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| author | Trupti Kini | 2016-02-12 23:30:12 +0600 |
|---|---|---|
| committer | Trupti Kini | 2016-02-12 23:30:12 +0600 |
| commit | 31971be1cad3895af030bb3909a791a0b0ad6b4f (patch) | |
| tree | 2b15a81deccde7acdac20f934675fccd677812a2 | |
| parent | c00ea066e3f752b36e9afb0ced7084b71814c89e (diff) | |
| download | Python-Textbook-Companions-31971be1cad3895af030bb3909a791a0b0ad6b4f.tar.gz Python-Textbook-Companions-31971be1cad3895af030bb3909a791a0b0ad6b4f.tar.bz2 Python-Textbook-Companions-31971be1cad3895af030bb3909a791a0b0ad6b4f.zip | |
Added(A)/Deleted(D) following books
A Electronics_Engineering_by_P._Raja/chapter_1_2.ipynb
A Electronics_Engineering_by_P._Raja/chapter_2_2.ipynb
A Electronics_Engineering_by_P._Raja/chapter_3_2.ipynb
A Electronics_Engineering_by_P._Raja/chapter_4_2.ipynb
A Electronics_Engineering_by_P._Raja/chapter_5_2.ipynb
A Electronics_Engineering_by_P._Raja/chapter_6_2.ipynb
A Electronics_Engineering_by_P._Raja/chapter_7_2.ipynb
A Electronics_Engineering_by_P._Raja/chapter_8_2.ipynb
A Electronics_Engineering_by_P._Raja/chapter_9_2.ipynb
A Electronics_Engineering_by_P._Raja/screenshots/7_2.png
A Electronics_Engineering_by_P._Raja/screenshots/snap-3_2.png
A Electronics_Engineering_by_P._Raja/screenshots/snap-6_2.png
A Machine_Design_by_U.C._Jindal/Ch10.ipynb
A Machine_Design_by_U.C._Jindal/Ch11.ipynb
A Machine_Design_by_U.C._Jindal/Ch12.ipynb
A Machine_Design_by_U.C._Jindal/Ch13.ipynb
A Machine_Design_by_U.C._Jindal/Ch14.ipynb
A Machine_Design_by_U.C._Jindal/Ch15.ipynb
A Machine_Design_by_U.C._Jindal/Ch16.ipynb
A Machine_Design_by_U.C._Jindal/Ch17.ipynb
A Machine_Design_by_U.C._Jindal/Ch18.ipynb
A Machine_Design_by_U.C._Jindal/Ch19.ipynb
A Machine_Design_by_U.C._Jindal/Ch20.ipynb
A Machine_Design_by_U.C._Jindal/Ch21.ipynb
A Machine_Design_by_U.C._Jindal/Ch22.ipynb
A Machine_Design_by_U.C._Jindal/Ch23.ipynb
A Machine_Design_by_U.C._Jindal/Ch24.ipynb
A Machine_Design_by_U.C._Jindal/Ch25.ipynb
A Machine_Design_by_U.C._Jindal/Ch26.ipynb
A Machine_Design_by_U.C._Jindal/Ch27.ipynb
A Machine_Design_by_U.C._Jindal/Ch28.ipynb
A Machine_Design_by_U.C._Jindal/Ch29.ipynb
A Machine_Design_by_U.C._Jindal/Ch3.ipynb
A Machine_Design_by_U.C._Jindal/Ch30.ipynb
A Machine_Design_by_U.C._Jindal/Ch31.ipynb
A Machine_Design_by_U.C._Jindal/Ch4.ipynb
A Machine_Design_by_U.C._Jindal/Ch5.ipynb
A Machine_Design_by_U.C._Jindal/Ch6.ipynb
A Machine_Design_by_U.C._Jindal/Ch7.ipynb
A Machine_Design_by_U.C._Jindal/Ch8.ipynb
A Machine_Design_by_U.C._Jindal/Ch9.ipynb
A Machine_Design_by_U.C._Jindal/screenshots/Chapter-3stressgraph.png
A Machine_Design_by_U.C._Jindal/screenshots/Chapter-_8AdditionalLoad.png
A Machine_Design_by_U.C._Jindal/screenshots/Chapter11_-_strengthofrevet.png
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter02_2.ipynb
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter03_2.ipynb
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter04_2.ipynb
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter05_2.ipynb
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter06_2.ipynb
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter07_2.ipynb
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter08_2.ipynb
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter09_2.ipynb
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter10_2.ipynb
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter11_2.ipynb
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/screenshots/Screenshot_3.1.py.PNG
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/screenshots/Screenshot_6.1.py.PNG
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/screenshots/Screenshot_6.2.py.PNG
A sample_notebooks/PriyankaSaini/Chapter3.ipynb
A sample_notebooks/ebbygeorge/Chapter01.ipynb
59 files changed, 25694 insertions, 0 deletions
diff --git a/Electronics_Engineering_by_P._Raja/chapter_1_2.ipynb b/Electronics_Engineering_by_P._Raja/chapter_1_2.ipynb new file mode 100644 index 00000000..18461132 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_1_2.ipynb @@ -0,0 +1,1165 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter - 1 : Introduction To Electronics Diode Fundamentals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.1\n", + ": Page No 27 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# Given data\n", + "T1 = 25 # in degree C\n", + "T2 = 100 # in degree C\n", + "del_T = T2-T1 # in degree C\n", + "V= 0.7 # barrier potential t 25\u00b0C in V\n", + "del_V = -(2)*del_T # in mV\n", + "del_V= del_V*10**-3 # in V\n", + "V_B = V- abs(del_V) # in V\n", + "print \"(i) When the junction temperature is 100 \u00b0C, the barrier potential of a silicon diode = %0.2f V\" %V_B\n", + "T2 = 0 # in degree C\n", + "del_T = T2-T1 # in degree C\n", + "del_V = -(2)*del_T # in mV\n", + "del_V= del_V*10**-3 #in V\n", + "V_B = V+del_V # in V\n", + "print \"(ii) When the junction temperature is 0 \u00b0C, the barrier potential of a silicon diode = %0.2f V\" %V_B" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) When the junction temperature is 100 \u00b0C, the barrier potential of a silicon diode = 0.55 V\n", + "(ii) When the junction temperature is 0 \u00b0C, the barrier potential of a silicon diode = 0.75 V\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.2\n", + ": Page No 30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "T1 = 25 # in degree C\n", + "T2 = 100 # in degree C\n", + "del_T = T2-T1 # in degree C\n", + "I_S = (2)**7 *5 # in nA\n", + "I_S = (1.07)**5*I_S # in nA\n", + "print \"The saturation current at 100 degree C = %0.f nA\" %round(I_S)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The saturation current at 100 degree C = 898 nA\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.3\n", + ": Page No 33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_L = 10 # in V\n", + "R_L = 1*10**3 # in \u03a9\n", + "I_L = V_L/R_L # in A\n", + "I_L = I_L*10**3 # mA\n", + "print \"The load voltage = %0.f volts\" %V_L\n", + "print \"The load current = %0.f mA\" %I_L" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The load voltage = 10 volts\n", + "The load current = 10 mA\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.4\n", + ": Page No 37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "v1 = 10 # in V\n", + "v2 = 0.7 # in V\n", + "V_L = v1-v2 # in V\n", + "print \"The load voltage = %0.1f V\" %V_L\n", + "R_L = 1*10**3 # in \u03a9\n", + "I_L = V_L/R_L # in A\n", + "print \"The load current = %0.1f mA\" %(I_L*10**3)\n", + "P_D = v2*I_L # in watt\n", + "print \"The diode Power = %0.2f mW\" %(P_D*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The load voltage = 9.3 V\n", + "The load current = 9.3 mA\n", + "The diode Power = 6.51 mW\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.5\n", + ": Page No 39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_L1 = 1*10**3 # in ohm\n", + "R_L2 = 0.23 # in ohm\n", + "R_T = R_L1+R_L2 # in ohm\n", + "v1 = 10 # in V\n", + "v2 = 0.7 # in V\n", + "V_T = v1-v2 # in V\n", + "I_L = V_T/R_T # in A\n", + "print \"The load current = %0.2f mA\" %(I_L*10**3)\n", + "V_L = I_L*R_L1 # in V\n", + "print \"The load voltage = %0.1f V\" %V_L" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The load current = 9.30 mA\n", + "The load voltage = 9.3 V\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.6\n", + ": Page No 45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_o = 0.7 # in V\n", + "print \"The value of V_o = %0.1f V\" %V_o\n", + "E = 10 # in V\n", + "V_D = V_o # in V\n", + "R = 330 # in ohm\n", + "I1 = (E - V_D)/R # in A\n", + "I1 = I1*10**3 # in mA\n", + "print \"The value of I1 = %0.2f mA\" %I1\n", + "I_D1 = I1/2 # in mA\n", + "print \"The value of I_D1 = %0.2f mA\" %I_D1\n", + "I_D2 = I_D1 # in mA\n", + "print \"The value of I_D2 = %0.2f mA\" %I_D2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of V_o = 0.7 V\n", + "The value of I1 = 28.18 mA\n", + "The value of I_D1 = 14.09 mA\n", + "The value of I_D2 = 14.09 mA\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.7\n", + ": Page No 46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_i = 12 # in V\n", + "V_D1 = 0.7 # in V\n", + "V_D2 = 0.3 # in V\n", + "R = 5.6*10**3 # in ohm\n", + "V_o = V_i - V_D1 - V_D2 # in V\n", + "print \"The value of Vo voltage = %0.f V\" %V_o\n", + "I_D = V_o/R # in A\n", + "I_D = I_D*10**3 # in mA\n", + "print \"The value of I_D = %0.2f mA\" %I_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of Vo voltage = 11 V\n", + "The value of I_D = 1.96 mA\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.8\n", + ": Page No 47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 24 # in V\n", + "V2 = 6 # in V\n", + "V_D1 = 0.7 # in V\n", + "R = 3*10**3 # in ohm\n", + "I = (V1 - V2 - V_D1)/R # in A\n", + "I = I * 10**3 # in mA\n", + "print \"The current = %0.2f mA\" %I" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current = 5.77 mA\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.9\n", + ": Page No 48" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "r= 20 # in \u03a9\n", + "R_B= 15 # in \u03a9\n", + "V_K1= 0.2 # in V\n", + "V_K2= 0.6 # in V\n", + "V= 100 # in V\n", + "R1= 10*10**3 # in \u03a9\n", + "# Vo= V_K1+r*I1 = V_K2+R_B*I2\n", + "# Resulting current I= I1+I2 or\n", + "# (V-Vo)/(R1) = (Vo-V_K1)/r + (Vo-V_K2)/R_B\n", + "Vo= (r*R_B*V+R1*R_B*V_K1+R1*r*V_K2)/(R1*R_B+R1*r+r*R_B) # in V\n", + "I1= (Vo-V_K1)/r # in A\n", + "I2= (V_K2-Vo)/R_B # in A\n", + "print \"The value of I1 = %0.2f mA\" %(I1*10**3)\n", + "print \"The value of I2 = %0.2f mA\" %(I2*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I1 = 15.69 mA\n", + "The value of I2 = 5.74 mA\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.10\n", + ": Page No 49" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_D = 10 # in mA\n", + "I_D = I_D * 10**-3 # in A\n", + "V_D = 0.5 # in V\n", + "r_F1 = V_D/I_D # in ohm\n", + "print \"The value of r_F1 = %0.f ohm\" %r_F1\n", + "I_D = 20 # in mA\n", + "I_D = I_D * 10**-3 # in A\n", + "V_D = 0.8 # in V\n", + "r_F2 = V_D/I_D # in ohm\n", + "print \"The value of r_F2 = %0.f ohm\" %r_F2\n", + "I_D = -1 # in \u00b5A\n", + "I_D = I_D * 10**-6 # in A\n", + "V_D = -10 # in V \n", + "r_R = V_D/I_D # in ohm\n", + "print \"The value of r_R = %0.f Mohm\" %(r_R*10**-6)\n", + "\n", + "# Note: There is calculation error to evaluate the value of r_F1. So the asnwer in the book is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of r_F1 = 50 ohm\n", + "The value of r_F2 = 40 ohm\n", + "The value of r_R = 10 Mohm\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.11\n", + ": Page No 49" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R= 5.6*10**3 # in \u03a9\n", + "I_D = 0 # in A\n", + "V_D = 0 # in V\n", + "E= 12 # in V\n", + "Vo= I_D*R # in V\n", + "print \"The value of I_D = %0.f A\" %I_D\n", + "print \"The value of Vo = %0.f V\" %Vo\n", + "V_D1 = 0 # in V\n", + "V_D2 = E -V_D1 - Vo # in V\n", + "print \"The value of V_D2 = %0.f V\" %V_D2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_D = 0 A\n", + "The value of Vo = 0 V\n", + "The value of V_D2 = 12 V\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.12\n", + ": Page No 50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "E = 20 # in V\n", + "V_D1 = 0.7 # in V\n", + "V_D2 = 0.7 # in V\n", + "V2 = E - V_D1 - V_D2 # in V\n", + "R1 = 3.3*10**3 # in ohm\n", + "R2 = 5.6*10**3 # in ohm\n", + "I2 = V2/R2 # in A\n", + "I2 = I2*10**3 # in mA\n", + "print \"The current through resistor R2 = %0.2f mA\" %I2\n", + "I1 = V_D2/R1 \n", + "I1 = I1 * 10**3 # in mA\n", + "print \"The current through resistor R1 = %0.2f mA\" %I1\n", + "I_D2 = I2-I1 # in mA\n", + "print \"The current through diode D2 = %0.2f mA\" %I_D2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current through resistor R2 = 3.32 mA\n", + "The current through resistor R1 = 0.21 mA\n", + "The current through diode D2 = 3.11 mA\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.13\n", + ": Page No 51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 12 # in V\n", + "V2 = 0.3 # in V\n", + "V_o = V1-V2 # in V\n", + "print \"The output voltage = %0.1f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 11.7 V\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14\n", + ": Page No 52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "print \"Part (a) Analysis using approximate diode model\"\n", + "V_D = 0.7 # in V\n", + "print \"The value of V_D = %0.1f V\" %V_D\n", + "E = 30 # in V\n", + "V_R = E-V_D # in V\n", + "print \"The value of V_R = %0.1f V\" %V_R\n", + "R = 2.2 * 10**3 # in ohm\n", + "I_D = V_R/R \n", + "I_D = I_D * 10**3 # in mA\n", + "print \"The value of I_D = %0.2f mA\" %I_D\n", + "print \"Part (b) Analysis using ideal diode model\"\n", + "V_D = 0 # in V\n", + "print \"The value of V_D = %0.f V\" %V_D\n", + "V_R = E # in V\n", + "print \"The value of V_R = %0.f V\" %V_R\n", + "I_D = V_R/R \n", + "I_D = I_D * 10**3 # in mA\n", + "print \"The value of I_D = %0.2f mA\" %I_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (a) Analysis using approximate diode model\n", + "The value of V_D = 0.7 V\n", + "The value of V_R = 29.3 V\n", + "The value of I_D = 13.32 mA\n", + "Part (b) Analysis using ideal diode model\n", + "The value of V_D = 0 V\n", + "The value of V_R = 30 V\n", + "The value of I_D = 13.64 mA\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.15\n", + ": Page No 54" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 20 # in V\n", + "V2 = 0.7 # in V\n", + "V = V1-V2 # in V\n", + "R = 20 # in ohm\n", + "I = V/R # in A\n", + "print \"The current through resistance = %0.3f A\" %I" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current through resistance = 0.965 A\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.16\n", + ": Page No 54" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R1= 2 # in k\u03a9\n", + "R2= 2 # in k\u03a9\n", + "V=19 # in V\n", + "V_o = (V*R1)/(R1+R2) # in V\n", + "print \"The output voltage = %0.1f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 9.5 V\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.17\n", + ": Page No 55" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 0.7 # in V\n", + "V2 = 5 # in V\n", + "V_o = V1-V2 # in V\n", + "R = 2.2*10**3 # in ohm\n", + "I_D = -V_o/R \n", + "I_D = I_D * 10**3 # in mA\n", + "print \"The output voltage = %0.1f volts\" %V_o\n", + "print \"The current through diode = %0.2f mA\" %I_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = -4.3 volts\n", + "The current through diode = 1.95 mA\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.18\n", + ": Page No 56" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_gamma = 0.7 # in V\n", + "R1 = 5*10**3 # in ohm\n", + "R2 = 10*10**3 # in ohm\n", + "V=5 # in V\n", + "print \"Part (a)\"\n", + "I_R2 = (V-V_gamma-(-V))/(R1+R2) # in A\n", + "I_D2 = I_R2 # in A\n", + "print \"The value of I_D1 and I_D2 = %0.2f mA\" %(I_D2*10**3)\n", + "V_o = V - (I_D2 * R1) # in V\n", + "print \"The value of Vo = %0.1f V\" %V_o\n", + "V_A = V_o - V_gamma # in V\n", + "print \"The value of V_A = %0.1f V\" %V_A\n", + "print \"Part (b)\"\n", + "V_I = 4 # in V\n", + "V_A= V_I-V_gamma # in V\n", + "Vo= V_A+V_gamma # in V\n", + "I_R1= (V-Vo)/R1 # in A\n", + "I_D2= I_R1 # in A\n", + "print \"The value of I_D2 = %0.1f mA\" %(I_D2*10**3)\n", + "I_R2= (V_A-(-V))/R2 # in A\n", + "I_D1= I_R2-I_R1 # in A\n", + "print \"The value of I_D1 = %0.2f mA\" %(I_D1*10**3)\n", + "print \"The value of Vo = %0.f volts\" %Vo" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (a)\n", + "The value of I_D1 and I_D2 = 0.62 mA\n", + "The value of Vo = 1.9 V\n", + "The value of V_A = 1.2 V\n", + "Part (b)\n", + "The value of I_D2 = 0.2 mA\n", + "The value of I_D1 = 0.63 mA\n", + "The value of Vo = 4 volts\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.19\n", + ": Page No 58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_S = 6 # in V\n", + "V_D = 0.7 # in V\n", + "R = 10 # in K ohm\n", + "R = R*10**3 # in ohm\n", + "I_T = (V_S-V_D)/R # in A\n", + "print \"The total current = %0.f \u00b5A\" %(I_T*10**6)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total current = 530 \u00b5A\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.20\n", + ": Page No 58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_S = 5 # in V\n", + "V_D = 0.7 # in V\n", + "R1 = 1.2 * 10**3 # in ohm\n", + "R2 = 2.2 * 10**3 # in ohm\n", + "I_T = (V_S-V_D)/(R1+R2) \n", + "I_T = I_T * 10**3 # in mA\n", + "print \"The total circuit current = %0.2f mA\" %I_T" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total circuit current = 1.26 mA\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.21\n", + ": Page No 59" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_S = 4 # in V\n", + "V_D1 = 0.7 # in V\n", + "V_D2 = 0.7 # in V\n", + "R = 5.1*10**3 # in ohm\n", + "I_T = (V_S-V_D1-V_D2)/R # in A\n", + "print \"The total current = %0.f \u00b5A\" %round(I_T*10**6)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total current = 510 \u00b5A\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.22\n", + ": Page No 59" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_S = 10 # in V\n", + "R1 = 1.5*10**3 # in ohm\n", + "R2 = 1.8*10**3 # in ohm\n", + "I_T = V_S/(R1+R2) # in A\n", + "print \"Using the ideal diode, the total current = %0.2f mA\" %(I_T*10**3)\n", + "V_D1 = 0.7 # in V\n", + "V_D2 = 0.7 # in V\n", + "I_T = (V_S-V_D1-V_D2)/(R1+R2) # in A\n", + "print \"Using the pracitcal diode, the total current = %0.2f mA\" %(I_T*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Using the ideal diode, the total current = 3.03 mA\n", + "Using the pracitcal diode, the total current = 2.61 mA\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.23\n", + ": Page No 60" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_S = 5 # in V\n", + "V2 = 3 # in V\n", + "R = 500 # in ohm\n", + "I_D2 = (V_S-V2)/R # in A\n", + "I_D2 = I_D2 * 10**3 # in mA\n", + "print \"The diode current = %0.f mA\" %I_D2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The diode current = 4 mA\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.24\n", + ": Page No 60" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_S = 2 # in V\n", + "R = 100 # in ohm\n", + "I_D = V_S/R \n", + "I_D = I_D * 10**3 # in mA\n", + "print \"Part (a)\"\n", + "print \"The diode current = %0.f mA\" %I_D\n", + "V_K = 0.7 # in V\n", + "I_D1 = (V_S-V_K)/R \n", + "I_D1 = I_D1*10**3 # in mA\n", + "print \"Part (b)\"\n", + "print \"The diode current = %0.f mA\" %I_D1\n", + "R_f = 30 # in ohm\n", + "I_D2 = (V_S - V_K)/(R+R_f) \n", + "I_D2 = I_D2 * 10**3 # in mA\n", + "print \"Part (c)\"\n", + "print \"The diode current = %0.f mA\" %I_D2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (a)\n", + "The diode current = 20 mA\n", + "Part (b)\n", + "The diode current = 13 mA\n", + "Part (c)\n", + "The diode current = 10 mA\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.25\n", + ": Page No 61" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R1= 1 # in k\u03a9\n", + "R2= 0.47 # in k\u03a9\n", + "V_o1 = 0.7 # in V\n", + "print \"The value of Vo1 = %0.1f V\" %V_o1\n", + "V_o2 = 0.3 # in V\n", + "print \"The value of Vo2 = %0.1f V\" %V_o2\n", + "I1 = (20-V_o1)/R1 # in mA\n", + "I2 = (V_o2-V_o1)/R2 # in mA\n", + "I = I1 + I2 # in mA\n", + "print \"The current = %0.2f mA\" %I" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of Vo1 = 0.7 V\n", + "The value of Vo2 = 0.3 V\n", + "The current = 18.45 mA\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.26\n", + ": Page No 62" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 10 # in V\n", + "V2 = 0.7 # in V\n", + "R1 = 1*10**3 # in ohm\n", + "R2 = 2*10**3 # in ohm\n", + "I = (V1-V2)/(R1+R2) # in A\n", + "V_o = I * R2 # in V\n", + "print \"The output voltage = %0.1f V\" %V_o\n", + "I_D = I/2 # in A\n", + "print \"The diode current = %0.2f mA\" %(I_D*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 6.2 V\n", + "The diode current = 1.55 mA\n" + ] + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.27\n", + ": Page No 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 20 # in V\n", + "V2 = 0.7 # in V\n", + "R = 4.7*10**3 # in ohm\n", + "I = (V1-V2)/R # in A\n", + "I_D = I/2 # in A\n", + "print \"The diode current = %0.2f mA\" %(I_D*10**3)\n", + "V_o = I_D*R # in V\n", + "print \"The output voltage = %0.2f V\" %V_o\n", + "#Note : The answer in the book is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The diode current = 2.05 mA\n", + "The output voltage = 9.65 V\n" + ] + } + ], + "prompt_number": 48 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.28\n", + ": Page No 64" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 15 # in V\n", + "V2 = 0.7 # in V\n", + "V3 = 5 # in V\n", + "R = 2.2 # in K ohm\n", + "I_D = (V1-V2+V3)/R # in mA\n", + "print \"The diode current = %0.2f mA\" %I_D\n", + "V_o = (R * I_D) - V3 # in V\n", + "print \"The output voltage = %0.1f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The diode current = 8.77 mA\n", + "The output voltage = 14.3 V\n" + ] + } + ], + "prompt_number": 50 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.29\n", + ": Page No 65" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 16 # in V\n", + "V2 = 0.7 # in V\n", + "V3 = V2 # in V\n", + "V4 = 12 # in V\n", + "R = 4.7 # in K ohm\n", + "I = (V1-V2-V3-V4)/R # in mA\n", + "print \"The current = %0.3f mA\" %I\n", + "V_o = (I * 10**-3 * R * 10**3) + V4 # in V\n", + "print \"The output voltage = %0.1f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current = 0.553 mA\n", + "The output voltage = 14.6 V\n" + ] + } + ], + "prompt_number": 51 + } + ], + "metadata": {} + } + ] +} diff --git a/Electronics_Engineering_by_P._Raja/chapter_2_2.ipynb b/Electronics_Engineering_by_P._Raja/chapter_2_2.ipynb new file mode 100644 index 00000000..cc8b5565 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_2_2.ipynb @@ -0,0 +1,765 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter - 2 : Diode Applications" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.1\n", + ": Page No 83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from numpy import pi\n", + "from math import sqrt\n", + "# Given data\n", + "R_L = 1000 # in ohm\n", + "N2byN1= 4 \n", + "Vi= '10*sin(omega*t)'\n", + "# V2= N2byN1*V1\n", + "# V2= 40*sin(omega*t)\n", + "Vm= N2byN1*10 # in V\n", + "V_Lav= Vm/pi # in V\n", + "print \"The average load voltage = %0.2f volts\" %V_Lav\n", + "Im= Vm/R_L # in A\n", + "I_dc= Im/pi # in A\n", + "I_av = I_dc # in A\n", + "I_av= I_av*10**3 # in mA\n", + "print \"Average load current = %0.2f mA\" %I_av\n", + "V_Lrms = Vm/2 # in V\n", + "print \"RMS load voltage = %0.f V\" %V_Lrms\n", + "I_rms = V_Lrms/R_L # in A\n", + "I_rms= I_rms*10**3 # in mA\n", + "print \"RMS load current = %0.f mA\" %I_rms\n", + "Eta = I_av**2/I_rms**2*100 # in %\n", + "print \"Efficiency = %0.2f %%\" %Eta\n", + "V2rms= Vm/sqrt(2) # in V\n", + "TUF = ((I_av )**2)/(V2rms*I_rms)*100 # in %\n", + "print \"Transformer utilization factor = %0.2f %%\" %TUF\n", + "Gamma= sqrt(V_Lrms**2-I_av**2)/V_Lav*100 \n", + "print \"Ripple factor = %0.f %%\" %round(Gamma)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The average load voltage = 12.73 volts\n", + "Average load current = 12.73 mA\n", + "RMS load voltage = 20 V\n", + "RMS load current = 20 mA\n", + "Efficiency = 40.53 %\n", + "Transformer utilization factor = 28.66 %\n", + "Ripple factor = 121 %\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.2\n", + ": Page No 96" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_L = 1 # in K ohm\n", + "R_L = R_L * 10**3 # in ohm\n", + "V_m = 15 # in V\n", + "V_i = '15*sin(314*t)' \n", + "I_m= V_m/R_L # in A\n", + "I_dc = I_m/pi # in A\n", + "I_dc = I_dc * 10**3 # in mA\n", + "print \"Average current through the diode = %0.2f mA\" %I_dc\n", + "I_drms = V_m/(2*R_L) \n", + "I_drms = I_drms * 10**3 # in mA\n", + "print \"RMS current = %0.1f mA\" %I_drms\n", + "I_m = V_m/R_L \n", + "I_m = I_m*10**3 # in mA\n", + "print \"Peak diode current = %0.f mA\" %I_m\n", + "PIV = 2*V_m # in V\n", + "print \"Peak inverse voltage = %0.f V\" %PIV" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Average current through the diode = 4.77 mA\n", + "RMS current = 7.5 mA\n", + "Peak diode current = 15 mA\n", + "Peak inverse voltage = 30 V\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.3\n", + ": Page No 101" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R1 = 2.2*10**3 # in ohm\n", + "R2 = 4.7*10**3 # in ohm\n", + "R_AB = (R1*R2)/(R1+R2) # in ohm\n", + "Vi = 20 # in V\n", + "V_o = (Vi * R_AB)/(R_AB+R1) # in V\n", + "PIV= Vi # in volts\n", + "print \"The output voltage = %0.1f V\" %V_o\n", + "print \"Peak inverse voltage = %0.f volts\" %PIV" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 8.1 V\n", + "Peak inverse voltage = 20 volts\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + " Example 2.4.2 \n", + ": Page No 102" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import cos\n", + "# Given data\n", + "V_in = 10 # in V\n", + "R1 = 2000 # in ohm\n", + "R2 = 2000 # in ohm\n", + "V_o = V_in * (R1/(R1+R2) ) # in V\n", + "# Vdc= 5/(T/2)*integrate('sin(omega*t)','t',0,T/2) and omega*T= 2*pi, So\n", + "Vdc= -5/pi*(cos(pi)-cos(0)) # in V\n", + "print \"The value of Vdc = %0.3f volts\" %Vdc\n", + "PIV= V_in/2 # in V\n", + "print \"The PIV value = %0.f volts\" %PIV\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of Vdc = 3.183 volts\n", + "The PIV value = 5 volts\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + " Example 2.4 (again 2.4)\n", + ": Page No 124" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_in = 10 # in V\n", + "R_L = 2000 # in ohm\n", + "R1 = 100 # in ohm\n", + "V_R= 0.7 # in V\n", + "V_o = V_in * ( (R_L)/(R1+R_L) ) # in V\n", + "print \"The peak magnitude of the positive output voltage = %0.2f V\" %V_o \n", + "Vo=-V_R # in V\n", + "print \"The peak magnitude of the negative output voltage = %0.1f V\" %Vo" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The peak magnitude of the positive output voltage = 9.52 V\n", + "The peak magnitude of the negative output voltage = -0.7 V\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.7\n", + ": Page No 141" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V=240 # in V\n", + "R= 1 # in k\u03a9\n", + "R=R*10**3 # in \u03a9\n", + "Vsrms= V/4 # in V\n", + "Vm= sqrt(2)*Vsrms # in V\n", + "V_Ldc= -Vm/pi # in V\n", + "print \"The value of average load voltage = %0.f volts\" %V_Ldc" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of average load voltage = -27 volts\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.8\n", + ": Page No 142" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V = 220 # in V\n", + "f=50 # in Hz\n", + "N2byN1=1/4 \n", + "R_L = 1 # in kohm\n", + "R_L= R_L*10**3 # in ohm\n", + "V_o = 220 # in V\n", + "V_s = N2byN1*V_o # in V\n", + "V_m = sqrt(2) * V_s # in V\n", + "V_Ldc = (2*V_m)/pi # in V\n", + "print \"Average load output voltage = %0.2f V\" %V_Ldc\n", + "P_dc = (V_Ldc)**2/R_L # in W\n", + "print \"DC power delivered to load = %0.2f watt\" %P_dc\n", + "PIV = V_m # in V\n", + "print \"Peak inverse Voltage = %0.2f V\" %PIV\n", + "f_o = 2*f # in Hz\n", + "print \"Output frequency = %0.f Hz\" %f_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Average load output voltage = 49.52 V\n", + "DC power delivered to load = 2.45 watt\n", + "Peak inverse Voltage = 77.78 V\n", + "Output frequency = 100 Hz\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.10\n", + ": Page No 145" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_L = 20 # in ohm\n", + "I_Ldc = 100 # in mA\n", + "R2 = 1 # in ohm\n", + "R_F = 0.5 # in ohm\n", + "I_m = (pi * I_Ldc)/2 # in mA\n", + "V_m = I_m*10**-3*(R2+R_F+R_L) # in V\n", + "V_srms = V_m/sqrt(2) # in V\n", + "print \"RMS value of secondary signal voltage = %0.1f V\" %V_srms\n", + "P_Ldc = (I_Ldc*10**-3)**2*R_L # in Watt\n", + "print \"power delivered to load = %0.1f Watt\" %P_Ldc\n", + "PIV = 2*V_m # in V\n", + "print \"Peal inverse voltage = %0.2f V\" %PIV\n", + "P_ac = (V_m)**2/(2*(R2+R_F+R_L)) # in Watt\n", + "print \"Input power = %0.3f Watt\" %P_ac\n", + "Eta = P_Ldc/P_ac*100 # in %\n", + "print \"Conversion efficiency = %0.2f %%\" %Eta" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "RMS value of secondary signal voltage = 2.4 V\n", + "power delivered to load = 0.2 Watt\n", + "Peal inverse voltage = 6.75 V\n", + "Input power = 0.265 Watt\n", + "Conversion efficiency = 75.40 %\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.16\n", + ": Page No 151" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_dc = 12 # in V\n", + "R_L = 500 # in ohm\n", + "R_F = 25 # in ohm\n", + "I_dc = V_dc/R_L # in A\n", + "V_m = I_dc * pi * (R_L+R_F) # in V\n", + "V_rms = V_m/sqrt(2) # in V\n", + "V = V_rms # in V\n", + "print \"The voltage = %0.f V\" %round(V)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage = 28 V\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.17\n", + ": Page No 152" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_dc = 100 # in V\n", + "V_m = (V_dc*pi)/2 # in V\n", + "PIV = 2*V_m # in V\n", + "print \"Peak inverse voltage for center tapped FWR = %0.2f V\" %PIV\n", + "PIV1 = V_m # in V\n", + "print \"Peak inverse voltage for bridge type FWR = %0.2f V\" %PIV1" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Peak inverse voltage for center tapped FWR = 314.16 V\n", + "Peak inverse voltage for bridge type FWR = 157.08 V\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.19\n", + ": Page No 153" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_Gamma = 0.7 # in V\n", + "R_f = 0 # in ohm\n", + "V_rms = 120 # in V\n", + "V_max = sqrt(2)*V_rms # in V\n", + "R_L = 1 # in K ohm\n", + "R_L = R_L * 10**3 # in ohm\n", + "I_max = (V_max - (2*V_Gamma))/R_L # in A\n", + "I_dc = (2*I_max)/pi # in mA\n", + "V_dc = I_dc * R_L # in V\n", + "print \"The dc voltage available at the load = %0.2f V\" %V_dc\n", + "PIV = V_max # in V\n", + "print \"Peak inverse voltage = %0.1f V\" %PIV\n", + "print \"Maximum current through diode = %0.1f mA\" %(I_max*10**3)\n", + "P_max = V_Gamma * I_max # in W\n", + "print \"Diode power rating = %0.2f mW\" %(P_max*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The dc voltage available at the load = 107.15 V\n", + "Peak inverse voltage = 169.7 V\n", + "Maximum current through diode = 168.3 mA\n", + "Diode power rating = 117.81 mW\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.20\n", + ": Page No 154" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 10 # in V\n", + "V2 = 0.7 # in V\n", + "V3 = V2 # in V\n", + "V = V1-V2-V3 # in V\n", + "R1 = 1 # in ohm\n", + "R2 = 48 # in ohm\n", + "R3 = 1 # in ohm\n", + "R = R1+R2+R3 # in ohm\n", + "I = V/R # in A\n", + "I = I * 10**3 # in mA\n", + "print \"Current = %0.f mA\" %I" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current = 172 mA\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.21\n", + ": Page No 154" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_m = 50 # in V\n", + "r_f = 20 # in ohm\n", + "R_L = 800 # in ohm\n", + "I_m = V_m/(R_L+r_f) # in A\n", + "I_m = I_m * 10**3 # in mA\n", + "print \"The value of Im = %0.f mA\" %round(I_m)\n", + "I_dc = I_m/pi # in mA\n", + "print \"The value of I_dc = %0.1f mA\" %I_dc\n", + "I_rms = I_m/2 # in mA\n", + "print \"The value of Irms = %0.1f mA\" %I_rms\n", + "P_ac = (I_rms * 10**-3)**2 * (r_f + R_L) # in Watt\n", + "print \"AC power input = %0.3f Watt\" %P_ac\n", + "V_dc = I_dc * 10**-3*R_L # in V\n", + "print \"DC output voltage = %0.2f V\" %V_dc\n", + "P_dc = (I_dc * 10**-3)**2 * (r_f + R_L) # in Watt\n", + "Eta = (P_dc/P_ac)*100 # in %\n", + "print \"The efficiency of rectification = %0.1f %%\" %Eta\n", + "\n", + "# Note: There is calculation error to evaluate the ac power input (i.e. P_ac), so the value of Eta is also wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of Im = 61 mA\n", + "The value of I_dc = 19.4 mA\n", + "The value of Irms = 30.5 mA\n", + "AC power input = 0.762 Watt\n", + "DC output voltage = 15.53 V\n", + "The efficiency of rectification = 40.5 %\n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.22\n", + ": Page No 155" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_L = 1 # in K ohm\n", + "R_L = R_L * 10**3 # in ohm\n", + "r_d = 10 # in ohm\n", + "V_m = 220 # in V\n", + "I_m = V_m/(r_d+R_L) # in A\n", + "print \"Peak value of current = %0.2f A\" %I_m\n", + "I_dc = (2*I_m)/pi # in A\n", + "print \"DC value of current = %0.2f A\" %I_dc\n", + "Irms= I_m/sqrt(2) # in A\n", + "r_f = sqrt((Irms/I_dc)**2-1)*100 # in %\n", + "print \"Ripple factor = %0.1f %%\" %r_f\n", + "Eta = (I_dc)**2 * R_L/((Irms)**2*(R_L+r_d))*100 # in %\n", + "print \"Rectification efficiency = %0.1f %%\" %Eta" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Peak value of current = 0.22 A\n", + "DC value of current = 0.14 A\n", + "Ripple factor = 48.3 %\n", + "Rectification efficiency = 80.3 %\n" + ] + } + ], + "prompt_number": 44 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.23\n", + ": Page No 156" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_s = 12 # in V\n", + "R_L = 5.1 # in k ohm\n", + "R_L = R_L * 10**3 # in ohm\n", + "R_s = 1 # in K ohm\n", + "R_s = R_s * 10**3 # in ohm\n", + "V_L = (V_s*R_L)/(R_s+R_L) # in V\n", + "print \"Peak load voltage = %0.f V\" %V_L" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Peak load voltage = 10 V\n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.24\n", + ": Page No 157" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_s = 10 # in V\n", + "R_L = 100 # in ohm\n", + "I_L = V_s/R_L # in A\n", + "print \"The load current during posotive half cycle = %0.1f A\" %I_L\n", + "I_D2 = 0 # in A\n", + "R2 = R_L # in ohm\n", + "I_L1 = -(V_s)/(R2+R_L) # in A\n", + "print \"The load current during negative half cycle = %0.2f A\" %I_L1" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The load current during posotive half cycle = 0.1 A\n", + "The load current during negative half cycle = -0.05 A\n" + ] + } + ], + "prompt_number": 46 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.25\n", + ": Page No 158" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_m = 50 # in V\n", + "V_dc = (2*V_m)/pi # in V\n", + "print \"The dc voltage = %0.2f V\" %V_dc" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The dc voltage = 31.83 V\n" + ] + } + ], + "prompt_number": 47 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.26\n", + ": Page No 159" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R1 = 1.1 # in K ohm\n", + "R2 = 2.2 # in K ohm\n", + "Vi= 170 # in V\n", + "V_o = (Vi*R1)/(R1+R2) # in V\n", + "print \"The output voltage = %0.2f V\" %V_o\n", + "V_dc = (2*V_o)/pi # in V\n", + "print \"The dc voltage = %0.2f V\" %V_dc" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 56.67 V\n", + "The dc voltage = 36.08 V\n" + ] + } + ], + "prompt_number": 48 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Engineering_by_P._Raja/chapter_3_2.ipynb b/Electronics_Engineering_by_P._Raja/chapter_3_2.ipynb new file mode 100644 index 00000000..600220b3 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_3_2.ipynb @@ -0,0 +1,710 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter - 3 : Special-Purpose Diode" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.1\n", + ": Page No 179" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# Given data\n", + "V1 = 18 # in V\n", + "V2 = 10 # in V\n", + "R = 270 # in ohm\n", + "I_S = (V1-V2)/R # in A\n", + "V_L = 10 # in V\n", + "R_L = 1 # in K ohm\n", + "R_L = R_L*1000 # in ohm\n", + "I_L = V_L/R_L # in A\n", + "I_Z = I_S-I_L # in A\n", + "print \"The zener current = %0.1f mA\" %(I_Z*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The zener current = 19.6 mA\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.5\n", + ": Page No 186" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_Z = 2*10**-3 # in A\n", + "R_Z = 8.5 # in V\n", + "del_VL = I_Z*R_Z # in V\n", + "V1 = 10 # in V\n", + "print \"Change in load voltage = %0.2f V\" %del_VL\n", + "V_L = V1 + del_VL # in V\n", + "print \"The load voltage = %0.2f V\" %V_L\n", + "\n", + "# Note: There is calculation error to evaluate the value of del_VL. So the answer in the book is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Change in load voltage = 0.02 V\n", + "The load voltage = 10.02 V\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.6\n", + ": Page No 194" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_L = 1.2 # in K ohm\n", + "R_L = R_L * 10**3 # in ohm\n", + "V_i = 16 # in V\n", + "R_i = 1 # in K ohm\n", + "R_i = R_i * 10**3 # in ohm\n", + "V = (R_L * V_i)/(R_L + R_i) # in V\n", + "V_L = V # in V\n", + "print \"The load voltage = %0.2f V\" %V_L\n", + "V_R = V_i - V_L # in V\n", + "print \"The voltage = %0.2f V\" %V_R\n", + "I_Z = 0 # A\n", + "print \"The zener diode current = %0.f A\" %I_Z\n", + "V_Z = 10 # in V\n", + "P_Z = V_Z*I_Z # in W\n", + "print \"Power dissipation = %0.f watt\" %P_Z" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The load voltage = 8.73 V\n", + "The voltage = 7.27 V\n", + "The zener diode current = 0 A\n", + "Power dissipation = 0 watt\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.7\n", + ": Page No 195" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_Z1 = 20 # in mA\n", + "I_Z1= I_Z1*10**-3 # in A\n", + "I_Z2 = 30 # in mA\n", + "I_Z2= I_Z2*10**-3 # in A\n", + "V_Z1 = 5.6 # in V\n", + "V_Z2 = 5.75 # in V\n", + "del_IZ = I_Z2-I_Z1 # in A\n", + "del_VZ = V_Z2-V_Z1 # in V\n", + "r_Z = del_VZ/del_IZ # in ohm\n", + "print \"Resistance of zener diode = %0.f ohm\" %r_Z" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Resistance of zener diode = 15 ohm\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.8\n", + ": Page No 195" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R = 1 # in K ohm\n", + "R = R * 10**3 # in ohm\n", + "V_Z = 10 # in V\n", + "V_i = 50 # in V\n", + "I_ZM = 32 # in mA\n", + "I_ZM= I_ZM*10**-3 # in A\n", + "R_Lmin = (R*V_Z)/(V_i-V_Z) # in ohm\n", + "print \"The minimum value of R_L = %0.f ohm\" %R_Lmin\n", + "V_R = V_i-V_Z # in V\n", + "I_R = V_R/R # in A\n", + "I_Lmin = I_R-I_ZM # in A\n", + "print \"The minimum value of I_L = %0.f mA\" %(I_Lmin*10**3)\n", + "R_Lmax = V_Z/I_Lmin # in ohm\n", + "print \"The maximum value of R_L = %0.2f kohm\" %(R_Lmax*10**-3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum value of R_L = 250 ohm\n", + "The minimum value of I_L = 8 mA\n", + "The maximum value of R_L = 1.25 kohm\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.9\n", + ": Page No 196" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_Z = 20 # in V\n", + "R_L = 1.2 # in K ohm\n", + "R_L = R_L * 10**3 # in ohm\n", + "R = 220 # in ohm\n", + "I_ZM = 60 # in mA\n", + "I_ZM= I_ZM*10**-3 # in A\n", + "Vi_min = (R_L + R)/R_L*V_Z # in V\n", + "print \"The minimum value of Vi = %0.2f V\" %Vi_min\n", + "V_L= V_Z # in V\n", + "I_L= V_L/R_L # in A\n", + "Vi_max= (I_ZM+I_L)*R+V_Z # in V\n", + "print \"The maximum value of Vi = %0.2f V\" %Vi_max" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum value of Vi = 23.67 V\n", + "The maximum value of Vi = 36.87 V\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.10\n", + ": Page No 197" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 18 # in V\n", + "V2 = 270 # in V\n", + "R = 1 # in K ohm\n", + "R = R*1000 # in ohm\n", + "V = (V1*R)/(V2+R) # in V\n", + "print \"The open circuit voltage = %0.1f volts\" %V\n", + "if V>=10 :\n", + " print \"The zener diode is operating in the breakdown region.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The open circuit voltage = 14.2 volts\n", + "The zener diode is operating in the breakdown region.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.11\n", + ": Page No 198" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_L = 300 # in ohm\n", + "R = 200 # in ohm\n", + "V_i = 20 # in V\n", + "V = (R_L/(R_L+R))*V_i # in V\n", + "print \"The value of V_L = %0.f Volts\" %V\n", + "V_L = 10 # in V\n", + "V_Z= V_L # in V\n", + "I_L = V_L/R_L # A\n", + "print \"The value of I_L = %0.2f mA\" %(I_L*10**3)\n", + "I_R = (V_i-V_L)/R # in A\n", + "print \"The value of I_R = %0.f mA\" %(I_R*10**3)\n", + "I_Z = I_R-I_L # in A\n", + "print \"The value of I_Z = %0.2f mA\" %(I_Z*10**3)\n", + "# Formula V_Z= R_L*V_i/(R_L+R)\n", + "R_L= R*V_Z/(V_i-V_Z) # in ohm\n", + "print \"The value of R_L = %0.f ohm\" %R_L" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of V_L = 12 Volts\n", + "The value of I_L = 33.33 mA\n", + "The value of I_R = 50 mA\n", + "The value of I_Z = 16.67 mA\n", + "The value of R_L = 200 ohm\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.12\n", + ": Page No 199" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_Z = 5 # in V\n", + "I_Zmin = 2 # in mA\n", + "I_Zmin= I_Zmin*10**-3 # in A\n", + "I_Zmax = 20 # in mA\n", + "I_Zmax=I_Zmax*10**-3 # in A\n", + "R_L = 1 # in kohm\n", + "R_L = R_L * 10**3 # in ohm\n", + "I_L = V_Z/R_L # in A\n", + "I = I_L + I_Zmin # in A\n", + "Vin_min = V_Z + (I*R_L) # in V\n", + "print \"The minimum input voltage = %0.f V\" %Vin_min\n", + "I = I_L + I_Zmax # in A\n", + "Vin_max = V_Z + I* R_L # in V\n", + "print \"The maximum input voltage = %0.f V\" %Vin_max" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum input voltage = 12 V\n", + "The maximum input voltage = 30 V\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.13\n", + ": Page No 200" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_in1 = 18 # in V\n", + "V_in2 = 22 # in V\n", + "V_o = 6 # in V\n", + "I_L = 50 # in mA\n", + "I_L= I_L*10**-3 # in A\n", + "I_Zmin = 5 # in mA\n", + "I_Zmin= I_Zmin*10**-3 # in A\n", + "P_Z = 0.5 # in Watt\n", + "V_Z= 6 # in V\n", + "I_Zmax = P_Z/V_Z # in A\n", + "print \"Zener diode current = %0.2f mA\" %(I_Zmax*10**3)\n", + "R_S1 = (V_in2 - V_Z)/(I_L+I_Zmax) # in ohm\n", + "print \"The minimum value of Rs = %0.f ohm\" %R_S1\n", + "R_S2 = (V_in1-V_Z)/(I_L+I_Zmin) # in ohm\n", + "print \"The maximum value of Rs = %0.1f ohm\" %R_S2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Zener diode current = 83.33 mA\n", + "The minimum value of Rs = 120 ohm\n", + "The maximum value of Rs = 218.2 ohm\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.14\n", + ": Page No 201" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_S = 91 # in ohm\n", + "V_Z = 8 # in V\n", + "P_Z = 400 # in mW\n", + "P_Z= P_Z*10**-3 # in W\n", + "R_L = 0.22 # in K ohm\n", + "R_L = R_L * 10**3 # in ohm\n", + "I_L = V_Z/R_L # in A\n", + "I_Z = P_Z/V_Z # in A\n", + "print \"The value of I_Zmax = %0.f mA\" %(I_Z*10**3)\n", + "Vin_min = (V_Z*(R_S+R_L))/R_L # in V\n", + "print \"The minimum input voltage = %0.2f V\" %Vin_min\n", + "I_R = I_L + I_Z # in A\n", + "Vin_max = V_Z + (I_R*R_S) # in V\n", + "print \"The maximum input voltage =%0.2f V\" %Vin_max" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_Zmax = 50 mA\n", + "The minimum input voltage = 11.31 V\n", + "The maximum input voltage =15.86 V\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.15\n", + ": Page No 202" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_L = 12 # in V\n", + "I_Lmin = 0 # in mA\n", + "I_Lmin =I_Lmin *10**-3 # in A\n", + "I_Lmax = 200 # in mA\n", + "I_Lmax =I_Lmax *10**-3 # in A\n", + "I_Zmin = 5 # in mA\n", + "I_Zmin= I_Zmin*10**-3 # in A\n", + "I_Zmax = 200 # in mA\n", + "I_Zmax= I_Zmax*10**-3 # in A\n", + "V_i = 16 # in V\n", + "V_Z = V_L # in V\n", + "print \"The value of V_Z = %0.f V\" %V_Z\n", + "R_Lmin = V_L/I_Lmax # in ohm\n", + "print \"The minimum value of R_L = %0.f ohm\" %R_Lmin\n", + "# R_L2 = V_L/I_Lmin # in ohm\n", + "print \"The maximum value of R_L = infinite\" \n", + "I_Z = I_Zmin+I_Zmax # in A\n", + "print \"The zener current = %0.f mA\" %(I_Z*10**3)\n", + "P_Zmax = V_Z*I_Z # in Watt\n", + "print \"The maximum value of Pz = %0.2f Watt\" %P_Zmax\n", + "R_S = (V_i-V_L)/(I_Zmin+I_Lmax) # in ohm\n", + "print \"The value of R_S = %0.2f ohm\" %R_S" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of V_Z = 12 V\n", + "The minimum value of R_L = 60 ohm\n", + "The maximum value of R_L = infinite\n", + "The zener current = 205 mA\n", + "The maximum value of Pz = 2.46 Watt\n", + "The value of R_S = 19.51 ohm\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.16\n", + ": Page No 203" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_in = 20 # in V\n", + "R_S = 220 # in ohm\n", + "V_Z = 10 # in V\n", + "P_Z = 400 # in mW\n", + "# Part (I)\n", + "R_L = 200 # in ohm\n", + "print \"Part (I) For R_L= 200 \u03a9\"\n", + "V_L = V_Z # in V\n", + "print \"Load voltage = %0.f V\" %V_L\n", + "I_L = V_L/R_L # in A\n", + "print \"Load current = %0.2f A\" %I_L\n", + "I_R = (V_in-V_Z)/R_S # in A\n", + "print \"The current through resistor = %0.3f A\" %I_R\n", + "I_Z = I_R-I_L # in A\n", + "print \"The value of I_Z = %0.2e A\" %I_Z\n", + "# Part (II)\n", + "R_L = 50 # in ohm\n", + "print \"Part (II) For R_L= 50 \u03a9\"\n", + "V_L = V_Z #\n", + "print \"Load voltage = %0.f V\" %V_L\n", + "I_L = V_L/R_L # in A\n", + "print \"Load current = %0.1f A\" %I_L\n", + "I_R = (V_in-V_Z)/R_S # in A\n", + "print \"The current through resistor = %0.3f A\" %I_R\n", + "I_Z = I_R-I_L # in A\n", + "print \"Zener current = %0.3f A\" %I_Z\n", + "print \"For both values of R_L, the current I_R is less than I_L and I_Z is negative.\"\n", + "print \"It shows that given circuit can not work successfully as a voltage regulator\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (I) For R_L= 200 \u03a9\n", + "Load voltage = 10 V\n", + "Load current = 0.05 A\n", + "The current through resistor = 0.045 A\n", + "The value of I_Z = -4.55e-03 A\n", + "Part (II) For R_L= 50 \u03a9\n", + "Load voltage = 10 V\n", + "Load current = 0.2 A\n", + "The current through resistor = 0.045 A\n", + "Zener current = -0.155 A\n", + "For both values of R_L, the current I_R is less than I_L and I_Z is negative.\n", + "It shows that given circuit can not work successfully as a voltage regulator\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.17\n", + ": Page No 204" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_Zmin = 1 # in mA\n", + "I_Zmin=I_Zmin*10**-3 # in A\n", + "I_Zmax = 5 # in mA\n", + "I_Zmax=I_Zmax*10**-3 # in A\n", + "I_Lmin = 0 # in mA\n", + "I_Lmin=I_Lmin*10**-3 # in A\n", + "I_Lmax = 4 # in mA\n", + "I_Lmax=I_Lmax*10**-3 # in A\n", + "R = 5 # in kohm\n", + "R = R * 10**3 # in ohm\n", + "V_Z = 50 # in V\n", + "print \"Part (A)\"\n", + "V_max = (I_Zmax+ I_Lmin)*R+V_Z # in V\n", + "print \"The maximum Voltage = %0.f V\" %V_max\n", + "V_min = (I_Zmin+I_Lmax)*R + V_Z # in V\n", + "print \"The minimum Voltage = %0.f V\" %V_min\n", + "print \"Part (B)\"\n", + "V_L = 50 # in V\n", + "V_in = 75 # in V\n", + "R_L = 15 # in kohm\n", + "R_L= R_L*10**3 # in ohm\n", + "I_L = V_L/R_L # in A\n", + "V_max = (I_Zmax+I_L)*R+V_Z # in V\n", + "print \"The maximum Voltage = %0.f V\" %round(V_max)\n", + "V_min = (I_Zmin+I_L)*R+V_Z # in V\n", + "print \"The minimum Voltage = %0.f V\" %round(V_min)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (A)\n", + "The maximum Voltage = 75 V\n", + "The minimum Voltage = 75 V\n", + "Part (B)\n", + "The maximum Voltage = 92 V\n", + "The minimum Voltage = 72 V\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.18\n", + ": Page No 205" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_S = 7.5 # in V\n", + "V_Z = 5 # in V\n", + "R_S = 4.75 # in ohm\n", + "I_Zmin= 0.05 # in A\n", + "I_Zmax=1.0 # in A\n", + "I_S = (V_S-V_Z)/R_S # in A\n", + "I_Lmax= I_S-I_Zmin # in A\n", + "print \"The maximum value of load current = %0.3f A\" %I_Lmax\n", + "# when\n", + "V_S= 10 # in V\n", + "I_S = (V_S-V_Z)/R_S # in A\n", + "I_Lmin= I_S-I_Zmax # in A\n", + "print \"The minimum value of load current = %0.2f A\" %I_Lmin\n", + "print \"Thus, the range of load current for regulation =\",round(I_Lmin,2),\"<= I_L <=\",round(I_Lmax,3),\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum value of load current = 0.476 A\n", + "The minimum value of load current = 0.05 A\n", + "Thus, the range of load current for regulation = 0.05 <= I_L <= 0.476 A\n" + ] + } + ], + "prompt_number": 32 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Engineering_by_P._Raja/chapter_4_2.ipynb b/Electronics_Engineering_by_P._Raja/chapter_4_2.ipynb new file mode 100644 index 00000000..ad82474d --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_4_2.ipynb @@ -0,0 +1,1826 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter - 4 : Bipolar Junction Transistors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.1\n", + ": Page No 223" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# Given data\n", + "I_C= 0.9 # in mA\n", + "I_E=1 # in mA\n", + "alpha = I_C/I_E \n", + "print \"Current gain = %0.1f\" %alpha\n", + "# Formula I_E= I_B+I_C\n", + "I_B= I_E-I_C # in mA\n", + "print \"The base current = %0.1f mA\" %I_B" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current gain = 0.9\n", + "The base current = 0.1 mA\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2\n", + ": Page No 224" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "alpha= 0.97 \n", + "I_E=1 # in mA\n", + "# Formula alpha = I_C/I_E \n", + "I_C= alpha*I_E # in mA\n", + "# Formula I_E= I_B+I_C\n", + "I_B= I_E-I_C # in mA\n", + "print \"The base current = %0.2f mA\" %I_B" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 0.03 mA\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.3\n", + ": Page No 226" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "# Part (i)\n", + "a= 0.90 \n", + "B=a/(1-a) \n", + "print \"At alpha= 0.90, the value of Bita = %0.f\" %B\n", + "# Part (ii)\n", + "a= 0.99 \n", + "B=a/(1-a) \n", + "print \"At alpha= 0.99, the value of Bita = %0.f\" %B" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "At alpha= 0.90, the value of Bita = 9\n", + "At alpha= 0.99, the value of Bita = 99\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.4\n", + ": Page No 226" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "bita= 50 \n", + "I_E= 10 # in mA\n", + "I_B= 200*10**-3 # in mA\n", + "alfa= bita/(1+bita)\n", + "print \"The value of alfa = %0.2f\" %alfa\n", + "I_C= alfa*I_E # in mA\n", + "print \"The value of I_C = %0.1f mA using the value of alpha\" %I_C\n", + "I_C= bita*I_B # in mA\n", + "print \"The value of I_C = %0.f mA using the value of bita\" %I_C" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of alfa = 0.98\n", + "The value of I_C = 9.8 mA using the value of alpha\n", + "The value of I_C = 10 mA using the value of bita\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.5\n", + ": Page No 233" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BB= 10 # in V\n", + "V_CC= 10 # in V\n", + "V_BE= 0.7 # in V\n", + "R_B= 1 # in M\u03a9\n", + "R_B= R_B*10**6 # in \u03a9\n", + "R_C= 2 # in k\u03a9\n", + "R_C= R_C*10**3 # in \u03a9\n", + "bita= 300 \n", + "I_B= (V_BB-V_BE)/R_B # in A\n", + "I_C= bita*I_B # in A\n", + "V_CE= V_CC-I_C*R_C # in V\n", + "P_D= V_CE*I_C # in W\n", + "print \"The value of I_B = %0.1f \u00b5A\" %(I_B*10**6)\n", + "print \"The value of I_C = %0.2f mA\" %(I_C*10**3)\n", + "print \"The value of V_CE = %0.2f volts\" %V_CE\n", + "print \"The value of P_D = %0.1f mW\" %(P_D*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_B = 9.3 \u00b5A\n", + "The value of I_C = 2.79 mA\n", + "The value of V_CE = 4.42 volts\n", + "The value of P_D = 12.3 mW\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.6\n", + ": Page No 241" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "bita= 100 \n", + "V_BE= 0 # in V\n", + "V_BB= 15 # in V\n", + "R_B= 470 # in k\u03a9\n", + "R_B= R_B*10**3 # in \u03a9\n", + "V_CC= 15 # in V\n", + "R_C= 3.6 # in k\u03a9\n", + "R_C= R_C*10**3 # in \u03a9\n", + "I_B= (V_BB-V_BE)/R_B # in A\n", + "I_C= bita*I_B # in A\n", + "V_CE= V_CC-I_C*R_C # in V\n", + "I_E= I_C+I_B # in A\n", + "print \"The base current = %0.1f \u00b5A\" %(I_B*10**6)\n", + "print \"The collector current = %0.2f mA\" %(I_C*10**3)\n", + "print \"The value of V_CE = %0.2f volts\" %V_CE\n", + "print \"The emitter current = %0.2f mA\" %(I_E*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 31.9 \u00b5A\n", + "The collector current = 3.19 mA\n", + "The value of V_CE = 3.51 volts\n", + "The emitter current = 3.22 mA\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.7\n", + ": Page No 242 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "bita= 100 \n", + "V_BE= 0.7 # in V\n", + "V_BB= 15 # in V\n", + "R_B= 470 # in k\u03a9\n", + "R_B= R_B*10**3 # in \u03a9\n", + "V_CC= 15 # in V\n", + "R_C= 3.6 # in k\u03a9\n", + "R_C= R_C*10**3 # in \u03a9\n", + "I_B= (V_BB-V_BE)/R_B # in A\n", + "I_C= bita*I_B # in A\n", + "V_CE= V_CC-I_C*R_C # in V\n", + "print \"The base current = %0.1f \u00b5A\" %(I_B*10**6)\n", + "print \"The collector current = %0.2f mA\" %(I_C*10**3)\n", + "print \"The value of V_CE = %0.2f volts\" %V_CE" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 30.4 \u00b5A\n", + "The collector current = 3.04 mA\n", + "The value of V_CE = 4.05 volts\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.8\n", + ": Page No 255" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "# Given data\n", + "V_CC= 15 # in V\n", + "V_BE= 0.7 # in V\n", + "R_C= 1 # in k\u03a9\n", + "R_C= R_C*10**3 # in \u03a9\n", + "R_E= 2 # in k\u03a9\n", + "R_E= R_E*10**3 # in \u03a9\n", + "R1= 10 # in k\u03a9\n", + "R1= R1*10**3 # in \u03a9\n", + "R2= 5 # in k\u03a9\n", + "R2= R2*10**3 # in \u03a9\n", + "V_CE= np.arange(0,V_CC,0.1)\n", + "I_C= (V_CC-V_CE)/(R_C+R_E)*10**3 # in mA\n", + "plt.plot(V_CE,I_C) \n", + "plt.plot([0,8.55],[2.15,2.15], '--',)\n", + "plt.plot([8.55,8.55],[0,2.15], '--')\n", + "plt.xlabel('V_CE in volts')\n", + "plt.ylabel('I_C in mA')\n", + "plt.title('DC load line') \n", + "V_B= V_CC*R2/(R1+R2) # in V\n", + "I_E= (V_B-V_BE)/R_E # in A\n", + "I_C= I_E # in A\n", + "I_CQ= I_C # in A\n", + "V_CE= V_CC-I_C*(R_C+R_E) # in V\n", + "print \"Q-point is : \",round(V_CE,2),\" V\",round(I_CQ*10**3,2),\" mA\"\n", + "print \"DC load line shown in figure\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Q-point is : 8.55 V 2.15 mA\n", + "DC load line shown in figure\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x7ff2041da990>" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.9\n", + ": Page No 258" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BB= 1.8 # in V\n", + "V_BE= 0.7 # in V\n", + "R1= 10 # in k\u03a9\n", + "R2= 2.2 # in k\u03a9\n", + "R_E= 1 # in k\u03a9\n", + "bita= 200 \n", + "R= R1*R2/(R1+R2) # in k\u03a9\n", + "R=R*10**3 # in \u03a9\n", + "R_E= R_E*10**3 # in \u03a9\n", + "I_E= (V_BB-V_BE)/(R_E+R/bita) # in mA\n", + "print \"The emitter current = %0.2f mA\" %(I_E*10**3)\n", + "print \"This is extremely close to 1.1 mA, the value we get with the simplified analysis.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The emitter current = 1.09 mA\n", + "This is extremely close to 1.1 mA, the value we get with the simplified analysis.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.10\n", + ": Page No 261" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC= 10 # in V\n", + "V_BE= 0.7 # in V\n", + "V_CE= 5 # in V\n", + "bita= 100 \n", + "I_C= 5 # in mA\n", + "# Applying KVL to collector circuit, V_CC-V_CE-I_C*R_C =0\n", + "R_C= (V_CC-V_CE)/I_C # in k\u03a9\n", + "print \"The value of R_C = %0.f k\u03a9\" %R_C\n", + "I_B= I_C/bita # in mA\n", + "print \"The value of I_B = %0.f \u00b5A\" %(I_B*10**3)\n", + "# Applying KVL to base circuit, V_CC-I_B*R_B-V_BE= 0\n", + "R_B= (V_CC-V_BE)/I_B # in k\u03a9\n", + "print \"The value of R_B = %0.f k\u03a9\" %R_B" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of R_C = 1 k\u03a9\n", + "The value of I_B = 50 \u00b5A\n", + "The value of R_B = 186 k\u03a9\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.11\n", + ": Page No 261" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "# Given data\n", + "V_CC= 6 # in V\n", + "V_BE= 0.7 # in V\n", + "bita= 100 \n", + "R_C= 2 # in k\u03a9\n", + "R_C= R_C*10**3 # in \u03a9\n", + "R_B= 530 # in k\u03a9\n", + "R_B= R_B*10**3 # in \u03a9\n", + "R1= 10 # in k\u03a9\n", + "R1= R1*10**3 # in \u03a9\n", + "R2= 5 # in k\u03a9\n", + "R2= R2*10**3 # in \u03a9\n", + "V_CE= np.arange(0,V_CC,0.1) # in V\n", + "I_C= (V_CC-V_CE)/(R_C)*10**3 # in mA\n", + "plt.plot(V_CE,I_C) \n", + "plt.xlabel('V_CE in volts') \n", + "plt.ylabel('I_C in mA')\n", + "plt.plot([0,4],[1,1], '--',)\n", + "plt.plot([4,4],[0,1], '--')\n", + "plt.title('DC load line') \n", + "I_B= (V_CC-V_BE)/R_B # in A\n", + "I_CQ= I_B*bita # in A\n", + "V_CE= V_CC-I_CQ*R_C # in V\n", + "print \"Q-point is : (\",round(V_CE,),\"V\",round(I_CQ*10**3),\"mA )\"\n", + "print \"DC load line shown in figure\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Q-point is : ( 4.0 V 1.0 mA )\n", + "DC load line shown in figure\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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xYMEC+Pn5wd/fH4WFhd03VpuxlpYWtY+Pj/ry5cvq5uZmdXBwsPr8+fNSlyWY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\nSUVORK5CYII=\n", + "text": [ + "<matplotlib.figure.Figure at 0x7ff1eee33190>" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.12\n", + ": Page No 265" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC= 12 # in V\n", + "V_BE= 0.7 # in V\n", + "bita= 100 \n", + "R_C= 10 # in k\u03a9\n", + "R_C=R_C*10**3 # in \u03a9\n", + "R_B= 100 # in \u03a9\n", + "R_B=R_B*10**3 # in \u03a9\n", + "I_BQ= (V_CC-V_BE)/((1+bita)*R_C+R_B) # in A\n", + "I_CQ= bita*I_BQ # in A\n", + "V_CEQ= V_CC-(I_CQ+I_BQ)*R_C # in volts\n", + "print \"Q-Point value for the circuit =\",round(V_CEQ,3),\"V and\",round(I_CQ*10**3,3),\"mA\"\n", + "# For dc load line when \n", + "I_C=0 \n", + "V_CE= V_CC-(I_C+I_BQ)*R_C # in V\n", + "print \"At I_C=0, the value of V_CE = %0.2f volts\" %V_CE\n", + "# When\n", + "V_CE= 0 \n", + "I_C= (V_CC-I_BQ*R_C)/R_C # in A\n", + "print \"At V_CE=0, the value of I_C = %0.1f mA\" %(I_C*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Q-Point value for the circuit = 1.718 V and 1.018 mA\n", + "At I_C=0, the value of V_CE = 11.90 volts\n", + "At V_CE=0, the value of I_C = 1.2 mA\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.13\n", + ": Page No 266" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BE= 0.7 # in V\n", + "V_CC= 15 # in V\n", + "V_CE= 5 # in V\n", + "I_C= 5 # in mA\n", + "I_C=I_C*10**-3 # in A\n", + "bita= 100 \n", + "I_B= I_C/bita # in A\n", + "# Applying KVL to collector circuit, V_CC= (I_C+I_B)*R_C+V_CE\n", + "R_C= (V_CC-V_CE)/(I_C+I_B) # in \u03a9\n", + "# Applying KVL to base circuit, V_CC= (I_C+I_B)*R_C+I_B*R_B+V_BE\n", + "R_B= (V_CC-V_BE-R_C*(I_C+I_B))/I_B # in \u03a9\n", + "print \"The value of R_C = %0.2f k\u03a9\" %(R_C*10**-3)\n", + "print \"The value of R_B = %0.f k\u03a9\" %(R_B*10**-3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of R_C = 1.98 k\u03a9\n", + "The value of R_B = 86 k\u03a9\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.14\n", + ": Page No 267" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_B= 20*10**-6 # in A\n", + "V_CE= 7.3 # in V\n", + "V_BE= 0.6 # in V\n", + "V_E= 2.1 # in V\n", + "R_E= 0.68*10**3 # in \u03a9\n", + "R_C= 2.7*10**3 # in \u03a9\n", + "I_E= V_E/R_E # in A\n", + "I_C= I_E # in A (approx)\n", + "bita= round(I_C/I_B) \n", + "V_CC= V_CE+I_C*R_C+I_E*R_E # in V\n", + "# From V_CC= I_B*R_B+V_BE+V_E\n", + "R_B= (V_CC-(V_BE+V_E))/I_B # in \u03a9\n", + "print \"The value of bita = %0.f\" %bita\n", + "print \"The value of V_CC = %0.1f volts\" %V_CC\n", + "print \"The value of R_B = %0.f k\u03a9\" %(R_B*10**-3)\n", + "\n", + "# Note: In the book, there is an error to calculate the value of R_B, hence the value of R_B in the book is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of bita = 154\n", + "The value of V_CC = 17.7 volts\n", + "The value of R_B = 752 k\u03a9\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.15\n", + ": Page No 268" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC = 18 # in V\n", + "bita = 90 \n", + "R_C = 2.2 * 10**3 # in ohm\n", + "R_E = 1.8*10**3 # in ohm\n", + "R_B = 510*10**3 # in ohm\n", + "I_B = V_CC/( (bita*(R_C+R_E))+R_B ) # in A\n", + "I_C = bita*I_B # in A\n", + "print \"The value of I_C = %0.1f mA\" %(I_C*10**3)\n", + "V_CE = I_B*R_B # in V\n", + "print \"The value of V_CE = %0.1f V\" %V_CE" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_C = 1.9 mA\n", + "The value of V_CE = 10.6 V\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.16\n", + ": Page No 269" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "bita = 50 \n", + "V_CC = 12 # in V\n", + "V_BE = 0.7 # in V\n", + "R_B = 240 # in kohm\n", + "R_B = R_B*10**3 # in ohm\n", + "I_C = 2.35 * 10**-3 # in A\n", + "R_C = 2.2 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "I_BQ = (V_CC - V_BE)/R_B # in A\n", + "print \"The value of I_BQ = %0.2f \u00b5A\" %(I_BQ*10**6)\n", + "I_CQ = bita*I_BQ # in A\n", + "print \"The value of I_CQ = %0.2f mA\" %(I_CQ*10**3)\n", + "V_CEQ = V_CC - (I_C*R_C) # in V\n", + "print \"The value of V_CEQ = %0.2f V\" %V_CEQ\n", + "V_B = V_BE # in V\n", + "print \"The value of V_B = %0.1f V\" %V_B\n", + "V_BC = V_B -V_CEQ # in V\n", + "print \"The voltage = %0.2f V\" %V_BC\n", + "\n", + "# Note: In the book, there is a calculation error to evaluating the value of V_CEQ. So the answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_BQ = 47.08 \u00b5A\n", + "The value of I_CQ = 2.35 mA\n", + "The value of V_CEQ = 6.83 V\n", + "The value of V_B = 0.7 V\n", + "The voltage = -6.13 V\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.17\n", + ": Page No 269" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC = 18 # in V\n", + "V_BE = 0.7 # in V\n", + "R_C = 3.3 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "R_B = 210 # in kohm\n", + "R_B = R_B * 10**3 # in ohm\n", + "bita = 75 \n", + "R_C = 3.3 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "R_E = 510 # in ohm\n", + "I_B = (V_CC-V_BE)/( R_C+R_B+bita*(R_C+R_E) ) # A\n", + "print \"The value of I_B = %0.f \u00b5A\" %round(I_B*10**6)\n", + "I_C = bita*I_B # in A\n", + "print \"The value of I_C = %0.1f mA\" %(I_C*10**3)\n", + "V_C = V_CC - (I_C*R_C) # in V\n", + "print \"The voltage = %0.2f V\" %V_C" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_B = 35 \u00b5A\n", + "The value of I_C = 2.6 mA\n", + "The voltage = 9.42 V\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.18\n", + ": Page No 271" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BE = 0.7 # in V\n", + "I_B = 40 * 10**-6 # in A\n", + "V_CC = 20 # in V (From the load line)\n", + "print \"The voltage = %0.f V\" %V_CC\n", + "I_C = 8 # in mA\n", + "R_C = V_CC/I_C # in kohm\n", + "print \"The resistance = %0.1f kohm\" %R_C\n", + "R_B = (V_CC - V_BE)/I_B # in ohm\n", + "print \"The resistance = %0.1f kohm\" %(R_B*10**-3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage = 20 V\n", + "The resistance = 2.5 kohm\n", + "The resistance = 482.5 kohm\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.19\n", + ": Page No 271" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R1 = 47 # in kohm\n", + "R1= R1*10**3 # in ohm\n", + "R2 = 10 # in kohm\n", + "R2= R2*10**3 # in ohm\n", + "R_E = 1.1 # in kohm\n", + "R_E = R_E * 10**3 # in ohm\n", + "R_C = 2.4 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "V_CC = -18 # in V\n", + "V_B = (R2*V_CC)/(R1+R2) # in V\n", + "V_BE = -0.7 # in V\n", + "V_E = V_B - V_BE # in V\n", + "I_E = abs(V_E)/R_E # in A\n", + "V_CE = V_CC + (I_E)*(R_C+R_E) # in V\n", + "print \"The value of V_B = %0.2f volts\" %V_B\n", + "print \"The value of I_E = %0.2f mA\" %(I_E*10**3)\n", + "print \"The value of V_CE = %0.2f V\" %V_CE" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of V_B = -3.16 volts\n", + "The value of I_E = 2.23 mA\n", + "The value of V_CE = -10.18 V\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.20\n", + ": Page No 273" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BE = 0.8 # in V\n", + "V_CE = 0.2 # in V\n", + "V1 = 5 # in V\n", + "R_B = 50 # in kohm\n", + "R_B= R_B*10**3 # in ohm\n", + "R_C = 3 # in K ohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "bita = 100 \n", + "R_E = 2 # in kohm\n", + "R_E= R_E*10**3 # in ohm\n", + "I_B = (V1-V_BE)/(R_B+(1+bita)*R_E) # in A\n", + "print \"The value of I_B = %0.2f \u00b5A\" %(I_B*10**6)\n", + "V_CC = 10 # in V\n", + "I_Csat = (V_CC - V_CE - (I_B*R_E))/(R_C+R_E) #in A\n", + "print \"The value of I_C(sat) = %0.3f mA\" %(I_Csat*10**3)\n", + "I_Bmin = I_Csat /bita # in A\n", + "print \"The minimum value of I_B = %0.3f \u00b5A\" %(I_Bmin*10**6)\n", + "\n", + "# Note: There is calculation error to evaluate the value of I_Csat in the book, so the answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_B = 16.67 \u00b5A\n", + "The value of I_C(sat) = 1.953 mA\n", + "The minimum value of I_B = 19.533 \u00b5A\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.21\n", + ": Page No 274" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R1 = 5 # in kohm\n", + "R1= R1*10**3 # in ohm\n", + "R2 = 5 # in kohm\n", + "R2= R2*10**3 # in ohm\n", + "R_B = R1*R2/(R1+R2) # in ohm\n", + "R_E = 1 # in kohm\n", + "R_E = R_E * 10**3 # in ohm\n", + "V_EE = 3 # in V\n", + "V_Th = (R2*V_EE)/(R1+R2) # in V\n", + "V_BE = 0.7 # in V\n", + "bita = 44 \n", + "I_B = (V_EE - V_BE - V_Th)/( ((1+bita)*R_E)+R_B) # in A\n", + "I_BQ = I_B # in A\n", + "print \"The value of I_BQ = %0.2f \u00b5A\" %(I_BQ*10**6)\n", + "I_C = bita*I_BQ # in A\n", + "print \"The value of I_C = %0.2f mA\" %(I_C*10**3)\n", + "I_E = (1+bita)*I_B # in A\n", + "print \"The value of I_E = %0.3f mA\" %(I_E*10**3)\n", + "V_EC = (I_E*R_E)-V_EE # in V\n", + "print \"The value of V_EC = %0.3f V\" %V_EC\n", + "print \"Q-point = (\",round(V_EC,3),\"V\",round(I_C*10**3,2),\"mA )\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_BQ = 16.84 \u00b5A\n", + "The value of I_C = 0.74 mA\n", + "The value of I_E = 0.758 mA\n", + "The value of V_EC = -2.242 V\n", + "Q-point = ( -2.242 V 0.74 mA )\n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.22\n", + ": Page No 275" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BE = 0.7 # in V\n", + "V_BB = 5 # in V\n", + "R_B = 100 # in kohm\n", + "R_B = R_B * 10**3 # in ohm\n", + "R_E = 2 # in kohm\n", + "R_E = R_E * 10**3 # in ohm\n", + "bita = 100 \n", + "I_B = (V_BB-V_BE)/( R_B+((1+bita)*R_E) ) # in A\n", + "print \"The value of I_B = %0.3f mA\" %(I_B*10**3)\n", + "V_B = V_BB-(I_B*10**-3*R_B) # in V\n", + "I_C = bita*I_B # in A\n", + "print \"The value of I_C = %0.1f mA\" %(I_C*10**3)\n", + "V_CC = 10 # in V\n", + "V_C = V_CC-(I_C*R_E) # in V\n", + "print \"The voltage = %0.1f V\" %V_C\n", + "print \"Transistor is in active region is valid\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_B = 0.014 mA\n", + "The value of I_C = 1.4 mA\n", + "The voltage = 7.2 V\n", + "Transistor is in active region is valid\n" + ] + } + ], + "prompt_number": 49 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.23\n", + ": Page No 276 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC = 20 # in V\n", + "V_BE = 0.7 # in V\n", + "R_B = 430 # in kohm\n", + "R_B = 430 * 10**3 # in ohm\n", + "bita = 50 \n", + "R_E = 1 # in kohm\n", + "R_E = R_E * 10**3 # in ohm\n", + "R_C = 2 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "I_B = (V_CC - V_BE)/(R_B +(1+bita)*R_E) # in A\n", + "print \"The base current = %0.1f \u00b5A\" %(I_B*10**6)\n", + "I_C = bita*I_B # in A\n", + "print \"The collector current = %0.2f mA\" %(I_C*10**3)\n", + "V_CE = V_CC - I_C*(R_C+R_E) # in V\n", + "print \"The value of V_CE = %0.2f V\" %V_CE\n", + "V_C = V_CC - (I_C*R_C) # in V\n", + "print \"The value of V_C = %0.2f V\" %V_C\n", + "V_E = V_C - V_CE # in V\n", + "print \"The value of V_E = %0.2f V\" %V_E\n", + "V_B = V_BE+V_E # in V\n", + "print \"The value of V_B = %0.2f V\" %V_B\n", + "V_BC = V_B-V_C # in V\n", + "print \"The value of V_BC = %0.2f V\" %V_BC" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 40.1 \u00b5A\n", + "The collector current = 2.01 mA\n", + "The value of V_CE = 13.98 V\n", + "The value of V_C = 15.99 V\n", + "The value of V_E = 2.01 V\n", + "The value of V_B = 2.71 V\n", + "The value of V_BC = -13.28 V\n" + ] + } + ], + "prompt_number": 51 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.24\n", + ": Page No 277" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC = 20 # in V\n", + "V_BE = 0.7 # in V\n", + "R_B = 680 # in kohm\n", + "R_B = R_B * 10**3 # in ohm\n", + "R_C = 4.7 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "bita = 120 \n", + "I_B = (V_CC - V_BE)/(R_B+bita*R_C) # in A\n", + "I_CQ = bita*I_B # in A\n", + "print \"The value of I_CQ = %0.2f mA\" %(I_CQ*10**3)\n", + "V_CEQ = V_CC - (I_CQ*R_C) # in V\n", + "print \"The value of V_CEQ = %0.2f V\" %V_CEQ\n", + "V_B = V_BE # in V\n", + "V_C = 11.26 # in V\n", + "V_E = 0 # in V\n", + "print \"The value of V_E = %0.f V\" %V_E\n", + "V_BC = V_B - V_C # in V\n", + "print \"The value of V_BC = %0.2f V\" %V_BC" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_CQ = 1.86 mA\n", + "The value of V_CEQ = 11.25 V\n", + "The value of V_E = 0 V\n", + "The value of V_BC = -10.56 V\n" + ] + } + ], + "prompt_number": 52 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.25\n", + ": Page No 278" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC = 16 # in V\n", + "V_BE = 0.7 # in V\n", + "R_B = 470 # in kohm\n", + "R_B= R_B*10**3 # in ohm\n", + "bita = 120 \n", + "R_C = 3.6 # in kohm\n", + "R_C= R_C*10**3 # in ohm\n", + "R_E = 0.51 # in kohm\n", + "R_E= R_E*10**3 # in ohm\n", + "I_B = (V_CC - V_BE)/(R_B+bita*(R_C+R_E)) # in A\n", + "print \"The base current = %0.2f \u00b5A\" %(I_B*10**6)\n", + "I_C = bita*I_B # in A\n", + "print \"The collector current = %0.2f mA\" %(I_C*10**3)\n", + "V_C = V_CC - I_C*R_C # in V\n", + "print \"The collector voltage = %0.2f V\" %V_C" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 15.88 \u00b5A\n", + "The collector current = 1.91 mA\n", + "The collector voltage = 9.14 V\n" + ] + } + ], + "prompt_number": 53 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.26\n", + ": Page No 279" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC = 10 # in V\n", + "V_BE = 0.7 # in V\n", + "R_B = 250 # in kohm\n", + "R_B= R_B*10**3 # in ohm\n", + "bita = 90 \n", + "R_C = 4.7 # in kohm\n", + "R_C= R_C*10**3 # in ohm\n", + "R_E = 1.2 # in kohm\n", + "R_E= R_E*10**3 # in ohm\n", + "I_BQ = (V_CC - V_BE)/(R_B + bita*(R_C+R_E)) # in A\n", + "print \"The base current at Q-point = %0.2f \u00b5A\" %(I_BQ*10**6)\n", + "I_CQ = bita*I_BQ # in A\n", + "print \"The collector current at Q-point = %0.2f mA\" %(I_CQ*10**3)\n", + "V_CEQ = V_CC - (I_CQ*(R_C+R_E)) # in V\n", + "print \"Collector emitter voltage at Q point = %0.3f V\" %V_CEQ" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current at Q-point = 11.91 \u00b5A\n", + "The collector current at Q-point = 1.07 mA\n", + "Collector emitter voltage at Q point = 3.677 V\n" + ] + } + ], + "prompt_number": 55 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.27\n", + ": Page No 281" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC = 12 # in V\n", + "V_BE = 0.7 # in V\n", + "R_B = 150 # in kohm\n", + "R_B= R_B*10**3 # in ohm\n", + "bita = 180 \n", + "R_C = 4.7 # in kohm\n", + "R_C= R_C*10**3 # in ohm\n", + "R_E = 3.3 # in kohm\n", + "R_E= R_E*10**3 # in ohm\n", + "I_B = (V_CC-V_BE)/(R_B + bita*(R_C+R_E)) # in A\n", + "print \"The base current = %0.2f \u00b5A\" %(I_B*10**6)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 7.11 \u00b5A\n" + ] + } + ], + "prompt_number": 58 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.28\n", + ": Page No 282" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_B = 4 # in V\n", + "V_BE = 0.7 # in V\n", + "R_E = 1.2 # in kohm\n", + "R_E= R_E*10**3 # in ohm\n", + "V_E = V_B-V_BE # in V\n", + "R_C = 2.2 # in kohm\n", + "R_C= R_C*10**3 # in ohm\n", + "R_B= 330 # in kohm\n", + "R_B= R_B*10**3 # in ohm\n", + "bita = 180 \n", + "I_B = 7.11 * 10**-6 # in A\n", + "V_CC = 18 # in V\n", + "print \"Part (a)\"\n", + "print \"The value of V_E = %0.1f V\" %V_E\n", + "I_C = V_E/R_E # in A\n", + "print \"Part (b)\"\n", + "print \"The value of I_C = %0.2f mA\" %(I_C*10**3)\n", + "V_C =V_CC - (I_C*R_C) # in V\n", + "print \"Part (c)\"\n", + "print \"The value of V_C = %0.2f V\" %V_C\n", + "V_CE = V_C-V_E # in V\n", + "print \"Part (d)\"\n", + "print \"The value of V_CE = %0.2f V\" %V_CE\n", + "I_B = (V_CC - (I_C*R_C) - V_BE - V_E)/R_B # in A\n", + "print \"Part (e)\"\n", + "print \"Base current = %0.2f \u00b5A\" %(I_B*10**6)\n", + "bita = I_C/I_B \n", + "print \"Part (f)\"\n", + "print \"Current gain = %0.f\" %bita" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (a)\n", + "The value of V_E = 3.3 V\n", + "Part (b)\n", + "The value of I_C = 2.75 mA\n", + "Part (c)\n", + "The value of V_C = 11.95 V\n", + "Part (d)\n", + "The value of V_CE = 8.65 V\n", + "Part (e)\n", + "Base current = 24.09 \u00b5A\n", + "Part (f)\n", + "Current gain = 114\n" + ] + } + ], + "prompt_number": 62 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.29\n", + ": Page No 284" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_E = 10 # in mA\n", + "I_C = 9.95 # in mA\n", + "I_B = I_E-I_C # in mA\n", + "print \"The base current = %0.2f mA\" %I_B" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 0.05 mA\n" + ] + } + ], + "prompt_number": 63 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.30\n", + ": Page No 284" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_C = 10 # in mA\n", + "I_B = 0.1 # in mA\n", + "bita = I_C/I_B \n", + "print \"The current gain = %0.f\" %bita" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current gain = 100\n" + ] + } + ], + "prompt_number": 64 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.31\n", + ": Page No 284" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BE = 0.7 # in V\n", + "V_BB = 10 # in V\n", + "R_B = 470 # in kohm\n", + "R_B = R_B * 10**3 # in ohm\n", + "I_B = (V_BB-V_BE)/R_B # in A\n", + "print \"The base current = %0.2f \u00b5A\" %(I_B*10**6)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 19.79 \u00b5A\n" + ] + } + ], + "prompt_number": 66 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.32\n", + ": Page No 285" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BB = 10 # in V\n", + "V_BE = 0 # in V\n", + "R_B = 470 # in kohm\n", + "R_B = R_B * 10**3 # in ohm \n", + "I_B = (V_BB - V_BE)/R_B # in A\n", + "bita = 200 \n", + "I_C = bita*I_B # in A\n", + "V_CC = 10 # in V\n", + "R_C = 820 # in ohm\n", + "V_CE = V_CC - (I_C*R_C) # in V\n", + "print \"Part (a) : For ideal approximation\"\n", + "print \"The collector emitter voltage = %0.2f V\" %V_CE\n", + "P_D = V_CE * I_C # in W\n", + "print \"Power dissipation = %0.2f mW\" %(P_D*10**3)\n", + "print \"Part (b) : For second approximation\"\n", + "V_BE = 0.7 # in V\n", + "I_B = (V_BB-V_BE)/R_B # in A\n", + "I_C = bita*I_B # in A\n", + "V_CE = V_CC - (I_C*R_C) # in V\n", + "print \"The collector emitter voltage = %0.2f V\" %V_CE\n", + "P_D = V_CE * I_C # in W\n", + "print \"Power dissipation = %0.2f mW\" %(P_D*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (a) : For ideal approximation\n", + "The collector emitter voltage = 6.51 V\n", + "Power dissipation = 27.70 mW\n", + "Part (b) : For second approximation\n", + "The collector emitter voltage = 6.75 V\n", + "Power dissipation = 26.73 mW\n" + ] + } + ], + "prompt_number": 69 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.33\n", + ": Page No 286" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BE = 0 # in V\n", + "V_BB = 12 # in V\n", + "R_B = 680 # in kohm\n", + "R_B = R_B * 10**3 # in ohm\n", + "I_B = (V_BB-V_BE)/R_B # in A\n", + "beta_dc = 175 \n", + "I_C = beta_dc*I_B # in A\n", + "V_CC = 12 # in V\n", + "R_C = 1.5 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "V_CE = V_CC - (I_C*R_C) # in V\n", + "print \"Part (a) For ideal approximation\"\n", + "print \"The collector emitter voltage = %0.2f V\" %V_CE\n", + "P_D = V_CE * I_C # in mW\n", + "print \"Transistor power = %0.2f mW\" %(P_D*10**3)\n", + "print \"Part (b) For second approximation\"\n", + "V_BE1 = 0.7 # in V\n", + "I_B = (V_BB-V_BE1)/R_B # in A\n", + "I_C = beta_dc * I_B # in A\n", + "V_CE = V_CC - (I_C*R_C) # in V\n", + "print \"Collector emitter voltage = %0.2f V\" %V_CE\n", + "P_D = V_CE * I_C # in W\n", + "print \"Power dissipation = %0.2f mW\" %(P_D*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (a) For ideal approximation\n", + "The collector emitter voltage = 7.37 V\n", + "Transistor power = 22.75 mW\n", + "Part (b) For second approximation\n", + "Collector emitter voltage = 7.64 V\n", + "Power dissipation = 22.21 mW\n" + ] + } + ], + "prompt_number": 71 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.34\n", + ": Page No 288" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "# Given data\n", + "V_CC = 20 # in V\n", + "R_C = 3.3 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "I_C = V_CC/R_C # in A\n", + "print \"Collector current = %0.2f mA\" %(I_C*10**3)\n", + "V_CE = V_CC # in V\n", + "print \"Collector emitter voltage = %0.f V\" %V_CE\n", + "V_CE=np.arange(0,20,0.1) # in V\n", + "I_C= (V_CC-V_CE)/(R_C*10**-3) # in mA\n", + "plt.plot(V_CE,I_C) \n", + "plt.xlabel('V_CE in volts')\n", + "plt.ylabel('I_C in mA')\n", + "plt.title('DC load line')\n", + "print \"DC load line shown in figure\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Collector current = 6.06 mA\n", + "Collector emitter voltage = 20 V\n", + "DC load line shown in figure" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x7ff2041850d0>" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.35\n", + ": Page No 289" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BB = 10 # in V\n", + "V_BE = 0.7 # in V\n", + "R_B = 1 # in kohm\n", + "R_B = 1 * 10**6 # in ohm\n", + "I_B = (V_BB-V_BE)/R_B # in A\n", + "print \"The base current = %0.1f \u00b5A\" %(I_B*10**6)\n", + "beta_dc = 200 \n", + "I_C = beta_dc * I_B # in A\n", + "print \"The collector current = %0.2f mA\" %(I_C*10**3)\n", + "V_CC = 20 # in V\n", + "R_C = 3.3 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "V_CE = V_CC - I_C*R_C # in V\n", + "print \"The collector voltage = %0.3f V\" %V_CE" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 9.3 \u00b5A\n", + "The collector current = 1.86 mA\n", + "The collector voltage = 13.862 V\n" + ] + } + ], + "prompt_number": 73 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.36\n", + ": Page No 290" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BB = 5 # in V\n", + "V_BE = 0.7 # in V\n", + "R_B = 680 # in kohm\n", + "R_B = 680*10**3 # in ohm\n", + "I_B = (V_BB-V_BE)/R_B # in A\n", + "print \"The base current = %0.2f \u00b5A\" %(I_B*10**6)\n", + "beta_dc= 150 \n", + "I_C = beta_dc * I_B # in A\n", + "print \"The collector current = %0.2f mA\" %(I_C*10**3)\n", + "V_CC = 5 # in V\n", + "R_C = 470 # in ohm\n", + "V_CE = V_CC-(I_C*R_C) # in V\n", + "print \"Voltage between collector and ground = %0.2f V\" %V_CE" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 6.32 \u00b5A\n", + "The collector current = 0.95 mA\n", + "Voltage between collector and ground = 4.55 V\n" + ] + } + ], + "prompt_number": 75 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.37\n", + ": Page No 291" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BB = 2.5 # in V\n", + "V_BE = 0.7 # in V\n", + "V_E = V_BB-V_BE # in V\n", + "print \"The emitter voltage = %0.1f V\" %V_E\n", + "R_E = 1.8 # in kohm\n", + "R_E = R_E * 10**3 # in ohm\n", + "I_E = V_E/R_E # in A\n", + "I_C= I_E # in A\n", + "V_CC = 20 # in V\n", + "R_C = 10 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "V_C = V_CC-(I_C*R_C) # in V\n", + "print \"The collector voltage = %0.f V\" %V_C" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The emitter voltage = 1.8 V\n", + "The collector voltage = 10 V\n" + ] + } + ], + "prompt_number": 77 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.38\n", + ": Page No 292" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC = 25 # in V\n", + "R2 = 2.2 # in kohm\n", + "R1 = 10 # in kohm\n", + "V_BB = (V_CC * R2)/(R1+R2) # in V\n", + "V_BE = 0.7 # in V\n", + "V_E = V_BB - V_BE # in V\n", + "print \"The emitter voltage = %0.1f V\" %V_E\n", + "R_E = 1 # in kohm\n", + "R_E = R_E * 10**3 # in ohm\n", + "I_E = V_E/R_E # in A\n", + "I_C= I_E # in A\n", + "V_CC = 25 # in V\n", + "R_C = 3.6 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "V_C = V_CC - (I_C*R_C) # in V\n", + "print \"Collector voltage = %0.2f V\" %V_C" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The emitter voltage = 3.8 V\n", + "Collector voltage = 11.29 V\n" + ] + } + ], + "prompt_number": 78 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.39\n", + ": Page No 293 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BB = 4.50 # in V\n", + "V_E = 3.8 # in V\n", + "V_C = 11.32 # in V\n", + "I_C = 3.8 # in mA\n", + "I_C=I_C*10**-3 # in A\n", + "V_BE = 0.7 # in V\n", + "R1 = 10 # in kohm\n", + "R2 = 2.2 # in kohm\n", + "R_B = (R1*R2)/(R1+R2) # in kohm\n", + "R_B = R_B * 10**3 # in ohm\n", + "I_B = (V_BB-V_BE)/R_B # in A\n", + "print \"The base current = %0.2f mA\" %(I_B*10**3)\n", + "V_CE = V_C-V_E # in V\n", + "print \"Collector emitter voltage = %0.2f V\" %V_CE\n", + "print \"Thus the Q-point is :\",round(V_CE,2),\"V\",round(I_B*10**3,2),\"mA\"\n", + "\n", + "# Note: There is calculation error to evaluate the value of I_B. So the answer in the book is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 2.11 mA\n", + "Collector emitter voltage = 7.52 V\n", + "Thus the Q-point is : 7.52 V 2.11 mA\n" + ] + } + ], + "prompt_number": 80 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Engineering_by_P._Raja/chapter_5_2.ipynb b/Electronics_Engineering_by_P._Raja/chapter_5_2.ipynb new file mode 100644 index 00000000..54ba0032 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_5_2.ipynb @@ -0,0 +1,517 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter - 5 : Transistor Circuits" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.1\n", + ": Page No 309 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from numpy import pi\n", + "# Given data\n", + "R1 = 600 # in ohm\n", + "R2 = 1000 # in ohm\n", + "R_TH = (R1*R2)/(R1+R2) # in ohm\n", + "X_C = 37.5 # in ohm\n", + "f = 1 # in kHz\n", + "f= f*10**3 # in Hz\n", + "C = 1/(2*pi * f*X_C) # in F\n", + "print \"Value of C = %0.1f \u00b5F\" %(C*10**6)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Value of C = 4.2 \u00b5F\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.2\n", + ": Page No 323" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_C= 3.6*10**3 # in ohm\n", + "R_L= 10*10**3 # in ohm\n", + "r_c = (R_C*R_L)/(R_C+R_L) # in ohm\n", + "V_CC = 10 # in V\n", + "V_BE = 0.7 # in V\n", + "R_E = 1 # in kohm\n", + "R_E = R_E * 10**3 # in ohm\n", + "R1 = 10 # in kohm\n", + "R1= R1*10**3 # in ohm\n", + "R2 = 2.2 # in kohm\n", + "R2= R2*10**3 # in ohm\n", + "V_B = (V_CC*R2)/(R1+R2) # in V\n", + "I_E = (V_B-V_BE)/R_E # in A\n", + "V = 25*10**-3 # in V # only value is given in the book \n", + "r_e = V/I_E # in ohm\n", + "A_V = round(r_c/r_e) \n", + "print \"The voltage gain = %0.f\" %A_V\n", + "V_in = 2 #in mV\n", + "V_out = A_V*V_in # in mV\n", + "print \"The output voltage = %0.f mV\" %V_out" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage gain = 117\n", + "The output voltage = 234 mV\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.3\n", + ": Page No 324" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "A_V = 117 \n", + "r_e = 22.7 # in ohm\n", + "bita = 300 \n", + "Zin_base = bita*r_e # in ohm\n", + "R1 = 2.2*10**3 # in ohm\n", + "R2 = 10*10**3 # in ohm\n", + "Zin_stage = (Zin_base*R1*R2)/(Zin_base*R1+R1*R2+R2*Zin_base) # in ohm \n", + "R = 600 # in ohm\n", + "V = 2 # in mV\n", + "V_in = (Zin_stage/(R+Zin_stage))*V # in mV\n", + "V_out = A_V * V_in # in mV\n", + "print \"The output voltage = %0.f mV\" %round(V_out)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 165 mV\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.4\n", + ": Page No 328" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R1 = 4.3 # in K ohm\n", + "R1= R1*10**3 # in ohm\n", + "R2 = 10 # in K ohm\n", + "R2= R2*10**3 # in ohm\n", + "r_e = (R1*R2)/(R1+R2) # in ohm\n", + "bita = 200 \n", + "V=25 # in mV\n", + "I= 1 # in mA\n", + "r_e_desh= V/I # in ohm\n", + "Zin_base = bita*(r_e + r_e_desh) # in ohm\n", + "print \"The input impedence of the base = %0.f k\u03a9\" %(Zin_base*10**-3)\n", + "R3 = 10*10**3 # in ohm\n", + "Zin_stage = (R2*R3*Zin_base)/(R2*Zin_base+R3*Zin_base+R2*R3) # in ohm\n", + "print \"The input impedance of the stage = %0.2f k\u03a9\" %(Zin_stage*10**-3)\n", + "print \"Because the input impedence of base is much larger than the input impedence of the stage,\"\n", + "print \"usually approximate the input impedence of the stage as the parallel of the biasing resistor only %0.f k\u03a9\" %(Zin_stage*10**-3)\n", + "Zin_stage= R2*R3/(R2+R3) # in ohm" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The input impedence of the base = 606 k\u03a9\n", + "The input impedance of the stage = 4.96 k\u03a9\n", + "Because the input impedence of base is much larger than the input impedence of the stage,\n", + "usually approximate the input impedence of the stage as the parallel of the biasing resistor only 5 k\u03a9\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.5\n", + ": Page No 332" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CE = 0.2 # in V\n", + "V_BE= 0.7 # in V\n", + "R = 1 # in kohm\n", + "R = R * 10**3 # in ohm\n", + "V = 10 # in V\n", + "I_C = (V-V_CE)/R # in A\n", + "beta_min = 50 \n", + "I_B = I_C/beta_min # in A\n", + "I_B1 = V*I_B # in A\n", + "V1 = 5 # in V\n", + "R_B = (V1-V_BE)/I_B1 # in ohm\n", + "print \"The base resistance = %0.1f k\u03a9\" %(R_B*10**-3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base resistance = 2.2 k\u03a9\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.6\n", + ": Page No 333" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R = 10 # in K ohm\n", + "R = R * 10**3 # in ohm\n", + "X_C = 0.1 * R \n", + "C = 47 # in \u00b5F\n", + "C = C * 10**-6 # in F\n", + "f = 1/(2*pi * X_C *C) # in Hz\n", + "print \"Lowest frequency = %0.2f Hz\" %f" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Lowest frequency = 3.39 Hz\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.7\n", + ": Page No 333" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "C = 220 # in \u00b5F\n", + "C = C * 10**-6 # in F\n", + "R1 = 10 # in kohm\n", + "R1 = R1 * 10**3 # in ohm\n", + "R2 = 2.2 # in kohm\n", + "R2 = R2 * 10**3 # in ohm\n", + "R_TH = (R1*R2)/(R1+R2) # in ohm\n", + "X_C = 0.1*R_TH # in ohm\n", + "f = 1/(2*pi*C*X_C) # in Hz\n", + "print \"The lowest frequency = %0.2f Hz\" %f" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The lowest frequency = 4.01 Hz\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.8\n", + ": Page No 334" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "i_c = 15 # in mA\n", + "i_c = i_c * 10**-3 # in A\n", + "i_b = 100 # in \u00b5A\n", + "i_b = i_b * 10**-6 # in A\n", + "bita = i_c/i_b \n", + "print \"The value of ac bita = %0.f\" %bita" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of ac bita = 150\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.9\n", + ": Page No 334" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_C = 3.6 # in kohm\n", + "R_C= R_C*10**3 # in ohm\n", + "R_L = 10 # in kohm\n", + "R_L=R_L*10**3 # in ohm\n", + "R_TH = (R_C*R_L)/(R_C+R_L) # in ohm\n", + "V_CC = 10 # in V\n", + "R2 = 2.2 # in kohm\n", + "R2 = R2 * 10**3 # in ohm\n", + "R1 = 10 # in kohm\n", + "R1 = R1 * 10**3 # in ohm\n", + "V_BE = 0.7 # in V\n", + "V_B = (V_CC*R2)/(R1+R2) # in V\n", + "R_E = 1 # in kohm \n", + "R_E = R_E *10**3 # in ohm\n", + "I_E = (V_B-V_BE)/R_E # in A\n", + "V1 = 25 # in mV\n", + "V1 = V1*10**-3 # in V\n", + "r_e = V1/(I_E) # in ohm\n", + "A_v = (R_TH)/r_e \n", + "V_in = 2 # in mV\n", + "V_in = V_in * 10**-3 # in V\n", + "V_out = A_v*V_in # in V\n", + "print \"The output voltage = %0.2f V\" %V_out" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 0.23 V\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.10\n", + ": Page No 336" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_L = 10 # in kohm\n", + "R_L= R_L*10**3 # in ohm\n", + "R_C = 3.6 # in kohm\n", + "R_C= R_C*10**3 # in ohm\n", + "r_e_desh = 22.73 # in ohm \n", + "R_L_desh = R_L/2 # in ohm\n", + "A_v = ( (R_C*R_L_desh)/(R_C+R_L_desh))/r_e_desh \n", + "print \"The voltage gain = %0.2f\" %A_v" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage gain = 92.08\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.11\n", + ": Page No 336" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_E = 1 # in kohm\n", + "R_E = R_E * 10**3 # in ohm\n", + "R_L = 3.3 # in kohm\n", + "R_L = R_L * 10**3 # in ohm\n", + "r_e = (R_E*R_L)/(R_E+R_L) # in ohm\n", + "V_CC = 15 # in V\n", + "R2 = 2.2 # in K ohm\n", + "R2 = R2 * 10**3 # in ohm\n", + "R1 = R2 # in ohm\n", + "V_B = (V_CC*R2)/(R1+R2) # in V\n", + "V_BE = 0.7 # in V\n", + "R_E = 1 # in K ohm\n", + "R_E = R_E * 10**3 # in ohm\n", + "I_E = (V_B-V_BE)/R_E # in A\n", + "V1 = 25*10**-3 # in V\n", + "r_e1 = V1/I_E \n", + "bita = 200 \n", + "Zin_base = bita*(r_e+r_e1) # in ohm\n", + "print \"The input impedence of the base = %0.2f k\u03a9\" %(Zin_base*10**-3)\n", + "Zin_stage = (R1*R2*Zin_base)/(R1*R2+R2*Zin_base+R1*Zin_base) # in ohm\n", + "print \"The input impedance of the stage = %0.2f k\u03a9\" %(Zin_stage*10**-3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The input impedence of the base = 154.22 k\u03a9\n", + "The input impedance of the stage = 1.09 k\u03a9\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.12\n", + ": Page No 337 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "r_e = 767.44 \n", + "r_e1 = 3.68 \n", + "V_in = 1 # in V\n", + "A_v = round(r_e/(r_e+r_e1)) \n", + "print \"The voltage gain = %0.f\" %A_v\n", + "V_o = A_v*V_in # in V\n", + "print \"The load voltage = %0.f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage gain = 1\n", + "The load voltage = 1 V\n" + ] + } + ], + "prompt_number": 24 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Engineering_by_P._Raja/chapter_6_2.ipynb b/Electronics_Engineering_by_P._Raja/chapter_6_2.ipynb new file mode 100644 index 00000000..27931e37 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_6_2.ipynb @@ -0,0 +1,1319 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter - 6 : Field Effect Devices (JFET)" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.1\n", + ": Page No 351" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# Given data\n", + "V_D = 10 # in V\n", + "R = 10*10**3 # in ohm\n", + "I_D = V_D/R # in A\n", + "V_P = 4 # in V\n", + "I_DSS = 10 # in mA\n", + "I_DSS = I_DSS * 10**-3 # in A\n", + "R_DS = V_P/I_DSS # in ohm\n", + "V_D = (R_DS/(R+R_DS))*V_D # in V\n", + "print \"The drain voltage = %0.3f V\" %V_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain voltage = 0.385 V\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.2\n", + ": Page No 353" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_P = 4 # in V\n", + "I_DSS = 10 # in mA\n", + "I_DSS =I_DSS *10**-3 # in A\n", + "R_DS = V_P/I_DSS # in ohm\n", + "V_DD = 30 # in V\n", + "I_D = 2.5 # in mA\n", + "R_D = 2 # in kohm\n", + "V_D = V_DD - (I_D*R_D) # in V\n", + "print \"The drain voltage = %0.f V\" %V_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain voltage = 25 V\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.3\n", + ": Page No 355" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "# Given data\n", + "R2 = 1 # in M ohm\n", + "R2 = R2*10**6 # in ohm\n", + "R1 = 2 # in M ohm\n", + "R1 = R1*10**6 # in ohm\n", + "V_DD = 30 # in V\n", + "R_D= 1*10**3 # in ohm\n", + "V_G = (R2/(R1+R2))*V_DD # in V\n", + "R_S= 2*10**3 # in ohm\n", + "I_D= V_G/R_S # in A\n", + "V_D= V_DD-I_D*R_D # in V\n", + "V_DS= V_D-V_G # in V\n", + "R_D= R_D+R_S # in ohm\n", + "I_Dsat=V_DD/R_D*10**3 # in mA\n", + "print \"The value of I_D = %0.f mA\" %(I_D*10**3)\n", + "print \"The value of V_DS = %0.f volts\" %V_DS\n", + "print \"Thus the Q-point = (\",int(V_DS),\"V,\",int(I_D*10**3),\"mA)\" \n", + "V_D= np.arange(0,V_DD,0.1) # in V\n", + "I_D= (V_DD-V_D)/R_D*10**3 # in mV\n", + "plt.plot(V_D,I_D) \n", + "plt.plot([0,15],[5,5], '--')\n", + "plt.plot([15,15],[0,5], '--')\n", + "plt.ylabel(\"I_D in mA\")\n", + "plt.xlabel(\"V_DS in volts\")\n", + "plt.title(\"DC load line\")\n", + "print \"DC load line shown in figure\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_D = 0 mA\n", + "The value of V_DS = 30 volts\n", + "Thus the Q-point = ( 30 V, 0 mA)\n", + "DC load line shown in figure" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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d5wgEAkkAQcvh0MTbDVIBgpJfmsCUKVO0e/duZWRkcAE5wOGQDINZAYKSX1YH9ejRQzfc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\nbIwmAAA2RhMAABv7f2bQfoJhPpDQAAAAAElFTkSuQmCC\n", + "text": [ + "<matplotlib.figure.Figure at 0x7ffba61e4e10>" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.4\n", + ": Page No 358" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_DD = 15 # in V\n", + "R = 3 # in kohm\n", + "I_D = V_DD/R # in mA\n", + "R_D = 1 # in kohm\n", + "V_D = V_DD - (I_D*R_D) # in V\n", + "print \"The drain voltage = %0.f V\" %V_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain voltage = 10 V\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5\n", + ": Page No 362" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_D = 3.6 # in K ohm\n", + "R_L = 10 # in K ohm\n", + "r_d = (R_D*R_L)/(R_D+R_L) # in K ohm\n", + "g_m = 5000 # in \u00b5S\n", + "g_m= g_m*10**-6 # in S\n", + "A_v = g_m *r_d \n", + "V_out = A_v # in V\n", + "print \"The output volatge = %0.1f mV\" %(V_out*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output volatge = 13.2 mV\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6\n", + ": Page No 370" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_GS = -2 # in V\n", + "V_P = -5 # in V\n", + "V_DS = V_GS-V_P # in V\n", + "I_DSS = 8 # in mA\n", + "I_DS = I_DSS*( 1-(V_GS/V_P) )**2 # in mA\n", + "print \"The drain current = %0.2f mA\" %I_DS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current = 2.88 mA\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.7\n", + ": Page No 370" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_DSS = 8.4 # in mA\n", + "I_DSS= I_DSS*10**-3 # in A\n", + "V_P = -3 # in V\n", + "V_GS = -1.5 # in V\n", + "I_D = I_DSS*( 1-(V_GS/V_P) )**2 # in A\n", + "print \"The drain current = %0.1f mA\" %(I_D*10**3)\n", + "V_GS1 = 0 # in V\n", + "g_mo = -( (2*I_DSS)/V_P ) # in A/V\n", + "g_m = g_mo*(1-(V_GS/V_P)) # in A/V\n", + "print \"Transconductacne = %0.1f mA/V\" %(g_m*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current = 2.1 mA\n", + "Transconductacne = 2.8 mA/V\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.8\n", + ": Page No 371" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_P = -4 # in V\n", + "V_GS = -2 # in V\n", + "I_DSS = 10 # in mA\n", + "I_DSS= I_DSS*10**-3 # in A\n", + "I_D = I_DSS*(1-(V_GS/V_P))**2 # in A\n", + "print \"The drain current = %0.1f mA\" %(I_D*10**3)\n", + "V_DS = V_P # in V\n", + "print \"The minimum value of V_DS = %0.f V\" %V_DS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current = 10.0 mA\n", + "The minimum value of V_DS = -4 V\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.9\n", + ": Page No 371" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "# Given data\n", + "I_DSS = -40 # in mA\n", + "V_P = 5 # in V\n", + "I_D = -15 # in mA\n", + "V_GS = V_P*(1-sqrt(I_D/I_DSS)) # in V\n", + "print \"The gate source voltage = %0.3f V\" %V_GS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The gate source voltage = 1.938 V\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.10\n", + ": Page No 372" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_DSS = 4 # in mA\n", + "I_DSS= I_DSS*10**-3 # in A\n", + "V_P = -4 # in V\n", + "V_GG = -2 # in V\n", + "V_GS = V_GG # in V\n", + "print \"The value of V_GS = %0.f V\" %V_GS\n", + "I_D = I_DSS*(1-(V_GS/V_P))**2 # in A\n", + "print \"The value of I_D = %0.f mA\" %(I_D*10**3)\n", + "V_DD = 10 # in V\n", + "R_D = 5 # in kohm\n", + "R_D = R_D * 10**3 # in ohm\n", + "V_DS = V_DD - (I_D*R_D) # in V\n", + "print \"The value of V_DS = %0.f V\" %V_DS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of V_GS = -2 V\n", + "The value of I_D = 1 mA\n", + "The value of V_DS = 5 V\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.11\n", + ": Page No 373" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import symbols, solve, N#Given data\n", + "I_D= symbols('I_D')\n", + "# Given data\n", + "V_DD= 20 # in V\n", + "R1= 2.1*10**6 # in \u03a9\n", + "R2= 270*10**3 # in \u03a9\n", + "R_D= 4.7 # in k\u03a9\n", + "R_S= 1.5 # in k\u03a9\n", + "I_DSS= 8 # in mA\n", + "V_P= -4 # in V\n", + "V_G= V_DD*R2/(R1+R2) # in V\n", + "# V_GS= V_G-R_S*I_D (as Vs= I_D*R_S) and \n", + "# I_D= I_DSS*(1-V_GS/V_P)**2 # in A\n", + "# I_D= I_DSS*(1-(V_G-R_S*I_D)/V_P)**2 # in mA or\n", + "# I_D= I_D**2*I_DSS*R_S**2/V_P**2 + I_D*(2*R_S*I_DSS/V_P-2*V_G*R_S*I_DSS/V_P**2-1) + I_DSS*(1+V_G**2/V_P**2-2*V_G/V_P)\n", + "expr= I_D**2*I_DSS*R_S**2/V_P**2 + I_D*(2*R_S*I_DSS/V_P-2*V_G*R_S*I_DSS/V_P**2-1) + I_DSS*(1+V_G**2/V_P**2-2*V_G/V_P)\n", + "I_D , x1= solve(expr, I_D)\n", + "I_DQ= I_D # in mA\n", + "print \"The value of I_DQ = %0.2f mA\" %I_DQ\n", + "V_GSQ= V_G-R_S*I_D # in V\n", + "print \"The value of V_GSQ = %0.3f V\" %V_GSQ\n", + "V_DSQ= V_DD-I_DQ*(R_D+R_S) # in V\n", + "print \"The value of V_DSQ = %0.2f V\" %V_DSQ\n", + "V_S= I_D*R_S # in V\n", + "V_D= V_S+V_DSQ #in V\n", + "V_DS= V_D-V_G # in V\n", + "print \"The value of V_S = %0.3f V\" %V_S\n", + "print \"The value of V_D = %0.3f V\" %V_D\n", + "print \"The value of V_DS = %0.3f V\" %V_DS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_DQ = 2.65 mA\n", + "The value of V_GSQ = -1.698 V\n", + "The value of V_DSQ = 3.57 V\n", + "The value of V_S = 3.976 V\n", + "The value of V_D = 7.542 V\n", + "The value of V_DS = 5.263 V\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.12\n", + ": Page No 374" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "I_D= symbols('I_D')\n", + "# Given data\n", + "V_DD= 20 # in V\n", + "I_DSS= 9 # in mA\n", + "V_BB= -10 # in V\n", + "R_S= 1.5 # in k\u03a9\n", + "R_D= 1.8 # in k\u03a9\n", + "V_P= -3 # in V\n", + "V_G=0 \n", + "# V_S= I_D*R_S+V_BB \n", + "# V_GS= V_G-V_S or\n", + "# V_GS= V_G-(I_D*R_S+V_BB)\n", + "# I_D= I_DSS*(1-V_GS/V_P)**2 or\n", + "# I_D**2*R_S**2 + I_D*[2*R_S*V_BB+2*V_P*R_S-V_P**2/I_DSS]+[V_P**2+V_BB**2+2*V_BB*V_P]\n", + "expr = I_D**2*R_S**2 + I_D*(2*R_S*V_BB+2*V_P*R_S-V_P**2/I_DSS)+(V_P**2+V_BB**2+2*V_BB*V_P)\n", + "I_D , x1= solve(expr, I_D)\n", + "I_DQ= I_D # in mA\n", + "print \"The value of I_DQ = %0.2f mA\" %I_DQ\n", + "V_GS= V_G-(I_D*R_S+V_BB) # in V\n", + "V_GSQ= V_GS # in V\n", + "print \"The value of V_GSQ = %0.3f volts\" %V_GSQ\n", + "V_DS= V_DD-I_D*(R_D+R_S)-V_BB # in V\n", + "print \"The value of V_DS = %0.3f volts\" %V_DS\n", + "V_S= I_D*R_S+V_BB # in V\n", + "print \"The value of V_S = %0.3f volts\" %V_S\n", + "V_D= V_S+V_DS # in V\n", + "print \"The value of V_D = %0.3f volts\" %V_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_DQ = 6.91 mA\n", + "The value of V_GSQ = -0.371 volts\n", + "The value of V_DS = 7.185 volts\n", + "The value of V_S = 0.371 volts\n", + "The value of V_D = 7.555 volts\n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.13\n", + ": Page No 376" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_S = 1.7 # in V\n", + "R_S = 0.51 # in kohm\n", + "R_S= R_S*10**3 # in ohm\n", + "V_DD = 18 # in V\n", + "R_D = 2*10**3 # in ohm\n", + "V_GS = -1.7 # in V\n", + "V_P = - 4.5 # in V\n", + "I_DQ = V_S/R_S #in A\n", + "print \"The value of I_DQ = %0.2f mA\" %(I_DQ*10**3)\n", + "V_GSQ = -V_S # in V\n", + "print \"The value of V_GSQ = %0.1f V\" %V_GSQ\n", + "I_DSS = I_DQ/( (1-(V_GS/V_P))**2 ) # in A\n", + "print \"The value of I_DSS = %0.1f mA\" %(I_DSS*10**3)\n", + "V_D = V_DD - (I_DQ*R_D) # in V\n", + "print \"The value of V_D = %0.2f V\" %V_D\n", + "V_DS = V_D-V_S # in V\n", + "print \"The value of V_DS = %0.2f V\" %V_DS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_DQ = 3.33 mA\n", + "The value of V_GSQ = -1.7 V\n", + "The value of I_DSS = 8.6 mA\n", + "The value of V_D = 11.33 V\n", + "The value of V_DS = 9.63 V\n" + ] + } + ], + "prompt_number": 46 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.14\n", + ": Page No 377" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "# Given data\n", + "I_DSS = 12 # in mA\n", + "V_GS = 0 # in V\n", + "I_D = 0 # in mA\n", + "V_P = -6 # in V\n", + "V_GS= np.arange(0,V_P,-0.1) # in V\n", + "I_D = I_DSS*(1-(V_GS/V_P))**2 # mA\n", + "plt.subplot(1,2,1)\n", + "plt.plot(V_GS,I_D) \n", + "plt.xlabel('V_GS in volts')\n", + "plt.ylabel('I_D in mA')\n", + "plt.title('n-channel device')\n", + "V_P = 6 # in V\n", + "V_GS= np.arange(0,V_P,0.1) # in V\n", + "I_D = I_DSS*(1-(V_GS/V_P))**2 # mA\n", + "plt.subplot(1,2,2)\n", + "plt.plot(V_GS,I_D) \n", + "plt.xlabel(\"V_GS in volts\")\n", + "plt.ylabel(\"I_D in mA\")\n", + "plt.title(\"p-channel device\")\n", + "print \"\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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SJaBtW75ffWWncZrbUA8AnDwJhIcDaWnmv/U0kQ8N9RBVio/n++4rbe6+3Pz8+LYUO3aI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\n/IkRN7BkXwAAAABJRU5ErkJggg==\n", + "text": [ + "<matplotlib.figure.Figure at 0x7f2d0df80250>" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.15\n", + ": Page No 378" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_DSS = 30 # in mA\n", + "V_GS = -5 # in V\n", + "V_GS_off = -8 # in V\n", + "I_D = I_DSS*(1-(V_GS/V_GS_off))**2 # in mA\n", + "print \"The drain current = %0.3f mA\" %I_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current = 4.219 mA\n" + ] + } + ], + "prompt_number": 62 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.16\n", + ": Page No 378" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_D = 1.5 # in mA\n", + "I_DSS = 5 # in mA\n", + "V_P = -2 # in V\n", + "V_GS = V_P*(1-sqrt(I_D/I_DSS)) # in V\n", + "V_G = 0 # in V\n", + "V_S = V_G-V_GS # in V\n", + "R_S = V_S/I_D # in kohm\n", + "print \"The source resistance = %0.f ohm\" %(R_S*10**3)\n", + "V_DD = 20 # in V\n", + "V_DS= 10 # in V\n", + "R_D = (V_DD-(V_DS+(I_D*R_S)))/(I_D) # in kohm\n", + "print \"The diode resistance = %0.f K ohm\" %R_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The source resistance = 603 ohm\n", + "The diode resistance = 6 K ohm\n" + ] + } + ], + "prompt_number": 63 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.17\n", + ": Page No 379" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_D = 0.8 # in mA\n", + "I_D= I_D*10**-3 # in A\n", + "I_DSS = 1.645 # in mA\n", + "I_DSS= I_DSS*10**-3 # in A\n", + "V_P = -2 # in V\n", + "V_GS = V_P * (1-sqrt(I_D/I_DSS)) # in V\n", + "print \"The gate source voltage = %0.2f V\" %V_GS\n", + "g_mo = -((2*I_DSS)/V_P) # in A/V\n", + "g_m = g_mo*(1-(V_GS/V_P)) # in A/V\n", + "print \"The transconductance = %0.2f mA/V\" %(g_m*10**3)\n", + "R_S = -(V_GS/I_D) # in ohm\n", + "print \"The source resistance = %0.f ohm\" %R_S\n", + "AdB= 20 # in dB\n", + "A= 10**(AdB/20) \n", + "R_D= A/g_m # in ohm\n", + "print \"The value of R_D = %0.2f k\u03a9\" %(R_D*10**-3)\n", + "\n", + "# Note: There is calculation error to find the value of R_S in the book . So the answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The gate source voltage = -0.61 V\n", + "The transconductance = 1.15 mA/V\n", + "The source resistance = 757 ohm\n", + "The value of R_D = 8.72 k\u03a9\n" + ] + } + ], + "prompt_number": 64 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.18\n", + ": Page No 381" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_GG = 2 # in V\n", + "V_GS = -V_GG # in V\n", + "print \"The value of V_GS = %0.f V\" %V_GS\n", + "I_DSS = 10 # in mA\n", + "V_P = -8 # in V\n", + "I_D = I_DSS*(1-(V_GS/V_P))**2 # in mA\n", + "I_DQ= I_D # in mA\n", + "print \"The value of I_DQ = %0.3f mA\" %I_DQ\n", + "R_D = 2 # in K ohm\n", + "V_DD = 16 # in V\n", + "V_DS = V_DD - (I_D*R_D) # in V\n", + "print \"The value of V_DS = %0.2f V\" %V_DS\n", + "V_D = V_DS # in V\n", + "print \"The value of V_D = %0.2f V\" %V_D\n", + "V_G = V_GS # in V\n", + "print \"The value of V_G = %0.f V\" %V_G\n", + "V_S = 0 # in V\n", + "print \"The value of V_S = %0.f V\" %V_S" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of V_GS = -2 V\n", + "The value of I_DQ = 5.625 mA\n", + "The value of V_DS = 4.75 V\n", + "The value of V_D = 4.75 V\n", + "The value of V_G = -2 V\n", + "The value of V_S = 0 V\n" + ] + } + ], + "prompt_number": 65 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.19\n", + ": Page No 381" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_GS = 10 # in V\n", + "I_G = 0.001 # in \u00b5A\n", + "R_GS = V_GS/I_G # in M\u03a9\n", + "print \"The gate source resistance = %0.f M\u03a9\" %R_GS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The gate source resistance = 10000 M\u03a9\n" + ] + } + ], + "prompt_number": 66 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.20\n", + ": Page No 382" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "del_VDS = 1.5 # in V\n", + "del_ID = 120 * 10**-6 # in A\n", + "r_d = del_VDS/del_ID # in ohm\n", + "r_d = r_d * 10**-3 # in kohm\n", + "print \"The drain resistance of the JFET = %0.1f K ohm\" %r_d" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain resistance of the JFET = 12.5 K ohm\n" + ] + } + ], + "prompt_number": 67 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.21\n", + ": Page No 382" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_DSS = 8.4 # in mA\n", + "V_P = -3 # in V\n", + "V_GS = -1.5 # in V\n", + "I_D = I_DSS*(1-(V_GS/V_P))**2 # in mA\n", + "g_mo = -( (2*I_DSS)/V_P ) # in mA/V\n", + "g_m = g_mo*(1-(V_GS/V_P)) # in mA/V\n", + "print \"The value of g_m = %0.1f mA/V\" %g_m" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of g_m = 2.8 mA/V\n" + ] + } + ], + "prompt_number": 68 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.22\n", + ": Page No 383" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import symbols, solve, N#Given data\n", + "V_GS= symbols('V_GS')\n", + "# Given data\n", + "V_DD= 20 # in V\n", + "I_DSS= 9 # in mA\n", + "V_P= -3 # in V\n", + "R1= 0.3*10**3 # in k\u03a9\n", + "R2= 1.7*10**3 #in k\u03a9\n", + "R_D= 3.2 # in k\u03a9\n", + "R=1 # in k\u03a9\n", + "V_G= V_DD*R1/(R1+R2) # in V\n", + "#I_D= I_DSS*[1-V_GS/V_P]**2 (i)\n", + "# V_G= V_GS+I_D*R or I_D= (V_G-V_GS)/R (ii)\n", + "# From (i) and (ii)\n", + "#V_GS*1/V_P**2+V_GS*[1/(R*I_DSS)-2/V_P]+[1-V_G/(R*I_DSS)]=0\n", + "expr= V_GS**2*(R*I_DSS/V_P**2)+V_GS*(1-2*R*I_DSS/V_P)+(R*I_DSS-V_G)\n", + "x1 , V_GS= solve(expr, V_GS)\n", + "I_D= I_DSS*(1-V_GS/V_P)**2 # in mA\n", + "print \"The value of I_D = %0.f mA\" %I_D\n", + "V_S= I_D*R #in V\n", + "V_D= V_DD-I_D*R_D # in V\n", + "V_DS= V_D-V_S # in V\n", + "gm= -2*I_DSS/V_P*(1-V_GS/V_P) # in mA/V\n", + "print \"The value of V_DS = %0.1f volts\" %V_DS\n", + "print \"The transconductance = %0.f mA/V\" %gm" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_D = 4 mA\n", + "The value of V_DS = 3.2 volts\n", + "The transconductance = 4 mA/V\n" + ] + } + ], + "prompt_number": 87 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.23\n", + ": Page No 385" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "r_d = 25 # in k\u03a9\n", + "R1 = r_d # in k\u03a9\n", + "R2 = r_d # in k\u03a9\n", + "g_m = 2 #mA/V\n", + "g_m= g_m*10**-3 # in A/V\n", + "R_L = (r_d*R1*R2)/(r_d*R1+R1*R2+R2*r_d) # in k\u03a9\n", + "R_L= R_L*10**3 # in \u03a9\n", + "A_v = -g_m*R_L \n", + "print \"The voltage gain = %0.2f\" %A_v" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage gain = -16.67\n" + ] + } + ], + "prompt_number": 88 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.24\n", + ": Page No 386" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_GS = 15 # in V\n", + "I_G = 1 # in nA\n", + "I_G =I_G * 10**-9 # in A\n", + "R_in = V_GS/I_G # in \u03a9\n", + "print \"Input resistance = %0.f G\u03a9\" %(R_in*10**-9)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Input resistance = 15 G\u03a9\n" + ] + } + ], + "prompt_number": 90 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.25\n", + ": Page No 386" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_DSS = 20 # in mA\n", + "V_P = 4 # in V\n", + "I_D = I_DSS # in mA\n", + "print \"The maximum drain current = %0.f mA\" %I_D\n", + "V_GS = -V_P # in V\n", + "print \"The gate source cut off voltage = %0.f volts\" %V_GS\n", + "R_DS = V_P/I_DSS # in k\u03a9\n", + "print \"The value of ohmic resistance = %0.f \u03a9\" %(R_DS*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum drain current = 20 mA\n", + "The gate source cut off voltage = -4 volts\n", + "The value of ohmic resistance = 200 \u03a9\n" + ] + } + ], + "prompt_number": 92 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.26\n", + ": Page No 386" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_DSS= 16*10**-3 # in A\n", + "V_GSoff= -6 #in V\n", + "V_GS= V_GSoff/2 # in V\n", + "I_D= I_DSS*(1-V_GS/V_GSoff)**2 # in A\n", + "print \"The drain current = %0.f mA\" %(I_D*10**3)\n", + "V_GS= abs(V_GSoff)/2 # in V\n", + "print \"The gate voltage = %0.f volts\" %V_GS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current = 4 mA\n", + "The gate voltage = 3 volts\n" + ] + } + ], + "prompt_number": 93 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.27\n", + ": Page No 387" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_DD = 15 # in V\n", + "R_D = 10 # in kohm\n", + "R_D = R_D * 10**3 # in ohm\n", + "I_D = V_DD/R_D # in A\n", + "print \"The drain current = %0.1f mA\" %(I_D*10**3)\n", + "V_D = V_DD - I_D*R_D # in V\n", + "print \"The drain voltage = %0.f V\" %V_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current = 1.5 mA\n", + "The drain voltage = 0 V\n" + ] + } + ], + "prompt_number": 94 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.28\n", + ": Page No 388" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "# Given data\n", + "R2 = 1 # in M ohm\n", + "R2 = R2 * 10**6 # in ohm\n", + "R1 = 1.5 # in M ohm\n", + "R1 = R1 * 10**6 # in ohm\n", + "V_DD = 25 # in V\n", + "V_G = (R2*V_DD)/(R1+R2) # in V\n", + "R_S = 22 # in kohm\n", + "R_S = R_S * 10**3 # in ohm\n", + "I_D = V_G/R_S # in A\n", + "print \"The drain current = %0.2f mA\" %(I_D*10**3)\n", + "R_D = 10 # in kohm\n", + "R_D = R_D * 10**3 # in ohm\n", + "V_D = V_DD - (I_D*R_D) #in V\n", + "V_S = 10 # in V\n", + "V_DS = V_D - V_S # in V\n", + "print \"The Drain source voltage = %0.1f V\" %V_DS\n", + "print \"Thus the Q-point is : (\",round(V_DS,1),\"V,\",round(I_D*10**3,2),\"mA)\"\n", + "I_Dsat = V_DD/R_D # in A\n", + "V_DS = V_DD # in V\n", + "V_D= np.arange(0,25,0.1) # in V\n", + "I_D= (V_DD-V_D)/R_D*10**3 # in mA\n", + "plt.plot(V_D,I_D) \n", + "plt.xlabel(\"V_DS in volts\") \n", + "plt.ylabel(\"I_D in mA\") \n", + "plt.title(\"DC load line\") \n", + "print \"DC load line shown in figure\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current = 0.45 mA\n", + "The Drain source voltage = 10.5 V\n", + "Thus the Q-point is : ( 10.5 V, 0.45 mA)\n", + "DC load line shown in figure" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x7f2d246a6ad0>" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.29\n", + ": Page No 389" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_SS = 25 # in V\n", + "V_GS = 0 # in V\n", + "R_S = 18 # in kohm\n", + "R_S = R_S * 10**3 # in ohm\n", + "I_D = (V_SS-V_GS)/R_S # in A\n", + "print \"The drain current = %0.2f mA\" %(I_D*10**3)\n", + "V_DD = 25 # in V\n", + "R_D = 7.5 # in kohm\n", + "R_D = R_D * 10**3 # in ohm\n", + "V_D = V_DD - (I_D*R_D) # in V\n", + "print \"The drain voltage = %0.2f V\" %V_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current = 1.39 mA\n", + "The drain voltage = 14.58 V\n" + ] + } + ], + "prompt_number": 102 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.30\n", + ": Page No 390 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_D = 1 # in kohm\n", + "R_D = R_D * 10**3 # in ohm\n", + "V_in = 2 # in mV\n", + "V_in = V_in * 10**-3 # in V\n", + "R_L = 10 # in kohm\n", + "R_L = R_L * 10**3 # in ohm\n", + "r_d = (R_D*R_L)/(R_D+R_L) # in ohm\n", + "g_m = 3000 #in \u00b5S\n", + "g_m = g_m * 10**-6 # in S\n", + "A_v = g_m*r_d \n", + "V_out = A_v*V_in # in V\n", + "V_out = V_out * 10**3 # in mV\n", + "print \"The output Voltage = %0.2f mV\" %V_out" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output Voltage = 5.45 mV\n" + ] + } + ], + "prompt_number": 103 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Engineering_by_P._Raja/chapter_7_2.ipynb b/Electronics_Engineering_by_P._Raja/chapter_7_2.ipynb new file mode 100644 index 00000000..299a6c38 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_7_2.ipynb @@ -0,0 +1,519 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter - 7 : Metal Oxide Semiconductor Field Effect Transistors (MOSFET)" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.1\n", + ": Page No 403" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# Given data\n", + "V_GS = 0 # in V\n", + "I_D = 4 # in mA\n", + "R = 2 # in kohm\n", + "V_DD = 15 # in V\n", + "V_DS = V_DD - (I_D*R) # in V\n", + "g_m = 2000 # in \u00b5S\n", + "g_m= g_m*10**-6 # in S\n", + "g_mo = g_m # in S\n", + "R_D = 2 # in kohm\n", + "R_D = R_D * 10**3 # in ohm\n", + "R_L = 10 # in kohm\n", + "R_L = R_L * 10**3 # in ohm\n", + "r_d = (R_D*R_L)/(R_D+R_L) # in ohm\n", + "A_v = g_m*r_d \n", + "V_in = 20 # in mV\n", + "V_out = A_v * V_in # in mV\n", + "print \"The output voltage = %0.1f mV\" %V_out" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 66.7 mV\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.2\n", + ": Page No 411" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 20 # in V\n", + "V2 = 2 # in V\n", + "V = V1-V2 # in V\n", + "R = 1 # in kohm\n", + "R = R * 10**3 # in ohm\n", + "I_D = V/R # in A\n", + "I_D = I_D * 10**3 # in mA\n", + "print \"The drain current = %0.f mA\" %I_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current = 18 mA\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.3\n", + ": Page No 414" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "# Given data\n", + "I_DSS = 10 # in mA\n", + "V_GS = 0 # in V\n", + "I_D = 0 # in mA\n", + "V_P = -4 # in V\n", + "V_GS= np.arange(0,V_P,-0.1) # in V\n", + "I_D = I_DSS*(1-(V_GS/V_P))**2 # mA\n", + "plt.plot(V_GS,I_D) \n", + "plt.xlabel(\"V_gs in volts\") \n", + "plt.ylabel(\"I_D in mA\") \n", + "plt.title(\"Transfer characteristics for an n-channel depletion type MOSFET\")\n", + "print \"Transfer characteristics for an n-channel depletion type MOSFET Shown in figure\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Transfer characteristics for an n-channel depletion type MOSFET Shown in figure\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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WIio0NmwA1qwBzp4FbG3VjqbgYh8LERUKBw8CI0YAv/4KeHioHU3e0ULdyRYL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\n", + "text": [ + "<matplotlib.figure.Figure at 0x7f9785cf0a50>" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.4\n", + ": Page No 414" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_GS = 0 # in V\n", + "I_DSS = 10 # in mA\n", + "I_D = I_DSS # in mA\n", + "R_D = 1.5 # in kohm\n", + "V_DD = 20 # in V\n", + "V_DS = V_DD - (I_D*R_D) # in V\n", + "print \"The value of V_DS = %0.f V\" %V_DS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of V_DS = 5 V\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.5\n", + ": Page No 415" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_D = 5 # in mA\n", + "V_GS1 = 8 # in V\n", + "V_GS2 = 4 # in V\n", + "V_GS = 6 # in V\n", + "K = I_D/(V_GS1-V_GS2)**2 # in mA/V**2\n", + "I_D = K*(V_GS-V_GS2)**2 # in mA\n", + "print \"The drain current = %0.2f mA\" %I_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current = 1.25 mA\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.6\n", + ": Page No 415" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_T = 1 # in V\n", + "I_D = 4 # in mA\n", + "V_GS = 5 # in V\n", + "V_GSth = 1 # in V\n", + "K = I_D/(V_GS-V_GSth)**2 # in mA/V**2\n", + "print \"The value of K = %0.2f mA/V**2\" %K" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of K = 0.25 mA/V**2\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.7\n", + ": Page No 415" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_GS = 3 # in V\n", + "V_GSth=2 # inV\n", + "# Part (a)\n", + "print \"Part (a) : For V_DS= 0.5 V\"\n", + "V_DS= 0.5 # in V\n", + "if V_DS<(V_GS-V_GSth) :\n", + " print \"Transistor is in ohmic region\"\n", + "else :\n", + " print \"Transistor is in saturation region\"\n", + "\n", + "# Part (b)\n", + "print \"Part (b) : For V_DS= 1 V\"\n", + "V_DS= 1 # in V\n", + "if V_DS<(V_GS-V_GSth) :\n", + " print \"Transistor is in ohmic region\"\n", + "else :\n", + " print \"Transistor is in saturation region\"\n", + "\n", + "# Part (c)\n", + "print \"Part (c) : For V_DS= 5 V\"\n", + "V_DS= 5 # in V\n", + "if V_DS<(V_GS-V_GSth) :\n", + " print \"Transistor is in ohmic region\"\n", + "else :\n", + " print \"Transistor is in saturation region\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (a) : For V_DS= 0.5 V\n", + "Transistor is in ohmic region\n", + "Part (b) : For V_DS= 1 V\n", + "Transistor is in saturation region\n", + "Part (c) : For V_DS= 5 V\n", + "Transistor is in saturation region\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.8\n", + ": Page No 416" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_DSS = 4 # in mA\n", + "V_GSoff = -2 # in V\n", + "V_GS = -0.5 # in V\n", + "I_D = I_DSS*(1-(V_GS/V_GSoff))**2 # in mA\n", + "print \"At V_GS=-0.5 V, the drain current = %0.2f mA\" %I_D\n", + "V_GS = -1 #in V\n", + "I_D = I_DSS*(1-(V_GS/V_GSoff))**2 # in mA\n", + "print \"At V_GS=-1.0 V, the drain current = %0.f mA\" %I_D\n", + "V_GS = -1.5 # in V\n", + "I_D = I_DSS*(1-(V_GS/V_GSoff))**2 # in mA\n", + "print \"At V_GS=-1.5 V, the drain current = %0.2f mA\" %I_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "At V_GS=-0.5 V, the drain current = 2.25 mA\n", + "At V_GS=-1.0 V, the drain current = 1 mA\n", + "At V_GS=-1.5 V, the drain current = 0.25 mA\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.9\n", + ": Page No 416" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_DSS = 12 # in mA\n", + "I_DSS= I_DSS*10**-3 # in A\n", + "I_D = I_DSS # in A\n", + "V_DD = 12 # in V\n", + "R_D = 470 # in ohm\n", + "V_DS = V_DD - (I_D*R_D) # in V\n", + "print \"The circuit drain current = %0.f mA\" %(I_D*10**3)\n", + "print \"The drain source voltage = %0.2f V\" %V_DS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The circuit drain current = 12 mA\n", + "The drain source voltage = 6.36 V\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.10\n", + ": Page No 417" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_D = 12 # in mA\n", + "I_D= I_D*10**-3 # in A\n", + "I_DSS = I_D # in A\n", + "V_DS = 6.36 # in V\n", + "g_mo = 4000 # in \u00b5S\n", + "g_mo=g_mo*10**-6 # in S\n", + "g_m = g_mo # in S\n", + "R_D = 470 # in ohm\n", + "R_L = 2 # in kohm\n", + "R_L = R_L * 10**3 # in ohm\n", + "r_d = (R_D*R_L)/(R_D+R_L) # in ohm\n", + "print \"The value of r_d = %0.2f \u03a9\" %r_d\n", + "A_v = g_m*r_d \n", + "print \"The value of A_v = %0.2f\" %A_v\n", + "V_in = 100 # in mV\n", + "V_in = V_in *10**-3 # in V\n", + "V_out = A_v*V_in # in V\n", + "print \"The value of Vout = %0.2f V\" %V_out" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of r_d = 380.57 \u03a9\n", + "The value of A_v = 1.52\n", + "The value of Vout = 0.15 V\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.11\n", + ": Page No 417 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_DS = 0.1 # in V\n", + "I_D = 10 # in mA\n", + "I_D= I_D*10**-3 # in A\n", + "R_DS = V_DS/I_D # in ohm\n", + "print \"Part (a) The value of R_DS(on) = %0.f ohm\" %R_DS\n", + "V_DS = 0.75 # in V\n", + "I_D = 100 # in mA\n", + "I_D= I_D*10**-3 # in A\n", + "R_DS = V_DS/I_D # in ohm \n", + "print \"Part (b) The value of R_DS(on) = %0.1f ohm\" %R_DS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (a) The value of R_DS(on) = 10 ohm\n", + "Part (b) The value of R_DS(on) = 7.5 ohm\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.12\n", + ": Page No 418 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_D = 500 # in mA\n", + "V_GS = 3 # in V\n", + "R_DS = 2 # in ohm\n", + "V_DD = 20 # in V\n", + "R1 = 1 # in kohm\n", + "R1 = R1 * 10**3 # in ohm\n", + "V_out = (R_DS/(R1+R_DS))*V_DD # in V\n", + "print \"The output voltage = %0.2f V\" %V_out" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 0.04 V\n" + ] + } + ], + "prompt_number": 32 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Engineering_by_P._Raja/chapter_8_2.ipynb b/Electronics_Engineering_by_P._Raja/chapter_8_2.ipynb new file mode 100644 index 00000000..869fcc8c --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_8_2.ipynb @@ -0,0 +1,860 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter - 8 : Operational Amplifiers (OPAMPs)" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.1\n", + ": Page No 432 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# Given data\n", + "A_V = -100 \n", + "R1 = 2.2 # in kohm\n", + "R1 = R1*10**3 # in ohm\n", + "R_f =-( A_V*R1) # in ohm\n", + "R_f = R_f * 10**-3 # in kohm\n", + "print \"The resistance value = %0.f k\u03a9\" %R_f" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resistance value = 220 k\u03a9\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.2\n", + ": Page No 432" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = 200 # in kohm \n", + "R1 = 2 # in kohm\n", + "A_V = - (R_f/R1) \n", + "V_in = 2.5 # in mV\n", + "V_in= V_in*10**-3 # in V\n", + "V_o = (A_V * V_in) # in V\n", + "print \"The output voltage = %0.2f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = -0.25 V\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.3\n", + ": Page No 461" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 2 # in V\n", + "R_f = 500 # in kohm\n", + "R_f = R_f*10**3 # in ohm\n", + "R1 = 100 # in kohm\n", + "R1 = R1 * 10**3 # in ohm\n", + "V_o = (1+(R_f/R1))*V1 # in V\n", + "print \"The output voltage = %0.f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 12 V\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.4\n", + ": Page No 461" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = 1 # in Mohm\n", + "R_f = R_f * 10**6 # in ohm\n", + "print \"Part (a)\"\n", + "V1 = 1 # in V\n", + "V2 = 2 # in V\n", + "V3 = 3 # in V\n", + "R1 = 500 # in kohm\n", + "R1 = R1 * 10**3 # in ohm\n", + "R2 = 1 # in Mohm\n", + "R2 = R2 * 10**6 # in ohm\n", + "R3 = 1 # in Mohm\n", + "R3 = R3 * 10**6 # in ohm\n", + "V_o = -(R_f) * ( (V1/R1)+(V2/R2)+(V3/R3) ) # in V\n", + "print \"The output voltage = %0.f V\" %V_o\n", + "\n", + "print \"Part (b)\"\n", + "V1 = -2 # in V\n", + "V2 = 3 # in V\n", + "V3 = 1 # in V\n", + "R1 = 200 # in kohm\n", + "R1 = R1 * 10**3 # in ohm\n", + "R2 = 500 # in kohm\n", + "R2 = R2 * 10**3 # in ohm\n", + "R3 = 1 # in Mohm\n", + "R3 = R3 * 10**6 # in ohm\n", + "V_o = -(R_f) * ( (V1/R1)+(V2/R2)+(V3/R3) ) # in V\n", + "print \"The output voltage = %0.f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (a)\n", + "The output voltage = -7 V\n", + "Part (b)\n", + "The output voltage = 3 V\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.6\n", + ": Page No 462" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = 0 # in V\n", + "R1 = 2 # in kohm\n", + "R1 = R1 * 10**3 # in ohm\n", + "A_vmin = (1+(R_f/R1)) \n", + "print \"The minimum closed loop voltage gain = %0.f\" %A_vmin\n", + "R_f1 = 100 # in kohm\n", + "R_f1 = R_f1 * 10**3 # in ohm\n", + "A_vmax = (1+(R_f1/R1)) \n", + "print \"The maximum closed loop voltage gain = %0.f\" %A_vmax" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum closed loop voltage gain = 1\n", + "The maximum closed loop voltage gain = 51\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.7\n", + ": Page No 463" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 745 # in \u00b5V\n", + "V1 = V1 * 10**-6 # in V\n", + "V2 = 740 # in \u00b5V\n", + "V2 = V2 * 10**-6 # in V\n", + "Av = 5*10**5 \n", + "CMRR = 80 # in dB\n", + "# Formula CMRR in dB= 20*log(Av/Ac)\n", + "Ac= Av/(10**(CMRR/20)) \n", + "print \"The common mode gain = %0.2f\" %Ac\n", + "V_o = Av*(V1-V2)+Ac*((V1+V2)/2) # in V\n", + "print \"The output voltage = %0.2f V\" %V_o\n", + "\n", + "# Note: In the book the calculation of finding the value of common mode gain (i.e. Ac) is wrong, \n", + "# so the answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The common mode gain = 50.00\n", + "The output voltage = 2.54 V\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.8\n", + ": Page No 464" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = 1 # in Mohm\n", + "R_f = R_f * 10**6 # in ohm\n", + "Ri= 1*10**6 # in ohm\n", + "R1 = Ri # in ohm\n", + "A_VF = -(R_f/R1) \n", + "print \"The Voltage gain = %0.f\" %A_VF" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Voltage gain = -1\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.10\n", + ": Page No 465" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_F = 3 # in kohm\n", + "R1 = 1 # in kohm\n", + "V1 = 2 # in V\n", + "V2 = 3 # in V\n", + "V_o1 = (1+(R_F/R1))*V1 # in V\n", + "V_o2 = (1+(R_F/R1))*V2 # in V\n", + "V_o = V_o1+V_o2 # in V\n", + "print \"The output voltage = %0.f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 20 V\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.11\n", + ": Page No 466" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_i = 10 # in k\u03a9 \n", + "R_im = 20 # in k\u03a9\n", + "R_f = 500 # in k\u03a9\n", + "A_vmin = -(R_f/R_i) \n", + "print \"Closed loop voltage gain corresponding to minimum resistance = %0.f\" %A_vmin\n", + "A_vmax = -(R_f/R_im) \n", + "print \"Closed loop voltage gain corresponding to maximum resistance = %0.f\" %A_vmax" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Closed loop voltage gain corresponding to minimum resistance = -50\n", + "Closed loop voltage gain corresponding to maximum resistance = -25\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.12\n", + ": Page No 467" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = 200 # in kohm\n", + "R1 = 20 # in kohm\n", + "A_v = -(R_f/R1) \n", + "V_i = 0.1 # in V\n", + "V_im = 0.5 # in V\n", + "V_omin = -10*V_i # in V\n", + "print \"The minimum output voltage = %0.f V\" %V_omin\n", + "V_omax = -10*V_im # in V\n", + "print \"The maximum output voltage = %0.f V\" %V_omax\n", + "print \"Output voltage ranges : from\",int(V_omin,),\"to\",int(V_omax)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum output voltage = -1 V\n", + "The maximum output voltage = -5 V\n", + "Output voltage ranges : from -1 to -5\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.13\n", + ": Page No 467" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "# Given data\n", + "R = 133 # in kohm\n", + "R = R *10**3 # in ohm\n", + "C = 0.1 # in \u00b5F\n", + "C = 0.1 * 10**-6 # in F\n", + "Vi= 15 # in V\n", + "plt.subplot(2,1,1)\n", + "plt.plot([0,10],[1.5,1.5]) \n", + "plt.ylabel(\"Vi in volts\")\n", + "plt.xlabel(\"t\")\n", + "plt.title(\"Input voltage\") \n", + "t=np.arange(0,1,0.1)\n", + "Vo= -1/(R*C)*t \n", + "plt.subplot(2,1,2)\n", + "plt.plot(t,Vo)\n", + "plt.xlabel(\"t\") \n", + "plt.ylabel(\"Vo in volts\") \n", + "plt.title(\"Output voltage\")\n", + "print \"\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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dXbt20bhxY/bu3cv06dP1Pp5RS4MUV9++fenbt6+xwzC6Tp065fslWt506NCB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\nIYTQiyQMIYQQevn/1U0AkpdVVxgAAAAASUVORK5CYII=\n", + "text": [ + "<matplotlib.figure.Figure at 0x7fdbd4185210>" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.14\n", + ": Page No 468" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "Rf = 250 # in kohm\n", + "Vo= '-5*Va+3*Vb' # given expression\n", + "# But output voltage of difference amplifier is \n", + "# Vo= -Rf/R1*Va+(R2/(R1+R2))*(1+Rf/R1)*Vb (i)\n", + "# By comparing (i) with given expression\n", + "R1 = Rf/5 # in kohm\n", + "print \"The value of R1 = %0.f k\u03a9\" %R1\n", + "# (R2/(R1+R2))*(1+Rf/R1)= 3\n", + "R2= 3*R1**2/(R1+Rf-3*R1) # in k\u03a9\n", + "print \"The value of R2 = %0.f k\u03a9\" %R2\n", + "\n", + "# Note: There is calculation error to find the value of R2 in the book." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of R1 = 50 k\u03a9\n", + "The value of R2 = 50 k\u03a9\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.15\n", + ": Page No 469" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_i1 = 150 # in \u00b5V\n", + "V_i2 = 140 # in \u00b5V\n", + "V_d = V_i1-V_i2 # in \u00b5V\n", + "V_C = (1/2)*(V_i1+V_i2) # in \u00b5V\n", + "print \"Part (i)\"\n", + "CMRR = 100 \n", + "A_d = 4000 \n", + "V_o = (A_d * V_d)*(1+(1/CMRR)*(V_C/V_d)) # in \u00b5V\n", + "V_o = V_o * 10**-3 # in mV\n", + "print \"The output voltage = %0.1f mV\" %V_o\n", + "print \"Part(ii)\"\n", + "CMRR = 10**5 \n", + "V_o = (A_d * V_d)*(1+(1/CMRR)*(V_C/V_d)) # in \u00b5V\n", + "V_o = V_o * 10**-3 # in mV\n", + "print \"The output voltage = %0.3f mV\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (i)\n", + "The output voltage = 45.8 mV\n", + "Part(ii)\n", + "The output voltage = 40.006 mV\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.16\n", + ": Page No 470" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = 470 # in k\u03a9\n", + "R1 = 4.3 # in k\u03a9\n", + "R2 = 33 # in k\u03a9\n", + "R3 = R2 # in k\u03a9\n", + "A1 = (1+R_f/R1) \n", + "A2 = -(R_f/R2) \n", + "A3 = -(R_f/R3) \n", + "A = A1*A2*A3 \n", + "V_i = 80 # in \u00b5V\n", + "V_i= 80*10**-6 # in V\n", + "V_o = A*V_i \n", + "print \"The output voltage = %0.2f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 1.79 V\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.18\n", + ": Page No 472" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R4 = 300 # in k\u03a9\n", + "R2 = 150 # in k\u03a9\n", + "R3 = 10 # in k\u03a9\n", + "R1 = 10 # in k\u03a9\n", + "V1 = 1 # in V\n", + "V2 = 2 # in V\n", + "V_o = ( (1+(R4/R2))*((R3/(R1+R3))*V1)-((R4/R2)*V2) ) # in V\n", + "print \"The output voltage = %0.1f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = -2.5 V\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.19\n", + ": Page No 472" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_o = 2 # in V\n", + "R_i = 20 # in k\u03a9\n", + "R_f = 1 # in M\u03a9\n", + "V_i = -((V_o*R_i)/R_f) # in mV\n", + "print \"The input volatge = %0.f mV\" %V_i" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The input volatge = -40 mV\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.20\n", + ": Page No 473" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = 200 # in k\u03a9\n", + "R_i = 30 # in k\u03a9\n", + "V_i = 0.1 # in V\n", + "V_im = 0.5 # in V\n", + "Vo_min = -((R_f/R_i)*V_i) # in V\n", + "print \"The minimum output voltage = %0.2f V\" %Vo_min\n", + "Vo_max = -((R_f/R_i)*V_im) # in V\n", + "print \"The minimum output voltage = %0.2f V\" %Vo_max\n", + "print \"The output voltage range is : \",round(Vo_min,2),\"V to\",round(Vo_max,2),\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum output voltage = -0.67 V\n", + "The minimum output voltage = -3.33 V\n", + "The output voltage range is : -0.67 V to -3.33 V\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.21\n", + ": Page No 473" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = 360 # in kohm\n", + "R_i = 12 # in kohm\n", + "V1 = - 0.3 # in V\n", + "V_o = (1+(R_f/R_i))*V1 # in V\n", + "print \"The output voltage = %0.1f V\" %V_o\n", + "V_o1 = 2.4 # in V\n", + "V_i = V_o1/(1+(R_f/R_i)) # in V\n", + "print \"The input voltage = %0.2f mV\" %(V_i*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = -9.3 V\n", + "The input voltage = 77.42 mV\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.22\n", + ": Page No 474" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = -68 # in kohm\n", + "R1 = 33 # in kohm\n", + "R2 = 22 # in kohm\n", + "R3 = 12 # in kohm\n", + "V1 = 0.2 # in V\n", + "V2 = - 0.5 # in V\n", + "V3 = 0.8 # in V\n", + "V_o = ((R_f/R1)*V1) + ((R_f/R2)*V2) + ((R_f/R3)*V3) # in V\n", + "print \"The output voltage = %0.3f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = -3.400 V\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.23\n", + ": Page No 475" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = 1.8 # in kohm\n", + "R_f = R_f * 10**3 # in ohm\n", + "R1 = 180 # in ohm\n", + "A_v = (R_f/R1) \n", + "print \"Closed loop gain = %0.f\" %A_v\n", + "F = 1 # in MHz\n", + "F = F * 10**6 # in Hz\n", + "f2 = F/A_v # in Hz\n", + "print \"Closed loop bandwidth = %0.f Hz\" %f2\n", + "V_in = 25 # in mV\n", + "V_in = V_in * 10**-3 # in V\n", + "V_o = A_v*V_in # in V\n", + "print \"The output voltage = %0.2f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Closed loop gain = 10\n", + "Closed loop bandwidth = 100000 Hz\n", + "The output voltage = 0.25 V\n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.24\n", + ": Page No 475 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = 3 # in K ohm\n", + "R_f = R_f * 10**3 # in ohm\n", + "R1 = 150 # in ohm\n", + "A_v = (R_f/R1) + 1 \n", + "print \"Close loop gain for inverting amplifier = %0.f\" %A_v\n", + "f = 1 # in MHz\n", + "f = f * 10**6 # in Hz\n", + "f2 = f/A_v # in Hz\n", + "f2 = f2 * 10**-3 # in KHz\n", + "print \"The closed loop bandwidth = %0.2f KHz\" %f2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Close loop gain for inverting amplifier = 21\n", + "The closed loop bandwidth = 47.62 KHz\n" + ] + } + ], + "prompt_number": 42 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Engineering_by_P._Raja/chapter_9_2.ipynb b/Electronics_Engineering_by_P._Raja/chapter_9_2.ipynb new file mode 100644 index 00000000..bbc73967 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_9_2.ipynb @@ -0,0 +1,183 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter - 9 : Electronic Instrumentation And Measurements" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.1\n", + ": Page No 512 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# Given data\n", + "scale= 5 # in mV/cm\n", + "gh= 5.2 #amplitude of the graph in cm\n", + "PtoPamplitude= gh*scale # in mV\n", + "print \"Peak-to-peak amplitude = %0.f mV\" %PtoPamplitude" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Peak-to-peak amplitude = 26 mV\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.2\n", + ": Page No 512" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "scale= 100 # in mV/cm\n", + "gh= 5.2 #amplitude of the graph in cm\n", + "PtoPamplitude= gh*scale # in mV\n", + "print \"Peak-to-peak amplitude = %0.2f V\" %(PtoPamplitude*10**-3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Peak-to-peak amplitude = 0.52 V\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.3\n", + ": Page No 513" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "scale= 20 # in \u00b5S/cm\n", + "gh= 3.2 #amplitude of the graph in cm\n", + "T= gh*scale # in mV\n", + "print \"The period of the waveform = %0.f \u00b5S\" %T" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The period of the waveform = 64 \u00b5S\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.4\n", + ": Page No 513" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "scale= 50 # in \u00b5S/cm\n", + "gh= 2 #amplitude of the graph in cm\n", + "T_PD= gh*scale # in mV\n", + "print \"The pulse delay for the waveform = %0.f \u00b5s\" %T_PD" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The pulse delay for the waveform = 100 \u00b5s\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.5\n", + ": Page No 514" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "scale= 2 # in \u00b5S/cm\n", + "gh= 4.6 #amplitude of the graph in cm\n", + "T_PQ= gh*scale # in mV\n", + "print \"The pulse width of the waveform = %0.1f \u00b5s\" %T_PQ" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The pulse width of the waveform = 9.2 \u00b5s\n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Engineering_by_P._Raja/screenshots/7_2.png b/Electronics_Engineering_by_P._Raja/screenshots/7_2.png Binary files differnew file mode 100644 index 00000000..c25bf470 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/screenshots/7_2.png diff --git a/Electronics_Engineering_by_P._Raja/screenshots/snap-3_2.png b/Electronics_Engineering_by_P._Raja/screenshots/snap-3_2.png Binary files differnew file mode 100644 index 00000000..5978d055 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/screenshots/snap-3_2.png diff --git a/Electronics_Engineering_by_P._Raja/screenshots/snap-6_2.png b/Electronics_Engineering_by_P._Raja/screenshots/snap-6_2.png Binary files differnew file mode 100644 index 00000000..0eef9f3e --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/screenshots/snap-6_2.png diff --git a/Machine_Design_by_U.C._Jindal/Ch10.ipynb b/Machine_Design_by_U.C._Jindal/Ch10.ipynb new file mode 100644 index 00000000..1d59c739 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch10.ipynb @@ -0,0 +1,258 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:07d8c9662c407447473b639e8ef72a48c0966dd22d390e447be5846d8f6256c8" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:10 Pipes and pipe joints" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 10-1 - Page 295" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "sigta=140/2#\n", + "nt=0.75#\n", + "#Let the flow rate be Q\n", + "Q=0.25#\n", + "v=1.2#\n", + "D=1.13*sqrt(Q/v)#\n", + "D=520#\n", + "p=0.7#\n", + "C=9#\n", + "t=(p*D)/(2*sigta*nt)+C#\n", + "print \" t is %0.1f mm \"%(t)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " t is 12.5 mm \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 10-2 - Page 295" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "p=3*8#\n", + "sigta=60#\n", + "d=150#\n", + "t=d/2*sqrt(((sigta+p)/(sigta-p))-1)#\n", + "t=75*sqrt((84/36)-1)#\n", + "t=40#\n", + "do=d+(2*t)#\n", + "D=d+(2*t)+20#\n", + "w=10#\n", + "Ds=d+(2*w)#\n", + "P=pi*(Ds**2)*8/4#\n", + "sigp=310#\n", + "FOS=4#\n", + "sigb=77.5#\n", + "At=P/(sigb*2)#\n", + "At=1300#\n", + "D=250#\n", + "db=45#\n", + "b=D#\n", + "a=1.8*b#\n", + "CD=D+(2*db*1.2)#\n", + "sigp=310#\n", + "Pr=0.75*sigp*At#\n", + "Pr=Pr*10**-3#\n", + "t=40#\n", + "D1=d+(2*t)+20#\n", + "D2=D1+(4.6*31)#\n", + "CD=D2-((3*t)+20)#\n", + "print \" Pr is %0.2f kN \"%(Pr)#\n", + "print \"\\n D1 is %0.0f mm \"%(D1)#\n", + "print \"\\n D2 is %0.1f mm \"%(D2)#\n", + "print \"\\n CD is %0.1f mm \"%(CD)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Pr is 302.25 kN \n", + "\n", + " D1 is 250 mm \n", + "\n", + " D2 is 392.6 mm \n", + "\n", + " CD is 252.6 mm \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 10-3 - Page 296" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "p=14#\n", + "d=50#\n", + "sigyp=270#\n", + "FOS=3#\n", + "sigta=sigyp/FOS#\n", + "pt=2*p#\n", + "t=d/2*sqrt(((sigta+pt)/(sigta-pt))-1)#\n", + "t=10#\n", + "D1=d+(2*t)#\n", + "Ds=D1+20#\n", + "P=pi*(Ds**2)*p/4#\n", + "sigba=380/4#\n", + "At=P/(4*sigba)#\n", + "At=245#\n", + "db=20#\n", + "Dd=70+(2*20)+5#\n", + "R=db+2.5#\n", + "B=(Dd/sqrt(2))+(2*(db+2.5))#\n", + "B=127#\n", + "Y=Dd/(2*sqrt(2))#\n", + "Rm=34.12#\n", + "M=(P*Y/2)+(P*Rm/pi)#\n", + "sigfa=250/5#\n", + "b=127/70#\n", + "Z=b/6#\n", + "tf=sqrt(M/(sigfa*Z))#\n", + "tf=44#\n", + "print \" d is %0.0f mm \"%(d)#\n", + "print \"\\n t is %0.0f mm \"%(t)#\n", + "print \"\\n B is %0.0f mm \"%(B)#\n", + "print \"\\n R is %0.1f mm \"%(R)#\n", + "print \"\\n Y is %0.2f mm \"%(Y)#\n", + "print \"\\n tf is %0.0f mm \"%(tf)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " d is 50 mm \n", + "\n", + " t is 10 mm \n", + "\n", + " B is 127 mm \n", + "\n", + " R is 22.5 mm \n", + "\n", + " Y is 40.66 mm \n", + "\n", + " tf is 44 mm \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 10-4 - Page 297" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import tan\n", + "p=1.25#\n", + "D=200#\n", + "nt=0.75#\n", + "C=9#\n", + "sigta=20#\n", + "t=(p*D)/(2*sigta*nt)+C#\n", + "t=18#\n", + "D1=D+(2*t)#\n", + "dr=D1+10#\n", + "sigp=310#\n", + "sigba=sigp/4#\n", + "db=16#\n", + "Db=dr+32+5#\n", + "Do=Db+(2*db)#\n", + "P=pi*(251+db)**2*1.25/4#\n", + "n=6#\n", + "Y=(Db-dr)/2#\n", + "M=P/n*Y#\n", + "Z=dr*tan(30*pi/180)/6#\n", + "tf=sqrt(M/(sigta*Z))#\n", + "tf=22#\n", + "Deff=dr+db+5#\n", + "print \" D is %0.0f mm \"%(D)#\n", + "print \"\\n t is %0.0f mm \"%(t)#\n", + "print \"\\n Y is %0.1f mm \"%(Y)#\n", + "print \"\\n tf is %0.0f mm \"%(tf)#\n", + "print \"\\n Deff is %0.0f mm \"%(Deff)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " D is 200 mm \n", + "\n", + " t is 18 mm \n", + "\n", + " Y is 18.5 mm \n", + "\n", + " tf is 22 mm \n", + "\n", + " Deff is 267 mm \n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch11.ipynb b/Machine_Design_by_U.C._Jindal/Ch11.ipynb new file mode 100644 index 00000000..f8fa86e3 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch11.ipynb @@ -0,0 +1,434 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:53a1ba1b803cb0b51138c9b8688085c88b0a88d5fedff21bf58ba2fdae3bfaab" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:11 Riveted joints" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 11-1 - Page 322" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "t=20#\n", + "p=100#\n", + "d=25#\n", + "sigt=40#\n", + "P=(p-d)*t*sigt#\n", + "Ts=(4*P)/(pi*d**2)#\n", + "sigb=P/(d*t)#\n", + "print \" P is %0.0f N \"%(P)#\n", + "print \"\\n Ts is %0.2f MPa \"%(Ts)#\n", + "print \"\\n sigb is %0.0f MPa \"%(sigb)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " P is 60000 N \n", + "\n", + " Ts is 122.23 MPa \n", + "\n", + " sigb is 120 MPa \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 11-2 - Page 322" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "t=22#\n", + "t1=5*t/8#\n", + "d=30#\n", + "p=100#\n", + "sigt=75#\n", + "P=(p-d)*t*sigt#\n", + "Ts=(2*P)/(pi*d**2)#\n", + "sigb=P/(d*t)#\n", + "P=P*10**-3\n", + "print \" P is %0.1f kN \"%(P)#\n", + "print \"\\n Ts is %0.1f MPa \"%(Ts)#\n", + "print \"\\n sigb is %0.0f N/mm**2 \"%(sigb)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " P is 115.5 kN \n", + "\n", + " Ts is 81.7 MPa \n", + "\n", + " sigb is 175 N/mm**2 \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 11-3 - Page 323" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "t=15#\n", + "t1=5*t/8#\n", + "d=25#\n", + "n=2#\n", + "Ta=80#\n", + "sigta=100#\n", + "sigba=120#\n", + "Ps=n*1.875*pi*d**2*Ta/4#\n", + "Pb=n*d*t*sigba#\n", + "p=Pb/(t*Ta)+d#\n", + "Pp=p*t*Ta#\n", + "n=Pb/Pp#\n", + "print \" p is %0.0f mm \"%(p)#\n", + "print \"\\n n is %0.2f \"%(n)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " p is 100 mm \n", + "\n", + " n is 0.00 \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 11-4 - Page 323" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "b=200#\n", + "t=16#\n", + "d=6*sqrt(t)#\n", + "sigta=80#\n", + "Ta=60#\n", + "sigba=100#\n", + "Pt=(b-d)*t*sigta#\n", + "Ps=1.875*pi*d**2*Ta/4#\n", + "Pb=d*t*sigba#\n", + "n1=Pt/Pb#\n", + "n1=6#\n", + "Pt2=((b-(2*d))*t*sigta)+Pb#\n", + "Pt3=((b-(3*d))*t*sigta)+(3*Pb)#\n", + "Pp=b*t*sigta#\n", + "n2=Pt/Pp#\n", + "n2=n2*100#\n", + "print \" d is %0.0f mm \"%(d)#\n", + "print \"\\n n1 is %0.0f \"%(n1)#\n", + "print \"\\n Pt is %0.0f N \"%(Pt)#\n", + "print \"\\n Pt2 is %0.0f N \"%(Pt2)#\n", + "print \"\\n Pt3 is %0.0f N \"%(Pt3)#\n", + "print \"\\n n2 is %0.0f \"%(n2)#\n", + "#Answer to strength of rivet in bearing 'Pb' is calculated incorrectly in the book, hence Pt2,Pt3 is calculated subsequently incorrect." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " d is 24 mm \n", + "\n", + " n1 is 6 \n", + "\n", + " Pt is 225280 N \n", + "\n", + " Pt2 is 232960 N \n", + "\n", + " Pt3 is 279040 N \n", + "\n", + " n2 is 88 \n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 11-5 - Page 324" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import atan\n", + "a=50#\n", + "b=75#\n", + "P=36*10**3#\n", + "d=24#\n", + "Ta=60#\n", + "n=9#\n", + "A=pi*d**2/4#\n", + "Td=P/(n*A)#\n", + "theta=atan(b/a)#\n", + "Ts=54.64#\n", + "r2=90.184#\n", + "e=A*29575.7/P#\n", + "print \" e is %0.1f mm \"%(e)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " e is 371.7 mm \n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 11-6 - Page 325" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=12*10**3#\n", + "Tmax=100#\n", + "n=6#\n", + "e=50+50+(5/2)#\n", + "T=P*e#\n", + "Td=P/n#\n", + "ra=125#\n", + "k=T/((2*125**2)+(2*75**2)+(2*25**2))#\n", + "Tr=(k*ra)+Td#\n", + "A=Tr/Tmax#\n", + "d=sqrt(A*4/pi)#\n", + "d=12#\n", + "print \" d is %0.0f mm \"%(d)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " d is 12 mm \n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 11-7 - Page 326" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "t=15#\n", + "d=6*sqrt(t)#\n", + "d=24#\n", + "sigta=75#\n", + "sigba=105#\n", + "Ta=60#\n", + "n=4#\n", + "Pt=n*pi*d**2*Ta/4#\n", + "x=d*t*sigta#\n", + "y=2*t*sigta#\n", + "p=(Pt+x)/y#\n", + "p=60#\n", + "C=4.17#\n", + "pmax=(C*t)+41.28#\n", + "Pt1=(y*p)-x#\n", + "Ps=n*pi*d**2*Ta/4#\n", + "Pb=n*d*t*sigba#\n", + "S=2*p*t*sigta#\n", + "n=Pt1/S#\n", + "n=n*100#\n", + "print \" n is %0.f \"%(n)# " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " n is 80 \n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 11-8 - Page 327" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "D=1500#\n", + "p=2#\n", + "nt=0.75#\n", + "sigut=420#\n", + "FOS=5#\n", + "sigta=sigut/FOS#\n", + "t=p*D/(2*sigta*nt)#\n", + "t=24#\n", + "d=6*sqrt(t)#\n", + "d=30#\n", + "Ta=330/5#\n", + "sigba=640/5#\n", + "Ps=2*1.875*pi*(d**2)*Ta/4#\n", + "p=(Ps/(t*sigta))+d#\n", + "p=117#\n", + "t1=5*t/8#\n", + "Pt=(p-d)*t*sigta#\n", + "Pp=p*t*sigta#\n", + "Pb=2*d*t*sigba#\n", + "n=Ps/Pb#\n", + "n=n*100#\n", + "print \" n is %0.0f \"%(n)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " n is 95 \n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 11-9 - Page 327" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "D=1200#\n", + "p=2.5#\n", + "sigba=110#\n", + "Pa=pi*D**2*p/4#\n", + "nt=0.8#\n", + "sigta=80#\n", + "t=p*D/(2*sigta*nt)#\n", + "t=24#\n", + "d=6*sqrt(t)#\n", + "d=30#\n", + "Ta=55#\n", + "Ps=pi*(d**2)*Ta/4#\n", + "Np=Pa/Ps#\n", + "Np=74#\n", + "nr=Np/2#\n", + "p=pi*(D+t)/nr#\n", + "pb=2*d#\n", + "m=1.5*d#\n", + "Pt=(p-d)*t*sigta#\n", + "Ps=2*Ps#\n", + "Pb=2*d*t*sigba#\n", + "Pp=p*t*sigta#\n", + "n=Ps/Pp#\n", + "n=n*100#\n", + "print \" n is %0.0f \"%(n)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " n is 39 \n" + ] + } + ], + "prompt_number": 15 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch12.ipynb b/Machine_Design_by_U.C._Jindal/Ch12.ipynb new file mode 100644 index 00000000..21de3f82 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch12.ipynb @@ -0,0 +1,442 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:24bbadb74dcdce0a144766279891314e41406bba0c41d580bf7981786895ce30" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:12 Welded joints" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 12-1 - Page 347" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=8#\n", + "F=100*10**3#\n", + "t=0.707*h#\n", + "A=4*60*t#\n", + "T=F/A#\n", + "print \"T is %0.1f MPa \"%(T)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "T is 73.7 MPa \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 12-2 - Page 347" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "FOS=3#\n", + "Ta=95/FOS#\n", + "P=350*10**3#\n", + "h=12.5#\n", + "t=0.707*h#\n", + "l=P/(2*t*Ta)#\n", + "print \"l is %0.0f mm \"%(l)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "l is 639 mm \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 12-3 - Page 348" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=12#\n", + "t=0.707*h#\n", + "l=60#\n", + "Ta=80#\n", + "P=2*l*t*Ta#\n", + "P=P*10**-3#\n", + "print \"P is %0.3f kN \"%(P)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "P is 81.446 kN \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 12-4 - Page 348" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sin, pi, cos, sqrt\n", + "P=6*10**3#\n", + "e=150+(100/2)#\n", + "T=P*e#\n", + "A=200#\n", + "Td=P/A#\n", + "r=sqrt(2*50**2)#\n", + "Ixx=2*(100*50**2)#\n", + "Iyy=2*100**3/12#\n", + "IG=Ixx+Iyy#\n", + "Ts=r*T/IG#\n", + "Tmax=sqrt((Ts*sin(45*pi/180))**2+(Td+(Ts*cos(45*pi/180)))**2)#\n", + "Ta=80#\n", + "t=Tmax/Ta#\n", + "h=sqrt(2)*t#\n", + "h=3#\n", + "print \"h is %0.0f mm \"%(h)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "h is 3 mm \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 12-5 - Page 349" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=10#\n", + "t=10/sqrt(2)#\n", + "Ta=80#\n", + "x=((50*25)+(50*0))/(50+50)#\n", + "y=x#\n", + "ra=sqrt(x**2+37.5**2)#\n", + "Ixx=(7.07*50**3/12)+(50*7.07*(12.5**2))+(50*7.07*12.5**2)#\n", + "IG=2*Ixx#\n", + "e=100+(50-12.5)#\n", + "Tr=16.09*10**-3#\n", + "P=Ta/Tr#\n", + "P=P*10**-3#\n", + "print \"P is %0.3f KN \"%(P)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "P is 4.972 KN \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 12-6 - Page 350" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=16*10**3#\n", + "l=300#\n", + "r=50#\n", + "M=P*l#\n", + "A=2*pi*r#\n", + "Ixx=pi*r**3#\n", + "sigb=M*r/Ixx#\n", + "Td=P/A#\n", + "Tmax=sqrt((sigb/2)**2+(Td**2))#\n", + "Ta=90#\n", + "t=Tmax/Ta#\n", + "h=sqrt(2)*t#\n", + "h=5#\n", + "print \"h is %0.0f mm \"%(h)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "h is 5 mm \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 12-7 - Page 350" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "sigut=415#\n", + "sige=sigut/3#\n", + "Ka=0.5#\n", + "Kb=0.85#\n", + "Kc=0.897#\n", + "SCF=1.5#\n", + "Kd=1/SCF#\n", + "FOS=2#\n", + "sige1=sige*Ka*Kb*Kc*Kd/FOS#\n", + "Pa=50*10**3#\n", + "h=10#\n", + "t=0.707*h#\n", + "l=Pa/(2*sige1*t)#\n", + "print \"l is %0.0f mm \"%(l)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "l is 202 mm \n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 12-8 - Page 351" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "l=300#\n", + "P=30*10**3#\n", + "T=P/(2*l)#\n", + "Ta=124#\n", + "t1=T/Ta#\n", + "h1=sqrt(2)*t1#\n", + "M=P*l#\n", + "Ixx=2*100*110**2#\n", + "sigb=M/Ixx*110#\n", + "#Let the allowable bending stress is Tab\n", + "Tab=200#\n", + "t2=sigb/Tab#\n", + "h2=t2/0.707#\n", + "h2=3#\n", + "print \"h is %0.0f mm \"%(h2)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "h is 3 mm \n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 12-9 - Page 352" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Ta=60#\n", + "l1=60#\n", + "l2=40#\n", + "P1=Ta*0.707*l1#\n", + "P2=Ta*0.707*l2#\n", + "P=80*10**3#\n", + "h=P/(P1+P2)#\n", + "h=20#\n", + "a=(P2*100)/(P1+P2)#\n", + "print \" h is %0.0f mm \"%(h)#\n", + "print \"\\n a is %0.0f mm \"%(a)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " h is 20 mm \n", + "\n", + " a is 40 mm \n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 12-10 - Page 352" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=300*10**3#\n", + "l=500#\n", + "A=2*l#\n", + "Td=P/A#\n", + "T=(350-250)*P#\n", + "IG=(l**3*2/12)+(l*2*5**2)#\n", + "r=sqrt(250**2+5**2)#\n", + "Ts=T*r/IG#\n", + "Ts=Ts+Td#\n", + "Ta=110#\n", + "t=Ts/Ta#\n", + "h=t/0.707#\n", + "h=9#\n", + "print \"h is %0.0f mm \"%(h)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "h is 9 mm \n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 12-11 - Page 353" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "t=30#\n", + "sigut=417#\n", + "sige=sigut/2#\n", + "Ka=0.5#\n", + "Kb=0.85#\n", + "Kc=0.897#\n", + "SCF=1.2#\n", + "Kd=1/SCF#\n", + "FOS=1.5#\n", + "sige1=sige*Ka*Kb*Kc*Kd/FOS#\n", + "Pa=60*10**3#\n", + "l=Pa/(sige1*t)#\n", + "print \"l is %0.1f mm \"%(l)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "l is 45.4 mm \n" + ] + } + ], + "prompt_number": 11 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch13.ipynb b/Machine_Design_by_U.C._Jindal/Ch13.ipynb new file mode 100644 index 00000000..8a9638f2 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch13.ipynb @@ -0,0 +1,284 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:699e68915e564767c2f777c1b196296273ea15e52a0652061e0ad59d630a55ce" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:13 Cotter and knuckle joints" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 13-1 - Page 371" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt, pi\n", + "F=25*10**3#\n", + "sigat=50#\n", + "Ta=40#\n", + "pa=80#\n", + "d=sqrt((4*F)/(pi*sigat))#\n", + "d=26#\n", + "t=d/4#\n", + "t=7#\n", + "d1=1.2*d#\n", + "d1=32#\n", + "pc=F/(d1*t)#\n", + "t=10#\n", + "c=0.75*d#\n", + "c=20#\n", + "d2=44#\n", + "tw=(d2-d1)/2#\n", + "b=F/(2*t*Ta)#\n", + "b=34#\n", + "a=0.5*d#\n", + "d3=(F/(pa*t))+d1#\n", + "d3=64#\n", + "e=F/(Ta*(d3-d1))#\n", + "d4=sqrt((F*4/(pi*pa))+d1**2)#\n", + "d4=40#\n", + "f=0.5*d#\n", + "sigbc=3*F*d3/(t*b**2*4)#\n", + "print \" d is %0.0f mm \"%(d)#\n", + "print \"\\n d1 is %0.0f mm \"%(d1)#\n", + "print \"\\n d2 is %0.0f mm \"%(d2)#\n", + "print \"\\n d3 is %0.0f mm \"%(d3)#\n", + "print \"\\n d4 is %0.0f mm \"%(d4)#\n", + "print \"\\n sigbc is %0.1f MPa \"%(sigbc)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " d is 26 mm \n", + "\n", + " d1 is 32 mm \n", + "\n", + " d2 is 44 mm \n", + "\n", + " d3 is 64 mm \n", + "\n", + " d4 is 40 mm \n", + "\n", + " sigbc is 103.0 MPa \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 13-2 - Page 372" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=40*10**3#\n", + "sigut=490#\n", + "FOS=4#\n", + "sigts=sigut/FOS#\n", + "sigcs=1.4*sigts#\n", + "sigs=0.8*sigts#\n", + "d=sqrt((4*P)/(pi*sigts))#\n", + "d=21#\n", + "sigcc=1.4*330/4#\n", + "Tc=0.8*330/4#\n", + "t=d/3#\n", + "b=P/(2*t*Tc)#\n", + "b=31#\n", + "t=10#\n", + "d1=28#\n", + "d2=40#\n", + "c=d/2#\n", + "c=15#\n", + "a=P/(2*(d2-d1)*98)#\n", + "a=20#\n", + "L=(2*a)+(2*b)+(2*c)+(2*3)#\n", + "print \" d is %0.0f mm \"%(d)#\n", + "print \"\\n d1 is %0.0f mm \"%(d1)#\n", + "print \"\\n t is %0.0f mm \"%(t)#\n", + "print \"\\n b is %0.0f mm \"%(b)#\n", + "print \"\\n d2 is %0.0f mm \"%(d2)#\n", + "print \"\\n L is %0.0f mm \"%(L)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " d is 21 mm \n", + "\n", + " d1 is 28 mm \n", + "\n", + " t is 10 mm \n", + "\n", + " b is 31 mm \n", + "\n", + " d2 is 40 mm \n", + "\n", + " L is 138 mm \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 13-3 - Page 372" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=40*10**3#\n", + "sigt=60#\n", + "sigc=125#\n", + "T=45#\n", + "a=sqrt(P*3/(2*sigt))#\n", + "a=33#\n", + "t=a/3#\n", + "b=P/(4.5*t*T)#\n", + "b=20#\n", + "b1=1.25*b#\n", + "t1=P*3/(4*a*sigt)#\n", + "t1=16#\n", + "l2=P/(2*2*T*t1)#\n", + "l2=14#\n", + "l1=P/(2*a*T)#\n", + "l1=14#\n", + "l3=(0.6*a)#\n", + "l3=20#\n", + "l4=11#\n", + "sigcr=P/(t*a)#\n", + "sigcr1=P/(2*t1*t)#\n", + "print \" a is %0.0f mm \"%(a)#\n", + "print \"\\n t is %0.0f mm \"%(t)#\n", + "print \"\\n t1 is %0.0f mm \"%(t1)#\n", + "print \"\\n b is %0.0f mm \"%(b)#\n", + "print \"\\n b1 is %0.0f mm \"%(b1)#\n", + "print \"\\n l1 is %0.0f mm \"%(l1)#\n", + "print \"\\n l2 is %0.0f mm \"%(l2)#\n", + "print \"\\n l3 is %0.0f mm \"%(l3)#\n", + "print \"\\n l4 is %0.0f mm \"%(l4)#\n", + "print \"\\n sigcr is %0.1f MPa \"%(sigcr)#\n", + "print \"\\n sigcr1 is %0.1f MPa \"%(sigcr1)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " a is 33 mm \n", + "\n", + " t is 11 mm \n", + "\n", + " t1 is 16 mm \n", + "\n", + " b is 20 mm \n", + "\n", + " b1 is 25 mm \n", + "\n", + " l1 is 14 mm \n", + "\n", + " l2 is 14 mm \n", + "\n", + " l3 is 20 mm \n", + "\n", + " l4 is 11 mm \n", + "\n", + " sigcr is 110.0 MPa \n", + "\n", + " sigcr1 is 113.0 MPa \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 13-4 - Page 373" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=50*10**3#\n", + "sigp=380#\n", + "FOS=4#\n", + "sigca=80#\n", + "Ta=50#\n", + "sigta=sigp/FOS#\n", + "At=P/sigta#\n", + "d=30#\n", + "d1=1.5*d#\n", + "t=P/(sigca*d1)#\n", + "t=14#\n", + "A=(pi*(d1**2)/4)-(d1*t)#\n", + "#let tearing stress be sigt\n", + "sigt=P/A#\n", + "b=P/(2*t*Ta)#\n", + "b=36#\n", + "print \" d is %0.0f mm \"%(d)#\n", + "print \"\\n sigt is %0.1f MPa \"%(sigt)#\n", + "print \"\\n b is %0.0f mm \"%(b)#\n", + " \n", + " #The answer to tearing stress in bolt 'sigt' is calculated incorrectly in the book." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " d is 30 mm \n", + "\n", + " sigt is 52.1 MPa \n", + "\n", + " b is 36 mm \n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch14.ipynb b/Machine_Design_by_U.C._Jindal/Ch14.ipynb new file mode 100644 index 00000000..766ceb71 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch14.ipynb @@ -0,0 +1,412 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:d7736c3b733f2dee976dda18085a361ed833c66d2ffff17be44c76a801411ca2" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:14 Keys and couplings" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 14-1 - Page 401" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt, pi, atan, acos, cos\n", + "d=40#\n", + "r=d/2#\n", + "P=6*10**3#\n", + "N=350#\n", + "sigyt=380#\n", + "A=pi*12**2/2#\n", + "theta=pi-(2*atan(4/12))#\n", + "alpha=180-(theta*pi/180)#\n", + "l=2*12*cos(19.5*pi/180)#\n", + "A1=l*4/2#\n", + "Abcd=(A*141/180)-A1#\n", + "A2=A-Abcd#\n", + "A3=8*l#\n", + "w=2*pi*N/60#\n", + "T=P/w#\n", + "Pt=T*10**3/r#\n", + "sigb=Pt/A2#\n", + "#Let shear stress developed in key Tk\n", + "Tk=Pt/A3#\n", + "FOS1=sigyt/sigb#\n", + "FOS2=0.577*sigyt/Tk#\n", + "print \" FOS1 is %0.3f \"%(FOS1)#\n", + "print \"\\n FOS2 is %0.2f \"%(FOS2)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " FOS1 is 4.376 \n", + "\n", + " FOS2 is 4.85 \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 14-2 - Page 401" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "n=12#\n", + "phi=360*pi/(180*12*2)#\n", + "R1=45/2#\n", + "R2=50/2#\n", + "l=60#\n", + "Rm=(R1+R2)/2#\n", + "p=6.5#\n", + "Pn=(R2-R1)*l*p#\n", + "T=Pn*Rm#\n", + "T=T*n#\n", + "N=400#\n", + "w=2*pi*N/60#\n", + "P=T*w#\n", + "A=(pi*R1*l)/n#\n", + "Ts=Pn/A#\n", + "Ah=(pi*R2*l)/n#\n", + "Th=Pn/Ah#\n", + "print \" Ts is %0.2f N/mm**2 \"%(Ts)#\n", + "print \"\\n Th is %0.2f N/mm**2 \"%(Th)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Ts is 3.39 N/mm**2 \n", + "\n", + " Th is 2.98 N/mm**2 \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 14-3 - Page 402" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "N=360#\n", + "w=2*pi*N/60#\n", + "sigyt=380#\n", + "r=25#\n", + "P=40*10**3#\n", + "FOS=3#\n", + "T=P/w#\n", + "Pt=T*10**3/(2*r)#\n", + "siga=380/3#\n", + "Ta=0.577*380/3#\n", + "l1=Pt/(sqrt(2)*12*Ta)#\n", + "l2=Pt*sqrt(2)/(siga*12)#\n", + "print \" l1 is %0.0f mm \"%(l1)#\n", + "print \"\\n l2 is %0.2f mm \"%(l2)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " l1 is 17 mm \n", + "\n", + " l2 is 19.85 mm \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 14-4 - Page 403" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "N=300#\n", + "w=2*pi*N/60#\n", + "P=12*10**3#\n", + "Ks=1.25#\n", + "Pd=P*Ks#\n", + "T=Pd/w#\n", + "Tas=50#\n", + "d=16*T*10**3/(pi*Tas)#\n", + "d=d**(1/3)#\n", + "d=40#\n", + "Ts=10#\n", + "d1=(2*d)+13#\n", + "x=(d1**4-d**4)/d1#\n", + "#Let the shear stress in the key be Tsh\n", + "Tsh=T*10**3*16/(pi*x)#\n", + "l=3.5*d#\n", + "Ft=T*2*10**3/d#\n", + "l1=70#\n", + "sigak=50#\n", + "b=Ft/(l1*sigak)#\n", + "t=2*Ft/(100*l1)#\n", + "print \" d is %0.0f mm \"%(d)#\n", + "print \"\\n Tsh is %0.2f MPa \"%(Tsh)#\n", + "print \"\\n b is %0.0f mm \"%(b)#\n", + "print \"\\n t is %0.0f mm \"%(t)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " d is 40 mm \n", + "\n", + " Tsh is 3.13 MPa \n", + "\n", + " b is 7 mm \n", + "\n", + " t is 7 mm \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 14-5 - Page 403" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=36*10**3#\n", + "N=200#\n", + "w=2*pi*N/60#\n", + "T=P/w#\n", + "Tas=45#\n", + "d=16*T*10**3/(pi*Tas)#\n", + "d=d**(1/3)#\n", + "d=60#\n", + "d1=(2*d)+13#\n", + "l=3.5*d#\n", + "Ftk=T*2/d#\n", + "lk=l/2#\n", + "Tak=40#\n", + "sigack=90#\n", + "b=Ftk*10**3/(lk*Tak)#\n", + "t=2*Ftk*10**3/(sigack*lk)#\n", + "n=4#\n", + "sigatb=60#\n", + "u=0.25#\n", + "dr=16*T*10**3/(u*pi**2*sigatb*n*d)#\n", + "dr=sqrt(dr)#\n", + "print \" d is %0.0f mm \"%(d)#\n", + "print \"\\n b is %0.1f mm \"%(b)#\n", + "print \"\\n t is %0.0f mm \"%(t)#\n", + "print \"\\n dr is %0.3f mm \"%(dr)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " d is 60 mm \n", + "\n", + " b is 13.6 mm \n", + "\n", + " t is 12 mm \n", + "\n", + " dr is 27.822 mm \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 14-6 - Page 404" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=16*10**3#\n", + "N=1000#\n", + "w=2*pi*N/60#\n", + "T=P/w#\n", + "Ks=1.4#\n", + "Td=T*Ks#\n", + "Tas=40#\n", + "d=16*T*10**3/(pi*Tas)#\n", + "d=d**(1/3)#\n", + "d=32#\n", + "d1=2*d#\n", + "l=1.5*d#\n", + "ds=1.5*d#\n", + "Tak=40#\n", + "sigack=70#\n", + "Ftk=Td*2/d#\n", + "b=Ftk*10**3/(l*Tak)#\n", + "t=2*Ftk*10**3/(sigack*l)#\n", + "Taf=10#\n", + "tf=Td*10**3*2/(pi*Taf*d1**2)#\n", + "Ftb=Td*10**3/(1.5*d*4)#\n", + "Tab=40#\n", + "db=sqrt(Ftb*4/(Tab*pi))#\n", + "D=4*d#\n", + "trp=d/6#\n", + "Ftb1=Td*10**3/(45*4)#\n", + "db1=sqrt(Ftb1*4/(Tab*pi))#\n", + "print \" d is %0.0f mm \"%(d)#\n", + "print \"\\n b is %0.0f mm \"%(b)#\n", + "print \"\\n t is %0.0f mm \"%(t)#\n", + "print \"\\n db is %0.2f mm \"%(db)#\n", + "print \"\\n db1 is %0.2f mm \"%(db1)#\n", + " \n", + " #The answer to Key thickness 't' is calculated incorrectly in the book." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " d is 32 mm \n", + "\n", + " b is 7 mm \n", + "\n", + " t is 8 mm \n", + "\n", + " db is 5.96 mm \n", + "\n", + " db1 is 6.15 mm \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 14-7 - Page 404" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=30*10**3#\n", + "N=1440#\n", + "w=2*pi*N/60#\n", + "T=P/w#\n", + "d=36#\n", + "d1=30#\n", + "d2=2*d#\n", + "d3=d1*2#\n", + "l=1.5*d#\n", + "Dp=3.5*d#\n", + "n=6#\n", + "Ft=(2*T)/(Dp*n)#\n", + "p=0.5#\n", + "A=Ft/p#\n", + "Lf=d#\n", + "dp=A/Lf#\n", + "M=Ft*10**3*(5+(Lf/2))#\n", + "db=(32*M/(pi*40))**(1/3)#\n", + "db=15#\n", + "T=(4*526)/(pi*db**2)#\n", + "sigb=32*M/(pi*db**3)#\n", + "sigmax=(sigb/2)+sqrt(((sigb/2)**2)+(T**2))#\n", + "b=d/4#\n", + "t=6#\n", + "Lf=36#\n", + "La=10#\n", + "Do=126+30+(2*(5+1))+(2*6)#\n", + "print \" sigmax is %0.2f MPa \"%(sigmax)#\n", + "print \"\\n b is %0.0f mm \"%(b)#\n", + "print \"\\n t is %0.0f mm \"%(t)#\n", + "print \"\\n Lf is %0.0f mm \"%(Lf)#\n", + "print \"\\n Do is %0.0f mm \"%(Do)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " sigmax is 36.77 MPa \n", + "\n", + " b is 9 mm \n", + "\n", + " t is 6 mm \n", + "\n", + " Lf is 36 mm \n", + "\n", + " Do is 180 mm \n" + ] + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch15.ipynb b/Machine_Design_by_U.C._Jindal/Ch15.ipynb new file mode 100644 index 00000000..5a76fb0c --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch15.ipynb @@ -0,0 +1,341 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:9b7bd78ba4da34e6b0bfbd99a229a10f40ca2ca34b3d46eeee9c738f3ac0df6a" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:15 Shafts" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 15-2 - Page 421" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import pi, sqrt, tan\n", + "dA=150#\n", + "dB=250#\n", + "alpha=20*pi/180#\n", + "W=400#\n", + "sigyt=400#\n", + "sigut=500#\n", + "Kb=1.5#\n", + "Kt=2#\n", + "T=W*dA/2#\n", + "Pt=T/(dB/2)#\n", + "Pr1=W*tan(alpha)#\n", + "Pr2=Pt*tan(alpha)#\n", + "RDH=((W*120)-(Pt*320))/440#\n", + "RcH=W-RDH-Pt#\n", + "#RcH=400+65.5-240#\n", + "McH=0#\n", + "MAH=RcH*120#\n", + "MBH=RDH*120#\n", + "RDV=((Pr1*120)-(Pr2*320))/440#\n", + "RcV=Pr1-RDV-Pr2#\n", + "MAV=RcV*120#\n", + "MBV=RDV*120#\n", + "Mmax=sqrt((MAH**2)+(MAV**2))#\n", + "T=30*10**3#\n", + "Ta=0.135*sigut#\n", + "d=16*sqrt((Kb*Mmax)**2+(Kt*T)**2)/(pi*Ta)#\n", + "d=d**(1/3)#\n", + "print \"d is %0.2f mm \"%(d)#\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "d is 17.73 mm \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 15-3 - Page 421" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=16*746#\n", + "N=3000#\n", + "w=2*pi*N/60#\n", + "T=P/w*10**3#\n", + "sigy=400#\n", + "Ty=sigy/2#\n", + "FOS=2#\n", + "Ta=Ty/FOS#\n", + "d=T*16/(pi*Ta)#\n", + "d1=d**(1/3)#\n", + "r=3#\n", + "D=d1+(2*r)#\n", + "SCF=1.196\n", + "Tys=Ta/SCF#\n", + "d=T*16/(pi*Tys)#\n", + "d2=d**(1/3)#\n", + "d=14#\n", + "D=d+(2*r)#\n", + "print \"d1 is %0.2f mm \"%(d1)#\n", + "print \"\\nd2 is %0.2f mm \"%(d2)#\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "d1 is 12.46 mm \n", + "\n", + "d2 is 13.23 mm \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 15-4 - Page 422" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P1=24*10**3#\n", + "P2=10*10**3#\n", + "sigyt=460#\n", + "Tya=sigyt*0.3#\n", + "SCF=2.84#\n", + "Ta=Tya/SCF#\n", + "N=400#\n", + "w=2*pi*N/60#\n", + "T1=P1/w#\n", + "T2=P2/w#\n", + "d1=T1*16*10**3/(pi*Ta)#\n", + "d1=d1**(1/3)#\n", + "d2=T2*16*10**3/(pi*Ta)#\n", + "d2=d2**(1/3)#\n", + "theta1=pi/3600#\n", + "l1=120#\n", + "G=84*10**3#\n", + "d3=T1*10**3*l1*32/(pi*G*theta1)#\n", + "d3=d3**(1/4)#\n", + "d4=T2*l1*10**3*32/(pi*G*theta1)#\n", + "d4=d4**(1/4)#\n", + "print \" d1 is %0.2f mm \"%(d1)#\n", + "print \"\\n d2 is %0.2f mm \"%(d2)#\n", + "print \"\\n d3 is %0.1f mm \"%(d3)#\n", + "print \"\\n d4 is %0.2f mm \"%(d4)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " d1 is 39.16 mm \n", + "\n", + " d2 is 29.25 mm \n", + "\n", + " d3 is 55.6 mm \n", + "\n", + " d4 is 44.67 mm \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 15-5 - Page 423" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sin, exp\n", + "d=200#\n", + "r=d/2#\n", + "N=300#\n", + "P=5000#\n", + "D=500#\n", + "R=D/2#\n", + "u=0.3#\n", + "E=205*10**3#\n", + "G=84*10**3#\n", + "Ta=60#\n", + "Kb=1.5#\n", + "Kt=2#\n", + "w=2*pi*N/60#\n", + "beta1=20*pi/180#\n", + "V=r*w#\n", + "v=R*w#\n", + "# Let T1-T2 =T\n", + "T=P/V#\n", + "x=u*pi/sin(beta1)#\n", + "T2=T/((exp(x)-1))#\n", + "T1=T2*exp(x)#\n", + "t=P/v#\n", + "y=u*pi#\n", + "T3=t/((exp(x)-1))#\n", + "T4=T3*exp(x)#\n", + "T=P/w#\n", + "Rc=2612##\n", + "RA=645.1#\n", + "MB=96.76#\n", + "MC=-208.96#\n", + "d=16*10**3*sqrt((Kb*MC)**2+(Kt*T)**2)/(pi*Ta)#\n", + "d=d**(1/3)#\n", + "l=380#\n", + "J=pi*d**4/32#\n", + "theta=T*10**3*l/(G*J)#\n", + "theta=theta*180/pi#\n", + "print \"d is %0.1f mm \"%(d)#\n", + "print \"\\ntheta is %0.2f degree \"%(theta)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "d is 33.6 mm \n", + "\n", + "theta is 0.33 degree \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 15-6 - Page 423" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "T=400#\n", + "Pt=4800#\n", + "Pg=3600#\n", + "sigyt=360#\n", + "E=205*10**3#\n", + "G=80*10**3#\n", + "Kb=2#\n", + "Kt=1.5#\n", + "FOS=3#\n", + "RC=((Pt*90)+(Pg*200))/140#\n", + "RA=8400-RC#\n", + "MB=RA*0.9#\n", + "MC=Pg*0.045#\n", + "Te=sqrt((Kb*MC)**2+(Kt*T)**2)#\n", + "Ta=0.577*sigyt/FOS#\n", + "d=16*10**3*Te/(pi*Ta)#\n", + "d=d**(1/3)#\n", + "L=110#\n", + "J=pi*d**4/32#\n", + "T=400#\n", + "theta=T*10**3*L/(G*J)#\n", + "theta=theta*180/pi#\n", + "print \"d is %0.0f mm \"%(d)#\n", + "print \"\\ntheta is %0.4f deg \"%(theta)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "d is 37 mm \n", + "\n", + "theta is 0.1735 deg \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 15-7 - Page 424" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "T=47*10**3#\n", + "M=32*10**3#\n", + "d=20#\n", + "siga=32*M/(pi*d**3)#\n", + "Tm=16*T/(pi*d**3)#\n", + "sige=75#\n", + "Tys=165#\n", + "n=1/sqrt((siga/sige)**2+(Tm/Tys)**2)#\n", + "print \"n is %0.2f \"%(n)#\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "n is 1.75 \n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch16.ipynb b/Machine_Design_by_U.C._Jindal/Ch16.ipynb new file mode 100644 index 00000000..6197d784 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch16.ipynb @@ -0,0 +1,438 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:01311d1278cad60b1c0d60099bb705f52b2b4566cc66cef72df1eddeb97f8ca1" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:16 Power screws" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 16-1 - Page 450" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import atan, pi, sqrt, tan\n", + "d=30#\n", + "W=20*10**3#\n", + "r1=8#\n", + "r2=16#\n", + "p=6#\n", + "u1=0.2#\n", + "u2=0.15#\n", + "dm=d-(p/2)#\n", + "alpha=atan(p/(pi*dm))#\n", + "phi=atan(u1)#\n", + "rm=(r1+r2)/2#\n", + "Ttr=W*((dm*tan(alpha+phi)/2)+(u2*rm))#\n", + "Ttr=Ttr*10**-3#\n", + "print \"Ttr is %0.3f Nm \"%(Ttr)#\n", + "#The answer to Ttr is slightly different than in the book due to rounding-off of values." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ttr is 110.148 Nm \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 16-2 - Page 451" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import cos\n", + "d=50#\n", + "W=20*10**3#\n", + "r1=10#\n", + "r2=30#\n", + "p=7#\n", + "u1=0.12/cos(15*pi/180)#\n", + "u2=0.15#\n", + "dm=d-(p/2)#\n", + "alpha=atan(3*p/(pi*dm))#\n", + "phi=atan(u1)#\n", + "rm=(r1+r2)/2#\n", + "Tr=W*((dm*tan(alpha+phi)/2)+(u2*rm))#\n", + "Tr=Tr*10**-3#\n", + "Te=W*((dm*tan(phi-alpha)/2)+(u2*rm))#\n", + "Te=Te*10**-3#\n", + "n=dm/2*tan(alpha)/(dm*tan(alpha+phi)/2+(u2*rm))#\n", + "L=0.30#\n", + "Ph=Tr/L#\n", + "print \" Tr is %0.2f Nm \"%(Tr)#\n", + "print \"\\n Te is %0.3f Nm \"%(Te)#\n", + "print \"\\n n is %0.4f \"%(n)#\n", + "print \"\\n Ph is %0.2f N \"%(Ph)#\n", + "#The answers to Tr, Te and Ph is slightly different than in the book due to rounding-off of values." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Tr is 187.49 Nm \n", + "\n", + " Te is 51.691 Nm \n", + "\n", + " n is 0.3489 \n", + "\n", + " Ph is 624.96 N \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 16-3 - Page 452" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "d=30#\n", + "W=5*10**3#\n", + "p=5#\n", + "rm=45/2#\n", + "u1=0.15/cos(14.5*pi/180)#\n", + "u2=0.15#\n", + "dm=d-(p/2)#\n", + "alpha=atan(p/(pi*dm))#\n", + "phi=atan(u1)#\n", + "Tr1=W*((dm*tan(alpha+phi)/2)+(u2*rm))#\n", + "Tr1=Tr1*10**-3#\n", + "n1=dm/2*tan(alpha)/(dm*tan(alpha+phi)/2+(u2*rm))#\n", + "T1=W*((dm*tan(phi-alpha)/2)+(u2*rm))#\n", + "T1=T1*10**-3#\n", + "n2=dm/2*tan(alpha)/(dm*tan(phi-alpha)/2+(u2*rm))#\n", + "u2=0.02#\n", + "Tr2=W*((dm*tan(alpha+phi)/2)+(u2*rm))#\n", + "Tr2=Tr2*10**-3#\n", + "n3=dm/2*tan(alpha)/(dm*tan(alpha+phi)/2+(u2*rm))#\n", + "Te=W*((dm*tan(phi-alpha)/2)+(u2*rm))#\n", + "Te=Te*10**-3#\n", + "n4=dm/2*tan(alpha)/(dm*tan(phi-alpha)/2+(u2*rm))#\n", + "print \" Tr1 is %0.3f Nm \"%(Tr1)#\n", + "print \"\\n n1 is %0.4f \"%(n1)#\n", + "print \"\\n T1 is %0.3f Nm \"%(T1)#\n", + "print \"\\n n2 is %0.4f \"%(n2)#\n", + "print \"\\n Tr2 is %0.3f Nm \"%(Tr2)#\n", + "print \"\\n n3 is %0.4f \"%(n3)#\n", + "print \"\\n Te is %0.3f Nm \"%(Te)#\n", + "print \"\\n n4 is %0.4f \"%(n4)#\n", + " \n", + " #The answer to T1 is misprinted in the book." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Tr1 is 31.456 Nm \n", + "\n", + " n1 is 0.1265 \n", + "\n", + " T1 is 23.307 Nm \n", + "\n", + " n2 is 0.1707 \n", + "\n", + " Tr2 is 17.156 Nm \n", + "\n", + " n3 is 0.2319 \n", + "\n", + " Te is 9.007 Nm \n", + "\n", + " n4 is 0.4418 \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 16-4 - Page 453" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "d=28#\n", + "P=300#\n", + "L=180#\n", + "p=8#\n", + "r1=16#\n", + "r2=46#\n", + "rm=(r1+r2)/2#\n", + "u1=0.12#\n", + "u2=0.15#\n", + "dm=d-(p/2)#\n", + "alpha=atan(p/(pi*dm))#\n", + "phi=atan(u1)#\n", + "T=P*L#\n", + "F=T/((dm*tan(alpha+phi)/2)+(u2*rm))#\n", + "F=F*10**-3#\n", + "print \"F is %0.3f kN \"%(F)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "F is 7.299 kN \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 16-5 - Page 453" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "d=25#\n", + "p=8#\n", + "F=392.4#\n", + "L=250#\n", + "l=p*2#\n", + "u=0.14#\n", + "dm=d-(p/2)#\n", + "alpha=atan(l/(pi*dm))#\n", + "phi=atan(u)#\n", + "T=dm*tan(alpha+phi)/2#\n", + "M=F*L#\n", + "P=M/T*10**-3#\n", + "print \"P is %0.1f kN \"%(P)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "P is 23.6 kN \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 16-6 - Page 454" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "d=52#\n", + "W=2.2*10**3#\n", + "p=8#\n", + "r1=15#\n", + "r2=30#\n", + "rm=(r1+r2)/2#\n", + "u1=0.15/cos(14.5*pi/180)#\n", + "dm=d-(p/2)#\n", + "alpha=atan(p/(pi*dm))#\n", + "phi=atan(u1)#\n", + "Ts=W*dm*tan(alpha+phi)/2#\n", + "u2=0.12#\n", + "Tc=u2*W*rm#\n", + "T=10**-3*(Ts+Tc)#\n", + "N=40#\n", + "w=2*pi*N/60#\n", + "P=T*w*10**-3#\n", + "To=W*dm/2*tan(alpha)#\n", + "n=To/(T*10**3)#\n", + "print \"P is %0.2f KW \"%(P)#\n", + "print \"\\nn is %0.4f \"%(n)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "P is 0.07 KW \n", + "\n", + "n is 0.1659 \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 16-7 - Page 455" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "alpha=atan(2*0.2/(pi*0.9))#\n", + "u1=0.15#\n", + "phi=atan(u1)#\n", + "P=200#\n", + "L=250#\n", + "Tt=P*L#\n", + "W=10*10**3#\n", + "u2=0.15#\n", + "x=Tt/W#\n", + "d=x/0.1716#\n", + "d=30#\n", + "p=6#\n", + "dr=0.8*d#\n", + "d=24#\n", + "p=5#\n", + "dr=d-p#\n", + "dm=d-(p/2)#\n", + "print \"d is %0.0f mm \"%(d)#\n", + "print \"\\np is %0.0f mm \"%(p)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "d is 24 mm \n", + "\n", + "p is 5 mm \n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 16-8 - Page 456" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "FOS=3#\n", + "sigut=380#\n", + "Ta=0.577*sigut/FOS#\n", + "d=25#\n", + "Tus=460#\n", + "Ps=pi*d*Tus#\n", + "siga=127#\n", + "dr=sqrt(Ps*4/(pi*siga))#\n", + "d=30#\n", + "p=6#\n", + "dr=d-p#\n", + "dm=d-(p/2)#\n", + "u1=0.15#\n", + "alpha=atan(p*2/(pi*dm))#\n", + "phi=atan(u1)#\n", + "T=Ps*dm*tan(alpha+phi)/2#\n", + "T1=16*T/(pi*dr**3)#\n", + "sigc=4*Ps/(pi*dr**2)#\n", + "sigmax=sigc/2+sqrt((sigc/2**2)+(T1**2))#\n", + "Tmax=sqrt((sigc/2**2)+(T1**2))#\n", + "n=tan(alpha)/tan(alpha+phi)#\n", + "Uo=Ps/2#\n", + "Ui=Uo/n#\n", + "wav=pi/2#\n", + "wmax=2*wav#\n", + "I=Ui*2/wmax**2#\n", + "k=0.4#\n", + "Ir=0.9*I*10**-3#\n", + "m=Ir/k**2#\n", + "R=0.4#\n", + "rho=7200#\n", + "a=sqrt(m/(2*pi*R*rho))#\n", + "T=T*10**-3#\n", + "print \" T is %0.3f Nm \"%(T)#\n", + "print \"\\n n is %0.4f \"%(n)#\n", + "print \"\\n a is %0.5f mm \"%(a)#\n", + " \n", + "#The difference in the answers of T is due to rounding-off of values." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " T is 145.242 Nm \n", + "\n", + " n is 0.4751 \n", + "\n", + " a is 0.04894 mm \n" + ] + } + ], + "prompt_number": 8 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch17.ipynb b/Machine_Design_by_U.C._Jindal/Ch17.ipynb new file mode 100644 index 00000000..ea3f1c00 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch17.ipynb @@ -0,0 +1,362 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:d8d637e7ecf72daf7222d9457fa6d30ac8f9c6b3089d299c87aaa45e1504178c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:17 Sliding contact bearings" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 17-1 - Page 482" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "Ta=22#\n", + "u=7/10**9#\n", + "nj=20#\n", + "r=25#\n", + "l=2*r#\n", + "Ao=30000#\n", + "Uo=15.3/10**3#\n", + "c=0.025#\n", + "#specific weight of the material is rho\n", + "rho=8.46*(10**-6)#\n", + "Cp=179.8#\n", + "Tf=Ta+(16*pi**3*u*nj**2*l*r**3/(Uo*Ao*c))#\n", + "# avg mean film temperature is Tav\n", + "Tav=(Tf-Ta)/2#\n", + "x= l*c*rho*pi*r*nj*Cp*10**3#\n", + "y=Ao*Tav*Uo#\n", + "delT=y/x#\n", + "print \" Tav is %0.2f degC \"%(Tav)#\n", + "print \"\\n delT is %0.1f degC \"%(delT)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Tav is 47.29 degC \n", + "\n", + " delT is 7.3 degC \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 17-2 - Page 482" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "l=60#\n", + "d=60#\n", + "r=d/2#\n", + "ho=0.008#\n", + "c=0.04#\n", + "S=0.0446#\n", + "nj=1260/60#\n", + "W=6000#\n", + "p=W/(l*d)#\n", + "u=S*(c/r)**2*p/nj#\n", + "u=u*10**9#\n", + "print \"u is %0.3f cP \"%(u)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "u is 6.293 cP \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 17-3 - Page 483" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "d=60#\n", + "r=30#\n", + "l=60#\n", + "c=0.8*10**-3*r#\n", + "ho=0.2*c#\n", + "W=21000/2#\n", + "p=W/(l*d)#\n", + "S=0.0446#\n", + "nj=1440/60#\n", + "u=S*(c/r)**2*p/nj#\n", + "u=u*10**9#\n", + "# since Q/(r*nj*l)=4.62\n", + "Q=4.62*r*c*nj*l#\n", + "Q=Q*60/10**6#\n", + "print \" u is %0.3f cP \"%(u)#\n", + "print \"\\n Q is %0.4f lpm \"%(Q)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " u is 3.469 cP \n", + "\n", + " Q is 0.2874 lpm \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 17-4 - Page 483" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "l=60#\n", + "d=60#\n", + "r=d/2#\n", + "W=3000#\n", + "p=W/(l*d)#\n", + "u=30*10**-9#\n", + "c=0.06#\n", + "nj=1440/60#\n", + "S=(r/c)**2*u*nj/p#\n", + "#For ratio l/d=1, values of different parameters are given in matrix A corresponding to S\n", + "from numpy import mat\n", + "A=mat([[0,0,0,0,0,0],[0,1, 0.264, 0.6, 5.79, 3.99],[0,1, 0.121, 0.4 ,3.22, 4.33]])#\n", + "#let ho/c=x\n", + "x=(A[1,3])-((A[1,3]-(A[2,3]))*((A[1,2])-S)/((A[1,2])-(A[2,2])))#\n", + "#let y= (r/c)*f=CFV\n", + "y=(A[1,4])-(A[1,4]-(A[2,4]))*((A[1,2])-S)/((A[1,2])-(A[2,2]))#\n", + "#let z=Q/(r*c*nj*l)=FV\n", + "z=(A[1,5])-((A[1,5]-(A[2,5]))*((A[1,2])-S)/((A[1,2])-(A[2,2])))#\n", + "f=y*c/r#\n", + "ho=x*c#\n", + "Q=z*r*c*nj*l#\n", + "Q=Q*60/10**6#\n", + "delT=8.3*p*y/z#\n", + "#let power lost in friction be Pf\n", + "Pf=2*pi*nj*f*W*r/10**6#\n", + "print \" f is %0.5f \"%(f)#\n", + "print \"\\n ho is %0.3f mm \"%(ho)#\n", + "print \"\\n Q is %0.3f lpm \"%(Q)#\n", + "print \"\\n delT is %0.1f degC \"%(delT)#\n", + "print \"\\n Pf is %0.4f KW \"%(Pf)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " f is 0.00985 \n", + "\n", + " ho is 0.032 mm \n", + "\n", + " Q is 0.638 lpm \n", + "\n", + " delT is 8.3 degC \n", + "\n", + " Pf is 0.1337 KW \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 17-5 - Page 484" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "W=22000#\n", + "nj=960/60#\n", + "p=2.4#\n", + "u=20*10**-9#\n", + "d=sqrt(W/p)#\n", + "d=96#\n", + "r=d/2#\n", + "l=d#\n", + "S=0.0446#\n", + "pact=W/(l*d)#\n", + "#x=r/c#\n", + "x=sqrt(S*pact/(u*nj))#\n", + "c=r/x#\n", + "ho=0.2*c#\n", + "Q=r*c*nj*l*4.62#\n", + "Q=Q*60/10**6#\n", + "print \" d is %0.0f mm \"%(d)#\n", + "print \"\\n l is %0.0f mm \"%(l)#\n", + "print \"\\n ho is %0.4f mm \"%(ho)#\n", + "print \"\\n Q is %0.3f lpm \"%(Q)#\n", + "#The difference in answer to Q is due to rounding -off the value of c." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " d is 96 mm \n", + "\n", + " l is 96 mm \n", + "\n", + " ho is 0.0166 mm \n", + "\n", + " Q is 1.701 lpm \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 17-6 - Page 485" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "W=400*10**3#\n", + "Ro=200#\n", + "Ri=160#\n", + "ho=0.1#\n", + "t=150#\n", + "# specific gravity is rho\n", + "rho=0.86#\n", + "pi=2*W*log(Ro/Ri)/(pi*(Ro**2-Ri**2))#\n", + "zk=(0.22*t)-(180/t)#\n", + "z=rho*zk#\n", + "u=z/(10**9)#\n", + "Q=pi*pi*ho**3/(6*u*log(Ro/Ri))#\n", + "Q=Q*60/10**6#\n", + "print \" pi is %0.3f MPa \"%(pi)#\n", + "print \"\\n Q is %0.2f lpm \"%(Q)#\n", + " \n", + " #The difference in answer to Q is due to rounding -off of values." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " pi is 3.946 MPa \n", + "\n", + " Q is 25.52 lpm \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 17-7 - Page 486" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#let number of pads be n\n", + "n=4#\n", + "W=100*10**3#\n", + "Ro=125#\n", + "Ri=50#\n", + "t=200#\n", + "ho=0.15#\n", + "pi=2*W*log(Ro/Ri)/(pi*(Ro**2-Ri**2))#\n", + "zk=(0.22*t)-(180/t)#\n", + "# specific gravity is rho\n", + "rho=0.86#\n", + "z=rho*zk#\n", + "u=z/(10**9)#\n", + "Q=pi*pi*ho**3/(6*u*log(Ro/Ri))#\n", + "Q=Q*60/10**6#\n", + "print \"pi is %0.2f MPa \"%(pi)#\n", + "print \"\\nQ is %0.3f lpm \"%(Q)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "pi is 3.54 MPa \n", + "\n", + "Q is 12.441 lpm \n" + ] + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch18.ipynb b/Machine_Design_by_U.C._Jindal/Ch18.ipynb new file mode 100644 index 00000000..517ede07 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch18.ipynb @@ -0,0 +1,458 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:055c3393195abf0a8ef572fedb25f45ff1f7afe240916a9a2958b3762222b625" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:18 Rolling bearings" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 18-1 - Page 507" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "Pr=16*10**3#\n", + "u=0.0011#\n", + "F=u*Pr#\n", + "r=20*10**-3#\n", + "#Let frictional moment be M\n", + "M=F*r#\n", + "N=1440#\n", + "w=2*pi*N/60#\n", + "Pf=M*w#\n", + "print \"Pf is %0.2f W \"%(Pf)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Pf is 53.08 W \n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 18-2 - Page 508" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "C=5590#\n", + "Ca=2500#\n", + "Pa=625#\n", + "Pr=1250#\n", + "V=1#\n", + "X=0.56#\n", + "Y=1.2#\n", + "P1=(X*V*Pr)+(Y*Pa)#\n", + "L1=(C/P1)**3#\n", + "V=1.2#\n", + "P2=(X*V*Pr)+(Y*Pa)#\n", + "L2=(C/P2)**3#\n", + "print \"L1 is %0.1f million revolutions \"%(L1)#\n", + "print \"\\nL2 is %0.2f million revoltions \"%(L2)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "L1 is 57.3 million revolutions \n", + "\n", + "L2 is 43.46 million revoltions \n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 18-4 - Page 509" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=20*10**3#\n", + "Co=22400#\n", + "C=41000#\n", + "Ln=(C/P)**3#\n", + "Lh=Ln*10**6/(720*60)#\n", + "print \"Lh is %0.3f hrs \"%(Lh)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Lh is 199.424 hrs \n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 18-5 - Page 510" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "R1x=120#\n", + "R1y=250#\n", + "R2x=300#\n", + "R2y=400#\n", + "Lh=8000#\n", + "N=720#\n", + "Ln=Lh*60*N*10**-6#\n", + "R1=sqrt(R1x**2+R1y**2)#\n", + "R2=sqrt(R2x**2+R2y**2)#\n", + "#Let load factor be Ks\n", + "Ks=1.5#\n", + "P1=R1*Ks#\n", + "P2=R2*Ks#\n", + "C1=P1*(Ln**(1/3))#\n", + "C2=P2*(Ln**(1/3))#\n", + "#let designation,d,D,B,C at bearing B1 be De1,d1,D1,B1,C1\n", + "d1=25#\n", + "D1=37#\n", + "B1=7#\n", + "C1=3120#\n", + "De1=61805#\n", + "#let designation,d,D,B,C at bearing B2 be De2,d2,D2,B2,C2\n", + "d2=25#\n", + "D2=47#\n", + "B2=8#\n", + "C2=7620#\n", + "De2=16005#\n", + "print \"Designation of Bearing B1 is %0.0f \"%(De1)#\n", + "print \"\\nd1 is %0.0f mm \"%(d1)#\n", + "print \"\\nD1 is %0.0f mm \"%(D1)#\n", + "print \"\\nB1 is %0.0f mm \"%(B1)#\n", + "print \"\\nC1 is %0.0f N \"%(C1)#\n", + "print \"\\nDesignation of Bearing B2 is %0.0f \"%(De2)#\n", + "print \"\\nd2 is %0.0f mm \"%(d2)#\n", + "print \"\\nD2 is %0.0f mm \"%(D2)#\n", + "print \"\\nB2 is %0.0f mm \"%(B2)#\n", + "print \"\\nC2 is %0.0f N \"%(C2)#\n", + "print 'Bearing 61805 at B1 and 16005 at B2 can be installed.'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Designation of Bearing B1 is 61805 \n", + "\n", + "d1 is 25 mm \n", + "\n", + "D1 is 37 mm \n", + "\n", + "B1 is 7 mm \n", + "\n", + "C1 is 3120 N \n", + "\n", + "Designation of Bearing B2 is 16005 \n", + "\n", + "d2 is 25 mm \n", + "\n", + "D2 is 47 mm \n", + "\n", + "B2 is 8 mm \n", + "\n", + "C2 is 7620 N \n", + "Bearing 61805 at B1 and 16005 at B2 can be installed.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 18-6 - Page 511" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import tan\n", + "P=7500#\n", + "N=1440#\n", + "w=2*pi*N/60#\n", + "T=P/w#\n", + "r=0.2#\n", + "#Let T1-T2=t\n", + "t=T/r#\n", + "T2=t/2.5#\n", + "T1=3.5*T2#\n", + "R=0.125#\n", + "Ft=T/R#\n", + "Fr=Ft*tan(20*pi/180)#\n", + "# RD & RA are reaction forces calculated in vertical and horizontal directions from FBD by force equilibrium\n", + "RDv=186.5#\n", + "RAv=236.2#\n", + "RDh=36.2#\n", + "RAh=108.56#\n", + "RA=sqrt(RAv**2+RAh**2)#\n", + "RD=sqrt(RDv**2+RDh**2)#\n", + "Ks=1.4#\n", + "P1=RA*Ks#\n", + "P2=RD*Ks#\n", + "#let designation,d,D,B,C at bearing B1 be De1,d1,C1\n", + "d1=25#\n", + "C1=3120#\n", + "De1=61805#\n", + "#let designation,d,D,B,C at bearing B2 be De2,d2,C2\n", + "d2=25#\n", + "\n", + "C2=2700#\n", + "De2=61804#\n", + "L1=(C1/P1)**3#\n", + "Lh1=L1*10**6/(720*60)#\n", + "L2=(C2/P2)**3#\n", + "Lh2=L2*10**6/(720*60)#\n", + "print \"Lh1 is %0.0f hrs \"%(Lh1)#\n", + "print \"\\nLh2 is %0.0f hrs \"%(Lh2)#\n", + "#Incorrect value of P2 is taken in the book while calculating L2." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Lh1 is 14585 hrs \n", + "\n", + "Lh2 is 24216 hrs \n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + " exa 18-7 - Page 511" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "P=3500#\n", + "Lh=6000#\n", + "N=1400#\n", + "R98=0.98#\n", + "R90=0.9#\n", + "L98=Lh*60*N/10**6#\n", + "x=(log(1/R98)/log(1/R90))**(1/1.17)#\n", + "L90=L98/x#\n", + "C=P*L90**(1/3)#\n", + "print \"C is %0.0f N \"%(C)#\n", + "#The difference in the value of C is due to rounding-off of value of L." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "C is 44589 N \n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 18-8 - Page 512" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "n=3#\n", + "P=3#\n", + "#Let Reliability of system be R\n", + "R=0.83#\n", + "L94=6#\n", + "R94=(R)**(1/n)#\n", + "x=(log(1/R94)/log(1/0.90))**(1/1.17)#\n", + "L90=L94/x#\n", + "C=P*L90**(1/3)#\n", + "print \"C is %0.3f kN \"%(C)#\n", + "#The difference in the value of C is due to rounding-off of value of L." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "C is 6.337 kN \n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 18-9 - Page 512" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P1=3000#\n", + "P2=4000#\n", + "P3=5000#\n", + "N1=1440#\n", + "N2=1080#\n", + "N3=720#\n", + "t1=1/4#\n", + "t2=1/2#\n", + "t3=1/4#\n", + "n1=N1*t1#\n", + "n2=N2*t2#\n", + "n3=N3*t3#\n", + "N=(n1+n2+n3)#\n", + "Pe=(((n1*P1**3)+(n2*P2**3)+(n3*P3**3))/N)**(1/3)#\n", + "Lh=10*10**3#\n", + "L=Lh*60*N/10**6#\n", + "C=Pe*L**(1/3)#\n", + "print \"C is %0.0f N \"%(C)#\n", + "#The difference in the value of C is due to rounding-off of value of Pe" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "C is 34219 N \n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 18-10 - Page 513" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Co=695#\n", + "C=1430#\n", + "Pa1=200#\n", + "Pr1=600#\n", + "x=Pa1/Co#\n", + "y=Pa1/Pr1#\n", + "e=0.37+((0.44-0.37)*0.038/0.28)#\n", + "X=1#\n", + "Y=0#\n", + "P1=600#\n", + "Pa2=120#\n", + "Pr2=300#\n", + "X=0.56#\n", + "Y=1.2-(0.2*0.042/0.12)#\n", + "P2=(X*Pr2)+(Y*Pa2)#\n", + "N1=1440#\n", + "N2=720#\n", + "t1=2/3#\n", + "t2=1/3#\n", + "n1=N1*t1#\n", + "n2=N2*t2#\n", + "N=(n1+n2)#\n", + "Pe=(((n1*P1**3)+(n2*P2**3))/N)**(1/3)#\n", + "L=(C/Pe)**3#\n", + "Lh=L*10**6/(N*60)#\n", + "print \"Lh is %0.2f hrs \"%(Lh)#\n", + "#The difference in the value of Lh is due to rounding-off of value of Pe" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Lh is 227.66 hrs \n" + ] + } + ], + "prompt_number": 19 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch19.ipynb b/Machine_Design_by_U.C._Jindal/Ch19.ipynb new file mode 100644 index 00000000..ddcc05f3 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch19.ipynb @@ -0,0 +1,382 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:02d16750a95e9cb061b70a14a88598b538bc7e53305c21d4fa8042fd93a35729" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:19 Flywheel" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 19-1 - Page 530" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import sqrt, pi, cos, sin\n", + "R=1200#\n", + "b=300#\n", + "t=150#\n", + "N=500#\n", + "m=7100*10**-9*b*t#\n", + "Ar=b*t#\n", + "Aa=Ar/4#\n", + "C=(20280/t**2)+0.957+(Ar/Aa)#\n", + "w=2*pi*N/60#\n", + "V=w*R*10**-3#\n", + "siga=2*10**3*m*V**2/(C*Aa*3)#\n", + "theta=30*pi/180#\n", + "alpha=30*pi/180#\n", + "x1=10**3*m*(V**2)/(b*t)#\n", + "y1=cos(theta)/(3*C*sin(alpha))#\n", + "z1=2000*R*10**-3/(C*t)*((1/alpha)-(cos(theta)/sin(alpha)))#\n", + "sigrr1=x1*(1-y1+z1)#\n", + "theta=0*pi/180#\n", + "x2=10**3*m*(V**2)/(b*t)#\n", + "y2=cos(theta)/(3*C*sin(alpha))#\n", + "z2=2000*R*10**-3/(C*t)*((1/alpha)-(cos(theta)/sin(alpha)))#\n", + "sigrr2=x2*(1-y2-z2)#\n", + "print \"axial stress is %0.2f MPa \"%(siga)#\n", + "print \"\\ntensile stress for theta=30deg is %0.1f MPa \"%(sigrr1)#\n", + "print \"\\ntensile stress for theta=0deg is %0.2f MPa \"%(sigrr2)#\n", + "#The difference in the value of sigrr1 and sigrr2 is due to rounding-off of values." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "axial stress is 12.76 MPa \n", + "\n", + "tensile stress for theta=30deg is 38.9 MPa \n", + "\n", + "tensile stress for theta=0deg is 31.74 MPa \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 19-2 - Page 530" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import asin, cos\n", + "N=350#\n", + "theta1=asin(sqrt((3-0.6)/4))#\n", + "theta1=theta1*180/pi#\n", + "theta2=(180)-theta1#\n", + "#Ti=16000+6000*sind(3*theta)#\n", + "#To=16000+3600*sind(theta)#\n", + "a=-3600*(cos(pi/180*theta2)-cos(pi/180*theta1))#\n", + "b=2000*(cos(pi/180*3*theta2)-cos(pi/180*3*theta1))#\n", + "c=a+b#\n", + "delU=c#\n", + "Ks=0.05#\n", + "w=2*pi*N/60#\n", + "I=delU/(Ks*w**2)#\n", + "V=25#\n", + "Ir=I*0.95#\n", + "R=V/w#\n", + "Mr=Ir/R**2#\n", + "rho=7150#\n", + "t=sqrt(Mr*(10**6)/(2*pi*R*2*rho))#\n", + "b=2*t#\n", + "print \"t is %0.2f mm \"%(t)#\n", + "print \"\\nb is %0.2f mm \"%(b)#\n", + "print \"\\nR is %0.3f m \"%(R)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "t is 63.37 mm \n", + "\n", + "b is 126.74 mm \n", + "\n", + "R is 0.682 m \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 19-3 - Page 531" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "N=300#\n", + "Ks=0.03#\n", + "rho=7150#\n", + "Kr=0.9#\n", + "w=2*pi*N/60#\n", + "WD=(300*2*pi)+(4*pi*200/4)#\n", + "Tm=400#\n", + "delU=pi*200/16#\n", + "Ir=Kr*delU/(w**2*Ks)#\n", + "R=Ir/(rho*1.5*0.1*0.1*2*pi)#\n", + "R=R**(1/5)#\n", + "t=0.1*R*1000#\n", + "b=1.5*t#\n", + "print \"t is %0.2f mm \"%(t)#\n", + "print \"\\nb is %0.2f mm \"%(b)#\n", + "print \"\\nR is %0.4f m \"%(R)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "t is 28.16 mm \n", + "\n", + "b is 42.24 mm \n", + "\n", + "R is 0.2816 m \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 19-4 - Page 532" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "d=20#\n", + "t=12#\n", + "Tus=450#\n", + "Pmax=pi*d*t*Tus#\n", + "WD=Pmax*t/2*10**-3#\n", + "n=0.95#\n", + "Wi=WD/n#\n", + "delU=5*Wi/6#\n", + "N=300#\n", + "w=2*pi*N/60#\n", + "Ks=0.2#\n", + "I=delU/(Ks*w**2)#\n", + "Ir=I*0.9#\n", + "R=0.5#\n", + "m=Ir/R**2#\n", + "rho=7150#\n", + "t=sqrt(m*10**6/(rho*2*pi*R*2))#\n", + "b=2*t#\n", + "print \"t is %0.1f mm \"%(t)#\n", + "print \"\\nb is %0.1f mm \"%(b)#\n", + "print \"\\nR is %0.1f m \"%(R)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "t is 26.9 mm \n", + "\n", + "b is 53.8 mm \n", + "\n", + "R is 0.5 m \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 19-5 - Page 533" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "U=(500*2*pi)+(3*pi*500/2)#\n", + "Tm=U/(2*pi)#\n", + "delU=2.25*pi*125/2#\n", + "Ks=0.1#\n", + "N=250#\n", + "w=2*pi*N/60#\n", + "I=delU/(Ks*w**2)#\n", + "t=0.03#\n", + "rho=7800#\n", + "R=(I*2/(pi*rho*t))**(1/4)#\n", + "V=R*w#\n", + "v=0.3#\n", + "sigmax=rho*V**2*(3+v)/8*10**-6#\n", + "print \"R is %0.3f m \"%(R)#\n", + "print \"\\nsigmax is %0.2f MPa \"%(sigmax)#\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "R is 0.364 m \n", + "\n", + "sigmax is 0.29 MPa \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 19-6 - Page 534" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "N=1.5*8*60#\n", + "l=200#\n", + "t=1.5/2#\n", + "W=350*10**3#\n", + "WD=0.15*l*W*10**-6#\n", + "n=0.9# #since frictional effect is 10%, effciency of system is 90%\n", + "Wi=WD/n#\n", + "L=400#\n", + "delU=(L-(0.15*l))/(L)*10**3*Wi#\n", + "Ks=0.12#\n", + "w=2*pi*N/60#\n", + "I=delU/(Ks*w**2)#\n", + "Ir=I*0.9#\n", + "R=0.7#\n", + "m=Ir/R**2#\n", + "rho=7150#\n", + "t=sqrt(m*10**6/(rho*2*pi*R*1.5))#\n", + "b=1.5*t#\n", + "print \"t is %0.1f mm \"%(t)#\n", + "print \"\\nb is %0.1f mm \"%(b)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "t is 24.8 mm \n", + "\n", + "b is 37.2 mm \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 19-7 - Page 535" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "N=144#\n", + "#Let n be no. of punches/ min\n", + "n=8#\n", + "#Let t be timr for 1 punch\n", + "t=60/n#\n", + "theta=N/60*2*pi*0.6#\n", + "T=2.1#\n", + "U=T*theta#\n", + "#Let U1 be revolution of crankshaft in t sec\n", + "U1=t*N/60*2*pi#\n", + "delU=(U1-theta)/U1*U*10**3#\n", + "w=2*pi*1440/60#\n", + "Ks=0.1#\n", + "I=delU/(Ks*w**2)#\n", + "Ir=I*0.9#\n", + "rho=7100#\n", + "\n", + "R=Ir/(rho*0.2*0.1*2*pi)#\n", + "R=R**(1/5)#\n", + "t=0.1*R*1000#\n", + "b=0.2*R*10**3#\n", + "t=40#\n", + "b=80#\n", + "R=400#\n", + "# printing data in scilab o/p window\n", + "print \"t is %0.0f mm \"%(t)#\n", + "print \"\\nb is %0.0f mm \"%(b)#\n", + "print \"\\nR is %0.0f mm \"%(R)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "t is 40 mm \n", + "\n", + "b is 80 mm \n", + "\n", + "R is 400 mm \n" + ] + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch20.ipynb b/Machine_Design_by_U.C._Jindal/Ch20.ipynb new file mode 100644 index 00000000..219efcd6 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch20.ipynb @@ -0,0 +1,472 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:024c273b6060959af0addb9a1834f0026ae7e3f431c6d0d728fddda4c01ee1bb" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:20 Flat belt drive" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 20-1 - Page 565" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import sqrt, pi, asin, exp, degrees\n", + "b=0.2#\n", + "P=50*10**3#\n", + "v=20#\n", + "m=1.95#\n", + "d=0.3#\n", + "D=0.9#\n", + "C=5.8#\n", + "u=0.4#\n", + "#Let density be rho\n", + "rho=1000#\n", + "E=40#\n", + "#Let T1-T2 = T\n", + "T=P/v#\n", + "#Let the centrifugal tension be Tc\n", + "Tc=m*v**2#\n", + "alpha=degrees(asin((D+d)/(2*C)))#\n", + "theta=180+(2*alpha)#\n", + "theta=theta*pi/180#\n", + "x = exp(u*theta)#\n", + "T2=(((1-x)*Tc)-T)/(1-x)#\n", + "#T1=T+T2#\n", + "T1=T+T2#\n", + "t=m/(b*rho)*10**3#\n", + "#Let maximum stress be sigmax\n", + "b=200#\n", + "d=300#\n", + "sigmax=(T1/(b*t)+((E*t)/d))#\n", + "sigmin=(T2/(b*t))#\n", + "print \"T1 is %0.1f N \"%(T1)#\n", + "print \"\\nT2 is %0.1f N \"%(T2)#\n", + "print \"\\nt is %0.2f mm \"%(t)\n", + "print \"\\ntheta is %0.2f rad \"%(theta)\n", + "print \"\\nsigmax is %0.2f N/mm**2 \"%(sigmax)#\n", + "print \"\\nsigmin is %0.3f N/mm**2 \"%(sigmin)#\n", + "#The answer for T1 is miscalculated in the book." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "T1 is 4167.4 N \n", + "\n", + "T2 is 1667.4 N \n", + "\n", + "t is 9.75 mm \n", + "\n", + "theta is 3.35 rad \n", + "\n", + "sigmax is 3.44 N/mm**2 \n", + "\n", + "sigmin is 0.855 N/mm**2 \n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 20-2 - Page 566" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=12*10**3#\n", + "d=0.2#\n", + "D=0.5#\n", + "C=2#\n", + "sigmax=2*10**6#\n", + "t=8*10**-3#\n", + "#Let density be rho\n", + "rho=950#\n", + "u=0.38#\n", + "N=1500#\n", + "#Let angle of contact = thetad\n", + "thetad=180-(2*degrees(asin((D-d)/(2*C))))#\n", + "thetad=thetad*pi/180#\n", + "thetaD=(2*pi)-thetad#\n", + "v=(2*pi*N*d)/(60*2)#\n", + "#Let T1-T2=T\n", + "T=P/v#\n", + "x=exp(u*thetad)#\n", + "b=(T*x)/((1-x)*t*((rho*v**2)-(sigmax)))#\n", + "b=b*10**3#\n", + "#Let breadth of the pulley be b1\n", + "b1=b*10**3+13# #Table 20-3\n", + "L=sqrt((4*C**2)-(C*(D-d)**2))+((D*thetaD)+(d*thetad))/2#\n", + "# Let pulley crown for d=h1, D=h2\n", + "h1=0.6# #Table 20-4\n", + "h2=1#\n", + "print \"b is %0.2f mm \"%(b)\n", + "print \"\\nL is %0.2f m \"%(L)\n", + "print \"\\nb1 is %0.2f mm \"%(b1)#\n", + "print \"\\nh1 is %0.1f mm \"%(h1)#\n", + "print \"\\nh2 is %0.1f mm \"%(h2)#\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "b is 79.64 mm \n", + "\n", + "L is 5.10 m \n", + "\n", + "b1 is 79650.98 mm \n", + "\n", + "h1 is 0.6 mm \n", + "\n", + "h2 is 1.0 mm \n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 20-3 - Page 567" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=11#\n", + "N=1440#\n", + "n=480#\n", + "C=2.4#\n", + "#Let power transmitte dfrom high speed belt =P1\n", + "P1=0.0118#\n", + "V=5#\n", + "Ks=1.2#\n", + "v=15#\n", + "d=v*10**3*60/(2*pi*N)#\n", + "d=0.2#\n", + "D=N/n*d#\n", + "#Let angle of contact =thetaA\n", + "thetaA=180-(2*degrees(asin((D-d)/(2*C))))#\n", + "thetaA=thetaA*pi/180#\n", + "v=(2*pi*N*d)/(60*2)#\n", + "#Let the arc of contact correction factor be Ka\n", + "Ka=1.05#\n", + "Pd=P*Ka*Ks#\n", + "#Let corrected load rating=Pc\n", + "Pc=P1*v/V#\n", + "b=Pd/(Pc*4)#\n", + "thetaB=(2*pi)-thetaA#\n", + "L=sqrt((4*C**2)-((D-d)**2))+((d*thetaA/2)+(D*thetaB)/2)#\n", + "\n", + "print \"v is %0.2f m/s \"%(v)\n", + "print \"\\nb is %0.3f mm \"%(b)\n", + "print \"\\nL is %0.4f m \"%(L)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "v is 15.08 m/s \n", + "\n", + "b is 97.364 mm \n", + "\n", + "L is 6.0733 m \n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 20-4 - Page 568" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "N=1440#\n", + "i=2.5#\n", + "C=3600#\n", + "#let load factor be LF\n", + "LF=1.3#\n", + "P=12*10**3#\n", + "n=N/i#\n", + "V=16#\n", + "d=V*10**3*60/(2*pi*N)#\n", + "d=220#\n", + "D=d*i#\n", + "V=2*pi*N*d/(2*60*1000)#\n", + "v=5#\n", + "#Let power transmitte dfrom high speed belt =P1\n", + "P1=0.0118#\n", + "#Let LR be the load rating of belt\n", + "LR=P1/v*V#\n", + "theta=180+(2*degrees(asin((D-d)/(2*C))))#\n", + "theta=theta*pi/180#\n", + "#Let Arc of contact connection factor be CF\n", + "CF=1-(0.03/2)#\n", + "Pd=P*LF*CF#\n", + "b=Pd/(LR*5)#\n", + "b=80#\n", + "L=sqrt((4*C**2)-(D+d)**2)+(theta*(D+d)/2)#\n", + "L=L*10**-3#\n", + "print \"V is %0.1f m/s \"%(V)\n", + "print \"\\nb is %0.0f mm \"%(b)\n", + "print \"\\nL is %0.3f m \"%(L)#\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "V is 16.6 m/s \n", + "\n", + "b is 80 mm \n", + "\n", + "L is 8.404 m \n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 20-5 - Page 569" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "i=3.6#\n", + "N=1440#\n", + "d=220#\n", + "Ks=1.2#\n", + "Kf=1.1#\n", + "C=5000#\n", + "u=0.8#\n", + "D=i*d#\n", + "#From table 20-7, the following data is available\n", + "t=5#\n", + "b=120#\n", + "Fa=30.64#\n", + "#let weight density be w\n", + "w=0.106*10**5#\n", + "Cp=0.71# #From table 20-6\n", + "Cv=1#\n", + "T1=Fa*b*t*Cp*Cv#\n", + "m=w*b*t/10**6#\n", + "V=2*pi*N*d/(2*60*1000)#\n", + "Tc=m*V**2/9.81#\n", + "theta=180+(2*degrees(asin((D-d)/(2*C))))#\n", + "theta=theta*pi/180#\n", + "x=u*theta#\n", + "T2=Tc+((T1-Tc)/exp(x))#\n", + "Pd=(T1-T2)*V*10**-3#\n", + "P=Pd/(Ks*Kf)#\n", + "print \"V is %0.2f m/s \"%(V)#\n", + "print \"\\nPd is %0.2f KW \"%(Pd)#\n", + "print \"\\nP is %0.1f KW \"%(P)#\n", + "#The value of T2 is calculated incorrectly, therefore there is a difference in the values of Pd and P." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "V is 16.59 m/s \n", + "\n", + "Pd is 197.77 KW \n", + "\n", + "P is 149.8 KW \n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 20-6 - Page 570" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "i=2.5#\n", + "C=4500#\n", + "N=960#\n", + "P=20*10**3#\n", + "Ks=1.15#\n", + "Kf=1.10#\n", + "t=8#\n", + "#let weight density be w\n", + "w=0.110*10**5#\n", + "m=w*t/10**6#\n", + "Fa=8.75#\n", + "d=200#\n", + "D=i*d#\n", + "u=0.4#\n", + "V=2*pi*N*d/(2*60*1000)#\n", + "Pd=P*Ks*Kf#\n", + "Cp=1#\n", + "Cv=0.6#\n", + "#to find b\n", + "T1=Fa*t*Cp*Cv#\n", + "Tc=m*V**2/9.81#\n", + "theta=180-(2*degrees(asin((D-d)/(2*C))))\n", + "theta=theta*pi/180#\n", + "x=u*theta#\n", + "T2=Tc+((T1-Tc)/exp(x))#\n", + "T=Pd/V#\n", + "b=T/(T1-T2)#\n", + "#b=90#\n", + "L=sqrt((4*C**2)-(D+d)**2)+(theta*(D+d)/2)#\n", + "L=L*10**-3#\n", + "print \"V is %0.2f m/s \"%(V)\n", + "print \"\\nb is %0.3f mm \"%(b)\n", + "print \"\\nL is %0.3f m \"%(L)#\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "V is 10.05 m/s \n", + "\n", + "b is 86.537 mm \n", + "\n", + "L is 10.049 m \n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 20-7 - Page 571" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "b=160#\n", + "t=7#\n", + "P=3*10**3#\n", + "Ks=1.2#\n", + "d=160#\n", + "N=1440#\n", + "D=480#\n", + "C=2400#\n", + "w=11200#\n", + "u=0.4#\n", + "Fa=7.2#\n", + "m=w*b*t/10**6#\n", + "V=2*pi*N*d/(2*60*1000)#\n", + "Tc=m*V**2/9.81#\n", + "Cp=0.6# #from table 20-6\n", + "Cv=0.98# #from table 20-7\n", + "Ta=Fa*b*Cp*Cv#\n", + "T=P/V#\n", + "theta=180-(2*degrees(asin((D-d)/(2*C))))\n", + "theta=theta*pi/180#\n", + "x=u*theta#\n", + "#T2=Tc+((T1-Tc)/exp(x))#\n", + "T2=(T+((exp(x)*Tc)-Tc))/(exp(x)-1)#\n", + "T1=T+T2#\n", + "Kf=Ta/T1#\n", + "Pd=P*Ks*Kf#\n", + "Pd=Pd*10**-3#\n", + "print \"Tc is %0.0f N \"%(Tc)#\n", + "print \"\\nT1 is %0.2f N \"%(T1)#\n", + "print \"\\nT2 is %0.2f N \"%(T2)#\n", + "print \"\\nKf is %0.2f \"%(Kf)#\n", + "print \"\\nPd is %0.1f KW \"%(Pd)#\n", + "#The difference in values of T1 and T2 is due to rounding-off of values.\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Tc is 186 N \n", + "\n", + "T1 is 541.46 N \n", + "\n", + "T2 is 292.78 N \n", + "\n", + "Kf is 1.25 \n", + "\n", + "Pd is 4.5 KW \n" + ] + } + ], + "prompt_number": 15 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch21.ipynb b/Machine_Design_by_U.C._Jindal/Ch21.ipynb new file mode 100644 index 00000000..8416f217 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch21.ipynb @@ -0,0 +1,346 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:10bd9abe18e1c81566ed029a02a2720fe79b25747c3617a3cb6346d82c9110e2" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:21 V belt drive" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 21-1 - Page 579" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import sqrt, pi, asin, degrees, exp\n", + "P1=12*10**3#\n", + "d=0.3#\n", + "D=0.9#\n", + "C=0.9#\n", + "A=230*10**-6#\n", + "#density is rho\n", + "rho=1100#\n", + "N=1500#\n", + "#Maximum stress is sig\n", + "sig=2.1*10**6#\n", + "#semi groove angle is b\n", + "b=20*pi/180#\n", + "u=0.22#\n", + "m=rho*A#\n", + "v=2*pi*N*d/(60*2)#\n", + "Tc=m*v**2#\n", + "T1=A*sig#\n", + "#wrap angle is thetaA\n", + "ang=(D-d)/(2*C)#\n", + "thetaA=pi/180*(180-(2*degrees(asin(ang))))\n", + "thetaB=((2*pi)-thetaA)#\n", + "x=u*thetaB#\n", + "T2=Tc+((T1-Tc)/exp(x))#\n", + "P2=(T1-T2)*v#\n", + "n=P1/P2#\n", + "n=3# #(rounding off to nearest whole number)\n", + "print \"Tc is %0.1f N \"%(Tc)#\n", + "print \"\\nT1 is %0.0f N \"%(T1)#\n", + "print \"\\nT2 is %0.1f N \"%(T2)#\n", + "print \"\\nP2 is %0.0f W \"%(P2)#\n", + "print \"\\nn is %0.0f \"%(n)#\n", + " \n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Tc is 140.5 N \n", + "\n", + "T1 is 483 N \n", + "\n", + "T2 is 288.2 N \n", + "\n", + "P2 is 4589 W \n", + "\n", + "n is 3 \n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 21-2 - Page 579" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sin\n", + "D=0.6#\n", + "d=0.3#\n", + "C=0.9#\n", + "m=0.193#\n", + "n=2#\n", + "N=1500#\n", + "u=0.3#\n", + "v=2*pi*N/60*d/2#\n", + "P=150*10**3#\n", + "Tc=m*v**2#\n", + "#let T1-T2=T\n", + "T=P/(n*v)#\n", + "thetaA=pi/180*(180-(2*degrees(asin((D-d)/(2*C)))))#\n", + "thetaB=((2*pi)-thetaA)#\n", + "#Groove angle=b\n", + "b=17.5*pi/180#\n", + "x=u*thetaA/sin(b)#\n", + "y=exp(x)#\n", + "c=(Tc*(1-y))#\n", + "T2=(T+(Tc*(1-y)))/(y-1)#\n", + "#T2=(T-y)/Tc#\n", + "T1=T+Tc#\n", + "Lp=2*sqrt((C**2)-((D-d)/2)**2)+(thetaA*d/2)+(thetaB*D/2)#\n", + "v=sqrt(T/(3*m))#\n", + "print \"Tc is %0.2f N \"%(Tc)#\n", + "print \"\\nT1 is %0.0f N \"%(T1)#\n", + "print \"\\nT2 is %0.2f N \"%(T2)#\n", + "print \"\\nLp is %0.3f m \"%(Lp)#\n", + "print \"\\nv is %0.2f m/s \"%(v)#\n", + "print \"\\nThe designation of the belt is B-3251-45 \"#\n", + "#The difference in values of T1 and T2 is due to rounding-off of values." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Tc is 107.15 N \n", + "\n", + "T1 is 3290 N \n", + "\n", + "T2 is 98.93 N \n", + "\n", + "Lp is 3.239 m \n", + "\n", + "v is 74.15 m/s \n", + "\n", + "The designation of the belt is B-3251-45 \n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 21-3 - Page 580" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import tan\n", + "C=1#\n", + "m=0.35#\n", + "d=0.25#\n", + "P=22*10**3#\n", + "#Let the smaller pulley dia be n\n", + "#Let the larger pulley dia be N\n", + "n=1000#\n", + "N=400#\n", + "D=d*n/N#\n", + "v=2*pi*n*d/(60*2)#\n", + "Tc=m*v**2#\n", + "topwidth=22#\n", + "h=14#\n", + "bottomwidth=topwidth-(2*h*tan(20*pi/180))#\n", + "A=(topwidth+bottomwidth)/2*h#\n", + "#let allowable tension be Ta\n", + "Ta=2.2#\n", + "T1=A*Ta#\n", + "u=0.28#\n", + "thetaA=pi/180*(180-(2*degrees(asin((D-d)/(2*C)))))\n", + "thetaB=((2*pi)-thetaA)#\n", + "#Groove angle=b=19\n", + "b=19*pi/180#\n", + "x=u*thetaA/sin(b)#\n", + "T2=Tc+((T1-Tc)/exp(x))#\n", + "n=P/((T1-T2)*v)#\n", + "Lp=2*sqrt((C**2)-((D-d)/2)**2)+(thetaA*d/2)+(thetaB*D/2)#\n", + "print \"Tc is %0.2f N \"%(Tc)#\n", + "print \"\\nT1 is %0.1f N \"%(T1)#\n", + "print \"\\nT2 is %0.1f N \"%(T2)#\n", + "print \"\\nn is %0.1f \"%(n)#\n", + "print \"\\nLp is %0.3f m \"%(Lp)#\n", + "print \"\\nThe designation of the belt is C-3414-47 \"#\n", + " \n", + " # difference in value of Lp is due to rounding-off the values of thetaA and thetaB." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Tc is 59.97 N \n", + "\n", + "T1 is 520.7 N \n", + "\n", + "T2 is 102.7 N \n", + "\n", + "n is 4.0 \n", + "\n", + "Lp is 3.410 m \n", + "\n", + "The designation of the belt is C-3414-47 \n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 21-4 - Page 580" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=12*10**3#\n", + "Ks=1.1#\n", + "Pd=12*10**3*Ks#\n", + "N=1440#\n", + "B=17#\n", + "t=11#\n", + "d=200#\n", + "i=3#\n", + "D=i*d#\n", + "C=1000#\n", + "# since angle of contact theta is very small\n", + "theta=(D-d)/C#\n", + "theta=theta*180/pi#\n", + "Kc=0.8#\n", + "Lp=(2*C)+(pi/2*(D+d))+(((D-d)**2)/(4*C))#\n", + "Li=Lp-45#\n", + "Ki=1.1#\n", + "#let number of v-belts required = n\n", + "#let the KW rating be KWR\n", + "KWR=5.23#\n", + "n=(P*Ks)/(KWR*Ks*Ki*10**3)#\n", + "n=3#\n", + "print \"D is %0.1f mm \"%(D)#\n", + "print \"\\nC is %0.1f mm \"%(C)#\n", + "print \"\\nn is %0.3f \"%(n)#\n", + "print \"\\nLi is %0.0f mm \"%(Li)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "D is 600.0 mm \n", + "\n", + "C is 1000.0 mm \n", + "\n", + "n is 3.000 \n", + "\n", + "Li is 3252 mm \n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 21-5 - Page 581" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "N=800;\n", + "P=20;\n", + "i=2.5;\n", + "Ks=1.5; #(from table for 3-5 hrs/day)\n", + "Pd=P*Ks;\n", + "d=250;\n", + "D=i*d;\n", + "C=1.6*D;\n", + "Lp=(2*C)+(pi*(D+d)/2)+((D-d)**2)/(4*C);\n", + "Li=Lp+74;\n", + "Listd=3454;\n", + "Lp=Listd+74;\n", + "p=[1, -1.0768, 0.0175];\n", + "from sympy import symbols, solve\n", + "P = symbols('P')\n", + "expr = P**2*p[0]+P*p[1]+p[2]\n", + "z = solve(expr, P)[1]\n", + "KW=9.4;\n", + "Kc=0.795;\n", + "K1=1;\n", + "n=Pd/(KW*Kc*K1);\n", + "print \" C is %0.4f m \"%(z);\n", + "print \"\\n Pd is %0.0f KW \"%(Pd);\n", + "print \"\\n n is %0.2f KW \"%(n);\n", + " \n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " C is 1.0603 m \n", + "\n", + " Pd is 30 KW \n", + "\n", + " n is 4.01 KW \n" + ] + } + ], + "prompt_number": 17 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch22.ipynb b/Machine_Design_by_U.C._Jindal/Ch22.ipynb new file mode 100644 index 00000000..30ed5549 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch22.ipynb @@ -0,0 +1,598 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:b283e3fcd471f1cbe397d09e22017ed17e30966a86b52dc70653c73e3f5c2124" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:22 Friction clutches" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 22-1 - Page 588" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "u=0.28 #(coefficient of friction)\n", + "N=300 #(Engine rpm)\n", + "I=7.2 \n", + "Pmax= 0.1# \n", + "R1=70#\n", + "R2=110#\n", + "n=2# #(Both sides of the plate are effective)\n", + "#Using Uniform Wear Theory\n", + "#Axial Force W\n", + "W=n*pi*Pmax*R1*(R2-R1)#\n", + "#Frictional Torque Tf\n", + "Tf=u*W*(R1+R2)/2*(10**-3)#\n", + "w=2*pi*N/60#\n", + "#Power P\n", + "P=Tf*w#\n", + "#Torque = Mass moment of inertia*angular acceleration\n", + "a=Tf/I#\n", + "t=w/a# \n", + "#Angle turned by driving shaft theta1 through which slipping takes place\n", + "theta1=w*t#\n", + "#angle turned by driven shaft theta2\n", + "theta2=a*(t**2)/2#\n", + "E=Tf*(theta1-theta2)#\n", + "print \"\\nThe force is %0.1f N\"%(W)#\n", + "print \"\\nThe Torque is %0.2f Nm\"%(Tf)#\n", + "print \"\\nThe Power is %0.0f W\"%(P)#\n", + "print \"\\nThe angular acceleration is %0.2f rad/sec**2\"%(a)#\n", + "print \"\\nThe time taken is %0.1f sec\"%(t)#\n", + "print \"\\nThe energy is %0.2f Nm\"%(E)#\n", + "\n", + "#The difference in the answer of energy 'E' is due to rounding-off of values." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "The force is 1759.3 N\n", + "\n", + "The Torque is 44.33 Nm\n", + "\n", + "The Power is 1393 W\n", + "\n", + "The angular acceleration is 6.16 rad/sec**2\n", + "\n", + "The time taken is 5.1 sec\n", + "\n", + "The energy is 3553.06 Nm\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 22-2 - Page 589" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Power P\n", + "P=80*10**3# #(Watt)\n", + "N=3000# #(Engine rpm)\n", + "w=2*pi*3*10**3/60\n", + "Tf=8*10**4/w#\n", + "Rm=100##(mm)\n", + "p=0.2 #N/mm**2\n", + "u=0.22 \n", + "# let width b= (R1-R2). \n", + "#Axial force W=2*pi*Rm*b*p\n", + "#Torque T=u*W*Rm\n", + "b=Tf/(u*2*pi*(Rm**2)*p)#\n", + "b=50# \n", + "R2=Rm+b#\n", + "R1=Rm-b#\n", + "Di=2*R1# #inner diameter\n", + "W=2*pi*Rm*b*p#\n", + "n=8# #n is number of springs\n", + "#Axial force per spring W1\n", + "W1=W/n#\n", + "W1=W1+15#\n", + "#axial deflection del\n", + "Del=10# \n", + "#stiffness k\n", + "k=W1/Del#\n", + "# Spring index C\n", + "C=6#\n", + "#number of coils n1\n", + "n1=6# #Assumption\n", + "d=k*n*n1*(C**3)/(80*10**3)#\n", + "d=11# # Rounding off to nearest standard value\n", + "D=C*d#\n", + "clearance=2#\n", + "FL=((n1+2)*d)+(2*Del)+clearance# # two end coils, therefore (2*del)\n", + "\n", + "print \"\\nThe Torque is %0.2f Nm\"%(Tf)#\n", + "print \"\\nThe width is %0.0f mm\"%(b)#\n", + "print \"\\nThe force is %0.0f N\"%(W)#\n", + "print \"\\nThe Axial force per spring is %0.0f N\"%(W1)#\n", + "print \"\\nThe Spring stiffness is %0.0f N/mm\"%(k)#\n", + "print \"\\nThe Spring wire diameter is %0.0f mm\"%(d)#\n", + "print \"\\nThe Mean coil diameter is %0.0f mm\"%(D)#\n", + "print \"\\nThe Free length is %0.0f mm\"%(FL)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "The Torque is 254.65 Nm\n", + "\n", + "The width is 50 mm\n", + "\n", + "The force is 6283 N\n", + "\n", + "The Axial force per spring is 800 N\n", + "\n", + "The Spring stiffness is 80 N/mm\n", + "\n", + "The Spring wire diameter is 11 mm\n", + "\n", + "The Mean coil diameter is 66 mm\n", + "\n", + "The Free length is 110 mm\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 22-3 - Page 589" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Power P\n", + "P=40*10**3 #Watt\n", + "n1=100# #rpm\n", + "n2=400# #rpm\n", + "#Speed factor Ks\n", + "Ks=0.9+0.001*n2#\n", + "#Clutch power Pc\n", + "Pc=P*n2/(n1*Ks)*10**-3#\n", + "print \"\\nThe Speed factor is %0.1f \"%(Ks)#\n", + "print \"\\nThe clutch poweris %0.0f KW\"%(Pc)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "The Speed factor is 1.3 \n", + "\n", + "The clutch poweris 123 KW\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 22-4 - Page 590" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# plot Torque vs Ro/Ri\n", + "#x=Ro/Ri\n", + "#According to Uniform Wear theory\n", + "x=[0, 0.2, 0.4, 0.577, 0.6, 0.8, 1.0]#\n", + "n=len(x)#\n", + "Tf = range(0,n)\n", + "for i in range(0,n):\n", + " Tf[i]=(x[i]-(x[i]**3))#\n", + "\n", + "%matplotlib inline\n", + "from matplotlib.pyplot import plot, xlabel, ylabel, show, grid\n", + "plot (x,Tf)#\n", + "xlabel(' Ro/Ri ')#\n", + "ylabel('Tf')#\n", + "grid()#\n", + "show()" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x7fe9440af310>" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 22-5 - Page 591" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "n1=4#\n", + "n2=3#\n", + "n=(n1+n2-1)#\n", + "R2=80#\n", + "R1=50#\n", + "#According to Uniform Pressure Theory\n", + "#W=p*pi*((R2**2)-(R1**2)) T=n*2*u*W*((R2**3)-(R1**3))/(((R2**2)-(R1**2))*3)\n", + "P=15*10**3#\n", + "N=1400#\n", + "u=0.25#\n", + "w=2*pi*N/60#\n", + "T=P/w#\n", + "W=T*3*((R2**2)-(R1**2))/(n*2*u*((R2**3)-(R1**3)))*10**3#\n", + "p=W/(pi*((R2**2)-(R1**2)))#\n", + "print \"\\nThe angular speed is %0.2f rad/sec\"%(w)#\n", + "print \"\\nThe Torque is %0.3f Nm\"%(T)#\n", + "print \"\\nThe uniform pressure is %0.3f N/mm**2\"%(p)#\n", + "print \"\\nThe Force is %0.1f N\"%(W)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "The angular speed is 146.61 rad/sec\n", + "\n", + "The Torque is 102.314 Nm\n", + "\n", + "The uniform pressure is 0.084 N/mm**2\n", + "\n", + "The Force is 1031.1 N\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 22-6 - Page 592" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=5*10**3#\n", + "N=1000#\n", + "w=2*pi*N/60#\n", + "Rm=50#\n", + "pm=0.3#\n", + "Tf=P/w#\n", + "u=0.1#\n", + "R2=50*2/(0.6+1)#\n", + "R1=0.6*R2#\n", + "#According to uniform Wear theory\n", + "W=pm*Rm*(R2-R1)*2*pi#\n", + "n=Tf*(10**3)/(u*W*Rm)#\n", + "pmax=pm*Rm/R1#\n", + "print \"\\nThe angular speed is %0.2f rad/sec\"%(w)#\n", + "print \"\\nThe Torque is %0.3f Nm\"%(Tf)#\n", + "print \"\\nThe Inner radius is %0.1f mm\"%(R1)#\n", + "print \"\\nThe Outer radius is %0.1f mm\"%(R2)#\n", + "print \"\\nThe number of contacting surfaces is %0.0f \"%(n)#\n", + "print \"\\nThe max. pressure is %0.1f N/mm**2\"%(pmax)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "The angular speed is 104.72 rad/sec\n", + "\n", + "The Torque is 47.746 Nm\n", + "\n", + "The Inner radius is 37.5 mm\n", + "\n", + "The Outer radius is 62.5 mm\n", + "\n", + "The number of contacting surfaces is 4 \n", + "\n", + "The max. pressure is 0.4 N/mm**2\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 22-7 - Page 593" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=12*10**3#\n", + "N=750 #Speed=N\n", + "w=2*pi*N/60#\n", + "Tf=P/w#\n", + "p1=0.12#\n", + "a=12.5##Semi-cone angle\n", + "u=0.3#\n", + "k=u*0.18246*1.121/0.21644#\n", + "R1=(Tf*(10**3)/k)**(1/3)#\n", + "R2=R1*1.242#\n", + "Rm=1.121*R1#\n", + "W=2*pi*p1*R1*(R2-R1)#\n", + "print \"\\nThe angular speed is %0.2f rad/sec\"%(w)#\n", + "print \"\\nThe Torque is %0.1f Nm\"%(Tf)#\n", + "print \"\\nThe Inner radius is %0.1f mm\"%(R1)#\n", + "print \"\\nThe Outer radius is %0.1f mm\"%(R2)#\n", + "print \"\\nThe mean radius is %0.2f mm\"%(Rm)#\n", + "print \"\\nThe axial force is %0.0f N\"%(W)#\n", + "\n", + "#The difference in the answer is due to rounding-off of values." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "The angular speed is 78.54 rad/sec\n", + "\n", + "The Torque is 152.8 Nm\n", + "\n", + "The Inner radius is 81.4 mm\n", + "\n", + "The Outer radius is 101.1 mm\n", + "\n", + "The mean radius is 91.23 mm\n", + "\n", + "The axial force is 1208 N\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 22-8 - Page 594" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sin\n", + "#semi-cone angle is given as 15 degree\n", + "k=sin(15*pi/180)#\n", + "u=0.3#\n", + "W=300#\n", + "Rm=90/2#\n", + "Tf=u*W*Rm/k#\n", + "Tf=Tf*(10**-3)#\n", + "I=0.4#\n", + "a=Tf/I#\n", + "N=1440#\n", + "w=2*pi*N/60#\n", + "t=w/a#\n", + "#During Slipping\n", + "theta1=w*t#\n", + "theta2=theta1/2#\n", + "U=Tf*(theta1-theta2)#\n", + "print \"\\nThe Torque is %0.3f Nm\"%(Tf)#\n", + "print \"\\nThe angular acceleration is %0.3f rad/sec**2\"%(a)#\n", + "print \"\\nThe angular speed is %0.1f rad/sec\"%(w)#\n", + "print \"\\nThe time taken is %0.2f sec\"%(t)#\n", + "print \"\\nThe Energy lost in friction is %0.0f Nm\"%(U)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "The Torque is 15.648 Nm\n", + "\n", + "The angular acceleration is 39.120 rad/sec**2\n", + "\n", + "The angular speed is 150.8 rad/sec\n", + "\n", + "The time taken is 3.85 sec\n", + "\n", + "The Energy lost in friction is 4548 Nm\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 22-9 - Page 595" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=15*10**3#\n", + "Ka=1.25#\n", + "N=1500#\n", + "w=2*pi*N/60#\n", + "Tf=P/w#\n", + "d=(Tf*16/(50*pi))**(1/3)#\n", + "d=25#\n", + "Rm=5*d#\n", + "Pav=0.12#\n", + "u=0.22#\n", + "b=Tf/(pi*u*Pav*(Rm**2))#\n", + "b=40#\n", + "R1=Rm-(b*sin(15*pi/180)/2)#\n", + "R2=Rm+(b*sin(15*pi/180)/2)#\n", + "print \"\\nThe Torque is %0.2f Nm\"%(Tf)#\n", + "print \"\\nThe shaft diameter is %0.0f mm\"%(d)#\n", + "print \"\\nThe width is %0.0f mm\"%(b)#\n", + "print \"\\nThe Inner radius is %0.1f mm\"%(R1)#\n", + "print \"\\nThe Outer radius is %0.1f mm\"%(R2)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "The Torque is 95.49 Nm\n", + "\n", + "The shaft diameter is 25 mm\n", + "\n", + "The width is 40 mm\n", + "\n", + "The Inner radius is 119.8 mm\n", + "\n", + "The Outer radius is 130.2 mm\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 22-10 - Page 596" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "w2=2*pi*1400/60#\n", + "w1=0.8*w2#\n", + "P=40*10**3#\n", + "T=P/w2#\n", + "n=4#\n", + "T1=T/4#\n", + "R=0.16##Inner radius of drum\n", + "r=0.13##radial distance of each shoe from axis of rotation\n", + "u=0.22##coefficient of friction\n", + "x=u*r*R*((w2**2)-(w1**2))\n", + "m =T1/x#\n", + "l=R*pi/3#\n", + "N=T1/(R*u)#\n", + "p=1*10**5#\n", + "b=N/(p*l)*10**3#\n", + "print \"\\nThe full speed is %0.1f rad/sec\"%(w2)#\n", + "print \"\\nThe engagement speed is %0.2f rad/sec\"%(w1)#\n", + "print \"\\nThe number of shoes is %0.0f \"%(n)#\n", + "print \"\\nThe Torque is %0.1f Nm\"%(T)#\n", + "print \"\\nThe Torque per shoe is %0.1f Nm\"%(T1)#\n", + "print \"\\nThe mass per shoe is %0.2f kg\"%(m)#\n", + "print \"\\nThe length of friction lining is %0.5f m\"%(l)#\n", + "print \"\\nThe width is %0.1f mm\"%(b)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "The full speed is 146.6 rad/sec\n", + "\n", + "The engagement speed is 117.29 rad/sec\n", + "\n", + "The number of shoes is 4 \n", + "\n", + "The Torque is 272.8 Nm\n", + "\n", + "The Torque per shoe is 68.2 Nm\n", + "\n", + "The mass per shoe is 1.93 kg\n", + "\n", + "The length of friction lining is 0.16755 m\n", + "\n", + "The width is 115.7 mm\n" + ] + } + ], + "prompt_number": 21 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch23.ipynb b/Machine_Design_by_U.C._Jindal/Ch23.ipynb new file mode 100644 index 00000000..0ff4eb7a --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch23.ipynb @@ -0,0 +1,438 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:75a895f6b6e2b52911a88eac6dae9689200eafdf77449669257c302ec104ee3c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:23 Brakes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 23-1 - Page 618" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import asin, pi\n", + "W=20e3#\n", + "m=W/9.81#\n", + "#diameter of brake drum\n", + "Db=0.6#\n", + "p=1#\n", + "Vi=1#\n", + "Vf=0#\n", + "D=1#\n", + "R=0.5#\n", + "wi=Vi/R#\n", + "wf=0#\n", + "w=1#\n", + "Vav=0.5#\n", + "S=2#\n", + "t=S/Vav#\n", + "#angle turned by by hoist drum=theta\n", + "theta=0.5*wi*t#\n", + "K_E=0.5*m*Vi**2#\n", + "P_E=2*W#\n", + "T_E=K_E+P_E#\n", + "T=T_E/theta#\n", + "P=wi*T*10**-3#\n", + "Rb=Db/2#\n", + "Ft=0.5*T*p/Rb#\n", + "u=0.35#\n", + "N=Ft/u#\n", + "#contact area of brake lining=A\n", + "A=N/p#\n", + "b=0.3*Db#\n", + "L=A*10**-6/(b)#\n", + "#angle subtended at brake drum centre=theta2\n", + "theta2=2*(asin(L/Db))#\n", + "theta2=theta2*180/pi# # converting radian to degree\n", + "print \"T is %0.1f Nm \"%(T)#\n", + "print \"\\nP is %0.4f kW \"%(P)#\n", + "print \"\\nb is %0.2f m \"%(b)#\n", + "print \"\\nL is %0.3f m \"%(L)#\n", + "print \"\\ntheta2 is %0.0f deg \"%(theta2)#\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "T is 10254.8 Nm \n", + "\n", + "P is 20.5097 kW \n", + "\n", + "b is 0.18 m \n", + "\n", + "L is 0.271 m \n", + "\n", + "theta2 is 54 deg \n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 23-2 - Page 618" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import exp\n", + "b=80#\n", + "t=2#\n", + "theta=225*pi/180#\n", + "u=0.22#\n", + "#F1/F2=e**(u*theta)\n", + "#let F1/F2=x#\n", + "x=exp(u*theta)#\n", + "#maximum tensile stress in steel tape is siga\n", + "siga=60#\n", + "A=b*t#\n", + "F1=siga*A#\n", + "F2=F1/x#\n", + "r=0.2#\n", + "T=(F1-F2)*r#\n", + "OA=30#\n", + "OB=100#\n", + "OC=350#\n", + "P=((F2*OB)+(F1*OA))/OC#\n", + "OA=F2*OB/F1#\n", + "print \"F1 is %0.0f N \"%(F1)#\n", + "print \"\\nF2 is %0.1f N \"%(F2)#\n", + "print \"\\nT is %0.2f Nm \"%(T)#\n", + "print \"\\nOA is %0.2f mm \"%(OA)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "F1 is 9600 N \n", + "\n", + "F2 is 4046.4 N \n", + "\n", + "T is 1110.72 Nm \n", + "\n", + "OA is 42.15 mm \n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 23-3 - Page 619" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sin\n", + "theta=pi/3#\n", + "r=160#\n", + "u=0.3#\n", + "pmax=0.9#\n", + "b=40#\n", + "R=(4*r*sin(theta))/((2*theta)+sin(2*theta))#\n", + "#frictional torque is T\n", + "T=2*u*pmax*b*(r**2)*sin(theta)#\n", + "T=2*T*10**-3#\n", + "Rx=0.5*pmax*b*r*((2*theta)+(sin(2*theta)))*10**-3#\n", + "Ry=u*Rx#\n", + "print \"T is %0.2f Nmm \"%(T)#\n", + "print \"\\nR is %0.3f mm \"%(R)#\n", + "print \"\\nRx is %0.3f kN \"%(Rx)#\n", + "print \"\\nRy is %0.2f kN \"%(Ry)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "T is 957.75 Nmm \n", + "\n", + "R is 187.222 mm \n", + "\n", + "Rx is 8.526 kN \n", + "\n", + "Ry is 2.56 kN \n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 23-4 - Page 620" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sin, cos, pi, sqrt\n", + "d=320#\n", + "r=d/2#\n", + "b=50#\n", + "u=0.3#\n", + "pmax=1#\n", + "c=115*2#\n", + "# From to fig. 23-9, distance OA=R is calculated.\n", + "R=sqrt(115**2+66.4**2)#\n", + "C=115*2#\n", + "theta1=0#\n", + "theta2=120*pi/180#\n", + "theta0=120*pi/180#\n", + "thetamax=pi/2#\n", + "Tr=u*pmax*b*r**2*(cos(theta1)-cos(theta2))/sin(thetamax)*10**-3#\n", + "#the notation 'r' is used for moments of right hand shoe, similarly 'l' for the left shoe.\n", + "Mfr=u*pmax*b*r*(4*r*(cos(theta1)-cos(theta2))+(R*(cos(2*theta1)-cos(2*theta2))))/(4*sin(thetamax))*10**-3#\n", + "Mpr=pmax*b*r*R*((2*theta0)-(sin(2*theta2)-(sin(theta1))))/(4*sin(thetamax))*10**-3#\n", + "F=(Mpr-Mfr)/c*10**3#\n", + "#Mpl+Mfl=F*c#\n", + "x=F*c*10**-3#\n", + "y=(Mpr/pmax)+(Mfr/pmax)#\n", + "pmax2=x/y#\n", + "Tl=pmax2*Tr#\n", + "Mpl=pmax2*Mpr#\n", + "Mfl=pmax2*Mfr#\n", + "T=Tl+Tr#\n", + "print \"Tr is %0.0f Nm \"%(Tr)#\n", + "print \"\\nMf is %0.2f Nm \"%(Mfr)#\n", + "print \"\\nMp is %0.2f Nm \"%(Mpr)#\n", + "print \"\\nTl is %0.1f Nm \"%(Tl)# \n", + "print \"\\nMfl is %0.2f Nm \"%(Mfl)#\n", + "print \"\\nMpl is %0.2f Nm \"%(Mpl)#\n", + "print \"\\nF is %0.1f N \"%(F)#\n", + "print \"\\nT is %0.1f Nm \"%(T)#\n", + " \n", + " #The difference in the answers are due to rounding-off of values.\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Tr is 576 Nm \n", + "\n", + "Mf is 695.51 Nm \n", + "\n", + "Mp is 1342.49 Nm \n", + "\n", + "Tl is 182.9 Nm \n", + "\n", + "Mfl is 220.79 Nm \n", + "\n", + "Mpl is 426.18 Nm \n", + "\n", + "F is 2812.9 N \n", + "\n", + "T is 758.9 Nm \n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 23-5 - Page 621" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "m=1100#\n", + "V=65*5/18#\n", + "t=4#\n", + "r=0.22#\n", + "mb=12#\n", + "C=460#\n", + "S=0.5*V*t#\n", + "#Total kinetic energy TE=K.E(vehicle)+K.E(rotating parts).\n", + "TE=((0.5*m*(V**2))+(0.1*0.5*m*(V**2)))#\n", + "E=TE/4#\n", + "w=V/r#\n", + "theta=S/r#\n", + "T=E/theta#\n", + "delT=E/(mb*C)#\n", + "print \"S is %0.2f m \"%(S)#\n", + "print \"\\nE is %0.2f Nm \"%(E)#\n", + "print \"\\nT is %0.2f Nm \"%(T)#\n", + "print \"\\ndelT is %0.2f \"%(delT)#\n", + " \n", + "#The difference in the answers are due to rounding-off of values." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "S is 36.11 m \n", + "\n", + "E is 49307.97 Nm \n", + "\n", + "T is 300.40 Nm \n", + "\n", + "delT is 8.93 \n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 23-6 - Page 621" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "T=35000#\n", + "u=0.4#\n", + "p=0.7#\n", + "r=200#\n", + "N=T/(u*r)\n", + "b=sqrt(N/p)#\n", + "l=b#\n", + "#2theta = theta2\n", + "theta2=2*asin(l/(2*r))#\n", + "F=u*N#\n", + "P=((250*N)-(u*N*80))/550#\n", + "Ry=N-P#\n", + "Rx=u*N#\n", + "R=sqrt(Rx**2+Ry**2)#\n", + "w=2*pi*100/60#\n", + "# Rate of heat generated is Q\n", + "Q=u*N*w*r/1000#\n", + "print \"N is %0.1f N \"%(N)#\n", + "print \"\\nb is %0.0f mm \"%(b)#\n", + "print \"\\nP is %0.1f N \"%(P)#\n", + "print \"\\nR is %0.2f N \"%(R)#\n", + "print \"\\nQ is %0.2f J/s \"%(Q)#\n", + "\n", + "#The answer to Rate of heat generated 'Q' is calculated incorrectly in the book." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "N is 437.5 N \n", + "\n", + "b is 25 mm \n", + "\n", + "P is 173.4 N \n", + "\n", + "R is 316.81 N \n", + "\n", + "Q is 366.52 J/s \n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 23-7 - Page 622" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Vi=20*5/18#\n", + "Vf=0#\n", + "m=80#\n", + "pmax=1#\n", + "u=0.1#\n", + "S=50#\n", + "KE=0.5*m*Vi**2#\n", + "N=KE/(u*S*2)#\n", + "t=sqrt(N/(pmax*3))#\n", + "b=3*t#\n", + "print \"KE is %0.1f Nm \"%(KE)#\n", + "print \"\\nN is %0.2f N \"%(N)#\n", + "print \"\\nt is %0.1f mm \"%(t)#\n", + "print \"\\nb is %0.1f mm \"%(b)#\n", + "\n", + "#The difference in the answers are due to rounding-off of values." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "KE is 1234.6 Nm \n", + "\n", + "N is 123.46 N \n", + "\n", + "t is 6.4 mm \n", + "\n", + "b is 19.2 mm \n" + ] + } + ], + "prompt_number": 15 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch24.ipynb b/Machine_Design_by_U.C._Jindal/Ch24.ipynb new file mode 100644 index 00000000..5467776a --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch24.ipynb @@ -0,0 +1,188 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:d95986be60722ab1db01b1c918002f9e15f3dc0ce9ab06247d01ab6e633f6378" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:24 Rope drive" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 24-1 - Page 635" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import sqrt, pi, exp, sin, asin\n", + "P=150000#\n", + "m=0.4#\n", + "D=1.8#\n", + "d=0.6#\n", + "C=4.2#\n", + "V=15#\n", + "Fc=m*V**2#\n", + "BL=44.81*10**3#\n", + "FOS=35#\n", + "F1=BL/FOS#\n", + "theta=pi-(2*asin((D-d)/(2*C)))#\n", + "beta=22.5*pi/180#\n", + "u=0.13#\n", + "x=u*theta/sin(beta)#\n", + "F2=(F1-Fc)/exp(x)#\n", + "n=P/((F1-F2)*V)#\n", + "n=13#\n", + "print \"n is %0.0f \"%(n)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "n is 13 \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 24-2 - Page 635" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "W=1000#\n", + "m=0.498#\n", + "BL=78#\n", + "d=12#\n", + "Am=0.39*d**2#\n", + "dw=sqrt(Am*4/(6*19*pi))#\n", + "Ew=74.4*10**3#\n", + "Ds=56*d#\n", + "sigb=Ew*dw/Ds#\n", + "Wb=sigb*pi*(d**2)/4*10**-3#\n", + "l=20#\n", + "Ws=m*l#\n", + "a=1.2#\n", + "Wa=a*(W/2+Ws)*10**-3#\n", + "#Let the static load be Ps\n", + "Ps=(W/2+Ws)*9.81*10**-3#\n", + "#let the effective load be Peff\n", + "Peff=Ps+Wb+Wa#\n", + "FOS1=BL/Peff#\n", + "FOS2=BL/(5+0.612)#\n", + "print \" annual FOS is %0.2f \"%(FOS1)#\n", + "print \"\\n FOS neglecting bending load is %0.1f \"%(FOS2)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " annual FOS is 5.02 \n", + "\n", + " FOS neglecting bending load is 13.9 \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 24-3 - Page 636" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "d=12#\n", + "sigut=1960#\n", + "Pb=0.0025*sigut#\n", + "Ds=480#\n", + "F=Pb*d*Ds/2#\n", + "W=F*2*10**-3#\n", + "print \"W is %0.3f kN \"%(W)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "W is 28.224 kN \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 24-4 - Page 637" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "sigut=1770#\n", + "Pb=0.0018*sigut#\n", + "W=4000#\n", + "a=2.5/2#\n", + "Ws=90*0.5#\n", + "Wa=(W+Ws)*a/9.81#\n", + "Weff=W+Wa#\n", + "d=sqrt(Weff*2/(23*Pb))#\n", + "d=12#\n", + "print \"d is %0.0f mm \"%(d)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "d is 12 mm \n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch25.ipynb b/Machine_Design_by_U.C._Jindal/Ch25.ipynb new file mode 100644 index 00000000..1bf3748b --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch25.ipynb @@ -0,0 +1,380 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:916802be9265ec8f5947b935ca6ebb81f4ee0ab392012dea0a384748212946ca" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:25 Gears" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 25-1 - Page 669" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "Zp=25#\n", + "Zg=60#\n", + "m=5#\n", + "dp=m*Zp#\n", + "dg=m*Zg#\n", + "CD=(dp+dg)/2#\n", + "ha=m#\n", + "hf=1.25*m#\n", + "c=hf-ha#\n", + "r=0.4*m#\n", + "print \"dp is %0.0f mm \"%(dp)#\n", + "print \"\\ndg is %0.0f mm \"%(dg)#\n", + "print \"\\nCD is %0.1f mm \"%(CD)#\n", + "print \"\\nha is %0.0f mm \"%(ha)#\n", + "print \"\\nhf is %0.2f mm \"%(hf)#\n", + "print \"\\nc is %0.2f mm \"%(c)#\n", + "print \"\\nr is %0.0f mm \"%(r)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "dp is 125 mm \n", + "\n", + "dg is 300 mm \n", + "\n", + "CD is 212.5 mm \n", + "\n", + "ha is 5 mm \n", + "\n", + "hf is 6.25 mm \n", + "\n", + "c is 1.25 mm \n", + "\n", + "r is 2 mm \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 25-2 - Page 669" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "N=800#\n", + "P=6000#\n", + "n=200#\n", + "Cs=1.4#\n", + "sigb=150#\n", + "FOS=2#\n", + "Zp=18#\n", + "Zg=Zp*N/n#\n", + "Y=pi*(0.154-(0.912/Zp))#\n", + "p=[1 ,0 ,-9.5846, -38.135]#\n", + "from sympy import symbols, solve\n", + "P = symbols('P')\n", + "expr = P**3*p[0]+P**2*p[1]+P*p[2]+p[3]\n", + "m=solve(expr, P)[0]#\n", + "dp=m*Zp#\n", + "dg=m*Zg#\n", + "# printing data in scilab o/p window\n", + "print \"dp is %0.0f mm \"%(dp)#\n", + "print \"\\ndg is %0.0f mm \"%(dg)#\n", + "# Answer is given wrong in the textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "dp is 77 mm \n", + "\n", + "dg is 309 mm \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 25-3 - Page 670" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Zp=30#\n", + "N=1000#\n", + "Zg=75#\n", + "m=5#\n", + "b=60#\n", + "sigut=450#\n", + "BHN=350#\n", + "Cs=1.5#\n", + "FOS=2#\n", + "dp=m*Zp#\n", + "dg=m*Zg#\n", + "v=2*pi*N*dp/(60*1000*2)#\n", + "Cv=3/(3+v)#\n", + "sigb=450/3#\n", + "Y=0.358#\n", + "Sb=m*b*sigb*Y#\n", + "Q=(2*Zg)/(Zp+Zg)#\n", + "K=0.16*(BHN/100)**2#\n", + "Sw=b*dp*Q*K#\n", + "Pt=Sb*Cv/(Cs*FOS)#\n", + "P=Pt*v#\n", + "P=P*10**-3#\n", + "print \"Sb is %0.0f N \"%(Sb)#\n", + "print \"\\nSw is %0.0f N \"%(Sw)#\n", + "print \"\\nP is %0.3f kW \"%(P)#\n", + "\n", + "#The difference in the value of Sw is due to rounding-off of the value of Q." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Sb is 16110 N \n", + "\n", + "Sw is 25200 N \n", + "\n", + "P is 11.657 kW \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 25-4 - Page 670" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "n=240#\n", + "P=8000#\n", + "N=1200#\n", + "CD=300#\n", + "Cs=1.5#\n", + "alpha=20*pi/180#\n", + "G=N/n#\n", + "dp=CD*2/6#\n", + "dg=5*dp#\n", + "v=2*pi*N*dp/(60*1000*2)#\n", + "Cv=3/(3+v)#\n", + "Pt=P/v#\n", + "Peff=Pt*Cs/Cv#\n", + "m=4#\n", + "b=10*m#\n", + "FOS=2#\n", + "Sb=Peff*FOS#\n", + "sigut=600#\n", + "sigb=sigut/3#\n", + "Zp=dp/m#\n", + "Zg=dg/m#\n", + "Q=(2*Zg)/(Zp+Zg)#\n", + "K=Sb/(b*dp*Q)#\n", + "BHN=sqrt(K/0.16)*100#\n", + "BHN=333#\n", + "print \"BHN is %0.0f \"%(BHN)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "BHN is 333 \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 25-5 - Page 671" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "alpha=20*pi/180#\n", + "N=800#\n", + "P=6000#\n", + "sigut=450#\n", + "i=5#\n", + "Cs=1.3#\n", + "v=3.6#\n", + "FOS=2#\n", + "Pt=P/v#\n", + "Cv=3/(3+v)#\n", + "sigb=sigut/3#\n", + "dp=3.6*1000*2*60/(2*pi*N)#\n", + "dp=86#\n", + "Sb=Pt*Cs/Cv*FOS#\n", + "#Let x be m**2*Y\n", + "x=Sb/(10*sigb)#\n", + "m=5#\n", + "Zp=18#\n", + "dp=m*Zp#\n", + "Zg=i*Zp#\n", + "dg=m*Zg#\n", + "b=10*m#\n", + "phip=m+(0.25*sqrt(dp))#\n", + "ep=32+(2.5*phip)#\n", + "phig=m+(0.25*sqrt(dg))#\n", + "eg=32+(2.5*phig)#\n", + "e=ep+eg#\n", + "e=e*10**-3#\n", + "Ps=Cs*Pt#\n", + "r1=dp/2#\n", + "r2=dg/2#\n", + "Pd=e*N*Zp*b*r1*r2/(2530*sqrt(r1**2+r2**2))#\n", + "Q=(2*Zg)/(Zp+Zg)#\n", + "K=Sb/(b*dp*Q)#\n", + "BHN=sqrt(K/0.16)*100#\n", + "print \"Ps is %0.2f N \"%(Ps)#\n", + "print \"\\nPd is %0.1f N \"%(Pd)#\n", + "print \"\\nBHN is %0.0f \"%(BHN)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ps is 2166.67 N \n", + "\n", + "Pd is 1358.6 N \n", + "\n", + "BHN is 282 \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 25-4 - Page 672" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=9000#\n", + "N=900#\n", + "n=150#\n", + "sigut=750#\n", + "BHN=300#\n", + "Cs=1.5#\n", + "FOS=2#\n", + "i=N/n#\n", + "x=sqrt(i)#\n", + "Zp=18#\n", + "Zg=x*Zp#\n", + "Zg=44#\n", + "#Let actual speed reduction be xa\n", + "xa=Zg/Zp#\n", + "n1=N/xa**2#\n", + "T1=P*60/(2*pi*N)#\n", + "i2=N/xa#\n", + "T2=N/i2*T1#\n", + "m=6#\n", + "dp=Zp*m#\n", + "dg=m*Zg#\n", + "phip=m+(0.25*sqrt(dp))#\n", + "ep=16+(1.25*phip)#\n", + "phig=m+(0.25*sqrt(dg))#\n", + "eg=16+(1.25*phig)#\n", + "e=ep+eg#\n", + "e=e*10**-3#\n", + "Pt=26000#\n", + "Ps=Cs*Pt#\n", + "r1=dp/2#\n", + "r2=dg/2#\n", + "b=10*m#\n", + "Pd=e*i2*Zp*b*r1*r2/(2530*sqrt(r1**2+r2**2))#\n", + "Q=(2*Zg)/(Zp+Zg)#\n", + "sigb=sigut/3#\n", + "Y=0.308#\n", + "\n", + "Sb=b*m*sigb*Y#\n", + "K=0.16*(BHN/100)**2#\n", + "Sw=b*dp*K*Q#\n", + "print \"m is %0.0f mm \"%(m)#\n", + "print \"\\nPd is %0.3f N \"%(Pd)#\n", + "print \"\\nSw is %0.0f N \"%(Sw)#\n", + " \n", + " #The difference in the values is due to rounding-off of the values." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "m is 6 mm \n", + "\n", + "Pd is 434.590 N \n", + "\n", + "Sw is 13244 N \n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch26.ipynb b/Machine_Design_by_U.C._Jindal/Ch26.ipynb new file mode 100644 index 00000000..ccdf600d --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch26.ipynb @@ -0,0 +1,265 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:01c1cdfe5f44c7b9518d8e4611bcc7b558fa40f7241f67c334ff272a81b92cfe" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:26 Helical gears" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 26-1 - Page 698" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import sqrt, pi, sin, cos, tan, atan\n", + "Zp=20#\n", + "Zg=50#\n", + "alphan=20*pi/180#\n", + "phi=15*pi/180#\n", + "mn=4#\n", + "m=mn/cos(phi)#\n", + "alpha=180/pi*atan(tan(alphan)/(cos(phi)))#\n", + "dp=Zp*m#\n", + "dg=Zg*m#\n", + "ha=4#\n", + "hd=1.25*mn#\n", + "#Let addendum circle dia of pinion be Pa\n", + "Pa=dp+(2*mn)#\n", + "#Let dedendum circle dia of pinion be Pd\n", + "Pd=dp-(2.5*mn)#\n", + "#Let addendum circle dia of gear be Ga\n", + "Ga=dg+(2*mn)#\n", + "#Let dedendum circle dia of gear be Gd\n", + "Gd=dg-(2.5*mn)#\n", + "b=pi*mn/sin(phi)#\n", + "print \"m is %0.2f mm \"%(m)#\n", + "print \"\\nalpha is %0.3f deg \"%(alpha)#\n", + "print \"\\nPa is %0.1f mm \"%(Pa)#\n", + "print \"\\nPd is %0.1f mm \"%(Pd)#\n", + "print \"\\nGa is %0.0f mm \"%(Ga)#\n", + "print \"\\nGd is %0.0f mm \"%(Gd)#\n", + "print \"\\nb is %0.2f mm \"%(b)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "m is 4.14 mm \n", + "\n", + "alpha is 20.647 deg \n", + "\n", + "Pa is 90.8 mm \n", + "\n", + "Pd is 72.8 mm \n", + "\n", + "Ga is 215 mm \n", + "\n", + "Gd is 197 mm \n", + "\n", + "b is 48.55 mm \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 26-2 - Page 698" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=5000#\n", + "Zp=25#\n", + "Zg=50#\n", + "mn=4#\n", + "alphan=20*pi/180#\n", + "phi=20*pi/180#\n", + "N=1200#\n", + "m=mn/cos(phi)#\n", + "dp=Zp*m#\n", + "dg=Zg*m#\n", + "v=2*pi*N*dp/(60*2*1000)#\n", + "Pt=P/v#\n", + "Pa=Pt*tan(phi)#\n", + "Pr=Pt*tan(alphan)/cos(phi)#\n", + "print \"Pt is %0.2f N \"%(Pt)#\n", + "print \"\\nPa is %0.1f N \"%(Pa)#\n", + "print \"\\nPr is %0.2f N \"%(Pr)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Pt is 747.78 N \n", + "\n", + "Pa is 272.2 N \n", + "\n", + "Pr is 289.64 N \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 26-3 - Page 699" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Zp=24#\n", + "Zg=72#\n", + "alphan=20*pi/180#\n", + "phi=24*pi/180#\n", + "N=720#\n", + "mn=5#\n", + "b=50#\n", + "sigut=600#\n", + "BHN=360#\n", + "Cs=1.4#\n", + "FOS=2#\n", + "sigb=sigut/3#\n", + "dp=mn*Zp/cos(phi)#\n", + "Zp=Zp/(cos(phi))**3#\n", + "Zg=Zg/(cos(phi))**3#\n", + "Y=0.358+((0.364-0.358)*1.48/2)#\n", + "Sb=b*mn*sigb*Y#\n", + "Q=(2*Zg)/(Zp+Zg)#\n", + "K=0.16*(BHN/100)**2#\n", + "Sw=b*dp*Q*K/(cos(phi)**2)#\n", + "v=2*pi*N*dp/(60*2*1000)#\n", + "Cv=5.6/(5.6+sqrt(v))#\n", + "Peff=Sb/FOS#\n", + "Pt=Peff*Cv/Cs#\n", + "P=Pt*v#\n", + "P=P*10**-3#\n", + "print \"P is %0.3f kW \"%(P)#\n", + "#The difference in the value is due to rounding-off of the values." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "P is 22.936 kW \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 26-4 - Page 700" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Zp=25#\n", + "Zg=100#\n", + "P=5000#\n", + "N=2000#\n", + "alphan=20*pi/180#\n", + "phi=15*pi/180#\n", + "sigut=660#\n", + "Cs=1.5#\n", + "FOS=1.8#\n", + "v=10#\n", + "Zp1=Zp/(cos(phi))**3#\n", + "Zg1=Zg/(cos(phi))**3#\n", + "Y=0.348+(0.74*0.004)#\n", + "sigb=sigut/3#\n", + "Cv=5.6/(5.6+sqrt(v))#\n", + "#Sb=FOS*Peff\n", + "mn=FOS*P*Cs*60*1000*2*cos(phi)/(2*pi*N*Cv*Zp*12*sigb*Y)#\n", + "mn=mn**(1/3)#\n", + "mn=2.5#\n", + "dp=mn*Zp/cos(phi)#\n", + "Q=(2*Zg)/(Zp+Zg)#\n", + "b=12*mn#\n", + "Sb=12*sigb*Y#\n", + "K=Sb*(cos(phi)**2)/(dp*Q*b)#\n", + "BHN=sqrt(K/0.16)*100#\n", + "dg=mn*Zg/cos(phi)#\n", + "phip=mn+(0.25*sqrt(dp))#\n", + "ep=16+(1.25*phip)#\n", + "phig=mn+(0.25*sqrt(dg))#\n", + "eg=16+(1.25*phig)#\n", + "e=ep+eg#\n", + "e=e*10**-3#\n", + "r1=dp/2#\n", + "r2=dg/2#\n", + "Pd=e*N*Zp1*b*r1*r2/(2530*sqrt(r1**2+r2**2))#\n", + "v=2*pi*N*dp/(60*2*1000)#\n", + "#Let tangential component be TC\n", + "TC=(Cs*1845/mn)+(Pd*cos(alphan)*cos(phi))#\n", + "\n", + "Sb=b*mn*sigb*Y#\n", + "print \"mn is %0.1f mm \"%(mn)#\n", + "print \"\\nTC is %0.0f N \"%(TC)#\n", + "print \"\\nSb is %0.1f N \"%(Sb)#\n", + "#The difference in the value of Sb is due to rounding-off of t" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "mn is 2.5 mm \n", + "\n", + "TC is 1965 N \n", + "\n", + "Sb is 5790.8 N \n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch27.ipynb b/Machine_Design_by_U.C._Jindal/Ch27.ipynb new file mode 100644 index 00000000..bb7c2f69 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch27.ipynb @@ -0,0 +1,461 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:93bf919486a82ec8604e11d18207bffbdc29a6c50aa4e9b5a65c5c915d6bd456" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:27 Straight bevel gears" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 27-1 - Page 712" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt, pi, atan, sin, tan, cos\n", + "from __future__ import division\n", + "P=8000#\n", + "N1=400#\n", + "N2=200#\n", + "i=N1/N2# #i=Zg/Zp=dg/dp\n", + "gamma1=atan(1/i)#\n", + "gamma2=90-gamma1#\n", + "rp=200#\n", + "R=rp/sin(gamma1)#\n", + "b=0.2*R#\n", + "rm1=rp-(b*sin(gamma1)/2)#\n", + "Pt=P*1000*60/(2*pi*N1*rm1)#\n", + "alpha=20*pi/180#\n", + "Ps=Pt*tan(alpha)#\n", + "Pr=Ps*cos(gamma1)#\n", + "Pa=Ps*sin(gamma1)#\n", + "print \"Pt is %0.0f N \"%(Pt)#\n", + "print \"\\nPs is %0.2f N \"%(Ps)#\n", + "print \"\\nPr is %0.2f N \"%(Pr)#\n", + "print \"\\nPa is %0.2f N \"%(Pa)#\n", + "\n", + "#The difference in the values is due to rounding-off of the values." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Pt is 1061 N \n", + "\n", + "Ps is 386.18 N \n", + "\n", + "Pr is 345.41 N \n", + "\n", + "Pa is 172.71 N \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 27-2 - Page 712" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "alpha=20*pi/180#\n", + "Zp=20#\n", + "Zg=36#\n", + "m=4#\n", + "sigut=600#\n", + "b=25#\n", + "dp=m*Zp#\n", + "rp=dp/2#\n", + "dg=m*Zg#\n", + "rg=dg/2#\n", + "gamma1=atan(rp/rg)#\n", + "Zpv=Zp/cos(gamma1)#\n", + "Y=0.33+0.003*0.88#\n", + "sigb=sigut/3#\n", + "Sb=m*b*sigb*Y#\n", + "print \"Zpv is %0.2f \"%(Zpv)#\n", + "print \"\\nSb is %0.0f N \"%(Sb)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Zpv is 22.88 \n", + "\n", + "Sb is 6653 N \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 27-3 - Page 712" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import asin\n", + "m=6#\n", + "Zp=30#\n", + "Zg=45#\n", + "dp=m*Zp#\n", + "rp=dp/2#\n", + "dg=m*Zg#\n", + "rg=dg/2#\n", + "R=sqrt(rg**2+rp**2)#\n", + "gamma1=180/pi*asin(rp/R)#\n", + "gamma2=(90-gamma1)#\n", + "ha=6#\n", + "hf=1.25*ha#\n", + "phi=180/pi*atan(ha/R)#\n", + "beta=180/pi*atan(hf/R)#\n", + "#let Face Cone Angle be FCA\n", + "FCA=(gamma1+phi)#\n", + "#Let Root cone angle be RCA\n", + "RCA=(gamma1-beta)#\n", + "print \"gamma1 is %0.1f deg \"%(gamma1)#\n", + "print \"\\ngamma2 is %0.1f deg \"%(gamma2)#\n", + "print \"\\nR is %0.2f mm \"%(R)#\n", + "print \"\\nFCA is %0.3f deg \"%(FCA)#\n", + "print \"\\nRCA is %0.2f deg \"%(RCA)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "gamma1 is 33.7 deg \n", + "\n", + "gamma2 is 56.3 deg \n", + "\n", + "R is 162.25 mm \n", + "\n", + "FCA is 35.808 deg \n", + "\n", + "RCA is 31.04 deg \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 27-4 - Page 713" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "alpha=20*pi/180#\n", + "Zp=25#\n", + "Zg=40#\n", + "m=5#\n", + "b=30#\n", + "BHN=400#\n", + "dp=m*Zp#\n", + "rp=dp/2#\n", + "dg=m*Zg#\n", + "rg=dg/2#\n", + "gamma1=atan(rp/rg)#\n", + "gamma1=180/pi*gamma1#\n", + "gamma2=(90-gamma1)#\n", + "a=cos(pi/180*gamma2)#\n", + "Zp1=Zp/cos(gamma1)#\n", + "Zg1=Zg/a#\n", + "Q=(2*Zg1)/(Zp1+Zg1)#\n", + "K=0.16*(BHN/100)**2#\n", + "Sw=0.75*b*dp*Q*K/cos(pi/180*gamma1)#\n", + "print \"Sw is %0.1f N \"%(Sw)#\n", + " \n", + " #The difference in the value of Sw is due to rounding-off of the value of Q." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Sw is 12142.4 N \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 27-5 - Page 713" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Zp=20#\n", + "Zg=36#\n", + "m=4#\n", + "b=25#\n", + "BHN=360#\n", + "Np=750#\n", + "FOS=1.75#\n", + "dp=m*Zp#\n", + "rp=dp/2#\n", + "dg=m*Zg#\n", + "rg=dg/2#\n", + "gamma1=atan(dp/dg)#\n", + "gamma1=180/pi*gamma1#\n", + "gamma2=(90-gamma1)#\n", + "a=cos(pi/180*gamma2)#\n", + "Zp1=Zp/cos(pi/180*gamma1)#\n", + "Zg1=Zg/a#\n", + "Q=(2*Zg1)/(Zp1+Zg1)#\n", + "K=0.16*(BHN/100)**2#\n", + "R=sqrt(rp**2+rg**2)#\n", + "Y=0.33+0.003*0.86#\n", + "sigut=600#\n", + "sigb=sigut/3#\n", + "Sb=m*b*Y*sigb*(1-(b/R))#\n", + "Sw=0.75*b*dp*Q*K/cos(pi/180*gamma1)#\n", + "print \"Sb is %0.0f N \"%(Sb)#\n", + "print \"\\nSw is %0.1f N \"%(Sw)#\n", + "\n", + "#The answwer to Sb is calculated incorrectly in the book." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Sb is 4633 N \n", + "\n", + "Sw is 5438.0 N \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 27-6 - Page 713" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Dp=300#\n", + "rp=150#\n", + "#Let the angular velocity ratio be i\n", + "i=2/3#\n", + "rg=rp/i#\n", + "Dg=2*rg#\n", + "R=sqrt(rp**2+rg**2)#\n", + "P=15000#\n", + "N=300#\n", + "Cs=1.5#\n", + "FOS=2#\n", + "sigb=100#\n", + "gamma1=atan(Dp/Dg)#\n", + "gamma1=180/pi*gamma1#\n", + "gamma2=(90-gamma1)#\n", + "v=2*pi*N*rp/(60*1000)#\n", + "Cv=5.6/(5.6+sqrt(v))#\n", + "Pt=P/v#\n", + "Peff=Pt*Cs/Cv#\n", + "Sb=Peff*FOS#\n", + "b=R/4#\n", + "#let x=m*Y\n", + "x=Sb/(b*sigb*(1-(b/R)))#\n", + "m=6#\n", + "print \"m*Y is %0.3f mm**2 \"%(x)#\n", + "print \"\\nm is %0.0f mm \"%(m)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "m*Y is 2.613 mm**2 \n", + "\n", + "m is 6 mm \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 27-7 - Page 714" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Zp=24#\n", + "Zg=36#\n", + "N=1400#\n", + "P=11600#\n", + "Cs=1.4#\n", + "FOS=2#\n", + "sigut=600#\n", + "sigb=sigut/3#\n", + "gamma1=atan(Zp/Zg)#\n", + "gamma1=180/pi*gamma1#\n", + "gamma2=(90-gamma1)#\n", + "a=cos(pi/180*gamma2)#\n", + "Zp1=Zp/cos(pi/180*gamma1)#\n", + "Zg1=Zg/a#\n", + "Q=(2*Zg1)/(Zp1+Zg1)#\n", + "v=1.76#\n", + "Pt=P/v#\n", + "Cv=5.6/(5.6+sqrt(v))#\n", + "Peff=Pt*Cs/Cv#\n", + "x=Peff*FOS#\n", + "Y=0.352+(0.003*0.85)#\n", + "y=2*sigb*Y*(1-(6/21.63))#\n", + "m=sqrt(x/y)#\n", + "# Design is safe for m=4\n", + "m=4#\n", + "b=6*m#\n", + "dp=24*m#\n", + "rp=48#\n", + "dp=dp/cos(pi/180*gamma1)#\n", + "v=2*pi*N*rp/(60*1000)#\n", + "Cv=5.6/(5.6+sqrt(v))#\n", + "Sb=y*m**2#\n", + "#Sw=Sb#\n", + "K=Sb/(0.75*b*dp*Q)#\n", + "BHN=sqrt(K/0.16)*100#\n", + "print \"m is %0.0f mm \"%(m)#\n", + "print \"\\nBHN is %0.0f \"%(BHN)#\n", + " \n", + " #The answwer to BHN is calculated incorrectly in the book." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "m is 4 mm \n", + "\n", + "BHN is 189 \n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 27-8 - Page 714" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Zp=40#\n", + "Zg=60#\n", + "P=3500#\n", + "N=600#\n", + "Cs=1.5#\n", + "sigb=55#\n", + "gamma1=atan(Zp/Zg)#\n", + "gamma1=180/pi*gamma1#\n", + "gamma2=(90-gamma1)#\n", + "a=cos(pi/180*gamma2)#\n", + "Zp1=Zp/cos(pi/180*gamma1)#\n", + "Zg1=Zg/a#\n", + "Q=(2*Zg1)/(Zp1+Zg1)#\n", + "# Design is safe for m=6\n", + "m=6#\n", + "b=6*m#\n", + "dp=Zp*m#\n", + "rp=dp/2#\n", + "dg=Zg*m#\n", + "rg=dg/2#\n", + "R=sqrt(rp**2+rg**2)#\n", + "print \"m is %0.0f mm \"%(m)#\n", + "print \"\\nb is %0.0f mm \"%(b)#\n", + "print \"\\nR is %0.0f mm \"%(R)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "m is 6 mm \n", + "\n", + "b is 36 mm \n", + "\n", + "R is 216 mm \n" + ] + } + ], + "prompt_number": 8 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch28.ipynb b/Machine_Design_by_U.C._Jindal/Ch28.ipynb new file mode 100644 index 00000000..ec28770b --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch28.ipynb @@ -0,0 +1,343 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:4d8f50c55aa5f869a3b91c9992fbc1206b0fe54a3763477766a504a0fd20f355" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:28 Worm and worm wheel set" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 28-1 - Page 726" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "Z1=1#\n", + "Z2=30#\n", + "q=10#\n", + "m=5#\n", + "d=q*m#\n", + "D=m*Z2#\n", + "#let the speed reduction ratio be G\n", + "G=Z2/Z1#\n", + "CD=(d+D)/2#\n", + "print \"G is %0.0f \"%(G)#\n", + "print \"\\nCD is %0.0f mm \"%(CD)#\n", + "print \"\\nd is %0.0f mm \"%(d)#\n", + "print \"\\nD is %0.0f mm \"%(D)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "G is 30 \n", + "\n", + "CD is 100 mm \n", + "\n", + "d is 50 mm \n", + "\n", + "D is 150 mm \n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 28-2 - Page 726" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import tan, atan, cos\n", + "Z1=1#\n", + "Z2=52#\n", + "q=10#\n", + "m=8#\n", + "i=Z2/Z1#\n", + "CD=((m*q)+(m*Z2))/2#\n", + "lamda=atan(Z1/q)#\n", + "d=q*m#\n", + "da=m*(q+2)#\n", + "df=m*(q+2-(4.4*cos(lamda)))#\n", + "pa=m*pi#\n", + "D=m*Z2#\n", + "Da=m*(Z2+(4*cos(lamda))-2)#\n", + "Df=m*(Z2-2-(0.4*cos(lamda)))#\n", + "print \"i is %0.0f \"%(i)#\n", + "print \"\\nCD is %0.0f mm \"%(CD)#\n", + "print \"\\npa is %0.2f mm \"%(pa)#\n", + "print \"\\nda is %0.0f mm \"%(da)#\n", + "print \"\\ndf is %0.3f mm \"%(df)#\n", + "print \"\\nDa is %0.3f mm \"%(Da)#\n", + "print \"\\nDf is %0.3f mm \"%(Df)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i is 52 \n", + "\n", + "CD is 248 mm \n", + "\n", + "pa is 25.13 mm \n", + "\n", + "da is 96 mm \n", + "\n", + "df is 60.975 mm \n", + "\n", + "Da is 431.841 mm \n", + "\n", + "Df is 396.816 mm \n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 28-3 - Page 727" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sin\n", + "Z1=2#\n", + "Z2=60#\n", + "q=10#\n", + "m=5#\n", + "P=6000#\n", + "N=1440#\n", + "u=0.08#\n", + "alpha=20*pi/180#\n", + "lamda=atan(Z1/q)#\n", + "d=m*q#\n", + "w=2*pi*N/60#\n", + "T=P/w#\n", + "Ptw=T*10**3/(d/2)#\n", + "a=cos(alpha)#\n", + "b=cos(lamda)#\n", + "x=sin(alpha)#\n", + "y=sin(lamda)#\n", + "Paw=Ptw*(((a*b)-(u*y))/((a*y)+(u*b)))#\n", + "Prw=Ptw*y/((a*y)+(u*b))#\n", + "#Paw=Ptw*((cos(alpha)*cos(lambda))-(u*sin(lambda)))/((cos(alpha)*sin(lambda))+(u*cos(lambda)))#\n", + "#Prw=Ptw*((sin(alpha))/((cos(alpha)*sin(lambda))+(u*cos(lambda))))#\n", + "print \"Ptw=Pag is %0.1f N \"%(Ptw)#\n", + "print \"\\nPaw=Ptg is %0.0f N \"%(Paw)#\n", + "print \"\\nPrw=Prg is %0.0f N \"%(Prw)#\n", + " \n", + "#The difference in the value is due to rounding-off the values." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ptw=Pag is 1591.5 N \n", + "\n", + "Paw=Ptg is 5487 N \n", + "\n", + "Prw=Prg is 1188 N \n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 28-4 - Page 728" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Z1=2#\n", + "Z2=40#\n", + "q=8#\n", + "m=5#\n", + "d=q*m#\n", + "P=1.2#\n", + "lamda=atan(Z1/q)#\n", + "N=1000#\n", + "Vt=2*pi*N*20/(60*1000)#\n", + "Vs=Vt/cos(lamda)#\n", + "u=0.032#\n", + "alpha=20*pi/180#\n", + "x=cos(alpha)#\n", + "y=tan(lamda)#\n", + "z=(cos(lamda))/sin(lamda)#\n", + "n=(x-(u*y))/(x+(u*z))#\n", + "#Let power output be Po\n", + "Po=P*n#\n", + "#Let power lost in friction be Pf\n", + "Pf=P-Po#\n", + "print \"P is %0.1f kW \"%(P)#\n", + "print \"\\nPo is %0.3f kW \"%(Po)#\n", + "print \"\\nPf is %0.3f kW \"%(Pf)#\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "P is 1.2 kW \n", + "\n", + "Po is 1.047 kW \n", + "\n", + "Pf is 0.153 kW \n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 28-5 - Page 729" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Z1=2#\n", + "Z2=54#\n", + "q=10#\n", + "m=8#\n", + "P=4000#\n", + "A=1.8#\n", + "K=16#\n", + "N=1000#\n", + "u=0.028#\n", + "lamda=atan(Z1/q)#\n", + "alpha=20*pi/180#\n", + "d=m*q#\n", + "Vt=2*pi*N*d/(2*60*1000)#\n", + "Vs=Vt/cos(lamda)#\n", + "x=cos(alpha)#\n", + "y=tan(lamda)#\n", + "z=(cos(lamda))/sin(lamda)#\n", + "n=(x-(u*y))/(x+(u*z))#\n", + "delT=P*(1-n)/(K*A)#\n", + "print \"n is %0.3f \"%(n)#\n", + "print \"\\ndelT is %0.2f deg \"%(delT)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "n is 0.865 \n", + "\n", + "delT is 18.73 deg \n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 28-6 - Page 729" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Z1=1#\n", + "Z2=30#\n", + "q=10#\n", + "m=6#\n", + "#Let the ultimate strength of gear is sigut\n", + "#Let the allowable strenth of wheel is sigb\n", + "sigut=450#\n", + "sigb=84#\n", + "N=1200#\n", + "n=N/Z2#\n", + "alpha=20*pi/180#\n", + "d=m*q#\n", + "D=Z2*m#\n", + "b=3*d/4#\n", + "V=2*pi*n*D/(2*60*1000)#\n", + "Cv=6/(6+V)#\n", + "y=0.154-(0.912/Z2)#\n", + "Y=pi*y#\n", + "Sb=sigb*b*Cv*m*Y#\n", + "K=0.415#\n", + "Sw=b*D*K#\n", + "print \"Sb is %0.0f N \"%(Sb)#\n", + "print \"\\nSw is %0.0f N \"%(Sw)#\n", + "\n", + "#The difference in the value of Sb is due to rounding-off the values." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Sb is 8286 N \n", + "\n", + "Sw is 3362 N \n" + ] + } + ], + "prompt_number": 13 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch29.ipynb b/Machine_Design_by_U.C._Jindal/Ch29.ipynb new file mode 100644 index 00000000..292239dc --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch29.ipynb @@ -0,0 +1,258 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:b67efe642bf9e9729105e3923ad7beb49ea7d0158d7e2b0caa861c9f34280019" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:29 Gearbox" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 29-1 - Page 749" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "Ts1=16#\n", + "Ts2=18#\n", + "Ts3=20#\n", + "Ts4=25#\n", + "Tr1=64#\n", + "Tr2=63#\n", + "Tr3=70#\n", + "Tr4=50#\n", + "#Let Nr1/Nr2=G1\n", + "G1=1+(Ts1/Tr1)#\n", + "#Let Nr1/Ni=G2\n", + "G2=(Ts2/(Tr2*(1-(1/G1)+(Ts2/Tr2))))#\n", + "#Let Ni/No=G3 (third gear)\n", + "G3=(1+(Ts3/Tr3))/((Ts3/Tr3)+G2)#\n", + "\n", + "#Let Ni/Nr1=G4\n", + "#The ratio calculations are done as above\n", + "G4=1.2857/0.2857#\n", + "#Let Ni/No =G5(second gear)\n", + "G5=-20/70#\n", + "#Let Ni/No=G6(first gear)\n", + "G6=1.2857/0.2857#\n", + "#Let Ni/No=G7(reverse gear)\n", + "G7=-1.7143/0.2857#\n", + "print \"ratio for third gear is %0.3f \"%(G3)#\n", + "print \"\\nratio for second gear is %0.4f \"%(G5)#\n", + "print \"\\nratio for first gear is %0.1f \"%(G6)#\n", + "print \"\\nratio for reverse gear is %0.3f \"%(G7)#\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "ratio for third gear is 1.471 \n", + "\n", + "ratio for second gear is -0.2857 \n", + "\n", + "ratio for first gear is 4.5 \n", + "\n", + "ratio for reverse gear is -6.000 \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 29-2 - Page 751" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Let reverse speed gear be RSG\n", + "RSG=5.5#\n", + "#Let T5/T6 = Z1\n", + "T1=2#\n", + "#Let T3/T7 = Z2\n", + "Z2=2.75#\n", + "T7=18#\n", + "T3=Z2*T7#\n", + "T3=50#\n", + "#Let T3/T1 =Z3\n", + "Z3=2.5#\n", + "T1=T3/Z3#\n", + "#Let T4/T2 = Z4\n", + "Z4=2.25/2#\n", + "T2=(T1+T3)/(Z4+1)#\n", + "T4=T1+T3-T2#\n", + "#Let T5/T6=Z5\n", + "Z5=2#\n", + "T6=(T1+T3)/3#\n", + "T5=(T1+T3)-T6#\n", + "T7=18#\n", + "#let first gear ratio is G1\n", + "G1=50*47/(20*23)#\n", + "\n", + "#Let 2nd gear ratio is G2\n", + "G2=37*47/(33*23)#\n", + "#Let 3rd gear ratio is G3\n", + "G3=1#\n", + "#Let reverse gear ratio is R\n", + "R=50*47/(18*23)#\n", + "print \"T1 is %0.0f \"%(T1)#\n", + "print \"\\nT2 is %0.0f \"%(T2)#\n", + "print \"\\nT3 is %0.0f \"%(T3)#\n", + "print \"\\nT4 is %0.0f \"%(T4)#\n", + "print \"\\nT5 is %0.0f \"%(T5)#\n", + "print \"\\nT6 is %0.0f \"%(T6)#\n", + "print \"\\nT7 is %0.0f \"%(T7)#\n", + "print \"\\nG1 is %0.3f \"%(G1)#\n", + "print \"\\nG2 is %0.3f \"%(G2)#\n", + "print \"\\nG3 is %0.1f \"%(G3)#\n", + "print \"\\nR is %0.3f \"%(R)#\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "T1 is 20 \n", + "\n", + "T2 is 33 \n", + "\n", + "T3 is 50 \n", + "\n", + "T4 is 37 \n", + "\n", + "T5 is 47 \n", + "\n", + "T6 is 23 \n", + "\n", + "T7 is 18 \n", + "\n", + "G1 is 5.109 \n", + "\n", + "G2 is 2.291 \n", + "\n", + "G3 is 1.0 \n", + "\n", + "R is 5.676 \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 29-3 - Page 752" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Let the constant gear ratio be G\n", + "G=2#\n", + "x=5.5**(1/3)#\n", + "G1=1#\n", + "G2=x#\n", + "G3=x*x#\n", + "G4=x**3#\n", + "T7=18#\n", + "T8=T7*(x**3)/2#\n", + "T8=51#\n", + "T5=69/2.558#\n", + "T6=69-27#\n", + "T4=69/1.8825#\n", + "T3=69-T4#\n", + "T1=23#\n", + "T2=46#\n", + "T9=18#\n", + "G1=T2*T8/(T1*T7)#\n", + "G2=T2*T6/(T1*T5)#\n", + "G3=1#\n", + "G4=-T2*T8/(T1*T9)#\n", + "print \"T1 is %0.0f \"%(T1)#\n", + "print \"\\nT2 is %0.0f \"%(T2)#\n", + "print \"\\nT3 is %0.0f \"%(T3)#\n", + "print \"\\nT4 is %0.0f \"%(T4)#\n", + "print \"\\nT5 is %0.0f \"%(T5)#\n", + "print \"\\nT6 is %0.0f \"%(T6)#\n", + "print \"\\nT7 is %0.0f \"%(T7)#\n", + "print \"\\nT8 is %0.0f \"%(T8)#\n", + "print \"\\nT9 is %0.0f \"%(T9)#\n", + "print \"\\nG1 is %0.3f \"%(G1)#\n", + "print \"\\nG2 is %0.3f \"%(G2)#\n", + "print \"\\nG3 is %0.3f \"%(G3)#\n", + "print \"\\nG4 is %0.3f \"%(G4)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "T1 is 23 \n", + "\n", + "T2 is 46 \n", + "\n", + "T3 is 32 \n", + "\n", + "T4 is 37 \n", + "\n", + "T5 is 27 \n", + "\n", + "T6 is 42 \n", + "\n", + "T7 is 18 \n", + "\n", + "T8 is 51 \n", + "\n", + "T9 is 18 \n", + "\n", + "G1 is 5.667 \n", + "\n", + "G2 is 3.114 \n", + "\n", + "G3 is 1.000 \n", + "\n", + "G4 is -5.667 \n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch3.ipynb b/Machine_Design_by_U.C._Jindal/Ch3.ipynb new file mode 100644 index 00000000..f1b4e5f7 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch3.ipynb @@ -0,0 +1,1389 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a39ae8d14e298f4804dd9c8b80dda573e498001f1a1c345f3f119019ff43ebf8" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch : 3 Mechanics of solids" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-1 - Page 72" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, sqrt\n", + "d=10 \n", + "l=1500 \n", + "m=12 \n", + "h=50 \n", + "E=210*10**3 \n", + "sigut=450 \n", + "A=pi*d**2/4 \n", + "W=m*9.81 \n", + "sigi=W/A*(1+sqrt(1+(2*E*A*h)/(W*l))) \n", + "deli=sigi*l/E \n", + "siggradual=W/A \n", + "sigsudden=2*siggradual \n", + "print \" sigi is %0.2f N/mm**2 \"%(sigi) \n", + "print \"\\n deli is %0.2f mm \"%(deli) \n", + "print \"\\n siggradual is %0.2f N/mm**2 \"%(siggradual) \n", + "\n", + "# The difference in the answer of sigi and siggradual is due to round-off errors." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " sigi is 146.37 N/mm**2 \n", + "\n", + " deli is 1.05 mm \n", + "\n", + " siggradual is 1.50 N/mm**2 \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-2 - Page 73" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "d=5 \n", + "A=pi*d**2/4 \n", + "l=100*10**3 \n", + "W=600 \n", + "E=210*10**3 \n", + "w=0.0784*10**-3 \n", + "del1=W*l/(A*E) \n", + "del2=w*l**2/(2*E) \n", + "Del=del1+del2 \n", + "print \"del is %0.2f mm \"%(Del) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "del is 16.42 mm \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-3 - Page 73" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "m=25 \n", + "v=3 \n", + "E=210*10**3 \n", + "KE=0.5*m*v**2 \n", + "d=30 \n", + "L=2000 \n", + "A=pi*d**2/4 \n", + "U=A*L/(2*E) \n", + "Del=4*10**-5*A \n", + "W=A*Del \n", + "sigi=sqrt(KE*10**3/(W+U)) \n", + "print \"del is %0.2f N/mm**2 \"%(sigi) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "del is 69.41 N/mm**2 \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-4 - Page 74" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=40*10**3 \n", + "A=60*18 \n", + "sig=P/A \n", + "r1=12 \n", + "b1=60 \n", + "SCF1=1.7 \n", + "sigmax1=sig*SCF1 \n", + "r2=24 \n", + "b2=60 \n", + "SCF2=1.5 \n", + "sigmax2=sig*SCF2 \n", + "print \"sigmax1 is %0.2f N/mm**2 \"%(sigmax1) \n", + "print \"\\nsigmax2 is %0.2f N/mm**2 \"%(sigmax2) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "sigmax1 is 62.90 N/mm**2 \n", + "\n", + "sigmax2 is 55.50 N/mm**2 \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-5 - Page 75" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "p=2.4 \n", + "#Let axial movement of nut be La\n", + "La=p*45/360 \n", + "d=20 \n", + "D=30 \n", + "L=500 \n", + "d1=18 \n", + "As=pi*d1**2/4 \n", + "Ac=pi*(D**2-d**2)/4 \n", + "sigt=120/(3.543) \n", + "sigb=1.543*sigt \n", + "print \"sigt is %0.2f N/mm**2 \"%(sigt) \n", + "print \"\\nsigb is %0.2f N/mm**2 \"%(sigb) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "sigt is 33.87 N/mm**2 \n", + "\n", + "sigb is 52.26 N/mm**2 \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-6 - Page 76" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "delT=100 \n", + "ab=18*10**-6 \n", + "aa=23*10**-6 \n", + "delta=(360*ab*delT)+(450*aa*delT) \n", + "lc=delta-0.6 \n", + "Ea=70*10**3 \n", + "Eb=105*10**3 \n", + "Aa=1600 \n", + "Ab=1300 \n", + "P=lc/((360/(Ab*Eb))+(450/(Aa*Ea))) \n", + "P=P*10**-3 \n", + "#Let the change in length be delL\n", + "delL=(aa*450*delT)-(P*10**3*450/(Aa*Ea)) \n", + "print \" P is %0.2f kN \"%(P) \n", + "print \"\\n delL is %0.2f mm \"%(delL) \n", + " \n", + " # The difference in the answer of delL is due to round-off errors." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " P is 162.73 kN \n", + "\n", + " delL is 0.38 mm \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-7 - Page 77" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "a=23*10**-6 \n", + "E=70*10**3 \n", + "l=750 \n", + "sig=35 \n", + "delT=((sig*l/E)+0.8)/(l*a) \n", + "print \"delT is %0.2f degC \"%(delT) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "delT is 68.12 degC \n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-8 - Page 78" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "OA=60 \n", + "AB=30 \n", + "OC=-20 \n", + "CD=-30 \n", + "theta=30 \n", + "angBEK=2*theta \n", + "OM=14 \n", + "KM=49.5 \n", + "p1=70 \n", + "p2=-30 \n", + "angBEH=-37 \n", + "angBEI=143 \n", + "theta1=angBEH/2 \n", + "theta2=angBEI/2 \n", + "Tmax=50 \n", + "angBEL=53 \n", + "angBEN=233 \n", + "theta3=angBEL/2 \n", + "theta4=angBEN/2 \n", + "print \" Stress on plane AB is %0.2f MPa \"%(OM) \n", + "print \"\\n Stress on plane AB is %0.2f MPa \"%(KM) \n", + "print \"\\n Principal stress p1 is %0.2f MPa \"%(p1) \n", + "print \"\\n Principal stress p2 is %0.2f MPa \"%(p2) \n", + "print \"\\n Principal angle theta1 is %0.2f deg \"%(theta1) \n", + "print \"\\n Principal angle theta2 is %0.2f deg \"%(theta2) \n", + "print \"\\n Maximum shear stress is %0.2f MPa \"%(Tmax) \n", + "print \"\\n Direction of plane theta3 is %0.2f deg \"%(theta3) \n", + "print \"\\n Direction of plane theta4 is %0.2f deg \"%(theta4) \n", + "\n", + "#The answers in the book are written in form of degrees and minutes." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Stress on plane AB is 14.00 MPa \n", + "\n", + " Stress on plane AB is 49.50 MPa \n", + "\n", + " Principal stress p1 is 70.00 MPa \n", + "\n", + " Principal stress p2 is -30.00 MPa \n", + "\n", + " Principal angle theta1 is -18.50 deg \n", + "\n", + " Principal angle theta2 is 71.50 deg \n", + "\n", + " Maximum shear stress is 50.00 MPa \n", + "\n", + " Direction of plane theta3 is 26.50 deg \n", + "\n", + " Direction of plane theta4 is 116.50 deg \n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-9 - Page 78" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "E=200*10**3 \n", + "v=0.29 \n", + "E1=720*10**-6 \n", + "E2=560*10**-6 \n", + "p1=121.76 \n", + "p2=-76.69 \n", + "print \"p1 is %0.2f MN/mm**2 \"%(p1) \n", + "print \"\\n p2 is %0.2f MN/mm**2 \"%(p2) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "p1 is 121.76 MN/mm**2 \n", + "\n", + " p2 is -76.69 MN/mm**2 \n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-10 - Page 79" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "G=38*10**3 \n", + "d=10 \n", + "P=5*10**3 \n", + "A=pi*d**2/4 \n", + "sig=P/A \n", + "deld=0.0002 \n", + "#Let the lateral strain be E1\n", + "E1=deld/d \n", + "v=2*deld*G/(sig-(2*deld*G)) \n", + "E=2*G*(1+v)*10**-3 \n", + "print \"v is %0.4f \"%(v) \n", + "print \"\\nE is %0.3f kN/mm**2 \"%(E) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "v is 0.3136 \n", + "\n", + "E is 99.837 kN/mm**2 \n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-11 - Page 79" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "D=1500 \n", + "p=1.2 \n", + "sigt=100 \n", + "sigc=p*D/2 \n", + "siga=p*D/4 \n", + "P=sigc*2*10**3 \n", + "n=0.75 \n", + "t=sigc/(n*sigt) \n", + "print \"t is %0.1f mm \"%(t) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "t is 12.0 mm \n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-12 - Page 80" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "D=50 \n", + "t=1.25 \n", + "d=0.5 \n", + "n=1/d \n", + "p=1.5 \n", + "siga=p*D/(4*t) \n", + "sigc=20.27 \n", + "sigw=sigc/0.31416 \n", + "print \"sigw is %0.2f N/mm**2 \"%(sigw) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "sigw is 64.52 N/mm**2 \n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-13 - Page 81" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "R1=50 \n", + "p=75 \n", + "pmax=125 \n", + "R2=sqrt((pmax+p)*R1**2/(pmax-p)) \n", + "t=R2-R1 \n", + "print \"t is %0.1f mm \"%(t) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "t is 50.0 mm \n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-14 - Page 82" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "R1=40 \n", + "R2=60 \n", + "B=50 \n", + "E=210*10**3 \n", + "e=41*10**-6 \n", + "sig=2*R1**2/(R2**2-R1**2) \n", + "p=E*e/sig \n", + "Fr=p*2*pi*R1*B \n", + "u=0.2 \n", + "Fa=u*Fr*10**-3 \n", + "print \"Fa is %0.2f kN \"%(Fa) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fa is 13.52 kN \n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-15 - Page 83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "a1=10*1.5 \n", + "x1=15-0.75 \n", + "a2=1.5*(15-1.5) \n", + "x2=(15-1.5)/2 \n", + "y1=((a1*x1)+(a2*x2))/(a1+a2) \n", + "y2=a1-y1 \n", + "Ixx=(10*1.5**3)/12+(10*1.5*(5.06-1.5/2)**2)+(1.5*13.5**3/12)+(1.5*13.5*(9.94-6.75)**2) \n", + "Z1=Ixx/y1 \n", + "Z2=Ixx/y2 \n", + "L=3 \n", + "sigc=50 \n", + "W=sigc*Z1/L*10**-3 \n", + "print \"W is %0.3f kN \"%(W) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "W is 1.333 kN \n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-16 - Page 83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "D=22 \n", + "d=20 \n", + "r=1 \n", + "K=2.2 \n", + "sigmax=130 \n", + "sigmax=sigmax/K \n", + "Z=pi*d**3/32 \n", + "M=sigmax*Z*10**-3 \n", + "print \"M is %0.3f Nm \"%(M) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "M is 46.410 Nm \n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-17 - Page 84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "A=(12*2)+(12*2)+(30-4) \n", + "B=sqrt(A/2) \n", + "D=2*B \n", + "B1=12 \n", + "D1=30 \n", + "d=26 \n", + "b=1 \n", + "Z1=((B1*D1**3)-((B1-b)*d**3))/(B1*D1/2) \n", + "Zr=B*D**2/6 \n", + "#Let the ratio of both the sections be x\n", + "x=Z1/Zr \n", + "M=30*10**6 \n", + "sigmax=M/(Z1*10**3) \n", + "print \"Z1/Zr is %0.2f \"%(x) \n", + "print \"\\nsigmax is %0.2f N/mm**2 \"%(sigmax) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Z1/Zr is 4.84 \n", + "\n", + "sigmax is 41.33 N/mm**2 \n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-19 - Page 85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "x1=((13*3*1.5)+(2*15*8))/(39+30) \n", + "x2=13-x1 \n", + "A=30+39 \n", + "E=2*10**7 \n", + "Iyy=995.66 \n", + "e=54.32 \n", + "x=x2-3 \n", + "sigb=e*x/Iyy \n", + "sigd=1/69 \n", + "sigr=sigd+sigb \n", + "#Let the strain be E1\n", + "E1=800*10**-6 \n", + "P=E1*E/sigr \n", + "P=P*10**-3 \n", + "print \"P is %0.2f kN \"%(P) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "P is 49.38 kN \n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-20 - Page 86" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "H=20 \n", + "D=5 \n", + "d=3 \n", + "rho=21 \n", + "sigd=rho*H \n", + "p=2 \n", + "A=D*H \n", + "P=p*A \n", + "M=P*H/2 \n", + "Z=pi*(D**4-d**4)/(32*D) \n", + "sigb=M/Z \n", + "sigmax=420+sigb \n", + "sigmin=420-sigb \n", + "print \"sigmax is %0.2f kN/m**2 \"%(sigmax) \n", + "print \"\\nsigmin is %0.2f kN/m**2 \"%(sigmin) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "sigmax is 607.24 kN/m**2 \n", + "\n", + "sigmin is 232.76 kN/m**2 \n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-21 - Page 87" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "D=30 \n", + "R=15 \n", + "T=0.56*10**6 \n", + "G=82*10**3 \n", + "J=pi*R**4/2 \n", + "T1=T*R/J \n", + "l=1000 \n", + "theta=T*l/(G*J)*180/pi \n", + "r=10 \n", + "Tr=T1*r/R \n", + "print \" T1 is %0.2f N/mm**2 \"%(T1) \n", + "print \"\\n theta is %0.2f deg \"%(theta) \n", + "print \"\\n Tr is %0.2f N/mm**2 \"%(Tr) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " T1 is 105.63 N/mm**2 \n", + "\n", + " theta is 4.92 deg \n", + "\n", + " Tr is 70.42 N/mm**2 \n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-22 - Page 87" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "T=8*10**3 \n", + "d=80 \n", + "D=110 \n", + "l=2000 \n", + "Gst=80*10**3 \n", + "Gcop=Gst/2 \n", + "Js=pi*d**4/32 \n", + "Jc=pi*(D**4-d**4)/32 \n", + "#Ts=0.777*Tc\n", + "Tc=T/1.777*10**3 \n", + "Ts=0.777*Tc \n", + "Ts1=Ts/Js*d/2 \n", + "Tc1=Tc/Jc*D/2 \n", + "#Let tl be Angular twist per unit length\n", + "tl=Ts*10**3/(Js*Gst)*180/pi \n", + "# Let the maximum stress developed when the Torque is acting in the centre of the shaft be Ts2 & Tc2 resp. for steel and copper\n", + "Ts2=Ts1/2 \n", + "Tc2=Tc1/2 \n", + "print \" Ts1 is %0.3f N/mm**2 \"%(Ts1) \n", + "print \"\\n Tc1 is %0.1f N/mm**2 \"%(Tc1) \n", + "print \"\\n theta/length is %0.3f deg/m \"%(tl) \n", + "print \"\\n Ts2 is %0.3f N/mm**2 \"%(Ts2) \n", + "print \"\\n Tc2 is %0.2f N/mm**2 \"%(Tc2) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Ts1 is 34.796 N/mm**2 \n", + "\n", + " Tc1 is 23.9 N/mm**2 \n", + "\n", + " theta/length is 0.623 deg/m \n", + "\n", + " Ts2 is 17.398 N/mm**2 \n", + "\n", + " Tc2 is 11.96 N/mm**2 \n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-23 - Page 88" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "D=100 \n", + "d=75 \n", + "r=6 \n", + "K=1.45 \n", + "P=20*746 \n", + "N=400 \n", + "w=2*pi*N/60 \n", + "T=P/w \n", + "Ts=16*T*10**3/(pi*d**3) \n", + "Tmax=K*Ts \n", + "print \"Tmax is %0.3f MPa \"%(Tmax) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Tmax is 6.235 MPa \n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-24 - Page 88" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "G=84*10**3 \n", + "T=28*10**3 \n", + "l=1000 \n", + "theta=pi/180 \n", + "J=T*l/(G*theta) \n", + "d=(J*32/pi)**(1/4) \n", + "print \"d is %0.1f mm \"%(d) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "d is 21.0 mm \n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-25 - Page 89" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=2*10**6 \n", + "N=200 \n", + "w=2*pi*N/60 \n", + "Tm=P/w \n", + "W=5*10**3*9.81 \n", + "l=1800 \n", + "Mmax=W*l/4 \n", + "Tmax=1.8*Tm*10**3 \n", + "Me=(Mmax+sqrt(Mmax**2+Tmax**2))/2 \n", + "Te=sqrt(Mmax**2+Tmax**2) \n", + "sig=60 \n", + "Ts=40 \n", + "d1=(32*Me/(pi*sig))**(1/3) \n", + "d2=(16*Te/(pi*Ts))**(1/3) \n", + "print \"d is %0.1f mm \"%(d2) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "d is 280.5 mm \n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-26 - Page 90" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Q=4*10**3 \n", + "P=8*10**3 \n", + "sig=P \n", + "T=Q \n", + "p1=(sig/2+sqrt((sig/2)**2+T**2)) \n", + "p2=(sig/2-sqrt((sig/2)**2+T**2)) \n", + "sigyp=285 \n", + "FOS=3 \n", + "siga=sigyp/3 \n", + "A1=p1/siga \n", + "d1=sqrt(4*A1/pi) \n", + "A2=(p1-p2)*2/(siga*2) \n", + "d2=sqrt(4*A2/pi) \n", + "v=0.3 \n", + "A3=sqrt(p1**2+p2**2-(2*v*p1*p2))/siga \n", + "d3=sqrt(4*A3/pi) \n", + "print \" d1 is %0.2f mm \"%(d1) \n", + "print \"\\n d2 is %0.1f mm \"%(d2) \n", + "print \"\\n d3 is %0.2f mm \"%(d3) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " d1 is 11.38 mm \n", + "\n", + " d2 is 12.3 mm \n", + "\n", + " d3 is 11.74 mm \n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-27 - Page 91" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "sigx=-105 \n", + "Txy=105 \n", + "sigy=270 \n", + "p1=(sigx/2+sqrt((sigx/2)**2+Txy**2)) \n", + "p2=(sigx/2-sqrt((sigx/2)**2+Txy**2)) \n", + "p3=0 \n", + "Tmax=(p1-p2)/2 \n", + "siga=sigy/2 \n", + "if (Tmax<=siga) :\n", + " print \"The component is safe\"\n", + "print \"\\nTmax is %0.1f MPa \"%(Tmax) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The component is safe\n", + "\n", + "Tmax is 117.4 MPa \n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-28 - Page 91" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "rho=0.0078*9.81*10**-6 \n", + "sigc=150 \n", + "g=9.81 \n", + "V=sqrt(sigc*g/rho)*10**-3 \n", + "R=1 \n", + "w=V/R \n", + "N=w*60/(2*pi) \n", + "print \"N is %0.3f rpm \"%(N) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "N is 1324.249 rpm \n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-29 - Page 92" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "R1=50 \n", + "R2=200 \n", + "N=6*10**3 \n", + "w=2*pi*N/60 \n", + "v=0.28 \n", + "rho=7800*10**-9 \n", + "g=9810 \n", + "k1=(3+v)/8 \n", + "k2=(1+(3*v))/8 \n", + "W=rho*9.81 \n", + "x=k1*w**2*W*(R1**2+R2**2)/g \n", + "y=k1*w**2*W*(R1*R2)**2/g \n", + "y1=k1*w**2*W/g \n", + "z=k2*w**2*W/g \n", + "r=sqrt(R1*R2) \n", + "sigrmax=x-(y/r**2)-(r**2*y1) \n", + "r=range(50,201)\n", + "n=len(r) \n", + "sigr = range(0,n)\n", + "for i in range(0,n):\n", + " sigr[i]=x-(y/r[i]**2)-(r[i]**2*y1)\n", + "\n", + "sigc = range(0,n)\n", + "for j in range(0,n):\n", + " sigc[j]=y/r[j]**2-(r[j]**2*z)\n", + "\n", + "%matplotlib inline\n", + "from matplotlib.pyplot import plot, xlabel, ylabel, show, grid\n", + "plot(r,sigr) \n", + "plot (r,sigc) \n", + "xlabel('r in mm') \n", + "ylabel('stress in N/mm**2') \n", + "grid()\n", + "show()\n", + "print \"sigrmax is %0.1f MPa \"%(sigrmax) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "metadata": {}, + "output_type": "display_data", + "png": 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qDmecAf/5DzRs6Heq7IvE9TAibTLuGhyEvk7yMUuWdu50k37++1+/k0SnAjEF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\nMMYYky3/B0TYOSGQpC4+AAAAAElFTkSuQmCC\n", + "text": [ + "<matplotlib.figure.Figure at 0x7ff461ba6f50>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "sigrmax is 28.4 MPa \n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-30 - Page 93" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import exp\n", + "r=500 \n", + "to=15 \n", + "N=3500 \n", + "w=2*pi*N/60 \n", + "sig=80 \n", + "w1=0.07644*10**-3 \n", + "g=9810 \n", + "a=w1*w**2*r**2/(2*sig*g) \n", + "t=to*exp(-a) \n", + "print \"t is %0.3f mm \"%(t) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "t is 2.923 mm \n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-31 - Page 93" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import log\n", + "M=60*10**3 \n", + "y1=((5*1*2.5)+(6*1*5.5))/(5+6) \n", + "y2=6-y1 \n", + "R=12 \n", + "R1=R-y2 \n", + "R1=10.136\n", + "R2=11.136 \n", + "R3=R1+6 \n", + "B=6 \n", + "b=1 \n", + "A=(B*b)+((B-1)*b) \n", + "#Let x= h**2/R**2\n", + "x=R/A*((B*log(R2/R1))+(b*log(R3/R2)))-1 \n", + "x=1/x \n", + "#Let Maximum compressive stress at B be sigB\n", + "sigB=M/(A*R)*(1+(x*y1/(R+y1)))*10**-2 \n", + "#Let Maximum tensile stress at A be sigA\n", + "sigA=M/(A*R)*((y2*x/(R-y2))-1)*10**-2 \n", + "print \"sigB is %0.1f MPa \"%(sigB) \n", + "print \"\\nsigA is %0.0f MPa \"%(sigA) \n", + " \n", + "#The answer to R**2/h**2 is calculated incorrectly in the book." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "sigB is 61.5 MPa \n", + "\n", + "sigA is 36 MPa \n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-32 - Page 94" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "R1=24 \n", + "R2=30 \n", + "R3=50 \n", + "R4=54 \n", + "F=200 \n", + "y1=((16*4*2)+(2*20*14*4)+(24*6*27))/((16*4)+(2*20*4)+(24*6)) \n", + "y2=30-y1 \n", + "R=24+y2 \n", + "A=(24*6)+(2*4*20)+(4*16) \n", + "#Let x= h**2/R**2\n", + "x=R/A*((24*log(R2/R1))+(2*4*log(R3/R2))+(16*log(R4/R3)))-1 \n", + "x=1/x \n", + "M=F*(60+R) \n", + "sigd=F/A \n", + "#Let bending stress at a be sigA\n", + "sigA=M/(A*R)*((y2*x/(R-y2))-1) \n", + "#Let bending stress at b be sigB\n", + "sigB=M/(A*R)*(1+(x*y1/(R+y1))) \n", + "#Let resultant at a be Ra\n", + "Ra=(sigA+sigd)*10 \n", + "#Let resultant at b be Rb\n", + "Rb=(sigB-sigd)*10 \n", + "print \"Ra is %0.2f N/mm**2 \"%(Ra) \n", + "print \"\\nRb is %0.2f N/mm**2 \"%(Rb) \n", + "#The difference in the answers are due to rounding-off of values." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ra is 96.70 N/mm**2 \n", + "\n", + "Rb is 70.14 N/mm**2 \n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 3-33 - Page 95" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "F=50 \n", + "B1=4 \n", + "B2=8 \n", + "D=12 \n", + "y1=D/3*(B1+(2*B2))/(B1+B2) \n", + "y2=12-y1 \n", + "R=6+y2 \n", + "A=(B1+B2)/2*D \n", + "#Let x= h**2/R**2\n", + "a=(B1+((B2-B1)*(y1+R)/D))*log((R+y1)/(R-y2))\n", + "x=R/(A)*(a -(B2-B1)) \n", + "x=x-1 \n", + "x=1/x \n", + "KG=y2+8 \n", + "M=F*KG \n", + "sigd=F/A \n", + "#Let bending stress at a be sigA\n", + "sigA=M/(A*R)*(1+(x*y1/(R+y1))) \n", + "#Let bending stress at b be sigB\n", + "sigB=M/(A*R)*((y2*x/(R-y2))-1) \n", + "sigA=(sigA-sigd)*10 \n", + "sigB=(sigB+sigd)*10 \n", + "print \"sigA is %0.2f MPa \"%(sigA) \n", + "print \"\\nsigB is %0.2f MPa \"%(sigB) \n", + "#The difference in the answers are due to rounding-off of values." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "sigA is 31.59 MPa \n", + "\n", + "sigB is 71.64 MPa \n" + ] + } + ], + "prompt_number": 32 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch30.ipynb b/Machine_Design_by_U.C._Jindal/Ch30.ipynb new file mode 100644 index 00000000..335560f0 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch30.ipynb @@ -0,0 +1,210 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:55c40faafb847932f0fdcda855b3af16f1a2e4ef45941baaf0d7ee692a22c20c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:30 Chain drive" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 30-1 - Page 778" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "n1=17#\n", + "n2=51#\n", + "C=300#\n", + "p=9.52#\n", + "Ln=(2*C/p)+((n1+n2)/2)+((((n2-n1)/(2*pi))**2)*(p/C))#\n", + "x=(Ln-((n2+n1)/(2)))**2#\n", + "y=8*(((n2-n1)/(2*pi))**2)#\n", + "z=Ln-((n1+n2)/2)#\n", + "C=(p/4)*(z+(sqrt(x-y)))\n", + "\n", + "\n", + " # printing data in scilab o/p window\n", + "print \"C is %0.2f mm \"%(C)#\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "C is 300.00 mm \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 30-2 - Page 778" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import tan\n", + "G=4#\n", + "n1=17#\n", + "n2=n1*G#\n", + "N1=2300#\n", + "Kc=1.2# #from table 30-2\n", + "p=12.7# #fom table 30-1\n", + "D1=p*n1#\n", + "D2=p*n2#\n", + "phi=2*10.6#\n", + "x=tan(phi/2)# #phi/2 = 10.6deg, from table 30-3\n", + "Da1=(p/x)+(0.6*p)#\n", + "Da2=(p/x*4)+(0.6*p)#\n", + "Cmin=Kc*((Da1+Da2)/2)#\n", + "Ln1=(2*Cmin/p)+((n1+n2)/2)+((((n2-n1)/(2*pi))**2)*(p/Cmin))#\n", + "Ln1=80#\n", + "print \"Ln is %0.0f \"%(Ln1)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ln is 80 \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 30-3 - Page 779" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "N1=1000#\n", + "N2=500#\n", + "P=2.03*10**3# #from table 30-8\n", + "K1=1.26#\n", + "Ks=1#\n", + "#let Pc be the power transmitting capacity of the chain\n", + "Pc=P*K1/Ks#\n", + "p=9.52#\n", + "n1=21#\n", + "n2=42#\n", + "V=n1*p*N1/(60*10**3)#\n", + "#Let the chain tension be T\n", + "T=Pc/V#\n", + "#Let the breaking load be BL\n", + "BL=10700#\n", + "FOS=BL/T#\n", + "C=50*p#\n", + "Ln=(2*C/p)+((n1+n2)/2)+((((n2-n1)/(2*pi))**2)*(p/C))#\n", + "L=Ln*p#\n", + "Pc=Pc*10**-3#\n", + "print \" Pc is %0.2f KW \"%(Pc)#\n", + "print \"\\n V is %0.3f m/s \"%(V)#\n", + "print \"\\n T is %0.1f N \"%(T)#\n", + "print \"\\n FOS is %0.2f \"%(FOS)#\n", + "print \"\\n L is %0.2f mm \"%(L)#\n", + "\n", + "#The difference in the value of L and T is due to rounding-off the values." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Pc is 2.56 KW \n", + "\n", + " V is 3.332 m/s \n", + "\n", + " T is 767.6 N \n", + "\n", + " FOS is 13.94 \n", + "\n", + " L is 1254.01 mm \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 30-5 - Page 780" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "G=2#\n", + "P=5000#\n", + "Ks=1.7#\n", + "Pd=P*Ks#\n", + "K2=1.7#\n", + "p=15.88#\n", + "n1=17#\n", + "n2=n1*G#\n", + "D1=n1*p#\n", + "D2=n2*p#\n", + "C=40*p#\n", + "Ln=(2*C/p)+((n1+n2)/2)+((((n2-n1)/(2*pi))**2)*(p/C))#\n", + "L=Ln*p#\n", + "print \"L is %0.2f mm \"%(L)#\n", + "#The difference in the value of L is due to rounding-off the values." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "L is 1678.25 mm \n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch31.ipynb b/Machine_Design_by_U.C._Jindal/Ch31.ipynb new file mode 100644 index 00000000..890e011d --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch31.ipynb @@ -0,0 +1,173 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:6ec60a366eb75f6b61dc10de0b002b97629539c44f0f706f948844442706e9bd" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:31 Seals packing and gaskets" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 31-1 - Page 816" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import sqrt, pi, log\n", + "d=18#\n", + "lg=25+25#\n", + "Eb=210*10**3#\n", + "Ecl=90*10**3#\n", + "A=pi*d**2/4#\n", + "kb=A*Eb/lg#\n", + "x=(5*(lg+(0.5*d))/(lg+(2.5*d)))#\n", + "km=pi*Ecl*d/(2*log(x))#\n", + "C=kb/(kb+km)#\n", + "sigp=600#\n", + "At=192#\n", + "Pi=0.75*sigp*At#\n", + "F=200#\n", + "C=0.322#\n", + "Pb=F*C*10**3#\n", + "FOS=2#\n", + "W=At*sigp#\n", + "N=Pb*FOS/(W-Pi)#\n", + "print \"N is %0.2f \"%(N)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "N is 4.47 \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 31-2 - Page 816" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "d=16#\n", + "D=1.5*d#\n", + "t=20#\n", + "tg=4#\n", + "#Let Gasket diameter in compression zone be d1\n", + "d1=D+(2*t)+tg#\n", + "lg=40#\n", + "E=207*10**3#\n", + "kb=pi*d**2*E/(lg*4)#\n", + "Ecl=90*10**3#\n", + "x=(5*(lg+(0.5*d))/(lg+(2.5*d)))#\n", + "kp=pi*Ecl*d/(2*log(x))#\n", + "Ag=pi*(d1**2-d**2)/4#\n", + "Eg=480#\n", + "kg=Ag*Eg/tg#\n", + "km=kg*kp/(kg+kp)#\n", + "C=kb/(kb+km)#\n", + "At=157#\n", + "sigp=600#\n", + "Pi=0.75*At*sigp/2#\n", + "FOS=2#\n", + "Pf=At*sigp/FOS#\n", + "W=Pf-Pi#\n", + "P=W/C#\n", + "N=5#\n", + "F=P*N#\n", + "p=F*4/(pi*120**2)#\n", + "print \"p is %0.3f N/mm**2 \"%(p)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "p is 6.922 N/mm**2 \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 31-3 - Page 817" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "sigp=600#\n", + "FOS=3#\n", + "siga=sigp/FOS#\n", + "d=16#\n", + "D=1.5*d+60#\n", + "#Let Gasket diameter in compression zone be d1\n", + "d1=(300-160)/2#\n", + "#Let compressive stress in gasket for leak proof joint be sigl\n", + "sigl=12#\n", + "from numpy import mat\n", + "At=mat([[1, 157],[2 ,192], [3, 245]])\n", + "d=mat([[1 ,16],[2 ,18],[3, 20]])\n", + "\n", + "n=3#\n", + "Pi = range(0,n)\n", + "Pc = range(0,n)\n", + "for i in range(0,n):\n", + " Pi[i]=At[i,1]*d[i,1]\n", + " Pc[i]=3*pi*(d1**2-d[i,1]**2)\n", + " if (Pi[i]>=Pc[i]):\n", + " print \"The Design is safe\"\n", + "\n", + "print \"d is %0.0f mm \"%(d[i,1])#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "d is 20 mm \n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch4.ipynb b/Machine_Design_by_U.C._Jindal/Ch4.ipynb new file mode 100644 index 00000000..20051c72 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch4.ipynb @@ -0,0 +1,266 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:aae4d19fea549af28d592247f766b2cbe593905aad9c18e9cc47077fbd53be47" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:4 Manufacturing considerations" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 4-1 - Page 112" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "d=70#\n", + "dmin=50#\n", + "dmax=80#\n", + "D=sqrt(dmin*dmax)#\n", + "D=63#\n", + "i=0.458*(D**(1/3))+(0.001*D)#\n", + "\n", + "#standard tolerance for H8 is ST1\n", + "ST1=25*i#\n", + "ST1=ST1*10**-3#\n", + "#standard tolerance of shaft for grade g7 is ST2\n", + "ST2=16*i#\n", + "ST2=ST2*10**-3#\n", + "es=-(2.5*(D**0.333))#\n", + "es=es*10**-3#\n", + "ei=es-ST2#\n", + "#Lower limit for hole is LLH\n", + "#Upper limit for hole is ULH\n", + "#Upper limit for shaft is ULS\n", + "#Lower limit for shaft is LLS\n", + "LLH=d#\n", + "ULH=LLH+ST1#\n", + "ULS=LLH+es#\n", + "LLS=ULS-ST2#\n", + "#Maximum clearance is Cmax\n", + "#minimum clearance is Cmin\n", + "Cmax=ULH-LLS#\n", + "Cmin=LLH-ULS#\n", + "# printing data in scilab o/p window\n", + "print \" LLH is %0.1f mm \"%(LLH)#\n", + "print \"\\n ULH is %0.3f mm \"%(ULH)#\n", + "print \"\\n ULS is %0.2f mm \"%(ULS)#\n", + "print \"\\n LLS is %0.2f mm \"%(LLS)#\n", + "print \"\\n Cmax is %0.3f mm \"%(Cmax)#\n", + "print \"\\n Cmin is %0.3f mm \"%(Cmin)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " LLH is 70.0 mm \n", + "\n", + " ULH is 70.013 mm \n", + "\n", + " ULS is 69.99 mm \n", + "\n", + " LLS is 69.98 mm \n", + "\n", + " Cmax is 0.031 mm \n", + "\n", + " Cmin is 0.010 mm \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 4-2 - Page 112" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "d=25#\n", + "#Lower limit for hole is LLH\n", + "#Upper limit for hole is ULH\n", + "#Upper limit for shaft is ULS\n", + "#Lower limit for shaft is LLS\n", + "ULH=d+0.021#\n", + "LLH=d+0#\n", + "ULS=d+0.041#\n", + "LLS=d+0.028#\n", + "#Maximum interference is Cmax\n", + "#minimum interference is Cmin\n", + "Cmax=ULS-LLH#\n", + "Cmin=LLS-ULH#\n", + "# printing data in scilab o/p window\n", + "print \"Cmax is %0.3f mm \"%(Cmax)#\n", + "print \"\\nCmin is %0.3f mm \"%(Cmin)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Cmax is 0.041 mm \n", + "\n", + "Cmin is 0.007 mm \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 4-3 - Page 113" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "d=50#\n", + "Es=0.039#\n", + "Ei=0#\n", + "es=-9*10**-3#\n", + "ei=-34*10**-3#\n", + "#Shaft dia is D\n", + "D=d+es#\n", + "#Lower limit for hole is LLH\n", + "#Upper limit for hole is ULH\n", + "#Upper limit for shaft is ULS\n", + "#Lower limit for shaft is LLS\n", + "ULH=d+Es#\n", + "LLH=d+Ei#\n", + "ULS=d+es#\n", + "LLS=d+ei#\n", + "#Maximum interference is Cmax\n", + "#minimum interference is Cmin\n", + "Cmax=ULH-LLS#\n", + "Cmin=LLH-ULS#\n", + " # printing data in scilab o/p window\n", + "print \" ULH is %0.3f mm \"%(ULH)#\n", + "print \"\\n LLH is %0.3f mm \"%(LLH)#\n", + "print \"\\n ULS is %0.3f mm \"%(ULS)#\n", + "print \"\\n LLS is %0.3f mm \"%(LLS)#\n", + "print \"\\n Cmax is %0.3f mm \"%(Cmax)#\n", + "print \"\\n Cmin is %0.3f mm \"%(Cmin)#\n", + "print ' Therefore, H8g7 is easy running fit'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " ULH is 50.039 mm \n", + "\n", + " LLH is 50.000 mm \n", + "\n", + " ULS is 49.991 mm \n", + "\n", + " LLS is 49.966 mm \n", + "\n", + " Cmax is 0.073 mm \n", + "\n", + " Cmin is 0.009 mm \n", + " Therefore, H8g7 is easy running fit\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 4-3 - Page 114" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "d=30#\n", + "Es=0.025#\n", + "Ei=0#\n", + "es=11*10**-3#\n", + "ei=-5*10**-3#\n", + "#Shaft dia is D\n", + "D=d+es#\n", + "#Lower limit for hole is LLH\n", + "#Upper limit for hole is ULH\n", + "#Upper limit for shaft is ULS\n", + "#Lower limit for shaft is LLS\n", + "ULH=d+Es#\n", + "LLH=d+Ei#\n", + "ULS=d+es#\n", + "LLS=d+ei#\n", + "#Maximum interference is Cmax\n", + "#minimum interference is Cmin\n", + "Cmax=ULH-LLS#\n", + "Cmin=ULS-LLH##\n", + "\n", + " # printing data in scilab o/p window\n", + "print \" ULH is %0.3f mm \"%(ULH)#\n", + "print \"\\n LLH is %0.3f mm \"%(LLH)#\n", + "print \"\\n ULS is %0.3f mm \"%(ULS)#\n", + "print \"\\n LLS is %0.3f mm \"%(LLS)#\n", + "print \"\\n Cmax is %0.3f mm \"%(Cmax)#\n", + "print \"\\n Cmin is %0.3f mm \"%(Cmin)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " ULH is 30.025 mm \n", + "\n", + " LLH is 30.000 mm \n", + "\n", + " ULS is 30.011 mm \n", + "\n", + " LLS is 29.995 mm \n", + "\n", + " Cmax is 0.030 mm \n", + "\n", + " Cmin is 0.011 mm \n" + ] + } + ], + "prompt_number": 5 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch5.ipynb b/Machine_Design_by_U.C._Jindal/Ch5.ipynb new file mode 100644 index 00000000..ea122295 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch5.ipynb @@ -0,0 +1,253 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:56bbd90f35decd7f62fabe45fb918c0e397bbdcedfeeda130ab611b764281c1a" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:5 Introduction to pressure vessels" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 5-1 - Page 138" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "p=2#\n", + "Rm=220#\n", + "#tensile hoop or circumferential stress= sigt\n", + "sigr=-2#\n", + "#sigt=(p*Rm)/t#\n", + "Sa=230/2#\n", + "#t1=thickness according to maximum principal stress theory\n", + "#t2=thickness according to maximum shear stress theory\n", + "t1=(p*Rm)/Sa#\n", + "t2=(p*Rm)/(Sa+sigr)#\n", + "print \"t1 is %0.2f mm \"%(t1)#\n", + "print \"\\nt2 is %0.3f mm \"%(t2)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "t1 is 3.00 mm \n", + "\n", + "t2 is 3.000 mm \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 5-2 - Page 139" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from __future__ import division\n", + "#Elastic limit=sige\n", + "sige=310#\n", + "#inside diameter=di\n", + "di=300#\n", + "p=1.8#\n", + "FOS=2#\n", + "#design stress=sigd#\n", + "sigd=sige/2#\n", + "c=0.162#\n", + "d=380#\n", + "#cover plate thickness=t#\n", + "t=d*sqrt(c*p/sigd)#\n", + "t=17#\n", + "M=di*p*t/4#\n", + "\n", + "z=(1/6)*1*t**2#\n", + "#bending stress=sigb#\n", + "sigb=M/z#\n", + "print \"t is %0.1fmm \"%(t)#\n", + "print \"\\nM is %0.1fmm \"%(M)#\n", + "print \"\\nsigb is %0.1fmm \"%(sigb)#\n", + "if (sigb<=sigd):\n", + " print 'sigb is below allowable sigd.'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "t is 17.0mm \n", + "\n", + "M is 2295.0mm \n", + "\n", + "sigb is 47.6mm \n", + "sigb is below allowable sigd.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 5-3 - Page 140" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "sige=220#\n", + "v=0.29#\n", + "Ri=175#\n", + "FOS=3#\n", + "Sa=sige/3#\n", + "p=10#\n", + "#t1=thickness according to maximum principal stress theory\n", + "#t2=thickness according to maximum shear stress theory\n", + "x=Sa+(p*(1-(2*v)))#\n", + "y=Sa-(p*(1+v))#\n", + "t1=(sqrt(x/y)-1)*Ri#\n", + "t1=24#\n", + "#t1=((sqrt((Sa+(p*(1-(2*v)))))/(Sa-(p*(1+v))))-1)*Ri#\n", + "t2=Ri*((sqrt(Sa/(Sa-(2*p))))-1)#\n", + "# printing data in scilab o/p window\n", + "print \"t1 is %0.1fmm \"%(t1)#\n", + "print \"\\nt2 is %0.3fmm \"%(t2)#\n", + "#The answer to t2 is not calculated in the book." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "t1 is 24.0mm \n", + "\n", + "t2 is 30.206mm \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 5-4 - Page 141" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "p=16#\n", + "Ri=250#\n", + "#Yield strength =sigy#\n", + "sigy=330#\n", + "v=0.3#\n", + "FOS=3#\n", + "Sa=sigy/3#\n", + "t=Ri*((sqrt(Sa/(Sa-(2*p))))-1)#\n", + "t=50#\n", + "\n", + "print \"t is %0.1fmm \"%(t)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "t is 50.0mm \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 5-5 - Page 141" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "d=15#\n", + "Eg=480#\n", + "t=3#\n", + "#flange thickness=ft#\n", + "ft=12#\n", + "A=pi*d**2/4#\n", + "l=d+t+(ft/2)#\n", + "E=210#\n", + "kb=A*E/l#\n", + "#effective area of gasket=Ag#\n", + "Ag=pi*(((ft+t+d)**2)-(d**2))/4#\n", + "kg=Ag*Eg/t#\n", + "# printing data in scilab o/p window\n", + "print \"kb is %0.3f N/mm \"%(kb)#\n", + "kb=kb*10**-3#\n", + "kg=kg*10**-3#\n", + "if (kb<=kg):\n", + " print \"\\nThe combines stiffness of bolt and gasket is %0.3f kN/mm\"%(kg)\n", + "\n", + "\n", + "#The difference in the value of kb is due to rounding-off the value of A \n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "kb is 1546.253 N/mm \n", + "\n", + "The combines stiffness of bolt and gasket is 84.823 kN/mm\n" + ] + } + ], + "prompt_number": 5 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch6.ipynb b/Machine_Design_by_U.C._Jindal/Ch6.ipynb new file mode 100644 index 00000000..b43adfe9 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch6.ipynb @@ -0,0 +1,370 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:e63b98d14884c3c025b7d2e036ef9d475b0841b82b0611aec18600a0b7443dfe" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:6 Levers" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 6-1 - Page 171" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt, pi\n", + "Del=10#\n", + "k=500#\n", + "W=k*Del#\n", + "#Let load arm be l1\n", + "l1=200#\n", + "#Let effort arm be l2\n", + "l2=500#\n", + "P=W*l1/l2#\n", + "Ro=sqrt(W**2+P**2)#\n", + "Ta=40#\n", + "d=sqrt(Ro*4/(2*pi*Ta))#\n", + "d=10#\n", + "pb=10#\n", + "d1=sqrt(Ro/(pb*1.5))#\n", + "d1=20#\n", + "l=1.5*d#\n", + "t=10#\n", + "T=Ro*4/(2*pi*d1**2)#\n", + "M=(Ro/2*(l/2+t/3))-(Ro/2*l/4)#\n", + "sigb=32*M/(pi*d1**3)#\n", + "sigmax=(sigb/2)+sqrt((sigb/2)**2+T**2)#\n", + "P=Ro/(l*d1)#\n", + "D=2*d1#\n", + "print \" d1 is %0.1f mm \"%(d1)#\n", + "print \"\\n D is %0.1f mm \"%(D)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " d1 is 20.0 mm \n", + "\n", + " D is 40.0 mm \n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 6-2 - Page 172" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "d1=80#\n", + "p=0.981#\n", + "Ta=40#\n", + "siga=80#\n", + "pa=15#\n", + "W=pi*(d1**2)*p/4#\n", + "P=W/8#\n", + "Ws=W-P#\n", + "d=sqrt(W*4/(pi*2*Ta))#\n", + "l=1.5*d#\n", + "D=2*d#\n", + "T=W/(2*pi*pa**2/4)#\n", + "M1=P*(700-87.5-(D/2))#\n", + "h=50#\n", + "b=h/4#\n", + "Z=b*h**2/6#\n", + "sigb=M1/Z#\n", + "pmax=80#\n", + "T=2465.6/h**2#\n", + "pmax=(sigb/2)+sqrt((sigb/2)**2+T**2)#\n", + "print \" h is %0.2f mm \"%(h)#\n", + "print \"\\n pmax is %0.2f MPa \"%(pmax)#\n", + " \n", + "#The difference in the value of pmax is due to rounding-off the digits." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " h is 50.00 mm \n", + "\n", + " pmax is 74.43 MPa \n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 6-3 - Page 173" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=((4*360)+(2*360))/900#\n", + "Fv=4-2#\n", + "Fh=P#\n", + "Fr=sqrt(Fv**2+Fh**2)#\n", + "P1=4*0.36/0.9#\n", + "Rf=sqrt(4**2+1.6**2)#\n", + "d=sqrt(Rf*10**3/(15*1.25))#\n", + "d=16#\n", + "l=1.25*d#\n", + "T=Rf*10**3*4/(2*pi*d**2)#\n", + "D=2*d#\n", + "M1=Rf*10**3*(360-(D/2))#\n", + "pa=15#\n", + "h=80#\n", + "b=h/4#\n", + "Z=b*h**2/6#\n", + "sigb=M1/Z#\n", + "T=4310/(b*h)#\n", + "pmax=(sigb/2)+sqrt((sigb/2)**2+T**2)#\n", + "print \"P is %0.1f KN \"%(P)#\n", + "print \"\\npmax is %0.2f MPa \"%(pmax)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "P is 2.0 KN \n", + "\n", + "pmax is 69.53 MPa \n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 6-4 - Page 174" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "l=360#\n", + "P=400#\n", + "Mh=2*P*l/3#\n", + "sigb=50#\n", + "l1=60#\n", + "d=(Mh*32/(pi*l1))**(1/3)#\n", + "d=30#\n", + "L=420#\n", + "siga=60#\n", + "H=20#\n", + "B=H/3#\n", + "Mx=P*(L-H/2)#\n", + "Tx=2*P*l/3#\n", + "sigb1=Mx*18/H**3#\n", + "Td=P/(B*H)#\n", + "Tr=17.17*Tx/H**4#\n", + "T=Tr+Td#\n", + "sigmax=(sigb1/2)+sqrt((sigb1/2)**2+T**2)#\n", + "Tmax=sqrt((sigb1/2)**2+T**2)#\n", + "T=P*L#\n", + "M=P*(l1+(2/3*l))#\n", + "Te=sqrt(T**2+M**2)#\n", + "Ta=40#\n", + "D=(Te*16/(pi*Ta))**(1/3)#\n", + "D=30# #Rounding off to nearest whole number\n", + "print \"d is %0.1f mm \"%(d)#\n", + "print \"\\nD is %0.1f mm \"%(D)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "d is 30.0 mm \n", + "\n", + "D is 30.0 mm \n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 6-5 - Page 175" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "l2=300#\n", + "l=450#\n", + "P=400#\n", + "Mx=2*P*l2/3#\n", + "siga=80#\n", + "dh=(Mx*32/(pi*siga))**(1/3)#\n", + "dh=22#\n", + "L=(2*l2/3)+l#\n", + "T=P*L#\n", + "Ta=40#\n", + "d=(T*16/(pi*Ta))**(1/3)#\n", + "d=35#\n", + "d1=1.6*d#\n", + "Th=T*16*d1/(pi*(d1**4-d**4))#\n", + "l1=1.5*d#\n", + "My=P*(L-(d1/2))#\n", + "B=dh#\n", + "H=sqrt(3.66*75)#\n", + "H=30#\n", + "Mz=P*l1/2#\n", + "Te=sqrt(T**2+Mz**2)#\n", + "d2=(Te*16/(pi*Ta))**(1/3)#\n", + "d2=32#\n", + "b=d/4#\n", + "b=9# #Rounding off to nearest whole number\n", + "t=d/6#\n", + "t=6# #Rounding off to nearest whole number\n", + "print \" d is %0.1f mm \"%(d)#\n", + "print \"\\n dh is %0.1f mm \"%(dh)#\n", + "print \"\\n d1 is %0.1f mm \"%(d1)#\n", + "print \"\\n l1 is %0.1f mm \"%(l1)#\n", + "print \"\\n d2 is %0.1f mm \"%(d2)#\n", + "print \"\\n b is %0.1f mm \"%(b)#\n", + "print \"\\n t is %0.1f mm \"%(t)#\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " d is 35.0 mm \n", + "\n", + " dh is 22.0 mm \n", + "\n", + " d1 is 56.0 mm \n", + "\n", + " l1 is 52.5 mm \n", + "\n", + " d2 is 32.0 mm \n", + "\n", + " b is 9.0 mm \n", + "\n", + " t is 6.0 mm \n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 6-6 - Page 175" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "L=450#\n", + "P=700#\n", + "T=P*L#\n", + "Ta=50#\n", + "d=(T*16/(pi*Ta))**(1/3)#\n", + "d=32#\n", + "d1=1.6*d#\n", + "d1=52# #Rounding off to nearest whole number\n", + "l1=1.25*d#\n", + "My=P*(L-d1/2)#\n", + "sigb=65#\n", + "H=(My*18/sigb)**(1/3)#\n", + "H=45#\n", + "B=H/3#\n", + "T1=P/(B*H)#\n", + "sigmax=(sigb/2)+sqrt((sigb/2)**2+T**2)#\n", + "Mx=P*l1/2#\n", + "Te=sqrt((T)**2+(Mx**2))#\n", + "d2=(Te*16/(pi*Ta))**(1/3)#\n", + "d2=d2+6#\n", + "d2=38# #Rounding off to nearest whole number\n", + "print \" d is %0.1f mm \"%(d)#\n", + "print \"\\n d1 is %0.1f mm \"%(d1)#\n", + "print \"\\n l1 is %0.1f mm \"%(l1)#\n", + "print \"\\n B is %0.1f mm \"%(B)#\n", + "print \"\\n H is %0.1f mm \"%(H)#\n", + "print \"\\n d2 is %0.1f mm \"%(d2)#\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " d is 32.0 mm \n", + "\n", + " d1 is 52.0 mm \n", + "\n", + " l1 is 40.0 mm \n", + "\n", + " B is 15.0 mm \n", + "\n", + " H is 45.0 mm \n", + "\n", + " d2 is 38.0 mm \n" + ] + } + ], + "prompt_number": 13 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch7.ipynb b/Machine_Design_by_U.C._Jindal/Ch7.ipynb new file mode 100644 index 00000000..3a83b942 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch7.ipynb @@ -0,0 +1,303 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:f63edcd22a3bf57c69f1f76aca0a01ff1dd288f9c7a8d1b01585ac7b29af9fbe" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:7 Struts and Columns" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 7-1 - Page 192" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "sigc=550#\n", + "FOS=4#\n", + "sigw=sigc/FOS#\n", + "l=4000#\n", + "le=l/2#\n", + "A=pi*(1-0.7**2)/4#\n", + "K=(1+0.7**2)/16#\n", + "Pr=800*10**3#\n", + "a=1/1600#\n", + "D=130# #Rounding off to nearest whole number\n", + "d=D*0.7#\n", + "print \"D is %0.1f mm \"%(D)#\n", + "print \"\\nd is %0.1f mm \"%(d)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "D is 130.0 mm \n", + "\n", + "d is 91.0 mm \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 7-2 - Page 192" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "l=500#\n", + "E=70*10**3#\n", + "P=20*10**3#\n", + "FOS=2#\n", + "d=P*2*12*4*l**2/((pi)**2*E)#\n", + "d=(sqrt(8)*d)**0.25#\n", + "b=d/sqrt(8)#\n", + "print \"d is %0.2f mm \"%(d)#\n", + "print \"\\nb is %0.2f mm \"%(b)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "d is 37.44 mm \n", + "\n", + "b is 13.24 mm \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 7-3 - Page193" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Ixx=(2*1696.6)+115.4#\n", + "Iyy=1696.6+(2*115.4)+(2*25.27*10.27**2)#\n", + "A=3*25.27#\n", + "Kmin=sqrt(Ixx/75.81)#\n", + "L=600#\n", + "k=L/Kmin#\n", + "sigc=110#\n", + "c=1/200#\n", + "sigw=sigc*(1-(c*k))#\n", + "Pw=sigw*A#\n", + "a=1/7500#\n", + "sigc1=320#\n", + "Pr=(sigc1*A)/(1+(a*(L/Kmin)**2))#\n", + "FOS=Pr/Pw#\n", + "print \"FOS is %0.2f \"%(FOS)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "FOS is 2.91 \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 7-4 - Page 194" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Iyy=193.4+(2*1.2*1.5**3/12)#\n", + "E=200*10**3#\n", + "l=500#\n", + "Pe=(pi**2)*E*Iyy*10**5/(l**2)#\n", + "A=35.53+(2*1.2*15)#\n", + "sige=Pe/7530#\n", + "k=sqrt(Iyy/A)#\n", + "xc=75#\n", + "sig=80#\n", + "sigo=20.875#\n", + "A=A*100#\n", + "P=sigo*A#\n", + "P=P*10**-3#\n", + "print \"P is %0.1f kN \"%(P)#\n", + " \n", + " #The difference in the value of P is due to rounding-off the digits." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "P is 149.3 kN \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 7-5 - Page 195" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "sigc=330#\n", + "a=1/7500#\n", + "t=4#\n", + "A=14.5*t**2#\n", + "l=300#\n", + "Kx=sqrt(1.4626*t**2)#\n", + "Pr=sigc*A/(1+(a*(l/Kx)**2))#\n", + "FOS=2#\n", + "P=Pr/FOS*10**-3#\n", + "print \"P is %0.4f KN \"%(P)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "P is 38.2800 KN \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 7-6 - Page 195" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P=1500#\n", + "FOS=2#\n", + "Pd=FOS*P#\n", + "l=280#\n", + "E=207*10**3#\n", + "I=Pd*l**2/(pi**2*E)#\n", + "D=(64*I/(pi*(1-0.8**4)))**(1/4)#\n", + "D=8#\n", + "d=6.4#\n", + "print \"D is %0.1f mm \"%(D)#\n", + "print \"\\nd is %0.1f mm \"%(d)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "D is 8.0 mm \n", + "\n", + "d is 6.4 mm \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 7-7 - Page 196" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "D=500#\n", + "p=0.3#\n", + "E=208*10**3#\n", + "sigc=320#\n", + "a=1/7500#\n", + "l=2000#\n", + "le=l/2#\n", + "W=pi*D**2*p/4#\n", + "FOS=4#\n", + "Wd=W*FOS#\n", + "I=Wd*l**2/(pi**2*E)#\n", + "d=(64*I/pi)**(1/4)#\n", + "A=pi*d**2/4#\n", + "k=d/4#\n", + "d=45# #Rounding off to nearest whole number\n", + "print \"d is %0.1f mm \"%(d)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "d is 45.0 mm \n" + ] + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch8.ipynb b/Machine_Design_by_U.C._Jindal/Ch8.ipynb new file mode 100644 index 00000000..520e04b8 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch8.ipynb @@ -0,0 +1,832 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:e4245d57622666881114047950aef2cecdad4d23a29878d669ecae5aa1b95679" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:8 Springs" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 8-1 - Page 224" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "d=5#\n", + "D=30#\n", + "G=84*(10**3)#\n", + "Na=15#\n", + "#Axial Load W\n", + "W=300#\n", + "#Spring index C\n", + "C=30/5#\n", + "#Shear stress Augmentation factor Ks\n", + "Ks=((2*C)+1)/(2*C)#\n", + "#Wahl's factor Kw\n", + "Kw=(((4*C)-1)/((4*C)-4))+(0.615/C)#\n", + "#Curvature correction factor Kc\n", + "Kc=Kw/Ks#\n", + "#Spring stiffness k\n", + "k=(G*(d**4))/(8*(D**3)*Na)#\n", + "#Axial deflection delta\n", + "delta=W/k#\n", + "print \" Ks is %0.4f \"%(Ks)#\n", + "print \"\\n Kw is %0.4f \"%(Kw)#\n", + "print \"\\n Kc is %0.3f \"%(Kc)#\n", + "print \"\\n The Spring Stiffness is %0.1f N/mm\"%(k)#\n", + "print \"\\n The Axial deflection is %0.3f mm\"%(delta)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Ks is 1.0000 \n", + "\n", + " Kw is 1.1025 \n", + "\n", + " Kc is 1.103 \n", + "\n", + " The Spring Stiffness is 16.0 N/mm\n", + "\n", + " The Axial deflection is 18.000 mm\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 8-2 - Page 224" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "W=196.2#\n", + "lenthofscale=50#\n", + "k=196.2/50#\n", + "C=8#\n", + "Ks=(1+(0.5/C))#\n", + "\n", + "# Let us choose oil tempered wire 0.6-0.7 %C. Refer to Table 8-4 for constants A and m, relating strength wire \n", + "#diameter.\n", + "G=77.2*(10**3)#\n", + "A=1855#\n", + "m=0.187#\n", + "# equating Tmax=0.5*sig(ut).\n", + "# Ks*(8*W*D/(pi*(d**3)))=0.5*A/(d**2)\n", + "d1=(Ks*(8*W*C/(pi*A*0.5)))#\n", + "d=d1**(1/1.813)#\n", + "D=C*d#\n", + "Na=G*(d**4)/(8*(D**3)*k)#\n", + "#Solid length = SL\n", + "SL=(Na-1)*d\n", + "\n", + "print \" wire diameter is %0.3f mm \"%(d)#\n", + "print \"\\n mean diameter is %0.3f mm \"%(D)#\n", + "print \"\\n Number of acting coils are %0.3f \"%(Na)#\n", + "\n", + "#The difference in the values of d,D and Na is due to rounding-off the digits." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " wire diameter is 2.314 mm \n", + "\n", + " mean diameter is 18.516 mm \n", + "\n", + " Number of acting coils are 11.117 \n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 8-3 - Page 225" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "d=1.626#\n", + "A=2211#\n", + "m=0.145#\n", + "rm=3#\n", + "ri=(rm-(d/2))#\n", + "sigma=A/(d**m)#\n", + "W=(sigma*pi*(d**3)*ri)/(32*(rm**2))#\n", + "print \" Ultimate tensile Strength is %0.1f MPa \"%(sigma)#\n", + "print \"\\n Force at which the spring hook fails is %0.1f N \"%(W)#\n", + "\n", + "#The difference in the values of sigma and W is due to rounding-off the digits." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Ultimate tensile Strength is 2060.5 MPa \n", + "\n", + " Force at which the spring hook fails is 211.3 N \n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 8-4 - Page 226" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Do=25#\n", + "# mean coil diameter D=25-d\n", + "W=150#\n", + "T=800#\n", + "G=81000#\n", + "# Substituting values in equation T=8*W*D/(pi*(d**3))\n", + "# therefore, the equation becomes d**3 + 0.477*d = 11.936\n", + "#consider d=2.2mm, (d can be taken between 2.2-2.3mm)\n", + "d=2.337# #(nearest available wire gauge)\n", + "C=9.5#\n", + "D=22.2# \n", + "Do=D+d#\n", + "Ks=1+(0.5/C)#\n", + "Tmax=Ks*8*W*D/(pi*(d**3))#\n", + "# check for safety- Tmax<T#\n", + "Lo=100#\n", + "Ls=40#\n", + "#Lo=Ls+delta+0.15*delta\n", + "delta=(Lo-Ls)/1.15#\n", + "delta=50#\n", + "k=150/50#\n", + "Na=(G*d**4)/(8*(D**3)*k)#\n", + "\n", + "N=Na+2#\n", + "Ls=N*d#\n", + "Lo=Ls+(1.15*delta)#\n", + "print \" d is %0.3fmm \"%(d)#\n", + "print \"\\n D is %0.2f mm\"%(D)#\n", + "print \"\\n Ls is %0.2f mm\"%(Ls)#\n", + "print \"\\n Lo is %0.2f mm\"%(Lo)#\n", + "if (Do<=25):\n", + " print '\\nThe diameter is within space constraints'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " d is 2.337mm \n", + "\n", + " D is 22.20 mm\n", + "\n", + " Ls is 26.18 mm\n", + "\n", + " Lo is 83.68 mm\n", + "\n", + "The diameter is within space constraints\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 8-5A - Page 227" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Di=15#\n", + "Do=20#\n", + "d=2.3#\n", + "D=17.5#\n", + "C=D/d#\n", + "Ks=1+(0.5/C)#\n", + "Wmax=100#\n", + "Tmax=Ks*8*Wmax*D/(pi*(d**3))#\n", + "G=81000#\n", + "delmax=67.7/2.366#\n", + "k=100/28#\n", + "Na=G*(d**4)/(8*k*(D**3))#\n", + "Ls=Na+1# #(for plain ends)\n", + "delmax=28#\n", + "#TL= total working length\n", + "TL=Ls+delmax+(0.15*delmax)#\n", + "print \" d is %0.1fmm \"%(d)#\n", + "print \"\\n C is %0.1f \"%(C)#\n", + "print \"\\n Na is %0.1f \"%(Na)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " d is 2.3mm \n", + "\n", + " C is 7.6 \n", + "\n", + " Na is 17.6 \n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 8-6 - Page- 228" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# 18 SWG=1.219MM in dia\n", + "d=1.219#\n", + "E=198.6*10**3#\n", + "G=80.7*10**3#\n", + "m=0.19#\n", + "A=1783#\n", + "sig=A/(d**m)#\n", + "Tys=(0.4*sig)#\n", + "Do=12.5#\n", + "D=Do-d#\n", + "C=D/d#\n", + "Ks=((2*C)+1)/(2*C)#\n", + "W=(Tys*pi*(d**3))/(8*D*Ks)#\n", + "Nt=13.5#\n", + "Na=Nt-2#\n", + "Del=(8*W*(D**3)*Na)/(G*(d**4))#\n", + "Ls=(Nt-1)*d#\n", + "Lo=Ls+Del+(0.15*Del)#\n", + "print \" Tys is %0.1f MPa \"%(Tys)#\n", + "print \"\\n W is %0.1f N \"%(W)#\n", + "print \"\\n del is %0.3f mm \"%(Del)#\n", + "print \"\\n Ls is %0.4f mm \"%(Ls)#\n", + "print \"\\n Lo is %0.2f mm \"%(Lo)#\n", + " \n", + "#Answers in the book for Torsional yeild strength have been rounded-off to the nearest whole number." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Tys is 686.9 MPa \n", + "\n", + " W is 41.1 N \n", + "\n", + " del is 30.457 mm \n", + "\n", + " Ls is 15.2375 mm \n", + "\n", + " Lo is 50.26 mm \n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 8-7 - Page 228" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "d=1.016#\n", + "A=2211#\n", + "m=0.145#\n", + "G=81000#\n", + "Nt=16#\n", + "Na=16-2#\n", + "sig=A/(d**m)#\n", + "Tys=0.45*sig#\n", + "Do=12.6#\n", + "D=Do-d#\n", + "C=D/d#\n", + "Ks=1+(0.5/C)#\n", + "W=(Tys*pi*(d**3))/(8*D*Ks)#\n", + "k=(G*(d**4))/(8*(D**3)*Na)#\n", + "Del=W/k#\n", + "Ls=(Nt-1)*d#\n", + "Lo=Ls+(1.15*Del)#\n", + "\n", + "print \"Tys is %0.1f MPa \"%(Tys)#\n", + "print \"\\n Do is %0.1f N \"%(Do)#\n", + "print \"\\n W is %0.1f N \"%(W)#\n", + "print \"\\n k is %0.3f N \"%(k)#\n", + "print \"\\n del is %0.2f mm \"%(Del)#\n", + "print \"\\n Ls is %0.2f mm \"%(Ls)#\n", + "print \"\\n Lo is %0.3f mm \"%(Lo)#\n", + " \n", + "if ((Lo/D)>=5.26):\n", + " print 'The spring will fail under buckling'\n", + "\n", + "\n", + "#Values after the decimal point has not been considered for answer of Torsional yeild strength in the book, whereas answers for deflection and free-length is different as entire value of variables is taken for calculation in the code." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Tys is 992.7 MPa \n", + "\n", + " Do is 12.6 N \n", + "\n", + " W is 33.8 N \n", + "\n", + " k is 0.496 N \n", + "\n", + " del is 68.20 mm \n", + "\n", + " Ls is 15.24 mm \n", + "\n", + " Lo is 93.669 mm \n", + "The spring will fail under buckling\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 8-8 - Page 229" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "d=2#\n", + "Do=20#\n", + "D=Do-d#\n", + "C=D/d#\n", + "Na=9#\n", + "#Material hard drawn spring steel\n", + "A=1783#\n", + "m=0.19#\n", + "G=81000#\n", + "sig=A/(d**m)#\n", + "Tys=0.45*sig\n", + "Kf=1.5#\n", + "Ta=Tys/Kf#\n", + "Ks=1+(0.5/C)#\n", + "W=(Ta*pi*(d**3))/(8*D*Ks)#\n", + "k=(G*(d**4))/(8*(D**3)*Na)#\n", + "Del=W/k#\n", + "Lo=((Na+1)*d)+(1.15*Del)#\n", + "p=(Lo-d)/Na#\n", + "print \"k is %0.3f N/mm \"%(k)#\n", + "print \"\\n W is %0.1f N \"%(W)#\n", + "print \"\\n Lo is %0.3f mm \"%(Lo)#\n", + "print \"\\n p is %0.3f mm \"%(p)#\n", + " \n", + " \n", + "if ((Lo)>=47.34):\n", + " print 'The spring will fail under buckling'\n", + "\n", + "#The answer for value of spring rate 'k' is misprinted in the book. Due to this all subsequent values of del,Lo,p is calucated incorrectly in the book." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "k is 3.000 N/mm \n", + "\n", + " W is 77.5 N \n", + "\n", + " Lo is 49.720 mm \n", + "\n", + " p is 5.302 mm \n", + "The spring will fail under buckling\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 8-9 - Page 230" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# for music wire\n", + "d1=11.5#\n", + "A=2211#\n", + "d=1.5#\n", + "m=0.145#\n", + "sigut=A/(d**m)#\n", + "sigy=0.78*sigut#\n", + "Do=16#\n", + "E=2*(10**5)#\n", + "Nb=4.25#\n", + "D=Do-d#\n", + "C=D/d#\n", + "Ki=((4*(C**2))-C-1)/(4*C*(C-1))#\n", + "Mmax=(sigy*pi*(d**3))/(32*Ki)#\n", + "kc=((d**4)*E)/(10.8*D*Nb)#\n", + "theta3=Mmax/kc#\n", + "l1=20#\n", + "l2=20#\n", + "Ne=(l1+l2)/(3*pi*D)#\n", + "Na=Nb+Ne#\n", + "k=((d**4)*E)/(10.8*Na*D)#\n", + "thetat=Mmax/k#\n", + "ke=(3*pi*(d**4)*E)/(10.8*(l1+l2))#\n", + "# angdisp=theta1+theta2=Mmax/ke#\n", + "angdisp=Mmax/ke#\n", + "#D1 is final coil diameter\n", + "D1=(Nb*D)/(Nb+theta3)#\n", + "#IRC=Initial radial clearance\n", + "IRC=((D-d)-d1)/2#\n", + "#FRC=Final radial clearance\n", + "FRC=((D1-d)-d1)/2#\n", + "\n", + "\n", + "\n", + "print \" Maximum Torque is %0.2f Nmm \"%(Mmax)#\n", + "print \"\\n theta3 is %0.3f turns \"%(theta3)#\n", + "print \"\\n Ne is %0.3f turns \"%(Ne)#\n", + "print \"\\n ke is %0.1f N/mm \"%(ke)#\n", + "print \"\\n theta1+theta2 is %0.4f turns \"%(angdisp)#\n", + "print \"\\n D1 is %0.2f mm \"%(D1)#\n", + "print \"\\n IRC is %0.2f mm \"%(IRC)#\n", + "print \"\\n FRC is %0.2f mm \"%(FRC)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Maximum Torque is 497.25 Nmm \n", + "\n", + " theta3 is 0.327 turns \n", + "\n", + " Ne is 0.293 turns \n", + "\n", + " ke is 22089.3 N/mm \n", + "\n", + " theta1+theta2 is 0.0225 turns \n", + "\n", + " D1 is 13.46 mm \n", + "\n", + " IRC is 0.75 mm \n", + "\n", + " FRC is 0.23 mm \n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 8-10 - Page 231" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "A=1783#\n", + "m=0.190#\n", + "d=1.5#\n", + "D=15#\n", + "M=300#\n", + "E=20800#\n", + "k=30#\n", + "#sigult= ultimate strength of the material\n", + "# sigy= yield strength of the material\n", + "sigult=A/(d**m)#\n", + "sigy=0.7*sigult#\n", + "#siga= allowable yield strength of the material\n", + "siga=sigy/2#\n", + "C=D/d#\n", + "Ki=(4*(C**2)-C-1)/(4*C*(C-1))#\n", + "Z=pi*(d**3)/32#\n", + "#sigb=bending strength of the material#\n", + "sigb=Ki*M/Z#\n", + "while (sigb>=siga) :\n", + " d=d+0.15#\n", + " D=15#\n", + " C=D/d#\n", + " sigult=A/(d**m)#\n", + " sigy=0.7*sigult#\n", + " siga=sigy/2#\n", + " Ki=(4*(C**2)-C-1)/(4*C*(C-1))#\n", + " Z=pi*(d**3)/32#\n", + " sigb=Ki*M/Z#\n", + "\n", + "d=2## rounding off the value of the diameter.\n", + "Na=(d**4)*E/(64*D*k)#\n", + "print \" d is %0.1f mm \"%(d)#\n", + "print \"\\n D is %0.1f mm \"%(D)#\n", + "print \"\\n Na is %0.2f mm \"%(Na)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " d is 2.0 mm \n", + "\n", + " D is 15.0 mm \n", + "\n", + " Na is 11.00 mm \n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 8-11 - Page 231" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "L=1180#\n", + "W=40*(10**3)#\n", + "Nf=2#\n", + "Ng=8#\n", + "E=207*(10**3)#\n", + "#sigut is ultimate strength\n", + "sigut=1400#\n", + "FOS=2#\n", + "#siga= allowable yield strength of the material\n", + "siga=1400/2#\n", + "#sigbf=bending strength in full length\n", + "sigbf=700#\n", + "b=75#\n", + "t=((4.5*W*L)/(((3*Nf)+(2*Ng))*sigbf))**(0.5)#\n", + "t=14#\n", + "I=(Nf*b*(t**3))/12#\n", + "Wf=(3*Nf*W)/((3*Nf)+(2*Ng))#\n", + "Del=(Wf*(L**3))/(48*E*I)#\n", + "print \" t is %0.0f mm \"%(t)#\n", + "print \"\\n Wf is %0.0f N \"%(Wf)#\n", + "print \"\\n I is %0.0f mm**4 \"%(I)#\n", + "print \"\\n del is %0.1f mm \"%(Del)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " t is 14 mm \n", + "\n", + " Wf is 10909 N \n", + "\n", + " I is 34300 mm**4 \n", + "\n", + " del is 52.0 mm \n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 8-12A - Page 232" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "W=80000#\n", + "sigbfr=500#\n", + "L=1100#\n", + "Nf=3#\n", + "Ng=10#\n", + "N=Nf+Ng#\n", + "t=((1.5*W*L)/(N*6*sigbfr))**(1/3)#\n", + "t=15#\n", + "b=6*t#\n", + "E=207*10**3#\n", + "deli=(W*(L**3))/(8*E*N*b*(t**3))#\n", + "Wi=(W*Nf*Ng)/(N*((3*Nf)+(2*Ng)))#\n", + "print \" t is %0.1f mm \"%(t)#\n", + "print \"\\n deli is %0.1f mm \"%(deli)#\n", + "print \"\\n Wi is %0.0f N \"%(Wi)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " t is 15.0 mm \n", + "\n", + " deli is 16.0 mm \n", + "\n", + " Wi is 6366 N \n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 8-13 - Page 233" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#ultimate strength=sigut\n", + "sigut=1500#\n", + "C=7#\n", + "d=3#\n", + "D=C*d#\n", + "Ks=1+(0.5/C)#\n", + "Kw=(((4*C)-1)/((4*C)-4))+(0.615/C)#\n", + "Pmax=120#\n", + "Pmin=40#\n", + "Pm=80#\n", + "Tm=(Ks*8*Pm*D)/(pi*(d**3))#\n", + "Ta=(Kw*8*Pmin*D)/(pi*(d**3))#\n", + "Tse=0.22*sigut#\n", + "Tys=0.45*sigut#\n", + "x=(Tys-(0.5*Tse))/(0.5*Tse)#\n", + "y=((x)*Ta)+Tm#\n", + "FOS=(Tys/y)#\n", + "print \" Tm is %0.2f MPa \"%(Tm)#\n", + "print \"\\n Ta is %0.1f MPa \"%(Ta)#\n", + "print \"\\n FOS is %0.3f \"%(FOS)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Tm is 169.77 MPa \n", + "\n", + " Ta is 86.2 MPa \n", + "\n", + " FOS is 1.548 \n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 8-14 - Page 234" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Tse=360#\n", + "Tys=660#\n", + "d=25#\n", + "P=0.03#\n", + "m=40#\n", + "Pmin=((pi*(d**2)*P)/4)+(m*9.81/1000)#\n", + "k=6#\n", + "#Additional load= Padd=k*further compression in spring\n", + "Padd=k*10#\n", + "Pmax=Padd+Pmin#\n", + "Pm=(Pmax+Pmin)/2#\n", + "Pa=(Pmax-Pmin)/2#\n", + "d=2#\n", + "D=12#\n", + "C=6#\n", + "Ks=1+(0.5/C)#\n", + "Ks=1.083#\n", + "Kw=(((4*C)-1)/((4*C)-4))+(0.615/C)#\n", + "Ta=(Kw*8*Pa*D)/(pi*(d**3))#\n", + "Tm=(Ks*8*Pm*D)/(pi*(d**3))#\n", + "x=(Tys-(0.5*Tse))/(0.5*Tse)#\n", + "y=((x)*Ta)+Tm#\n", + "FOS=(Tys/y)#\n", + "print \" Tm is %0.2f MPa \"%(Tm)#\n", + "print \"\\n Ta is %0.3f MPa \"%(Ta)#\n", + "print \"\\n FOS is %0.2f \"%(FOS)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Tm is 186.64 MPa \n", + "\n", + " Ta is 126.337 MPa \n", + "\n", + " FOS is 1.26 \n" + ] + } + ], + "prompt_number": 28 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/Ch9.ipynb b/Machine_Design_by_U.C._Jindal/Ch9.ipynb new file mode 100644 index 00000000..0e704ca7 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/Ch9.ipynb @@ -0,0 +1,358 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:4885451d1eb84f1c7ede3570bb9a4c180b0235da374d3be406dcc799b426e236" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch:9 Threaded Fasteners" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 9-1 - Page 266" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "p1=2#\n", + "d=16#\n", + "dt1=d-(0.93825*p1)#\n", + "At1=pi*dt1**2/4#\n", + "p2=1.5#\n", + "d=16#\n", + "dt2=d-(0.93825*p2)#\n", + "At2=pi*dt2**2/4#\n", + "\n", + "print \" At1 is %0.1f mm**2 \"%(At1)#\n", + "print \"\\n At2 is %0.1f mm**2 \"%(At2)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " At1 is 156.7 mm**2 \n", + "\n", + " At2 is 167.2 mm**2 \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 9-2 - Page 266" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "W=20*10**3#\n", + "n=4#\n", + "#Let the load on each bolt be W1\n", + "W1=W/n#\n", + "At=W1/80#\n", + "\n", + "print \"At is %0.1f mm**2 \"%(At)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "At is 62.0 mm**2 \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 9-3 - Page 267" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import atan, tan, cos\n", + "d=18#\n", + "p=2.5#\n", + "dr=d-(1.2268*p)#\n", + "dm=(d+dr)/2#\n", + "alpha=atan(p/(pi*dm))#\n", + "theta=pi*30/180#\n", + "u1=0.15#\n", + "u2=0.13#\n", + "x=(tan(alpha)+(u1/cos(theta)))/(1-(tan(alpha)*u1/cos(theta)))#\n", + "K=dm*x/(2*d)+(0.625*u2)#\n", + "print \"K is %0.5f \"%(K)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "K is 0.18343 \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 9-4 - Page 267" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "d=20#\n", + "t=4#\n", + "Lg=84#\n", + "Ad=pi*d**2/4#\n", + "Eb=205*10**3#\n", + "Ed=105*10**3#\n", + "kb=Ad*Eb/Lg#\n", + "lg=80#\n", + "x=5*(lg+(0.5*d))/(lg+(2.5*d))#\n", + "kp=pi*Ed*d/(2*log(x))#\n", + "At=245#\n", + "sigb=105#\n", + "Pe=20*10**3#\n", + "Pb=Pe*kb/(kb+kp)#\n", + "sigad=Pb/At#\n", + "finalst=sigb+sigad#\n", + "\n", + "print \"final stress is %0.2f N/mm**2 \"%(finalst)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "final stress is 123.28 N/mm**2 \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 9-5 - Page 268" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Eb=207*10**3#\n", + "Ec=105*10**3#\n", + "sigp=650#\n", + "At=115#\n", + "Pi=0.75*sigp*At#\n", + "F=sigp*At#\n", + "#Let the additional load Fadd\n", + "Padd=F-Pi#\n", + "d=14#\n", + "Ad=pi*d**2/4#\n", + "Lg=63#\n", + "kb=Ad*Eb/Lg#\n", + "lg=60#\n", + "x=5*(lg+(0.5*d))/(lg+(2.5*d))#\n", + "km=pi*Ec*d/(2*log(x))#\n", + "C=kb/(kb+km)#\n", + "Pe=Padd/C#\n", + "K=0.2#\n", + "Ti=Pi*K*d*10**-3#\n", + "\n", + "print \"Ti is %0.2f Nm \"%(Ti)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ti is 156.97 Nm \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 9-6 - Page 269" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "d=20#\n", + "sigp=600#\n", + "At=245#\n", + "Pi=120*10**3#\n", + "Pe=30*10**3#\n", + "C=0.35#\n", + "Pb=C*Pe#\n", + "P=Pi+Pb#\n", + "sigi=Pi/At#\n", + "sigf=P/At#\n", + "K=0.18#\n", + "T=K*d*Pi*10**-3#\n", + "E1=sigi/sigp#\n", + "E2=sigf/sigp#\n", + "\n", + "print \" sigi is %0.1f MPa \"%(sigi)#\n", + "print \"\\n sigi is %0.2f MPa \"%(sigf)#\n", + "print \"\\n T is %0.0f Nm \"%(T)#\n", + "print \"\\n E1 is %0.3f \"%(E1)#\n", + "print \"\\n E2 is %0.3f \"%(E2)#\n", + " \n", + " #Value upto tenthth place is considered in the book for value of final stress in bolt, 'sigf'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " sigi is 489.0 MPa \n", + "\n", + " sigi is 532.65 MPa \n", + "\n", + " T is 432 Nm \n", + "\n", + " E1 is 0.000 \n", + "\n", + " E2 is 0.888 \n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 9-7 - Page 269" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "p=2#\n", + "d=16#\n", + "dt=d-(0.938*p)#\n", + "At=pi*dt**2/4#\n", + "r=60*sqrt(2)#\n", + "Td=1/(4*At)#\n", + "Ta=120#\n", + "T=8.722*10**-3#\n", + "P=Ta/T*10**-3#\n", + "\n", + "print \"P is %0.3f kN \"%(P)#\n", + " \n", + " #Value upto hundredth place is considered in the book for value of permissible load, 'P'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "P is 13.758 kN \n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exa 9-8 - Page 270" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "sigyp=460#\n", + "FOS=2#\n", + "Ts=0.577*sigyp/FOS#\n", + "At=245#\n", + "r=100#\n", + "P=Ts*At/1.453*10**-3#\n", + "print \"The eccentric load is %f N \"%(P)\n", + "print \"P is %0.3f kN \"%(P)#\n", + "#Value of thousandth place of eccentric load, 'P' is misprinted in the book. " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The eccentric load is 22.377116 N \n", + "P is 22.377 kN \n" + ] + } + ], + "prompt_number": 9 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Machine_Design_by_U.C._Jindal/screenshots/Chapter-3stressgraph.png b/Machine_Design_by_U.C._Jindal/screenshots/Chapter-3stressgraph.png Binary files differnew file mode 100644 index 00000000..8b1f0efc --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/screenshots/Chapter-3stressgraph.png diff --git a/Machine_Design_by_U.C._Jindal/screenshots/Chapter-_8AdditionalLoad.png b/Machine_Design_by_U.C._Jindal/screenshots/Chapter-_8AdditionalLoad.png Binary files differnew file mode 100644 index 00000000..e2ec2d6d --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/screenshots/Chapter-_8AdditionalLoad.png diff --git a/Machine_Design_by_U.C._Jindal/screenshots/Chapter11_-_strengthofrevet.png b/Machine_Design_by_U.C._Jindal/screenshots/Chapter11_-_strengthofrevet.png Binary files differnew file mode 100644 index 00000000..15578bc7 --- /dev/null +++ b/Machine_Design_by_U.C._Jindal/screenshots/Chapter11_-_strengthofrevet.png diff --git a/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter02_2.ipynb b/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter02_2.ipynb new file mode 100644 index 00000000..aa1ce937 --- /dev/null +++ b/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter02_2.ipynb @@ -0,0 +1,1029 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c74fb10129132335df7b12f8b742cc9cb4132c6756929f4670fe8146e4f5aa90"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 02 - BIPOLAR JUNCTION TRANSISTORS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 - Pg 20"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.1\n",
+ "# Given data\n",
+ "I_C = 15.;# in mA\n",
+ "I_CbyI_E= 0.995;\n",
+ "I_E= I_C/I_CbyI_E;# in mA\n",
+ "I_B= 0.005*I_E;# in mA\n",
+ "I_CBO = 3.;# in uA\n",
+ "I_CBO = I_CBO * 10.**-3.;# in mA\n",
+ "alpha_dc= I_C/I_E;\n",
+ "print '%s %.2f' %(\"The value of Alpha_dc is\",alpha_dc);\n",
+ "# I_C = Alpha_dc*I_E + I_CBO;\n",
+ "I_E = (I_C-I_CBO)/alpha_dc;# in mA\n",
+ "print '%s %.2f' %(\"The value of I_E in mA is\",I_E);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Alpha_dc is 0.99\n",
+ "The value of I_E in mA is 15.07\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 - Pg 21"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.2\n",
+ "# Given data\n",
+ "Alpha_dc = 0.99;\n",
+ "I_CBO = 10.;# in uA\n",
+ "I_CBO = I_CBO * 10.**-3.;# in mA\n",
+ "I_E = 10.;# in mA\n",
+ "# To calculate I_C : \n",
+ "I_C = (Alpha_dc*I_E) + I_CBO;# in mA\n",
+ "print '%s %.2f' %(\"The value of I_C in mA is\",I_C);\n",
+ "# To calculate I_B : \n",
+ "I_B = I_E-I_C;# in mA\n",
+ "I_B = I_B * 10.**3.;# in uA\n",
+ "print '%s %.2f' %(\"The value of I_B in uA is\",I_B);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_C in mA is 9.91\n",
+ "The value of I_B in uA is 90.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 - Pg 22"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.3\n",
+ "# Given data\n",
+ "Alpha_dc = 0.99;\n",
+ "I_C = 6.;# in mA\n",
+ "I_CBO = 15.;# in uA\n",
+ "I_CBO = I_CBO * 10.**-3.;# in mA\n",
+ "# I_C = Alpha_dc*I_E + I_CBO;\n",
+ "I_E = (I_C - I_CBO)/Alpha_dc;# in mA\n",
+ "I_B = I_E - I_C;# in mA\n",
+ "I_B = I_B * 10.**3.;# in uA\n",
+ "print '%s %.2f' %(\"The value of I_B in uA is\",I_B);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_B in uA is 45.45\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E05 - Pg 24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.5\n",
+ "# Given data\n",
+ "Alpha_dc = 0.98;\n",
+ "I_CBO = 12.;# in uA\n",
+ "I_CBO = I_CBO * 10.**-6.;# in A\n",
+ "I_B = 120.;# in uA\n",
+ "I_B = I_B * 10**-6;# in A\n",
+ "# Calculation of Beta_dc\n",
+ "Beta_dc = Alpha_dc/(1-Alpha_dc);\n",
+ "I_E = (1+Beta_dc)*I_B + (1+Beta_dc)*I_CBO;# in A\n",
+ "I_E = I_E * 10.**3.;# in mA\n",
+ "print '%s %.2f' %(\"The value of I_E in mA is\",I_E);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_E in mA is 6.60\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E06 - Pg 24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.6\n",
+ "# Given data\n",
+ "V_BEsat = 0.8;# in V\n",
+ "V_BEact = 0.7;# in V\n",
+ "V_CEsat = 0.2;# in V\n",
+ "V_CC = 10.;# in V\n",
+ "Beta = 100.;\n",
+ "V = 5.;# in V \n",
+ "R_B = 50.* 10.**3.;# in ohm\n",
+ "R_E = 2.* 10.**3.;# in ohm\n",
+ "R_C = 3.* 10.**3.;# in ohm\n",
+ "# Applying KVL to input loop, V = R_B*I_B + V_BEact + I_C*R_E and I_C = Beta*I_B;\n",
+ "I_B =17.2;# (V-V_BEact)/(R_B+R_E*Beta);# in A\n",
+ "# Applying KVL to collector circuit, V_CC= I_C*R_C+V_CEsat+I_E*R_E and I_E=I_C+I_B\n",
+ "#I_C = (V_CC-V_CEsat-I_B*R_E)/(R_C+R_E);# in A\n",
+ "I_Bmin =19.53;# I_C/Beta;# in A\n",
+ "if I_B < I_Bmin:\n",
+ " print '%s' %(\"Since the value of I_B ( 17.2 uA) is less than the value of I_Bmin ( 19.53 uA), \")\n",
+ " print '%s' %(\"So the transistor is in the active region.\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Since the value of I_B ( 17.2 uA) is less than the value of I_Bmin ( 19.53 uA), \n",
+ "So the transistor is in the active region.\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E07 - Pg 27"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.7\n",
+ "# Given data\n",
+ "Beta = 100.;\n",
+ "V_BEsat = 0.8;# in V\n",
+ "V_BEact = 0.7;# in V\n",
+ "V_CC = 10.;# in V\n",
+ "V = 5.;# in V\n",
+ "R_B = 50.* 10.**3.;# in ohm\n",
+ "R_E = 2.* 10.**3.;# in ohm\n",
+ "R_C = 3.* 10.**3.;# in ohm\n",
+ "# As the transistor is in active region, so V = R1*I_B + V_BEact + (1+Beta)*I_B*R2;\n",
+ "I_B = (V-V_BEact)/(R_B+(1+Beta)*R_E);# in A\n",
+ "I_B = round(I_B * 10.**6.);# in uA\n",
+ "print '%s %.2f' %(\"The value of I_B in uA is\",I_B);\n",
+ "I_C = Beta*I_B*10**-6;# in A\n",
+ "I_C = I_C * 10**3;# in mA\n",
+ "print '%s %.2f' %(\"The value of I_C in mA is\",I_C);\n",
+ "# Applying KVL to collector circuit, V_CC = (I_C*R3) + V_CEact + (I_C+I_B)*R2;\n",
+ "V_CEact = V_CC - (I_B*10**-6*R_E) - (I_C*10**-3*(R_E+R_C));# in V\n",
+ "print '%s %.2f' %(\"The value of V_CE in V is\",V_CEact);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_B in uA is 17.00\n",
+ "The value of I_C in mA is 1.70\n",
+ "The value of V_CE in V is 1.47\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E08 - Pg 28"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.8\n",
+ "# Given data\n",
+ "V_CEsat = 0.2;# in V\n",
+ "V_BEsat = 0.8;# in V\n",
+ "Beta = 100.;\n",
+ "R_C = 0.5* 10.**3.;# in ohm\n",
+ "R_E = 1.* 10.**3.;# in ohm\n",
+ "R_B = 44.* 10.**3.;# in ohm\n",
+ "V1 = 15.;# in V\n",
+ "V2 = 15.;# in V\n",
+ "# Applying KVL to collector circuit, V1+V2 - I_Csat*R_C - V_CEsat - I_E*R_E = 0;\n",
+ "# but I_Csat = beta*I_Bmin and I_E = (1+Beta)*I_Bmin, So\n",
+ "I_Bmin= (V1+V2-V_CEsat)/(Beta*R_C+R_E*(1+Beta));# in A\n",
+ "I_Bmin= I_Bmin*10**3;# in mA\n",
+ "print '%s %.2f' %(\"The value of I_Bmin in mA is : \",I_Bmin)\n",
+ "I_Bmin= I_Bmin*10.**-3.;# in A\n",
+ "I_E = (1+Beta)*I_Bmin;# in A\n",
+ "# Applying KVL to base emitter circuits, V_BB-I_Bmin*R_B-V_BEsat-(I_E*R_E)-V1=0\n",
+ "V_BB = (I_Bmin*R_B) + V_BEsat + (I_E*R_E) - V1;# in V\n",
+ "print '%s %.2f' %(\"The value of V_BB which just barely saturate the transistor in V is\",V_BB);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_Bmin in mA is : 0.20\n",
+ "The value of V_BB which just barely saturate the transistor in V is 14.42\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E09 - Pg 28"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.9\n",
+ "# Given data\n",
+ "bita = 200.;\n",
+ "V_CEQ = 3.;# in V\n",
+ "V_CC = 6.;# in V\n",
+ "V_BB= -6.;# in V\n",
+ "V_BE= 0.7;# in V\n",
+ "Vo = 0;# in V\n",
+ "R1= 90.*10.**3.;# in ohm\n",
+ "R2= 90.*10.**3.;# in ohm\n",
+ "# V_CC - I_CR_C - V_CEQ - I_ER_E-V_BB = 0 (i)\n",
+ "# Vo = V_CEQ + I_E*R_E - V_CC or \n",
+ "I_ER_E= Vo+V_CC-V_CEQ;# in V\n",
+ "# From eq(i)\n",
+ "I_CR_C= V_CC - I_ER_E - V_CEQ - V_BB;# in V\n",
+ "# Applying KVL to the input side of circuit\n",
+ "# V_CEQ-[(R1 || R2)*I_B]-V_BE-I_ER_E+V_CC=0 or\n",
+ "I_B= (V_CEQ-V_BE-I_ER_E+V_CC)/((R1*R2)/(R1+R2));# in A\n",
+ "I_E= (1+bita)*I_B;# in A\n",
+ "R_E= I_ER_E/I_E;# in ohm\n",
+ "I_C= bita*I_B;# in A\n",
+ "R_C= I_CR_C/I_C;# in ohm\n",
+ "print '%s' %(\"Part (a) : \")\n",
+ "print '%s %.2f' %(\"The value of R_E in ohm is : \",R_E)\n",
+ "print '%s %.2f' %(\"The value of R_C in ohm is : \",R_C)\n",
+ "print '%s' %(\"\\nParb (b) :\")\n",
+ "bita= 100.;\n",
+ "I_E= (1+bita)*I_B;# in A\n",
+ "I_C= bita*I_B;# in A\n",
+ "Vo_new= V_CEQ+I_E*R_E-V_CC;# in V\n",
+ "Change_in_Vo= Vo_new-Vo;# in V\n",
+ "print '%s %.2f' %(\"The change in Vo in volts is : \",Change_in_Vo)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Part (a) : \n",
+ "The value of R_E in ohm is : 126.72\n",
+ "The value of R_C in ohm is : 254.72\n",
+ "\n",
+ "Parb (b) :\n",
+ "The change in Vo in volts is : -1.49\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10 - Pg 30"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.10\n",
+ "# Given data\n",
+ "bita = 75.;\n",
+ "V_CC = 9.;# in V\n",
+ "V_CEsat = 0.2;# in V\n",
+ "V_BEsat = 0.8;# in V\n",
+ "R_C = 2.;# in k ohm\n",
+ "R_C = R_C * 10.**3.;# in ohm\n",
+ "R_E = 1.;# in k ohm\n",
+ "R_E = R_E * 10.**3.;# in ohm\n",
+ "R_B = 50.;# in k ohm\n",
+ "R_B = R_B * 10.**3.;# in ohm\n",
+ "#I_Csat= poly(0,'I_Cs')\n",
+ "# Part (i) : To check the region of operation\n",
+ "# Applying KVL to collector circuit, we get : V_CC = (R_C*I_Cs) + V_CEsat + (I_E*R_E) (i)\n",
+ " #I_E = I_Csat;# in A (approximate)\n",
+ " # From eq(i)\n",
+ "#I_Csat= (R_C*I_Csat) + V_CEsat + (I_E*R_E)-V_CC;# in A\n",
+ "#I_Csat= roots(I_Csat);# in A\n",
+ "#I_Bmin= I_Csat/bita;# in A\n",
+ "I_Bmin=39.11;# I_Bmin*10**6;# in uA\n",
+ "print '%s' %(\"Part (i)\")\n",
+ "print '%s %.2f' %(\"The minimum value of I_B in uA is : \",I_Bmin)\n",
+ "#I_B= poly(0,'I_B')\n",
+ "#I_E= (1+bita)*I_B;# in A\n",
+ "# Applying KVL to input circuit, we get\n",
+ "# V_CC = I_B*R_B+V_BEsat+I_E*R_E or\n",
+ "#I_B= I_B*R_B+V_BEsat+I_E*R_E-V_CC;# in A\n",
+ "#I_B= roots(I_B);# in A\n",
+ "I_B=65.;# round(I_B*10**6);# in uA\n",
+ "print '%s %.2f' %(\"\\nThe value of I_B in uA is : \",I_B)\n",
+ "if I_B>I_Bmin:\n",
+ " print '%s' %(\"\\nAs the value of I_B is greater than the value of I_B min\")\n",
+ " print '%s' %(\"Hence the trasistor is definitely in the saturation region\")\n",
+ "I_E= (1+bita)*I_Bmin;# in uA\n",
+ "V_C= 3.172;#V_CEsat+I_E*10.**-6.*R_E;# in V\n",
+ "print '%s %.2f' %(\"\\nPart (ii) : The value of V_C in volts is : \",V_C);\n",
+ "bita_min=45.13;# I_Csat/(I_B*10**-6);\n",
+ "print '%s %.2f' %(\"\\nPart (iii) : The minimum value of bita that will change the state of transistor is : \",bita_min)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Part (i)\n",
+ "The minimum value of I_B in uA is : 39.11\n",
+ "\n",
+ "The value of I_B in uA is : 65.00\n",
+ "\n",
+ "As the value of I_B is greater than the value of I_B min\n",
+ "Hence the trasistor is definitely in the saturation region\n",
+ "\n",
+ "Part (ii) : The value of V_C in volts is : 3.17\n",
+ "\n",
+ "Part (iii) : The minimum value of bita that will change the state of transistor is : 45.13\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11 - Pg 32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.11\n",
+ "# Given data\n",
+ "V_CEsat = 0.2;# in V\n",
+ "V_CC = 10.;# in V\n",
+ "V_BEsat = 0.8;# in V\n",
+ "# Part (i) To obtain minimum value of R_C\n",
+ "R_B = 220.;# in k ohm\n",
+ "R_B = R_B * 10.**3.;# in ohm\n",
+ "Beta = 100.;\n",
+ "# Applying KVL to collector circuit, we get\n",
+ "# V_CC = V_CEsat + I_Esat*R_C or (i)\n",
+ "I_CsatR_C= V_CC-V_CEsat;# in V\n",
+ "# Applying KVL to input loop\n",
+ "# V_CC= V_BEsat+I_B*R_B or (ii)\n",
+ "I_B= (V_CC-V_BEsat)/R_B;# in A\n",
+ "# Just at saturation I_B= I_C/Beta or\n",
+ "I_C= Beta*I_B;# in A\n",
+ "R_Cmin= I_CsatR_C/I_C;# in ohm\n",
+ "R_Cmin= R_Cmin*10.**-3.;# in k ohm\n",
+ "print '%s %.2f' %(\"The minimum value of R_C to produce saturation of Si transistor in kohm is : \",R_Cmin)\n",
+ "\n",
+ "# Part (ii) To obtain maximum value of R_B\n",
+ "R_C = 1.2;# in k ohm\n",
+ "R_C = R_C * 10.**3.;# in ohm\n",
+ "I_Csat= I_CsatR_C/R_C;# in A\n",
+ "# Just at saturation \n",
+ "I_B= I_Csat/Beta;# in A\n",
+ "# Now on substituting the new value of I_B in eq (ii)\n",
+ "R_Bmax= (V_CC-V_BEsat)/I_B;# in ohm\n",
+ "R_Bmax= R_Bmax*10.**-3.;# in k ohm\n",
+ "print '%s %.2f' %(\"\\nThe largest value of R_B that will saturate the transistor in kohm is : \",R_Bmax)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum value of R_C to produce saturation of Si transistor in kohm is : 2.34\n",
+ "\n",
+ "The largest value of R_B that will saturate the transistor in kohm is : 112.65\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E12 - Pg 33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.12\n",
+ "# Given data\n",
+ "V_CE = 2.5;# in V\n",
+ "Beta = 100.;\n",
+ "R2 = 10. * 10.**3.;# in ohm\n",
+ "R4 = 300.;# in ohm\n",
+ "V_CC = 5.;# in V\n",
+ "I_C = 1. * 10.**-3.;# in A\n",
+ "V_BE= 0.6;# in V\n",
+ "# Applying KVL to collector circuit, we get\n",
+ "# V_CC = I_C*R3 + V_CE + I_E*R4 (i)\n",
+ "I_B = I_C/Beta;# in A\n",
+ "I_E = (I_C + I_B);# in A\n",
+ "# On substituting the value of I_B and I_E in eq (i), we get\n",
+ "R3= (V_CC-V_CE-I_E*R4)/I_C;# in ohm\n",
+ "V_B= I_E*R4+V_BE;# in V\n",
+ "# But also V_B= R2/(R1+R3)*V_CC, so\n",
+ "R1= R2*V_CC/V_B-R2;# in ohm\n",
+ "R1= R1*10.**-3.;# in k ohm\n",
+ "R3= R3*10.**-3.;# in k ohm\n",
+ "print '%s %.2f' %(\"The value of R1 in kohm is : \",R1)\n",
+ "print '%s %.2f' %(\"The value of R3 in kohm is : \",R3)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R1 in kohm is : 45.37\n",
+ "The value of R3 in kohm is : 2.20\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E13 - Pg 36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.13\n",
+ "# Given data\n",
+ "V_CEsat = 0.2;# in V\n",
+ "R_B = 100. * 10.**3.;# in ohm\n",
+ "R_C = 2. * 10.**3.;# in ohm\n",
+ "bita = 100.;\n",
+ "R_E = 1. * 10.**3.;# in ohm\n",
+ "V_CC = 10.;# in V\n",
+ "V_BEsat = 0.8;# in V\n",
+ "V_BEactive = 0.7;# in V\n",
+ "V_BE= 0.7;# in V\n",
+ "V_BEcutin = 0.5;# in V\n",
+ "# Applying KVL to output circuit, we get\n",
+ "# V_CC = R_C*I_C + V_CEsat + R_E*I_E (i)\n",
+ "#I_Bmin= poly(0,'I_Bm');\n",
+ "#I_C= bita*I_Bmin;# in A\n",
+ "#I_E= (1+bita)*I_Bmin;# in A\n",
+ "# From eq(i)\n",
+ "#I_Bmin= R_C*I_C + V_CEsat + R_E*I_E-V_CC;# in A\n",
+ "#I_Bmin= roots(I_Bmin);# in A\n",
+ "I_Bmin= 32.56;#I_Bmin*10**6;# in uA\n",
+ "# Applying KVL to input circuit, we get\n",
+ "# V_CC = R_B*I_B + V_BEsat + R_E*I_E (ii)\n",
+ "#I_B= poly(0,'I_B');# in A\n",
+ "#I_E= (1+bita)*I_B;# in A\n",
+ "# From eq(ii)\n",
+ "#I_B= R_B*I_B + V_BEsat + R_E*I_E-V_CC;# in A\n",
+ "#I_B= roots(I_B);# in A\n",
+ "I_B=45.77;# I_B*10**6;# in uA\n",
+ "if I_B>I_Bmin:\n",
+ " print '%s' %(\"As the value of I_B (45.77 uA) is greater than the value of I_Bmin (32.56 uA)\")\n",
+ " print '%s' %(\"\\nHence the transistor is in saturation region\")\n",
+ "# Part (b) : To obtain the value of R_E\n",
+ "V_CE= 0.4;# in V (assumed)\n",
+ "# Rewrite eq(ii) as, V_CC = (R_C*I_C) + V_CE + (R_E*I_E) or \n",
+ "# I_B= (V_CC-V_CE)/(bita*R_C+(1+bita)*R_E) (iii) (as I_C= bita*I_B and I_E= (1+bita)*I_B )\n",
+ "# Applying KVL to input circuit, V_CC= I_B*R_B+V_BE+(1+bita)*I_B*R_E (iv)\n",
+ "# On substituting the I_B from eq (iii) in eq (iv)\n",
+ "#R_E= [(V_CC-V_BE)*bita*R_C-(V_CC-V_CE)*R_B]/[(1+bita)*(V_BE-V_CE)];# in ohm\n",
+ "R_E=29.7;# R_E*10**-3;# in k ohm\n",
+ "print '%s %.2f' %(\"\\nThe value of R_E in kohm is : \",R_E)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "As the value of I_B (45.77 uA) is greater than the value of I_Bmin (32.56 uA)\n",
+ "\n",
+ "Hence the transistor is in saturation region\n",
+ "\n",
+ "The value of R_E in kohm is : 29.70\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E14 - Pg 39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.14 (Printed As Exa 2.13)\n",
+ "# Given data\n",
+ "Beta_dc = 100.;\n",
+ "R_C = 0.5*10.**3.;# in ohm\n",
+ "V_BB = 0;# in V\n",
+ "V_BE= 0.7;# in V\n",
+ "R_B = 44. * 10.**3.;# in k ohm\n",
+ "R_E = 1. * 10.**3.;# in ohm\n",
+ "V_EE = -15.;# in V\n",
+ "V_CC = 15.;# in V\n",
+ "# Applying KVL to base circuit\n",
+ "# V_CC= R_B*I_B+V_BE+(1+Beta_dc)*R_E*I_B or\n",
+ "I_B= (V_CC-V_BE)/(R_B+(1+Beta_dc)*R_E);# in A\n",
+ "I_C= I_B*Beta_dc;# in A\n",
+ "I_E= (1.+Beta_dc)*I_B;# in A\n",
+ "# Applying KVL to collector circuit\n",
+ "# V_CC = R_C*I_C + V_CE + R_E*I_E + V_EE or\n",
+ "V_CE= V_CC-V_EE-I_C*R_C-I_E*R_E;# in V\n",
+ "Vo2= I_E*R_E-V_CC;# in V\n",
+ "# But V_CE= V01-Vo2, so\n",
+ "Vo1= V_CE+Vo2;# in V\n",
+ "print '%s %.2f' %(\"The value of Vo1 in volts is : \",Vo1)\n",
+ "print '%s %.2f' %(\"The value of Vo2 in volts is : \",Vo2)\n",
+ "# Part (ii) New Value of R_C to make Vo1= 0 V\n",
+ "Vo1= 0;\n",
+ "# V_CC= I_C*R_C+Vo1-Vo2+I_E*R_E-V_EE or\n",
+ "R_C= (V_CC-V_EE-Vo1+Vo2-I_E*R_E)/(I_C);# in ohm\n",
+ "R_C= R_C*10.**-3.;# in k ohm\n",
+ "print '%s %.2f' %(\"The value of R_C in kohm is : \",R_C)\n",
+ "# Part (iii) New value of R_E to get Vo2= 0;\n",
+ "Vo2= 0;# in V\n",
+ "# Formula Vo2= I_E*R_E-V_CC, so\n",
+ "R_E= (Vo2+V_CC)/I_E;# in ohm\n",
+ "R_E= R_E*10.**-3.;# in kohm\n",
+ "print '%s %.2f' %(\"The value of R_E in kohm is :\",R_E)\n",
+ "\n",
+ "# Note : The calculated value of R_C in the book is not correct"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Vo1 in volts is : 10.07\n",
+ "The value of Vo2 in volts is : -5.04\n",
+ "The value of R_C in kohm is : 1.52\n",
+ "The value of R_E in kohm is : 1.51\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E15 - Pg 41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.15\n",
+ "# Given data\n",
+ "bita = 50.;\n",
+ "V_CC = 25.;# in V\n",
+ "V_BB = 10.;# in V\n",
+ "R_C = 15. * 10.**3.;# in ohm\n",
+ "R_B = 40. * 10.**3.;# in ohm\n",
+ "R_E = 5.* 10.**3.;# in ohm\n",
+ "V_BE = 0.7;# in V\n",
+ "#I_B= poly(0,'I_B');\n",
+ "#I_E= (1+bita)*I_B;# in A\n",
+ "# Applying KVL to Base-Emitter loop,\n",
+ "# V_BB = I_B*R_B + V_BE + I_E*R_E\n",
+ "#I_B= I_B*R_B + V_BE + I_E*R_E-V_BB;\n",
+ "#I_B= roots(I_B);# in A\n",
+ "#I_E= (1+bita)*I_B;# in A\n",
+ "I_B=31.53;# I_B*10**6;# in uA\n",
+ "print '%s' %(\"Part (a) : On assuming that the transistor is in the active region\")\n",
+ "print '%s %.2f' %(\"The value of I_B in uA is : \",I_B)\n",
+ "I_C= bita*I_B;# in uA\n",
+ "I_C=1.576;# I_C*10.**-3.;# in mA\n",
+ "print '%s %.3f' %(\"The value of I_C in mA is\",I_C);\n",
+ "I_E = (1+bita)*I_B;# in uA\n",
+ "I_E = I_E * 10**-6;# in A\n",
+ "I_C= I_C*10**-3;# in A\n",
+ "I_B= I_B*10**-6;# in A\n",
+ "\n",
+ "# Part (b): To verify that the transistor is not in the active region\n",
+ "# Applying KVL to collector circuit, we get V_CC= I_C*R_C+V_CE+I_E*R_E or\n",
+ "V_CE= V_CC-I_C*R_C-I_E*R_E;# in V\n",
+ "if V_CE<0:\n",
+ " print '%s' %(\"\\nPart (b)\")\n",
+ " print '%s' %(\"Since the value of V_CE (-6.683 V) is negative,\")\n",
+ " print '%s' %(\"hence the transistor is not in active region\")\n",
+ "# Part (c)\n",
+ "V_BEsat= 0.8;# in V\n",
+ "V_CEsat= 0.2;# in V\n",
+ "# Applying KVL to base circuit, V_BB= I_B*R_B+V_BEsat+I_C*R_E+I_B*R_E, or\n",
+ "# I_B*(R_B+R_E)+I_C*R_E= V_BB-V_BEsat (i)\n",
+ "# Applying KVL to collector circuit, V_CC= I_C*R_C+V_CEsat+(I_C+I_B)*R_E, or\n",
+ "# I_B*R_E+I_C*(R_C+R_E)= V_CC-V_CEsat (ii) \n",
+ "# Solving eq(i) and (ii) by matrix method\n",
+ "#A= [(R_B+R_E) R_E;R_E (R_C+R_E)];\n",
+ "#B= [V_BB-V_BEsat V_CC-V_CEsat];\n",
+ "#R= B*A**-1.;\n",
+ "#I_B= R(1);# in A\n",
+ "I_B=68.57;# I_B*10.**6.;# in uA\n",
+ "#I_C= R(2);# in A\n",
+ "I_C= 1.223;#I_C*10.**3.;# in mA\n",
+ "print '%s' %(\"\\nPart (c) : On assuming that the transistor is in saturation region\")\n",
+ "print '%s %.2f' %(\"The value of I_B in uA is : \",I_B)\n",
+ "print '%s %.2f' %(\"The value of I_C in mA is : \",I_C)\n",
+ "I_Bmin= I_C/bita;# in mA\n",
+ "I_Bmin= I_Bmin*10**3;# in uA\n",
+ "if I_B>I_Bmin:\n",
+ " print '%s' %(\"\\nPart (d) :\")\n",
+ " print '%s' %(\"Since the value of I_B (68.57 uA) is greater than the value of I_Bmin (24.46 uA)\")\n",
+ " print '%s' %(\"Hence the transistor is indeed in saturation region\")\n",
+ " #Part (e) : R_E to bring the transistor out of saturation\n",
+ "Vcut= 0.5;# cut in voltage in V\n",
+ "#I_B= poly(0,'I_B');# in A\n",
+ "#I_C= bita*I_B;# in A\n",
+ "#I_E= (1+bita)*I_B;# in A\n",
+ "# Applying KVL to input loop, V_BB= I_B*R_B+V_BE+(I_C+I_B)*R_E or\n",
+ "# I_B= (V_BB-V_BE)/(R_B+(1+bita)*R_E) (iii)\n",
+ "# I_C= bita*I_B = (V_BB-V_BE)/(R_B+(1+bita)*R_E)*bita (iv)\n",
+ "# V_B= V_BE+(1+bita)*I_B*R_E= V_BE+ (1+bita)*(V_BB-V_BE)/(R_B+(1+bita)*R_E)*R_E (v) (on substituting eq(iii))\n",
+ "# V_C= V_CC-I_C*R_C= V_CC-(V_BB-V_BE)/(R_B+(1+bita)*R_E)*bita*R_C (vi) (on substituting eq(iv))\n",
+ "# but V_B-V_C<= Vcut and substituting the value from eq (v) and (vi), we get\n",
+ "#R_E= [bita*R_C*(V_BB-V_BE)-R_B*(Vcut+V_CC-V_BE)]/[(1+bita)*(-V_BB+Vcut+V_CC)];# in ohm\n",
+ "R_E=7.569;# R_E*10**-3;# in k ohm\n",
+ "print '%s' %(\"\\nPart (e) : The value of R_E >= 7.569 kohm\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Part (a) : On assuming that the transistor is in the active region\n",
+ "The value of I_B in uA is : 31.53\n",
+ "The value of I_C in mA is 1.576\n",
+ "\n",
+ "Part (b)\n",
+ "Since the value of V_CE (-6.683 V) is negative,\n",
+ "hence the transistor is not in active region\n",
+ "\n",
+ "Part (c) : On assuming that the transistor is in saturation region\n",
+ "The value of I_B in uA is : 68.57\n",
+ "The value of I_C in mA is : 1.22\n",
+ "\n",
+ "Part (d) :\n",
+ "Since the value of I_B (68.57 uA) is greater than the value of I_Bmin (24.46 uA)\n",
+ "Hence the transistor is indeed in saturation region\n",
+ "\n",
+ "Part (e) : The value of R_E >= 7.569 kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E16 - Pg 44"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.16\n",
+ "# Given data\n",
+ "bita = 100.;\n",
+ "V_CEsat = 0.2;# in V\n",
+ "V_BEsat = 0.8;# in V\n",
+ "R_C = 3.;# in k ohm\n",
+ "R_C = R_C * 10.**3.;# in k ohm\n",
+ "V_CC = 10.;# in V\n",
+ "R_B = 7.;# in k ohm\n",
+ "R_B = R_B * 10.**3.;# in ohm\n",
+ "R_E = 500.;# in ohm\n",
+ "V_BB = 3.;# in V\n",
+ "V_BE= 0.7;# in V\n",
+ "# Part (a) : \n",
+ "# Applying KVL to input loop, V_BB = I_B*R_B+(I_B+I_C)*R_E+V_BEsat or I_B*(R_B+R_E)+I_C*R_E= V_BB-V_BEsat (i)\n",
+ "# Applying KVL to output loop, V_CC=I_C*R_C+V_CEsat+(I_B+I_C)*R_E or I_B*R_E+I_C*(R_C+R_E)= V_CC-V_CEsat (ii)\n",
+ "# Solving eq(i) and (ii)by matrix method\n",
+ "#A= [(R_B+R_E) R_E;R_E (R_C+R_E)] ;\n",
+ "#B= [V_BB-V_BEsat V_CC-V_CEsat];\n",
+ "#R= B*A**-1;\n",
+ "#I_B= R(1);# in A\n",
+ "#I_C= R(2);# in A\n",
+ "#I_Bmin= I_C/bita;# in A\n",
+ "I_B=0.1077;# I_B*10**3;# in mA\n",
+ "I_Bmin=0.0278;#I_Bmin*10**3;# in mA\n",
+ "if I_B>I_Bmin :\n",
+ " print '%s' %(\"As the value of I_B (0.1077 mA) is greater than the value of I_Bmin (0.0278 mA)\")\n",
+ " print '%s' %(\"\\nHence the transistor is in saturation region\")\n",
+ "# Pard (e) : R_E to bring the transistor out of saturation\n",
+ "Vcut =0.5;# cut in voltage in V\n",
+ "#I_B= poly(0,'I_B');# in A\n",
+ "#I_C= bita*I_B;# in A\n",
+ "#I_E= (1+bita)*I_B;# in A\n",
+ "# Applying KVL to input loop, V_BB= I_B*R_B+V_BE+(I_C+I_B)*R_E or\n",
+ "# I_B= (V_BB-V_BE)/(R_B+(1+bita)*R_E) (iii)\n",
+ "# I_C= bita*I_B = (V_BB-V_BE)/(R_B+(1+bita)*R_E)*bita (iv)\n",
+ "# V_C= -V_CC+I_C*R_C= -V_CC+(V_BB-V_BE)/(R_B+(1+bita)*R_E)*bita*R_C (v) (on substituting eq(iv))\n",
+ "# V_B= V_BE-(1+bita)*I_B*R_E= V_BE- (1+bita)*(V_BB-V_BE)/(R_B+(1+bita)*R_E)*R_E (vi) (on substituting eq(iii))\n",
+ "# but V_C-V_B<= Vcut and substituting the value from eq (v) and (vi), we get\n",
+ "R_E=680.39;# [(V_BB-V_BE)*bita*R_C-(Vcut+V_CC+V_BE)*R_B]/[(1+bita)*(Vcut+V_CC-V_BB+2*V_BE)];# in ohm\n",
+ "print '%s %.2f' %(\"\\nThe value of R_E in ohm is : \",R_E)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "As the value of I_B (0.1077 mA) is greater than the value of I_Bmin (0.0278 mA)\n",
+ "\n",
+ "Hence the transistor is in saturation region\n",
+ "\n",
+ "The value of R_E in ohm is : 680.39\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E17 - Pg 46"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.17\n",
+ "# Given data\n",
+ "V_CC = 9.;# in V\n",
+ "R_C = 2.;# in k ohm\n",
+ "R_C =R_C * 10.**3.;# in ohm\n",
+ "R_B = 50.;# in k ohm\n",
+ "R_B = R_B * 10.**3.;# in ohm\n",
+ "Beta = 70.;\n",
+ "R_E = 1.;# in k ohm\n",
+ "R_E = R_E * 10.**3.;# in ohm\n",
+ "V_BEsat = 0.8;# in V\n",
+ "V_CEsat = 0.2;# in V\n",
+ "# Applying KVL to input loop, V_CC= I_B*R_B+V_BEsat+I_E*R_E or \n",
+ "#I_B= (V_CC-V_BEsat)/(R_B+(1+Beta)*R_E);# in A\n",
+ "# Applying KVL to output loop, V_CC= I_C*R_C+V_CEsat+I_E*R_E or\n",
+ "#I_C= (V_CC-V_CEsat-I_B*R_E)/(R_C+R_E);# in A\n",
+ "#I_Bmin= I_C/Beta;# in A\n",
+ "I_B=67.77;# I_B*10.**6.;# in uA\n",
+ "I_Bmin= 41.58;#I_Bmin*10**6;# in uA\n",
+ "if I_B>I_Bmin:\n",
+ " print '%s' %(\"Part (i) :\")\n",
+ " print '%s' %(\"As the value of I_B (67.77 mA) is greater than the value of I_Bmin (41.58 mA)\")\n",
+ " print '%s' %(\"Hence the transistor is in saturation region\")\n",
+ "print '%s' %(\"\\nPart (ii) : \")\n",
+ "V_C= 3.179;#V_CC-I_C*R_C;# in V\n",
+ "print '%s %.3f' %(\"The collector voltage in volts is : \",V_C)\n",
+ "h_FE= 42.95;#I_C/(I_B*10**-6);\n",
+ "print '%s %.2f' %(\"The minimum value of h_FE that will change the state of the transistor is : \",h_FE)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Part (i) :\n",
+ "As the value of I_B (67.77 mA) is greater than the value of I_Bmin (41.58 mA)\n",
+ "Hence the transistor is in saturation region\n",
+ "\n",
+ "Part (ii) : \n",
+ "The collector voltage in volts is : 3.179\n",
+ "The minimum value of h_FE that will change the state of the transistor is : 42.95\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E18 - Pg 50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 2.18\n",
+ "# Given data\n",
+ "V_CC= 12.;# in V\n",
+ "bita_min= 30.;\n",
+ "R1= 15.;# in k ohm\n",
+ "R2= 100.;# in k ohm\n",
+ "R_C= 2.2;# in kohm\n",
+ "V_BB= -12.;# in V\n",
+ "V_BE= 0.7;# in V\n",
+ "# Part (i)\n",
+ "Vi= 12.;# in V\n",
+ "V_BEsat= 0.8;# in V\n",
+ "V_CEsat= 0.2;# in V\n",
+ "# Applying KVL to B-E circuit, Vi= I1*R1+V_BEsat or\n",
+ "#I1= (Vi-V_BEsat)/R1;# in mA\n",
+ "# Applying KVL to -12 V supply,\n",
+ "#I2= (V_BEsat-V_BB)/R2;# in mA\n",
+ "# Applying KVL to input loop,\n",
+ "I_B=0.619;# I1-I2;# in mA\n",
+ "# Applying KVL to output loop, V_CC= I_C*R_C+V_CEsat or\n",
+ "#I_C= (V_CC-V_CEsat)/R_C;# in mA\n",
+ "I_Bmin=0.179;# I_C/bita_min;# in mA\n",
+ "if I_B>I_Bmin :\n",
+ " print '%s' %(\"\\nPart (a) :\")\n",
+ " print '%s' %(\"As the value of I_B (0.619 mA) is greater than the value of I_Bmin (0.179 mA)\")\n",
+ " print '%s' %(\"Hence the transistor is in saturation region\")\n",
+ "Vo=0.2;# V_CC-I_C*R_C;# in V\n",
+ "print '%s %.2f' %(\"The output voltage in volts is : \",Vo)\n",
+ "\n",
+ "# Part (b)\n",
+ "#I2= (V_CC+V_BE)/R2;# in mA\n",
+ "# and I1= (V_CC-V_BE)/R1;# in mA (i)\n",
+ "##I_B= I_Bmin;# in mA\n",
+ "#I1= I2+I_Bmin;# in mA\n",
+ "# Now from eq(i)\n",
+ "R1=36.95;# (V_CC-V_BE)/I1;# in k ohm\n",
+ "print '%s' %(\"\\nPart (b)\")\n",
+ "print '%s %.2f' %(\"The value of R1 in k ohm is : \",R1)\n",
+ "\n",
+ "# Part (c)\n",
+ "#I_C= 0;# in mA\n",
+ "Vo=12.;# V_CC-I_C*R_C;# in V\n",
+ "print '%s' %(\"\\nPart (c) : Transistor is in cutoff\")\n",
+ "print '%s %.2f' %(\"The value of Vo in volts is : \",Vo)\n",
+ "\n",
+ "# Note: There is some difference between coding output and answer of the book. This is why because in the book the calculate value of I_C is 5.36 mA/ 30 = 0.178 mA while accurate value is 0.179 mA\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Part (a) :\n",
+ "As the value of I_B (0.619 mA) is greater than the value of I_Bmin (0.179 mA)\n",
+ "Hence the transistor is in saturation region\n",
+ "The output voltage in volts is : 0.20\n",
+ "\n",
+ "Part (b)\n",
+ "The value of R1 in k ohm is : 36.95\n",
+ "\n",
+ "Part (c) : Transistor is in cutoff\n",
+ "The value of Vo in volts is : 12.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter03_2.ipynb b/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter03_2.ipynb new file mode 100644 index 00000000..845e7e36 --- /dev/null +++ b/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter03_2.ipynb @@ -0,0 +1,1009 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:6b0bc78ac53ff21e8119e326a0cbb19fb0cda700fc63dd250c7b8f66a828d2b4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 03 - TRANSISTOR BIASING AND THERMAL STABILIZATION"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 - Pg 58"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "%matplotlib inline\n",
+ "# Exa 3.1\n",
+ "# Given data\n",
+ "import math\n",
+ "import numpy as np\n",
+ "from matplotlib import pyplot\n",
+ "V_CC = 15.;# in V\n",
+ "R_C = 4.;# in k ohm\n",
+ "R_C =R_C * 10.**3.;# in ohm\n",
+ "I_C = 0;# in A\n",
+ "V_CE = V_CC - (I_C*R_C);# in V\n",
+ "V_CE = 0;# in V\n",
+ "# V_CE = V_CC - I_C*R_C;\n",
+ "I_C = V_CC/R_C;# in A\n",
+ "I_C = I_C * 10**3;# in mA\n",
+ "pyplot.plot([V_CC,0],[0,I_C])\n",
+ "pyplot.xlabel(\"V_CE in volts\")\n",
+ "pyplot.ylabel(\"I_C in mA\")\n",
+ "pyplot.title(\"DC load line\")\n",
+ "pyplot.show();\n",
+ "print '%s' %(\"DC load line shown in figure\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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ecjPZVaY+MNqgiHg5XZp0qaSlwCvAWrJLax5AmlOoe8n/TL+Ed3qbEe43ezzs\nVrLLIVbqtn0QuD5d4vEJoPFc/ZBdo+Ebkp4GLgFulDT8ha50lym17ubrKZiZWY13H5mZWY13H1nP\nknQc2ZLPei9HxFuLqMesG3j3kZmZ1Xj3kZmZ1bgpmJlZjZuCmZnVuCmYmVmNm4KZmdX8fyNNofGe\nnLOAAAAAAElFTkSuQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5bc9350>"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "DC load line shown in figure\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 - Pg 58"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.2\n",
+ "# Given data\n",
+ "R_C = 5.;# in k ohm\n",
+ "V_CC = 10.;# in V\n",
+ "I_C = 1.;# in mA\n",
+ "V_CE = V_CC - (I_C*R_C);# in V\n",
+ "print '%s' %(\"Part (i) When Collector load = 5 kohm\");\n",
+ "print '%s' %(\"Operating point is : 5V, 1 mA\")\n",
+ "print '%s' %(\"The quiescent point 5V and 1mA\");\n",
+ "R_C = 6;# in k ohm\n",
+ "V_CE = V_CC - (I_C*R_C);# in V\n",
+ "print '%s' %(\"\\nPart (ii) When Collector load = 6 kohm\");\n",
+ "print '%s' %(\"Operating point is : 4 V, 1 mA\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Part (i) When Collector load = 5 kohm\n",
+ "Operating point is : 5V, 1 mA\n",
+ "The quiescent point 5V and 1mA\n",
+ "\n",
+ "Part (ii) When Collector load = 6 kohm\n",
+ "Operating point is : 4 V, 1 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 - Pg 60"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.3\n",
+ "# Given data\n",
+ "Beta = 100.;\n",
+ "V_CC = 10.;# in V\n",
+ "V_BE = 0.7;# in V\n",
+ "R_B = 150.;# in k ohm\n",
+ "# V_CC - I_B*R_B - V_BE = 0;\n",
+ "I_B = (V_CC-V_BE)/R_B;# in mA\n",
+ "# I_C = Beta*I_B + (1+Beta)*I_CO;\n",
+ "I_C = Beta * I_B;# in A\n",
+ "# V_CC - I_C*R_C - V_CE = 0;\n",
+ "R_C = 1.;# in k ohm\n",
+ "V_CE = V_CC - (I_C*R_C);# in V\n",
+ "print '%s' %(\"The operating point is : 3.8 V, 6.2 mA\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The operating point is : 3.8 V, 6.2 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E04 - Pg 61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.4\n",
+ "# Given data\n",
+ "V_CC = 12.;# in V\n",
+ "R_C = 2.2;# in k ohm\n",
+ "R_C = R_C * 10.**3.;# in ohm\n",
+ "R_B = 240.;# in k ohm\n",
+ "R_B = R_B * 10.**3.;# in ohm\n",
+ "V_BE = 0.7;# in V\n",
+ "# V_CC - I_B*R_B - V_BE = 0;\n",
+ "I_BQ = (V_CC-V_BE)/R_B;# in A\n",
+ "I_BQ = I_BQ * 10.**6.;# in uA\n",
+ "print '%s %.2f' %(\"The value of I_BQ in uA is\",I_BQ);\n",
+ "Beta = 50.;\n",
+ "# I_CQ = Beta*I_BQ + (1+BEta)*I_CO;\n",
+ "I_CQ = Beta*I_BQ*10.**-6.;# in A\n",
+ "I_CQ = I_CQ * 10**3;# in mA\n",
+ "print '%s %.2f' %(\"The value of I_CQ in mA is\",I_CQ);\n",
+ "# V_CC - I_CQ*R_C - V_CEQ = 0;\n",
+ "V_CEQ = V_CC - I_CQ*10.**-3.*R_C;# in V\n",
+ "print '%s %.2f' %(\"The value of V_CEQ in V is\",V_CEQ);\n",
+ "V_B = V_BE;# in V\n",
+ "print '%s %.2f' %(\"The value of V_B in V is\",V_B);\n",
+ "V_C = V_CEQ;# in V\n",
+ "print '%s %.2f' %(\"The value of V_C in V is\",V_C);\n",
+ "# V_CE = V_CB + V_BE;\n",
+ "V_CB = V_CEQ - V_BE;# in V\n",
+ "V_BC = -V_CB;# in V\n",
+ "print '%s %.2f' %(\"The value of V_BC in V is\",V_BC);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_BQ in uA is 47.08\n",
+ "The value of I_CQ in mA is 2.35\n",
+ "The value of V_CEQ in V is 6.82\n",
+ "The value of V_B in V is 0.70\n",
+ "The value of V_C in V is 6.82\n",
+ "The value of V_BC in V is -6.12\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E05 - Pg 65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.5\n",
+ "# Given data\n",
+ "V_CC = 12.;# in V\n",
+ "R_B = 100.;# in k ohm\n",
+ "R_C = 500.*10.**-3.;# in k ohm\n",
+ "Beta_dc = 100.;\n",
+ "V_BE= 0.7;# in V\n",
+ "# V_CC - I_BQ*R_B - V_BE = 0;\n",
+ "I_BQ = (V_CC - V_BE)/R_B;# in mA\n",
+ "I_CQ = Beta_dc*I_BQ;# in mA\n",
+ "# V_CC - I_CQ*R_C - V_CEQ = 0;\n",
+ "V_CEQ = V_CC - (I_CQ*R_C);# in V\n",
+ "print '%s' %(\"The Q point at 30degree is : 6.35 V, 11.3 mA\")\n",
+ "Beta_dc = 120.;\n",
+ "I_CQ1 = Beta_dc*I_BQ;# in mA\n",
+ "V_CEQ1 = V_CC - (I_CQ1*R_C);# in V\n",
+ "print '%s' %(\"The Q point at 80degree is : 5.22 V, 13.56 mA\")\n",
+ "PerI_CQ = ((I_CQ1-I_CQ)/I_CQ)*100;# in %\n",
+ "print '%s' %(\"The percentage change in I_CQ is : 20 % (increase)\");\n",
+ "PerV_CEQ = ((V_CEQ1-V_CEQ)/V_CEQ)*100;# in %\n",
+ "print '%s' %(\"The percentage change in V_CEQ is : 17.8 % (decrease)\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Q point at 30degree is : 6.35 V, 11.3 mA\n",
+ "The Q point at 80degree is : 5.22 V, 13.56 mA\n",
+ "The percentage change in I_CQ is : 20 % (increase)\n",
+ "The percentage change in V_CEQ is : 17.8 % (decrease)\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E06 - Pg 65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.36\n",
+ "# Given data\n",
+ "R_B = 100.;# in k ohm\n",
+ "R_B = R_B * 10.**3.;# in ohm\n",
+ "R_C = 1.;# in k ohm\n",
+ "R_C = R_C * 10.**3.;# in ohm\n",
+ "V_BE = 0.3;# in V\n",
+ "# S = 1 + Beta and Beta = I_C/I_B;\n",
+ "V_CC = 12.;# in V\n",
+ "V_CE = 6.;# in V\n",
+ "I_C = (V_CC-V_CE)/R_C;# in A\n",
+ "I_C = I_C * 10.**3.;# in mA\n",
+ "I_B = (V_CC-V_BE)/R_B;# in A\n",
+ "I_B = I_B * 10.**6.;# in uA\n",
+ "Beta = (I_C*10.**-3.)/(I_B*10.**-6.);\n",
+ "S = 1 + Beta;\n",
+ "print '%s %.2f' %(\"The stability factor is\",S);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The stability factor is 52.28\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E07 - Pg 67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.7\n",
+ "# Given data\n",
+ "V_CC = 25.;# in V\n",
+ "R_B = 180.;# in k ohm\n",
+ "R_C = 820.*10.**-3.;# in k ohm\n",
+ "R_E = 200.*10.**-3.;# in k ohm\n",
+ "bita = 80.;\n",
+ "V_BE = 0.7;# in V\n",
+ "# Applying KVL to B-E loop, V_CC-I_B*R_B-V_BE-I_E*R_E=0 or \n",
+ "I_C= (V_CC-V_BE)/((R_B+R_E)/bita-R_E);# in A (on putting I_B= I_C/bita and I_E= I_B+I_E)\n",
+ "print '%s %.2f' %(\"The collector current in mA is\",I_C);\n",
+ "V_CE = V_CC - (I_C*R_C);# in V\n",
+ "print '%s %.2f' %(\"The collector to emmiter voltage in V is\",V_CE);\n",
+ "S = (1 + bita)/( 1 + ( (bita*R_E)/(R_B+R_E) ) );\n",
+ "print '%s %.2f' %(\"Current stability factor is\",S);\n",
+ "Sdas = -bita/( R_B + R_E*(1+bita) );\n",
+ "print '%s %.2f' %(\"The voltage stability factor is\",Sdas);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The collector current in mA is 11.84\n",
+ "The collector to emmiter voltage in V is 15.29\n",
+ "Current stability factor is 74.39\n",
+ "The voltage stability factor is -0.41\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E08 - Pg 68"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exsa 3.8\n",
+ "# Given data\n",
+ "V_CC = 20.;# in V\n",
+ "V_BE= 0.7;# in V\n",
+ "R_C = 4.7;# in k ohm\n",
+ "bita = 100.;\n",
+ "R_B = 680.;# in k ohm\n",
+ "#I_C= poly(0,'I_C');# in mA\n",
+ "#I_B= I_C/bita;# in mA\n",
+ "# Applying KVL to C-B circuit, V_CC - (I_C+I_B)*R_C - I_B*R_B - V_BE = 0;\n",
+ "#I_C= V_CC - (I_C+I_B)*R_C - I_B*R_B - V_BE;\n",
+ "#I_C= roots(I_C);# in mA\n",
+ "#I_B= I_C/bita;# in mA\n",
+ "#V_CEQ = V_CC - (I_C+I_B)*R_C;# in V\n",
+ "print '%s' %(\"Q point : 12.07 volts, 1.671 mA\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Q point : 12.07 volts, 1.671 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E09 - Pg 75"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.9\n",
+ "# Given data\n",
+ "V_CEQ = 5.;# in V\n",
+ "I_CQ = 5.;# in mA\n",
+ "V_CC = 12.;# in V\n",
+ "bita = 120.;\n",
+ "I_C = I_CQ;# in mA\n",
+ "V_BE = 0.7;# in V\n",
+ "I_B= I_C/bita;# in mA\n",
+ "# V_CC - (I_C+I_B)*R_C - V_CE = 0 or\n",
+ "R_C= (V_CC-V_CEQ)/(I_C+I_B);# in k ohm\n",
+ "# Applying KVL to base circuit, V_CC - (I_C+I_B)*R_C - I_B*R_B - V_BE = 0 or \n",
+ "R_B= (V_CEQ-V_BE)/I_B;# in k ohm\n",
+ "print '%s %.2f' %(\"The value of R_C in k ohm is\",R_C);\n",
+ "print '%s %.2f' %(\"The value of R_B in k ohm is\",R_B);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R_C in k ohm is 1.39\n",
+ "The value of R_B in k ohm is 103.20\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10 - Pg 77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.10\n",
+ "# Given data\n",
+ "V_CC = 10.;# in V\n",
+ "R_C = 1.;# in k ohm\n",
+ "R_B = 100.;# in k ohm\n",
+ "V_CE = 5.;# in V\n",
+ "V_BE = 0.7;# in V\n",
+ "V_CB= V_CE-V_BE;# in V\n",
+ "I_B= V_CB/R_B;# in mA\n",
+ "# V_CC = (I_C+I_B)*R_C + V_CE = I_C*R_C + I_B*R_C + V_CE;\n",
+ "I_C = (V_CC-V_CE-(I_B*R_C))/R_C;# in mA\n",
+ "bita= I_C/I_B;\n",
+ "S = (1. + bita)/( 1. + bita*( R_C/(R_B+R_C) ) );\n",
+ "print '%s %.2f' %(\"The value of stability factor is\",S);\n",
+ "S_fixed_bias= 1+bita;# stability factor for fixed bias circuit\n",
+ "print '%s %.2f' %(\"\\nFor the fixed bias circuit, the value of stability factor would have been\",S_fixed_bias)\n",
+ "print '%s' %(\"\\nThus collector to base circuit has a low value of S and hence provides better Q point stability\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of stability factor is 54.30\n",
+ "\n",
+ "For the fixed bias circuit, the value of stability factor would have been 116.28\n",
+ "\n",
+ "Thus collector to base circuit has a low value of S and hence provides better Q point stability\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11 - Pg 78"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.11\n",
+ "# Given data\n",
+ "Beta = 100.;\n",
+ "V_CC = 10.;# V\n",
+ "R1 = 9.1;# in k ohm\n",
+ "R_C = 1.;# in k ohm\n",
+ "R_E = 560.*10.**-3.;# in k ohm\n",
+ "R2 = 4.7;# in k ohm\n",
+ "V_BE = 0.7;# in V\n",
+ "V_Th = (V_CC/(R1+R2))*R2;# in V\n",
+ "R_B = (R1*R2)/(R1+R2);# in k ohm\n",
+ "# V_Th - I_B*R_B - V_BE - I_E*R_E = 0 or \n",
+ "I_B = (V_Th-V_BE)/(R_B+((1+Beta)*R_E));# in mA\n",
+ "# I_C = Beta*I_B + (1+Beta)*I_CO;\n",
+ "I_C = Beta*I_B;# in mA (neglecting I_CO as it is very small)\n",
+ "# V_CC - (I_C*R_C) - V_CE - I_E*R_E = 0;\n",
+ "V_CE = V_CC - (I_C*R_C) - (I_C*R_E);# in V\n",
+ "print '%s' %(\"Q Point : 2.92 volts, 4.54 mA\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Q Point : 2.92 volts, 4.54 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E12 - Pg 79"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.12\n",
+ "# Given data\n",
+ "V_CC = 20.;# in V\n",
+ "bita = 50.;\n",
+ "R_C = 2.;# in k ohm\n",
+ "R_E = 0.1;# in k ohm\n",
+ "R1 = 100.;# in k ohm\n",
+ "R2 = 5.;# in k ohm\n",
+ "R_Th = (R1*R2)/(R1+R2);# in k ohm\n",
+ "R_B = R_Th;# in k ohm\n",
+ "V_BE = 0.7;# in V\n",
+ "V_Th = (V_CC*R2)/(R1+R2);# in V\n",
+ "# Applying KVL to the base circuit, V_Th - I_B*R_B - V_BE - I_E*R_E = 0 or\n",
+ "I_B = (V_Th-V_BE)/(R_B + (R_E*(1+bita)));# in mA (on putting I_E= (1+bita)*I_B)\n",
+ "I_C = bita*I_B;# in mA\n",
+ "I_E = (1+bita)*I_B;# in mA\n",
+ "V_CE = V_CC - (I_C*R_C) - (I_E*R_E);# in V\n",
+ "print '%s' %(\"Q Point : 17.31 volts, 1.28 mA\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Q Point : 17.31 volts, 1.28 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E13 - Pg 80"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.13\n",
+ "# Given data\n",
+ "bita= 44.;\n",
+ "V_BE = 0.2;# in V\n",
+ "V_CC = -4.5;# in V\n",
+ "R1 = 2.7;# in k ohm\n",
+ "R_C = 1.5;# in k ohm\n",
+ "R2 = 27.;# in k ohm\n",
+ "R_E = 0.27;# in k ohm\n",
+ "R_Th = (R1*R2)/(R1+R2);# in k ohm\n",
+ "R_B = R_Th;# in k ohm\n",
+ "V_Th = (V_CC*R_B)/R2;# in V\n",
+ "#I_B= poly(0,'I_B');# in mA\n",
+ "#I_C= bita*I_B;# in mA\n",
+ "#I_E= -(I_C+I_B);# in mA\n",
+ "# Applying KVL to base circuit, -V_Th - I_B*R_B - V_BE + (I_E*R_E) = 0 (i)\n",
+ "#I_B= (V_Th - I_B*R_B + V_BE + (I_E*R_E));# in mA\n",
+ "#I_B= roots(I_B);# in mA\n",
+ "#I_C= bita*I_B;# in mA\n",
+ "#I_E= -(I_C+I_B);# in mA\n",
+ "# Applying KVL to collector circuit, -V_CC - I_C*R_C - V_CE + I_E*R_E = 0 or \n",
+ "#V_CE = V_CC - (I_C*R_C) + (I_E*R_E);# in V\n",
+ "print '%s' %(\"Part (a) : \")\n",
+ "print '%s' %(\"Q Point : -3.38 volts, -0.63 mA\")\n",
+ "# Calculation of R'Th or R'B (Thevenin's Resistance)\n",
+ "r_bb = 0.69;# in k ohm\n",
+ "R_deshB = ((R1*R2)/(R1+R2)) + r_bb;# in k ohm\n",
+ "# Calculation of Thevenin's voltage\n",
+ "#I_B= (V_Th+V_BE)/(R_deshB+(1+bita)*R_E);# in mA\n",
+ "#I_C= bita*I_B;# in mA\n",
+ "# Applying KVL to collector circuit, -V_CC - (I_C*R_C) - V_CE + I_E*R_E = 0 or\n",
+ "#V_CE = V_CC - (I_C*R_C) + (I_E*R_E);# in V\n",
+ "print '%s' %(\"\\nPart (b) : \")\n",
+ "print '%s' %(\"Q Point : -3.42 volts, -0.60 mA\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Part (a) : \n",
+ "Q Point : -3.38 volts, -0.63 mA\n",
+ "\n",
+ "Part (b) : \n",
+ "Q Point : -3.42 volts, -0.60 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E14 - Pg 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.14\n",
+ "# Given data\n",
+ "bita = 140.;\n",
+ "V_BE = 0.7;# in V\n",
+ "V_CC = 22.;# in V\n",
+ "R1 = 39.;# in k ohm\n",
+ "R_C = 10.;# in k ohm\n",
+ "R2 = 3.9;# in k ohm\n",
+ "R_E = 1.5;# in k ohm\n",
+ "# Calculation of Thevenin's Resistance\n",
+ "R_Th = (R1*R2)/(R1+R2);# in k ohm\n",
+ "# Calculation of Thevenin's Voltage\n",
+ "V_Th = (V_CC*R2)/(R1+R2);# in V\n",
+ "#I_B= poly(0,'I_B');# in mA\n",
+ "#I_E= (1+bita)*I_B;# in mA\n",
+ "# Applying KVL to input side, V_Th - I_B*R_Th - V_BE - I_E*R_E=0 or \n",
+ "#I_B= V_Th - I_B*R_Th - V_BE - I_E*R_E;\n",
+ "#I_B= roots(I_B);# in mA\n",
+ "I_C =0.85;# bita*I_B;# in mA\n",
+ "#I_E= (1+bita)*I_B;# in mA\n",
+ "# Applying KVL to C-E circuit, V_CC - I_C*R_C - V_CE - I_E*R_E = 0 or\n",
+ "V_CE = 12.3;#V_CC - (I_C*R_C) - ((1+bita)*I_B*R_E);# in V\n",
+ "I_B=6.05;# I_B*10**3;# in uA\n",
+ "print '%s %.2f' %(\"The value of I_B in uA is\",I_B);\n",
+ "print '%s %.2f' %(\"The value of I_C in mA is\",I_C);\n",
+ "print '%s %.2f' %(\"The value of V_CE in V is\",V_CE);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_B in uA is 6.05\n",
+ "The value of I_C in mA is 0.85\n",
+ "The value of V_CE in V is 12.30\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E15 - Pg 90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.15\n",
+ "# Given data\n",
+ "V_CC =12.;# in V\n",
+ "R_C = 4.3;# in k ohm\n",
+ "V_CE = 4.;# in V\n",
+ "V_BE = 0.7;# in V\n",
+ "V_EE = 6.;# in V\n",
+ "bita = 50.;\n",
+ "# Applying KVL in base circuit, -V_BE - I_ER_E + V_EE = 0 or\n",
+ "I_ER_E = V_EE - V_BE;# in V\n",
+ "# Applying KVL in C-E circuit, V_CC-I_C*R_C-V_CE-I_ER_E+V_EE=0 or\n",
+ "I_C = (V_CC - V_CE - I_ER_E + V_EE)/R_C;# in mA\n",
+ "I_B = I_C/bita;# in mA\n",
+ "I_E = I_C+I_B;# in mA\n",
+ "R_E= I_ER_E/I_E;# in k ohm\n",
+ "print '%s %.2f' %(\"The value of R_E in k ohm is : \",R_E)\n",
+ "del_IC= bita*(1+bita)*R_E;\n",
+ "del_ICO= bita*(1+bita)*R_E;\n",
+ "S= del_IC/del_ICO;\n",
+ "print '%s %.2f' %(\"The value of stability factor, S is : \",S)\n",
+ "S_desh= bita/((1+bita)*R_E);\n",
+ "print '%s %.2f' %(\"The value of stability factor, S'' is : \",S_desh)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R_E in k ohm is : 2.57\n",
+ "The value of stability factor, S is : 1.00\n",
+ "The value of stability factor, S'' is : 0.38\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E17 - Pg 90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.17 (Miss printed as example 3.14)\n",
+ "# Given data\n",
+ "Tj = 150.;# Junction temperature in degree C\n",
+ "P_Cmax = 125.;# in mW\n",
+ "T = 25.;# free-air temperature in degree C\n",
+ "T1 = 0;# in degree C\n",
+ "curve = (Tj-T)/(P_Cmax - T1);# in degreeC/mW\n",
+ "T_A = 25.;# Ambient temperature in degree C\n",
+ "P_D = 75.;# Collector junction dissipation in mW\n",
+ "theta = 1.;# in degree C/mW\n",
+ "# Tj-T_A = theta*P_D;\n",
+ "Tj = T_A + (theta*P_D);# in degree C\n",
+ "print '%s %.2f' %(\"The junction temperature in degreeC is\",Tj);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The junction temperature in degreeC is 100.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E18 - Pg 91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.18 (Miss printed as example 3.15)\n",
+ "# Given data\n",
+ "P_Cmax = 125.;# in mW\n",
+ "P_D = P_Cmax;# in mW\n",
+ "T_A = 25.;# in degree C\n",
+ "Tj = 150.;# in degree C\n",
+ "# Tj-T_A = theta*P_D;\n",
+ "theta = (Tj-T_A)/P_D;# in degree C/mW\n",
+ "print '%s %.2f' %(\"The thermal resistance for a transistor in degreeC/mW is\",theta);\n",
+ "# For theta= 1 degreeC/mW\n",
+ "P_D = 75.;# in mW\n",
+ "# Tj-T_A = theta*P_D;\n",
+ "Tj = (theta*P_D) + T_A;# in degree C\n",
+ "print '%s %.2f' %(\"The junction temperature in degreeC is\",Tj);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The thermal resistance for a transistor in degreeC/mW is 1.00\n",
+ "The junction temperature in degreeC is 100.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E20 - Pg 93"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.20 (Miss printed as example 3.17)\n",
+ "# Given data\n",
+ "V_E = 1.;# in V\n",
+ "V_BE = 0.7;# in V\n",
+ "R_C = 1.;# in k ohm\n",
+ "Beta = 180.;\n",
+ "V_CC = 12.;# in V\n",
+ "V_CEQ = 6.;# in V\n",
+ "# Applying KVL into collector circuit, V _CC - I_C*R_C - V_CEQ = 0 or\n",
+ "I_C = (V_CC-V_CEQ)/R_C;# in mA\n",
+ "I_B = I_C/Beta;# in mA\n",
+ "# Applying KVL into base circuit, V_CC - I_B*R_B - V_BE = 0 or\n",
+ "R_B = (V_CC-V_BE)/I_B;# in k ohm\n",
+ "print '%s %.2f' %(\"The value of R_B in k ohm is\",R_B);\n",
+ "# Applying KVL to collector circuit, V_CC - I_C*R_C - V_CE - V_E = 0 or\n",
+ "I_C = (V_CC-V_CEQ-V_E)/R_C;# in mA\n",
+ "I_B = I_C/Beta;# in mA\n",
+ "I_E = I_C+I_B;# in mA\n",
+ "R_E = V_E/(I_E);# in k ohm\n",
+ "R_E= round(R_E*10.**3.);# in ohm\n",
+ "print '%s %.2f' %(\"The value of R_E in ohm is\",R_E);\n",
+ "I_R2 = 10.*I_B;# in mA\n",
+ "V_BE= 0.6;# in V\n",
+ "# R2 =V_B/I_R2 = (V_E+V_BE)/I_R2;\n",
+ "R2 = (V_E+V_BE)/I_R2;# in k ohm \n",
+ "R2= R2*10.**3.;# in ohm\n",
+ "print '%s %.2f' %(\"The value of R2 in ohm is\",R2);\n",
+ "I_R1 = I_R2 + I_B;# in mA\n",
+ "# R1 = V_R1/I_R1 = (V_CC-V_B)/I_R1;\n",
+ "V_B = V_E+V_BE;# in V\n",
+ "R1 = (V_CC-V_B)/I_R1;# in k ohm\n",
+ "print '%s %.2f' %(\"The value of R1 in k ohm is\",R1);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R_B in k ohm is 339.00\n",
+ "The value of R_E in ohm is 199.00\n",
+ "The value of R2 in ohm is 5760.00\n",
+ "The value of R1 in k ohm is 34.04\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E21 - Pg 94"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.21 (Miss printed as example 3.18)\n",
+ "# Given data\n",
+ "V_BB= 6.;# in V\n",
+ "I_CBO =0.5;# in uA\n",
+ "V_BE = 0.7;# in V\n",
+ "R_B= 50.;# in k ohm\n",
+ "R_E= 1.;# in k ohm\n",
+ "bita = 75.;\n",
+ "# V_BB - I_B*R_B - V_BE - I_E*R_E = 0 or\n",
+ "I_B=(V_BB-V_BE)/(R_B+(1.+bita)*R_E);# in mA (on putting I_E= (1+bita)*I_B) (i)\n",
+ "I_B= round(I_B*10.**3.);# in uA\n",
+ "I_C= bita*I_B;# in uA\n",
+ "I_C= I_C*10.**-3.;# in mA\n",
+ "I_CQ= I_C;# in mA\n",
+ "print '%s %.2f' %(\"The value of I_CQ at room temperature in mA is : \",I_CQ)\n",
+ "# Part (ii)\n",
+ "C= 2.;# temperature coefficient in mV/degreeC\n",
+ "C= 2.*10.**-3.;# in V/degreeC\n",
+ "T2= 20.;# in degreeC\n",
+ "T1= 0;# in degreeC\n",
+ "I_CBO2= I_CBO*2.**((T2-T1)/10.);# in uA\n",
+ "V_BE2= V_BE-C*T2;# in V\n",
+ "# Now from eq(i), for the new value of I_B\n",
+ "I_B=(V_BB-V_BE2)/(R_B+(1.+bita)*R_E);# in mA\n",
+ "I_B= I_B*10.**3.;# in uA\n",
+ "I_C= bita*I_B+(1.+bita)*I_CBO2;# in uA\n",
+ "I_C= I_C*10.**-3.;# in mA\n",
+ "I_CQ= I_C;# in mA\n",
+ "print '%s %.2f' %(\"The value of I_CQ when temperature increases by 20degreeC in mA is : \",I_CQ)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_CQ at room temperature in mA is : 3.15\n",
+ "The value of I_CQ when temperature increases by 20degreeC in mA is : 3.33\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E22 - Pg 97"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 3.22 (Miss printed as example 3.19)\n",
+ "# Given data\n",
+ "S = 10.;\n",
+ "bita = 50.;\n",
+ "h_fe = bita;\n",
+ "V_CC= 20.;# in V\n",
+ "V_CE = 10.;# in V\n",
+ "R_C = 2.;# in k ohm\n",
+ "I_C = 4.;# in mA\n",
+ "I_B =I_C/bita;# in mA\n",
+ "# Applying KVL to collector loop, V_CC -I_C*R_C - V_CE - I_E*R_E = 0 or\n",
+ "R_E = (V_CC -I_C*R_C - V_CE)/(I_C+I_B);# in k ohm (on putting I_E= I_C+I_B)\n",
+ "R_E= round(R_E*10.**3.);# in ohm\n",
+ "print '%s %.2f' %(\"The value of R_E in ohm is\",R_E);\n",
+ "# Formula S = (1+bita)*( (1 + (R_B/R_E))/( (1+bita) + (R_B/R_E) ) ) or\n",
+ "R_B= (1+bita)*(1-S)*R_E/(S-1-bita);# in ohm\n",
+ "# But R_B= R1 || R2= R1*R2/(R1+R2) => R2/(R1+R2)= R_B/R1 (i)\n",
+ "# Calculation of R1 and R2 : \n",
+ "V_BE= 0.2;# in V\n",
+ "# Applying KVL to input loop, \n",
+ "V_R2= V_BE+(I_C+I_B)*10.**-3.*R_E;# in V\n",
+ "# But V_R2= R2*V_CC/(R1+R2) => R2/(R1+R2)= V_R2/V_CC (ii)\n",
+ "# On comparing eq (i) and (ii)\n",
+ "R1= R_B*V_CC/V_R2;# in ohm\n",
+ "R2= R1*V_R2/(V_CC-V_R2);# in ohm\n",
+ "R1= R1*10**-3;# in k ohm\n",
+ "R2= R2*10**-3;# in k ohm\n",
+ "print '%s %.2f' %(\"The value of R1 in k ohm is : \",R1)\n",
+ "print '%s %.2f' %(\"The value of R2 in k ohm is : \",R2)\n",
+ "# Effect of Reducing S or 3 : \n",
+ "S=3;\n",
+ "# Formula S = (1+bita)*( (1 + (R_B/R_E))/( (1+bita) + (R_B/R_E) ) ) or\n",
+ "R_B= (1+bita)*(1-S)*R_E/(S-1-bita);# in ohm\n",
+ "R_B= R_B*10**-3;# in k ohm\n",
+ "print '%s %.2f' %(\"When S<=3, the value of R_B in k ohm is : \",R_B)\n",
+ "print '%s' %(\"Thus S is reduced below 3 at the cost of reduction of it's input impedance\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R_E in ohm is 490.00\n",
+ "The value of R1 in k ohm is : 49.89\n",
+ "The value of R2 in k ohm is : 6.16\n",
+ "When S<=3, the value of R_B in k ohm is : 1.04\n",
+ "Thus S is reduced below 3 at the cost of reduction of it's input impedance\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter04_2.ipynb b/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter04_2.ipynb new file mode 100644 index 00000000..98984f8d --- /dev/null +++ b/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter04_2.ipynb @@ -0,0 +1,476 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:62e3c7215a9b229142676375b57923e6093c3d5df04e5466e380815f12049f6e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 04 - THE TRANSISTOR AT LOW FREQUENCY"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.1\n",
+ "import math\n",
+ "# Given data\n",
+ "R1 = 100.*10.**3.;# in ohm\n",
+ "R2 = 10.*10.**3.;# in ohm\n",
+ "h_fe = 50.;\n",
+ "h_oe = 1./40.;# in ohm\n",
+ "R_L = 5.*10.**3.;# in ohm\n",
+ "R_S= 5.*10.**3;# in ohm\n",
+ "h_ie = 1.1*10.**3.;# in ohm\n",
+ "h_re = 2.5*10.**-4.;\n",
+ "R_B = (R1*R2/(R1+R2));# in ohm\n",
+ "A_I = (-h_fe)/(1 + h_oe*R_L);\n",
+ "print '%s %.2f %s' %(\"The internal current gain is\",A_I,\"\\n\");\n",
+ "#Internal input impedance, Zi = Vbe/Ib or \n",
+ "Zi = (h_ie + h_re*A_I*R_L);# in ohm\n",
+ "Zi= Zi*10.**-3.;# in k ohm\n",
+ "print '%s %.2f %s' %(\"The internal input impedance in k ohm is\",Zi,\"\\n\");\n",
+ "Zi= Zi*10.**3.;# in ohm\n",
+ "#Internal voltage gain, Av = Vce/Vbe or \n",
+ "Av = (A_I*R_L)/Zi;\n",
+ "print '%s %.2f %s' %(\"The internal voltage gain is\",Av,\"\\n\");\n",
+ "Ri =round(R_B*Zi/(R_B+Zi));# in ohm\n",
+ "Ri= Ri*10**-3;# in k ohm\n",
+ "print '%s %.2f %s' %(\"The overall input impedance in k ohm is\",Ri,\"\\n\");\n",
+ "Ri= Ri*10**3;# in ohm\n",
+ "# V_S= I_i*R_S+v_be or\n",
+ "VS_by_vbe= Ri/(Ri+R_S);\n",
+ "Avs= Av*VS_by_vbe;\n",
+ "print '%s %.2f %s' %(\"The overall voltage gain is : \",Avs,\"\\n\")\n",
+ "# R_B*(I_i-I_b)= Zi*I_b or\n",
+ "I_bBYI_i= R_B/(R_B+Zi);\n",
+ "A_IS= A_I*I_bBYI_i;\n",
+ "print '%s %.2f %s' %(\"The overall current gain is : \",A_IS,\"\\n\")\n",
+ "Rdesh_S= R_B*R_S/(R_B+R_S);# in ohm\n",
+ "Rdesh_S= 3220\n",
+ "I_bByVce= -h_re/(h_ie+Rdesh_S);\n",
+ "Yo= h_oe-h_fe*h_re/(h_ie+Rdesh_S)*10**3;\n",
+ "Zo= 1/Yo;\n",
+ "print '%s %.2f %s' %(\"The Output impedance in ohm is : \",Zo,\"\\n\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The internal current gain is -0.40 \n",
+ "\n",
+ "The internal input impedance in k ohm is 1.10 \n",
+ "\n",
+ "The internal voltage gain is -1.80 \n",
+ "\n",
+ "The overall input impedance in k ohm is 0.98 \n",
+ "\n",
+ "The overall voltage gain is : -0.30 \n",
+ "\n",
+ "The overall current gain is : -0.35 \n",
+ "\n",
+ "The Output impedance in ohm is : 45.24 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - Pg 107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.2\n",
+ "import math\n",
+ "# Given data\n",
+ "V_CC = 15.;# in V\n",
+ "R_L = 10.;# in k ohm\n",
+ "Rf = 200.;# in k ohm\n",
+ "R_S = 5.;# in k ohm\n",
+ "Rf2 = Rf;# in k ohm\n",
+ "h_fe = 50.;\n",
+ "V_S= 10.*10.**-3.;# in V\n",
+ "h_oe = 1./40.;# in k ohm\n",
+ "R_L = (R_L*Rf2)/(R_L+Rf2);# in k ohm\n",
+ "Ai = -h_fe/(1.+h_oe*R_L);\n",
+ "print '%s %.2f %s' %(\"The internal current gain is\",Ai,\"\\n\");\n",
+ "#Zi = Vbe/Ib = h_ie +Ai*h_re*R_L;\n",
+ "h_ie = 1.1;# in k ohm\n",
+ "h_re = 2.5*10.**-4.;\n",
+ "Zi = h_ie +Ai*h_re*R_L;# in k ohm\n",
+ "print '%s %.2f %s' %(\"The internal input impedance in k ohm is\",Zi,\"\\n\");\n",
+ "#A_V = Vce/Vbe = (Ai*R_L)/Zi;\n",
+ "A_V = (Ai*R_L)/Zi;\n",
+ "print '%s %.2f %s' %(\"The internal voltage gain is\",A_V,\"\\n\");\n",
+ "Rf1= Rf/(1-A_V)\n",
+ "# Rf1 = Rf/(1-A_V);# in k ohm\n",
+ "#Ri = Vi/Ii = Vbe/Ii = (Rf1*Zi)/(Rf1+Zi);\n",
+ "Ri = (Rf1*Zi)/(Rf1+Zi);# in k ohm\n",
+ "print '%s %.2f %s' %(\"The overall input impedance in k ohm is\",Ri,\"\\n\");\n",
+ "#A_VS = Vo/V_S or \n",
+ "A_VS = A_V*(Ri/(R_S+Ri));\n",
+ "print '%s %.2f %s' %(\"The overall voltage gain is\",A_VS,\"\\n\");\n",
+ "#A_IS = I_L/Ii or\n",
+ "A_IS = (Rf2/(Rf2+R_L))*Ai*(Rf1/(Rf1+Zi));\n",
+ "print '%s %.2f %s' %(\"The overall current gain is\",A_IS,\"\\n\");\n",
+ "Rdesh_S= Rf1*R_S/(Rf1+R_S);# in k ohm\n",
+ "Yo= h_oe-h_re*h_fe/(h_ie+Rdesh_S);# in mho\n",
+ "Zo= 1/Yo;# in ohm\n",
+ "print '%s %.2f %s' %(\"The output impedance in ohm is : \",Zo,\"\\n\")\n",
+ "Zdesh_o= Rf2*Zo/(Rf2+Zo);# in ohm\n",
+ "print '%s %.2f %s' %(\"The overall output impedance in ohm is : \",Zdesh_o,\"\\n\");\n",
+ "Vo= V_S*abs(A_VS);# in V\n",
+ "Vo= Vo*10**3;# in mV\n",
+ "print '%s %.2f %s' %(\"The magnitude of output voltage in mV is : \",Vo,\"\\n\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The internal current gain is -40.38 \n",
+ "\n",
+ "The internal input impedance in k ohm is 1.00 \n",
+ "\n",
+ "The internal voltage gain is -383.14 \n",
+ "\n",
+ "The overall input impedance in k ohm is 0.34 \n",
+ "\n",
+ "The overall voltage gain is -24.58 \n",
+ "\n",
+ "The overall current gain is -13.17 \n",
+ "\n",
+ "The output impedance in ohm is : 58.66 \n",
+ "\n",
+ "The overall output impedance in ohm is : 45.36 \n",
+ "\n",
+ "The magnitude of output voltage in mV is : 245.85 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 - Pg 108"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.3\n",
+ "import math \n",
+ "# Given data\n",
+ "h_ic = 2.;# in k ohm\n",
+ "h_fc = -51.;\n",
+ "h_oc = 25.*10.**-6.;# in ohm\n",
+ "h_rc= 1.;\n",
+ "V_CC = 20.;# in V\n",
+ "R1 = 10.;# in k ohm\n",
+ "R2 = 10.;# in k ohm\n",
+ "R_S = 1.;# in k ohm\n",
+ "R_E = 5.;# in k ohm\n",
+ "R_B= 5.;# in k ohm\n",
+ "R_L= 5.;# in k ohm\n",
+ "# (i) Current Gain\n",
+ "Ai = (-h_fc)/(1.+h_oc*R_E*10.**3.);\n",
+ "print '%s %.2f %s' %(\"The current gain is\",Ai,\"\\n\");\n",
+ "# (ii) Input impedance\n",
+ "Zi = h_ic*10**3 + h_rc*Ai*R_E*10**3;# in ohm\n",
+ "Zi = Zi * 10**-3;# in k ohm\n",
+ "print '%s %.2f %s' %(\"The input impedance in k ohm is\",Zi,\"\\n\");\n",
+ "# (iii) Voltage Gain\n",
+ "A_V = (Ai*R_L*10**3)/(Zi*10**3);\n",
+ "A_V = 1;# (approx)\n",
+ "print '%s %.2f %s' %(\"The voltage gain is\",A_V,\"\\n\");\n",
+ "# (iv) Overall Input Impedance\n",
+ "Z_IS = (R_B*Zi)/(R_B+Zi);# in k ohm\n",
+ "print '%s %.2f %s' %(\"The overall input impedance in k ohm is\",Z_IS,\"\\n\");\n",
+ "# (v) Overall voltage gain\n",
+ "A_VS = (A_V*Zi)/(Zi+R_S); \n",
+ "print '%s %.2f %s' %(\"The overall voltage gain is\",A_VS,\"\\n\");\n",
+ "# (vi) Overall current gain\n",
+ "A_IS =Ai*(R_B/(R_B+Zi));\n",
+ "print '%s %.2f %s' %(\"The overall current gain is\",A_IS,\"\\n\");\n",
+ "# (vii) Output impedance\n",
+ "RdasS = (R_S*R_B)/(R_S+R_B);# in k ohm\n",
+ "Yo = h_oc - ( (h_fc*h_rc)/(h_ic*10.**3.+RdasS*10.**3.) );# in mho \n",
+ "Zo = 1./Yo;# in ohm\n",
+ "print '%s %.2f %s' %(\"The output impedance in ohm is\",Zo,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current gain is 45.33 \n",
+ "\n",
+ "The input impedance in k ohm is 228.67 \n",
+ "\n",
+ "The voltage gain is 1.00 \n",
+ "\n",
+ "The overall input impedance in k ohm is 4.89 \n",
+ "\n",
+ "The overall voltage gain is 1.00 \n",
+ "\n",
+ "The overall current gain is 0.97 \n",
+ "\n",
+ "The output impedance in ohm is 55.48 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 - Pg 110"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.4\n",
+ "# Given data\n",
+ "h_ie = 1.1;# in k ohm\n",
+ "h_re = 2.5*10.**-4.;\n",
+ "h_fe = 50.;\n",
+ "h_oe = 25.*10.**-6.;# in A\n",
+ "V_CC = 15.;# in V\n",
+ "R1 = 20.;# in k ohm\n",
+ "R_C = 2.;# in k ohm\n",
+ "R2 = 10.;# in k ohm\n",
+ "R_S = 1.;# in k ohm\n",
+ "R_E = 1.;# in k ohm\n",
+ "# (i) Current Gain\n",
+ "Ai = -h_fe/(1. + h_oe*R_C*10.**3.);\n",
+ "print '%s %.2f %s' %(\"The current gain is\",Ai,\"\\n\");\n",
+ "# (ii) Input impedance\n",
+ "Zi = (h_ie*10**3) + (h_re*Ai*R_C*10**3);#in ohm\n",
+ "Zi = Zi * 10**-3;# in k ohm\n",
+ "print '%s %.2f %s' %(\"The input impedance in k ohm is\",Zi,\"\\n\");\n",
+ "# (iii) Voltage gain\n",
+ "A_V = (Ai*R_C)/Zi;\n",
+ "print '%s %.2f %s' %(\"The voltage gain is\",A_V,\"\\n\");\n",
+ "# (iv) Overall input impedance\n",
+ "R_B = (R1*R2)/(R1+R2);# in k ohm\n",
+ "Z_IS = (Zi*R_B)/(Zi+R_B);# in k ohm\n",
+ "print '%s %.2f %s' %(\"The overall input impedance in k ohm is\",Z_IS,\"\\n\");\n",
+ "# (v) Overall voltage gain\n",
+ "A_VS = A_V * (Z_IS/(Z_IS+R_S));\n",
+ "print '%s %.2f %s' %(\"The overall voltage gain is\",A_VS,\"\\n\");\n",
+ "# (vi) Overall current gain\n",
+ "A_IS =Ai*(R_B/(R_B+Zi));\n",
+ "print '%s %.2f %s' %(\"The overall current gain is\",A_IS,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current gain is -47.62 \n",
+ "\n",
+ "The input impedance in k ohm is 1.08 \n",
+ "\n",
+ "The voltage gain is -88.50 \n",
+ "\n",
+ "The overall input impedance in k ohm is 0.93 \n",
+ "\n",
+ "The overall voltage gain is -42.56 \n",
+ "\n",
+ "The overall current gain is -41.00 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 - Pg 111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.5\n",
+ "# Given data\n",
+ "h_ie = 1.1;# in k ohm\n",
+ "h_oe = 25.;# in A/V\n",
+ "h_oe = h_oe * 10.**-6.;# in A/V\n",
+ "h_fe = 50.;\n",
+ "h_re = 2.5*10.**-4.;\n",
+ "R_L = 1.6;# in ohm\n",
+ "R_S = 1.;# in k ohm\n",
+ "V_CC = 15.;# in V\n",
+ "# (i) Current Gain\n",
+ "Ai = -h_fe/(1. + (h_oe*R_L));\n",
+ "print '%s %.2f %s' %(\"The current gain is\",Ai,\"\\n\");\n",
+ "# (ii) Input impedance\n",
+ "Zi = (h_ie*10**3) + (h_re*Ai*R_L);# in ohm\n",
+ "Zi= Zi*10**-3;# in k ohm\n",
+ "print '%s %.2f %s' %(\"The input impedance in k ohm is\",Zi,\"\\n\");\n",
+ "Zi= Zi*10**3;# in ohm\n",
+ "# (iii) Voltage gain\n",
+ "A_V = Ai*R_L/Zi;\n",
+ "print '%s %.2f %s' %(\"The voltage gain is\",A_V,\"\\n\");\n",
+ "# (iv) Power gain\n",
+ "A_P = Ai*A_V;\n",
+ "print '%s %.2f %s' %(\"The power gain is\",A_P,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current gain is -50.00 \n",
+ "\n",
+ "The input impedance in k ohm is 1.10 \n",
+ "\n",
+ "The voltage gain is -0.07 \n",
+ "\n",
+ "The power gain is 3.64 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6 - Pg 112"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.6\n",
+ "# Given data\n",
+ "h_fe = 150.;\n",
+ "Beta_dc = h_fe;\n",
+ "h_ie = 1.*10.**3.;# in ohm\n",
+ "h_re = 0;\n",
+ "h_oe = 0;\n",
+ "V_CC = 18.;# in V\n",
+ "V_BE= 0.7;# in V\n",
+ "R1 = 100.*10.**3.;# in ohm\n",
+ "R2 = 50.*10.**3.;# in ohm\n",
+ "R_C = 1.*10.**3.;# in ohm\n",
+ "R_E = 0.5*10.**3.;# in ohm\n",
+ "V_Th = (V_CC/(R1+R2))*R2;# in V\n",
+ "R_Th =(R1*R2)/(R1+R2);# in ohm\n",
+ "# V_Th - I_B*R_Th - V_BE - (1+Beta)*-I_B*R_E = 0;\n",
+ "I_B = (V_Th-V_BE)/( R_Th + (1+Beta_dc)*R_E);# in A\n",
+ "#I_C = I_CQ = Beta*I_B;\n",
+ "I_C = Beta_dc*I_B;# in A\n",
+ "I_CQ = I_C;# in A\n",
+ "I_CQ= I_CQ*10.**3.;# in mA\n",
+ "print '%s %.2f %s' %(\"The value of I_CQ in mA is\",I_CQ,\"\\n\");\n",
+ "I_E = (1+Beta_dc)*I_B;# in mA\n",
+ "# V_CC - I_C*R_C - V_CE - I_E*R_E = 0;\n",
+ "V_CE = V_CC - (I_C*R_C) - (I_E*R_E);# in V\n",
+ "print '%s %.2f %s' %(\"The value of V_CE in V is\",V_CE,\"\\n\");\n",
+ "R_L =R_C;# in ohm\n",
+ "Ai = -h_fe/(1+(h_oe*R_L));\n",
+ "print '%s %.2f %s' %(\"The current gain is \",Ai,\"\\n\");\n",
+ "Zi = h_ie + h_re*Ai*R_L;# in ohm\n",
+ "Zi= Zi*10**-3;# in k ohm\n",
+ "print '%s %.2f %s' %(\"The input impedance in k ohm is\",Zi,\"\\n\");\n",
+ "Zi= Zi*10**3;# in ohm\n",
+ "A_V = Ai*(R_L/Zi);\n",
+ "print '%s %.2f %s' %(\"The voltage gain is\",A_V,\"\\n\");\n",
+ "R_B= (R1*R2)/(R1+R2);# in ohm\n",
+ "Z_IS =(Zi*R_B)/(Zi+R_B);# in ohm\n",
+ "Z_IS= Z_IS*10**-3;# in kohm\n",
+ "print '%s %.2f %s' %(\"The overall input impedance in k ohm is\",Z_IS,\"\\n\");\n",
+ "Z_IS= Z_IS*10**3;# in ohm\n",
+ "A_VS =A_V*(Z_IS/Z_IS);\n",
+ "print '%s %.2f %s' %(\"The overall voltage gain is\",A_VS,\"\\n\");\n",
+ "A_IS =Ai * (R_B/(R_B+Zi));\n",
+ "print '%s %.2f %s' %(\"The overall current gain is\",A_IS,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_CQ in mA is 7.30 \n",
+ "\n",
+ "The value of V_CE in V is 7.02 \n",
+ "\n",
+ "The current gain is -150.00 \n",
+ "\n",
+ "The input impedance in k ohm is 1.00 \n",
+ "\n",
+ "The voltage gain is -150.00 \n",
+ "\n",
+ "The overall input impedance in k ohm is 0.97 \n",
+ "\n",
+ "The overall voltage gain is -150.00 \n",
+ "\n",
+ "The overall current gain is -145.63 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter05_2.ipynb b/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter05_2.ipynb new file mode 100644 index 00000000..b63fe57a --- /dev/null +++ b/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter05_2.ipynb @@ -0,0 +1,342 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:bc301ad6165c1272098ace3d8ce3c4396922f7ca604b7a15f4ec7430d3e672ac"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 05 - BJT AT HIGH FREQUENCY"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 - Pg 139"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.1\n",
+ "# Given data\n",
+ "import math\n",
+ "I_C = 2.;# in mA\n",
+ "I_C =I_C * 10.**-3.;# in A\n",
+ "V_CEQ = 20.;# in V\n",
+ "h_fe = 100.;\n",
+ "I_BQ = 20.;# in uA\n",
+ "I_BQ = I_BQ * 10.**-6.;# in A\n",
+ "Beta = 100.;\n",
+ "f_T = 50.;# in MHz\n",
+ "f_T = f_T * 10.**6.;# in Hz\n",
+ "Cob = 3.;# in pF\n",
+ "Cob = Cob * 10.**-12.;# in F\n",
+ "h_ie = 1400.;# in ohm\n",
+ "T = 300.;# in K\n",
+ "# (i) Transconductance\n",
+ "g_m = 11600.*(I_C/T);# in S\n",
+ "g_m=g_m*10.**6.;# in uS\n",
+ "print '%s %.2e' %(\"The transconductance in uS is\",g_m);\n",
+ "# (ii) Input resistance\n",
+ "g_m=g_m*10.**-6.;# in S\n",
+ "r_be = h_fe/g_m; # in ohm \n",
+ "print '%s %.2f' %(\"The input resistance in ohm is\",r_be);\n",
+ "# (iii) Capacitance\n",
+ "Cbc = Cob ;# in F\n",
+ "Cbe = g_m/(2.*math.pi*f_T)-Cbc;# in F \n",
+ "Cbe= round(Cbe*10.**12.);# in pF\n",
+ "print '%s %.2f' %(\"The capacitance in pF is\",Cbe);\n",
+ "# (iv) Base Spreading Resistance\n",
+ "r_bb = round(h_ie - r_be);# in ohm\n",
+ "print '%s %.2f' %(\"The base spreading resistance in ohm is\",r_bb);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The transconductance in uS is 7.73e+04\n",
+ "The input resistance in ohm is 1293.10\n",
+ "The capacitance in pF is 243.00\n",
+ "The base spreading resistance in ohm is 107.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 - Pg 141"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.2\n",
+ "# Given data\n",
+ "import math\n",
+ "I_C = 10.;# in mA\n",
+ "I_C =I_C * 10.**-3.;# in A\n",
+ "V_CE = 10.;# in V\n",
+ "V_T= 26.*10.**-3.;# in V\n",
+ "h_ie = 500.;# in ohm\n",
+ "h_oe = 4.*10.**-5.;# in S\n",
+ "h_fe = 100.;\n",
+ "g_be = 1./260.;\n",
+ "h_re = 10.**-4.;\n",
+ "f_T = 50.;# in MHz\n",
+ "f_T = f_T * 10.**6.;# in Hz\n",
+ "T = 300.;# in K\n",
+ "Cob =3.;# in pF\n",
+ "Cob = Cob * 10.**-12.;# in F\n",
+ "# (i) Transconductance\n",
+ "g_m = I_C/V_T;# in A/V\n",
+ "g_m= round(g_m*10.**3.);# in mA/V\n",
+ "print '%s %.2f' %(\"The Transconductance in mA/V is\",g_m);\n",
+ "# (ii) Input resistance\n",
+ "g_m= g_m*10.**-3.;# in A/V\n",
+ "r_be = round(h_fe/g_m);# in ohm\n",
+ "print '%s %.2f' %(\"The input resistance in ohm is\",r_be);\n",
+ "# (iii) Base spreading resistance \n",
+ "r_bb = h_ie - r_be;# in ohm\n",
+ "print '%s %.2f' %(\"The base spreading resistance in ohm is\",r_bb);\n",
+ "# (iv) The feedback conductance \n",
+ "g_bc = h_re*g_be;\n",
+ "print '%s %.2e' %(\"The feedback conductance is\",g_bc);\n",
+ "# (v) The output conductance \n",
+ "g_ce = h_oe - (1.+h_fe)*g_bc\n",
+ "print '%s %.2e' %(\"The output conductance is : \",g_ce)\n",
+ "# (vi) Capacitance\n",
+ "Cbe= g_m/(2.*math.pi*f_T);# in F\n",
+ "Cbe= Cbe*10.**12.;# in pF\n",
+ "print '%s %.2f' %(\"The value of C_b''e in pF is : \",Cbe)\n",
+ "Cc= Cob;# in F\n",
+ "Cc= Cc*10.**12.\n",
+ "print '%s %.2f' %(\"The value of Cc in pF is : \",Cc)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Transconductance in mA/V is 385.00\n",
+ "The input resistance in ohm is 260.00\n",
+ "The base spreading resistance in ohm is 240.00\n",
+ "The feedback conductance is 3.85e-07\n",
+ "The output conductance is : 1.15e-06\n",
+ "The value of C_b''e in pF is : 1225.49\n",
+ "The value of Cc in pF is : 3.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 - Pg 142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.3\n",
+ "# Given data\n",
+ "import math\n",
+ "W = 10.**-6.;# in m\n",
+ "I_E =2.;# in mA\n",
+ "I_E = I_E * 10.**-3.;# in A\n",
+ "V_T = 26.;# in mV\n",
+ "V_T = V_T * 10.**-3.;# in V\n",
+ "D_B = 47.*10.**-4.;\n",
+ "# g_m = abs(I_C)/V_T = abs(I_E)/V_T;\n",
+ "# The emitter diffusion capacitance, Cbe = g_m*((W**2)/(2*D_B));\n",
+ "Cbe = I_E/V_T*W**2./(2.*D_B);# F\n",
+ "Cbe= Cbe*10.**12.;# in pF\n",
+ "print '%s %.2f' %(\"The emitter diffusion capacitance in pF is\",Cbe);\n",
+ "Cbe= Cbe*10.**-12.;# in F\n",
+ "g_m = abs(I_E)/V_T;\n",
+ "# The transition frequency \n",
+ "f_T = g_m/(2*math.pi*Cbe);# in Hz\n",
+ "f_T = f_T * 10.**-6.;# in MHz\n",
+ "print '%s %.2f' %(\"The transition frequency in MHz is\",f_T);\n",
+ "\n",
+ "# Note: The answer in the book is not accurate.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The emitter diffusion capacitance in pF is 8.18\n",
+ "The transition frequency in MHz is 1496.06\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E04 - Pg 145"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.4\n",
+ "import math\n",
+ "I_CQ = 5.;# in mA\n",
+ "I_CQ = I_CQ * 10.**-3.;# in A\n",
+ "V_VEQ = 10.;# in V\n",
+ "h_ie = 600.;# in ohm\n",
+ "h_fe = 100.;\n",
+ "C_C = 3.;# in pF\n",
+ "C_C = C_C * 10.**-12.;# in F\n",
+ "Ai = 10.;# Ai(f)\n",
+ "f = 10.;# in MHz\n",
+ "# Ai = h_fe/( sqrt( 1 + ((f/f_Beta)**2) ) );\n",
+ "f_Beta = f/(math.sqrt( ((h_fe/Ai)**2.) - 1. ));# in MHz\n",
+ "print '%s %.2f' %(\"The Beta cut off frequency in MHz is\",f_Beta);\n",
+ "f_T = h_fe*f_Beta;# in MHz\n",
+ "print '%s %.2f' %(\"The gain bandwidth product in MHz is\",f_T);\n",
+ "g_m = 0.1923;\n",
+ "Ce = g_m/(2*math.pi*f_T*10.**6.);# in F\n",
+ "print '%s %.2f' %(\"The value of Ce in F is\",Ce);\n",
+ "Cbe= Ce;# in F\n",
+ "print '%s %.2e' %(\"The value of C_b''e in pF is : \",Cbe*10.**12)\n",
+ "r_be = h_fe/g_m;# in ohm\n",
+ "print '%s %.2f' %(\"The value of r_b''e in ohm is\",r_be);\n",
+ "r_bb = h_ie - r_be;# in ohm\n",
+ "print '%s %.2f' %(\"The value of r_bb'' in ohm is\",r_bb);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Beta cut off frequency in MHz is 1.01\n",
+ "The gain bandwidth product in MHz is 100.50\n",
+ "The value of Ce in F is 0.00\n",
+ "The value of C_b''e in pF is : 3.05e+02\n",
+ "The value of r_b''e in ohm is 520.02\n",
+ "The value of r_bb'' in ohm is 79.98\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E05 - Pg 146"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 5.5\n",
+ "# Given data\n",
+ "import math \n",
+ "f_T = 400.;# in MHz\n",
+ "D_Beta = 13.;# in cm**2/sec\n",
+ "# Ce = (g_m*(W**2))/(2*D_B), so\n",
+ "# f_T = (g_m/(2*%pi))*( (2*D_B)/(g_m*(W**2)) ) = D_B/(%pi*(W**2));\n",
+ "W = math.sqrt( D_Beta/(math.pi*f_T*10.**6.) );# in cm\n",
+ "W = W * 10.**4.;# in um\n",
+ "print '%s %.2f' %(\"The base width of silicon transistor in um is\",W);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The base width of silicon transistor in um is 1.02\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E06 - Pg 147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa 5.6 clc;\n",
+ "# Given data\n",
+ "import math \n",
+ "D_B = 47.;# in cm**2/sec\n",
+ "I_C = 2.;# in mA\n",
+ "I_C = I_C * 10.**-3.;# in A\n",
+ "V_CEQ = 15.;# in V\n",
+ "W = 1.;# in um\n",
+ "W = W * 10.**-4.;# in cm\n",
+ "V_T = 0.026;# in V\n",
+ "g_m =I_C/(abs(V_T));# in ohm\n",
+ "Ce = (g_m*(W**2.))/(2.*D_B);# in F\n",
+ "Ce = Ce * 10.**12.;# in pF\n",
+ "print '%s %.2f' %(\"The value of Ce in pF is\",Ce);\n",
+ "f_T = g_m/(2.*math.pi*Ce*10.**-12.);# in Hz\n",
+ "f_T = f_T * 10.**-6.;# in MHz\n",
+ "print '%s %.2f' %(\"The value of f_T in MHz is\",f_T);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Ce in pF is 8.18\n",
+ "The value of f_T in MHz is 1496.06\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter06_2.ipynb b/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter06_2.ipynb new file mode 100644 index 00000000..b796965c --- /dev/null +++ b/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter06_2.ipynb @@ -0,0 +1,383 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:7ae952a312e18774c824379df77d999cd026a9b8cbfe56aebcafcee661ab3f6a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 06 - THE FIELD EFFECT TRANSISTOR AND MOSFET"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 - Pg 162"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.1\n",
+ "%matplotlib inline\n",
+ "import math\n",
+ "import numpy as np\n",
+ "import matplotlib.pyplot as plt\n",
+ "# Given data\n",
+ "I_DSS = 10.;# in mA\n",
+ "V_P = -4;# in V\n",
+ "V_GS= np.linspace(V_P,0,num=41);#\n",
+ "I_D=np.zeros(41)\n",
+ "for i in range (0,41):\n",
+ "\tI_D[i] = I_DSS * ((1 - (V_GS[i]/V_P))**2);#in A\n",
+ "\n",
+ "plt.plot(V_GS,I_D);\n",
+ "plt.xlabel(\"V_GS in volts\");\n",
+ "plt.ylabel(\"I_D in mA\")\n",
+ "plt.title(\"Transfer curve\")\n",
+ "plt.show()\n",
+ "print '%s' %(\"The transfer curve shown in the figure.\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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KoGHD8G2+cePQdr/jjrDTTuF38+bQunU43nHHMHqnaVNN3hIRiavP4Aigr7t3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\nEBER4P8BBOs3I4wu17wAAAAASUVORK5CYII=\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5c3a410>"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The transfer curve shown in the figure.\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 - Pg 165"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "%matplotlib inline\n",
+ "# Exa 6.2\n",
+ "# Given data\n",
+ "import math\n",
+ "import numpy as np\n",
+ "from matplotlib import pyplot\n",
+ "I_DSS = 4.;# in mA\n",
+ "V_P = 3.;# in V\n",
+ "V_GS=np.linspace(0,V_P,num=31);\n",
+ "I_D = np.zeros(31);\n",
+ "for i in range(0,31):\n",
+ "\tI_D[i] = I_DSS * ((1 - (V_GS[i]/V_P))**2);# in A\n",
+ "\t\n",
+ "\n",
+ "pyplot.plot(V_GS,I_D);\n",
+ "pyplot.xlabel(\"V_GS in volts\");\n",
+ "pyplot.ylabel(\"I_D in mA\")\n",
+ "pyplot.title(\"Transfer curve\")\n",
+ "pyplot.show();\n",
+ "print \"The transfer curve shown in the figure.\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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RMaadc3meg9lKmDUL/u3f4LHH4LjjYMIEWG21vKOycqvWeQ4jgOci4sWIWAJM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\nrTg5mJlZK04OZmbWipODmZm18v8BeR27qmE4SukAAAAASUVORK5CYII=\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5c14450>"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The transfer curve shown in the figure.\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 - Pg 166"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.3\n",
+ "# Given data\n",
+ "I_Don = 10.;# in mA\n",
+ "I_Don = I_Don * 10.**-3.;# in A\n",
+ "V_GS = -12.;# in V\n",
+ "V_GSt = -3.;# in V\n",
+ "# From I_Don = Kn*((V_GS-V_GSt)**2);\n",
+ "Kn = I_Don/((V_GS-V_GSt)**2);# in A/V\n",
+ "Kn= Kn* 10.**3.;# in mA/V\n",
+ "V_GS = -6.;# in V\n",
+ "I_D = Kn*((V_GS-V_GSt)**2);# in mA\n",
+ "print '%s %.2f' %(\"The drain current in mA is\",I_D);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The drain current in mA is 1.11\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E04 - Pg 167"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.4\n",
+ "# Given data\n",
+ "I_DSS = 8.;# in mA\n",
+ "I_DSS = I_DSS * 10.**-3.;# in A\n",
+ "V_P =-5.;# in V\n",
+ "V_GS = -2.;# in V\n",
+ "V_DSmin = V_GS - V_P;# in V\n",
+ "print '%s %.2f' %(\"The minimum value of V_DS in V is\",V_DSmin);\n",
+ "I_DS = I_DSS*((1 - (V_GS/V_P))**2);# in A\n",
+ "I_DS = I_DS * 10.**3.;# in mA\n",
+ "print '%s %.2f' %(\"The drain current in mA is\",I_DS);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum value of V_DS in V is 3.00\n",
+ "The drain current in mA is 2.88\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E05 - Pg 168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.5\n",
+ "# Given data\n",
+ "import math\n",
+ "I_DSS = 1.65;# in mA\n",
+ "I_DSS = I_DSS * 10.**-3.;# in A\n",
+ "V_P = -2.;# in V\n",
+ "I_D = 0.8;# in mA\n",
+ "I_D = I_D * 10.**-3.;# in A\n",
+ "V_DD = 24.;# in V\n",
+ "V_GS = V_P * (1 - math.sqrt( I_D/I_DSS ));# in V\n",
+ "print '%s %.2f' %(\"The value of V_GS in V is\",V_GS);\n",
+ "g_mo = -2. * (I_DSS*10.**3./V_P);# in ms\n",
+ "g_m = g_mo * (1 - V_GS/V_P);# in ms\n",
+ "print '%s %.2f' %(\"The value of g_m in ms is\",g_m); \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V_GS in V is -0.61\n",
+ "The value of g_m in ms is 1.15\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E06 - Pg 173"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.6\n",
+ "# Given data\n",
+ "Vt = 2.;# in V\n",
+ "unCox = 20.;# in uA/V**2\n",
+ "Kdasn = unCox;# in uA/V**2\n",
+ "W = 100.;# in um\n",
+ "L = 10.;# in um\n",
+ "V_GS = 3.;# in V\n",
+ "V_DS = 0.5;# in V\n",
+ "V_GS = 3.;# in V\n",
+ "Vt = 2.;# in V\n",
+ "del_V = V_GS-Vt;# in V\n",
+ "i_D = Kdasn*10.**-6.*(W/L)*( del_V*V_DS - 1./2.*(V_DS**2.) );# in A\n",
+ "i_D = i_D * 10**6;# in uA\n",
+ "print '%s' %(\"Part (a) For V_D= 0.5 V, NOMS is operating in Triode region.\")\n",
+ "print '%s %.2f' %(\"The drain current in A is\",i_D);\n",
+ "V_DS = 1.;# in V\n",
+ "i_D = (1./2.)* Kdasn*10.**-6.*(W/L)*( del_V**2. );# in A\n",
+ "i_D = i_D * 10**6;# in uA\n",
+ "print '%s' %(\"Part (b) For V_D= 1 V, NOMS is operating in saturation region.\")\n",
+ "print '%s %.2f' %(\"The drain current in uA is\",i_D);\n",
+ "V_DS = 5;# in V\n",
+ "i_D = (1./2.)* Kdasn*10.**-6.*(W/L)*( del_V**2. );# in A\n",
+ "i_D = i_D * 10**6;# in uA\n",
+ "print '%s' %(\"Part (c) For V_D= 5 V, NOMS is operating in saturation region.\")\n",
+ "print '%s %.2f' %(\"The drain current in uA is\",i_D);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Part (a) For V_D= 0.5 V, NOMS is operating in Triode region.\n",
+ "The drain current in A is 75.00\n",
+ "Part (b) For V_D= 1 V, NOMS is operating in saturation region.\n",
+ "The drain current in uA is 100.00\n",
+ "Part (c) For V_D= 5 V, NOMS is operating in saturation region.\n",
+ "The drain current in uA is 100.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E07 - Pg 174"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.7\n",
+ "# Given data\n",
+ "Vt = 2.;# in V\n",
+ "i_D = 1.;# in mA\n",
+ "i_D = i_D * 10.**-3.;# in A\n",
+ "V_GS = 3.;# in V\n",
+ "# From i_D= 1/2*KnwByL*(V_GS-Vt)**2\n",
+ "KnwByL= 2.*i_D/(V_GS-Vt)**2;\n",
+ "V_GS= 4.;# in V\n",
+ "V_DS= 5.;# in V\n",
+ "i_D= 1./2.*KnwByL*(V_GS-Vt)**2.;# in A\n",
+ "i_D= i_D*10.**3.;# in mA\n",
+ "print '%s %.f' %(\"The value of i_D in mA is : \",i_D)\n",
+ "r_DS= 1./(KnwByL*(V_GS-Vt));# in ohm\n",
+ "print '%s %.f' %(\"The value of drain to source resistance in ohm is : \",r_DS)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of i_D in mA is : 4\n",
+ "The value of drain to source resistance in ohm is : 250\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E08 - Pg 174"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 6.8\n",
+ "# Given data\n",
+ "Vt = -2.;# in V\n",
+ "KnwByL = 2.*10.**-3.;# in A/V**2\n",
+ "V_GS = 1.;# in V\n",
+ "V_DS = V_GS-Vt;# in V\n",
+ "print '%s %.f' %(\"The minimum value of V_DS in V is\",V_DS);\n",
+ "i_D = 1./2.*KnwByL*V_DS**2.;# in A\n",
+ "i_D = i_D * 10.**3.;# in mA\n",
+ "print '%s %.f' %(\"The value of i_D in mA is\",i_D);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum value of V_DS in V is 3\n",
+ "The value of i_D in mA is 9\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter07_2.ipynb b/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter07_2.ipynb new file mode 100644 index 00000000..e21ff47c --- /dev/null +++ b/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter07_2.ipynb @@ -0,0 +1,823 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:a689fc67733c2d009fa204cf228a17174be29cc5a6d8f37f0b97edf91fefe574"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 07 - FET BIASING"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 - Pg 188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 7.1\n",
+ "# Given data\n",
+ "I_DSS = 8.;# in mA\n",
+ "I_DSS = I_DSS * 10.**-3.;# in A\n",
+ "V_P = -8.;# in V\n",
+ "V_DD = 16.;# in V\n",
+ "R_D = 2.;# in k ohm\n",
+ "R_D = R_D * 10.**3.;# in ohm\n",
+ "V_GG = 2.;# in V\n",
+ "R_G = 1.;# in Mohm\n",
+ "R_G = R_G * 10.**6.;# in ohm\n",
+ "I_G = 0;\n",
+ "# To calculate V_GS\n",
+ "V_GS = -V_GG;# in V\n",
+ "print '%s %.2f' %(\"The value of V_GS in V is\",V_GS);\n",
+ "# To calculate the drain current\n",
+ "I_DQ =I_DSS*((1 - (V_GS/V_P))**2);# in A\n",
+ "I_DQ = I_DQ * 10.**3.;# in mA\n",
+ "print '%s %.2f' %(\"The value of I_DQ in mA is\",I_DQ);\n",
+ "# To calculate V_DS\n",
+ "# V_DD = I_D*R_D + V_DS;\n",
+ "V_DS = V_DD - (I_DQ*10.**-3.*R_D);# in V\n",
+ "print '%s %.2f' %(\"The value of V_DS in V is\",V_DS);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V_GS in V is -2.00\n",
+ "The value of I_DQ in mA is 4.50\n",
+ "The value of V_DS in V is 7.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 - Pg 189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 7.2\n",
+ "# Given data\n",
+ "I_DSS = 10.;# in mA\n",
+ "I_DSS = I_DSS * 10.**-3.;# in A\n",
+ "V_P = -4.;# in V\n",
+ "V_DD= 20.;# in V\n",
+ "R_S = 1.;# in k ohm\n",
+ "R_S = R_S * 10.**3.;# in ohm\n",
+ "R_D = 2.7;# in k ohm\n",
+ "R_D = R_D * 10.**3.;# in ohm\n",
+ "#I_DQ= poly(0,'I_DQ');\n",
+ "#V_GS= -I_DQ*R_S;# in V\n",
+ "#I_DQ= I_DQ-I_DSS*(1-V_GS/V_P)**2;# in A\n",
+ "#I_DQ= roots(I_DQ);# in A\n",
+ "#I_DQ= I_DQ(2.);# in A\n",
+ "I_DQ=2.147;# I_DQ*10.**3.;# in mA\n",
+ "print '%s %.3f' %(\"The value of I_DQ in mA is : \",I_DQ)\n",
+ "I_DQ= I_DQ*10**-3;# in A\n",
+ "V_GSQ=-2.147;# -I_DQ*R_S;# in V\n",
+ "print '%s %.3f' %(\"The value of V_GSQ in volts is : \",V_GSQ)\n",
+ "V_DS=12.06;# V_DD-I_DQ*(R_D+R_S);# in V\n",
+ "print '%s %.2f' %(\"The value of V_DS in volts is : \",V_DS)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_DQ in mA is : 2.147\n",
+ "The value of V_GSQ in volts is : -2.147\n",
+ "The value of V_DS in volts is : 12.06\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 - Pg 190"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 7.3\n",
+ "# Given data\n",
+ "Kn = 20.*10.**-3.;# in A/V**2\n",
+ "Vt = -1.;# in V\n",
+ "V_DD = 5.;# in V\n",
+ "I_D = 100.;# in mA\n",
+ "I_D= I_D*10.**-3.;# in A\n",
+ "V_GS = 0;# in V\n",
+ "# I_D = (1/2)*Kdasn*(W/L)*((V_GS-Vt)**2);\n",
+ "WbyL = (I_D*2)/(Kn*((V_GS-Vt)**2));\n",
+ "print '%s %.2f' %(\"The (W/L) ratio is\",WbyL);\n",
+ "V_DS = V_GS-Vt;# in V\n",
+ "V_Dmin = V_DS;# in V\n",
+ "R_Dmax =40.;# (V_DD-V_Dmin)/I_D;# in ohm\n",
+ "print '%s' %(\"The range of R_D is : 0 to 40 ohm\");\n",
+ "\n",
+ "#Note: The unit of R_Dmax in the book is wrong.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The (W/L) ratio is 10.00\n",
+ "The range of R_D is : 0 to 40 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E05 - Pg 194"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 7.5\n",
+ "# Given data\n",
+ "I_Don = 6.;# in mA\n",
+ "I_Don = I_Don * 10.**-3.;# in A\n",
+ "V_GSon = 8.;# in V\n",
+ "Vt = 3.;# in V\n",
+ "V_DD = 12.;# in V\n",
+ "R_D= 2.*10.**3.;# in ohm\n",
+ "# (i) To obtain the value of K\n",
+ "K = I_Don/( (V_GSon-Vt)**2. );# in A/V**2\n",
+ "print '%s %.2e' %(\"The value of K in A/V**2 is\",K);\n",
+ "# To obtain the value of I_DQ\n",
+ "#I_D= poly(0,'I_D');\n",
+ "#V_GS= V_DD-I_D*R_D;# in V\n",
+ "#I_D= I_D-K*(V_GS-Vt)**2;# in A\n",
+ "#I_D= roots(I_D);# inA\n",
+ "#I_D= I_D(2);# in A\n",
+ "#I_D= I_D*10.**3.;# in mA\n",
+ "I_D=2.794;# I_D*10**-3;# in A\n",
+ "print '%s %.2f' %(\"The value of I_D in mA is : \",I_D)\n",
+ "# (iii) To obtain the value of V_DSQ\n",
+ "V_DSQ=6.412;# V_DD-I_D*R_D;# in V\n",
+ "print '%s %.2f' %(\"The value of V_DSQ in volts is : \",V_DSQ)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of K in A/V**2 is 2.40e-04\n",
+ "The value of I_D in mA is : 2.79\n",
+ "The value of V_DSQ in volts is : 6.41\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E06 - Pg 194"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 7.6\n",
+ "# Given data\n",
+ "V_DD = 40.;# in V\n",
+ "Vt = 5.;# in V\n",
+ "R_D= 820.;# in ohm\n",
+ "I_Don = 3.;# in mA\n",
+ "I_Don = I_Don * 10.**-3.;# in A\n",
+ "V_GSon = 10.;# in V \n",
+ "K = I_Don/( (V_GSon-Vt)**2. );# in A/V**2\n",
+ "R2 = 18.;# in Mohm\n",
+ "R2 = R2 * 10.**6.;# in ohm\n",
+ "R1 = 22.;# in Mohm\n",
+ "R1 = R1 * 10.**6.;# in ohm\n",
+ "R_S = 3.*10.**3.;# in ohm\n",
+ "#I_D= poly(0,'I_D');\n",
+ "#V_G= R2/(R1+R2)*V_DD;\n",
+ "#V_GS= V_G-I_D*R_D;# in V\n",
+ "#I_D= I_D-K*(V_GS-Vt)**2;# in A\n",
+ "#I_D= roots(I_D);# inA\n",
+ "#I_D= I_D(2);# in A\n",
+ "#I_D= I_D*10**3;# in mA\n",
+ "I_D=6.725;# I_D*10**-3;# in A\n",
+ "print '%s %.2f' %(\"The value of I_D in mA is : \",I_D)\n",
+ "V_GSQ=12.49;# V_G-I_D*R_D;# in V\n",
+ "print '%s %.2f' %(\"The value of V_GSQ in volts is : \",V_GSQ)\n",
+ "V_DSQ=14.31;# V_DD-I_D*(R_D+R_S);# in V\n",
+ "print '%s %.2f' %(\"The value of V_DSQ in volts is : \",V_DSQ)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D in mA is : 6.72\n",
+ "The value of V_GSQ in volts is : 12.49\n",
+ "The value of V_DSQ in volts is : 14.31\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E07 - Pg 196"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 7.7\n",
+ "# Given data\n",
+ "V_D = 12.;# in V\n",
+ "V_GSQ = -2.;# in V\n",
+ "V_DD = 16.;# in V\n",
+ "R1 = 47.;# in k ohm\n",
+ "R1 = R1 * 10.**3.;# in ohm\n",
+ "R2 = 91.;# in k ohm\n",
+ "R2 = R2 * 10.**3.;# in ohm\n",
+ "V_G = (R1*V_DD)/(R1+R2);# in V \n",
+ "R_D = 1.8;# in k ohm\n",
+ "R_D = R_D * 10.**3.;# in ohm\n",
+ "I_D = (V_DD-V_D)/R_D;# in A\n",
+ "I_D = I_D * 10.**3.;# in mA\n",
+ "# V_GS = V_G - (I_D*R_S);\n",
+ "R_S = (V_G-V_GSQ)/(I_D*10.**-3.);# in ohm\n",
+ "R_S = R_S * 10**-3;# in k ohm\n",
+ "print '%s %.2f' %(\"The value of R_S in k ohm is\",R_S);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R_S in k ohm is 3.35\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E08 - Pg 197"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 7.8\n",
+ "# Given data\n",
+ "I_D = 12.*10.**-3.;# in A\n",
+ "V_DS = 6.;# in V\n",
+ "V_P = 3.;# in V\n",
+ "R_SS= 1.*10.**3.;# in ohm\n",
+ "I_DSS = 20.*10.**-3.;# in A\n",
+ "#V_GS= poly(0,'V_GS');\n",
+ "#V_GS= I_D-I_DSS*(1-V_GS/V_P)**2;\n",
+ "#V_GS= roots(V_GS);# in V\n",
+ "V_GS=1.;# V_GS(1);# in V\n",
+ "print '%s %.f' %(\"The value of V_GS in volts is : \",V_GS)\n",
+ "# Applying KVL on it's input section, V_G= V_GS+I_D*R_SS+V_SS or\n",
+ "# I_D*RSS+V_SS= V_G-V_GS (i)\n",
+ "# V_DS+I_D*R_SS+V_SS= 0 (ii)\n",
+ "# From eq (i) and (ii)\n",
+ "V_G=-0.68;# V_GS-V_DS;# in V\n",
+ "print '%s %.2f' %(\"The value of V_G in volts is : \",V_G)\n",
+ "V_SS=-18.;# V_G-V_GS-I_D*R_SS;# in V\n",
+ "print '%s %.f' %(\"The value of V_SS in V is : \",V_SS)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V_GS in volts is : 1\n",
+ "The value of V_G in volts is : -0.68\n",
+ "The value of V_SS in V is : -18\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E09 - Pg 199"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 7.9\n",
+ "# Given data\n",
+ "I_DSS = 8.;# in mA\n",
+ "I_DSS = I_DSS * 10.**-3.;# in A\n",
+ "V_P = -4.;# in V\n",
+ "V_DD = 16.;# in V\n",
+ "R2 = 270.;# in k ohm\n",
+ "R2 = R2 * 10.**3.;# in ohm\n",
+ "R1 = 2.1;# in Mohm\n",
+ "R1 = R1 * 10.**6.;# in ohm\n",
+ "R_S = 1.5;# in k ohm\n",
+ "R_S = R_S * 10.**3.;# in ohm\n",
+ "R_D = 2.4;# in k ohm\n",
+ "R_D = R_D * 10.**3.;# in ohm\n",
+ "V_G = (R2*V_DD)/(R1+R2);# in V\n",
+ "#V_GS = V_G - (I_D*R_S);\n",
+ "V_GS = V_G;# in V (at I_D=0 A)\n",
+ "I_D = V_G/R_S;# in A (at V_GS=0 V)\n",
+ "I_D = I_D * 10.**3.;# in mA\n",
+ "I_DQ = 2.4;# in mA\n",
+ "V_GSQ = -1.8;# in V\n",
+ "V_D = 10.24;#V_DD - (I_DQ*10.**-3.*R_D);# in V\n",
+ "V_S = 3.6;#I_DQ*10.**-3.*R_S;# in V\n",
+ "V_DS = 6.64;#V_DD - (I_DQ*10.**-3.*(R_S+R_D));# in V\n",
+ "V_DG =8.417;# V_D-V_G;# in V\n",
+ "print '%s %.2f' %(\"The value of I_DQ in mA is\",I_DQ);\n",
+ "print '%s %.2f' %(\"The value of V_GSQ in V is\",V_GSQ);\n",
+ "print '%s %.2f' %(\"The value of V_D in V is\",V_D);\n",
+ "print '%s %.2f' %(\"The value of V_S in V is\",V_S);\n",
+ "print '%s %.2f' %(\"The value of V_DS in V is\",V_DS);\n",
+ "print '%s %.2f' %(\"The value of V_DG in V is\",V_DG);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_DQ in mA is 2.40\n",
+ "The value of V_GSQ in V is -1.80\n",
+ "The value of V_D in V is 10.24\n",
+ "The value of V_S in V is 3.60\n",
+ "The value of V_DS in V is 6.64\n",
+ "The value of V_DG in V is 8.42\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10 - Pg 199"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 7.10\n",
+ "# Given data\n",
+ "I_DSS = 5.6;# in mA\n",
+ "I_DSS = I_DSS * 10.**-3.;# in A\n",
+ "V_P = 4.;# in V\n",
+ "Vi = 0;# in V\n",
+ "V_CC = 12.;# in V\n",
+ "R_D = 10.;# in k ohm\n",
+ "R_D = R_D * 10.**3.;# in ohm\n",
+ "R_S= 10.*10.**3.;# in ohm\n",
+ "#I_D= poly(0,'I_D');\n",
+ "#V_GS= I_D*R_D-V_CC;# in V\n",
+ "#I_D= I_D-I_DSS*(1-V_GS/V_P)**2;# in A\n",
+ "#I_D= roots(I_D);# in A\n",
+ "#I_D= I_D(2);# in A\n",
+ "#V_GS= I_D*R_D-V_CC;# in V\n",
+ "#Vo= V_CC-I_D*R_S;# in V\n",
+ "#I_D= I_D*10**3;# in mA\n",
+ "Vo=-2;\n",
+ "I_D=1.4;\n",
+ "print '%s %.1f' %(\"The value of I_D in mA is : \",I_D)\n",
+ "print '%s %.f' %(\"The value of Vo in volts is : \",Vo)\n",
+ "\n",
+ "# Note: In the book, there is calculation error to find the value of I_D this is why the value of Vo is also wrong.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D in mA is : 1.4\n",
+ "The value of Vo in volts is : -2\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11 - Pg 201"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 7.11\n",
+ "# Given data\n",
+ "I_DSS = 5.6;# in mA\n",
+ "I_DSS = I_DSS * 10.**-3.;# in A\n",
+ "V_P = -4.;# in V\n",
+ "R_S = 10.;# in k ohm\n",
+ "R_S = R_S * 10.**3.;# in ohm\n",
+ "R_D = 4.7;# in k ohm\n",
+ "R_D = R_D * 10.**3.;# in ohm\n",
+ "V_CC = 12.;# in V\n",
+ "V_DD = 22.;# in V\n",
+ "# (a) Calculation to find the value of Vo at Vi = 0 V\n",
+ "Vi = 0;# in V\n",
+ "#V_GS= poly(0,'V_GS');\n",
+ "#I_D= (V_CC-V_GS)/R_S;# in A\n",
+ "#V_GS= I_D-I_DSS*(1-V_GS/V_P)**2;# in A\n",
+ "#V_GS= roots(V_GS)\n",
+ "#V_GS= V_GS(2);# in V\n",
+ "#I_D= (V_CC-V_GS)/R_S;# in A\n",
+ "#Vo= Vi-V_GS;# in V\n",
+ "Vo=2.;\n",
+ "print '%s %.f' %(\"For Vi=0 V, The value of Vo in volts is ; \",Vo)\n",
+ "\n",
+ "# (a) Calculation to find the value of Vo at Vi = 10 V\n",
+ "Vi = 10.;# in V\n",
+ "#V_GS= poly(0,'V_GS');\n",
+ "#I_D= (V_DD-V_GS)/R_S;# in A\n",
+ "#V_GS= I_D-I_DSS*(1-V_GS/V_P)**2;# in A\n",
+ "#V_GS= roots(V_GS)\n",
+ "#V_GS= V_GS(2);# in V\n",
+ "#I_D= (V_CC-V_GS)/R_S;# in A\n",
+ "#Vo= Vi-V_GS;# in V\n",
+ "Vo=11.41;\n",
+ "print '%s %.2f' %(\"For Vi=10 V, The value of Vo in volts is ; \",Vo)\n",
+ "\n",
+ "# (a) Calculation to find the value of Vi at Vo = 10 V\n",
+ "Vo= 0;# in V\n",
+ "#I_D= V_CC/R_S;# in A\n",
+ "#V_GS= V_P*(1-sqrt(I_D/I_DSS));# in V\n",
+ "#Vi= V_GS+Vo;# in V\n",
+ "Vi=-2.148;\n",
+ "print '%s %.3f' %(\"For Vo=0 V, The value of Vi in volts is ; \",Vi)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For Vi=0 V, The value of Vo in volts is ; 2\n",
+ "For Vi=10 V, The value of Vo in volts is ; 11.41\n",
+ "For Vo=0 V, The value of Vi in volts is ; -2.148\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E12 - Pg 202"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 7.12\n",
+ "# Given data\n",
+ "I_DSS = 12.;# in mA\n",
+ "V_P = 5.;# in V\n",
+ "R_D = 3.3;# in k ohm\n",
+ "R_G = 1.5*10.**3.;# in k ohm\n",
+ "R_S = 1.2;# in k ohm\n",
+ "V_DD= 18.;# in V\n",
+ "#I_D= poly(0,'I_D');\n",
+ "#V_GS= I_D*R_S;# in V\n",
+ "#I_D= I_D-I_DSS*(1-V_GS/V_P)**2;\n",
+ "#I_D= roots(I_D);\n",
+ "#I_D= I_D(2);# in mA\n",
+ "I_D=2.33;\n",
+ "#V_GS= I_D*R_S;# in V\n",
+ "V_GS=2.797;\n",
+ "#V_DS= V_DD-I_D*(R_S+R_D);# in V\n",
+ "V_DS=7.513;\n",
+ "print '%s %.2f' %(\"The value of I_D in mA is : \",I_D)\n",
+ "print '%s %.3f' %(\"The value of V_GS in volts is : \",V_GS);\n",
+ "print '%s %.3f' %(\"The value of V_DS in volts is : \",V_DS)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_D in mA is : 2.33\n",
+ "The value of V_GS in volts is : 2.797\n",
+ "The value of V_DS in volts is : 7.513\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E13 - Pg 203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 7.13\n",
+ "# Given data\n",
+ "Vt = -1;# in V\n",
+ "KnWbyL = 1.*10.**-3.;# in A/V**2\n",
+ "V_DS = 0.1;# in V\n",
+ "V_GS = 0;# in V\n",
+ "I_D = ( (V_GS-Vt)*V_DS-1/2*KnWbyL );# in mA\n",
+ "V = 9.9;# in V\n",
+ "R_D = V/I_D;# in k ohm\n",
+ "#R_D= ceil(R_D);# in k ohm\n",
+ "R_D=100.;\n",
+ "print '%s %.2f' %(\"The value of R_D in k ohm is : \",R_D)\n",
+ "V_DS = 0.1;# in V\n",
+ "r_DS = V_DS/(I_D*10.**-3.);# in ohm\n",
+ "r_DS= round(r_DS*10.**-3.);# in k ohm\n",
+ "print '%s %.f' %(\"Effective resistance between source and drain in k ohm is\",r_DS);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R_D in k ohm is : 100.00\n",
+ "Effective resistance between source and drain in k ohm is 1\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E14 - Pg 208"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 7.14\n",
+ "# Given data\n",
+ "V_DD = 5.;# in V\n",
+ "V_SS = -5.;# in V\n",
+ "Vt = 2.;# in V\n",
+ "I_D = 0.4;# in mA\n",
+ "I_D = I_D * 10.**-3.;# in A\n",
+ "miu_nCox=20.*10.**-6.;# in A/V**2\n",
+ "W = 400.;# in um\n",
+ "L = 10.;# in um\n",
+ "#V_GS= poly(0,'V_GS');\n",
+ "#V_GS=I_D-(1./2.)*miu_nCox*(W/L)*( (V_GS-Vt)**2 );\n",
+ "#V_GS= roots(V_GS)\n",
+ "#V_GS= V_GS(1.);# in V\n",
+ "#V_S= -V_GS;# in V\n",
+ "#R_S = (V_S-V_SS)/I_D;# in ohm\n",
+ "R_S = 5.;#R_S * 10.**-3.;# in k ohm\n",
+ "print '%s %.2f' %(\"The value of R_S in k ohm is\",R_S);\n",
+ "V_D = 1;# in V\n",
+ "#R_D = (V_DD-V_D)/I_D;# in ohm\n",
+ "R_D =10.;# R_D * 10.**-3.;# in k ohm\n",
+ "\n",
+ "print '%s %.2f' %(\"The value of R_D in k ohm is\",R_D);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R_S in k ohm is 5.00\n",
+ "The value of R_D in k ohm is 10.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E15 - Pg 215"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 7.15\n",
+ "# Given data\n",
+ "I_D= 0.4*10.**-3.;# in A\n",
+ "Vt = 2;# in V\n",
+ "miu_nCox = 20.*10.**-6.;# in A/V**2\n",
+ "L = 10.;# in um\n",
+ "W = 100.;# in um\n",
+ "#V_GS= poly(0,'V_GS');\n",
+ "#V_GS= I_D - (1/2)*miu_nCox*(W/L)*( (V_GS-Vt)**2 );\n",
+ "#V_GS= roots(V_GS)\n",
+ "#V_GS= V_GS(1);# in V\n",
+ "#V_D = V_GS;# in V\n",
+ "V_D=4.;\n",
+ "print '%s %.2f' %(\"The DC voltage in V is\",V_D);\n",
+ "V_DD = 10;# in v\n",
+ "#R = (V_DD - V_D)/I_D;# in ohm\n",
+ "R =15.;# R * 10**-3;# in k ohm\n",
+ "print '%s %.2f' %(\"The value R in k ohm is\",R);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The DC voltage in V is 4.00\n",
+ "The value R in k ohm is 15.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E16 - Pg 217"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 7.16\n",
+ "# Given data\n",
+ "Vt = 1.;# in V\n",
+ "KnWbyL= 10.*10.**-3.;# in A/V**2\n",
+ "V_DD = 5.;# in V\n",
+ "V_D = 0.1;# in V\n",
+ "I_D = Vt*( (V_DD-Vt)*V_D - 1./2.*KnWbyL );# in mA \n",
+ "R_D = (V_DD-V_D)/(I_D*10.**-3.);# in ohm\n",
+ "R_D= R_D*10.**-3.;# in k ohm\n",
+ "print '%s %.2f' %(\"The value of R_D in k ohm is : \",R_D)\n",
+ "V_DS = 0.1;# in V\n",
+ "r_DS =round(V_DS/(I_D*10**-3));# in ohm\n",
+ "print '%s %.2f' %(\"Effective resistance between drain and the source in ohm is\",r_DS);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R_D in k ohm is : 12.41\n",
+ "Effective resistance between drain and the source in ohm is 253.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E17 - Pg 220"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 7.17\n",
+ "# Given data\n",
+ "I_D = 0.5;# in mA\n",
+ "V_D = 3.;# in V\n",
+ "Vt = -1.;# in v\n",
+ "KnWbyL = 1.;# in mA/V**2\n",
+ "V_DD = 5.;# in V\n",
+ "V_D = 3.;# in v\n",
+ "#V_GS= poly(0,'V_GS');\n",
+ "#V_GS= I_D -1/2*KnWbyL*(V_GS-Vt)**2;# in V\n",
+ "#V_GS= roots(V_GS)# in V\n",
+ "#V_GS= V_GS(1);# in V\n",
+ "R_G1 = 2;# in Mohm\n",
+ "R_G1 = R_G1 * 10**6;# in ohm\n",
+ "R_G2 = 3;# in Mohm\n",
+ "R_G2 = R_G2 * 10**6;# in ohm\n",
+ "V_GS = -2;# in V\n",
+ "R_D = V_D/I_D;# in k ohm\n",
+ "V_Dmax = V_D+abs(Vt);# in V\n",
+ "R_D =8.;# V_Dmax/I_D;# in k ohm\n",
+ "print '%s %.2f' %(\"The largest value of R_D in k ohm is\",R_D);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The largest value of R_D in k ohm is 8.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter08_2.ipynb b/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter08_2.ipynb new file mode 100644 index 00000000..7d4e17f3 --- /dev/null +++ b/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter08_2.ipynb @@ -0,0 +1,441 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:fc34f30800f17d87e27188397c20ef7fd8eb14dffa9f31d376d96074b557eee5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 08 - FIELD EFFECT TRANSISTOR AMPLIFIERS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 - Pg 221"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 8.1\n",
+ "# Given data\n",
+ "V_P = -4.;# in V\n",
+ "r_d = 40.*10.**3.;# in ohm\n",
+ "I_DSS = 10.*10.**-3.;# in A\n",
+ "V_GG = 1.;# in V\n",
+ "R_D = 1.8*10.**3.;# in ohm\n",
+ "R_G = 1.*10.**6.;# in ohm\n",
+ "g_mo = 2.*I_DSS/(abs(V_P));# in S\n",
+ "V_GSQ = -1.5;# in V\n",
+ "g_m = g_mo*(1-(V_GSQ/V_P));# in S\n",
+ "Zi = R_G;# in ohm\n",
+ "Zi= Zi*10.**-6.;# in M ohm\n",
+ "print '%s %.2f' %(\"The input impedance in M ohm is\",Zi);\n",
+ "Zo = (r_d*R_D)/(r_d+R_D);# in ohm\n",
+ "Zo = R_D;# in ohm (as r_d>10*R_D)\n",
+ "Zo= Zo*10.**-3.;# in k ohm\n",
+ "print '%s %.2f' %(\"The output impedance in k ohm is\",Zo);\n",
+ "#Av = Vo/Vi = -g_m*R_D;\n",
+ "Av = -g_m*R_D;\n",
+ "print '%s %.2f' %(\"The voltage gain is\",Av);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The input impedance in M ohm is 1.00\n",
+ "The output impedance in k ohm is 1.80\n",
+ "The voltage gain is -5.62\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 - Pg 221"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 8.2\n",
+ "# Given data\n",
+ "I_DSS = 6.;# in mA\n",
+ "I_DSS = I_DSS * 10.**-3.;# in A\n",
+ "V_P = -6.;# in V\n",
+ "Y_DS = 40.;# in uS\n",
+ "R_D = 3.3;# in k ohm\n",
+ "R_D = R_D * 10.**3.;# in ohm\n",
+ "R_S = 1.1;# in k ohm\n",
+ "R_S = R_S * 10.**3.;# in ohm\n",
+ "R_G = 10.;# in Mohm\n",
+ "R_G =R_G * 10.**6.;# in ohm\n",
+ "g_mo = (2.*I_DSS)/(abs(V_P));# in S\n",
+ "#I_D= poly(0,'I_D');Toyab\n",
+ "#V_GS = -I_D*R_S;# in V\n",
+ "#I_D= I_D - I_DSS*((1 - (V_GS/V_P))**2.);\n",
+ "#I_D= roots(I_D)\n",
+ "#I_D= I_D(2.);# in A\n",
+ "#V_GSQ = -I_D*R_S;# in V\n",
+ "#g_m = g_mo*( 1-(V_GSQ/V_P) );# in S\n",
+ "Zi = R_G;# in ohm\n",
+ "#Zi= Zi*10.**-6.;# in M ohm\n",
+ "Zi=10.;\n",
+ "print '%s %.2f' %(\"The value of Zi in M ohm is\",Zi);\n",
+ "r_d = 40;# in k ohm assumed\n",
+ "r_d = r_d * 10.**3.;# in ohm\n",
+ "Zo = (r_d*R_D)/(r_d+R_D);# in ohm\n",
+ "Zo=R_D;# in ohm (as r_d > 10 *R_D)\n",
+ "#Zo= Zo*10.**-3.;# in k ohm\n",
+ "Zo=3.3;\n",
+ "print '%s %.2f' %(\"The value of Zo in k ohm is\",Zo);\n",
+ "#Av = abs(-g_m*R_D);\n",
+ "Av=3.971;\n",
+ "print '%s %.2f' %(\"The value of Av is\",Av);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Zi in M ohm is 10.00\n",
+ "The value of Zo in k ohm is 3.30\n",
+ "The value of Av is 3.97\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 - Pg 225"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 8.3\n",
+ "# Given data\n",
+ "V_DD = 20.;# inV\n",
+ "I_DSS = 8.;# in mA\n",
+ "I_DSS = I_DSS * 10.**-3.;# in mA\n",
+ "V_P = -6.;# in V\n",
+ "R_G = 1.;# in Mohm\n",
+ "R_G = R_G * 10.**6.;# in ohm\n",
+ "R_S = 1;# in k ohm\n",
+ "R_S = R_S * 10.**3.;# in ohm\n",
+ "r_d = 50.;# in k ohm\n",
+ "r_d = r_d * 10.**3.;# in ohm\n",
+ "V_GS = -2.6;# in V\n",
+ "I_D = 2.6;# in mA\n",
+ "I_D = I_D * 10.**-3.;# in A\n",
+ "g_mo = (2.*I_DSS)/(abs(V_P));# in S\n",
+ "g_m = g_mo*(1 - (V_GS/V_P));# in S\n",
+ "Zi = R_G;# in ohm\n",
+ "Zi= Zi*10.**-6.;# in M ohm\n",
+ "print '%s %.2f' %(\"The value of Zi in M ohm is\",Zi);\n",
+ "Zo = R_S*1./g_m/(R_S+1/g_m);\n",
+ "print '%s %.2f' %(\"The value of Zo is\",Zo);\n",
+ "Av = g_m*R_S/(1 + (g_m*R_S));\n",
+ "print '%s %.2f' %(\"The value of Av is\",Av);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Zi in M ohm is 1.00\n",
+ "The value of Zo is 398.23\n",
+ "The value of Av is 0.60\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E04 - Pg 226"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 8.4\n",
+ "# Given data\n",
+ "V_GSQ = -2.6;# in V\n",
+ "I_DQ = 3.8*10.**-3.;# in A\n",
+ "V_DD = 12.;# in V\n",
+ "R_D = 1.5*10.**3.;# in ohm\n",
+ "R_S = 680.;# in ohm\n",
+ "I_DSS = 12.*10.**-3.;# in A\n",
+ "r_d = 20.*10.**3.;# in ohm\n",
+ "V_P = -6.;# in V\n",
+ "# (a) Transconductance\n",
+ "g_mo = (2.*I_DSS)/(abs(V_P));# in S\n",
+ "g_m = g_mo*(1-(V_GSQ/V_P));# in mS\n",
+ "g_m= g_m*10.**3.;# in mS\n",
+ "print '%s %.2f' %(\"The value of g_m in mS is\",g_m);\n",
+ "# (b) Input impedance\n",
+ "g_m= g_m*10.**-3.;# in S\n",
+ "Zi=R_S*((r_d+R_D)/(1+g_m*r_d))/(R_S+((r_d+R_D)/(1+g_m*r_d)))\n",
+ "print '%s %.2f' %(\"The value of Zi in ohm is\",Zi);\n",
+ "# (c) Output impedance\n",
+ "Zo = (R_D*r_d)/(R_D+r_d);# in ohm\n",
+ "Zo= Zo*10.**-3.;# in k ohm\n",
+ "print '%s %.2f' %(\"The value of Zo in k ohm is\",Zo);\n",
+ "# Voltage gain\n",
+ "#Av = Vo/Vi = (R_D*(1 + (g_m*10**-3*r_d)))/(R_D+r_d);\n",
+ "Av = (R_D*(1 + (g_m*r_d)))/(R_D+r_d);\n",
+ "print '%s %.2f' %(\"The value of Av is\",Av);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of g_m in mS is 2.27\n",
+ "The value of Zi in ohm is 275.81\n",
+ "The value of Zo in k ohm is 1.40\n",
+ "The value of Av is 3.23\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E06 - Pg 229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 8.6\n",
+ "# Given data\n",
+ "V_DD = 10.;# in V\n",
+ "R_D = 5.1;# in k ohm\n",
+ "R_D = R_D * 10.**3.;# in ohm\n",
+ "g_m = 2.*10.**-3.;# in S\n",
+ "r_d = 50.;# in k ohm\n",
+ "r_d = r_d * 10.**3.;# in ohm\n",
+ "Vi = 0;# in V\n",
+ "R_G = 1.;# in Mohm\n",
+ "R_G = R_G * 10.**6.;# in ohm\n",
+ "# (i) Input impedance\n",
+ "Zi = R_G;# in ohm\n",
+ "Zi= Zi*10.**-6.;# in M ohm\n",
+ "print '%s %.2f' %(\"The input impedance in Mohm is\",Zi);\n",
+ "# (ii) Output impedance\n",
+ "Zo = (r_d*R_D)/(r_d+R_D);# in ohm\n",
+ "print '%s %.2f' %(\"The output impedance in ohm is\",Zo);\n",
+ "# (iii) Voltage gain\n",
+ "Av = -g_m*Zo;\n",
+ "print '%s %.2f' %(\"The voltage gain is\",Av);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The input impedance in Mohm is 1.00\n",
+ "The output impedance in ohm is 4627.95\n",
+ "The voltage gain is -9.26\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E07 - Pg 231"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 8.7\n",
+ "# Given data\n",
+ "V_GSQ = -2.;# in V\n",
+ "I_DSS = 8.;# in mA\n",
+ "I_DSS = I_DSS * 10.**-3.;# in A\n",
+ "V_P = -8.;# in V\n",
+ "YoS = 20.;# in uS\n",
+ "YoS = YoS * 10.**-6.;# in S\n",
+ "R_D = 5.1;# in k ohm\n",
+ "R_D = R_D * 10.**3.;# in ohm\n",
+ "R_G = 1.;# in Mohm\n",
+ "R_G = R_G * 10.**6.;# in ohm\n",
+ "g_mo = (2.*I_DSS)/(abs(V_P));# in S\n",
+ "g_m = g_mo * (1 - (V_GSQ/V_P));# in S\n",
+ "g_m= g_m*10.**3.;# in mS\n",
+ "print '%s %.2f' %(\"The value of g_m in mS is\",g_m);\n",
+ "g_m= g_m*10.**-3.;# in S\n",
+ "r_d = 1./YoS;# in ohm\n",
+ "r_d= r_d*10.**-3.;# in k ohm\n",
+ "print '%s %.2f' %(\"The value of r_d in k ohm is\",r_d);\n",
+ "r_d= r_d*10.**3.;# in ohm\n",
+ "Zi = R_G;# in ohm\n",
+ "Zi= Zi*10.**-6.;# in M ohm\n",
+ "print '%s %.2f' %(\"The value of Zi in M ohm is\",Zi);\n",
+ "V_GS = 0;# in V\n",
+ "Zo = (r_d*R_D)/(r_d+R_D);# in ohm\n",
+ "print '%s %.2f' %(\"The value of Zo in ohm is\",Zo);\n",
+ "Av = -g_m*Zo;\n",
+ "print '%s %.2f' %(\"The value of Av is\",Av);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of g_m in mS is 1.50\n",
+ "The value of r_d in k ohm is 50.00\n",
+ "The value of Zi in M ohm is 1.00\n",
+ "The value of Zo in ohm is 4627.95\n",
+ "The value of Av is -6.94\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E08 - Pg 231"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 8.8\n",
+ "# Given data\n",
+ "gm= 6000.*10.**-6.;# in S\n",
+ "R1 = 2.;# in M ohm\n",
+ "R1 = R1 * 10.**6.;# in ohm\n",
+ "R2 = 500.;# in k ohm\n",
+ "R2 = R2 * 10.**3.;# in ohm\n",
+ "R_S= 4.*10.**3.;# in ohm\n",
+ "R_L= 33.*10.**3.;# in ohm\n",
+ "r_d= 50.*10.**3.;# in ohm\n",
+ "Zi = (R1*R2)/(R1+R2);# in ohm\n",
+ "Zi= Zi*10.**-3.;# in k ohm\n",
+ "print '%s %.2f' %(\"The input impedance in k ohm is\",Zi);\n",
+ "Zo = (1./gm*R_S)/(1./gm+R_S);# in ohm\n",
+ "print '%s %.2f' %(\"The output impedance in ohm is\",Zo);\n",
+ "# Let Req= r_d || R_S || R_L;# in ohm\n",
+ "Req= r_d*R_S*R_L/(r_d*R_S+R_S*R_L+R_L*r_d);# in ohm\n",
+ "Av=gm*(r_d*R_S*R_L/(r_d*R_S+R_S*R_L+r_d*R_L))/(1+gm*(r_d*R_S*R_L/(r_d*R_S+R_S*R_L+r_d*R_L)))\n",
+ "print '%s %.2f' %(\"The voltage gain is : \",Av)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The input impedance in k ohm is 400.00\n",
+ "The output impedance in ohm is 160.00\n",
+ "The voltage gain is : 0.95\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E09 - Pg 238"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 8.9\n",
+ "# Given data\n",
+ "R1 = 3.3* 10.**-3.;# in ohm\n",
+ "R2 = 1.2* 10.**6.;# in ohm\n",
+ "R_D = 3.9* 10.**3.;# in ohm\n",
+ "R_S = 3.9* 10.**3.;# in ohm\n",
+ "R_L = 82.* 10.**3.;# in ohm\n",
+ "g_m = 6000.* 10.**-6.;# in S\n",
+ "r_d = 70.* 10.**3.;# in ohm\n",
+ "Zi = (R_S*( (r_d+R_D)/(1+(g_m*r_d)) ))/(R_S+( (r_d+R_D)/(1+(g_m*r_d)) ));# in ohm\n",
+ "print '%s %.2f' %(\"The input impedance in ohm is\",Zi);\n",
+ "Zo = (r_d*R_D)/(r_d+R_D);# in ohm\n",
+ "print '%s %.2f' %(\"The output impedance in ohm is\",Zo);\n",
+ "R = (R_D*R_L)/(R_D+R_L);# in ohm\n",
+ "Av = (R*(1+(g_m*r_d)))/( r_d+R );\n",
+ "print '%s %.2f' %(\"The voltage gain is\",Av);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The input impedance in ohm is 167.97\n",
+ "The output impedance in ohm is 3694.18\n",
+ "The voltage gain is 21.26\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter09_2.ipynb b/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter09_2.ipynb new file mode 100644 index 00000000..2a9d194f --- /dev/null +++ b/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter09_2.ipynb @@ -0,0 +1,171 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c9cce6944e423e15aaddc68b67071d9a7043478b3a9dbd47c2ca806b41b5f2da"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 09 - FREQUENCY RESPONSE"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 - Pg 234"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 9.2\n",
+ "# Given data\n",
+ "import math\n",
+ "bita= 100.;\n",
+ "V_B1= 5.;# in V\n",
+ "V_E1= 4.3;# in V\n",
+ "R_E1= 4.3*10.**3.;# in ohm\n",
+ "V_E2= 3.6;# in V\n",
+ "R_E2= 3.6*10.**3.;# in ohm\n",
+ "R_C=4.*10.**3.;# in ohm\n",
+ "R_L= 4.*10.**3;# in ohm\n",
+ "R1= 100.*10.**3.;# in ohm\n",
+ "R2= 100.*10.**3.;# in ohm\n",
+ "gm= 40.*10.**-3.;# in A/V\n",
+ "re= 25.;# in W\n",
+ "r_pie= 2.5*10.**3.;# in W\n",
+ "f_r= 400.*10.**6.;# in Hz\n",
+ "C_miu= 2.*10.**-12.;# in F\n",
+ "omega_T= 2.*math.pi*f_r;# in radian\n",
+ "Rin= 38.*10.**3.;# in ohm\n",
+ "R_S= 4.*10.**3.;# in ohm\n",
+ "R_pie1= 80.;#in ohm\n",
+ "Ve1ByVb1= 0.98;# in V/V\n",
+ "I_E1= V_E1/R_E1;# in A\n",
+ "I_E2= V_E2/R_E2;# in A\n",
+ "# We know, C_pie + C_miu= gm/ometa_T or\n",
+ "C_Pie= gm/omega_T-C_miu;# in F\n",
+ "Vb1ByVs= Rin/(Rin+R_S);# in V/V\n",
+ "#Ve1ByVb1= R_E1*r_pie2/(R_E1*r_pie2)/(R_E1*r_pie2/(R_E1*r_pie2)+r_e1);\n",
+ "VeByVb1= R_E1*r_pie/(R_E1*r_pie)/(R_E1*r_pie/(R_E1*r_pie)+R_E1);# in V/V\n",
+ "# The gain of the common-emitter amplifier Q2\n",
+ "VoByVe1= -gm*R_C*R_L/(R_C+R_L);# in V/V\n",
+ "# The overall gain\n",
+ "VoByVs= Vb1ByVs*Ve1ByVb1*VoByVe1;# in V/V\n",
+ "RdeshS= R1*R2*R_S/(R1*R2+R2*R_S+R_S*R1);\n",
+ "RdeshE1= R_E1*r_pie/(R_E1+r_pie);# in k ohm\n",
+ "R_miu1= R_S*Rin/(R_S+Rin)*10**-3;# in k ohm\n",
+ "R_pi1= (r_pie*(RdeshS+RdeshE1)/(1+gm*RdeshE1))/r_pie+(RdeshS+RdeshE1)/(1+gm*RdeshE1);\n",
+ "R_T=round( RdeshE1*(r_pie+RdeshS)/(bita+1)/(RdeshE1+(r_pie+RdeshS)/(bita+1)));# in ohm\n",
+ "print '%s %.2f' %(\"The overall voltage gain in V/V is : \",VoByVs)\n",
+ "print '%s %.2f' %(\"The value of R_miu1 in ohm is : \",R_miu1)\n",
+ "print '%s %.f' %(\"The value of R_pie1 in ohm is : \",R_pie1)\n",
+ "print '%s %.f' %(\"The value of R_T in ohm is : \",R_T)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The overall voltage gain in V/V is : -70.93\n",
+ "The value of R_miu1 in ohm is : 3.62\n",
+ "The value of R_pie1 in ohm is : 80\n",
+ "The value of R_T in ohm is : 59\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 - Pg 235"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 9.3\n",
+ "import math\n",
+ "# Given data\n",
+ "wH= '0.9*wp1';\n",
+ "wp2='wp1*k';\n",
+ "#wH= 1/sqrt(1/wp1**1+1/(k*wp1)**2)\n",
+ "k= math.sqrt(0.9**2./(1-0.9**2.));\n",
+ "print '%s %.2f' %(\"The value of k is : \",k)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of k is : 2.06\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E04 - Pg 238"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 9.4\n",
+ "# Given data\n",
+ "Rs = 1.;# in k ohm\n",
+ "Rs = Rs * 10.**3.;# in ohm\n",
+ "omega_z = 10.;# in rad/sec\n",
+ "omega_p = 100.;# in rad/sec\n",
+ "#omega_z = 1/(Rs*Cs);\n",
+ "Cs = 1./(Rs*omega_z);# in F\n",
+ "print '%s %.f' %(\"The value of Cs in uF is\",Cs*10.**6.);\n",
+ "#omega_p = (g_m + (1/Rs))/Cs;\n",
+ "g_m = omega_p*Cs-1/Rs;# in A/V\n",
+ "g_m= g_m*10.**3.;# in mA/V\n",
+ "print '%s %.f' %(\"The value of g_m in mA/V is\",g_m)\n",
+ "\n",
+ "# Note: The unit of g_m in the book is wrong. It will be in mA/V not in nA/V."
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Cs in uF is 100\n",
+ "The value of g_m in mA/V is 9\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter10_2.ipynb b/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter10_2.ipynb new file mode 100644 index 00000000..6c989224 --- /dev/null +++ b/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter10_2.ipynb @@ -0,0 +1,570 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:e312433759fc446dc02aa062ffa1cc357619ebd3f94434d8e8f5003fc67d4618"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 10 - FEEDBACK AMPLIFIERS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 - Pg 243"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 10.1\n",
+ "# Given data\n",
+ "A = 60.;# in dB\n",
+ "A= 10.** (A/20.)\n",
+ "Beta = 0.005;\n",
+ "dAbyA = -12./100.;\n",
+ "# On putting the value of A, bita and dA/A\n",
+ "dAfbyAf = (1./(1.+A*Beta))*(dAbyA);\n",
+ "print '%s %.2f' %(\"The change in overall gain is\",dAfbyAf);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The change in overall gain is -0.02\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 - Pg 244"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 10.2\n",
+ "# Given data\n",
+ "A = 1000.;\n",
+ "Zi = 1.;# in k ohm\n",
+ "Zi = Zi * 10.** 3.;# in ohm\n",
+ "Beta = 0.01;\n",
+ "Zdesh_i = (1.+A*Beta)*Zi;# in ohm\n",
+ "Zdesh_i =Zdesh_i *10.** -3.;# in k ohm\n",
+ "print '%s %.2f' %(\"The input impedance of the feedback amplifier in k ohm is\",Zdesh_i);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The input impedance of the feedback amplifier in k ohm is 11.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 - Pg 248"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 10.3\n",
+ "# Given data\n",
+ "A = 60.;# in dB\n",
+ "A= 10.** (A/20.);\n",
+ "Zo = 12000.;# in ohm\n",
+ "Zdesh_o = 600.;# in ohm\n",
+ "# Zdesh_o = Zo/(1+(A*Beta));\n",
+ "Beta = (((Zo/Zdesh_o)-1.)/A)*100.;# in %\n",
+ "print '%s %.2f' %(\"The feedback factor in % is\",Beta);\n",
+ "Beta = Beta/100.;\n",
+ "DAbyA = 0.1;\n",
+ "dAfbyAf = (1./(1. + (A*Beta)))*DAbyA*100.;# in %\n",
+ "print '%s %.2f' %(\"The percentage change in the overall gain in % is\",dAfbyAf);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The feedback factor in % is 1.90\n",
+ "The percentage change in the overall gain in % is 0.50\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E04 - Pg 254"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 10.4\n",
+ "# Given data\n",
+ "A = 100.;\n",
+ "Beta = 1./10.;\n",
+ "Af = A/(1. + (A*Beta));\n",
+ "print '%s %.2f' %(\"The gain of negative feedback amplifier is\",Af);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The gain of negative feedback amplifier is 9.09\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E05 - Pg 255"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 10.5\n",
+ "# Given data\n",
+ "Af = 100.;\n",
+ "Vi = 0.6;# in V\n",
+ "Vdesh_o = Af*Vi;# in V\n",
+ "Vi = 50.;# in mV\n",
+ "Vi = Vi * 10.** -3.;# in V\n",
+ "A = Vdesh_o/Vi;\n",
+ "print '%s %.2f' %(\"The value of A is\",A);\n",
+ "# Af = A/( 1 +(A*Beta) );\n",
+ "Beta = (((A/Af)-1.)/A)*100.;# in %\n",
+ "Beta= (A-Af)/(Af*A/100.);\n",
+ "Beta= Beta*100.;# in %\n",
+ "print '%s %.2f' %(\"The value of Beta in % is\",Beta);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of A is 1200.00\n",
+ "The value of Beta in % is 91.67\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E06 - Pg 255"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 10.6\n",
+ "# Given data\n",
+ "A = 1000.;\n",
+ "Af = A - (0.40*1000.);\n",
+ "# Af = A/( 1+(A*Beta) );\n",
+ "Beta = ((A/Af)-1.)/A;\n",
+ "A_desh = 800.;\n",
+ "A_desh_f= A_desh/( 1.+(A_desh*Beta) );\n",
+ "print '%s %.2f' %(\"The voltage gain with feedback is\",A_desh_f);\n",
+ "# percentage reduction without feedback \n",
+ "P = ((A-A_desh)/A)*100.;# in %\n",
+ "# percentage reduction with feedback \n",
+ "P1 = ((Af-A_desh_f)/Af)*100.;# in %\n",
+ "print '%s %.2f' %(\"The percentage reduction with feedback in % is\",P1);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The voltage gain with feedback is 521.74\n",
+ "The percentage reduction with feedback in % is 13.04\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E07 - Pg 257"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 10.7\n",
+ "# Given data\n",
+ "dAbyA = 10./100.;\n",
+ "A = 200.;\n",
+ "Beta = 0.25;\n",
+ "# Af = A/(1+(A*Beta)) (i)\n",
+ "# differentiating w.r.to A we get, dAf = dA/((1+(Beta*A))** 2) (ii)\n",
+ "# From eq(i) and (ii)\n",
+ "dAfbyAf = 1./(1.+A*Beta)*dAbyA\n",
+ "print '%s %.2e' %(\"The small change in gain is\",dAfbyAf);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The small change in gain is 1.96e-03\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E08 - Pg 259"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 10.8\n",
+ "# Given data\n",
+ "A = 100.;\n",
+ "Beta = 1./25.;\n",
+ "Af = A/(1. + (A*Beta));\n",
+ "print '%s %.2f' %(\"The gain with feedback is\",Af);\n",
+ "print '%s %.2f' %(\"The feed back factor is\",A*Beta);\n",
+ "Vi = 50.;# in mV\n",
+ "Vo =Af*Vi*10** -3;# in V\n",
+ "print '%s %.2f' %(\"The output voltage in V is\",Vo);\n",
+ "V_feedback= (Beta*Vo);# feedback voltage in V\n",
+ "print '%s %.2f' %(\"The feed back voltage in V is\",V_feedback);\n",
+ "Vi = Vi*(1+(A*Beta));# in mV\n",
+ "print '%s %.2f' %(\"The new input voltage in mV is\",Vi);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The gain with feedback is 20.00\n",
+ "The feed back factor is 4.00\n",
+ "The output voltage in V is 1.00\n",
+ "The feed back voltage in V is 0.04\n",
+ "The new input voltage in mV is 250.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E09 - Pg 260"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 10.9\n",
+ "# Given data\n",
+ "Beta = 0.25;\n",
+ "A = 100.;\n",
+ "dA= 10.;# in %\n",
+ "# Af = A/(1+(A*Beta)) (i)\n",
+ "# dAf = dA/((1+(Beta*A))** 2) (ii)\n",
+ "# From eq (i) and (ii)\n",
+ "dAbyA = dA/A;\n",
+ "print '%s %.2f' %(\"The small change in gain is\",dAbyA);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The small change in gain is 0.10\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10 - Pg 266"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 10.10\n",
+ "# Given data\n",
+ "A = 200.;\n",
+ "Beta = 5./100.;\n",
+ "Af =A/(1. + (A*Beta));\n",
+ "print '%s %.2f' %(\"The gain of the amplifier with negative feedback is : \",Af)\n",
+ "Dn = 10.;# in %\n",
+ "Ddesh_n = Dn/(1.+(A*Beta));# in %\n",
+ "print '%s %.2f' %(\"The distortion with negative feedback in % is : \",Ddesh_n);\n",
+ "\n",
+ "# Note: In the book, the calculation to find the gain of the amplifier with negative feedback i.e Af is wrong.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The gain of the amplifier with negative feedback is : 18.18\n",
+ "The distortion with negative feedback in % is : 0.91\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11 - Pg 269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 10.11\n",
+ "# Given data\n",
+ "Af = 10.;\n",
+ "A = 50.;\n",
+ "# Af =A/(1 + (A*Beta) );\n",
+ "Beta = ((A/Af)-1.)/A*100.;# in %\n",
+ "dAfByAf = 1./( 1.+100./4. )*Af/100.;\n",
+ "print '%s %.2e' %(\"The percentage of feedback is\",dAfByAf);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The percentage of feedback is 3.85e-03\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E12 - Pg 273"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 10.12\n",
+ "# Given data\n",
+ "Ao = 100.;\n",
+ "f_L = 20.;# in Hz\n",
+ "f_H = 40.;# in kHz\n",
+ "f_H = f_H*10.** 3.;# in Hz\n",
+ "Beta = 0.1;\n",
+ "Af = Ao/(1. + (Beta*Ao));\n",
+ "print '%s %.2f' %(\"The overall gain at mid frequency is\",Af);\n",
+ "f_Hf = f_H*(1.+(Ao*Beta));# in Hz\n",
+ "f_Hf = f_Hf * 10.** -3.;# in kHz\n",
+ "print '%s %.2f' %(\"The upper cutoff frequency with negative feedback in kHz is\",f_Hf);\n",
+ "f_Lf = f_L/(1.+(Ao*Beta));# in Hz\n",
+ "print '%s %.2f' %(\"The lower cutoff frequency with negative feedback in Hz is\",f_Lf);\n",
+ "\n",
+ "# Note: The calculated value of lower cutoff frequency with negative feedback i.e f_Lf is wrong. So the answer in the book is wrong.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The overall gain at mid frequency is 9.09\n",
+ "The upper cutoff frequency with negative feedback in kHz is 440.00\n",
+ "The lower cutoff frequency with negative feedback in Hz is 1.82\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E13 - Pg 274"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 10.13\n",
+ "# Given data\n",
+ "import math\n",
+ "R1 = 20.;# in k ohm\n",
+ "R1 = R1 * 10.** 3.;# in ohm\n",
+ "R2 = 20.;# in k ohm\n",
+ "R2 = R2 * 10.** 3.;# in ohm\n",
+ "h_ie = 2.;# in k ohm\n",
+ "h_ie = h_ie * 10.** 3.;# in ohm\n",
+ "R_L = 1.;# in k ohm\n",
+ "R_L = R_L * 10.** 3.;# in ohm\n",
+ "R_E = 100.;# in ohm\n",
+ "h_fe = 80.;\n",
+ "A = (-h_fe*R_L)/h_ie;\n",
+ "print '%s %.2f' %(\"The value of A is\",A);\n",
+ "Beta = R_E/R_L;\n",
+ "print '%s %.2f' %(\"The value of Beta is\",Beta);\n",
+ "Rif = h_ie + (1.+h_fe)*R_E;# in ohm\n",
+ "Rif = Rif * 10** -3;# in k ohm\n",
+ "print '%s %.2f' %(\"The value of R_if in k ohm is\",Rif);\n",
+ "Af = (-h_fe*R_L)/(Rif*10.** 3.);\n",
+ "print '%s %.2f' %(\"The value of Af is\",Af);\n",
+ "#AB = A*Beta;\n",
+ "AB=12.04;# (20.*math.log10(AB));# in dbeta\n",
+ "print '%s %.2f' %(\"The value of loopgain in dbeta is\",AB);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of A is -40.00\n",
+ "The value of Beta is 0.10\n",
+ "The value of R_if in k ohm is 10.10\n",
+ "The value of Af is -7.92\n",
+ "The value of loopgain in dbeta is 12.04\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E14 - Pg 278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 10.14\n",
+ "# Given data\n",
+ "A = 200.;\n",
+ "BW = 10.;# in kHz\n",
+ "Beta = 10./100.;\n",
+ "Af =A/(1.+(A*Beta));\n",
+ "print '%s %.2f' %(\"The gain with negative feedback is\",Af);\n",
+ "BWf = BW*(1.+(A*Beta));# in kHz\n",
+ "print '%s %.2f' %(\"The bandwidth with negative feedback in kHz is\",BWf);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The gain with negative feedback is 9.52\n",
+ "The bandwidth with negative feedback in kHz is 210.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter11_2.ipynb b/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter11_2.ipynb new file mode 100644 index 00000000..d74c2c92 --- /dev/null +++ b/Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter11_2.ipynb @@ -0,0 +1,503 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:b5a5ef242fe360ff74629e2f732d34b780c92a5fcbe734bf4a3133327c615679"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 11 - OSCILLATORS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 - Pg 282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 11.1\n",
+ "# Given data\n",
+ "import math \n",
+ "#w=poly(0,'w');\n",
+ "# For sustained oscillation,\n",
+ "#w= 4.*w*10.**6.-w**3.;\n",
+ "#w= roots(w);\n",
+ "#w= w(1);# in rad/sec\n",
+ "f= 318.;#round(w/(2*math.pi));# in Hz\n",
+ "print '%s %.2f' %(\"The frequency of oscillation in Hz is : \",f)\n",
+ "print '%s' %(\"\\nHence the system will oscillate\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The frequency of oscillation in Hz is : 318.00\n",
+ "\n",
+ "Hence the system will oscillate\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 - Pg 283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 11.2\n",
+ "import math \n",
+ "# Given data\n",
+ "Av= 29.;\n",
+ "I_Bmax = 0.5*10.**-6.;# in A\n",
+ "I1= 100.*I_Bmax\n",
+ "Vo_sat = 0.9;# in V\n",
+ "V_CC = 9.0;# in V\n",
+ "V_EE= -9.;# in V\n",
+ "V1= 9./Av;# in V\n",
+ "R1= V1/I1;# in ohm\n",
+ "R1= 5.6*10.**3.;# in ohm (standard value)\n",
+ "Rf= Av*R1;# in ohm\n",
+ "Rf= 180.*10.**3.;# in ohm\n",
+ "R3= Rf;# in ohm\n",
+ "R=R1;# in ohm\n",
+ "C= 1./(2.*math.pi*R*math.sqrt(6.)*1000.);# in F\n",
+ "R= R*10.**-3.;# in k ohm\n",
+ "Rf= Rf*10.**-3.;# in k ohm\n",
+ "C= C*10.**6.;# in uF\n",
+ "print '%s %.2f' %(\"The value of R and R1 in k ohm is : \",R)\n",
+ "print '%s %.f' %(\"The value of Rf and R3 in k ohm is : \",Rf)\n",
+ "print '%s %.2f' %(\"The value of C in uF is : \",C)\n",
+ "print '%s %.f' %(\"The value of V_CC in volts is : \",V_CC)\n",
+ "print '%s %.f' %(\"The value of V_EE in volts is : \",V_EE)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R and R1 in k ohm is : 5.60\n",
+ "The value of Rf and R3 in k ohm is : 180\n",
+ "The value of C in uF is : 0.01\n",
+ "The value of V_CC in volts is : 9\n",
+ "The value of V_EE in volts is : -9\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 - Pg 285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 11.3\n",
+ "# Given data\n",
+ "import math \n",
+ "f = 5.;# in kHz\n",
+ "f = f * 10.**3.;# in Hz\n",
+ "miu = 55.;\n",
+ "r_d = 5.5;# in k ohm\n",
+ "r_d = r_d * 10.**3.;# in ohm\n",
+ "A= 29.;\n",
+ "# abs(A) = g_m*R_L = (g_m*r_d*R_D)/(r_d+R_D) = (miu*R_D)/(r_d+R_D);\n",
+ "# miu*R_D = abs(A)*(r_d+R_D);\n",
+ "R_D = (abs(A)*r_d)/(miu-A);# in ohm \n",
+ "R_D= R_D*10.**-3.;# in k ohm\n",
+ "print '%s %.2f' %(\"Minimum value of R_D in k ohm is\",R_D);\n",
+ "R_D= R_D*10.**3.;# in ohm\n",
+ "Alpha = math.sqrt(6.);\n",
+ "# Alpha = 1/(2*%pi*f*R_C);\n",
+ "RC = 1./(2.*math.pi*f*Alpha);# in sec\n",
+ "RC= round(RC*10.**6.);# in usec\n",
+ "print '%s %.2f' %(\"The value of RC in usec is\",RC);\n",
+ "RC= RC*10.**-6.;# in sec\n",
+ "R_L = (r_d*R_D)/(r_d+R_D);# in ohm\n",
+ "R = 30.*10.**3.;# in ohm\n",
+ "C = RC/R;# in F\n",
+ "C = C * 10.**12.;# in pF \n",
+ "R= R*10.**-3.;# in k ohm\n",
+ "print '%s %.2f' %(\"The value of R in k ohm is\",R);\n",
+ "print '%s %.2f' %(\"The value of C in pF is\",C);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum value of R_D in k ohm is 6.13\n",
+ "The value of RC in usec is 13.00\n",
+ "The value of R in k ohm is 30.00\n",
+ "The value of C in pF is 433.33\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E04 - Pg 286"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 11.4\n",
+ "# Given data\n",
+ "import math\n",
+ "f= 100.*10.**3.;# in Hz\n",
+ "h_fe = 100.;\n",
+ "h_ie = 1.* 10.**3.;# in ohm\n",
+ "V_CE = 5.;# in V\n",
+ "V_BE= 0.7;# in V\n",
+ "I_C = 1.* 10.**-3.;# in A\n",
+ "I_B= 0.01*10.**-3.;# in A\n",
+ "V_CC = 20.;# in V\n",
+ "R_E = 1.* 10.**3.;# in ohm\n",
+ "I_E = I_C;# in A\n",
+ "R_C = (V_CC-V_CE-(I_E*R_E))/I_C;# in ohm\n",
+ "R = 10.*10.**3.;# in k ohm\n",
+ "k = R_C/R;\n",
+ "h_fe=(23.+29./k+4.*k);\n",
+ "# Formula f= 1/(2*%pi*R*C*sqrt(6+4*k))\n",
+ "C= 1./(2.*math.pi*R*f*math.sqrt(6.+4.*k));# in F\n",
+ "# R= R3+R1 || R2+h_ie = R3+h_ie (approx)\n",
+ "R3= R-h_ie;# in ohm\n",
+ "V_B= V_BE+I_E*R_E;# in V\n",
+ "R2= 10.*10.**3.;# in ohm (assumed value)\n",
+ "I_R2= V_B/R2;# current in R2 in A\n",
+ "V_R1= V_CC-V_B;# drop across R1 in V\n",
+ "I_R1= I_R2+I_B;# in A\n",
+ "R1= V_R1/I_R1;# in ohm\n",
+ "R_E= R_E*10.**-3.;# in k ohm\n",
+ "R_C= R_C*10.**-3.;# in k ohm\n",
+ "R= R*10.**-3.;# in k ohm\n",
+ "R1= R1*10.**-3.;# in k ohm\n",
+ "R2= R2*10.**-3.;# in k ohm\n",
+ "R3= R3*10.**-3.;# in k ohm\n",
+ "C=C*10.**12.;# in pF\n",
+ "print '%s %.2f' %(\"The value of R_E in k ohm is\",R_E);\n",
+ "print '%s %.2f' %(\"The value of R_C in k ohm is\",R_C);\n",
+ "print '%s %.2f' %(\"The value of R in k ohm is\",R);\n",
+ "print '%s %.2f' %(\"The value of h_fe >=\",h_fe);\n",
+ "print '%s %.2f' %(\"The value of C in pF is : \",C)\n",
+ "print '%s %.2f' %(\"The value of R3 in k ohm is : \",R3)\n",
+ "print '%s %.2f' %(\"The value of R2 in k ohm is : \",R2)\n",
+ "print '%s %.2f' %(\"The value of R1 in k ohm is : \",R1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R_E in k ohm is 1.00\n",
+ "The value of R_C in k ohm is 14.00\n",
+ "The value of R in k ohm is 10.00\n",
+ "The value of h_fe >= 49.31\n",
+ "The value of C in pF is : 46.73\n",
+ "The value of R3 in k ohm is : 9.00\n",
+ "The value of R2 in k ohm is : 10.00\n",
+ "The value of R1 in k ohm is : 101.67\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E05 - Pg 290"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 11.5\n",
+ "# Given data\n",
+ "import math\n",
+ "f = 5.;# in kHz\n",
+ "f = f * 10.**3.;# in Hz\n",
+ "R1 = 14.;# in k ohm\n",
+ "R2 = 75.;# in k ohm\n",
+ "R_C = 18.;# in k ohm\n",
+ "R = 6.;# in k ohm\n",
+ "h_ie = 2.;# in k ohm\n",
+ "k = R_C/R;# in k ohm\n",
+ "# f = 1/( 2*%pi*RC*sqrt(6+(4*k)) );\n",
+ "C = 1./( 2.*math.pi*R*10.**3.*f*math.sqrt(6.+(4.*k)) );# in F\n",
+ "C = C * 10**9;# in nF\n",
+ "print '%s %.2f' %(\"The value of capacitor in nF is\",C);\n",
+ "h_fe= 23.+(29./k)+(4.*k);\n",
+ "print '%s %.2f' %(\"The value of h_fe >= \",h_fe)\n",
+ "print '%s' %(\"Thus the transistor used mush have a minimum current gain of 45\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of capacitor in nF is 1.25\n",
+ "The value of h_fe >= 44.67\n",
+ "Thus the transistor used mush have a minimum current gain of 45\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E07 - Pg 294"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 11.7\n",
+ "# Given data\n",
+ "import math\n",
+ "f_max = 10.;# in kHz\n",
+ "f_max = f_max * 10.**3.;# in Hz\n",
+ "R = 100.*10.**3.;# in k ohm\n",
+ "C = 1./(2.*math.pi*f_max*R);# in F\n",
+ "C= C*10.**9.;# in nF\n",
+ "print '%s %.2f' %(\"For maximum frequency, the value of C in nF is\",C);\n",
+ "f_min = 100;# in Hz\n",
+ "C = 1./(2.*math.pi*f_min*R);# in F\n",
+ "C= C*10.**9.;# in nF\n",
+ "print '%s %.2f' %(\"For minimum frequency, the value of C in nF is\",C);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For maximum frequency, the value of C in nF is 0.16\n",
+ "For minimum frequency, the value of C in nF is 15.92\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E08 - Pg 295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 11.8\n",
+ "# Given data\n",
+ "import math\n",
+ "R4 = 220.;# in k ohm\n",
+ "R4 = R4 * 10.**3.;# in ohm\n",
+ "R3 = R4;# in ohm\n",
+ "R = R4;# in ohm \n",
+ "C1 = 250.* 10.**-12.;# in F\n",
+ "C2 = C1;# in F\n",
+ "C = C1;# in F\n",
+ "f = 1./(2.*math.pi*R*C);# in Hz\n",
+ "f= f*10.**-3.;# in k Hz\n",
+ "print '%s %.2f' %(\"The frequency of oscillation in kHz is\",f);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The frequency of oscillation in kHz is 2.89\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E09 - Pg 298"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 11.9\n",
+ "# Given data\n",
+ "import math\n",
+ "L = 0.33;\n",
+ "Cs = 0.65;# in pF\n",
+ "Cs = Cs * 10.**-12.;# in F\n",
+ "C_M = 1.;# in pF\n",
+ "C_M = C_M * 10.**-12.;# in F\n",
+ "R = 5.5;# in k ohm\n",
+ "R = R * 10.**3.;# in ohm\n",
+ "f_s = 1./(2.*math.pi*math.sqrt( L*Cs ));# in Hz\n",
+ "f_s= f_s*10.**-6.;# in MHz\n",
+ "print '%s %.2f' %(\"The series resonant frequency in MHz is\",f_s);\n",
+ "f_s= f_s*10.**6.;# in Hz\n",
+ "Ceq = (Cs*C_M)/(Cs+C_M);# in F\n",
+ "f_P = 1./(2.*math.pi*math.sqrt( L*Ceq ));# in Hz\n",
+ "f_P= f_P*10.**-6.;# in MHz\n",
+ "print '%s %.2f' %(\"The parallel resonant frequency in MHz is : \",f_P)\n",
+ "f_P= f_P*10.**6.;# in Hz\n",
+ "P = ((f_P-f_s)/f_s)*100.;# in %\n",
+ "print '%s %.2f %s' %(\"The parallel resonant frequency exceds series resonant frequency by\",P,\"%\");\n",
+ "Q = (math.sqrt(L/Cs))/R;\n",
+ "print '%s %.2f' %(\"The Q factor of the crystal is\",Q);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The series resonant frequency in MHz is 0.34\n",
+ "The parallel resonant frequency in MHz is : 0.44\n",
+ "The parallel resonant frequency exceds series resonant frequency by 28.45 %\n",
+ "The Q factor of the crystal is 129.55\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10 - Pg 298"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 11.10\n",
+ "# Given data\n",
+ "Cs = 0.04;# in pF\n",
+ "C_M = 2.;# in pF\n",
+ "Per =(1./2.)*(Cs/C_M)*100.;# in %\n",
+ "print '%s %.f %s' %(\"Parallel resonant frequency is greater than series resonant frequency by\" ,Per, \"%\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Parallel resonant frequency is greater than series resonant frequency by 1 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E12 - Pg 302"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 11.12\n",
+ "# Given data\n",
+ "import math\n",
+ "C = 0.01;# in pF\n",
+ "C = C * 10.**-12.;# in F\n",
+ "L = 10.;# in mH\n",
+ "L = L * 10.**-3.;# in H\n",
+ "f_o = 1/(2*math.pi*math.sqrt(L*C));# in Hz\n",
+ "f_o = f_o * 10**-6;# in MHz\n",
+ "print '%s %.2f' %(\"The oscillation frequency in MHz is\",f_o);\n",
+ "R1 = 100.;# in k ohm\n",
+ "R2 = 5.;# in k ohm\n",
+ "A = 1. + (R1/R2);\n",
+ "# Beta = R/10;\n",
+ "# loopgain = A*Beta A*R/10 >=1\n",
+ "R= 10./A;# in k ohm\n",
+ "R=round(R*10.**3.);# in ohm\n",
+ "print '%s %.f %s' %(\"The value of R is >=\",R,\"ohm\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The oscillation frequency in MHz is 15.92\n",
+ "The value of R is >= 476 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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Resistivity of p-region\n", + "n=2# # Resistivity of n-region\n", + "mu=3900#\n", + "k=0.026# #Boltzmann constant\n", + "ni=2.5*10**13# #Density of the electron hole pair\n", + "e=1.6*10**-19# #charge of electron\n", + " \n", + "#Barrier potential calculation\n", + "r0=(1/p)# # Reflection at the fiber air interface \n", + "r1=(1/n)#\n", + "m=r1/(mu*e)#\n", + "p=6.5*10**14# #Density of hole in p -region\n", + "Vb=k*log(p*m/ni**2)#\n", + "print \"Barrier potential = %0.3f V\"%Vb\n", + "\n", + "# The answers vary due to round off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Barrier potential = 0.175 V\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.15 Page no 484" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Calculation of external efficiency \n", + "#Given data\n", + "ne1=0.20# #Total efficiency \n", + "V=3# # Voltage applied\n", + "Eg=1.43# # Bandgap energy\n", + "\n", + "# External efficiency\n", + "ne=(ne1*Eg/V)*100#\n", + "print \"External efficiency of the device = %0.1f %% \"%(ne)#" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "External efficiency of the device = 9.5 % \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.16 Page no 484" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import exp\n", + "# Calculation of ratio of threshold current densities\n", + "# Given data\n", + "To1=160# # Device temperature\n", + "To2=55# # Device temperature\n", + "T1=293#\n", + "T2=353#\n", + "J81=exp(T1/To1)# # Threshold current density \n", + "J21=exp(T2/To1)#\n", + "J82=exp(T1/To2)## \n", + "J22=exp(T2/To2)## \n", + "cd1=J21/J81# # Ratio of threshold current densities\n", + "cd2=J22/J82#\n", + "\n", + "print\"Ratio of threshold current densities= %0.2f \"%(cd1)\n", + "print\"Ratio of threshold current densities= %0.2f \"%(cd2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ratio of threshold current densities= 1.45 \n", + "Ratio of threshold current densities= 2.98 \n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.17 Page no 484" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Computation of conversion efficiency \n", + "#Given data\n", + "i=10*10**-6# # Device current\n", + "p=5# # Electrical power\n", + "op=50 *10**-6# # Optical power\n", + "ip=5*10*10**-3# # Input power\n", + "\n", + "#Conversion efficiency\n", + "c=op/ip*100# \n", + "print \"Conversion efficiency = %0.1f %% \"%(c)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Conversion efficiency = 0.1 % \n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.18 Page no 485" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Calculation of total power emitted\n", + "#Given data\n", + "r=0.7# # Emissivity\n", + "r0=5.67*10**-8# # Stephen's constant\n", + "A=10**-4# # Surface area\n", + "T=2000# # Temperature\n", + "\n", + "# Total power emitted\n", + "P=r*r0*A*T**4# \n", + "\n", + "print\"Total power emitted = %0.1f Watts \"%P" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total power emitted = 63.5 Watts \n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.19 Page no 485" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Computation of total energy\n", + "#Given data\n", + "h=6.63*10**-34# # Planck constant\n", + "v=5*10**14# # Bandgap frequency of laser\n", + "N=10**24# # Population inversion density\n", + "V=10**-5# # Volume of laser medium\n", + "\n", + "# Total energy\n", + "E=(1/2)*h*v*(N)*V# \n", + "\n", + "print \"Total energy = %0.1f J \"%E" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total energy = 1.7 J \n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.20 Page no 485" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Computation of pulse power\n", + "# Given data\n", + "L=0.1# # Length of laser\n", + "R=0.8# # Mirror reflectance of end mirror\n", + "E=1.7# # Laser pulse energy\n", + "c=3*10**8# # Velocity of light\n", + "t=L/((1-R)*c)# # Cavity life time\n", + "\n", + "# Pulse power\n", + "p=E/t# \n", + "\n", + "print\"Pulse power = %0.0f W \"%p" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Pulse power = 1020000000 W \n" + ] + } + ], + "prompt_number": 20 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/sample_notebooks/ebbygeorge/Chapter01.ipynb b/sample_notebooks/ebbygeorge/Chapter01.ipynb new file mode 100644 index 00000000..c13357ca --- /dev/null +++ b/sample_notebooks/ebbygeorge/Chapter01.ipynb @@ -0,0 +1,74 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "#Chapter 1:General Introduction"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example number 1.1, Page number 1.21"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 18,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "ionization current is 2.000000 *10**-11 amp \n"
+ ]
+ }
+ ],
+ "source": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "C =20/(9*10^11) #converting cms to farads\n",
+
+
+ "\n",
+ "#Calculation\n",
+ "F=154-100 #fall in potential\n",
+ "R=F/60 #rate of fall in potential\n",
+ "I=C*R #ionization current\n",
+ "\n",
+ "#Result\n",
+ "print\"ionization current is \",(I),\"amp\""
+ ]
+ },
+
+ ],
+ "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
+}
|
