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diff --git a/Electronic_Devices_and_Circuits/Chapter3_2.ipynb b/Electronic_Devices_and_Circuits/Chapter3_2.ipynb deleted file mode 100755 index 8f8a6a3b..00000000 --- a/Electronic_Devices_and_Circuits/Chapter3_2.ipynb +++ /dev/null @@ -1,1201 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 03 : Diode applications"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.1, Page No 73"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#initialisation of variables\n",
- "Vf=.7\n",
- "Rl=500.0\n",
- "Vi=22.0\n",
- "Vpi=1.414*Vi\n",
- "\n",
- "#Calculations\n",
- "Vpo=Vpi-Vf\n",
- "print(\" peak vouput voltage is %3.2fV \" %Vpo)\n",
- "Ip=Vpo/Rl\n",
- "\n",
- "#Results\n",
- "print(\"peak load current is %3.4fA \" %Ip)\n",
- "PIV=Vpi\n",
- "print(\"diode paek reverse voltage %3.2fV \" %PIV)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " peak vouput voltage is 30.41V \n",
- "peak load current is 0.0608A \n",
- "diode paek reverse voltage 31.11V \n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.2, Page No 79"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#initialisation of variables\n",
- "Vi=30.0\n",
- "Rl=300.0\n",
- "Vf=0.7\n",
- "\n",
- "#Calculations\n",
- "Vpi=1.414*Vi\n",
- "Vpo=Vpi-2*Vf\n",
- "print(\" peak output voltage %.3f V \" %Vpo)\n",
- "Ip=Vpo/Rl\n",
- "\n",
- "#Results\n",
- "print(\" current bridge is %.1f mA \" %(Ip*1000))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " peak output voltage 41.020 V \n",
- " current bridge is 136.7 mA \n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3 Page No 83"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#initialisation of variables\n",
- "C1=680.0*10**-6\n",
- "Eo=28.0\n",
- "Rl=200.0\n",
- "f=60.0\n",
- "\n",
- "#Calculations\n",
- "Il=Eo/Rl\n",
- "T=1/f\n",
- "t1=T\n",
- "Vr=(Il*t1)/C1\n",
- "\n",
- "#Results\n",
- "print(\"peak to peak ripple voltage is %.2f V \" %Vr)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "peak to peak ripple voltage is 3.43 V \n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.4, Page No 84"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#initialisation of variables\n",
- "Eo=20.0\n",
- "Rl=500.0\n",
- "f=60.0\n",
- "\n",
- "#Calculations\n",
- "Vr=(10*Eo)/100#10% of Eo\n",
- "Il=Eo/Rl\n",
- "T=1/f\n",
- "t1=T\n",
- "C1=((Il*t1)/Vr)*10**6\n",
- "\n",
- "#Results\n",
- "print(\"Reservior capacitance is %.2f uF \" %C1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Reservior capacitance is 333.33 uF \n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.5 Page No 85"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#initialisation of variables\n",
- "Eo=20.0\n",
- "f=60.0\n",
- "Rl=500.0\n",
- "Il=Eo/Rl\n",
- "\n",
- "#Calculations\n",
- "Vr=(10.0*Eo)/100\n",
- "print(\"10percent of Eo is %.2f V \" %Vr)\n",
- "Eomin=Eo-0.5*Vr\n",
- "Eomax=Eo+0.5*Vr\n",
- "Q1=math.asin(Eomin/Eomax)\n",
- "Q1=65\n",
- "Q2=90-Q1\n",
- "T=1/f\n",
- "t2=(Q2*T)/360\n",
- "print(\" charging time is %.2fs \" %t2)\n",
- "t1=T-t2\n",
- "print(\"discharging time is %.2fs \" %t1)\n",
- "C1=((Il*t1)/Vr)*10**6\n",
- "\n",
- "#Results\n",
- "print(\"reservior capacitance is %.2f uF \" %C1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "10percent of Eo is 2.00 V \n",
- " charging time is 0.00s \n",
- "discharging time is 0.02s \n",
- "reservior capacitance is 310.19 uF \n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.6 Page No 88"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#initialisation of variables\n",
- "Eo=21.0\n",
- "Vf=0.7\n",
- "\n",
- "#Calculations\n",
- "t1=1.16*10**-3\n",
- "t2=15.54*10**-3\n",
- "Vp=Eo+Vf\n",
- "Vr=2*Vp\n",
- "Il=40*10**-4\n",
- "Ifrm=(Il*(t1+t2))/t2\n",
- "Ifsm=30.0\n",
- "Rs=Vp/Ifsm\n",
- "\n",
- "#Results\n",
- "print(\" surge limiting resistance is %3.2fohm \" %Rs)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " surge limiting resistance is 0.72ohm \n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.7, Page No 89 "
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#initialisation of variables\n",
- "Vf=.7\n",
- "Eo=21.0\n",
- "\n",
- "#Calculations\n",
- "Il=40*10**-3\n",
- "Vp=115.0\n",
- "Vs=.707*(Vf+Eo)\n",
- "print(\" Vrms voltage is %3.3fV \" %Vs)\n",
- "Is=3.6*Il\n",
- "print(\" rms current is %.2f mA \" %(Is*1000))\n",
- "Ip=(Vs*Is)/Vp\n",
- "\n",
- "#Results\n",
- "print(\"primary current is %.2f mA \" %(Ip*1000))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " Vrms voltage is 15.342V \n",
- " rms current is 144.00 mA \n",
- "primary current is 19.21 mA \n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.8 Page No 92"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math \n",
- "\n",
- "#initialisation of variables\n",
- "Vr=2.0\n",
- "T=16.7*10**-3\n",
- "t2=1.16*10**-3\n",
- "\n",
- "#Calculations\n",
- "Il=40.0*10**-3#from example 3.5\n",
- "t1=(T/2.0)-t2\n",
- "C1=(Il*t1)/Vr\n",
- "\n",
- "#Results\n",
- "print(\" resrvior capacitor is %.2f mF \" %(C1*10**6))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " resrvior capacitor is 143.80 mF \n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.9 Page No 93"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#initialisation of variables\n",
- "Vr=2.0\n",
- "T=16.7*10**-3\n",
- "Il=40.0*10**-3\n",
- "\n",
- "#Calculations\n",
- "t1=T/2\n",
- "C1=(Il*t1)/Vr\n",
- "\n",
- "#Results\n",
- "print(\" reservior capacitance is %.1fF \" %(C1*10**6))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " reservior capacitance is 167.0F \n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.10 Page No 93"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#initialisation of variables\n",
- "Eo=21.0\n",
- "Vf=0.7\n",
- "Il=40.0*10**-3\n",
- "t1=7.19*10**-3\n",
- "t2=1.16*10**-3\n",
- "\n",
- "#Calculations\n",
- "Vp=Eo+(2*Vf)\n",
- "Vr=Vp\n",
- "If=Il/2\n",
- "Ifrm=Il*(t1+t2)/t2\n",
- "Ifsm=30\n",
- "Rs=Vp/Ifsm\n",
- "\n",
- "#Results\n",
- "print(\"surge limiting resistance is %.3fohm \" %Rs)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "surge limiting resistance is 0.747ohm \n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.11, Page No 93"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#initialisation of variables\n",
- "Eo=21.0\n",
- "Vf=0.7\n",
- "Il=40*10**-3\n",
- "Vp=115.0\n",
- "\n",
- "#Calculations\n",
- "Vs=0.707*(Eo+2*Vf)\n",
- "Is=1.6*Il\n",
- "Ip=(Vs*Is)/Vp\n",
- "\n",
- "#Results\n",
- "print(\" supply current is %.1f mA \" %(Ip*10**3))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " supply current is 8.8 mA \n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.12, Page No 97"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#initialisation of variables\n",
- "Eo=20.0\n",
- "Il=40.0*10**-3\n",
- "R1=22.0\n",
- "Vr=2.0\n",
- "C1=150*10**-6\n",
- "C2=C1\n",
- "fr=120\n",
- "\n",
- "#Calculations\n",
- "Vo=Eo-Il*R1\n",
- "vi=Vr/3.14\n",
- "Xc2=1/(2*3.14*fr*C2)\n",
- "vo=(vi*Xc2)/math.sqrt((R1**2) + (Xc2**2))\n",
- "print(\" dc output voltage is %.3fV \" %vo)\n",
- "Vpp=2*vo\n",
- "\n",
- "#Results\n",
- "print(\" peak to peak voltage is %.1fV \" %(Vpp*10**3))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " dc output voltage is 0.238V \n",
- " peak to peak voltage is 475.3V \n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.13, Page No 98"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#initialisation of variables\n",
- "C1=150*10**-6\n",
- "C2=C1\n",
- "vi=4.0\n",
- "vo=1.0\n",
- "f=120.0\n",
- "\n",
- "#Calculations\n",
- "Xc2=8.84 #FROM EXAMPLE 3.12\n",
- "Xl=Xc2*((vi/vo)+1)\n",
- "L1=Xl/(2*3.14*f)\n",
- "\n",
- "#Results\n",
- "print(\" suitable value of L1 is %.3fH \" %(L1*10**3))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " suitable value of L1 is 58.652H \n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.14, Page No 101"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#initialisation of variables\n",
- "Edc=20.0\n",
- "vo=0.24\n",
- "Vo=20.0\n",
- "Il=40*10**-3\n",
- "fr=120.0\n",
- "\n",
- "#Calculations\n",
- "Eomax=(3.14*Edc)/2\n",
- "Epeak=(4*Eomax)/(3*3.14)\n",
- "vi=Epeak\n",
- "Rl=Vo/Il\n",
- "Xlc=(2*Rl)/3\n",
- "Lc=Xlc/(2*3.14*fr)\n",
- "L=1.25*Lc\n",
- "Xl=2*3.14*fr*L\n",
- "Xc=Xl/((vi/vo)+1)\n",
- "C1=1/(2*3.14*fr*Xc)\n",
- "\n",
- "#Results\n",
- "print(\"The value of c1 = %.2f mF \" %(C1*10**6))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of c1 = 180.11 mF \n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.15, Page No 105"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#initialisation of variables\n",
- "Eo=20.0\n",
- "E0=20-19.7 #load effect\n",
- "\n",
- "#Calculations\n",
- "loadregulation =(E0*100)/Eo#percentage\n",
- "sourceeffect=20.2-20\n",
- "lineregulation =(sourceeffect*100)/Eo\n",
- "\n",
- "#Results\n",
- "print(\"Line regulation = %.1f percent \" %lineregulation)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Line regulation = 1.0 percent \n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.16, Page No 108"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#initialisation of variables\n",
- "Vz=9.1\n",
- "Izt=20*10**-3\n",
- "Es=30.0\n",
- "\n",
- "#Calculations\n",
- "R1=(Es-Vz)/Izt\n",
- "Pr1=(Izt**2)*R1\n",
- "Es=27\n",
- "Iz=(Es-Vz)/R1\n",
- "\n",
- "#Results\n",
- "print(\"The circuit current is %.2f mA \" %(Iz*10**3))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The circuit current is 17.13 mA \n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.17, Page No 110"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#initialisation of variables\n",
- "Vz=6.2\n",
- "Pd=400.0*10**-3\n",
- "Es=16.0\n",
- "\n",
- "#Calculations\n",
- "Izm=Pd/Vz\n",
- "R1=(Es-Vz)/Izm\n",
- "Pr1=(Izm**2)*R1\n",
- "Izmin=5.0*10**-3\n",
- "Izmax=Izm-Izmin\n",
- "\n",
- "#Results\n",
- "print(\"maximum current is %3.2f mA \" %(Izmax*10**3))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "maximum current is 59.52 mA \n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.18, Page No 112"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#initialisation of variables\n",
- "Zz=7.0\n",
- "Es=16.0\n",
- "Vo=6.2\n",
- "Il=59.5*10**-3\n",
- "\n",
- "#Calculations\n",
- "es=(10*Es)/100.0 #10% os Es\n",
- "Rl=Vo/Il\n",
- "print(\"es*Zz||Rl/R1+Zz||Rl\")\n",
- "V0=es*((Zz*Rl)/(Zz+Rl))/(R1+((Zz*Rl)/(Zz+Rl)))\n",
- "lineregulation=(V0*100)/Vo\n",
- "print(\"line regulation voltage is %3.3fpercentage \" %lineregulation)\n",
- "V0=Il*((Zz*R1)/(Zz+R1))\n",
- "loadregulation=(V0*100)/Vo\n",
- "print(\"loadregulation voltage is %3.3fpercentage \" %loadregulation)\n",
- "Rr=((Zz*Rl)/(Zz+Rl))/(R1+(Zz*Rl)/(Zz+Rl))\n",
- "\n",
- "#Results\n",
- "print(\"ripple rejection is %3.2f X 10^-2 \" %(Rr*10**2))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "es*Zz||Rl/R1+Zz||Rl\n",
- "line regulation voltage is 1.068percentage \n",
- "loadregulation voltage is 6.422percentage \n",
- "ripple rejection is 4.14 X 10^-2 \n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.19, Page No 114"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#initialisation of variables\n",
- "\n",
- "E=9.0\n",
- "Vf=.7\n",
- "\n",
- "#Calculations\n",
- "If=1.0*10**-3\n",
- "Vo=E-Vf\n",
- "R1=Vo/If\n",
- "Vr=E\n",
- "\n",
- "#Results\n",
- "print(\"diode forward voltage is %3.2fohm \" %Vr)\n",
- "print(\"diode forward current is %3.1fA \" %(If*10**3))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "diode forward voltage is 9.00ohm \n",
- "diode forward current is 1.0A \n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.20, Page No 117"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#initialisation of variables\n",
- "E=5.0\n",
- "Vo=4.5\n",
- "Il=2.0*10**-3\n",
- "\n",
- "#Calculations\n",
- "R1=(E-Vo)/Il\n",
- "print(\" suitable resistance is %dohm \" %R1)\n",
- "Vr=E\n",
- "print(\"when diode is forward baised\")\n",
- "If=(E-Vf)/R1\n",
- "\n",
- "#Results\n",
- "print(\" diode forward current is %3.2fA \" %(If*10**3))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " suitable resistance is 250ohm \n",
- "when diode is forward baised\n",
- " diode forward current is 17.20A \n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.21, Page No 119"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#initialisation of variables\n",
- "Vo=2.7\n",
- "Vf=.7\n",
- "E=9.0\n",
- "If=1*10**-3\n",
- "\n",
- "#Calculations\n",
- "Il=If\n",
- "Vb=Vo-Vf\n",
- "R1=(E-Vo)/(Il+If)\n",
- "\n",
- "#Results\n",
- "print(\"resistance is %.2f kOhm \" %(R1/10**3))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "resistance is 3.15 kOhm \n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.22, Page No 120"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#initialisation of variables\n",
- "Vo=5.0\n",
- "Vf=0.7\n",
- "Iz=5.0\n",
- "Il=1.0\n",
- "E=20.0\n",
- "\n",
- "#Calculations\n",
- "Vz=Vo-Vf\n",
- "R1=(E-Vo)/(Il+Iz)\n",
- "\n",
- "#Results\n",
- "print(\"zener diode resistance si %.2f ohm \" %R1)\n",
- "#Answer in the book is wrong"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "zener diode resistance si 2.50 ohm \n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.23, Page No 122"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#initialisation of variables\n",
- "E=10.0\n",
- "R1=56.0*10**3\n",
- "f=1000.0\n",
- "C1=1.0*10**-6\n",
- "\n",
- "#Calculations\n",
- "Vo=2*E\n",
- "Ic=Vo/R1\n",
- "t=1/(2*f)\n",
- "Vc=(Ic*t)/C1\n",
- "\n",
- "#Results\n",
- "print(\" tilt output voltage is %3.2fV \" %(Vc*10**3))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " tilt output voltage is 178.57V \n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.24, Page No 124"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#initialisation of variables\n",
- "f=500.0\n",
- "Rs=600.0\n",
- "E=8.0\n",
- "\n",
- "#Calculations\n",
- "t=1.0/(2*f)\n",
- "PW=t\n",
- "C1=PW/Rs\n",
- "Vo=2.0*E\n",
- "Vc=(1*Vo)/100#1% of the Vo\n",
- "Ic=(Vc*C1)/t\n",
- "R1=(2*E)/(Ic*1000)\n",
- "\n",
- "#Results\n",
- "print(\"suitable value of R1 is %.2f ohm \" %R1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "suitable value of R1 is 60.00 ohm \n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.25, Page No 125"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#initialisation of variables\n",
- "Vf=0.7\n",
- "E=6.0\n",
- "Vb1=3.0\n",
- "\n",
- "#Calculations\n",
- "Vc=Vb1-Vf-(-E)\n",
- "Vo=Vb1-Vf\n",
- "print(\"when input is -E\")\n",
- "Vo=E+Vc\n",
- "Vo=Vb1+Vf\n",
- "print(\"Capicitor voltage is %.2f ohm \" %Vc)\n",
- "print(\"when input is +E\")\n",
- "Vo=E+(Vc)\n",
- "\n",
- "#Results\n",
- "print(\"Capicitor voltage is %.2f ohm \" %Vo)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "when input is -E\n",
- "Capicitor voltage is 8.30 ohm \n",
- "when input is +E\n",
- "Capicitor voltage is 14.30 ohm \n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.26, Page No 130"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#initialisation of variables\n",
- "E=12.0\n",
- "Vf=0.7\n",
- "Rl=47*10**3\n",
- "f=5000.0\n",
- "\n",
- "#Calculations\n",
- "Vo=2*(E-Vf)\n",
- "Il=Vo/Rl\n",
- "print(\" capacitor discharge time\")\n",
- "t=1.0/(2*f)\n",
- "print(\" for 1% ripple allow .5% due to discharge of C2 %.5%due to discharge of C1\")\n",
- "Vc=(.5*Vo)/100\n",
- "C2=((Il*t)/Vc)*10**6\n",
- "print(\" value of capacitor C2 is %3.2fuF \" %C2)\n",
- "C1=2*C2\n",
- "\n",
- "#Results\n",
- "print(\"value of capacitor C1 is %3.2fuF \" %C1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " capacitor discharge time\n",
- " for 1% ripple allow .5% due to discharge of C2 %.5%due to discharge of C1\n",
- " value of capacitor C2 is 0.43uF \n",
- "value of capacitor C1 is 0.85uF \n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.27, Page No 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#initialisation of variables\n",
- "\n",
- "Vcc=5.0\n",
- "Vf=.7\n",
- "R1=3.3*10**3\n",
- "\n",
- "#Calculations\n",
- "print(\"A)\")\n",
- "Ir1=(Vcc-Vf)/R1\n",
- "print(\"diode forward current when all input are low is %3.4fA \" %Ir1)\n",
- "print(\"for each diode\")\n",
- "If=Ir1/3\n",
- "print(\"B)\")\n",
- "If2=Ir1/2\n",
- "If3=If2\n",
- "print(\" forward current when input A is high is %3.5fA \" %If3)\n",
- "print(\"C)\")\n",
- "If3=Ir1\n",
- "\n",
- "#Results\n",
- "print(\" forward current when input A and B are high and C is low %3.2fA \" %(If3*10**3))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "A)\n",
- "diode forward current when all input are low is 0.0013A \n",
- "for each diode\n",
- "B)\n",
- " forward current when input A is high is 0.00065A \n",
- "C)\n",
- " forward current when input A and B are high and C is low 1.30A \n"
- ]
- }
- ],
- "prompt_number": 27
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file |