{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 2 : Special Purpose Diodes" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.1, Page No. 68" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# maximum current\n", "\n", "import math\n", "#Variable declaration\n", "Pmax=364.0 #dissipation in milliwatt\n", "Vz=9.1 #in V\n", "\n", "#Calculations\n", "Izmax=Pmax/Vz #in mA\n", "\n", "#Result\n", "print(\"maximum current the diode can handle is ,(mA)= %.f\"%Izmax)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "maximum current the diode can handle is ,(mA)= 40\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.2, Page No. 68" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# resistance\n", "\n", "import math\n", "#Vaariable declaration\n", "mip=15.0 #in volt\n", "op=6.8 #output potential in volt\n", "pd=mip-op #potential difference across series resistor\n", "Il=5 #load current in mA\n", "nmip=20 #new maximum input voltage in volt\n", "pd1=nmip-op #new potential difference across series resistor\n", "Il1=20 #new load current in mA\n", "\n", "#Calculations\n", "R=((pd1-pd)/((Il1-Il)*10**-3))\n", "print(\"value of series resistance is,(ohm)= %.1f\"%R)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "value of series resistance is,(ohm)= 333.3\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2.3, Page No.69" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Current\n", "\n", "import math\n", "#VAriable declaration\n", "V=120.0 #in V\n", "Vz=50.0 #in V\n", "R=5.0 #in ohm\n", "Rl=10.0 #in k-ohm\n", "\n", "#Calculations\n", "vd5=V-Vz #voltage drop across 5 ohm resistor\n", "I5=vd5/R #current through 5 ohm resistor\n", "Il=Vz/(Rl*10**3) #current through load resistor\n", "Iz=I5-Il #in A\n", "\n", "#Result\n", "print(\"current through zener diode is ,(A)= %.3f\"%Iz)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "current through zener diode is ,(A)= 13.995\n" ] } ], "prompt_number": 8 } ], "metadata": {} } ] }