge = 5.32 # density of Ge in gm/c.c\n",
+      "Mu = 3800 # in cm**2/v.s.\n",
+      "C_Ge = (AvagadroNumber/at_Ge) * D_Ge # concentration of Ge atoms in per cm**3\n",
+      "n_d = C_Ge/10**8 # in per cc\n",
+      "Sigma = n_d * Mu * e # in mho/cm\n",
+      "print \"The conductivity = %0.3f mho/cm \" %Sigma"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The conductivity = 0.268 mho/cm \n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.17.4 - Page No : 2-43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "Rho = 0.3623 * 10**-3 # in Ohm m\n",
+      "Sigma = 1/Rho #in mho/m\n",
+      "D = 4.42 * 10**28 # Ge density in atom/m**3\n",
+      "n_d = D / 10**6 # in atom/m**3\n",
+      "e = 1.6 * 10**-19 # in C\n",
+      "Mu = Sigma/(n_d * e) # in m**2/V.sec\n",
+      "print \"The mobility of electron in germanium = %0.2f m**2/V.sec \" %Mu"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mobility of electron in germanium = 0.39 m**2/V.sec \n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.17.5 - Page No : 2-44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "AvagadroNumber = 6.025 * 10**26 # in kg.Mole\n",
+      "W = 72.59 # atomic weight of Ge\n",
+      "D = 5.36 * 10**3 #density of Ge in kg/m**3\n",
+      "Rho = 0.42 # resistivity in Ohm m\n",
+      "e = 1.6 * 10**-19 # in C\n",
+      "Sigma = 1/Rho # in mho/m\n",
+      "n = (AvagadroNumber/W) * D # number of Ge atoms present per unit volume\n",
+      "# Holes per unit volume, H = n*10**-6%\n",
+      "H= n*10**-8 \n",
+      "a=H \n",
+      "# Formula sigma= a*e*Mu_h\n",
+      "Mu_h = Sigma/(a * e) # in m**2/V.sec\n",
+      "print \"Mobility of holes = %0.4f m**2/V.sec \" %Mu_h"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mobility of holes = 0.0334 m**2/V.sec \n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.17.6 - Page No : 2-45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "e = 1.6 * 10**-19 # in C\n",
+      "n_i = 2 * 10**19 # in /m**3\n",
+      "Mu_e = 0.36 # in m**2/v.s\n",
+      "Mu_h = 0.17 # in m**2/v.s\n",
+      "A = 1 * 10**-4 # in  m**2\n",
+      "V = 2 #in volts\n",
+      "l = 0.3 # in mm\n",
+      "l = l * 10**-3 # in m\n",
+      "E=V/l # in volt/m\n",
+      "Sigma = n_i * e * (Mu_e + Mu_h) # in mho/m\n",
+      "# J = I/A = Sigma * E\n",
+      "I= Sigma*E*A \n",
+      "print \"The current produced in a small germanium plate = %0.2f amp \" %I\n",
+      "\n",
+      " "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current produced in a small germanium plate = 1.13 amp \n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.17.7 - Page No : 2-45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "D = 4.2 * 10**28 #density of Ge atoms in atoms/m**3\n",
+      "N_d = D / 10**6 # in atoms/m**3\n",
+      "e = 1.6 * 10**-19 # in C\n",
+      "Mu_e = 0.36 # in m**2/vs\n",
+      "Sigma_n = N_d *  e * Mu_e # in mho/m\n",
+      "Rho_n = 1/Sigma_n # in ohm m\n",
+      "print \"The resistivity of drop Ge = %0.3e ohm m \" %Rho_n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistivity of drop Ge = 4.134e-04 ohm m \n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.17.8 - Page No : 2-46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "e = 1.6 * 10**-19 # in C\n",
+      "n_i = 1 * 10**19 # in per m**3\n",
+      "Mu_e = 0.36 # in m**2/volt.sec\n",
+      "Mu_h = 0.17 # in m**2/volt.sec \n",
+      "A = 2 # in cm**2\n",
+      "A = A * 10**-4 # im m**2\n",
+      "t = 0.1 # in mm\n",
+      "t = t * 10**-3 # in m\n",
+      "V = 4 # in volts\n",
+      "Sigma_i = n_i * e * (Mu_e + Mu_h) # in mho/m\n",
+      "J = Sigma_i * (V/t) # in Amp/m**2\n",
+      "I = J * A # in Amp\n",
+      "print \"The current produced in a Ge sample = %0.3f Amp \" %I"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current produced in a Ge sample = 6.784 Amp \n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.17.9 - Page No : 2-46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "e = 1.6 * 10**-19 # in C\n",
+      "Mu_h = 500 # in cm**2/V.s.\n",
+      "Mu_e = 1500 # in cm**2/V.s.\n",
+      "n_i = 1.6 * 10**10 # in per cm**3\n",
+      "Sigma_i = n_i * e * ( Mu_h + Mu_e) # in mho/cm\n",
+      "print \"Conductivity of pure silicon at room temperature = %0.2e mho/cm \" %Sigma_i"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Conductivity of pure silicon at room temperature = 5.12e-06 mho/cm \n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.19.1 - Page No : 2-50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given data\n",
+      "l= 0.50*10**-2 # width of ribbon in m\n",
+      "d= 0.10*10**-3 # thickness of ribbon in m\n",
+      "A= l*d # area of ribbon in m**2\n",
+      "B = 0.8 # in Tesla\n",
+      "D = 10.5 #density in gm/cc\n",
+      "I = 2 # in amp\n",
+      "q = 1.6 * 10**-19 # in C\n",
+      "n=6*10**28 # number of elec. per m**3\n",
+      "V_H = ( I * B * d)/(n * q * A) # in volts\n",
+      "print \"The hall Voltage produced = %0.2e volts \" %V_H"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The hall Voltage produced = 3.33e-08 volts \n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.19.2 - Page No : 2-51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "l = 1 # in m\n",
+      "d = 1 # in cm\n",
+      "d = d * 10**-2 # in m\n",
+      "W = 1 # in mm\n",
+      "W = W * 10**-3 # in m\n",
+      "A = d * W # in m**2\n",
+      "I= 1 # in Amp\n",
+      "B = 1 # Tesla\n",
+      "V_H = 0.074 * 10**-6 # in Volts\n",
+      "Sigma = 5.8 * 10**7 # in mho/m\n",
+      "R_H = (V_H * A)/(B*I*d) # in m**3/c\n",
+      "print \"The hall coefficient = %0.1e m**3/c \" %R_H\n",
+      "Mu = Sigma * R_H # in m**2/volt.sec\n",
+      "print \"The mobility  of electrons in copper = %0.1e m**2/volt-sec \" %Mu"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The hall coefficient = 7.4e-11 m**3/c \n",
+        "The mobility  of electrons in copper = 4.3e-03 m**2/volt-sec \n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.20.1 - Page No : 2-52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "n_i = 1.4 * 10**18 # in /m**3\n",
+      "n_D = 1.4 * 10**24 # in /m**3\n",
+      "n=n_D # in /m**3\n",
+      "p = n_i**2/n # in /m**3\n",
+      "R = n/p \n",
+      "print \"The ratio of electrons to hole concentration = %0.1e\" %R"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The ratio of electrons to hole concentration = 1.0e+12\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.20.2 - Page No : 2-53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import atan, pi\n",
+      "# Given data\n",
+      "B = 0.48 # in wb/m**2\n",
+      "R_H = 3.55 * 10**-4 # in m**3/c\n",
+      "Rho = 0.00912 # in ohm-m\n",
+      "Sigma = 1/Rho # in (ohm-m)**-1\n",
+      "theta_H = atan( Sigma * B * R_H)*180/pi # in degree\n",
+      "print \"The hall angle for a hall coefficient = %0.4f\u00b0\" %theta_H"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The hall angle for a hall coefficient = 1.0704\u00b0\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.20.3 - Page No : 2-53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given data\n",
+      "R = 9 * 10**-3 # in ohm-m\n",
+      "R_H = 3.6 * 10**-4 # in m**3\n",
+      "e = 1.6 * 10**-19 # in C\n",
+      "Sigma = 1/R # in (ohm-m)**-1\n",
+      "Rho = 1/R_H # in coulomb/m**3\n",
+      "n = Rho/e # in /m**3\n",
+      "print \"Density of charge carriers = %0.5e per m**3 \" %n\n",
+      "Mu = Sigma * R_H # in m**2/v-s\n",
+      "print \"Mobility of charge carriers = %0.2f m**2/V-s \" %Mu"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Density of charge carriers = 1.73611e+22 per m**3 \n",
+        "Mobility of charge carriers = 0.04 m**2/V-s \n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.24.1 - Page No : 2-60"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "e = 1.6 * 10**-19 # in C\n",
+      "R_H = 0.0145 # in m**3/coulomb\n",
+      "Mu_e = 0.36 # in m**2/v-s\n",
+      "E = 100 # in V/m\n",
+      "n = 1/(e * R_H) # in /m**3\n",
+      "J = n * e * Mu_e * E # in A/m**2\n",
+      "print \"The current density of specimen = %0.3e A/m**2 \" %J"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current density of specimen = 2.483e+03 A/m**2 \n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.24.2 - Page No : 2-60"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "#Given data\n",
+      "Mu_e = 7.04 * 10**-3 # in m**2/v-s\n",
+      "m = 9.1 * 10**-31 \n",
+      "E_F = 5.5 # in eV\n",
+      "n = 5.8 * 10**28 \n",
+      "e = 1.6 * 10**-19 # in C\n",
+      "Torque = (Mu_e/e) * m # in sec\n",
+      "print \"Relaxation Time = %0.3e sec \" %Torque\n",
+      "Rho = 1 /(n * e * Mu_e) # in ohm-m\n",
+      "print \"Resistivity of conductor = %0.3e ohm-m \" %Rho\n",
+      "V_F = sqrt((2 * E_F * e)/m) # in m/s\n",
+      "print \"Velocity of electrons with fermi-energy = %0.4e m/s \" %V_F"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Relaxation Time = 4.004e-14 sec \n",
+        "Resistivity of conductor = 1.531e-08 ohm-m \n",
+        "Velocity of electrons with fermi-energy = 1.3907e+06 m/s \n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.24.3 - Page No : 2-61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "# Given data\n",
+      "E= 5.95 # in eV\n",
+      "EF= 6.25 # in eV\n",
+      "delE= 0.01 \n",
+      " # delE= 1-1/(1+exp((E-EF)/KT))\n",
+      "K=1.38*10**-23 # Boltzman Constant in J/K\n",
+      "T = ((E-EF)/log(1/(1-delE) -1)*1.6*10**-19)/K # in K\n",
+      "print \"The temperature =\" ,int(T),\"K\"\n",
+      " "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The temperature = 756 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.24.4 - Page No : 2-61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "N_V = 1.04 * 10**19 # in cm**-3\n",
+      "T1 = 300 # in K\n",
+      "T2 = 400 # in K\n",
+      "del_E = 0.27 # in eV\n",
+      "N_V = N_V * (T2/T1)**1.5 # in cm**-3\n",
+      "KT = (0.0259) * (T2/T1) # in eV\n",
+      "P_o = N_V * exp(-(del_E)/KT) # in cm**-3\n",
+      "print \"The thermal equilibrium hole concentration in silicon = %0.2e cm**-3 \" %P_o"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The thermal equilibrium hole concentration in silicon = 6.44e+15 cm**-3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.24.5 - Page No : 2-62"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given data\n",
+      "N_c = 2.8 * 10**19 \n",
+      "N_V = 1.04 *10**19 \n",
+      "T1 = 550 # in K\n",
+      "T2 = 300 # in K\n",
+      "E_g = 1.12 \n",
+      "KT = (0.0259)  \n",
+      "n_i = sqrt(N_c *N_V *(T1/T2)**3* exp(-(E_g)/KT*T2/T1)) # in cm**-3\n",
+      "# n_o = N_d/2 + sqrt((N_d/2)**2 + (n_i)**2)\n",
+      "# 1.05*N_d -N_d/2= sqrt((N_d/2)**2 + (n_i)**2)\n",
+      "N_d=sqrt((n_i)**2/((0.55)**2-1/4)) \n",
+      "print \"Minimum donor concentration required = %0.3e cm**-3 \"  %N_d"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Minimum donor concentration required = 1.396e+15 cm**-3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 2.24.6 - Page No : 2-62"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given data\n",
+      "T = 300 # in K\n",
+      "n_o = 10**15 # in cm**-3\n",
+      "n_i = 10**10 # in cm**-3\n",
+      "p_o = 10**5 # in cm**-3\n",
+      "del_n = 10**13 # in cm**-3\n",
+      "del_p = del_n # in cm**-3\n",
+      "KT = 0.0259 # in eV\n",
+      "delta_E1= KT*log(n_o/n_i) # value of E_F-E_Fi in eV\n",
+      "delta_E2= KT*log((n_o+del_n)/n_i) # value of E_Fn-E_Fi in eV\n",
+      "delta_E3= KT*log((p_o+del_p)/n_i) # value of E_Fi-E_Fp in eV\n",
+      "print \"The Fermi level for thermal equillibrium = %0.4f eV \" %delta_E1\n",
+      "print \"The quase-Fermi level for electrons in non equillibrium = %0.4f eV \" %delta_E2\n",
+      "print \"The quasi-Fermi level for holes in non equillibrium = %0.3f eV \" %delta_E3\n",
+      "print \"The quasi-Fermi level for electrons is above E_Fi while the quasi-Fermi level for holes is below E_Fi\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Fermi level for thermal equillibrium = 0.2982 eV \n",
+        "The quase-Fermi level for electrons in non equillibrium = 0.2984 eV \n",
+        "The quasi-Fermi level for holes in non equillibrium = 0.179 eV \n",
+        "The quasi-Fermi level for electrons is above E_Fi while the quasi-Fermi level for holes is below E_Fi\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_3_1.ipynb b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_3_1.ipynb
new file mode 100644
index 00000000..aedb74be
--- /dev/null
+++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_3_1.ipynb
@@ -0,0 +1,898 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:764d1c60d9fefbb17ec0ffb4d8cdbf1cddd8802a6e86f8e004d6805017f0ecae"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter : 3 - Excess Carriers in Semiconductor"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.1 - Page No : 3-38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# Given data\n",
+      "N_d = 10**17 # atoms/cm**3\n",
+      "n_i = 1.5 * 10**10 # in /cm**3\n",
+      "n_o = 10**17 # in cm**3\n",
+      "# p_o * n_o = (n_i)**2\n",
+      "p_o = (n_i)**2 / n_o #in holes/cm**3\n",
+      "print \"The holes concentration at equilibrium = %0.2e holes/cm**3 \" %p_o"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The holes concentration at equilibrium = 2.25e+03 holes/cm**3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.3 - Page No : 3-40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "# Given data\n",
+      "n_i = 1.5 * 10 **10 # in /cm**3 for silicon\n",
+      "N_d = 10**17 # in atoms/cm**3\n",
+      "n_o = 10**17 # electrons/cm**3\n",
+      "KT = 0.0259 \n",
+      "# E_r - E_i = KT * log(n_o/n_i)\n",
+      "del_E = KT * log(n_o/n_i) # in eV\n",
+      "print \"The energy band for this type material, E_F = Ei +\",round(del_E,3),\" eV\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The energy band for this type material, E_F = Ei + 0.407  eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.4 - Page No : 3-41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "K = 1.38 * 10**-23 # in J/K\n",
+      "T = 27 # in degree\n",
+      "T = T + 273 # in K\n",
+      "e = 1.6 * 10**-19 # in C\n",
+      "Mu_e = 0.17 # in m**2/v-s\n",
+      "Mu_e1 = 0.025 # in m**2/v-s\n",
+      "D_n = ((K * T)/e) * Mu_e # in m**2/s\n",
+      "print \"The diffusion coefficient of electrons = %0.1e m**2/s \" %D_n\n",
+      "D_p = ((K * T)/e) * Mu_e1 # in m**2/s\n",
+      "print \"The diffusion coefficient of  holes = %0.2e m**2/s \" %D_p"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The diffusion coefficient of electrons = 4.4e-03 m**2/s \n",
+        "The diffusion coefficient of  holes = 6.47e-04 m**2/s \n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.5 - Page No : 3-41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "# Given data\n",
+      "Mu_n = 0.15 # in m**2/v-s\n",
+      "K = 1.38 * 10**-23  # in J/K\n",
+      "T = 300 # in K\n",
+      "del_n = 10**20 # in per m**3\n",
+      "Toh_n = 10**-7 # in s\n",
+      "e = 1.6 * 10**-19 # in C\n",
+      "D_n = Mu_n * ((K * T)/e) # in m**2/s\n",
+      "print \"The diffusion coefficient = %0.2e m**2/s \" %D_n\n",
+      "L_n = sqrt(D_n * Toh_n) # in m \n",
+      "print \"The Diffusion length = %0.2e m \" %L_n\n",
+      "J_n = (e * D_n * del_n)/L_n # in A/m**2\n",
+      "print \"The diffusion current density = %0.2e A/m**2 \"  %J_n\n",
+      "# Note : The value of diffusion coefficient in the book is wrong."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The diffusion coefficient = 3.88e-03 m**2/s \n",
+        "The Diffusion length = 1.97e-05 m \n",
+        "The diffusion current density = 3.15e+03 A/m**2 \n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.6 - Page No : 3-42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "Sigma = 0.1 # in (ohm-m)**-1\n",
+      "Mu_n = 1300 \n",
+      "n_i = 1.5 * 10**10 \n",
+      "q = 1.6 * 10**-19 # in C\n",
+      "n_n = Sigma/(Mu_n * q) # in electrons/cm**3\n",
+      "print \"The concentration of electrons = %0.2e per m**3 \" %(n_n*10**6)\n",
+      "p_n = (n_i)**2/n_n # in per cm**3\n",
+      "p_n = p_n * 10**6 # in perm**3\n",
+      "print \"The concentration of holes = %0.2e per m**3 \" %p_n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The concentration of electrons = 4.81e+20 per m**3 \n",
+        "The concentration of holes = 4.68e+11 per m**3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.7 - Page No : 3-43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "Mu_e = 0.13 # in m**2/v-s\n",
+      "Mu_h = 0.05 # in m**2/v-s\n",
+      "Toh_h = 10**-6 # in s\n",
+      "L = 100 # in \u00b5m\n",
+      "L = L * 10**-6 # in m\n",
+      "V = 2 # in V\n",
+      "t_n =L**2/(Mu_e * V) # in s\n",
+      "print \"Electron transit time = %0.1e seconds \" %t_n\n",
+      "p_g = (Toh_h/t_n) * (1 + Mu_h/Mu_e) #photo conductor gain \n",
+      "print \"Photo conductor gain = %0.2f\" %p_g\n",
+      "\n",
+      "# Note: There is a calculation error to evaluate the value of t_n. So the answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Electron transit time = 3.8e-08 seconds \n",
+        "Photo conductor gain = 36.00\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.8 - Page No : 3-43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "n_i = 2.5 * 10**13 \n",
+      "Mu_n = 3800 \n",
+      "Mu_p = 1800 \n",
+      "q = 1.6 * 10**-19 # in C\n",
+      "Sigma = n_i * (Mu_n + Mu_p) * q # in (ohm-cm)**-1\n",
+      "Rho = 1/Sigma # in ohm-cm\n",
+      "Rho= round(Rho) \n",
+      "print \"The resistivity of intrinsic germanium = %0.f ohm-cm \" %Rho\n",
+      "N_D = 4.4 * 10**22/(1*10**8) # in atoms/cm**3\n",
+      "Sigma_n = N_D * Mu_n * q # in (ohm-cm)**-1\n",
+      "Rho_n = 1/Sigma_n # in ohm-cm\n",
+      "print \"If a donor type impurity is added to the extent of 1 atom per 10**8 Ge atoms, then the resistivity drops = %0.2f ohm-cm \" %Rho_n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistivity of intrinsic germanium = 45 ohm-cm \n",
+        "If a donor type impurity is added to the extent of 1 atom per 10**8 Ge atoms, then the resistivity drops = 3.74 ohm-cm \n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.9 - Page No : 3-44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "n_i = 10**16 # in /m3\n",
+      "N_D = 10**22 # in /m**3\n",
+      "n = N_D # in /m**3\n",
+      "print \"Electron concentration = %0.1e per m**3 \" %n\n",
+      "p = (n_i)**2/n # in /m**3\n",
+      "print \"Hole concentration = %0.1e per m**3 \" %p"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Electron concentration = 1.0e+22 per m**3 \n",
+        "Hole concentration = 1.0e+10 per m**3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.10 - Page No : 3-44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "Rho = 9.6 * 10**-2 # in ohm-m\n",
+      "Sigma_n = 1/Rho # in (ohm-m)**-1\n",
+      "q = 1.6 * 10**-19 # in C\n",
+      "Mu_n = 1300 * 10**-4 # in m**2/v-s\n",
+      "N_D = Sigma_n / (Mu_n * q) # in atoms/m**3\n",
+      "A_D = N_D # Atom density in atoms/cm**3\n",
+      "A_D = A_D * 10**6 # atoms/m**3\n",
+      "R_si = N_D/A_D # ratio\n",
+      "print \"The ratio of donor atom to silicon atom = %0.1e\" %R_si\n",
+      "\n",
+      "# Note: In the book the wrong value of N_D (5*10**22) is putted to evaluate the value of \n",
+      "#       Atom Density (A_D) whereas the value of N_D is calculated as 5*10**20.So the answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The ratio of donor atom to silicon atom = 1.0e-06\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.11 - Page No : 3-45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "n_i = 1.5 * 10**10 # in per cm**3\n",
+      "n_n = 2.25 * 10**15 # in per cm**3\n",
+      "p_n = (n_i)**2/n_n # in per cm**3\n",
+      "print \"The equilibrium electron = %0.1e per cm**3 \" %p_n\n",
+      "h_n = n_n # in cm**3\n",
+      "print \"Hole densities = %0.2e per cm**3 \" %h_n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The equilibrium electron = 1.0e+05 per cm**3 \n",
+        "Hole densities = 2.25e+15 per cm**3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.12 - Page No : 3-45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "N_A = 2 * 10**16 # in atoms/cm**3\n",
+      "N_D = 10**16 # in atoms/cm**3\n",
+      "C_c = N_A-N_D # C_c stands for Carrier concentration in /cm**3\n",
+      "print \"Carrier concentration = %0.1e holes/cm**3 \" %C_c"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Carrier concentration = 1.0e+16 holes/cm**3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.13 - Page No : 3-46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "del_n = 10**15 # in cm**3\n",
+      "Torque_p = 10 * 10**-6 # in sec\n",
+      "R_g = del_n/Torque_p # in hole pairs/sec/cm**3\n",
+      "print \"The rate of generation of minority carrier = %0.1e electron hole pairs/sec/cm**3 \" %R_g"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The rate of generation of minority carrier = 1.0e+20 electron hole pairs/sec/cm**3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.14 - Page No : 3-46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "v = 1/(20 * 10**-6) # in cm/sec\n",
+      "E = 10 # in V/cm\n",
+      "Mu= v/E # in cm**2/V-sec\n",
+      "print \"The mobility of minority charge carrier = %0.f cm**2/V-sec \" %Mu"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mobility of minority charge carrier = 5000 cm**2/V-sec \n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.15 - Page No : 3-47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "q = 1.6 * 10**-19 # in C\n",
+      "N_D = 4.5 * 10**15 # in /cm**3\n",
+      "del_p = 10**21 \n",
+      "e=10 # in cm\n",
+      "A = 1 # in mm**2\n",
+      "A = A * 10**-14 # cm**2\n",
+      "l = 10 # in cm\n",
+      "Torque_p = 1 # in microsec\n",
+      "Torque_p = Torque_p * 10**-6 # in sec\n",
+      "Torque_n = 1 # in microsec\n",
+      "Torque_n = Torque_n * 10**-6 # in  sec\n",
+      "n_i = 1.5 * 10**10 # in /cm**3\n",
+      "D_n = 30 # in cm**2/sec\n",
+      "D_p = 12 # in cm**2/sec\n",
+      "n_o = N_D # in /cm**3\n",
+      "p_o = (n_i)**2/n_o # in /cm**3\n",
+      "print \"Hole concentration at thermal equilibrium = %0.1e per cm**3 \" %p_o\n",
+      "l_n = sqrt(D_n * Torque_n) # in cm\n",
+      "print \"Diffusion length of electron = %0.2e cm \" %l_n\n",
+      "l_p = sqrt(D_p * Torque_p) # in cm\n",
+      "print \"Diffusion length of holes = %0.1e cm \" %l_p\n",
+      "x=34.6*10**-4 # in cm\n",
+      "dpBYdx = del_p *e # in cm**4\n",
+      "print \"Concentration gradient of holes = %0.1e cm**4 \" %dpBYdx\n",
+      "e1 = 1.88 * 10**1 # in cm\n",
+      "dnBYdx = del_p * e1 # in cm**4 \n",
+      "print \"Concentration gradient of electrons = %0.2e per cm**4 \" %dnBYdx\n",
+      "J_P = -(q) * D_p * dpBYdx # in A/cm**2\n",
+      "print \"Current density of holes due to diffusion = %0.2e A/cm**2 \" %J_P\n",
+      "J_n = q * D_n * dnBYdx # in A/cm**2\n",
+      "print \"Current density of electrons due to diffusion = %0.1e A/cm**2 \" %J_n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hole concentration at thermal equilibrium = 5.0e+04 per cm**3 \n",
+        "Diffusion length of electron = 5.48e-03 cm \n",
+        "Diffusion length of holes = 3.5e-03 cm \n",
+        "Concentration gradient of holes = 1.0e+22 cm**4 \n",
+        "Concentration gradient of electrons = 1.88e+22 per cm**4 \n",
+        "Current density of holes due to diffusion = -1.92e+04 A/cm**2 \n",
+        "Current density of electrons due to diffusion = 9.0e+04 A/cm**2 \n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.16 - Page No : 3-49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "e= 1.6*10**-19 # electron charge in C\n",
+      "h = 6.626 * 10**-34 # in J-s\n",
+      "h= h/e # in eV\n",
+      "c = 3 * 10**8 # in m/s\n",
+      "lembda = 5490 * 10**-10 # in m\n",
+      "f = c/lembda \n",
+      "E = h * f # in eV\n",
+      "print \"The energy band gap of the semiconductor material = %0.2f eV \" %E"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The energy band gap of the semiconductor material = 2.26 eV \n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.17 - Page No : 3-49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "y2 = 6 * 10**16 # in /cm**3\n",
+      "y1 = 10**17 # in /cm**3\n",
+      "x2 = 2 # in \u00b5m\n",
+      "x1 = 0 # in \u00b5m\n",
+      "D_n = 35 # in cm**2/sec\n",
+      "q = 1.6 *10**-19 # in C\n",
+      "dnBYdx = (y2 - y1)/((x2-x1) * 10**-4) \n",
+      "J_n = q * D_n * dnBYdx # in A/cm**2\n",
+      "print \"The current density in silicon = %0.f A/cm**2 \" %J_n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current density in silicon = -1120 A/cm**2 \n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.18 - Page No : 3-50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "q = 1.6 * 10**-19 # in C\n",
+      "n_n = 5 * 10**20 # in /m**3\n",
+      "n_n = n_n * 10**-6 # in cm**3\n",
+      "Mu_n = 0.13 # in m**2/V-sec\n",
+      "Mu_n = Mu_n * 10**4 # in cm**2/V-sec\n",
+      "Sigma_n = q * n_n * Mu_n # in (ohm-cm)**-1\n",
+      "Rho = 1/Sigma_n # in \u03a9-cm\n",
+      "l = 0.1 # in cm\n",
+      "A = 100 # \u00b5m**2\n",
+      "A = A * 10**-8 # in cm**2\n",
+      "R = Rho * (l/A) # in Ohm\n",
+      "R=round(R*10**-6) # in M\u03a9\n",
+      "print \"The resistance of the bar = %0.f M\u03a9 \"  %R"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistance of the bar = 1 M\u03a9 \n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.19 - Page No : 3-51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "t_d = 3 # total depletion in \u00b5m\n",
+      "D = t_d/9 # in \u00b5m\n",
+      "print \"Depletion width = %0.1f \u00b5m \" %D"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Depletion width = 0.3 \u00b5m \n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.20 - Page No : 3-51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "n_i = 1.5 * 10**16 # in /m**3\n",
+      "n_n = 5 * 10**20 # in /m**3\n",
+      "p_n = (n_i)**2/n_n # in /m**3\n",
+      "print \"The majority carrier density = %0.2e per m**3 \" %p_n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The majority carrier density = 4.50e+11 per m**3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.21 - Page No : 3-52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "D_n = 25 # in cm**2/sec\n",
+      "q = 1.6 * 10**-19 # in C\n",
+      "y2 = 10**14 # in /cm**3\n",
+      "y1 = 0 # in /cm**3\n",
+      "x2 = 0 #in  \u00b5m\n",
+      "x1 = 0.5 # in \u00b5m\n",
+      "x1 = x1 * 10**-4 # in cm\n",
+      "dnBYdx = abs((y2-y1)/(x2-x1)) # in /cm**4 \n",
+      "J_n = q * D_n * (dnBYdx) # in /cm**4\n",
+      "J_n = J_n * 10**-1 # in A/cm**2\n",
+      "print \"The collector current density = %0.f A/cm**2 \" %J_n\n",
+      "\n",
+      "# Note: In the book, the calculated value of dn by dx (2*10**19) is wrong. Correct value is 2*10**18\n",
+      "#       so the answer in the book is wrong."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The collector current density = 1 A/cm**2 \n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 3.21.22 - Page No : 3-53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "h = 6.64 * 10**-34 # in J-s\n",
+      "e= 1.6*10**-19 # electron charge in C\n",
+      "c= 3 * 10**8 # in m/s\n",
+      "lembda = 0.87 # in \u00b5m\n",
+      "lembda = lembda * 10**-6 # in m\n",
+      "E_g = (h * c)/lembda # in J-s\n",
+      "E_g= E_g/e # in eV\n",
+      "print \"The band gap of the material = %0.3f eV \" %E_g"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The band gap of the material = 1.431 eV \n"
+       ]
+      }
+     ],
+     "prompt_number": 46
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.23 - Page No : 3-53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import exp\n",
+      "# Given data\n",
+      "I_o = 10 # in mW\n",
+      "e = 1.6 * 10**-19 # in J/eV\n",
+      "hv = 2 # in eV\n",
+      "hv1=1.43 # in eV\n",
+      "alpha = 5 * 10**4 # in cm**-1\n",
+      "l = 46 # in \u00b5m\n",
+      "l = l * 10**-6 # in m\n",
+      "I_t = round(I_o * exp(-(alpha) * l)) # in mW\n",
+      "AbsorbedPower= I_o-I_t # in mW\n",
+      "AbsorbedPower=AbsorbedPower*10**-3 # in W or J/s\n",
+      "print \"The absorbed power = %0.1e watt or J/s \" %AbsorbedPower\n",
+      "F= (hv-hv1)/hv # fraction of each photon energy unit\n",
+      "EnergyConToHeat= AbsorbedPower*F # in J/s\n",
+      "print \"The amount of energy converted to heat per second = %0.2e in J/s \" %EnergyConToHeat\n",
+      "A= (AbsorbedPower-EnergyConToHeat)/(e*hv1) \n",
+      "print \"The number of photon per sec given off from recombination events = %0.2e photons/s \" %A"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The absorbed power = 9.0e-03 watt or J/s \n",
+        "The amount of energy converted to heat per second = 2.57e-03 in J/s \n",
+        "The number of photon per sec given off from recombination events = 2.81e+16 photons/s \n"
+       ]
+      }
+     ],
+     "prompt_number": 47
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.24 - Page No : 3-54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "Mu_p = 500 # in cm**2/v-s\n",
+      "kT = 0.0259 \n",
+      "Toh_p = 10**-10 # in sec\n",
+      "p_o = 10**17 # in cm**-3\n",
+      "q= 1.6*10**-19 # in C\n",
+      "A=0.5 # in square meter\n",
+      "del_p = 5 * 10**16 # in cm**-3\n",
+      "n_i= 1.5*10**10 # in cm**-3    \n",
+      "D_p = kT * Mu_p # in cm/s\n",
+      "L_p = sqrt(D_p * Toh_p) # in cm\n",
+      "x = 10**-5 # in cm\n",
+      "p = p_o+del_p* exp(x/L_p) # in cm**-3\n",
+      "# p= n_i*%e**(Eip)/kT where Eip=E_i-F_p\n",
+      "Eip= log(p/n_i)*kT # in eV\n",
+      "Ecp= 1.1/2-Eip # value of E_c-E_p in eV\n",
+      "Ip= q*A*D_p/L_p*del_p*exp(x/L_p) # in A\n",
+      "print \"The hole current = %0.2e A \" %Ip\n",
+      "Qp= q*A*del_p*L_p # in C\n",
+      "print \"The value of Qp = %0.2e C \" %Qp\n",
+      "\n",
+      "# Note: There is a calculation error or miss print to evalaute the value of hole current but they putted correct \n",
+      "# value of it to evaluate the value of Qp.Hence the value of hole current in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The hole current = 1.90e+03 A \n",
+        "The value of Qp = 1.44e-07 C \n"
+       ]
+      }
+     ],
+     "prompt_number": 48
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_4_1.ipynb b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_4_1.ipynb
new file mode 100644
index 00000000..c6818489
--- /dev/null
+++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_4_1.ipynb
@@ -0,0 +1,948 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:05740426f73b9605877bfd0b7708432621382112013a8d4f8751c47309a06737"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter : 4 - Junction Properties"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  4.16.1 - Page No : 4-69"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from math import log\n",
+      "# Given data\n",
+      "t = 4.4 * 10**22 # total number of Ge atoms/cm**3\n",
+      "n = 1 * 10**8 # number of impurity atoms\n",
+      "N_A = t/n # in atoms/cm**3\n",
+      "N_A = N_A * 10**6 # in atoms/m**3\n",
+      "N_D = N_A * 10**3 # in atoms/m**3\n",
+      "n_i = 2.5 * 10**13 # in atoms/cm**3\n",
+      "n_i = n_i * 10**6 # in atoms/m**3\n",
+      "V_T = 26 #in mV\n",
+      "V_T= V_T*10**-3 # in V\n",
+      "V_J = V_T * log((N_A * N_D)/(n_i)**2) # in V\n",
+      "print \"Part (a) : The contact potential = %0.3f V \" %V_J\n",
+      "# Part (b)\n",
+      "t = 5* 10**22 # total number of Si atoms/cm**3\n",
+      "N_A = t/n # in atoms/cm**3\n",
+      "N_A = N_A * 10**6 # in atoms/m**3\n",
+      "N_D = N_A * 10**3 # in atoms/m**3\n",
+      "n_i = 1.5 * 10**10 # in atoms/cm**3\n",
+      "n_i = n_i * 10**6 # in atoms/m**3\n",
+      "V_T = 26 #in mV\n",
+      "V_T= V_T*10**-3 # in V\n",
+      "V_J = V_T * log((N_A * N_D)/(n_i)**2) # in V\n",
+      "print \"Part (b) : The contact potential  = %0.3f V \" %V_J"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part (a) : The contact potential = 0.329 V \n",
+        "Part (b) : The contact potential  = 0.721 V \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  4.16.2 - Page No : 4-71"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "V_T = 26 # in mV\n",
+      "V_T=V_T*10**-3 # in V\n",
+      "n_i = 2.5 * 10**13 \n",
+      "Sigma_p = 1 \n",
+      "Sigma_n = 1 \n",
+      "Mu_n = 3800 \n",
+      "q = 1.6 * 10**-19 # in C\n",
+      "Mu_p = 1800 \n",
+      "N_A = Sigma_p/(2* q * Mu_p) # in /cm**3\n",
+      "N_D = Sigma_n /(q * Mu_n) # in /cm**3\n",
+      "V_J = V_T * log((N_A * N_D)/(n_i)**2) # in V\n",
+      "print \"For Ge the height of the energy barrier = %0.2f V \" %V_J\n",
+      "# For Si p-n juction\n",
+      "n_i = 1.5 * 10**10 \n",
+      "Mu_n = 1300 \n",
+      "Mu_p = 500 \n",
+      "N_A = Sigma_p/(2* q * Mu_p) # in /cm**3\n",
+      "N_D = Sigma_n /(q * Mu_n) # in /cm**3\n",
+      "V_J = V_T * log((N_A * N_D)/(n_i)**2) # in V\n",
+      "print \"For Si p-n junction  the height of the energy barrier = %0.3f V \" %V_J"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For Ge the height of the energy barrier = 0.22 V \n",
+        "For Si p-n junction  the height of the energy barrier = 0.666 V \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.16.3 - Page No : 4-72"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import exp\n",
+      "# Given data\n",
+      "Eta = 1 \n",
+      "V_T = 26 # in mV\n",
+      "V_T= V_T*10**-3 # in V\n",
+      "# I = I_o * (exp(V/(Eta*V_T)) - 1) and I = -(0.9) * I_o \n",
+      "V= log(1-0.9)*V_T # in V\n",
+      "print \"The voltage = %0.2f volts \" %V\n",
+      "# Part (ii)\n",
+      "V1=0.05 # in V\n",
+      "V2= -0.05 # in V\n",
+      "ratio= (exp(V1/(Eta*V_T))-1)/(exp(V2/(Eta*V_T))-1)\n",
+      "print \"The ratio of the current for a forward bias to reverse bias = %0.2f\" %ratio\n",
+      "# Part (iii)\n",
+      "Io= 10 # in \u00b5A\n",
+      "Io=Io*10**-3 # in mA\n",
+      "#For \n",
+      "V=0.1 # in V\n",
+      "I = Io * (exp(V/(Eta*V_T)) - 1) # in mA\n",
+      "print \"For v=0.1 V , the value of I = %0.3f mA \" %I\n",
+      "#For \n",
+      "V=0.2 # in V\n",
+      "I = Io * (exp(V/(Eta*V_T)) - 1) # in mA\n",
+      "print \"For v=0.2 V , the value of I = %0.1f mA \" %I\n",
+      "#For \n",
+      "V=0.3 # in V\n",
+      "I = Io * (exp(V/(Eta*V_T)) - 1) # in mA\n",
+      "print \"For v=0.3 V , the value of I = %0.2f A \" %(I*10**-3)\n",
+      "print \"From three value of I, for small rise in forward voltage, the diode current increase rapidly\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The voltage = -0.06 volts \n",
+        "The ratio of the current for a forward bias to reverse bias = -6.84\n",
+        "For v=0.1 V , the value of I = 0.458 mA \n",
+        "For v=0.2 V , the value of I = 21.9 mA \n",
+        "For v=0.3 V , the value of I = 1.03 A \n",
+        "From three value of I, for small rise in forward voltage, the diode current increase rapidly\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.16.4 - Page No : 4-73"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "# Part (i)\n",
+      "T1= 25 # in \u00b0C\n",
+      "T2= 80 # in \u00b0C\n",
+      "# Formula Io2= Io1*2**((T2-T1)/10)\n",
+      "AntiFactor= 2**((T2-T1)/10) \n",
+      "print \"Anticipated factor for Ge = %0.f \" %AntiFactor\n",
+      "# Part (ii)\n",
+      "T1= 25 # in \u00b0C\n",
+      "T2= 150 # in \u00b0C\n",
+      "AntiFactor= 2**((T2-T1)/10) \n",
+      "print \"Anticipated factor for Si = %0.f \" %AntiFactor"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Anticipated factor for Ge = 45 \n",
+        "Anticipated factor for Si = 5793 \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.16.5 - Page No : 4-74"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "I=5 # in \u00b5A\n",
+      "V=10 # in V\n",
+      "T1= 0.11 # in \u00b0C**-1\n",
+      "T2= 0.07 # in \u00b0C**-1\n",
+      "# Io+I_R=I                            (i)\n",
+      "# dI_by_dT= dIo_by_dT      (ii)\n",
+      "# 1/Io*dIo_by_dT = T1 and 1/I*dI_by_dT = T2, So\n",
+      "Io= T2*I/T1 # in \u00b5A\n",
+      "I_R= I-Io # in \u00b5A\n",
+      "R= V/I_R # in M\u03a9\n",
+      "print \"The leakage resistance = %0.1f M\u03a9 \" %R"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The leakage resistance = 5.5 M\u03a9 \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.16.6 - Page No : 4-75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "Eta = 1 \n",
+      "T = 125 # in \u00b0C\n",
+      "T = T + 273 # in K\n",
+      "V_T = 8.62 * 10**-5 * 398 # in V\n",
+      "I_o = 30 # in \u00b5A\n",
+      "I_o= I_o*10**-6 # in A\n",
+      "v = 0.2 # in V\n",
+      "r_f = (Eta * V_T)/(I_o * exp(v/(Eta* V_T))) # in ohm\n",
+      "print \"The dynamic resistance in the forward direction = %0.2f ohm \" %r_f\n",
+      "r_r = (Eta * V_T)/(I_o * exp(-v/(Eta* V_T))) # in ohm\n",
+      "print \"The dynamic resistance in the reverse direction = %0.2f kohm \" %(r_r*10**-3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The dynamic resistance in the forward direction = 3.36 ohm \n",
+        "The dynamic resistance in the reverse direction = 389.08 kohm \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  4.16.7 - Page No : 4-76"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# Given data\n",
+      "epsilon = 16/(36 * pi * 10**11) # in F/cm\n",
+      "A = 1 * 10**-2 \n",
+      "W = 2 * 10**-4 \n",
+      "C_T = (epsilon * A)/W # in F\n",
+      "print \"The barrier capacitance = %0.2f pF \" %(C_T*10**12)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The barrier capacitance = 70.74 pF \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.16.8 - Page No : 4-76"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "# Given data\n",
+      "A = 1 # in mm**2\n",
+      "A = A * 10**-6 # in m**2\n",
+      "N_A = 3 * 10**20 # in atoms/m**3\n",
+      "q = 1.6 *10**-19 # in C\n",
+      "V_o = 0.2 # in V\n",
+      "epsilon_r=16 \n",
+      "epsilon_o= 8.854*10**-12 # in F/m\n",
+      "epsilon=epsilon_r*epsilon_o \n",
+      "# Part (a)\n",
+      "V=-10 # in V\n",
+      "# V_o - V = 1/2*((q * N_A )/epsilon) * W**2\n",
+      "W = sqrt(((V_o - V) * 2 * epsilon)/(q * N_A)) # m\n",
+      "C_T1 = (epsilon * A)/W # in F\n",
+      "print \"The width of the depletion layer  for an applied reverse voltage of 10V     = %0.2f \u00b5m \" %(W*10**6)\n",
+      "# Part (b)\n",
+      "V=-0.1 # in V\n",
+      "W = sqrt(((V_o - V) * 2 * epsilon)/(q * N_A)) # m\n",
+      "C_T2 = (epsilon * A)/W # in F\n",
+      "print \"The width of the depletion layer  for an applied reverse voltage of 0.1V    = %0.2f \u00b5m \" %(W*10**6)\n",
+      "# Part (c)\n",
+      "V=0.1 # in V\n",
+      "W = sqrt(((V_o - V) * 2 * epsilon)/(q * N_A)) # m\n",
+      "print \"The width of the depletion layer  for an applied for a forward bias of 0.1V = %0.3f \u00b5m \" %(W*10**6)\n",
+      "# Part (d)\n",
+      "print \"The space charge capacitance for an applied reverse voltage of 10V          = %0.2f pF \" %(C_T1*10**12)\n",
+      "print \"The space charge capacitance for an applied reverse voltage of 0.1V         = %0.2f pF \" %(C_T2*10**12)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The width of the depletion layer  for an applied reverse voltage of 10V     = 7.76 \u00b5m \n",
+        "The width of the depletion layer  for an applied reverse voltage of 0.1V    = 1.33 \u00b5m \n",
+        "The width of the depletion layer  for an applied for a forward bias of 0.1V = 0.768 \u00b5m \n",
+        "The space charge capacitance for an applied reverse voltage of 10V          = 18.26 pF \n",
+        "The space charge capacitance for an applied reverse voltage of 0.1V         = 106.46 pF \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  4.16.9 - Page No : 4-78"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "I_o = 1.8 * 10**-9 # A\n",
+      "v = 0.6 # in V\n",
+      "Eta = 2 \n",
+      "V_T = 26 # in mV\n",
+      "V_T=V_T*10**-3 # in V\n",
+      "I = I_o *(exp(v/(Eta * V_T))) # in A\n",
+      "print \"The current in the junction = %0.3f mA \" %(I*10**3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current in the junction = 0.185 mA \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  4.16.10 - Page No : 4-78"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "I_o = 2.4 * 10**-14 \n",
+      "I = 1.5 # in mA\n",
+      "I=I*10**-3 # in A\n",
+      "Eta = 1\n",
+      "V_T = 26 # in mV\n",
+      "V_T= V_T*10**-3 # in V\n",
+      "v =log((I + I_o)/I_o) * V_T # in V\n",
+      "print \"The forward biasing voltage across the junction = %0.4f V \" %v"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The forward biasing voltage across the junction = 0.6463 V \n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  4.16.11 - Page No : 4-79"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "I_o = 10 # in nA\n",
+      "# I = I_o * ((e**(v/(Eta * V_T))) - 1)\n",
+      "# e**(v/(Eta * V_T)<< 1, so neglecting it\n",
+      "I = I_o * (-1) # in nA\n",
+      "print \"The Diode current = %0.f nA \" %I"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Diode current = -10 nA \n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  4.16.12 - Page No : 4-79"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "R = 4.5 # in ohm\n",
+      "I = 44.4 # in mA\n",
+      "I=I*10**-3 # in A\n",
+      "V = R * I # in V\n",
+      "Eta = 1 \n",
+      "V_T = 26 #in mV\n",
+      "V_T=V_T*10**-3 # in V\n",
+      "I_o = I/((exp(V/(Eta * V_T))) -1) # in A\n",
+      "# At\n",
+      "V = 0.1 # in V\n",
+      "r_f = (Eta * V_T)/(I_o * ((exp(V/(Eta * V_T)))-1)) # in ohm\n",
+      "print \"The diode dynamic resistance = %0.2f \u03a9 \" %r_f"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The diode dynamic resistance = 27.78 \u03a9 \n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  4.16.13 - Page No : 4-80"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%matplotlib inline\n",
+      "import matplotlib.pyplot as plt\n",
+      "import numpy as np\n",
+      "# Given data\n",
+      "V_D = 10 # in V\n",
+      "# V_S = i*R_L + V_D\n",
+      "V_S = V_D # in V (i * R_L = 0)\n",
+      "print \"when diode is OFF, the voltage = %0.f volts \" %V_S\n",
+      "R_L = 250 # in ohm\n",
+      "I = V_S/R_L # in A\n",
+      "print \"when diode is ON, the current = %0.f mA \" %(I*10**3)\n",
+      "V_D= np.arange(0,10,0.1) # in V\n",
+      "I= (V_S-V_D)/R_L*1000 # in mA\n",
+      "plt.plot(V_D,I)\n",
+      "plt.xlabel('V_D in volts') \n",
+      "plt.ylabel('Current in mA')\n",
+      "plt.title('DC load line')\n",
+      "plt.axis([0, 12, 0, 50])\n",
+      "plt.show()\n",
+      "print \"DC load line shown in figure\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when diode is OFF, the voltage = 10 volts \n",
+        "when diode is ON, the current = 40 mA \n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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MjFgxEJGyabVaaLXaBu9f5z2Bbt264cKFC0b/xX7t2jU0bdoUjo6OuHPnDoYPH45Fixbh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\nCsYkQESkYP8HqYZH+eX0mvAAAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7f0e680b52d0>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "DC load line shown in figure\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  4.16.14 - Page No : 4-81"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "V = 0.25 # in V\n",
+      "I_o = 1.2 # in \u00b5A\n",
+      "I_o = I_o * 10**-6 # in A\n",
+      "V_T = 26 # in mV\n",
+      "V_T = V_T * 10**-3 # in V\n",
+      "Eta = 1 \n",
+      "r = (Eta * V_T)/(I_o * (exp(V/(Eta * V_T)))) # in ohm\n",
+      "print \"The ac resistance of the diode = %0.3f ohm \" %r"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The ac resistance of the diode = 1.445 ohm \n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  4.16.15 - Page No : 4-82"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "t = 4.4 * 10**22 # in total number of atoms/cm**3\n",
+      "n = 1 * 10**8 # number of impurity\n",
+      "N_A = t/n # in atoms/cm**3\n",
+      "N_A = N_A * 10**6 # in atoms/m**3\n",
+      "N_D = N_A * 10**3 # in atoms/m**3\n",
+      "V_T = 26 # in mV\n",
+      "V_T = V_T * 10**-3 # in V\n",
+      "n_i = 2.5 * 10**19 # in /cm**3\n",
+      "V_J = V_T * log((N_A * N_D)/(n_i)**2) # in V\n",
+      "print \"The junction potential = %0.3f V \"%V_J"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The junction potential = 0.329 V \n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  4.16.16 - Page No : 4-82"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "Eta = 1 \n",
+      "I_o = 30 # in MuA\n",
+      "I_o = I_o * 10**-6 # in A\n",
+      "v = 0.2 # in V\n",
+      "K = 1.381 * 10**-23 # in J/degree K \n",
+      "T = 125 # in \u00b0C\n",
+      "T = T + 273 # in K\n",
+      "q = 1.6 * 10**-19 # in C\n",
+      "V_T = (K*T)/q # in V\n",
+      "r_f = (Eta * V_T)/(I_o * (exp(v/(Eta * V_T)))) # in ohm\n",
+      "print \"The forward dynamic resistance = %0.3f ohm \" %r_f\n",
+      "r_f1 = (Eta * V_T)/(I_o * (exp(-(v)/(Eta * V_T)))) # in ohm\n",
+      "print \"The Reverse dynamic resistance = %0.2f k\u03a9 \" %(r_f1*10**-3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The forward dynamic resistance = 3.391 ohm \n",
+        "The Reverse dynamic resistance = 386.64 k\u03a9 \n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  4.16.17 - Page No : 4-83"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "q = 1.6 * 10**-19 # in C\n",
+      "N_A = 3 * 10**20 # in /m**3\n",
+      "A = 1 # in \u00b5m**2\n",
+      "A = A * 10**-6 # in m**2\n",
+      "V = -10 # in V\n",
+      "V_J = 0.25 # in V\n",
+      "V_B = V_J - V # in V\n",
+      "epsilon_o = 8.854 # in pF/m\n",
+      "epsilon_o = epsilon_o * 10**-12 # in F/m\n",
+      "epsilon_r = 16 \n",
+      "epsilon = epsilon_o * epsilon_r \n",
+      "W = sqrt((V_B * 2 * epsilon)/(q * N_A)) # in m \n",
+      "print \"The width of depletion layer = %0.2f \u00b5m \" %(W*10**6)\n",
+      "C_T = (epsilon * A)/W # in pF\n",
+      "print \"The space charge capacitance = %0.4f pF \" %(C_T*10**12)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The width of depletion layer = 7.78 \u00b5m \n",
+        "The space charge capacitance = 18.2127 pF \n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  4.16.18 - Page No : 4-84"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "W = 2 * 10**-4 # in cm\n",
+      "W = W * 10**-2 # in m\n",
+      "A = 1 # in mm**2\n",
+      "A = A * 10**-6 # in m**2\n",
+      "epsilon_r = 16 \n",
+      "epsilon_o = 8.854 * 10**-12 # in F/m\n",
+      "epsilon = epsilon_r * epsilon_o \n",
+      "C_T = (epsilon * A)/W # in F\n",
+      "print \"The barrier capacitance = %0.3f pF \" %(C_T*10**12)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The barrier capacitance = 70.832 pF \n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  4.16.19 - Page No : 4-85"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "C_T = 100 # in pF\n",
+      "C_T=C_T*10**-12 # in F\n",
+      "epsilon_r = 12 \n",
+      "epsilon_o = 8.854 * 10**-12 # in F/m\n",
+      "epsilon = epsilon_r * epsilon_o \n",
+      "Rho_p = 5 # in ohm-cm\n",
+      "Rho_p = Rho_p * 10**-2 # in ohm-m\n",
+      "V_j = 0.5 # in V\n",
+      "V = -4.5 # in V\n",
+      "Mu_p = 500 # in cm**2\n",
+      "Mu_p = Mu_p * 10**-4 # in m**2\n",
+      "Sigma_p = 1/Rho_p # in per ohm-m\n",
+      "qN_A = Sigma_p/ Mu_p \n",
+      "V_B = V_j - V \n",
+      "W = sqrt((V_B * 2 * epsilon)/qN_A) # in m\n",
+      "#C_T = (epsilon * A)/W \n",
+      "A = (C_T * W)/ epsilon # in m\n",
+      "D = sqrt(A * (4/pi)) # in m\n",
+      "D = D * 10**3 # in mm\n",
+      "print \"The value of diameter = %0.3f mm \" %D\n",
+      " "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of diameter = 1.398 mm \n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  4.16.20 - Page No : 4-20"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "q = 1.6 * 10**-19 # in C\n",
+      "Mu_p = 500 # in cm**2/V-sec\n",
+      "Rho_p = 3.5 # in ohm-cm\n",
+      "Mu_n = 1500 # in cm**2/V-sec\n",
+      "Rho_n = 10 # in ohm-cm\n",
+      "N_A = 1/(Rho_p * Mu_p * q) # in /cm**3\n",
+      "N_D = 1/(Rho_n * Mu_n * q) # in /cm**3\n",
+      "V_J = 0.56 # in V\n",
+      "n_i = 1.5 * 10**10 # in /cm**3\n",
+      "V_T = V_J/log((N_A * N_D)/(n_i)**2) # in V\n",
+      "# V_T = T/11600\n",
+      "T = V_T * 11600 # in K\n",
+      "T = T /19.78 # in \u00b0C   ( 1 degree K = 19.78 degree C)\n",
+      "print \"The Temperature of junction = %0.3f \u00b0C \" %T"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Temperature of junction = 14.524 \u00b0C \n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  4.16.21 - Page No : 4-88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "V_T = 26 # in mV\n",
+      "V_T = V_T * 10**-3 # in V\n",
+      "Eta = 1 \n",
+      "# I = -90% for Io, so\n",
+      "IbyIo= 0.1 \n",
+      "# I  = I_o * ((e**(v/(Eta * V_T)))-1)\n",
+      "V = log(IbyIo) * V_T # in V\n",
+      "print \"The reverse bias voltage = %0.5f volts \" %V"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The reverse bias voltage = -0.05987 volts \n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  4.16.22 - Page No : 4-89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "R = 5 # in ohm\n",
+      "I = 50 # in mA\n",
+      "I=I*10**-3 # in A\n",
+      "V = R * I # in V\n",
+      "Eta = 1 \n",
+      "V_T = 26 # in mV\n",
+      "V_T=V_T*10**-3 # in V\n",
+      "I_o = I/((exp(V/(Eta * V_T))) - 1) # in A\n",
+      "print \"Reverse saturation current = %0.2f \u00b5A \" %(I_o*10**6)\n",
+      "v1 = 0.2 # in V\n",
+      "r = (Eta * V_T)/(I_o * (exp(v1/(Eta * V_T)))) # in ohm\n",
+      "print \"Dynamic resistance of the diode = %0.3f \u03a9 \" %r"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reverse saturation current = 3.33 \u00b5A \n",
+        "Dynamic resistance of the diode = 3.558 \u03a9 \n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_5_1.ipynb b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_5_1.ipynb
new file mode 100644
index 00000000..ded28641
--- /dev/null
+++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_5_1.ipynb
@@ -0,0 +1,834 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:f580dc14cd164012c93f758d593e5fbb20692db2a4830c462cc2ecc139c62a4a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter : 5 - Bipolar Junction Transistors (BJTs)"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  5.9.1 - Page No : 5-22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# Given data\n",
+      "V_EE = 8 # in V\n",
+      "V_BE = 0.7 # in V\n",
+      "R_E = 1.5 # in k ohm\n",
+      "I_E = (V_EE - V_BE)/R_E # in mA\n",
+      "I_C = I_E # in mA\n",
+      "print \"The value of I_C = %0.2f mA \" %I_C\n",
+      "V_CC = 18 # in V\n",
+      "R_C = 1.2 # in k\u03a9\n",
+      "V_CB = V_CC - (I_C * R_C) # in V\n",
+      "print \"The value of V_CB = %0.2f V \" %V_CB"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_C = 4.87 mA \n",
+        "The value of V_CB = 12.16 V \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  5.9.2 - Page No : 5-23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "alpha = 0.9 \n",
+      "I_E = 1 # mA\n",
+      "I_C = alpha * I_E # in mA\n",
+      "I_B = I_E - I_C # in mA\n",
+      "print \"The value of base current = %0.1f mA \" %I_B"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of base current = 0.1 mA \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  5.11.1 - Page No : 5-41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "bita = 50 \n",
+      "I_B= 20 # in \u00b5A\n",
+      "I_B=I_B*10**-6 # in A\n",
+      "I_C= bita*I_B # in A\n",
+      "I_E= I_C+I_B # in A\n",
+      "I_E = I_E * 10**3 # in mA\n",
+      "print \"The Emitter current = %0.2f mA \" %I_E"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Emitter current = 1.02 mA \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  5.11.1(a) - Page No : 5-42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "beta_dc = 90 \n",
+      "I_C = 15 # in mA\n",
+      "I_C = I_C * 10**-3 # in A\n",
+      "I_B = I_C/beta_dc # in A\n",
+      "print \"The base current = %0.2f \u00b5A \" %(I_B*10**6)\n",
+      "I_E = I_C + I_B # in A\n",
+      "I_E = I_E * 10**3 # in mA\n",
+      "print \"The Emitter current = %0.3f mA \" %I_E\n",
+      "alpha_dc = beta_dc/(1+beta_dc) \n",
+      "print \"The value of alpha_dc = %0.3f\" %alpha_dc"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The base current = 166.67 \u00b5A \n",
+        "The Emitter current = 15.167 mA \n",
+        "The value of alpha_dc = 0.989\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  5.11.3 - Page No : 5-43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "del_ic = 1.8 # in mA\n",
+      "del_ie = 1.89 # in mA\n",
+      "alpha = del_ic / del_ie \n",
+      "bita = alpha/(1 - alpha) \n",
+      "del_ib = del_ic/bita # in mA\n",
+      "del_ib = del_ib * 10**3 # in \u00b5A\n",
+      "print \"The change in I_B = %0.f \u00b5A \"  %del_ib"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The change in I_B = 90 \u00b5A \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 5.11.4 - Page No : 5-43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "V_CC = 10 # in V\n",
+      "R_C = 3 # in k \u03a9\n",
+      "R_C= R_C*10**3 # in \u03a9\n",
+      "bita = 100 \n",
+      "I_CO = 20 # in nA\n",
+      "I_CO = I_CO * 10**-9 # in A\n",
+      "V_BB = 5 # in V\n",
+      "R_B = 200 # in k\u03a9\n",
+      "R_B= R_B*10**3 # in \u03a9\n",
+      "V_BE = 0.7 # in V\n",
+      "# Applying KVL to the base circuit, V_BB= I_B*R_B+V_BE\n",
+      "I_B = (V_BB - V_BE)/R_B # in A\n",
+      "print \"The base current = %0.1f \u00b5A \" %(I_B*10**6)\n",
+      "I_C = (bita * I_B) + I_CO # in A\n",
+      "print \"The collector current = %0.5f mA \" %(I_C*10**3)\n",
+      "I_E = I_C + I_B # in A\n",
+      "print \"Emitter current = %0.5f mA \" %(I_E*10**3)\n",
+      "V_CE = V_CC - (I_C * R_C) # in V\n",
+      "print \"Collector emitter voltage = %0.4f V \" %(V_CE)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The base current = 21.5 \u00b5A \n",
+        "The collector current = 2.15002 mA \n",
+        "Emitter current = 2.17152 mA \n",
+        "Collector emitter voltage = 3.5499 V \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.11.5 - Page No : 5-44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "bita = 100 \n",
+      "I_CBO = 4 # in \u00b5A\n",
+      "I_B = 40 # in \u00b5A\n",
+      "I_C = (bita * I_B) + ((1+bita) * I_CBO) # in \u00b5A\n",
+      "I_C = I_C * 10**-3 # in msA\n",
+      "print \"The collector current = %0.3f mA \" %I_C"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The collector current = 4.404 mA \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  5.11.6 - Page No : 5-45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%matplotlib inline\n",
+      "import matplotlib.pyplot as plt\n",
+      "import numpy as np\n",
+      "from __future__ import division\n",
+      "# Given data\n",
+      "del_IC = 1 * 10**-3 # in A\n",
+      "del_IB = 10 * 10**-6 # in A\n",
+      "CurrentGain= del_IC/del_IB \n",
+      "print \"The current gain = %0.f\" %CurrentGain\n",
+      "del_IC= del_IC*10**3 # in mA\n",
+      "del_IB= del_IB*10**6 # in \u00b5A\n",
+      "I_B=np.arange(0,50,0.1) # in \u00b5A\n",
+      "I_C= I_B/del_IB+del_IC # in mA\n",
+      "plt.plot(I_B,I_C)\n",
+      "plt.xlabel('Base current in micro A')\n",
+      "plt.ylabel('Collector current in mA')\n",
+      "plt.title('Transfer Characteristics')\n",
+      "plt.axis([0, 60, 0, 7])\n",
+      "plt.show()\n",
+      "print \"Transfer Characteristics is shown in figure\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current gain = 100\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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5IrJnNt/GYV+eiMiGw559eSKi39hkG4d9eSKi+mwq7NmXJyJqnE20cdiXJyJq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\nEdkBhj0RkR1g2BMR2YH/A2IOTkNSzG6GAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7f8cdc73de90>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Transfer Characteristics is shown in figure\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  5.14.1 - Page No : 5-51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "I_CBO = 3 #in \u00b5A\n",
+      "I_CBO= I_CBO*10**-3 # in mA \n",
+      "I_C= 15 # in mA\n",
+      "# But it is given that I_C= 99.5% of I_E, SO\n",
+      "I_E= I_C/99.5*100 # in mA\n",
+      "alpha_dc= I_C/I_E \n",
+      "print \"The value of alpha_dc = %0.3f\" %alpha_dc\n",
+      "print \"The value of I_E = %0.2f mA \" %I_E"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of alpha_dc = 0.995\n",
+        "The value of I_E = 15.08 mA \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 5.14.2 - Page No : 5-52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given data\n",
+      "alpha_dc = 0.99 \n",
+      "I_CBO = 10 # in \u00b5A\n",
+      "I_CBO= I_CBO*10**-6 # in A\n",
+      "I_E = 10 # in mA\n",
+      "I_E= I_E*10**-3 # in A\n",
+      "I_C = (alpha_dc * I_E) + I_CBO # in A\n",
+      "print \"The value of I_C = %0.2f mA \" %(I_C*10**3)\n",
+      "I_B = I_E - I_C # in A\n",
+      "I_B = I_B * 10**6 # in \u00b5A\n",
+      "print \"The value of I_B = %0.f \u00b5A \" %I_B"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_C = 9.91 mA \n",
+        "The value of I_B = 90 \u00b5A \n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  5.14.3 - Page No : 5-52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "alpha_dc = 0.99 \n",
+      "I_C = 6 # in mA\n",
+      "I_C= I_C*10**-3 # in A\n",
+      "I_CBO = 15 # in \u00b5A\n",
+      "I_CBO= I_CBO*10**-6 # in A\n",
+      "I_E = (I_C - I_CBO)/alpha_dc # in A\n",
+      "I_B = I_E - I_C # in A \n",
+      "print \"The value of I_B = %0.f \u00b5A \" %(I_B*10**6)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_B = 45 \u00b5A \n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 5.14.5 - Page No : 5-53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "alpha_dc = 0.98 \n",
+      "I_CBO = 12 # in \u00b5A\n",
+      "I_CBO = I_CBO * 10**-6 # in A\n",
+      "I_B = 120 # in \u00b5A\n",
+      "I_B = I_B * 10**-6 # in A\n",
+      "beta_dc = alpha_dc/(1-alpha_dc) \n",
+      "I_E = ((1 + beta_dc) * I_B) + ((1 + beta_dc) * I_CBO) #in A\n",
+      "I_E = I_E * 10**3 # in mA\n",
+      "print \"The value of I_E = %0.1f mA \" %I_E"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_E = 6.6 mA \n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 5.14.6 - Page No : 5-53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "bita= 100 \n",
+      "V_BEsat= 0.8 # in V\n",
+      "V_CEsat= 0.2 # in V\n",
+      "V_BEact= 0.7 # in V\n",
+      "V_CC = 10 # in V\n",
+      "V_BB=5 # in V\n",
+      "R_E = 2 # in k\u03a9\n",
+      "R_C = 3 # in k\u03a9\n",
+      "R_B= 50 # in k\u03a9\n",
+      "# Applying KVL to collector loop\n",
+      "# V_CC= I_Csat*R_C +V_CEsat +I_E*R_E and I_E= I_Csat+I_B, So\n",
+      "#I_B= ((V_CC-V_CEsat)-(R_C+R_E)*I_Csat)/R_E            (i)\n",
+      "# Applying KVL to base loop\n",
+      "# V_BB-I_B*R_B -V_BEsat-I_E*R_E =0 and I_E= I_Csat+I_B, So\n",
+      "#V_BB-V_BEsat= R_E*I_Csat + (R_B+R_E)*I_B              (ii)\n",
+      "# From eq (i) and (ii)\n",
+      "I_B = ((V_BB-V_BEsat)*5- (V_CC-V_CEsat)*2) / ((R_B+R_E)*5 - R_E*2)  # in mA\n",
+      "I_Csat= ((V_CC-V_CEsat)-R_E*I_B)/(R_C+R_E) # in mA\n",
+      "I_Bmin= I_Csat/bita # in mA\n",
+      "if I_B<I_Bmin :\n",
+      "    print \"Since the value of I_B (\",round(I_B*10**3,2),\"\u00b5A) is less than the value of I_Bmin (\",round(I_Bmin*10**3,1),\"\u00b5A)\" \n",
+      "    print \"So the transistor is not in the saturation region. But it is conducting hence it can not be in cutoff.\"\n",
+      "    print \"Therefore the transistor is in the active region\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Since the value of I_B ( 5.47 \u00b5A) is less than the value of I_Bmin ( 19.6 \u00b5A)\n",
+        "So the transistor is not in the saturation region. But it is conducting hence it can not be in cutoff.\n",
+        "Therefore the transistor is in the active region\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.14.7 - Page No : 5-55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "bita= 100 \n",
+      "V_BEsat= 0.8 # in V\n",
+      "V_CEsat= 0.2 # in V\n",
+      "V_BEact= 0.7 # in V\n",
+      "V_CC = 10 # in V\n",
+      "V_BB=5 # in V\n",
+      "R_E = 2 # in k\u03a9\n",
+      "R_C = 3 # in k\u03a9\n",
+      "R_B= 50 # in k\u03a9\n",
+      "# Applying KVL to input loop\n",
+      "# V_BB= I_B*R_B+(1+bita)*I_B*R_E+V_BEact or \n",
+      "I_B= (V_BB-V_BEact)/(R_B+(1+bita)*R_E) # in mA\n",
+      "I_C= bita*I_B # in mA\n",
+      "# Applying KVL to collector circuit\n",
+      "# V_CC= I_Csat*R_C +V_CEsat +(I_C+I_B)*R_E\n",
+      "V_CEact= V_CC-I_B*R_E-I_C*(R_C+R_E) # in V\n",
+      "print \"The value of I_B = %0.f \u00b5A \" %(I_B*10**3)\n",
+      "print \"The value of I_C = %0.1f mA \" %I_C\n",
+      "print \"The value of V_CE = %0.3f volts \" %V_CEact"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_B = 17 \u00b5A \n",
+        "The value of I_C = 1.7 mA \n",
+        "The value of V_CE = 1.434 volts \n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.14.8 - Page No : 5-55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "bita = 100 \n",
+      "V_CEsat = 0.2 # in V\n",
+      "R_B = 150 # in kohm\n",
+      "R_C = 2 # in kohm\n",
+      "V_CC  = 10 # in V\n",
+      "V_BEsat = 0.8 # in V\n",
+      "I_B = (V_CC - V_BEsat)/R_B # in mA\n",
+      "I_C = (V_CC - V_CEsat)/R_C # in mA\n",
+      "I_Bmin = I_C/bita # in mA\n",
+      "if I_B>I_Bmin :\n",
+      "    print \"Since the value of I_B (\",round(I_B*10**3,2),\"\u00b5A) is greater than the value of I_Bmin (\",int(I_Bmin*10**3),\"\u00b5A)\" \n",
+      "    print \"So the transistor is in the saturation region.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Since the value of I_B ( 61.33 \u00b5A) is greater than the value of I_Bmin ( 49 \u00b5A)\n",
+        "So the transistor is in the saturation region.\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.14.9 - Page No : 5-57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "bita = 100 \n",
+      "V_CE = 0.2 #in V\n",
+      "V_BE = 0.8 # in V\n",
+      "R_C= 500 # in \u03a9\n",
+      "R_B= 44*10**3 # in \u03a9\n",
+      "R_E= 1*10**3 # in \u03a9\n",
+      "V_CC= 15 # in V\n",
+      "V_GE= -15 # in V\n",
+      "# Applying KVL to collector circuit\n",
+      "# V_CC-V_GE - I_Csat*R_C-V_CE-I_E*R_E=0, but I_Csat= bita*I_Bmin and I_E= 1+bita\n",
+      "I_Bmin= (V_CC-V_GE-V_CE)/(R_C*bita+(1+bita)*R_E) # in A\n",
+      "# Applying KVL to the base emitter circuit\n",
+      "# V_BB-I_Bmin*R_B-V_BE-I_E*R_E + V_CC=0\n",
+      "V_BB= I_Bmin*R_B + V_BE + (1+bita)*I_Bmin*R_E-V_CC # in V\n",
+      "print \"The value of I_B(min) = %0.3f mA \" %(I_Bmin*10**3)\n",
+      "print \"The value of V_BB = %0.1f volts \" %V_BB"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_B(min) = 0.197 mA \n",
+        "The value of V_BB = 14.4 volts \n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  5.14.10 - Page No : 5-58"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "V_ECsat= 0.2 # in V\n",
+      "V_CC= 10 # in V\n",
+      "V_EBsat= 0.8 # in V\n",
+      "\n",
+      "# Part (i)\n",
+      "bita= 100 \n",
+      "R_B= 220 # in k\u03a9\n",
+      "# Applying KVL to collector circuit, V_CC= V_EC+ICRC\n",
+      "ICRC= V_CC-V_ECsat # in V\n",
+      "# Applying KVL to input loop, V_CC= V_EBsat+I_B*R_B        (i)\n",
+      "I_B= (V_CC-V_EBsat)/R_B # in mA\n",
+      "I_C= bita*I_B # in mA\n",
+      "R_Cmin= ICRC/I_C # in k\u03a9\n",
+      "print \"The minimum value of R_C = %0.3f k\u03a9 \" %R_Cmin\n",
+      "# Part (ii)\n",
+      "R_C= 1.2 # in k\u03a9\n",
+      "I_Csat= ICRC/R_C # in mA\n",
+      "I_B= I_Csat/bita # in mA\n",
+      "# From eq (i)\n",
+      "R_B= (V_CC-V_EBsat)/I_B # in k\u03a9\n",
+      "print \"The maximum value of R_B = %0.2f k\u03a9 \" %R_B"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum value of R_C = 2.343 k\u03a9 \n",
+        "The maximum value of R_B = 112.65 k\u03a9 \n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 5.14.11 - Page No : 5-60"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "bita= 100 \n",
+      "V_BEsat= 0.8 # in V\n",
+      "V_CEsat= 0.2 # in V\n",
+      "V_BEact= 0.7 # in V\n",
+      "V_CC = 10 # in V\n",
+      "R_E = 1 # in k\u03a9\n",
+      "R_C = 2 # in k\u03a9\n",
+      "R_B= 100 # in k\u03a9\n",
+      "bita=100 \n",
+      "alpha= bita/(1+bita) \n",
+      "# Applying KVL to collector circuit\n",
+      "# V_CC= I_Csat*R_C +V_CE +R_E*I_E\n",
+      "# but I_E= alpha*I_Csat\n",
+      "I_Csat= (V_CC-V_CEsat)/(R_C+R_E*alpha) # in mA\n",
+      "I_Bmin= I_Csat/bita # in mA\n",
+      "# Applying KVL to base loop\n",
+      "# V_CC= I_B*R_B +V_BEsat +I_E*R_E\n",
+      "# but I_E= I_Csat+I_B\n",
+      "I_B= (V_CC-V_BEsat-I_Csat*R_E)/(R_B+R_E) # in mA\n",
+      "print \"The value of I_B = %0.2f \u00b5A \" %(I_B*10**3)\n",
+      "print \"The minimum value of I_B = %0.1f \u00b5A \" %(I_Bmin*10**3)\n",
+      "if I_B>I_Bmin :\n",
+      "    print \"Since the value of I_B is greater than the value of I_Bmin\"\n",
+      "    print \"Hence the transistor is in saturation .\"\n",
+      "I_E= (1+bita)*I_Bmin # in mA\n",
+      "R_E= (V_CC-V_BEact-I_Bmin*R_B)/I_E # in k\u03a9\n",
+      "print \"The value of R_E = %0.3f k\u03a9 \" %R_E\n",
+      "print \"So R_E should be greater than this value in order to bring the transistor just out of saturation \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_B = 58.64 \u00b5A \n",
+        "The minimum value of I_B = 32.8 \u00b5A \n",
+        "Since the value of I_B is greater than the value of I_Bmin\n",
+        "Hence the transistor is in saturation .\n",
+        "The value of R_E = 1.819 k\u03a9 \n",
+        "So R_E should be greater than this value in order to bring the transistor just out of saturation \n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  5.14.12 - Page No : 5-62"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "V_CC = 9 # in V\n",
+      "V_BE = 0.8 # in V\n",
+      "V_CE = 0.2 # in V\n",
+      "R_B = 50 # in k\u03a9\n",
+      "R_C=2 # in k\u03a9\n",
+      "R_E = 1 # in k\u03a9\n",
+      "bita=70 \n",
+      "# Applying KVL to input loop, V_CC= I_B*R_B +V_BE +I_E*R_E\n",
+      "# V_CC- V_BE= (R_B+R_E)*I_B + R_E*I_C          (i)\n",
+      "# Applying KVL to output loop, V_CC= R_C*I_C +V_CE +I_C*R_E +I_B*R_E\n",
+      "#I_B = ((V_CC- V_CE)-(R_C+R_E)*I_C)/R_E         (ii)\n",
+      "# From eq (i) and (ii)\n",
+      "I_C= ( (V_CC- V_BE)-(R_B+R_E)* (V_CC- V_CE)/R_E)/(1-(R_B+R_E)*(R_C+R_E)) # in mA\n",
+      "I_B = ((V_CC- V_CE)-(R_C+R_E)*I_C)/R_E# in mA\n",
+      "I_Bmin= I_C/bita # in mA\n",
+      "if I_B>I_Bmin :\n",
+      "    print \"Since the value of I_B (\",round(I_B,3),\" mA) is greater than the value of I_Bmin (\",round(I_Bmin,4),\" mA)\"\n",
+      "    print \"So the transistor is in saturation \"\n",
+      "V_C= V_CC-I_C*R_C # in V\n",
+      "print \"The value of collector voltage = %0.3f volts \" %V_C\n",
+      "bita= I_C/I_B \n",
+      "print \"The minimum value of bita that will change the state of the trasistor = %0.3f\" %bita"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Since the value of I_B ( 0.104  mA) is greater than the value of I_Bmin ( 0.0414  mA)\n",
+        "So the transistor is in saturation \n",
+        "The value of collector voltage = 3.203 volts \n",
+        "The minimum value of bita that will change the state of the trasistor = 27.886\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_6_1.ipynb b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_6_1.ipynb
new file mode 100644
index 00000000..844651b3
--- /dev/null
+++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_6_1.ipynb
@@ -0,0 +1,208 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter : 6 - Field Effect Transistors"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 6.9.1 - Page No : 6-26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from numpy import pi\n",
+      "# Given data\n",
+      "q = 1.6 * 10**-19;# in C\n",
+      "N_D = 10**15;# in electrons/cm**3\n",
+      "N_D = N_D * 10**6;# in electrons/m**3\n",
+      "epsilon_r = 12;\n",
+      "epsilon_o = (36 * pi * 10**9)**-1;\n",
+      "epsilon = epsilon_o * epsilon_r;\n",
+      "a = 3 * 10**-4;# in cm\n",
+      "a = a * 10**-2;# in m\n",
+      "V_P = (q * N_D * a**2)/( 2 * epsilon);# in V\n",
+      "print \"The Pinch off voltage = %0.1f V\" %V_P\n",
+      "# V_GS = V_P * (1-(b/a))**2\n",
+      "b = (1-0.707) *a;# in m\n",
+      "print \"The value of b = %0.3f \u00b5m\" %(b*10**6)\n",
+      "print \"Hence the channel width has been reduced to about one third of its value for V_GS = 0\"\n",
+      "# Note : The unit of b in the book is wrong since the value of b is calculated in \u00b5m."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Pinch off voltage = 6.8 V\n",
+        "The value of b = 0.879 \u00b5m\n",
+        "Hence the channel width has been reduced to about one third of its value for V_GS = 0\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 6.9.2 - Page No : 6-26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "# Given data\n",
+      "I_DSS = 8;# in mA\n",
+      "V_P = -4;# in V\n",
+      "I_D = 3;# in mA\n",
+      "V_GS = V_P * (1 - sqrt(I_D/I_DSS));# in V\n",
+      "print \"The value of V_GS = %0.2f V\" %V_GS\n",
+      "V_DS = V_GS - V_P;# in V\n",
+      "print \"The value of V_DS = %0.2f V\" %V_DS"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of V_GS = -1.55 V\n",
+        "The value of V_DS = 2.45 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 6.9.3 - Page No : 6-27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "V_P = -4;# in V\n",
+      "I_DSS = 9;# in mA\n",
+      "I_DSS = I_DSS * 10**-3;# in A\n",
+      "V_GS = -2;# in V\n",
+      "I_D = I_DSS * ((1 - (V_GS/V_P))**2);# in A\n",
+      "print \"The drain current = %0.2f mA\" %(I_D*10**3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The drain current = 2.25 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 6.9.4 - Page No : 6-27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "I_DSS = 12;# in mA\n",
+      "I_DSS = I_DSS * 10**-3;# in A\n",
+      "V_P = -(6);# in V\n",
+      "V_GS = -(1);# in V\n",
+      "g_mo = (-2 * I_DSS)/V_P;# in A/V\n",
+      "g_m = g_mo * (1 - (V_GS/V_P));# in S\n",
+      "print \"The value of transconductance = %0.2f mS\" %(g_m*10**3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of transconductance = 3.33 mS\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 6.9.5 - Page No : 6-28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "I_DSS = 10;# in mA \n",
+      "I_DSS = I_DSS * 10**-3;# in A\n",
+      "V_P = -(5);# in V\n",
+      "V_GS = -(2.5);# in V\n",
+      "g_m = ((-2 * I_DSS)/V_P) * (1 -(V_GS/V_P));# in S\n",
+      "g_m = g_m * 10**3;# in mS\n",
+      "print \"The Transconductance = %0.f mS\" %g_m\n",
+      "I_D = I_DSS * ((1 - (V_GS/V_P))**2);# in A\n",
+      "print \"The drain current = %0.1f mA\" %(I_D*10**3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Transconductance = 2 mS\n",
+        "The drain current = 2.5 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_7_1.ipynb b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_7_1.ipynb
new file mode 100644
index 00000000..26297594
--- /dev/null
+++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_7_1.ipynb
@@ -0,0 +1,143 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter : 7 - Optoelectronic Devices"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 7.6.1 - Page No : 7-16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# Given data\n",
+      "O_V = 5;# output voltage in V\n",
+      "V_D = 1.5;#voltage drop in V\n",
+      "R = (O_V - V_D)/O_V;\n",
+      "R = R * 10**3;# in ohm\n",
+      "print \"The resistance value = %0.f \u03a9\" %R\n",
+      "print \"As this is not standard value, use R=680 \u03a9 which is a standard value\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistance value = 700 \u03a9\n",
+        "As this is not standard value, use R=680 \u03a9 which is a standard value\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 7.23.1 - Page No : 7-57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt, log\n",
+      "# Given data\n",
+      "N_A = 7.5*10**24;# in atoms/m**3\n",
+      "N_D = 1.5*10**22;# in atoms/m**3\n",
+      "I_lembda = 12.5*10**-3;# in A/cm**2\n",
+      "D_e = 25*10**-4;# in m**2/s\n",
+      "D_h = 1*10**-3;# in m**2/s\n",
+      "Torque_eo = 500;# in ns\n",
+      "Torque_ho = 100;# in ns\n",
+      "n_i = 1.5*10**16;# in /m**3\n",
+      "e = 1.6*10**-19;\n",
+      "P_C = 12.5;# in mA/cm**2\n",
+      "L_e = sqrt(D_e*Torque_ho*10**-9);# in m\n",
+      "L_e = L_e * 10**6;# in \u00b5m\n",
+      "L_h = sqrt(D_h*Torque_ho*10**-9);# in m\n",
+      "L_h = L_h * 10**6;# in \u00b5m\n",
+      "J_s = e*((n_i)**2)*( (D_e/(L_e*10**-6*N_A)) + (D_h/(L_h*10**-6*N_D)) );# in A/m**2\n",
+      "J_s = J_s * 10**-4;# in A/cm**2\n",
+      "V_T = 26;# in mV\n",
+      "V_OC = V_T*log( 1+(I_lembda/J_s) );# in mV\n",
+      "V_OC = V_OC * 10**-3;# in V\n",
+      "print \"Open circuit voltage = %0.3f V\" %V_OC"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Open circuit voltage = 0.522 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 7.23.2 - Page No : 7-57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import exp\n",
+      "# Given data\n",
+      "Phi_o = 1*10**21;# in m**-2s**-1\n",
+      "Alpha = 1*10**5;# in m**-1\n",
+      "W = 25;# in \u00b5m\n",
+      "W =W * 10**-6;# in m\n",
+      "e = 1.6*10**-19;# in C\n",
+      "G_L1 = Alpha*Phi_o;# in m**-3s**-1\n",
+      "G_L2 = Alpha*Phi_o*exp( (-Alpha*W) );# in m**-3s**-1\n",
+      "J_L = e*Phi_o*(1-exp(-Alpha*W));# in A/m**2\n",
+      "J_L = J_L * 10**-1;# in mA/cm**2\n",
+      "print \"Photo current density = %0.2f mA/cm**2\" %J_L"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Photo current density = 14.69 mA/cm**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_8_1.ipynb b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_8_1.ipynb
new file mode 100644
index 00000000..62a399cf
--- /dev/null
+++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_8_1.ipynb
@@ -0,0 +1,96 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter : 8 - Negative Conductance Microwave Devices And Power Devices"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 8.13.1 - Page No : 8-35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# Given data\n",
+      "R= 10;# in k\u03a9\n",
+      "R= R*10**3;# in \u03a9\n",
+      "# Part (i)\n",
+      "V=300;# in V\n",
+      "I_A= V/R;# in A\n",
+      "print \"Part (i) : For 300 V voltage : \"\n",
+      "print \"The anode current = %0.f mA\" %(I_A*10**3)\n",
+      "# Part (ii)\n",
+      "V=100;# in V\n",
+      "I_A= V/R;# in A\n",
+      "print \"Part (ii) : For 100 V voltage : \"\n",
+      "print \"The anode current = %0.f mA\" %(I_A*10**3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part (i) : For 300 V voltage : \n",
+        "The anode current = 30 mA\n",
+        "Part (ii) : For 100 V voltage : \n",
+        "The anode current = 10 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example : 8.14.1 - Page No : 8-43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "t_rr = 10;# in \u00b5s\n",
+      "Q_rr = 150;# in \u00b5c\n",
+      "I_rr = (2*Q_rr)/t_rr;# in A\n",
+      "print \"The peak reverse recovery current = %0.f A\" %I_rr"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The peak reverse recovery current = 30 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/1_1.png b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/1_1.png
new file mode 100644
index 00000000..0d53ffcc
--- /dev/null
+++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/1_1.png
diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/6_1.png b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/6_1.png
new file mode 100644
index 00000000..02ebc92e
--- /dev/null
+++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/6_1.png
diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/7_1.png b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/7_1.png
new file mode 100644
index 00000000..f87b5693
--- /dev/null
+++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/7_1.png