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+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 4 Diode Applications"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.1 Page No 83"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The dc voltage across the load resistor = 17.99 volts\n",
+ "The peak inverse voltage across the diode = 56.58 volts\n",
+ "The dc current through the diode = 0.90 A\n"
+ ]
+ }
+ ],
+ "source": [
+ "# given data\n",
+ "V2rms= 40.0## V\n",
+ "R_L= 20.0## Ω\n",
+ "V2peak= V2rms/0.707## V\n",
+ "Vout_peak= V2peak## V\n",
+ "# The dc voltage across the load resistor \n",
+ "Vdc=0.318*Vout_peak## V\n",
+ "#The peak inverse voltage across the diode \n",
+ "PIV= V2peak## V\n",
+ "Idc= Vdc/R_L## A\n",
+ "# The dc current through the diode \n",
+ "I_diode= Idc## A\n",
+ "print \"The dc voltage across the load resistor = %.2f volts\"%Vdc\n",
+ "print \"The peak inverse voltage across the diode = %.2f volts\"%PIV\n",
+ "print \"The dc current through the diode = %.2f A\"%I_diode"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.2 Page No 86"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The dc load voltage = 17.99 volts\n",
+ "The peak inverse voltage across each diode = 56.58 volts\n",
+ "The dc current through each diode = 0.45 A\n"
+ ]
+ }
+ ],
+ "source": [
+ "# given data\n",
+ "Vrms= 40.0## V\n",
+ "R_L= 20.0## Ω\n",
+ "V2peak= Vrms/0.707## V\n",
+ "Vout_peak= V2peak/2## V\n",
+ "# The dc load voltage \n",
+ "Vdc=0.636*Vout_peak## V\n",
+ "# The peak inverse voltage across each diode \n",
+ "PIV= V2peak## V\n",
+ "Idc= Vdc/R_L## A\n",
+ "# The dc current through each diode \n",
+ "I_diode= Idc/2## A\n",
+ "print \"The dc load voltage = %.2f volts\"%Vdc\n",
+ "print \"The peak inverse voltage across each diode = %.2f volts\"%PIV\n",
+ "print \"The dc current through each diode = %.2f A\"%I_diode"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.3 Page No 88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The dc load voltage = 35.98 volts\n",
+ "The peak inverse voltage across each diode = 56.58 volts\n",
+ "The dc current through each diode = 0.90 A\n"
+ ]
+ }
+ ],
+ "source": [
+ "# given data\n",
+ "Vrms= 40.0## V\n",
+ "R_L= 20.0## Ω\n",
+ "V2peak= Vrms/0.707## V\n",
+ "Vout_peak= V2peak## V\n",
+ "# The value of Vdc \n",
+ "Vdc=0.636*Vout_peak## V\n",
+ "# The value of PIV \n",
+ "PIV= V2peak## V\n",
+ "Idc= Vdc/R_L## A\n",
+ "#The value of I_diode\n",
+ "I_diode= Idc/2## A\n",
+ "print \"The dc load voltage = %.2f volts\"%Vdc\n",
+ "print \"The peak inverse voltage across each diode = %.2f volts\"%PIV\n",
+ "print \"The dc current through each diode = %.2f A\"%I_diode"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.6 Page No 94"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The peak-to-peak ripple = 4.72 volts\n",
+ "The dc load voltage = 54.24 volts\n"
+ ]
+ }
+ ],
+ "source": [
+ "# given data\n",
+ "Vdc= 56.6## V\n",
+ "R_L= 100.0## Ω\n",
+ "f=120.0## Hz\n",
+ "C= 1000.0## µF\n",
+ "C= C*10**-6## F\n",
+ "V2peak= Vdc## V\n",
+ "Idc= Vdc/R_L## A\n",
+ "# The peak-to-peak ripple \n",
+ "Vrip= Idc/(f*C)## V\n",
+ "# The dc load voltage \n",
+ "Vdc= V2peak-Vrip/2## V\n",
+ "print \"The peak-to-peak ripple = %.2f volts\"%Vrip\n",
+ "print \"The dc load voltage = %.2f volts\"%Vdc"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.8 Page No 98"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The zener current = 5.36 mA\n"
+ ]
+ }
+ ],
+ "source": [
+ "# given data\n",
+ "V2rms= 12.6## V\n",
+ "V_Z= 6.8## V\n",
+ "V2peak= V2rms/0.707## V\n",
+ "Vin= V2peak## V\n",
+ "Vout= V_Z## V\n",
+ "R_L= 1.2## kΩ\n",
+ "R_L= R_L*10**3##in Ω\n",
+ "Rs= 1## kΩ\n",
+ "Rs= Rs*10**3## Ω\n",
+ "Is= (Vin-Vout)/Rs## A\n",
+ "I_L= Vout/R_L## A\n",
+ "# The zener current \n",
+ "Iz= Is-I_L## A\n",
+ "Iz= Iz*10**3## mA\n",
+ "print \"The zener current = %.2f mA\"%Iz\n",
+ "# Note: The calculation = %.2f the book is not accurate.\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.9 Page No 98"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The ripple across the load resistance = 4.56 mV\n"
+ ]
+ }
+ ],
+ "source": [
+ "# given data\n",
+ "C= 100.0##in µF\n",
+ "C= C*10**-6## F\n",
+ "Rz= 5.0##in Ω\n",
+ "Rs= 1.0*10**3##in Ω\n",
+ "Idc= 11.0*10**-3##in A\n",
+ "f=120.0##in Hz\n",
+ "Vin_rip= Idc/(f*C)## V\n",
+ "# The ripple across the load resistance \n",
+ "Vout_rip= Rz*Vin_rip/(Rs+Rz)##in A\n",
+ "Vout_rip= Vout_rip*10**3## mV\n",
+ "print \"The ripple across the load resistance = %.2f mV\"%Vout_rip"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}