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| author | Trupti Kini | 2016-01-01 23:30:07 +0600 |
|---|---|---|
| committer | Trupti Kini | 2016-01-01 23:30:07 +0600 |
| commit | 3b01cb04f651e64f9df771578149f1dd90754676 (patch) | |
| tree | 1e1df5fc269201d5b78c3f84902c1af75cd69b66 | |
| parent | 9daed43fd5f272f8d29ae36f5d924070563ec279 (diff) | |
| download | Python-Textbook-Companions-3b01cb04f651e64f9df771578149f1dd90754676.tar.gz Python-Textbook-Companions-3b01cb04f651e64f9df771578149f1dd90754676.tar.bz2 Python-Textbook-Companions-3b01cb04f651e64f9df771578149f1dd90754676.zip | |
Added(A)/Deleted(D) following books
A Electronics_Devices_and_Circuits_by_G._S._N._Raju/README.txt
A Electronics_Engineering_by_P._Raja/chapter_1_1.ipynb
A Electronics_Engineering_by_P._Raja/chapter_2_1.ipynb
A Electronics_Engineering_by_P._Raja/chapter_3_1.ipynb
A Electronics_Engineering_by_P._Raja/chapter_4_1.ipynb
A Electronics_Engineering_by_P._Raja/chapter_5_1.ipynb
A Electronics_Engineering_by_P._Raja/chapter_6_1.ipynb
A Electronics_Engineering_by_P._Raja/chapter_7_1.ipynb
A Electronics_Engineering_by_P._Raja/chapter_8_1.ipynb
A Electronics_Engineering_by_P._Raja/chapter_9_1.ipynb
A Electronics_Engineering_by_P._Raja/screenshots/7_1.png
A Electronics_Engineering_by_P._Raja/screenshots/snap-3_1.png
A Electronics_Engineering_by_P._Raja/screenshots/snap-6_1.png
A Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_1_1.ipynb
A Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_2_1.ipynb
A Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_3_1.ipynb
A Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_4_1.ipynb
A Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_5_1.ipynb
A Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_6_1.ipynb
A Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_7_1.ipynb
A Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_8_1.ipynb
A Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/1_1.png
A Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/6_1.png
A Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/7_1.png
A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/README.txt
A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/README.txt
A Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/README.txt
A Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap_/README.txt
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_10_SILICON_CONTROLLED_RECTIFIER_1.ipynb
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_1_CRYSTAL_STRUCTURES_1.ipynb
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_6_ELECTRICAL_BREAKDOWN_IN_PN_JUNCTIONS_1.ipynb
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_2_ENERGY_BAND_THEORY_OF_SOLIDS_1.ipynb
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_3_CARRIER_TRANSPORT_IN_SEMICONDUCTOR_1.ipynb
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_4__EXCESS_CARRIER_IN_SEMICONDUCTOR_1.ipynb
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_5_PN_JUNCTION_DIODE_1.ipynb
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_7_BIPOLAR_JUNCTION_TRANSISTORB_1.ipynb
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_8_THE_FIELD_EFFECT_TRANSISTOR_1.ipynb
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2015-11-05_at_11.31.11_pm_1.png
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2015-11-05_at_11.32.51_pm_1.png
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2015-11-05_at_11.33.45_pm_1.png
40 files changed, 19298 insertions, 0 deletions
diff --git a/Electronics_Devices_and_Circuits_by_G._S._N._Raju/README.txt b/Electronics_Devices_and_Circuits_by_G._S._N._Raju/README.txt new file mode 100644 index 00000000..6ffdd616 --- /dev/null +++ b/Electronics_Devices_and_Circuits_by_G._S._N._Raju/README.txt @@ -0,0 +1,10 @@ +Contributed By: vivekkumar gupta +Course: btech +College/Institute/Organization: iitbombay +Department/Designation: aerospace engnieering +Book Title: Electronics Devices and Circuits +Author: G. S. N. Raju +Publisher: I. K. International, New Delhi +Year of publication: 2006 +Isbn: 81-89866-02-8 +Edition: 1
\ No newline at end of file diff --git a/Electronics_Engineering_by_P._Raja/chapter_1_1.ipynb b/Electronics_Engineering_by_P._Raja/chapter_1_1.ipynb new file mode 100644 index 00000000..18461132 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_1_1.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_1.ipynb b/Electronics_Engineering_by_P._Raja/chapter_2_1.ipynb new file mode 100644 index 00000000..cc8b5565 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_2_1.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_1.ipynb b/Electronics_Engineering_by_P._Raja/chapter_3_1.ipynb new file mode 100644 index 00000000..600220b3 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_3_1.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_1.ipynb b/Electronics_Engineering_by_P._Raja/chapter_4_1.ipynb new file mode 100644 index 00000000..346bf695 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_4_1.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 0x7f9b900aba50>" + ] + } + ], + "prompt_number": 109 + }, + { + "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 0x7f9ba804b910>" + ] + } + ], + "prompt_number": 110 + }, + { + "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 0x7f9b92fd8850>" + ] + } + ], + "prompt_number": 91 + }, + { + "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_1.ipynb b/Electronics_Engineering_by_P._Raja/chapter_5_1.ipynb new file mode 100644 index 00000000..54ba0032 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_5_1.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_1.ipynb b/Electronics_Engineering_by_P._Raja/chapter_6_1.ipynb new file mode 100644 index 00000000..a9fbbd80 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_6_1.ipynb @@ -0,0 +1,1316 @@ +{ + "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 = 2.5 mA\n", + "The minimum value of V_DS = -4 V\n" + ] + } + ], + "prompt_number": 31 + }, + { + "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\njoSYOyr8xKwUFPDhjuho0ZGIERPDh7kIqQoq/MSsbN8O2Nnx3q8W9e8PJCUBf/0lOhJizqjwE7Oy\nYAHv9WqVrS0/VzghQXQkxJzRzV1iNrKy+Fm0aWnAYxtjVpg53twtsmUL8PHHwNGjEgZFVINu7hJV\nWboU6Nq16kXf3L36KnDrFnD8uOhIiLmiwk/MAmPADz8Aw4aJjkQ8S0s+3DV/vuhIiLmioR5iFg4f\n5tsvX7gAWEjQXTHnoR4AuHKFb9l87RpQq5ZEgRFVoKEeoho//AAMHSpN0VcDZ2egTRu+dQUhFUU9\nfqJ4RRuynT4NODpKc01z7/EDwI8/ArNnAzt3ShAUUQ3q8RNVWLUKaN9euqKvFq+/zn8ZXrwoOhJi\nbqjwE8UrGuYhJVlZ8UPmf/hBdCTE3NBQD1G0U6eATp34oerVJDwoVA1DPQBw7hwQHMxv9lpZSXJJ\nYuZoqIeYvXnz+NRFKYu+mri7Ay1aAOvWiY6EmBPq8RPFysvjN3WPHAFcXKS9tlp6/ABf2LZ0KbBp\nk2SXJGaMevzErP30E9C6tfRFX20iIvgvx7Q00ZEQc0GFnyjW998DI0aIjkL5atUCBg6klbzEeDTU\nQxSp6Kbl1atA9erSX19NQz0An9YZFsZvgsvx8yLmg4Z6iNmaO5ff1KUiZhwvL75z6fr1oiMh5oB6\n/ERxim7qHj4MNG0qTxtq6/ED/AbvkiXA5s2SX5qYEerxE7O0ahUQGChf0Ver3r2BY8eA1FTRkRCl\no8JPFOe774CRI0VHYX5q1gSGDOE3xQkpDw31EEU5eRLo1o1PTZRz0ZYah3oA4Px5vq/RlStAjRqy\nNEEUjoZ6iNmZO5cftkIrdSvHzQ3w9gZWrxYdCVEy6vETxbh3D3juOSAlBWjcWN621NrjB/h2zbNm\nAb/+KlsTRMGox0/MytKlQGio/EVf7bp351s1nzolOhKiVMIK/7Rp0+Dl5QUfHx9ERUXh4cOHokIh\nCsAYMGcO8OaboiOpOtG5Xb06Hy777juTNkvMiJDCn56ejnnz5uHYsWNISUlBQUEBEhMTRYRCFGLv\nXkCv5z1+c6aU3B4+HFi+nA+fEfI4IYW/Tp06qF69OnJzc5Gfn4/c3Fw0pvf3mvbtt7y3r9OJjqRq\nlJLbTk5ASAiwbJnJmyZmQEjht7W1xQcffABnZ2c4OjqiXr166NChg4hQiAJkZgIbN/I56OZOSbn9\n1lv8FyrNkSCPEzJp7tKlS/j666+Rnp6OunXrok+fPli2bBkGDBhQ4nkTJ040fBwcHIzg4GDTBkpM\n4ocf+KrTevXkayM5ORnJycnyNfA3JeV2aCgfPtuzh8/tJ+pUmdwWMp1zxYoV2Lp1K374+7DQJUuW\n4MCBA5g9e/Y/gdF0Tk3Iz+dbM6xbB/j7m65dufJLabk9axYv/CtWmKQ5ogCKnc7p7u6OAwcOIC8v\nD4wxbNu2DZ6eniJCIYKtXcsPWjFl0ZeT0nJ7yBBg61bg+nVhIRAFElL4/fz8MHjwYLRu3Rq+vr4A\ngBF04oYmffMN8M47oqOQjtJyu04doH9/2r+HlEQrd4kwp04BHTsC6emAlZVp21bzyt3HnTkDvPoq\nP6TF1D9nYnqKHeohBABmz+ZHK1IxkpenJ+DhwbdyIASgHj8RJDub39Q9fRpwdDR9+1rq8QN807bp\n04F9+0zaLBGAevxEsebPB157TUzR16LwcODGDX6qGSFU+InJFRTwm7r/+pfoSLSjWjXg7beBmTNF\nR0KUgHY9Jya3fj1gbw8EBYmORFuGDeMHsmdkAI0aiY6GiEQ9fmJyM2dSb1+E+vWBfv34YTdE2+jm\nLjGplBSgUycxUziL09rN3SKnTwMdOvCfPx3NqE50c5coTlwc34WTpnCK4eXFj2akLRy0jXr8xGT+\n+ANo0YIfCN6wodhYtNrjB4CkJOCTT4Bjx8x/G2zyJOrxE0WZOxeIiBBf9LWuc2cgLw/YtUt0JESU\nShX+vLw8rFq1SupYiIo9fMj3hh81SnQk5dNCbltYAO+9B/zvf6IjIaIYXfgLCgqwYcMGDBw4EC4u\nLnRUIqmQFSsAHx8+vqw0WsztwYP5ds2XLomOhIhQ7hg/Ywy//vorEhISkJSUhKCgIOzevRtpaWmw\ntraWNzAa41cNxoCWLYEpU/hqXSXQ6XTYuXOnpnN73Dg+5BMXJzoSIiVj8qvcwu/k5ARPT0/ExMQg\nPDwczzzzDJo2bYq0tDTJg30iMIW8OEjV7djBV42ePs2HGZRAp9MhLCxM07l97Rrg68t7/fXri46G\nSKXKN3d79+6NixcvYsWKFVi3bh3u378vaYBEG2bMAEaPVk7RL6L13HZyArp1o736teip0zkLCwuR\nnJyMhIQEbNy4EdnZ2Zg/fz66du0KGxsb+QJTSK+IVM3Zs0BICF8wVLOm6Gj+odPpUFBQoPncPnkS\n6NoVSE2ltRVqUeWhnsfp9Xps3rwZCQkJ2Lx5M27fvl3lIMsMTEEvDlJ5w4fznuVnn4mOpKTH80vL\nud2hAz+icdAg0ZEQKUhe+IvLy8tDrVq1KhWYMZT24iAVl5kJuLsrY8HW48rLL63l9saN/EbviRO0\noEsNJFvAtW7dOgQEBKB+/fqoXbs2ateuDXt7e0mCJOo1ezbQt6/yin5xlNt8QVdBAbBtm+hIiKkY\n1eN3dXXF6tWr4e3tDQsT3aFTWq+IVMz9+/yErT17ADc30dE8qSi/KLe5+Hhg+XJgyxbRkZCqkqzH\n7+TkBC8vL5O9MIj5W7AAaN9emUW/OMptLiqKH8p+4oToSIgpGNXjP3DgAD799FOEhITA6u9b/zqd\nDqNHj5YvMAX2iohx8vOB5s2BhASgbVvR0ZSuKL8ot/8xfTov/MuWiY6EVIUx+WXUCVwTJkxA7dq1\n8eDBA+j1ekmCI+r1449AkybKLfrFUW7/Y8QIfkLX5cvAc8+JjobIyagev7e3N06dOmWKeAyU2isi\n5WMMaNUKmDyZLw5SqqL8otwuaexYvqEebeNgviQb43/ttdewefNmSYIi6rZ1Ky8cStmT52kot0sa\nNQpYsgS4dUt0JERORvX4bWxskJubCysrK1SvXp1/oU6Hu3fvyheYgntFpGyhoUB0tPIXAxXlF+X2\nk4YPBxo3BiZOFB0JqQxZF3DJTekvDvKkgwf5Yd4XLgB/11DF0vIJXE9z4QLQrh2QlgbIuHMFkQmd\nwEVMKjYWGDNG+UWflK95c76/0rx5oiMhchFW+LOzs9G7d294eHjA09MTBw4cEBUKkcCZM8D+/UBM\njOhIxFNDbo8bx3dVffhQdCREDsIK/3vvvYfXXnsNZ8+exW+//QYPDw9RoRAJxMYC774LyHyGiVlQ\nQ263bAl4efEbvUR9jB7jLygoQGZmJvLz8w2fc3Z2rlSjd+7cQUBAAFJTU8sOTOHjoOQfqalAYCBw\n8SJQr57oaIxTPL8ot0u3axd/B3fuHFDNqBU/RAkkW8A1a9YsTJo0CXZ2drC0tDR8PiUlpVKBpaWl\noWHDhoiOjsbJkyfRqlUrxMXFyX7kHZHHV18B//d/5lP0i6PcLtvLLwMODsDKlXxLB6IeRm/SdujQ\nITRo0ECSRo8cOYIXXngB+/btQ5s2bTBq1CjUqVMHkydP/icwnQ6fFdvEPTg4GMHBwZK0T6Rz/To/\nRP3335W9C2dycjKSk5MN/540aZJhkzbK7bJt2sRv2P/2m/JOUCNcWbldLmaE4OBgptfrjXmqUW7e\nvMlcXFwM/969ezfr2rVriecYGRoR7P33+R9zU5RflNvlKyxkrFUrxlavFh0JMZYx+WXUUE/Tpk0R\nEhKCrl27SrKRVaNGjdCkSROcP38ebm5u2LZtG7y8vCp1LSLOn3/y7XwrOSqiCJTb5dPpgE8+Ab74\nAujenQ5qUQujCr+zszOcnZ2h1+uh1+vBGIOuihkwa9YsDBgwAHq9Hq6urli4cGGVrkdMb8YMoH9/\nvsrTXFFuP1337vzozI0bzWcrDlI+WrlLKuXWLaBFC+D4caCSE2CEopW7FbNqFf9Fv38/9fqVrspb\nNrz33nuIi4tDeHh4qRf/5Zdfqh5lWYGZ4YtDSz75hBf/uXNFR1I5RflFuW2cwkJ+E/+//wU6dRId\nDSlPlQv/0aNH0apVqxJ3jItf/JVXXqlykGUGZoYvDq3IyuLL+o8eBVxcREdTOUX5RbltvIQEYNYs\nYO9e6vUrGW3SRmTx6ad8Guf8+aIjqTwa6qm4ggK+mvebb4AOHURHQ8pChZ9Irqi3f/gwP63JXFHh\nr5xly4BvvwX27KFev1LR7pxEcv/9L9Czp3kXfVJ5kZH8l//WraIjIVVBPX5itNu3ATc38x7bL0I9\n/spLTAS+/ppm+CiVJD3++Ph4tGzZEtbW1rC2tkbr1q2xaNEiyYIk5mPGDCAiwvyLfhHK7crp0we4\nd49v50DMVHnLeuPj45m/vz/bsWMH++uvv1hWVhbbvn07a9myJVu0aJERi4cr7ymhERP74w/GbG0Z\nS08XHYk0AFBuV8HKlYy1bs23dCDKYkx+lTvUExQUhMTERDRt2rTE59PT09GvXz8cPHhQtl9I5v52\nWG3GjAHy8oDZs0VHIg2dTofU1FTK7UoqLAQCAoDJk/nKXqIcVd6W+d69e0+8MADAxcUF9+7dq1p0\nxGzcuAEsWACcOiU6EmlRbleehQXw+efA+PFAeDjt3Gluyv3vqlmzZqUeI+oyZQo/kMPRUXQkpkG5\nbZzwcKBWLb5fPzEv5Q711KpVC82aNSv1sUuXLiE3N1e+wFTwdlgN0tOBVq34KUxK3m+/onQ6Hby9\nvUt9jHLbeNu2AW+/DZw+Tad0KUWVF3Clp6eX+8UuMk7vUNOLw5xFRwNOTvxtvZrodDqkpaWV+Tjl\ntnEYA0JCgMGD+btCIp7JVu6+8MIL2L9/f1UvU4KaXhzm6vRp/qK+cAGoW1d0NNIyNr8ot59u3z6+\nsOv8eYBGycQz2crdBw8eSHEZojDjxwNjx6qv6FcE5fbTtWvHZ/jMmSM6EmIsuhdPSnXgAHDkCB+/\nJeRppk4FYmOBu3dFR0KMQYWfPIExYNw4YOJEPmuDkKfx8gK6dOGru4nyUeEnT9i0CcjIAIYMER0J\nMScTJ/ItmzMyREdCnkaSm7spKSnw8fGRIh4Dtd0AMxcFBf+syOzRQ3Q08jE2vyi3K+aDD/gK72+/\nFR2JdlV5Vo+NjU2ZB0/rdDrclXFAT80vDiWLjwfmzVP/fus6nQ42NjZlPka5XTm3bwPu7jx/WrQQ\nHY020UEspELy8viLNTGRz9RQM9qWWT5ffcUnB/z8s+hItIkOYiEVMnMm0KaN+os+kde77/IZYXv3\nio6ElIV6/AQA8OefgIcHX4zj5iY6GvlRj19eixfzef379ql7yFCJqMdPjDZxIhAVpY2iT+Q3cCCg\n19MGbkpFPX6Cs2eBl1/mG7E1aCA6GtOgHr/8kpP5Xk9nz9JWDqZEPX5ilA8/5Au2tFL0iWkEBwN+\nfvzeEVEW6vFr3NatwMiRwJkzQI0aoqMxHerxm8b583yywJkzgJ2d6Gi0gXr8pFz5+cD77wP/+Y+2\nij4xHTc3vmXz+PGiIyHFCS38BQUFCAgIQHh4uMgwNGvuXMDeXt0rdEWgvC7p00+BX34BTpwQHQkp\nIrTwx8XFwdPTs8zVwUQ+WVnApEnA11/TdDupUV6XVK8ez7X33uMbABLxhBX+a9euISkpCcOGDdPM\neKeSTJwI9O4NSLwNjeZRXpdu2DAgOxv48UfRkRBAYOF///33MX36dFhY0G0GU0tJ4dsyTJ4sOhL1\nobwunaUln90zZgwg43HGxEhCjkdev3497OzsEBAQgOTk5DKfN3HiRMPHwcHBCA4Olj02tWOML6mf\nNAl49lnR0ZhOcnJyubkmBWPzGtBmbr/yCp/hM22a+s5wFqkyuS1kOue///1vLFmyBNWqVcODBw9w\n9+5dREREYPHixf8EpqEpb6aUmAh8+SXfS8XSUnQ04siRX8bktVxtm4tr1wB/f+DgQcDVVXQ06mQW\nu3P++uuv+M9//oN169aV+LyWXxxyycnh+/EkJgIvvig6GrHkzq+y8toUbSvdl1/ybZtL+dEQCZjN\nPH6a/WBknV/pAAAXGklEQVQakycDoaFU9E2F8rp0o0bxhV1U+MUR3uMvi9Z7RVI7dYoX/ZQUPndf\n62jlrljbtgHDhwOnTwPW1qKjURez6fETeTEGvPUWv6FLRZ8oQYcOQNu2wJQpoiPRJurxa8CiRfwQ\n7AMHtH1Dtzjq8Yt34wbg6wvs3s3vPRFpmMXN3bLQi0Mat28DXl7A+vVA69aio1EOKvzKEBcHrF4N\n7NxJK8ilQkM9BGPGAP36UdEnyvTOO3y2WXy86Ei0hXr8KrZzJzBkCL+BVru26GiUhXr8ynH8ONC5\nM5+A0LCh6GjMH/X4NezBA+D//o+P7VPRJ0oWEAAMGgSMHi06Eu2gHr9KjR/Pj7z76SfRkSgT9fiV\n5f59wNsb+O47oFMn0dGYN7q5q1EnTgAdOwInTwIODqKjUSYq/MqzZQswYgRfa0LvUiuPCr8G5ecD\nQUH8pll0tOholIsKvzJFR/OiT+f0Vh4Vfg368ku+KnLLFpoeVx4q/MqUlcXPiFi5krYWqSwq/Bpz\n7hzw0kvA4cNA06aio1E2KvzK9fPPwMcf8yHLWrVER2N+qPBrSH4+7yENGcK3ZyDlo8KvbP37A46O\nwIwZoiMxP1T4NeTLL/nwztatAB3+9HRU+JXt9u1/hnxeekl0NOaFCr9GnDnDTzc6fBhwcREdjXmg\nwq98a9YAH37IZ6fRDp7Go8KvAXo98MILfLHWiBGiozEfVPjNw6BBQN26fCEiMQ4Vfg2YMIEveV+3\njmbxVAQVfvOQnc138Jw3jxZ2GYsKv8rt3w/07MlnPzRqJDoa80KF33xs384nLZw8CTRoIDoa5aO9\nelQsJwcYPBj49lsq+kTdXn0V6NMHePNNfqgQqTrq8ZupmBj+94IFYuMwV9TjNy8PHgBt2vCN3GhF\nevmMya9qJoqFSGjFCmDPHuDYMdGREGIaNWsCCQlASAif3tm8ueiIzBv1+M3M5cu857NxI9Cqleho\nzBf1+M3T7Nn80Ja9ewErK9HRKBPd3FWZR4+A4GCge3dg7FjR0Zg3KvzmiTGe/y1aANOni45Gmajw\nq0zR/iUbNtDq3Kqiwm++bt/mh7fMmQN07So6GuWhwq8imzcDw4bxcX06nq7qqPCbt717gYgI4MgR\nwMlJdDTKQtM5VeL6deCNN4ClS6noEwLwDQlHjQIiI/kQKKkY6vErnF7Px/XDw/lQD5EG9fjNX2Eh\nf124u9MunsXRUI8KjBoFpKbyDatoXF86VPjVISuLz26bPh3o3Vt0NMpA8/jN3IoVfA+eI0eo6BNS\nGltbYNUqoEsXfli7u7voiMyDkHJy9epVhISEwMvLC97e3phJB2w+4bff+Lm5P/4I1K8vOhpiLMpt\n02vdGpg2je9bdfeu6GjMg5ChnoyMDGRkZMDf3x85OTlo1aoV1qxZAw8Pj38C0/Db4awsvkjr88+B\nqCjR0aiTXPlFuS3Om28CN24Aq1dr+x2yYmf1NGrUCP7+/gAAGxsbeHh44MaNGyJCUZz8fH7sXM+e\nVPTNEeW2OHFxfI7/55+LjkT5hI/xp6en4/jx4wgKChIdiiJ8+CFfnRgbKzoSUlWU26ZlZcWHRgMD\n+Xh/RIToiJRLaOHPyclB7969ERcXBxsbG5GhKML33wNJScCBA0A14b+SSVVQbovRqBGfAdepE9C0\nKdCypeiIlElYeXn06BEiIiIwcOBA9OjRo9TnTJw40fBxcHAwgoODTROcADt38tO09uyhm7lySE5O\nRnJysknaotwWq2VL4Lvv+J4+Bw8Cjo6iI5JXZXJbyM1dxhiGDBmCBg0a4H//+1+pz9HSDbCzZ/ki\nrYQEIDRUdDTaIFd+UW4rx9SpfOhn1y5AS2+6FLuAa8+ePXj55Zfh6+sL3d8HxU6bNg2dO3f+JzCN\nvDgyM/lh6Z99xo+XI6YhV35RbisHY8Dw4cDNm8DatdoZPlVs4TeGFl4c9+/znn7XrkCxd/7EBGjl\nrjY8egR068bH++fMAf7+XaxqVPgV7NEjPgZpb8+PT9RCQioJFX7tuHsXePllPstnwgTR0ciPtmxQ\nqMJCfmauhQWfyUNFnxD51KnDT6x76SXe0RoxQnRE4lHhNzHG+Fz91FRg61agenXRERGifg4O/EyL\nl18GGjSgOf5U+E1s8mRe8JOTAWtr0dEQoh3NmvHT6zp3Bp55hv+tVRre0cL0ZswAli/nhd/WVnQ0\nhGhPQADfy2fwYODXX0VHIw4VfhOZPRv45htg2zY+zkgIEaNdOyAxEejTB9i3T3Q0YlDhN4HZs/lB\nETt2AE2aiI6GEBIaCixeDPTooc3iT4VfZt9+y4v+zp18LjEhRBk6d9Zu8afCL6MZM/7p6VPRJ0R5\niop/9+68c6YVVPhlwBgwaRKfo79rF/D886IjIoSUpXNnYOVKoG9fvjuuFlDhl1hhIfD++8BPP/Gi\nT2P6hChfSAg/3zo6mm+WqHY0j19Cej3wxhvAtWt8qhhtr0yI+WjbFti+nR/cnpkJjBolOiL5UI9f\nInfu8M3W8vL4CkEq+oSYH29vfibG3LnA2LH8HbwaUeGXQHo68OKLQIsWfP/vWrVER0QIqaznnuPF\n/8ABPu6fmys6IulR4a+iAwf4gpARI/gCLUtL0RERQqqqQQO+wr5WLb51+s2boiOSFhX+KliwAHj9\ndT5751//Eh0NIURKNWrwqZ6vvw60acOPcVQL2o+/EvR64IMPgC1b+Mk+7u6iIyIVRfvxk4pYtw4Y\nOhSYNo3/rWR0EIsMLl/m436NGgGLFgH16omOiFQGFX5SUWfP8u2cg4L4NixK3V3XmPyioZ4K+OUX\nIDCQb+60Zg0VfUK0xMMDOHQIyM/ndeD0adERVR71+I1w/z4f2tm8GVi6lM/gIeaNevykshjj9/fG\njQM++wx4+21lnaJHPX4J7N8PtGoF5OQAJ05Q0SdE63Q6Ps6/dy8f7n3tNb5o05xQ4S9DXh4wZgzQ\nqxc/NWvpUqBuXdFREUKUws2N7+r5wgv8gJcffuDvBswBDfWUYuNG4J13+DjezJlAw4ZCwiAyoqEe\nIqWUFL7Pj40N34rd01NcLDTUU0Hp6UDv3sC77/K79gkJVPQJIU/n48Pn+ffuDbzyCh//v3dPdFRl\no8IPvs/OuHF8LN/Xl//21vJBzISQirO05CMFv/3GV/q6uQHz5gEFBaIje5KmC//9+8CXXwLNmwN/\n/MEL/qef0l47hJDKc3DgN33XrweWLeMbv61YoawN3zQ5xp+dDXz3HRAXB7z8MjBxIp+jS7SDxviJ\nKTDG9/yZMIFv9jZuHF8AWr26fG3Syt3HXLoEzJkDLFzIp2CNHcvH5oj2UOEnpsQYsGkT8NVXQFoa\n39srOlqe7dsVfXN306ZNcHd3R/PmzfHll1/K1s7Dh/w0rC5d+EELFhbAsWPAkiVU9Ik8TJXbxHzo\ndLwG7dzJj3k8fpwfyRoTA+zeLWAaKBMgPz+fubq6srS0NKbX65mfnx87c+ZMiedUJbSHDxnbtImx\n4cMZs7VlLDSUsUWLGMvNLfm8nTt3VrqNijBFO2ppw1TtyJX6cue2sSgflN9GZiZjX37JmJcXY02b\nMvbvfzN2+DBjhYVVa8eY/BLS4z906BCaNWsGFxcXVK9eHZGRkVi7dm2lr8cYcPEiPzWnTx++gdrk\nyfyu+vHj/Di1wYOfvGmbnJxctW/ESKZoRy1tmLIdOUid25VF+aD8Nuzs+HBzSgo/wKmgABgwgB8E\nM3w4f2fwxx/yxCLkzN3r16+jSbFTyJ2cnHDQyM2uc3P5GNmZM3yTpGPH+GEoVlZAaCgQHs4XXTk4\nyBU9IWWrSm4TbdLpgJYt+Z9p04Dff+dbvi9Zwg94atCAD1P7+fEZQu7uQJMmVbtBLKTw64zc0Ygx\nfgf81i0gK4vPjb17F3B25ivjvL2BQYP4SjknJ5mDJsQIxuY2IaXR6Xhhd3fnN4ALC4Fz53jnNiUF\n2LaN/2LIyODvGOzs+C+Gdu347ESjVW00qXL279/POnXqZPj31KlTWWxsbInnuLq6MgD0h/7I8sfV\n1ZVym/6o8o8xuS1kOmd+fj5atGiB7du3w9HREYGBgUhISIAHTaYnZo5ym5gDIUM91apVwzfffINO\nnTqhoKAAQ4cOpRcGUQXKbWIOFLuAixBCiDwUvVfPrFmz4OHhAW9vb3z00UeytDFx4kQ4OTkhICAA\nAQEB2LRpkyztAMCMGTNgYWGBrKwsWa4/YcIE+Pn5wd/fH6+++iquXr0qeRsffvghPDw84Ofnh169\neuHOnTuSt7Fq1Sp4eXnB0tISx44dk/z6ohZYmaLdmJgY2Nvbw0fG1YlXr15FSEgIvLy84O3tjZkz\nZ0rexoMHDxAUFAR/f394enri448/lryNIgUFBQgICEB4eLhsbbi4uMDX1xcBAQEIDAyUpY3s7Gz0\n7t0bHh4e8PT0xIEDB8p+sqR3tiS0Y8cO1qFDB6bX6xljjP3xxx+ytDNx4kQ2Y8YMWa5d3JUrV1in\nTp2Yi4sLu337tixt3L171/DxzJkz2dChQyVvY8uWLaygoIAxxthHH33EPvroI8nbOHv2LPv9999Z\ncHAwO3r0qKTXNmaBlRxM1e6uXbvYsWPHmLe3t+TXLnLz5k12/Phxxhhj9+7dY25ubrJ8L/fv32eM\nMfbo0SMWFBTEdu/eLXkbjDE2Y8YMFhUVxcLDw2W5PmNM1td9kcGDB7P58+czxvjPLDs7u8znKrbH\nP2fOHHz88ceo/vdk1YYybozPTDDaNXr0aHz11VeytlG7dm3Dxzk5OXj22WclbyMsLAwWFjxtgoKC\ncE2GM+fc3d3h5uYm+XUBcQusTNVu+/btUV+ODWCKadSoEfz9/QEANjY28PDwwI0bNyRvx9raGgCg\n1+tRUFAAW1tbydu4du0akpKSMGzYMNnrgJzXv3PnDnbv3o2YmBgA/F5T3XKODFRs4b9w4QJ27dqF\ntm3bIjg4GEeOHJGtrVmzZsHPzw9Dhw5Fdna25Ndfu3YtnJyc4OvrK/m1H/fJJ5/A2dkZixYtwrhx\n42Rta8GCBXjttddkbUNqpS2wun79umrblVt6ejqOHz+OoKAgya9dWFgIf39/2NvbIyQkBJ4yHGv1\n/vvvY/r06YbOjFx0Oh06dOiA1q1bY968eZJfPy0tDQ0bNkR0dDRatmyJ4cOHIzc3t8znC5nVUyQs\nLAwZGRlPfH7KlCnIz8/HX3/9hQMHDuDw4cPo27cvUlNTJW/nzTffxKeffgqAj5F/8MEHmD9/vqRt\nTJs2DVu2bDF8riq/+ctqZ+rUqQgPD8eUKVMwZcoUxMbG4v3338fChQslbwPg35eVlRWioqIq/k0Y\n2YYcRC2wUuPCrpycHPTu3RtxcXGwsbGR/PoWFhY4ceIE7ty5g06dOiE5ORnBwcGSXX/9+vWws7ND\nQECA7Fs27N27Fw4ODvjzzz8RFhYGd3d3tG/fXrLr5+fn49ixY/jmm2/Qpk0bjBo1CrGxsZg8eXKp\nzxda+Ldu3VrmY3PmzEGvXr0AAG3atIGFhQVu376NBg0aSNpOccOGDat00SmrjVOnTiEtLQ1+fn4A\n+FvLVq1a4dChQ7Czs5OsncdFRUVVujf+tDbi4+ORlJSE7du3V+r6xrQhl8aNG5e46X316lU4mWDZ\nt6h25fLo0SNERERg4MCB6NGjh6xt1a1bF127dsWRI0ckLfz79u3DL7/8gqSkJDx48AB3797F4MGD\nsXjxYsnaKOLw9x4yDRs2RM+ePXHo0CFJC7+TkxOcnJzQpk0bAEDv3r0RGxtb5vMVO9TTo0cP7Nix\nAwBw/vx56PX6ShX9p7l586bh49WrV0s+G8Lb2xuZmZlIS0tDWloanJyccOzYsUoV/ae5cOGC4eO1\na9ciICBA8jY2bdqE6dOnY+3atahZs6bk13+c1OOirVu3xoULF5Ceng69Xo8VK1bg9ddfl7QNJbUr\nB8YYhg4dCk9PT4waNUqWNm7dumUYds3Ly8PWrVslz+epU6fi6tWrSEtLQ2JiIkJDQ2Up+rm5ubj3\n9wG89+/fx5YtWySvM40aNUKTJk1w/vx5AMC2bdvg5eVV9hfIepu5CvR6PRs4cCDz9vZmLVu2lG3b\n1EGDBjEfHx/m6+vLunfvzjIyMmRpp0jTpk1lu7sfERHBvL29mZ+fH+vVqxfLzMyUvI1mzZoxZ2dn\n5u/vz/z9/dmbb74peRs///wzc3JyYjVr1mT29vasc+fOkl4/KSmJubm5MVdXVzZ16lRJry263cjI\nSObg4MCsrKyYk5MTW7BggeRt7N69m+l0Oubn52fIg40bN0raxm+//cYCAgKYn58f8/HxYV999ZWk\n139ccnKybLN6UlNTmZ+fH/Pz82NeXl6y/d+fOHGCtW7dmvn6+rKePXuWO6uHFnARQojGKHaohxBC\niDyo8BNCiMZQ4SeEEI2hwk8IIRpDhZ8QQjSGCj8hhGgMFX5CCNEYKvwSCQ0NLbEfDwB8/fXXeOut\nt8r8mgsXLqBbt25o1qwZWrdujdDQUOzevRsAkJmZiW7dusHf3x9eXl7o2rVrqdd48cUXpfsmShEc\nHGzYE3/q1KmytkWUh/JapWRZQqZB33//PYuOji7xubZt25a5h3heXh5r3rw5W7duneFzp06dYvHx\n8YwxxkaMGMFmzpxpeCwlJUWGqJ+u+J74NjY2QmIg4lBeqxP1+CUSERGBDRs2ID8/HwDfrvbGjRt4\n6aWXSn3+smXL8OKLL6Jbt26Gz3l5eWHIkCEAgIyMDDRu3NjwmLe3d6nXKdoVsWjnwj59+sDDwwMD\nBw584rnnzp0rsX1uenq6Yavo7du3o2XLlvD19cXQoUOh1+sNz2OMYdy4ccjLy0NAQAAGDRqE3Nxc\ndO3aFf7+/vDx8cHKlSuN+jkR80J5rc68psIvEVtbWwQGBiIpKQkAkJiYiH79+pX5/DNnzqBly5Zl\nPv72229j6NChCA0NxdSpU0tsJldc8e1+T5w4gbi4OJw5cwapqanYu3dviee6u7tDr9cjPT0dALBi\nxQpERkbiwYMHiI6OxsqVK/Hbb78hPz8fc+bMKdFGbGwsatWqhePHj2PJkiXYuHEjGjdujBMnTiAl\nJQWdO3d+6s+ImB/Ka3XmNRV+CfXv3x+JiYkAePL179+/3OezYtsk9ezZEz4+PoiIiAAAdOzYEamp\nqRg+fDjOnTuHgIAA3Lp1q9zrBQYGwtHRETqdDv7+/oYXQnF9+/bFihUrAAArV65Ev3798Pvvv6Np\n06Zo1qwZAGDIkCHYtWtXuW35+vpi69atGDduHPbs2YM6deqU+3xiviiv1YcKv4Ref/11bN++HceP\nH0dubm6528h6eXmVOEh89erViI+PL3EQe/369dG/f38sXrwYbdq0eWrS1qhRw/CxpaWl4e15cf36\n9cPKlStx4cIF6HQ6uLq6PvEcZsS+fc2bN8fx48fh4+OD8ePH4/PPP3/q1xDzRHmtPlT4JWRjY4OQ\nkBBER0c/9WSqqKgo7N27F+vWrTN87v79+4a3uDt37jQcnXbv3j1cunQJzz33XJVjfP7552FpaYnP\nP/8ckZGRAIAWLVogPT0dly5dAgAsWbKk1AMvqlevbnjR3bx5EzVr1sSAAQMwZsyYEi92oi6U1+oj\n9AQuNerfvz969er11JtCNWvWxPr16zF69GiMGjUK9vb2qF27NsaPHw8AOHr0KN555x1Uq1YNhYWF\nGD58OFq1avXEdYqPhT5+vF9Zx/3169cPY8eOxRdffGGIZeHChejTpw/y8/MRGBiIkSNHPvF1I0aM\ngK+vL1q1aoVBgwbhww8/hIWFBaysrEqMnRL1obxWF9qPnxBCNIaGegghRGNoqEdmKSkpGDx4cInP\n1axZE/v37xcUESFVR3lt3miohxBCNIaGegghRGOo8BNCiMZQ4SeEEI2hwk8IIRpDhZ8QQjTm/wFW\n/IkRN7BkXwAAAABJRU5ErkJggg==\n", + "text": [ + "<matplotlib.figure.Figure at 0x7ffba5748810>" + ] + } + ], + "prompt_number": 18 + }, + { + "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": [ + "# 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 0x7ffba566e950>" + ] + } + ], + "prompt_number": 12 + }, + { + "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_1.ipynb b/Electronics_Engineering_by_P._Raja/chapter_7_1.ipynb new file mode 100644 index 00000000..299a6c38 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_7_1.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_1.ipynb b/Electronics_Engineering_by_P._Raja/chapter_8_1.ipynb new file mode 100644 index 00000000..c1dea420 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_8_1.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 0x7f6a3b4fb510>" + ] + } + ], + "prompt_number": 43 + }, + { + "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_1.ipynb b/Electronics_Engineering_by_P._Raja/chapter_9_1.ipynb new file mode 100644 index 00000000..bbc73967 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_9_1.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_1.png b/Electronics_Engineering_by_P._Raja/screenshots/7_1.png Binary files differnew file mode 100644 index 00000000..c25bf470 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/screenshots/7_1.png diff --git a/Electronics_Engineering_by_P._Raja/screenshots/snap-3_1.png b/Electronics_Engineering_by_P._Raja/screenshots/snap-3_1.png Binary files differnew file mode 100644 index 00000000..5978d055 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/screenshots/snap-3_1.png diff --git a/Electronics_Engineering_by_P._Raja/screenshots/snap-6_1.png b/Electronics_Engineering_by_P._Raja/screenshots/snap-6_1.png Binary files differnew file mode 100644 index 00000000..0eef9f3e --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/screenshots/snap-6_1.png diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_1_1.ipynb b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_1_1.ipynb new file mode 100644 index 00000000..4377829c --- /dev/null +++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_1_1.ipynb @@ -0,0 +1,707 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter : 1 - Semiconductor Materials And Crystal Properties" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 1.18.3 - Page No : 1-43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# Given data\n", + "At = 28;# Atomic weight\n", + "n = 4;\n", + "N_A = 6.023*10**23;\n", + "a = 5.3;# in \u00c5\n", + "a = a * 10**-10;# in m\n", + "m = At/N_A;# in gm\n", + "m = m*10**-3;# in kg\n", + "V = (a)**3;# in m**3\n", + "Rho = (m*n)/V;# in gm/m**3\n", + "Rho = Rho * 10**-3;# in kg/m**3\n", + "print \"The volume density = %0.3f kg/m**3 \" %Rho" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The volume density = 1.249 kg/m**3 \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 1.18.4 - Page No : 1-43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "# Given data\n", + "r = 1.278;# in \u00c5\n", + "a = (4*r)/(sqrt(2));# in \u00c5\n", + "a = a * 10**-10;# in m\n", + "V = (a)**3;# in m**3\n", + "At = 63.5;# atomic weight\n", + "N_A = 6.023*10**23;\n", + "m = At/N_A;# in gm\n", + "m = m * 10**-3;# in kg\n", + "n = 4;\n", + "Rho = (m*n)/V;# in gm/m**3\n", + "Rho = Rho * 10**-3;# in kg/m**3\n", + "print \"The density of copper crystal = %0.3f kg/m**3\" %Rho" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The density of copper crystal = 8.929 kg/m**3\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 1.21.1 - Page No : 1-50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sin, pi\n", + "# Given data\n", + "d = 2.82;# in \u00c5\n", + "d = d * 10**-10;# in m\n", + "n = 1;\n", + "theta = 10;# in degree \n", + "#Formula 2*d*sin(theta) = n*lembda;\n", + "lembda = 2*d*sin(theta*pi/180);# in m\n", + "lembda = lembda * 10**10;# in \u00c5\n", + "print \"The wavelength of X-ray = %0.3f \u00c5\" %lembda" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The wavelength of X-ray = 0.979 \u00c5\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 1.21.2 - Page No : 1-51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "lembda = 1.6;# in \u00c5\n", + "lembda = lembda * 10**-10;# in m\n", + "theta = 14.2;# in degree\n", + "n = 1;\n", + "#Formula 2*d*sin(theta) = n*lembda;\n", + "d = (n*lembda)/(2*sin(theta*pi/180));# in m\n", + "d = d * 10**10;# in \u00c5\n", + "print \"The spacing of atomic layer in the crystal = %0.2f \u00c5\" %d\n", + "# Note : In the book the unit of the answer is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The spacing of atomic layer in the crystal = 3.26 \u00c5\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 1.21.3 - Page No : 1-51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "n = 1;\n", + "theta = 30;# in degree \n", + "lembda = 1.78;# in \u00c5\n", + "lembda = lembda * 10**-10;# in m\n", + "#Formula 2*d*sin(theta) = n*lembda;\n", + "d = (n*lembda)/(2*sin(theta*pi/180));# in m\n", + "d = d * 10**10;# in \u00c5\n", + "print \"The interplaner spacing = %0.2f \u00c5\" %d" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The interplaner spacing = 1.78 \u00c5\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 1.21.4 - Page No : 1-51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "lembda = 0.58;# in \u00c5\n", + "n = 1;\n", + "theta1 = 6.45;# in degree\n", + "d = (n*lembda)/(2*sin(theta1*pi/180));# in \u00c5 \n", + "print \"Part (i) : At angle of 6.45\u00b0, Interplaner spacing of the crystal = %0.3f \u00c5\" %d\n", + "theta2 = 9.15;# in degree\n", + "d1 = (n*lembda)/(2*sin(theta2*pi/180));# in \u00c5 \n", + "print \"Part(ii) : At angle of 9.15\u00b0, Interplaner spacing of the crystal = %0.3f \u00c5\" %d1\n", + "theta3 = 13;# in degree\n", + "n2 = 1;\n", + "d2 = (n2*lembda)/(2*sin(theta3*pi/180));# in \u00c5 \n", + "print \"Part(iii) : At angle of 13\u00b0, Interplaner spacing of the crystal = %0.3f \u00c5\" %d2\n", + "# For \n", + "n=2;\n", + "d2 = (n*lembda)/(2*sin(theta3*pi/180));# in \u00c5 \n", + "print \"Part (iv) : The interplaner spacing = %0.3f \u00c5\" %d2\n", + "print \"The interplaner spacing for some other set of reflecting = %0.3f \u00c5\" %d1" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (i) : At angle of 6.45\u00b0, Interplaner spacing of the crystal = 2.582 \u00c5\n", + "Part(ii) : At angle of 9.15\u00b0, Interplaner spacing of the crystal = 1.824 \u00c5\n", + "Part(iii) : At angle of 13\u00b0, Interplaner spacing of the crystal = 1.289 \u00c5\n", + "Part (iv) : The interplaner spacing = 2.578 \u00c5\n", + "The interplaner spacing for some other set of reflecting = 1.824 \u00c5\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 1.21.5 - Page No : 1-52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import asin\n", + "# Given data\n", + "a = 2.814;# in \u00c5\n", + "h = 1;\n", + "k = 0;\n", + "l = 0;\n", + "d = a/(sqrt( ((h)**2) + ((k)**2) + ((l)**2) ));# in \u00c5\n", + "n = 2;\n", + "lembda = 0.710;# in \u00c5\n", + "#Formula 2*d*sin(theta) = n*lembda;\n", + "theta = asin(n*lembda/(2*d) )*180/pi;# in degree\n", + "print \"The glacing angle = %0.2f degree\" %theta" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The glacing angle = 14.61 degree\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 1.21.6 - Page No : 1-52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "a = 3.65;# in \u00c5\n", + "h = 1;\n", + "k = 0;\n", + "l = 0;\n", + "d = a/(sqrt( ((h)**2) + ((k)**2) + ((l)**2) ));# in \u00c5\n", + "n = 1;\n", + "theta = 60;# in degree\n", + "#Formula 2*d*sin(theta) = n*lembda;\n", + "lembda = 2*d*sin(theta*pi/180);# in \u00c5\n", + "print \"Wavelength of X ray = %0.4f \u00c5\" %lembda" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of X ray = 6.3220 \u00c5\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 1.21.7 - Page No : 1-53" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "lembda = 1.54;# in \u00c5\n", + "lembda= lembda*10**-8;# in cm\n", + "At = 63.54;# atomic weight\n", + "density = 9.024;# in gm/cc\n", + "n = 1;\n", + "N_A = 6.023*10**23;\n", + "m = At/N_A;# mass\n", + "a =(density*m)**(1/3);# in cm\n", + "h = 1;\n", + "k = 0;\n", + "l = 0;\n", + "d = a/(sqrt( ((h)**2) + ((k)**2) + ((l)**2) ));# in cm\n", + "n = 1;\n", + "#Formula 2*d*sin(theta) = n*lembda;\n", + "theta = asin( (lembda)/(2*d) )*180/pi;# in degree\n", + "print \"The glancing angle = %0.4f degree\" %theta\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The glancing angle = 4.4893 degree\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 1.21.8 - Page No : 1-53" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "a = 3.615;# in \u00c5\n", + "theta = 22;# in degree\n", + "n = 1;\n", + "h = 1;\n", + "k = 1;\n", + "l = 1;\n", + "d = a/(sqrt( ((h)**2) + ((k)**2) + ((l)**2) ));# in \u00c5\n", + "# 2*d*sin(theta) = n*lembda;\n", + "lembda = 2*d*sin(theta*pi/180);# in \u00c5\n", + "print \"The wavelength = %0.3f \u00c5\" %lembda\n", + "n = 2;\n", + "#Formula 2*d*sin(theta) = n*lembda;\n", + "theta = asin(n*lembda/(2*d) )*180/pi;# in degree\n", + "print \"The glacing angle for second order = %0.2f degree\" %theta\n", + "print \"To get the 2nd order spectrum the position of the detector should be %0.2f \u00b0\" %(2*theta)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The wavelength = 1.564 \u00c5\n", + "The glacing angle for second order = 48.52 degree\n", + "To get the 2nd order spectrum the position of the detector should be 97.04 \u00b0\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 1.21.9 - Page No : 1-54" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "n = 1;\n", + "lembda = 1.54;# in \u00c5\n", + "theta = 21.7;# in degree\n", + "#Formula 2*d*sin(theta) = n*lembda;\n", + "d = (lembda*n)/(2*sin(theta*pi/180));# in \u00c5\n", + "h = 1;\n", + "k = 1;\n", + "l = 1;\n", + "# Formula d = a/(sqrt( ((h)**2) + ((k)**2) + ((l)**2) ));\n", + "a = d*(sqrt( ((h)**2) + ((k)**2) + ((l)**2) ));# in \u00c5\n", + "print \"Lattice constant = %0.3f \u00c5\" %a" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Lattice constant = 3.607 \u00c5\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 1.22.1 - Page No : 1-56" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "a = 4.8;# in \u00c5\n", + "h = 2;\n", + "k = 1;\n", + "l = 1;\n", + "d = a/(sqrt( ((h)**2) + ((k)**2) + ((l)**2) ));# in \u00c5\n", + "print \"The distance between d_211 plains = %0.2f \u00c5\" %d" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The distance between d_211 plains = 1.96 \u00c5\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 1.22.2 - Page No : 1-56" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "r = 1.28;# in \u00c5\n", + "#Formula r = (a*sqrt(2))/4;\n", + "a = (4*r)/(sqrt(2));# in \u00c5\n", + "a = a * 10**-8;# in cm\n", + "n = 4;\n", + "M = 63.5;\n", + "N_A = 6.023*10**23;\n", + "Rho = (n*M)/(N_A*((a)**3));# in gm/cc\n", + "print \"The density = %0.2f gm/cc\" %Rho" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The density = 8.89 gm/cc\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 1.22.3 - Page No : 1-57" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "a = 2.9;# in \u00c5\n", + "a = a * 10**-8;# in cm\n", + "Rho = 7.87;# in gm/cc\n", + "N_A = 6.023*10**23;\n", + "M = 55.85\n", + "# Rho = (n*M)/(N_A*((a)**3))\n", + "n = (Rho*N_A*((a)**3))/M;# in atoms per unit cell\n", + "print \"The number of atoms per unit cell = %0.f \" %n\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The number of atoms per unit cell = 2 \n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 1.22.4 - Page No : 1-57" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "M =60;# in gm/mole\n", + "Rho = 6.23;# in gm/cc\n", + "n = 4;\n", + "N_A = 6.023*10**23;\n", + "# Rho = (n*M)/(N_A*((a)**3));\n", + "a = ( (n*M)/(N_A*Rho) )**(1/3);# in cm\n", + "r = (a*sqrt(2))/4;# in cm\n", + "r = r * 10**8;# in \u00c5\n", + "print \"The radius = %0.3f \u00c5\" %r" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The radius = 1.414 \u00c5\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 1.22.5 - Page No : 1-57" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "Rho = 5.96;# in gm/cc\n", + "M = 50;\n", + "n = 2;\n", + "N_A = 6.023*10**23;\n", + "#Formula Rho = (n*M)/(N_A*((a)**3));\n", + "a = ( (n*M)/(N_A*Rho) )**(1/3);# in cm\n", + "r = (a*sqrt(3))/4;# in cm\n", + "P_F = (2*(4/3)*pi*((r)**3))/((a)**3);# Packing factor\n", + "print \"The Packing factor = %0.2f\" %P_F" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Packing factor = 0.68\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 1.22.6 - Page No : 1-58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "d = 5.2;# in gm/cc\n", + "n = 2;\n", + "M = 120;\n", + "N_A = 6.023*10**23;\n", + "m = M/N_A;#mass of 1 atom in gm\n", + "m = n*m;#mass of unit cell in gm\n", + "g = 20;# in gm\n", + "m = g/m;# in unit cells\n", + "print \"The number of unit cell in its 20 gm = %0.3e\" %m" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The number of unit cell in its 20 gm = 5.019e+22\n" + ] + } + ], + "prompt_number": 43 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 1.22.7 - Page No : 1-58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "Rho = 2.48;# in gm/cc\n", + "K = 39;# molecular weight of K\n", + "F = 19; # molecular weight of F\n", + "M = K+F;# molecular weight of KF\n", + "n = 4;\n", + "N_A = 6.023*10**23;\n", + "#Formula Rho = (n*M)/(N_A*((a)**3));\n", + "a = ( (n*M)/(N_A*Rho) )**(1/3);# in cm\n", + "a = a * 10**8;# in \u00c5\n", + "r = (a*sqrt(2))/4;# in \u00c5\n", + "r = 2*r;# in \u00c5\n", + "print \"The distance between ions = %0.1f \u00c5\" %r\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The distance between ions = 3.8 \u00c5\n" + ] + } + ], + "prompt_number": 44 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_2_1.ipynb b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_2_1.ipynb new file mode 100644 index 00000000..64ef8b00 --- /dev/null +++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_2_1.ipynb @@ -0,0 +1,1136 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:146b2a737d67bfa0a93cceee1f0df7461bc034ad0becba42baf98a47d6983e5d" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter : 2 - Energy Bands And Charge Carriers In Semiconductor" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.6.1 - Page No : 2-10" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# Given data\n", + "h = 6.625 * 10**-34 # in J\n", + "c = 3 * 10**8 # in J\n", + "lambda_Gr = 17760 * 10**-10 # in m\n", + "lambda_Si = 11000 # in A\u00b0\n", + "lambda_Si = lambda_Si * 10**-10 # in m\n", + "E_g = (h*c)/lambda_Si # in J\n", + "E_g = E_g /(1.6*10**-19) # in eV\n", + "print \"The energy gap of Si = %0.3f eV \" %E_g" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The energy gap of Si = 1.129 eV \n" + ] + } + ], + "prompt_number": 53 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.6.2 - Page No : 2-10" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "E_g = 0.75# in eV\n", + "E_g = 0.75 * 1.6 * 10**-19 # in J\n", + "h = 6.63 * 10**-34 # in J\n", + "c = 3 * 10**8 # in m/s \n", + "# hv = E_g\n", + "#E_g = (h*c)/lambda\n", + "Lambda=(h*c)/E_g # in m\n", + "Lambda=Lambda * 10**10 # in A\u00b0\n", + "print \"The wavelength at which germanium starts to absorb light = %0.f A\u00b0 \" %Lambda" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The wavelength at which germanium starts to absorb light = 16575 A\u00b0 \n" + ] + } + ], + "prompt_number": 54 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.15.1\n", + " - Page No : 2-30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "del_E = 0.3 # in eV\n", + "T1 = 300 # in K\n", + "T2 = 330 # in K\n", + "# del_E = K * T1 * log(N/N_c) where del_E= E_C-E_F\n", + "# del_E1 = K * T2 * log(N/N_c) where del_E1= E_C-E_F at T= 330 \u00b0K\n", + "del_E1 = del_E*(T2/T1) # in eV \n", + "print \"The Fermi level will be \",round(del_E1,2),\" eV below the conduction band\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Fermi level will be 0.33 eV below the conduction band\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.15.2\n", + " - Page No : 2-31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import exp\n", + "# Given data\n", + "N_c = 2.8 * 10**19 # in cm**-3\n", + "del_E = 0.25 # fermi energy in eV\n", + "KT = 0.0259 \n", + "f_F = exp(-(del_E)/KT) \n", + "print \"The probaility in the condition band is occupied by an electron = %0.2e\" %f_F\n", + "n_o = N_c * exp(-(del_E)/KT) # in cm**-3\n", + "print \"The thermal equilibrium electron concentration = %0.1e cm**-3 \" %n_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The probaility in the condition band is occupied by an electron = 6.43e-05\n", + "The thermal equilibrium electron concentration = 1.8e+15 cm**-3 \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.15.3\n", + " - Page No : 2-33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "T1 = 300 # in K\n", + "T2 = 400 # in K\n", + "del_E = 0.27 # Fermi level in eV\n", + "KT = (0.0259) * (T2/T1) # in eV\n", + "N_v = 1.04 * 10**19 # in cm**-3\n", + "N_v = N_v * (T2/T1)**(3/2) # in cm**-3 \n", + "p_o = N_v * exp(-(del_E)/KT) # in per cm**3\n", + "print \"The thermal equilibrium hole concentration = %0.2e per cm**3 \" %p_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The thermal equilibrium hole concentration = 6.44e+15 per cm**3 \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.16.1 - Page No : 2-37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "At = 63.5;# atomic weight\n", + "Rho = 1.7*10**-6;# in ohm cm\n", + "d = 8.96;# in gm/cc\n", + "N_A = 6.02*10**23;# in /gm.mole\n", + "e = 1.6*10**-19;# in C\n", + "n = (N_A/At)*d;\n", + "Miu_e = 1/(Rho*n*e);# in cm**2/volt.sec\n", + "print \"The electron mobility = %0.3f cm**2/volt-sec\" %Miu_e" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The electron mobility = 43.281 cm**2/volt-sec\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.16.2 - Page No : 2-38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "l = 0.1;# in m\n", + "A = 1.7;# in mm**2\n", + "A = A * 10**-6;# in m**2\n", + "R = 0.1;# in ohm\n", + "At = 63.5;# atomic weight\n", + "N_A = 6.02*10**23;\n", + "d = 8.96;# in gm/cc\n", + "n = (N_A/At)*d;# in /cc\n", + "n = n * 10**6;# in /m**3\n", + "#Formula R = Rho*(l/A);\n", + "Rho = (R*A)/l;# in ohm m\n", + "Sigma = 1/Rho;# in mho/m\n", + "e = 1.6*10**-19;\n", + "# Formula Sigma = n*e*Miu_e\n", + "Miu_e = Sigma/(n*e);# in m**2/V.sec\n", + "print \"The mobility = %0.2e m**2/V-sec\" %Miu_e" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The mobility = 4.33e-05 m**2/V-sec\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.16.3 - Page No : 2-39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# Given data\n", + "d = 10.5;# in gm/cc\n", + "At = 108;# atomic weight\n", + "N_A = 6.025*10**23;# in /gm mole\n", + "r = 10**-3;# in m\n", + "q = 1.6*10**-19;# in C\n", + "n = (N_A/At)*d;# in /cm**3\n", + "n = n * 10**6;# in /m**3\n", + "A = pi*((r)**2);# in m**2\n", + "I = 2;# in A\n", + "# I = q*n*A*v;\n", + "v = I/(n*q*A);# in m/s\n", + "print \"The drift velocity = %0.e m/s\" %v\n", + "\n", + "# Note: There is calculation error to find the value of drift velocity (i.e v), so the answer in the book is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drift velocity = 7e-05 m/s\n" + ] + } + ], + "prompt_number": 44 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.16.4 - Page No : 2-39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "d = 1.03;# in mm\n", + "d = d *10**-3;# in m\n", + "r = d/2;# in m\n", + "R = 6.51;# in ohm\n", + "l = 300;# in mm\n", + "e = 1.6*10**-19;\n", + "n = 8.4*10**28;# in /m**3\n", + "A = pi*r**2;\n", + "# R = Rho*(l/A);\n", + "Rho = (R*A)/l;#in ohm m\n", + "Sigma = 1/Rho;# in mho/m\n", + "print \"The conductivity of copper = %0.3e mho/m\" %Sigma\n", + "#Formula sigma = n*e*Miu_e\n", + "Miu_e = Sigma/(n*e);# in m**2/V.sec\n", + "print \"The mobility = %0.3e m**2/V-sec\" %Miu_e" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The conductivity of copper = 5.531e+07 mho/m\n", + "The mobility = 4.115e-03 m**2/V-sec\n" + ] + } + ], + "prompt_number": 49 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.16.5 - Page No : 2-40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "At = 63.5;# atomic weight\n", + "d = 8.96;# in gm/cc\n", + "Miu_e = 43.28;# in cm**2/V.sec\n", + "N_A = 6.02*10**23;# in /gm mole\n", + "e = 1.6*10**-19;# in C\n", + "n = (N_A/At)*d;# in /cc\n", + "Rho = 1/(n*e*Miu_e);# in ohm-cm\n", + "print \"The resistivity = %0.1e ohm-cm\" %Rho" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resistivity = 1.7e-06 ohm-cm\n" + ] + } + ], + "prompt_number": 52 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.17.1 - Page No : 2-42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "Mu_e = 1500 # in cm**2/volt sec\n", + "Mu_h = 500 # in cm**2/volt sec\n", + "n_i = 1.6 * 10**10 # in per cm**3\n", + "e = 1.6 * 10**-19 # in C\n", + "Sigma = n_i * (Mu_e + Mu_h) * e # in mho/cm\n", + "print \"The conductivity of pure semiconductor = %0.2e mho/cm \" %Sigma" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The conductivity of pure semiconductor = 5.12e-06 mho/cm \n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.17.2 - Page No : 2-42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "Rho = 10 # in \u03a9-cm\n", + "Mu_d = 500 # in cm**2/v.s.\n", + "e = 1.6*10**-19 \n", + "n_d = 1/(Rho * e * Mu_d) # in per cm**3\n", + "print \"The number of donor atom must be added to achieve = %0.2e per cm**3 \" %n_d" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The number of donor atom must be added to achieve = 1.25e+15 per cm**3 \n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.17.3 - Page No : 2-43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given data\n", + "AvagadroNumber = 6.02 * 10**23 # in atoms/gm.mole\n", + "at_Ge = 72.6 # atom weight of Ge\n", + "e = 1.6 * 10**-19 # in C\n", + "D_Ge = 5.32 # density of Ge in gm/c.c\n", + "Mu = 3800 # in cm**2/v.s.\n", + "C_Ge = (AvagadroNumber/at_Ge) * D_Ge # concentration of Ge atoms in per cm**3\n", + "n_d = C_Ge/10**8 # in per cc\n", + "Sigma = n_d * Mu * e # in mho/cm\n", + "print \"The conductivity = %0.3f mho/cm \" %Sigma" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The conductivity = 0.268 mho/cm \n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.17.4 - Page No : 2-43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "Rho = 0.3623 * 10**-3 # in Ohm m\n", + "Sigma = 1/Rho #in mho/m\n", + "D = 4.42 * 10**28 # Ge density in atom/m**3\n", + "n_d = D / 10**6 # in atom/m**3\n", + "e = 1.6 * 10**-19 # in C\n", + "Mu = Sigma/(n_d * e) # in m**2/V.sec\n", + "print \"The mobility of electron in germanium = %0.2f m**2/V.sec \" %Mu" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The mobility of electron in germanium = 0.39 m**2/V.sec \n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.17.5 - Page No : 2-44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "AvagadroNumber = 6.025 * 10**26 # in kg.Mole\n", + "W = 72.59 # atomic weight of Ge\n", + "D = 5.36 * 10**3 #density of Ge in kg/m**3\n", + "Rho = 0.42 # resistivity in Ohm m\n", + "e = 1.6 * 10**-19 # in C\n", + "Sigma = 1/Rho # in mho/m\n", + "n = (AvagadroNumber/W) * D # number of Ge atoms present per unit volume\n", + "# Holes per unit volume, H = n*10**-6%\n", + "H= n*10**-8 \n", + "a=H \n", + "# Formula sigma= a*e*Mu_h\n", + "Mu_h = Sigma/(a * e) # in m**2/V.sec\n", + "print \"Mobility of holes = %0.4f m**2/V.sec \" %Mu_h" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mobility of holes = 0.0334 m**2/V.sec \n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.17.6 - Page No : 2-45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "e = 1.6 * 10**-19 # in C\n", + "n_i = 2 * 10**19 # in /m**3\n", + "Mu_e = 0.36 # in m**2/v.s\n", + "Mu_h = 0.17 # in m**2/v.s\n", + "A = 1 * 10**-4 # in m**2\n", + "V = 2 #in volts\n", + "l = 0.3 # in mm\n", + "l = l * 10**-3 # in m\n", + "E=V/l # in volt/m\n", + "Sigma = n_i * e * (Mu_e + Mu_h) # in mho/m\n", + "# J = I/A = Sigma * E\n", + "I= Sigma*E*A \n", + "print \"The current produced in a small germanium plate = %0.2f amp \" %I\n", + "\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current produced in a small germanium plate = 1.13 amp \n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.17.7 - Page No : 2-45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "D = 4.2 * 10**28 #density of Ge atoms in atoms/m**3\n", + "N_d = D / 10**6 # in atoms/m**3\n", + "e = 1.6 * 10**-19 # in C\n", + "Mu_e = 0.36 # in m**2/vs\n", + "Sigma_n = N_d * e * Mu_e # in mho/m\n", + "Rho_n = 1/Sigma_n # in ohm m\n", + "print \"The resistivity of drop Ge = %0.3e ohm m \" %Rho_n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resistivity of drop Ge = 4.134e-04 ohm m \n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.17.8 - Page No : 2-46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "e = 1.6 * 10**-19 # in C\n", + "n_i = 1 * 10**19 # in per m**3\n", + "Mu_e = 0.36 # in m**2/volt.sec\n", + "Mu_h = 0.17 # in m**2/volt.sec \n", + "A = 2 # in cm**2\n", + "A = A * 10**-4 # im m**2\n", + "t = 0.1 # in mm\n", + "t = t * 10**-3 # in m\n", + "V = 4 # in volts\n", + "Sigma_i = n_i * e * (Mu_e + Mu_h) # in mho/m\n", + "J = Sigma_i * (V/t) # in Amp/m**2\n", + "I = J * A # in Amp\n", + "print \"The current produced in a Ge sample = %0.3f Amp \" %I" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current produced in a Ge sample = 6.784 Amp \n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.17.9 - Page No : 2-46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "e = 1.6 * 10**-19 # in C\n", + "Mu_h = 500 # in cm**2/V.s.\n", + "Mu_e = 1500 # in cm**2/V.s.\n", + "n_i = 1.6 * 10**10 # in per cm**3\n", + "Sigma_i = n_i * e * ( Mu_h + Mu_e) # in mho/cm\n", + "print \"Conductivity of pure silicon at room temperature = %0.2e mho/cm \" %Sigma_i" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Conductivity of pure silicon at room temperature = 5.12e-06 mho/cm \n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.19.1 - Page No : 2-50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given data\n", + "l= 0.50*10**-2 # width of ribbon in m\n", + "d= 0.10*10**-3 # thickness of ribbon in m\n", + "A= l*d # area of ribbon in m**2\n", + "B = 0.8 # in Tesla\n", + "D = 10.5 #density in gm/cc\n", + "I = 2 # in amp\n", + "q = 1.6 * 10**-19 # in C\n", + "n=6*10**28 # number of elec. per m**3\n", + "V_H = ( I * B * d)/(n * q * A) # in volts\n", + "print \"The hall Voltage produced = %0.2e volts \" %V_H" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The hall Voltage produced = 3.33e-08 volts \n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.19.2 - Page No : 2-51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "l = 1 # in m\n", + "d = 1 # in cm\n", + "d = d * 10**-2 # in m\n", + "W = 1 # in mm\n", + "W = W * 10**-3 # in m\n", + "A = d * W # in m**2\n", + "I= 1 # in Amp\n", + "B = 1 # Tesla\n", + "V_H = 0.074 * 10**-6 # in Volts\n", + "Sigma = 5.8 * 10**7 # in mho/m\n", + "R_H = (V_H * A)/(B*I*d) # in m**3/c\n", + "print \"The hall coefficient = %0.1e m**3/c \" %R_H\n", + "Mu = Sigma * R_H # in m**2/volt.sec\n", + "print \"The mobility of electrons in copper = %0.1e m**2/volt-sec \" %Mu" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The hall coefficient = 7.4e-11 m**3/c \n", + "The mobility of electrons in copper = 4.3e-03 m**2/volt-sec \n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.20.1 - Page No : 2-52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "n_i = 1.4 * 10**18 # in /m**3\n", + "n_D = 1.4 * 10**24 # in /m**3\n", + "n=n_D # in /m**3\n", + "p = n_i**2/n # in /m**3\n", + "R = n/p \n", + "print \"The ratio of electrons to hole concentration = %0.1e\" %R" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The ratio of electrons to hole concentration = 1.0e+12\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.20.2 - Page No : 2-53" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import atan, pi\n", + "# Given data\n", + "B = 0.48 # in wb/m**2\n", + "R_H = 3.55 * 10**-4 # in m**3/c\n", + "Rho = 0.00912 # in ohm-m\n", + "Sigma = 1/Rho # in (ohm-m)**-1\n", + "theta_H = atan( Sigma * B * R_H)*180/pi # in degree\n", + "print \"The hall angle for a hall coefficient = %0.4f\u00b0\" %theta_H" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The hall angle for a hall coefficient = 1.0704\u00b0\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.20.3 - Page No : 2-53" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given data\n", + "R = 9 * 10**-3 # in ohm-m\n", + "R_H = 3.6 * 10**-4 # in m**3\n", + "e = 1.6 * 10**-19 # in C\n", + "Sigma = 1/R # in (ohm-m)**-1\n", + "Rho = 1/R_H # in coulomb/m**3\n", + "n = Rho/e # in /m**3\n", + "print \"Density of charge carriers = %0.5e per m**3 \" %n\n", + "Mu = Sigma * R_H # in m**2/v-s\n", + "print \"Mobility of charge carriers = %0.2f m**2/V-s \" %Mu" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Density of charge carriers = 1.73611e+22 per m**3 \n", + "Mobility of charge carriers = 0.04 m**2/V-s \n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.24.1 - Page No : 2-60" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "e = 1.6 * 10**-19 # in C\n", + "R_H = 0.0145 # in m**3/coulomb\n", + "Mu_e = 0.36 # in m**2/v-s\n", + "E = 100 # in V/m\n", + "n = 1/(e * R_H) # in /m**3\n", + "J = n * e * Mu_e * E # in A/m**2\n", + "print \"The current density of specimen = %0.3e A/m**2 \" %J" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current density of specimen = 2.483e+03 A/m**2 \n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.24.2 - Page No : 2-60" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "#Given data\n", + "Mu_e = 7.04 * 10**-3 # in m**2/v-s\n", + "m = 9.1 * 10**-31 \n", + "E_F = 5.5 # in eV\n", + "n = 5.8 * 10**28 \n", + "e = 1.6 * 10**-19 # in C\n", + "Torque = (Mu_e/e) * m # in sec\n", + "print \"Relaxation Time = %0.3e sec \" %Torque\n", + "Rho = 1 /(n * e * Mu_e) # in ohm-m\n", + "print \"Resistivity of conductor = %0.3e ohm-m \" %Rho\n", + "V_F = sqrt((2 * E_F * e)/m) # in m/s\n", + "print \"Velocity of electrons with fermi-energy = %0.4e m/s \" %V_F" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Relaxation Time = 4.004e-14 sec \n", + "Resistivity of conductor = 1.531e-08 ohm-m \n", + "Velocity of electrons with fermi-energy = 1.3907e+06 m/s \n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.24.3 - Page No : 2-61" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "# Given data\n", + "E= 5.95 # in eV\n", + "EF= 6.25 # in eV\n", + "delE= 0.01 \n", + " # delE= 1-1/(1+exp((E-EF)/KT))\n", + "K=1.38*10**-23 # Boltzman Constant in J/K\n", + "T = ((E-EF)/log(1/(1-delE) -1)*1.6*10**-19)/K # in K\n", + "print \"The temperature =\" ,int(T),\"K\"\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The temperature = 756 K\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.24.4 - Page No : 2-61" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data \n", + "N_V = 1.04 * 10**19 # in cm**-3\n", + "T1 = 300 # in K\n", + "T2 = 400 # in K\n", + "del_E = 0.27 # in eV\n", + "N_V = N_V * (T2/T1)**1.5 # in cm**-3\n", + "KT = (0.0259) * (T2/T1) # in eV\n", + "P_o = N_V * exp(-(del_E)/KT) # in cm**-3\n", + "print \"The thermal equilibrium hole concentration in silicon = %0.2e cm**-3 \" %P_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The thermal equilibrium hole concentration in silicon = 6.44e+15 cm**-3 \n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.24.5 - Page No : 2-62" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given data\n", + "N_c = 2.8 * 10**19 \n", + "N_V = 1.04 *10**19 \n", + "T1 = 550 # in K\n", + "T2 = 300 # in K\n", + "E_g = 1.12 \n", + "KT = (0.0259) \n", + "n_i = sqrt(N_c *N_V *(T1/T2)**3* exp(-(E_g)/KT*T2/T1)) # in cm**-3\n", + "# n_o = N_d/2 + sqrt((N_d/2)**2 + (n_i)**2)\n", + "# 1.05*N_d -N_d/2= sqrt((N_d/2)**2 + (n_i)**2)\n", + "N_d=sqrt((n_i)**2/((0.55)**2-1/4)) \n", + "print \"Minimum donor concentration required = %0.3e cm**-3 \" %N_d" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum donor concentration required = 1.396e+15 cm**-3 \n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 2.24.6 - Page No : 2-62" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given data\n", + "T = 300 # in K\n", + "n_o = 10**15 # in cm**-3\n", + "n_i = 10**10 # in cm**-3\n", + "p_o = 10**5 # in cm**-3\n", + "del_n = 10**13 # in cm**-3\n", + "del_p = del_n # in cm**-3\n", + "KT = 0.0259 # in eV\n", + "delta_E1= KT*log(n_o/n_i) # value of E_F-E_Fi in eV\n", + "delta_E2= KT*log((n_o+del_n)/n_i) # value of E_Fn-E_Fi in eV\n", + "delta_E3= KT*log((p_o+del_p)/n_i) # value of E_Fi-E_Fp in eV\n", + "print \"The Fermi level for thermal equillibrium = %0.4f eV \" %delta_E1\n", + "print \"The quase-Fermi level for electrons in non equillibrium = %0.4f eV \" %delta_E2\n", + "print \"The quasi-Fermi level for holes in non equillibrium = %0.3f eV \" %delta_E3\n", + "print \"The quasi-Fermi level for electrons is above E_Fi while the quasi-Fermi level for holes is below E_Fi\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Fermi level for thermal equillibrium = 0.2982 eV \n", + "The quase-Fermi level for electrons in non equillibrium = 0.2984 eV \n", + "The quasi-Fermi level for holes in non equillibrium = 0.179 eV \n", + "The quasi-Fermi level for electrons is above E_Fi while the quasi-Fermi level for holes is below E_Fi\n" + ] + } + ], + "prompt_number": 29 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_3_1.ipynb b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_3_1.ipynb new file mode 100644 index 00000000..aedb74be --- /dev/null +++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_3_1.ipynb @@ -0,0 +1,898 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:764d1c60d9fefbb17ec0ffb4d8cdbf1cddd8802a6e86f8e004d6805017f0ecae" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter : 3 - Excess Carriers in Semiconductor" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.1 - Page No : 3-38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# Given data\n", + "N_d = 10**17 # atoms/cm**3\n", + "n_i = 1.5 * 10**10 # in /cm**3\n", + "n_o = 10**17 # in cm**3\n", + "# p_o * n_o = (n_i)**2\n", + "p_o = (n_i)**2 / n_o #in holes/cm**3\n", + "print \"The holes concentration at equilibrium = %0.2e holes/cm**3 \" %p_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The holes concentration at equilibrium = 2.25e+03 holes/cm**3 \n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.3 - Page No : 3-40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "# Given data\n", + "n_i = 1.5 * 10 **10 # in /cm**3 for silicon\n", + "N_d = 10**17 # in atoms/cm**3\n", + "n_o = 10**17 # electrons/cm**3\n", + "KT = 0.0259 \n", + "# E_r - E_i = KT * log(n_o/n_i)\n", + "del_E = KT * log(n_o/n_i) # in eV\n", + "print \"The energy band for this type material, E_F = Ei +\",round(del_E,3),\" eV\" " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The energy band for this type material, E_F = Ei + 0.407 eV\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.4 - Page No : 3-41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "K = 1.38 * 10**-23 # in J/K\n", + "T = 27 # in degree\n", + "T = T + 273 # in K\n", + "e = 1.6 * 10**-19 # in C\n", + "Mu_e = 0.17 # in m**2/v-s\n", + "Mu_e1 = 0.025 # in m**2/v-s\n", + "D_n = ((K * T)/e) * Mu_e # in m**2/s\n", + "print \"The diffusion coefficient of electrons = %0.1e m**2/s \" %D_n\n", + "D_p = ((K * T)/e) * Mu_e1 # in m**2/s\n", + "print \"The diffusion coefficient of holes = %0.2e m**2/s \" %D_p" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The diffusion coefficient of electrons = 4.4e-03 m**2/s \n", + "The diffusion coefficient of holes = 6.47e-04 m**2/s \n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.5 - Page No : 3-41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "# Given data\n", + "Mu_n = 0.15 # in m**2/v-s\n", + "K = 1.38 * 10**-23 # in J/K\n", + "T = 300 # in K\n", + "del_n = 10**20 # in per m**3\n", + "Toh_n = 10**-7 # in s\n", + "e = 1.6 * 10**-19 # in C\n", + "D_n = Mu_n * ((K * T)/e) # in m**2/s\n", + "print \"The diffusion coefficient = %0.2e m**2/s \" %D_n\n", + "L_n = sqrt(D_n * Toh_n) # in m \n", + "print \"The Diffusion length = %0.2e m \" %L_n\n", + "J_n = (e * D_n * del_n)/L_n # in A/m**2\n", + "print \"The diffusion current density = %0.2e A/m**2 \" %J_n\n", + "# Note : The value of diffusion coefficient in the book is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The diffusion coefficient = 3.88e-03 m**2/s \n", + "The Diffusion length = 1.97e-05 m \n", + "The diffusion current density = 3.15e+03 A/m**2 \n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.6 - Page No : 3-42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "Sigma = 0.1 # in (ohm-m)**-1\n", + "Mu_n = 1300 \n", + "n_i = 1.5 * 10**10 \n", + "q = 1.6 * 10**-19 # in C\n", + "n_n = Sigma/(Mu_n * q) # in electrons/cm**3\n", + "print \"The concentration of electrons = %0.2e per m**3 \" %(n_n*10**6)\n", + "p_n = (n_i)**2/n_n # in per cm**3\n", + "p_n = p_n * 10**6 # in perm**3\n", + "print \"The concentration of holes = %0.2e per m**3 \" %p_n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The concentration of electrons = 4.81e+20 per m**3 \n", + "The concentration of holes = 4.68e+11 per m**3 \n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.7 - Page No : 3-43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "Mu_e = 0.13 # in m**2/v-s\n", + "Mu_h = 0.05 # in m**2/v-s\n", + "Toh_h = 10**-6 # in s\n", + "L = 100 # in \u00b5m\n", + "L = L * 10**-6 # in m\n", + "V = 2 # in V\n", + "t_n =L**2/(Mu_e * V) # in s\n", + "print \"Electron transit time = %0.1e seconds \" %t_n\n", + "p_g = (Toh_h/t_n) * (1 + Mu_h/Mu_e) #photo conductor gain \n", + "print \"Photo conductor gain = %0.2f\" %p_g\n", + "\n", + "# Note: There is a calculation error to evaluate the value of t_n. So the answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Electron transit time = 3.8e-08 seconds \n", + "Photo conductor gain = 36.00\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.8 - Page No : 3-43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "n_i = 2.5 * 10**13 \n", + "Mu_n = 3800 \n", + "Mu_p = 1800 \n", + "q = 1.6 * 10**-19 # in C\n", + "Sigma = n_i * (Mu_n + Mu_p) * q # in (ohm-cm)**-1\n", + "Rho = 1/Sigma # in ohm-cm\n", + "Rho= round(Rho) \n", + "print \"The resistivity of intrinsic germanium = %0.f ohm-cm \" %Rho\n", + "N_D = 4.4 * 10**22/(1*10**8) # in atoms/cm**3\n", + "Sigma_n = N_D * Mu_n * q # in (ohm-cm)**-1\n", + "Rho_n = 1/Sigma_n # in ohm-cm\n", + "print \"If a donor type impurity is added to the extent of 1 atom per 10**8 Ge atoms, then the resistivity drops = %0.2f ohm-cm \" %Rho_n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resistivity of intrinsic germanium = 45 ohm-cm \n", + "If a donor type impurity is added to the extent of 1 atom per 10**8 Ge atoms, then the resistivity drops = 3.74 ohm-cm \n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.9 - Page No : 3-44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "n_i = 10**16 # in /m3\n", + "N_D = 10**22 # in /m**3\n", + "n = N_D # in /m**3\n", + "print \"Electron concentration = %0.1e per m**3 \" %n\n", + "p = (n_i)**2/n # in /m**3\n", + "print \"Hole concentration = %0.1e per m**3 \" %p" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Electron concentration = 1.0e+22 per m**3 \n", + "Hole concentration = 1.0e+10 per m**3 \n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.10 - Page No : 3-44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "Rho = 9.6 * 10**-2 # in ohm-m\n", + "Sigma_n = 1/Rho # in (ohm-m)**-1\n", + "q = 1.6 * 10**-19 # in C\n", + "Mu_n = 1300 * 10**-4 # in m**2/v-s\n", + "N_D = Sigma_n / (Mu_n * q) # in atoms/m**3\n", + "A_D = N_D # Atom density in atoms/cm**3\n", + "A_D = A_D * 10**6 # atoms/m**3\n", + "R_si = N_D/A_D # ratio\n", + "print \"The ratio of donor atom to silicon atom = %0.1e\" %R_si\n", + "\n", + "# Note: In the book the wrong value of N_D (5*10**22) is putted to evaluate the value of \n", + "# Atom Density (A_D) whereas the value of N_D is calculated as 5*10**20.So the answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The ratio of donor atom to silicon atom = 1.0e-06\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.11 - Page No : 3-45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "n_i = 1.5 * 10**10 # in per cm**3\n", + "n_n = 2.25 * 10**15 # in per cm**3\n", + "p_n = (n_i)**2/n_n # in per cm**3\n", + "print \"The equilibrium electron = %0.1e per cm**3 \" %p_n\n", + "h_n = n_n # in cm**3\n", + "print \"Hole densities = %0.2e per cm**3 \" %h_n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The equilibrium electron = 1.0e+05 per cm**3 \n", + "Hole densities = 2.25e+15 per cm**3 \n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.12 - Page No : 3-45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "N_A = 2 * 10**16 # in atoms/cm**3\n", + "N_D = 10**16 # in atoms/cm**3\n", + "C_c = N_A-N_D # C_c stands for Carrier concentration in /cm**3\n", + "print \"Carrier concentration = %0.1e holes/cm**3 \" %C_c" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Carrier concentration = 1.0e+16 holes/cm**3 \n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.13 - Page No : 3-46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "del_n = 10**15 # in cm**3\n", + "Torque_p = 10 * 10**-6 # in sec\n", + "R_g = del_n/Torque_p # in hole pairs/sec/cm**3\n", + "print \"The rate of generation of minority carrier = %0.1e electron hole pairs/sec/cm**3 \" %R_g" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The rate of generation of minority carrier = 1.0e+20 electron hole pairs/sec/cm**3 \n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.14 - Page No : 3-46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "v = 1/(20 * 10**-6) # in cm/sec\n", + "E = 10 # in V/cm\n", + "Mu= v/E # in cm**2/V-sec\n", + "print \"The mobility of minority charge carrier = %0.f cm**2/V-sec \" %Mu" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The mobility of minority charge carrier = 5000 cm**2/V-sec \n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.15 - Page No : 3-47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "q = 1.6 * 10**-19 # in C\n", + "N_D = 4.5 * 10**15 # in /cm**3\n", + "del_p = 10**21 \n", + "e=10 # in cm\n", + "A = 1 # in mm**2\n", + "A = A * 10**-14 # cm**2\n", + "l = 10 # in cm\n", + "Torque_p = 1 # in microsec\n", + "Torque_p = Torque_p * 10**-6 # in sec\n", + "Torque_n = 1 # in microsec\n", + "Torque_n = Torque_n * 10**-6 # in sec\n", + "n_i = 1.5 * 10**10 # in /cm**3\n", + "D_n = 30 # in cm**2/sec\n", + "D_p = 12 # in cm**2/sec\n", + "n_o = N_D # in /cm**3\n", + "p_o = (n_i)**2/n_o # in /cm**3\n", + "print \"Hole concentration at thermal equilibrium = %0.1e per cm**3 \" %p_o\n", + "l_n = sqrt(D_n * Torque_n) # in cm\n", + "print \"Diffusion length of electron = %0.2e cm \" %l_n\n", + "l_p = sqrt(D_p * Torque_p) # in cm\n", + "print \"Diffusion length of holes = %0.1e cm \" %l_p\n", + "x=34.6*10**-4 # in cm\n", + "dpBYdx = del_p *e # in cm**4\n", + "print \"Concentration gradient of holes = %0.1e cm**4 \" %dpBYdx\n", + "e1 = 1.88 * 10**1 # in cm\n", + "dnBYdx = del_p * e1 # in cm**4 \n", + "print \"Concentration gradient of electrons = %0.2e per cm**4 \" %dnBYdx\n", + "J_P = -(q) * D_p * dpBYdx # in A/cm**2\n", + "print \"Current density of holes due to diffusion = %0.2e A/cm**2 \" %J_P\n", + "J_n = q * D_n * dnBYdx # in A/cm**2\n", + "print \"Current density of electrons due to diffusion = %0.1e A/cm**2 \" %J_n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Hole concentration at thermal equilibrium = 5.0e+04 per cm**3 \n", + "Diffusion length of electron = 5.48e-03 cm \n", + "Diffusion length of holes = 3.5e-03 cm \n", + "Concentration gradient of holes = 1.0e+22 cm**4 \n", + "Concentration gradient of electrons = 1.88e+22 per cm**4 \n", + "Current density of holes due to diffusion = -1.92e+04 A/cm**2 \n", + "Current density of electrons due to diffusion = 9.0e+04 A/cm**2 \n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.16 - Page No : 3-49" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "e= 1.6*10**-19 # electron charge in C\n", + "h = 6.626 * 10**-34 # in J-s\n", + "h= h/e # in eV\n", + "c = 3 * 10**8 # in m/s\n", + "lembda = 5490 * 10**-10 # in m\n", + "f = c/lembda \n", + "E = h * f # in eV\n", + "print \"The energy band gap of the semiconductor material = %0.2f eV \" %E" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The energy band gap of the semiconductor material = 2.26 eV \n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.17 - Page No : 3-49" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "y2 = 6 * 10**16 # in /cm**3\n", + "y1 = 10**17 # in /cm**3\n", + "x2 = 2 # in \u00b5m\n", + "x1 = 0 # in \u00b5m\n", + "D_n = 35 # in cm**2/sec\n", + "q = 1.6 *10**-19 # in C\n", + "dnBYdx = (y2 - y1)/((x2-x1) * 10**-4) \n", + "J_n = q * D_n * dnBYdx # in A/cm**2\n", + "print \"The current density in silicon = %0.f A/cm**2 \" %J_n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current density in silicon = -1120 A/cm**2 \n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.18 - Page No : 3-50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "q = 1.6 * 10**-19 # in C\n", + "n_n = 5 * 10**20 # in /m**3\n", + "n_n = n_n * 10**-6 # in cm**3\n", + "Mu_n = 0.13 # in m**2/V-sec\n", + "Mu_n = Mu_n * 10**4 # in cm**2/V-sec\n", + "Sigma_n = q * n_n * Mu_n # in (ohm-cm)**-1\n", + "Rho = 1/Sigma_n # in \u03a9-cm\n", + "l = 0.1 # in cm\n", + "A = 100 # \u00b5m**2\n", + "A = A * 10**-8 # in cm**2\n", + "R = Rho * (l/A) # in Ohm\n", + "R=round(R*10**-6) # in M\u03a9\n", + "print \"The resistance of the bar = %0.f M\u03a9 \" %R" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resistance of the bar = 1 M\u03a9 \n" + ] + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.19 - Page No : 3-51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "t_d = 3 # total depletion in \u00b5m\n", + "D = t_d/9 # in \u00b5m\n", + "print \"Depletion width = %0.1f \u00b5m \" %D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Depletion width = 0.3 \u00b5m \n" + ] + } + ], + "prompt_number": 43 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.20 - Page No : 3-51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "n_i = 1.5 * 10**16 # in /m**3\n", + "n_n = 5 * 10**20 # in /m**3\n", + "p_n = (n_i)**2/n_n # in /m**3\n", + "print \"The majority carrier density = %0.2e per m**3 \" %p_n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The majority carrier density = 4.50e+11 per m**3 \n" + ] + } + ], + "prompt_number": 44 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.21 - Page No : 3-52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "D_n = 25 # in cm**2/sec\n", + "q = 1.6 * 10**-19 # in C\n", + "y2 = 10**14 # in /cm**3\n", + "y1 = 0 # in /cm**3\n", + "x2 = 0 #in \u00b5m\n", + "x1 = 0.5 # in \u00b5m\n", + "x1 = x1 * 10**-4 # in cm\n", + "dnBYdx = abs((y2-y1)/(x2-x1)) # in /cm**4 \n", + "J_n = q * D_n * (dnBYdx) # in /cm**4\n", + "J_n = J_n * 10**-1 # in A/cm**2\n", + "print \"The collector current density = %0.f A/cm**2 \" %J_n\n", + "\n", + "# Note: In the book, the calculated value of dn by dx (2*10**19) is wrong. Correct value is 2*10**18\n", + "# so the answer in the book is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The collector current density = 1 A/cm**2 \n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.22 - Page No : 3-53" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "h = 6.64 * 10**-34 # in J-s\n", + "e= 1.6*10**-19 # electron charge in C\n", + "c= 3 * 10**8 # in m/s\n", + "lembda = 0.87 # in \u00b5m\n", + "lembda = lembda * 10**-6 # in m\n", + "E_g = (h * c)/lembda # in J-s\n", + "E_g= E_g/e # in eV\n", + "print \"The band gap of the material = %0.3f eV \" %E_g" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The band gap of the material = 1.431 eV \n" + ] + } + ], + "prompt_number": 46 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.23 - Page No : 3-53" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import exp\n", + "# Given data\n", + "I_o = 10 # in mW\n", + "e = 1.6 * 10**-19 # in J/eV\n", + "hv = 2 # in eV\n", + "hv1=1.43 # in eV\n", + "alpha = 5 * 10**4 # in cm**-1\n", + "l = 46 # in \u00b5m\n", + "l = l * 10**-6 # in m\n", + "I_t = round(I_o * exp(-(alpha) * l)) # in mW\n", + "AbsorbedPower= I_o-I_t # in mW\n", + "AbsorbedPower=AbsorbedPower*10**-3 # in W or J/s\n", + "print \"The absorbed power = %0.1e watt or J/s \" %AbsorbedPower\n", + "F= (hv-hv1)/hv # fraction of each photon energy unit\n", + "EnergyConToHeat= AbsorbedPower*F # in J/s\n", + "print \"The amount of energy converted to heat per second = %0.2e in J/s \" %EnergyConToHeat\n", + "A= (AbsorbedPower-EnergyConToHeat)/(e*hv1) \n", + "print \"The number of photon per sec given off from recombination events = %0.2e photons/s \" %A" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The absorbed power = 9.0e-03 watt or J/s \n", + "The amount of energy converted to heat per second = 2.57e-03 in J/s \n", + "The number of photon per sec given off from recombination events = 2.81e+16 photons/s \n" + ] + } + ], + "prompt_number": 47 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 3.21.24 - Page No : 3-54" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "Mu_p = 500 # in cm**2/v-s\n", + "kT = 0.0259 \n", + "Toh_p = 10**-10 # in sec\n", + "p_o = 10**17 # in cm**-3\n", + "q= 1.6*10**-19 # in C\n", + "A=0.5 # in square meter\n", + "del_p = 5 * 10**16 # in cm**-3\n", + "n_i= 1.5*10**10 # in cm**-3 \n", + "D_p = kT * Mu_p # in cm/s\n", + "L_p = sqrt(D_p * Toh_p) # in cm\n", + "x = 10**-5 # in cm\n", + "p = p_o+del_p* exp(x/L_p) # in cm**-3\n", + "# p= n_i*%e**(Eip)/kT where Eip=E_i-F_p\n", + "Eip= log(p/n_i)*kT # in eV\n", + "Ecp= 1.1/2-Eip # value of E_c-E_p in eV\n", + "Ip= q*A*D_p/L_p*del_p*exp(x/L_p) # in A\n", + "print \"The hole current = %0.2e A \" %Ip\n", + "Qp= q*A*del_p*L_p # in C\n", + "print \"The value of Qp = %0.2e C \" %Qp\n", + "\n", + "# Note: There is a calculation error or miss print to evalaute the value of hole current but they putted correct \n", + "# value of it to evaluate the value of Qp.Hence the value of hole current in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The hole current = 1.90e+03 A \n", + "The value of Qp = 1.44e-07 C \n" + ] + } + ], + "prompt_number": 48 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_4_1.ipynb b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_4_1.ipynb new file mode 100644 index 00000000..c6818489 --- /dev/null +++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_4_1.ipynb @@ -0,0 +1,948 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:05740426f73b9605877bfd0b7708432621382112013a8d4f8751c47309a06737" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter : 4 - Junction Properties" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 4.16.1 - Page No : 4-69" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import log\n", + "# Given data\n", + "t = 4.4 * 10**22 # total number of Ge atoms/cm**3\n", + "n = 1 * 10**8 # number of impurity atoms\n", + "N_A = t/n # in atoms/cm**3\n", + "N_A = N_A * 10**6 # in atoms/m**3\n", + "N_D = N_A * 10**3 # in atoms/m**3\n", + "n_i = 2.5 * 10**13 # in atoms/cm**3\n", + "n_i = n_i * 10**6 # in atoms/m**3\n", + "V_T = 26 #in mV\n", + "V_T= V_T*10**-3 # in V\n", + "V_J = V_T * log((N_A * N_D)/(n_i)**2) # in V\n", + "print \"Part (a) : The contact potential = %0.3f V \" %V_J\n", + "# Part (b)\n", + "t = 5* 10**22 # total number of Si atoms/cm**3\n", + "N_A = t/n # in atoms/cm**3\n", + "N_A = N_A * 10**6 # in atoms/m**3\n", + "N_D = N_A * 10**3 # in atoms/m**3\n", + "n_i = 1.5 * 10**10 # in atoms/cm**3\n", + "n_i = n_i * 10**6 # in atoms/m**3\n", + "V_T = 26 #in mV\n", + "V_T= V_T*10**-3 # in V\n", + "V_J = V_T * log((N_A * N_D)/(n_i)**2) # in V\n", + "print \"Part (b) : The contact potential = %0.3f V \" %V_J" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (a) : The contact potential = 0.329 V \n", + "Part (b) : The contact potential = 0.721 V \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 4.16.2 - Page No : 4-71" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_T = 26 # in mV\n", + "V_T=V_T*10**-3 # in V\n", + "n_i = 2.5 * 10**13 \n", + "Sigma_p = 1 \n", + "Sigma_n = 1 \n", + "Mu_n = 3800 \n", + "q = 1.6 * 10**-19 # in C\n", + "Mu_p = 1800 \n", + "N_A = Sigma_p/(2* q * Mu_p) # in /cm**3\n", + "N_D = Sigma_n /(q * Mu_n) # in /cm**3\n", + "V_J = V_T * log((N_A * N_D)/(n_i)**2) # in V\n", + "print \"For Ge the height of the energy barrier = %0.2f V \" %V_J\n", + "# For Si p-n juction\n", + "n_i = 1.5 * 10**10 \n", + "Mu_n = 1300 \n", + "Mu_p = 500 \n", + "N_A = Sigma_p/(2* q * Mu_p) # in /cm**3\n", + "N_D = Sigma_n /(q * Mu_n) # in /cm**3\n", + "V_J = V_T * log((N_A * N_D)/(n_i)**2) # in V\n", + "print \"For Si p-n junction the height of the energy barrier = %0.3f V \" %V_J" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For Ge the height of the energy barrier = 0.22 V \n", + "For Si p-n junction the height of the energy barrier = 0.666 V \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.16.3 - Page No : 4-72" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import exp\n", + "# Given data\n", + "Eta = 1 \n", + "V_T = 26 # in mV\n", + "V_T= V_T*10**-3 # in V\n", + "# I = I_o * (exp(V/(Eta*V_T)) - 1) and I = -(0.9) * I_o \n", + "V= log(1-0.9)*V_T # in V\n", + "print \"The voltage = %0.2f volts \" %V\n", + "# Part (ii)\n", + "V1=0.05 # in V\n", + "V2= -0.05 # in V\n", + "ratio= (exp(V1/(Eta*V_T))-1)/(exp(V2/(Eta*V_T))-1)\n", + "print \"The ratio of the current for a forward bias to reverse bias = %0.2f\" %ratio\n", + "# Part (iii)\n", + "Io= 10 # in \u00b5A\n", + "Io=Io*10**-3 # in mA\n", + "#For \n", + "V=0.1 # in V\n", + "I = Io * (exp(V/(Eta*V_T)) - 1) # in mA\n", + "print \"For v=0.1 V , the value of I = %0.3f mA \" %I\n", + "#For \n", + "V=0.2 # in V\n", + "I = Io * (exp(V/(Eta*V_T)) - 1) # in mA\n", + "print \"For v=0.2 V , the value of I = %0.1f mA \" %I\n", + "#For \n", + "V=0.3 # in V\n", + "I = Io * (exp(V/(Eta*V_T)) - 1) # in mA\n", + "print \"For v=0.3 V , the value of I = %0.2f A \" %(I*10**-3)\n", + "print \"From three value of I, for small rise in forward voltage, the diode current increase rapidly\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage = -0.06 volts \n", + "The ratio of the current for a forward bias to reverse bias = -6.84\n", + "For v=0.1 V , the value of I = 0.458 mA \n", + "For v=0.2 V , the value of I = 21.9 mA \n", + "For v=0.3 V , the value of I = 1.03 A \n", + "From three value of I, for small rise in forward voltage, the diode current increase rapidly\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.16.4 - Page No : 4-73" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "# Part (i)\n", + "T1= 25 # in \u00b0C\n", + "T2= 80 # in \u00b0C\n", + "# Formula Io2= Io1*2**((T2-T1)/10)\n", + "AntiFactor= 2**((T2-T1)/10) \n", + "print \"Anticipated factor for Ge = %0.f \" %AntiFactor\n", + "# Part (ii)\n", + "T1= 25 # in \u00b0C\n", + "T2= 150 # in \u00b0C\n", + "AntiFactor= 2**((T2-T1)/10) \n", + "print \"Anticipated factor for Si = %0.f \" %AntiFactor" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Anticipated factor for Ge = 45 \n", + "Anticipated factor for Si = 5793 \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.16.5 - Page No : 4-74" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I=5 # in \u00b5A\n", + "V=10 # in V\n", + "T1= 0.11 # in \u00b0C**-1\n", + "T2= 0.07 # in \u00b0C**-1\n", + "# Io+I_R=I (i)\n", + "# dI_by_dT= dIo_by_dT (ii)\n", + "# 1/Io*dIo_by_dT = T1 and 1/I*dI_by_dT = T2, So\n", + "Io= T2*I/T1 # in \u00b5A\n", + "I_R= I-Io # in \u00b5A\n", + "R= V/I_R # in M\u03a9\n", + "print \"The leakage resistance = %0.1f M\u03a9 \" %R" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The leakage resistance = 5.5 M\u03a9 \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.16.6 - Page No : 4-75" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "Eta = 1 \n", + "T = 125 # in \u00b0C\n", + "T = T + 273 # in K\n", + "V_T = 8.62 * 10**-5 * 398 # in V\n", + "I_o = 30 # in \u00b5A\n", + "I_o= I_o*10**-6 # in A\n", + "v = 0.2 # in V\n", + "r_f = (Eta * V_T)/(I_o * exp(v/(Eta* V_T))) # in ohm\n", + "print \"The dynamic resistance in the forward direction = %0.2f ohm \" %r_f\n", + "r_r = (Eta * V_T)/(I_o * exp(-v/(Eta* V_T))) # in ohm\n", + "print \"The dynamic resistance in the reverse direction = %0.2f kohm \" %(r_r*10**-3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The dynamic resistance in the forward direction = 3.36 ohm \n", + "The dynamic resistance in the reverse direction = 389.08 kohm \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 4.16.7 - Page No : 4-76" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# Given data\n", + "epsilon = 16/(36 * pi * 10**11) # in F/cm\n", + "A = 1 * 10**-2 \n", + "W = 2 * 10**-4 \n", + "C_T = (epsilon * A)/W # in F\n", + "print \"The barrier capacitance = %0.2f pF \" %(C_T*10**12)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The barrier capacitance = 70.74 pF \n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.16.8 - Page No : 4-76" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "# Given data\n", + "A = 1 # in mm**2\n", + "A = A * 10**-6 # in m**2\n", + "N_A = 3 * 10**20 # in atoms/m**3\n", + "q = 1.6 *10**-19 # in C\n", + "V_o = 0.2 # in V\n", + "epsilon_r=16 \n", + "epsilon_o= 8.854*10**-12 # in F/m\n", + "epsilon=epsilon_r*epsilon_o \n", + "# Part (a)\n", + "V=-10 # in V\n", + "# V_o - V = 1/2*((q * N_A )/epsilon) * W**2\n", + "W = sqrt(((V_o - V) * 2 * epsilon)/(q * N_A)) # m\n", + "C_T1 = (epsilon * A)/W # in F\n", + "print \"The width of the depletion layer for an applied reverse voltage of 10V = %0.2f \u00b5m \" %(W*10**6)\n", + "# Part (b)\n", + "V=-0.1 # in V\n", + "W = sqrt(((V_o - V) * 2 * epsilon)/(q * N_A)) # m\n", + "C_T2 = (epsilon * A)/W # in F\n", + "print \"The width of the depletion layer for an applied reverse voltage of 0.1V = %0.2f \u00b5m \" %(W*10**6)\n", + "# Part (c)\n", + "V=0.1 # in V\n", + "W = sqrt(((V_o - V) * 2 * epsilon)/(q * N_A)) # m\n", + "print \"The width of the depletion layer for an applied for a forward bias of 0.1V = %0.3f \u00b5m \" %(W*10**6)\n", + "# Part (d)\n", + "print \"The space charge capacitance for an applied reverse voltage of 10V = %0.2f pF \" %(C_T1*10**12)\n", + "print \"The space charge capacitance for an applied reverse voltage of 0.1V = %0.2f pF \" %(C_T2*10**12)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The width of the depletion layer for an applied reverse voltage of 10V = 7.76 \u00b5m \n", + "The width of the depletion layer for an applied reverse voltage of 0.1V = 1.33 \u00b5m \n", + "The width of the depletion layer for an applied for a forward bias of 0.1V = 0.768 \u00b5m \n", + "The space charge capacitance for an applied reverse voltage of 10V = 18.26 pF \n", + "The space charge capacitance for an applied reverse voltage of 0.1V = 106.46 pF \n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 4.16.9 - Page No : 4-78" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_o = 1.8 * 10**-9 # A\n", + "v = 0.6 # in V\n", + "Eta = 2 \n", + "V_T = 26 # in mV\n", + "V_T=V_T*10**-3 # in V\n", + "I = I_o *(exp(v/(Eta * V_T))) # in A\n", + "print \"The current in the junction = %0.3f mA \" %(I*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current in the junction = 0.185 mA \n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 4.16.10 - Page No : 4-78" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_o = 2.4 * 10**-14 \n", + "I = 1.5 # in mA\n", + "I=I*10**-3 # in A\n", + "Eta = 1\n", + "V_T = 26 # in mV\n", + "V_T= V_T*10**-3 # in V\n", + "v =log((I + I_o)/I_o) * V_T # in V\n", + "print \"The forward biasing voltage across the junction = %0.4f V \" %v" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The forward biasing voltage across the junction = 0.6463 V \n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 4.16.11 - Page No : 4-79" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_o = 10 # in nA\n", + "# I = I_o * ((e**(v/(Eta * V_T))) - 1)\n", + "# e**(v/(Eta * V_T)<< 1, so neglecting it\n", + "I = I_o * (-1) # in nA\n", + "print \"The Diode current = %0.f nA \" %I" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Diode current = -10 nA \n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 4.16.12 - Page No : 4-79" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R = 4.5 # in ohm\n", + "I = 44.4 # in mA\n", + "I=I*10**-3 # in A\n", + "V = R * I # in V\n", + "Eta = 1 \n", + "V_T = 26 #in mV\n", + "V_T=V_T*10**-3 # in V\n", + "I_o = I/((exp(V/(Eta * V_T))) -1) # in A\n", + "# At\n", + "V = 0.1 # in V\n", + "r_f = (Eta * V_T)/(I_o * ((exp(V/(Eta * V_T)))-1)) # in ohm\n", + "print \"The diode dynamic resistance = %0.2f \u03a9 \" %r_f" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The diode dynamic resistance = 27.78 \u03a9 \n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 4.16.13 - Page No : 4-80" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "# Given data\n", + "V_D = 10 # in V\n", + "# V_S = i*R_L + V_D\n", + "V_S = V_D # in V (i * R_L = 0)\n", + "print \"when diode is OFF, the voltage = %0.f volts \" %V_S\n", + "R_L = 250 # in ohm\n", + "I = V_S/R_L # in A\n", + "print \"when diode is ON, the current = %0.f mA \" %(I*10**3)\n", + "V_D= np.arange(0,10,0.1) # in V\n", + "I= (V_S-V_D)/R_L*1000 # in mA\n", + "plt.plot(V_D,I)\n", + "plt.xlabel('V_D in volts') \n", + "plt.ylabel('Current in mA')\n", + "plt.title('DC load line')\n", + "plt.axis([0, 12, 0, 50])\n", + "plt.show()\n", + "print \"DC load line shown in figure\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "when diode is OFF, the voltage = 10 volts \n", + "when diode is ON, the current = 40 mA \n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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MjFgxEJGyabVaaLXaBu9f5z2Bbt264cKFC0b/xX7t2jU0bdoUjo6OuHPnDoYPH45Fixbh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\nCsYkQESkYP8HqYZH+eX0mvAAAAAASUVORK5CYII=\n", + "text": [ + "<matplotlib.figure.Figure at 0x7f0e680b52d0>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "DC load line shown in figure\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 4.16.14 - Page No : 4-81" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V = 0.25 # in V\n", + "I_o = 1.2 # in \u00b5A\n", + "I_o = I_o * 10**-6 # in A\n", + "V_T = 26 # in mV\n", + "V_T = V_T * 10**-3 # in V\n", + "Eta = 1 \n", + "r = (Eta * V_T)/(I_o * (exp(V/(Eta * V_T)))) # in ohm\n", + "print \"The ac resistance of the diode = %0.3f ohm \" %r" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The ac resistance of the diode = 1.445 ohm \n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 4.16.15 - Page No : 4-82" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "t = 4.4 * 10**22 # in total number of atoms/cm**3\n", + "n = 1 * 10**8 # number of impurity\n", + "N_A = t/n # in atoms/cm**3\n", + "N_A = N_A * 10**6 # in atoms/m**3\n", + "N_D = N_A * 10**3 # in atoms/m**3\n", + "V_T = 26 # in mV\n", + "V_T = V_T * 10**-3 # in V\n", + "n_i = 2.5 * 10**19 # in /cm**3\n", + "V_J = V_T * log((N_A * N_D)/(n_i)**2) # in V\n", + "print \"The junction potential = %0.3f V \"%V_J" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The junction potential = 0.329 V \n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 4.16.16 - Page No : 4-82" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "Eta = 1 \n", + "I_o = 30 # in MuA\n", + "I_o = I_o * 10**-6 # in A\n", + "v = 0.2 # in V\n", + "K = 1.381 * 10**-23 # in J/degree K \n", + "T = 125 # in \u00b0C\n", + "T = T + 273 # in K\n", + "q = 1.6 * 10**-19 # in C\n", + "V_T = (K*T)/q # in V\n", + "r_f = (Eta * V_T)/(I_o * (exp(v/(Eta * V_T)))) # in ohm\n", + "print \"The forward dynamic resistance = %0.3f ohm \" %r_f\n", + "r_f1 = (Eta * V_T)/(I_o * (exp(-(v)/(Eta * V_T)))) # in ohm\n", + "print \"The Reverse dynamic resistance = %0.2f k\u03a9 \" %(r_f1*10**-3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The forward dynamic resistance = 3.391 ohm \n", + "The Reverse dynamic resistance = 386.64 k\u03a9 \n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 4.16.17 - Page No : 4-83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "q = 1.6 * 10**-19 # in C\n", + "N_A = 3 * 10**20 # in /m**3\n", + "A = 1 # in \u00b5m**2\n", + "A = A * 10**-6 # in m**2\n", + "V = -10 # in V\n", + "V_J = 0.25 # in V\n", + "V_B = V_J - V # in V\n", + "epsilon_o = 8.854 # in pF/m\n", + "epsilon_o = epsilon_o * 10**-12 # in F/m\n", + "epsilon_r = 16 \n", + "epsilon = epsilon_o * epsilon_r \n", + "W = sqrt((V_B * 2 * epsilon)/(q * N_A)) # in m \n", + "print \"The width of depletion layer = %0.2f \u00b5m \" %(W*10**6)\n", + "C_T = (epsilon * A)/W # in pF\n", + "print \"The space charge capacitance = %0.4f pF \" %(C_T*10**12)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The width of depletion layer = 7.78 \u00b5m \n", + "The space charge capacitance = 18.2127 pF \n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 4.16.18 - Page No : 4-84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "W = 2 * 10**-4 # in cm\n", + "W = W * 10**-2 # in m\n", + "A = 1 # in mm**2\n", + "A = A * 10**-6 # in m**2\n", + "epsilon_r = 16 \n", + "epsilon_o = 8.854 * 10**-12 # in F/m\n", + "epsilon = epsilon_r * epsilon_o \n", + "C_T = (epsilon * A)/W # in F\n", + "print \"The barrier capacitance = %0.3f pF \" %(C_T*10**12)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The barrier capacitance = 70.832 pF \n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 4.16.19 - Page No : 4-85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "C_T = 100 # in pF\n", + "C_T=C_T*10**-12 # in F\n", + "epsilon_r = 12 \n", + "epsilon_o = 8.854 * 10**-12 # in F/m\n", + "epsilon = epsilon_r * epsilon_o \n", + "Rho_p = 5 # in ohm-cm\n", + "Rho_p = Rho_p * 10**-2 # in ohm-m\n", + "V_j = 0.5 # in V\n", + "V = -4.5 # in V\n", + "Mu_p = 500 # in cm**2\n", + "Mu_p = Mu_p * 10**-4 # in m**2\n", + "Sigma_p = 1/Rho_p # in per ohm-m\n", + "qN_A = Sigma_p/ Mu_p \n", + "V_B = V_j - V \n", + "W = sqrt((V_B * 2 * epsilon)/qN_A) # in m\n", + "#C_T = (epsilon * A)/W \n", + "A = (C_T * W)/ epsilon # in m\n", + "D = sqrt(A * (4/pi)) # in m\n", + "D = D * 10**3 # in mm\n", + "print \"The value of diameter = %0.3f mm \" %D\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of diameter = 1.398 mm \n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 4.16.20 - Page No : 4-20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "q = 1.6 * 10**-19 # in C\n", + "Mu_p = 500 # in cm**2/V-sec\n", + "Rho_p = 3.5 # in ohm-cm\n", + "Mu_n = 1500 # in cm**2/V-sec\n", + "Rho_n = 10 # in ohm-cm\n", + "N_A = 1/(Rho_p * Mu_p * q) # in /cm**3\n", + "N_D = 1/(Rho_n * Mu_n * q) # in /cm**3\n", + "V_J = 0.56 # in V\n", + "n_i = 1.5 * 10**10 # in /cm**3\n", + "V_T = V_J/log((N_A * N_D)/(n_i)**2) # in V\n", + "# V_T = T/11600\n", + "T = V_T * 11600 # in K\n", + "T = T /19.78 # in \u00b0C ( 1 degree K = 19.78 degree C)\n", + "print \"The Temperature of junction = %0.3f \u00b0C \" %T" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Temperature of junction = 14.524 \u00b0C \n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 4.16.21 - Page No : 4-88" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_T = 26 # in mV\n", + "V_T = V_T * 10**-3 # in V\n", + "Eta = 1 \n", + "# I = -90% for Io, so\n", + "IbyIo= 0.1 \n", + "# I = I_o * ((e**(v/(Eta * V_T)))-1)\n", + "V = log(IbyIo) * V_T # in V\n", + "print \"The reverse bias voltage = %0.5f volts \" %V" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The reverse bias voltage = -0.05987 volts \n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 4.16.22 - Page No : 4-89" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R = 5 # in ohm\n", + "I = 50 # in mA\n", + "I=I*10**-3 # in A\n", + "V = R * I # in V\n", + "Eta = 1 \n", + "V_T = 26 # in mV\n", + "V_T=V_T*10**-3 # in V\n", + "I_o = I/((exp(V/(Eta * V_T))) - 1) # in A\n", + "print \"Reverse saturation current = %0.2f \u00b5A \" %(I_o*10**6)\n", + "v1 = 0.2 # in V\n", + "r = (Eta * V_T)/(I_o * (exp(v1/(Eta * V_T)))) # in ohm\n", + "print \"Dynamic resistance of the diode = %0.3f \u03a9 \" %r" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Reverse saturation current = 3.33 \u00b5A \n", + "Dynamic resistance of the diode = 3.558 \u03a9 \n" + ] + } + ], + "prompt_number": 22 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_5_1.ipynb b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_5_1.ipynb new file mode 100644 index 00000000..ded28641 --- /dev/null +++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_5_1.ipynb @@ -0,0 +1,834 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:f580dc14cd164012c93f758d593e5fbb20692db2a4830c462cc2ecc139c62a4a" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter : 5 - Bipolar Junction Transistors (BJTs)" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 5.9.1 - Page No : 5-22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# Given data\n", + "V_EE = 8 # in V\n", + "V_BE = 0.7 # in V\n", + "R_E = 1.5 # in k ohm\n", + "I_E = (V_EE - V_BE)/R_E # in mA\n", + "I_C = I_E # in mA\n", + "print \"The value of I_C = %0.2f mA \" %I_C\n", + "V_CC = 18 # in V\n", + "R_C = 1.2 # in k\u03a9\n", + "V_CB = V_CC - (I_C * R_C) # in V\n", + "print \"The value of V_CB = %0.2f V \" %V_CB" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_C = 4.87 mA \n", + "The value of V_CB = 12.16 V \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 5.9.2 - Page No : 5-23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "alpha = 0.9 \n", + "I_E = 1 # mA\n", + "I_C = alpha * I_E # in mA\n", + "I_B = I_E - I_C # in mA\n", + "print \"The value of base current = %0.1f mA \" %I_B" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of base current = 0.1 mA \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 5.11.1 - Page No : 5-41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "bita = 50 \n", + "I_B= 20 # in \u00b5A\n", + "I_B=I_B*10**-6 # in A\n", + "I_C= bita*I_B # in A\n", + "I_E= I_C+I_B # in A\n", + "I_E = I_E * 10**3 # in mA\n", + "print \"The Emitter current = %0.2f mA \" %I_E" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Emitter current = 1.02 mA \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 5.11.1(a) - Page No : 5-42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "beta_dc = 90 \n", + "I_C = 15 # in mA\n", + "I_C = I_C * 10**-3 # in A\n", + "I_B = I_C/beta_dc # in A\n", + "print \"The base current = %0.2f \u00b5A \" %(I_B*10**6)\n", + "I_E = I_C + I_B # in A\n", + "I_E = I_E * 10**3 # in mA\n", + "print \"The Emitter current = %0.3f mA \" %I_E\n", + "alpha_dc = beta_dc/(1+beta_dc) \n", + "print \"The value of alpha_dc = %0.3f\" %alpha_dc" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 166.67 \u00b5A \n", + "The Emitter current = 15.167 mA \n", + "The value of alpha_dc = 0.989\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 5.11.3 - Page No : 5-43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "del_ic = 1.8 # in mA\n", + "del_ie = 1.89 # in mA\n", + "alpha = del_ic / del_ie \n", + "bita = alpha/(1 - alpha) \n", + "del_ib = del_ic/bita # in mA\n", + "del_ib = del_ib * 10**3 # in \u00b5A\n", + "print \"The change in I_B = %0.f \u00b5A \" %del_ib" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The change in I_B = 90 \u00b5A \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 5.11.4 - Page No : 5-43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC = 10 # in V\n", + "R_C = 3 # in k \u03a9\n", + "R_C= R_C*10**3 # in \u03a9\n", + "bita = 100 \n", + "I_CO = 20 # in nA\n", + "I_CO = I_CO * 10**-9 # in A\n", + "V_BB = 5 # in V\n", + "R_B = 200 # in k\u03a9\n", + "R_B= R_B*10**3 # in \u03a9\n", + "V_BE = 0.7 # in V\n", + "# Applying KVL to the base circuit, V_BB= I_B*R_B+V_BE\n", + "I_B = (V_BB - V_BE)/R_B # in A\n", + "print \"The base current = %0.1f \u00b5A \" %(I_B*10**6)\n", + "I_C = (bita * I_B) + I_CO # in A\n", + "print \"The collector current = %0.5f mA \" %(I_C*10**3)\n", + "I_E = I_C + I_B # in A\n", + "print \"Emitter current = %0.5f mA \" %(I_E*10**3)\n", + "V_CE = V_CC - (I_C * R_C) # in V\n", + "print \"Collector emitter voltage = %0.4f V \" %(V_CE)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 21.5 \u00b5A \n", + "The collector current = 2.15002 mA \n", + "Emitter current = 2.17152 mA \n", + "Collector emitter voltage = 3.5499 V \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.11.5 - Page No : 5-44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "bita = 100 \n", + "I_CBO = 4 # in \u00b5A\n", + "I_B = 40 # in \u00b5A\n", + "I_C = (bita * I_B) + ((1+bita) * I_CBO) # in \u00b5A\n", + "I_C = I_C * 10**-3 # in msA\n", + "print \"The collector current = %0.3f mA \" %I_C" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The collector current = 4.404 mA \n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 5.11.6 - Page No : 5-45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "from __future__ import division\n", + "# Given data\n", + "del_IC = 1 * 10**-3 # in A\n", + "del_IB = 10 * 10**-6 # in A\n", + "CurrentGain= del_IC/del_IB \n", + "print \"The current gain = %0.f\" %CurrentGain\n", + "del_IC= del_IC*10**3 # in mA\n", + "del_IB= del_IB*10**6 # in \u00b5A\n", + "I_B=np.arange(0,50,0.1) # in \u00b5A\n", + "I_C= I_B/del_IB+del_IC # in mA\n", + "plt.plot(I_B,I_C)\n", + "plt.xlabel('Base current in micro A')\n", + "plt.ylabel('Collector current in mA')\n", + "plt.title('Transfer Characteristics')\n", + "plt.axis([0, 60, 0, 7])\n", + "plt.show()\n", + "print \"Transfer Characteristics is shown in figure\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current gain = 100\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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5IrJnNt/GYV+eiMiGw559eSKi39hkG4d9eSKi+mwq7NmXJyJqnE20cdiXJyJq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\nEdkBhj0RkR1g2BMR2YH/A2IOTkNSzG6GAAAAAElFTkSuQmCC\n", + "text": [ + "<matplotlib.figure.Figure at 0x7f8cdc73de90>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Transfer Characteristics is shown in figure\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 5.14.1 - Page No : 5-51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_CBO = 3 #in \u00b5A\n", + "I_CBO= I_CBO*10**-3 # in mA \n", + "I_C= 15 # in mA\n", + "# But it is given that I_C= 99.5% of I_E, SO\n", + "I_E= I_C/99.5*100 # in mA\n", + "alpha_dc= I_C/I_E \n", + "print \"The value of alpha_dc = %0.3f\" %alpha_dc\n", + "print \"The value of I_E = %0.2f mA \" %I_E" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of alpha_dc = 0.995\n", + "The value of I_E = 15.08 mA \n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 5.14.2 - Page No : 5-52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given data\n", + "alpha_dc = 0.99 \n", + "I_CBO = 10 # in \u00b5A\n", + "I_CBO= I_CBO*10**-6 # in A\n", + "I_E = 10 # in mA\n", + "I_E= I_E*10**-3 # in A\n", + "I_C = (alpha_dc * I_E) + I_CBO # in A\n", + "print \"The value of I_C = %0.2f mA \" %(I_C*10**3)\n", + "I_B = I_E - I_C # in A\n", + "I_B = I_B * 10**6 # in \u00b5A\n", + "print \"The value of I_B = %0.f \u00b5A \" %I_B" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_C = 9.91 mA \n", + "The value of I_B = 90 \u00b5A \n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 5.14.3 - Page No : 5-52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "alpha_dc = 0.99 \n", + "I_C = 6 # in mA\n", + "I_C= I_C*10**-3 # in A\n", + "I_CBO = 15 # in \u00b5A\n", + "I_CBO= I_CBO*10**-6 # in A\n", + "I_E = (I_C - I_CBO)/alpha_dc # in A\n", + "I_B = I_E - I_C # in A \n", + "print \"The value of I_B = %0.f \u00b5A \" %(I_B*10**6)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_B = 45 \u00b5A \n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 5.14.5 - Page No : 5-53" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "alpha_dc = 0.98 \n", + "I_CBO = 12 # in \u00b5A\n", + "I_CBO = I_CBO * 10**-6 # in A\n", + "I_B = 120 # in \u00b5A\n", + "I_B = I_B * 10**-6 # in A\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 \"The value of I_E = %0.1f mA \" %I_E" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_E = 6.6 mA \n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 5.14.6 - Page No : 5-53" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "bita= 100 \n", + "V_BEsat= 0.8 # in V\n", + "V_CEsat= 0.2 # in V\n", + "V_BEact= 0.7 # in V\n", + "V_CC = 10 # in V\n", + "V_BB=5 # in V\n", + "R_E = 2 # in k\u03a9\n", + "R_C = 3 # in k\u03a9\n", + "R_B= 50 # in k\u03a9\n", + "# Applying KVL to collector loop\n", + "# V_CC= I_Csat*R_C +V_CEsat +I_E*R_E and I_E= I_Csat+I_B, So\n", + "#I_B= ((V_CC-V_CEsat)-(R_C+R_E)*I_Csat)/R_E (i)\n", + "# Applying KVL to base loop\n", + "# V_BB-I_B*R_B -V_BEsat-I_E*R_E =0 and I_E= I_Csat+I_B, So\n", + "#V_BB-V_BEsat= R_E*I_Csat + (R_B+R_E)*I_B (ii)\n", + "# From eq (i) and (ii)\n", + "I_B = ((V_BB-V_BEsat)*5- (V_CC-V_CEsat)*2) / ((R_B+R_E)*5 - R_E*2) # in mA\n", + "I_Csat= ((V_CC-V_CEsat)-R_E*I_B)/(R_C+R_E) # in mA\n", + "I_Bmin= I_Csat/bita # in mA\n", + "if I_B<I_Bmin :\n", + " print \"Since the value of I_B (\",round(I_B*10**3,2),\"\u00b5A) is less than the value of I_Bmin (\",round(I_Bmin*10**3,1),\"\u00b5A)\" \n", + " print \"So the transistor is not in the saturation region. But it is conducting hence it can not be in cutoff.\"\n", + " print \"Therefore the transistor is in the active region\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Since the value of I_B ( 5.47 \u00b5A) is less than the value of I_Bmin ( 19.6 \u00b5A)\n", + "So the transistor is not in the saturation region. But it is conducting hence it can not be in cutoff.\n", + "Therefore the transistor is in the active region\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.14.7 - Page No : 5-55" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "bita= 100 \n", + "V_BEsat= 0.8 # in V\n", + "V_CEsat= 0.2 # in V\n", + "V_BEact= 0.7 # in V\n", + "V_CC = 10 # in V\n", + "V_BB=5 # in V\n", + "R_E = 2 # in k\u03a9\n", + "R_C = 3 # in k\u03a9\n", + "R_B= 50 # in k\u03a9\n", + "# Applying KVL to input loop\n", + "# V_BB= I_B*R_B+(1+bita)*I_B*R_E+V_BEact or \n", + "I_B= (V_BB-V_BEact)/(R_B+(1+bita)*R_E) # in mA\n", + "I_C= bita*I_B # in mA\n", + "# Applying KVL to collector circuit\n", + "# V_CC= I_Csat*R_C +V_CEsat +(I_C+I_B)*R_E\n", + "V_CEact= V_CC-I_B*R_E-I_C*(R_C+R_E) # in V\n", + "print \"The value of I_B = %0.f \u00b5A \" %(I_B*10**3)\n", + "print \"The value of I_C = %0.1f mA \" %I_C\n", + "print \"The value of V_CE = %0.3f volts \" %V_CEact" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_B = 17 \u00b5A \n", + "The value of I_C = 1.7 mA \n", + "The value of V_CE = 1.434 volts \n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.14.8 - Page No : 5-55" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "bita = 100 \n", + "V_CEsat = 0.2 # in V\n", + "R_B = 150 # in kohm\n", + "R_C = 2 # in kohm\n", + "V_CC = 10 # in V\n", + "V_BEsat = 0.8 # in V\n", + "I_B = (V_CC - V_BEsat)/R_B # in mA\n", + "I_C = (V_CC - V_CEsat)/R_C # in mA\n", + "I_Bmin = I_C/bita # in mA\n", + "if I_B>I_Bmin :\n", + " print \"Since the value of I_B (\",round(I_B*10**3,2),\"\u00b5A) is greater than the value of I_Bmin (\",int(I_Bmin*10**3),\"\u00b5A)\" \n", + " print \"So the transistor is in the saturation region.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Since the value of I_B ( 61.33 \u00b5A) is greater than the value of I_Bmin ( 49 \u00b5A)\n", + "So the transistor is in the saturation region.\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.14.9 - Page No : 5-57" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "bita = 100 \n", + "V_CE = 0.2 #in V\n", + "V_BE = 0.8 # in V\n", + "R_C= 500 # in \u03a9\n", + "R_B= 44*10**3 # in \u03a9\n", + "R_E= 1*10**3 # in \u03a9\n", + "V_CC= 15 # in V\n", + "V_GE= -15 # in V\n", + "# Applying KVL to collector circuit\n", + "# V_CC-V_GE - I_Csat*R_C-V_CE-I_E*R_E=0, but I_Csat= bita*I_Bmin and I_E= 1+bita\n", + "I_Bmin= (V_CC-V_GE-V_CE)/(R_C*bita+(1+bita)*R_E) # in A\n", + "# Applying KVL to the base emitter circuit\n", + "# V_BB-I_Bmin*R_B-V_BE-I_E*R_E + V_CC=0\n", + "V_BB= I_Bmin*R_B + V_BE + (1+bita)*I_Bmin*R_E-V_CC # in V\n", + "print \"The value of I_B(min) = %0.3f mA \" %(I_Bmin*10**3)\n", + "print \"The value of V_BB = %0.1f volts \" %V_BB" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_B(min) = 0.197 mA \n", + "The value of V_BB = 14.4 volts \n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 5.14.10 - Page No : 5-58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_ECsat= 0.2 # in V\n", + "V_CC= 10 # in V\n", + "V_EBsat= 0.8 # in V\n", + "\n", + "# Part (i)\n", + "bita= 100 \n", + "R_B= 220 # in k\u03a9\n", + "# Applying KVL to collector circuit, V_CC= V_EC+ICRC\n", + "ICRC= V_CC-V_ECsat # in V\n", + "# Applying KVL to input loop, V_CC= V_EBsat+I_B*R_B (i)\n", + "I_B= (V_CC-V_EBsat)/R_B # in mA\n", + "I_C= bita*I_B # in mA\n", + "R_Cmin= ICRC/I_C # in k\u03a9\n", + "print \"The minimum value of R_C = %0.3f k\u03a9 \" %R_Cmin\n", + "# Part (ii)\n", + "R_C= 1.2 # in k\u03a9\n", + "I_Csat= ICRC/R_C # in mA\n", + "I_B= I_Csat/bita # in mA\n", + "# From eq (i)\n", + "R_B= (V_CC-V_EBsat)/I_B # in k\u03a9\n", + "print \"The maximum value of R_B = %0.2f k\u03a9 \" %R_B" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum value of R_C = 2.343 k\u03a9 \n", + "The maximum value of R_B = 112.65 k\u03a9 \n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 5.14.11 - Page No : 5-60" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "bita= 100 \n", + "V_BEsat= 0.8 # in V\n", + "V_CEsat= 0.2 # in V\n", + "V_BEact= 0.7 # in V\n", + "V_CC = 10 # in V\n", + "R_E = 1 # in k\u03a9\n", + "R_C = 2 # in k\u03a9\n", + "R_B= 100 # in k\u03a9\n", + "bita=100 \n", + "alpha= bita/(1+bita) \n", + "# Applying KVL to collector circuit\n", + "# V_CC= I_Csat*R_C +V_CE +R_E*I_E\n", + "# but I_E= alpha*I_Csat\n", + "I_Csat= (V_CC-V_CEsat)/(R_C+R_E*alpha) # in mA\n", + "I_Bmin= I_Csat/bita # in mA\n", + "# Applying KVL to base loop\n", + "# V_CC= I_B*R_B +V_BEsat +I_E*R_E\n", + "# but I_E= I_Csat+I_B\n", + "I_B= (V_CC-V_BEsat-I_Csat*R_E)/(R_B+R_E) # in mA\n", + "print \"The value of I_B = %0.2f \u00b5A \" %(I_B*10**3)\n", + "print \"The minimum value of I_B = %0.1f \u00b5A \" %(I_Bmin*10**3)\n", + "if I_B>I_Bmin :\n", + " print \"Since the value of I_B is greater than the value of I_Bmin\"\n", + " print \"Hence the transistor is in saturation .\"\n", + "I_E= (1+bita)*I_Bmin # in mA\n", + "R_E= (V_CC-V_BEact-I_Bmin*R_B)/I_E # in k\u03a9\n", + "print \"The value of R_E = %0.3f k\u03a9 \" %R_E\n", + "print \"So R_E should be greater than this value in order to bring the transistor just out of saturation \"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_B = 58.64 \u00b5A \n", + "The minimum value of I_B = 32.8 \u00b5A \n", + "Since the value of I_B is greater than the value of I_Bmin\n", + "Hence the transistor is in saturation .\n", + "The value of R_E = 1.819 k\u03a9 \n", + "So R_E should be greater than this value in order to bring the transistor just out of saturation \n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 5.14.12 - Page No : 5-62" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC = 9 # in V\n", + "V_BE = 0.8 # in V\n", + "V_CE = 0.2 # in V\n", + "R_B = 50 # in k\u03a9\n", + "R_C=2 # in k\u03a9\n", + "R_E = 1 # in k\u03a9\n", + "bita=70 \n", + "# Applying KVL to input loop, V_CC= I_B*R_B +V_BE +I_E*R_E\n", + "# V_CC- V_BE= (R_B+R_E)*I_B + R_E*I_C (i)\n", + "# Applying KVL to output loop, V_CC= R_C*I_C +V_CE +I_C*R_E +I_B*R_E\n", + "#I_B = ((V_CC- V_CE)-(R_C+R_E)*I_C)/R_E (ii)\n", + "# From eq (i) and (ii)\n", + "I_C= ( (V_CC- V_BE)-(R_B+R_E)* (V_CC- V_CE)/R_E)/(1-(R_B+R_E)*(R_C+R_E)) # in mA\n", + "I_B = ((V_CC- V_CE)-(R_C+R_E)*I_C)/R_E# in mA\n", + "I_Bmin= I_C/bita # in mA\n", + "if I_B>I_Bmin :\n", + " print \"Since the value of I_B (\",round(I_B,3),\" mA) is greater than the value of I_Bmin (\",round(I_Bmin,4),\" mA)\"\n", + " print \"So the transistor is in saturation \"\n", + "V_C= V_CC-I_C*R_C # in V\n", + "print \"The value of collector voltage = %0.3f volts \" %V_C\n", + "bita= I_C/I_B \n", + "print \"The minimum value of bita that will change the state of the trasistor = %0.3f\" %bita" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Since the value of I_B ( 0.104 mA) is greater than the value of I_Bmin ( 0.0414 mA)\n", + "So the transistor is in saturation \n", + "The value of collector voltage = 3.203 volts \n", + "The minimum value of bita that will change the state of the trasistor = 27.886\n" + ] + } + ], + "prompt_number": 30 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_6_1.ipynb b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_6_1.ipynb new file mode 100644 index 00000000..844651b3 --- /dev/null +++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_6_1.ipynb @@ -0,0 +1,208 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter : 6 - Field Effect Transistors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 6.9.1 - Page No : 6-26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from numpy import pi\n", + "# Given data\n", + "q = 1.6 * 10**-19;# in C\n", + "N_D = 10**15;# in electrons/cm**3\n", + "N_D = N_D * 10**6;# in electrons/m**3\n", + "epsilon_r = 12;\n", + "epsilon_o = (36 * pi * 10**9)**-1;\n", + "epsilon = epsilon_o * epsilon_r;\n", + "a = 3 * 10**-4;# in cm\n", + "a = a * 10**-2;# in m\n", + "V_P = (q * N_D * a**2)/( 2 * epsilon);# in V\n", + "print \"The Pinch off voltage = %0.1f V\" %V_P\n", + "# V_GS = V_P * (1-(b/a))**2\n", + "b = (1-0.707) *a;# in m\n", + "print \"The value of b = %0.3f \u00b5m\" %(b*10**6)\n", + "print \"Hence the channel width has been reduced to about one third of its value for V_GS = 0\"\n", + "# Note : The unit of b in the book is wrong since the value of b is calculated in \u00b5m." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Pinch off voltage = 6.8 V\n", + "The value of b = 0.879 \u00b5m\n", + "Hence the channel width has been reduced to about one third of its value for V_GS = 0\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 6.9.2 - Page No : 6-26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "# Given data\n", + "I_DSS = 8;# in mA\n", + "V_P = -4;# in V\n", + "I_D = 3;# in mA\n", + "V_GS = V_P * (1 - sqrt(I_D/I_DSS));# in V\n", + "print \"The value of V_GS = %0.2f V\" %V_GS\n", + "V_DS = V_GS - V_P;# 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 V_GS = -1.55 V\n", + "The value of V_DS = 2.45 V\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 6.9.3 - Page No : 6-27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_P = -4;# in V\n", + "I_DSS = 9;# in mA\n", + "I_DSS = I_DSS * 10**-3;# in A\n", + "V_GS = -2;# in V\n", + "I_D = I_DSS * ((1 - (V_GS/V_P))**2);# in A\n", + "print \"The drain current = %0.2f mA\" %(I_D*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current = 2.25 mA\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 6.9.4 - Page No : 6-27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_DSS = 12;# in mA\n", + "I_DSS = I_DSS * 10**-3;# in A\n", + "V_P = -(6);# in V\n", + "V_GS = -(1);# in V\n", + "g_mo = (-2 * I_DSS)/V_P;# in A/V\n", + "g_m = g_mo * (1 - (V_GS/V_P));# in S\n", + "print \"The value of transconductance = %0.2f mS\" %(g_m*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of transconductance = 3.33 mS\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 6.9.5 - Page No : 6-28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_DSS = 10;# in mA \n", + "I_DSS = I_DSS * 10**-3;# in A\n", + "V_P = -(5);# in V\n", + "V_GS = -(2.5);# in V\n", + "g_m = ((-2 * I_DSS)/V_P) * (1 -(V_GS/V_P));# in S\n", + "g_m = g_m * 10**3;# in mS\n", + "print \"The Transconductance = %0.f mS\" %g_m\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)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Transconductance = 2 mS\n", + "The drain current = 2.5 mA\n" + ] + } + ], + "prompt_number": 14 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_7_1.ipynb b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_7_1.ipynb new file mode 100644 index 00000000..26297594 --- /dev/null +++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_7_1.ipynb @@ -0,0 +1,143 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter : 7 - Optoelectronic Devices" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 7.6.1 - Page No : 7-16" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# Given data\n", + "O_V = 5;# output voltage in V\n", + "V_D = 1.5;#voltage drop in V\n", + "R = (O_V - V_D)/O_V;\n", + "R = R * 10**3;# in ohm\n", + "print \"The resistance value = %0.f \u03a9\" %R\n", + "print \"As this is not standard value, use R=680 \u03a9 which is a standard value\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resistance value = 700 \u03a9\n", + "As this is not standard value, use R=680 \u03a9 which is a standard value\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 7.23.1 - Page No : 7-57" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt, log\n", + "# Given data\n", + "N_A = 7.5*10**24;# in atoms/m**3\n", + "N_D = 1.5*10**22;# in atoms/m**3\n", + "I_lembda = 12.5*10**-3;# in A/cm**2\n", + "D_e = 25*10**-4;# in m**2/s\n", + "D_h = 1*10**-3;# in m**2/s\n", + "Torque_eo = 500;# in ns\n", + "Torque_ho = 100;# in ns\n", + "n_i = 1.5*10**16;# in /m**3\n", + "e = 1.6*10**-19;\n", + "P_C = 12.5;# in mA/cm**2\n", + "L_e = sqrt(D_e*Torque_ho*10**-9);# in m\n", + "L_e = L_e * 10**6;# in \u00b5m\n", + "L_h = sqrt(D_h*Torque_ho*10**-9);# in m\n", + "L_h = L_h * 10**6;# in \u00b5m\n", + "J_s = e*((n_i)**2)*( (D_e/(L_e*10**-6*N_A)) + (D_h/(L_h*10**-6*N_D)) );# in A/m**2\n", + "J_s = J_s * 10**-4;# in A/cm**2\n", + "V_T = 26;# in mV\n", + "V_OC = V_T*log( 1+(I_lembda/J_s) );# in mV\n", + "V_OC = V_OC * 10**-3;# in V\n", + "print \"Open circuit voltage = %0.3f V\" %V_OC" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Open circuit voltage = 0.522 V\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 7.23.2 - Page No : 7-57" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import exp\n", + "# Given data\n", + "Phi_o = 1*10**21;# in m**-2s**-1\n", + "Alpha = 1*10**5;# in m**-1\n", + "W = 25;# in \u00b5m\n", + "W =W * 10**-6;# in m\n", + "e = 1.6*10**-19;# in C\n", + "G_L1 = Alpha*Phi_o;# in m**-3s**-1\n", + "G_L2 = Alpha*Phi_o*exp( (-Alpha*W) );# in m**-3s**-1\n", + "J_L = e*Phi_o*(1-exp(-Alpha*W));# in A/m**2\n", + "J_L = J_L * 10**-1;# in mA/cm**2\n", + "print \"Photo current density = %0.2f mA/cm**2\" %J_L" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Photo current density = 14.69 mA/cm**2\n" + ] + } + ], + "prompt_number": 8 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_8_1.ipynb b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_8_1.ipynb new file mode 100644 index 00000000..62a399cf --- /dev/null +++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/chapter_8_1.ipynb @@ -0,0 +1,96 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter : 8 - Negative Conductance Microwave Devices And Power Devices" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 8.13.1 - Page No : 8-35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# Given data\n", + "R= 10;# in k\u03a9\n", + "R= R*10**3;# in \u03a9\n", + "# Part (i)\n", + "V=300;# in V\n", + "I_A= V/R;# in A\n", + "print \"Part (i) : For 300 V voltage : \"\n", + "print \"The anode current = %0.f mA\" %(I_A*10**3)\n", + "# Part (ii)\n", + "V=100;# in V\n", + "I_A= V/R;# in A\n", + "print \"Part (ii) : For 100 V voltage : \"\n", + "print \"The anode current = %0.f mA\" %(I_A*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (i) : For 300 V voltage : \n", + "The anode current = 30 mA\n", + "Part (ii) : For 100 V voltage : \n", + "The anode current = 10 mA\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example : 8.14.1 - Page No : 8-43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "t_rr = 10;# in \u00b5s\n", + "Q_rr = 150;# in \u00b5c\n", + "I_rr = (2*Q_rr)/t_rr;# in A\n", + "print \"The peak reverse recovery current = %0.f A\" %I_rr" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The peak reverse recovery current = 30 A\n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/1_1.png b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/1_1.png Binary files differnew file mode 100644 index 00000000..0d53ffcc --- /dev/null +++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/1_1.png diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/6_1.png b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/6_1.png Binary files differnew file mode 100644 index 00000000..02ebc92e --- /dev/null +++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/6_1.png diff --git a/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/7_1.png b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/7_1.png Binary files differnew file mode 100644 index 00000000..f87b5693 --- /dev/null +++ b/Fundamentals_Of_Electronics_Devices_by_Dr._K._C._Nandi/screenshots/7_1.png diff --git a/Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/README.txt b/Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/README.txt new file mode 100644 index 00000000..b67a8ea3 --- /dev/null +++ b/Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/README.txt @@ -0,0 +1,10 @@ +Contributed By: hemla rathod +Course: btech +College/Institute/Organization: Gokaraju Rangaraju Institute of Engineering and Technology +Department/Designation: mechanical engneering +Book Title: Introduction To Mechanical Engineering +Author: S. Chandra And O. Singh +Publisher: New Age International Pvt. Ltd, Delhi +Year of publication: 2001 +Isbn: 81-224-1342-0 +Edition: 1
\ No newline at end of file diff --git a/Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/README.txt b/Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/README.txt new file mode 100644 index 00000000..8c105aea --- /dev/null +++ b/Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/README.txt @@ -0,0 +1,10 @@ +Contributed By: pratik sonone +Course: btech +College/Institute/Organization: iitbombay +Department/Designation: metallurgical engineering +Book Title: Materials Science and Engineering - A First Course +Author: V. Raghavan +Publisher: Prentice Hall, India +Year of publication: 2007 +Isbn: 9788120324558 +Edition: 5
\ No newline at end of file diff --git a/Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/README.txt b/Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/README.txt new file mode 100644 index 00000000..6a76af40 --- /dev/null +++ b/Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/README.txt @@ -0,0 +1,10 @@ +Contributed By: mmn vijayadurga +Course: others +College/Institute/Organization: Manchukonda Srinivasa Rao & Co +Department/Designation: Special Telugu +Book Title: Measurement Systems +Author: E. O. Doebelin And D. N. Manik +Publisher: Tata McGraw - Hill Education +Year of publication: 2007 +Isbn: 978-0070173385 +Edition: 5
\ No newline at end of file diff --git a/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap_/README.txt b/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap_/README.txt new file mode 100644 index 00000000..5de12de7 --- /dev/null +++ b/Principles_of_Electrical_Engineering_Materials_by_S._O._Kasap_/README.txt @@ -0,0 +1,10 @@ +Contributed By: Jignesh Bhadani +Course: bca +College/Institute/Organization: Freelancer +Department/Designation: BCA +Book Title: Principles of Electrical Engineering Materials +Author: S. O. Kasap +Publisher: Mgh +Year of publication: 2000 +Isbn: 0072356448 +Edition: 1
\ No newline at end of file diff --git a/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_10_SILICON_CONTROLLED_RECTIFIER_1.ipynb b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_10_SILICON_CONTROLLED_RECTIFIER_1.ipynb new file mode 100644 index 00000000..6410798f --- /dev/null +++ b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_10_SILICON_CONTROLLED_RECTIFIER_1.ipynb @@ -0,0 +1,119 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 10 SILICON CONTROLLED RECTIFIER" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exmaple 10_2 pgno: 296" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nd = 100000000000000 /cmˆ−3\n", + "Er = 11.9\n", + "e = 1.6e-19 columns\n", + "Eo = 8.854e-14 F/cm\n", + "W = 0.01 cm\n", + " total permittivity ,E=Eo∗Er= 1.053626e-12 F/cm\n", + "Punch trough voltage ,Vpt=(e∗Nd∗Wˆ2)/(2∗E))= 759.282705628 V\n" + ] + } + ], + "source": [ + "#exa 10.2\n", + "Nd =10**14\n", + "print\"Nd = \",Nd,\" /cmˆ−3\" # initializing value of donor ion concentration .\n", + "Er =11.9\n", + "print\"Er = \",Er # initializing value of relative dielectric permittivity constant .\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\" columns\" # initializing value of charge of electrons .\n", + "Eo=8.854*10**-14\n", + "print\"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "W=100*10**-4\n", + "print\"W = \",W,\" cm\" # initializing value of width of SCR.\n", + "E=Eo*Er\n", + "print\" total permittivity ,E=Eo∗Er=\",E,\" F/cm\" # calculation\n", + "Vpt=(e*Nd*W**2)/(2*E)\n", + "print\"Punch trough voltage ,Vpt=(e∗Nd∗Wˆ2)/(2∗E))=\",Vpt,\" V\"# calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exmaple 10_3 pgno: 296" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ia = 0.002 A\n", + "(ap+an) = 0.9\n", + "a = 0.45\n", + "Ico=Ia∗(1−(2∗an))= 0.0002 A\n", + "(da/dt)=1/2∗Ico∗((Ia)ˆ−2))= 25.0 /A\n" + ] + } + ], + "source": [ + "#exa 10.3\n", + "Ia =2e-3\n", + "print\"Ia = \",Ia,\" A\" # initializing value of forward current of thyrsistor .\n", + "x=0.9\n", + "print\"(ap+an) = \",x # initializing value of sum of current gain of n,ptype semiconductor [ value is get in by variable x,but represented on console window through ap +an ] .\n", + "a=0.45\n", + "print\"a = \",a # initializing value of current gain of both n,p type semiconductor (as it is assume that ap[current gain of n type semiconductor]=an[ current gain of ptype semiconductor ] in the question ) .\n", + "Ico=Ia*(1-(2*a))\n", + "print\"Ico=Ia∗(1−(2∗an))=\",Ico,\" A\" # calculation\n", + "y=1./2.*Ico*((Ia)**-2)\n", + "print\"(da/dt)=1/2∗Ico∗((Ia)ˆ−2))=\",y,\" /A\" # calculation\n", + "#The answer for (da/dt) after doing calculation is provided wrong in the book ." + ] + } + ], + "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": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_1_CRYSTAL_STRUCTURES_1.ipynb b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_1_CRYSTAL_STRUCTURES_1.ipynb new file mode 100644 index 00000000..cd376de8 --- /dev/null +++ b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_1_CRYSTAL_STRUCTURES_1.ipynb @@ -0,0 +1,677 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1 CRYSTAL STRUCTURES" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1_4 pgno:10" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a= 1.0\n", + "r=a/2 = 0.5\n", + "Volume of one atom ,v=((4∗%pi∗(rˆ3))/3)= 0.523598775598\n", + "Total Volume of the cube ,V=aˆ3 = 1.0\n", + "Fp(S.C)=(v∗100/V)= 52.3598775598\n" + ] + } + ], + "source": [ + "#exa 1.4\n", + "from math import pi\n", + "a=1.\n", + "print \"a= \",a # initializing value of lattice constant(a)=1.\n", + "r=a/2.\n", + "print \"r=a/2 = \",r # initializing value of radius of atom for simple cubic .\n", + "v=((4*pi*(r**3))/3)\n", + "print \"Volume of one atom ,v=((4∗%pi∗(rˆ3))/3)= \",v # calcuation . \n", + "V=a**3\n", + "print \"Total Volume of the cube ,V=aˆ3 = \",V # calcuation .\n", + "Fp=(v*100/V)\n", + "print \"Fp(S.C)=(v∗100/V)= \",Fp,# calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1_5 pgno:11" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a= 1.0\n", + "Radius of the atoms,r=(sqrt(3)∗(aˆ2/4)) = 0.433012701892\n", + "Volume of two atom,v=((4∗pi∗(rˆ3))/3)∗2 = 0.680174761588\n", + "Total Volume of the cube ,V=aˆ3 = 1.0\n", + "Fp(B.C.C)=(v∗100/V)= 68.0174761588 %\n" + ] + } + ], + "source": [ + "#exa 1.5\n", + "from math import sqrt\n", + "a=1.\n", + "print \"a= \",a # initializing value of lattice constant(a)=1.\n", + "r=(sqrt(3)*(a**2/4))\n", + "print \"Radius of the atoms,r=(sqrt(3)∗(aˆ2/4)) = \",r # initializing value of radius of atom for BCC.\n", + "v=((4*pi*(r**3))/3)*2\n", + "print \"Volume of two atom,v=((4∗pi∗(rˆ3))/3)∗2 = \",v # calcuation \n", + "V=a**3\n", + "print \"Total Volume of the cube ,V=aˆ3 = \",V # calcuation .\n", + "Fp=(v*100/V)\n", + "print \"Fp(B.C.C)=(v∗100/V)= \",Fp,\"%\" # calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## example 1_6 pgno:12" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a= 1\n", + "Radius of the atom,r=(a/(2∗sqrt(2)))= 0.353553390593\n", + "Volume of the four atom,v=(((4∗pi∗(rˆ3))/3)∗4)= 0.740480489693\n", + "Total volume of the cube ,V=aˆ3= 2\n", + "Fp(F.C.C)=(v∗100/V)= 37.0240244847 %\n" + ] + } + ], + "source": [ + "#exa 1.6\n", + "a=1\n", + "print \"a= \",a # initializing value of lattice constant(a)=1.\n", + "r=(a/(2*sqrt(2)))\n", + "print \"Radius of the atom,r=(a/(2∗sqrt(2)))= \",r # initializing value of radius of atom for FCC .\n", + "v=(((4*pi*(r**3))/3)*4)\n", + "print \"Volume of the four atom,v=(((4∗pi∗(rˆ3))/3)∗4)= \",v # calcuation \n", + "V=a^3\n", + "print \"Total volume of the cube ,V=aˆ3= \",V # calcuation .\n", + "Fp=(v*100/V)\n", + "print \"Fp(F.C.C)=(v∗100/V)= \",Fp,\"%\" # calculation\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## example 1_8 pgno:14" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a= 1\n", + "Radius of the atom , r=(sqrt (3)∗a/8))= 0.216506350946\n", + "v=(((4∗pi∗(rˆ3))/3)∗8) = 0.340087380794\n", + "V=aˆ3= 2\n", + "Fp(Diamond)=(v∗100/V) = 17.0043690397 %\n" + ] + } + ], + "source": [ + "#Exa 1.8 \n", + "a=1\n", + "print \"a= \",a # initializing value of lattice constant(a)=1.\n", + "r=((sqrt(3)*a/8))\n", + "print \"Radius of the atom , r=(sqrt (3)∗a/8))= \",r # initializing value of radius of atom for diamond .\n", + "v=(((4*pi*(r**3))/3)*8)\n", + "print \"v=(((4∗pi∗(rˆ3))/3)∗8) = \",v # calcuation .\n", + "V=a^3\n", + "print \"V=aˆ3= \",V # calcuation .\n", + "Fp=(v*100/V)\n", + "print \"Fp(Diamond)=(v∗100/V) = \",Fp,\"%\" # calculation\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## example 1_9 pgno:14" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a = 5e-08 cm\n", + "Radius of the atom,r=(sqrt(3)∗(a/4))= 2.16506350946e-08\n", + "Volume of the two atoms ,v=((4∗pi∗(rˆ3))/3)∗2= 8.50218451985e-23\n", + "Total Volume of the cube ,V=aˆ3 = 1.25e-22\n", + "Fp(B.C.C)=(v∗100/V) = 68.0174761588 %\n" + ] + } + ], + "source": [ + "#exa 1.9\n", + "a=5*10**-8\n", + "print \"a = \",a,\" cm\" # initializing value of lattice constant .\n", + "r=(sqrt(3)*(a/4))\n", + "print \"Radius of the atom,r=(sqrt(3)∗(a/4))= \",r # initializing value of radius of atom for BCC.\n", + "v=((4*pi*(r**3))/3)*2\n", + "print \"Volume of the two atoms ,v=((4∗pi∗(rˆ3))/3)∗2= \",v # calcuation .\n", + "V=a**3\n", + "print \"Total Volume of the cube ,V=aˆ3 = \",V # calcuation .\n", + "Fp=(v*100/V)\n", + "print \"Fp(B.C.C)=(v∗100/V) = \",Fp,\"%\" # calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## example 1_10 pgno:" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "x intercept = 1\n", + "y intercept = inf\n", + "z intercept = inf\n", + "miller indices ,h=(1/x )= [1]\n", + "k=(1/y)= [0.0]\n", + "l=(1/z) = [0.0]\n" + ] + } + ], + "source": [ + "#exa 1.10\n", + "x=1\n", + "print \"x intercept = \",x # initializing value of x intercept .\n", + "y=float('inf')\n", + "print \"y intercept = \",y # initializing value of y intercept .\n", + "z=float('inf')\n", + "print \"z intercept = \",z # initializing value of z intercept .\n", + "h=[1/x]\n", + "print \"miller indices ,h=(1/x )= \",h # calculation\n", + "k=[1/y]\n", + "print \"k=(1/y)= \",k # calculation\n", + "l=[1/z]\n", + "print \"l=(1/z) = \",l # calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## example 1_11 pgno:15" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "x intercept = inf\n", + "y intercept = inf\n", + "z intercept = 1\n", + "miller indices ,h=[1/x] = [0.0]\n", + "k=[1/y] = [0.0]\n", + "l=[1/z] = [1]\n" + ] + } + ], + "source": [ + "#exa 1.11\n", + "x=float('inf')\n", + "print \"x intercept = \",x # initializing of x intercept .\n", + "y=float('inf') \n", + "print\"y intercept = \",y # initializing of Y intercept .\n", + "z=1\n", + "print \"z intercept = \",z # initializing of Z intercept .\n", + "h=[1/x]\n", + "print \"miller indices ,h=[1/x] = \",h # calculation\n", + "k=[1/y]\n", + "print \"k=[1/y] = \",k # calculation \n", + "l=[1/z]\n", + "print \"l=[1/z] = \",l # calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## example 1_12 pgno: 16" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "x intercept = inf\n", + "y intercept = 1\n", + "z intercept = inf\n", + "miller indices ,h=[1/x] = [0.0]\n", + "k=[1/y] = [1]\n", + "l=[1/z] = [0.0]\n" + ] + } + ], + "source": [ + "#exa 1.12\n", + "x=float('inf') \n", + "print \"x intercept = \",x # initializing of X intercept .\n", + "y=1\n", + "print \"y intercept = \",y # initializing of X intercept .\n", + "z=float('inf') \n", + "print \"z intercept = \",z # initializing of X intercept .\n", + "h=[1/x]\n", + "print \"miller indices ,h=[1/x] = \",h # calculation\n", + "k=[1/y]\n", + "print \"k=[1/y] = \",k # calculation \n", + "l=[1/z]\n", + "print \"l=[1/z] = \",l #calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## example 1_13 pgno:16" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "x intercept = 1\n", + "y intercept = 1\n", + "z intercept = inf\n", + "miller indices ,h=[1/x] = [1]\n", + "k=[1/y] = [1]\n", + "l=[1/z] = [0.0]\n" + ] + } + ], + "source": [ + "#exa 1.13\n", + "x=1\n", + "print \"x intercept = \",x # initializing of X intercept .\n", + "y=1\n", + "print \"y intercept = \",y # initializing of X intercept .\n", + "z=float('inf') \n", + "print \"z intercept = \",z # initializing of X intercept .\n", + "h=[1/x]\n", + "print \"miller indices ,h=[1/x] = \",h # calculation\n", + "k=[1/y]\n", + "print \"k=[1/y] = \",k # calculation \n", + "l=[1/z]\n", + "print \"l=[1/z] = \",l #calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## example 1_14 pgno:17" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "x intercept = inf\n", + "y intercept = 1\n", + "z intercept = 1\n", + "miller indices ,h=[1/x] = [0.0]\n", + "k=[1/y] = [1]\n", + "l=[1/z] = [1]\n" + ] + } + ], + "source": [ + "#exa 1.14\n", + "x=float('inf') \n", + "print \"x intercept = \",x # initializing of X intercept .\n", + "y=1\n", + "print \"y intercept = \",y # initializing of X intercept .\n", + "z=1\n", + "print \"z intercept = \",z # initializing of X intercept .\n", + "h=[1/x]\n", + "print \"miller indices ,h=[1/x] = \",h # calculation\n", + "k=[1/y]\n", + "print \"k=[1/y] = \",k # calculation \n", + "l=[1/z]\n", + "print \"l=[1/z] = \",l #calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## example 1_15 pgno:18" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "x intercept = 2\n", + "y intercept = 2\n", + "z intercept = 2\n", + "common factor of all the intercept= 2\n", + "miller indices ,h=[c/x] = [1]\n", + "k=[c/y] = [1]\n", + "l=[c/z] = [1]\n" + ] + } + ], + "source": [ + "x=2\n", + "print \"x intercept = \",x # initializing of X intercept .\n", + "y=2\n", + "print \"y intercept = \",y # initializing of X intercept .\n", + "z=2\n", + "print \"z intercept = \",z # initializing of X intercept .\n", + "c=2\n", + "print \"common factor of all the intercept= \",c # initializing value of common factor of all the intercepts .\n", + "h=[c/x]\n", + "print \"miller indices ,h=[c/x] = \",h # calculation\n", + "k=[c/y]\n", + "print \"k=[c/y] = \",k # calculation \n", + "l=[c/z]\n", + "print \"l=[c/z] = \",l #calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## example 1_16 pgno: 18" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Wa = 28.1\n", + "D = 2.33 ram/cmˆ3\n", + "Na = 6.02e+23 atoms/mole\n", + "na =(Na∗D)/(Wa)= 4.99167259786e+22 atoms/cmˆ3\n" + ] + } + ], + "source": [ + "#exa 1.16\n", + "Wa =28.1\n", + "print \"Wa = \",Wa # initializing value of atomic weight .\n", + "D=2.33\n", + "print \"D = \",D,\"ram/cmˆ3\" # initializing value of density .\n", + "Na=6.02*10**23\n", + "print \"Na = \",Na,\"atoms/mole\" # initializing value of avagadro number .\n", + "na =(Na*D)/(Wa)\n", + "print \"na =(Na∗D)/(Wa)= \",na,\" atoms/cmˆ3\" # calculation\n", + "# the value of na (number of atoms in 1 cmˆ3 of silicon ) , provided after calculation in the book is wrong." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## example 1_17 pgno: 18" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a= 5e-08 cm\n", + "N= 2\n", + "V=aˆ3 = 1.25e-22 cmˆ3\n", + "na=(no.of atoms in unit cell/Volume of theunit cell) =(N/(V))= 1.6e+22\n" + ] + } + ], + "source": [ + "#exa 1.17\n", + "a=5*10**-8\n", + "print \"a= \",a,\"cm\" # initializing value of lattice constant .\n", + "N=2\n", + "print \"N= \",N # initializing value of no. of atoms in unit cell .\n", + "V=a**3\n", + "print \"V=aˆ3 = \",V,\"cmˆ3\" # initializing value of total Volume of the unit cell.\n", + "na =(N/(V))\n", + "print \"na=(no.of atoms in unit cell/Volume of theunit cell) =(N/(V))= \",na # calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## example 1_18 pgno: 18" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a = 5.43e-08 cm\n", + "N = 8\n", + "Number of atom in the cmˆ3,ns =(N/(aˆ3))= 4.99678310227e+22\n" + ] + } + ], + "source": [ + "#exa 1.18\n", + "a=5.43*10**-8\n", + "print \"a = \",a,\"cm\" # initializing value of lattice constant .\n", + "N=8\n", + "print \"N = \",N # initializing value of no. of atoms in a unit cell .\n", + "ns =(N/(a**3))\n", + "print \"Number of atom in the cmˆ3,ns =(N/(aˆ3))= \",ns # calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## example 1_19 pgno: 18" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a = 5.43e-08 cm\n", + "Wa = 28.1\n", + "Na = 6.02e+23\n", + "ns = 50000000000000000000000 atoms/cmˆ3\n", + "Density of silicon ,D =(ns∗Wa)/(Na)= 2.33388704319 gm/cmˆ2\n" + ] + } + ], + "source": [ + "#exa 1.19\n", + "a=5.43*10**-8\n", + "print \"a = \",a,\"cm\" # initializing value of lattice constant .\n", + "Wa =28.1\n", + "print \"Wa = \",Wa # initializing value of atomic weight .\n", + "Na=6.02*10**23\n", + "print \"Na = \",Na # initializing value of avagdro number .\n", + "ns =5*10**22\n", + "print \"ns = \",ns,\"atoms/cmˆ3\" # initializing value of atoms/cmˆ3.\n", + "D =(ns*Wa)/(Na)\n", + "print \"Density of silicon ,D =(ns∗Wa)/(Na)= \",D,\" gm/cmˆ2\" # calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## example 1_20 pgno: 19" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a = 4.75e-08 cm\n", + "N = 4\n", + "na =(N/(aˆ3))= 3.73232249599e+22\n" + ] + } + ], + "source": [ + "#exa 1.20\n", + "a=4.75*10**-8\n", + "print \"a = \",a,\"cm\" # initializing value of lattice constant .\n", + "N=4\n", + "print \"N = \",N # initializing value of number of atoms in the unit cell .\n", + "na =(N/(a**3))\n", + "print \"na =(N/(aˆ3))=\",na # calculation" + ] + } + ], + "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": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_6_ELECTRICAL_BREAKDOWN_IN_PN_JUNCTIONS_1.ipynb b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_6_ELECTRICAL_BREAKDOWN_IN_PN_JUNCTIONS_1.ipynb new file mode 100644 index 00000000..95315558 --- /dev/null +++ b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_6_ELECTRICAL_BREAKDOWN_IN_PN_JUNCTIONS_1.ipynb @@ -0,0 +1,991 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6 ELECTRICAL BREAKDOWN IN PN JUNCTIONS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6_2 pgno: 183" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "X1 = 4.13 eV\n", + "X2 = 4.07 eV\n", + "Eg1 = 0.7 eV\n", + "Eg2 = 1.43 F/cm\n", + "Nv1 = 6e+18 cmˆ−3\n", + "Nv2 = 7e+18 cmˆ−3\n", + "Vt = 0.0259 eV\n", + "e = 1.6e-19 columbs\n", + "no = 2.5e+13 cmˆ−3\n", + "Pp = 1e+17 cmˆ−3\n", + "Nd = 1e+17 cmˆ−3\n", + "np= 6250000000.0 cmˆ−3\n", + "delta Eg=(Eg2−Eg1)= 0.73 eV\n", + "delta Ec=(X1−X2)= 0.06 eV\n", + "delta  Ev=(delta  Eg−delta  Ec )= 0.67 eV\n", + "Vbi=((delta Ev∗1.6∗10ˆ−19)/(e))+((Vt∗log((Nv1∗Nd) /(np∗Nv2) ) ) )= 1.09563926875 V\n" + ] + } + ], + "source": [ + "#exa 6.2\n", + "from math import log\n", + "X1 =4.13\n", + "print\"X1 = \",X1,\" eV\" # initializing value of eldelta Ectron effinity of germanium.\n", + "X2 =4.07\n", + "print\"X2 = \",X2,\" eV\" # initializing value of electron effinity of gallium arsenide .\n", + "Eg1 =0.7\n", + "print\"Eg1 = \",Eg1,\" eV\" # initializing value of energy gap of germanium .\n", + "Eg2 =1.43\n", + "print\"Eg2 = \",Eg2,\" F/cm\" # initializing value of energy gap of gallium arsenide..\n", + "Nv1 =6e18\n", + "print\"Nv1 = \",Nv1,\" cmˆ−3\" # initializing value of density of states in Valence band,Nv for germanium .\n", + "Nv2 =7e18\n", + "print\"Nv2 = \",Nv2,\" cmˆ−3\" # initializing value of density of states in Valence band,Nv for galliminum arsenide .\n", + "Vt=0.0259\n", + "print\"Vt = \",Vt,\" eV\" # initializing valueof thermal voltage . . . Vt = K∗T/e\n", + "e=1.6e-19\n", + "print\"e = \",e,\" columbs\" # initializing value of electronic charge .\n", + "no=2.5e13\n", + "print\"no = \",no,\" cmˆ−3\" # initializingvalue of intrinsic carrier concentration .\n", + "Pp=1e17\n", + "print\"Pp = \",Pp,\" cmˆ−3\" # initializing value of hole concentration on the depletion edge of the N region .\n", + "Nd=1e17\n", + "print\"Nd = \",Nd,\" cmˆ−3\" # initializing value of number of donor ions (which is equal to hole concentration on the depletion edge of the N region).\n", + "np=(no**2)/Pp\n", + "print\"np=\",np,\" cmˆ−3\"# calculation\n", + "delta_Eg=(Eg2-Eg1)\n", + "print\"delta Eg=(Eg2−Eg1)=\",delta_Eg,\" eV\"#calculation\n", + "delta_Ec=(X1-X2)\n", + "print\"delta Ec=(X1−X2)=\",delta_Ec,\" eV\"#calculation\n", + "delta_Ev=(delta_Eg-delta_Ec)\n", + "print\"delta  Ev=(delta  Eg−delta  Ec )=\",delta_Ev,\" eV\"# calculation\n", + "Vbi=((delta_Ev*1.6*10**-19)/(e))+((Vt*log((Nv1*Nd)/(np*Nv2))))\n", + "print\"Vbi=((delta Ev∗1.6∗10ˆ−19)/(e))+((Vt∗log((Nv1∗Nd) /(np∗Nv2) ) ) )=\",Vbi,\" V\"# calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6_4 pgno: 184" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nc = 2.8e+19 cmˆ−3\n", + "k = -4e+15 cmˆ4Fˆ−2Vˆ−1\n", + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + "e = 1.6e-19 columns\n", + "Vt = 0.0259 eV\n", + "VBI = 0.3 V\n", + " total permittivity ,E=Eo∗Er = 1.053626e-12 F/cm \n", + "Nd=((−2)/(e∗E)∗(1/k)))= 2.96594806886e+15 cmˆ−3\n", + "Vn=(Vt∗( log (Nc/Nd) ) )= 0.237056563109 V\n", + "VBn=(VBI+Vn)= 0.537056563109 V\n" + ] + } + ], + "source": [ + "#exa 6.4\n", + "from math import log\n", + "Nc=2.8e19\n", + "print\"Nc = \",Nc,\" cmˆ−3\" # initializing value of effective density of state in the conduction band .\n", + "k=-4e15\n", + "print\"k = \",k,\" cmˆ4Fˆ−2Vˆ−1\" # initializing value of slope of the (1/Cˆ2) versus V curve.\n", + "Er =11.9\n", + "print\"Er = \",Er # initializing value of relative dielectric permittivity constant .\n", + "Eo=8.854e-14\n", + "print\"Eo = \",Eo,\" F/cm\" # initializing value of dielectric constant of free space.\n", + "e=1.6e-19\n", + "print\"e = \",e,\" columns\" # initializing value of charge of electrons .\n", + "Vt=0.0259\n", + "print\"Vt = \",Vt,\" eV\" # initializing value of thermal voltage .\n", + "VBI=0.3\n", + "print\"VBI = \",VBI,\" V\" # initializing value of built in voltage .\n", + "E=Eo*Er\n", + "print\" total permittivity ,E=Eo∗Er =\",E,\" F/cm \"# calculation\n", + "Nd=((-2)/(e*E)*(1/k))\n", + "print\"Nd=((−2)/(e∗E)∗(1/k)))=\",Nd,\" cmˆ−3\" # c a l c u l a t i o n\n", + "Vn=(Vt*(log(Nc/Nd)))\n", + "print\"Vn=(Vt∗( log (Nc/Nd) ) )=\",Vn,\" V\"#calculation\n", + "VBn=(VBI+Vn)\n", + "print\"VBn=(VBI+Vn)=\",VBn,\" V\"# calculation\n", + "# taking ,... d(1/Cˆ2)/dV as k,... for simlification," + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6_5 pgno: 184" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nd = 2e+17 /cmˆ−3\n", + "Nc = 2.8e+19 /cmˆ−3\n", + "Js = 4e-05 A/cmˆ2\n", + "T = 300 K\n", + "R = 110 A/(K−cmˆ2)\n", + "Vt = 0.0259 eV\n", + "VBn = 0.679478119251 V\n", + "Vn = 0.127988538746 V\n", + "VBI=(VBn−Vn))= 0.551489580505 V\n" + ] + } + ], + "source": [ + "#exa 6.5\n", + "from math import log\n", + "Nd =2e17\n", + "print\"Nd = \",Nd,\"/cmˆ−3\" # initializing value of donor concentration .\n", + "Nc=2.8e19\n", + "print\"Nc = \",Nc,\"/cmˆ−3\" # initializing value of effective density of state in the conduction band .\n", + "Js =40e-6\n", + "print\"Js = \",Js,\"A/cmˆ2\" # initializing value of saturation current density .\n", + "T=300\n", + "print\"T = \",T,\"K\" # initializing value of absolute temperature .\n", + "R=110\n", + "print\"R = \",R,\" A/(K−cmˆ2)\" #initializing value of richardson ’ s constant .\n", + "Vt=0.0259\n", + "print\"Vt = \",Vt,\" eV\" # initializing value of thermal voltage .\n", + "VBn=(Vt*(log(R*T**2/Js)))\n", + "print\"VBn = \",VBn,\" V\" # calculation .\n", + "Vn=(Vt*(log(Nc/Nd)))\n", + "print\"Vn = \",Vn,\" V\" # calculation .\n", + "VBI=(VBn-Vn)\n", + "print\"VBI=(VBn−Vn))=\",VBI,\" V\"#calculation\n", + "#The value of Vn (after calculation ) is provided wrong in the book,due to which VBI also differ." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6_6 pgno: 186" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nd = 200000000000000000 /cmˆ−3\n", + "Dp = 30 cmˆ2/s\n", + "Nc = 2.8e+19 /cmˆ−3\n", + "Js = 4e-05 A/cmˆ2\n", + "no = 15000000000.0 cmˆ−3\n", + "tp = 1e-06 s\n", + "T = 300 K\n", + "R = 110 A/(K−cmˆ2)\n", + "Vt = 0.0259 eV\n", + "e = 1.6e-19 columbs\n", + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + " total permittivity ,E=Eo∗Er)= 1.053626e-12 F/cm\n", + "VBn = 0.679478119251 V\n", + "Vn = 0.127988538746 V\n", + "VBI=(VBn−Vn))= 0.551489580505 V\n", + "current density in a metal semiconductor junction ,W = 4.26124893939e-06 A\n", + "Diffusion length ,Lp=(sqrt(Dp∗tp)) = 0.00547722557505 cm\n", + " saturation hole current density , Jpo=(e∗Dp∗noˆ2) /(Lp∗Nd) ) = 9.85900603509e-13 A/cmˆ2\n" + ] + } + ], + "source": [ + "#exa 6.6\n", + "from math import sqrt\n", + "from math import log\n", + "Nd =2*10**17\n", + "print\"Nd = \",Nd,\" /cmˆ−3\" # initializing value of donor concentration .\n", + "Dp=30\n", + "print\"Dp = \",Dp,\" cmˆ2/s\" # initializing value of diffusion cofficient .\n", + "Nc=2.8*10**19\n", + "print\"Nc = \",Nc,\" /cmˆ−3\" # initializing value of effective density of state in the conduction band .\n", + "Js =40*10**-6\n", + "print\"Js = \",Js, \"A/cmˆ2\" # initializing value of saturation current density .\n", + "no=1.5*10**10\n", + "print\"no = \",no,\" cmˆ−3\" # initializing value of intrinsic concentration of electrons .\n", + "tp=10**-6\n", + "print\"tp = \",tp,\" s\" # initializing value of hole life−time.\n", + "T=300\n", + "print\"T = \",T,\" K\" # initializing value of absolute temperature .\n", + "R=110\n", + "print\"R = \",R,\" A/(K−cmˆ2)\" #initializing value of richardson ’ s constant .\n", + "Vt=0.0259\n", + "print\"Vt = \",Vt,\" eV\" # initializing value of thermal voltage .\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\" columbs\" # initializing value of charge of electrons .\n", + "Er=11.9\n", + "print\"Er = \",Er # initializing value of relative dielectric permittivity constant .\n", + "Eo=8.854*10**-14\n", + "print\"Eo = \",Eo,\" F/cm\" # initializing value of dielectric constant of free space.\n", + "E=Eo*Er\n", + "print\" total permittivity ,E=Eo∗Er)=\",E,\" F/cm\"# calculation\n", + "VBn=(Vt*(log(R*T**2/Js)))\n", + "print\"VBn = \",VBn,\" V\" # calculation .\n", + "Vn=(Vt*(log(Nc/Nd)))\n", + "print\"Vn = \",Vn,\" V\" # calculation .\n", + "VBI=(VBn-Vn)\n", + "print\"VBI=(VBn−Vn))=\",VBI,\" V\"#calculation\n", + "W=(sqrt((E*VBI)/(e*Nd)))\n", + "print\"current density in a metal semiconductor junction ,W = \",W,\" A\" # calculation .\n", + "Lp=(sqrt(Dp*tp))\n", + "print\"Diffusion length ,Lp=(sqrt(Dp∗tp)) = \", Lp,\" cm\" # calculation .\n", + "Jpo=(e*Dp*no**2)/(Lp*Nd)\n", + "print\" saturation hole current density , Jpo=(e∗Dp∗noˆ2) /(Lp∗Nd) ) = \",Jpo,\" A/cmˆ2\" # calculation .\n", + "#The value of Vn (after calculation ) is provided wrong in the book,due to which VBI differ and due to VBI ,current density in a metal semiconductor junction (W) gets changed .\n", + "#The value of Jpo ( saturation hole current density ),after calculation is also provided wrong in the book .," + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6_8 pgno:186" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + "VBD = 20 V\n", + "e = 1.6e-19 columns\n", + " total permittivity ,E=Eo∗Er)= 1.053626e-12 F/cm\n", + "Emax = 500000 V/cm\n", + "ND=(Eo∗Er∗(Emaxˆ2))/(2∗e∗VBD)= 4.1157265625e+16 cmˆ−3\n" + ] + } + ], + "source": [ + "#exa 6.8\n", + "Er =11.9\n", + "print\"Er = \",Er # initializing value of relative dielectric permittivity constant.\n", + "Eo=8.854*10**-14\n", + "print\"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "VBD =20\n", + "print\"VBD = \",VBD,\" V\" #initializing value of break down voltage .\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\" columns\" # initializing value of charge of electrons .\n", + "E=Eo*Er\n", + "print\" total permittivity ,E=Eo∗Er)=\",E,\" F/cm\"# calculation\n", + "Emax =5*10**5\n", + "print\"Emax = \",Emax,\" V/cm\" # initializing value of maximum critical electric field .\n", + "ND=(Eo*Er*(Emax**2))/(2*e*VBD)\n", + "print\"ND=(Eo∗Er∗(Emaxˆ2))/(2∗e∗VBD)=\",ND,\"cmˆ−3\"# calculation\n", + "#the formula given in the solution for VBD is somewhat written wrong.The correct formula is ( VBD=(E∗Emaxˆ2/2∗e∗ND)) ." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6_9 pgno: 187" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + "e = 1.6e-19 columns\n", + "no = 15000000000.0 cmˆ−3\n", + "Nd= 1e+16 cmˆ−3\n", + "Emax = 200000.0 V/cm\n", + "Na= 1e+16 cmˆ−3\n", + "Vt = 0.0259 eV\n", + " total permittivity ,E=Eo∗Er)= 1.053626e-12 F/cm\n", + "VBI=(Vt∗(log(Na∗Nd/noˆ2))) = 0.694640354303 V\n", + "breakdown voltage for symetrical abrupt junction ,VBD+VBI=(E∗Emaxˆ2) /( e∗Nd) )= 26.34065 V\n", + "VBD=V−VBI = 25.6460096457 V\n" + ] + } + ], + "source": [ + "#exa 6.9\n", + "Er =11.9\n", + "print\"Er = \",Er # initializing value of relative dielectric permittivity constant.\n", + "Eo=8.854e-14\n", + "print\"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "e=1.6e-19\n", + "print\"e = \",e,\" columns\" # initializing value of charge of electrons .\n", + "no=1.5e10\n", + "print\"no = \",no,\"cmˆ−3\" # initializing value of intrinsic concentration of electrons .\n", + "Nd =1e16\n", + "print\"Nd=\",Nd,\" cmˆ−3\"#initializing the value of donor concentration .\n", + "Emax =2e5\n", + "print\"Emax = \",Emax,\" V/cm\" # initializing value of maximum critical electric field .\n", + "Na =1e16\n", + "print\"Na=\",Na,\" cmˆ−3\"# initializing the value of acceptor concentration .\n", + "Vt=0.0259\n", + "print\"Vt = \",Vt,\" eV\" # initializing value of thermal voltage .\n", + "E=Eo*Er\n", + "print\" total permittivity ,E=Eo∗Er)=\",E,\" F/cm\"# calculation\n", + "VBI=(Vt*(log(Na*Nd/no**2)))\n", + "print\"VBI=(Vt∗(log(Na∗Nd/noˆ2))) = \",VBI,\" V\" # calculation .\n", + "V=(E*Emax**2)/(e*Nd)\n", + "print\"breakdown voltage for symetrical abrupt junction ,VBD+VBI=(E∗Emaxˆ2) /( e∗Nd) )=\",V,\"V\" # calculation \n", + "VBD=V-VBI\n", + "print\"VBD=V−VBI =\",VBD,\" V\"# calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6_10 pgno: 187" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + "e = 1.6e-19 columns\n", + "no = 15000000000.0 cmˆ−3\n", + "Emax = 1000000 V/cm\n", + "Nd= 1000000000000000000 cmˆ−3\n", + "Na= 1000000000000000000 cmˆ−3\n", + "Vt = 0.0259 eV\n", + "VBI=(Vt∗(log(Na∗Nd/noˆ2))) = 0.933188169937 V\n", + " total permittivity ,E=Eo∗Er)= 1.053626e-12 F/cm 99\n", + "breakdown voltage for symetrical abrupt junction ,VBD+VBI=(E∗Emaxˆ2) /( e∗Nd) )= 6.5851625 V\n", + "VBD=V−VBI)= 5.65197433006 V\n" + ] + } + ], + "source": [ + "#exa 6.10\n", + "from math import log\n", + "Er =11.9\n", + "print\"Er = \",Er # initializing value of relative dielectric permittivity constant.\n", + "Eo=8.854*10**-14\n", + "print\"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\" columns\" # initializing value of charge of electrons .\n", + "no=1.5*10**10\n", + "print\"no = \",no,\"cmˆ−3\" # initializing value of intrinsic concentration of electrons .\n", + "Emax=10**6\n", + "print\"Emax = \",Emax,\" V/cm\" # initializing value of maximum critical electric field ..\n", + "Nd =1*10**18\n", + "print\"Nd=\",Nd,\" cmˆ−3\"#initializing the value of donor concentration .\n", + "Na =1*10**18\n", + "print\"Na=\",Na,\" cmˆ−3\"# initializing the value of acceptor concentration .\n", + "Vt=0.0259\n", + "print\"Vt = \",Vt,\" eV\" # initializing value of thermal voltage .\n", + "VBI=(Vt*(log(Na*Nd/no**2)))\n", + "print\"VBI=(Vt∗(log(Na∗Nd/noˆ2))) = \",VBI,\" V\" # calculation .\n", + "E=Eo*Er\n", + "print\" total permittivity ,E=Eo∗Er)=\",E,\" F/cm 99\"# calculation\n", + "V=(E*Emax**2)/(e*Nd)\n", + "print\"breakdown voltage for symetrical abrupt junction ,VBD+VBI=(E∗Emaxˆ2) /( e∗Nd) )=\",V,\"V\"# calculation\n", + "VBD=V-VBI\n", + "print\"VBD=V−VBI)=\",VBD,\" V\"# calculation" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nd = 1e+18 cmˆ−3\n", + "Na = -1e+18 cmˆ3\n", + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + "e = 1.6e-19 columns\n", + "Vt = 0.0259 eV\n", + "Vbd = 15 eV\n", + "W = 0.0002 cm\n", + " total permittivity ,E=Eo∗Er= 1.053626e-12 F/cm\n", + "slope of doping profile curve ,a=((Nd−Na)/(W))= 1e+22 cmˆ−4\n", + "Emax=(((Vbd)ˆ2)∗9∗e∗a/(32∗E))ˆ(1/3)= 1.0 V/cm\n" + ] + } + ], + "source": [ + "#exa 6.11\n", + "Nd =1e18\n", + "print\"Nd = \",Nd,\" cmˆ−3\" # initializing value of donor concentration .\n", + "Na = -1e18\n", + "print\"Na = \",Na,\" cmˆ3\" # initializing value of acceptor concentration .\n", + "Er =11.9\n", + "print\"Er = \",Er # initializing value of relative dielectric permittivity constant .\n", + "Eo=8.854e-14\n", + "print\"Eo = \",Eo,\" F/cm\" # initializing value of dielectric constant of free space.\n", + "e=1.6e-19\n", + "print\"e = \",e,\" columns\" # initializing value of charge of electrons .\n", + "Vt=0.0259\n", + "print\"Vt = \",Vt,\" eV\" # initializing value of thermal voltage .\n", + "Vbd=15\n", + "print\"Vbd = \",Vbd,\" eV\" # initializing value of break down voltage .\n", + "W=2e-4\n", + "print\"W = \",W,\" cm\" # initializing value of the distance over which doping profile varies.\n", + "E=Eo*Er\n", + "print\" total permittivity ,E=Eo∗Er=\",E,\" F/cm\"# calculation\n", + "a=((Nd-Na)/(W))\n", + "print\"slope of doping profile curve ,a=((Nd−Na)/(W))= \",a,\" cmˆ−4\"# calculation\n", + "Emax=(((Vbd)**2)*9*e*a/(32*E))**(1/3)\n", + "print\"Emax=(((Vbd)ˆ2)∗9∗e∗a/(32∗E))ˆ(1/3)=\",Emax,\" V/cm\"# calculation\n", + "## calculation was given wrong in the book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6_12 pgno: 188" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ew = 4.55 V\n", + "X = 4.01 V\n", + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + "e = 1.6e-19 columns\n", + "Nc = 2.8e+19 /cmˆ−3\n", + "Nd = 100000000000000000 /cmˆ−3\n", + "Vt = 0.0259 eV\n", + " Barrier height ,VB=(Ew−X) = 0.54 V\n", + "Ec Ef=(Vt∗log(Nc/Nd))= 0.145941050722 V\n", + "VBI=(VB−(Ec  Ef ) )= 0.394058949278 V\n", + "Depletion width ,xn=sqrt(2∗Eo∗Er∗VBI/(e∗Nd))= 7.20408525154e-06 cm\n", + "maximum electric field ,Emax=(e∗Nd∗xn/(Eo∗Er))= 109398.746827 V/cm\n" + ] + } + ], + "source": [ + "#exa 6.12\n", + "from math import sqrt\n", + "from math import log\n", + "Ew =4.55\n", + "print\"Ew = \",Ew,\" V\" # initializing value of work function of tungusten .\n", + "X=4.01\n", + "print\"X = \",X,\"V\" # initializing value of electron effinity of silicon .\n", + "Er =11.9\n", + "print\"Er = \",Er # initializing value of relative dielectric permittivity constant.\n", + "Eo=8.854*10**-14\n", + "print\"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\" columns\" # initializing value of charge of electrons .\n", + "Nc=2.8*10**19\n", + "print\"Nc = \",Nc,\"/cmˆ−3\" # initializing value of effective density of state in the conduction band .\n", + "Nd=10**17\n", + "print\"Nd = \",Nd,\"/cmˆ−3\" # initializing value of donor concentration .\n", + "Vt=0.0259\n", + "print\"Vt = \",Vt,\" eV\" # initializing value of thermal voltage .\n", + "VB=(Ew-X)\n", + "print\" Barrier height ,VB=(Ew−X) = \",VB,\" V\" # calculation .\n", + "Ec_Ef=(Vt*log(Nc/Nd))\n", + "print\"Ec Ef=(Vt∗log(Nc/Nd))=\",Ec_Ef,\" V\"#calculation\n", + "VBI=(VB-(Ec_Ef))\n", + "print\"VBI=(VB−(Ec  Ef ) )=\",VBI,\" V\"# calculation\n", + "xn=sqrt(2*Eo*Er*VBI/(e*Nd))\n", + "print\"Depletion width ,xn=sqrt(2∗Eo∗Er∗VBI/(e∗Nd))=\",xn,\" cm\"# calculation\n", + "Emax=(e*Nd*xn/(Eo*Er))\n", + "print\"maximum electric field ,Emax=(e∗Nd∗xn/(Eo∗Er))=\",Emax,\" V/cm\"# calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6_13 pgno: 189" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ew = 4.5 V\n", + "X = 4.01 V\n", + "Er = 12\n", + "Eo = 8.854e-14 F/cm\n", + "Vr = 3 V\n", + "e = 1.6e-19 columns\n", + "Nc = 2.8e+19 /cmˆ−3\n", + "Nd = 100000000000000000 /cmˆ−3\n", + "Vt = 0.0259 eV\n", + " barrier height ,VB=(Ew−X) = 0.49 V\n", + "Ec Ef=(Vt∗log(Nc/Nd))= 0.145941050722 V\n", + "VBI=(VB−(Ec  Ef ) )= 0.344058949278 V\n", + "Depletion width ,xn=sqrt(2∗Eo∗Er∗(VBI+Vr)/(e∗Nd))= 2.10742608187e-05 cm\n", + "maximum electric field ,Emax=(e∗Nd∗xn/(Eo∗Er))= 317359.548508 V/cm\n", + "Capitance per unit area ,C=sqrt (( e∗Eo∗Er∗Nd)/(2∗(VBI+Vr) ) )= 5.04160031586e-08 F/cmˆ2\n" + ] + } + ], + "source": [ + "#exa 6.13\n", + "from math import sqrt\n", + "from math import log\n", + "Ew =4.5\n", + "print\"Ew = \",Ew,\" V\" # initializing value of work function of tungusten .\n", + "X=4.01\n", + "print\"X = \",X,\"V\" # initializing value of electron effinity of silicon .\n", + "Er=12\n", + "print\"Er = \",Er # initializing value of relative dielectric permittivity constant.\n", + "Eo=8.854*10**-14\n", + "print\"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "Vr=3\n", + "print\"Vr = \",Vr,\" V\" # initializing value of reverse voltage .\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\" columns\" # initializing value of charge of electrons .\n", + "Nc=2.8*10**19\n", + "print\"Nc = \",Nc,\"/cmˆ−3\" # initializing value of effective density of state in the conduction band .\n", + "Nd=10**17\n", + "print\"Nd = \",Nd,\"/cmˆ−3\" # initializing value of donor concentration .\n", + "Vt=0.0259\n", + "print\"Vt = \",Vt,\" eV\" # initializing value of thermal voltage .\n", + "VB=(Ew-X)\n", + "print\" barrier height ,VB=(Ew−X) = \",VB,\" V\"# calculation .\n", + "Ec_Ef=(Vt*log(Nc/Nd))\n", + "print\"Ec Ef=(Vt∗log(Nc/Nd))=\",Ec_Ef,\" V\"#calculation\n", + "VBI=(VB-(Ec_Ef))\n", + "print\"VBI=(VB−(Ec  Ef ) )=\",VBI,\" V\"#calculation\n", + "xn=sqrt((2*Eo*Er*(VBI+Vr))/(e*Nd))\n", + "print\"Depletion width ,xn=sqrt(2∗Eo∗Er∗(VBI+Vr)/(e∗Nd))=\",xn,\" cm\"#calculation\n", + "Emax=(e*Nd*xn/(Eo*Er))\n", + "print\"maximum electric field ,Emax=(e∗Nd∗xn/(Eo∗Er))=\",Emax,\" V/cm\"# calculation\n", + "C=sqrt((e*Eo*Er*Nd)/(2*(VBI+Vr)))\n", + "print\"Capitance per unit area ,C=sqrt (( e∗Eo∗Er∗Nd)/(2∗(VBI+Vr) ) )=\",C,\" F/cmˆ2\"# calculation\n", + "#the Value of reverse voltage(Vr) provided in the question is different than used in the solution . I have used the value provided in the solution ( i . e Vr=3).\n", + "#the value of C (Capitance per unit area) after calculation is provided wrong in the book." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6_13 pgno: 190" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ew = 4.28 V\n", + "X = 4.01 V\n", + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + "e = 1.6e-19 columns\n", + "Nc = 2.8e+19 /cmˆ−3\n", + "Nd = 1000000000000000 /cmˆ−3\n", + "Vt = 0.0259 eV\n", + " barrier height ,VB=(Ew−X) = 0.27 V\n", + "Ec Ef=(Vt∗log(Nc/Nd))= 0.265214958539 V\n", + "VBI=(VB−(Ec  Ef ) )= 0.00478504146083 V\n", + "Depletion width ,xn=sqrt(2∗Eo∗Er∗VBI/(e∗Nd))= 7.93854843013e-06 cm\n", + "maximum electric field ,Emax=(e∗Nd∗xn/(Eo∗Er))= 1205.52050616 V/cm\n" + ] + } + ], + "source": [ + "#exa 6.14\n", + "from math import log\n", + "Ew =4.28\n", + "print\"Ew = \",Ew,\" V\" # initializing value of work function of tungusten .\n", + "X=4.01\n", + "print\"X = \",X,\"V\" # initializing value of electron effinity of silicon .\n", + "Er =11.9\n", + "print\"Er = \",Er # initializing value of relative dielectric permittivity constant.\n", + "Eo=8.854*10**-14\n", + "print\"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\" columns\" # initializing value of charge of electrons .\n", + "Nc=2.8*10**19\n", + "print\"Nc = \",Nc,\"/cmˆ−3\" # initializing value of effective density of state in the conduction band .\n", + "Nd=10**15\n", + "print\"Nd = \",Nd,\"/cmˆ−3\" # initializing value of donor concentration .\n", + "Vt=0.0259\n", + "print\"Vt = \",Vt,\" eV\" # initializing value of thermal voltage .\n", + "VB=(Ew-X)\n", + "print\" barrier height ,VB=(Ew−X) = \",VB,\" V\"# calculation .\n", + "Ec_Ef=(Vt*log(Nc/Nd))\n", + "print\"Ec Ef=(Vt∗log(Nc/Nd))=\",Ec_Ef,\" V\"#calculation\n", + "VBI=(VB-(Ec_Ef))\n", + "print\"VBI=(VB−(Ec  Ef ) )=\",VBI,\" V\"#calculation\n", + "xn=sqrt(2*Eo*Er*VBI/(e*Nd))\n", + "print\"Depletion width ,xn=sqrt(2∗Eo∗Er∗VBI/(e∗Nd))=\",xn,\" cm\"# calculation\n", + "Emax=(e*Nd*xn/(Eo*Er))\n", + "print\"maximum electric field ,Emax=(e∗Nd∗xn/(Eo∗Er))=\",Emax,\" V/cm\"# calculation\n", + "#the Value of donor concentration (Nd) provided in the question is different than used in the solution . I have used the value provided in the question(i.e Nd=10ˆ15). ,i.e answer differs than provided in the book ." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6_15 pgno: 191" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ew = 5.1 V\n", + "X = 4.01 V\n", + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + "e = 1.6e-19 columns\n", + "Nc = 2.8e+19 /cmˆ−3\n", + "Nd = 5000000000000000 /cmˆ−3\n", + "Vt = 0.0259 eV\n", + "Vr = 5 V\n", + "A = 0.0001 cmˆ2\n", + " barrier height ,VB=(Ew−X) = 1.09 V\n", + "Ec Ef=(Vt∗log(Nc/Nd))= 0.223530516607 V\n", + "VBI=(VB−(Ec  Ef ) )= 0.866469483393 V\n", + "Capitance per unit area ,C1=sqrt (( e∗Eo∗Er∗Nd)/(2∗(VBI+Vr) ) )= 8.4758805431e-09 F/cmˆ2\n", + "total junction capatiance ,C=C1∗A= 8.4758805431e-13 F\n" + ] + } + ], + "source": [ + "#exa 6.15\n", + "from math import sqrt\n", + "from math import log\n", + "Ew =5.1\n", + "print\"Ew = \",Ew,\" V\" # initializing value of work function of tungusten .\n", + "X=4.01\n", + "print\"X = \",X,\"V\" # initializing value of electron effinity of silicon .\n", + "Er =11.9\n", + "print\"Er = \",Er # initializing value of relative dielectric permittivity constant.\n", + "Eo=8.854*10**-14\n", + "print\"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\" columns\" # initializing value of charge of electrons .\n", + "Nc=2.8*10**19\n", + "print\"Nc = \",Nc,\"/cmˆ−3\" # initializing value of effective density of state in the conduction band .\n", + "Nd =5*10**15\n", + "print\"Nd = \",Nd,\"/cmˆ−3\" # initializing value of donor concentration .\n", + "Vt=0.0259\n", + "print\"Vt = \",Vt,\" eV\" # initializing value of thermal voltage .\n", + "Vr=5\n", + "print\"Vr = \",Vr,\" V\" # initializing value of reverse voltage .\n", + "A=1*10**-4\n", + "print\"A = \",A,\" cmˆ2\" # initializing valueof area of the gold silicon junction diode..\n", + "VB=(Ew-X)\n", + "print\" barrier height ,VB=(Ew−X) = \",VB,\" V\"# calculation .\n", + "Ec_Ef=(Vt*log(Nc/Nd))\n", + "print\"Ec Ef=(Vt∗log(Nc/Nd))=\",Ec_Ef,\" V\"#calculation\n", + "VBI=(VB-(Ec_Ef))\n", + "print\"VBI=(VB−(Ec  Ef ) )=\",VBI,\" V\"#calculation\n", + "C1=sqrt((e*Eo*Er*Nd)/(2*(VBI+Vr)))\n", + "print\"Capitance per unit area ,C1=sqrt (( e∗Eo∗Er∗Nd)/(2∗(VBI+Vr) ) )=\",C1,\" F/cmˆ2\"#calculation\n", + "C=C1*A\n", + "print\"total junction capatiance ,C=C1∗A=\",C,\"F\"# calculation" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Er = 13.1\n", + "Eo = 8.854e-14 F/cm\n", + "e = 1.6e-19 columns\n", + "Emax = 30000 V/cm\n", + " total permittivity ,E=Eo∗Er)= 1.159874e-12 F/cm\n", + "lowering of the barrier height ,V=sqrt(e∗Emax/(4∗pi∗E) )= 0.0181472273453 V\n", + " position of the maximum barrier height ,Xmax=sqrt(e/(16∗%pi∗E∗Emax))= 3.02453789089e-07 cm\n" + ] + } + ], + "source": [ + "#exa 6.17\n", + "from math import pi\n", + "from math import sqrt\n", + "Er =13.1\n", + "print\"Er = \",Er # initializing value of relative dielectric permittivity constant.\n", + "Eo=8.854*10**-14\n", + "print\"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\" columns\" # initializing value of charge of electrons .\n", + "Emax =30*10**3\n", + "print\"Emax = \",Emax,\" V/cm\" # initializing value of maximum critical electric field ..\n", + "E=Eo*Er\n", + "print\" total permittivity ,E=Eo∗Er)=\",E,\" F/cm\"# calculation\n", + "V=sqrt(e*Emax/(4*pi*E))\n", + "print\"lowering of the barrier height ,V=sqrt(e∗Emax/(4∗pi∗E) )=\",V,\" V\"# calculation\n", + "Xmax=sqrt(e/(16*pi*E*Emax))\n", + "print\" position of the maximum barrier height ,Xmax=sqrt(e/(16∗%pi∗E∗Emax))=\",Xmax,\" cm\"# calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6_18 pgno: 192" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A = 0.0001 cmˆ−2\n", + "VBn = 0.55 V\n", + "T = 300 K\n", + "R = 110 A/(K−cmˆ2)\n", + "Vt = 0.0259 eV\n", + "V = 0.25 V\n", + "reverse saturation current , Io=A∗R∗Tˆ2∗exp(−VBn/Vt) = 5.93151320618e-07 A\n", + "diode current , I=Io(exp(V/Vt)−1)= 0.00922931077027 A\n" + ] + } + ], + "source": [ + "#exa 6.18\n", + "from math import exp\n", + "A=10**-4\n", + "print\"A = \",A,\" cmˆ−2\" # initializing value of cross sectional area .\n", + "VBn =0.55\n", + "print\"VBn = \",VBn,\"V\" # initializing value of barrier height .\n", + "T=300\n", + "print\"T = \",T,\"K\" # initializing value of absolute temperature .\n", + "R=110\n", + "print\"R = \",R,\" A/(K−cmˆ2)\" #initializing value of richardson ’ s constant .\n", + "Vt=0.0259\n", + "print\"Vt = \",Vt,\" eV\" # initializing value of thermal voltage .\n", + "V=0.25\n", + "print\"V = \",V,\" V\" # initializing value of forward bias voltage .\n", + "Io=A*R*T**2*exp(-VBn/Vt)\n", + "print\"reverse saturation current , Io=A∗R∗Tˆ2∗exp(−VBn/Vt) = \",Io,\" A\" # calculation .\n", + "I=Io*((exp(V/Vt))-1)\n", + "print\"diode current , I=Io(exp(V/Vt)−1)=\",I,\"A\"# calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6_19 pgno:192" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Io1 = 1e-09 A\n", + "Io2 = 1e-14 A\n", + "Vt = 0.0259 eV\n", + "I = 0.0001 A\n", + "forward Voltage for silicon SBD,VfSBD=Vt∗(( log(I/Io1+1)))= 0.298185028541 V\n", + "forward Voltage for silicon SBD,VfPN=Vt∗((log(I/Io2+1)))= 0.596369539088 V\n" + ] + } + ], + "source": [ + "#exa 6.19\n", + "from math import log\n", + "Io1 =10**-9\n", + "print\"Io1 = \",Io1,\" A\" # initializing value of reverse saturation current of silicon SBD.\n", + "Io2 =10**-14\n", + "print\"Io2 = \",Io2,\"A\" # initializing value of reverse saturation current of a PN junction .\n", + "Vt =0.0259\n", + "print\"Vt = \",Vt,\" eV\" # initializing value of thermal voltage .\n", + "I=100*10**-6\n", + "print\"I = \",I,\" A\" # initializing value of required current .\n", + "VfSBD=Vt*((log(I/Io1+1)))\n", + "print\"forward Voltage for silicon SBD,VfSBD=Vt∗(( log(I/Io1+1)))= \",VfSBD,\" V\" # calculation\n", + "VfPN=Vt*((log(I/Io2+1)))\n", + "print\"forward Voltage for silicon SBD,VfPN=Vt∗((log(I/Io2+1)))=\",VfPN,\" V\"#calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6_20 pgno: 193" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Io1 = 1e-06 A\n", + "Io2 = 1e-06 A\n", + "Vt = 0.0259 eV\n", + "I = 0.001 A\n", + "V = 0.25 V\n", + "forward Voltage for silicon SBD,VfSBD=Vt∗(( log(I/Io1+1)))= 0.178936748784 V\n", + "forward volage applied across the PN Diode ,VfPN=(V+VfSBD)= 0.428936748784 V\n", + "reverse saturation current of the PN junction Diode,Io=(I/((exp(VfPN/Vt))−1))= 6.41998882039e-11 A\n" + ] + } + ], + "source": [ + "#exa 6.20\n", + "from math import log\n", + "Io1 =10*10**-7\n", + "print\"Io1 = \",Io1,\" A\" # initializing value of reverse saturation current of silicon SBD.\n", + "Io2 =10*10**-7\n", + "print\"Io2 = \",Io2,\"A\" # initializing value of reverse saturation current of a PN junction .\n", + "Vt =0.0259\n", + "print\"Vt = \",Vt,\" eV\" # initializing value of thermal voltage .\n", + "I=1*10**-3\n", + "print\"I = \",I,\" A\" # initializing value of forward current .\n", + "V=0.25\n", + "print\"V = \",V,\" V\" # initializing value of difference in the forward voltage of the two diode .\n", + "VfSBD=Vt*((log(I/Io1+1)))\n", + "print\"forward Voltage for silicon SBD,VfSBD=Vt∗(( log(I/Io1+1)))= \",VfSBD,\" V\" # calculation 109\n", + "VfPN=(V+VfSBD)\n", + "print\"forward volage applied across the PN Diode ,VfPN=(V+VfSBD)=\",VfPN,\" V\"#calculation \n", + "Io=(I/((exp(VfPN/Vt))-1))\n", + "print\"reverse saturation current of the PN junction Diode,Io=(I/((exp(VfPN/Vt))−1))=\",Io,\" A\" # calculation" + ] + } + ], + "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": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_2_ENERGY_BAND_THEORY_OF_SOLIDS_1.ipynb b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_2_ENERGY_BAND_THEORY_OF_SOLIDS_1.ipynb new file mode 100644 index 00000000..7b0ddd98 --- /dev/null +++ b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_2_ENERGY_BAND_THEORY_OF_SOLIDS_1.ipynb @@ -0,0 +1,998 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2 ENERGY BAND THEORY OF SOLIDS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_1 pgno:49" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "no = 15000000000.0 /cmˆ3\n", + "n = 1000000000000000000 /cmˆ3\n", + "number of holes ,p=(noˆ2/n))= 225.0 /cmˆ3\n" + ] + } + ], + "source": [ + "#exa 2.1\n", + "no=1.5*10**10\n", + "print \"no = \",no,\"/cmˆ3\" # initializing value of electrons and hole per cmˆ3.\n", + "n=1*10**18\n", + "print \"n = \",n,\"/cmˆ3\" # initializing value of number of electrons per cmˆ3.\n", + "p=(no**2/n)\n", + "print \"number of holes ,p=(noˆ2/n))= \",p,\" /cmˆ3\" # calculation\n", + "#this is solved problem 2.1 of chapter 2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_2 pgno:49" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "n = 100000 /cmˆ3\n", + "p = 10000000000000000000 /cmˆ3\n", + "Value of intrinsic concentration ,no=sqrt(n∗p))= 1e+12 /cmˆ3\n" + ] + } + ], + "source": [ + "#exa 2.2\n", + "from math import sqrt\n", + "n=1*10**5\n", + "print\"n = \",n,\" /cmˆ3\" # initializing value of electrons and hole per cmˆ3.\n", + "p=1*10**19\n", + "print\"p = \",p,\" /cmˆ3\" # initializing value of number of hole per cmˆ3\n", + "no=sqrt(n*p)\n", + "print\"Value of intrinsic concentration ,no=sqrt(n∗p))= \",no,\" /cmˆ3\"# calculation\n", + "#this is solved problem 2.2 of chapter 2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_3 pgno:49" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "e = 1.6e-19 columb\n", + "Ef−Efi = 0.309 eV\n", + "no = 2.5e+13 /cmˆ3\n", + "T = 300 K\n", + "exp = 2.718\n", + "k = ”,k,” J/K\n", + "number of electrons per cmˆ3, n=no∗(exˆ((Ef−Efi)/kT)))= 3.83494867662e+18 /cmˆ3\n" + ] + } + ], + "source": [ + "#exa 2.3\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\"columb\" # initializing the value of electronic charge .\n", + "Ef_Efi =0.309\n", + "print\"Ef−Efi = \",Ef_Efi,\" eV\" # initializing the value of difference in the energy levels .\n", + "no=2.5*10**13\n", + "print\"no = \",no,\" /cmˆ3\" # initializing value of number of electrons per cmˆ3\n", + "T=300\n", + "print\"T = \",T,\" K\" # initializing value of temperature .\n", + "ex=2.718\n", + "print\"exp = \",ex # initializing the value of exponential .\n", + "k=1.38*10**-23\n", + "print\"k = ”,k,” J/K\" # initializing value of boltzmann constant .\n", + "n=no*(ex**((Ef_Efi*e)/(k*T)))\n", + "print\"number of electrons per cmˆ3, n=no∗(exˆ((Ef−Efi)/kT)))= \",n,\" /cmˆ3\" #calculation\n", + "#This is solved problem 2.3 of chapter 2.\n", + "#The value used for ”Ef−Efi” in the solution is different than provided in the question .\n", + "#I have used the value provided in the solution ( i .e Ef Efi =0.309)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_4 pgno:50" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "e = 1.6e-19 columb\n", + "Ef = 0.4065 eV\n", + "n = 100000000000000000 /cmˆ3\n", + "T = 300 K\n", + "exp = 2.718\n", + "k = 1.38e-23 J/K\n", + "Number of electrons per cmˆ3, no=n/(exˆ((Ef)/kT) ) )= 15061844796.9 electrons /cmˆ3\n" + ] + } + ], + "source": [ + "#exa 2.4\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\" columb\" # initializing the value of electronic charge .\n", + "Ef =0.4065\n", + "print \"Ef = \",Ef,\" eV\" # initializing the value of fermi level .\n", + "n=10**17\n", + "print\"n = \",n,\" /cmˆ3\" # initializing value of number of electrons per cmˆ3.\n", + "T=300\n", + "print\"T = \",T,\" K\" # initializing value of temperature .\n", + "ex=2.718\n", + "print\"exp = \",ex # initializing the value of exponential .\n", + "k=1.38*10**-23\n", + "print\"k = \",k,\" J/K\" # initializing value of boltzmann constant .\n", + "no=n/(ex**((Ef*e)/(k*T)))\n", + "print\"Number of electrons per cmˆ3, no=n/(exˆ((Ef)/kT) ) )= \",no,\" electrons /cmˆ3\" # calculation\n", + "#this is solved problem 2.4 of chapter 2.\n", + "#the value used for \"n\" in the solution is different than provided in the question .\n", + "#I have used the value provided in the solution ( i .e n=10ˆ17)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_5 pgno:50" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "e = 1.6e-19 columb\n", + "n = 10000000000000000000000 /mˆ3\n", + "u = 0.12 mˆ2/Vs\n", + "L = 0.001 m\n", + "A = 1e-10 mˆ2\n", + " conductivity , sigma=n∗e∗u)= 192.0 siemen/m\n", + "Resistivity ,p=(1/sigma))= 0.00520833333333 ohm metre\n", + " resistance ,R=(p∗L/A) )= 52083.3333333 ohm\n" + ] + } + ], + "source": [ + "#exa 2.5\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\"columb\" # initializing the value of electronic charge .\n", + "n=1*10**22\n", + "print\"n = \",n,\" /mˆ3\" # initializing value of number of electrons per cmˆ3\n", + "u=1200*10**-4\n", + "print\"u = \",u,\" mˆ2/Vs\" # initializing the value of mobility .\n", + "L=0.1*10**-2\n", + "print\"L = \",L,\" m\" # initializing the value of length .\n", + "A=100*10**-12\n", + "print\"A = \",A,\" mˆ2\" # initializing the value of area of cross section .\n", + "sigma=n*e*u\n", + "print\" conductivity , sigma=n∗e∗u)= \",sigma,\"siemen/m\" # calculation .\n", + "p=(1/sigma)\n", + "print\"Resistivity ,p=(1/sigma))= \",p,\" ohm metre\"#calculation .\n", + "R=(p*L/A)\n", + "print\" resistance ,R=(p∗L/A) )= \",R,\" ohm\" #calculation .\n", + "#this is solved problem 2.5 of chapter 2.\n", + "#the value used for \"A\" in the solution is different than provided in the question .\n", + "#I have used the value provided in the solution ( i .e A=100∗10ˆ−12)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_6 pgno:50" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R = 52080.0 ohm\n", + "V = 5 volt\n", + " Drift current , I=(V/R) )= 9.60061443932e-05 amphere\n" + ] + } + ], + "source": [ + "#exa 2.6\n", + "R=52.08*10**3\n", + "print\"R = \",R,\"ohm\" # initializing value of Resistance .\n", + "V=5\n", + "print\"V = \",V,\"volt\" # initializing value of voltage .\n", + "I=(V/R)\n", + "print\" Drift current , I=(V/R) )= \",I,\" amphere\" # calculation\n", + "#this is solved problem 2.6 of chapter 2." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_7 pgno:50" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Energy gap of GaAs = 1.43 eV\n", + " Energy gap of GaP = 2.43 eV\n", + " Plank constant = 6.624e-34 joule \n", + " Light speed = 300000000 m/s\n", + "Difference between the energy gap of GaAs and GaP ,x=(Eg2−Eg1) )= 1.0 eV\n", + "Excess energy gap added to GaAs to form GaAsP,(0.4∗x))= 0.4 eV \n", + "Band gap energy GaAsP,Eg=(Eg1+g))= 1.83 eV \n", + "wavelength of radiation emitted , lamda=(c∗h/Eg))= 6.7868852459e-07 metre \n" + ] + } + ], + "source": [ + "#exa 2.7\n", + "Eg1 =1.43\n", + "print\" Energy gap of GaAs = \",Eg1,\"eV\" # initializing the value of energy gap of GaAs.\n", + "Eg2 =2.43\n", + "print\" Energy gap of GaP = \",Eg2,\"eV\"# initializing the value of energy gap of Gap.\n", + "h=6.624*10**-34\n", + "print\" Plank constant = \",h,\" joule \"# initializing the value of plank constant .\n", + "c=3*10**8\n", + "print\" Light speed = \",c,\"m/s\" # initializing the value of speed of light.\n", + "x=(Eg2-Eg1)\n", + "print\"Difference between the energy gap of GaAs and GaP ,x=(Eg2−Eg1) )= \",x,\" eV\"# calculation\n", + "g=(0.4*x)\n", + "print\"Excess energy gap added to GaAs to form GaAsP,(0.4∗x))= \",g,\" eV \"#calculation\n", + "Eg=(Eg1+g)\n", + "print\"Band gap energy GaAsP,Eg=(Eg1+g))= \",Eg ,\" eV \"#calculation\n", + "lamda=(c*h/(Eg*1.6*10**-19))\n", + "print\"wavelength of radiation emitted , lamda=(c∗h/Eg))= \",lamda,\" metre \"\n", + "# calculation 19 #this is solved problem 2.7 of chapter 2." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_8 pgno:51" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Energy gap of GaAs = 1.43 eV\n", + " Energy gap of GaP = 2.43 eV\n", + " Plank constant = 6.624e-34 joule\n", + " Light speed = 300000000 m/s\n", + " lamda = 540000000 m\n", + "Difference between the energy gap of GaAs and GaP ,x=(Eg2−Eg1) )= 1.0 eV\n", + "Band gap energy ,Eg=(c∗h/lamda∗(1.6∗10ˆ−19)))= 2.3e-15 eV\n", + "X=Eg−(Eg1)= -1.43\n" + ] + } + ], + "source": [ + "#exa 2.8\n", + "Eg1 =1.43\n", + "print\" Energy gap of GaAs = \",Eg1,\" eV\" # initializing the value of energy gap of GaAs.\n", + "Eg2 =2.43\n", + "print\" Energy gap of GaP = \",Eg2,\" eV\"# initializing the value of energy gap of Gap.\n", + "h=6.624*10**-34\n", + "print\" Plank constant = \",h,\" joule\"# initializing the value of plank constant .\n", + "c=3*10**8\n", + "print\" Light speed = \",c,\" m/s\" # initializing the value of speed of light.\n", + "lamda =540*10**6\n", + "print\" lamda = \",lamda,\" m\" # initializing the value of wavelength .\n", + "x=(Eg2-Eg1)\n", + "print\"Difference between the energy gap of GaAs and GaP ,x=(Eg2−Eg1) )= \",x,\" eV\"# calculation\n", + "Eg=((c*h/(lamda*(1.6*10**-19))))\n", + "print\"Band gap energy ,Eg=(c∗h/lamda∗(1.6∗10ˆ−19)))=\",Eg,\" eV\"# calculation\n", + "X=Eg-(Eg1)\n", + "print\"X=Eg−(Eg1)= \",X # calculation \n", + "#this is solved problem 2.8 of chapter 2.\n", + "#the value of Eg(band gap energy )is provided wrong in the book after calculation.Due to this value ofX,alsodiffer." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_9 pgno:51" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Temperature 1 = 500 K\n", + " Nv = 2e+19 cmˆ−3\n", + " Temperature 2 = 300 K\n", + "NV at 500K=(Nv((500/300) ˆ(3/2) ) ) )= 2e+19 cmˆ−3 \n" + ] + } + ], + "source": [ + "#exa 2.9\n", + "T1 =500\n", + "print\" Temperature 1 = \",T1,\"K\" # initializing the value of temperature 1.\n", + "Nv =2*10**19\n", + "print\" Nv = \",round(Nv,3),\"cmˆ−3\"# initializing the value of effective density of state for valence band .\n", + "T2 =300\n", + "print\" Temperature 2 = \",T2,\"K\"# initializing the value of temperature 2.\n", + "NV=(Nv*((500/300)**(3/2)))\n", + "print\"NV at 500K=(Nv((500/300) ˆ(3/2) ) ) )= \",round(NV,3),\" cmˆ−3 \"#calculation\n", + "#this is solved problem 2.9 of chapter 2." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_10 pgno:52" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nd = 100000000000000000 cmˆ−3\n", + " Ec Ed = 0.045\n", + "Vt = 0.0259 eV \n", + " Nc = 2.8e+19 cmˆ−3\n", + "exp = 2.718\n", + "Fraction of electron still in the donor state,(nd/(nd+n)=(((Nc/Nd)∗eˆ((−Ec Ed)/Vt),1)ˆ−1)= 0.0198886296934\n" + ] + } + ], + "source": [ + "#exa 2.10\n", + "Nd =1*10**17\n", + "print\"Nd = \",Nd,\"cmˆ−3\" # initializing the value of effective energy density of state.\n", + "Ec_Ed =0.045\n", + "print\" Ec Ed = \",Ec_Ed # initializing the value of donor ionisation level .\n", + "Vt =0.0259\n", + "print\"Vt = \",Vt,\" eV \"# initializing the value of thermal voltage .\n", + "Nc=2.8*10**19\n", + "print\" Nc = \",Nc,\"cmˆ−3\"# initializing the value of effective density of state of conduction band .\n", + "e=2.718\n", + "print\"exp = \",e # initializing the value of exponential .\n", + "N=(((Nc/Nd)*e**((-(Ec_Ed))/Vt))+1)**-1\n", + "print \"Fraction of electron still in the donor state,(nd/(nd+n)=(((Nc/Nd)∗eˆ((−Ec Ed)/Vt),1)ˆ−1)= \",N # calculation\n", + "#this is solved problem 2.10 of chapter 2." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_11 pgno:52" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na = 10000000000000000 cmˆ−3\n", + "Ea Ev = 0.045\n", + "Nv = 1.04e+19 cmˆ−3\n", + "Vt = 0.0259 eV\n", + "Fraction of holes that are still in the acceptor state ,(pa/(pa+p))=(1+((Nv/4∗Na)∗exp(−(Ea −Ev)/Vt)))ˆ(−1)= 0.0213895767669\n" + ] + } + ], + "source": [ + "#exa 2.11\n", + "from math import exp\n", + "Na =1*10**16\n", + "print\"Na = \",Na,\" cmˆ−3\"# initializing the value of acceptor concentration \n", + "Ea_Ev =0.045\n", + "print\"Ea Ev = \",Ea_Ev # initializing the boron acceptor ionization energy .\n", + "Nv=(1.04*10**19)\n", + "print\"Nv = \",Nv,\" cmˆ−3\"# initializing the value of effective density of state for valence band .\n", + "Vt=(0.0259)\n", + "print\"Vt = \",Vt,\" eV\"# initializing the value of thermal voltage .\n", + "p=(1+((Nv/(4*Na))*exp(-(Ea_Ev)/Vt)))**(-1)\n", + "print\"Fraction of holes that are still in the acceptor state ,(pa/(pa+p))=(1+((Nv/4∗Na)∗exp(−(Ea −Ev)/Vt)))ˆ(−1)= \",p #calculation\n", + "#this is solved problem 2.11 of chapter 2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_12 pgno:52" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nd = 100000000000000000 cmˆ−3\n", + "Na = 0 cmˆ−3\n", + "ni = 15000000000.0 cmˆ−3\n", + "Electron concentration ,n=(−(Na−Nd)+sqrt ((Na−Nd)ˆ2+4∗no))/2)= 1e+17 cmˆ−3\n", + "Hole concentration ,p)= 2250.0 cmˆ−3\n" + ] + } + ], + "source": [ + "#exa 2.12\n", + "Nd =1*10**17\n", + "print\"Nd = \",Nd,\" cmˆ−3\" # initializing the value of donor concentration .\n", + "Na=0\n", + "print\"Na = \",Na,\" cmˆ−3\"# initializing the value of acceptor concentration .\n", + "no=1.5*10**10\n", + "print\"ni = \",no,\" cmˆ−3\"# initializing the value of electron hole per cmˆ3.\n", + "n=(-(Na-Nd)+sqrt((Na-Nd)**2+4*no))/2\n", + "print\"Electron concentration ,n=(−(Na−Nd)+sqrt ((Na−Nd)ˆ2+4∗no))/2)= \",n,\" cmˆ−3\"#calculation\n", + "p=(no**2/n)\n", + "print\"Hole concentration ,p)= \",p,\" cmˆ−3\" # calculation\n", + "#this is solved problem 2.13 of chapter 2." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_14 pgno:53" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nd = 60000000000000000 cmˆ−3\n", + "Na = 100000000000000000 cmˆ−3\n", + "no = 15000000000.0 cmˆ−3\n", + "Hole concentration ,n=(−(Na−Nd)+sqrt ((Na−Nd)ˆ2+4∗no))/2)= 4e+16 cmˆ−3\n", + "Electron concentration ,n=(noˆ2/p))= 5625.0\n" + ] + } + ], + "source": [ + "#exa 2.14\n", + "Nd =6*10**16\n", + "print\"Nd = \",Nd,\" cmˆ−3\" # initializing the value of donor concentration .\n", + "Na =10**17\n", + "print\"Na = \",Na,\" cmˆ−3\"# initializing the value of acceptor concentration .\n", + "no=1.5*10**10\n", + "print\"no = \",no,\" cmˆ−3\"# initializing the value of electron and hole per cmˆ3.\n", + "p=((Na-Nd)+sqrt((Na-Nd)**2+4*no))/2\n", + "print\"Hole concentration ,n=(−(Na−Nd)+sqrt ((Na−Nd)ˆ2+4∗no))/2)= \",p,\" cmˆ−3\"#calculation\n", + "n=(no**2/p)\n", + "print \"Electron concentration ,n=(noˆ2/p))= \",n # calculation\n", + "#this is solved problem 2.14 of chapter 2.\n", + "#the value of Na,Nd in the solution is different than provided in the question\n", + "#I have used the value used in the solution(i.e Na=10ˆ17 ,Nd=6∗10ˆ16)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_15 pgno:53" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nd = 60000000000000000 cmˆ−3\n", + "Na = 100000000000000000 cmˆ−3\n", + "no = 15000000000.0 cmˆ−3\n", + "Hole concentration ,n=(−(Na−Nd)+sqrt ((Na−Nd)ˆ2+4∗no))/2)= 4e+16 cmˆ−3\n", + "Electron concentration ,n=(noˆ2/p))= 5625.0\n" + ] + } + ], + "source": [ + "#exa 2.15\n", + "Nd =6*10**16\n", + "print\"Nd = \",Nd,\" cmˆ−3\" # initializing the value of donor concentration .\n", + "Na =10**17\n", + "print\"Na = \",Na,\" cmˆ−3\"# initializing the value of acceptor concentration .\n", + "no=1.5*10**10\n", + "print\"no = \",no,\" cmˆ−3\"# initializing the value of electron and hole per cmˆ3.\n", + "p=((Na-Nd)+sqrt((Na-Nd)**2+4*no))/2\n", + "print\"Hole concentration ,n=(−(Na−Nd)+sqrt ((Na−Nd)ˆ2+4∗no))/2)= \",p,\"cmˆ−3\"#calculation\n", + "n=(no**2/p)\n", + "print\"Electron concentration ,n=(noˆ2/p))= \",n # calculation\n", + "#this is solved problem 2.15 of chapter 2.\n", + "#the value of Na,Nd in the solution is different than provided in the question\n", + "#I have used the value used in the solution(i.e Na=10ˆ17 ,Nd=6∗10ˆ16)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_16 pgno:53" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nv = 1.04e+19 cmˆ−3\n", + "Ef Ev = 0.3 eV\n", + "T = 300 K\n", + "T = 500 K\n", + "Vt1 = 0.0259 eV\n", + "k = 1.38e-23 J/K\n", + "e = 1.6e-19 columb\n", + "Value of constant ,K1=(Nv/((T) ˆ(3/2) ) )= 3.46666666667e+16 cmˆ−3 K(−2/3)\n", + "Value of valence band concentration at 500K,Nv =K1∗T(3/2)= 1.73333333333e+19 cmˆ−3\n", + "Value of parameter VT at 500K,VT=(K∗T/e)= 0.043125 cmˆ−3\n", + "Hole concentration ,p=(Nv∗(exp(Ef Ev)/(VT)))= 1.65083278171e+16 cmˆ−3\n" + ] + } + ], + "source": [ + "#exa 2.16\n", + "from math import exp\n", + "Nv1 =1.04*10**19\n", + "print\"Nv = \",Nv1,\" cmˆ−3\"# initializing the value of valence band concentration at 300K.\n", + "Ef_Ev =0.3\n", + "print\"Ef Ev = \",Ef_Ev,\" eV\"# initializing the value of boron acceptor ionization energy.\n", + "T1 =300\n", + "print\"T = \",T1,\"K\"# initializing the value of temperature 1.\n", + "T2 =500\n", + "print\"T = \",T2,\"K\"# initializing the value of temperature 2.\n", + "Vt1 =0.0259\n", + "print\"Vt1 = \",Vt1,\"eV\"# initializing the value of thermal voltage at 300K.\n", + "k=1.38*10**-23\n", + "print\"k = \",k,\"J/K\" # initializing value of boltzmann constant .\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\"columb\" # initializing the value of electronic charge .\n", + "K1=(Nv1/((T1)**(3/2)))\n", + "print\"Value of constant ,K1=(Nv/((T) ˆ(3/2) ) )= \",K1,\" cmˆ−3 K(−2/3)\"# calculation\n", + "Nv2=K1*T2**(3/2)\n", + "print\"Value of valence band concentration at 500K,Nv =K1∗T(3/2)= \",Nv2,\" cmˆ−3\"# calculation\n", + "VT=(k*T2/e)\n", + "print\"Value of parameter VT at 500K,VT=(K∗T/e)= \",VT,\" cmˆ−3\"# calculation\n", + "p=(Nv2*(exp(-(Ef_Ev)/(VT))))\n", + "print\"Hole concentration ,p=(Nv∗(exp(Ef Ev)/(VT)))= \",p,\" cmˆ−3\"# calculation\n", + "#this is solved problem 2.16 of chapter 2.\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_17 pgno:54" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nv = 7000000000000000000 cmˆ−3\n", + "Nc = 4.7e+17 cmˆ−3\n", + "T = 300 K\n", + "T = 450 K\n", + "Vt1 = 0.0259 eV\n", + "Vt2 = 0.03881 eV\n", + "Eg = 1.42 eV\n", + "intrinsic concentration at 300K,no=(sqrt(Nc∗Nv∗(exp(−Eg/Vt1))))= 2255422.87974\n", + "Value of constant ,K1=(Nc/((T) ˆ(3/2) ) )= 1.56666666667e+15\n", + "Value of constant k1 at 450K ,k1=(K1∗T2ˆ(3/2))= 7.05e+17\n", + "Value of constant ,K2=(Nv/((T1) ˆ(3/2) ) )= 23333333333333333\n", + "Value of constant k2 at 450K ,k2=(K2∗T2ˆ(3/2))= 10499999999999999850\n", + "Value of constant K,= 7.4025e+36\n", + "intrinsic concentration at 450K,no=(sqrt(K∗(exp(−Eg/Vt2) ) ) )= 30874193378.4 cmˆ3\n" + ] + } + ], + "source": [ + "#exa 2.17\n", + "from math import sqrt\n", + "Nv =7*10**18\n", + "print\"Nv = \",Nv,\"cmˆ−3\"# initializing the value of valence band concentration at 300K.\n", + "Nc=4.7*10**17\n", + "print\"Nc = \",Nc,\"cmˆ−3\"# initializing the value of conduction band concentration at 300K.\n", + "T1 =300\n", + "print\"T = \",T1,\"K\"# initializing the value of temperature 1.\n", + "T2 =450\n", + "print\"T = \",T2,\"K\"# initializing the value of temperature 2.\n", + "Vt1 =0.0259\n", + "print\"Vt1 = \",Vt1,\"eV\"# initializing the value of thermal voltage at 300K.\n", + "Vt2 =0.03881\n", + "print\"Vt2 = \",Vt2,\"eV\"# initializing the value of thermal voltage at 450K.\n", + "Eg=1.42\n", + "print\"Eg = \",Eg,\"eV\"# initializing the value of thermal voltage .\n", + "no=(sqrt(Nc*Nv*(exp(-Eg/Vt1))))\n", + "print\"intrinsic concentration at 300K,no=(sqrt(Nc∗Nv∗(exp(−Eg/Vt1))))= \",no #calculation\n", + "K1=(Nc/((T1)**(3/2)))\n", + "print\"Value of constant ,K1=(Nc/((T) ˆ(3/2) ) )= \",K1 # calculation\n", + "k1=(K1*T2**(3/2))\n", + "print\"Value of constant k1 at 450K ,k1=(K1∗T2ˆ(3/2))= \",k1# calculation\n", + "K2=(Nv/((T1)**(3/2)))\n", + "print\"Value of constant ,K2=(Nv/((T1) ˆ(3/2) ) )= \",K2# calculation\n", + "k2=(K2*T2**(3/2))\n", + "print\"Value of constant k2 at 450K ,k2=(K2∗T2ˆ(3/2))= \",k2 # calculation\n", + "K=k1*k2\n", + "print\"Value of constant K,= \",K # calculation\n", + "no1=(sqrt(K*(exp(-Eg/Vt2))))\n", + "print\"intrinsic concentration at 450K,no=(sqrt(K∗(exp(−Eg/Vt2) ) ) )= \",no1,\" cmˆ3\"# calculation\n", + "#this is solved problem 2.17 of chapter 2." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_18 pgno:55" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nv = 1.04e+19 cmˆ−3\n", + "Nc = 2.8e+19 cmˆ−3\n", + "T = 300 K\n", + "T = 550 K\n", + "Vt1 = 0.0259 eV\n", + "Vt2 = 0.0474 eV\n", + "Eg1 = 1.12 eV\n", + "intrinsic concentration at 300K,no=(sqrt(Nc∗Nv∗(exp(−Eg1/Vt1))))= 6949358641.26\n", + "Value of constant ,K1=(Nc/((T) ˆ(3/2) ) )= 9.3023255814e+16\n", + "Value of constant k1 at 550K ,k1=(K1∗T2ˆ(3/2))= 5.11627906977e+19\n", + "Value of constant ,K2=(Nv/((T1) ˆ(3/2) ) )= 3.46666666667e+16\n", + "Value of constant k2 at 550K ,k2=(K2∗T2ˆ(3/2))= 1.90666666667e+19\n", + "Value of constant K,= 9.75503875969e+38\n", + "Intrinsic concentration at 550K,no=(sqrt(K∗(exp(−Eg1/Vt2))))= 2.31051731905e+14 cmˆ3\n", + "Donor concentration at which intrinsic concentration is 10% of the total electron concentration ,Nd=(4∗(no1ˆ2) /(1.2) )= 1.77949676054e+29 cmˆ3\n" + ] + } + ], + "source": [ + "#exa 2.18\n", + "from math import sqrt\n", + "from math import exp\n", + "Nv=1.04*10**19\n", + "print\"Nv = \",Nv,\"cmˆ−3\"# initializing the value of valence band concentration at 300K.\n", + "Nc=2.8*10**19\n", + "print\"Nc = \",Nc,\"cmˆ−3\"# initializing the value of conduction band concentration at 300K.\n", + "T1 =300\n", + "print\"T = \",T1,\"K\"# initializing the value of temperature 1.\n", + "T2 =550\n", + "print\"T = \",T2,\"K\"# initializing the value of temperature 2.\n", + "Vt1 =0.0259\n", + "print\"Vt1 = \",Vt1,\"eV\"# initializing the value of thermal voltage at 300K.\n", + "Vt2 =0.0474\n", + "print\"Vt2 = \",Vt2,\"eV\"# initializing the value of thermal voltage at 550K.\n", + "Eg1=1.12\n", + "print\"Eg1 = \",Eg1,\"eV\"# initializing the value of thermal voltage .\n", + "no=(sqrt(Nc*Nv*(exp(-Eg1/Vt1))))\n", + "print\"intrinsic concentration at 300K,no=(sqrt(Nc∗Nv∗(exp(−Eg1/Vt1))))= \",no #calculation\n", + "K1=(Nc/((T1)^(3/2)))\n", + "print\"Value of constant ,K1=(Nc/((T) ˆ(3/2) ) )= \",K1 # calculation\n", + "k1=(K1*T2**(3/2))\n", + "print\"Value of constant k1 at 550K ,k1=(K1∗T2ˆ(3/2))= \",k1 # calculation \n", + "K2=(Nv/((T1)**(3/2)))\n", + "print\"Value of constant ,K2=(Nv/((T1) ˆ(3/2) ) )= \",K2 # calculation\n", + "k2=(K2*T2**(3/2))\n", + "print\"Value of constant k2 at 550K ,k2=(K2∗T2ˆ(3/2))= \",k2 # calculation\n", + "K=k1*k2\n", + "print\"Value of constant K,= \",K # calculation\n", + "no1=(sqrt(K*(exp(-Eg1/Vt2))))\n", + "print\"Intrinsic concentration at 550K,no=(sqrt(K∗(exp(−Eg1/Vt2))))= \",no1,\" cmˆ3\"# calculation\n", + "Nd=(4*(no1**2)/(1.2))\n", + "print\"Donor concentration at which intrinsic concentration is 10% of the total electron concentration ,Nd=(4∗(no1ˆ2) /(1.2) )= \",Nd,\" cmˆ3\"# calculation\n", + "#this is solved problem 2.18 of chapter 2.\n", + "#the value of temperature and % of the intrinsic carrier concentration given in the question is different than used in the solution .\n", + "#I have used the value provided in the solution (i.e T2=550 and % of the intrinsic carrier concentration =10%)\n", + "#the value of Donor concentration at which intrinsic concentration is 10% of the total electron concentration (Nd) , is provided wrong in the book after calculation .\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_19 pgno:55" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ec Ef = 0.2 eV\n", + "Nc = 2.8e+19 cmˆ−3\n", + "Na = 30000000000000000 cmˆ−3\n", + "Vt = 0.0259 eV\n", + "Donor concentration ,Nd=(Nc∗(exp(−(Ec Ef)/(Vt))) )+(Na)= 4.24031697774e+16 cmˆ−3\n" + ] + } + ], + "source": [ + "#exa 2.19\n", + "Ec_Ef =0.2\n", + "print\"Ec Ef = \",Ec_Ef,\" eV\" # initializing the value of difference in the energy levels.\n", + "Nc=2.8*10**19\n", + "print\"Nc =\",Nc,\" cmˆ−3\"# initializing the conduction band concentration .\n", + "Na =3*10**16\n", + "print\"Na =\",Na,\" cmˆ−3\"# initializing the acceptor concentration .\n", + "Vt =0.0259\n", + "print\"Vt =\",Vt,\" eV\"# initializing the thermal voltage at 300K.\n", + "Nd=(Nc*(exp(-(Ec_Ef)/(Vt))))+(Na)\n", + "print\"Donor concentration ,Nd=(Nc∗(exp(−(Ec Ef)/(Vt))) )+(Na)= \",Nd,\" cmˆ−3\"# calculation\n", + "#this is solved problem 2.19 of chapter 2." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_20 pgno:56" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nv = 6000000000000000000 cmˆ−3\n", + "Nc = 1.04e+19 cmˆ−3\n", + "T1 = 300 K\n", + "T2 = 200 K\n", + "Vt1 = 0.0259 eV\n", + "Vt2 = 0.0173 eV\n", + "Eg1 = 0.6 eV\n", + "intrinsic concentration at 300K,no=(sqrt(Nc∗Nv∗(exp(−Eg1/Vt1))))= 7.36468677124e+13\n", + "Eg2 = 0.66 eV\n", + "Value of constant ,K1=(Nc/((T) ˆ(3/2) ) )= 3.46666666667e+16\n", + "Value of constant k1 at 200K ,k1=(K1∗T2ˆ(3/2))= 6.93333333333e+18\n", + "Value of constant ,K2=(Nv/((T1) ˆ(3/2) ) )= 20000000000000000\n", + "Value of constant k2 at 200K ,k2=(K2∗T2ˆ(3/2))= 4000000000000000000\n", + "Value of constant K,= 2.77333333333e+37\n", + "intrinsic concentration at 200K,no=(sqrt(K∗(exp(−Eg2/Vt2))))= 27369762834.6 cmˆ3\n" + ] + } + ], + "source": [ + "#exa 2.20\n", + "from math import sqrt\n", + "from math import exp\n", + "Nv =6*10**18\n", + "print\"Nv = \",Nv,\"cmˆ−3\"# initializing the value of valence band concentration at 300K.\n", + "Nc=1.04*10**19\n", + "print\"Nc = \",Nc,\"cmˆ−3\"# initializing the value of conduction band concentration at 300K.\n", + "T1 =300\n", + "print\"T1 = \",T1,\"K\"# initializing the value of temperature 1.\n", + "T2 =200\n", + "print\"T2 = \",T2,\"K\"# initializing the value of temperature 2.\n", + "Vt1 =0.0259\n", + "print\"Vt1 = \",Vt1,\"eV\"# initializing the value of thermal voltage at 300K.\n", + "Vt2 =0.0173\n", + "print\"Vt2 = \",Vt2,\"eV\"# initializing the value of thermal voltage at 200K.\n", + "Eg1=0.60\n", + "print\"Eg1 = \",Eg1,\"eV\"# initializing the value of thermal voltage used for 300K .\n", + "no=(sqrt(Nc*Nv*(exp(-Eg1/Vt1))))\n", + "print\"intrinsic concentration at 300K,no=(sqrt(Nc∗Nv∗(exp(−Eg1/Vt1))))= \",no #calculation\n", + "Eg2=0.66\n", + "print\"Eg2 = \",Eg2,\"eV\"# initializing the value of thermal voltage used for 200K.\n", + "K1=(Nc/((T1)**(3/2)))\n", + "print\"Value of constant ,K1=(Nc/((T) ˆ(3/2) ) )= \",K1 # calculation\n", + "k1=(K1*T2**(3/2))\n", + "print\"Value of constant k1 at 200K ,k1=(K1∗T2ˆ(3/2))= \",k1 # calculation\n", + "K2=(Nv/((T1)**(3/2)))\n", + "print\"Value of constant ,K2=(Nv/((T1) ˆ(3/2) ) )= \",K2 # calculation\n", + "k2=(K2*T2**(3/2))\n", + "print\"Value of constant k2 at 200K ,k2=(K2∗T2ˆ(3/2))= \",k2 # calculation\n", + "K=k1*k2\n", + "print\"Value of constant K,= \",K # calculation\n", + "no1=(sqrt(K*(exp(-Eg2/Vt2))))\n", + "print\"intrinsic concentration at 200K,no=(sqrt(K∗(exp(−Eg2/Vt2))))= \",round(no1,2),\" cmˆ3\"# calculation\n", + "#this is solved problem 2.20 of chapter 2.\n", + "#The answer of intrinsic concentration at 300K,(no) is provided wrong in the book." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_21 pgno:56" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Eg1 = 2 eV\n", + "Eg2 = 2.2 eV\n", + "Vt = 0.0259 eV\n", + "Ratio of their intrinsic concentration at 300K,(no1/no2)=sqrt(exp((−Eg1/Vt)−(−Eg2/Vt)))= 47.5130239084\n" + ] + } + ], + "source": [ + "#exa 2.21\n", + "Eg1=2\n", + "print\"Eg1 = \",Eg1,\" eV\" # initializing the value of band energy gap for semiconductor1.\n", + "Eg2 =2.2\n", + "print\"Eg2 = \",Eg2,\" eV\"# initializing the value of band energy gap for semiconductor2.\n", + "Vt =0.0259\n", + "print\"Vt = \",Vt,\" eV\"# initializing the value of thermal voltage at 300K.\n", + "No=sqrt(exp((-Eg1/Vt)-(-Eg2/Vt)))\n", + "print\"Ratio of their intrinsic concentration at 300K,(no1/no2)=sqrt(exp((−Eg1/Vt)−(−Eg2/Vt)))= \",No # calculation" + ] + } + ], + "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": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_3_CARRIER_TRANSPORT_IN_SEMICONDUCTOR_1.ipynb b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_3_CARRIER_TRANSPORT_IN_SEMICONDUCTOR_1.ipynb new file mode 100644 index 00000000..2cc9b53b --- /dev/null +++ b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_3_CARRIER_TRANSPORT_IN_SEMICONDUCTOR_1.ipynb @@ -0,0 +1,700 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3 CARRIER TRANSPORT IN SEMICONDUCTOR" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_1 pgno: 71" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "I = 0.005 amphere\n", + "B= 1e-06 Tesla\n", + "w = 0.0001 m\n", + "l = 0.001 m\n", + "t = 1e-05 m\n", + "p = 100000000000000000 atoms/mˆ3\n", + "e = 1.6e-19 columb\n", + "Hall electric field ,EH=(I∗B)/(w∗t∗p∗e)= 312.5 V/m\n", + "Hall electric field in centimeter ,EH=(I∗B)/(w∗ t∗p∗e)= 3.125 V/cm\n" + ] + } + ], + "source": [ + "#exa 3.1\n", + "I=5*10**-3\n", + "print \"I = \",I,\" amphere\" # initializing value of current flowing through the sample.\n", + "B=1*10**-6\n", + "print \"B= \",B,\" Tesla\" # initializing value of magnetic field .\n", + "w=0.01*10**-2\n", + "print \"w = \",w,\" m\" # initializing value of width of germanium sample .\n", + "l=0.1*10**-2\n", + "print \"l = \",l,\" m\" # initializing value of length of germanium sample .\n", + "t=0.001*10**-2\n", + "print \"t = \",t,\" m\" # initializing value of thickness of germanium sample .\n", + "p=10**17\n", + "print \"p = \",p,\" atoms/mˆ3\" # initializing value of doped acceptor atoms .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\" columb\" # initializing value of charge of electron .\n", + "EH=(I*B)/(w*t*p*e)\n", + "print \"Hall electric field ,EH=(I∗B)/(w∗t∗p∗e)= \",EH,\" V/m\" # calculation 18 \n", + "E=EH*10**-2\n", + "print \"Hall electric field in centimeter ,EH=(I∗B)/(w∗ t∗p∗e)= \",E,\" V/cm\" # calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_2 pgno: 72" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "I = 0.005 amphere\n", + "B= 1e-06 Tesla\n", + "w = 0.0001 m\n", + "l = 0.001 m\n", + "t = 1e-05 m\n", + "p = 100000000000000000 atoms/cmˆ3\n", + "e = 1.6e-19 columb\n", + "hall cofficient ,Rh=(1/(p∗e))= 62.5 cmˆ3/C\n" + ] + } + ], + "source": [ + "#exa 3.2\n", + "I=5*10**-3\n", + "print \"I = \",I,\" amphere\" # initializing value of current flowing through the sample.\n", + "B=1*10**-6\n", + "print \"B= \",B,\" Tesla\" # initializing value of magnetic field .\n", + "w=0.01*10**-2\n", + "print \"w = \",w,\" m\" # initializing value of width of germanium sample .\n", + "l=0.1*10**-2\n", + "print \"l = \",l,\" m\" # initializing value of length of germanium sample .\n", + "t=0.001*10**-2\n", + "print \"t = \",t,\" m\" # initializing value of thickness of germanium sample .\n", + "p=10**17\n", + "print \"p = \",p,\" atoms/cmˆ3\" # initializing value of doped acceptor atoms .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\" columb\" # initializing value of charge of electron .\n", + "Rh=(1/(p*e))\n", + "print \"hall cofficient ,Rh=(1/(p∗e))= \",Rh,\" cmˆ3/C\"# calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_3 pgno: 72" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "I = 0.01 amphere\n", + "B= 1e-05 Tesla\n", + "w = 0.0001 m\n", + "l = 0.001 m\n", + "t = 1e-05 m\n", + "n = 10000000000000000 atoms/cmˆ3\n", + "e = 1.6e-19 columb\n", + "Hall voltage ,Vh=((I∗B)/(n∗e∗t)))= 6.25 V\n" + ] + } + ], + "source": [ + "#exa 3.3\n", + "I=10*10**-3\n", + "print \"I = \",I,\" amphere\" # initializing value of current flowing through the sample.\n", + "B=10*10**-6\n", + "print \"B= \",B,\" Tesla\" # initializing value of magnetic field .\n", + "w=0.01*10**-2\n", + "print \"w = \",w,\" m\" # initializing value ofwidth of germanium sample .\n", + "l=0.1*10**-2\n", + "print \"l = \",l,\" m\" # initializing value of length of germanium sample .\n", + "t=0.001*10**-2\n", + "print \"t = \",t,\" m\" # initializing value of thickness of germanium sample .\n", + "n=10**16\n", + "print \"n = \",n,\" atoms/cmˆ3\" # initializing value of doped donor atoms .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\" columb\" # initializing value of charge of electron .\n", + "Vh=((I*B)/(n*e*t))\n", + "print \"Hall voltage ,Vh=((I∗B)/(n∗e∗t)))= \",Vh,\" V\" # calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_4 pgno: 72" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "I = 0.01 amphere\n", + "B= 1e-05 Tesla\n", + "w = 0.0001 m\n", + "l = 0.001 m\n", + "t = 1e-05 m\n", + "p = 1000000000000000000 atoms/cmˆ3\n", + "e = 1.6e-19 columb\n", + "Hall voltage ,Yh=((B)/(p∗e∗t)))= 6.25 ohm\n" + ] + } + ], + "source": [ + "#exa 3.4\n", + "I=10*10**-3\n", + "print \"I = \",I,\" amphere\" # initializing value of current flowing through the sample.\n", + "B=10*10**-6\n", + "print \"B= \",B,\" Tesla\" # initializing value of magnetic field .\n", + "w=0.01*10**-2\n", + "print \"w = \",w,\" m\" # initializing value of width of germanium sample .\n", + "l=0.1*10**-2\n", + "print \"l = \",l,\" m\" # initializing value of length of germanium sample .\n", + "t=0.001*10**-2\n", + "print \"t = \",t,\" m\" # initializing value of thickness of germanium sample .\n", + "p=10**18\n", + "print \"p = \",p,\" atoms/cmˆ3\" # initializing value of doped donor atoms .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\" columb\" # initializing value of charge of electron .\n", + "Yh=((B)/(p*e*t))\n", + "print \"Hall voltage ,Yh=((B)/(p∗e∗t)))= \",Yh,\"ohm\" # calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_8 pgno: 75" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "no = 15000000000.0\n", + "n= 20000000000000000\n", + "un = 1200\n", + "up = 500\n", + "e = 1.6e-19 columb\n", + "resistivity ,p=(1/(2∗e∗no∗(sqrt(un/up))))= 268957.17682 ohm\n", + "conductivity ,s=(1/p))= 3.71806401236e-06 S /cm\n", + "intrinsic conductivity ,sigma=e∗no∗(un+up))= 4.08e-06 S/cm\n" + ] + } + ], + "source": [ + "#exa 3.8\n", + "from math import sqrt\n", + "no=1.5*10**10\n", + "print \"no = \",no # initializing value of electron hole per cmˆ3.\n", + "n=2*10**16\n", + "print \"n= \",n # initializing value of number of electrons per cmˆ3.\n", + "un =1200\n", + "print \"un = \",un # initializing value of mobility of n−type carrier .\n", + "up =500\n", + "print \"up = \",up # initializing value of mobility of p−type carrier .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\" columb\" # initializing value of charge of electron .\n", + "p=(1/(2*e*no*(sqrt(un*up))))\n", + "print \"resistivity ,p=(1/(2∗e∗no∗(sqrt(un/up))))= \",p,\" ohm\" # calculation\n", + "sigmamin=(1/p)\n", + "print \"conductivity ,s=(1/p))= \",sigmamin,\" S /cm\" # calculation\n", + "sigma=e*no*(un+up)\n", + "print \"intrinsic conductivity ,sigma=e∗no∗(un+up))= \",sigma,\" S/cm\" # calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_10 pgno: 76" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "po = 1000000000000000000 cmˆ−3\n", + "no = 15000000000.0 /cmˆ−3\n", + "P(o)= 100000000000000000 cmˆ−3\n", + "A = 0.1 cmˆ−2\n", + "up = 300 cmˆ2/Vs\n", + "t = 7e-09 sec\n", + "T = 300 K\n", + "Vt = 0.0259 eV\n", + "x = 5e-06 cm\n", + "Diffusion cofficient ,Dp=(Vt∗up))= 7.77 cmˆ2/s\n", + "Diffusion length ,Lp=(sqrt(Dp∗t)))= 0.000233216637485 cm\n", + "Excess charge generated ,p(x)=(po+(P(o)∗exp(−x/Lp) ) )= 1.09787888943e+18 cmˆ−3\n", + "Fermi level ,Efi Efp=(Vt∗log(p(x)/no)))= 0.469012627899 eV\n" + ] + } + ], + "source": [ + "#exa 3.10\n", + "from math import sqrt\n", + "from math import exp\n", + "from math import log\n", + "po =10**18\n", + "print \"po = \",po,\" cmˆ−3\" # initializing value of N type doping level .\n", + "no=1.5*10**10\n", + "print \"no = \",no,\" /cmˆ−3\" # initializing value of electron and hole concentration per cmˆ3.\n", + "Po =10**17\n", + "print \"P(o)= \",Po,\" cmˆ−3\" # initializing value of excess hole concentration .\n", + "A=0.1\n", + "print \"A = \",A,\" cmˆ−2\" # initializing the value of area .\n", + "up=300\n", + "print \"up = \",up,\" cmˆ2/Vs\" # initializing value of mobility of p−type carrier .\n", + "t=7*10**-9\n", + "print \"t = \",t,\" sec\" # initializing value of transit time.\n", + "T=300\n", + "print \"T = \",T,\"K\" # initializing value of temperature .\n", + "Vt=0.0259\n", + "print \"Vt = \",Vt,\" eV\" # initializing value of thermal voltage at 300K.\n", + "x=500*10**-8\n", + "print \"x = \",x,\" cm\" # initializing value of distance at which difference in fermi level is to calculated .\n", + "Dp=(Vt*up)\n", + "print \"Diffusion cofficient ,Dp=(Vt∗up))= \",Dp,\" cmˆ2/s\" #calculation \n", + "Lp=(sqrt(Dp*t))\n", + "print \"Diffusion length ,Lp=(sqrt(Dp∗t)))= \",Lp,\" cm\" # calculation\n", + "px=(po+(Po*exp(-x/Lp)))\n", + "print \"Excess charge generated ,p(x)=(po+(P(o)∗exp(−x/Lp) ) )= \",px,\" cmˆ−3\" # calculation\n", + "Efi_Efp=(Vt*log(px/no))\n", + "print \"Fermi level ,Efi Efp=(Vt∗log(p(x)/no)))= \",Efi_Efp,\" eV\" # calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_11 pgno: 77" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A = 1e-05 cmˆ2\n", + "Dp= 0.000777 cmˆ2/s\n", + "Lp = 2.33e-06 cm\n", + "x = 5e-06 cm\n", + "P(O)−po = 100000000000000000000000\n", + "e = 1.6e-19 column\n", + "Hole current ,I=(((e∗A∗Dp∗[P(O)−po])/Lp)∗exp(−x/Lp))= 6.24054720884 amphere\n", + " stored excess hole ,Q=(e∗A∗Dp∗Lp∗P))= 2.896656e-10 C\n" + ] + } + ], + "source": [ + "#exa 3.11\n", + "from math import exp\n", + "A=0.1*10**-4\n", + "print \"A = \",A,\" cmˆ2\" # initializing value of area .\n", + "Dp =7.77*10** -4\n", + "print \"Dp= \",Dp,\" cmˆ2/s\" # initializing value of diffusion cofficient .\n", + "Lp =0.233*10** -5\n", + "print \"Lp = \",Lp,\" cm\" # initializing value of diffusion length .\n", + "x=500*10**-8\n", + "print \"x = \",x,\" cm\" # initializing value of distance\n", + "P=10**17*10**6\n", + "print \"P(O)−po = \",P # initializing value of P(O)−po\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\"column\" # initializing value of charge of electron .\n", + "I=(((e*A*Dp*P)/Lp)*exp(-x/Lp))\n", + "print \"Hole current ,I=(((e∗A∗Dp∗[P(O)−po])/Lp)∗exp(−x/Lp))= \",I,\"amphere\" # calculation\n", + "Q=(e*A*Dp*Lp*P)\n", + "print \" stored excess hole ,Q=(e∗A∗Dp∗Lp∗P))= \",Q,\"C\" # calculation\n", + "# the value of current(I) given after calculation inthe book is wrong, (as the value of Lp used in the formula while finding value of hole current ( I)at two places is used different).\n", + "# I have used the value Lp=0.233∗10ˆ−5 cm" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_12 pgno: 77" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "I = 0.002 amphere\n", + "B= 0.1 Tesla\n", + "w = 0.0002 mm\n", + "l = 0.002 m\n", + "t = 2e-05 m\n", + "Vaa = 10 V\n", + "Vh = -0.01 V\n", + "e = 1.6e-19 columb\n", + "electron concentration ,n=((I∗B)/(e∗t∗Vh))= -6.25e+21 mˆ−3\n", + "mobility ,un=(I∗L/(e∗n∗Vaa∗w∗t))= 0.1 mˆ2/Vs\n" + ] + } + ], + "source": [ + "#exa 3.12\n", + "I=2*10**-3\n", + "print \"I = \",I,\" amphere\" # initializing value of current flowing through the sample.\n", + "B=1000*10**-4\n", + "print \"B= \",B,\" Tesla\" # initializing value of magnetic field .\n", + "w=0.2*10**-3\n", + "print \"w = \",w,\" mm\" # initializing value of width of sample .\n", + "l=2*10**-3\n", + "print \"l = \",l,\" m\" # initializing value of length of sample .\n", + "t=0.02*10**-3\n", + "print \"t = \",t,\" m\" # initializing value of thickness of sample .\n", + "Vaa=10\n", + "print \"Vaa = \",Vaa,\" V\" # initializing value of applied voltage .\n", + "Vh = -10*10** -3\n", + "print \"Vh = \",Vh,\" V\" # initializing value of hall voltage .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\" columb\" # initializing value of charge of electron .\n", + "n=((I*B)/(e*t*Vh))\n", + "print \"electron concentration ,n=((I∗B)/(e∗t∗Vh))= \",n,\" mˆ−3\" # calculation\n", + "un=(I*l/(e*abs(n)*Vaa*w*t))\n", + "print \"mobility ,un=(I∗L/(e∗n∗Vaa∗w∗t))= \",un,\" mˆ2/Vs\" # calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_14 pgno: 78" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ND x = ((10ˆ17) −(10ˆ18∗x))\n", + "differentiating above equation with resprct to x\n", + "d[ND x]/dx = (−10ˆ18) cmˆ−4\n", + "now, electric field is given by \n", + "E x = −(VT/ND x)∗(d[ND x]/dx) = (0.0259∗10ˆ18)/((10ˆ15) −(10ˆ18∗x))\n", + "for x = 0\n", + "E x = 25.9 V/cm\n", + "for x = 1∗10ˆ−4 cm\n", + "E x = 28.7777777778 V/cm\n" + ] + } + ], + "source": [ + "#exa 3.14\n", + "print \"ND x = ((10ˆ17) −(10ˆ18∗x))\" # donor concentration in an N type semiconductor\n", + "print \"differentiating above equation with resprct to x\"\n", + "print \"d[ND x]/dx = (−10ˆ18) cmˆ−4\"\n", + "print \"now, electric field is given by \"\n", + "print \"E x = −(VT/ND x)∗(d[ND x]/dx) = (0.0259∗10ˆ18)/((10ˆ15) −(10ˆ18∗x))\" # equation for electric field\n", + "print \"for x = 0\" \n", + "x=0\n", + "E_x = (0.0259*10**18)/((10**15) -(10**18*x))\n", + "print \"E x = \",E_x,\"V/cm\"\n", + "print \"for x = 1∗10ˆ−4 cm\"\n", + "x = 1*10**-4\n", + "E_x = (0.0259*10**18)/((10**15) -(10**18*x))\n", + "print \"E x = \",E_x,\"V/cm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_17 pgno: 81" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nd = 100000000000000000 /cmˆ3\n", + "Na= 0 /cmˆ3\n", + "no = 1800000.0 /cmˆ3\n", + "E = 5 V/cm\n", + "un = 7500 cmˆ2/s\n", + "n1= 100000000000000000 cmˆ−3\n", + "e = 1.6e-19 columb\n", + "Electron concentration ,n=(−(Na−Nd)+sqrt ((Na−Nd)ˆ2+4∗no))/2)= 1e+17 cmˆ−3\n", + "Hole concentration ,p=(noˆ2/n))= 3.24e-05 cmˆ−3\n", + "Drift current density , Jdrift=n1∗un∗e∗E)= 600.0 A/cmˆ2\n" + ] + } + ], + "source": [ + "#exa 3.17\n", + "from math import sqrt\n", + "Nd =10**17\n", + "print \"Nd = \",Nd,\" /cmˆ3\" # initializing value of donor concentration .\n", + "Na=0\n", + "print \"Na= \",Na,\"/cmˆ3\" # initializing value of acceptor concentration .\n", + "no=1.8*10**6\n", + "print \"no = \",no,\" /cmˆ3\" # initializing value of electron and hole concentration per cmˆ3.\n", + "E=5\n", + "print \"E = \",E,\" V/cm\" # initializing value of electric field .\n", + "un=7500\n", + "print \"un = \",un,\" cmˆ2/s\" # initializing value of mobility .\n", + "n1=10**17\n", + "print \"n1= \",n1,\" cmˆ−3\" # initializing value of impurity concentration .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\" columb\" # initializing value of charge of electron .\n", + "n=(-(Na-Nd)+sqrt((Na-Nd)**2+4*no))/2\n", + "print \"Electron concentration ,n=(−(Na−Nd)+sqrt ((Na−Nd)ˆ2+4∗no))/2)= \",n,\" cmˆ−3\" #calculation\n", + "p=(no**2/n)\n", + "print \"Hole concentration ,p=(noˆ2/n))= \",p,\" cmˆ−3\" # calculation\n", + "Jdrift=n1*un*e*E\n", + "print \"Drift current density , Jdrift=n1∗un∗e∗E)= \",Jdrift,\" A/cmˆ2\" # calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_18 pgno: 82" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nd = 0 /cmˆ3\n", + "Na= 100000000000000000 /cmˆ3\n", + "no = 1800000.0 /cmˆ3\n", + "E = 10 V/cm\n", + "un = 200 cmˆ2/s\n", + "p1= 100000000000000000 cmˆ−3\n", + "e = 1.6e-19 columb\n", + "Electron concentration ,p=−(−(Na−Nd)−sqrt ((Na−Nd)**2+4∗(no**2)))/2= 1e+17 cmˆ−3\n", + "Hole concentration ,n=(noˆ2/p))= 3.24e-05 cmˆ−3\n", + "Drift current density , Jdrift=n1∗un∗e∗E)= 32.0 A/cmˆ2\n" + ] + } + ], + "source": [ + "#exa 3.18\n", + "from math import sqrt\n", + "Nd=0\n", + "print \"Nd = \",Nd,\" /cmˆ3\" # initializing value of donor concentration .\n", + "Na =10**17\n", + "print \"Na= \",Na,\" /cmˆ3\" # initializing value of acceptor concentration .\n", + "no=1.8*10**6\n", + "print \"no = \",no,\" /cmˆ3\" # initializing value of electron and hole concentration per cmˆ3.\n", + "E=10\n", + "print \"E = \",E,\" V/cm\" # initializing value of electric field .\n", + "un=200\n", + "print \"un = \",un,\" cmˆ2/s\" # initializing value of mobility \n", + "p1=10**17\n", + "print \"p1= \",p1,\" cmˆ−3\" # initializing value of impurity concentration .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\" columb\" # initializing value of charge of electron .\n", + "p=-(-(Na-Nd)-sqrt((Na-Nd)**2+4*(no**2)))/2\n", + "print \"Electron concentration ,p=−(−(Na−Nd)−sqrt ((Na−Nd)**2+4∗(no**2)))/2= \",p,\" cmˆ−3\" # calculation\n", + "n=(no**2/p)\n", + "print \"Hole concentration ,n=(noˆ2/p))= \",n,\"cmˆ−3\" # calculation\n", + "Jdrift=p1*un*e*E\n", + "print \"Drift current density , Jdrift=n1∗un∗e∗E)= \",Jdrift,\" A/cmˆ2\" # calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_19 pgno: 82" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "D = 120 A/cmˆ2\n", + "E = 5 V/cm\n", + "e = 1.6e-19 columb\n", + "thermal equilibrium value of hole concentration ,p=(D/(450∗ e∗E)))= 3.33333333333e+17 /cmˆ3\n" + ] + } + ], + "source": [ + "#exa 3.19\n", + "D=120\n", + "print \"D = \",D,\" A/cmˆ2\" # initializing value of drift current density .\n", + "E=5\n", + "print \"E = \",E,\" V/cm\" # initializing value of electric field .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\"columb\" # initializing value of charge of electron .\n", + "p=(D/(450*e*E))\n", + "print \"thermal equilibrium value of hole concentration ,p=(D/(450∗ e∗E)))= \",p,\" /cmˆ3\" # calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_20 pgno: 83" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nd = 50000000000000000 /cmˆ3\n", + "A= 5e-07 cmˆ2\n", + "l = 0.2 /cm\n", + "E = 10 V\n", + "un = 1100 cmˆ2/s\n", + "p= 50000000000000000 /cmˆ−3\n", + "e = 1.6e-19 columb\n", + "Current through the bar,I=(p∗up∗e∗E∗A)/l)= 0.00022 A\n" + ] + } + ], + "source": [ + "#exa 3.20\n", + "Nd =5*10**16\n", + "print \"Nd = \",Nd,\" /cmˆ3\" # initializing value of donor concentration .\n", + "A=50*10**-8\n", + "print \"A= \",A,\" cmˆ2\" # initializing value of area .\n", + "l=0.2\n", + "print \"l = \",l,\" /cm\" # initializing value of length .\n", + "E=10\n", + "print \"E = \",E,\" V\" # initializing value of electric field .\n", + "up=1100\n", + "print \"un = \",up,\" cmˆ2/s\" # initializing value of mobility .\n", + "p=5*10**16\n", + "print \"p= \",p,\" /cmˆ−3\" # initializing value of impurity concentration .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\"columb\" # initializing value of charge of electron .\n", + "I=(p*up*e*E*A)/l\n", + "print \"Current through the bar,I=(p∗up∗e∗E∗A)/l)= \",I,\"A\" # calculation" + ] + } + ], + "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": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_4__EXCESS_CARRIER_IN_SEMICONDUCTOR_1.ipynb b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_4__EXCESS_CARRIER_IN_SEMICONDUCTOR_1.ipynb new file mode 100644 index 00000000..4b62aae8 --- /dev/null +++ b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_4__EXCESS_CARRIER_IN_SEMICONDUCTOR_1.ipynb @@ -0,0 +1,266 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4 EXCESS CARRIER IN SEMICONDUCTOR" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4_2 pgno: 100" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nd = 200000000000000000 cmˆ−3\n", + "Er = 11.9\n", + "e = 1.6e-19 columns\n", + "eo = 8.854e-14\n", + "un = 1350 cm2/Vs\n", + " conducitivity , sigma=e∗un∗Nd)= 43.2 S/cm\n", + "Dielectric releaxation time ,td=((Er∗Eo)/sigma))= 2.43894907407e-14 s\n" + ] + } + ], + "source": [ + "#exa 4.2\n", + "Nd =2*10**17\n", + "print\"Nd = \",Nd,\"cmˆ−3\" # initializing value of donor concentration .\n", + "Er =11.9\n", + "print\"Er = \",Er # initializing value of relative dielectric constant.\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\"columns\" # initializing value of charge of electrons .\n", + "Eo=8.854*10**-14\n", + "print\"eo = \",Eo # initializing value of permittivity of free space .\n", + "un=1350\n", + "print\"un = \",un,\"cm2/Vs\" # initializing value of mobility .\n", + "sigma=e*un*Nd\n", + "print\" conducitivity , sigma=e∗un∗Nd)=\",sigma,\"S/cm\"# calculation\n", + "td=((Er*Eo)/sigma)\n", + "print\"Dielectric releaxation time ,td=((Er∗Eo)/sigma))=\",td,\"s\"# calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4_3 pgno: 101" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "n = 1000000000000000 cmˆ−3\n", + "no = 10000000000 cmˆ−3\n", + "t = 1e-06 s\n", + "Excess electron concentration = 100000000000000 cmˆ−3\n", + "electron hole recombination ,R=(c/t))= 1e+20 /cmˆ3s\n" + ] + } + ], + "source": [ + "#exa 4.3\n", + "n=10**15\n", + "print\"n = \",n,\"cmˆ−3\" # initializing value of concentration of electrons/cmˆ3.\n", + "no =10**10\n", + "print\"no = \",no,\"cmˆ−3\" # initializing value of intrinsic concentration of electron .\n", + "t=10**-6\n", + "print\"t = \",t,\"s\" # initializing value of carrier lifetime .\n", + "c=1*10**14\n", + "print\"Excess electron concentration = \",c,\"cmˆ−3\" # initializing value of excess electrons concentration .\n", + "R=(c/t)\n", + "print\"electron hole recombination ,R=(c/t))=\",R,\" /cmˆ3s\"# calculation," + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nd = 1000000000000000 cmˆ−3\n", + "minority carrier lifetime = 1e-05 s\n", + "no = 15000000000.0 cmˆ−3\n", + "excess carrier concentration ,p=(noˆ2/Nd))= 225000.0 /cmˆ3\n", + "electron hole generation and recombination rate ,R=(p/t))= 22500000000.0 /cmˆ3s\n", + "majority carrier concentration ,t=Nd/R)= 44444.4444444 s\n" + ] + } + ], + "source": [ + "#exa 4.4\n", + "Nd =10**15\n", + "print\"Nd = \",Nd,\"cmˆ−3\" # initializing value of donor concentration ..\n", + "tn =10*10**-6\n", + "print\"minority carrier lifetime = \",tn,\"s\" #initializing value of minority carrier lifetime\n", + "no=1.5*10**10\n", + "print\"no = \",no,\"cmˆ−3\" # initializing value of electron and hole concentration per cmˆ3.\n", + "p=(no**2/Nd)\n", + "print\"excess carrier concentration ,p=(noˆ2/Nd))=\",p,\"/cmˆ3\"# calculation\n", + "R=(p/tn)\n", + "print\"electron hole generation and recombination rate ,R=(p/t))=\",R,\"/cmˆ3s\"#calculation\n", + "t=Nd/R\n", + "print\"majority carrier concentration ,t=Nd/R)=\",t,\"s\"# calculation .\n", + "#the value of majority carrier concentration,t=Nd/R( after calculation ) , is provided wrong in the solution .\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4_5 pgno: 101" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nd = 10000000000000000 cmˆ−3\n", + "p = 1000000 cmˆ−3\n", + "no = 10000000000 cmˆ−3\n", + "n∗ = 1000000000000000 cmˆ−3\n", + "p∗ = 1000000000000000 cmˆ−3\n", + "KT = 0.0259 eV\n", + "T = 300 K\n", + "Thermal equilibirium fermi level ,( Ef Efi )=(KT∗log(n/no)))= 0.357821723451 eV\n", + "Quasi−fermi levels for n−type dopant ,( Efn Efi )=(KT∗log ((n+n∗)/no))= 0.360290257108 eV\n", + "Quasi−fermi levels for p−type dopant ,( Efi Efp )=(KT∗log ((p+p∗)/no))= 0.360290257108 eV\n" + ] + } + ], + "source": [ + "#exa 4.5\n", + "from math import log\n", + "Nd =10**16\n", + "print\"Nd = \",Nd,\" cmˆ−3\" # initializing value of donor concentration .\n", + "p=10**6\n", + "print\"p = \",p,\" cmˆ−3\" # initializing value of minority hole concentration .\n", + "no =10**10\n", + "print\"no = \",no,\" cmˆ−3\" # initializing value of electron and hole concentration per cm ˆ3..\n", + "n1 =10**15\n", + "print\"n∗ = \",n1,\" cmˆ−3\" # initializing value of excess electron carrier concentration(denoted by n∗).\n", + "p1=10**15\n", + "print\"p∗ = \",p1,\" cmˆ−3\" # initializing value of excess hole carrier concentration( denoted by p∗).\n", + "KT=0.0259\n", + "print\"KT = \",KT,\" eV\" # initializing value of multipication of temperature and bolzmann constant .\n", + "T=300\n", + "print\"T = \",T,\" K\" # initializing value of temperature .\n", + "Ef_Efi=(log(Nd/no)*KT)\n", + "print\"Thermal equilibirium fermi level ,( Ef Efi )=(KT∗log(n/no)))=\",Ef_Efi,\" eV\"#calculation .\n", + "Efn_Efi=log((Nd+n1)/no)*KT\n", + "print\"Quasi−fermi levels for n−type dopant ,( Efn Efi )=(KT∗log ((n+n∗)/no))=\",Efn_Efi,\" eV\"# calculation .\n", + "Efi_Efp=log((Nd+p1)/no)*KT\n", + "print\"Quasi−fermi levels for p−type dopant ,( Efi Efp )=(KT∗log ((p+p∗)/no))=\",Efi_Efp,\" eV\"# calculation .\n", + "#the answer for Efn Efi , Efi Efp is provided wrong in the book.\n", + "#In this question,Nd=(n(used in the formula))." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4_6 pgno: 102" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nd = 50000000000000000 cmˆ−3\n", + "Na = 0 cmˆ−3\n", + "no = 15000000000.0 cmˆ−3\n", + "n∗ = 500000000000000 cmˆ−3\n", + "p∗ = 500000000000000 cmˆ−3\n", + "KT = 0.0259\n", + "thermal equilibrium fermi level ,( Ef Efi )=(KT∗log(n/no)))= 0.389004619083 eV\n", + "Excess carrier concentration ,(Efn Efi)=(KT∗log ((n+n∗)/no))= 0.389262332652 eV\n", + "(Ef Efi)=(KT∗log((p+p∗)/no))= 0.269730711266 eV\n" + ] + } + ], + "source": [ + "#exa 4.6\n", + "from math import log\n", + "Nd =5*10**16\n", + "print\"Nd = \",Nd,\"cmˆ−3\" # initializing value of donor ion concentration .\n", + "Na=0\n", + "print\"Na = \",Na,\"cmˆ−3\" # initializing value of value of acceptor ion concentration .\n", + "no=1.5*10**10\n", + "print\"no =\",no,\"cmˆ−3\" # initializing electron and hole concentration per cmˆ3.\n", + "n1 =5*10**14\n", + "print\"n∗ =\",n1,\"cmˆ−3\" # initializing excess electron carrier concentration .\n", + "p1 =5*10**14\n", + "print\"p∗ =\",p1,\"cmˆ−3\" # initializing excess hole carrier concentration .\n", + "KT=0.0259\n", + "print\"KT =\",KT #initializing value of voltage \n", + "Ef_Efi=(KT*log(Nd/no))\n", + "print\"thermal equilibrium fermi level ,( Ef Efi )=(KT∗log(n/no)))=\",Ef_Efi,\"eV\" #calculation .\n", + "Efn_Efi=log((Nd+n1)/no)*KT\n", + "print\"Excess carrier concentration ,(Efn Efi)=(KT∗log ((n+n∗)/no))=\",Efn_Efi,\"eV\" # calculation .\n", + "Efi_Efp=log((Na+p1)/no)*KT\n", + "print\"(Ef Efi)=(KT∗log((p+p∗)/no))=\",Efi_Efp,\"eV\"# calculation ." + ] + } + ], + "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": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_5_PN_JUNCTION_DIODE_1.ipynb b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_5_PN_JUNCTION_DIODE_1.ipynb new file mode 100644 index 00000000..f92767c3 --- /dev/null +++ b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_5_PN_JUNCTION_DIODE_1.ipynb @@ -0,0 +1,1364 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# chapter 5 PN JUNCTION DIODE" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_5 pgno: 142" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na 100000000000000000 /cmˆ3\n", + "Nd= 1000000000000000 /cmˆ3\n", + "no = 15000000000.0 cmˆ−3\n", + "e = 1.6e-19 columns\n", + "K = 1.38e-23 J/k\n", + "T = 300 K\n", + "(a)fermi level in the P region , Efi Efp=((KT/e) ∗ log (Na/no ) ) )= 0.406564315296 eV\n", + "fermi level in the n region,Efn Efi=((KT/e)∗log (Nd/no) ) )= 0.287405536734 eV\n", + "(b) junction potential at room temperature ,Efn Efp=(Efi Efp)+(Efn Efi))= 0.69396985203 eV\n" + ] + } + ], + "source": [ + "#exa 5.5\n", + "from math import log\n", + "Na =10**17\n", + "print \"Na\",Na,\"/cmˆ3\" # initializing value of medium p doping concentration .\n", + "Nd =10**15\n", + "print \"Nd= \",Nd,\"/cmˆ3\" # initializing value of light n doping .\n", + "no=1.5*10**10\n", + "print \"no = \",no,\"cmˆ−3\" # initializing value of intrinsic carrier concentration .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\"columns\" # initializing value of charge of electrons .\n", + "K=1.38*10**-23\n", + "print \"K = \",K,\"J/k\" #initializing value of boltzmann constant .\n", + "T=300\n", + "print \"T = \",T,\"K\" # initializing value of temperature .\n", + "Efi_Efp=((K*T/e)*log(Na/no))\n", + "print \"(a)fermi level in the P region , Efi Efp=((KT/e) ∗ log (Na/no ) ) )= \",Efi_Efp,\"eV\" # calculation .\n", + "Efn_Efi=((K*T/e)*log(Nd/no))\n", + "print \"fermi level in the n region,Efn Efi=((KT/e)∗log (Nd/no) ) )=\",Efn_Efi,\" eV\" # calculation\n", + "Efn_Efp=(Efi_Efp)+(Efn_Efi)\n", + "print\"(b) junction potential at room temperature ,Efn Efp=(Efi Efp)+(Efn Efi))=\",Efn_Efp,\" eV\" #calculation ." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_7 pgno: 143" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Pp= 1000000000000000000 /cmˆ3\n", + "Nn= 1000000000000000 /cmˆ3\n", + "tp = 7e-06 s\n", + "tn = 2e-07 s\n", + "up= 800 cm2/Vs\n", + "un= 300 cm2/Vs\n", + "no = 15000000000.0 cmˆ−3\n", + "Vf = 0.6 V\n", + "A = 0.0001 mˆ2\n", + "e = 1.6e-19 columns\n", + "K = 1.38e-23 J/k\n", + "T = 300 K\n", + "Vt=(K∗T/e))= 0.025875 eV\n", + "Dp=Vt∗un= 7.7625 cmˆ−3\n", + "Dn=Vt∗up= 20.7 cmˆ−3\n", + "Lp=(sqrt(Dp∗tp))= 0.00737139742518 cm\n", + "Ln=(sqrt(Dn∗tn))= 0.00203469899494 cm\n", + "npo=(no^2/Pp)= 225.0 cmˆ−3\n", + "Ppo=(noˆ2/Nn)= 225000.0 cmˆ−3\n", + "Reverse saturation current ,Io=(((Dp∗Ppo)/(Lp)) + ( ( D n ∗ n p o ) / ( L n ) ) ) ∗ e ∗ A= 3.827628972e-15 A \n", + "Diode forward current , If=Io∗((exp(Vf/Vt))−1)= 4.5032547414e-05 A\n" + ] + } + ], + "source": [ + "#exa 5.7\n", + "from math import sqrt\n", + "from math import exp\n", + "Pp =10**18\n", + "print \"Pp= \",Pp,\"/cmˆ3\" # initializing value of doping concentration in p region.\n", + "Nn =10**15\n", + "print \"Nn= \",Nn,\"/cmˆ3\" # initializing value of doping concentration in n region.\n", + "tp =7*10** -6\n", + "print \"tp = \",tp,\"s\" # initializing value of hole lifetime .\n", + "tn =0.2*10** -6\n", + "print \"tn = \",tn,\"s\" # initializing value of electron lifetime .\n", + "up=800\n", + "print \"up= \",up,\"cm2/Vs\" # initializing value of P side mobility .\n", + "un=300\n", + "print \"un= \",un,\"cm2/Vs\" # initializing value of n side mobility .\n", + "no=1.5*(10**10)\n", + "print \"no = \",no,\"cmˆ−3\" # initializing value of intrinsic concentration .\n", + "Vf=0.6\n", + "print \"Vf = \",Vf,\"V\" # initializing value of forward bias voltage .\n", + "A=100*(10**-6)\n", + "print \"A = \",A,\"mˆ2\"# initializing value of diode cross−sectional area .\n", + "e=1.6*(10**-19)\n", + "print \"e = \",e,\"columns\" # initializing value of charge of electrons .\n", + "K=1.38*(10**-23)\n", + "print \"K = \",K,\"J/k\" # initializing value of boltzmann constant .\n", + "T=300\n", + "print \"T = \",T,\"K\" # initializing value of temperature .\n", + "Vt=(K*T/e)\n", + "print \"Vt=(K∗T/e))=\",Vt,\"eV\" #calculation .\n", + "Dp=Vt*un\n", + "print \"Dp=Vt∗un=\",Dp,\"cmˆ−3\" # calculation .\n", + "Dn=Vt*up\n", + "print \"Dn=Vt∗up=\",Dn,\"cmˆ−3\" # calculation .\n", + "Lp=sqrt(Dp*tp)\n", + "print \"Lp=(sqrt(Dp∗tp))=\",Lp,\"cm\" # calculation .\n", + "Ln=(sqrt(Dn*tn))\n", + "print \"Ln=(sqrt(Dn∗tn))=\",Ln,\"cm\" # calculation .\n", + "npo=(no**2/Pp)\n", + "print \"npo=(no^2/Pp)=\",npo,\"cmˆ−3\" # calculation .\n", + "Ppo=(no**2/Nn)\n", + "print \"Ppo=(noˆ2/Nn)=\",Ppo,\"cmˆ−3\" #calculation .\n", + "Io=(((Dp*Ppo)/(Lp))+((Dn*npo)/(Ln)))*e*A\n", + "print \"Reverse saturation current ,Io=(((Dp∗Ppo)/(Lp)) + ( ( D n ∗ n p o ) / ( L n ) ) ) ∗ e ∗ A= \",Io,\" A \" #calculation .\n", + "If=Io*((exp(Vf/Vt))-1)\n", + "print \"Diode forward current , If=Io∗((exp(Vf/Vt))−1)=\",If,\"A\" # calculation .\n", + "#//the value of Io(reverse saturation current ),after calculation is provided wrong in the book.Due to which If (diode forward current )also differ.\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_8 pgno: 144" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na = 40000000000000000 cmˆ−3\n", + "Nd = 20000000000000000000 cmˆ−3\n", + "no = 15000000000.0 cmˆ−3\n", + "e = 1.6e-19 columns\n", + "K = 1.38e-23 J/k\n", + "T = 300 K\n", + "Built in potential potential ,Vbi=((K∗T/e)∗log((Na∗Nd) /(no) ˆ2) )= 0.926513569765 V\n" + ] + } + ], + "source": [ + "#exa 5.8\n", + "Na =4*10**16\n", + "print \"Na = \",Na,\"cmˆ−3\" # initializing value of acceptor concentration .\n", + "Nd =2*10**19\n", + "print \"Nd = \",Nd,\"cmˆ−3\" # initializing value of donor concentration .\n", + "no=1.5*10**10\n", + "print \"no = \",no,\"cmˆ−3\" # initializing value of intrinsic carrier concentration .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\"columns\" # initializing value of charge of electrons .\n", + "K=1.38*10**-23\n", + "print \"K = \",K,\"J/k\" # initializing value of boltzmann constant .\n", + "T=300\n", + "print \"T = \",T,\"K\" # initializing value of temperature .\n", + "Vbi=((K*T/e)*log((Na*Nd)/(no)**2))\n", + "print \"Built in potential potential ,Vbi=((K∗T/e)∗log((Na∗Nd) /(no) ˆ2) )=\",Vbi,\"V\" # calculation\n", + "#The value used for Nd in the book for solution is different than provided in the question .\n", + "#I have used the value provided in the solution(i.eNd=2∗10ˆ19)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_9 pgno: 144" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na = 40000000000000000 cmˆ−3\n", + "Nd = 20000000000000000000 cmˆ−3\n", + "no = 1800000.0 cmˆ−3\n", + "e = 1.6e-19 columns\n", + "K = 1.38e-23 J/k\n", + "T = 300 K\n", + "Built in potential potential ,Vbi=((K∗T/e)∗log((Na∗Nd) /(no) ˆ2) )= 1.39371354345 V\n" + ] + } + ], + "source": [ + "#exa 5.9\n", + "Na =4*10**16\n", + "print \"Na = \",Na,\"cmˆ−3\" # initializing value of acceptor concentration .\n", + "Nd =2*10**19\n", + "print \"Nd = \",Nd,\"cmˆ−3\" # initializing value of donor concentration .\n", + "no=1.8*10**6\n", + "print \"no = \",no,\"cmˆ−3\" # initializing value of intrinsic carrier concentration .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\"columns\" # initializing value of charge of electrons .\n", + "K=1.38*10**-23\n", + "print \"K = \",K,\"J/k\" # initializing value of boltzmann constant .\n", + "T=300\n", + "print \"T = \",T,\"K\" # initializing value of temperature .\n", + "Vbi=((K*T/e)*log((Na*Nd)/(no)**2))\n", + "print \"Built in potential potential ,Vbi=((K∗T/e)∗log((Na∗Nd) /(no) ˆ2) )= \",Vbi,\"V\" # calculation .\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_10 pgno: 144" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na= 1e+17 /cmˆ3\n", + "Nd= 1e+19 /cmˆ3\n", + "Vbi = 0.64 V\n", + "e = 1.6e-19 columns\n", + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + " total permittivity ,E=Eo∗Er= 1.053626e-12 F/cm\n", + "W=sqrt((2∗E∗Vbi/e)∗((Nd+Na)/(Na∗Nd))))= 9.22675353524e-06 cm\n", + "xn=((W∗Na) /(Nd+Na) ) )= 9.13539953984e-08 cm\n", + "xp=((W∗Nd) /(Nd+Na) ) )= 9.13539953984e-06 cm\n", + "Emax=(−e∗Nd∗xn)/E)= -138727.017592 V/cm\n" + ] + } + ], + "source": [ + "# exa 5.10\n", + "from math import sqrt\n", + "Na =10e16\n", + "print \"Na= \",Na,\"/cmˆ3\" # initializing value of medium p doping concentration .\n", + "Nd =10e18\n", + "print \"Nd= \",Nd,\"/cmˆ3\" # initializing value of light n doping .\n", + "Vbi =0.64\n", + "print \"Vbi = \",Vbi,\"V\" # initializing value of built in voltage .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\"columns\" # initializing value of charge of electrons .\n", + "Er=11.9\n", + "print \"Er = \",Er # initializing value of relative dielectric permittivity constant .\n", + "Eo=8.854*10**-14\n", + "print \"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "E=Eo*Er\n", + "print \" total permittivity ,E=Eo∗Er= \",E,\" F/cm\" #calculation .\n", + "W=sqrt((2*E*Vbi/e)*((Nd+Na)/(Na*Nd)))\n", + "print \"W=sqrt((2∗E∗Vbi/e)∗((Nd+Na)/(Na∗Nd))))=\",W,\" cm\" # calculation .\n", + "xn=((W*Na)/(Nd+Na))\n", + "print \"xn=((W∗Na) /(Nd+Na) ) )=\",xn,\"cm\" # calculation .\n", + "xp=((W*Nd)/(Nd+Na))\n", + "print \"xp=((W∗Nd) /(Nd+Na) ) )=\",xp,\"cm\" # calculation .\n", + "Emax=(-e*Nd*xn)/E\n", + "print \"Emax=(−e∗Nd∗xn)/E)=\",Emax,\"V/cm\" #calculation .\n", + "# The value and unit of W(depletion width) ,provided after calculation in the book is wrong.Due to this xn,xp ,Emax also differ.\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_12 pgno: 145" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na = 10000000000000000 /cmˆ3\n", + "Nd = 1000000000000000000 /cmˆ3\n", + "Vbi = 0.64 V\n", + "Vr = 20 V\n", + "e = 1.6e-19 columns\n", + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + " total permittivity ,E=Eo∗Er= 1.053626e-12 F/cm\n", + "Emax=−(sqrt(((2∗e∗(Vbi+Vr))/(E))∗((Nd∗Na)/(Na+Nd)))))= -249129.931857 V/cm\n" + ] + } + ], + "source": [ + "#exa 5.12\n", + "Na =10**16\n", + "print \"Na = \",Na,\"/cmˆ3\" # initializing value of medium p doping concentration .\n", + "Nd =10**18\n", + "print \"Nd = \",Nd,\"/cmˆ3\" # initializing value of light n doping .\n", + "Vbi =0.64\n", + "print \"Vbi = \",Vbi,\"V\" # initializing value of built in voltage .\n", + "Vr=20\n", + "print \"Vr = \",Vr,\"V\" # initializing value of applied reverse voltage .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\"columns\" # initializing value of charge of electrons .\n", + "Er=11.9\n", + "print \"Er = \",Er # initializing value of relative dielectric permittivity constant .\n", + "Eo=8.854*10**-14\n", + "print \"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "E=Eo*Er\n", + "print \" total permittivity ,E=Eo∗Er= \",E,\" F/cm\" #calculation .\n", + "Emax=-(sqrt(((2*e*(Vbi+Vr))/(E))*((Nd*Na)/(Na+Nd))))\n", + "print \"Emax=−(sqrt(((2∗e∗(Vbi+Vr))/(E))∗((Nd∗Na)/(Na+Nd)))))= \",Emax,\"V/cm\" #calculation .\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_13 pgno: 146" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Emax = 200000 V/cm\n", + "Nd= 1000000000000000000 /cmˆ3\n", + "Vbi = 0.54 V\n", + "Vr = 20 V \n", + "e = 1.6e-19 columns\n", + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + "total permittivity ,E=Eo∗Er= 1.053626e-12 F/cm\n", + "Na=(Emaxˆ2∗E∗Nd)/((2∗e∗(Vbi+Vr)∗Nd)−(Emaxˆ2∗E)) )= 6.45341703981e+15 cmˆ−3\n" + ] + } + ], + "source": [ + "#exa 5.13\n", + "Emax =2*10**5\n", + "print \"Emax = \",Emax,\"V/cm\" # initializing value of maximum electric field .\n", + "Nd=1*10**18\n", + "print \"Nd= \",Nd,\"/cmˆ3\" # initializing value of donor concentration .\n", + "Vbi=0.54\n", + "print \"Vbi = \",Vbi,\"V\" # initializing value of built in voltage .\n", + "Vr=20\n", + "print \"Vr = \",Vr,\"V \" #initializing value of applied reverse voltage .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\"columns\" # initializing value of charge of electrons .\n", + "Er=11.9\n", + "print \"Er = \",Er # initializing value of relative dielectric permittivity constant .\n", + "Eo=8.854*10**-14\n", + "print \"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "E=Eo*Er\n", + "print \"total permittivity ,E=Eo∗Er=\",E,\" F/cm\" #calculation .\n", + "Na=((Emax**2)*E*Nd)/((2*e*(Vbi+Vr)*Nd)-((Emax**2)*E))\n", + "print \"Na=(Emaxˆ2∗E∗Nd)/((2∗e∗(Vbi+Vr)∗Nd)−(Emaxˆ2∗E)) )= \",Na,\"cmˆ−3\" # calculation .\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_14 pgno: 146" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na = 10000000000000000 /cmˆ3\n", + "Nd = 1000000000000000000 /cmˆ3\n", + "A = 1 cmˆ2\n", + "Vj = 0.54 V\n", + "Va = 10 V\n", + "e = 1.6e-19 columns\n", + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + "total permittivity ,E=Eo∗Er= 1.053626e-12 F/cm\n", + "Cj=sqrt((e∗E∗Aˆ2/(2∗(Va+Vj))∗((Na∗Nd)/(Na+Nd))))= 8.89830403817e-09 f/cmˆ2\n" + ] + } + ], + "source": [ + "#exa 5.14\n", + "Na =10**16\n", + "print \"Na = \",Na,\"/cmˆ3\" # initializing value of acceptor concentration .\n", + "Nd =10**18\n", + "print \"Nd = \",Nd,\"/cmˆ3\" # initializing value of donor concentration .\n", + "A=1\n", + "print \"A = \",A,\"cmˆ2\" # initializing value of area for finding junction capacitance per unit area.\n", + "Vj =0.54\n", + "print \"Vj =\",Vj,\"V\" # initializing value of built in voltage .\n", + "Va=10\n", + "print \"Va = \",Va,\"V\" # initializing value of applied reverse voltage .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\"columns\" # initializing value of charge of electrons .\n", + "Er=11.9\n", + "print \"Er = \",Er # initializing value of relative dielectric permittivity constant .\n", + "Eo=8.854*10**-14\n", + "print \"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "E=Eo*Er\n", + "print \"total permittivity ,E=Eo∗Er= \",E,\" F/cm\" # calculation .\n", + "Cj=sqrt((e*E*A**2/(2*(Va+Vj)))*((Na*Nd)/(Na+Nd)))\n", + "print \"Cj=sqrt((e∗E∗Aˆ2/(2∗(Va+Vj))∗((Na∗Nd)/(Na+Nd))))=\",Cj,\"f/cmˆ2\" # calculation .\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_15 pgno: 146" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na= 1000000000000000 cmˆ−3\n", + "Nd= 1000000000000000000 cmˆ−3\n", + "no = 1800000.0 cmˆ−3\n", + "e = 1.6e-19 columbs\n", + "K = 1.38e-23 J/k\n", + "T = 300 K\n", + "Built in potential potential ,Vbi=((K∗T/e)∗log((Na∗Nd)/(no)ˆ2))= 1.220749215 V\n" + ] + } + ], + "source": [ + "#exa 5.15\n", + "Na =10**15\n", + "print \"Na= \",Na,\"cmˆ−3\" # initializing value of acceptor concentration .\n", + "Nd =10**18\n", + "print \"Nd= \",Nd,\" cmˆ−3\" # initializing value of donor concentration .\n", + "no=1.8*10**6\n", + "print \"no = \",no,\" cmˆ−3\" # initializing value of intrinsic carrier concentration .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\"columbs\" # initializing value of charge of electrons .\n", + "K=1.38*10**-23\n", + "print \"K = \",K,\" J/k\" # initializing value of boltzmann constant .\n", + "T=300\n", + "print \"T = \",T,\" K\" # initializing value of temperature .\n", + "Vbi=((K*T/e)*log((Na*Nd)/(no)**2))\n", + "print \"Built in potential potential ,Vbi=((K∗T/e)∗log((Na∗Nd)/(no)ˆ2))= \",Vbi,\"V\" # calculation .\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_16 pgno: 146" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na= 1000000000000000000 cmˆ−3\n", + "Nd= 1000000000000000000 cmˆ−3\n", + "Vbi = 1.4\n", + "e = 1.6e-19 columbs\n", + "K = 1.38e-23 J/k\n", + "T = 300 K\n", + "Vt = 0.0259 eV\n", + "no=sqrt ((Na∗Nd)/(exp(Vbi/Vt)))= 1829411.05814 cmˆ−3\n" + ] + } + ], + "source": [ + "#5.16\n", + "Na =10**18\n", + "print \"Na= \",Na, \"cmˆ−3\" # initializing value of acceptor concentration .\n", + "Nd =10**18\n", + "print \"Nd= \",Nd,\" cmˆ−3\" # initializing value of donor concentration .\n", + "Vbi =1.4\n", + "print \"Vbi = \",Vbi # initializing value of built in voltage .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\" columbs\" # initializing value of charge of electrons .\n", + "K=1.38*10**-23\n", + "print \"K = \",K,\"J/k\" # initializing value of boltzmann constant .\n", + "T=300\n", + "print \"T = \",T,\" K\" # initializing value of temperature .\n", + "Vt=0.0259\n", + "print \"Vt = \",Vt,\" eV\" # initializing value of thermal voltage .\n", + "no=sqrt((Na*Nd)/(exp(Vbi/Vt)))\n", + "print \"no=sqrt ((Na∗Nd)/(exp(Vbi/Vt)))=\",no,\"cmˆ−3\" #calculation ." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_18 pgno: 147" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na = 100000000000000000 cmˆ−3\n", + "Nd = 50000000000000000 cmˆ−3\n", + "e = 1.6e-19 columbs\n", + "no = 15000000000.0 cmˆ3\n", + "T = 300 K\n", + "Vt = 0.0259 eV\n", + "(a)Vbi=(Vt∗(log(Na∗Nd/(noˆ2))))= 0.795961750143 V\n", + "(b)value of fermi level on each side of junction ,Efi Efp=(Vt∗log(Na/(no)))= 0.40695713106 V\n", + "Efn Efi=(Vt∗log(Nd/(no)))= 0.389004619083 V\n", + "(c)The energy band digram is similar to Fig P5 .3\n", + "(d)Vbi=((Efi Efp)+(Efn Efi))/(e)=Vj= 0.795961750143 V\n" + ] + } + ], + "source": [ + "#exa 5.18\n", + "Na =10**17\n", + "print \"Na = \",Na,\" cmˆ−3\" # initializing value of acceptor concentration .\n", + "Nd =5*10**16\n", + "print \"Nd = \",Nd,\" cmˆ−3\" # initializing value of donor concentration .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\" columbs\" # initializing value of charge of electrons .\n", + "no=1.5*10**10\n", + "print \"no = \",no,\" cmˆ3\" # initializing value of intrinsic carrier concentration .\n", + "T=300\n", + "print \"T = \",T,\" K\" # initializing value of temperature .\n", + "Vt=0.0259\n", + "print \"Vt = \",Vt,\" eV\" # initializing value of thermal voltage .\n", + "Vbi=(Vt*(log(Na*Nd/(no**2))))\n", + "print \"(a)Vbi=(Vt∗(log(Na∗Nd/(noˆ2))))= \",Vbi,\" V\" #calculation .\n", + "Efi_Efp=(Vt*log(Na/(no)))\n", + "print \"(b)value of fermi level on each side of junction ,Efi Efp=(Vt∗log(Na/(no)))=\",Efi_Efp,\" V\" # calculation .\n", + "Efn_Efi=(Vt*log(Nd/(no)))\n", + "print \"Efn Efi=(Vt∗log(Nd/(no)))=\",Efn_Efi,\" V\" #calculation .\n", + "print \"(c)The energy band digram is similar to Fig P5 .3\"\n", + "Vbi=((Efi_Efp)+(Efn_Efi))\n", + "print \"(d)Vbi=((Efi Efp)+(Efn Efi))/(e)=Vj=\",Vbi,\" V\" # calculation .\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_19 pgno: 148" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na = 500000000000000000 /cmˆ3\n", + "Nd = 500000000000000000 /cmˆ3\n", + "no = 15000000000.0 cmˆ−3\n", + "e = 1.6e-19 columbs\n", + "K = 1.38e-23 J/k\n", + "T = 300 K\n", + "(a)Built in potential potential ,Vbi=((K∗T/e)∗log ((Na∗Nd)/(no)ˆ2))= 0.896417042561 eV\n", + "(b)fermi level in the P region and N region , Efi Efp=((KT/e)∗log(Na/no)))= 0.44820852128 eV\n", + "(c)VBI from the fermi level ,VBI=2∗(Efi Efp))= 0.896417042561 V\n" + ] + } + ], + "source": [ + "#exa 5.19\n", + "from math import log\n", + "Na =5*10**17\n", + "print \"Na = \",Na,\"/cmˆ3\" # initializing value of medium p doping concentration .\n", + "Nd =5*10**17\n", + "print \"Nd = \",Nd,\"/cmˆ3\" # initializing value of light n doping .\n", + "no=1.5*10**10\n", + "print \"no = \",no,\"cmˆ−3\" # initializing value of intrinsic concentration .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\"columbs\" # initializing value of charge of electrons .\n", + "K=1.38*10**-23\n", + "print \"K = \",K,\"J/k\" # initializing value of boltzmann constant .\n", + "T=300\n", + "print \"T = \",T,\"K\" # initializing value of temperature .\n", + "Vbi=((K*T/e)*log((Na*Nd)/(no)**2))\n", + "print \"(a)Built in potential potential ,Vbi=((K∗T/e)∗log ((Na∗Nd)/(no)ˆ2))=\",Vbi,\"eV\" #calculation .\n", + "Efi_Efp=((K*T/e)*log(Na/no))\n", + "print \"(b)fermi level in the P region and N region , Efi Efp=((KT/e)∗log(Na/no)))= \",Efi_Efp,\"eV\" # calculation .\n", + "VBI=2*(Efi_Efp)\n", + "print \"(c)VBI from the fermi level ,VBI=2∗(Efi Efp))=\",VBI,\"V\" # calculation ." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_20 pgno: 148" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nc = 2.8e+19 /cmˆ3\n", + "Nv = 1.04e+19 /cmˆ3\n", + "no = 15000000000.0 cmˆ−3\n", + "e = 1.6e-19 columbs\n", + "K = 8.62e-05 J/k\n", + "T = 300 K\n", + "Vt = 0.0259 eV\n", + "Ec Ef = 0.21 eV\n", + "Ef Ev = 0.18 eV\n", + "Nd=(Nc/exp((Ec−Ef)/(K∗T))))= 8.32539212771e+15 cmˆ−3\n", + "Na=(Nv/exp (( Ef−Ev) /(K∗T) ) ) )= 9.86510951303e+15 cmˆ−3\n", + "Built in potential ,Vbi=(Vt∗(log(Na∗Nd/(noˆ2))))= 0.68954178887 V\n" + ] + } + ], + "source": [ + "#exa 5.20\n", + "Nc=2.8*10**19\n", + "print \"Nc = \",Nc,\" /cmˆ3\" # initializing value of number of electron in the conduction band .\n", + "Nv=1.04*10**19\n", + "print \"Nv = \",Nv,\" /cmˆ3\" # initializing value of number of electron in the valence band..\n", + "no=1.5*10**10\n", + "print \"no = \",no,\" cmˆ−3\" # initializing value of intrinsic carrier concentration .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\" columbs\" # initializing value of charge of electrons .\n", + "K=8.62*10**-5\n", + "print \"K = \",K,\" J/k\" # initializing value of boltzmann constant .\n", + "T=300\n", + "print \"T = \",T,\" K\" # initializing value of temperature .\n", + "Vt=0.0259\n", + "print \"Vt = \",Vt,\" eV\" # initializing value of thermal voltage .\n", + "Ec_Ef =0.21\n", + "print \"Ec Ef = \",Ec_Ef,\" eV\" # initializing value of energy difference between conduction band and fermi level.\n", + "Ef_Ev =0.18\n", + "print \"Ef Ev = \",Ef_Ev,\" eV\" # initializing value of energy difference between fermi level and valence band .\n", + "Nd=(Nc/exp((Ec_Ef)/(K*T)))\n", + "print \"Nd=(Nc/exp((Ec−Ef)/(K∗T))))= \",Nd,\" cmˆ−3\" #calculation .\n", + "Na=(Nv/exp((Ef_Ev)/(K*T)))\n", + "print \"Na=(Nv/exp (( Ef−Ev) /(K∗T) ) ) )=\",Na,\"cmˆ−3\" # calculation .\n", + "Vbi=(Vt*(log(Na*Nd/(no**2))))\n", + "print \"Built in potential ,Vbi=(Vt∗(log(Na∗Nd/(noˆ2))))= \",Vbi,\" V\" #calculation ." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_21 pgno: 149" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vbi = 1.2 /cmˆ3\n", + "no = 1800000.0 cmˆ−3”\n", + "Vt = 0.0259 eV\n", + "Er = 13.1\n", + "Eo = 8.854e-14 F/cm\n", + "e = 1.6e-19 columbs\n", + " total permittivity ,E=Eo∗Er= 1.159874e-12 F/cm\n", + "(a)NaNd=((noˆ2)∗(exp(Vbi/Vt)))= 4.28841757806e+32 /cmˆ6\n", + "Na=(sqrt(NaNd/(4)))= 1.03542474112e+16 /cmˆ3\n", + "(b)Nd=4∗Na= 4.14169896446e+16 /cmˆ3\n", + "(c)W=sqrt((2∗E∗Vbi/e)∗((Nd+Na)/(Na∗Nd))))= 4.58296745959e-05 cm\n", + "(d)xn=0.2∗W= 9.16593491919e-06 cm\n", + "xp=0.8∗W= 3.66637396767e-05 cm\n", + "(e)Emax=(−e∗Nd∗xn)/E)= -52367.8167292 V/cm\n" + ] + } + ], + "source": [ + "#exa 5.21\n", + "from math import sqrt\n", + "from math import exp\n", + "Vbi =1.2\n", + "print \"Vbi = \",Vbi,\"/cmˆ3\" # initializing value of built in voltage .\n", + "no=1.8*10**6\n", + "print \"no = \",no,\"cmˆ−3”\" # initializing value of intrinsic concentration .\n", + "Vt =0.0259\n", + "print \"Vt = \",Vt,\"eV\" # initializing value of thermal voltage .\n", + "Er =13.1\n", + "print \"Er = \",Er # initializing value of relative dielectric permittivity constant .\n", + "Eo=8.854*10**-14\n", + "print \"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "e =1.6*(10**-19)\n", + "print \"e = \",e,\" columbs\" # initializing value of charge of electrons .\n", + "E=Eo*Er\n", + "print \" total permittivity ,E=Eo∗Er=\",E,\" F/cm\" # calculation .\n", + "NaNd=((no**2)*(exp(Vbi/Vt)))\n", + "print \"(a)NaNd=((noˆ2)∗(exp(Vbi/Vt)))= \",NaNd,\" /cmˆ6\" # calculation .\n", + "Na=(sqrt(NaNd/(4)))\n", + "print \"Na=(sqrt(NaNd/(4)))=\",Na,\" /cmˆ3\" # calculation .\n", + "Nd=4*Na\n", + "print \"(b)Nd=4∗Na= \",Nd,\" /cmˆ3\" # calculation.\n", + "W=sqrt((2*E*Vbi/e)*((Nd+Na)/(Na*Nd)))\n", + "print \"(c)W=sqrt((2∗E∗Vbi/e)∗((Nd+Na)/(Na∗Nd))))= \",W,\" cm\" # calculation .\n", + "xn=0.2*W\n", + "print \"(d)xn=0.2∗W= \",xn,\" cm\" # calculation .\n", + "xp=0.8*W\n", + "print \"xp=0.8∗W= \",xp,\" cm\" # calculation .\n", + "Emax=(-e*Nd*xn)/E\n", + "print \"(e)Emax=(−e∗Nd∗xn)/E)= \",Emax,\"V/cm\" # calculation .\n", + "#The value of Na after calculation is provided wrong in the book.Due to which value of W,xn,xp and Emax differ ,than the answer provided in the book ." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_22 pgno: 150" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na = 10000000000000000 cmˆ−3\n", + "Nd = 5000000000000000 cmˆ−3\n", + "no = 15000000000.0 cmˆ−3\n", + "Vbi = 0.676 V\n", + "e = 1.6e-19 columns\n", + "K = 1.38e-23 J/k\n", + "T=(Vbi∗(e/K)∗(1/(log((Na∗Nd)/(noˆ2))))))= 299.984615257 K\n" + ] + } + ], + "source": [ + "#exa 5.22\n", + "from math import log\n", + "Na =10**16\n", + "print \"Na = \",Na,\"cmˆ−3\" # initializing value of acceptor concentration .\n", + "Nd =5*10**15\n", + "print \"Nd = \",Nd,\"cmˆ−3\" # initializing value of donor concentration .\n", + "no=1.5*10**10\n", + "print \"no = \",no,\"cmˆ−3\" # initializing value of intrinsic concentration .\n", + "Vbi =0.676\n", + "print \"Vbi = \",Vbi,\"V\" # initializing value of built in voltage .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\"columns\" # initializing value of charge of electrons .\n", + "K=1.38*10**-23\n", + "print \"K = \",K,\"J/k\" # initializing value of boltzmann constant .\n", + "T=(Vbi*(e/K)*(1/(log((Na*Nd)/(no**2)))))\n", + "print \"T=(Vbi∗(e/K)∗(1/(log((Na∗Nd)/(noˆ2))))))= \",T,\"K\" # calculation ." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_23 pgno: 150" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na = 500000000000000000 cmˆ−3\n", + "Nd = 100000000000000000 cmˆ−3\n", + "no = 15000000000.0 cmˆ−3\n", + "e = 1.6e-19 columns\n", + "K = 1.38e-23 J/k\n", + "T = 300 K\n", + "VBI = 0.847 V\n", + "(a)Built in potential potential ,Vbi=((K∗T/e)∗log ((Na∗Nd)/(no)ˆ2))= 0.854772836577 V\n", + "(b)T=(VBI∗(e/K)∗(1/(log((Na∗Nd)/(noˆ2))))))= 297.271964113 K\n" + ] + } + ], + "source": [ + "#exa 5.23\n", + "Na =5*10**17\n", + "print \"Na = \",Na,\"cmˆ−3\" # initializing value of acceptor concentration .\n", + "Nd =10**17\n", + "print \"Nd = \",Nd,\"cmˆ−3\" # initializing value of donor concentration .\n", + "no=1.5*10**10\n", + "print \"no = \",no,\"cmˆ−3\" # initializing value of intrinsic concentration .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\"columns\" # initializing value of charge of electrons .\n", + "K=1.38*10**-23\n", + "print \"K = \",K,\"J/k\" # initializing value of boltzmann constant .\n", + "T=300\n", + "print \"T = \",T,\"K\" # initializing value of temperature .\n", + "VBI=0.847\n", + "print \"VBI = \",VBI,\"V\" # initializing value of VBI when VBI is reduced by 1%.\n", + "Vbi=((K*T/e)*log((Na*Nd)/(no)**2))\n", + "print \"(a)Built in potential potential ,Vbi=((K∗T/e)∗log ((Na∗Nd)/(no)ˆ2))= \",Vbi,\"V\" # calculation .\n", + "T=(e*VBI/K)*((log(Na*Nd/(no**2)))**-1)\n", + "print \"(b)T=(VBI∗(e/K)∗(1/(log((Na∗Nd)/(noˆ2))))))= \",T,\"K\" # calculation .\n", + "#the answer for part (b) is not provided in the book ." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_24 pgno: 150" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false, + "slideshow": { + "slide_type": "subslide" + } + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na = 4e+12 /cmˆ3\n", + "Nd = 4e+16 /cmˆ3\n", + "no = 1.5e+11 /cmˆ3\n", + "K = 1.38e-23 J/k\n", + "T = 300 K\n", + "e = 1.6e-19 columbs\n", + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + "total permittivity ,E=Eo∗Er= 1.053626e-12 F/cm\n", + "Built in potential potential ,Vbi=((K∗T/e)∗log((Na∗Nd)/(no)ˆ2))= 0.408234249531 eV\n", + "W=sqrt((2∗E∗Vbi/e)∗((Nd+Na)/(Na∗Nd))))= 0.00115943039897 cm\n", + "xn=((W∗Na) /(Nd+Na) ) )= 1.15931446752e-07 cm\n", + "xp=((W∗Nd) /(Nd+Na) ) )= 0.00115931446752 cm\n", + "Emax=(e∗Nd∗xn)/E)= 704.19794046 V/cm\n" + ] + } + ], + "source": [ + "#exa 5.24\n", + "from math import log\n", + "from math import sqrt\n", + "Na =4e12\n", + "print \"Na = \",Na,\" /cmˆ3\" # initializing value of medium p doping concentration .\n", + "Nd =4e16\n", + "print \"Nd = \",Nd,\" /cmˆ3\" # initializing value of light n doping.\n", + "no=1.5*10e10\n", + "print \"no = \",no,\" /cmˆ3\" # initializing value of intrinsic carrier concentration .\n", + "K=1.38e-23\n", + "print \"K = \",K,\" J/k\" # initializing value of boltzmann constant .\n", + "T=300\n", + "print \"T = \",T,\" K\" # initializing value of temperature .\n", + "e=1.6e-19\n", + "print \"e = \",e,\" columbs\" # initializing value of charge of electrons .\n", + "Er=11.9\n", + "print \"Er = \",Er # initializing value of relative dielectric permittivity constant .\n", + "Eo=8.854e-14\n", + "print \"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "E=Eo*Er\n", + "print \"total permittivity ,E=Eo∗Er=\",E,\" F/cm\" # calculation .\n", + "Vbi=((K*T/e)*log((Na*Nd)/(no)**2))\n", + "print \"Built in potential potential ,Vbi=((K∗T/e)∗log((Na∗Nd)/(no)ˆ2))=\",Vbi,\" eV\" #calculation .\n", + "W=sqrt((2.*E*Vbi/e)*((Nd+Na)/(Na*Nd)))\n", + "print\"W=sqrt((2∗E∗Vbi/e)∗((Nd+Na)/(Na∗Nd))))=\",W,\" cm\" # calculation .\n", + "xn=((W*Na)/(Nd+Na))\n", + "print \"xn=((W∗Na) /(Nd+Na) ) )=\",xn,\"cm\" # calculation .\n", + "xp=((W*Nd)/(Nd+Na))\n", + "print \"xp=((W∗Nd) /(Nd+Na) ) )=\",xp,\"cm\" #calculation .\n", + "Emax=(e*Nd*xn)/E\n", + "print \"Emax=(e∗Nd∗xn)/E)=\",Emax,\" V/cm\" #calculation .\n", + "#the value of W( depletion width) , after calculation is provided wrong in the book,due to this xn,xp ,Emax also differ.(also,the value of Nd+Na substitute in the formula for for xn,xp is wrong )" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_25 pgno: 151" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na = 400000000000000000 /cmˆ3\n", + "Nd = 4000000000000000 /cmˆ3\n", + "no = 15000000000.0 cmˆ−3\n", + "Emax = 300000 /cmˆ3\n", + "K = 1.38e-23 J/k\n", + "T = 300 K\n", + "e = 1.6e-19 columns\n", + " Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + " total permittivity ,E=Eo∗Er= 1.053626e-12 F/cm\n", + "Built in potential potential ,Vbi=((K∗T/e)∗log ((Na∗Nd)/(no)ˆ2))= 0.765710585218 V\n", + "xn=(E∗Emax) /( e∗Nd) )= 0.0004938871875 cm\n", + "W=(xn(Nd+Na)/Na))= 0.000498826059375 cm\n", + "Vr=(Wˆ2∗e/(2∗E))∗((Na∗Nd)/(Na+Nd))−(Vbi))= 74.058198321 V\n" + ] + } + ], + "source": [ + "#exa 5.25\n", + "Na =4*10**17\n", + "print \"Na = \",Na,\"/cmˆ3\" # initializing value of donor concentration .\n", + "Nd =4*10**15\n", + "print \"Nd = \",Nd,\"/cmˆ3\" # initializing value of light n doping.\n", + "no=1.5*10**10\n", + "print \"no = \",no,\"cmˆ−3\" # initializing value of intrinsic concentration .\n", + "Emax =300*10**3\n", + "print \"Emax = \",Emax,\"/cmˆ3\" # initializing value of maximum electric field .\n", + "K=1.38*10**-23\n", + "print \"K = \",K,\"J/k\" # initializing value of boltzmann constant .\n", + "T=300\n", + "print \"T = \",T,\"K\" # initializing value of temperature .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\"columns\" # initializing value of charge of electrons .\n", + "Er=11.9\n", + "print \" Er = \",Er # initializing value of relative dielectric permittivity constant .\n", + "Eo=8.854*10**-14\n", + "print \"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "E=Eo*Er\n", + "print \" total permittivity ,E=Eo∗Er=\",E,\" F/cm\" # calculation .\n", + "Vbi=((K*T/e)*log((Na*Nd)/(no)**2))\n", + "print \"Built in potential potential ,Vbi=((K∗T/e)∗log ((Na∗Nd)/(no)ˆ2))=\",Vbi,\" V\" # calculation .\n", + "xn=(E*Emax/(Nd*e))\n", + "print \"xn=(E∗Emax) /( e∗Nd) )=\",xn,\" cm\" #calculation .\n", + "W=(xn*(Nd+Na)/Na)\n", + "print \"W=(xn(Nd+Na)/Na))=\",W,\" cm\" #calculation .\n", + "Vr=((W**2*e/(2*E))*((Na*Nd)/(Na+Nd)))-(Vbi)\n", + "print \"Vr=(Wˆ2∗e/(2∗E))∗((Na∗Nd)/(Na+Nd))−(Vbi))=\",Vr,\" V\" # calculation ." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_26 pgno: 152" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na = 5000000000000000 cmˆ−3\n", + "Nd = 1000000000000000000 cmˆ−3\n", + "e = 1.6e-19 columns\n", + "Vr = 10 V\n", + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + " total permittivity ,E=Eo∗Er= 1.053626e-12 F/cm\n", + "Emax = 1000000 V/cm\n", + "W = 2e-05 cm\n", + "Nd=(Emax∗e)/(W∗e))= 3.29258125e+17 cmˆ−3\n" + ] + } + ], + "source": [ + "#exa 5.26\n", + "Na =5*10**15\n", + "print \"Na = \",Na,\"cmˆ−3\" # initializing value of acceptor concentration .\n", + "Nd =10**18\n", + "print \"Nd = \",Nd,\"cmˆ−3\" # initializing value of donor concentration .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\"columns\" # initializing value of charge of electrons .\n", + "Vr=10\n", + "print \"Vr = \",Vr,\"V\" # initializing value reverse voltage .\n", + "Er=11.9\n", + "print \"Er = \",Er # initializing value of relative dielectric permittivity constant .\n", + "Eo=8.854*10**-14\n", + "print \"Eo = \",Eo,\"F/cm\" # initializing value of permittivity of free space .\n", + "E=Eo*Er\n", + "print \" total permittivity ,E=Eo∗Er=\",E,\" F/cm\" # calculation .\n", + "Emax=10**6\n", + "print \"Emax = \",Emax,\"V/cm\" # initializing value of maximum electric field .\n", + "W=(2.*Vr/(Emax))\n", + "print \"W = \",W,\"cm\" # calculation .\n", + "Nd=(Emax*E)/(W*e)\n", + "print \"Nd=(Emax∗e)/(W∗e))=\",Nd,\"cmˆ−3\" # calculation " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_27 pgno: 152" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na = 5000000000000000 cmˆ3\n", + "Nd = 1000000000000000000 cmˆ3\n", + "no = 15000000000.0 cmˆ−3\n", + "Vr1 = 0 V\n", + "Vr2 = 5 V\n", + "A = 3e-05 cmˆ2\n", + "e = 1.6e-19 columns\n", + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + "total permittivity ,E=Eo∗Er= 1.053626e-12 F/cm\n", + "Vt= 0.0259 V\n", + "Built in potential ,Vbi=(Vt∗log ((Na∗Nd)/(no)ˆ2) )= 0.795961750143 V\n", + "Cj1=sqrt((e∗E∗(Aˆ2)/(2∗(Vr1+Vbi))∗((Na∗Nd)/(Na+Nd))))= 6.88597370389e-13 F\n", + "Cj2=sqrt((e∗E∗(Aˆ2)/(2∗(Vr2+Vbi))∗((Na∗Nd)/(Na+Nd))))= 2.55181216611e-13 F\n" + ] + } + ], + "source": [ + "#exa 5.27\n", + "from math import sqrt\n", + "from math import log\n", + "Na =5*10**15\n", + "print \"Na = \",Na,\"cmˆ3\" # initializing value of acceptor concentration .\n", + "Nd =10**18\n", + "print \"Nd = \",Nd,\"cmˆ3\" # initializing value of donor concentration .\n", + "no=1.5*10**10\n", + "print \"no = \",no,\"cmˆ−3\" # initializing value of intrinsic carrier concentration .\n", + "Vr1=0\n", + "print \"Vr1 = \",Vr1,\"V\" # initializing value of built in voltage .\n", + "Vr2=5\n", + "print \"Vr2 = \",Vr2,\"V\" # initializing value of applied reverse voltage .\n", + "A=3*10**-5\n", + "print \"A = \",A,\"cmˆ2\" # initializing value of cross sectional area .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\"columns\" # initializing value of charge of electrons .\n", + "Er=11.9\n", + "print \"Er = \",Er # initializing value of relative dielectric permittivity constant .\n", + "Eo=8.854*10**-14\n", + "print \"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "E=Eo*Er\n", + "print \"total permittivity ,E=Eo∗Er=\",E,\" F/cm\" # calculation .\n", + "Vt=0.0259\n", + "print \"Vt=\",Vt,\" V\" # initializing the value of thermal voltage .\n", + "Vbi=((Vt)*log((Na*Nd)/(no)**2))\n", + "print \"Built in potential ,Vbi=(Vt∗log ((Na∗Nd)/(no)ˆ2) )= \",Vbi,\" V\" # calculation .\n", + "Cj1=sqrt((e*E*(A**2)/(2*(Vr1+Vbi)))*((Na*Nd)/(Na+Nd)))\n", + "print \"Cj1=sqrt((e∗E∗(Aˆ2)/(2∗(Vr1+Vbi))∗((Na∗Nd)/(Na+Nd))))=\",Cj1,\" F\" #calculation .\n", + "Cj2=sqrt((e*E*(A**2)/(2*(Vr2+Vbi)))*((Na*Nd)/(Na+Nd)))\n", + "print \"Cj2=sqrt((e∗E∗(Aˆ2)/(2∗(Vr2+Vbi))∗((Na∗Nd)/(Na+Nd))))=\",Cj2,\" F\" #calculation .\n", + "# the value of Vr2 use for calculating answer of Cj2 is different than provided in question .\n", + "# I have used the value provided in the solution ( i .e. Vr2=5)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_28 pgno: 153" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na = 10000000000000000 cmˆ3\n", + "Nd = 50000000000000000 cmˆ3\n", + "no = 15000000000.0 cmˆ−3\n", + "Dn = 25 cmˆ2/sec\n", + "Dp = 10 cmˆ2/sec\n", + "tn = 5e-07 s\n", + "tp = 5e-07 s\n", + "e = 1.6e-19 columns\n", + "Pno=(noˆ2/Nd))= 4500.0 cmˆ−3\n", + "Npo=(noˆ2/Na))= 22500.0 cmˆ−3\n", + "Lp=(sqrt(Dp∗tp)))= 0.0022360679775 cm\n", + "Ln=(sqrt(Dn∗tn)))= 0.00353553390593 cm\n", + "Jo=((e ∗((Dp∗Pno/(Lp) )+(Dn∗Npo) /(Ln) ) ) )= 2.86757820103e-11 A/cmˆ2\n" + ] + } + ], + "source": [ + "#exa 5.28\n", + "from math import sqrt\n", + "Na =1*10**16\n", + "print \"Na = \",Na,\"cmˆ3\" # initializing value of acceptor concentration. \n", + "Nd = 5*10**16\n", + "print \"Nd = \",Nd,\"cmˆ3\" # initializing value of donor concentration .\n", + "no=1.5*10**10\n", + "print \"no = \",no,\"cmˆ−3\" # initializing value of intrinsic concentration .\n", + "Dn=25\n", + "print \"Dn = \",Dn,\"cmˆ2/sec\" # initializing value of diffusion cofficient on the P side.\n", + "Dp=10\n", + "print \"Dp = \",Dp,\"cmˆ2/sec\" # initializing value of diffusion cofficient on the N side .\n", + "tp =5*10** -7\n", + "print \"tn = \",tp,\"s\" # initializing value of hole lifetime .\n", + "tn =5*10** -7\n", + "print \"tp = \",tn,\"s\" # initializing value of electron lifetime .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\"columns\" # initializing value of charge of electrons .\n", + "Pno=(no**2/Nd)\n", + "print \"Pno=(noˆ2/Nd))= \",Pno,\"cmˆ−3\" # calculation .\n", + "Npo=(no**2/Na)\n", + "print \"Npo=(noˆ2/Na))= \",Npo,\"cmˆ−3\" # calculation .\n", + "Lp=(sqrt(Dp*tp))\n", + "print \"Lp=(sqrt(Dp∗tp)))= \",Lp,\"cm\" # calculation .\n", + "Ln=(sqrt(Dn*tn))\n", + "print \"Ln=(sqrt(Dn∗tn)))= \",Ln,\"cm\" # calculation .\n", + "Jo=((e*((Dp*Pno/(Lp))+(Dn*Npo)/(Ln))))\n", + "print \"Jo=((e ∗((Dp∗Pno/(Lp) )+(Dn∗Npo) /(Ln) ) ) )= \",Jo,\" A/cmˆ2\" # calculation ." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_29 pgno: 154" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na = 1000000000000000 cmˆ3\n", + "Nd = 1000000000000000 cmˆ3\n", + "no = 15000000000.0 cmˆ−3\n", + "Dn = 50 cmˆ2/sec\n", + "Dp = 20 cmˆ2/sec\n", + "tn = 5e-07 s\n", + "tp = 5e-07 s\n", + "e = 1.6e-19 columns\n", + "Pno=(noˆ2/Nd))= 225000.0 cmˆ−3\n", + "Npo=(noˆ2/Na))= 225000.0 cmˆ−3\n", + "Lp=(sqrt(Dp∗tp)))= 0.00316227766017 cm\n", + "Ln=(sqrt(Dn∗tn)))= 0.005 cm\n", + "Jo=((e ∗((Dp∗Pno/(Lp) )+(Dn∗Npo) /(Ln))))= 5.87683991532e-10 A/cmˆ2\n" + ] + } + ], + "source": [ + "#exa 5.29\n", + "from math import sqrt\n", + "Na =10**15\n", + "print \"Na = \",Na,\"cmˆ3\" # initializing value of acceptor concentration .\n", + "Nd =10**15\n", + "print \"Nd = \",Nd,\"cmˆ3\" # initializing value of donor concentration .\n", + "no=1.5*10**10\n", + "print \"no = \",no,\"cmˆ−3\" # initializing value of intrinsic carrier concentration .\n", + "Dn=50\n", + "print \"Dn = \",Dn,\"cmˆ2/sec\" # initializing value of built in voltage .\n", + "Dp=20\n", + "print \"Dp = \",Dp,\"cmˆ2/sec\" # initializing value of applied reverse voltage .\n", + "tp =5*10** -7\n", + "print \"tn = \",tp,\"s\" # initializing value of hole lifetime .\n", + "tn =5*10** -7\n", + "print \"tp = \",tn,\"s\" # initializing value of electrons lifetime .\n", + "e=1.6*10**-19\n", + "print \"e = \",e,\"columns\" # initializing value of charge of electrons .\n", + "Pno=(no**2/Nd)\n", + "print \"Pno=(noˆ2/Nd))= \",Pno,\"cmˆ−3\" # calculation .\n", + "Npo=(no**2/Na)\n", + "print \"Npo=(noˆ2/Na))= \",Npo,\"cmˆ−3\" # calculation .\n", + "Lp=(sqrt(Dp*tp))\n", + "print \"Lp=(sqrt(Dp∗tp)))= \",Lp,\"cm\" # calculation .\n", + "Ln=(sqrt(Dn*tn))\n", + "print \"Ln=(sqrt(Dn∗tn)))= \",Ln,\"cm\" # calculation .\n", + "Jo=((e*((Dp*Pno/(Lp))+(Dn*Npo)/(Ln))))\n", + "print \"Jo=((e ∗((Dp∗Pno/(Lp) )+(Dn∗Npo) /(Ln))))=\",Jo,\"A/cmˆ2\" # calculation .\n", + "# the value of tp , tn provided in the question , is different than that provided in the solution.\n", + "# I have used the value ,provided in the solution(i. e . tp=tn =5∗10ˆ7)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_30 pgno: 154" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Eg = -1.1 V\n", + "Vf1 = 0.6 V\n", + "T1 = 300 K\n", + "T2 = 310 K\n", + "Forward voltage for case 2,Vf2=((Eg+Vf1)∗T2)/( T1)+Eg)= 0.583333333333 V\n" + ] + } + ], + "source": [ + "#exa 5.30\n", + "Eg = -1.1\n", + "print \"Eg = \",Eg,\"V\" # initializing value of energy gap .\n", + "Vf1 =0.6\n", + "print \"Vf1 = \",Vf1,\"V\" # initializing value of forward voltage for case 1.\n", + "T1 =300\n", + "print \"T1 = \",T1,\"K\" # initializing value of temperature for case 1.\n", + "T2 =310\n", + "print \"T2 = \",T2,\"K\" # initializing value of temperature for case 2 .\n", + "Vf2=(((Eg+Vf1)*T2)/(T1))-Eg\n", + "print \"Forward voltage for case 2,Vf2=((Eg+Vf1)∗T2)/( T1)+Eg)= \",Vf2,\" V\" # calculation .\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_7_BIPOLAR_JUNCTION_TRANSISTORB_1.ipynb b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_7_BIPOLAR_JUNCTION_TRANSISTORB_1.ipynb new file mode 100644 index 00000000..bbe984b6 --- /dev/null +++ b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_7_BIPOLAR_JUNCTION_TRANSISTORB_1.ipynb @@ -0,0 +1,496 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# chapter 7 BIPOLAR JUNCTION TRANSISTORB" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7_1 pgno: 220" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Dnb = 20 cmˆ2/s\n", + "nB = 10000 /cmˆ3\n", + "xB = 1e-06 m\n", + "AB = 0.0001 cmˆ2\n", + "e = 1.6e-19 columns\n", + "Vbe = 0.5 V\n", + "VT = 0.0259 V\n", + "WB = 0.0001 cm\n", + "Magnitude of Io , Io=((e∗AB∗Dnb∗nB)/(WB)))= 3.2e-14 A\n", + "Collector current ,Ic=Io((exp(Vbe/VT))−1))= 7.7484232166e-06 A\n" + ] + } + ], + "source": [ + "#exa 7.1\n", + "from math import exp\n", + "Dnb =20\n", + "print\"Dnb = \",Dnb,\" cmˆ2/s\" #initializiation the value of one of base parametre of NPN transistor .\n", + "nB =10**4\n", + "print\"nB = \",nB,\" /cmˆ3\" # initializiation the value of one of base parametre of NPN transistor .\n", + "xB =1*10**-6\n", + "print\"xB = \",xB,\" m\" # initializiation the value of one of base parametre of NPN transistor .\n", + "AB =10**-4\n", + "print\"AB = \",AB,\" cmˆ2\" #initializiation the value of one of base parametre of NPN transistor .\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\" columns\" # initializiation the value of electronic charge .\n", + "Vbe=0.5\n", + "print\"Vbe = \",Vbe,\" V\" # initializiation the value of base emitter voltage of NPN transistor ..\n", + "VT=0.0259\n", + "print\"VT = \",VT,\" V\" # initializiation the value of threshold voltage .\n", + "WB=10**-4\n", + "print\"WB = \",WB,\" cm\" # initializiation the value of base width of NPN transistor .\n", + "Io=((e*AB*Dnb*nB)/(WB))\n", + "print\"Magnitude of Io , Io=((e∗AB∗Dnb∗nB)/(WB)))=\",Io,\" A\"# calculation\n", + "Ic=Io*(exp(Vbe/VT)-1)\n", + "print\"Collector current ,Ic=Io((exp(Vbe/VT))−1))=\",Ic,\" A\"# calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7_2 pgno: 221" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "NE = 500000000000000000 /cmˆ3\n", + "NB = 10000000000000000 /cmˆ3\n", + "NC = 1000000000000000 /cmˆ3\n", + "WB = 8e-05 cm\n", + "no = 15000000000.0 cmˆ−3\n", + "Number of Majority holes in the emitter ,pEO=(noˆ2/NE) )= 450.0 /cmˆ3\n", + "Number of Majority holes in the base,nBO=(no ˆ2/NB))= 22500.0 /cmˆ3\n", + "Number of Majority holes in the collector ,pCO=(noˆ2/NC) )= 225000.0 /cmˆ3\n" + ] + } + ], + "source": [ + "#exa 7.2\n", + "NE =5*10**17\n", + "print\"NE = \",NE,\" /cmˆ3\" # initializiation the value of doping concentration in the emitter\n", + "NB =10**16\n", + "print\"NB = \",NB,\" /cmˆ3\" # initializiation the value of doping concentration in the base.\n", + "NC =10**15\n", + "print\"NC = \",NC,\" /cmˆ3\" # initializiation the value of doping concentration in the collector .\n", + "WB =0.8*10**-4\n", + "print\"WB = \",WB,\" cm\" # initializiation the value of base width of NPN transistor .\n", + "no=1.5*10**10\n", + "print\"no = \",no,\"cmˆ−3\" # initializing the intrinsic carrier concentration .\n", + "pEO=(no**2/NE)\n", + "print\"Number of Majority holes in the emitter ,pEO=(noˆ2/NE) )=\",pEO,\" /cmˆ3\"# calculationnBO=(no^2/NB)\n", + "nBO=(no**2/NB)\n", + "print\"Number of Majority holes in the base,nBO=(no ˆ2/NB))=\",nBO,\" /cmˆ3\"#calculation\n", + "pCO=(no**2/NC)\n", + "print\"Number of Majority holes in the collector ,pCO=(noˆ2/NC) )=\",pCO,\" /cmˆ3\"# calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7_3 pgno: 221" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "NE = 500000000000000000 /cmˆ3\n", + "NB = 10000000000000000 /cmˆ3\n", + "NC = 1000000000000000 /cmˆ3\n", + "WB = 8e-05 cm\n", + "no = 15000000000.0 cmˆ−3\n", + "VT = 0.0259 V\n", + "VJ=Vbe = 0.6258 V\n", + "Number of Majority holes in the emitter ,pEO=(noˆ2/NE) )= 450.0 /cmˆ3\n", + "Number of Majority holes in the base,nBO=(no ˆ2/NB))= 22500.0 /cmˆ3\n", + "Number of Majority holes in the collector ,pCO=(noˆ2/NC) )= 225000.0 /cmˆ3\n", + "pE(O)=pEO∗( exp (VJ/VT) ) )= 1.40186506034e+13 /cmˆ3 \n", + "nB=(nBO∗( exp (VJ/VT) ) ) )= 7.00932530169e+14 /cmˆ3\n" + ] + } + ], + "source": [ + "#exa 7.3\n", + "from math import exp\n", + "NE =5*10**17\n", + "print\"NE = \",NE,\" /cmˆ3\" # initializiation of doping concentration in the emitter .\n", + "NB =10**16\n", + "print\"NB = \",NB,\" /cmˆ3\" # initializiation of doping concentration in the base .\n", + "NC =10**15\n", + "print\"NC = \",NC,\" /cmˆ3\" # initializiation of doping concentration in the collector .\n", + "WB =0.8*10**-4\n", + "print\"WB = \",WB,\" cm\" # initializiation the value of base width of NPN transistor .\n", + "no=1.5*10**10\n", + "print\"no = \",no,\"cmˆ−3\" # initializing the value of intrinsic carrier concentration .\n", + "VT=0.0259\n", + "print\"VT = \",VT,\" V\" # initializiation the value of threshold voltage .\n", + "VJ=0.6258\n", + "print\"VJ=Vbe = \",VJ,\" V\" # initializiation the value of base emitter voltage .\n", + "pEO=(no**2/NE)\n", + "print\"Number of Majority holes in the emitter ,pEO=(noˆ2/NE) )=\",pEO,\" /cmˆ3\"# calculation\n", + "nBO=(no**2/NB)\n", + "print\"Number of Majority holes in the base,nBO=(no ˆ2/NB))=\",nBO,\" /cmˆ3\"#calculation\n", + "pCO=(no**2/NC)\n", + "print\"Number of Majority holes in the collector ,pCO=(noˆ2/NC) )=\",pCO,\" /cmˆ3\"# calculation\n", + "pE=pEO*(exp(VJ/VT))\n", + "print\"pE(O)=pEO∗( exp (VJ/VT) ) )=\",pE, \"/cmˆ3 \" # calculation\n", + "nB=nBO*(exp(VJ/VT))\n", + "print\"nB=(nBO∗( exp (VJ/VT) ) ) )=\",nB,\"/cmˆ3\" # calculation\n", + "#the answer provided in the book for pE,nB is some what different than actual calculated ." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7_1 pgno: 222" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Db = 10 cmˆ2/s\n", + "Bt = 0.95\n", + "tb = 1e-07 s\n", + "Lp=(sqrt(Db∗tb)))= 0.001 cm\n", + "WB=(Lp∗( acosh (1/Bt) ) )= 0.000323036439272 cm\n" + ] + } + ], + "source": [ + "#exa 7.5\n", + "from math import sqrt\n", + "from math import acosh\n", + "Db=10\n", + "print\"Db = \",Db,\" cmˆ2/s\" # initializiation the value of one of parametere of the transistor .\n", + "Bt =0.95\n", + "print\"Bt = \",Bt # initializiation the value of base transport factor of the transistor.\n", + "tb =10**-7\n", + "print\"tb = \",tb,\" s\" # initializiation the value of one of parametere of the transistor.\n", + "Lp=(sqrt(Db*tb))\n", + "print\"Lp=(sqrt(Db∗tb)))=\",Lp,\" cm\"# calculation\n", + "WB=(Lp*(acosh(1/Bt)))\n", + "print\"WB=(Lp∗( acosh (1/Bt) ) )=\",WB,\"cm\" #calculation," + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7_7 pgno: 224" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Jro = 1e-09 A/cmˆ2\n", + "Jo = 1e-12 A/cmˆ2\n", + "Vbe = 0.5 V\n", + "VT = 0.0259 V\n", + "delta (recombination factor)=(1+((Jro/Jo)∗(exp((−Vbe) /(2∗VT) ) ) ) )ˆ−1)= 0.939616412003\n" + ] + } + ], + "source": [ + "#exa 7.7\n", + "from math import exp\n", + "Jro =10**-9\n", + "print\"Jro = \",Jro,\" A/cmˆ2\" # initializiation the value of recombination current density .\n", + "Jo =10**-12\n", + "print\"Jo = \",Jo,\" A/cmˆ2\" # initializiation the value of reverse saturation current density.\n", + "Vbe =0.5\n", + "print\"Vbe = \",Vbe,\" V\" # initializiation the value of base emitter voltage .\n", + "VT =0.0259\n", + "print\"VT = \",VT,\" V\" # initializiation the value of threshold voltage .\n", + "delta=(1+((Jro/Jo)*(exp((-Vbe)/(2*VT)))))**-1\n", + "print\"delta (recombination factor)=(1+((Jro/Jo)∗(exp((−Vbe) /(2∗VT) ) ) ) )ˆ−1)=\",delta # calculation ." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7_8 pgno: 224" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "NE = 100000000000000000 /cmˆ3\n", + "NB = 1000000000000000 /cmˆ3\n", + "WE = 6e-05 cm\n", + "WB = 8e-05 cm\n", + "no = 15000000000.0 cmˆ−3\n", + "e = 1.6e-19 columns\n", + "DE = 15 cmˆ2/s\n", + "DB = 20 cmˆ2/s\n", + "tE = 2e-07 s\n", + "tB = 1e-07 s\n", + "Vbe = 0.6 V\n", + "VT = 0.0259 V\n", + "Jro = 2e-08 A/cmˆ2\n", + "LE=(sqrt(DE∗tE)))= 0.00173205080757 cm\n", + "LB=(sqrt(DB∗tB)))= 0.00141421356237 cm\n", + "Number of Majority holes in the emitter ,pEO=(noˆ2/NE) )= 2250.0 /cmˆ3\n", + "Number of Majority holes in the base,nBO=(no ˆ2/NB))= 225000.0 /cmˆ3\n", + "Emitter injection efficiency ,Y=(1+((NB∗DE∗LB) /(NE∗DB∗LE)∗(tanh(WB/LB)/tanh(WE/LE)))) )= 0.990105536375\n", + "Base transport factor ,Bt=(cosh(WB/LB))ˆ−1)= 0.998402130561\n", + "Reverse saturation current Density , Jro=((e∗DB∗n BO)/(LB∗tanh(WB/LB)))) = 9.00959795262e-09 A/cmˆ2\n", + "delta(recombination factor)=(1+((Jro/Jo)∗(exp((−Vbe)/(2∗VT)))))ˆ−1)= 0.999979304421 A\n", + "common base current amplification factor ,(alpha=Bt∗delta∗Y)= 0.988503018931\n", + "common emitter current amplification factor ,Beta=(a/(1−a) ) )= 85.9793551861\n" + ] + } + ], + "source": [ + "#exa 7.8\n", + "from math import sqrt\n", + "from math import cosh\n", + "from math import tanh\n", + "NE =1*10**17\n", + "print\"NE = \",NE,\" /cmˆ3\" # initializiation the value of doping concentration of emitter in the NPN transistor .\n", + "NB =10**15\n", + "print\"NB = \",NB,\" /cmˆ3\" # initializiation the value of doping concentration of base in the NPN transistor .\n", + "WE =0.6*10**-4\n", + "print\"WE = \",WE,\" cm\" # initializiation the value of one of parametre of the transistor.\n", + "WB =0.8*10**-4\n", + "print\"WB = \",WB,\" cm\" # initializiation the value of one of parametre of the transistor.\n", + "no=1.5*10**10\n", + "print\"no = \",no,\"cmˆ−3\" # initializing the value of intrinsic carrier concentration .\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\" columns\" # initializiation the value of electronic charge\n", + "DE=15\n", + "print\"DE = \",DE,\" cmˆ2/s\" # initializiation the value of one of parametere of the transistor .\n", + "DB=20\n", + "print\"DB = \",DB,\" cmˆ2/s\" #initializiation the value of one of parametere of the transistor .\n", + "tE =0.2*10**-6\n", + "print\"tE = \",tE,\" s\" # initializiation the value of one of parametere of the transistor.\n", + "tB =0.1*10**-6\n", + "print\"tB = \",tB,\" s\" # initializiation the value of one of parametere of the transistor.\n", + "Vbe=0.60\n", + "print\"Vbe = \",Vbe,\" V\" # initializiation the value of base emitter voltage .\n", + "VT=0.0259\n", + "print\"VT = \",VT,\" V\" # initializiation the value of threshold voltage .\n", + "Jro =2*10**-8\n", + "print\"Jro = \",Jro,\" A/cmˆ2\" # initializiation the value of recombination current density .\n", + "LE=(sqrt(DE*tE))\n", + "print\"LE=(sqrt(DE∗tE)))=\",LE,\" cm\"#calculation\n", + "LB=(sqrt(DB*tB))\n", + "print\"LB=(sqrt(DB∗tB)))=\",LB,\" cm\"#calculation\n", + "pEO=(no**2/NE)\n", + "print\"Number of Majority holes in the emitter ,pEO=(noˆ2/NE) )=\",pEO,\" /cmˆ3\"# calculation\n", + "nBO=(no**2/NB)\n", + "print\"Number of Majority holes in the base,nBO=(no ˆ2/NB))=\",nBO,\" /cmˆ3\"#calculation\n", + "Y=(1+(((NB*DE*LB)/(NE*DB*LE))*((tanh(WB/LB)/tanh(WE/ LE)))))**(-1)\n", + "print\"Emitter injection efficiency ,Y=(1+((NB∗DE∗LB) /(NE∗DB∗LE)∗(tanh(WB/LB)/tanh(WE/LE)))) )=\",Y # calculation\n", + "Bt=(cosh(WB/LB))**-1\n", + "print\"Base transport factor ,Bt=(cosh(WB/LB))ˆ−1)=\",Bt# calculation\n", + "Jo=((e*DB*nBO)/(LB*tanh(WB/LB)))\n", + "print\"Reverse saturation current Density , Jro=((e∗DB∗n BO)/(LB∗tanh(WB/LB)))) = \",Jo, \"A/cmˆ2\" # calculation\n", + "delta=(1+((Jro/Jo)*(exp((-Vbe)/(2*VT)))))**-1\n", + "print\"delta(recombination factor)=(1+((Jro/Jo)∗(exp((−Vbe)/(2∗VT)))))ˆ−1)=\",delta,\" A\"# calculation\n", + "a=Bt*delta*Y\n", + "print\"common base current amplification factor ,(alpha=Bt∗delta∗Y)=\",a # calculation\n", + "B=(a/(1-a))\n", + "print\"common emitter current amplification factor ,Beta=(a/(1−a) ) )=\",B # calculation\n", + "#the value of NE provided in the question is different than used in the solution .\n", + "#I have used the value (while solving) provided in the question ( i . e NE=10ˆ17/cmˆ3) " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7_9 pgno: 225" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "NB = 5e+16 /cmˆ3\n", + "NC = 2e+15 /cmˆ3\n", + "WBm = 6e-05 cm\n", + "e = 1.6e-19 columns\n", + "VCB1 = 1 V\n", + "VCB2 = 4 V\n", + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + "no = 15000000000.0 cmˆ−3\n", + "VT = 0.0259 V\n", + " VBI=VT∗(log((NB∗NC)/noˆ2))= 0.694640354303 V\n", + "WBS=((2∗Eo∗Er∗(VBI+VCB1)/e)∗(NC/NB)∗(1/(NC+NB)))ˆ(1/2))= 4.14348090604e-06 cm\n", + "Neutral base width for VCB1,WB( neutral )=WBm− WBS1= 5.5856519094e-05 cm\n", + "WBS=((2∗Eo∗Er∗(VBI+VCB2)/e)∗(NC/NB)∗(1/(NC+NB) ))ˆ(1/2))= 1.0 cm\n", + "Neutral base width for VCB2,WB( neutral )=WBm−WBS2= -0.99994 cm\n", + "change in the neutal base width ,deltaWb(neutral )=Wb1−Wb2= 0.999995856519 cm\n" + ] + } + ], + "source": [ + "#exa 7.9\n", + "from math import log\n", + "NB =5e16\n", + "print\"NB = \",NB,\" /cmˆ3\" # initializiation the doping concentration in the base .\n", + "NC =2e15\n", + "print\"NC = \",NC,\" /cmˆ3\" # initializiation the doping concentration in the collector .\n", + "WBm =0.6e-4\n", + "print\"WBm = \",WBm,\" cm\" # initializiation the value of actual base width .\n", + "e=1.6e-19\n", + "print\"e = \",e,\" columns\" # initializiation the value of electronic charge .\n", + "VCB1=1\n", + "print\"VCB1 = \",VCB1,\" V\" # initializiation the initial value of collector base voltage .\n", + "VCB2=4\n", + "print\"VCB2 = \",VCB2,\" V\" # initializiation the final value of collector base voltage.\n", + "Er=11.9\n", + "print\"Er = \",Er # initializing value of relative dielectric permittivity constant .\n", + "Eo=8.854e-14\n", + "print\"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "no=1.5e10\n", + "print\"no = \",no,\"cmˆ−3\" # initializing the value of intrinsic charge carriers\n", + "VT=0.0259\n", + "print\"VT = \",VT,\" V\" # initializiation the value of threshold voltage .\n", + "VBI=VT*(log((NB*NC)/no**2))\n", + "print\" VBI=VT∗(log((NB∗NC)/noˆ2))=\",VBI,\" V\" # calculation\n", + "WBS1=((2*Eo*Er*(VBI+VCB1)/e)*(NC/NB)*(1/(NC+NB)))**(1./2.)\n", + "print\"WBS=((2∗Eo∗Er∗(VBI+VCB1)/e)∗(NC/NB)∗(1/(NC+NB)))ˆ(1/2))=\",WBS1,\" cm\"#calculation\n", + "Wb1=WBm-WBS1\n", + "print\"Neutral base width for VCB1,WB( neutral )=WBm− WBS1=\",Wb1,\" cm\"# calculation\n", + "WBS2=((2*Eo*Er*(VBI+VCB2)/e)*(NC/NB)*(1/(NC+NB)))**(1/2)\n", + "print\"WBS=((2∗Eo∗Er∗(VBI+VCB2)/e)∗(NC/NB)∗(1/(NC+NB) ))ˆ(1/2))=\",WBS2,\" cm\"#calculation\n", + "Wb2=WBm-WBS2\n", + "print\"Neutral base width for VCB2,WB( neutral )=WBm−WBS2=\",Wb2,\" cm\"# calculation\n", + "deltaWbneutral=Wb1-Wb2\n", + "print\"change in the neutal base width ,deltaWb(neutral )=Wb1−Wb2=\",deltaWbneutral,\" cm\" # calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7_10 pgno: 226" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ro = 5000000.0 ohm\n", + "Vce1 = 7 V\n", + "Vce2 = 1 V\n", + "change in the collector −emitter voltage , Vce1−Vce2 = 6 V\n", + "change in the collector current , Ic=(Vce/ro))= 1.2e-06 A\n" + ] + } + ], + "source": [ + "#exa 7.10\n", + "ro=500*10e3\n", + "print\"ro = \",ro,\" ohm\" # initializiation the value of output resistance .\n", + "Vce1 =7\n", + "print\"Vce1 = \",Vce1,\" V\" # initializiation the initial value of collector emitter voltage .\n", + "Vce2 =1\n", + "print\"Vce2 = \",Vce2,\" V\" # initializiation the final value of collector emitter voltage .\n", + "Vce=6\n", + "print\"change in the collector −emitter voltage , Vce1−Vce2 = \",Vce,\" V\" # calculation .\n", + "Ic=(Vce/ro)\n", + "print\"change in the collector current , Ic=(Vce/ro))=\" ,Ic,\" A\"# calculation," + ] + } + ], + "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": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_8_THE_FIELD_EFFECT_TRANSISTOR_1.ipynb b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_8_THE_FIELD_EFFECT_TRANSISTOR_1.ipynb new file mode 100644 index 00000000..0c2e0420 --- /dev/null +++ b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_8_THE_FIELD_EFFECT_TRANSISTOR_1.ipynb @@ -0,0 +1,806 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8 THE FIELD EFFECT TRANSISTOR" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_1 pgno: 267" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nd= 10000000000000000 /cmˆ3\n", + "Er= 3.9\n", + "Eo = 8.854e-14 F/cm\n", + "W = 5e-05 cm\n", + "L = 0.0001 cm\n", + "tox = 4e-06 cm\n", + " total permittivity ,E=Eo∗Er= 3.45306e-13 F/cm\n", + "Oxide capacitance ,Cox=(E∗W∗L)/tox)= 4.316325e-16 F\n", + "Capacitance per unit area ,Co=(Cox/(W∗L)))= 8.63265e-08 F/cmˆ2\n" + ] + } + ], + "source": [ + "#exa 8.1\n", + "Nd =10**16\n", + "print\"Nd= \",Nd,\" /cmˆ3\" # initializing value of donor ion concentration .\n", + "Er =3.9\n", + "print\"Er= \",Er # initializing value of relative dielectric permittivity constant .\n", + "Eo=8.854*10**-14\n", + "print\"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "W=0.5*10**-4\n", + "print\"W = \",W,\" cm\" # initializing value of width of p−substrate .\n", + "L=10**-4\n", + "print\"L = \",L,\" cm\" # initializing value of length of p−substrate .\n", + "tox =400*10**-8\n", + "print\"tox = \",tox,\" cm\" # initializing value of thickness of p−substrate .\n", + "E=Eo*Er\n", + "print\" total permittivity ,E=Eo∗Er=\",E,\" F/cm\"# calculation\n", + "Cox=(E*W*L)/tox\n", + "print\"Oxide capacitance ,Cox=(E∗W∗L)/tox)=\",Cox,\" F\"# calculation\n", + "Co=(Cox/(W*L))\n", + "print\"Capacitance per unit area ,Co=(Cox/(W∗L)))=\",Co,\" F/cmˆ2\"# calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_2 pgno: 267" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na = 100000000000000000 /cmˆ3\n", + "Vt = 0.0259 V\n", + "e = 1.6e-19 columns\n", + "ni = 15000000000.0 /cmˆ3\n", + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + "Vs=Vt∗log(Na/ni))= 0.40695713106 V\n", + " total permittivity ,E=Eo∗Er= 1.053626e-12 F/cm\n", + "maximum depletion width ,Wd(max)=Sqrt(4∗E∗Vs/(e∗Na)))= 1.03535092381e-05 cm\n" + ] + } + ], + "source": [ + "#exa 8.2\n", + "from math import log\n", + "from math import sqrt\n", + "Na =10**17\n", + "print\"Na = \",Na,\" /cmˆ3\" # initializing value of acceptor ion concentration .\n", + "Vt =0.0259\n", + "print\"Vt = \",Vt,\"V\" # initializing value of thermal voltage .\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\" columns\" # initializing value of charge of electrons .\n", + "ni=1.5*10**10\n", + "print\"ni = \",ni,\"/cmˆ3\" #initializing value of intrinsic carrier concentration .\n", + "Er=11.9\n", + "print\"Er = \",Er # initializing value of relative dielectric permittivity constant\n", + "Eo=8.854*10**-14\n", + "print\"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "Vs=Vt*log(Na/ni)\n", + "print\"Vs=Vt∗log(Na/ni))=\",Vs,\" V\"#calculation\n", + "E=Eo*Er\n", + "print\" total permittivity ,E=Eo∗Er=\",E,\" F/cm\" # calculation\n", + "Wd=sqrt(4*E*Vs/(e*Na))\n", + "print\"maximum depletion width ,Wd(max)=Sqrt(4∗E∗Vs/(e∗Na)))=\",Wd,\" cm\"#calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_3 pgno: 268" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nd = 3000000000000000000 /cmˆ3\n", + "Vt = 0.0259 V\n", + "e = 1.6e-19 columns\n", + "ni = 15000000000.0 /cmˆ3\n", + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + "Vs=Vt∗log(Nd/ni))= 0.495048143245 V\n", + " total permittivity ,E=Eo∗Er= 1.053626e-12 F/cm\n", + "maximum depletion width ,Wd(max)=Sqrt(4∗E∗Vs/(e∗Nd)))= 2.08485729922e-06 cm\n" + ] + } + ], + "source": [ + "#exa 8.3\n", + "from math import sqrt\n", + "from math import log\n", + "Nd =3*10**18\n", + "print\"Nd = \",Nd,\" /cmˆ3\" # initializing value of acceptor ion concentration .\n", + "Vt =0.0259\n", + "print\"Vt = \",Vt,\"V\" # initializing value of thermal voltage .\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\" columns\" # initializing value of charge of electrons .\n", + "ni=1.5*10**10\n", + "print\"ni = \",ni,\"/cmˆ3\" #initializing value of intrinsic carrier concentration .\n", + "Er=11.9\n", + "print\"Er = \",Er # initializing value of relative dielectric permittivity constant\n", + "Eo=8.854*10**-14\n", + "print\"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "Vs=Vt*log(Nd/ni)\n", + "print\"Vs=Vt∗log(Nd/ni))=\",Vs,\" V\"# calculation\n", + "E=Eo*Er\n", + "print\" total permittivity ,E=Eo∗Er=\",E,\" F/cm\" # calculation\n", + "Wd=sqrt(4*E*Vs/(e*Nd))\n", + "print\"maximum depletion width ,Wd(max)=Sqrt(4∗E∗Vs/(e∗Nd)))=\",Wd,\" cm\"#calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_5 pgno: 269" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vm = 3.2 V\n", + "X = 3.25 V\n", + "Nd = 20000000000000000 /cmˆ3\n", + "ni = 15000000000.0 V\n", + "Vt = 0.0259 V\n", + "Eg = 1.12 V\n", + "Vfp=(Vt∗log(Nd/ni))= 0.365272689128 V\n", + "Vms=−(Vm+(Eg/2)+Vfp−Vm)= -0.925272689128 V\n" + ] + } + ], + "source": [ + "#exa 8.5\n", + "from math import log\n", + "Vm =3.2\n", + "print\"Vm = \",Vm,\" V\" # initializing value of modified metal work function .\n", + "X=3.25\n", + "print\"X = \",X,\" V\" # initializing value of modified electron affinity .\n", + "Nd =2*10**16\n", + "print\"Nd = \",Nd,\" /cmˆ3\" # initializing value of donor concentration .\n", + "ni=1.5*10**10\n", + "print\"ni = \",ni,\" V\" # initializing value of intrinsic carrier concentration .\n", + "Vt=0.0259\n", + "print\"Vt = \",Vt,\"V\" # initializing value of thermal voltage .\n", + "Eg=1.12\n", + "print\"Eg = \",Eg,\"V\" # initializing value of energy gap .\n", + "Vfp=(Vt*log(Nd/ni))\n", + "print\"Vfp=(Vt∗log(Nd/ni))=\",Vfp,\" V\" # calculation .\n", + "Vms=-(Vm+(Eg/2)+Vfp-Vm)\n", + "print\"Vms=−(Vm+(Eg/2)+Vfp−Vm)=\",Vms,\" V\" # calculation ." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_7 pgno: 270" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nd = 10000000000000000 /cmˆ3\n", + "Vms = -1.12 V\n", + "Er = 3.9\n", + "Eo = 8.854e-14 F/cm\n", + "tox = 2e-06 cm\n", + "Qss = 2.5e-08 columbs/cmˆ2\n", + "Total permittivity ,Eox=Eo∗Er= 3.45306e-13 F/cm\n", + "Capacitance per unit area ,Co=(E/tox))= 1.72653e-07 F/cmˆ2\n", + " Flat band voltage , Vfb=(Vms−(Qss/Co) ) )= -1.26479910572 V\n" + ] + } + ], + "source": [ + "#exa 8.7\n", + "Nd =10**16\n", + "print\"Nd = \",Nd,\" /cmˆ3\" # initializing value of donor ion concentration .\n", + "Vms = -1.12\n", + "print\"Vms = \",Vms,\" V\" # initializing value of metal semiconductor work function difference .\n", + "Er =3.9\n", + "print\"Er = \",Er # initializing value of relative dielectric permittivity constant .\n", + "Eo=8.854*10**-14\n", + "print\"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "tox =200*10**-8\n", + "print\"tox = \",tox,\" cm\" # initializing value of thickness of p−type substrate .\n", + "Qss =2.5*10**-8\n", + "print\"Qss = \",Qss,\" columbs/cmˆ2\" # initializing value of charge density on semiconductor surface .\n", + "Eox=Eo*Er\n", + "print\"Total permittivity ,Eox=Eo∗Er=\",Eox,\" F/cm\"# calculation\n", + "Co=(Eox/tox)\n", + "print\"Capacitance per unit area ,Co=(E/tox))=\",Co,\" F/cmˆ2\"# calculation\n", + "Vfb=(Vms-(Qss/Co))\n", + "print\" Flat band voltage , Vfb=(Vms−(Qss/Co) ) )=\",Vfb,\" V\"# calculation\n", + "#the answer for Co after calculation is provided wrong in the book " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_9 pgno: 271" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na = 30000000000000000 /cmˆ3\n", + "Vms = -1.12 V\n", + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + "ni = 15000000000.0 cmˆ−3\n", + "e = 1.6e-19 columns\n", + "tox = 3e-06 cm\n", + "Vfb = -1.12 V\n", + "Qss = 100000000000 electronic charge columns/cmˆ2\n", + "Vt = 0.0259 eV\n", + "er = 3.9\n", + "total permittivity ,Eox=Eo∗Er= 1.053626e-12 F/cm\n", + "Potential ,Vfp=Vt∗(log(Na/(ni))))= 0.375774235428 V\n", + "Maximum depletion width ,Wd=sqrt ((4∗E∗Vs)/(e∗Nd)))= 1.81641933617e-05 cm\n", + "Over all maximum depletion width ,QDmax=(e∗Na∗ Wd) )= 8.71881281361e-08 columns/cmˆ2\n", + "Threshold Voltage ,VT=(((QDmax−1.6∗10ˆ−8)∗tox)/(er∗Eo),(2∗Vfp+Vfb)= 0.250027108378 V\n" + ] + } + ], + "source": [ + "#exa 8.9\n", + "from math import log\n", + "Na =3*10**16\n", + "print\"Na = \",Na,\" /cmˆ3\" # initializing value of acceptor ion concentration .\n", + "Vms = -1.12\n", + "print\"Vms = \",Vms,\"V\" # initializing value of metal semiconductor work function difference .\n", + "Er =11.9\n", + "print\"Er = \",Er # initializing value of relative dielectric permittivity constant .\n", + "Eo=8.854*10**-14\n", + "print\"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "ni=1.5*10**10\n", + "print\"ni = \",ni,\"cmˆ−3\" # initializing value of intrinsic concentration of electrons .\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\" columns\" # initializing value of charge of electrons .\n", + "tox =300*10**-8\n", + "print\"tox = \",tox,\" cm\" # initializing value of thickness of p−type substrate .\n", + "Vfb=-1.12\n", + "print\"Vfb = \",Vfb,\" V\" # initializing value of flat band voltage .\n", + "Qss=10**11\n", + "print\"Qss = \",Qss,\" electronic charge columns/cmˆ2\" # initializing value of charge density on semiconductor surface .\n", + "Vt=0.0259\n", + "print\"Vt = \",Vt,\" eV\" # initializing value of thermal voltage .\n", + "er=3.9\n", + "print\"er = \",er # initializing value of relative dielectric permittivity constant\n", + "Eox=Eo*Er\n", + "print\"total permittivity ,Eox=Eo∗Er=\",Eox,\" F/cm\"# calculation\n", + "Vfp=Vt*(log(Na/(ni)))\n", + "print\"Potential ,Vfp=Vt∗(log(Na/(ni))))=\",Vfp,\" V\"#calculation\n", + "Wd=sqrt((4*Eox*Vfp)/(e*Na))\n", + "print\"Maximum depletion width ,Wd=sqrt ((4∗E∗Vs)/(e∗Nd)))=\",Wd,\" cm\"#calculation\n", + "QDmax=(e*Na*Wd)\n", + "print\"Over all maximum depletion width ,QDmax=(e∗Na∗ Wd) )=\",QDmax,\" columns/cmˆ2\" # calculation\n", + "VT=(((QDmax -1.6*10**-8)*tox)/(er*Eo))+(2*Vfp+Vfb)\n", + "print \"Threshold Voltage ,VT=(((QDmax−1.6∗10ˆ−8)∗tox)/(er∗Eo),(2∗Vfp+Vfb)=\",VT,\" V\" # calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_10 pgno: 271" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "L = 0.000125 cm\n", + "un = 600 cmˆ2/V−s\n", + "Co = 6.9e-09 F/cmˆ2\n", + "VT = 0.6 V\n", + "Vgs = 4 V\n", + "W = 0.0012 cm\n", + "Drain current ,Id=((Co∗un∗W)/(L)∗((Vgs−VT)ˆ2/(2)))= 0.00022972032 A\n" + ] + } + ], + "source": [ + "#exa 8.10\n", + "L=1.25*10**-4\n", + "print\"L = \",L,\" cm\" # initializing value of length of channel .\n", + "un =600\n", + "print\"un = \",un,\"cmˆ2/V−s\" # initializing value of mobility of n−channel MOS transistor .\n", + "Co =6.9*10**-9\n", + "print\"Co = \",Co,\"F/cmˆ2\" # initializing value of capacitance per unit area .\n", + "VT =0.60\n", + "print\"VT = \",VT,\" V\" # initializing value of threshold Voltage .\n", + "Vgs=4\n", + "print\"Vgs = \",Vgs,\" V\" # initializing value of gate to source voltage .\n", + "W=12*10**-4\n", + "print\"W = \",W,\"cm\" # initializing value of width of channel .\n", + "Id=((Co*un*W)/(L)*((Vgs-VT)**2/(2)))\n", + "print\"Drain current ,Id=((Co∗un∗W)/(L)∗((Vgs−VT)ˆ2/(2)))=\",Id,\" A\"#calculation .\n", + "#The answer provided in the book (for Id) is wrong as the value of mobility used for solution is different than provided in the question and also value of (Vgs−Vt) is put wrong in the solution than given in the book .\n", + "#I have used the value given in the question i.e. answer differ ." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_13 pgno: 273" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Na = 200000000000000000 /cmˆ3\n", + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + "ni = 15000000000.0 cmˆ−3\n", + "e = 1.6e-19 columns\n", + "tox = 4e-06 cm\n", + "Vt = 0.0259 eV\n", + "er = 3.9\n", + "Potential ,Vfp=Vt∗(log(Na/(ni))))= 0.424909643036 V\n", + "Depletion width ,Wd=sqrt ((4∗Er∗Eo∗Vs)/(e∗Nd)))= 7.48077408723e-06 cm\n", + "Minimum Capacitance,CTmin=(er∗Eo/((er/Er)∗(Wd)+(tox)))= 5.35218545918e-08 F/cmˆ2\n", + "Flat band capacitance ,CFB=((er∗Eo) /((( er/Er)∗sqrt(Vt∗Er∗Eo/(e∗Na))))+(tox))= 8.02543256028e-08 F/ cmˆ2\n" + ] + } + ], + "source": [ + "#exa 8.13\n", + "from math import sqrt\n", + "from math import log\n", + "Na =2*10**17\n", + "print\"Na = \",Na,\" /cmˆ3\" # initializing value of acceptor ion concentration .\n", + "Er =11.9\n", + "print\"Er = \",Er # initializing value of relative dielectric permittivity constant .\n", + "Eo=8.854*10**-14\n", + "print\"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "ni=1.5*10**10\n", + "print\"ni = \",ni,\"cmˆ−3\" # initializing value of intrinsic concentration of electrons .\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\" columns\" # initializing value of charge of electrons .\n", + "tox =400*10**-8\n", + "print\"tox = \",tox,\" cm\" # initializing value of thickness of p−type substrate .\n", + "Vt=0.0259\n", + "print\"Vt = \",Vt,\" eV\" # initializing value of thermal voltage .\n", + "er=3.9\n", + "print\"er = \",er # initializing value of relative dielectric permittivity constant\n", + "Vfp=Vt*(log(Na/(ni)))\n", + "print\"Potential ,Vfp=Vt∗(log(Na/(ni))))=\",Vfp,\" V\"#calculation\n", + "Wd=sqrt((4*Er*Eo*Vfp)/(e*Na))\n", + "print\"Depletion width ,Wd=sqrt ((4∗Er∗Eo∗Vs)/(e∗Nd)))=\",Wd,\" cm\"# calculation\n", + "CTmin=(er*Eo/(((er/Er)*(Wd))+(tox)))\n", + "print\"Minimum Capacitance,CTmin=(er∗Eo/((er/Er)∗(Wd)+(tox)))=\",CTmin,\" F/cmˆ2\"#calculation\n", + "CFB=((er*Eo)/((((er/Er)*sqrt(Vt*Er*Eo/(e*Na))))+(tox)))\n", + "print\"Flat band capacitance ,CFB=((er∗Eo) /((( er/Er)∗sqrt(Vt∗Er∗Eo/(e∗Na))))+(tox))=\",CFB,\" F/ cmˆ2\"# calculation\n", + "#the value of Na (acceptor ion concentration) and tox ( thickness of p−type substrate ) is provided different in the question than used in the solution .\n", + "#I have used the value provided in the solution .( i . e Na=2∗10ˆ17 and tox =400∗10ˆ8cm)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_14 pgno: 274" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vfb = -1.0 V\n", + "Vms = -0.9 V\n", + "tox = 2e-06 cm\n", + "et = 3.9\n", + "eo = 8.85e-14 F/cm\n", + "e = 1.6e-19 columns\n", + "eox=(eo∗et))= 3.4515e-13 F/cmˆ2\n", + "Oxide capacitance ,Cox=(eox/tox))= 1.72575e-07 F/cmˆ2\n", + "charge density on semiconductor surface ,Qss=(( Vms−Vfb)∗Cox))= 1.72575e-08 C/cmˆ2\n", + "charge density on semiconductor surface (in terms of number of charges) ,Qss∗=Qss/e)= 1.07859375e+11 electrons/cmˆ2\n" + ] + } + ], + "source": [ + "#exa 8.14\n", + "Vfb = -1.0\n", + "print\"Vfb = \",Vfb,\" V\" # initializing value of flat band voltage .\n", + "Vms = -0.9\n", + "print\"Vms = \",Vms,\"V\" # initializing value of metal semiconductor work function difference .\n", + "tox =200*10**-8\n", + "print\"tox = \",tox,\" cm\" # initializing value of gate oxide thickness .\n", + "et =3.9\n", + "print\"et = \",et # initializing value of relative permittivity .\n", + "eo =8.85*10**-14\n", + "print\"eo = \",eo,\"F/cm\" # initializing value of free space permittivity .\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\" columns\" # initializing value of charge of electrons .\n", + "eox=(eo*et)\n", + "print\"eox=(eo∗et))=\",eox,\" F/cmˆ2\" # calculation\n", + "Cox=(eox/tox)\n", + "print\"Oxide capacitance ,Cox=(eox/tox))=\",Cox,\" F/cmˆ2\"# calculation\n", + "Qss=((Vms-Vfb)*Cox)\n", + "print\"charge density on semiconductor surface ,Qss=(( Vms−Vfb)∗Cox))=\",Qss,\" C/cmˆ2\" # calculation\n", + "Qss1=Qss/e\n", + "print\"charge density on semiconductor surface (in terms of number of charges) ,Qss∗=Qss/e)=\",Qss1,\" electrons/cmˆ2\" #calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_15 pgno: 274" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "L = 3e-06 meter\n", + "un = 800.0 cmˆ2/V−s\n", + "VT = 1.0 V\n", + "Vgs = 0 V\n", + "tox = 5e-06 cm\n", + "et = 3.9\n", + "eo = 8.85e-14 F/cm\n", + "W = 3e-05 m\n", + "eox=(eo∗et))= 3.4515e-13 F/cmˆ2\n", + "Drain current ,Id=((eox∗un∗W)/(tox∗L)∗((Vgs−VT)ˆ2/(2))))= 276120.0 A\n" + ] + } + ], + "source": [ + "#exa 8.15\n", + "L=3e-6\n", + "print\"L = \",L,\" meter\" # initializing value of length of channel .\n", + "un =800.\n", + "print\"un = \",un,\"cmˆ2/V−s\" # initializing value of mobility of n−channel MOS transistor .\n", + "VT=1.\n", + "print\"VT = \",VT,\" V\" # initializing value of threshold Voltage .\n", + "Vgs=0\n", + "print\"Vgs = \",Vgs,\" V\" # initializing value of gate to source voltage .\n", + "tox =500e-8\n", + "print\"tox = \",tox,\" cm\" # initializing value of gate oxide thickness .\n", + "et=3.9\n", + "print\"et = \",et # initializing value of relative permittivity .\n", + "eo =8.85e-14\n", + "print\"eo = \",eo,\"F/cm\" # initializing value of free space permittivity .\n", + "W=30e-6\n", + "print\"W = \",W,\"m\" # initializing value of width of channel .\n", + "eox=(eo*et)\n", + "print\"eox=(eo∗et))=\",eox,\" F/cmˆ2\"# calculation\n", + "Id=((eox*un*W)/(tox*L)*((Vgs-VT)**2/(2)))*(1e9)\n", + "print\"Drain current ,Id=((eox∗un∗W)/(tox∗L)∗((Vgs−VT)ˆ2/(2))))=\",Id,\" A\"#calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_16 pgno: 274" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "L = 2.5e-06 meter\n", + "un = 800 cmˆ2/V−s\n", + "VT = 0.8 V\n", + "Vgs = 1 V\n", + "tox = 4e-06 cm\n", + "et = 3.9\n", + "eo = 8.85e-14 F/cm\n", + "eox=(eo∗et))= 3.4515e-13 F/cmˆ2\n", + "W = 2.5e-05 m\n", + "Drain current ,Id=((eox∗un∗W)/(tox∗L)∗((Vgs−VT)ˆ2/(2))))= 1.3806e-05 A\n" + ] + } + ], + "source": [ + "#exa 8.16\n", + "L=2.5*10**-6\n", + "print\"L = \",L,\" meter\" # initializing value of length of channel .\n", + "un =800\n", + "print\"un = \",un,\"cmˆ2/V−s\" # initializing value of mobility of n−channel MOS transistor .\n", + "VT =0.8\n", + "print\"VT = \",VT,\" V\" # initializing value of threshold Voltage .\n", + "Vgs=1\n", + "print\"Vgs = \",Vgs,\" V\" # initializing value of gate to source voltage .\n", + "tox =400*10**-8\n", + "print\"tox = \",tox,\" cm\" # initializing value of gate oxide thickness .\n", + "et=3.9\n", + "print\"et = \",et # initializing value of relative permittivity .\n", + "eo =8.85*10**-14\n", + "print\"eo = \",eo,\"F/cm\" # initializing value of free space permittivity .\n", + "eox=(eo*et)\n", + "print\"eox=(eo∗et))=\",eox,\" F/cmˆ2\" # calculation\n", + "W=25*10**-6\n", + "print\"W = \",W,\"m\" # initializing value of width of channel . .\n", + "Id=((eox*un*W)/(tox*L)*((Vgs-VT)**2/(2)))\n", + "print\"Drain current ,Id=((eox∗un∗W)/(tox∗L)∗((Vgs−VT)ˆ2/(2))))=\",Id,\" A\"#calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_17 pgno: 274" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "un = 525 cmˆ2/V−s\n", + "VT = 0.75 V\n", + "Vgs = 2 V\n", + "tox = 4e-06 cm\n", + "et = 3.9\n", + "eo = 8.85e-14 F/cm\n", + "eox=(eo∗et))= 3.4515e-13 F/cmˆ2\n", + "Id = 0.006 A\n", + "width to length ratio ,W/L=((Id∗tox∗2)/(eox∗un∗((Vgs−VT)ˆ2)))= 169.532915296\n" + ] + } + ], + "source": [ + "#8.17\n", + "un =525\n", + "print\"un = \",un,\"cmˆ2/V−s\" # initializing value of mobility of n−channel MOS transistor .\n", + "VT =0.75\n", + "print\"VT = \",VT,\" V\" # initializing value of threshold Voltage .\n", + "Vgs=2\n", + "print\"Vgs = \",Vgs,\" V\" # initializing value of gate to source voltage .\n", + "tox =400*10**-8\n", + "print\"tox = \",tox,\" cm\" # initializing value of gate oxide thickness .\n", + "et=3.9\n", + "print\"et = \",et # initializing value of relative permittivity .\n", + "eo =8.85*10**-14\n", + "print\"eo = \",eo,\"F/cm\" # initializing value of free space permittivity .\n", + "eox=(eo*et)\n", + "print\"eox=(eo∗et))=\",eox,\" F/cmˆ2\" # calculation\n", + "Id =6*10**-3\n", + "print\"Id = \",Id,\"A\" # initializing value of width of channel . .\n", + "X=((Id*tox*2)/(eox*un*((Vgs-VT)**2)))\n", + "print\"width to length ratio ,W/L=((Id∗tox∗2)/(eox∗un∗((Vgs−VT)ˆ2)))= \",X # calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_18 pgno: 275" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nd = 20000000000000000 /cmˆ3\n", + "a = 0.0002 cm\n", + "e = 1.6e-19 columns\n", + "Er = 11.9\n", + "Eo = 8.85e-14 F/cm\n", + "E=(Eo∗Er))= 1.05315e-12 F/cmˆ2\n", + "Pinch off Voltage ,Vp=((e∗Nd∗aˆ2)/(2∗E)))= 60.7700707402 V\n" + ] + } + ], + "source": [ + "#exa 8.18\n", + "Nd =2*10**16\n", + "print\"Nd = \",Nd,\" /cmˆ3\" # initializing value of donor ion concentration .\n", + "a=2*10**-4\n", + "print\"a = \",a,\" cm\" # initializing value of height of channel at pinch off .\n", + "e=1.6*10**-19\n", + "print\"e = \",e,\" columns\" # initializing value of charge of electrons .\n", + "Er =11.9\n", + "print\"Er = \",Er # initializing value of relative permittivity .\n", + "Eo =8.85*10**-14\n", + "print\"Eo = \",Eo,\"F/cm\" # initializing value of free space permittivity .\n", + "E=(Eo*Er)\n", + "print\"E=(Eo∗Er))=\",E,\" F/cmˆ2\"#calculation\n", + "Vp=((e*Nd*a**2)/(2*E))\n", + "print\"Pinch off Voltage ,Vp=((e∗Nd∗aˆ2)/(2∗E)))=\",Vp,\" V\"# calculation," + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_20 pgno: 275" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a = 0.0002 cm\n", + "Er = 11.9\n", + "Eo = 8.854e-14 F/cm\n", + "un = 1350 cmˆ2/V−s\n", + "W = 0.0008 cm\n", + "L = 0.001 cm\n", + "e = 1.6e-19 columns\n", + "Vp = 4 V\n", + "Vgs = 0 V\n", + " total permittivity ,E=Eo∗Er= 1.053626e-12 F/cm\n", + "Donor ion concentration ,Nd=((Vp∗2∗E)/(e∗aˆ2)))= 1.3170325e+15 /cmˆ3\n", + "On Drain resistance ,rds=(L/(W∗a∗e∗un∗Nd)))= 21969.9856953 ohm\n" + ] + } + ], + "source": [ + "#exa 8.20\n", + "a=2*10**-4\n", + "print\"a = \",a,\" cm\" # initializing value of height of channel at pinch off .\n", + "Er =11.9\n", + "print\"Er = \",Er # initializing value of relative dielectric permittivity constant .\n", + "Eo=8.854*10**-14\n", + "print\"Eo = \",Eo,\" F/cm\" # initializing value of permittivity of free space .\n", + "un =1350\n", + "print\"un = \",un,\"cmˆ2/V−s\" # initializing value of mobility of n−type silicon Mosfet.\n", + "W=8*10**-4\n", + "print\"W = \",W,\" cm\" # initializing value of width of p−substrate .\n", + "L=10*10**-4\n", + "print\"L = \",L,\" cm\" # initializing value of length of p−substrate \n", + "e=1.6*10**-19\n", + "print\"e = \",e,\" columns\" # initializing value of charge of electrons .\n", + "Vp=4\n", + "print\"Vp = \",Vp,\" V\" # initializing value of thickness of p−substrate . \n", + "Vgs=0\n", + "print\"Vgs = \",Vgs,\" V\" # initializing value of gate to source voltage .\n", + "E=Eo*Er\n", + "print\" total permittivity ,E=Eo∗Er=\",E,\" F/cm\" # calculation\n", + "Nd=((Vp*2*E)/(e*a**2))\n", + "print\"Donor ion concentration ,Nd=((Vp∗2∗E)/(e∗aˆ2)))=\",Nd,\" /cmˆ3\"# calculation\n", + "rds=(L/(W*a*e*un*Nd))\n", + "print\"On Drain resistance ,rds=(L/(W∗a∗e∗un∗Nd)))=\",rds,\" ohm\"# calculation" + ] + } + ], + "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": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2015-11-05_at_11.31.11_pm_1.png b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2015-11-05_at_11.31.11_pm_1.png new file mode 100644 index 00000000..e69de29b --- /dev/null +++ b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2015-11-05_at_11.31.11_pm_1.png diff --git a/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2015-11-05_at_11.32.51_pm_1.png b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2015-11-05_at_11.32.51_pm_1.png new file mode 100644 index 00000000..e69de29b --- /dev/null +++ b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2015-11-05_at_11.32.51_pm_1.png diff --git a/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2015-11-05_at_11.33.45_pm_1.png b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2015-11-05_at_11.33.45_pm_1.png new file mode 100644 index 00000000..e69de29b --- /dev/null +++ b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2015-11-05_at_11.33.45_pm_1.png |
