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diff --git a/Basic_Electronics_and_Linear_Circuits/ch4.ipynb b/Basic_Electronics_and_Linear_Circuits/ch4.ipynb deleted file mode 100755 index 701f4aaf..00000000 --- a/Basic_Electronics_and_Linear_Circuits/ch4.ipynb +++ /dev/null @@ -1,188 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:ae9612ced78590c195456849014e70cbd1cdee113840d747c19ee249579f79a3" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 4:Semiconductor Diode" - ] - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 4.1 Page No.85" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "Vrms=220 #Volts, power supply\n", - "n2=1 #Assumption\n", - "n1=12*n2 #Turns Ratio\n", - "\n", - "#Calculation\n", - "import math\n", - "Vp=math.sqrt(2)*Vrms #Maximum(Peak) Primary Voltage\n", - "Vm=n2*Vp/n1 #Maximum Secondary Voltage\n", - "Vdc=Vm/math.pi #DC load Voltage \n", - "# Results \n", - "print \"The DC load Voltage is = \",round(Vdc,2),\"V\"\n", - "print \"The Peak Inverse Voltage(PIV) is = \",round(Vm,1),\"V\"" - ], - "language": "python", - "metadata": {}, - "outputs": [] - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 4.2 Page No.90" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "Vrms=220 #Volts, power supply rms voltage\n", - "n2=1 #Assumption\n", - "n1=12*n2 #Turns Ratio\n", - "\n", - "#Calculation\n", - "import math\n", - "Vp=math.sqrt(2)*Vrms #Maximum(Peak Primary Voltage\n", - "Vm=n2*Vp/n1 #Maximum Secondary Voltage\n", - "Vdc=2*Vm/math.pi #DC load Voltage \n", - "\n", - "# Results \n", - "print \"The DC load Voltage is = \",round(Vdc,1),\"V\"\n", - "print \"The Peak Inverse Voltage(PIV of Bridge Rectifier is = \",round(Vm,1),\"V\"\n", - "print \"The Peak Inverse Voltage(PIV of Centre-tap Rectifier is = \",round(2*Vm,1),\"v\"" - ], - "language": "python", - "metadata": {}, - "outputs": [] - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 4.3 Page No.95" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "import math\n", - "Rl=1000.0 #Ohms, load resistance\n", - "rd=10.0 #Ohms forward base dynamic resistance\n", - "Vm=220.0 #Volts(Peak Value of Voltage)\n", - "#Calculation\n", - "Im=Vm/(rd+Rl) #Peak Value of Current\n", - "\n", - "# Result\n", - "print \"The Peak Value of Current is = \",round(Im*1000,1),\"mA\"\n", - "\n", - "#(b) dc or av value of current\n", - "\n", - "Idc=2*Im/math.pi #DC Value of Current\n", - "# Results \n", - "print \"The DC or Average Value of Current is \",round(Idc*1000,2),\"mA\"\n", - "\n", - "#(c)\n", - "Irms=Im/math.sqrt(2) #RMS Value of Current\n", - "# Results \n", - "print \"The RMS Value of Current is = \",round(Irms*1000,1),\"mA\"\n", - "\n", - "#(d)\n", - "r=math.sqrt((Irms/Idc)**2-1) #Ripple Factor\n", - "# Results i\n", - "print \" The Ripple Factor r = \",round(r,3)\n", - "\n", - "#(e)\n", - "Pdc=Idc**2*Rl\n", - "Pac=Irms**2*(rd+Rl)\n", - "n=Pdc/Pac #Rectification Efficiency\n", - "# Results\n", - "print \"The Rectification EFficiency n(eeta) = percent.\",round(n*100,2)" - ], - "language": "python", - "metadata": {}, - "outputs": [] - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 4.4 Page No.103" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "Vz=9.1 #Volts\n", - "P=0.364 #Watts\n", - "#Calculation\n", - "Iz=P/Vz\n", - "#Result\n", - "print \" The Maximum permissible Current is \",Iz*1000,\"mA\"" - ], - "language": "python", - "metadata": {}, - "outputs": [] - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 4.5 Page No.105" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "Ci=18*10**(-12) #i.e. 18 pF, capacitance of diode\n", - "Vi=4 #volt, initial voltage\n", - "Vf=8 #v, final voltage\n", - "\n", - "#Calculation\n", - "import math\n", - "Vf=8 \n", - "K=Ci*math.sqrt(Vi)\n", - "Cf=K/math.sqrt(Vf)\n", - "#Result\n", - "print \" The Final Value of Capacitance is C = \",round(Cf/10**(-12),3),\"pF\"" - ], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
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