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{
"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": {}
}
]
}
|
