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diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter1.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter1.ipynb new file mode 100644 index 00000000..dff05eac --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter1.ipynb @@ -0,0 +1,214 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1 : Electricity" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 1_5 Page No. 24" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Output Voltage of a Battery = 7.20 Volts\n" + ] + } + ], + "source": [ + "# What is the output voltage of a battery that expends 3.6 J of energy in moving 0.5C of charge?\n", + "\n", + "# Given data\n", + "\n", + "W = 3.6# # Work=3.6 Jouls\n", + "Q = 0.5# # Charge=0.5 Columb\n", + "\n", + "V = W/Q#\n", + "print 'The Output Voltage of a Battery = %0.2f Volts'%V" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 1_6 Page No. 24" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Intensity of Charge Flow = 12.00 Amps\n" + ] + } + ], + "source": [ + "#The charge of 12 C moves past a given point every second. How much is the intensity of charge flow?\n", + "\n", + "# Given data\n", + "\n", + "Q = 12# # Charge=12 Columb\n", + "T = 1# # Time=1 Sec i.e every second\n", + "\n", + "I = Q/T#\n", + "print 'The Intensity of Charge Flow = %0.2f Amps'%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 1_7 Page No. 24" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current = 5.00 Amps\n" + ] + } + ], + "source": [ + "# The charge of 5 C moves past a given point in 1 s. How much is the current?\n", + "\n", + "# Given data\n", + "\n", + "Q = 5# # Charge=5 Columb\n", + "T = 1# # Time=1 Sec\n", + "\n", + "I = Q/T#\n", + "print 'The Current = %0.2f Amps'%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 1_8 Page No. 25" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistance for Conductance value 0.05 S = 20.00 Ohms\n", + "The Resistance for Conductance value 0.1 S = 10.00 Ohms\n" + ] + } + ], + "source": [ + "# Calculate the resistance for the following conductance values: (a) 0.05 S (b) 0.1 S\n", + "\n", + "# Given data\n", + "\n", + "G1 = 0.05# # G1=0.05 Siemins\n", + "G2 = 0.1# # G1=0.1 Siemins\n", + "\n", + "R1 = 1/G1#\n", + "print 'The Resistance for Conductance value 0.05 S = %0.2f Ohms'%R1\n", + "\n", + "R2 = 1/G2#\n", + "print 'The Resistance for Conductance value 0.1 S = %0.2f Ohms'%R2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 1_9 Page No. 26" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Conductance for Resistance value 1 kohms = 1e-03 Siemens\n", + "OR 1 mS\n", + "The Conductance for Resistance value 5 kohms = 2e-04 Siemens\n", + "OR 200 uS\n" + ] + } + ], + "source": [ + "#Calculate the conductance for the following resistance values: (a) 1 kOhms \b(b)5 kOhms\n", + "\n", + "# Given data\n", + "\n", + "R1 = 1.0*10**3# # R1=1k Ohms\n", + "R2 = 5.0*10**3# # R2=5k Ohms\n", + "\n", + "G1 = 1/R1#\n", + "print 'The Conductance for Resistance value 1 kohms = %0.e Siemens'%G1\n", + "print 'OR 1 mS'\n", + "\n", + "G2 = 1/R2#\n", + "print 'The Conductance for Resistance value 5 kohms = %0.e Siemens'%G2\n", + "print 'OR 200 uS'" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter11.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter11.ipynb new file mode 100644 index 00000000..4e938ea1 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter11.ipynb @@ -0,0 +1,246 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11 : Conductors and Insulators" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 11_1 Page No. 329" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Circular Area = 25.00 cmils\n" + ] + } + ], + "source": [ + "# What is the area in circular mils of a wire with a diameter of 0.005 in.?\n", + "\n", + "# Given data\n", + "\n", + "Din = 0.005# # Diameter in Inches=0.005 in.\n", + "Dmil = 5# # Diameter in Mils=5 mil.\n", + "\n", + "# 0.005 in. = 5 mil\n", + "# Therefore: Din == Dmil\n", + "\n", + "A = Dmil*Dmil#\n", + "print 'The Circular Area = %0.2f cmils'%A" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 11_2 Page No. 335" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Total Area = 3224.00 cmils\n" + ] + } + ], + "source": [ + "# A stranded wire is made up of 16 individual strands of No. 27 gage wire. What is its equivalent gage size in solid wire?\n", + "\n", + "# Given data\n", + "\n", + "N = 16# # No. of strands=16\n", + "A27 = 201.5 # Circular area of No. 27 Guage wire=201.5 cmils\n", + "\n", + "A = N*A27#\n", + "print 'The Total Area = %0.2f cmils'%A" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 11_3 Page No. 340" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistance of 100 ft of No. 20 gage Copper Wire = 1.02 ohms\n" + ] + } + ], + "source": [ + "# How much is the resistance of 100 ft of No. 20 gage copper wire?\n", + "\n", + "# Given data\n", + "\n", + "roh = 10.4# # roh or specific resistance=10.4 (for Copper)\n", + "l = 100.# # Lenght=100 feet\n", + "A = 1022# # Area of No. 20 Gage=1022 cmil\n", + "\n", + "R = roh*(l/A)#\n", + "print 'The Resistance of 100 ft of No. 20 gage Copper Wire = %0.2f ohms'%R" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 11_4 Page No. 342" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistance of 100 ft of No. 20 gage Copper Wire = 2.04 ohms\n" + ] + } + ], + "source": [ + "# How much is the resistance of 100 ft of No. 23 gage copper wire?\n", + "\n", + "# Given data\n", + "\n", + "roh = 10.4# # roh or specific resistance=10.4 (for Copper)\n", + "l = 100# # Lenght=100 feet\n", + "A = 509.5# # Area of No. 23 Gage=509.5 cmil\n", + "\n", + "R = roh*(l/A)#\n", + "print 'The Resistance of 100 ft of No. 20 gage Copper Wire = %0.2f ohms'%R" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 11_5 Page No. 344" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistance of a Slab of Germanium = 11.00 ohms\n" + ] + } + ], + "source": [ + "# How much is the resistance of a slab of germanium 0.2 cm long with a crosssectional area of 1 sqcm?\n", + "\n", + "# Given data\n", + "\n", + "roh = 55.0# # roh or specific resistance=55 (for Germanium)\n", + "l = 0.2*10**-2# # Lenght=100 feet\n", + "A = 1.0*10**-2# # Area=1 sqcm\n", + "\n", + "R = roh*(l/A)#\n", + "print 'The Resistance of a Slab of Germanium = %0.2f ohms'%R" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 11_6 Page No. 344" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistance at 120°C = 21.00 ohms\n" + ] + } + ], + "source": [ + "# A tungsten wire has a 14-Ohms\u0002 R at 20°C. Calculate its resistance at 120°C.\n", + "\n", + "# Given data\n", + "\n", + "Tmax = 120.# # Temp(max)=120 degree Centigrates\n", + "Tmin = 20.# # Temp(min)=20 degree Centigrates\n", + "Ro = 14.# # Wire Resistance=14 Ohms\n", + "alpha = 0.005# # Aplha=0.005 (for Tungsten)\n", + "\n", + "delta = Tmax-Tmin#\n", + "\n", + "Rt = Ro+Ro*(alpha*delta)#\n", + "print 'The Resistance at 120°C = %0.2f ohms'%Rt" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter12.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter12.ipynb new file mode 100644 index 00000000..8876ae94 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter12.ipynb @@ -0,0 +1,67 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 12 : Batteries" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 12_1 Page No. 368" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistance ri = 10 Ohms\n" + ] + } + ], + "source": [ + "# Calculate ri if the output of a generator drops from 100 V with zero load current to 80 V when Il is 2 A.\n", + "\n", + "# Given data\n", + "\n", + "Vo0 = 100# # Vo at zero load current=100 Volts\n", + "Vo1 = 80# # Vo at 2 A load current=80 Volts\n", + "Il = 2# # Load current=2 Amps\n", + "\n", + "Ri = (Vo0-Vo1)/Il#\n", + "print 'The Resistance ri = %0.f Ohms'%Ri" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter13.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter13.ipynb new file mode 100644 index 00000000..89a82ed0 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter13.ipynb @@ -0,0 +1,182 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 13 : Magnetism" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 13_1 Page No. 291" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The 25000 Maxwell = 2.50e-04 Wabers\n", + "i.e 250*10**-6 Wb or 250 uWb\n", + "The 0.005 Wabers = 500000.00 Maxwell\n", + "i.e 5.0*10**5 Mx\n" + ] + } + ], + "source": [ + "# Make the following conversions: (a) 25,000 Mx to Wb# (b) 0.005 Wb to Mx.\n", + "\n", + "# Given data\n", + "\n", + "A = 25000.0# # A=25000 Maxwell\n", + "B = 0.005# # B=0.005 Wabers\n", + "C = 1*10**8# # Conversion Factor\n", + "\n", + "Wb = A*(1.0/C)#\n", + "print 'The 25000 Maxwell = %0.2e Wabers'%Wb\n", + "print 'i.e 250*10**-6 Wb or 250 uWb'\n", + "\n", + "Mx = B*C#\n", + "print 'The 0.005 Wabers = %0.2f Maxwell'%Mx\n", + "print 'i.e 5.0*10**5 Mx'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 13_2 Page No. 392" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Flux Density = 2000.00 Guass (G)\n" + ] + } + ], + "source": [ + "# With a flux of 10,000 Mx through a perpendicular area of 5 sqcm, what is the flux density in gauss?\n", + "\n", + "# Given data\n", + "\n", + "A = 5.# # Area=5 sqcm\n", + "flux = 10000.0# # Total Flux=10000 Mx\n", + "\n", + "B = flux/A#\n", + "print 'The Flux Density = %0.2f Guass (G)'%B" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 13_3 Page No. 394" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Flux Density = 0.80 Tesla (T)\n" + ] + } + ], + "source": [ + "# With a flux of 400 u\u0002Wb through an area of 0.0005 sqm, what is the flux density B in tesla units?\n", + "\n", + "# Given data\n", + "\n", + "A = 0.0005# # Area=0.0005 sqm\n", + "flux = 400*10**-6# # Total Flux=400 uWb\n", + "\n", + "B = flux/A#\n", + "print 'The Flux Density = %0.2f Tesla (T)'%B" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 13_4 Page No. 394" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The 0.003 Tesla = 30.00 Guass\n", + "The 15,000 Guass = 1.50 Tesla\n" + ] + } + ], + "source": [ + "# Make the following conversions: (a) 0.003 T to G# (b) 15,000 G to T.\n", + "\n", + "# Given data\n", + "\n", + "A = 0.003# # A=0.003 Tesla\n", + "B = 15000.# # B=15000 Guass\n", + "C = 1.*10**4# # Conversion Factor\n", + "\n", + "G = A*C#\n", + "print 'The 0.003 Tesla = %0.2f Guass'%G\n", + "\n", + "T = B*(1/C)#\n", + "print 'The 15,000 Guass = %0.2f Tesla'%T" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter14.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter14.ipynb new file mode 100644 index 00000000..c8957744 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter14.ipynb @@ -0,0 +1,248 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter14 : Electromagnetism" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 14_1 Page No. 421" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Amps-Turn (A.t) of Magneto-Motive Force (mmf) = 10.00 A.t\n" + ] + } + ], + "source": [ + "# Calculate the ampere-turns of mmf for a coil with 2000 turns and a 5-mA current.\n", + "\n", + "# Given data\n", + "\n", + "I = 5*10**-3# # Current=5 mAmps\n", + "N = 2000# # No. of Turns=2000\n", + "\n", + "mmf = I*N#\n", + "print 'The Amps-Turn (A.t) of Magneto-Motive Force (mmf) = %0.2f A.t'%mmf" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 14_2 Page No. 421" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Turns necessary are : 150\n" + ] + } + ], + "source": [ + "# A coil with 4 A is to provide a magnetizing force of 600 A\u0002 t. How many turns are necessary?\n", + "\n", + "# Given data\n", + "\n", + "I = 4# # Current=4 Amps\n", + "mmf = 600# # Magnetizing Force=600 A.t\n", + "\n", + "N = mmf/I#\n", + "print 'The Turns necessary are : ',N" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 14_3 Page No. 424" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current necessary = 2 Amps\n" + ] + } + ], + "source": [ + "# A coil with 400 turns must provide 800 A\u0002 t of magnetizing force. How much current is necessary?\n", + "\n", + "# Given data\n", + "\n", + "mmf = 800# # Magnetizing Force=800 A.t\n", + "N = 400# # No. of Turns=400\n", + "\n", + "I = mmf/N#\n", + "print 'The Current necessary = %0.f Amps'%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 14_4 Page No. 426" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current necessary when a wire is connected to 6-V Battery = 2 Amps\n", + "The Amps-Turn (A.t) of Magneto-Motive Force (mmf) = 500 A.t\n" + ] + } + ], + "source": [ + "# The wire in a solenoid of 250 turns has a resistance of 3 Ohms. (a)How much is the current when the coil is connected to a 6-V battery? (b) Calculate the ampereturns of mmf.\n", + "\n", + "# Given data\n", + "\n", + "V = 6# # Voltage=6 Volts\n", + "R = 3# # Resistance=3 Ohms\n", + "N = 250# # No. of Turns=250\n", + "\n", + "I = V/R#\n", + "print 'The Current necessary when a wire is connected to 6-V Battery = %0.f Amps'%I\n", + "\n", + "mmf = I*N#\n", + "print 'The Amps-Turn (A.t) of Magneto-Motive Force (mmf) = %0.f A.t'%mmf" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 14_5 Page No. 426" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Absolute u as B/H in CGS = 500 (G/Oe)\n", + "The Absolute u as B/H in SI = 6.300e-04 (T/(A.t/m))\n", + "i.e 630*10**-6 T/(A.t/m)\n" + ] + } + ], + "source": [ + "# A magnetic material has a \u0003ur of 500. Calculate the absolute u\u0003 as B/H (a) in CGS units and (b) in SI units.\n", + "\n", + "# Given data\n", + "\n", + "ur = 500# # ur=500\n", + "uoa = 1# # uo for CGS Units=1\n", + "uob = 1.26*10**-6# # uo for SI Units=1.26 u\n", + "\n", + "ua = ur*uoa#\n", + "print 'The Absolute u as B/H in CGS = %0.f (G/Oe)'%ua\n", + "\n", + "ub = ur*uob#\n", + "print 'The Absolute u as B/H in SI = %0.3e (T/(A.t/m))'%ub\n", + "print 'i.e 630*10**-6 T/(A.t/m)'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 14_6 Page No. 427" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Flux density = 0.63 Tesla\n" + ] + } + ], + "source": [ + "# u = 630*10**-\u00056 in SI units, calculate the flux density B that will be produced by the field intensity H equal to 1000 A.t/m.\n", + "\n", + "# Given data\n", + "\n", + "u = 630*10**-6# # u=630 micro T/(A.t/m)\n", + "H = 1000# # H=1000 A.t/m\n", + "\n", + "B = u*H#\n", + "print 'The Flux density = %0.2f Tesla'%B" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter15.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter15.ipynb new file mode 100644 index 00000000..682677a5 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter15.ipynb @@ -0,0 +1,354 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 15 : Alternating voltage & current" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 15_1 Page No: 451" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage at 30° = 50.00 Volts\n", + "The Voltage at 45° = 70.71 Volts\n", + "The Voltage at 90° = 100.00 Volts\n", + "The Voltage at 270° = -100.00 Volts\n" + ] + } + ], + "source": [ + "from numpy import sin, pi\n", + "# A sine wave of voltage varies from zero to a maximum of 100 V. How much is the voltage at the instant of 30° of the cycle? 45°? 90°? 270°?\n", + "\n", + "# Given data\n", + "\n", + "Vm = 100# # Vm=100 Volts\n", + "t1 = 30# # Theta 1=30°.\n", + "t2 = 45# # Theta 2=45°.\n", + "t3 = 90# # Theta 3=90°.\n", + "t4 = 270# # Theta 4=270°.\n", + "\n", + "v1 = Vm*sin(t1*pi/180)\n", + "print 'The Voltage at 30° = %0.2f Volts'%v1\n", + "\n", + "v2 = Vm*sin(t2*pi/180)\n", + "print 'The Voltage at 45° = %0.2f Volts'%v2\n", + "\n", + "v3 = Vm*sin(t3*pi/180)\n", + "print 'The Voltage at 90° = %0.2f Volts'%v3\n", + "\n", + "v4 = Vm*sin(t4*pi/180)\n", + "print 'The Voltage at 270° = %0.2f Volts'%v4" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 15_2 Page No: 456" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Time period = 1e-03 Seconds\n", + "i.e 1/1000 sec\n", + "The Frequency = 1000.00 Hertz\n", + "OR 1 kHz\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# An alternating current varies through one complete cycle in 1 ⁄ 1000 s. Calculate the period and frequency.\n", + "\n", + "# Given data\n", + "\n", + "tc = 1/1000# # One Complete Cycle=1 ⁄ 1000 sec.\n", + "\n", + "T = tc#\n", + "print 'The Time period = %0.e Seconds'%T\n", + "print 'i.e 1/1000 sec'\n", + "\n", + "f = 1/tc#\n", + "print 'The Frequency = %0.2f Hertz'%f\n", + "print 'OR 1 kHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 15_3 Page No: 476" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Time period = 1e-06 Seconds of 1 MHz freq.\n", + "i.e 1*10**-6 sec = 1 usec\n", + "The Time period = 5e-07 Seconds of 2 MHz freq.\n", + "i.e 0.5*10**-6 sec = 0.5 usec\n" + ] + } + ], + "source": [ + "# Calculate the period for the two frequencies of 1 MHz and 2 MHz.Calculate the period for the two frequencies of 1 MHz and 2 MHz.\n", + "\n", + "# Given data\n", + "\n", + "f1 = 1*10**6# # Freq=1 MHz\n", + "f2 = 2*10**6# # Freq=2 MHz\n", + "\n", + "t1 = 1/f1#\n", + "print 'The Time period = %0.e Seconds of 1 MHz freq.'%t1\n", + "print 'i.e 1*10**-6 sec = 1 usec'\n", + "\n", + "t2 = 1/f2#\n", + "print 'The Time period = %0.e Seconds of 2 MHz freq.'%t2\n", + "print 'i.e 0.5*10**-6 sec = 0.5 usec'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 15_4 Page No: 478" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Lamda or Wavelenght = 1.00 cm\n" + ] + } + ], + "source": [ + "# Calculate lamda for a radio wave witf f of 30 GHz.\n", + "\n", + "# Given data\n", + "\n", + "c = 3*10**10# # Speed of light=3*10**10 cm/s\n", + "f = 30*10**9# # Freq=30 GHz\n", + "\n", + "l = c/f#\n", + "print 'The Lamda or Wavelenght = %0.2f cm'%l" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 15_5 Page No: 480" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Height = 250.00 cm\n", + "The Height = 8.20 feet\n" + ] + } + ], + "source": [ + "# The length of a TV antenna is lamda\u0005/2 for radio waves with f of 60 MHz. What is the antenna length in centimeters and feet?\n", + "\n", + "# Given data\n", + "\n", + "c = 3*10**10# # Speed of light=3*10**10 cm/s\n", + "f = 60*10**6# # Freq=60 MHz\n", + "In = 2.54# # 2.54 cm = 1 in\n", + "ft = 12# # 12 in = 1 ft\n", + "\n", + "l1 = c/f#\n", + "l = l1/2#\n", + "print 'The Height = %0.2f cm'%l\n", + "\n", + "li = l/In\n", + "lf = li/ft#\n", + "print 'The Height = %0.2f feet'%lf" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 15_6 Page No: 481" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Frequency = 50000000 Hertz\n", + "i.e 50*10**6 Hz OR 50 MHz\n" + ] + } + ], + "source": [ + "# For the 6-m band used in amateur radio, what is the corresponding frequency?\n", + "\n", + "# Given data\n", + "\n", + "v = 3*10**10# # Speed of light=3*10**10 cm/s\n", + "l = 6*10**2# # lamda=6 meter\n", + "\n", + "f = v/l\n", + "print 'The Frequency = %0.f Hertz'%f\n", + "print 'i.e 50*10**6 Hz OR 50 MHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 15_7 Page No: 481" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Lamda or Wavelenght = 11.30 ft\n" + ] + } + ], + "source": [ + "# What is the wavelength of the sound waves produced by a loudspeaker at a frequency of 100 Hz?\n", + "\n", + "# Given data\n", + "\n", + "c = 1130# # Speed of light=1130 ft/s\n", + "f = 100# # Freq=100 Hz\n", + "\n", + "l = c/f#\n", + "print 'The Lamda or Wavelenght = %0.2f ft'%l" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 15_8 Page No: 482" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Lamda or Wavelength = 0.03 ft\n", + "The Lamda or Wavelength = 1.00 cm\n" + ] + } + ], + "source": [ + "# For ultrasonic waves at a frequency of 34.44 kHz, calculate the wavelength in feet and in centimeters.\n", + "\n", + "# Given data\n", + "\n", + "c = 1130# # Speed of light=1130 ft/s\n", + "f = 34.44*10**3# # Freq=100 Hz\n", + "In = 2.54# # 2.54 cm = 1 in\n", + "ft = 12# # 12 in = 1 ft\n", + "\n", + "l = c/f#\n", + "print 'The Lamda or Wavelength = %0.2f ft'%l\n", + "\n", + "a = l*ft#\n", + "\n", + "l1 = a*In#\n", + "print 'The Lamda or Wavelength = %0.2f cm'%l1" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter16.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter16.ipynb new file mode 100644 index 00000000..94af71c5 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter16.ipynb @@ -0,0 +1,285 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 16 : Capacitance" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 16_1 Page No. 492" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Charge Stored = 1.00e-04 Columb\n", + "i.e 100*10**-6 Columbs\n" + ] + } + ], + "source": [ + "# How much charge is stored in a 2 uF capacitor connected across a 50-V supply?\n", + "\n", + "# Given data\n", + "\n", + "V = 50# # Voltage=50 Volts\n", + "C = 2*10**-6# # Capacitor=2 uFarad\n", + "\n", + "Q = C*V#\n", + "print 'The Charge Stored = %0.2e Coulomb'%Q\n", + "print 'i.e 100*10**-6 Coulombs'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 16_2 Page No. 492" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Charge Stored = 2.00e-03 Columb\n", + "i.e 2000*10**-6 Columbs\n" + ] + } + ], + "source": [ + "# How much charge is stored in a 40 uF capacitor connected across a 50-V supply?\n", + "\n", + "# Given data\n", + "\n", + "V = 50# # Voltage=50 Volts\n", + "C = 40*10**-6# # Capacitor=2 uFarad\n", + "\n", + "Q = C*V#\n", + "print 'The Charge Stored = %0.2e Coulomb'%Q\n", + "print 'i.e 2000*10**-6 Coulombs'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 16_3 Page No. 493" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Charge Stored = 4.00e-05 Columb\n", + "i.e 40*10**-6 Columbs OR 40 uColumb\n" + ] + } + ], + "source": [ + "# A constant current of 2 uA charges a capacitor for 20 s. How much charge is stored? Remember I=Q/t or Q=I*t.\n", + "\n", + "# Given data\n", + "\n", + "I = 2*10**-6# # Current=2 uAmps\n", + "t = 20# # Time=20 Sec\n", + "\n", + "Q = I*t\n", + "print 'The Charge Stored = %0.2e Coulomb'%Q\n", + "print 'i.e 40*10**-6 Coulombs OR 40 uCoulomb'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 16_4 Page No. 494" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Capacitance = 2.00e-06 Farad\n", + "i.e 2 uF\n" + ] + } + ], + "source": [ + "# The voltage across the charged capacitor is 20 V. Calculate C.\n", + "\n", + "#Given data\n", + "\n", + "V = 20# # Voltage=20 Volts\n", + "Q = 40*10**-6# # Charge=40 uCoulomb\n", + "\n", + "C = Q/V\n", + "print 'The Capacitance = %0.2e Farad'%C\n", + "print 'i.e 2 uF'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 16_5 Page No. 495" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage across Capacitor = 500.00 Volts\n" + ] + } + ], + "source": [ + "# A constant current of 5 mA charges a 10 uF capacitor for 1 s. How much is the voltage across the capacitor?\n", + "\n", + "# Given data\n", + "\n", + "I = 5*10**-3# # Current=5 mAmps\n", + "t = 1# # Time=1 Sec\n", + "C = 10*10**-6# # Cap=10 uFarad\n", + "\n", + "Q = I*t#\n", + "\n", + "V = Q/C#\n", + "print 'The Voltage across Capacitor = %0.2f Volts'%V" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 16_6 Page No. 496" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Capacitance = 1.77e-09 Farad\n", + "i.e 1700*10**-12 F OR 1770 pF\n" + ] + } + ], + "source": [ + "# Calculate C for two plates, each with an area 2 sqm, separated by 1 cm with a dielectric of air.\n", + "\n", + "# Given data\n", + "\n", + "c = 8.85*10**-12# # Constant=8.85 p\n", + "A = 2# # Area=2 sqm\n", + "d = 1*10**-2# # Distance=1 cm\n", + "K = 1 # Permeability=1\n", + "\n", + "C = K*c*(A/d)#\n", + "print 'The Capacitance = %0.2e Farad'%C\n", + "print 'i.e 1700*10**-12 F OR 1770 pF'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 16_11 Page No. 514" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Energy Stored = 0.23 Joules\n" + ] + } + ], + "source": [ + "# The high-voltage circuit for a color picture tube can have 30 kV across 500 pF of C . Calculate the stored energy.\n", + "\n", + "# Given data\n", + "\n", + "V = 30*10**3# # Voltage=30 kVolts\n", + "C = 500*10**-12# # Cap=500 pFarad\n", + "\n", + "E = 0.5*C*V*V\n", + "print 'The Energy Stored = %0.2f Joules'%E" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter17.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter17.ipynb new file mode 100644 index 00000000..a7a1d312 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter17.ipynb @@ -0,0 +1,307 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter17 : Capacitive Reactance" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 17_1 Page No. 530" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Capacitive Reactance = 1136.82 Ohms\n", + "approx 1140 Ohms\n", + "The Capacitive Reactance = 113.68 Ohms\n", + "approx 114 Ohms\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# How much is Xc for (a) 0.1 u\u0003F of C at 1400 Hz? (b) 1 u\u0003F of C at the same frequency?\n", + "\n", + "# Given data\n", + "\n", + "f = 1400# # Frequency=1400 Hz\n", + "C1 = 0.1*10**-6# # Cap1=0.1 uF\n", + "C2 = 1*10**-6# # Cap2=1 uF\n", + "\n", + "Xc1 = 1./(2.*pi*f*C1)#\n", + "print 'The Capacitive Reactance = %0.2f Ohms'%Xc1\n", + "print 'approx 1140 Ohms'\n", + "\n", + "Xc2 = 1./(2.*pi*f*C2)#\n", + "print 'The Capacitive Reactance = %0.2f Ohms'%Xc2\n", + "print 'approx 114 Ohms'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 17_2 Page No. 530" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Capacitive Reactance = 3386.28 Ohms\n", + "approx 3388 Ohms\n", + "The Capacitive Reactance = 338.63 Ohms\n" + ] + } + ], + "source": [ + "from math import pi\n", + "#How much is the Xc of a 47-pF value of C at (a) 1 MHz? (b) 10 MHz?\n", + "\n", + "# Given data\n", + "\n", + "f1 = 1*10**6# # Frequency1=1 MHz\n", + "f2 = 10*10**6# # Frequency2=10 MHz\n", + "C = 47*10**-12# # Cap=47 pF\n", + "\n", + "# For 1 MHz\n", + "\n", + "Xc1 = 1./(2.*pi*f1*C)#\n", + "print 'The Capacitive Reactance = %0.2f Ohms'%Xc1\n", + "print 'approx 3388 Ohms'\n", + "\n", + "# For 10 MHz\n", + "\n", + "Xc2 = 1./(2.*pi*f2*C)#\n", + "print 'The Capacitive Reactance = %0.2f Ohms'%Xc2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 17_3 Page No. 532" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Capacitance = 4.68e-10 Farads\n", + "approx 468 pF\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# What C is needed for Xc of 100 Ohms\u0003 at 3.4 MHz?\n", + "\n", + "# Given data\n", + "\n", + "f = 3.4*10**6# # Frequency=3.4 MHz\n", + "Xc = 100# # Capacitive Reactance=100 Ohms\n", + "\n", + "C = 1./(2.*pi*f*Xc)#\n", + "print 'The Capacitance = %0.2e Farads'%C\n", + "print 'approx 468 pF'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 17_4 Page No. 533" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Frequency = 159.15 Hertz\n", + "approx 159 Hz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# At what frequency will a 10 uF capacitor have Xc equal to 100 Ohms\u0003?\n", + "\n", + "# Given data\n", + "\n", + "Xc = 100# # Capacitive Reactance=100 Ohms\n", + "C = 10*10**-6# # Cap=10 uF\n", + "\n", + "f = 1./(2.*pi*C*Xc)#\n", + "print 'The Frequency = %0.2f Hertz'%f\n", + "print 'approx 159 Hz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 17_5 Page No. 534" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Instantaneous Value of Charging Current ic produced = 3.00e-04 Amps\n", + "i.e 300 uAmps\n" + ] + } + ], + "source": [ + "# Calculate the instantaneous value of charging current ic produced by a 6 u\u0003F C when its potential difference is increased by 50 V in 1 s.\n", + "\n", + "# Given data\n", + "\n", + "C = 6*10**-6# # Cap=6 uF\n", + "dv = 50.# # differential voltage increased by 50 Volts\n", + "dt = 1.# # differectial time is 1 sec\n", + "\n", + "ic = C*(dv/dt)#\n", + "print 'The Instantaneous Value of Charging Current ic produced = %0.2e Amps'%ic\n", + "print 'i.e 300 uAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 17_6 Page No. 535" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Instantaneous Value of Discharging Current ic produced = -3.00e-04 Amps\n", + "i.e -300 uAmps\n" + ] + } + ], + "source": [ + "# Calculate the instantaneous value of charging current ic produced by a 6 u\u0003F C when its potential difference is decreased by 50 V in 1 s.\n", + "\n", + "# Given data\n", + "\n", + "C = 6*10**-6# # Cap=6 uF\n", + "dv = -50.# # differential voltage decreased by 50 Volts\n", + "dt = 1.# # differectial time is 1 sec\n", + "\n", + "ic = C*(dv/dt)#\n", + "print 'The Instantaneous Value of Discharging Current ic produced = %0.2e Amps'%ic\n", + "print 'i.e -300 uAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 17_7 Page No. 536" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Instantaneous Value of ic produced = 1.25e-02 Amps\n", + "12500 uAmps or 12.5 mAmps\n" + ] + } + ], + "source": [ + "# Calculate ic produced by a 250-pF capacitor for a change of 50 V in 1 u\u0002s.\n", + "\n", + "# Given data\n", + "\n", + "C = 250*10**-12# # Cap=250 pF\n", + "dv = 50.# # differential voltage increased by 50 Volts\n", + "dt = 1.*10**-6# # differectial time is 1 usec\n", + "\n", + "ic = C*(dv/dt)#\n", + "print 'The Instantaneous Value of ic produced = %0.2e Amps'%ic\n", + "print '12500 uAmps or 12.5 mAmps'" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter18.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter18.ipynb new file mode 100644 index 00000000..55e5731f --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter18.ipynb @@ -0,0 +1,144 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 18 : Capacitive Circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 18_1 Page No. 545" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Zt = 50.00 Ohms\n", + "I = 2.00 Ampers\n", + "Voltage Across Resistor = 60 Volts\n", + "Voltage Across Capacitive Reactance = 80.00 Volts\n", + "Theta z =-53.13 degree\n", + "Sum of Voltage Drop is Equal to Applied Voltage of 100V = 100.00 Volts\n" + ] + } + ], + "source": [ + "from math import sqrt,pi,atan\n", + "# If a R=30ohms and Xc=40ohms are in series with 100V applied, find the following: Zt, I, Vr, Vc and Theta z. What is the phase angle between Vc and Vr with respect to I? Prove that the sum of the series voltage drop equals the applied voltage Vt\n", + "\n", + "# Given data\n", + "\n", + "R = 30.# # Resistance=30 Ohms\n", + "Xc = 40.# # Capacitive Reactance=40 Ohms\n", + "Vt = 100.# # Applied Voltage=100 Volts\n", + "\n", + "R1 = R*R#\n", + "Xc1 = Xc*Xc#\n", + "\n", + "Zt = sqrt(R1+Xc1)#\n", + "print 'Zt = %0.2f Ohms'%Zt\n", + "\n", + "I = (Vt/Zt)#\n", + "print 'I = %0.2f Ampers'%I\n", + "\n", + "Vr = I*R#\n", + "print 'Voltage Across Resistor = %02.f Volts'%Vr\n", + "\n", + "Vc = I*Xc#\n", + "print 'Voltage Across Capacitive Reactance = %0.2f Volts'%Vc\n", + "\n", + "Oz = atan(-(Xc/R))*180/pi\n", + "print 'Theta z =%0.2f degree'%Oz\n", + "\n", + "#Prove that the sum of the series voltage drop equals the applied voltage Vt\n", + "\n", + "Vt = sqrt((Vr*Vr)+(Vc*Vc))#\n", + "print 'Sum of Voltage Drop is Equal to Applied Voltage of 100V = %0.2f Volts'%Vt" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 18_2 Page No. 548" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Total Current = 0.05 Amps\n", + "i.e 50 mAmps\n", + "The Equivqlent Impedence = 1440.00 Ohms\n", + "i.e 1.44 kohms\n", + "The Value of Theta I = 53.13 degrees\n" + ] + } + ], + "source": [ + "from math import sqrt,pi,atan\n", + "# A 30-mA Ir is in parallel with another branch current of 40 mA for Ic. The applied voltage Va is 72 V. Calculate It, Zeq and Theta \u0002I.\n", + "\n", + "# Given data\n", + "\n", + "Ir = 30.*10**-3# # Current Ir=30 mA\n", + "Ic = 40.*10**-3# # Current Ic=40 mA\n", + "Va = 72.# # Applied Voltage=72 Volts\n", + "\n", + "A = Ir*Ir#\n", + "B = Ic*Ic#\n", + "\n", + "It = sqrt(A+B)#\n", + "print 'The Total Current = %0.2f Amps'%It\n", + "print 'i.e 50 mAmps'\n", + "\n", + "Zeq = Va/It#\n", + "print 'The Equivqlent Impedence = %0.2f Ohms'%Zeq\n", + "print 'i.e 1.44 kohms'\n", + "\n", + "Oi = atan(Ic/Ir)*180/pi\n", + "print 'The Value of Theta I = %0.2f degrees'%Oi" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter19.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter19.ipynb new file mode 100644 index 00000000..5e3f9929 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter19.ipynb @@ -0,0 +1,830 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 19 : Inductance" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_1 Page No. 580" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The di/dt Rate of Current change = 4 A/s\n" + ] + } + ], + "source": [ + "# The current in an inductor changes from 12 to 16 A in 1s. How much is the di/\u0002dt rate of current change in amperes per second?\n", + "\n", + "# Given data\n", + "\n", + "di = 4# # Differential current=16-12=4 Amps\n", + "dt = 1# # Differential time=1 sec\n", + "\n", + "A = di/dt#\n", + "print 'The di/dt Rate of Current change = %0.f A/s'%A" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_2 Page No. 580" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The di/dt Rate of Current change = 2.50e+04 A/s\n" + ] + } + ], + "source": [ + "# The current in an inductor changes by 50 mA in 2 us. How much is the di/\u0002dt rate of current change in amperes per second?\n", + "\n", + "# Given data\n", + "\n", + "di = 50.*10**-3# # Differential current=50 mAmps\n", + "dt = 2.*10**-6# # Differential time=2 usec\n", + "\n", + "A = di/dt#\n", + "print 'The di/dt Rate of Current change = %0.2e A/s'%A" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_3 Page No. 581" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Inductance = 10 Henry\n" + ] + } + ], + "source": [ + "# How much is the inductance of a coil that induces 40 V when its current changes at the rate of 4 A/\u0002s?\n", + "\n", + "# Given data\n", + "\n", + "Vl = 40# # Induced voltage=40 Volts\n", + "R = 4 # Current changing rate=di/dt=4 A/s\n", + "\n", + "L = Vl/R#\n", + "print 'The Value of Inductance = %0.f Henry'%L" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_4 Page No. 582" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Inductance = 4.00e-02 Henry\n", + "OR 40 mH\n" + ] + } + ], + "source": [ + "# How much is the inductance of a coil that induces 1000 V when its current changes at the rate of 50 mA in 2\u0002us?\n", + "\n", + "# Given data\n", + "\n", + "Vl = 1000# # Induced voltage=1000 Volts\n", + "di = 50*10**-3# # differential current=50 mAmps\n", + "dt = 2*10**-6# # differential time=2 usec\n", + "\n", + "A = di/dt#\n", + "\n", + "L = Vl/A#\n", + "print 'The Value of Inductance = %0.2e Henry'%L\n", + "print 'OR 40 mH'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_5 Page No. 582" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Self-Induced Voltage = 48 Volts\n" + ] + } + ], + "source": [ + "# How much is the self-induced voltage across a 4-H inductance produced by a current change of 12 A/\u0002s?\n", + "\n", + "# Given data\n", + "\n", + "L = 4# # Inductor=4 H\n", + "R = 12# # current change=di/dt=12 A/s\n", + "\n", + "Vl = L*R#\n", + "print 'The Value of Self-Induced Voltage = %0.f Volts'%Vl" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_6 Page No. 583" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Self-Induced Voltage = 1.00e+04 Volts\n", + "OR 10 kVolts\n" + ] + } + ], + "source": [ + "# The current through a 200-mH L changes from 0 to 100 mA in 2 u\u0002s. How much is Vl ?\n", + "\n", + "# Given data\n", + "\n", + "L = 200*10**-3# # Inductor=200 mH\n", + "di = 100*10**-3# # differential current=100 mAmps\n", + "dt = 2*10**-6# # differectial time=2 usec\n", + "\n", + "A = di/dt#\n", + "\n", + "Vl = L*A#\n", + "print 'The Value of Self-Induced Voltage = %0.2e Volts'%Vl\n", + "print 'OR 10 kVolts'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_7 Page No. 583" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Coefficient of Coupling k between Coil L1 and Coil L2 = 0.75\n" + ] + } + ], + "source": [ + "# A coil L1 produces 80 u\u0002Wb of magnetic flux. Of this total flux, 60 u\u0002Wb arelinked with L2. How much is k between L1 and L2?\n", + "\n", + "# Given data\n", + "\n", + "lf1 = 80.*10**-6# # Magnetic flux of coil L1=80 uWb\n", + "lf2 = 60.*10**-6# # Magnetic flux of coil L2=60 uWb\n", + "\n", + "k = lf2/lf1#\n", + "print 'The Coefficient of Coupling k between Coil L1 and Coil L2 = %0.2f'%k" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_8 Page No. 584" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Coefficient of Coupling k between Coil L1 and Coil L2 = 1\n" + ] + } + ], + "source": [ + "# A 10-H inductance L1 on an iron core produces 4 Wb of magnetic flux. Another coil L2 is on the same core. How much is k between L1 and L2?\n", + "\n", + "# Given data\n", + "\n", + "lf1 = 4# # Magnetic flux of coil L1=4 Wb\n", + "lf2 = 4# # Magnetic flux of coil L2=4 Wb\n", + "\n", + "k = lf2/lf1#\n", + "print 'The Coefficient of Coupling k between Coil L1 and Coil L2 = %0.f'%k" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_9 Page No. 585" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The mutual inductance = 0.08 Henry\n", + "i.e 80*10**-3 H OR 80 mH\n" + ] + } + ], + "source": [ + "\n", + "from math import sqrt\n", + "# Two 400-mH coils L1 and L2 have a coefficient of coupling k equal to 0.2. Calculate Lm.\n", + "\n", + "# Given data\n", + "\n", + "L1 = 400*10**-3# # L1=400 mH\n", + "L2 = 400*10**-3# # L2=400 mH\n", + "k = 0.2# # Coupling coefficient=0.2\n", + "\n", + "Lm = k*sqrt(L1*L2)#\n", + "print 'The mutual inductance = %0.2f Henry'%Lm\n", + "print 'i.e 80*10**-3 H OR 80 mH'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_10 Page No. 586" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Coupling Coefficient k = 0.10\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "# If the two coils had a mutual inductance LM of 40 mH, how much would k be?\n", + "\n", + "# Given data\n", + "\n", + "L1 = 400.*10**-3# # Coil Inductance 1=400 mH\n", + "L2 = 400.*10**-3# # Coil Inductance 2=400 mH\n", + "Lm = 40.*10**-3# # Mutual inductance=40 mH\n", + "\n", + "lt = sqrt(L1*L2)#\n", + "\n", + "k = Lm/lt#\n", + "print'The Coupling Coefficient k = %0.2f'% k" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_11 Page No. 590" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Turns Ratio = 0.1667\n", + "OR 1:6\n", + "The Secondary Voltage = 720.00 Volts\n" + ] + } + ], + "source": [ + "# A power transformer has 100 turns for Np and 600 turns for Ns. What is the turns ratio? How much is the secondary voltage Vs if the primary voltage Vp is 120 V?\n", + "\n", + "# Given data\n", + "\n", + "np = 100.# # Turns in primary coil=100\n", + "ns = 600.# # Turns in secondary coil=600\n", + "vp = 120.# # Primary voltage=120 Volts\n", + "\n", + "Tr = np/ns#\n", + "print 'The Turns Ratio = %0.4f'%Tr\n", + "print 'OR 1:6'\n", + "\n", + "vs = vp*(ns/np)#\n", + "print 'The Secondary Voltage = %0.2f Volts'%vs" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_12 Page No. 590" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Turns Ratio 20:1 or 20.00\n", + "The Secondary Voltage = 6.00 Volts\n" + ] + } + ], + "source": [ + "# A power transformer has 100 turns for Np and 5 turns for Ns. What is the turns ratio? How much is the secondary voltage Vs with a primary voltage of 120 V?\n", + "\n", + "# Given data\n", + "\n", + "np = 100.# # Turns in primary coil=100\n", + "ns = 5.# # Turns in secondary coil=5\n", + "vp = 120.# # Primary voltage=120 Volts\n", + "\n", + "Tr = np/ns#\n", + "print 'The Turns Ratio 20:1 or %0.2f'%Tr\n", + "\n", + "vs = vp*(ns/np)#\n", + "print 'The Secondary Voltage = %0.2f Volts'%vs" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_13 Page No. 591" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Secondary Current = 0.10 Amps\n", + "The Primary Current = 0.60 Amps\n" + ] + } + ], + "source": [ + "# A transformer with a 1:6 turns ratio has 720 V across 7200 Ohms\u0006 in the secondary. (a) How much is Is? (b) Calculate the value of Ip.\n", + "\n", + "# Given data\n", + "\n", + "vs = 720.# # Secondary voltage=720 Volts\n", + "Rl = 7200.# # Secondary load=7200 Ohms\n", + "tr = 1./6# # Turns ratio=1:6\n", + "\n", + "Is = vs/Rl#\n", + "print 'The Secondary Current = %0.2f Amps'%Is\n", + "\n", + "Ip = Is/tr#\n", + "print 'The Primary Current = %0.2f Amps'%Ip" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_14 Page No. 591" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Secondary Current = 10.00 Amps\n", + "The Primary Current = 0.50 Amps\n" + ] + } + ], + "source": [ + "# A transformer with a 20:1 voltage step-down ratio has 6 V across 0.6 \u0006 in the secondary. (a) How much is Is? (b) How much is Ip?\n", + "\n", + "# Given data\n", + "\n", + "vs = 6.# # Secondary voltage=6 Volts\n", + "Rl = 0.6# # Secondary load=0.6 Ohms\n", + "tr = 20./1# # Turns ratio=20:1\n", + "\n", + "Is = vs/Rl#\n", + "print 'The Secondary Current = %0.2f Amps'%Is\n", + "\n", + "Ip = Is/tr#\n", + "print 'The Primary Current = %0.2f Amps'%Ip" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_15 Page No. 593" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Primary current = 0.42 Amps\n", + "OR 420 mAmps\n" + ] + } + ], + "source": [ + "# Calculate the primary current I P if the secondary current Is equals its rated value of 2 A.\n", + "\n", + "# Given data\n", + "\n", + "vs = 25.2# # Secondary voltage=25.2 Volts\n", + "vp = 120.# # Primary voltage=120 Volts\n", + "Is = 2.# # Secondary current=2 Amps\n", + "\n", + "Ip = Is*(vs/vp)#\n", + "print 'The Primary current = %0.2f Amps'%Ip\n", + "print 'OR 420 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_16 Page No. 593" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Primary Impedence = 128.00 Ohms by Method 1\n", + "Primary Impedence = 128.00 Ohms by Method 2\n" + ] + } + ], + "source": [ + "# Determine the Primary Impedence Zo\n", + "\n", + "# Method 1\n", + "# Given data\n", + "\n", + "Vp = 32.# # Primary Voltage = 32 Volts\n", + "Rl = 8.# # Load Resistance = 8 Ohms\n", + "TR = 4.# # Turns Ratio Np/Ns = 4/1\n", + "\n", + "Vs = Vp/TR#\n", + "\n", + "Is = Vs/Rl#\n", + "\n", + "Ip = ((Vs/Vp)*Is)#\n", + "\n", + "Zp = Vp/Ip#\n", + "print 'Primary Impedence = %0.2f Ohms by Method 1'%Zp\n", + "\n", + "# Method 2\n", + "\n", + "Zp = TR*TR*Rl#\n", + "print 'Primary Impedence = %0.2f Ohms by Method 2'%Zp" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_17 Page No. 594" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Turns ratio Np/Ns = 0.50\n", + "OR 1/2\n", + "The Turns ratio Np/Ns is 1.414/1 or 1.41\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "# Calculate the turns ratio Np/Ns that will produce a reflected primary impedance Zp of (a) 75 Ohms# (b) 600 Ohms.\n", + "\n", + "# Given data\n", + "\n", + "Zs = 300.# # Secondary impedence=300 Ohms\n", + "Zp1 = 75.# # Primary impedence=75 Ohms\n", + "Zp2 = 600.# # Primary impedence=600 Ohms\n", + "\n", + "tra = sqrt (Zp1/Zs)#\n", + "print 'The Turns ratio Np/Ns = %0.2f'%tra\n", + "print 'OR 1/2'\n", + "\n", + "trb = sqrt (Zp2/Zs)#\n", + "print 'The Turns ratio Np/Ns is 1.414/1 or %0.2f'%trb" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_18 Page No. 595" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Total Inductance = 1.50e-02 Henry\n", + "i.e 15 mH\n" + ] + } + ], + "source": [ + "# Inductance L1 is 5 mH and L2 is 10 mH. How much is Lt?\n", + "\n", + "# Given data\n", + "\n", + "l1 = 5*10**-3# # Inductor 1=5 mH\n", + "l2 = 10*10**-3# # Inductor 2=10 mH\n", + "\n", + "Lt = l1+l2#\n", + "print 'The Total Inductance = %0.2e Henry'%Lt\n", + "print 'i.e 15 mH'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_19 Page No. 597" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Equivalent Inductance = 0.04 Henry\n", + "i.e 4 mH\n" + ] + } + ], + "source": [ + "# Inductances L1 and L2 are each 8 mH. How much is Leq?\n", + "\n", + "# Given data\n", + "\n", + "l1 = 8*10**-3# # Inductor 1=8 mH\n", + "l2 = 8*10**-3# # Inductor 2=8 mH\n", + "\n", + "a = 1./l1#\n", + "b = 1./l2#\n", + "\n", + "Leq = 10/(a+b)#\n", + "print 'The Equivalent Inductance = %0.2f Henry'%Leq\n", + "print 'i.e 4 mH'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_20 Page No. 598" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Mutual Inductance = 2.50e-05 Henry\n", + "i.e 25 uH\n", + "The Coupling coefficient k = 0.10\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "# Two series coils, each with an L of 250 u\u0002H, have a total inductance of 550 u\u0002H connected series-aiding and 450 uH series-opposing. (a) How much is the mutual inductance Lm between the two coils? (b) How much is the coupling coefficient k?\n", + "\n", + "# Given data\n", + "\n", + "l1 = 250*10**-6# # Coil Inductance 1=250 uH\n", + "l2 = 250*10**-6# # Coil Inductance 2=250 uH\n", + "Lts = 550*10**-6# # Inductance series-aiding=550 uH\n", + "Lto = 450*10**-6# # Inductance series-opposing=450 uH\n", + "\n", + "Lm = (Lts-Lto)/4.\n", + "print 'The Mutual Inductance = %0.2e Henry'%Lm\n", + "print 'i.e 25 uH'\n", + "\n", + "lt = sqrt(l1*l2)#\n", + "\n", + "k = Lm/lt#\n", + "print 'The Coupling coefficient k = %0.2f'%k" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_21 Page No. 608" + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Energy Stored in the Magnetic Field = 0.29 Joules\n" + ] + } + ], + "source": [ + "# A current of 1.2 A flows in a coil with an inductance of 0.4 H. How much energy is stored in the magnetic field?\n", + "\n", + "# Given data\n", + "\n", + "l1 = 0.4# # Coil Inductance 1=0.4 H\n", + "I = 1.2# # Current=1.2 Amps\n", + "\n", + "E = (l1*I*I)/2#\n", + "print 'The Energy Stored in the Magnetic Field = %0.2f Joules'%E" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter20.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter20.ipynb new file mode 100644 index 00000000..01adabd9 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter20.ipynb @@ -0,0 +1,296 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 20 : Inductive Reactance" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 20_1 Page No. 625" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Inductive Reactance = 15709.22 Ohms\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# How much is Xl of a 6-mH L at 41.67 kHz?\n", + "\n", + "# Given data\n", + "\n", + "f = 41.67*10**3# # Frequency=41.67 kHz\n", + "L = 6.*10**-3# # Inductor=6 mH\n", + "\n", + "Xl = 20*pi*f*L#\n", + "print 'The Inductive Reactance = %0.2f Ohms'%Xl" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 20_2 Page No. 627" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Inductive Reactance = 37680 Ohms\n", + "The Inductive Reactance = 1884 Ohms\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# Calculate the Xl of (a) a 10-H L at 60 Hz and (b) a 5-H L at 60 Hz.\n", + "\n", + "# Given data\n", + "\n", + "f = 60.# # Frequency=60 Hz\n", + "L1 = 10.# # Inductor 1=10 H\n", + "L2 = 5.# # Inductor 2=5 H\n", + "pi = 3.14\n", + "\n", + "Xl1 = 20*pi*f*L1#\n", + "print 'The Inductive Reactance = %0.f Ohms'%Xl1\n", + "\n", + "Xl2 = 2*pi*f*L2#\n", + "print 'The Inductive Reactance = %0.f Ohms'%Xl2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 20_3 Page No. 629" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Inductive Reactance = 1570 Ohms\n", + "The Inductive Reactance = 15700 Ohms\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# Calculate the Xl of a 250-u\u0003H coil at (a) 1 MHz and (b) 10 MHz.\n", + "\n", + "# Given data\n", + "\n", + "f1 = 1.*10**6# # Frequency1=1 MHz\n", + "f2 = 10.0*10**6# # Frequency2=10 MHz\n", + "L = 250.0*10**-6# # Inductor=250 uH\n", + "pi = 3.14#\n", + "\n", + "# For 1 Mhz\n", + "\n", + "Xl1 = 2*pi*f1*L#\n", + "print 'The Inductive Reactance = %0.f Ohms'%Xl1\n", + "\n", + "# For 10 Mhz\n", + "\n", + "Xl2 = 2*pi*f2*L#\n", + "print 'The Inductive Reactance = %0.f Ohms'%Xl2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 20_4 Page No. 631" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Inductive Reactance = 6280 Ohms\n" + ] + } + ], + "source": [ + "# A coil with negligible resistance has 62.8 V across it with 0.01 A of current. How much is Xl?\n", + "\n", + "# Given data\n", + "\n", + "Vl = 62.8# # Voltage across coil=62.8 Volts\n", + "Il = 0.01# # Current in coil=0.01 Amps\n", + "\n", + "Xl = Vl/Il#\n", + "print 'The Inductive Reactance = %0.f Ohms'%Xl" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 20_5 Page No. 632" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of Inductor = 1.00 Henry\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# Calculate L of the coil when the frequency is 1000 Hz.\n", + "\n", + "# Given data\n", + "\n", + "Xl = 6280.# # Inductive reactance=6280 Ohms\n", + "f = 1000.# # Frequency=1000 Hz\n", + "pi = 3.14#\n", + "\n", + "L = Xl/(2.*pi*f)#\n", + "print'The value of Inductor = %0.2f Henry'% L" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 20_6 Page No. 633" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of Inductor = 2.08e-04 Henry\n", + "i.e Approx 208.8*10**-6 OR 208.8 uH\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# Calculate L of a coil that has 15,700 Ohms\u0003 of Xl at 12 MHz.\n", + "\n", + "# Given data\n", + "\n", + "Xl = 15700.# # Inductive reactance=15700 Ohms\n", + "f = 12.0*10**6# # Frequency=12 MHz\n", + "pi = 3.14#\n", + "\n", + "L = Xl/(2*pi*f)#\n", + "print'The value of Inductor = %0.2e Henry'% L\n", + "print 'i.e Approx 208.8*10**-6 OR 208.8 uH'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 20_7 Page No. 634" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Frequency = 159.15 Hertz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# At what frequency will an inductance of 1 H have a reactance of 1000 \u0003?\n", + "\n", + "# Given data\n", + "\n", + "Xl = 1000.# # Inductive reactance=1000 Ohms\n", + "L = 1.# # Inductor=1 H\n", + "\n", + "f = Xl/(2.*pi*L)#\n", + "print 'The Frequency = %0.2f Hertz'%f" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter21.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter21.ipynb new file mode 100644 index 00000000..59eb0bda --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter21.ipynb @@ -0,0 +1,209 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 21 : Inductive Circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 21_1 Page No. 650" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Zt = 50.00 ohms\n", + "I = 2.00 Ampers\n", + "Vr = 60.00 Volts\n", + "Vl = 80.00 Volts\n", + "Theta z = 53.13 degree\n", + "Sum of Voltage Drop = 100 which is Equal to Applied Voltage 100V \n" + ] + } + ], + "source": [ + "from math import sqrt,atan,pi\n", + "# If a R=30 ohms and Xl=40 ohms are in series with 100V applied, find the following: Zt, I, Vr, Vl and Theta z. What is the phase angle between Vl and Vr with respect to I? Prove that the sum of the series voltage drop equals the applied voltage Vt\n", + "\n", + "# Given data\n", + "\n", + "R = 30.# # Resistance=30 Ohms\n", + "Xl = 40.# # Inductive reactance=40 Ohms\n", + "Vt = 100.# # Applied voltage=100 Volts\n", + "\n", + "\n", + "R1 = R*R#\n", + "Xl1 = Xl*Xl#\n", + "\n", + "Zt = sqrt(R1+Xl1)#\n", + "print 'Zt = %0.2f ohms'%Zt\n", + "\n", + "I = (Vt/Zt)#\n", + "print 'I = %0.2f Ampers'%I\n", + "\n", + "Vr = I*R#\n", + "print 'Vr = %0.2f Volts'%Vr\n", + "\n", + "Vl = I*Xl#\n", + "print 'Vl = %0.2f Volts'%Vl\n", + "\n", + "Oz = atan(Xl/R)*180/pi\n", + "print 'Theta z = %0.2f degree'%Oz\n", + "\n", + "#Prove that the sum of the series voltage drop equals the applied voltage Vt\n", + "\n", + "Vt = sqrt((Vr*Vr)+(Vl*Vl))#\n", + "print 'Sum of Voltage Drop = %0.f which is Equal to Applied Voltage 100V '%Vt" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 21_2 Page No. 654" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Total Impedence = 268.33 Ohms\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "# What is the total Z of a 600-Ohms\u0005 R in parallel with a 300 Ohms\u0005 Xl? Assume 600 V for the applied voltage.\n", + "\n", + "# Given data\n", + "\n", + "R = 600.# # Resistance=600 Ohms\n", + "Xl = 300.# # Inductive reactance=300 Ohms\n", + "V = 600.# # Applied voltage=600 Volts\n", + "\n", + "Ir = V/R#\n", + "Il = V/Xl#\n", + "A = Ir*Ir#\n", + "B = Il*Il#\n", + "It = sqrt(A+B)#\n", + "\n", + "Zeq = V/It#\n", + "print 'The Total Impedence = %0.2f Ohms'%Zeq" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 21_3 Page No. 656" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Q of Coil = 350.00\n" + ] + } + ], + "source": [ + "# An air-core coil has an Xl of 700 Ohms\u0005 and an Re of 2 Ohms\u0005. Calculate the value of Q for this coil.\n", + "\n", + "# Given data\n", + "\n", + "Xl = 700# # Inductive reactance=700 Ohms\n", + "Re = 2# # AC effective resistance=2 Ohms\n", + "\n", + "Q = Xl/Re#\n", + "print 'The Q of Coil = %0.2f'%Q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 21_4 Page No. 658" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The AC Effective Resistance = 1.57 Ohms\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# A 200 uH coil has a Q of 40 at 0.5 MHz. Find Re.\n", + "\n", + "# Given data\n", + "\n", + "L = 200.*10**-6# # L of coil=200 uHenry\n", + "Q = 400# # Q=40\n", + "f = 0.5*10**6# # Frequency=0.5 MHz\n", + "pi = 3.14#\n", + "\n", + "Xl = 2*pi*L*f#\n", + "\n", + "Re = Xl/Q#\n", + "print 'The AC Effective Resistance = %0.2f Ohms'%Re" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter22.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter22.ipynb new file mode 100644 index 00000000..cd340733 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter22.ipynb @@ -0,0 +1,418 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 22 : RC and L/R Time Constants" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 22_1 Page No. 674" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Time Constant = 0.20 Seconds\n" + ] + } + ], + "source": [ + "# What is the time constant of a 20-H coil having 100 Ohms\u0002 of series resistance?\n", + "\n", + "# Given data\n", + "\n", + "L = 20.# # Inductor=20 Henry\n", + "R = 100.# # Resistor=100 Ohms\n", + "\n", + "T = L/R#\n", + "print 'The Time Constant = %0.2f Seconds'%T" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 22_2 Page No. 674" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Since 0.2 sec is one time constant, I is 63% of 100 mA\n", + "The current at 0.2 sec time constant = 0.06 A\n", + "After 1 sec the current reaches its steady state value of 100 mAmps \n" + ] + } + ], + "source": [ + "# An applied dc voltage of 10 V will produce a steady-state current of 100 mA in the 100-Ohms coil. How much is the current after 0.2 s? After 1 s?\n", + "\n", + "# Given data\n", + "\n", + "L = 20.# # Inductor=20 Henry\n", + "R = 100.# # Resistor=100 Ohms\n", + "I = 100.*10**-3# # Steady-state current=100 mAmps\n", + "\n", + "print 'Since 0.2 sec is one time constant, I is 63% of 100 mA'\n", + "I1 = 0.63*I#\n", + "print 'The current at 0.2 sec time constant = %0.2f A'%I1\n", + "\n", + "print 'After 1 sec the current reaches its steady state value of 100 mAmps '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 22_3 Page No. 675" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Time Constant = 2.00e-05 Seconds\n", + "i.e 20 us\n" + ] + } + ], + "source": [ + "# If a 1-M Ohms\u0002 R is added in series with the coil, how much will the time constant be for the higher resistance RL circuit?\n", + "\n", + "# Given data\n", + "\n", + "L = 20.# # Inductor=20 Henry\n", + "R = 1.*10**6# # Resistor=1 MOhms\n", + "\n", + "T = L/R#\n", + "print 'The Time Constant = %0.2e Seconds'%T\n", + "print 'i.e 20 us'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 22_4 Page No. 676" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Time Constant = 1.00e-02 Seconds\n" + ] + } + ], + "source": [ + "# What is the time constant of a 0.01-u\u0003F capacitor in series with a 1-M\u0002 Ohmsresistance?\n", + "\n", + "# Given data\n", + "\n", + "C = 0.01*10**-6# # Capacitor=0.01 uFarad\n", + "R = 1*10**6# # Resistor=1 MOhms\n", + "\n", + "T = C*R#\n", + "print 'The Time Constant = %0.2e Seconds'%T" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 22_5 Page No. 677" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Time Constant = 1.00e-02 Seconds\n", + "Since 0.01 sec is one time constant, the voltage across C then is 63% of 300 V,\n", + "The Capacitor voltage at 0.01 Sec = 189.00 Volts\n", + "After 5 time constants or 0.05 Sec Capacitor voltage = 300.00 volts \n", + "After 2 hours or 2 days the C will be still charged to 300 V if the supply is still connected\n" + ] + } + ], + "source": [ + "# With a dc voltage of 300 V applied, how much is the voltage across C in Example 22–4 after 0.01 s of charging? After 0.05 s? After 2 hours? After 2 days?\n", + "\n", + "# Given data\n", + "\n", + "C = 0.01*10**-6# # Capacitor=0.01 uFarad\n", + "R = 1.*10**6# # Resistor=1 MOhms\n", + "V = 300.# # Applied DC=300 Volts\n", + "\n", + "T = C*R#\n", + "print 'The Time Constant = %0.2e Seconds'%T\n", + "\n", + "print 'Since 0.01 sec is one time constant, the voltage across C then is 63% of 300 V,'\n", + "\n", + "T1 = 0.63*V#\n", + "print 'The Capacitor voltage at 0.01 Sec = %0.2f Volts'%T1\n", + "\n", + "T2 = V\n", + "print 'After 5 time constants or 0.05 Sec Capacitor voltage = %0.2f volts '%V\n", + "\n", + "print 'After 2 hours or 2 days the C will be still charged to 300 V if the supply is still connected'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 22_6 Page No. 678" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "In one time constant, C discharges to 37% of its initial voltage\n", + "The Capacitor voltage after 0.01 sec start of discharge = 1110 volts\n" + ] + } + ], + "source": [ + "# If the capacitor is allowed to charge to 300 V and then discharged, how much is the capacitor voltage 0.01 s after the start of discharge? The series resistance is the same on discharge as on charge.\n", + "\n", + "# Given data\n", + "\n", + "C = 0.01*10**-6# # Capacitor=0.01 uFarad\n", + "R = 1.*10**6# # Resistor=1 MOhms\n", + "V = 3000# # Applied DC=300 Volts\n", + "\n", + "print 'In one time constant, C discharges to 37% of its initial voltage'\n", + "\n", + "V1 = 0.37*V#\n", + "print 'The Capacitor voltage after 0.01 sec start of discharge = %0.f volts'%V1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 22_7 Page No. 680" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "In one time constant, C discharges to 37% of its initial voltage\n", + "The Capacitor voltage after 0.01 sec start of discharge = 74 volts\n" + ] + } + ], + "source": [ + "# Assume the capacitor is discharging after being charged to 200 V. How much will the voltage across C be 0.01 s after the beginning of discharge? The series resistance is the same on discharge as on charge.\n", + "\n", + "# Given data\n", + "\n", + "C = 0.01*10**-6# # Capacitor=0.01 uFarad\n", + "R = 1*10**6# # Resistor=1 MOhms\n", + "V = 200# # Capacitor voltage=200 Volts\n", + "\n", + "print 'In one time constant, C discharges to 37% of its initial voltage'\n", + "\n", + "V1 = 0.37*V#\n", + "print 'The Capacitor voltage after 0.01 sec start of discharge = %0.f volts'%V1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 22_8 Page No. 681" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Time Constant = 2.00e-02 Seconds\n" + ] + } + ], + "source": [ + "# If a 1-M Ohms resistance is added in series with the capacitor 0.01-u\u0003F and resistor 1-M Ohms in, how much will the time constant be?\n", + "\n", + "# Given data\n", + "\n", + "C = 0.01*10**-6# # Capacitor=0.01 uFarad\n", + "R = 2*10**6# # Resistor= 2 MOhms \n", + "\n", + "T = C*R#\n", + "print 'The Time Constant = %0.2e Seconds'%T" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 22_9 Page No. 682" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Vr = 5.42 Volts\n" + ] + } + ], + "source": [ + "from math import log10\n", + "# An RC circuit has a time constant of 3 s. The capacitor is charged to 40 V. Then C is discharged. After 6 s of discharge, how much is Vr?\n", + "\n", + "# Given data\n", + "\n", + "RC = 3# # RC time constant=3 Sec\n", + "t = 6# # Discharge time=6 Sec\n", + "Vc = 40# # Capacitor voltage=40 Volts\n", + "\n", + "A = t/RC# # constant factor\n", + "B = log10(Vc)#\n", + "\n", + "Vr = 10**(B-(A*0.434))#\n", + "print 'The Value of Vr = %0.2f Volts'%Vr" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 22_10 Page No. 682" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Time Constant = 5.00e-04 Seconds\n", + "i.e 0.5*10**-3 Sec OR 0.5 mSec\n", + "Time required to charge Capacitor upto 24 Volts = 5.49e-04 Seconds\n", + "i.e approx 0.549*10**-3 Sec OR 0.549 mSec\n" + ] + } + ], + "source": [ + "from math import log10\n", + "# An RC circuit has an R of 10 k Ohms\u0002 and a C of 0.05 u\u0003F. The applied voltage for charging is 36 V. (a) Calculate the time constant. (b) How long will it take C to charge to 24 V?\n", + "\n", + "C = 0.05*10**-6# # Capacitor=0.05 uFarad\n", + "R = 10*10**3# # Resistor=10 kOhms\n", + "V = 36# # Applied voltage=36 Volts\n", + "v = 12# # Voltage drops from 36 to 12 Volts\n", + "A = 2.3# # Specific factor\n", + "\n", + "T = C*R#\n", + "print 'The Time Constant = %0.2e Seconds'%T\n", + "print 'i.e 0.5*10**-3 Sec OR 0.5 mSec'\n", + "\n", + "t = A*T*log10(V/v)#\n", + "print 'Time required to charge Capacitor upto 24 Volts = %0.2e Seconds'%t\n", + "print 'i.e approx 0.549*10**-3 Sec OR 0.549 mSec'" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter23.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter23.ipynb new file mode 100644 index 00000000..59551e45 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter23.ipynb @@ -0,0 +1,136 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 23 : Alternating Current Circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 23_1 Page No. 715" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total Impedence Zt = 38.18 Ohms\n", + "Current I = 0.00 Ampers\n", + "i.e 1.31 mAmps\n", + "Theta z = 45.00 Degree\n" + ] + } + ], + "source": [ + "from math import sqrt,atan,pi\n", + "# A 27-Ohms R is in series with 54 Ohms\u0004 of Xl and 27 Ohms\u0004 of Xc. The applied voltage Vt is 50 mV. Calculate ZT, I, and Theta z.\n", + "\n", + "# Given data\n", + "\n", + "R = 27.# # Resistance=27 Ohms\n", + "Xl = 54.# # Inductive reactance=54 Ohms\n", + "Vt = 50.*10**-3# # Applied voltage=100 Volts\n", + "Xc = 27.# # Capacitive reactance=27 Ohms\n", + "\n", + "nXl = Xl-Xc# # Net Inductive reactance\n", + "R1 = R*R#\n", + "nXl1 = nXl*nXl#\n", + "\n", + "Zt = sqrt(R1+nXl1)#\n", + "print 'Total Impedence Zt = %0.2f Ohms'%Zt\n", + "\n", + "I = (Vt/Zt)#\n", + "print 'Current I = %0.2f Ampers'%I\n", + "print 'i.e 1.31 mAmps'\n", + "\n", + "Oz = atan(Xc/R)*180/pi\n", + "print 'Theta z = %0.2f Degree'%Oz" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 23_2 Page No. 717" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Total Current It = 0.00 Amps\n", + "i.e 2.55 mAmps\n", + "The Equivqlent Impedence Zeq = 19.64 Ohms\n", + "approx 19.61 Ohms\n", + "Theta z = -45.00 Degree\n" + ] + } + ], + "source": [ + "from math import atan,pi,sqrt\n", + "# The following branch currents are supplied from a 50-mV source: Ir=1.8 mA# Il=2.8 mA# Ic=1 mA. Calculate It, Zeq, and Theta I.\n", + "\n", + "# Given data\n", + "\n", + "Va = 50.*10**-3# # Applied voltage=50m Volts\n", + "Ir = 1.8*10**-3# # Ir=1.8 mAmps\n", + "Il = 2.8*10**-3# # Ir=2.8 mAmps\n", + "Ic = 1.*10**-3# # Ic=1 mAmps\n", + "\n", + "nI = Il-Ic# # net current\n", + "Ir1 = Ir*Ir#\n", + "nI1 = nI*nI#\n", + "\n", + "It = sqrt(Ir1+nI1)#\n", + "print 'The Total Current It = %0.2f Amps'%It\n", + "print 'i.e 2.55 mAmps'\n", + "\n", + "Zeq = Va/It#\n", + "print 'The Equivqlent Impedence Zeq = %0.2f Ohms'%Zeq\n", + "print 'approx 19.61 Ohms'\n", + "\n", + "Oz = atan(-(nI/Ir))*180/pi\n", + "print 'Theta z = %0.2f Degree'%Oz" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter25.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter25.ipynb new file mode 100644 index 00000000..fc3c2316 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter25.ipynb @@ -0,0 +1,441 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 25 : Resonance" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 25_1 Page No. 775" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resonant frequency = 12.58 Hertz\n", + "approx 12.6 Hertz\n" + ] + } + ], + "source": [ + "from math import pi,sqrt\n", + "# Calculate the resonant frequency for an 8-H inductance and a 20-u\u0003F capacitance.\n", + "\n", + "# Given data\n", + "\n", + "L = 8.# # L=8 Henry\n", + "C = 20.*10**-6# # C=20 uFarad\n", + "\n", + "fr = 1./(2.*pi*sqrt(L*C))#\n", + "print 'The resonant frequency = %0.2f Hertz'%fr\n", + "print 'approx 12.6 Hertz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 25_2 Page No. 776" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resonant frequency = 65007689.56 Hertz\n", + "i.e 65 MHz\n" + ] + } + ], + "source": [ + "# Calculate the resonant frequency for a 2-uH inductance and a 3-pF capacitance.\n", + "\n", + "# Given data\n", + "\n", + "L = 2.*10**-6# # Inductor=2 uHenry\n", + "C = 3.*10**-12# # Capacitor=3 pFarad\n", + "pi = 3.14#\n", + "\n", + "fr = 1./(2.*pi*sqrt(L*C))#\n", + "print 'The resonant frequency = %0.2f Hertz'%fr\n", + "print 'i.e 65 MHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 25_3 Page No. 778" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of Capacitor = 1.06e-10 Farads\n", + "i.e 106 pF\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# What value of C resonates with a 239-u\u0003H L at 1000 kHz?\n", + "\n", + "# Given data\n", + "\n", + "L = 239.*10**-6# # Inductor=239 uHenry\n", + "fr = 1000.*10**3# # Resonant frequency=1000 kHertz\n", + "\n", + "A = pi*pi# # pi square\n", + "B = fr*fr# # Resonant frequency square\n", + "\n", + "C = 1./(4.*A*B*L)#\n", + "print 'The value of Capacitor = %0.2e Farads'%C\n", + "print 'i.e 106 pF'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 25_4 Page No. 781" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of Inductor = 2.39e-04 Henry\n", + "i.e 239 uF\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# What value of L resonates with a 106-pF C at 1000 kHz, equal to 1 MHz?\n", + "\n", + "# Given data\n", + "\n", + "C = 106.*10**-12# # Capacitor=106 pFarad\n", + "fr = 1.*10**6# # Resonant frequency=1 MHertz\n", + "\n", + "A = pi*pi# # pi square\n", + "B = fr*fr# # Resonant frequency square\n", + "\n", + "C = 1./(4*A*B*C)#\n", + "print 'The value of Inductor = %.2e Henry'%C\n", + "print 'i.e 239 uF'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 25_5 Page No. 782" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Magnification factor Q =50.00\n" + ] + } + ], + "source": [ + "# A series circuit resonant at 0.4 MHz develops 100 mV across a 250-u\u0003H L with a 2-mV input. Calculate Q .\n", + "\n", + "# Given data\n", + "\n", + "Vo = 100.*10**-3# # Output voltage=100 mVolts\n", + "Vi = 2*10**-3# # Input voltage=2 mVolts\n", + "L = 250*10**-6# # Inductor=250 uHenry\n", + "f = 0.4*10**6# # Frequency=0.4 MHertz\n", + "\n", + "Q = Vo/Vi#\n", + "print 'The Magnification factor Q =%0.2f'%Q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 25_6 Page No. 784" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Ac Resistance of Coil = 12.57 Ohms\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# What is the ac resistance of the coil in A series circuit resonant at 0.4 MHz develops 100 mV across a 250-u\u0003H L with a 2-mV input.\n", + "\n", + "# Given data\n", + "\n", + "Vo = 100.*10**-3# # Output voltage=100 mVolts\n", + "Vi = 2.0*10**-3# # Input voltage=2 mVolts\n", + "L = 250.0*10**-6# # Inductor=250 uHenry\n", + "f = 0.4*10**6# # Frequency=0.4 MHertz\n", + "\n", + "Q = Vo/Vi#\n", + "Xl = 2*pi*f*L#\n", + "\n", + "rs = Xl/Q#\n", + "print 'The Ac Resistance of Coil = %0.2f Ohms'%rs" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 25_7 Page No. 785" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Because they divide Vt equally\n", + "The Equivalent Impedence = 225000 Ohms\n", + "i.e 225 kOhms\n", + "The Q =150.00\n" + ] + } + ], + "source": [ + "# In Fig. 25–9, assume that with a 4-mVac input signal for VT, the voltage across R1 is 2 mV when R1 is 225-kOhms\u0003. Determine Zeq and Q.\n", + "\n", + "# Given data\n", + "\n", + "vin = 4.*10**-3# # Input AC signal=4 mVac\n", + "R1 = 225.*10**3# # Resistance1=225 kOhms\n", + "vR1 = 2.*10**-3# # Voltage across Resistor1=2 mVac\n", + "xl = 1.5*10**3# # Inductive Reactance=1.5 kOhms\n", + "\n", + "print 'Because they divide Vt equally'\n", + "\n", + "Zeq = R1#\n", + "print 'The Equivalent Impedence = %0.f Ohms'%Zeq\n", + "print 'i.e 225 kOhms'\n", + "\n", + "Q = Zeq/xl#\n", + "print 'The Q =%0.2f'%Q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 25_8 Page No. 786" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Magnification factor Q = 40.02\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# A parallel LC circuit tuned to 200 kHz with a 350-u\u0003H L has a measured ZEQ of 17,600. Calculate Q.\n", + "\n", + "# Given data\n", + "\n", + "L = 350.*10**-6# # Inductor=350 uHenry\n", + "f = 200.*10**3# # Frequency=200 kHertz\n", + "Zeq = 17600.# # Equivalent Impedence=17600 Ohms\n", + "\n", + "Xl = 2*pi*f*L#\n", + "\n", + "Q = Zeq/Xl#\n", + "print 'The Magnification factor Q = %0.2f'%Q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 25_9 Page No. 788" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Bandwidth BW or Delta f = 20000 Hertz\n", + "i.e 20 kHz\n", + "The Edge Frequency f1 = 1990000 Hertz\n", + "i.e 1990 kHz\n", + "The Edge Frequency f2 = 2010000 Hertz\n", + "i.e 2010 kHz\n" + ] + } + ], + "source": [ + "# An LC circuit resonant at 2000 kHz has a Q of 100. Find the total bandwidth delta f and the edge frequencies f1 and f2.\n", + "\n", + "# Given data\n", + "\n", + "fr = 2000.*10**3# # Resonant frequency=2000 kHertz\n", + "Q = 100.# # Magnification factor=100\n", + "\n", + "Bw = fr/Q#\n", + "print 'The Bandwidth BW or Delta f = %0.f Hertz'%Bw\n", + "print 'i.e 20 kHz'\n", + "\n", + "f1 = fr-Bw/2#\n", + "print 'The Edge Frequency f1 = %0.f Hertz'%f1\n", + "print 'i.e 1990 kHz'\n", + "\n", + "f2 = fr+Bw/2#\n", + "print 'The Edge Frequency f2 = %0.f Hertz'%f2\n", + "print 'i.e 2010 kHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 25_10 Page No. 789" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Bandwidth BW or Delta f = 60000 Hertz\n", + "i.e 60 kHz\n", + "The Edge Frequency f1 = 5970000 Hertz\n", + "i.e 5970 kHz\n", + "The Edge Frequency f2 = 6030000 Hertz\n", + "i.e 6030 kHz\n" + ] + } + ], + "source": [ + "# An LC circuit resonant at 6000 kHz has a Q of 100. Find the total bandwidth delta f and the edge frequencies f1 and f2.\n", + "\n", + "# Given data\n", + "\n", + "fr = 6000.*10**3# # Resonant frequency=6000 kHertz\n", + "Q = 100.# # Magnification factor=100\n", + "\n", + "Bw = fr/Q#\n", + "print 'The Bandwidth BW or Delta f = %0.f Hertz'%Bw\n", + "print 'i.e 60 kHz'\n", + "\n", + "f1 = fr-Bw/2#\n", + "print 'The Edge Frequency f1 = %0.f Hertz'%f1\n", + "print 'i.e 5970 kHz'\n", + "\n", + "f2 = fr+Bw/2.0#\n", + "print 'The Edge Frequency f2 = %0.f Hertz'%f2\n", + "print 'i.e 6030 kHz'" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter26.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter26.ipynb new file mode 100644 index 00000000..471f38d0 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter26.ipynb @@ -0,0 +1,425 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 26 : Filters" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 26_1 Page No. 824" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Cutoff Frequency = 1591.55 Hertz\n", + "i.e 1.592 kHz\n", + "The Output Voltage = 7.07 Vpp\n", + "The Phase angle (Theta z) = -45.00 Degree\n" + ] + } + ], + "source": [ + "from math import pi,sqrt,atan\n", + "# Calculate (a)the cutoff frequency fc# (b)Vout at fc# (c)Theta at fc (Assume Vin = 10 Vpp for all frequencies)\n", + "\n", + "# Given data\n", + "\n", + "R = 10.*10**3# # Resistor=10 kOhms\n", + "C = 0.01*10**-6# # Capacitor=0.01 uFarad\n", + "Vin = 10.# # Input Voltage=10Vpp\n", + "# To calculate fc\n", + "\n", + "fc = 1./(2*pi*R*C)#\n", + "print 'The Cutoff Frequency = %0.2f Hertz'%fc\n", + "print 'i.e 1.592 kHz'\n", + "\n", + "# To calculate Vout at fc\n", + "\n", + "Xc = 1./(2*pi*fc*C)#\n", + "\n", + "Zt = sqrt((R*R)+(Xc*Xc))#\n", + "\n", + "Vout = Vin*(Xc/Zt)#\n", + "print 'The Output Voltage = %0.2f Vpp'%Vout\n", + "\n", + "# To calculate Theta\n", + "\n", + "Theta = atan(-(R/Xc))*180/pi\n", + "print 'The Phase angle (Theta z) = %0.2f Degree'%Theta" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 26_2 Page No. 825" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Cutoff Frequency = 3183.10 Hertz i.e 3.183 kHz\n", + "The Output Voltage = 9.54 Vpp\n", + "approx 9.52 Volts(p-p)\n", + "The Phase angle (Theta z) = -17.44 Degree\n" + ] + } + ], + "source": [ + "from math import pi,sqrt,atan\n", + "# Calculate (a)the cutoff frequency fc# (b)Vout at 1 kHz# (c)Theta at 1 kHz (Assume Vin = 10 Vpp for all frequencies)\n", + "\n", + "# Given data\n", + "\n", + "R = 1.*10**3# # Resistor=1 kOhms\n", + "L = 50.*10**-3 # Inductor=50 mHenry\n", + "Vin = 10.# # Input Voltage=10Vpp\n", + "f = 1.*10**3# # Frequency=1 kHz\n", + "# To calculate fc\n", + "\n", + "fc = R/(2*pi*L)#\n", + "print 'The Cutoff Frequency = %0.2f Hertz'%fc,\n", + "print 'i.e 3.183 kHz'\n", + "\n", + "# To calculate Vout at fc\n", + "\n", + "Xl = 2*pi*f*L#\n", + "\n", + "Zt = sqrt((R*R)+(Xl*Xl))#\n", + "\n", + "Vout = Vin*(R/Zt)#\n", + "print 'The Output Voltage = %0.2f Vpp'%Vout\n", + "print 'approx 9.52 Volts(p-p)'\n", + "\n", + "# To calculate Theta\n", + "\n", + "Theta = atan(-(Xl/R))*180/pi\n", + "print 'The Phase angle (Theta z) = %0.2f Degree'%Theta" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 26_3 Page No. 826" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Cutoff Frequency for RC High-Pass Filter = 10610.33 Hertz\n", + "i.e 10.61 kHz\n", + "The Cutoff Frequency for RL High-Pass Filter = 2387.32 Hertz\n", + "approx 2.39 kHz\n" + ] + } + ], + "source": [ + "from math import pi,sqrt\n", + "# Calculate the cutoff frequency for (a) the RC high-pass filter# (b) the RL high-pass filter\n", + "\n", + "# Given data\n", + "\n", + "R = 1.5*10**3# # Resistor=1.5 kOhms\n", + "L = 100.*10**-3 # Inductor=100 mHenry\n", + "C = 0.01*10**-6# # Capacitor=0.01 uFarad\n", + "\n", + "# To calculate fc for RC high-pass filter\n", + "\n", + "fc = 1./(2*pi*R*C)#\n", + "print 'The Cutoff Frequency for RC High-Pass Filter = %0.2f Hertz'%fc\n", + "print 'i.e 10.61 kHz'\n", + "\n", + "# To calculate fc for RL high-pass filter\n", + "\n", + "fc1 = R/(2*pi*L)#\n", + "print 'The Cutoff Frequency for RL High-Pass Filter = %0.2f Hertz'%fc1\n", + "print 'approx 2.39 kHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 26_4 Page No. 827" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Cutoff Frequency for RC High-Pass filter = 159.15 Hertz\n", + "i.e 159 Hz\n", + "The Cutoff Frequency for RC High-Pass filter = 1591.55 Hertz\n", + "i.e 1.59 kHz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# Calculate the cutoff frequencies fc1 and fc2.\n", + "\n", + "#Given data\n", + "\n", + "R1 = 1.*10**3# # Resistor 1=1 kOhms\n", + "C1 = 1.*10**-6# # Capacitor 1=1 uFarad\n", + "R2 = 100.*10**3# # Resistor 2=100 kOhms\n", + "C2 = 0.001*10**-6# # Capacitor 2=0.001 uFarad\n", + "\n", + "# To calculate fc1 for RC high-pass filter\n", + "\n", + "fc1 = 1/(2*pi*R1*C1)#\n", + "print 'The Cutoff Frequency for RC High-Pass filter = %0.2f Hertz'%fc1\n", + "print 'i.e 159 Hz'\n", + "\n", + "# To calculate fc2 for RC high-pass filter\n", + "\n", + "fc2 = 1/(2*pi*R2*C2)#\n", + "print 'The Cutoff Frequency for RC High-Pass filter = %0.2f Hertz'%fc2\n", + "print 'i.e 1.59 kHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 26_5 Page No. 828" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Notch Frequency for RC Low-Pass filter = 7957.75 Hertz\n", + "i.e 7.96 kHz\n", + "The Required Value of 2R1 = 2000.00 Ohms\n", + "i.e 2 kohms\n", + "The Required Value of 2C1 = 2.00e-08 Ohms\n", + "0.02 uF\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# Calculate the notch frequency fn if R1 is 1 kOhms\u0004 and C1 is\u0005 0.01 \u0002uF. Also, calculate the required values for 2R1 and 2C1 in the low-pass filter.\n", + "\n", + "# Given data\n", + "\n", + "R1 = 1.*10**3# # Resistor 1=1 kOhms\n", + "C1 = 0.01*10**-6# # Capacitor 1=0.01 uFarad\n", + "\n", + "# To calculate Notch frequency fn for RC low-pass filter\n", + "\n", + "fn = 1/(4*pi*R1*C1)#\n", + "print 'The Notch Frequency for RC Low-Pass filter = %0.2f Hertz'%fn\n", + "print 'i.e 7.96 kHz'\n", + "\n", + "A = 2*R1#\n", + "print 'The Required Value of 2R1 = %0.2f Ohms'%A\n", + "print 'i.e 2 kohms'\n", + "\n", + "B = 2*C1#\n", + "print 'The Required Value of 2C1 = %0.2e Ohms'%B\n", + "print '0.02 uF'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 26_6 Page No. 829" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Power Gain of Amplifier = 20.00 dB\n" + ] + } + ], + "source": [ + "from math import log10\n", + "# A certain amplifier has an input power of 1 W and an output power of 100 W.Calculate the dB power gain of the amplifier.\n", + "\n", + "# Given data\n", + "\n", + "Pi = 1.# # Input power=1 Watts\n", + "Po = 100.# # Output power=100 Watts\n", + "\n", + "N = 10*log10(Po/Pi)#\n", + "print 'The Power Gain of Amplifier = %0.2f dB'%N" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 26_7 Page No. 830" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Attenuation offered by the Filter = -13.01 dB\n" + ] + } + ], + "source": [ + "from math import log10\n", + "# The input power to a filter is 100 mW, and the output power is 5 mW. Calculate the attenuation, in decibels, offered by the filter.\n", + "\n", + "# Given data\n", + "\n", + "Pi = 100.*10**-3# # Input power=1 Watts\n", + "Po = 5.*10**-3# # Output power=100 Watts\n", + "\n", + "N = 10*log10(Po/Pi)#\n", + "print 'The Attenuation offered by the Filter = %0.2f dB'%N" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 26_8 Page No. 832" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Attenuation at 0 Hz = 0 dB\n", + "The Attenuation at 1.592 kHz = -3.01 dB\n", + "The Attenuation at 15.92 kHz = -20.05 dB\n" + ] + } + ], + "source": [ + "from math import pi,log10,sqrt\n", + "# Calculate the attenuation, in decibels, at the following frequencies: (a) 0 Hz# (b) 1.592 kHz# (c) 15.92 kHz. (Assume that Vin is 10 V p-p at all frequencies.)\n", + "\n", + "# Given data\n", + "\n", + "f1 = 0# # Frequency 1=0 Hz\n", + "f2 = 1.592*10**3# # Frequency 2=1.592 kHz (cutoff frequency)\n", + "f3 = 15.92*10**3# # Frequency 3=15.92 kHz\n", + "Vi = 10.# # Voltage input=10 Volts(p-p)\n", + "R = 10.*10**3# # Resistor 1=10 kOhms\n", + "C = 0.01*10**-6# # Capacitor 1=0.01 uFarad \n", + "\n", + "Vo1 = Vi#\n", + "Vo2 = 0.707*Vi#\n", + "\n", + "# At 0 Hz\n", + "\n", + "N1 = 20*log10(Vo1/Vi)#\n", + "print 'The Attenuation at 0 Hz = %0.f dB'%N1\n", + "\n", + "#At 1.592 kHz (cutoff frequency)\n", + "\n", + "N2 = 20*log10(Vo2/Vi)#\n", + "print 'The Attenuation at 1.592 kHz = %0.2f dB'%N2\n", + "\n", + "# At 15.92 kHz\n", + "\n", + "Xc = 1./(2*pi*f3*C)#\n", + "\n", + "A = R*R#\n", + "B = Xc*Xc#\n", + "\n", + "Zt = sqrt (A+B)#\n", + "\n", + "N3 = 20*log10(Xc/Zt)#\n", + "print 'The Attenuation at 15.92 kHz = %0.2f dB'%N3" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter27.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter27.ipynb new file mode 100644 index 00000000..77d3bb82 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter27.ipynb @@ -0,0 +1,712 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 27 : Diodes and Diodes Applications" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_1 Page No. 864" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Forward Resistance at Point A = 59.09 Ohms\n", + "Approx 59.1 Ohms\n", + "The Forward Resistance at Point B = 31.11 Ohms\n" + ] + } + ], + "source": [ + "# For the diode curve, calculate the dc resistance, RF, at points A and B.\n", + "\n", + "# Given data\n", + "\n", + "Vf1 = 0.65# # Forward votage 1=0.65 Volts\n", + "If1 = 11*10**-3 # Forward current 1=11 mAmps\n", + "Vf2 = 0.7# # Forward votage 2=0.7 Volts\n", + "If2 = 22.5*10**-3 # Forward current 2=22.5 mAmps\n", + "\n", + "Rf1 = Vf1/If1#\n", + "print 'The Forward Resistance at Point A = %0.2f Ohms'%Rf1\n", + "print 'Approx 59.1 Ohms'\n", + "\n", + "Rf2 = Vf2/If2#\n", + "print 'The Forward Resistance at Point B = %0.2f Ohms'%Rf2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_2 Page No. 865" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Bulk Resistance = 0.40 Ohms\n" + ] + } + ], + "source": [ + "# A silicon diode has a forward voltage drop of 1.1 V for a forward diode current, If, of 1 A. Calculate the bulk resistance, Rb.\n", + "\n", + "# Given data\n", + "\n", + "Vf1 = 1.1# # Forward votage 1=1.1 Volts\n", + "If1 = 1. # Forward current 1=1 Amps\n", + "Vf2 = 0.7# # Fwd. vltg. 2=0.7 Volts (min working vltg of diode is 0.7 V)\n", + "If2 = 0 # Forward current=0 Amps\n", + "\n", + "delV = Vf1-Vf2# # diff. between max. min. Voltages\n", + "delI = If1-If2# # diff. between max. min. Currents\n", + "\n", + "Rb = delV/delI#\n", + "print 'The Bulk Resistance = %0.2f Ohms'%Rb" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_3 Page No. 868" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Load Voltage of First Approximation = 10.00 Volts(dc)\n", + "The Load Current of First Approximation = 0.10 Amps\n", + "i.e 100 mAmps\n", + "The Load Voltage of Second Approximation = 9.30 Volts\n", + "The Load Current of Second Approximation = 0.09 Amps\n", + "i.e 93 mAmps\n", + "The Load Current of Third Approximation = 0.09 Amps\n", + "i.e 90.73 mAmps\n", + "The Load Voltage of Third Approximation 9.07 Volts\n" + ] + } + ], + "source": [ + "# Solve for the load voltage and current using the first, second, and third diode approximations.\n", + "\n", + "# Given data\n", + "\n", + "Rl = 100.# # Load resistance=100 Ohms\n", + "Rb = 2.5# # Resistance=2.5 Ohms\n", + "Vin = 10.# # Input voltage=10 Volts\n", + "Vb = 0.7# # Voltage=0.7 Volts\n", + "\n", + "\n", + "# Using first approximation\n", + "\n", + "Vl1 = Vin\n", + "print 'The Load Voltage of First Approximation = %0.2f Volts(dc)'%Vl1\n", + "\n", + "Il1 = Vl1/Rl#\n", + "print 'The Load Current of First Approximation = %0.2f Amps'%Il1\n", + "print 'i.e 100 mAmps'\n", + "\n", + "# Using second approximation\n", + "\n", + "Vl2 = Vin-Vb\n", + "print 'The Load Voltage of Second Approximation = %0.2f Volts'%Vl2\n", + "\n", + "Il2 = Vl2/Rl#\n", + "print 'The Load Current of Second Approximation = %0.2f Amps'%Il2\n", + "print 'i.e 93 mAmps'\n", + "\n", + "# Using third approximation\n", + "\n", + "Il3 = (Vin-Vb)/(Rl+Rb)#\n", + "print 'The Load Current of Third Approximation = %0.2f Amps'%Il3\n", + "print 'i.e 90.73 mAmps'\n", + "\n", + "Vl3 = Il3*Rl#\n", + "print 'The Load Voltage of Third Approximation %0.2f Volts'%Vl3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_4 Page No. 875" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Secondary Voltage = 40.00 Volts(ac)\n", + "The DC Voltage = 17.76 Volts\n", + "The Load Current = 0.18 Amps\n", + "The DC Diode Current = 0.18 Amps\n", + "The PIV for Diode-1 = 56.56 Volts\n", + "The Output Frequency = 60.00 Hertz\n" + ] + } + ], + "source": [ + "# If the turns ratio Np:Ns is 3:1, calculate the following: Vs, Vdc, Il, Idiode, PIV for D1, and fout.\n", + "\n", + "# Given data\n", + "\n", + "Vp = 120.# # Primary voltage=120 Vac\n", + "A = 3/1.# # Turns ratio Np:Ns=3:1\n", + "B = 1/3.# # Turns ratio Ns:Np=1:3\n", + "Rl = 100.# # Load resistance=100 Ohms\n", + "fi = 60.# # Input frequency=60\n", + "\n", + "Vs = B*Vp#\n", + "print 'The Secondary Voltage = %0.2f Volts(ac)'%Vs\n", + "\n", + "Vspk = (Vs*1.414)#\n", + "\n", + "C = Vspk-0.7#\n", + "\n", + "Vdc = 0.318*C#\n", + "print 'The DC Voltage = %0.2f Volts'%Vdc\n", + "\n", + "Il = Vdc/Rl#\n", + "print 'The Load Current = %0.2f Amps'%Il\n", + "\n", + "Idiode = Il#\n", + "print 'The DC Diode Current = %0.2f Amps'%Idiode\n", + "\n", + "PIV = Vspk#\n", + "print 'The PIV for Diode-1 = %0.2f Volts'%PIV\n", + "\n", + "fo =fi#\n", + "print 'The Output Frequency = %0.2f Hertz'%fo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_5 age No. 878" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The DC Voltage = 17.54 Volts\n", + "The Load Current = 0.18 Amps\n", + "i.e 175.4 mAmps\n", + "The DC Diode Current = 0.09 Amps\n", + "i.e 87.7 mAmps\n", + "The PIV for Diode-1 = 55.86 Volts\n", + "The Output Frequency = 120.00 Hertz\n" + ] + } + ], + "source": [ + "# If the turns ratio Np:Ns is\u0006 3:1, calculate the following: Vdc, Il, Idiode, PIV for D1, and fout.\n", + "\n", + "# Given data\n", + "\n", + "Vp = 120. # Primary voltage=120 Vac\n", + "A = 3/1. # Turns ratio Np:Ns = 3:1\n", + "B = 1/3. # Turns ratio Ns:Np = 1:3\n", + "Rl = 100. # Load resistance=100 Ohms\n", + "\n", + "Vs = B*Vp#\n", + "Vspk = 1.414*(Vs/2)#\n", + "Vopk = Vspk-0.7#\n", + "\n", + "Vdc = 0.636*Vopk#\n", + "print 'The DC Voltage = %0.2f Volts'%Vdc\n", + "\n", + "Il = Vdc/Rl#\n", + "print 'The Load Current = %0.2f Amps'%Il\n", + "print 'i.e 175.4 mAmps'\n", + "\n", + "Idiode = Il/2#\n", + "print 'The DC Diode Current = %0.2f Amps'%Idiode\n", + "print 'i.e 87.7 mAmps'\n", + "\n", + "C = (Vspk*2)-0.7#\n", + "\n", + "PIV = C#\n", + "print 'The PIV for Diode-1 = %0.2f Volts'%PIV\n", + "\n", + "f =120#\n", + "print 'The Output Frequency = %0.2f Hertz'%f" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_6 Page No. 880" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The DC Voltage = 35.08 Volts\n", + "The Load Current = 0.35 Amps\n", + "i.e 350.8 mAmps\n", + "The DC Diode Current = 0.1754 Amps\n", + "i.e 175.4 mAmps\n", + "The PIV for each Diode = 55.86 Volts\n", + "The Output Frequency = 120.00 Hertz\n" + ] + } + ], + "source": [ + "# If the turns ratio Np:Ns is\u0006 3:1, calculate the following: Vdc, Il, Idiode, PIV for each diode, and fout.\n", + "\n", + "# Given data\n", + "\n", + "Vp = 120.# # Primary voltage=120 Vac\n", + "A = 3./1# # Turns ratio Np:Ns = 3:1\n", + "B = 1./3# # Turns ratio Ns:Np = 1:3\n", + "Rl = 100.# # Load resistance=100 Ohms\n", + "\n", + "Vs = B*Vp#\n", + "Vspk = 1.414*(Vs)#\n", + "Vopk = Vspk-1.4#\n", + "\n", + "Vdc = 0.636*Vopk#\n", + "print 'The DC Voltage = %0.2f Volts'%Vdc\n", + "\n", + "Il = Vdc/Rl#\n", + "print 'The Load Current = %0.2f Amps'%Il\n", + "print 'i.e 350.8 mAmps'\n", + "\n", + "Idiode = Il/2#\n", + "print 'The DC Diode Current = %0.4f Amps'%Idiode\n", + "print 'i.e 175.4 mAmps'\n", + "\n", + "C = Vspk-0.7#\n", + "\n", + "PIV = C#\n", + "print 'The PIV for each Diode = %0.2f Volts'%PIV\n", + "\n", + "f =120#\n", + "print 'The Output Frequency = %0.2f Hertz'%f" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_7 Page No. 883" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Ripple Voltage for Turns Ratio Np:Ns=4:1 = 5.21 Volts(p-p)\n", + "Approx 5.21 Volts(p-p)\n", + "The DC Voltage for Turns Ratio Np:Ns=4:1 = 39.12 Volts\n", + "Approx 39.12 Volts\n", + "The Ripple Voltage for Turns Ratio Np:Ns=2:1 = 2.69 Volts(p-p)\n", + "Approx 2.69 Volts(p-p)\n", + "The DC Voltage for Turns Ratio Np:Ns=2:1 = 40.38 Volts\n", + "Approx 40.38 Volts\n" + ] + } + ], + "source": [ + "from math import exp\n", + "# Assume the transformer turns ratio Np:Ns = 4:1 in Fig. 27–21 a and 2:1 in Fig. 27–22a. Compare Vripple and Vdc if C =\u0006 500 uF and Rl =\u0006 250\u0007.\n", + " \n", + "# Given data\n", + "\n", + "A1 = 4./1# # Turns ratio Np:Ns=4:1\n", + "B1 = 1./4# # Turns ratio Ns:Np=1:4\n", + "A2 = 2./1# # Turns ratio Np:Ns=2:1\n", + "B2 = 1./2# # Turns ratio Ns:Np=1:2\n", + "Vp = 120.# # Primary voltage=120 Vac\n", + "Vb = 0.7# # \n", + "t1 = 16.67*10**-3# # Charging Time of Capacitor of Turns ratio Np:Ns=4:1=16.67 mSec\n", + "t2 = 8.33*10**-3# # Charging Time of Capacitor of Turns ratio Np:Ns=4:1=8.33 mSec\n", + "Rl = 250.# # Load resistance=250 Ohms\n", + "C = 500.*10**-6# # Capacitor=500 uFarad\n", + "\n", + "# Calculations for Turns Ratio Np:Ns=4:1\n", + "\n", + "Vs1 = B1*Vp#\n", + "Vspk1 = Vs1*1.414#\n", + "Vopk1 = Vspk1 - Vb#\n", + "D = -t1/(Rl*C)#\n", + "\n", + "Vrp1 = Vopk1*(1-exp(D))#\n", + "print 'The Ripple Voltage for Turns Ratio Np:Ns=4:1 = %0.2f Volts(p-p)'%Vrp1\n", + "print 'Approx 5.21 Volts(p-p)'\n", + "\n", + "Vdc1 = Vopk1-(Vrp1/2)#\n", + "print 'The DC Voltage for Turns Ratio Np:Ns=4:1 = %0.2f Volts'%Vdc1\n", + "print 'Approx 39.12 Volts'\n", + "\n", + "# Calculations for Turns Ratio Np:Ns = 2:1\n", + "\n", + "Vs2 = B2*Vp#\n", + "V2 = Vs2/2#\n", + "V2pk2 = V2*1.414\n", + "Vopk2 = V2pk2 - Vb#\n", + "E = -t2/(Rl*C)#\n", + "\n", + "Vrp2 = Vopk2*(1-(exp(E)))\n", + "print 'The Ripple Voltage for Turns Ratio Np:Ns=2:1 = %0.2f Volts(p-p)'%Vrp2\n", + "print 'Approx 2.69 Volts(p-p)'\n", + "\n", + "Vdc2 = Vopk2-(Vrp2/2)#\n", + "print 'The DC Voltage for Turns Ratio Np:Ns=2:1 = %0.2f Volts'%Vdc2\n", + "print 'Approx 40.38 Volts'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_8 Page No. 885" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The LED Current = 0.01 Amps\n", + "i.e 10 mAmps\n" + ] + } + ], + "source": [ + "# Calculate the LED current.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 24.# # Input voltage=24 Volts\n", + "Vled = 2.# # Voltage drop at LED=2 Volts\n", + "Rs = 2.2*10**3# # Source Resistance=2.2 kOhms\n", + "\n", + "Iled = (Vin-Vled)/Rs#\n", + "print 'The LED Current = %0.2f Amps'%Iled\n", + "print 'i.e 10 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_9 Page No. 888" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistance Rs, Required to Provide an LED Current of 25 mA = 880.00 Ohms\n" + ] + } + ], + "source": [ + "# Calculate the resistance Rs, required to provide an LED current of 25 mA.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 24.# # Input voltage=24 Volts\n", + "Vled = 2.# # Voltage drop at LED=2 Volts\n", + "Iled = 25.*10**-3# # LED Current=25 mAmps\n", + "\n", + "Rs = (Vin-Vled)/Iled#\n", + "print 'The Resistance Rs, Required to Provide an LED Current of 25 mA = %0.2f Ohms'%Rs" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_10 Page No. 889" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Maximum Rated Current of Zener = 0.10 Amps\n", + "i.e 100 mAmps\n" + ] + } + ], + "source": [ + "# Calculate the maximum rated zener current for a 1 W, 10 V zener.\n", + "\n", + "# Given data\n", + "\n", + "Pzm = 1.# # Power rating of zener= 1 Watts\n", + "Vz = 10.# # Voltage rating of zener= 10 Volts\n", + "\n", + "Izm = Pzm/Vz#\n", + "print 'The Maximum Rated Current of Zener = %0.2f Amps'%Izm\n", + "print 'i.e 100 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_11 Page No. 890" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Zener Current = 0.01 Amps\n", + "i.e 15 mAmps\n" + ] + } + ], + "source": [ + "# If Vz=10 V, calculate Iz.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 25.# # Input voltage=25 Volts\n", + "Vz = 10.# # Zener voltage=10 Volts\n", + "Rs = 1.*10**3# # Source Resistance=1 kOhms\n", + "\n", + "Iz = (Vin-Vz)/Rs#\n", + "print 'The Zener Current = %0.2f Amps'%Iz\n", + "print 'i.e 15 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_12 Page No. 891" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Source Current = 0.075 Amps\n", + "i.e 75 mAmps\n", + "The Load Current = 0.03 Amps\n", + "i.e 30 mAmps\n", + "The Zener Current = 0.045 Amps\n", + "i.e 45 mAmps\n", + "The Power Dissipation of Zener = 0.3375 Watts\n", + "i.e 337.5 mWatts\n" + ] + } + ], + "source": [ + "# If R L increases to 250 Ohms, calculate the following: Is, Il, Iz, and Pz.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 25# # Input voltage=25 Volts\n", + "Vz = 7.5# # Zener voltage=7.5 Volts\n", + "Rl = 250# # Load Resistance=250 Ohms\n", + "Is = 75*10**-3# # Source current=75 mAmps\n", + "\n", + "print 'The Source Current = %0.3f Amps'%Is\n", + "print 'i.e 75 mAmps'\n", + "\n", + "Il = Vz/Rl#\n", + "print 'The Load Current = %0.2f Amps'%Il\n", + "print 'i.e 30 mAmps'\n", + "\n", + "Iz = Is-Il#\n", + "print 'The Zener Current = %0.3f Amps'%Iz\n", + "print 'i.e 45 mAmps'\n", + "\n", + "Pz = Vz*Iz#\n", + "print 'The Power Dissipation of Zener = %0.4f Watts'%Pz\n", + "print 'i.e 337.5 mWatts'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_13 Page No. 892" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Source Current = 0.06 Amps.\n", + "i.e 60 mAmps\n", + "The Load Current for 200 ohms Load = 0.05 Amps.\n", + "i.e 50 mAmps\n", + "The Zener Current for 200 ohms Load = 0.01 Amps.\n", + "i.e 10 mAmps\n", + "The Load Current for 500 ohms Load = 0.02 Amps.\n", + "i.e 20 mAmps\n", + "The Zener Current for 500 ohms load = 0.04 Amps.\n", + "i.e 40 mAmps\n" + ] + } + ], + "source": [ + "# Calculate Is, Il and Iz for (a)Rl=200 ohms# (b)Rl=500 ohms.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 16.# # Vin=16 Volts given\n", + "Vz = 10.# # Vz=10 Volts given\n", + "Rs = 100.# # Source Resistance = 100 ohms given\n", + "Rla = 200.# # Load Resistance A = 200 ohms given\n", + "Rlb = 500.# #Load Resistance B = 500 ohms given\n", + "\n", + "# For Rl 200 ohms\n", + "\n", + "Is = (Vin-Vz)/Rs#\n", + "print 'The Source Current = %0.2f Amps.'%Is\n", + "print 'i.e 60 mAmps'\n", + "\n", + "Ila = Vz/Rla#\n", + "print 'The Load Current for 200 ohms Load = %0.2f Amps.'%Ila\n", + "print 'i.e 50 mAmps'\n", + "\n", + "Iza = Is-Ila\n", + "print 'The Zener Current for 200 ohms Load = %0.2f Amps.'%Iza\n", + "print 'i.e 10 mAmps'\n", + "\n", + "# For Rl 500 ohms\n", + "\n", + "Ilb = Vz/Rlb#\n", + "print 'The Load Current for 500 ohms Load = %0.2f Amps.'%Ilb\n", + "print 'i.e 20 mAmps'\n", + "\n", + "Izb = Is-Ilb\n", + "print 'The Zener Current for 500 ohms load = %0.2f Amps.'%Izb\n", + "print 'i.e 40 mAmps'" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter28.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter28.ipynb new file mode 100644 index 00000000..67d92401 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter28.ipynb @@ -0,0 +1,723 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 28 : Bipolar Junction Transistors" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_1 Page No. 910" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Emitter Current Ie = 5.00 Amps\n" + ] + } + ], + "source": [ + "# A transistor has the following currents: Ib is\u0002 20 mA and Ic is 4.98 A. Calculate Ie.\n", + "\n", + "# Given data\n", + "\n", + "Ib = 20*10**-3# # Base current=20 mAmps\n", + "Ic = 4.98# # Collector current=4.98 Amps\n", + "\n", + "Ie = Ic+Ib#\n", + "print 'The Emitter Current Ie = %0.2f Amps'%Ie" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_2 Page No. 912" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Collector Current Ic = 0.09804 Amps\n", + "i.e 98.04 mAmps\n" + ] + } + ], + "source": [ + "# A transistor has the following currents: Ie is\u0002 100 mA, Ib is\u0002 1.96 mA. Calculate Ic.\n", + "\n", + "# Given data\n", + "\n", + "Ie = 100.0*10**-3# # Emitter current=100 mAmps\n", + "Ib = 1.96*10**-3# # Base current=4.98 Amps\n", + "\n", + "Ic = Ie-Ib#\n", + "print 'The Collector Current Ic = %0.5f Amps'%Ic\n", + "print 'i.e 98.04 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_3 Page No. 913" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Base Current Ib = 0.0010 Amps\n", + "i.e 1 mAmps\n" + ] + } + ], + "source": [ + "# A transistor has the following currents: Ie is\u0002 50 mA, Ic is\u0002 49 mA. Calculate Ib.\n", + "\n", + "# Given data\n", + "\n", + "Ie = 50.0*10**-3# # Emitter current=50 mAmps\n", + "Ic = 49.0*10**-3# # Collector current=20 mAmps\n", + "\n", + "Ib = Ie-Ic#\n", + "print 'The Base Current Ib = %0.4f Amps'%Ib\n", + "print 'i.e 1 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_4 Page No. 914" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Alpha(dc) = 0.9960\n" + ] + } + ], + "source": [ + "# A transistor has the following currents: Ie is\u0002 15 mA, Ib is\u0002 60 u\u0004A. Calculate \u0002Alpha(dc).\n", + "\n", + "# Given data\n", + "\n", + "Ie = 15.*10**-3# # Emitter current=15 mAmps\n", + "Ib = 60.*10**-6# # Base current=60 uAmps\n", + "\n", + "Ic = Ie-Ib#\n", + "\n", + "Adc = Ic/Ie#\n", + "print 'The Value of Alpha(dc) = %0.4f'%Adc" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_5 Page No. 916" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Beta(dc) = 200\n" + ] + } + ], + "source": [ + "# A transistor has the following currents: Ic is\u0002 10 mA and Ib is 50 uA. Calculate Beta(dc).\n", + "\n", + "# Given data\n", + "\n", + "Ic = 10.*10**-3# # Collector current=10 mAmps\n", + "Ib = 50.*10**-6# # Base current=50 uAmps\n", + "\n", + "Bdc = Ic/Ib#\n", + "print 'The Value of Beta(dc) = %0.f'%Bdc" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_6 Page No. 918" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Collector Current Ic = 0.01125 Amps\n", + "i.e 11.25 mAmps\n" + ] + } + ], + "source": [ + "# A transistor has Beta(dc) of 150 and Ib of 75 uAmps. Calculate Ic.\n", + "\n", + "# Given data\n", + "\n", + "Bdc = 150.# # Beta(dc)=150\n", + "Ib = 75.*10**-6# # Base current=75 uAmps\n", + "\n", + "Ic = Bdc*Ib#\n", + "print 'The Collector Current Ic = %0.5f Amps'%Ic\n", + "print 'i.e 11.25 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_7 Page No. 920" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Alpha(dc) = 0.9901\n" + ] + } + ], + "source": [ + "# A transistor has Beta(dc) of 100. Calculate Alpha(dc).\n", + "\n", + "# Given data\n", + "\n", + "Bdc = 100.0# # Beta(dc)=100\n", + "\n", + "Adc = Bdc/(1+Bdc)#\n", + "print 'The Value of Alpha(dc) = %0.4f'%Adc" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_8 Page No. 922" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Beta(dc) =199.00\n" + ] + } + ], + "source": [ + "# A transistor has Alpha(dc) of 0.995. Calculate Beta(dc).\n", + "\n", + "# Given data\n", + "\n", + "Adc = 0.995# # Alpha(dc)=100\n", + "\n", + "Bdc = Adc/(1-Adc)#\n", + "print 'The Value of Beta(dc) =%0.2f'%Bdc" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_9 Page No. 922" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Power Dissipation = 0.05 Watts\n", + "i.e 50 mWatts\n" + ] + } + ], + "source": [ + "# Calculate Pd if Vcc is 10 V and Ib is 50 uAmps. Assume Beta(dc) is 100.\n", + "\n", + "# Given data\n", + "\n", + "Bdc = 100.# # Beta(dc)=100\n", + "Ib = 50.*10**-6# # Base current=50 uAmps\n", + "Vcc = 10.# # Supply voltage=10 Volts\n", + "\n", + "Vce = Vcc\n", + "\n", + "Ic = Bdc*Ib#\n", + "\n", + "Pd = Vce*Ic#\n", + "print 'The Power Dissipation = %0.2f Watts'%Pd\n", + "print 'i.e 50 mWatts'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_10 Page No. 922" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Maximum Allowable Collector Current Ic(max) = 0.025 Amps\n", + "i.e 25 mAmps\n" + ] + } + ], + "source": [ + "# The transistor has a power rating of 0.5 W. If Vce is\u0002 20 V, calculate the maximum allowable collector current, Ic, that can exist without exceeding the transistor’s power rating.\n", + "\n", + "# Given data\n", + "\n", + "Pdmax = 0.5# # Power dissipation(max)=0.5 Watts\n", + "Vce = 20.# # Voltage (collector to emitter)=20 Volts\n", + "\n", + "Ic = Pdmax/Vce#\n", + "print 'The Maximum Allowable Collector Current Ic(max) = %0.3f Amps'%Ic\n", + "print 'i.e 25 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_11 Page No. 923" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Power Rating at 50°C = 0.28 Watts\n", + "i.e 280 mWatts\n" + ] + } + ], + "source": [ + "# Assume that a transistor has a power rating Pd(max) of 350 mW at an ambient temperature Ta of 25°C. The derating factor is 2.8 mW/°C. Calculate the power rating at 50°C.\n", + "\n", + "# Given data\n", + "\n", + "f = 2.8*10**-3# # Derating factor=2.8 mW/°C\n", + "Pd = 350.*10**-3# # Power dissipation(max)=350 mWatts\n", + "Ta = 25.# # Ambient Temperature=25°C\n", + "Tp = 50.# # Power rating at 50°C\n", + "\n", + "delT = Tp-Ta# # Difference between max and min temp\n", + "\n", + "delPd = delT*f#\n", + "\n", + "Prat = Pd-delPd#\n", + "print 'The Power Rating at 50°C = %0.2f Watts'%Prat\n", + "print 'i.e 280 mWatts'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_12 Page No. 923" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Base Current = 0.0000 Amps.\n", + "Approx 28.97 mAmps\n", + "The Collector Current = 0.0043 Amps\n", + "Approx 4.35 mAmps\n", + "The Voltage Collector-Emitter = 5.48 Volts\n", + "Q(5.480769,0.004346)\n", + "\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fbd5f1197d0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "from matplotlib.pyplot import plot,xlabel,ylabel,show,title\n", + "# Solve for Ib, Ic, Vce. Also, Construct a dc load line showing the valuse of Ic(sat), Vce(off), Icq, Vceq.\n", + "\n", + "# Given data\n", + "Vcc = 12.# # Supply voltage=12 Volts\n", + "Vbe = 0.7# # Base-Emitter Voltage=0.7 Volts\n", + "Rb = 390.*10**3# # Base Resistor=390K Ohms\n", + "Rc = 1.5*10**3# # Collector Resistor=1.5K Ohms\n", + "B = 150.# # Beta(dc)=150\n", + "\n", + "Ib = (Vcc-Vbe)/Rb#\n", + "print 'The Base Current = %0.4f Amps.'%Ib\n", + "print 'Approx 28.97 mAmps'\n", + "\n", + "Icq = B*Ib#\n", + "print 'The Collector Current = %0.4f Amps'%Icq\n", + "print 'Approx 4.35 mAmps'\n", + "\n", + "Vceq = Vcc-(Icq*Rc)#\n", + "print 'The Voltage Collector-Emitter = %0.2f Volts'%Vceq\n", + "\n", + "# For DC load line\n", + "\n", + "Icsat = (Vcc/Rc)#\n", + "Vceoff = Vcc#\n", + "\n", + "Vce1=[Vceoff, Vceq ,0]\n", + "Ic1=[0 ,Icq ,Icsat]\n", + "\n", + "#To plot DC load line\n", + "\n", + "print \"Q(%f,%f)\\n\"%(Vceq,Icq)\n", + "plot(Vce1, Ic1)\n", + "plot(Vceq,Icq)\n", + "plot(0,Icq)\n", + "plot(Vceq,0)\n", + "plot(0,Icsat)\n", + "plot(Vceoff,0)\n", + "xlabel(\"Vce in volt\")\n", + "ylabel(\"Ic in Ampere\")\n", + "title(\"DC Load-line for Base-Biased Transistor Circuit\")\n", + "show() " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_13 Page No. 924" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Base Voltage = 2.61 Volts\n", + "The Emmiter Voltage = 1.91 Volts\n", + "The Collector Voltage = 10.65 Volts\n", + "Approx 10.65 Volts\n", + "The Collector-Emitter Voltage = 8.74 Volts\n", + "Approx 8.74 Volts\n", + "The Current Ic(sat) = 0.01 Amps\n", + "i.e 9.52 mAmps\n", + "The Voltage Vce(off) = 18.00 Volts\n", + "Q(8.737067,0.004901)\n", + "\n" + ] + }, + { + "data": { + "image/png": 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jH4NmhI5Xk9KWmcTWITH6Ap+lFFVVZBKh/T3x758qWtCrMf5LXbx+/etfJx6DYiqsuBST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\nMkOBRwDcfTawq5ntUcW6W9a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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fbd5f078490>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "from matplotlib.pyplot import plot,xlabel,ylabel,show,title\n", + "\n", + "# Solve for Vb, Ve, Ic, Vc, and Vce. Also, calculate Ic(sat) and Vce(off). Finally, construct a dc load line showing the values of Ic(sat), Vce(off), Icq, and Vceq.\n", + "\n", + "# Given data\n", + "\n", + "R1 = 33.*10**3# # Resistor 1=33 kOhms\n", + "R2 = 5.6*10**3# # Resistor 2=5.6 kOhms\n", + "Rc = 1.5*10**3# # Collector resistance=1.5 kOhms\n", + "Re = 390.# # Emitter resistance=390 Ohms\n", + "Bdc = 200.# # Beta(dc)= 200\n", + "Vcc = 18.# # Supply voltage = 18 Volts\n", + "Vbe = 0.7# # Base-Emmiter Voltage=0.7 Volts\n", + "\n", + "Vb = Vcc*(R2/(R1+R2))#\n", + "print 'The Base Voltage = %0.2f Volts'%Vb\n", + "\n", + "Ve = Vb-Vbe#\n", + "print 'The Emmiter Voltage = %0.2f Volts'%Ve\n", + "\n", + "Ie = Ve/Re# # Emitter current\n", + "\n", + "Ic = Ie#\n", + "\n", + "Vc = Vcc-(Ic*Rc)#\n", + "print 'The Collector Voltage = %0.2f Volts'%Vc\n", + "print 'Approx 10.65 Volts'\n", + "\n", + "Vce = Vcc-(Ic*(Rc+Re))#\n", + "print 'The Collector-Emitter Voltage = %0.2f Volts'%Vce\n", + "print 'Approx 8.74 Volts'\n", + "\n", + "Icsat = Vcc/(Rc+Re)#\n", + "print 'The Current Ic(sat) = %0.2f Amps'%Icsat\n", + "print 'i.e 9.52 mAmps'\n", + "\n", + "Vceoff = Vcc#\n", + "print 'The Voltage Vce(off) = %0.2f Volts'%Vceoff\n", + "\n", + "Icq = Ic\n", + "Vceq = Vce\n", + "\n", + "Vce1=[Vcc, Vceq, 0]\n", + "Ic1=[0, Icq, Icsat]\n", + "\n", + "#To plot DC load line\n", + "\n", + "print \"Q(%f,%f)\\n\"%(Vceq,Icq)\n", + "plot(Vce1, Ic1)\n", + "plot(Vceq,Icq)\n", + "plot(0,Icq)\n", + "plot(Vceq,0)\n", + "plot(0,Icsat)\n", + "plot(Vceoff,0)\n", + "xlabel(\"Vce in Volt\")\n", + "ylabel(\"Ic in mAmps\")\n", + "title(\"DC Load-line for Voltage Divider-Biased Transistor Circuit\")\n", + "show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_14 Page No. 925" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Base Voltage = -1.90 Volts\n", + "Approx -1.9 Volts\n", + "The Emitter Voltage = -1.20 Volts\n", + "Approx -1.2 Volts\n", + "The Collector Current = 0.00 Amps\n", + "Approx 2.4 mAmps\n", + "The Collector Voltage = -7.21 Volts\n", + "The Collector-Emitter Voltage = -6.01 Volts\n" + ] + } + ], + "source": [ + "# For the pnp transistor, solve for Vb, Ve, Ic, Vc, and Vce.\n", + "\n", + "# Given data\n", + "\n", + "R1 = 33.*10**3# # Resistor1=33 kOhms\n", + "R2 = 6.2*10**3# # Resistor2=6.2 kOhms\n", + "Rc = 2.*10**3# # Collector resistance=2 kOhms\n", + "Re = 500.# # Emitter resistance=500 Ohms\n", + "Vcc = 12.# # Supply voltage=12 Volts\n", + "Vbe = 0.7# # Base-Emmiter Voltage=0.7 Volts\n", + "\n", + "\n", + "Vb = -Vcc*(R2/(R1+R2))#\n", + "print 'The Base Voltage = %0.2f Volts'%Vb\n", + "print 'Approx -1.9 Volts'\n", + "\n", + "Ve = Vb-(-Vbe)#\n", + "print 'The Emitter Voltage = %0.2f Volts'%Ve\n", + "print 'Approx -1.2 Volts'\n", + "\n", + "Ic = -(Ve/Re)# # Ic =~ Ie\n", + "print 'The Collector Current = %0.2f Amps'%Ic\n", + "print 'Approx 2.4 mAmps'\n", + "\n", + "Vc = -Vcc+(Ic*Rc)\n", + "print 'The Collector Voltage = %0.2f Volts'%Vc\n", + "\n", + "Vce = -Vcc+(Ic*(Rc+Re))#\n", + "print 'The Collector-Emitter Voltage = %0.2f Volts'%Vce" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_15 Page No. 926" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Emitter current = 0.0053 Amps\n", + "i.e 5.3 mAmps\n", + "The Collector voltage = 7.05 Volts\n" + ] + } + ], + "source": [ + "# Calculate Ie and Vc\n", + "\n", + "# Given data\n", + "\n", + "Vee = 6.# # Supply voltage at emitter=6 Volts\n", + "Vcc = 15.# # Supply voltage at collector=15 Volts\n", + "Vbe = 0.7# # Base-Emmiter Voltage=0.7 Volts\n", + "Rc = 1.5*10**3# # Collector resistance=1.5 kOhms\n", + "Re = 1.*10**3# # Emitter resistance=1 kOhms\n", + "\n", + "Ie = (Vee-Vbe)/Re#\n", + "print 'The Emitter current = %0.4f Amps'%Ie\n", + "print 'i.e 5.3 mAmps'\n", + "\n", + "Ic = Ie# # Ic =~ Ie\n", + "\n", + "Vc = Vcc-Ic*Rc#\n", + "print 'The Collector voltage = %0.2f Volts'%Vc" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter29.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter29.ipynb new file mode 100644 index 00000000..040d35f9 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter29.ipynb @@ -0,0 +1,595 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 29 : Transistor Amplifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 29_1 Page No. 940" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Ac Resistance with R=10 kOhms = 26.88 Ohms\n", + "The Ac Resistance with R=5 kOhms = 13.44 Ohms\n", + "The Ac Resistance with R=1 kOhms = 2.69 Ohms\n", + "Approx 2.69 Ohms\n" + ] + } + ], + "source": [ + "# For the diode circuit, calculate the ac resistance, rac, for the following values of R: (a) 10 kOhms, (b) 5 kOhms, and (c) 1 kOhms. Use the second approximation of a diode.\n", + "\n", + "# Given data\n", + "\n", + "R1 = 10.*10**3# # Resistance 1=10 kOhms\n", + "R2 = 5.*10**3# # Resistance 2=5 kOhms\n", + "R3 = 1.*10**3# # Resistance 3=1 kOhms\n", + "Vdc = 10.# # DC supply=10 Volts\n", + "V = 0.7# # Starting voltage of diode=0.7 Volts\n", + "A = 25.*10**-3# # Constant\n", + "\n", + "# For R=10 kOhms\n", + "\n", + "Id1 = (Vdc-V)/R1#\n", + "\n", + "rac1 = A/Id1#\n", + "print 'The Ac Resistance with R=10 kOhms = %0.2f Ohms'%rac1\n", + "\n", + "# For R=5 kOhms\n", + "\n", + "Id2 = (Vdc-V)/R2#\n", + "\n", + "rac2 = A/Id2#\n", + "print 'The Ac Resistance with R=5 kOhms = %0.2f Ohms'%rac2\n", + "\n", + "# For R=1 kOhms\n", + "\n", + "Id3 = (Vdc-V)/R3#\n", + "\n", + "rac3 = A/Id3#\n", + "print 'The Ac Resistance with R=1 kOhms = %0.2f Ohms'%rac3\n", + "print 'Approx 2.69 Ohms'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 29_2 Page No. 941" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Gain Av =200.00\n" + ] + } + ], + "source": [ + "#A common-emitter amplifier circuit has an input of 25 mVp-p and an output of 5 Vp-p. Calculate Av.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 25*10**-3# # Input voltage=25 mVolts(p-p)\n", + "Vo = 5# # Output voltage=5 Volts(p-p).\n", + "\n", + "Av = Vo/Vin#\n", + "print 'The Voltage Gain Av =%0.2f'%Av" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 29_3 Page No. 942" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Output Voltage = 1.50 Volts(p-p)\n" + ] + } + ], + "source": [ + "# assume Av still equals 300. If vin is\u0003 5 mVp-p, calculate Vout.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 5*10**-3# # Input voltage=5 mVolts(p-p)\n", + "Av = 300# # Voltage gain=300\n", + "\n", + "Vo = Av*Vin#\n", + "print 'The Output Voltage = %.2f Volts(p-p)'%Vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 29_4 Page No. 948" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Gain Av =89.96\n", + "Approx 90\n" + ] + } + ], + "source": [ + "# Assume that re varies from 3.33 Ohms to 6.67 Ohms as the temperature of the transistor changes. Calculate the variation in the voltage gain, Av.\n", + "\n", + "# Given data\n", + "\n", + "rl = 600# # Load resistance=600 Ohms\n", + "re = 6.67# # Internal emitter resistance=6.67 Ohms\n", + "\n", + "Av = rl/re#\n", + "print 'The Voltage Gain Av =%0.2f'%Av\n", + "print 'Approx 90'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 29_5 Page No. 949" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Gain Av(max) when r`e=3.33 Ohms: 9.47\n", + "The Voltage Gain Av(min) when r`e=6.67 Ohms: 9.00\n" + ] + } + ], + "source": [ + "# Assume that r'e varies from 3.33 Ohms\u0003 to 6.67 Ohms. Calculate the minimum and maximum values for Av.\n", + "\n", + "# Given data\n", + "\n", + "rl = 600.# # Load resistance=600 Ohms\n", + "re1 = 3.33# # Internal emitter resistance=3.33 Ohms\n", + "re2 = 6.67# # Internal emitter resistance=6.67 Ohms\n", + "rE = 60.# # Emitter resistance=60 Ohms\n", + "\n", + "Av1 = rl/(re1+rE)#\n", + "print \"The Voltage Gain Av(max) when r`e=3.33 Ohms: %0.2f\"%Av1\n", + "\n", + "Av2 = rl/(re2+rE)#\n", + "print 'The Voltage Gain Av(min) when r`e=6.67 Ohms: %0.2f'%Av2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 29_6 Page No. 950" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Gain Av =0.996\n", + "i.e 996 mVolts(p-p)\n", + "The Output Voltage = 0.700 Volts(p-p)\n" + ] + } + ], + "source": [ + "# Find the exact value of Av. Also, find Vout.\n", + "\n", + "# Given data\n", + "\n", + "rl = 909.# # Load resistance=909 Ohms\n", + "re = 3.35# # Internal emitter resistance=3.35 Ohms\n", + "Vin = 1.# # Input voltage=1 Volts(p-p)\n", + "\n", + "Av = rl/(re+rl)#\n", + "print 'The Voltage Gain Av =%0.3f'%Av\n", + "print 'i.e 996 mVolts(p-p)'\n", + "Vo = Av*Vin#\n", + "print 'The Output Voltage = %0.3f Volts(p-p)'%V\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 29_7 Page No. 951" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Input impedence = 2485.72 Ohms\n", + "i.e 2.48 kOhms\n" + ] + } + ], + "source": [ + "# Calculate Zin.\n", + "\n", + "# Given data\n", + "\n", + "rl = 909.# # Load resistance=909 Ohms\n", + "re = 3.35# # Internal emitter resistance=3.35 Ohms\n", + "B = 100.# # Beta=100\n", + "R1 = 4.7*10**3# # Resistance1=4.7 kOhms\n", + "R2 = 5.6*10**3# # Resistance2=5.6 kOhms\n", + "\n", + "Zibase = B*(re+rl)#\n", + "A = (R1*R2)/(R1+R2)#\n", + "\n", + "Zin = (Zibase*A)/(A+Zibase)#\n", + "print 'The Input impedence = %0.2f Ohms'%Zin\n", + "print 'i.e 2.48 kOhms'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 29_8 Page No. 952" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Base Voltage = 9.00 Volts\n", + "The Emitter Voltage = 8.30 Volts\n", + "The Collector current = 0.01 Amps\n", + "i.e 5.53 mAmps\n", + "The Collector Voltage = 20.00 Volts\n", + "The Collector-Emmiter Voltage = 11.70 Volts\n", + "The AC emmiter resistance = 4.52 Ohms\n", + "Approx 4.52 Ohms\n", + "The Voltage gain =0.99\n", + "The Input Base Impedence = 120903.61 Ohms\n", + "i.e 120.9 kOhms\n", + "The Input Impedence = 9150.71 Ohms\n", + "i.e 9.15 kOhms\n", + "The AC base voltage = 4.69 Volts(p-p)\n", + "The AC output voltage = 4.66 Volts(p-p)\n", + "Q(11.700000,0.005533)\n", + "\n" + ] + }, + { + "data": { + "image/png": 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+/TV06hSu9S6ST9JJIs+Y2fUZLMDYGliZsL0qeiydffasomw7oLeZvWBm5WbW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\nQUREqlawy56IiEjuFdyM9XQ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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fb8844408d0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "from matplotlib.pyplot import plot,xlabel,title,ylabel,show\n", + "# Calculate the following quantities: Vb, Ve, Ic, Vc, Vce, r'e, Zin(base), Zin, Av, vb, and vout. Also, plot the dc load line.\n", + "\n", + "# Given data\n", + "\n", + "R1 = 22.*10**3# # Resistance1=22 kOhms\n", + "R2 = 18.*10**3# # Resistance2=18 kOhms\n", + "Rg = 600.# # Generator resistance=600 Ohms\n", + "Re = 1.5*10**3# # Emitter resistance=1.5 kOhms\n", + "Rl = 1.*10**3# # Load resistance=1 kOhms\n", + "Vcc = 20.# # Supply Voltage=20 Volts\n", + "Vbe = 0.7# # Voltage Base-Emitter=0.7 Volts\n", + "B = 200.# # Beta=200\n", + "vin = 5.# # Input Voltage=5 Volts(p-p)\n", + "\n", + "# Calculate the DC quantities first:\n", + "\n", + "Vb = Vcc*(R2/(R1+R2))#\n", + "print 'The Base Voltage = %0.2f Volts'%Vb\n", + "\n", + "Ve = Vb-Vbe#\n", + "print 'The Emitter Voltage = %0.2f Volts'%Ve\n", + "\n", + "Ie = Ve/Re#\n", + "Ic = Ie# # Ic =~ Ie\n", + "print 'The Collector current = %0.2f Amps'%Ic\n", + "print 'i.e 5.53 mAmps'\n", + "\n", + "Vc = Vcc# # Since the collector is tied directly to Vcc\n", + "print 'The Collector Voltage = %0.2f Volts'%Vc\n", + "\n", + "Vce = Vcc-Ve#\n", + "print 'The Collector-Emmiter Voltage = %0.2f Volts'%Vce\n", + "\n", + "Icsat = Vcc/Re#\n", + "\n", + "Vceoff = Vcc#\n", + "\n", + "# Now, calculate AC quantities:\n", + "\n", + "a = 25.*10**-3#\n", + "\n", + "re = a/Ie#\n", + "print 'The AC emmiter resistance = %0.2f Ohms'%re\n", + "print 'Approx 4.52 Ohms'\n", + "\n", + "b = Re*Rl#\n", + "c = Re+Rl#\n", + "rl = b/c#\n", + "\n", + "Av = rl/(rl+re)#\n", + "print 'The Voltage gain =%0.2f'%Av\n", + "\n", + "Zinbase = B*(re+rl)#\n", + "print 'The Input Base Impedence = %0.2f Ohms'%Zinbase\n", + "print 'i.e 120.9 kOhms'\n", + "\n", + "d = 1./Zinbase#\n", + "e = 1./R1#\n", + "f = 1./R2#\n", + "\n", + "Zin = (d+e+f)**-1\n", + "print 'The Input Impedence = %0.2f Ohms'%Zin\n", + "print 'i.e 9.15 kOhms'\n", + "\n", + "vb = vin*(Zin/(Zin+Rg))#\n", + "print 'The AC base voltage = %0.2f Volts(p-p)'%vb\n", + "\n", + "vout = Av*vb#\n", + "print 'The AC output voltage = %0.2f Volts(p-p)'%vout\n", + "\n", + "Icq = Ic\n", + "Vceq = Vce\n", + "\n", + "Vce1=[Vcc, Vceq, 0]\n", + "Ic1=[0 ,Icq ,Icsat]\n", + "\n", + "#To plot DC load line\n", + "\n", + "print \"Q(%f,%f)\\n\"%(Vceq,Icq)\n", + "plot(Vce1, Ic1)\n", + "plot(Vceq,Icq)\n", + "plot(0,Icq)\n", + "plot(Vceq,0)\n", + "plot(0,Icsat)\n", + "plot(Vceoff,0)\n", + "xlabel(\"Vce in Volt\")\n", + "ylabel(\"Ic in mAmps\")\n", + "title(\"DC Load-line for Emitter Follower Circuit\")\n", + "show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 29_9 Page No. 963" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Emmiter current = 0.0046 Amps\n", + "i.e 4.61 mApms\n", + "The Collector-Base Voltage = 8.08 Volts\n", + "Approx 8.09 Volts\n", + "The AC emmiter resistance = 5.42 Ohms\n", + "The Voltage gain =138.33\n", + "The AC output voltage = 3.46 Volts(p-p)\n", + "Approx 3.46 Volts(p-p)\n", + "The Input Impedence = 5.41 Ohms\n" + ] + } + ], + "source": [ + "# Calculate the following: Ie, Vcb, r'\u0005e, Av, vout and zin.\n", + "\n", + "# Given data\n", + "\n", + "Rc = 1.5*10**3# # Collector resistance=1.5 kOhms\n", + "Re = 1.8*10**3# # Emitter resistance=1.8 kOhms\n", + "Rl = 1.5*10**3# # Load resistance=1.5 kOhms\n", + "Vcc = 15.# # +ve Supply Voltage=15 Volts\n", + "Vee = 9.# # -ve Supply Voltage=9 Volts\n", + "Vbe = 0.7# # Voltage Base-Emitter=0.7 Volts\n", + "vin = 25.*10**-3# # Input Voltage=25 mVolts(p-p)\n", + "\n", + "\n", + "Ie = (Vee-Vbe)/Re#\n", + "print 'The Emmiter current = %0.4f Amps'%Ie\n", + "print 'i.e 4.61 mApms'\n", + "\n", + "Ic = Ie# # Ic =~ Ie\n", + "\n", + "Vcb = Vcc-(Ic*Rc)#\n", + "print 'The Collector-Base Voltage = %0.2f Volts'%Vcb\n", + "print 'Approx 8.09 Volts'\n", + "\n", + "a = 25.*10**-3#\n", + "\n", + "re = a/Ie#\n", + "print 'The AC emmiter resistance = %0.2f Ohms'%re\n", + "\n", + "b = Rc*Rl#\n", + "c = Rc+Rl#\n", + "\n", + "rl = b/c#\n", + "\n", + "Av = rl/re#\n", + "print 'The Voltage gain =%0.2f'%Av\n", + "\n", + "vout = Av*vin#\n", + "print 'The AC output voltage = %0.2f Volts(p-p)'%vout\n", + "print 'Approx 3.46 Volts(p-p)'\n", + "\n", + "d = Re*re\n", + "e = Re+re\n", + "\n", + "Zin = d/e#\n", + "print 'The Input Impedence = %0.2f Ohms'%Zin" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 29_10 Page No. 968" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The AC output voltage = 1.45 Volts(p-p)\n" + ] + } + ], + "source": [ + "# Calculate the ac output voltage, vout.\n", + "\n", + "# Given data\n", + "\n", + "Rc = 1.2*10**3# # Collector resistance=1.2 kOhms\n", + "Re = 2.2*10**3# # Emitter resistance=2.2 kOhms\n", + "Rl = 3.3*10**3# # Load resistance=3.3 kOhms\n", + "Rg = 600.# # Generator Resistance=600 Ohms\n", + "Vcc = 12.# # +ve Supply Voltage=15 Volts\n", + "Vee = 12.# # -ve Supply Voltage=9 Volts\n", + "Vbe = 0.7# # Voltage Base-Emitter=0.7 Volts\n", + "vin = 1.# # Input Voltage=1 Volts(p-p)\n", + "\n", + "Ie = (Vee-Vbe)/Re#\n", + "\n", + "a = 25*10**-3#\n", + "re = a/Ie#\n", + "\n", + "b = Rc*Rl#\n", + "c = Rc+Rl#\n", + "rl = b/c#\n", + "\n", + "Av = rl/re#\n", + "\n", + "d = Re*re\n", + "e = Re+re\n", + "Zin = d/e#\n", + "\n", + "ve = vin*(Zin/(Zin+Rg))#\n", + "\n", + "vout = Av*ve#\n", + "print 'The AC output voltage = %0.2f Volts(p-p)'%vout" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter3.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter3.ipynb new file mode 100644 index 00000000..099cbe37 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter3.ipynb @@ -0,0 +1,636 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3 : Ohm's Law" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_1 Page No. 88" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current I = 15.00 Amps\n" + ] + } + ], + "source": [ + "# A heater with the resistance of 8 Ohms\u0003 is connected across the 120-V power line. How much is current I?\n", + "\n", + "# Given data\n", + "\n", + "V = 120# # Voltage of Power line=120 Volts\n", + "R = 8# # Heater Resistance=8 Ohms\n", + "\n", + "I = V/R#\n", + "print 'The Current I = %0.2f Amps'%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_2 Page No. 88" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current I = 0.05 Amps\n" + ] + } + ], + "source": [ + "# A small lightbulb with a resistance of 2400 Ohms\u0003 is connected across the 120-V power line. How much is current I?\n", + "\n", + "# Given data\n", + "\n", + "V = 120.# # Voltage of Power line=120 Volts\n", + "R = 2400# # Lightbulb Resistance=2400 Ohms\n", + "\n", + "I = V/R#\n", + "print 'The Current I = %0.2f Amps'%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_3 Page No. 88" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage = 30.00 Volts\n" + ] + } + ], + "source": [ + "# If a 12-Ohms\u0003 resistor is carrying a current of 2.5 A, how much is its voltage?\n", + "\n", + "# Given data\n", + "\n", + "I = 2.5# # Current=2.5 Amps\n", + "R = 12# # Resistance=12 Ohms\n", + "\n", + "V = I*R#\n", + "print 'The Voltage = %0.2f Volts'%V" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_4 Page No. 89" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistance = 75.00 ohms\n" + ] + } + ], + "source": [ + "# How much is the resistance of a lightbulb if it draws 0.16 A from a 12-V battery?\n", + "\n", + "# Given data\n", + "\n", + "V = 12# # Voltage of Battery=12 Volts\n", + "I = 0.16# # Current drawn form Battery=0.16 Amps\n", + "\n", + "R = V/I\n", + "print 'The Resistance = %0.2f ohms'%R" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_5 Page No. 89" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage = 40.00 Volts\n" + ] + } + ], + "source": [ + "# The I of 8 mA flows through a 5-kOhms Resistor. How much is the IR voltage?\n", + "\n", + "# Given data\n", + "\n", + "I = 8*10**-3# # Current flowing through Resistor=8m Amps\n", + "R = 5*10**3# # Resistance=5k Ohms\n", + "\n", + "V = I*R#\n", + "print 'The Voltage = %0.2f Volts'%V" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_6 Page No. 90" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current I = 0.005 Amps\n", + "i.e 5 mAmps\n" + ] + } + ], + "source": [ + "# How much current is produced by 60 V across 12 kOhms?\n", + "\n", + "# Given data\n", + "\n", + "V = 60.0# # Voltage=60 Volts\n", + "R = 12*10**3# # Resistance=12k Ohms\n", + "\n", + "I = V/R#\n", + "print 'The Current I = %0.3f Amps'%I\n", + "print 'i.e 5 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_7 Page No. 91" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Power used = 1200.00 Watts\n", + "OR 1.2 kW\n" + ] + } + ], + "source": [ + "# A toaster takes 10 A from the 120-V power line. How much power is used?\n", + "\n", + "# Given data\n", + "\n", + "V = 120# # Voltage of Power line=120 Volts\n", + "I = 10# # Current drawn from Powerline=10 Amps\n", + "\n", + "P = V*I#\n", + "print 'The Power used = %0.2f Watts'%P\n", + "print 'OR 1.2 kW'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_8 Page No. 91" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current I = 2.50 Amps\n" + ] + } + ], + "source": [ + "# How much current flows in the filament of a 300-W bulb connected to the 120-V power line?\n", + "\n", + "# Given Data\n", + "\n", + "V = 120.0# # Voltage of Power line=120 Volts\n", + "P = 300# # Power of Bulb=300 Watts\n", + "\n", + "I = P/V#\n", + "print 'The Current I = %0.2f Amps'%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_9 Page No. 91" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current I = 0.50 Amps\n", + "OR 500 mA\n" + ] + } + ], + "source": [ + "# How much current flows in the filament of a 60-W bulb connected to the 120-V power line?\n", + "\n", + "# Given Data\n", + "\n", + "V = 120# # Voltage of Power line=120 Volts\n", + "P = 60.0# # Power of Bulb=60 Watts\n", + "\n", + "I = P/V#\n", + "print 'The Current I = %0.2f Amps'%I\n", + "print 'OR 500 mA'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_10 Page No. 93" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Cost = 4.32 $\n" + ] + } + ], + "source": [ + "# Asuming that the cost of electricity is 6 cent per kWh, how much will it cost to light a 100-W lightbulb for 30 days?\n", + "\n", + "h = 24*30# # Total hours = 24 hrs * 30 days\n", + "\n", + "kWh = 0.1*h# # 100W=0.1kW\n", + "\n", + "Cost = kWh*0.06# # 6 cent = $0.06\n", + "\n", + "print 'Cost = %0.2f $'%Cost" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_11 Page No. 95" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Power = 200.00 Watts\n" + ] + } + ], + "source": [ + "# Calculate the power in a circuit where the source of 100 V produces 2 A in a 50 Ohms Resistor.\n", + "\n", + "# Given data\n", + "\n", + "I = 2# # Current=2 Amps\n", + "R = 50# # Resistance=50 Ohms\n", + "V = 100# # Voltage Source=100 Volts\n", + "\n", + "P = I*I*R#\n", + "print 'The Power = %0.2f Watts'%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_12 Page No. 95" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Power = 400.00 Watts\n" + ] + } + ], + "source": [ + " \n", + "# Calculate the power in a circuit where the source of 100 V produces 4 A in a 25 Ohms Resistor.\n", + "\n", + "# Given data\n", + "\n", + "I = 4# # Current=4 Amps\n", + "R = 25# # Resistance=25 Ohms\n", + "V = 100# # Voltage Source=100 Volts\n", + "\n", + "P = I*I*R#\n", + "print 'The Power = %0.2f Watts'%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_13 Page No. 96" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current I = 5.00 Amps\n" + ] + } + ], + "source": [ + "# How much current is needed for a 600-W, 120-V toaster?\n", + "\n", + "# Given data\n", + "\n", + "V = 120.0# # Applied Voltage=120 Volts\n", + "P = 600# # Power of toaster=600 Watts\n", + "\n", + "I = P/V#\n", + "print 'The Current I = %0.2f Amps'%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_14 Page No. 97" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistance = 24.00 Ohms\n" + ] + } + ], + "source": [ + "# How much is the resistance of a 600-W, 120-V toaster?\n", + "\n", + "# Given data\n", + "\n", + "V = 120.0# # Applied Voltage=120 Volts\n", + "P = 600# # Power of toaster=600 Watts\n", + "\n", + "R = (V*V)/P#\n", + "print 'The Resistance = %0.2f Ohms'%R" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_15 Page No. 97" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current I = 5.00 Amps\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "# How much current is needed for a 24 Ohms\u0003 Resistor that dissipates 600 W?\n", + "\n", + "# Given data\n", + "\n", + "R = 24.0# # Resistance=24 Ohms\n", + "P = 600.0# # Power=600 Watts\n", + "\n", + "I = sqrt(P/R)#\n", + "print 'The Current I = %0.2f Amps'%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_16 Page No. 98" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistor value = 1500.00 Ohms\n", + "i.e 1.5 kohms\n", + "The Power = 0.60 Watts\n", + "OR 600 mW\n" + ] + } + ], + "source": [ + "# Determine the required resistance and appropriate wattage rating of a resistor to meet the following requirements: The resistor must have a 30-V IR drop when its current is 20 mA. The resistors available have the following wattage ratings: 1⁄8, 1⁄4, 1⁄2, 1, and 2 W.\n", + "\n", + "# Given data\n", + "\n", + "I = 20.0*10**-3# # Current=20m Amps\n", + "V = 30.0# # Voltage Drop=30 Volts\n", + "\n", + "R = V/I#\n", + "print 'The Resistor value = %0.2f Ohms'%R\n", + "print 'i.e 1.5 kohms'\n", + "\n", + "P = I*I*R#\n", + "print 'The Power = %0.2f Watts'%P\n", + "print 'OR 600 mW'\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_17 Page No. 99" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistor value = 1500000 Ohms\n", + "i.e 1.5 Mohms\n", + "The Power = 0.03375 Watts\n", + "i.e 33.75 mW\n" + ] + } + ], + "source": [ + "#Determine the required resistance and appropriate wattage rating of a carbonfilm resistor to meet the following requirements: The resistor must have a 225-V IR drop when its current is 150 uA. The resistors available have the following wattage ratings: 1⁄8, 1⁄4, 1⁄2, 1, and 2 W.\n", + "\n", + "# Given data\n", + "\n", + "I = 150.0*10**-6# # Current=150 uAmps\n", + "V = 225# # Voltage Drop=225 Volts\n", + "\n", + "R = V/I#\n", + "print 'The Resistor value = %0.f Ohms'%R\n", + "print 'i.e 1.5 Mohms'\n", + "\n", + "P = I*I*R#\n", + "print 'The Power = %0.5f Watts'%P\n", + "print 'i.e 33.75 mW'" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter30.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter30.ipynb new file mode 100644 index 00000000..5b2dbf97 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter30.ipynb @@ -0,0 +1,578 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 30 : Field Effect Transistors" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 30_1 Page No. 984" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Id for Vgs is 0 Volts = 0.01 Amps\n", + "i.e 10 mAmps\n", + "The Value of Id for Vgs is -0.5 Volts = 0.0077 Amps\n", + "i.e 7.65 mAmps\n", + "The Value of Id for Vgs is -1 Volts = 0.0056 Amps\n", + "i.e 5.62 mAmps\n", + "The Value of Id for Vgs is -2 Volts = 0.0025 Amps\n", + "i.e 2.5 mAmps\n", + "The Value of Id for Vgs is -3 Volts = 0.0006 Amps\n", + "i.e 0.625 mAmps\n" + ] + } + ], + "source": [ + "# Determine Id for each value of Vgs (a) 0V# (b) -0.5V# (c) -1V (d) -2V (e) -3V\n", + "\n", + "# Given Data\n", + "\n", + "Vgs1 = 0# # Voltage Gate-Source 1=0 Volts\n", + "Vgs2 = -0.5# # Voltage Gate-Source 2=-0.5 Volts\n", + "Vgs3 = -1.# # Voltage Gate-Source 3=-1 Volts\n", + "Vgs4 = -2.# # Voltage Gate-Source 4=-2 Volts\n", + "Vgs5 = -3.# # Voltage Gate-Source 5=-3 Volts\n", + "Vgsoff = -4.# # Voltage Gate-Source(off)=-4 Volts\n", + "Idss = 10.*10**-3 # Idss = 10m Amps\n", + "\n", + "a = (1-(Vgs1/Vgsoff))\n", + "b = (1-(Vgs2/Vgsoff))\n", + "c = (1-(Vgs3/Vgsoff))\n", + "d = (1-(Vgs4/Vgsoff))\n", + "e = (1-(Vgs5/Vgsoff))\n", + "\n", + "# Vgs = 0 Volts\n", + "\n", + "Id1 = Idss*a*a\n", + "print 'The Value of Id for Vgs is 0 Volts = %0.2f Amps'%Id1\n", + "print 'i.e 10 mAmps'\n", + "\n", + "# Vgs = -0.5 Volts\n", + "\n", + "Id2 = Idss*b*b\n", + "print 'The Value of Id for Vgs is -0.5 Volts = %0.4f Amps'%Id2\n", + "print 'i.e 7.65 mAmps'\n", + "\n", + "# Vgs = -1 Volts\n", + "\n", + "Id3 = Idss*c*c\n", + "print 'The Value of Id for Vgs is -1 Volts = %0.4f Amps'%Id3\n", + "print 'i.e 5.62 mAmps'\n", + "\n", + "# Vgs = -2 Volts\n", + "\n", + "Id4 = Idss*d*d\n", + "print 'The Value of Id for Vgs is -2 Volts = %0.4f Amps'%Id4\n", + "print 'i.e 2.5 mAmps'\n", + "\n", + "# Vgs = -3 Volts\n", + "\n", + "Id5 = Idss*e*e\n", + "print 'The Value of Id for Vgs is -3 Volts = %0.4f Amps'%Id5\n", + "print 'i.e 0.625 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 30_2 Page No. 985" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Id = 1.2500e-04 Amps using Minimum Values\n", + "i.e 125 uAmps\n", + "The Value of Vds = 1.88 Volts using Minimum Values\n", + "The Value of Id = 0.0132 Amps using Maximum Values\n", + "i.e 13.2 mAmps\n", + "The Value of Vds = -11.20 Volts using Maximun Values\n", + "The Value of Vds(p) = 6.50 Volts using Maximun Values\n" + ] + } + ], + "source": [ + "# Find the minimim and maximum value of Id and Vds if Vgs=-1.5 Volts\n", + "\n", + "# Given Data\n", + "\n", + "Idssmin = 2.*10**-3# # Idss(min)=2m Amp\n", + "Idssmax = 20.*10**-3# # Idss(max)=20m Amp\n", + "Vgs = -1.5# # Voltage Gate-Source=-1.5V\n", + "Vgsoffmin = -2.# # Voltage Gate-Source(off)(min)=-2 Volts\n", + "Vgsoffmax = -8.# # Voltage Gate-Source(off)(max)=-8 Volts\n", + "Vdd = 2.0# # Supply Voltage(Drain)=20 Volts\n", + "Rd = 1.*10**3# # Drain Resistance=1k Ohms\n", + "\n", + "a = 1-(Vgs/Vgsoffmin)#\n", + "b = 1-(Vgs/Vgsoffmax)#\n", + "\n", + "# Calculation using Minimum Values\n", + "\n", + "Id1 = Idssmin*a*a#\n", + "print 'The Value of Id = %0.4e Amps using Minimum Values'%Id1\n", + "print 'i.e 125 uAmps'\n", + "\n", + "Vds1 = Vdd-Id1*Rd#\n", + "print 'The Value of Vds = %0.2f Volts using Minimum Values'%Vds1\n", + "\n", + "# Calculation using Maximum Values\n", + "\n", + "Id2 = Idssmax*b*b#\n", + "print 'The Value of Id = %0.4f Amps using Maximum Values'%Id2\n", + "print 'i.e 13.2 mAmps'\n", + "\n", + "Vds2 = Vdd-Id2*Rd#\n", + "print 'The Value of Vds = %0.2f Volts using Maximun Values'%Vds2\n", + "\n", + "Vp = -Vgsoffmax#\n", + "\n", + "Vdsp = Vp+Vgs#\n", + "print 'The Value of Vds(p) = %0.2f Volts using Maximun Values'%Vdsp" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 30_3 Page No. 989" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Drain Voltage Vd = 5.00 Volts\n" + ] + } + ], + "source": [ + "# Calculate the value of Vd\n", + "\n", + "# Given Data\n", + "\n", + "Vs = 1.# # Voltage at Resistor Rs=1 Volts\n", + "Rs = 200.# # Source Resistor=200 Ohms\n", + "Vdd = 10.# # Supply Voltage(Drain)=10 Volts\n", + "Rd = 1.*10**3# # Drain Resistor=1k Ohms\n", + "\n", + "Is=Vs/Rs#\n", + "\n", + "Id = Is#\n", + "\n", + "Vd = Vdd-Id*Rd#\n", + "print 'The Drain Voltage Vd = %0.2f Volts'%Vd," + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 30_4 Page No. 991" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Vg = 3.06 Volts\n", + "i.e 3 Volts\n", + "The Value of Vs = 4.06 Volts\n", + "i.e 4 Volts\n", + "The Value of Id = 5.08e-03 Amps.\n", + "i.e 5 mAmps\n", + "The Value of Vd = 9.92 Volts\n", + "Approx 10 Volts\n" + ] + } + ], + "source": [ + "# Calculate Vg, Vs, Id, Vd.\n", + "\n", + "# Given Data\n", + "\n", + "R1 = 390.*10**3# # Resistor 1=390k Ohms\n", + "R2 = 100.*10**3# # Resistor 2=100k Ohms\n", + "Rd = 1.*10**3# # Drain Resistor=1k Ohms\n", + "Vdd = 15.# # Supply Voltage(Drain)=15 Volts\n", + "Vgs = -1.# # Voltage Gate-Source=-1 Volts\n", + "Rs = 800.# # Source Resistor=800 Ohms\n", + "\n", + "Vg = (R2/(R1+R2))*Vdd#\n", + "print 'The Value of Vg = %0.2f Volts'%Vg\n", + "print 'i.e 3 Volts'\n", + "\n", + "Vs = Vg-Vgs#\n", + "print 'The Value of Vs = %0.2f Volts'%Vs\n", + "print 'i.e 4 Volts'\n", + "\n", + "Id = Vs/Rs#\n", + "print 'The Value of Id = %0.2e Amps.'%Id\n", + "print 'i.e 5 mAmps'\n", + "\n", + "Vd = Vdd-Id*Rd\n", + "print 'The Value of Vd = %0.2f Volts'%Vd\n", + "print 'Approx 10 Volts'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 30_5 Page No. 992" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Drain Current Id = 6.50e-03 Amps\n", + "i.e 6.5 mAmps\n", + "The Drain Voltage Vd = 8.50 Voltage\n" + ] + } + ], + "source": [ + "# Calculate the value Drain Current Id and Drain Voltage Vd.\n", + "\n", + "# Given Data\n", + "\n", + "Vdd = 15# # Supply Voltage(Drain)=15 Volts\n", + "Vbe = 0.7# # Voltage Base-Emitter=0.7 Volts\n", + "Re = 2.2*10**3# # Emitter Resistor=2.2 kOhms\n", + "Rd = 1*10**3# # Drain Resistor=1 kOhms\n", + "Vee = 15# # Supply Voltage(Emitter)=15 Volts\n", + "\n", + "\n", + "Ic = (Vee-Vbe)/Re#\n", + "\n", + "Id = Ic#\n", + "print 'The Drain Current Id = %0.2e Amps'%Id\n", + "print 'i.e 6.5 mAmps'\n", + "\n", + "Vd = Vdd-Id*Rd#\n", + "print 'The Drain Voltage Vd = %0.2f Voltage'%Vd" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 30_6 Page No. 997" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Gain Av =4.89\n", + "Approx 4.875\n", + "The Output Voltage Vo = 0.978 Volts(p-p)\n" + ] + } + ], + "source": [ + "# Calculate the Voltage Gain Av and Output Voltage Vo\n", + "\n", + "# Given Data\n", + "\n", + "Rd = 1.5*10**3# # Drain Resistor=1.5 kOhms\n", + "Rl = 10*10**3# # Load Resistor=10 kOhms\n", + "Idss = 10*10**-3# # Idss=10 mAmps\n", + "Vgs = -1# # Voltage Gate-Source=-1 Volts\n", + "Vgsoff = -4.# # Voltage Gate-Source(off)=-4 Volts\n", + "Vin = 0.2# # Input Voltage=0.2 Volts(p-p)\n", + "\n", + "gmo = 2*Idss/(-Vgsoff)#\n", + "\n", + "gm = gmo*(1-(Vgs/Vgsoff))#\n", + "\n", + "rl = (Rd*Rl)/(Rd+Rl)#\n", + "\n", + "Av = gm*rl#\n", + "print 'The Voltage Gain Av =%0.2f'%Av\n", + "print 'Approx 4.875'\n", + "\n", + "Vo = Av*Vin\n", + "print 'The Output Voltage Vo = %0.3f Volts(p-p)'%Vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 30_7 Page No. 998" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Gain Av =0.37\n", + "The Output Voltage Vo = 0.37 Volts(p-p)\n", + "The Output Impedence Zo = 143.28 Ohms\n", + "Approx 143.5 Ohms\n" + ] + } + ], + "source": [ + "# Calculate Av, Vo & Zo.\n", + "\n", + "# Given Data\n", + "\n", + "Rs = 240.# # Source Resistor=240 Ohms\n", + "Rl = 1.8*10**3# # Load Resistor=1.8 kOhms\n", + "Vgsoff = -8.# # Voltage Gate-Source(off)=-8 Volts\n", + "Vgs = -2.# # Voltage Gate-Source=-2 Volts\n", + "Idss = 15.*10**-3 # Idss=15 mAmps.\n", + "Vin = 1.# # Input Voltage=1 Volts(p-p)\n", + "\n", + "rl = ((Rs*Rl)/(Rs+Rl))#\n", + "gmo = 2*Idss/-Vgsoff#\n", + "gm = gmo*(1-(Vgs/Vgsoff))#\n", + "\n", + "Av = gm*rl/(1+gm*rl)#\n", + "print 'The Voltage Gain Av =%0.2f'%Av\n", + "\n", + "Vo = Av*Vin#\n", + "print 'The Output Voltage Vo = %0.2f Volts(p-p)'%Vo\n", + "\n", + "A = (1/gm)#\n", + "Zo = ((Rs*A)/(Rs+A))#\n", + "print 'The Output Impedence Zo = %0.2f Ohms'%Zo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 30_8 Page No. 1000" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Gain Av =4.17\n", + "The Output Voltage = 4.17e-02 Volts(p-p)\n", + "Approx 41.6 mVolts(p-p)\n", + "The Output Impedence Zi = 114.29 Ohms\n", + "Approx 114 Ohms\n" + ] + } + ], + "source": [ + "#Calculate Av, Vo, Zin.\n", + "\n", + "# Given Data\n", + "\n", + "Rd = 1.2*10**3# # Drain Resistor=1.2 kOhms\n", + "Rl = 15.*10**3# # Load Resistor=15 kOhms\n", + "gm = 3.75*10**-3# # Transconductance=3.75 mSiemens\n", + "Vin = 10.*10**-3# # Input Voltage=10 mVpp\n", + "Rs = 200.# # Source Resistor=200 Ohms\n", + "\n", + "rl = ((Rd*Rl)/(Rd+Rl))#\n", + "\n", + "Av = gm*rl#\n", + "print 'The Voltage Gain Av =%0.2f'%Av\n", + "\n", + "Vo = Av*Vin#\n", + "print 'The Output Voltage = %0.2e Volts(p-p)'%Vo\n", + "print 'Approx 41.6 mVolts(p-p)'\n", + "\n", + "A = (1/gm)#\n", + "\n", + "Zi = ((Rs*A)/(Rs+A))#\n", + "print 'The Output Impedence Zi = %0.2f Ohms'%Zi\n", + "print 'Approx 114 Ohms'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 30_9 Page No. 1001" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Id for Vgs is 2 Volts = 2.25e-02 Amps\n", + "i.e 22.5 mAmps\n", + "The Value of Id for Vgs is -2 Volts = 2.50e-03 Amps\n", + "i.e 2.5 mAmps\n", + "The Value of Id for Vgs is 0 Volts = 1.00e-02 Amps\n", + "i.e 10 mAmps\n" + ] + } + ], + "source": [ + "#Determine Id for each value of Vgs (a) 2V# (b) -2V# (c) 0V\n", + "\n", + "# Given Data\n", + "Vgs1 = 2.# # Voltage Gate-Source 1=2 Volts\n", + "Vgs2 = -2.# # Voltage Gate-Source 2=-2 Volts\n", + "Vgs3 = 0# # Voltage Gate-Source 3=0 Volts\n", + "Vgsoff = -4.# # Voltage Gate-Source(off)=-4 Volts\n", + "Idss = 10.*10**-3# # Idss = 10m Amps\n", + "\n", + "a = (1-(Vgs1/Vgsoff))#\n", + "b = (1-(Vgs2/Vgsoff))#\n", + "c = (1-(Vgs3/Vgsoff))#\n", + "\n", + "# Vgs = 2 Volts\n", + "\n", + "Id1 = Idss*a*a#\n", + "print 'The Value of Id for Vgs is 2 Volts = %0.2e Amps'%Id1\n", + "print 'i.e 22.5 mAmps'\n", + "\n", + "# Vgs = -2 Volts\n", + "\n", + "Id2 = Idss*b*b#\n", + "print 'The Value of Id for Vgs is -2 Volts = %0.2e Amps'%Id2\n", + "print 'i.e 2.5 mAmps'\n", + "\n", + "# Vgs = 0 Volts\n", + "\n", + "Id3 = Idss*c*c#\n", + "print 'The Value of Id for Vgs is 0 Volts = %0.2e Amps'%Id3\n", + "print 'i.e 10 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 30_10 Page No. 1002" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Drain Resistance = 500.00 Ohms\n", + "A 470 Ohms resistor would provide the proper biasing voltage at the gate\n" + ] + } + ], + "source": [ + "# Calculate the value of Rd to provide an Id(on) of 10m Amps.\n", + "\n", + "# Given Data\n", + "\n", + "Vdd = 15.# # Suppy Voltage(Drain)=15 Volts\n", + "Vgson = 10.# # Voltage Gate-Source(on)=10 Volts\n", + "Idon = 10.*10**-3# # Drain Current(on)=10m Amps\n", + "\n", + "Rd = (Vdd-Vgson)/Idon#\n", + "print 'The Drain Resistance = %0.2f Ohms'%Rd\n", + "\n", + "print 'A 470 Ohms resistor would provide the proper biasing voltage at the gate'" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter31.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter31.ipynb new file mode 100644 index 00000000..6cbfbe90 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter31.ipynb @@ -0,0 +1,472 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 31 : Power Amplifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 31_1 Page No. 1019" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of Icq = 7.95e-03 Amps\n", + "i.e Approx 7.91 mAmps\n", + "The value of Vceq = 10.14 Volts\n", + "The Power Dissipation = 8.06e-02 Watts\n", + "i.e 80.6 mWatts\n", + "The value of Ic(sat) = 1.61e-02 Amps\n", + "i.e 16.1 mAmps\n", + "The value of Vce(off) = 20.00 Volts\n", + "Q(10.138406,0.007953)\n", + "\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f383ca8da10>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "from matplotlib.pyplot import plot,show,title,xlabel,ylabel\n", + "# Calculate the following dc quantities Icq, Vceq, Pd, Ic(sat) and Vce(off). Also draw the dc load line\n", + "\n", + "# Given Data\n", + "\n", + "R1 = 18.*10**3# # Resistor 1=18k Ohms\n", + "R2 = 2.7*10**3# # Resistor 2=2.7k Ohms\n", + "Vcc = 20.# # Supply Voltage(Collector)=20 Volts\n", + "Vbe = 0.7# # Voltage Base-Emitter=0.7 Volts\n", + "Re = 240.# # Emitter Resistor=240 Ohms\n", + "Rc = 1.*10**3# # Collector Resistor=1k Ohms\n", + "\n", + "Vb = Vcc*(R2/(R1+R2))#\n", + "\n", + "Ve = Vb-Vbe#\n", + "\n", + "#Ie = Ic#\n", + "\n", + "Icq = Ve/Re#\n", + "print 'The value of Icq = %0.2e Amps'%Icq\n", + "print 'i.e Approx 7.91 mAmps'\n", + "\n", + "Vceq = Vcc-Icq*(Rc+Re)#\n", + "print 'The value of Vceq = %0.2f Volts'%Vceq\n", + "\n", + "Pd = Vceq*Icq#\n", + "print 'The Power Dissipation = %0.2e Watts'%Pd\n", + "print 'i.e 80.6 mWatts'\n", + "\n", + "Icsat = Vcc/(Rc+Re)#\n", + "print 'The value of Ic(sat) = %0.2e Amps'%Icsat\n", + "print 'i.e 16.1 mAmps'\n", + "\n", + "Vceoff = Vcc#\n", + "print 'The value of Vce(off) = %0.2f Volts'%Vceoff\n", + "\n", + "# For DC load line\n", + "\n", + "Vce1=[Vceoff ,Vceq, 0]\n", + "Ic1=[0, Icq, Icsat]\n", + "\n", + "#To plot DC load line\n", + "\n", + "print \"Q(%f,%f)\\n\"%(Vceq,Icq)\n", + "plot(Vce1, Ic1)\n", + "plot(Vceq,Icq)\n", + "plot(0,Icq)\n", + "plot(Vceq,0)\n", + "plot(0,Icsat)\n", + "plot(Vceoff,0)\n", + "xlabel(\"Vce in volt\")\n", + "ylabel(\"Ic in Ampere\")\n", + "title(\"DC Load-line for Common-Emitter Class A Amplifier Circuit\")\n", + "show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 31_2 Page No. 1024" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Gain Av =189.84\n", + "Approx 190\n", + "The Output Voltage = 4.75 Volts\n", + "The Load Power = 1.88e-03 Watts\n", + "i.e Approx 1.88 mWatts\n", + "The Dc Input Power = 1.78e-01 Watts\n", + "i.e Approx 177.4 mWatts\n", + "The Efficiency in % =1.06\n", + "Approx 1%\n", + "The Y-axis Value of AC Load-line is ic(sat) = 2.49e-02 Amps\n", + "i.e 24.89 mAmps\n", + "The X-axis value of AC Load-line is vce(off) = 14.94 Volts\n", + "Q(10.190000,0.007910)\n", + "\n" + ] + }, + { + "data": { + "image/png": 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8hAwPu7vPrc8Bm4oKE2kqGzfC+PFhed68MLy9SDHKazWXu89JvQFPABpoQiTq\n0QOWL4eRI0OflIULk45IpOnlNJyKmX0K+CZwJtANWOzuV+Q5tgZRZiJJqKgIk28de2wYNLJjx6Qj\nEsldXjITM/u4mU0ws6XAn4DPAL3c/TOFXpCIJOWYY2DNGmjbNky+tWpV0hGJNI2smYmZvQ88DvzY\n3VfFxza4e68mjK/elJlI0hYvhosvhksugWuvhdata99GJEn5ajO5htA2coeZXW1mn61XdCIt1Kmn\nwrPPhkEjjz8+9KIXaa6yFibufou7f5nQVtIKWAIcbGZXmdnhTRWgSDHr1g2WLoXTTguzOc6fn3RE\nIvlRp/lMzOwLhEb4M9y9oDMVVXNJoamsDI3z/frBjBnQqVPSEYnsLd894Hdz9+fd/dpCL0hEClFp\nabjaq3PnsLxyZdIRiTQezbQokoCHH4YLL4QLLoDJk6FNm6QjEmnCzEREGsfo0aHa65lnYPDgMP+8\nSDFTYSKSkJISeOSRMD3woEEwezYomZZilcvYXIOBH7DvHPCfyW9oDaNqLikm69aFxvk+fWDmTOjS\nJemIpCXKdzXX3cA0YDDwpXgbUJ+DiUhmRx4Jq1eHQSJLS2HFiqQjEqmbXDKTp2N/k6KizESK1dKl\ncN55YZrgG24IQ7OINIV8D0F/I6HT4m/Zew74Z+tzwKaiwkSK2Ztvhiu9Nm+Ge++Fvn2TjkhagnwX\nJuXAPiu5+9D6HLCpqDCRYucOd90F//Vf8KMfwcSJYPU6zUVyk9fCpCHMbCRwCyGzmeXuN2VY51bg\nROA9YIK7r4mP3wOcBLzh7l9IWb8L8ADwaeAV4HR3355hvypMpFmoqgqN8927w6xZcOCBSUckzVW+\nhqAfF/9eYWaXp9yuMLPLcwiqFXA7MBL4HHCmmR2Rts4ooLe7HwZMBO5MeXp23Dbd1cDj7n44sDze\nF2m2+vYNQ9n37Rsa55ctSzoikX3VdDVX+/i3Y9rtY/FvbQYA6939FXffCdwPnJK2zsnAXAB3fxro\nZGYHxfsrgbcz7Hf3NvHvN3KIRaSotW0LN90UpgU+/3yYNAk++CDpqET2yDrDgrvPjH+n1HPfhwCb\nUu5vBtKvCsu0ziHA6zXst8Tdt8blrWgKYWlBhg2DtWtD+8mAAbBgQbisWCRp+ewBn2uDRXr9XM4N\nHbFRRA0j0qJ06QKLFoXsZOjQMD2wmgclafmc++1VoHvK/e6EzKOmdQ6Nj9Vkq5kd5O6vm9nBwBvZ\nVpwyZcrURcQvAAARnklEQVTu5bKyMsrKymqPWqQImMG558KQITB2LPzud2E4lhLl6VIH5eXllJeX\nN8q+8nY1l5m1Bv4CDAe2AKuBM939pZR1RgGXuvsoMxsI3OLuA1Oe7wk8lHY110+At9z9JjO7Gujk\n7vs0wutqLmkpdu6E668PV3rNmgUnnZR0RFKs8jqciplNNbPOKfc7m9mPatvO3T8ELgWWAi8CD7j7\nS2Z2kZldFNd5FPibma0HZgKXpBznPuCPwOFmtsnMzo1P3Qh8zcz+CgyL90VarDZtQk/5hQvh29+G\nSy+F999POippaXLptFjp7qVpj61x9355jayBlJlIS7R9O1xySRjefsGCcCmxSK7yPdDjfmbWLuVg\nBwAaLUikAHXqFAqR666DESNg2jTYtSvpqKQlyCUzuYrQt+MewpVX5wIPZurNXkiUmUhLt2FDGCyy\nfXuYOxe6dUs6Iil0eR9OxcxOBL5KuAz3cXdfWp+DNSUVJiLw4YcwdSrMmAF33gmnnpp0RFLICnZs\nriSpMBHZY9WqcAnx8OEwfTp06JB0RFKI8jU217/N7F9Zbu/UP1wRaWoDB4ZG+Z07oX9/qKhIOiJp\nbpSZiLQwCxfCZZeFHvRXXgmtWiUdkRQKVXNloMJEJLtNm2DcuLA8b14Y3l4k35cGi0gz0707LF8O\nJ54IxxwTshWRhlBmItLCPfNMmHxr0KAwaGTHXCaYkGZJmYmI1NvRR8Ozz8L++4ce86tWJR2RFCNl\nJiKy25Il8B//EYZkufZaaJ3PccWl4KgBPgMVJiL1s2ULTJgA774L8+dDr15JRyRNRdVcItJounWD\nxx6DMWPgy18OBYpIbZSZiEhWa9eGxvmjjoI77ggDSUrzpcxERPLiqKNCb/kuXULj/MqVSUckhUqZ\niYjk5JFH4MIL4fzzYfLkMCmXNC/KTEQk7046CdasCf1SBg+G9euTjkgKiQoTEclZSUnIUMaNC50c\nZ88GVQAIqJpLROrphRfgzDOhTx+YOTO0q0hxUzWXiDS5z38eVq8O43yVlsKKFUlHJElSZiIiDbZ0\nKZx3Xpgm+IYboG3bpCOS+lBmIiKJOuGEMPlWVVVoS6mqSjoiaWoqTESkURx4YBjba+JEGDIktKOo\ncqDlUDWXiDS6qqrQc757d5g1KxQ0UvhUzSUiBaVv3zCUfd++oXF+2bKkI5J8U2YiInn1xBNwzjlh\n4MipU6Fdu6QjkmyUmYhIwRo2LAwYuWkTDBgA69YlHZHkgwoTEcm7Ll1g0SKYNAmGDg3TA6vioHlR\nNZeINKn162HsWOjaNQzHUlKSdERSTdVcIlI0eveGp54Kc8+XloaxvqT4KTMRkcSsXBkGjRw9Gm6+\nGQ44IOmIWjZlJiJSlIYMCT3nt20LmUplZdIRSX2pMBGRRHXqBAsWwHXXwYgRMG0a7NqVdFRSV6rm\nEpGCsWFDGCyyfXuYOxe6dUs6opZF1Vwi0iz06gVPPgnHHQf9+8PixUlHJLlSZiIiBWnVqnAJ8fDh\nMH06dOiQdETNX8FmJmY20syqzOxlM7sqyzq3xufXmlm/2rY1sylmttnM1sTbyHy+BhFJxsCBoUF+\nx46QpVRUJB2R1CRvhYmZtQJuB0YCnwPONLMj0tYZBfR298OAicCdOWzrwDR37xdvj+XrNYhIsjp2\nhDlz4PrrYdQouPFG+OijpKOSTPKZmQwA1rv7K+6+E7gfOCVtnZOBuQDu/jTQycwOymHbeqVhIlKc\nzjgjZCaPPRaqvTZtSjoiSZfPwuQQIPUt3xwfy2WdbrVse1msFrvbzDo1XsgiUqh69IDly2HkyNAn\nZeHCpCOSVK3zuO9cW7/rmmXcCVwfl28Afgacn2nFKVOm7F4uKyujrKysjocSkULSqhVcfTV89ath\n8q1HHw2DRnbsmHRkxam8vJzy8vJG2VferuYys4HAFHcfGe9fA+xy95tS1vkFUO7u98f7VcDxQK/a\nto2P9wQecvcvZDi+ruYSacbefTeMQrx8Odx7b2iwl4Yp1Ku5KoDDzKynmbUFzgAeTFvnQWA87C58\ntrv71pq2NbODU7Y/FXg+j69BRApUhw5w113w05/CN74RGuk//DDpqFquvPYzMbMTgVuAVsDd7j7V\nzC4CcPeZcZ3qq7beBc5192ezbRsf/xVQSqhG2wBcFAug9GMrMxFpIbZsCbM5vvcezJ8fOj9K3TUk\nM1GnRRFpFnbtgltuCVMDT58ehmWRulFhkoEKE5GWqbIyNM736wczZoSBJCU3hdpmIiLS5EpLQ5+U\nzp3D8sqVSUfUMigzEZFm6+GH4cIL4YILYPJkaNMm6YgKmzITEZEMRo8O1V7PPAODB4f55yU/VJiI\nSLNWUhLmmR83DgYNgtmzQZUWjU/VXCLSYqxbFxrn+/SBmTOhS5ekIyosquYSEcnBkUfC6tXQvXto\nnF+xIumImg9lJiLSIi1dCuedF/qj3HADtG2bdETJU2YiIlJHJ5wQGuerqkJbSlVV0hEVNxUmItJi\nHXggLFkCEyfCkCGhHUUVGvWjai4REUJmctZZoT1l1qxQ0LQ0quYSEWmgvn1h1arwt7QUli1LOqLi\nosxERCTNE0+EUYjHjAkDR7Zrl3RETUOZiYhIIxo2DNauDXPNDxgQ+qdIzVSYiIhk0KULLFoUZnMc\nOjRMD6zKjuxUzSUiUov162HsWOjaNQzHUlKSdET5oWouEZE86t0bnnoKjj46NM4/8kjSERUeZSYi\nInWwcmUYNHL0aLj5ZjjggKQjajzKTEREmsiQIaHn/LZtIVOprEw6osKgwkREpI46dYIFC+C662DE\nCJg2LcxB35KpmktEpAE2bAiDRbZvD3PnQrduSUdUf6rmEhFJSK9e8OSTcNxx0L8/LF6cdETJUGYi\nItJIVq0KlxAPHw7Tp0OHDklHVDfKTERECsDAgaFBfseOkKVUVCQdUdNRYSIi0og6doQ5c+D662HU\nKLjxRvjoo6Sjyj9Vc4mI5MnGjTB+fFieNy8Mb1/IVM0lIlKAevSA5cth5MjQJ2XhwqQjyh9lJiIi\nTaCiIky+deyxYdDIjh2TjmhfykxERArcMcfAmjXQtm0Y32vVqqQjalzKTEREmtjixXDxxXDJJXDt\ntdC6ddIRBQ3JTFSYiIgkYMuWMJvje+/B/Pmh82PSVM0lIlJkunWDpUvhtNPCbI7z5ycdUcMoMxER\nSVhlZWic79cPZswIA0kmQZmJiEgRKy0NV3t17hyWV65MOqK6U2YiIlJAHn4YLrwQLrgAJk+GNm2a\n7tgFm5mY2UgzqzKzl83sqizr3BqfX2tm/Wrb1sy6mNnjZvZXM1tmZgklhCIijW/06FDt9cwzMHhw\nmH++GOStMDGzVsDtwEjgc8CZZnZE2jqjgN7ufhgwEbgzh22vBh5398OB5fF+USovL086hJwozsal\nOBtXc4yzpCTMMz9uHAwaBLNnQ6FXtOQzMxkArHf3V9x9J3A/cEraOicDcwHc/Wmgk5kdVMu2u7eJ\nf7+Rx9eQV83xJEiS4mxcirNx1TVOM7j0UlixIgxnf/rpYargQpXPwuQQYFPK/c3xsVzW6VbDtiXu\nvjUubwVKGitgEZFCc+SRsHp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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f383c6dc490>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "from matplotlib.pyplot import plot,show,title,xlabel,ylabel\n", + "\n", + "# Claculate the following AC quantities Av, Vout, Pl, Pcc and percent efficiency. Also calculate the endpoints of ac loadline\n", + "\n", + "# Given data\n", + "\n", + "Icq = 7.91*10**-3# # Collector Currect(Q-point)=7.91 mAmps\n", + "Rl = 1.5*10**3# # Load Resistor=1.5 kOhms\n", + "Rc = 1.*10**3# # Collector Resistor=1 kOhms\n", + "Vin = 25.*10**-3# # Input Voltage=25 mVolts(p-p)\n", + "R1 = 18.*10**3# # Resistor 1=18 kOhms\n", + "R2 = 2.7*10**3# # Resistor 2=2.7 kOhms\n", + "Vcc = 20.# # Supply Voltage(Collector)=20 Volts\n", + "Vceq = 10.19# # Voltage Colector-Emitter(Q-point)=10.19 Volts\n", + "\n", + "rc = (25.*10**-3)/Icq#\n", + "rl = (Rc*Rl)/(Rc+Rl)\n", + "\n", + "Av = rl/rc#\n", + "print 'The Voltage Gain Av =%0.2f'%Av\n", + "print 'Approx 190'\n", + "\n", + "Vout = Av*Vin#\n", + "print 'The Output Voltage = %0.2f Volts'%Vout\n", + "\n", + "Pl = (Vout*Vout)/(8*Rl)#\n", + "print 'The Load Power = %0.2e Watts'%Pl\n", + "print 'i.e Approx 1.88 mWatts'\n", + "\n", + "Ivd = Vcc/(R1+R2)#\n", + "# Ic = Icq\n", + "Icc = Ivd+Icq#\n", + "\n", + "Pcc = Vcc*Icc#\n", + "print 'The Dc Input Power = %0.2e Watts'%Pcc\n", + "print 'i.e Approx 177.4 mWatts'\n", + "\n", + "efficiency = ((Pl/Pcc)*100)#\n", + "print 'The Efficiency in %% =%0.2f'%efficiency\n", + "print 'Approx 1%'\n", + "\n", + "# Endpoints of AC load line\n", + "\n", + "icsat = Icq+(Vceq/rl)#\n", + "print 'The Y-axis Value of AC Load-line is ic(sat) = %0.2e Amps'%icsat\n", + "print 'i.e 24.89 mAmps'\n", + "\n", + "vceoff = Vceq+Icq*rl#\n", + "print 'The X-axis value of AC Load-line is vce(off) = %0.2f Volts'%vceoff\n", + "\n", + "# For AC load line\n", + "\n", + "Vce1=[vceoff, Vceq, 0]\n", + "Ic1=[0 ,Icq ,icsat]\n", + "\n", + "#To plot AC load line\n", + "\n", + "print \"Q(%f,%f)\\n\"%(Vceq,Icq)\n", + "plot(Vce1, Ic1)\n", + "plot(Vceq,Icq)\n", + "plot(0,Icq)\n", + "plot(Vceq,0)\n", + "plot(0,icsat)\n", + "plot(vceoff,0)\n", + "xlabel(\"Vce in volt\")\n", + "ylabel(\"Ic in Ampere\")\n", + "title(\"AC Load-line for Common-Emitter Class A Amplifier Circuit\")\n", + "show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 31_3 Page No. 1025" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Load Power = 6.25 Watts\n", + "The DC Input Power = 9.54 Watts\n", + "The Efficiency in % =65.51\n", + "The Maximum Power Dissipation = 1.80 Watts\n" + ] + } + ], + "source": [ + "#Calculate the following quantities: Pl, Pcc, Pdmax & percent efficiency\n", + "\n", + "# Given data\n", + "\n", + "Vin = 20.# # Input Voltage=20 Volts(p-p)\n", + "Vopp = 20.# # Output Voltage(p-p)=20 Volts(p-p)\n", + "Vcc = 24.# # Supply Voltage(Collector)=24 Volts\n", + "Vop = 10.# # Output Voltage(peak)=10 Volts\n", + "Rl = 8.# # Load Resistor=8 Ohms\n", + "\n", + "Vopp1 = Vopp*Vopp#\n", + "Pl = (Vopp1/(8*Rl))#\n", + "print 'The Load Power = %0.2f Watts'%Pl\n", + "\n", + "Icc = ((Vop/Rl)*0.318)#\n", + "\n", + "Pcc = Vcc*Icc\n", + "print 'The DC Input Power = %0.2f Watts'%Pcc\n", + "\n", + "eff = ((Pl/Pcc)*100)#\n", + "print 'The Efficiency in %% =%0.2f'%eff\n", + "\n", + "Pd = (Vcc*Vcc)/(40*Rl)#\n", + "print 'The Maximum Power Dissipation = %0.2f Watts'%Pd" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 31_4 Page No. 1037" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Load Power = 39.00 Watts\n", + "The DC Input Power = 57.24 Watts\n", + "The Efficiency in % = 68.13\n" + ] + } + ], + "source": [ + "# Calculate the following quantities Pl, Pcc & percent efficiency\n", + "\n", + "# Given data\n", + "\n", + "Rl = 8# # Load Resistor=8 Ohms\n", + "Vopp = 50# # Output Voltage(p-p)=50 Volts(p-p)\n", + "Vcc = 30# # Supply Voltage(Collector)=30 Volts\n", + "Vopk = Vopp/2# # Output Voltage(peak)\n", + "\n", + "Pl = (Vopp*Vopp)/(8*Rl)#\n", + "print 'The Load Power = %0.2f Watts'%Pl\n", + "\n", + "Pcc = Vcc*0.636*(Vopk/Rl)#\n", + "print 'The DC Input Power = %0.2f Watts'%Pcc\n", + "\n", + "efficiency = ((Pl/Pcc)*100)#\n", + "print 'The Efficiency in %% = %0.2f'%efficiency" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 31_5 Page No. 1038" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resonant Frequency = 2.00e+06 Hertz\n", + "i.e 2 MHz\n", + "The DC Bias Voltage at Base = 0.80 Volts\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "# Calculate the fr of LC tank circuit and dc bias voltage at base\n", + "\n", + "# Given data\n", + "\n", + "L = 100*10**-6# # Inductor=100 uHenry\n", + "C = 63.325*10**-12# # Capacitor=63.325 pFarad\n", + "Vin = 1.5# # Input Voltage(peak)=1.5 Volts\n", + "Vbe = 0.7# # Voltage Base-Emitter=0.7 Volts\n", + "\n", + "A = sqrt(L*C)#\n", + "fr = 1./(2*3.14*A)#\n", + "print 'The Resonant Frequency = %0.2e Hertz'%fr\n", + "print 'i.e 2 MHz'\n", + "\n", + "Vdc = (Vin-Vbe)#\n", + "print 'The DC Bias Voltage at Base = %0.2f Volts'%Vdc" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 31_6 Page No. 1039" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Minimum Base Reisitance Rb to Provide Clamping Action = 500 Ohms\n" + ] + } + ], + "source": [ + "# Calculate the minimum base reisitance Rb, necessary to provide clamping action\n", + "\n", + "# Given data\n", + "\n", + "C = 0.01*10**-6# # Capacitor=0.01 uFarad\n", + "fr = 2.*10**6# # Resonant Frequency=2 MHertz\n", + "\n", + "fin = fr\n", + "T = 1/fin\n", + "\n", + "Rb = 10*T/C\n", + "print 'The Minimum Base Reisitance Rb to Provide Clamping Action = %0.f Ohms'%Rb" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 31_7 Page No. 1040" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Bandwidth = 45120 Hertz\n", + "i.e Approx 45 kHz\n" + ] + } + ], + "source": [ + "# Calculate the Bandwidth\n", + "\n", + "# Given data\n", + "\n", + "L = 100*10**-6# # Inductor=100 uHenry\n", + "fr = 2*10**6# # Resonant Frequency=2 MHertz\n", + "ri = 12.56# # Resistance of Coil=12.56 Ohms\n", + "Rp = 100*10**3# # Rp=100 kOhms\n", + "\n", + "Xl = 2*3.14*fr*L#\n", + "Qcoil = Xl/ri#\n", + "Ztank = Qcoil*Xl#\n", + "\n", + "A = Ztank#\n", + "B = Rp#\n", + "C = (A*B)/(A+B)#\n", + "Qckt = C/Xl#\n", + "\n", + "BW = fr/Qckt#\n", + "print 'The Bandwidth = %0.f Hertz'%BW\n", + "print 'i.e Approx 45 kHz'" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter32.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter32.ipynb new file mode 100644 index 00000000..bc93187e --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter32.ipynb @@ -0,0 +1,71 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 32 : Thyristors" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 32_1 Page No. 1058" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Frequency of the Emmiter Voltage Waveform = 49.61 Hertz\n" + ] + } + ], + "source": [ + "from math import log\n", + "# Calculate the frequency of the emmiter voltage waveform. Assume n=0.6\n", + "\n", + "# Given data\n", + "\n", + "Rt = 220*10**3# # Resistor Rt=220k Ohms\n", + "Ct = 0.1*10**-6# # Capacitor Ct=0.1u Farad\n", + "n = 0.6# # Constant\n", + "\n", + "A = 1./(1-n)#\n", + "T = Rt*Ct*log(A)#\n", + "\n", + "f = 1./T#\n", + "print 'The Frequency of the Emmiter Voltage Waveform = %0.2f Hertz'%f" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter33.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter33.ipynb new file mode 100644 index 00000000..c663815f --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter33.ipynb @@ -0,0 +1,1000 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 33 : Operational Amplifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_1 Page No. 1072" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Differential Voltage Gain =143.00\n", + "The Ac Output Voltage = 1.43 Volts(p-p)\n" + ] + } + ], + "source": [ + "# Calculate the differential voltage gain, Ad, and the ac output voltage, Vout.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 10*10**-3# # Input voltage=10 mVolts(p-p)\n", + "Rc = 10*10**3# # Collector resistance=10 kOhms\n", + "Ie = 715.*10**-6# # Emitter current=715 uAmps\n", + "\n", + "re = (25*10**-3)/Ie#\n", + "\n", + "Ad = Rc/(2*re)#\n", + "print 'The Differential Voltage Gain =%0.2f'%Ad\n", + "\n", + "Av = Ad\n", + "\n", + "Vo = Av*Vin#\n", + "print 'The Ac Output Voltage = %0.2f Volts(p-p)'%Vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_2 Page No. 1073" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Common-Mode Voltage Gain Acm = 0.50\n", + "The Commom-Mode Rejection Ratio = 49.12 dB\n" + ] + } + ], + "source": [ + "from math import log10\n", + "# calculate the common-mode voltage gain, ACM, and the CMRR (dB).\n", + "\n", + "# Given data\n", + "\n", + "Rc = 10*10**3# # Collector resistance=10 kOhms\n", + "Re = 10.*10**3# # Emitter resistance=10 kOhms\n", + "Ad = 142.86# # Differential gain=142.86\n", + "\n", + "Acm = Rc/(2*Re)#\n", + "print 'The Common-Mode Voltage Gain Acm = %0.2f'%Acm\n", + "\n", + "CMRR = 20*log10(Ad/Acm)#\n", + "print 'The Commom-Mode Rejection Ratio = %0.2f dB'%CMRR" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_3 Page No. 1074" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Output Frequency = 7.96e+04 Hertz\n", + "i.e 79.6 kHz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# Calculate fmax for an op amp that has an Sr of 5 V/u\u0002s and a peak output voltage of 10 V.\n", + "\n", + "# Given data\n", + "\n", + "Vpk = 10.# # Peak output voltage=10 Volts\n", + "Sr = 5./10**-6# # Slew rate=5 V/us\n", + "\n", + "\n", + "fo = Sr/(2*pi*Vpk)#\n", + "print 'The Output Frequency = %0.2e Hertz'%fo\n", + "print 'i.e 79.6 kHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_4 Page No. 1075" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Closed-Loop Voltage Gain Acl =-10.00\n", + "The Output Voltage = 10.00 Volts(p-p)\n", + "The -ve sign indicates that input and output voltages are 180° out-of-phase\n" + ] + } + ], + "source": [ + "# calculate the closed-loop voltage gain, Acl, and the output voltage, Vout.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 1.# # Input voltage=1 Volts(p-p)\n", + "Rf = 10.*10**3# # Feedback resistance=10 kOhms\n", + "Ri = 1.*10**3# # Input resistance=1 kOhms\n", + "\n", + "Acl = -(Rf/Ri)#\n", + "print 'The Closed-Loop Voltage Gain Acl =%0.2f'%Acl\n", + "\n", + "Vo = -Vin*Acl#\n", + "print 'The Output Voltage = %0.2f Volts(p-p)'%Vo\n", + "print 'The -ve sign indicates that input and output voltages are 180° out-of-phase'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_5 Page No. 1076" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Differential Input Voltage = 1.00e-04 Volts(p-p)\n", + "i.e 100 uVolts(p-p)\n" + ] + } + ], + "source": [ + "#If Avol equals 100,000, calculate the value of Vid.\n", + "\n", + "# Given data\n", + "\n", + "Avol = 100000.# # Open loop voltage gain=100,000\n", + "Vo = 10.# # Output voltage=10 Volts(p-p)\n", + "\n", + "Vid = Vo/Avol#\n", + "print 'The Differential Input Voltage = %0.2e Volts(p-p)'%Vid\n", + "print 'i.e 100 uVolts(p-p)'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_6 Page No. 1078" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Input Impedence = 1.00e+03 Ohms\n", + "i.e 1 kOhms\n", + "The Closed Loop Output Impedence = 0.01 Ohms\n" + ] + } + ], + "source": [ + "# calculate Zin and Zout(CL). Assume AVOL is\u0004 100,000 and Zout(OL) is\u0004 75 Ohms.\n", + "\n", + "# Given data\n", + "\n", + "Avol = 100000.# # Open loop voltage gain=100,000\n", + "Rf = 10.*10**3# # Feedback resistance=10 kOhms\n", + "Ri = 1.*10**3# # Input resistance=1 kOhms\n", + "Zool = 75.# # Output impedence (open-loop)=75 Ohms\n", + "\n", + "Zi = Ri#\n", + "print 'The Input Impedence = %0.2e Ohms'%Zi\n", + "print 'i.e 1 kOhms'\n", + "\n", + "Beta = Ri/(Ri+Rf)#\n", + "\n", + "A = Avol*Beta#\n", + "\n", + "Zocl = Zool/(1+A)#\n", + "print 'The Closed Loop Output Impedence = %0.2f Ohms'%Zocl" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_7 Page No. 1083" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Output Frequency = 1.59e+04 Hertz\n", + "i.e 15.915 kHz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# Calculate the 5-V power bandwidth.\n", + "\n", + "# Given data\n", + "\n", + "Vo = 10.# # Output voltage=10 Volts(p-p)\n", + "Sr = 0.5/10**-6# # Slew rate=0.5 V/us\n", + "\n", + "Vpk = Vo/2#\n", + "\n", + "fo = Sr/(2*pi*Vpk)#\n", + "print 'The Output Frequency = %0.2e Hertz'%fo\n", + "print 'i.e 15.915 kHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_8 Page No. 1085" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Closed-Loop Voltage Gain Acl =11.00\n", + "The Output Voltage = 11.00 Volts(p-p)\n" + ] + } + ], + "source": [ + "# Calculate the closed-loop voltage gain, Acl, and the output voltage, Vout.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 1# # Input voltage=1 Volts(p-p)\n", + "Rf = 10.*10**3# # Feedback resistance=10 kOhms\n", + "Ri = 1.*10**3# # Input resistance=1 kOhms\n", + "\n", + "Acl = 1+(Rf/Ri)#\n", + "print 'The Closed-Loop Voltage Gain Acl =%0.2f'%Acl\n", + "\n", + "Vo = Vin*Acl#\n", + "print 'The Output Voltage = %0.2f Volts(p-p)'%Vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_9 Page No. 1089" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Input Impedence Closed-Loop = 1.82e+10 Ohms\n", + "i.e 18 GOhms\n", + "The Closed-Loop Output Impedence = 0.01 Ohms\n" + ] + } + ], + "source": [ + "# Calculate Zin(CL) and Zout(CL). Assume Rin is\u0004 2 MOhms\u0006, Avol is 100,000, and Zout(OL) is 75 Ohms.\n", + "\n", + "# Given data\n", + "\n", + "Avol = 100000.# # Open loop voltage gain=100,000\n", + "Ri = 2.*10**6# # Input resistance=2 MOhms\n", + "B = 0.0909# # Beta=0.0909\n", + "Zool = 75.# # Output impedence (open-loop)=75 Ohms\n", + "\n", + "Zicl = Ri*(1+Avol*B)#\n", + "print 'The Input Impedence Closed-Loop = %0.2e Ohms'%Zicl\n", + "print 'i.e 18 GOhms'\n", + "\n", + "A = Avol*B#\n", + "\n", + "Zocl = Zool/(1+A)#\n", + "print 'The Closed-Loop Output Impedence = %0.2f Ohms'%Zocl" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_10 Page No. 1090" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Input impedence closed-loop = 2.00e+11 Ohms\n", + "i.e 200 GOhms\n", + "The Closed loop Output Impedence = 0.001 Ohms\n" + ] + } + ], + "source": [ + "# Assume Rin is 2 MOhms, Avol is 100,000, and Zout(OL) is 75 Ohms. Calculate Zin(CL) and Zout(CL)\n", + "\n", + "# Given data\n", + "\n", + "Avol = 100000.# # Open loop voltage gain=100,000\n", + "Ri = 2.0*10**6# # Input resistance=2 MOhms\n", + "B = 1.0# # Beta=1\n", + "Zool = 75.# # Output impedence (open-loop)=75 Ohms\n", + "\n", + "Zicl = Ri*(1+Avol*B)#\n", + "print 'The Input impedence closed-loop = %0.2e Ohms'% Zicl\n", + "print 'i.e 200 GOhms'\n", + "\n", + "A = Avol*B#\n", + "\n", + "Zocl = Zool/(1+A)#\n", + "print 'The Closed loop Output Impedence = %0.3f Ohms'%Zocl" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_11 Page No. 1091" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Closed-Loop Voltage Gain Acl =-10.00\n", + "The Output Voltage = 7.50 Volts\n" + ] + } + ], + "source": [ + "# Calculate the closed-loop voltage gain, Acl, and the dc voltage at the op-amp output terminal.\n", + "\n", + "# Given data\n", + "\n", + "V = 15.# # Voltage at +ve terminal of op-amp=15 Volts\n", + "Rf = 10.*10**3# # Feedback resistance=10 kOhms\n", + "Ri = 1.*10**3# # Input resistance=1 kOhms\n", + "R1 = 10.*10**3# # Resistance1=10 kOhms\n", + "R2 = 10.*10**3# # Rsistance2=10 kOhms\n", + "\n", + "Acl = -(Rf/Ri)#\n", + "print 'The Closed-Loop Voltage Gain Acl =%0.2f'%Acl\n", + "\n", + "Vo = V*(R2/(R1+R2))#\n", + "print 'The Output Voltage = %0.2f Volts'%Vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_12 Page No. 1095" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Output Voltage 1 Volts\n" + ] + } + ], + "source": [ + "# Calculate the output voltage, Vout.\n", + "\n", + "# Given data\n", + "\n", + "V1 = 1# # Input voltage1=1 Volts\n", + "V2 = -5# # Input voltage2=-5 Volts\n", + "V3 = 3# # Input voltage3=3 Volts\n", + "\n", + "Vo = -(V1+V2+V3)#\n", + "print 'The Output Voltage %0.f Volts'%Vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_13 Page No. 1097" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Output Voltage = 3.00 Volts\n" + ] + } + ], + "source": [ + "# Calculate the output voltage, Vout.\n", + "\n", + "# Given data\n", + "\n", + "V1 = 0.5# # Input voltage1=0.5 Volts\n", + "V2 = -2.0# # Input voltage2=-2 Volts\n", + "Rf = 10.*10**3# # Feedback resistance=10 kOhms\n", + "R1 = 1.*10**3# # Resistance1=1 kOhms\n", + "R2 = 2.5*10**3# # Rsistance2=2.5 kOhms\n", + "\n", + "A = Rf/R1#\n", + "B = Rf/R2#\n", + "\n", + "Vo = -(A*V1+B*V2)#\n", + "print 'The Output Voltage = %0.2f Volts'%Vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_14 Page No. 1101" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Output Voltage of Case A = -12.50 Volts\n", + "The Output Voltage of Case B = 10.00 Volts\n", + "The Output Voltage of Case C = -0.00 Volts\n" + ] + } + ], + "source": [ + "# Calculate the output voltage, Vout, if (a) Vx is 1 Vdc and Vy is -0.25 Vdc, (b) -Vx is 0.5 Vdc and Vy is 0.5 Vdc, (c) Vx is 0.3 V and Vy is 0.3 V.\n", + "\n", + "# Given data\n", + "\n", + "Rf = 10.*10**3# # Feedback resistance=10 kOhms\n", + "R1 = 1.*10**3# # Resistance1=1 kOhms\n", + "Vx1 = 1.# # Input voltage Vx1 at -ve terminal of op-amp=1 Volts\n", + "Vy1 = -0.25# # Input voltage Vy1 at +ve terminal of op-amp=-0.25 Volts\n", + "Vx2 = -0.5# # Input voltage Vx2 at -ve terminal of op-amp=-0.5 Volts\n", + "Vy2 = 0.5# # Input voltage Vy2 at +ve terminal of op-amp=0.5 Volts\n", + "Vx3 = 0.3# # Input voltage Vx3 at -ve terminal of op-amp=0.3 Volts\n", + "Vy3 = 0.3# # Input voltage Vy3 at +ve terminal of op-amp=0.3 Volts\n", + "\n", + "A = -Rf/R1#\n", + "\n", + "# Case A\n", + "\n", + "Voa = A*(Vx1-Vy1)#\n", + "print 'The Output Voltage of Case A = %0.2f Volts'%Voa\n", + "\n", + "# Case B\n", + "\n", + "Voa = A*(Vx2-Vy2)#\n", + "print 'The Output Voltage of Case B = %0.2f Volts'%Voa\n", + "\n", + "# Case C\n", + "\n", + "Voa = A*(Vx3-Vy3)#\n", + "print 'The Output Voltage of Case C = %0.2f Volts'%Voa" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_15 Page No. 1102" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Output of Differential Amplifier = 5.00 Volts\n" + ] + } + ], + "source": [ + "# Assume that Rd increases to 7.5 k\u0006 due to an increase in the ambient temperature. Calculate the output of the differential amplifier. Note: Rb is 5 kOhms\u0006.\n", + "\n", + "# Given data\n", + "\n", + "Vi = 5.# # Voltage input=5 Volts(dc)\n", + "Rf = 10.*10**3# # Feedback resistance=10 kOhms\n", + "R1 = 1.*10**3# # Resistance1=1 kOhms\n", + "Ra = 5.*10**3# # Resistance A at wein bridge=5 kOhms\n", + "Rb = 10.*10**3# # Resistance B at wein bridge=10 kOhms\n", + "Rc = 5.*10**3# # Resistance C at wein bridge=5 kOhms\n", + "Rd = 7.5*10**3# # Resistance D at wein bridge=7.5 kOhms\n", + "\n", + "Vx = Vi*(Ra/Rb)#\n", + "Vy = Vi*(Rd/(Rd+Rc))#\n", + "A = -Rf/R1\n", + "\n", + "Vo = A*(Vx-Vy)#\n", + "print 'The Output of Differential Amplifier = %0.2f Volts'%Vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_16 Page No. 1103" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Cutoff Frequency = 1.59e+03 Hertz\n", + "i.e 1.591 kHz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# Calculate the cutoff frequency, fc.\n", + "\n", + "# Given data\n", + "\n", + "Rf = 10.*10**3# # Feedback resistance=10 kOhms\n", + "Cf = 0.01*10**-6# # Feedback capacitance=0.01 uFarad\n", + "\n", + "fc = 1./(2.*pi*Rf*Cf)#\n", + "print 'The Cutoff Frequency = %0.2e Hertz'%fc\n", + "print 'i.e 1.591 kHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_17 Page No. 1104" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Closed-Loop Voltage Gain at 0 Hz =-10.00\n", + "The Closed-Loop Voltage Gain at 1 MHz =-0.02\n" + ] + } + ], + "source": [ + "from math import pi,sqrt\n", + "# Calculate the Voltage gain, Acl at (a)0 Hz and (b) 1 MHz\n", + "\n", + "# Given data\n", + "\n", + "f1 = 1.*10**6# # Frequency=1 MHertz\n", + "Rf = 10.*10**3# # Feedback resistance=10 kOhms\n", + "R1 = 1.*10**3# # Resistance1=1 kOhms\n", + "Cf = 0.01*10**-6# # Feedback capacitance=0.01 uFarad\n", + "\n", + "# At 0 Hz, Xcf = infinity ohms, So, Zf=Rf \n", + "\n", + "Acl = -Rf/R1#\n", + "print 'The Closed-Loop Voltage Gain at 0 Hz =%0.2f'%Acl\n", + "\n", + "# At 1 MHz\n", + "\n", + "Xcf = 1/(2*pi*f1*Cf)#\n", + "\n", + "A = (Rf*Rf)#\n", + "B = (Xcf*Xcf)#\n", + "\n", + "Zf = ((Xcf*Rf)/sqrt(A+B))#\n", + "\n", + "Acl1 = -Zf/R1#\n", + "print 'The Closed-Loop Voltage Gain at 1 MHz =%0.2f'%Acl1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_18 Page No. 1105" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Gain at 0 Hz = 20.00 dB\n", + "The Voltage Gain at 1.591 kHz = 16.99 dB\n", + "approx 17dB\n" + ] + } + ], + "source": [ + "from math import log10,pi,sqrt\n", + "# Calculate the dB voltage gain, at (a)0 Hz and (b) 1.591 kHz\n", + "\n", + "# Given data\n", + "\n", + "f1 = 1.591*10**3# # Frequency=1.591 kHertz\n", + "Rf = 10.*10**3# # Feedback resistance=10 kOhms\n", + "Ri = 1.*10**3# # Input resistance=1 kOhms\n", + "Cf = 0.01*10**-6# # Feedback capacitance=0.01 uFarad\n", + "\n", + "# At 0 Hz, Xcf = infinity ohms, So, Zf=Rf \n", + "\n", + "A = Rf/Ri\n", + "\n", + "Acl = 20*log10(A)#\n", + "print 'The Voltage Gain at 0 Hz = %0.2f dB'%Acl\n", + "\n", + "# At 1.591 kHz\n", + "\n", + "Xcf = 1/(2*pi*f1*Cf)#\n", + "B = (Rf*Rf)#\n", + "C = (Xcf*Xcf)#\n", + "Zf = (Xcf*Rf/sqrt(B+C))#\n", + "D = Zf/Ri#\n", + "\n", + "Acl1 = 20*log10(D)#\n", + "print 'The Voltage Gain at 1.591 kHz = %0.2f dB'%Acl1\n", + "print 'approx 17dB'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_19 Page No. 1106" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Cutoff Frequency = 1591.55 Hertz\n", + "i.e 1.591 kHz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# Calculate the cutoff frequency, fc.\n", + "\n", + "# Given data\n", + "\n", + "Ri = 1.*10**3# # Input resistance=10 kOhms\n", + "Ci = 0.1*10**-6# # Input capacitance=0.01 uFarad\n", + "\n", + "fc = 1/(2*pi*Ri*Ci)#\n", + "print 'The Cutoff Frequency = %0.2f Hertz'%fc\n", + "print 'i.e 1.591 kHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_20 Page No. 1118" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Output Current = 5.00e-03 Amps\n", + "i.e 5 mAmps\n" + ] + } + ], + "source": [ + "# Vin is 5 V, R is 1 kOhms , and Rl is 100 Ohms . Calculate the output current, Iout.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 5.# # Input votage=5 Volts\n", + "Ri = 1.*10**3# # Input resistance=1 kOhms\n", + "Rl = 100.# # Load resistance=100 Ohms\n", + "\n", + "Io = Vin/Ri#\n", + "print 'The Output Current = %0.2e Amps'%Io\n", + "print 'i.e 5 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_21 Page No. 1120" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Output Voltage = 1.50 Volts\n" + ] + } + ], + "source": [ + "# Iin is 1.5 mA, R is 1 kOhms, and Rl is 10 kOhms. Calculate Vout.\n", + "\n", + "# Given data\n", + "\n", + "Iin = 1.5*10**-3# # Input votage=5 Volts\n", + "Ri = 1.*10**3# # Input resistance=1 kOhms\n", + "Rl = 100.# # Load resistance=100 Ohms\n", + "\n", + "Vo = Iin*Ri#\n", + "print 'The Output Voltage = %0.2f Volts'%Vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_22 Page No. 1121" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Upper Trigger Point = 0.129 Volts\n", + "i.e 128.7 mVolts\n", + "The Lower Trigger Point = -0.129 Volts\n", + "i.e -128.7 mVolts\n", + "The Hysterisis Voltage = 0.257 Volts\n", + "i.e 257.4 mVolts\n" + ] + } + ], + "source": [ + "# R1 is 1 kOhms and R2 is 100 kOhms . Calculate UTP, LTP, and VH.\n", + "\n", + "# Given data\n", + "\n", + "R1 = 1.*10**3# # Resistance1=1 kOhms\n", + "R2 = 100.*10**3# # Resistance2=100 kOhms\n", + "Vcc = 15.# # Applied votage=15 Volts\n", + "Vsat = 13.# # Assume Saturation voltage=13 Volts\n", + "\n", + "Beta = R1/(R1+R2)#\n", + "\n", + "Utp = Beta*Vsat#\n", + "print 'The Upper Trigger Point = %0.3f Volts'%Utp\n", + "print 'i.e 128.7 mVolts'\n", + "\n", + "Ltp = -Beta*Vsat#\n", + "print 'The Lower Trigger Point = %0.3f Volts'%Ltp\n", + "print 'i.e -128.7 mVolts'\n", + "\n", + "Vh = Utp-Ltp#\n", + "print 'The Hysterisis Voltage = %0.3f Volts'%Vh\n", + "print 'i.e 257.4 mVolts'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_23 Page No. 1124" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Minimum value of required Capacitor = 1.00e-04 Farads\n", + "i.e 100 uFarad\n" + ] + } + ], + "source": [ + "# Rl is 1 kOhms and the frequency of the input voltage equals 100 Hz. Calculate the minimum value of C required.\n", + "\n", + "# Given data\n", + "\n", + "f = 100.# # Applied frequency=100 Hertz\n", + "Rl = 1.*10**3# # Load resistance=1 kOhms\n", + "\n", + "T = 1./f#\n", + "\n", + "C = (10*T)/Rl#\n", + "print 'The Minimum value of required Capacitor = %0.2e Farads'%C\n", + "print 'i.e 100 uFarad'" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter4.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter4.ipynb new file mode 100644 index 00000000..b71eae51 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter4.ipynb @@ -0,0 +1,336 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4 : Series Circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 4_1 Page No. 117" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Combined Series Resistance = 20 Ohms\n" + ] + } + ], + "source": [ + "# Two resistances R1 and R2 of 5 Ohms\u0004 each and R3 of 10 Ohms\u0004 are in series. How much is Rt?\n", + "\n", + "# Given data\n", + "\n", + "R1 = 5# # Resistor 1=5 Ohms\n", + "R2 = 5# # Resistor 2=5 Ohms\n", + "R3 = 10# # Resistor 3=10 Ohms\n", + "\n", + "Rt = R1+R2+R3#\n", + "print 'The Combined Series Resistance = %0.f Ohms'%Rt" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 4_2 Page No. 117" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current in Resistor R3 connected in Series = 4 Amps\n" + ] + } + ], + "source": [ + "#With 80 V applied across the series string, how much is the current in R3?\n", + "\n", + "# Given data\n", + "\n", + "Rt = 20# # Total Resistance=20 Ohms\n", + "Vt = 80# # Applied Voltage=80 Volts\n", + "\n", + "I = Vt/Rt#\n", + "print 'The Current in Resistor R3 connected in Series = %0.f Amps'%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 4_3 Page No. 119" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The combined series resistance = 60 ohms\n", + "The current = 0.20 Amps\n", + "i.e 200 mA\n", + "The Voltage Drop of Resistor R1 = 2.00 Volts\n", + "The Voltage Drop of Resistor R2 = 4.00 Volts\n", + "The Voltage Drop of Resistor R3 = 6.00 Volts\n" + ] + } + ], + "source": [ + "# Solve for Rt, I and the individual resistor voltage drops at R1, R2, R3.\n", + "\n", + "# Given data\n", + "\n", + "R1 = 10.# # Resistor 1=10 Ohms\n", + "R2 = 20.# # Resistor 2=20 Ohms\n", + "R3 = 30.# # Resistor 3=30 Ohms\n", + "Vt = 12.0# # Applied Voltage=12 Volts\n", + "\n", + "Rt = R1+R2+R3#\n", + "print 'The combined series resistance = %0.f ohms'%Rt\n", + "\n", + "I = Vt/Rt#\n", + "print 'The current = %0.2f Amps'%I\n", + "print 'i.e 200 mA'\n", + "\n", + "V1 = I*R1\n", + "print 'The Voltage Drop of Resistor R1 = %0.2f Volts'%V1\n", + "\n", + "V2 = I*R2\n", + "print 'The Voltage Drop of Resistor R2 = %0.2f Volts'%V2\n", + "\n", + "V3 = I*R3\n", + "print 'The Voltage Drop of Resistor R3 = %0.2f Volts'%V3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 4_4 Page No. 123" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Applied Voltage Vt = 280 Volts\n" + ] + } + ], + "source": [ + "# A voltage source produces an IR drop of 40 V across a 20 Ohms R1, 60 V across a 30 Ohms\u0004 R2, and 180 V across a 90 Ohms\u0004 R3, all in series. According to Kirchhoff’s voltage law, how much is the applied voltage Vt ?\n", + "\n", + "# Given data\n", + "\n", + "V1 = 40# # Voltage drop at R1=40 Volts\n", + "V2 = 60# # Voltage drop at R2=60 Volts\n", + "V3 = 180# # Voltage drop at R3=180 Volts\n", + "\n", + "Vt = V1+V2+V3#\n", + "print 'The Applied Voltage Vt = %0.f Volts'%Vt" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 4_5 Page No. 123" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Drop across Resistor R2 = 80 Volts\n" + ] + } + ], + "source": [ + "# An applied Vt of 120 V produces IR drops across two series resistors R 1 and R 2 If the voltage drop across R1 is 40 V, how much is the voltage drop across R2?\n", + "\n", + "# Given data\n", + "\n", + "V1 = 40# # Voltage drop at R1=40 Volts\n", + "Vt = 120# # Applied Voltage=120 Volts\n", + "\n", + "V2 = Vt-V1#\n", + "print 'The Voltage Drop across Resistor R2 = %0.f Volts'%V2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 4_6 Page No. 131" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Drop of Resistor R1 = 6.00 Volts\n", + "The Voltage Drop of Resistor R2 = 4.80 Volts\n", + "The Voltage Drop of Resistor R3 = 7.20 Volts\n", + "The Voltage Drop of Resistor R4 = 6.00 Volts\n", + "The Resistor R3 is defective since it is open circuit and drops all the voltage arround it\n" + ] + } + ], + "source": [ + "# Assume that the series circuit in Fig. 4–20 has failed. A technician troubleshooting the circuit used a voltmeter to record the following resistor voltage drops. V1=0 V# V2=0 V# V3=24 V# V4=0 V. Based on these voltmeter readings, which component is defective and what type of defect is it? (Assume that only one component is defective.)\n", + "\n", + "# Given data\n", + "\n", + "R1 = 150.# # Resistor 1=150 Ohms\n", + "R2 = 120.# # Resistor 2=120 Ohms\n", + "R3 = 180.# # Resistor 3=180 Ohms\n", + "R4 = 150.# # Resistor 4=150 Ohms\n", + "Vt = 24.# # Applied Voltage=24 Volts\n", + "\n", + "Rt = R1+R2+R3+R4#\n", + "\n", + "I = Vt/Rt#\n", + "\n", + "V1 = I*R1\n", + "print 'The Voltage Drop of Resistor R1 = %0.2f Volts'%V1\n", + "\n", + "V2 = I*R2\n", + "print 'The Voltage Drop of Resistor R2 = %0.2f Volts'%V2\n", + "\n", + "V3 = I*R3\n", + "print 'The Voltage Drop of Resistor R3 = %0.2f Volts'%V3\n", + "\n", + "V4 = I*R4\n", + "print 'The Voltage Drop of Resistor R4 = %0.2f Volts'%V4\n", + "\n", + "print 'The Resistor R3 is defective since it is open circuit and drops all the voltage arround it'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 4_7 Page No. 133" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Calculated from the Circuit\n", + "The Voltage Drop of Resistor R1 = 6.00 Volts\n", + "The Voltage Drop of Resistor R2 = 4.80 Volts\n", + "The Voltage Drop of Resistor R3 = 7.20 Volts\n", + "The Voltage Drop of Resistor R4 = 6.00 Volts\n" + ] + } + ], + "source": [ + "# Assume that the series circuit has failed. A technician troubleshooting the circuit used a voltmeter to record the following resistor voltage drops: V1 \u0005 8 V#V2 \u0005 6.4 V#V3 \u0005 9.6 V#V4 \u0005 0 V. Based on the voltmeter readings, which component is defective and what type of defect is it? (Assume that only one component is defective.)\n", + "\n", + "# Given data\n", + "\n", + "R1 = 150.# # Resistor 1=150 Ohms\n", + "R2 = 120.# # Resistor 2=120 Ohms\n", + "R3 = 180.# # Resistor 3=180 Ohms\n", + "R4 = 150.# # Resistor 4=150 Ohms\n", + "Vt = 24.# # Applied Voltage=24 Volts\n", + "\n", + "print 'Calculated from the Circuit'\n", + "\n", + "Rt = R1+R2+R3+R4#\n", + "\n", + "I = Vt/Rt#\n", + "\n", + "V1 = I*R1\n", + "print 'The Voltage Drop of Resistor R1 = %0.2f Volts'%V1\n", + "\n", + "V2 = I*R2\n", + "print 'The Voltage Drop of Resistor R2 = %0.2f Volts'%V2\n", + "\n", + "V3 = I*R3\n", + "print 'The Voltage Drop of Resistor R3 = %0.2f Volts'%V3\n", + "\n", + "V4 = I*R4\n", + "print 'The Voltage Drop of Resistor R4 = %0.2f Volts'%V4\n" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter5.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter5.ipynb new file mode 100644 index 00000000..64665511 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter5.ipynb @@ -0,0 +1,291 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5 : Parallel Circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 5_1 Page No. 149" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current Resistor R1 = 0.015 Amps\n", + "i.e 15 mAmps\n", + "The Current Resistor R2 = 0.03 Amps\n", + "i.e 25 mAmps\n" + ] + } + ], + "source": [ + "# Solve for branch currents I1 and I2.\n", + "\n", + "R1 = 1.*10**3# # Resistor 1=1*10**3 Ohms\n", + "R2 = 600.# # Resistor 2=600 Ohms\n", + "Va = 15.# # Applied Voltage=15 Volts\n", + "\n", + "I1 = Va/R1#\n", + "print 'The Current Resistor R1 = %0.3f Amps'%I1\n", + "print 'i.e 15 mAmps'\n", + "\n", + "I2 = Va/R2#\n", + "print 'The Current Resistor R2 = %0.2f Amps'%I2\n", + "print 'i.e 25 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 5_2 Page No. 150" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Total Current in the Mainline = 11 Amps\n" + ] + } + ], + "source": [ + "# An R1 of 20 Ohms\u0002, an R2 of 40 Ohms\u0002, and an R3 of 60 Ohms\u0002 are connected in parallel across the 120-V power line. Using Kirchhoff’s current law, determine the total current It.\n", + "\n", + "# Given data\n", + "\n", + "R1 = 20.# # Resistor 1=20 Ohms\n", + "R2 = 40.# # Resistor 2=40 Ohms\n", + "R3 = 60.# # Resistor 3=60 Ohms\n", + "Va = 120.# # Applied Voltage=120 Volts\n", + "\n", + "I1 = Va/R1#\n", + "I2 = Va/R2#\n", + "I3 = Va/R3#\n", + "\n", + "It = I1+I2+I3\n", + "print 'The Total Current in the Mainline = %0.f Amps'%It" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 5_3 Page No. 151" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current in R2 branch = 5 Amps\n" + ] + } + ], + "source": [ + "# Two branches R1 and R2 across the 120-V power line draw a total line current It of 15 A. The R1 branch takes 10 A. How much is the current I2 in the R2 branch?\n", + "\n", + "# Given data\n", + "\n", + "I1 = 10# # Current in R1 branch=10 Amps\n", + "It = 15# # Total Current=15 Amps\n", + "\n", + "I2 = It-I1#\n", + "print 'The Current in R2 branch = %0.f Amps'%I2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 5_4 Page No. 152" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Total Current = 0.6008 Amps\n", + "i.e 600.8 mAmps\n" + ] + } + ], + "source": [ + "# Three parallel branch currents are 0.1 A, 500 mA, and 800 \u0002A. Using Kirchhoff’s current law, calculate It.\n", + "\n", + "\n", + "# Given data\n", + "\n", + "I1 = 0.1# # Branch Current 1=0.1 Amps\n", + "I2 = 0.5# # Branch Current 2=500m Amps\n", + "I3 = 800*10**-6# # Branch Current 3=800u Amps\n", + "\n", + "It = I1+I2+I3#\n", + "print 'The Total Current = %0.4f Amps'%It\n", + "print 'i.e 600.8 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 5_5 Page No. 153" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Equivalent Resistance Req = 9 Ohms\n" + ] + } + ], + "source": [ + "# Two branches, each with a 5-A current, are connected across a 90-V source. How much is the equivalent resistance Req?\n", + "\n", + "# Given data\n", + "\n", + "I1 = 5# # Branch Current 1=5 Amps\n", + "I2 = 5# # Branch Current 2=5 Amps\n", + "Va = 90# # Applied Voltage=90 Volts\n", + "\n", + "It = I1+I2#\n", + "Req = Va/It#\n", + "print 'The Equivalent Resistance Req = %0.f Ohms'%Req" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 5_6 Page No. 158" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Rx = 60.00 Ohms\n" + ] + } + ], + "source": [ + "# What Rx in parallel with 40 Ohms\u0002 will provide an Req of 24 Ohms?\n", + "\n", + "# Given data\n", + "\n", + "R = 40.0# # Resistance=40 Ohms\n", + "Req = 24.0# # Equivqlent Resistance=24 Ohms\n", + "\n", + "Rx = (R*Req)/(R-Req)#\n", + "print 'The Value of Rx = %0.2f Ohms'%Rx" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 5_7 Page No. 158" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of R = 50000 Ohms\n", + "i.e 50 kohms\n" + ] + } + ], + "source": [ + "# What R in parallel with 50 kOhms will provide an Req of 25 kOhms\u0002\n", + "\n", + "# Given data\n", + "\n", + "R1 = 50*10**3# # R1=50k Ohms\n", + "Req = 25*10**3# # Req=25k Ohms\n", + "\n", + "R = (R1*Req)/(R1-Req)#\n", + "print 'The value of R = %0.f Ohms'%R\n", + "print 'i.e 50 kohms'" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter6.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter6.ipynb new file mode 100644 index 00000000..5ad41975 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter6.ipynb @@ -0,0 +1,141 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6 : Series-Parallel Crcuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 6_1 Page No. 194" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Unknown Resistance Rx = 564.20 ohms\n" + ] + } + ], + "source": [ + "# The Current in M1 reads 0 \u0002A with the standard resistor RS adjusted to 5642 Ohms\u0002. What is the value of the unknown resistor Rx?\n", + "\n", + "# Given data\n", + "\n", + "Rs = 5642.# # Standard Resistor=5642 Ohms\n", + "R1 = 1.*10**3# # Resistor 1=1k Ohms\n", + "R2 = 10.*10**3# # Resistor 2=10k Ohms\n", + "\n", + "Rx = Rs*(R1/R2)#\n", + "print 'The Unknown Resistance Rx = %0.2f ohms'%Rx" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 6_2 Page No. 195" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Unknown Resistance Rx(max) = 999.90 Ohms\n" + ] + } + ], + "source": [ + "# what is the maximum unknown resistance Rx that can be measured for the ratio arm values shown?\n", + "\n", + "# Given data\n", + "\n", + "Rsmax = 9999.# # Standard Resistor(max)=9999 Ohms\n", + "R1 = 1.*10**3# # Resistor 1=1k Ohms\n", + "R2 = 10.*10**3# # Resistor 2=10k Ohms\n", + "\n", + "Rxmax = Rsmax*(R1/R2)#\n", + "print 'The Unknown Resistance Rx(max) = %0.2f Ohms'%Rxmax" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 6_3 Page No. 197" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistance R(AB) = 100.00 ohms\n" + ] + } + ], + "source": [ + "# Assume that the series-parallel circuit in Fig. 6–15a has failed. A technician troubleshooting the circuit has measured the following voltages: V1 = 10.8 V# VAB = 9 V# V4 = 16.2 V. These voltage readings are shown in Fig. 6–15b. Based on the voltmeter readings shown, which component is defective and what type of defect does it have?\n", + "\n", + "# Given data\n", + "\n", + "V1 = 10.8# # Voltage at R1=10.8 Volts\n", + "Vab = 9.0# # Voltage at point (AB)=9 Volts\n", + "V4 = 16.2# # Voltage at R4=16.2 Volts\n", + "R1 = 120.0# # Resistor 1=120 Ohms\n", + "\n", + "\n", + "It = V1/R1#\n", + "Rab = Vab/It#\n", + "print 'The Resistance R(AB) = %0.2f ohms'%Rab\n", + "\n" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter7.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter7.ipynb new file mode 100644 index 00000000..b7757e4c --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter7.ipynb @@ -0,0 +1,70 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7 : Voltage Dividers and Current Dividers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 7_1 Page No. 199" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Drop across each Resistor = 60.00 Volts\n" + ] + } + ], + "source": [ + "# Three 50 Ohms resistors R1, R2 and R3 are in series across an applied voltage of 180 V. How much is the IR voltage drop across each resistor?\n", + "\n", + "# Given data\n", + "\n", + "R1 = 50*10**3# # Resistor 1=50k Ohms\n", + "R2 = 50*10**3# # Resistor 2=50k Ohms\n", + "R3 = 50.*10**3# # Resistor 3=50k Ohms\n", + "Vt = 180.# # Applied Voltage=180 Volts\n", + "\n", + "R = R1 # R = R1 = R2 = R3\n", + "Rt = R1+R2+R3#\n", + "V = Vt*(R/Rt)#\n", + "print 'The Voltage Drop across each Resistor = %0.2f Volts'%V" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter8.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter8.ipynb new file mode 100644 index 00000000..95d4e266 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter8.ipynb @@ -0,0 +1,106 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8 : Analog & Digital Multimeters" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 8_1 Page No. 251" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current through Shunt at Full Scale Deflection = 0.000950 Amps\n", + "i.e 950 uAmps\n" + ] + } + ], + "source": [ + "# A shunt extends the range of a 50-u\u0002A meter movement to 1 mA. How much is the current through the shunt at full-scale deflection?\n", + "\n", + "# Given data\n", + "\n", + "It = 1*10**-3# # Total Current=1 mAmps\n", + "Im = 50*10**-6# # Current (cause of meter movement)=50 uAmps\n", + "\n", + "Is = It-Im#\n", + "print 'The Current through Shunt at Full Scale Deflection = %0.6f Amps'%Is\n", + "print 'i.e 950 uAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 8_2 Page No. 252" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Shunt Resistance Rs needed to extend the range to 500 uA = 111.11 Ohms\n" + ] + } + ], + "source": [ + "# A 50 u\u0002A meter movement has an Rm of 1000 Ohms\u0002. What Rs is needed to extend the range to 500 uA?\n", + "\n", + "# Given data\n", + "\n", + "It = 500*10**-6# # Total Current=500u Amps\n", + "Im = 50*10**-6# # Current (cause of meter movement)=50 uAmps\n", + "rm = 1000# # Resistance of moving coil=1000 Ohms\n", + "\n", + "Is = It-Im#\n", + "Vm = Im*rm#\n", + "\n", + "Rs = Vm/Is#\n", + "print 'The Shunt Resistance Rs needed to extend the range to 500 uA = %0.2f Ohms'%Rs\n" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter9.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter9.ipynb new file mode 100644 index 00000000..c836f96b --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter9.ipynb @@ -0,0 +1,126 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9 : Kirchoff's Laws" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 9_1 Page No. 268" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Branch 3 Current I3 = 4.5 Amps\n" + ] + } + ], + "source": [ + "# Apply Kirchhoff’s current law to solve for the unknown current, I3.\n", + "\n", + "# Given data\n", + "\n", + "I1 = 2.5# # Branch 1 Current=2.5 Amps\n", + "I2 = 8# # Branch 2 Current=8 Amps\n", + "I4 = 6# # Branch 3 Current=6 Amps\n", + "I5 = 9# # Branch 4 Current=9 Amps\n", + "\n", + "# I1+I2+I3-I4-I5 = 0 Sum of all currents at node is ZERO\n", + "# I1+I2+I3 = I4+I5 Total Incomming Current = Total Outgoing Current\n", + "\n", + "I3 = I4+I5-I1-I2#\n", + "print 'The Branch 3 Current I3 = %0.1f Amps'%I3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 9_2 Page No. 275" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage V(AG) = 7.20 Volts\n", + "The Voltage V(BG) = -1.80 Volts\n" + ] + } + ], + "source": [ + "# Apply Kirchhoff’s voltage law to solve for the voltages V(AG) & V(BG).\n", + "\n", + "# Given data\n", + "\n", + "V1 = 18.# # Source Voltage 1=18 Volts\n", + "V2 = 18.# # Source Voltage 2=18 Volts\n", + "R1 = 120.# # Resistor 10=120 Ohms\n", + "R2 = 100.# # Resistor 2=100 Ohms\n", + "R3 = 180.# # Resistor 3=180 Ohms\n", + "\n", + "Vt = V1+V2#\n", + "Rt = R1+R2+R3#\n", + "\n", + "I = Vt/Rt#\n", + "\n", + "VR1 = I*R1#\n", + "VR2 = I*R2#\n", + "VR3 = I*R3#\n", + "\n", + "# V1+V2-VR1-VR2-VR3=0 Sum of all Voltages in loop is ZERO\n", + "# V1+V2 = VR1+VR2+VR3 Total Applied Voltage = Total Dropped Voltage in Resistors\n", + "\n", + "Vt = VR1+VR2+VR3#\n", + "\n", + "VAG = VR2+VR3-V2#\n", + "print 'The Voltage V(AG) = %0.2f Volts'%VAG\n", + "\n", + "VBG = V1-VR1-VR2#\n", + "print 'The Voltage V(BG) = %0.2f Volts'%VBG" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/ChapterI.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/ChapterI.ipynb new file mode 100644 index 00000000..3f57bc74 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/ChapterI.ipynb @@ -0,0 +1,347 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter - I : Introduction to powers of 10" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. I_9 Page No. 9" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The addition of 170*10**3 and 23*10**4 =4.00E+05\n" + ] + } + ], + "source": [ + "# Add 170*10**3 and 23*10**4. Express the final answer in scientific notation.\n", + "\n", + "# Given data\n", + "\n", + "A = 170*10**3# # Variable 1\n", + "B = 23*10**4# # Variable 2\n", + "\n", + "C = A+B#\n", + "print 'The addition of 170*10**3 and 23*10**4 =%0.2E'%C" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. I_10 Page No. 9" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The substraction of 250*10**3 and 1.5*10**6 =1.25E+06\n" + ] + } + ], + "source": [ + "# Substract 250*10**3 and 1.5*10**6. Express the final answer in scientific notation.\n", + "\n", + "# Given data\n", + "\n", + "A = 1.5*10**6# # Variable 1\n", + "B = 250*10**3# # Variable 2\n", + "\n", + "C = A-B#\n", + "print 'The substraction of 250*10**3 and 1.5*10**6 =%0.2E'%C" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. I_11 Page No. 10" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The multiplication of 3*10**6 by 150*10**2 =4.50E+10\n" + ] + } + ], + "source": [ + "# Multiply 3*10**6 by 150*10**2. Express the final answer in scientific notation.\n", + "\n", + "# Given data\n", + "\n", + "A = 3*10**6# # Variable 1\n", + "B = 150*10**2# # Variable 2\n", + "\n", + "C = A*B#\n", + "print 'The multiplication of 3*10**6 by 150*10**2 =%0.2E'%C" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. I_12 Page No. 10" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The division of 5.0*10**7 by 2.0*10**4 =2.50E+03\n" + ] + } + ], + "source": [ + "# Divide 5.0*10**7 by 2.0*10**4. Express the final answer in scientific notation.\n", + "\n", + "# Given data\n", + "\n", + "A = 5.0*10**7# # Variable 1\n", + "B = 2.0*10**4# # Variable 2\n", + "\n", + "C = A/B#\n", + "print 'The division of 5.0*10**7 by 2.0*10**4 =%0.2E'%C" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. I_13 Page No. 10" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The reciprocal of 10**5 =1.00e-05\n", + "i.e 10**-5\n", + "The reciprocal of 10-3 = 1.00e+03\n", + "i.e 10**3\n" + ] + } + ], + "source": [ + "# Find the reciprocals for the following powers of 10: (a) 10**5 (b) 10**-3.\n", + "\n", + "# Given data\n", + "\n", + "A = 10**5# # Variable 1\n", + "B = 10**-3# # Variable 2\n", + "\n", + "C = 1./A#\n", + "print 'The reciprocal of 10**5 =%0.2e'%C\n", + "print 'i.e 10**-5'\n", + "\n", + "D = 1./B#\n", + "print 'The reciprocal of 10-3 = %0.2e'%D\n", + "print 'i.e 10**3'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. I_14 Page No. 11" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The square of 3.0*10**4 =9.00E+08\n" + ] + } + ], + "source": [ + "# Square 3.0*10**4. Express the answer in scientific notation.\n", + "\n", + "# Given data\n", + "\n", + "A = 3.0*10**4# # Variable 1\n", + "\n", + "B = A*A#\n", + "print 'The square of 3.0*10**4 =%0.2E'%B" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. I_15 Page No. 11" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The squareroot of 4*10**6 = 2.00E+03\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "# Find the squareroot of 4*10**6. Express the answer in scientific notation.\n", + "\n", + "# Given data\n", + "\n", + "A = 4*10**6# # Variable 1\n", + "\n", + "B = sqrt(A)#\n", + "print 'The squareroot of 4*10**6 = %0.2E'%B" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. I_16 Page No. 12" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The squareroot of 90*10**5 =3.00E+03\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "# Find the squareroot of 90*10**5. Express the answer in scientific notation.\n", + "\n", + "# Given data\n", + "\n", + "A = 90*10**5# # Variable 1\n", + "\n", + "B = sqrt(A)#\n", + "print 'The squareroot of 90*10**5 =%0.2E'% B" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. I_17 Page No. 12" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The multiplication of 40*10**-3 by 5*10**6 =200000.00\n", + "i.e 200.000*10**03 OR 200E03\n" + ] + } + ], + "source": [ + "# Show the keystrokes for multiplying 40*10**-3 by 5*10**6.\n", + "\n", + "# Given data\n", + "\n", + "A = 40*10**-3# # Variable 1\n", + "B = 5*10**6# # Variable 2\n", + "\n", + "C = A*B#\n", + "print 'The multiplication of 40*10**-3 by 5*10**6 =%0.2f'%C\n", + "print 'i.e 200.000*10**03 OR 200E03'" + ] + } + ], + "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 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/screenshots/DCLoadLine.png b/Grob's_Basic_Electronics_by_M._E._Schultz/screenshots/DCLoadLine.png Binary files differnew file mode 100644 index 00000000..214d7c48 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/screenshots/DCLoadLine.png diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/screenshots/EmitterCurrent.png b/Grob's_Basic_Electronics_by_M._E._Schultz/screenshots/EmitterCurrent.png Binary files differnew file mode 100644 index 00000000..0f981bd1 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/screenshots/EmitterCurrent.png diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/screenshots/freqNperiod.png b/Grob's_Basic_Electronics_by_M._E._Schultz/screenshots/freqNperiod.png Binary files differnew file mode 100644 index 00000000..e4989287 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/screenshots/freqNperiod.png |
