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@@ -1,206 +1,188 @@
-{
- "metadata": {
- "name": "Ch 4"
- },
- "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": [
- "#Example 4.1\n",
- "# Assuming the diode resistance to be zero \n",
- "#Determine DC Voltage across the load and PIV of the Diode\n",
- "\n",
- "#Given Circuit Data\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": [
- "#Example 4.2\n",
- "#Determine DC Voltage across the load and \n",
- "#PIV of the \n",
- "#Centre Tap Rectifier and Bridge Rectifier\n",
- "\n",
- "#Given Circuit Data\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": [
- "#Example 4.3(a\n",
- "# determine the Peak Value of Current\n",
- "\n",
- "#Given Circuit Data\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": [
- "#Example 4.4\n",
- "# determine Maximum Current the Given Zener Diode can handle\n",
- "\n",
- "#Given Circuit Data\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": [
- "#Example 4.5\n",
- "# determine Capacitance of Varactor Diode if the\n",
- "#Reverse-Bias Voltage is increased from 4V to 8V \n",
- "\n",
- "#Given Circuit Data\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": {}
- }
- ]
+{
+ "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": {}
+ }
+ ]
} \ No newline at end of file
generated by cgit (git 2.34.1) at 2025-02-10 06:32:00 +0000