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diff --git a/sample_notebooks/DeepTrambadia/Diode_Applications.ipynb b/sample_notebooks/DeepTrambadia/Diode_Applications.ipynb
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+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:90aab70b55d4896f0f22aa66f516161405cb3435a2adc68d32d5e3787e2d9de5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2:Diode Applications\n"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1 page : 53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#(a)\n",
+      "#initialisation of variables\n",
+      "\n",
+      "E=10      #E in V\n",
+      "R=1     #R in Kohm\n",
+      "\n",
+      "\n",
+      "#Calculations\n",
+      " \n",
+      "Id=E/R    #Eq.(2.2)\n",
+      "Vd=E\n",
+      "print \"(a) \\nThe current Ic is = %fmA \"%(Id),\";Vd=0V\"\n",
+      "print \"The diode voltage is = %fV\"%(Vd),\";Id=0A\"\n",
+      "print \"\\nThe resulting load line appears in Fig. 2.4. The intersection between the load line \\nand the characteristic curve defines the Q-point as\"\n",
+      "print \"\\nThe level of VD is certainly an estimate, and the accuracy of ID is limited by the chosenscale. \\nA higher degree of accuracy would require a plot that would be much large and perhaps unwieldy\"\n",
+      "\n",
+      "\n",
+      "#(B)\n",
+      "print \"\\n(B)\\n\"\n",
+      "Ir=9.25 #Ir in mA\n",
+      "Vdq=0.78 #Vdq in v\n",
+      "Vr=Ir*R\n",
+      "print \"Vr = Ir*R = Idq*R  = %dV\"%(Vr),\"or\"\n",
+      "Vr = E-Vdq\n",
+      "print \"Vr = E-Vdq = %fV\" %(Vr)\n",
+      "print \"\\nThe difference in results is due to the accuracy with which the graph can be read. \\nIdeally,the results obtained either way should be the same.\"\n",
+      "\n",
+      "#Graph solution to example 2.1\n",
+      "\n",
+      "import numpy as np\n",
+      "import matplotlib.pyplot as plt\n",
+      "\n",
+      "Vd = np.linspace(0.0,10.0)\n",
+      "Id = np.linspace(0.0,10.0)\n",
+      "Id= -Vd + 10\n",
+      "plt.plot(Vd, Id)\n",
+      "Vd = [0,0,0.1,0.1,0.2,0.2,0.3,0.3,0.3,0.3,0.4,0.5,0.6,0.7]\n",
+      "Id = [0,0,0,0,0,0,0,0,0.1,0.1,0.3,0.7,2.0,10.0]\n",
+      "\n",
+      "plt.plot(Vd, Id,'yo-')\n",
+      "\n",
+      "plt.xlabel('Voltage (v)')\n",
+      "plt.ylabel('current (mA)')\n",
+      "plt.title('Characteristics of diode')\n",
+      "plt.grid(True)\n",
+      "plt.savefig(\"test.png\")\n",
+      "\n",
+      "plt.show()\n",
+      "\n",
+      "print \"example 2.2:\"\n",
+      "print \"repeat the example 2.1 for R =2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) \n",
+        "The current Ic is = 10.000000mA  ;Vd=0V\n",
+        "The diode voltage is = 10.000000V ;Id=0A\n",
+        "\n",
+        "The resulting load line appears in Fig. 2.4. The intersection between the load line \n",
+        "and the characteristic curve defines the Q-point as\n",
+        "\n",
+        "The level of VD is certainly an estimate, and the accuracy of ID is limited by the chosenscale. \n",
+        "A higher degree of accuracy would require a plot that would be much large and perhaps unwieldy\n",
+        "\n",
+        "(B)\n",
+        "\n",
+        "Vr = Ir*R = Idq*R  = 9V or\n",
+        "Vr = E-Vdq = 9.220000V\n",
+        "\n",
+        "The difference in results is due to the accuracy with which the graph can be read. \n",
+        "Ideally,the results obtained either way should be the same.\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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mN0BE9kRTM4DU1JlwdGyOdu1m1flYnA0Qka3hDKAKcn4mMLsBIlI7jRUA+T8T2NKzAa5v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\nASAi0igWACIijWIBICLSKBYAIiKN+j91tdZGQm+jrgAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7f6e4daea050>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "example 2.2:\n",
+        "repeat the example 2.1 for R =2\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6 Page : 60"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 2.6\n",
+      "#For the series diode configuration of Fig. 2.16, determine VD, VR, and ID.\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "\n",
+      "\n",
+      "E=8       #E in V\n",
+      "R=2.2     #R in Kohm\n",
+      "Vd=0.7    #Vd in V \n",
+      "\n",
+      "#Calculations\n",
+      "\n",
+      "Vr=E-Vd \n",
+      "Id=Vr/R    \n",
+      "print \"Since the applied voltage establishes a current in the clockwise direction to match thearrow of the symbol \\nand the diode is in the 'on' state,\\n\"\n",
+      "print \"The diode voltage is = %.1fV\"%(Vd),\";Id=0A\"\n",
+      "print \"The voltage Vr is = %.1fV\"%(Vr)\n",
+      "print \"The current Id is = %.2fmA \"%(Id)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Since the applied voltage establishes a current in the clockwise direction to match thearrow of the symbol \n",
+        "and the diode is in the 'on' state,\n",
+        "\n",
+        "The diode voltage is = 0.7V ;Id=0A\n",
+        "The voltage Vr is = 7.3V\n",
+        "The current Id is = 3.32mA \n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7  Page : 60"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 2.7\n",
+      "#Repeat Example 2.6 with the diode reversed\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "\n",
+      "\n",
+      "E=8       #E in V\n",
+      "R=2.2     #R in Kohm\n",
+      "I=0       #For open circuit\n",
+      "\n",
+      "#Calculations\n",
+      "\n",
+      "Vr=I*R \n",
+      "Vd=E-Vr    \n",
+      "print \"Removing the diode, we find that the direction of I is opposite to the arrow in the diode symbol \\nand the diode equivalent is the open circuit no matter which model isemployed.\"\n",
+      "print \"The result is the network of Fig. 2.17, where ID = 0A due to the open circuit.\\n\"\n",
+      "print \"The voltage Vr is = %.1fV\"%(Vr)\n",
+      "print \"The diode voltage is = %.1fV\"%(Vd)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Removing the diode, we find that the direction of I is opposite to the arrow in the diode symbol \n",
+        "and the diode equivalent is the open circuit no matter which model isemployed.\n",
+        "The result is the network of Fig. 2.17, where ID = 0A due to the open circuit.\n",
+        "\n",
+        "The voltage Vr is = 0.0V\n",
+        "The diode voltage is = 8.0V\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.8 Page : 61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 2.8\n",
+      "#For the series diode configuration of Fig. 2.19, determine VD, VR, and ID.\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "\n",
+      "\n",
+      "E=0.5     #E in volt\n",
+      "R=1.2     #R in Kohm\n",
+      "Id=0      #For open circuit\n",
+      "\n",
+      "\n",
+      "#calculation\n",
+      "\n",
+      "Vr=Id*R\n",
+      "Vd=E\n",
+      "\n",
+      "print \"The voltage Vr is = %.1fV\"%(Vr)\n",
+      "print \"The diode voltage Vd is = %.1fV\"%(Vd)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The voltage Vr is = 0.0V\n",
+        "The diode voltage Vd is = 0.5V\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.9 Page : 62"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 2.9      Page no 62\n",
+      "\n",
+      "#initialisation of variables\n",
+      "\n",
+      "R=5.6             # resistance in kilo ohm\n",
+      "E=12\t\t\t  # supply voltage in volt\n",
+      "Vt1=0.7           # threshold voltage of siicon in volt\n",
+      "Vt2=0.3            # threshold voltage of germanium in volt\n",
+      "\n",
+      "print \"Applying KVL rule in fig 2.2,\"\n",
+      "\n",
+      "Vo=E-(Vt1+Vt2)    # resulting voltage in volt\n",
+      "\n",
+      "Id=(Vo/R)\n",
+      "\n",
+      "print \"The resulting voltage is = %dV\"%(Vo)\n",
+      "\n",
+      "print \"The current through diode is = %.2fmA\"%(Id)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Applying KVL rule in fig 2.2,\n",
+        "The resulting voltage is = 11V\n",
+        "The current through diode is = 1.96mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.12 Page : 64"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "\n",
+      "E=10      #supply voltage in vol't\n",
+      "R=0.33    #resistance in kilo ohms\n",
+      "Vd=0.7    # voltage across silicon diode\n",
+      "\n",
+      "print\"From figure 2.31 it can be said that both diodes are opened so\"\n",
+      "\n",
+      "Vo=0.7    # resulting voltage in volt\n",
+      "\n",
+      "I1=(E-Vd)/R\n",
+      "print\"the value of Id1 is = %.2fmA\"%(I1)\n",
+      "print\"\\nDiodes are of similar characteristics so\"\n",
+      "\n",
+      "Id2=(I1/2)\n",
+      "print\"the value of Id2 is = %.2fmA\"%(Id2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "From figure 2.31 it can be said that both diodes are opened so\n",
+        "the value of Id1 is = 28.18mA\n",
+        "\n",
+        "Diodes are of similar characteristics so\n",
+        "the value of Id2 is = 14.09mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.13 Page : 65"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialization of variables\n",
+      "\n",
+      "E1=20       #supply voltage in V\n",
+      "E2=4        #second port voltage in V\n",
+      "Vd=0.7      #thresold voltage\n",
+      "R=2.2       #R in Kohm\n",
+      "\n",
+      "\n",
+      "#calculation\n",
+      "\n",
+      "I = (E1-E2-Vd)/R\n",
+      "\n",
+      "print \"Diode D1 turn on and Diode D2 turn off\"\n",
+      "print \"the resultant current I is = %.2fmA\" %(I)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diode D1 turn on and Diode D2 turn off\n",
+        "the resultant current I is = 6.95mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.14 Page :66"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialization of variables\n",
+      "E=12           #supply Voltage in V\n",
+      "Vd=0.3         #thresold voltage in V\n",
+      "\n",
+      "\n",
+      "#calculation\n",
+      "\n",
+      "V0 = E-Vd\n",
+      "\n",
+      "print \"If initially both were 'on,'' the 0.7-V drop across the silicon diode would not match the 0.3 V \"\n",
+      "print \"Across the germanium diode as required by the fact that the voltage across parallel elements must be the same.\"\n",
+      "print \"\\nThe silicon diode will never have the opportunity to capture its required 0.7 V \\nand therefore remains in its open-circuit state.\"\n",
+      "print \"the resultant Voltage V0 is = %.1fV\" % (V0)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "If initially both were 'on,'' the 0.7-V drop across the silicon diode would not match the 0.3 V \n",
+        "Across the germanium diode as required by the fact that the voltage across parallel elements must be the same.\n",
+        "\n",
+        "The silicon diode will never have the opportunity to capture its required 0.7 V \n",
+        "and therefore remains in its open-circuit state.\n",
+        "the resultant Voltage V0 is = 11.7V\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.15 Page :66"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialization of variables\n",
+      "\n",
+      "E=20        #supply voltage in V\n",
+      "VT1=0.7      #thresold voltage\n",
+      "VT2=0.7      #thresold voltage\n",
+      "R1=3.3       #R in Kohm\n",
+      "R2=5.6       #R in Kohm\n",
+      "\n",
+      "#calculation\n",
+      "\n",
+      "print \"Both Diodes will turn 'on'\"\n",
+      "print \"So diode voltage will appear over the resistance\"\n",
+      "\n",
+      "I1 = (VT2)/R1\n",
+      "\n",
+      "print \"the resultant current I2 is = %.3fmA\" %(I1)\n",
+      "\n",
+      "print \"\\nApplying Kirchhoff's voltage law around the indicated loop in the clockwise direction yields\"\n",
+      "\n",
+      "V2 = E-VT1-VT2\n",
+      "I2 = V2/R2\n",
+      "\n",
+      "print \"the voltage V2 = %.1fV\" %(V2)\n",
+      "print \"the current I2 = %.2fmA\"%(I2)\n",
+      "\n",
+      "#At hte bottom node (a)\n",
+      "\n",
+      "ID2=I2-I1\n",
+      "\n",
+      "print \"the current I2 = %.3fmA\" %(ID2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Both Diodes will turn 'on'\n",
+        "So diode voltage will appear over the resistance\n",
+        "the resultant current I2 is = 0.212mA\n",
+        "\n",
+        "Applying Kirchhoff's voltage law around the indicated loop in the clockwise direction yields\n",
+        "the voltage V2 = 18.6V\n",
+        "the current I2 = 3.32mA\n",
+        "the current I2 = 3.109mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.18 Page :71"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialization of variables\n",
+      "\n",
+      "Vm=20          #peak voltage in V\n",
+      "VT=0.7         #thresold voltage in V\n",
+      "\n",
+      "\n",
+      "#calculation\n",
+      "#(a)\n",
+      "print \"(a)\\nIdeal Diode:\"\n",
+      "\n",
+      "Vdc1= -(0.318*Vm)             #Vdc=-0.318(Vm-VT)\n",
+      "\n",
+      "print \" In this situation the diode will conduct during the negative part of the input\"\n",
+      "print \"For the full period DC level is = %.2fV\" %(Vdc1)\n",
+      "print \"\\nThe negative sign indicates that the polarity of the output is opposite to the defined polarity\"\n",
+      "\n",
+      "\n",
+      "#(b)\n",
+      "\n",
+      "print \"\\n(b)\\nsilicon Diode:\"\n",
+      "\n",
+      "Vdc2= -0.318*(Vm-0.7)\n",
+      "\n",
+      "print \"Vdc2 = %.2fV\" %(Vdc2)\n",
+      "\n",
+      "#(c)\n",
+      "\n",
+      "print \"\\n(c)\\nIf Vm is increased to 200V:\"\n",
+      "\n",
+      "Vm = 200         #new peak voltage\n",
+      "Vdc1= -(0.318*Vm)\n",
+      "Vdc2= -0.318*(Vm-0.7)\n",
+      "\n",
+      "print \"using (a), Vdc = %.2fV\" %(Vdc1)\n",
+      "print \"using (b), Vdc = %.2fV\" %(Vdc2)\n",
+      "\n",
+      "\n",
+      "\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)\n",
+        "Ideal Diode:\n",
+        " In this situation the diode will conduct during the negative part of the input\n",
+        "For the full period DC level is = -6.36V\n",
+        "\n",
+        "The negative sign indicates that the polarity of the output is opposite to the defined polarity\n",
+        "\n",
+        "(b)\n",
+        "silicon Diode:\n",
+        "Vdc2 = -6.14V\n",
+        "\n",
+        "(c)\n",
+        "If Vm is increased to 200V:\n",
+        "using (a), Vdc = -63.60V\n",
+        "using (b), Vdc = -63.38V\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.19 Page :75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#example 2.19 page no.75\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "#initialization of variables\n",
+      "Vi=10          #input voltage in V\n",
+      "\n",
+      "#calculation\n",
+      "\n",
+      "print \"After redrawing the network configuration,\"\n",
+      "\n",
+      "V0 = 0.5*Vi\n",
+      "\n",
+      "print \"voltage across the resistance V0 = %dV\" %(V0)\n",
+      "print \"\\nFor the negative part of the input the roles of the diodes will be interchanged\"\n",
+      "\n",
+      "#The effect of removing diodes\n",
+      "\n",
+      "Vdc = 0.636*(V0)\n",
+      "print \"The effect of removing diodes:\"\n",
+      "print \"\\tReduced available DC level = %.2fV\" %(Vdc)\n",
+      "print \"\\tPIV = the maximum voltage across R is = %dV\" %(V0)\n",
+      "print \"or half of that required for a half-wave rectifier with the same input\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "After redrawing the network configuration,\n",
+        "voltage across the resistance V0 = 5V\n",
+        "\n",
+        "For the negative part of the input the roles of the diodes will be interchanged\n",
+        "The effect of removing diodes:\n",
+        "\tReduced available DC level = 3.18V\n",
+        "\tPIV = the maximum voltage across R is = 5V\n",
+        "or half of that required for a half-wave rectifier with the same input\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file