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+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:7bf8d28c6c82b7e899c3ac6b15267e576e5304a1722b176d625beb68170b13b4"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "Chapter03:Semiconductor Diodes"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E1 - Page 60"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#find the value of threshold voltage\n",

+      "#given\n",

+      "t1=25.;#degrees C#initial temperature\n",

+      "t2=100.;#degrees C#final temperature\n",

+      "V=2.*10.**-3.;#V per celsius degree#decrease in barrier potential per degree\n",

+      "V0=0.7#V#Potential at normal temperature\n",

+      "Vd=(t2-t1)*V;#decrease in barrier potential\n",

+      "Vt=V0-Vd;#threshold volatge at 100degree C\n",

+      "print '%s %.2f %s' %(\"Threshold volatge at 100 degrees C =\",Vt,'V');\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Threshold volatge at 100 degrees C = 0.55 V\n"

+       ]

+      }

+     ],

+     "prompt_number": 2

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E2 - Pg 62"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#detrenmine dc resistance of silicon diode\n",

+      "#given\n",

+      "#At Id = 2 mA\n",

+      "Id=2.*10.**-3.;#Ampere#diode current\n",

+      "Vd=0.5;#V#voltage(from given curve)\n",

+      "Rf=(Vd/Id);\n",

+      "print '%s %.f %s' %(\"The dc resistance is =\",Rf,\"ohm\\n\");\n",

+      "\n",

+      "#At Id = 20 mA\n",

+      "Id=20.*10.**-3.;#Ampere#diode current\n",

+      "Vd=0.75;#V#voltage(from given curve)\n",

+      "Rf=(Vd/Id);\n",

+      "print '%s %.1f %s' %(\"The dc resistance is =\",Rf,\"ohm\\n\");\n",

+      "\n",

+      "#At Vd = - 10 V \n",

+      "Id=-2.*10.**-6.;#Ampere#diode current(from given curve)\n",

+      "Vd=-10.;#V#voltage\n",

+      "Rf=(Vd/Id);\n",

+      "print '%s %.f %s' %(\"The dc resistance is =\",Rf/10**6,\"M ohm\\n\");\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The dc resistance is = 250 ohm\n",

+        "\n",

+        "The dc resistance is = 37.5 ohm\n",

+        "\n",

+        "The dc resistance is = 5 M ohm\n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 4

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E3 - Pg 63"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#determine dc & ac resistance of silicon diode\n",

+      "#given\n",

+      "Id=20.*10.**-3.;#A#diode current\n",

+      "Vd=0.75;#V# as given in the V-I graph\n",

+      "Rf=Vd/Id;\n",

+      "print '%s %.1f %s' %(\"The dc resistance of diode is =\",Rf,\"ohm\\n\");\n",

+      "\n",

+      "#From Graph the values of dynamic voltage and current are\n",

+      "#which is equal to MN and NL repectively (in graph)\n",

+      "del_Vd=(0.8-0.68);#V\n",

+      "del_Id=(40-0)*10.**-3.;#A\n",

+      "rf=del_Vd/del_Id;\n",

+      "print '%s %.f %s' %(\"The ac resistance of the diode is =\",rf,\"ohm\")\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The dc resistance of diode is = 37.5 ohm\n",

+        "\n",

+        "The ac resistance of the diode is = 3 ohm\n"

+       ]

+      }

+     ],

+     "prompt_number": 5

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E4 - Pg 65"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#determine ac resistance of silicon diode\n",

+      "#given\n",

+      "#At Id =10mA\n",

+      "Id=10.;#mA\n",

+      "rf=25./Id;\n",

+      "print '%s %.1f %s' %(\"The ac resistance of the diode is(At Id= 10mA) =\",rf,\"ohm\\n\")\n",

+      "\n",

+      "#At Id =20mA\n",

+      "Id=20.;#mA\n",

+      "rf=25./Id;\n",

+      "print '%s %.2f %s' %(\"The ac resistance of the diode is(At Id= 20mA) =\",rf,\"ohm\")\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The ac resistance of the diode is(At Id= 10mA) = 2.5 ohm\n",

+        "\n",

+        "The ac resistance of the diode is(At Id= 20mA) = 1.25 ohm\n"

+       ]

+      }

+     ],

+     "prompt_number": 7

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E5 - Pg 67"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#Find current through diode\n",

+      "#given\n",

+      "Vt=0.3;#V#Threshold voltage\n",

+      "rf=25.;#ohm# average resistance\n",

+      "\n",

+      "#assuming it to be ideal\n",

+      "#from fig 3.19\n",

+      "Vaa=10.;#V#supply\n",

+      "R1=45.;#ohm\n",

+      "R2=5.;#ohm\n",

+      "Vab=Vaa*R2/(R1+R2);\n",

+      "#Vab>Vt therefore diode is forward bias and no current flow through R2\n",

+      "Idi=Vaa/R1;       #for ideal\n",

+      "print '%s %.f %s' %(\"The diode current (for ideal) is =\",Idi*1000,\"mA\\n\");\n",

+      "\n",

+      "#assuming it to be real\n",

+      "#Thevenins equivalent circuit parameters of fig 3.19\n",

+      "Vth=Vaa*R2/(R1+R2);\n",

+      "Rth=R1*R2/(R1+R2);\n",

+      "Idr=(Vth-Vt)/(Rth+rf);          #for real\n",

+      "print '%s %.1f %s' %(\"The diode current (for real) is =\",Idr*1000,\"mA\");\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The diode current (for ideal) is = 222 mA\n",

+        "\n",

+        "The diode current (for real) is = 23.7 mA\n"

+       ]

+      }

+     ],

+     "prompt_number": 16

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E6 - Pg 68"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#Find current through resistance in given figure\n",

+      "#From fig\n",

+      "Vaa=20.;#V#supply\n",

+      "Vt=0.7;#V#threshold voltage of diode\n",

+      "rf=5.;#ohm #forward resistance\n",

+      "R=90.;#ohm#given resistor\n",

+      "\n",

+      "#Diode D1 and D4 are forward bias and D2 and D3 are reverse biased\n",

+      "\n",

+      "Vnet=Vaa-Vt-Vt;\n",

+      "Rt=R+rf+rf;\n",

+      "I=Vnet/Rt;\n",

+      "print '%s %.f %s' %(\"Current through 90 ohm resistor is =\",I*1000,\"mA\");\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Current through 90 ohm resistor is = 186 mA\n"

+       ]

+      }

+     ],

+     "prompt_number": 14

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E7 - Pg 68"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#Find current drawn by the battery\n",

+      "#From fig\n",

+      "Vaa=10.;#V#supply\n",

+      "R1=100.;#ohm\n",

+      "R2=100.;#ohm\n",

+      "\n",

+      "#Forward Bias\n",

+      "Id=Vaa/R1;\n",

+      "print '%s %.1f %s' %(\"Current drawn from battery (forward bias) =\",Id,\"A\\n\");\n",

+      "\n",

+      "#Reverse Bias\n",

+      "Rnet=R1+R2;\n",

+      "Id=Vaa/Rnet;\n",

+      "print '%s %.2f %s' %(\"Current drawn from battery (reverse bias) =\",Id,\"A\");\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Current drawn from battery (forward bias) = 0.1 A\n",

+        "\n",

+        "Current drawn from battery (reverse bias) = 0.05 A\n"

+       ]

+      }

+     ],

+     "prompt_number": 17

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E8 - Pg 79"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#determine dc current through load and rectification efficiency and peak inverse voltage\n",

+      "#given\n",

+      "import math\n",

+      "TR=31./2.;#Turn ratio of the transformer\n",

+      "rf=20.;#ohm#Dynamic forward resistance\n",

+      "Rl=1000.;#ohm#Load resistance\n",

+      "Vt=0.66;#V#Threshold voltage of diode\n",

+      "V=220.;#V#input voltage of transformer\n",

+      "Vp=math.sqrt(2.)*220.#V#peak value of primary voltage\n",

+      "Vm=(1./TR)*Vp;\n",

+      "Im=(Vm-Vt)/(rf+Rl);\n",

+      "Idc=Im/math.pi;\n",

+      "n=40.6/(1.+rf/Rl);\n",

+      "print '%s %.f %s' %(\"The dc current through load is =\",Idc*1000,\"mA\\n\");\n",

+      "print '%s %.1f %s' %(\"The rectification efficiency is =\",n,\"percent\\n\");\n",

+      "print '%s %.2f %s' %(\"Peak inverse voltage =Vm = \",Vm,\"V\\n\")\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The dc current through load is = 6 mA\n",

+        "\n",

+        "The rectification efficiency is = 39.8 percent\n",

+        "\n",

+        "Peak inverse voltage =Vm =  20.07 V\n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 18

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E9 - Pg 79"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#determine dc voltage across load and peak inverse voltage across each diode\n",

+      "#given\n",

+      "import math\n",

+      "TR=12./1.##Turn ratio of the transformer\n",

+      "V=220.##V#input voltage of transformer\n",

+      "Vp=math.sqrt(2.)*220.#V#peak value of primary voltage\n",

+      "Vm=(1./TR)*Vp#\n",

+      "Vdc=(2.*Vm)/math.pi#\n",

+      "print '%s %.1f %s' %(\"The dc voltage across load =\",Vdc,\"V\\n\")#\n",

+      "print '%s %.1f %s' %(\"Peak inverse voltage (for bridge rectifier) =\",Vm,\"V\\n\")#\n",

+      "print '%s %.1f %s' %(\"Peak inverse voltage (for centre tap rectifier) =\",2*Vm,\"V\\n\")#\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The dc voltage across load = 16.5 V\n",

+        "\n",

+        "Peak inverse voltage (for bridge rectifier) = 25.9 V\n",

+        "\n",

+        "Peak inverse voltage (for centre tap rectifier) = 51.9 V\n",

+        "\n"

+       ]

+      }

+     ],

+     "prompt_number": 19

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E10 - Pg 80"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#find dc power supplied to load and efficiency and PIV rating of the diode\n",

+      "#given\n",

+      "import math\n",

+      "rf=2.;#ohm#Dynamic forward resistance\n",

+      "Rs=5.;#ohm#resistaqnce of secondary\n",

+      "Rl=25.;#ohm#Load resistance\n",

+      "Idc=0.1;#A#dc current to a load\n",

+      "Pdc=Idc**2.*Rl;      #dc power\n",

+      "n=(81.2*Rl)/(Rl+rf+Rs);        #efficiency\n",

+      "Im=(math.pi*Idc)/2.;              #peak value current\n",

+      "Vm=Im*(Rl+rf+Rs);              #peak voltage\n",

+      "Vlm=Vm-Im*(rf+Rs);              #peak voltage across load\n",

+      "PIV=Vm+Vlm;\n",

+      "print '%s %.2f %s' %(\"The dc power supplied to the load is =\",Pdc,'W\\n');\n",

+      "print '%s %.2f %s' %(\"Efficiency =\",n,'percent\\n');\n",

+      "print '%s %.3f %s' %(\"The peak inverse voltage is =\",PIV,'V');\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The dc power supplied to the load is = 0.25 W\n",

+        "\n",

+        "Efficiency = 63.44 percent\n",

+        "\n",

+        "The peak inverse voltage is = 8.954 V\n"

+       ]

+      }

+     ],

+     "prompt_number": 21

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E11 - Pg 87"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#Calculate output voltage and current through load and voltage across series resistor and current and power dissipated in zener diode\n",

+      "#given\n",

+      "Vi=110.;#V       #input voltage\n",

+      "Rl=6.*10.**3.;# ohm    #load resistance\n",

+      "Rs=2.*10.**3.;#ohm       #series resistance\n",

+      "Vz=60.;#V        #Zener voltage\n",

+      "V=Vi*Rl/(Rs+Rl);\n",

+      "\n",

+      "#This V>Vz therefore Zener diode is ON\n",

+      "\n",

+      "Vo=Vz;          #output voltage\n",

+      "Il=Vo/Rl;       #Current through load resistance\n",

+      "Vs=Vi-Vo;       #Voltage drop across the series resistor\n",

+      "Is=Vs/Rs         #current through the series resistor\n",

+      "Iz=Is-Il        #/By applying kirchhoffs law\n",

+      "Pz=Vz*Iz        #Power dissipated accross zener diode\n",

+      "\n",

+      "print '%s %.f %s' %(\"The output voltage is =\",Vo,\"V\\n\");\n",

+      "print '%s %.f %s' %(\"The current through load resistance is =\",Il*1000,\"mA\\n\");\n",

+      "print '%s %.f %s' %(\"Voltage across series resistor is =\",Vs,\"V\\n\")\n",

+      "print '%s %.f %s' %(\"Current in zener diode is =\",Iz*1000,\"mA\\n\")\n",

+      "print '%s %.f %s' %(\"Power dissipated by zener diode =\",Pz*1000,'mW');\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The output voltage is = 60 V\n",

+        "\n",

+        "The current through load resistance is = 10 mA\n",

+        "\n",

+        "Voltage across series resistor is = 50 V\n",

+        "\n",

+        "Current in zener diode is = 15 mA\n",

+        "\n",

+        "Power dissipated by zener diode = 900 mW\n"

+       ]

+      }

+     ],

+     "prompt_number": 20

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E12 - Pg 88"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "\n",

+      "#Calculate max and min values of zener diode current\n",

+      "#given\n",

+      "Vimin=80.;#V       #minimum input voltage\n",

+      "Vimax=120.;#V       #maximum input voltage\n",

+      "Rl=10.*10.**3.;# ohm    #load resistance\n",

+      "Rs=5.*10.**3.;#ohm       #series resistance\n",

+      "Vz=50.;#V        #Zener voltage\n",

+      "V=Vimin*Rl/(Rs+Rl);\n",

+      "\n",

+      "#This V>Vz therefore Zener diode is ON\n",

+      "\n",

+      "#For minimum value of zener diode\n",

+      "\n",

+      "Vo=Vz;          #output voltage\n",

+      "Vs=Vimin-Vo;       #Voltage drop across the series resistor\n",

+      "Is=Vs/Rs         #current through the series resistor\n",

+      "Il=Vo/Rl;       #Current through load resistance\n",

+      "Izmin=Is-Il;\n",

+      "print '%s %.f %s' %(\"Minimum values of zener diode current is =\",Izmin*1000,\"mA\\n\");\n",

+      "\n",

+      "#For maximum value of zener diode\n",

+      "\n",

+      "Vo=Vz;          #output voltage\n",

+      "Vs=Vimax-Vo;       #Voltage drop across the series resistor\n",

+      "Is=Vs/Rs         #current through the series resistor\n",

+      "Il=Vo/Rl;       #Current through load resistance\n",

+      "Izmax=Is-Il;\n",

+      "print '%s %.f %s' %(\"Maximum values of zener diode current is =\",Izmax*1000,\"mA\");\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Minimum values of zener diode current is = 1 mA\n",

+        "\n",

+        "Maximum values of zener diode current is = 9 mA\n"

+       ]

+      }

+     ],

+     "prompt_number": 22

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E13 - Pg 88"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#determine value of the series resistor and wattage rating\n",

+      "#given\n",

+      "Vi=12.##V       #input voltage\n",

+      "Vz=7.2##V        #Zener voltage\n",

+      "Izmin=10.*10.**-3.##A             #min current through zener diode\n",

+      "Ilmax=100*10.**-3.##A            #max current through load\n",

+      "Ilmin=12.*10.**-3.##A            #min current through load\n",

+      "Vs=Vi-Vz#       #Voltage drop across the series resistor\n",

+      "Is=Izmin+Ilmax#     #Current through the series resistor\n",

+      "Rs=Vs/Is#\n",

+      "print '%s %.1f %s' %(\"The series resistor so that 10mA current flow through zener diode is =\",Rs,\"ohm\\n\")#\n",

+      "Izmax=Is-Ilmin#max zener through zener diode\n",

+      "Pmax=Izmax*Vz#\n",

+      "print '%s %.1f %s' %(\"The maximum wattage rating is =\",Pmax*1000,\"mW\")#\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "The series resistor so that 10mA current flow through zener diode is = 43.6 ohm\n",

+        "\n",

+        "The maximum wattage rating is = 705.6 mW\n"

+       ]

+      }

+     ],

+     "prompt_number": 23

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E14 - Pg 90"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#Find the capacitance of a varactor diode\n",

+      "#given\n",

+      "import math\n",

+      "C=5.;#pf#capcitance of varactor diode at V=4V\n",

+      "V=4.;#V\n",

+      "K=C*math.sqrt(4.);\n",

+      "#When bias voltage is increased upto 6 V\n",

+      "Vn=6.;#V#new bias voltage\n",

+      "Cn=K/(math.sqrt(Vn));\n",

+      "print '%s %.3f %s' %(\"Capacitance (At 6 V ) =\",Cn,\"pf\");\n"

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Capacitance (At 6 V ) = 4.082 pf\n"

+       ]

+      }

+     ],

+     "prompt_number": 24

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
\ No newline at end of file