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| author | hardythe1 | 2015-06-03 15:27:17 +0530 |
|---|---|---|
| committer | hardythe1 | 2015-06-03 15:27:17 +0530 |
| commit | df60071cf1d1c18822d34f943ab8f412a8946b69 (patch) | |
| tree | ab059cf19bad4a1233a464ccf5d72cf8b3fb323c /Integrated_Electronics__Analog_And_Digital_Circuits_and_Systems/Chapter4.ipynb | |
| parent | fba055ce5aa0955e22bac2413c33493b10ae6532 (diff) | |
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diff --git a/Integrated_Electronics__Analog_And_Digital_Circuits_and_Systems/Chapter4.ipynb b/Integrated_Electronics__Analog_And_Digital_Circuits_and_Systems/Chapter4.ipynb new file mode 100755 index 00000000..353ad61d --- /dev/null +++ b/Integrated_Electronics__Analog_And_Digital_Circuits_and_Systems/Chapter4.ipynb @@ -0,0 +1,262 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 04 : Diode Circuits"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.3, Page No 91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Vr=0.6 #v\n",
+ "\n",
+ "#Calculations\n",
+ "i=(1-Vr)/2.01\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print(\"The current i is= %.2f mA \" %(i))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current i is= 0.20 mA \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4, Page No 92"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Vr=0.7 #v\n",
+ "\n",
+ "#Calculations\n",
+ "i2=((1-Vr)/300.0*10**3)\n",
+ "i1=3-i2\n",
+ "\n",
+ "#Results\n",
+ "print(\"The current i is= %.2f mA \" %(i1))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current i is= 2.00 mA \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.5, Page No 95"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Rl=9.0 # kohm\n",
+ "R=1.0 # kohm\n",
+ "Vr=1.0 #v\n",
+ "\n",
+ "#Calculations\n",
+ "v0=(Rl/(R+Rl))*Vr\n",
+ "\n",
+ "#Results\n",
+ "print(\"The output voltage is= %.2f V \" %(v0))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The output voltage is= 0.90 V \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.7, Page No 101"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Vs=1.0 #v\n",
+ "Rc=100.0 #ohm\n",
+ "RL=1000 #ohm\n",
+ "\n",
+ "#Calculations\n",
+ "Vcmin=Vs*(2+(Rc/RL))\n",
+ "\n",
+ "#Results\n",
+ "print(\"The concentration of germanium atom is= %.2f X 10^22 atom/cm^3 \" %(Vcmin))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The concentration of germanium atom is= 2.10 X 10^22 atom/cm^3 \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.8, Page No 105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "Vrms=220.0 #v\n",
+ "Rf=10.0\n",
+ "Rl=500.0\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "Im=(math.sqrt(2.0)*Vrms)/(Rf+Rl)\n",
+ "Idc=Im/math.pi\n",
+ "Irms=Im/2\n",
+ "Vdc=(-Im*Rl)/math.pi\n",
+ "pd=Vrms*Irms\n",
+ "reg=Rf/Rl*100\n",
+ "\n",
+ "\n",
+ "#Required Formula\n",
+ "print(\"Peak load current is = %.2f A \" %Im)\n",
+ "print(\"DC load current is = %.2f A \" %Idc)\n",
+ "print(\"RMS load current is = %.2f A \" %Irms)\n",
+ "print(\"The dc diode voltage is = %.2f V \" %Vdc)\n",
+ "print(\"The total input power to the curcuit is = %.2f A \" %pd)\n",
+ "print(\"percentage regulation will go from 0 percent to %.f percent \" %reg)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Peak load current is = 0.61 A \n",
+ "DC load current is = 0.19 A \n",
+ "RMS load current is = 0.31 A \n",
+ "The dc diode voltage is = -97.09 V \n",
+ "The total input power to the curcuit is = 67.11 A \n",
+ "percentage regulation will go from 0 percent to 2 percent \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.9 Page No 107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "RL=100.0 #ohm\n",
+ "Rf=10.0 #ohm\n",
+ "Vm=5.0 #v\n",
+ "Vr=0.6 #v\n",
+ "\n",
+ "#Calculations\n",
+ "Vdc=2*(((Vm/math.pi)*(RL/(RL+Rf)))-Vr)\n",
+ "\n",
+ "#Results\n",
+ "print(\"The dc output voltage is = %.2f v \" %Vdc)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The dc output voltage is = 1.69 v \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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