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{
"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": {}
}
]
}
|
