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| author | Trupti Kini | 2016-01-07 23:30:03 +0600 |
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| committer | Trupti Kini | 2016-01-07 23:30:03 +0600 |
| commit | 09dec1387f25ee7463ceb1b4c1b641cb3e022c3b (patch) | |
| tree | 20a0e68b2fffd6040422b2a91df8082a8491f24e | |
| parent | 2dae62c25a4f05d6de7d2b0d3af33a8abd744c4c (diff) | |
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Added(A)/Deleted(D) following books
A Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/README.txt
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch10.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch11.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch12.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch14.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch15.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch16.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch17.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch18.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch19.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch2.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch20.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch21.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch22.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch23.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch26.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch3.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch4.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch5.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch6.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch7.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch8.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch9.ipynb
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/screenshots/14_fraction_of_load.png
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/screenshots/7_atom_conc.png
A Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/screenshots/ch4no_of_eff_atoms.png
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_10_SILICON_CONTROLLED_RECTIFIER_2.ipynb
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_1_CRYSTAL_STRUCTURES_2.ipynb
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_6_ELECTRICAL_BREAKDOWN_IN_PN_JUNCTIONS_2.ipynb
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_2_ENERGY_BAND_THEORY_OF_SOLIDS_2.ipynb
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_3_CARRIER_TRANSPORT_IN_SEMICONDUCTOR_2.ipynb
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_4__EXCESS_CARRIER_IN_SEMICONDUCTOR_2.ipynb
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_5_PN_JUNCTION_DIODE_2.ipynb
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_7_BIPOLAR_JUNCTION_TRANSISTORB_2.ipynb
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_8_THE_FIELD_EFFECT_TRANSISTOR_2.ipynb
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2016-01-07_at_8.46.13_pm.png
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2016-01-07_at_8.46.31_pm.png
A Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2016-01-07_at_8.46.51_pm.png
38 files changed, 13902 insertions, 0 deletions
diff --git a/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/README.txt b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/README.txt new file mode 100644 index 00000000..9de55847 --- /dev/null +++ b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/README.txt @@ -0,0 +1,10 @@ +Contributed By: kartik sankhla +Course: btech +College/Institute/Organization: iitbombay +Department/Designation: aerospace engnieering +Book Title: Introduction to Nuclear Engineering +Author: J. R. Lamarsh and A. J. Baratta +Publisher: Prentice Hall, New Jersey +Year of publication: 2001 +Isbn: 0-201-82498-1 +Edition: 3
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch10.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch10.ipynb new file mode 100644 index 00000000..47f6c5a5 --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch10.ipynb @@ -0,0 +1,716 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:4899a3f156bb3463e97f6b375e9947501acc46cec2332bed6d0ada61bec0b1d0" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-10, Mechanical tests and factors affecting mechanical properties" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-10.1 : page no-298" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#initial gauge length of the specimen\n", + "l0=50*10**-3 #m\n", + "#initia l gauge diameter of the specimen\n", + "d0=12*10**-3 #m\n", + "#extended gauge length of fracture\n", + "lf=58*10**-3 #m\n", + "#reduced gauge diameter\n", + "df=7*10**-3 #m\n", + "#initial an final cross sectional areas are\n", + "A_i=3.14/4*d0**2 #m**2\n", + "A_f=3.14/4*df**2 #m**2\n", + "#various applied loads are in k N\n", + "P1=0\n", + "P2=5\n", + "P3=10\n", + "P4=15\n", + "P5=20\n", + "P6=25\n", + "P7=30\n", + "P8=32\n", + "P9=33\n", + "P10=32\n", + "P11=31\n", + "P12=35\n", + "P13=40\n", + "P16=130\n", + "#corresponding to these load we have recorded elongation as\n", + "delta1=0\n", + "delta2=0.011\n", + "delta3=0.022\n", + "delta4=0.035\n", + "delta5=0.048\n", + "delta6=0.059\n", + "delta7=0.073\n", + "delta8=0.088\n", + "delta9=0.100\n", + "delta10=0.125\n", + "delta11=0.150\n", + "delta12=0.230\n", + "delta13=0.400\n", + "delta16=8.000\n", + "#stress and strain corresponding to these loads and elongations are\n", + "sigma1=P1/A_i\n", + "strain1=delta1/l0\n", + "sigma2=P2/A_i\n", + "strain2=delta2/l0\n", + "sigma3=P3/A_i\n", + "strain3=delta3/l0\n", + "sigma4=P4/A_i\n", + "strain4=delta4/l0\n", + "sigma5=P5/A_i\n", + "strain5=delta5/l0\n", + "sigma6=P6/A_i\n", + "strain6=delta6/l0\n", + "sigma7=P7/A_i\n", + "strain7=delta7/l0\n", + "sigma8=P8/A_i\n", + "strain8=delta8/l0\n", + "sigma9=P9/A_i\n", + "strain9=delta9/l0\n", + "sigma10=P10/A_i\n", + "strain10=delta10/l0\n", + "sigma11=P11/A_i\n", + "strain11=delta11/l0\n", + "sigma12=P12/A_i\n", + "strain12=delta12/l0\n", + "sigma13=P13/A_i\n", + "strain13=delta13/l0\n", + "sigma16=P16/A_i\n", + "strain16=delta16/l0\n", + "#part(a)\n", + "#modulus of elasticity\n", + "E=(sigma4-sigma1)/(strain4-strain1)/10**3 #G Pa\n", + "#part(b)\n", + "#ultimate stress (maximum)\n", + "ultimate_sigma=P16/A_i*10**3/10**6 #M N/m**3\n", + "#part (c)\n", + "#upper yield point at C (shown in fig 10.3)\n", + "u_yield=P9/A_i*1000/10**6 #M N/m**2\n", + "#lower yield point at D (shown in fig 10.3)\n", + "l_yield=P11/A_i*1000/10**6 #M N/m**2\n", + "#part(d)\n", + "#percentage reduction in area\n", + "percent_A=(d0**2-df**2)/d0**2*100 #%\n", + "#part (e)\n", + "#percentage elongation\n", + "percent_l=(lf-l0)/l0*100 #%\n", + "#part(f)\n", + "#apparent breaking stress\n", + "app_breaking_stress=P16*1000/A_i/10**6 #M N/m**2\n", + "#actual breaking stress\n", + "actual_breaking_stress=P16*1000/A_f/10**6 #M N/m**2\n", + "print \"\"\"the modulus of elasticity is %0.3f G Pa\n", + "the ultimate (maximum) stress is %0.3f M n/m**2 \n", + "upper yield point is %0.3f M N/m**2 \n", + "lower yield point is %0.3f M n/m**2\n", + "percentage reduction in area is %0.3f \n", + "percentage elongation in length %0.3f\n", + "apparent breaking stress is %0.3f M n/mm**2\n", + "actual breaking point is %0.3f M n/m**2\"\"\"%(E,ultimate_sigma,u_yield,l_yield,percent_A,percent_l,app_breaking_stress,actual_breaking_stress)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the modulus of elasticity is 189.566 G Pa\n", + "the ultimate (maximum) stress is 1150.035 M n/m**2 \n", + "upper yield point is 291.932 M N/m**2 \n", + "lower yield point is 274.239 M n/m**2\n", + "percentage reduction in area is 65.972 \n", + "percentage elongation in length 16.000\n", + "apparent breaking stress is 1150.035 M n/mm**2\n", + "actual breaking point is 3379.696 M n/m**2\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-10.2 : page no-303" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#initial length of the specimen\n", + "h0=24.02*10**-3 #m\n", + "#initial gauge diameter of the specimen\n", + "d0=18.74*10**-3 #m\n", + "#final length of specimen\n", + "hf=18.70*10**-3 #m\n", + "#final diameter\n", + "df=21.54*10**-3 #m\n", + "#initial an final cross sectional areas are\n", + "A_i=3.14/4*d0**2 #m**2\n", + "A_f=3.14/4*df**2 #m**2\n", + "#various applied loads are in k N\n", + "P1=0\n", + "P2=5\n", + "P3=10\n", + "P4=15\n", + "P5=20\n", + "P6=25\n", + "P7=30\n", + "P8=35\n", + "P9=40\n", + "P10=45\n", + "P11=50\n", + "P12=55\n", + "P13=60\n", + "P14=65\n", + "P15=70\n", + "P16=75\n", + "P17=80\n", + "P18=85\n", + "P19=131\n", + "#corresponding to these load we have recorded contraction as\n", + "delta1=0\n", + "delta2=0.004\n", + "delta3=0.008\n", + "delta4=0.012\n", + "delta5=0.015\n", + "delta6=0.017\n", + "delta7=0.020\n", + "delta8=0.023\n", + "delta9=0.025\n", + "delta10=0.028\n", + "delta11=0.032\n", + "delta12=0.036\n", + "delta13=0.040\n", + "delta14=0.044\n", + "delta15=0.049\n", + "delta16=0.054\n", + "delta17=0.061\n", + "delta18=0.069\n", + "#stress and strain corresponding to these loads and elongations are\n", + "sigma1=P1/A_i\n", + "strain1=delta1/h0\n", + "sigma2=P2/A_i\n", + "strain2=delta2/h0\n", + "sigma3=P3/A_i\n", + "strain3=delta3/h0\n", + "sigma4=P4/A_i\n", + "strain4=delta4/h0\n", + "sigma5=P5/A_i\n", + "strain5=delta5/h0\n", + "sigma6=P6/A_i\n", + "strain6=delta6/h0\n", + "sigma7=P7/A_i\n", + "strain7=delta7/h0\n", + "sigma8=P8/A_i\n", + "strain8=delta8/h0\n", + "sigma9=P9/A_i\n", + "strain9=delta9/h0\n", + "sigma10=P10/A_i\n", + "strain10=delta10/h0\n", + "sigma11=P11/A_i\n", + "strain11=delta11/h0\n", + "sigma12=P12/A_i\n", + "strain12=delta12/h0\n", + "sigma13=P13/A_i\n", + "strain13=delta13/h0\n", + "sigma14=P14/A_i\n", + "strain14=delta14/h0\n", + "sigma15=P15/A_i\n", + "strain15=delta15/h0\n", + "sigma16=P16/A_i\n", + "strain16=delta16/h0\n", + "sigma17=P17/A_i\n", + "strain17=delta17/h0\n", + "sigma18=P18/A_i\n", + "strain18=delta18/h0\n", + "#part(a)\n", + "#modulus of elasticity\n", + "E=(sigma13-sigma1)/(strain13-strain1)/10**3 #G Pa\n", + "#part(b)\n", + "# yield stress at D (shown in fig 10.6)\n", + "Yield=P15/A_i*1000/10**6 #M N/m**2\n", + "#part (c)\n", + "#ultimate stress (maximum)\n", + "ultimate_sigma=P19/A_i*10**3/10**6 #M N/m**3\n", + "#part (d)\n", + "#percentage contraction\n", + "percent_l=(h0-hf)/h0*100 #%\n", + "#part(e)\n", + "#percentage increase in area\n", + "percent_A=(df**2-d0**2)/d0**2*100 #%\n", + "#part(f)\n", + "#apparent breaking stress\n", + "app_breaking_stress=P19*1000/A_i/10**6 #M N/m**2\n", + "#actual breaking stress\n", + "actual_breaking_stress=P19*1000/A_f/10**6 #M N/m**2\n", + "print \"\"\"The modulus of elasticity is %0.3f G Pa \n", + "Yield stress is %0.3f M n/m**2\n", + "The ultimate (maximum) stress is %0.3f M n/m**2 \n", + "Percentage contraction in length %0.3f\n", + "Percentage increase in area is %0.3f \n", + "Apparent breaking stress is %0.3f M n/mm**2\n", + "Actual breaking point is %0.3f M n/m**2\"\"\"%(E,Yield,ultimate_sigma,percent_l,percent_A,app_breaking_stress,actual_breaking_stress)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The modulus of elasticity is 130.694 G Pa \n", + "Yield stress is 253.915 M n/m**2\n", + "The ultimate (maximum) stress is 475.185 M n/m**2 \n", + "Percentage contraction in length 22.148\n", + "Percentage increase in area is 32.115 \n", + "Apparent breaking stress is 475.185 M n/mm**2\n", + "Actual breaking point is 359.675 M n/m**2\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-10.3 page no-307" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#length of glass piece\n", + "l=1.1*10**3 #mm\n", + "#width of glass piece\n", + "b=225 #mm\n", + "#height or thicness of plate\n", + "h=10 #mm\n", + "#load\n", + "P=250 #N\n", + "#for a simply supported beam subjected to concentrated load in the middle of its span,\n", + "M=P*l/4 #N mm\n", + "#and force\n", + "F=P/2 #N\n", + "#part(a)\n", + "#flexure strength\n", + "sigma=6*M/b/h**2 #N/mm**2\n", + "#part (b)\n", + "#shear strength\n", + "tau=3*F/2/b/h #N/mm**2\n", + "#part (c)\n", + "P1=350 #N\n", + "M1=P1*l/4\n", + "#ineria\n", + "I=b*h**3/12 #mm**4\n", + "y=h/2 #mm\n", + "#the modulus of rupture is given by\n", + "sigmar=M1*y/I\n", + "print \"The flexture strength, shear strength and modulus of rupture are :\\n%0.3f N/mm**2, %0.3f N/mm**2 and %0.3f N/mm**2 resp\"%(sigma, tau, sigmar)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The flexture strength, shear strength and modulus of rupture are :\n", + "18.333 N/mm**2, 0.083 N/mm**2 and 25.667 N/mm**2 resp\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-10.4 page no-313" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, sqrt\n", + "#given\n", + "#diameter of ball\n", + "D=0.5*10 #mm\n", + "#indentation diameter\n", + "d=32.5/10 #mm (diveided by 10 because it is 10 times magnified)\n", + "#from table- 10.3 of book , the load for steel specimen is\n", + "P=30*D**2 #kg f\n", + "#hardness\n", + "BHN=P/((pi)*D/2*(D-sqrt(D**2-d**2))) \n", + "print \"The hardness is %0.3f\" %(BHN)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The hardness is 79.556\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-10.5 page no-321" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sin, cos, acos\n", + "#given\n", + "#dimension of steel specimen\n", + "l=75 #mm\n", + "b=10 #mm\n", + "t=10 #mm\n", + "#depth of V-notch is t/5\n", + "#in the absence of specimen, frictional and windage loss\n", + "L1=0.1 #kg f.m\n", + "#in the presence ofspecimen, which is placed on support breaks\n", + "L2=5.9 #kg f m\n", + "#rupture energy\n", + "U=L2-L1 #kg f m\n", + "#since the depth of V-notch is t/5\n", + "#so t/5=2\n", + "te=t-2 #mm\n", + "#volume of specimen\n", + "Ve=l*b*te*10**-9 #m**3\n", + "#modulus of rupture\n", + "Ur=U/Ve #kg f/m**2\n", + "#effective area of cross section\n", + "Ae=b*te*10**-6 #m**2\n", + "#notch impact strength\n", + "Is=U/Ae #kg/m\n", + "#given that \n", + "Ui=30 #kg f.m\n", + "alpha=160*(pi)/180 #radians\n", + "#swing diameter\n", + "D=1600 #mm\n", + "R=D/2*10**-3 #m\n", + "#weight of hammer\n", + "#as we know that\n", + "#Ut=W*R(1-cos(aplha))\n", + "#so\n", + "W=Ui/R/(1-cos(alpha)) #kg f.m\n", + "#capacity of izod impact testing machine\n", + "L3=30 #kg f.m\n", + "#so Uf will be\n", + "Uf=L3-L2 #kbg f.m\n", + "#we know that energy after rupture\n", + "#Uf=W*R(1-cos(beta))\n", + "bet=acos(1-Uf/W/R) #radins\n", + "#beta in degrees\n", + "Beta=bet*180/(pi) #degrees\n", + "#also we know that Uf=W*hf\n", + "#so hf will be\n", + "hf=Uf/W #m\n", + "print \"\"\"rupture energy is %0.3f kg f.m \n", + "modulus of rupture %0.3f kg f/m**2 \n", + "notch impact strength %0.3f kg/m \n", + "angle of hammer after striking %0.3f degrees\n", + "height risen by hammer after breaking %0.3f m\"\"\"%(U,Ur,Is,Beta,hf)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "rupture energy is 5.800 kg f.m \n", + "modulus of rupture 966666.667 kg f/m**2 \n", + "notch impact strength 72500.000 kg/m \n", + "angle of hammer after striking 123.933 degrees\n", + "height risen by hammer after breaking 1.247 m\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-10.6 page no-323" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#diameter of circular section of bean\n", + "d=60 #mm\n", + "#length of circular section of beam\n", + "l=500 #mm\n", + "#maximum and minimum load \n", + "Pmin= 20 #kN\n", + "Pmax= 50 #kN\n", + "#ultimate strength\n", + "sigmau=650 #MPa\n", + "#yoeld strength\n", + "sigmay=520 #MPa\n", + "#factor of safety\n", + "fos=1.8\n", + "#maximum bending moment\n", + "Mmax=Pmax*l/4 #kN mm\n", + "#minimum bending moment\n", + "Mmin=Pmin*l/4 #kN mm\n", + "#mean bending moment\n", + "Mm=(Mmax+Mmin)/2 #kN mm\n", + "#alerting (variable) bending moment\n", + "Ma=(Mmax-Mmin)/2 #kN mm\n", + "#section modulus of beam\n", + "Z=(pi)*d**3/32 #mm**3\n", + "#mean bending stress\n", + "sigmam=Mm/Z*1000 #MPa\n", + "#variable bending stress\n", + "sigmaa=Ma/Z*1000 #MPa\n", + "#endurance stress from\n", + "#(i) gerber's parabolic relation\n", + "sigmae1=sigmaa/(1/fos-(sigmam/sigmau)**2*fos) #MPa\n", + "#(ii) goodman's straight line relation\n", + "sigmae2=sigmaa/(1/fos-sigmam/sigmau) #MPa\n", + "#(iii) soderberg's straight line realtion\n", + "sigmae3=sigmaa/(1/fos-sigmam/sigmay) #MPa\n", + "print \"\"\"Endurance strength of the material are :\n", + "Gerbers parabolic fomula %0.3f MPa\n", + "goodmans straight line formula %0.3f MPa\n", + "sodergerbs straight line relation %0.3f MPa \"\"\"%(sigmae1,sigmae2,sigmae3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Endurance strength of the material are :\n", + "Gerbers parabolic fomula 236.280 MPa\n", + "goodmans straight line formula 371.272 MPa\n", + "sodergerbs straight line relation 556.791 MPa \n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-10.7 page no-329" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#yeild strength of polycrystalline material increases from sigmay1 to sigmay2\n", + "sigmay1=118 #MPa\n", + "sigmay2=207 #MPa\n", + "#decreasing grain diameter from d1 to d2\n", + "d1=0.253*10**-3 #m\n", + "d2=0.0224*10**-3 #m\n", + "#to find the yield strngth at \n", + "d=0.095*10**-3 #m\n", + "#as we know that according to hall and petch equation,\n", + "#sigmay=sigma0+C/sqrt(d)\n", + "#putting sigmay1,sigmay2,d1 and d2.. we get 2 equations\n", + "#sigma0+C/sqrt(d1)=sigmay1 -------(1)\n", + "#and sigma0+C/sqrt(d2)=sigmay2 -----(2)\n", + "#solving equation 1 and 2 we get\n", + "sigma0=80.3 #MPa\n", + "#and\n", + "C=0.1896 #MN/m**(3/2)\n", + "#so the yield stress for the grain size\n", + "d=0.095*10**-3 #m\n", + "sigma=sigma0+C/sqrt(d) #MPa\n", + "print \"The yield stress for a grainof size of 0.095 mm is %0.3f MPa\"%(sigma)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The yield stress for a grainof size of 0.095 mm is 99.753 MPa\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-10.8 page no-330" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#ASTM number\n", + "n=12\n", + "#as we know that\n", + "N=2**(n-1) \n", + "#1 square inch=645 mm**2\n", + "#so grain diameter for ASTM number 12 will be\n", + "d=1/sqrt((N/645)*10**4) #mm\n", + "print \"The grain diameter of ASTM number 12 is %0.4f mm\"%(d)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The grain diameter of ASTM number 12 is 0.0056 mm\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-10.9 page no-330" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#ASTM number of grain\n", + "n=5\n", + "#as we know that\n", + "N=2**(n-1) #grains/inche**2 at magnification 100*\n", + "# as lineal and areal magnifications are related as *100=10,000 areal\n", + "N1=N/0.01/0.01 #grains/inch**2 at 1*\n", + "#average area of one grain\n", + "A=2.54*2.54/N1 #cm**2\n", + "#now N1 grains/ inch**2 of surface is = sqrt(160,000)=400 grain/inch of length and this is equal to =(400)**3=6.4*10**7 grains/m**3 of volume\n", + "#surface area of eaxh cubic surface\n", + "S=(1/400)**2 #inch**2\n", + "#there are 6 surfaces in a cubic grain\n", + "#so total surface area of each grain\n", + "TS=6*S #inch**2\n", + "#each boundary is composed of two grain surfaces, therefore , total boundary in the volume is\n", + "TotS=1/2*TS*(400)**3 #inch**2 boundary per cubic of steel\n", + "#total suface area in cm**2\n", + "TotalS=TotS/(2.54) #cm**2 boundary per cubic cm of steel\n", + "print \"total boundary in the volume is %0.3f cm**2 per cm**3 of steel\"%(TotalS)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "total boundary in the volume is 472.441 cm**2 per cm**3 of steel\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-10.16 page no-332" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#ASTM number\n", + "n=4\n", + "#as we know that\n", + "N=2**(n-1) #per inch**2 at a magnification of 100\n", + "#let r be the radius of grain\n", + "#so\n", + "#N*A=1/100 inch**2 where A=(pi)*r**2\n", + "#so\n", + "r=sqrt(1/100/N/(pi)) #inch\n", + "#radius of grain in mm\n", + "R=r*25.4 #mm\n", + "print \"the radius of grain is %0.3f mm\"%(R)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the radius of grain is 0.507 mm\n" + ] + } + ], + "prompt_number": 24 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch11.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch11.ipynb new file mode 100644 index 00000000..f77066eb --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch11.ipynb @@ -0,0 +1,485 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:832a91ae4051faf82e71f1eeca9ef1ab61ad84ac2a8a6a3e0aeb6be4b007f372" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-11, Theories of inelastic, enelastic and viscoelastic deformation and fracture" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-11.1 page no- 343" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import sqrt\n", + "#given\n", + "#applied stress\n", + "sigmax=3.5 #MPa\n", + "#aluminium crystal slips from (111) plane in the direction (110) when the stess is applied to (1-11)\n", + "#so\n", + "h1=1\n", + "k1=1\n", + "l1=1\n", + "h2=1\n", + "k2=-1\n", + "l2=1\n", + "#magnitude of plane (111)\n", + "M1=sqrt(h1**2+k1**2+l1**2)\n", + "#magnitude of (1-11)\n", + "M2=sqrt(h2**2+k2**2+l2**2)\n", + "#direction (110)\n", + "h3=1\n", + "k3=1\n", + "l3=0\n", + "#magnitude of direction(110)\n", + "M3=sqrt(h3**2+k3**2+l3**2)\n", + "#the angle between the planes (111) and (1-11) is \n", + "cosphie=((h1*h2+k1*k2+l1*l2)/(M1*M2))\n", + "sinphie=sqrt(1-(cosphie)**2)\n", + "#similarly angle between the plane (111) and the direction (110) is given by\n", + "costheta=((h1*h3+k1*k3+l1*l3)/(M1*M3)) \n", + "#critical resolved shear stress\n", + "taucr=sigmax*2*sinphie*cosphie*costheta/2 #MPa\n", + "print \"the critical resolved shear stress is %0.3f MPa\"%(taucr)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the critical resolved shear stress is 0.898 MPa\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exaplme-11.2 page no-351" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#shera stress\n", + "tau=715*10**6 #Pa\n", + "#shear modulus\n", + "G=25*10**9 #Pa\n", + "#atomic radius\n", + "b=4.05*10**-10 #m\n", + "#as we know that \n", + "#tau=G*b/l\n", + "#so\n", + "l=G*b/tau #m\n", + "print \"the length of frank- read source in aluminium crystal is %.10f m\"%(l)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the length of frank- read source in aluminium crystal is 0.0000000142 m\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-11.2 page no-351" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#shear modulous\n", + "G=25*10**9 #Pa\n", + "#shear stress\n", + "tau=50*10**6 #Pa\n", + "#lattice constant of aluminium is\n", + "a=4.05*10**-10 #m\n", + "#burger's vector for aluminium is 1/2(110)\n", + "h=1\n", + "k=1\n", + "l=0\n", + "#atomic radius of aluminium \n", + "b=a/sqrt(h**2+k**2+l**2) # m\n", + "#as we know that\n", + "#tau=G*b/L\n", + "#so\n", + "L=G*b/tau #m\n", + "#disloaction density is rhoD (let)\n", + "rhoD =1/L**2 # per m**2 (calculation mistake, there in book formula written is correct but calculation is wrong)\n", + "print \"the dislocation density is %0.3e per m**2\"%(rhoD)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the dislocation density is 4.877e+13 per m**2\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-11.4 page no-352" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#burger vector\n", + "b=4*10**-10 #m\n", + "#density\n", + "N=10**13 #lines/m**2\n", + "#velocity\n", + "v=10**-5 #m/s\n", + "#rate of straining\n", + "epsilonP=N*b*v #lines/sec\n", + "print \"the rate of starining is %0.3f lines/sec\"%(epsilonP)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the rate of starining is 0.040 lines/sec\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-11.5 page no-357" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log, exp\n", + "#given\n", + "#conventional stress\n", + "sigmac=98.9 #MPa\n", + "#conventional strain\n", + "epsilonc=0.35 #mm/mm\n", + "#as we know that\n", + "epsilont=log(1+epsilonc) #mm/mm\n", + "#also\n", + "#sigmac=sigmat*exp(-epsilont) \n", + "#so\n", + "sigmat=sigmac/exp(-epsilont) #MPa\n", + "print \"the true strees is %0.3f MPa\\n and true strain is %0.3f mm/mm\"%(sigmat,epsilont)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the true strees is 133.515 MPa\n", + " and true strain is 0.300 mm/mm\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-11.6 page no-357" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#material constant\n", + "K=500 #MPa\n", + "#strain hardenening coefficient\n", + "n=0.20\n", + "#according to true stress-strain relation\n", + "#sigmat=d sigmat/d epsilont=d(Kepsilont**n)/d epsilont\n", + "#so we get\n", + "#sigmat=n*K*epsilont**(n-1) -----(1)\n", + "#also we know that\n", + "#sigmat=K*epsilont**n ------------(2)\n", + "#equating (1) and (2)\n", + "epsilont=n\n", + "#also we know that\n", + "#epsilont=log (1+epsilonc)\n", + "#so\n", + "epsilonc=exp(epsilont)-1 \n", + "#now\n", + "sigmat=K*epsilont**n #MPa\n", + "print \"\"\"the maximum tensile strength %0.3f\n", + "the linear strain are %0.3f MPa\"\"\"%(epsilonc,sigmat)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the maximum tensile strength 0.221\n", + "the linear strain are 362.390 MPa\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exapmle-11.7 page no-363" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#applied strain\n", + "epsilon=0.4\n", + "#immediate stress\n", + "sigmai=10*10**9 #Pa\n", + "#after 42 days stress is\n", + "sigma=5*10**9 #Pa\n", + "t=42 #days\n", + "#as we know that\n", + "#sigma=sigmai*exp(-t/tr)\n", + "#so\n", + "tr=t/log(sigmai/sigma) #days\n", + "#stress after 90 days\n", + "t90=90 #days\n", + "sigma90=sigmai*exp(-t90/tr)*10**-9 #MPa\n", + "print \"\"\"relaxation time for the tensile stress to decrease from 10 to 5 MPa is %0.3f days\n", + "the stress after 90 days is %0.3f MPa\"\"\"%(tr,sigma90)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "relaxation time for the tensile stress to decrease from 10 to 5 MPa is 60.593 days\n", + "the stress after 90 days is 2.264 MPa\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-11.8 page no-370" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "#given\n", + "#crack length of glass piece\n", + "l=3*10**-6/2 #m divided by 2 because min general we consider 2*l be the length of crack\n", + "#young modulus\n", + "E=70*10**9 #Pa\n", + "#specific surface energy\n", + "gammae=1.05 #/J/m**2\n", + "#fracture strength\n", + "sigmaf=sqrt(2*E*gammae/(pi)/l) #N/m**2\n", + "#ratio of strength and young's modulus\n", + "R=sigmaf/E \n", + "print \"\"\"fracture strength is %0.3e N/m**2\n", + "the ratio of strength and youngs modulus is %0.3f\"\"\"%(sigmaf,R)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "fracture strength is 1.766e+08 N/m**2\n", + "the ratio of strength and youngs modulus is 0.003\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "exaple-11.9 page no-370" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#young's modulus\n", + "E=71*10**9 #Pa\n", + "#fracture strength\n", + "sigmaf=115*10**6 #Pa\n", + "#lenght of crack\n", + "#2l=6 micro m\n", + "#so\n", + "l=6/2*10**-6 #m\n", + "#as we know that\n", + "#sigmaf=sqrt(2*E*sigmac/(pi)/l)\n", + "#so\n", + "sigmac=(sigmaf)**2*(pi)*l/2/E #J/m**2 \n", + "print \"surface energy of the etched glass is %0.3f J/m**2\"%(sigmac)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "surface energy of the etched glass is 0.878 J/m**2\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-11.10 page no-370" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "E=70*10**9 #Pa\n", + "gammae=0.85 #J/m**2\n", + "l=2*10**-6 #m\n", + "#necessary stress to satisfy the griffith's energy\n", + "sigmaf=sqrt(2*E*gammae/(pi)/l)/10**6 #Pa\n", + "print \"the necessary stress is %0.3f MPa\"%(sigmaf)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the necessary stress is 137.621 MPa\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-11.11 page 371" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "E=72*10**9 #Pa\n", + "gammae=0.9 #J/m**2\n", + "sigmaf=17.5*10**6 #Pa\n", + "#as we know that\n", + "#sigmaf=sqrt(2*E*gammae/(pi)/l) #N/m**2\n", + "#so\n", + "l=2*E*gammae/(pi)/(sigmaf)**2*1000 #mm\n", + "#length of intenal crack\n", + "L=2*l #m\n", + "print \"\"\"length of crack on the outer surface is %0.3f mm\n", + "length of crack internally %0.3f mm\"\"\"%(l,L)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "length of crack on the outer surface is 0.135 mm\n", + "length of crack internally 0.269 mm\n" + ] + } + ], + "prompt_number": 26 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch12.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch12.ipynb new file mode 100644 index 00000000..fcc3aca3 --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch12.ipynb @@ -0,0 +1,178 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:6fb610a5c61b7f5f5a11cdaca7365589bd559a2226305d387dc828099826367f" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-12, Organic materials, plastic, rubber, elastomers and wood" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-12.1 page no-382" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#degree of polymerization of styrene\n", + "DOP=10000\n", + "#formula of styrene= C8H8\n", + "#molecular weight of styrene monomer\n", + "Mm=12*8+1*8\n", + "#molecular weight of polymer\n", + "Mp=DOP*Mm\n", + "print \"the molecular weight of styrene polymer is %d\"%(Mp)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the molecular weight of styrene polymer is 1040000\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-12.2 page no-382" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#degree of polymerization of teflon\n", + "DOP=100000\n", + "#chemical formula of teflon is C2F4\n", + "#molecular weight of monomer teflon\n", + "Mm=2*12+4*19\n", + "#molecular weight of teflon polymer\n", + "Mp=DOP*Mm\n", + "#molecular weight of polythene monomer i.ee C2H4\n", + "MmP=2*12+4*1\n", + "#molecular weight of polythene polymer\n", + "MpP=DOP*MmP\n", + "print \"molecular weight of PTFE anf Polythene are %d and %d\"%(Mm,MpP)\n", + "#ratio of molecular weight of PTFE and Polythene\n", + "R=Mp/MpP\n", + "print \"the ration of molecular weight of PTFE and Polythene having same DOP is %0.3f\"%(R)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "molecular weight of PTFE anf Polythene are 100 and 2800000\n", + "the ration of molecular weight of PTFE and Polythene having same DOP is 3.571\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-12.3 page no-396" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#the molecular weight of polyisoprene monomer\n", + "Mm=68 #gm\n", + "#after vulcanisation with sulphur, it is observedthat the 2 molecules of isoprene monimer require 2 molecules of sulphur\n", + "#hence for full cross linking ,(68*2) gm of isoprene requires (32*2)gm of sulphur. therefore 68kg of isoprene requires\n", + "M=32*2*68/68/2 #kg of sulphur\n", + "print \"the weight of sulphur required for cross link polymerization of polyisoprene is %d of sulphur\"%(M)\n", + "print \"the fully cross linked product will be EBONITE\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the weight of sulphur required for cross link polymerization of polyisoprene is 32 of sulphur\n", + "the fully cross linked product will be EBONITE\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-12.4 page no-396" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#molecular weight of butdiene,isoprene,sulhur and carbon black are\n", + "MB=4*12+6*1\n", + "MI=68\n", + "MS=32\n", + "MC=12\n", + "#percentages of diiferent constituents in rubber\n", + "PB=27/MB\n", + "PI=51/MI\n", + "PS=16/MS\n", + "PC=6/MC\n", + "#percentage of coss linking\n", + "percent=PS/(PI+PB)*100\n", + "print \"the percentage of cross linking is %d\"%(percent)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the percentage of cross linking is 40\n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch14.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch14.ipynb new file mode 100644 index 00000000..f656ae27 --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch14.ipynb @@ -0,0 +1,306 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:00cefcd555e0826aa9214fd45f5432d45db966e939f0000eade5d36dd64962a8" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-14, Composite Material" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-14.1 page no-429" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#part(a)\n", + "#thickness of GRPS sheet\n", + "ts=3 #mm\n", + "#depth of this skin\n", + "tc=24 #mm\n", + "#breadth of skin\n", + "b=100 #mm\n", + "d=(ts+tc)/2 #mm\n", + "#moduli of polyster skin and foamare\n", + "Es=7000 #N/mm**2\n", + "Ec=20 #N/mm**2\n", + "#values of Is and Ic are \n", + "Is=2*(b*ts**3/12+b*ts*d**2) #mm**4\n", + "Ic=b*tc**3/12 #mmm**4\n", + "#as we know that flexure rigidity is D, flexure rigidity of skin is Ds and that of core is Dc\n", + "#D=Ds+Dc \n", + "D=2*Es*Is+Ec*Ic #mm**2 for two skins ---------------(1)\n", + "#part(b)\n", + "ts1=6 #mm\n", + "#Ds1=Es*Is1 \n", + "Ds1=Es*b*ts1**2/12 #Nmm**2 --------------(2) (there is calculation mistake I myself have checked from calculator too)\n", + "#ratio of (1) and (2) is R (let)\n", + "R=D/Ds1 \n", + "print \"flexure rigidity of sandwich beam is %d times more\"%(R)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "flexure rigidity of sandwich beam is 733 times more\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-14.2 page no-441" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#young's modulus of aluminium, iron and boron are resp\n", + "#here we are representing a,b and c for aluminum, iron and boron resp.\n", + "Ea=71*10**9 #Pa\n", + "Eb=210*10**9 #Pa\n", + "Ec=440*10**9 #Pa\n", + "#as we know that Ec=Ef*Vf+Em*Vm where Ef and Em are the young's modulus of fibre and matrix resp.\n", + "#so we get\n", + "#210=71*Va+440*Vb --------------(1)\n", + "#assumin void volume is zero and we know that\n", + "#Va+Vb=1\n", + "#so Vb=1-Va ---------------(2)\n", + "#on solving 1 and 2 we get\n", + "Va=31.8 \n", + "Vb=68.2\n", + "#ratio of Va and Vb\n", + "R=Va/Vb\n", + "print \"the volume ratio of aluminium and boron in aluminium boron composite %0.2f\"%(R)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the volume ratio of aluminium and boron in aluminium boron composite 0.47\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-14.3 page no-442" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#elastic moduli of carbon fibre and epoxy resin\n", + "Ef=430 #GPa\n", + "E=3.6 #/GPa\n", + "#modular ratio\n", + "Pf=Ef/E\n", + "#part(a)\n", + "Vf1=0.15\n", + "Vm1=1-Vf1\n", + "R1=Vm1/Vf1\n", + "Pc1=Pf+R1\n", + "volume_fibre1=Pf/Pc1\n", + "#part(b)\n", + "Vf2=0.65\n", + "Vm2=1-Vf2\n", + "R2=Vm2/Vf2\n", + "Pc2=Pf+R2\n", + "volume_fibre2=Pf/Pc2\n", + "print \"\"\"the fraction of load carried by fibre in carbon-epoxy matrix composite containing 15 percent fibre by volume is %0.3f\n", + "that by 65 percent fibre by volume is %0.3f\"\"\"%(volume_fibre1,volume_fibre2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the fraction of load carried by fibre in carbon-epoxy matrix composite containing 15 percent fibre by volume is 0.955\n", + "that by 65 percent fibre by volume is 0.996\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-14.4 page no-442" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "Vf=.65\n", + "Vm=1-Vf\n", + "#part(a)\n", + "#longitudinal strength is determined by\n", + "#sigmac=sigmaf*Vf+sigmam*Vm\n", + "#here according to table given in question we have\n", + "sigmaf=2.8\n", + "sigmam=0.0025\n", + "sigmac=sigmaf*Vf+sigmam*Vm #GPa\n", + "#part(b)\n", + "#longitudinal modulus is given by\n", + "#Ec=Ef*Vf+Em*Vm\n", + "#here according to table\n", + "Ef=130\n", + "Em=3.5\n", + "Ec=(Ef*Vf+Em*Vm) #GPa\n", + "#part(c)\n", + "#transverse modulus is given by\n", + "#1/EC=Vf/Ef+Vm/Em\n", + "EC=(Vf/Ef+Vm/Em)**-1 #GPa\n", + "#part(d)\n", + "#poisson ratio\n", + "#nuLT=nuf*Vf+num*Vm\n", + "#here according to the table\n", + "nuf=0.34\n", + "num=0.36\n", + "nuLT=nuf*Vf+num*Vm \n", + "#part(e)\n", + "#shear modulus\n", + "#1/Glt=Vf/Gf+Vm/Gm\n", + "# here according to the table \n", + "Gf=2.2\n", + "Gm=1.2\n", + "GLT=(Vf/Gf+Vm/Gm)**-1 #GPa\n", + "print \"\"\"the longitudinal strength is %0.3f GPa\n", + "the logitudinal modulus is %0.3f GPa\n", + "the transvrse modulus is %0.3f GPa\n", + "the poissons ratio is %0.3f\n", + "shear modulus is %0.3f GPa\"\"\"%(sigmac,Ec,EC,nuLT,GLT)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the longitudinal strength is 1.821 GPa\n", + "the logitudinal modulus is 85.725 GPa\n", + "the transvrse modulus is 9.524 GPa\n", + "the poissons ratio is 0.347\n", + "shear modulus is 1.703 GPa\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-14.5 page no-444" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#content of polyster\n", + "Vm=0.45\n", + "Vf=1-Vm\n", + "#longitudinal strength and modulus are calculated using following formulas\n", + "#sigmac=nu*sigmaf*Vf+singmam*Vm\n", + "#Ec=nu*Ef*Vf+Em*Vm\n", + "#according to the table given in the ques\n", + "Ef=240 #GPa\n", + "Em=4.5 #GPa\n", + "sigmaf=1.7 #GPa\n", + "sigmam=0.0029 #GPa\n", + "Gf=1.7\n", + "Gm=1.4\n", + "nuf=0.28\n", + "num=0.32\n", + "#values of nu\n", + "nu1=1\n", + "nu2=1/2\n", + "nu3=1/4\n", + "nu4=3/8\n", + "nu5=1/6\n", + "#now longitudinal strength for differnt nu\n", + "sigmac1=nu1*(sigmaf*Vf+sigmam*Vm) #GPa\n", + "sigmac2=nu2*(sigmaf*Vf+sigmam*Vm) #GPa \n", + "sigmac3=nu3*(sigmaf*Vf+sigmam*Vm) #GPa\n", + "sigmac4=nu4*(sigmaf*Vf+sigmam*Vm) #GPa\n", + "sigmac5=nu5*(sigmaf*Vf+sigmam*Vm) #GPa\n", + "#liongitudina modulus for differnt nu\n", + "EC1=nu1*(Ef*Vf+Em*Vf) #GPa\n", + "EC2=nu2*(Ef*Vf+Em*Vf) #GPa\n", + "EC3=nu3*(Ef*Vf+Em*Vf) #GPa\n", + "EC4=nu4*(Ef*Vf+Em*Vf) #GPa\n", + "EC5=nu5*(Ef*Vf+Em*Vf) #GPa\n", + "print \"\"\"longitudinal strength and longitudinal modulus for nu=1 is %0.3f GPa and %0.3f GPa\n", + "longitudinal strength and longitudinal modulus for nu=1/2 is %0.3f GPa and %0.3f GPa\n", + "longitudinal strength and longitudinal modulus for nu=1/4 is %0.3f GPa and %0.3f GPa\n", + "longitudinal strength and longitudinal modulus for nu=3/8 is %0.3f GPa and %0.3f GPa\n", + "longitudinal strength and longitudinal modulus for nu=1/6 is %0.3f GPa and %0.3f GPa\"\"\"%(sigmac1,EC1,sigmac2,EC2,sigmac3,EC3,sigmac4,EC4,sigmac5,EC5)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "longitudinal strength and longitudinal modulus for nu=1 is 0.936 GPa and 134.475 GPa\n", + "longitudinal strength and longitudinal modulus for nu=1/2 is 0.468 GPa and 67.237 GPa\n", + "longitudinal strength and longitudinal modulus for nu=1/4 is 0.234 GPa and 33.619 GPa\n", + "longitudinal strength and longitudinal modulus for nu=3/8 is 0.351 GPa and 50.428 GPa\n", + "longitudinal strength and longitudinal modulus for nu=1/6 is 0.156 GPa and 22.412 GPa\n" + ] + } + ], + "prompt_number": 12 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch15.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch15.ipynb new file mode 100644 index 00000000..ad133b85 --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch15.ipynb @@ -0,0 +1,240 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:8e6683b0ce6857dd42d958479a1ed2d6f82135436c1b895c7b5905b161f18985" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-15, Phase transformation" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-15.1 page no-461" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import log, exp, pi\n", + "#given\n", + "#reaction comlete in 500 mins at 10 degree celcius and in 1 min at 80 degrees celcius\n", + "#so\n", + "t1=1 #min\n", + "t2=500 #min\n", + "T1=273+80 #kelvin\n", + "T2=273+10 #Kelvin\n", + "#gas constant \n", + "R=8.314\n", + "#as we know that phie=1/t*\n", + "#so\n", + "#log(phie1/phie2)=log(t2/t1)\n", + "#so we get\n", + "#log(t2/t1)=Ea/2.303*(1/T1-1/T2)\n", + "#so Ea is given by\n", + "Ea=2.303*log(t2/t1)*(1/(1/T1-1/T2))*R\n", + "#to find the reaction completion time when temperature is 40 degrees celcius\n", + "#so\n", + "T3=273+40 #Kelvin\n", + "t3=t1*exp(Ea/2.303/R*(1/T1-1/T3)) #min\n", + "print \" the reaction completion tym at 40 degree celcius is %0.2f min\"%(t3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the reaction completion tym at 40 degree celcius is 24.80 min\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-15.2 page no-463" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#temp\n", + "T=983+273 #K\n", + "Tm=1083+273 #K\n", + "deltaT=Tm-T #K\n", + "#given that\n", + "#latent heat of fusion of copper\n", + "deltaHm=1.88*10**9 #J/m**3\n", + "#interface energy/unit area\n", + "gammasL=0.144 #J/m**2\n", + "#change in free energy of vapour\n", + "deltaGv=deltaHm*deltaT/Tm #J/m**3\n", + "#critical radius of copper during solidification\n", + "#for notation only we are using r*=R\n", + "R=2*gammasL/deltaGv*10**10 #A\n", + "print \" the critical radius of copper during solidification is %0.3f A\"%(R)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the critical radius of copper during solidification is 20.773 A\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-15.3 page no-463" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#latent heat of fusion of pure gold\n", + "deltaHf=-1.16*10**9 #J/m**2\n", + "#surface free energy\n", + "gama=0.132 #J/m**2\n", + "#melting point of gold\n", + "Tm=1064+273 #K\n", + "#supercooling value\n", + "deltaT=230 #K\n", + "#critical radius is given by R intead of r* just for notation\n", + "R=(-2*gama*Tm/deltaHf)*(1/deltaT)*10**9 #nm\n", + "print \" critical radius of pure gold at 230 degree celcius is %0.3f nm\"%(R)\n", + "#change in free energy of vapours is\n", + "deltaGv=deltaHf*deltaT/Tm #J/m**3\n", + "#activation free energy\n", + "#for notation deltaG* we use deltaG\n", + "deltaG=16*(pi)*(gama)**3/3/(deltaGv)**2 #J\n", + "print \" the activation free energy is %0.3e J\"%(deltaG)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " critical radius of pure gold at 230 degree celcius is 1.323 nm\n", + " the activation free energy is 9.677e-19 J\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-15.5 page no-472" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#diameter of hard inert particle\n", + "D=2 #m\n", + "#average centre to centre distance between the particles\n", + "l=20*10**-6 #m becaus 1 micro m=10**-6 m\n", + "#shear modulus of copper\n", + "G=41*10**9 #Pa\n", + "#burgers vector\n", + "b=0.64*10**-9 #m\n", + "#contribution of these particles on the yeild strength of material\n", + "tau=G*b/l/10**6 #MPa\n", + "print \" the contribution of these particles on the yeild strength of material is %0.3f MPa\"%(tau)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the contribution of these particles on the yeild strength of material is 1.312 MPa\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-15.7 page no-474" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#heat for recovery of yeild point in zinc\n", + "Q=83.14*10**3 #J/mole\n", + "#gas constant\n", + "R=8.314*10**3 #J/mol K\n", + "#temperature\n", + "T1=0+273 #K\n", + "T2=27+273 #K\n", + "#recovery time at 0 degree celcius\n", + "t2=5 #min\n", + "#the recovery time for the two different temp is given by\n", + "#1/t1=A*exp(-Q/R/T1) --------------(1)\n", + "#1/t2=A*exp(-Q/R/T2) -----------(2)\n", + "#taking ratio of (1) to (2)\n", + "#we get\n", + "t1=t2*exp(Q/R*(1/T1-1/T2)) #min (there is calculation mistake I myself have checked from calculator too)\n", + "print \" the recovery time at 27 degree celcius is %0.2f min\"%(t2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the recovery time at 27 degree celcius is 5.00 min\n" + ] + } + ], + "prompt_number": 12 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch16.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch16.ipynb new file mode 100644 index 00000000..5d1affc4 --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch16.ipynb @@ -0,0 +1,106 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:f92e9700573aa24c5004333de60aedbbc7f2ae80bd6c9e8b75276c45115c8416" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-16, Heat treatment" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-16.1 page no- 484" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import sqrt\n", + "#given\n", + "#atomic radiii of gamma-iron having FCC lattice\n", + "rFCC=1.26 #A\n", + "#atomic radius of alpha-iron having BCC lattica\n", + "rBCC=1.24 #A\n", + "#as we know that FCC and BCC has effective no of atoms 4 and 2 resp\n", + "#so\n", + "aBCC=4/sqrt(3)*rBCC #A\n", + "aFCC=2*sqrt(2)*rFCC #A\n", + "#volume of lattice for FCC and BCC\n", + "VFCC=(aFCC)**3 #A**3\n", + "VBCC=(aBCC)**3 #A**3\n", + "#percentage change in volume during phase transformation of gamma-iron to alpha-iron is given by\n", + "percent_vol_change=(VFCC/4-VBCC/2)/(VFCC/4)*100\n", + "print \"the percentage volume change during phase transformation of gamma-iron to alpha-iron is %0.3f\"%(percent_vol_change)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the percentage volume change during phase transformation of gamma-iron to alpha-iron is -3.764\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-16.3 page no-495" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#young's modulus \n", + "E=51*10**9 #Pa\n", + "#poisson's ratio\n", + "nu=0.22\n", + "#magnitude of burger's vector\n", + "b=2*10**-10 #m\n", + "#we know that shear modulus is given by\n", + "G=E/2/(1-nu) #Pa\n", + "#elastic strain energy of dislocation in cold worked tin\n", + "Ue=1/2*G*b**2*10**12 #Pa m\n", + "#the strain energy in tin before cold working may be neglected as it is smaller by three orders of magnitude.\n", + "print \" the change in energy during recrystallization is %0.3f Pa m\"%(Ue)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the change in energy during recrystallization is 653.846 Pa m\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch17.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch17.ipynb new file mode 100644 index 00000000..65767cdc --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch17.ipynb @@ -0,0 +1,188 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:8f783907d07720e4a586c63392d5e7f018b540c022c382705c66c64148aba8e2" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-17, Corrosion & Oxidation" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-17.1 page no-511" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#density of nickel and nickel oxide (NiO)\n", + "rhom=8900 #kg/m**3\n", + "rhoc=7080 #kg/m**3\n", + "#molecular weight of nickel and nickel oxide\n", + "Mm=58.51\n", + "Mc=74.71\n", + "#molar volumes are\n", + "Vm=Mm/rhom #m**3/mol\n", + "Vc=Mc/rhoc #m**3/mol\n", + "if Vc>Vm:\n", + " print \" it is clear that Vc>Vm, so a protective film will form over nickel\"\n", + "else :\n", + " print \" it is clear that Vc<Vm, so a protective film will not form over nickel\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " it is clear that Vc>Vm, so a protective film will form over nickel\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-17.2 page no-513" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "#given\n", + "#oxidation loss on the copper surface is 0.1 mm in 25 hours\n", + "t1=25 #hours\n", + "x1=0.1 #mm\n", + "#to find the loss in 300 hours\n", + "t2=300 #hours\n", + "# as we know that oxidation loss is given by\n", + "#x**2=Cp*t\n", + "#for t1 and t2 we have following equations\n", + "#x1**2=Cp*t1 -------(1)\n", + "#x2**2=Cp*t2 -----------(2)\n", + "#dividing equation 1 by 2\n", + "#we get\n", + "x2=x1*sqrt(t2/t1) #mm\n", + "print \"Loss in 300 hours is %0.3f mm\"%(x2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Loss in 300 hours is 0.346 mm\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-17.3 page no-515" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#atomic weight and density of magnesium\n", + "MMg=24.3 #gm\n", + "rhoMg=1.740 #g/cm**3\n", + "#molecular weight and density of MgO\n", + "MMgO=40.3 #gm\n", + "rhoMgO=3.650 #g/cm**3\n", + "#volume of Mg and Mgo\n", + "VMg=MMg/rhoMg #cm**3\n", + "VMgO=MMgO/rhoMgO #cm**3\n", + "#we know that pilling-bedworth ratio is given by\n", + "PBR=VMgO/VMg \n", + "if PBR<1:\n", + " print \"Since PBR is less than 1 so the film formed is not protective \"\n", + "else:\n", + " print \"Since PBR is greater than 1 so the film formed is protective \"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Since PBR is less than 1 so the film formed is not protective \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-17.4 page no-526" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#one mole of magnesium weighs\n", + "MMg=0.0243 #Kg\n", + "#current density\n", + "I=20*10**-3 #A/m**2\n", + "#design life of 15 years\n", + "t=15*365*24*3600 #sec\n", + "#charge= 2 faradya\n", + "Q=2*96490 #C\n", + "#charge needed per m**2 of structure's for design life\n", + "q=I*t #A s\n", + "#the amount of magnesium needed\n", + "AMg=MMg*q/Q #kg/m**2\n", + "print \"the amount of magnesiun meeded is %0.3f kg/m**2\"%(AMg)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the amount of magnesiun meeded is 1.191 kg/m**2\n" + ] + } + ], + "prompt_number": 8 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch18.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch18.ipynb new file mode 100644 index 00000000..08313513 --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch18.ipynb @@ -0,0 +1,274 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:7b265e751ff1e7a83221e62eaaf1fb3cb58beb0eac8456c9e59af597a832c4e3" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-18, Electron theories & conducting material properties, behaviour and application" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-18.1 page no-535" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#side of cube\n", + "l=20*10**-3 #m\n", + "#plank's constant\n", + "h=6.626*10**-34 #Js\n", + "#mass of electron\n", + "m=9.109*10**-31 #kg\n", + "# lower energy level\n", + "nx1=1\n", + "ny1=1\n", + "nz1=1\n", + "#higher energy level\n", + "nx2=2\n", + "ny2=1\n", + "nz2=1\n", + "#energy in the lower level\n", + "E1=h**2*(nx1**2+ny1**2+nz1**2)/(8*m*l**2) #J\n", + "#energy in the higher level\n", + "E2=h**2*(nx2**2+ny2**2+nz2**2)/(8*m*l**2) #J\n", + "#difference in energies of the two levels\n", + "deltaE=E2-E1 #J\n", + "print \"deltaE = %0.2e\"%deltaE\n", + "print \"since deltaE is very small, so the assumption that E varies continuously with k, is justified\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "deltaE = 4.52e-34\n", + "since deltaE is very small, so the assumption that E varies continuously with k, is justified\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-18.2 page no-537" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#distance between the plates\n", + "d=5*10**-3 #m\n", + "#voltage difference between the plates\n", + "V=230 #Volts\n", + "#electric field is given by\n", + "E=-V/d #V/m\n", + "print \"electric field between the plates is %.f V/m\"%(E)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "electric field between the plates is -46000 V/m\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-18.3 page no- 537" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#conductivity of material\n", + "sigma=0.018 # per ohm m\n", + "#no of electrons\n", + "n=10**19 #electrons/m**3\n", + "#voltage\n", + "V=0.16 #V\n", + "#thickness of material\n", + "t=0.29*10**-3 #m\n", + "#mass and charge of electron\n", + "m=9.1*10**-31 #kg\n", + "e=1.602*10**-19 #C\n", + "#electric feilf gradient\n", + "kie=V/t #V/m\n", + "#as we know that\n", + "#sigma=n*e*vd/kie\n", + "#so v\n", + "vd=sigma*kie/n/e #m/s\n", + "print \" drift velocity of carriers is %0.3f m/s\"%(vd)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " drift velocity of carriers is 6.199 m/s\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-18.4 page no-547" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "#given\n", + "#resistance of wire\n", + "R=21 #ohm\n", + "#length of wire\n", + "l=200 #m\n", + "#diameter of wire\n", + "d=0.44*10**-3 #m\n", + "#area of cross section of wire\n", + "A=(pi)/4*d**2 #m**2\n", + "#as we know that \n", + "#R=rho*l/A\n", + "#so\n", + "rho=R*A/l #ohm m\n", + "print \"the specific resistance of wire is %0.2e ohm m\"%(rho)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the specific resistance of wire is 1.60e-08 ohm m\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-18.5 page no-548" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#resistance of wire at 70 degree celcius is 57.2 ohm and at 25 degrees celcius 50 ohm\n", + "R25=50 #ohm\n", + "R70=57.2 #ohm\n", + "T1=25 #degree C\n", + "T2=70 #degree C\\\n", + "#as we know that\n", + "#Rt=R0*(1+t*alpha)\n", + "#putting above values in the above given equation we get\n", + "#R25=R0*(1+25*alpha) -------(1)\n", + "#R70=R0*(1+70*alpha) --------(2)\n", + "#applying R70/R25\n", + "#we get\n", + "#R70/R25=(1+70*alpha)/(1+25*alpha)\n", + "#solving for alpha we get equation as follows\n", + "alpha=(R70-R25)/(T2*R25-T1*R70) #K**-1\n", + "print \"the temperature coefficient of resistance is %f K**-1\"%(alpha)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the temperature coefficient of resistance is 0.003478 K**-1\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-18.6 page no-550" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#resistivity of copper , nickel and silver are\n", + "rhoCu=0.015*10**-6 #ohm m\n", + "rhoNi=0.012*10**-6 #ohm m\n", + "rhoAg=0.016*10**-6 #ohm m\n", + "#atomic percent of nickle and silver\n", + "CNi=0.25\n", + "CAg=0.40\n", + "#the resistivity of Cu-Ni-Ag alloy at 300 K\n", + "rho=rhoCu+(rhoNi*CNi)+(rhoAg*CAg) #ohm m\n", + "print \" the resistivity of Cu=Ni-Ag alloy is %.2e ohm m\"%(rho)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the resistivity of Cu=Ni-Ag alloy is 2.44e-08 ohm m\n" + ] + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch19.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch19.ipynb new file mode 100644 index 00000000..88f8962f --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch19.ipynb @@ -0,0 +1,648 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:67f7218fdc52f5680eb92fd41ae852facfa629b4691b8574101a958a20b2627c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-19, Semiconducting material property, behaviour & application" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-19.1 page no-561" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import exp, log\n", + "from __future__ import division\n", + "#given\n", + "pE=0.05\n", + "#let E-Ef=EE\n", + "EE=0.4 #eV\n", + "k=8.614*10**-5\n", + "#we know that\n", + "#p(E)=1/exp((E-EF)/k/T)\n", + "#putting above values in above equation we get\n", + "T=EE/k/log(1/pE-1) #K\n", + "print \"\"\"The temperature at which there is a probability of 5 percent \n", + "for an electron to occupy energy state which is 0.4 eV above the fermi level is %0.3f K\"\"\"%(T)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The temperature at which there is a probability of 5 percent \n", + "for an electron to occupy energy state which is 0.4 eV above the fermi level is 1577.076 K\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-19.2 page no-563" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#resistivity of pure silicon is \n", + "rho= 3000 #ohm m\n", + "#conductivity of pure silicon is reciproca of resistivity\n", + "sigma=1/rho #(ohm m)**-1\n", + "#mobility of electrons and holes\n", + "muh=0.05 #m**2/V s\n", + "mue=0.14 #m**2/V s\n", + "#charge on elctron\n", + "e=1.602*10**-19 #C\n", + "# we know that\n", + "#sigme=ne*ee*mue+nh*eh*muh\n", + "#here we have ne=nh=n and ee=eh=e\n", + "#so we have sigma=n*e*muh+n*e*mue\n", + "#so \n", + "n=sigma/(e*muh+e*mue) #ohm m**2/(C V s)\n", + "print \" the density of intrensic carriers is %0.3e /m**3\"%(n)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the density of intrensic carriers is 1.095e+16 /m**3\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-19.3 page no-563" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#mobility of silicon\n", + "mue=0.17 #m**2/V s\n", + "#mobility of holes\n", + "muh=0.035 #m**2/V s\n", + "#carrier density\n", + "n=1.1*10**16 #per m**3 (here ne=nh=n)\n", + "#electronic charge\n", + "e=1.602*10**-19 #C (here ee=eh)\n", + "#as we know that\n", + "#sigma=ne*ee*mue+nh*eh*muh\n", + "#so we get\n", + "sigma=n*e*(mue+muh) #per ohm m\n", + "#resistivity\n", + "rho=1/sigma #ohm m\n", + "print \" the resistivity of silicon is %0.3f ohm m approx.\"%(rho)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the resistivity of silicon is 2768.158 ohm m approx.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-19.4 page no-565" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#resistivity\n", + "rho=2*10**-3 #ohm m\n", + "#conductivity\n", + "sigma=1/rho #per ohm m\n", + "#electrons and holes mobility\n", + "mue=0.3 #m**2/V s\n", + "muh=0.1 #m**2/V s\n", + "#charge on holes and electrons are same so\n", + "e=1.602*10**-19 #C\n", + "#we know that\n", + "#sigma=ne*e*mue+nh*e*muh\n", + "#here ne=nh=n\n", + "#so\n", + "n=sigma/(e*(mue+muh)) #per m**3 (some approximation is done in book)\n", + "print \"the carrier density is %0.3e per m**3\"%(n)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the carrier density is 7.803e+21 per m**3\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-19.5 page no-566" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#current density\n", + "Id=1000 #A/m**2\n", + "#resistivity\n", + "rho=0.05 #ohm m\n", + "#conductivity\n", + "sigma=1/rho #per ohm m\n", + "#electron mobility\n", + "mue=0.4 #m**2/V m\n", + "#length of crystal\n", + "l=100*10**-6 #m\n", + "#charge on electron\n", + "e=1.602*10**-19 #C\n", + "#in n-type semiconductor ne>>>nh\n", + "#so\n", + "n=sigma/e/mue #per m**3\n", + "#also we know that\n", + "#Ie=n*e*vd\n", + "#so\n", + "vd=Id/n/e #m/s\n", + "#the distance l travelled in time t at drift velocity vd, by an electron is given by\n", + "#l=t*vd\n", + "#so\n", + "t=l/vd*10**6 #micro sec\n", + "print \"the velocity and time taken by the electron to travel 100 m in the crystal is %d m/s and %d micro sec\"%(vd,t)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the velocity and time taken by the electron to travel 100 m in the crystal is 20 m/s and 5 micro sec\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-19.6 page no-567" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "#given\n", + "#electronic charge\n", + "e=1.602*10**-19 #C\n", + "#lengthof rod\n", + "l=10*10**-3 #m\n", + "#diameter of rod \n", + "d=1*10**-3 #m\n", + "#area of cross section\n", + "A=(pi)/4*d**2 #m**2\n", + "#resistance of wire\n", + "R=100 #ohm\n", + "#mobility of holes \n", + "muh=0.19 #m**2/V s\n", + "#resistivity of wire\n", + "rho=R*A/l #ohm m\n", + "#conductivity\n", + "sigma=1/rho #per ohm m\n", + "#we know that i p-type semiconductore nh>>>ne and eh=e\n", + "#so\n", + "#sigma=nh*e*muh\n", + "#so\n", + "nh=sigma/e/muh #per m**3\n", + "print \"the impurity conc in the rod is %0.3e per m**3\"%(nh)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the impurity conc in the rod is 4.183e+21 per m**3\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-19.7 page no 569" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#hole density \n", + "n=10**19 #per m**3\n", + "#intrinsic carriers concentration\n", + "ni=1.5*10**16 # per m**3\n", + "#no of conduction electrons are given by\n", + "p=(ni)**2/n #per m**3\n", + "print \"The no of intrinsic carrier are %0.3e per m**3\"%(p)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The no of intrinsic carrier are 2.250e+13 per m**3\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-19.8 page no-569" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#hole density in silicon\n", + "ND=10**17 #per cm**3\n", + "#intrinsic carrier concentretion\n", + "ni=1.5*10**10 #per cm**3\n", + "#since ND>>>> ni, so ne=ND\n", + "ne=ND\n", + "#the holes concentration\n", + "nh=(ni)**2/ne #per cm**3\n", + "print \" the hole concentration is %.f per cm**3\"%(nh)\n", + "#relative location of EF and Ei are found from\n", + "#EF-Ei=k*Tlog (ne/ni)\n", + "#let us assume for notation\n", + "#EF-Ei=EE\n", + "#temp\n", + "T=300 #K\n", + "k=8.614*10**-5 #eV\n", + "#so now \n", + "EE=k*T*log(ne/ni) #eV\n", + "print \" EF is located at %0.3f eV away from Ei, toeards Ec side as denoted in book\"%(EE)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the hole concentration is 2250 per cm**3\n", + " EF is located at 0.406 eV away from Ei, toeards Ec side as denoted in book\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-19.9 page no-582" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#deimension of aluminium piece\n", + "t=15*10**-3 #m (thickness)\n", + "b=60*10**-3 #m (width)\n", + "l=180*10**-3 #m (length)\n", + "#magnetic field\n", + "betaz=0.6 #T (tesla)\n", + "#current\n", + "I=25 #A\n", + "#hole mobility\n", + "sigmah=0.0012\n", + "#electrical conductivity\n", + "sigma=3.8*10**7 # per ohm m\n", + "#part(a)\n", + "#hall coefficient\n", + "HC=sigmah/sigma #Vm/AT\n", + "#part(b)\n", + "#hall voltage\n", + "VAB=HC*I*betaz/t #V\n", + "#part(c)\n", + "#resistance\n", + "R=l/sigma/b/t #ohm\n", + "print \" the hall coefficient is %.12f Vm/AT\\n, hall voltage is %.9f V\\n and resistance is %.6f ohm\"%(HC,VAB,R)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the hall coefficient is 0.000000000032 Vm/AT\n", + ", hall voltage is 0.000000032 V\n", + " and resistance is 0.000005 ohm\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-19.13 page no-583" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#electron mobility\n", + "mue=1.065*10**-3 #m**2/V s\n", + "#relaxation time\n", + "tau=6*10**-15 #s\n", + "#charge on electron\n", + "e=1.6*10**-19 #C\n", + "#no of electrons\n", + "n=1 \n", + "#mass of electron\n", + "me=9.1*10**-31 #kg\n", + "#as we know that \n", + "#mue=sigma*HC and sigma=n*e**2*tau/me\n", + "#so \n", + "#mue=n*e**2*tau*HC/me\n", + "#from above equation we can get\n", + "HC=mue*me/n/e**2/tau #V m**3/A Wb \n", + "#condustivity\n", + "sigma=mue/HC #per ohm m (calculation mistake in book)\n", + "print \" the hall coefficient is %.2e V m**3/A Wb and conductivity is %.2e per ohm m\"%(HC,sigma)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the hall coefficient is 6.31e+18 V m**3/A Wb and conductivity is 1.69e-22 per ohm m\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-19.14 page no-583" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#electron mobility\n", + "mue=1.065*10**-3 #m**2/Vs\n", + "#relaxation time\n", + "tau=6*10**-15 #sec\n", + "#charge on an electron\n", + "e=1.6*10**-19 #C\n", + "#mass of electron\n", + "me=9.1*10**-31 #Kg\n", + "n=1\n", + "#as we know that\n", + "#mue=sigma*Hc and sigma=n*e**2*tau/me\n", + "#so \n", + "#mue=n*e**2*tau*Hc/me\n", + "#upon rearrangement\n", + "Hc=mue*me/n/e**2/tau \n", + "#and \n", + "sigma=mue/Hc #(ohm m)**-1 #(ohm m)**-1 (calulation mistake is there in book)\n", + "print \" the hall coefficent is %0.3e \\n and conductivity of Al sample is %0.3e (ohm m)**-1 \"%(Hc,sigma)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the hall coefficent is 6.310e+18 \n", + " and conductivity of Al sample is 1.688e-22 (ohm m)**-1 \n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-19.15 page no-584" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import atan\n", + "#given\n", + "#magnetic flux density\n", + "betaz=0.5 #T or Wb/m**2\n", + "#hall coefficient\n", + "HC=3.66*10**-4 #m**3/C\n", + "#resistivity of semiconductors \n", + "rho=0.00893 #ohm m\n", + "#hall angle\n", + "thetaH=(atan(HC*betaz/rho))*180/(pi) #degrees\n", + "print \" the hall angle is %0.3f degrees\"%(thetaH)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the hall angle is 1.174 degrees\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-19.18 page no-585" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "#given\n", + "#fermi velocity of electron\n", + "VF=0.85*10**6 #m/s\n", + "#rest mass of electron\n", + "m0=9.109*10**-31 #kg\n", + "#charge on an electron\n", + "e=1.602*10**-19 #C\n", + "#fermi energy is\n", + "EF=1/2*m0*(VF)**2 # J\n", + "#energy in eV\n", + "E=EF/e #eV\n", + "print \"the fermi energy value = %0.2f ev\"%E\n", + "print \"it indicate that the metal is potassium\\n\"\n", + "#speed of light\n", + "c=3*10**8 #m/s\n", + "#varying mass of electron can bee calculated as\n", + "m=m0*sqrt(1-(VF/c)**2) #kg\n", + "#fermi energy \n", + "EF=1/2*m*(VF)**2 #J\n", + "#energy in eV\n", + "E=EF/e #eV\n", + "print \"the fermi energy value = %0.2f ev\"%E\n", + "print \"it indicate that the metal is potassium\\n\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the fermi energy value = 2.05 ev\n", + "it indicate that the metal is potassium\n", + "\n", + "the fermi energy value = 2.05 ev\n", + "it indicate that the metal is potassium\n", + "\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-19.19 page no-586" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#wavelength of the edge of the absorption edge of a semiconductor material is \n", + "lamda=1771*10**-9 #m\n", + "#plank's constant\n", + "h=6.626*10**-34\n", + "#speed of light\n", + "c=3*10**8 #m/s\n", + "#band gap energy\n", + "Eg=h*c/lamda #J\n", + "#charge on electron\n", + "e=1.602*10**-19 #C\n", + "#energy in eV\n", + "E=Eg/e #eV\n", + "print \" the band gap energy is %0.3f eV\"%(E)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the band gap energy is 0.701 eV\n" + ] + } + ], + "prompt_number": 28 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch2.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch2.ipynb new file mode 100644 index 00000000..f023bdf3 --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch2.ipynb @@ -0,0 +1,389 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:5006cf0b5772f0a5da2f92f1c244fa4b7e9f85dd25f91fb0cd299755ca106f4d" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-2 : Atomic model & bonding in solids" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-2.1 page no-28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import pi\n", + "#given\n", + "#atomic no. of gold\n", + "Z=79\n", + "#kinetic energy of alpha particle\n", + "E=7.68*1.6*(10)**(-13) #J because [1MeV=1.6*(10)**(-13)]\n", + "e=1.6*10**(-19) #C\n", + "E0=8.854*10**(-12) #F/m\n", + "#the distance of closest approach is given by:\n", + "d0=2*e*Z*e/(4*(pi)*E0*E) #m\n", + "print \"the closest approach of alpha particle is %.2e m\"%d0 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the closest approach of alpha particle is 2.96e-14 m\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-2.2 page no-29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, sin\n", + "#given\n", + "#IN THE RUTHERFORD SCATTERING EXPERIMENT\n", + "#the no of particles scattered at\n", + "theta1=(pi)/2 #radians\n", + "#is\n", + "N90=44 #per minute\n", + "#the number of particles scattered particales N is given by\n", + "#N=C*(1/(sin(theta/2))**4) where C is propotionality constant\n", + "#solving above equation for C\n", + "C=N90*(sin(theta1/2))**4 \n", + "# now to find the no of particles scatering at 75 and 135 degrees\n", + "theta2=75*(pi)/180 #radians\n", + "N75=C*(1/(sin(theta2/2))**4) #per minute\n", + "theta3=135*(pi)/180 #radians\n", + "N135=C*(1/(sin(theta3/2))**4) #per minute\n", + "print \"the no of particles scattered at 75 and 135 degrees are %d per minute and %d per minutes\"%(N75,N135)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the no of particles scattered at 75 and 135 degrees are 80 per minute and 15 per minutes\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-2.3 page no-32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#mass of electron\n", + "m=9.11*10**(-31) #kg\n", + "#charge on an electron\n", + "e=1.6*10**(-19) #C\n", + "#plank's constant\n", + "h=6.62*10**(-34)\n", + "E0=8.85*10**(-12) \n", + "#NO OF ELECTRONS SHELLS IN HYDROZEN ATOm\n", + "n=1\n", + "#atomic number of hydrogen\n", + "Z=1\n", + "#radius of first orbit of hydrogen is given by\n", + "r1=n**2*E0*h**2/((pi)*m*Z*e**2) #m\n", + "print \"the radius of the first orbit of the electron in the hydrogen atom %.2e\"%(r1)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the radius of the first orbit of the electron in the hydrogen atom 5.29e-11\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-2.4 page no-32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#mass of electron\n", + "m=9.11*10**(-31) #kg\n", + "#charge on an electron\n", + "e=1.6*10**(-19) #C\n", + "#plank's constant\n", + "h=6.62*10**(-34)\n", + "E0=8.85*10**(-12) \n", + "#NO OF ELECTRONS SHELLS IN HYDROZEN ATOm\n", + "n=1\n", + "#atomic number of hydrogen\n", + "Z=1\n", + "#ionization potential energy of hydrogen atom is given by\n", + "E=m*Z**2*e**4/(8*(E0)**2*h**2*n**2) #J\n", + "#energy in eV\n", + "EV=E/e #eV\n", + "print \"the ionization potential for hydrogen atom is %0.2f V\"%(EV)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the ionization potential for hydrogen atom is 13.59 V\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-2.5 page no-34" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "n=4\n", + "#the geometery of elliptical path is obtained by\n", + "#b/a=(l+1)/n\n", + "#let b/a=t just for notation\n", + "#for s suborbit\n", + "l1=0\n", + "t1=(l1+1)/n\n", + "#so b=a/4\n", + "#for p orbit\n", + "l2=1\n", + "t2=(l2+1)/n\n", + "#so b=a/2\n", + "#for d suborbit\n", + "l3=2\n", + "t3=(l3+1)/n\n", + "#so b=3*a/4\n", + "#for f suborbit\n", + "l4=3\n", + "t4=(l4+1)/n\n", + "#so b=a\n", + "print \"The dimension of elliptical locus of different suborbits are :\",\n", + "print t1,',',t2,',',t3,',',t4" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The dimension of elliptical locus of different suborbits are : 0.25 , 0.5 , 0.75 , 1.0\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-2.6 page no-36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#uncertainity in the momentum\n", + "deltap=10**-27 #kg ms**-1\n", + "#according to uncertainity principle\n", + "#deltap* deltax >=h/(2*(pi))\n", + "#we know that \n", + "h=6.626*10**-34 #Js\n", + "#here instead of inequality we are using only equality just for notation otherwise it is greater than equal to as mentioned above\n", + "#now deltax is given by\n", + "deltax=h/(2*(pi)*deltap) #m\n", + "print \"the minimum uncertainity is %.2e m\"%(deltax)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the minimum uncertainity is 1.05e-07 m\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-2.10 page no- 57" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#ionization potential of hydrogen\n", + "E1=13.6 #eV\n", + "#when \n", + "n=3\n", + "E3=-E1/n**2 #eV\n", + "#when \n", + "n=5\n", + "E5=-E1/n**2 #eV\n", + "print \"energy of 3rd and 5th orbits are %0.2f eV and %0.2f eV\"%(E3,E5)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "energy of 3rd and 5th orbits are -1.51 eV and -0.54 eV\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-2.11 page no-59" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#dipole moment og HF is\n", + "DM=6.375*10**(-30) #Cm\n", + "#intermolecular distance\n", + "r=0.9178*10**(-10) #m\n", + "#charge on an electron\n", + "e=1.67*10**(-19) #C\n", + "#since the HF posses ionic characters\n", + "#so\n", + "#Hf in fully ionic state has dipole moment as\n", + "DM2=r*e #Cm\n", + "#percentage ionic characters\n", + "percentage=DM/DM2*100 #%\n", + "print \"the percentage ionic character is %0.2f approx.\"%(percentage)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the percentage ionic character is 41.59 approx.\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-2.12 page no-60" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import exp\n", + "#given\n", + "#elctronegativity of In\n", + "EnIn=1.5\n", + "#elctronegativity of As\n", + "EnAs=2.2\n", + "#elctronegativity of Ga\n", + "EnGa=1.8\n", + "#for InAs\n", + "ionic_charater1=(1-exp((-0.25)*(EnAs-EnIn)**2))*100\n", + "#for GaAs\n", + "ionic_charater2=(1-exp((-0.25)*(EnAs-EnGa)**2))*100\n", + "print \"percent ionic character in InAs and GaAs are %0.2f and %0.2f\"%(ionic_charater1, ionic_charater2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "percent ionic character in InAs and GaAs are 11.53 and 3.92\n" + ] + } + ], + "prompt_number": 26 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch20.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch20.ipynb new file mode 100644 index 00000000..72766378 --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch20.ipynb @@ -0,0 +1,437 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:b9f5d7f0a48b46e1babebe6bd43ebb6898c49a7f38ee85ae4b3cf64f271cc1c8" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-20, Dielectric, Ferroelectric, piezoelastic & pyroelastic material" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-20.1 page no-593" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#dielecric strength of natural rubber\n", + "DS=40000 #volts/mm\n", + "#current \n", + "I=33*10**3 #V\n", + "#required thickness of insulation\n", + "t=I/DS #mm\n", + "print \"the thickness of wire required for insulation is %0.3f mm\"%(t)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the thickness of wire required for insulation is 0.825 mm\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-20.2 page no-594" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "#given\n", + "#capacitance of capacitor\n", + "C=0.025*10**-6 #F\n", + "#power factor\n", + "#tan delta=0.0005\n", + "#for notation let tan delta=delta\n", + "delta=0.0005\n", + "#current\n", + "I=200 #A\n", + "#frequency\n", + "f=25*10**3 #Hz\n", + "#volatge across a capacitor is\n", + "V=I/(2*(pi)*f*C) #V\n", + "#dielectric loss\n", + "P=V*I*delta # W\n", + "print \" the dielectric loss is %0.3f Watt\"%(P)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the dielectric loss is 5092.958 Watt\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-20.3 page no-596" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#electric field\n", + "E=600 #V/m\n", + "#dielectric constant\n", + "Er=6.1\n", + "E0=8.85*10**-12 \n", + "#polarization is given by\n", + "P=E*E0*(Er-1) #V/m\n", + "print \"The polarization produced is %0.3e V/m\"%(P)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The polarization produced is 2.708e-08 V/m\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-20.4 page no-596" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#charge\n", + "Q=10*10**-6 #C\n", + "#voltage\n", + "V=10*10**3 #V\n", + "#seperation betweemn the plates\n", + "d=5*10**-4 #m\n", + "#dielectric eonstant\n", + "Er=10\n", + "E0=8.854*10**-12\n", + "#we know that\n", + "#Q=C*V\n", + "#so\n", + "C=Q/V #F\n", + "#also we know that\n", + "#C=Er*E0*A/d\n", + "#so\n", + "A=C*d/Er/E0 #m**2\n", + "print \"area between the plates is %0.3f m**2\"%(A)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "area between the plates is 0.006 m**2\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-20.5 page no-597" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#area of plate\n", + "A=10*10*10**-6 #m**2\n", + "#capacitance\n", + "C=10**-9 #F\n", + "#distance between the plates\n", + "d=2*10**-3 #m\n", + "#contant\n", + "E0=8.854*10**-12 #F/m\n", + "#dielectric constant\n", + "Er=C*d/(E0*A) \n", + "print \"the cielectric constant of the crystal is %0.3f\"%(Er)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the cielectric constant of the crystal is 2258.866\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-20.6 page no-597" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#dielectric constant\n", + "Er1=6.0\n", + "Er2=3.0\n", + "#thickness of plates\n", + "d1=0.25*10**-3 #m\n", + "d2=0.1*10**-3 #m\n", + "#taking A1=A2\n", + "#we know that \n", + "#C=Er*E0*A/d\n", + "#for plate1\n", + "#C1=Er1*E0*A1/d1 -----------(1)\n", + "#for plate 2\n", + "#C2=Er1*E0*A2/d2 --------------(2)\n", + "#dividing 1 and 2\n", + "#we get\n", + "#C1/C2=Er1*d2/(Er2*d1)\n", + "#let C1/C2=c\n", + "C=Er1*d2/(Er2*d1) \n", + "#so we get\n", + "#C1=0.8*C2\n", + "print \"The plastic film wil hold more charge\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The plastic film wil hold more charge\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-20.7 page no-609" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#thickness of BaTiO3 wafer\n", + "t=0.15*10**-3 #m\n", + "#compressive strength\n", + "sigma=25*10**6 #N/m**2\n", + "#young's modulus of elasticity\n", + "Y=70*10**9 #N/m**2\n", + "#electric field E produced by the stress sigma is related as\n", + "#E=sigma*lambda where lambda is constant known as voltage output coefficient\n", + "lamda=1*10**-10 #m/V\n", + "#and modulus of elasticity is gven by\n", + "#Y=1/(lambda*t)\n", + "#so we get from 1 and 2\n", + "#E=sigma/(Y*t)\n", + "#also E=V/t\n", + "#so\n", + "V=sigma*t/(lamda*Y) #V\n", + "print \"Potential difference producd across tha wafer is %0.3f V\"%(V)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Potential difference producd across tha wafer is 535.714 V\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-20.8 page no-613" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "#given\n", + "#vibration frequency\n", + "f=434*10**3 #Hz\n", + "#young's modulus of elasticity\n", + "E=80*10**9 #Pa\n", + "#density\n", + "rho=2655 #kg/m**3\n", + "#and fundamental overtones may be 1,2,3.......\n", + "#so\n", + "n=1\n", + "#we konow tha\n", + "#f=n/(2*t)*sqrt(E/rho) wher t is the thickness of crystal\n", + "#so\n", + "t=n/(2*f)*sqrt(E/rho)*10**3 #mm\n", + "print \" the thickness of crystal is %0.3f mm\"%(t)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the thickness of crystal is 6.324 mm\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-20.9 page no-613" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#capacitance of paper capacitance\n", + "C=0.02*10**-6 #F\n", + "#thickness of capacitor\n", + "d=1*10**-3 #m\n", + "#relative permitivity\n", + "Er=2.6\n", + "E0=8.85*10**-12\n", + "#dielectric strength\n", + "k=1.8*10**7 #V/m\n", + "#area of capacitor is given by\n", + "A=C*d/(Er*E0)*10**4 #cm**2\n", + "#breakdown volatage\n", + "Vbreakdown=k*d #V\n", + "print \"The area of capacitor is %0.3f am**2 \\n and breakdown volatage is %0.3f V\"%(A,Vbreakdown)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The area of capacitor is 8691.873 am**2 \n", + " and breakdown volatage is 18000.000 V\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-20.10 page no-614" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#capacitance of capacitor\n", + "C=0.2*10**-6 #F\n", + "#loss factor\n", + "#tan delta=0.004\n", + "# for notation let us use tan delta=delta\n", + "delta=0.004\n", + "#voltage\n", + "V=240 #V\n", + "#frequency\n", + "f=50 #Hz\n", + "#and \n", + "omega=2*(pi)*f\n", + "#power loss is given by\n", + "P=V**2*omega*C*delta #W \n", + "print \" power loss in the capacitor is %0.3f W\"%(P)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " power loss in the capacitor is 0.014 W\n" + ] + } + ], + "prompt_number": 20 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch21.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch21.ipynb new file mode 100644 index 00000000..402b7276 --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch21.ipynb @@ -0,0 +1,444 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a30f79ff459b2d070e97267af2c0b53a7ca4c8f26e341ccc7fe8201d7c4afe5e" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-21, Magnetic material properties, behaviour and application" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-21.1 page no-621" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import pi\n", + "#given\n", + "#magnetic field\n", + "H=2400 #A/m\n", + "#susceptibilty\n", + "kie=1500\n", + "#part(a)\n", + "#relative permeability is given by\n", + "mur=1+kie\n", + "#part(b)\n", + "#intensity of magnetization \n", + "M=kie*H #A/m\n", + "#part(c)\n", + "#permeability\n", + "mu0=4*(pi)*10**-7\n", + "#remanance\n", + "B=mu0*mur*H #T\n", + "print \" the realative permeability is %0.f \\n, the intensity of magnetisation is %.f A/m \\n and the remanance is %0.3f\"%(mur,M,B)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the realative permeability is 1501 \n", + ", the intensity of magnetisation is 3600000 A/m \n", + " and the remanance is 4.527\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-21.2 page no-621" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#relative permeability of superalloy\n", + "mur=200000\n", + "mu0=4*(pi)*10**-7 #henry/m\n", + "#intensity of magnetisation\n", + "M=6000 #A/m\n", + "#magnetic field is given by\n", + "H=M/(mur-1) #A/m\n", + "#strength of magnet\n", + "B=mu0*mur*H #tesla\n", + "print \"The strength of magnet is %0.3f T\"%(B)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The strength of magnet is 0.008 T\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-21.3 page no-621" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#magnetic moment is 0.6 times bohr magneton and we know that beta is 9.27*10**-24 Am**2\n", + "beta=9.27*10**-24 #A/m**2\n", + "M=0.6*beta #A/m**2\n", + "#attice constant \n", + "a=0.35*10**-9 #m\n", + "#no of atoms per unit cell is given by\n", + "Ne=4\n", + "#saturation magnetisation for FCC unit cell is given by\n", + "Ms=Ne*M/a**3 #A/m\n", + "print \"he saturation magnetisation is %0.3f A/m\"%(Ms)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "he saturation magnetisation is 518903.790 A/m\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-21.4 page no-632" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#refer to fig-21.6\n", + "#width of loop\n", + "#W=(OA+OB) \n", + "#so \n", + "W=80*2 #A/m\n", + "#height of loop\n", + "#H=(OC+OD) #wb/m**2\n", + "#so\n", + "H=0.15*2 #Wb/m**2\n", + "#area of loop\n", + "A=W*H #T A/m or J\n", + "print \"The energy loss per ubitvolume of the magnetic material during one cycle is %.f J\"%(A)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The energy loss per ubitvolume of the magnetic material during one cycle is 48 J\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-21.5 page no-633" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#volume\n", + "V=0.01 #m**3\n", + "#frequency\n", + "f=50 #Hz\n", + "#area of loop\n", + "A=600 #J/m**2\n", + "#as we know that\n", + "#A=mu*Bmax**16\n", + "#also power loss is given by\n", + "#P=mu*Bmax**16*f*V\n", + "#so \n", + "P=A*f*V #watt\n", + "print \" the power loss due to hysteresis is %.f W\"%(P)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the power loss due to hysteresis is 300 W\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-21.6 page no-633" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#hysteresis loss is\n", + "W1=300 #W\n", + "#max flux density is \n", + "Bmax1=0.9 #Wb/m**2\n", + "Bmax2=1.1 #Wb/m**2\n", + "#frequency\n", + "f1=50 #Hz\n", + "f2=40 #Hz\n", + "#we know that\n", + "#W=nu*(Bmax)**1.7*f*V\n", + "#s0 \n", + "#W1=nu*(Bmax1)**1.7*f1*V ---------(1)\n", + "#W2=nu*(Bmax2)**1.7*f*V -------(2)\n", + "#from one and 2 we get\n", + "W2=W1*(Bmax2)**1.7*f2/((Bmax1)**1.7*f1) #W\n", + "print \"The hysteresis loss at 40 Hz frequency is %0.3f W\"%(W2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The hysteresis loss at 40 Hz frequency is 337.572 W\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-21.7 page no 634" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#flux density\n", + "Bm=1.10 #Wb/m**2\n", + "#frequency\n", + "f=50 #Hz\n", + "#thickness of sheet\n", + "t=0.5*10**-3 #m\n", + "#resistivity\n", + "rho=30*10**-8 #per ohm m\n", + "#density\n", + "rhodash=7800 #kg/m**3\n", + "#mass\n", + "m=1 #kg\n", + "#volume of material\n", + "V=m/rhodash #m**3\n", + "#and \n", + "k=1.11\n", + "#hysteresis loss in each cycle\n", + "Wh=380 #W s/m**3\n", + "#loss per kg of specimen is given by\n", + "We=4/3*(Bm*f*t*k)**2*V/(rho) #watt/kg\n", + "print \" the loss is %0.3f watt/kg\"%(We)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the loss is 0.531 watt/kg\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-21.8 page no-634" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#frequency\n", + "f=50 #Hz\n", + "#mass\n", + "m=50 #kg\n", + "#density\n", + "rho=7500 #kg/m**3\n", + "#volume of material\n", + "V=m/rho #m**3\n", + "#hysteresis loop area\n", + "A=150 #m**2\n", + "#scale factor\n", + "#1 cm=0.008 Wb/m**2 on y-axis and 1cm=20 A/m on x-axis\n", + "#energy lost during each cycle\n", + "E=A*0.008*20*10**4 #J/m**3\n", + "#poer loss due to hysteresis\n", + "P=E*f*V #J/s\n", + "#energy lost in one hour\n", + "Wh=P*(60*60) #J\n", + "print \" the energy lost in one hour is %0.3e J\"%(Wh)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the energy lost in one hour is 2.880e+08 J\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-21.11 page no-650" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#frequency\n", + "f=50 #Hz\n", + "#eddy current loss in transformer\n", + "We=100 #W\n", + "#to find eddy current loss at frequencies\n", + "f1=60 #Hz\n", + "f2=100 #Hz\n", + "#as we know that\n", + "#We is directly proportional to f**2\n", + "We1=f1**2*We/f**2 #W\n", + "#similarly\n", + "We2=f2**2*We/f**2 #W\n", + "print \" the eddy current loss at 60 Hz is %0.f W \\n and at 100 Hz is %.f W\"%(We1,We2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the eddy current loss at 60 Hz is 144 W \n", + " and at 100 Hz is 400 W\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-21.15 page no-651" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#length of wire\n", + "l=250*10**-3 #m\n", + "#no of turns\n", + "N=400\n", + "#current\n", + "I=15 #A\n", + "#permeability in vaccum\n", + "mu0=1.2457*10**-6 #H/m\n", + "#relative permeability\n", + "mur=1\n", + "#magnetic field strength \n", + "H=N*I/l #AT/m\n", + "#flux density is\n", + "B=mu0*mur*H #Wb/m**2\n", + "print \" the magnetic field strength is %.f AT/m and flux density is %0.3f Wb/m\"%(H,B)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the magnetic field strength is 24000 AT/m and flux density is 0.030 Wb/m\n" + ] + } + ], + "prompt_number": 23 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch22.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch22.ipynb new file mode 100644 index 00000000..74bd9c9f --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch22.ipynb @@ -0,0 +1,294 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:9b85f177bb11398b4f5d182de9fe8e9e406f1c070a747a1b0c1a3320e4f20360" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-22, Superconduting material properties, behaviour and application" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-22.2 page no-660" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#critical temp of Pb\n", + "T0=7.17 #K\n", + "#critical field\n", + "H0=0.0803 #A/m\n", + "#to find the critical field at \n", + "T1=3 #K\n", + "T2=10 #K\n", + "#critical field at T1\n", + "Hc1=H0*(1-T1**2/T0**2) #A/m\n", + "#critical field at T2\n", + "Hc2=H0*(1-T2**2/T0**2) #A/m\n", + "print \" the critical field at 3K temp is %0.3f A/m and at 10K ia %0.3f A/m\"%(Hc1,Hc2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the critical field at 3K temp is 0.066 A/m and at 10K ia -0.076 A/m\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-22.3 page no-660" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "#given\n", + "#critical magnetic field\n", + "Hc=7.9*10**3 #A/m\n", + "#diameter of aluminium wire\n", + "d=1*10**-3 #m\n", + "#critical current is give by\n", + "Ic=2*(pi)*d*Hc #A\n", + "print \" the critical current which can pass through a long thin superconducting wire of aluminium is %0.3f A\"%(Ic)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the critical current which can pass through a long thin superconducting wire of aluminium is 49.637 A\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-22.4 page no-667" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "#given\n", + "#specific density of lead\n", + "sd=11.4\n", + "#density of lead\n", + "rho=sd*10**3 #kg/m**3\n", + "#atomic weight of lead\n", + "Aw=207.2 #kg/kg-mol\n", + "#velocity of sound in lead\n", + "v=1200 #m/s\n", + "NA=6.023*10**26 #particles/kg-mol (avogadro's number)\n", + "e=1.602*10**-19 #C/electrons (charge on an electron)\n", + "m=9.1*10**-31 #kg (mass of electron)\n", + "mu0=4*(pi)*10**-7 #H/m (permeability)\n", + "#since lead is type I superconductor, so London's theory of superconductivity is applicable\n", + "#so\n", + "ne=2*rho*NA/Aw #electrons/m**3\n", + "#critical current density\n", + "Ied=ne*e*v #A/m**2\n", + "#depth of penetration at the surface of lead\n", + "dp=sqrt(m/(mu0*ne*e**2))*10**10 #A\n", + "print \"\"\"the electron density is %0.3e electrons/m**3\n", + "the critical current density is %0.3e A/m**2\n", + "the depth of penetration is %0.3e A\"\"\"%(ne,Ied,dp)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the electron density is 6.628e+28 electrons/m**3\n", + "the critical current density is 1.274e+13 A/m**2\n", + "the depth of penetration is 2.063e+02 A\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-22.9 page no-673" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#magnetic field at 0K temp \n", + "H0=65*10**3 #A/m\n", + "#critical temp \n", + "Tc=7.18 #K\n", + "#diameter of wire\n", + "d=1*10**-3 #m\n", + "#radius of wire\n", + "r=d/2 #m\n", + "#area of cross section\n", + "A=(pi)*r**2 #m**2\n", + "#to find the current density at 4.2 K\n", + "#since it is given that Hc is parabolicaly dependent on T, so\n", + "T=4.2 #K\n", + "Hc=H0*(1-T**2/Tc**2) #A/m\n", + "#critical current\n", + "Ic=2*(pi)*r*Hc #A\n", + "#critical current density Jc\n", + "Jc=Ic/A #A/m**2\n", + "print \" the critical current density of lead is %0.3e A/m**2\"%(Jc)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the critical current density of lead is 1.710e+08 A/m**2\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-22.10 page no-673" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "#given\n", + "#critical field at 3K and 14 K are 21 A/m and 10 A/m\n", + "T1=7 #K\n", + "T2=14 #K\n", + "Hc1=21 #A/m\n", + "Hc2=10 #A/m\n", + "#DETERMINING CRITICAL TEMP\n", + "#as we know that H=H0*(1-T**2/Tc**2)\n", + "#so we get\n", + "#71=H0*(1-7**2/Tc**2) ----(1)\n", + "#10=H0*(1-14**2/Tc**2) --(2)\n", + "#dividing 1 and 2 we get\n", + "#71/10=(Tc**2-7**2)/(Tc**2-14**2)\n", + "#on solving we get\n", + "Tc=sqrt(3626/11) #K\n", + "#DETERMINING CRITICAL FIELD AT 0K\n", + "H0=Hc1/(1-T1**2/Tc**2) #A/m\n", + "#DETERMINING CRITICAL FIELD AT \n", + "T=4.2 #K\n", + "Hc=H0*(1-T**2/Tc**2) #A/m\n", + "print \" the critical temp is %0.3f K\\n, the critical field at 0K is %0.3f A/m and critical field at 4.2 K is %0.3f A/m\"%(Tc,H0,Hc)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the critical temp is 18.156 K\n", + ", the critical field at 0K is 24.667 A/m and critical field at 4.2 K is 23.347 A/m\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-22.11 page no-674" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#depth of penetration at 3K is 39.8 nm and at 7.1 K is 1730 A\n", + "T1=3 #K\n", + "T2=7.1 #K\n", + "dp1=39.6*10**-9 #m\n", + "dp2=1730*10**-10 #m\n", + "#as we know that depth of penetration and temp are related as\n", + "#(dp(T)/dp(T0))=1/(1-t**4/Tc**4)\n", + "#so we get\n", + "#at 3K\n", + "#let dp(T0)=dp0\n", + "#dp0=sqrt(dp1**2*(1-T1**4/Tc**4)) -(1)\n", + "#also\n", + "#dp0=sqrt(dp2**2*(1-T2**4/Tc**4)) .----(2)\n", + "#solving 1 and 2 we get\n", + "#((Tc**4-81)/(Tc**4-(7.1)**4))=(173)**2/(39.6)**2\n", + "#so we get\n", + "Tc=(48417.9/18.085)**(1/4) #K\\\n", + "#depth of penetration at absolute zero willbe\n", + "dp0=sqrt(dp1**2*(1-T1**4/Tc**4))*10**9 #nm\n", + "print \" critica temp is %0.3f K\\n and depth of penetration at critica zero is %0.3f nm\"%(Tc,dp0)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " critica temp is 7.193 K\n", + " and depth of penetration at critica zero is 38.996 nm\n" + ] + } + ], + "prompt_number": 14 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch23.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch23.ipynb new file mode 100644 index 00000000..17895a18 --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch23.ipynb @@ -0,0 +1,130 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:5b9459ecae57c08b61ec14610fd69508bc529ffb99e434161f835cad507f2c18" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-23, Thermal properties & materials" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-23.1 page no-681" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#length of glass rod and steel rod is equal at 273 K and differ by 1.2 mm at 373 K\n", + "T1=273 #K\n", + "T2=373 #K\n", + "#coefficients of linear expansion af glass and steel are\n", + "alphaG=8*10**-6 #per degree C\n", + "alphaS=12*10**-6 #per degrees C\n", + "#we know that\n", + "#lT2=lT1*(1+alpha(T2-T1))\n", + "#so for glass rod\n", + "#l100G=l0*(1+(alpha1)*(T2-T1))\n", + "#similarly for steel rod\n", + "#l100G=1.0008*l0 ----(1)\n", + "#l100S=lo*(1+(alpha2)*(T2-T1))\n", + "#l100S=1.0012*l0 ----------(2)\n", + "#we have given that\n", + "#l100S-l100G=1.2 mm ---(3)\n", + "#from 1 and 2 put in 3, we get\n", + "#1.0012*l0-1.0008*l0=1.2\n", + "#so\n", + "l0=1.2*10**-3/(0.0012-.0008) #m\n", + "print \" the length of rod at 0 degrees celcius is %.f m\"%(l0)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the length of rod at 0 degrees celcius is 3 m\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-23.2 page no-684" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#coefficient of linear expansion of Cu and steel are\n", + "alphaCu=18*10**-6 #cm/cm/degree C\n", + "alphaSteel=14*10**-6 #cm/cm/degree C\n", + "#young's modulus of elasticity\n", + "ECu=106*10**9 #Pa\n", + "ESteel=200*10**9 #Pa\n", + "#part(a)\n", + "#since alphaCu>alphaSteel\n", + "#so steel will contract less. \n", + "print \" Hence strip will bend in the direction of copper\"\n", + "#part(b)\n", + "#annealing temp\n", + "T2=530 #degrees celcius\n", + "#room temp\n", + "T1=30 #degrees celcius\n", + "#difference in temp\n", + "T=T2-T1 #degrees celcius\n", + "#differnce in values of coefficient of linear expansion\n", + "alpha=alphaCu-alphaSteel #cm/cm/degrees celcius\n", + "#differential contraction\n", + "contraction=T*alpha #cm/cm\n", + "#since the two metals in the strip have equal dimension\n", + "#sigmaCu=sigmaSteel=sigma\n", + "#strain in copper\n", + "#eCu=sigma/Ecu -----(1)\n", + "#strain in steel\n", + "#eSteel=sigma/ESteel ---(2)\n", + "#sum of strains given by equation 1 and 2\n", + "sigma=contraction*ESteel*ECu/(ESteel+ECu)*10**-6 # MPa\n", + "print \" the stresses in each metal is %0.3f MPa\"%(sigma)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Hence strip will bend in the direction of copper\n", + " the stresses in each metal is 138.562 MPa\n" + ] + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch26.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch26.ipynb new file mode 100644 index 00000000..6a6b9761 --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch26.ipynb @@ -0,0 +1,272 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:0c5a5b237d1d9ee2de03397e0c595a70fe68a27ab5ea34fc9d5e4643a075bb47" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-26, Optical properties of materials and materials for opto electronic devices" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-26.1 page no-769" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#wavelength of light\n", + "lamda=5.893*10**-7 #m\n", + "#plank's constant\n", + "h=6.626*10**-34 #J s\n", + "#velocity of light \n", + "c=3*10**8 #m/s\n", + "#energy of photon\n", + "Ephoton=h*c/lamda #J\n", + "#we know that 1eV=1.6*10**-19 J\n", + "#so\n", + "EPhoton=Ephoton/(1.6*10**-19) #eV\n", + "print \" the energy of photon is %.23f J or %f eV\"%(Ephoton,EPhoton)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the energy of photon is 0.00000000000000000033732 J or 2.108222 eV\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-26.2 page no-769" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import exp\n", + "#given\n", + "##thickness of sample\n", + "t=0.45*10**-4 #cm\n", + "#energy of light\n", + "E1=3 #eV\n", + "#absorption coefficient\n", + "alpha=50000 #per cm\n", + "#incident power on the sample\n", + "I0=15*10**-3 #W\n", + "#we know that\n", + "#intensity of transmitted light is given by\n", + "It=I0*exp(-alpha*t) #W or J/s\n", + "#thus total energy absorbed is \n", + "Iabsorbed=I0-It #W or J/s\n", + "print \" total energy absorbed is %0.3e J/s\"%(Iabsorbed)\n", + "#plank's constant\n", + "h=6.626*10**-34 #J s\n", + "#energy of outgoing radiation\n", + "E2=2.35 #eV\n", + "#fraction of each photon energy unit which is converted ton heat\n", + "E=(E1-E2)/E1 \n", + "#therefore total amount of energy converted to heat per second is\n", + "EC=E*Iabsorbed #J/s\n", + "print \" total amount of energy coverted to heat is %0.3e J/s\"%(EC)\n", + "#charge on an electron\n", + "e=1.6*10**-19 #C\n", + "#no of photons = nphoton \n", + "nphoton=Iabsorbed/(e*E1) #photons/sec (calculation mistake is there in book)\n", + "print \" the no of photon given off from recombination is %0.3e photons/sec\"%(nphoton)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " total energy absorbed is 1.342e-02 J/s\n", + " total amount of energy coverted to heat is 2.907e-03 J/s\n", + " the no of photon given off from recombination is 2.796e+16 photons/sec\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-26.3 page no-783" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#energy of photon\n", + "E=1.5*10**-19 #J\n", + "#quantum eficiency\n", + "muquantam=0.6\n", + "#photon current\n", + "Iopc=3*10**-6 #A\n", + "#speed of light\n", + "c=3*10**8 #m/s\n", + "#plank's constant\n", + "h=6.626*10**-34 #J s\n", + "#wavelength at which the photodiode is operatng\n", + "lamda=h*c/E*10**6 #micro m\n", + "#responsivity of diode\n", + "R=0.64\n", + "#incident optical power is given by\n", + "Piop=Iopc/R*10**6 #micro W\n", + "print \"wavelength at which photodiode is operating is %0.3f micro m \\n and incident optical power is %0.3f micro W\"%(lamda,Piop)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "wavelength at which photodiode is operating is 1.325 micro m \n", + " and incident optical power is 4.688 micro W\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-26.4 page no-784" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import asin, pi\n", + "#given\n", + "#refractive index of core and cladding\n", + "mucladding=1.47\n", + "mucore=1.50\n", + "#critocal angle at the core cladding interface\n", + "thetac=asin(mucladding/mucore) *180/(pi) #degrees\n", + "print \"the critical angle at core cladding interface is %0.3f degrees\"%(thetac)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the critical angle at core cladding interface is 78.522 degrees\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-26.5 page no-784" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#energy band gap\n", + "Eg=0.75*1.6*10**-19 #J\n", + "#plank's constant\n", + "h=6.626*10**-34 #Js\n", + "#speed of light\n", + "c=3*10**8 #m/s\n", + "#wavelength of light\n", + "lamda=h*c/Eg*10**10 #A\n", + "print \" wavelength of light is %.f A\"%(lamda)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " wavelength of light is 16565 A\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-26.7 page no-784" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#frequency of light\n", + "f=1.5*10**9*10**6 #Hz\n", + "#pank's constant\n", + "h=6.626*10**-34 #J s\n", + "#threshold frequency is\n", + "f0=1.2*10**9*10**6 #Hz\n", + "#maximum energy of emitted photoelectron is\n", + "Emax=h*(f-f0)/(1.6*10**-19) #eV\n", + "print \"The maximum enery of the emitted photoe lectron is %0.3f eV\"%(Emax)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum enery of the emitted photoe lectron is 1.242 eV\n" + ] + } + ], + "prompt_number": 8 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch3.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch3.ipynb new file mode 100644 index 00000000..5413f744 --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch3.ipynb @@ -0,0 +1,559 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:bc212e8bce28ec0d7c268cd20b237b2f8a7e58b06860a7ec51f691263671de76" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-3, Crystal geometry of metal & miller indices" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-3.3 page no-78" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "#given\n", + "#ionia radius of Na and cl are resp.\n", + "r1=0.98*10**(-10) #m\n", + "r2=1.81*10**(-10) #m\n", + "#no of atons in FCC\n", + "N=4\n", + "#total volume of Na and cl ions are resp\n", + "v1=N*4/3*(pi)*(r1)**3 #m**3\n", + "v2=N*4/3*(pi)*(r2)**3 #m**3\n", + "#the lattice constant\n", + "a=2*r1+2*r2\n", + "#average packing fraction\n", + "APF=(v1+v2)/a**3 \n", + "print \"the average packing fraction is %0.2f\"%(APF)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the average packing fraction is 0.62\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-3.4 page no-80" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "#given\n", + "#atomic radius of copper\n", + "r=1.278*10**(-8) #cm\n", + "#lattice structure of copper is FCC\n", + "a=4*r/sqrt(2) #cm\n", + "#atomic weight of copper\n", + "Aw=63.54\n", + "N=4\n", + "#avogadro's number\n", + "Na=6.023*10**(23) \n", + "#density of copper atoms in lattice\n", + "rho=Aw*N/(Na*a**3)*1000 #kg/m**3\n", + "print \"the density of copper is %d kg/m**3\"%(rho)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the density of copper is 8934 kg/m**3\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-3.5 page no-80" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#density of NaCl is\n", + "rho=2.18 #g/cm**3\n", + "#no of effective atoms in FCC\n", + "N=4\n", + "#molecular weight of NaCl\n", + "Aw=23+35.5\n", + "#avogadro's number\n", + "Na=6.023*10**(23)\n", + "#lenght of sice of the unit cell\n", + "a=(Aw*N/(Na*rho))**(1/3)*10**8 #A\n", + "#distance between two adjacent atoms\n", + "r=a/2 #A\n", + "print \"the distance b/w the two adjacent atoms is %0.1e A\"%(r) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the distance b/w the two adjacent atoms is 5.0e+07 A\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-3.6 page no-80" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#density of iron\n", + "rho=7.86 #gm/cm**3\n", + "#atomic weight of iron\n", + "Aw=55.85\n", + "#iron has BCC unit structure\n", + "Ne=2\n", + "#avogadros no.\n", + "Na=6.023*10**(23) \n", + "#side of the unit cell\n", + "a=(Aw*Ne/(Na*rho))**(1/3) #cm\n", + "#atomic radius\n", + "r=3**(1/3)*a/4*10**8 #A\n", + "print \"the atomic radius of iron is %0.1e A\"%(r)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the atomic radius of iron is 2.5e+07 A\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-3.7 page no-85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#the intercepts \n", + "#c1(along x-axis)=a\n", + "#c2(along y-axis)=b/2\n", + "#c3(along z-axis)=3*c\n", + "#therefore\n", + "p=1\n", + "q=1.0/2\n", + "r=3\n", + "#so \n", + "h=1/p\n", + "k=1/q\n", + "l=1/r\n", + "#hence (hkl)=(121/3)=1/3*(361) or (361)\n", + "print \"miller indices for such a simple cubic unit cell is\",\n", + "print str('%0.f')%(h/l)+str('%0.f')%(k/l)+str('%0.f')%(l/l)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "miller indices for such a simple cubic unit cell is 361\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-3.8 page no-85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#the ratio of intercepts of orthorombic unit cell is a:b:c=0.429:1:0.379\n", + "#to find:- the miller indices of faces of the following intecepts\n", + "#(I) 0.214:1:0.188\n", + "#(II) 0.858:1:0.754\n", + "#(III) 0.429:1:0.126\n", + "#solution\n", + "#let\n", + "x=0.429\n", + "y=1\n", + "z=0.379\n", + "x1=0.214\n", + "y1=1\n", + "z1=0.188\n", + "p1=x1/x\n", + "q1=y1/y\n", + "r1=z1/z\n", + "#intercepts will be 1/2:1:1/2\n", + "#reciprocal is 2:1:2\n", + "#so miller indices is (212)\n", + "x2=0.858\n", + "y2=1\n", + "z2=0.754\n", + "p2=x2/x\n", + "q2=y2/y\n", + "r2=z2/z\n", + "#intercepts willbe 2:1:2\n", + "#reciprocal is 1/2:1:1/2\n", + "#so miller inces is (121)\n", + "x3=0.429\n", + "y3=1\n", + "z3=0.126\n", + "p3=x3/x\n", + "q3=y3/y\n", + "r3=z3/z\n", + "#th e intercept will be 1:1:1/3\n", + "#reciprocal is 1:1:3\n", + "#so miller indices is (113)\n", + "print \"the miller indices for part (I),(II) and (III) are :\"\n", + "print str('%0.f')%(1/p1)+str('%0.f')%(1/q1)+str('%0.f')%(1/r1)+'',\n", + "print str('%0.f')%(1/p2*p2)+str('%0.f')%(1/q2*p2)+str('%0.f')%(1/r2*p2)+'',\n", + "print str('%0.f')%(1/p3)+str('%0.f')%(1/q3)+str('%0.f')%(1/r3),\n", + "print 'respectively'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the miller indices for part (I),(II) and (III) are :\n", + "212 121 113 respectively\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-3.9 page no-85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#a:b:c=4:3:2\n", + "#so\n", + "a=4\n", + "b=3\n", + "c=2\n", + "#intercepts are\n", + "x=2 #armstrong\n", + "y=3 #armstrong\n", + "z=4 #armstrong\n", + "p=x/a\n", + "q=y/b\n", + "r=z/c\n", + "#reciprocals will be\n", + "h=1/p\n", + "k=1/q\n", + "l=1/r\n", + "#(hkl)=(2 1 1/2)=1/2*(421)=(421)\n", + "print \"the miller indices is\",\n", + "print str('%0.f')%(h/l)+str('%0.f')%(k/l)+str('%0.f')%(l/l)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the miller indices is 421\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-3.11 page no-93" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "#given\n", + "#atomic radius in a FCC crystal is\n", + "r=1.246 #A\n", + "#side of the unit cell\n", + "a=4*r/sqrt(2) #A\n", + "#part(I)\n", + "h1=2\n", + "k1=0\n", + "l1=0\n", + "d1=a/sqrt((h1)**2+(k1)**2+(l1)**2) #A\n", + "#part (II)\n", + "h2=2\n", + "k2=2\n", + "l2=0\n", + "d2=a/sqrt((h2)**2+(k2)**2+(l2)**2) #A\n", + "#part (III)\n", + "h3=1\n", + "k3=1\n", + "l3=1\n", + "d3=a/sqrt((h3)**2+(k3)**2+(l3)**2) #A\n", + "print \"the inteplaners spacing will be d1,d2 and d3 are %0.3f A, %0.3f A and %0.3f A \"%(d1,d2,d3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the inteplaners spacing will be d1,d2 and d3 are 1.762 A, 1.246 A and 2.035 A \n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-3.12 page no-94" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#lattice parameter of an orthorombic unit cell are\n", + "a=0.82 #nm\n", + "b=0.94 #nm\n", + "c=0.75 #nm\n", + "#part(I)\n", + "h1=1\n", + "k1=2\n", + "l1=3\n", + "d1=sqrt(1/((h1/a)**2+(k1/b)**2+(l1/c)**2)) #nm\n", + "#part(II)\n", + "h2=2\n", + "k2=4\n", + "l2=6\n", + "d2=sqrt(1/((h2/a)**2+(k2/b)**2+(l2/c)**2)) #nm\n", + "#aprt(III)- comment\n", + "#we note that\n", + "d=d2/d1\n", + "#i.e d2=1/2*d1\n", + "print \"IT MEANS THAT INCREASING THE MILLER INDICES FREOM (123) TO (246) I.E BY 2 TIMES , DECREASES THE SEPERATION B/W THE PLANES BY 2 TIMES.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "IT MEANS THAT INCREASING THE MILLER INDICES FREOM (123) TO (246) I.E BY 2 TIMES , DECREASES THE SEPERATION B/W THE PLANES BY 2 TIMES.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-3.14 page no-96" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#lattice constant for copper unit cell is \n", + "a=3.61*10**(-10) #m\n", + "#since the lattice structure of copper is FCC\n", + "#so the length of lattice structure (110)\n", + "r110=a*sqrt(2) #m because r110=sqrt(a**2+a**2)\n", + "#effective no of atoms present on vector OA is\n", + "Ne110=(1/2+1+1/2) #1/2 is for the atoms at corners of diagonal OA and 1 is for the atom at the centre of diagonal OA\n", + "#therefore linear density of atoms along (110) in copper unit cell is\n", + "invr110=Ne110/r110 #atoms per m\n", + "#so the length of lattice structure (111)\n", + "r111=a*sqrt(3) #m because r111=sqrt(a**2+a**2+a**2)\n", + "#effective no of atoms present on vector OB is\n", + "Ne111=(1/2+1/2) #1/2 is for each atoms at corners of diagonal OB\n", + "#therefore linear density of atoms along (111) in copper unit cell is\n", + "invr111=Ne111/r111 #atoms per m\n", + "print \"the linear density of atoms along (110) and (111) are %0.2e atoms/mm and %0.2e atoms/mm resp\"%(invr110,invr111)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the linear density of atoms along (110) and (111) are 3.92e+09 atoms/mm and 1.60e+09 atoms/mm resp\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-3.15 page no-101" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# to find the linear density of (100) direction in BCC\n", + "#density is given by the ratio of toatal line length of the intersecting atoms to the distance between the centres of atom\n", + "#rho=2*r/(4*r/sqrt(3))\n", + "#so\n", + "rho=2*sqrt(3)/4 \n", + "print \"the linear density of (100) direction in BCC structure is %0.3f\"%(rho)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the linear density of (100) direction in BCC structure is 0.866\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-3.17 page no-101" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#atomic radii of polonium,rhodium and chromium are\n", + "rPo=1.7*10**(-10) #m\n", + "rRh=1.34*10**(-10) #m\n", + "rCr=1.25*10**(-10) #m\n", + "#latice strucrure of polonium , rhodiun and chromium are SC, FCC and BCC resp\n", + "#so lattice constants are\n", + "aPo=2*rPo #m\n", + "aRh=2*sqrt(2)*rRh #m\n", + "aCr=4/sqrt(3)*rCr #m\n", + "#planer density on (100) in polonium is given by\n", + "rhoPo=1/(aPo)**2 #per m**2\n", + "#planer denity on (110) in rhodium\n", + "rhoRh=sqrt(2)/(aRh)**2 #per m**2\n", + "#planer density on (111) in chromium \n", + "rhoCr=sqrt(3)/(aCr)**2 #per m**2\n", + "print \"\"\"the planer density of polonium in (100) is %0.2e, \n", + "rhodium in (110) is %0.2e per m**2 and chromium in (111) is %0.2e per m**2\"\"\"%(rhoPo,rhoRh,rhoCr)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the planer density of polonium in (100) is 8.65e+18, \n", + "rhodium in (110) is 9.84e+18 per m**2 and chromium in (111) is 2.08e+19 per m**2\n" + ] + } + ], + "prompt_number": 12 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch4.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch4.ipynb new file mode 100644 index 00000000..f4fd8151 --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch4.ipynb @@ -0,0 +1,237 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c3acc1fab1ab8a2f10940ad0f121e6fed364ac2fa1b9f49347a7b95bd54bf0dd" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-4, Solid structure & allotropy" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-4.1 page no-110" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import sqrt\n", + "#given\n", + "#atomic radii of iron in BCC and FCC at 910 degrees celcius temp are\n", + "rBCC=1.258 #A\n", + "rFCC=1.292 #A\n", + "#latice side\n", + "aBCC=4*rBCC/sqrt(3)\n", + "aFCC=2*sqrt(2)*rFCC\n", + "#effective number of atoms in BCC and FCC \n", + "nBCC=2\n", + "nFCC=4\n", + "#volume\n", + "VBCC=nBCC/2*(aBCC)**3\n", + "VFCC=nFCC/4*(aFCC)**3\n", + "#change in volume percent\n", + "deltaV=(-1+VFCC/VBCC)*100\n", + "print \"the volume change = %0.2f %% and here +ve sign indicates that the volume increases during polymorphism transformation\"%deltaV" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the volume change = 99.01 % and here +ve sign indicates that the volume increases during polymorphism transformation\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-4.2 page no-115" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import tan, pi\n", + "#given\n", + "#height of zinc unit cell\n", + "h=4.935*10**-10 #m\n", + "#side of the lattice\n", + "a=2.66*10**-10 #m\n", + "#as we know that zinc has HCP unit cell. \n", + "#the number of effective atoms\n", + "Ne=6 \n", + "#as we know \n", + "#tan(pi/3)=x/(a/2)\n", + "#so\n", + "x=a/2*tan (pi/3) #m\n", + "#area of basal plane\n", + "Ar=6*a*x/2 #m**2\n", + "#volume of the unit cell\n", + "V=Ar*h #m**3\n", + "#atomic weight of zinc\n", + "Aw=65.37\n", + "#avogadro's number\n", + "NA=6.023*10**23\n", + "#density of zinc\n", + "rho=Aw*Ne/(NA*V)/1000 #kg/m**3\n", + "print \"the no of effective atoms, the volume of unit cell and density of zinc are %d, %0.2e and %d kg/m**3 resp\"%(Ne,V,rho)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the no of effective atoms, the volume of unit cell and density of zinc are 6, 9.07e-29 and 7178 kg/m**3 resp\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-4.3 page no-121" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#ionic radius of anion is \n", + "ra=2.11*10**-10 #m\n", + "#given that the atom has eight fold ligancy with the anion\n", + "#so, we know that for this\n", + "#rc/ra=0.732\n", + "rc=0.732*ra #m\n", + "print \"the radius of smallest cation is 1.54 A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the radius of smallest cation is 1.54 A\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-4.6 page no-125" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#radii of cation and anion in CaO\n", + "rc=0.94*10**-10 #m\n", + "ra=1.32*10**-10 #m\n", + "#so the lattice side will be\n", + "a=2*(rc+ra) #m\n", + "#effective no of atoms in FCC lattice structure\n", + "Ne=4 #because CuO has FCC cubic structure\n", + "#molecular weight of CuO\n", + "Aw=40.08+16\n", + "#atomic weight unit\n", + "amu=1.66*10**-27 #amu\n", + "#mass of atom per unit cell\n", + "M=Aw*amu\n", + "#density\n", + "rho=M*Ne/a**3 #kg/m**3\n", + "print \"density of CuO is %0.f Kg/m**3\"%rho" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "density of CuO is 4032 Kg/m**3\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-4.7 page no-126" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#radii of cation and anion in MgO\n", + "rc=0.78*10**-10 #m\n", + "ra=1.32*10**-10 #m\n", + "#so the lattice side will be\n", + "a=2*(rc+ra) #m\n", + "#effective no of atoms in FCC lattice structure\n", + "Ne=4 #because MgO has FCC cubic structure\n", + "#molecular weight of MgO\n", + "Aw=24.3+16\n", + "#atomic weight unit\n", + "amu=1.66*10**-27 #amu\n", + "#mass of atom per unit cell\n", + "M=Aw*amu\n", + "#density\n", + "rho=M*Ne/a**3 #kg/m**3\n", + "print \"density of MgO is %0.3f Kg/m**3\" %rho" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "density of MgO is 3611.813 Kg/m**3\n" + ] + } + ], + "prompt_number": 5 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch5.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch5.ipynb new file mode 100644 index 00000000..78100746 --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch5.ipynb @@ -0,0 +1,362 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:b4c4e8b4e1c16c859c0b2970e90b60e284d6f5febb43327aebce670fd8dbf006" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-5,Material Characterization" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-5.1 page no-136" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, sin, asin\n", + "#given\n", + "#wavelength of X-rays beams of light\n", + "lamda=0.824*10**-10 #m\n", + "#glancing angle of the incident light\n", + "theta1=(8+35/60)*(pi)/180 #radians\n", + "n1=1\n", + "#to find theta3 i.e at\n", + "n3=3\n", + "#as we know that\n", + "#2*d*sin(theta)=n*lambda\n", + "#so for n1 and n3 we get in the same way and solving together we get \n", + "theta3=asin(3*sin(theta1))\n", + "#so \n", + "d=lamda/2/sin(theta1)\n", + "print \"the galncing angle for the third order diffraction is and interplanar spacing of the crystal is 2.761 A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the galncing angle for the third order diffraction is and interplanar spacing of the crystal is 2.761 A\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-5.2 page no-141" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt, pi\n", + "#given\n", + "#bragg's angle of reflection\n", + "theta1=17.03*(pi)/180 #radians\n", + "#wavelength of light\n", + "lamda=0.71 #A\n", + "#according to bragg's equation \n", + "#n*lambda=2*d*sin(theta)\n", + "#for n=1\n", + "d=lamda/2/sin(theta1) #A\n", + "#given that h**2+k**2+l**2=8\n", + "#let (h**2+k**2+l**2)**1/2=H\n", + "#we get\n", + "H=sqrt(8)\n", + "a=d*H #A\n", + "print \"since h**2+k**2+l**2=8 ,hence the reflecting planes will be (220). family of planes (220) include (220), (202), (022) ,etc.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "since h**2+k**2+l**2=8 ,hence the reflecting planes will be (220). family of planes (220) include (220), (202), (022) ,etc.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-5.3 page n0-141" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#lattice constant\n", + "a=1.54 #A\n", + "#wavelength of beam of light\n", + "lamda=1.54 #A\n", + "#according to bragg's equation\n", + "#n*lambda=2*d*sin(theta)\n", + "#following angles are given\n", + "theta1=20.3*(pi)/180\n", + "theta2=29.2*(pi)/180\n", + "theta3=36.7*(pi)/180\n", + "theta4=43.6*(pi)/180\n", + "#interplaner spadcing is \n", + "d1=lamda/(2*sin(theta1)) #A\n", + "d2=lamda/(2*sin(theta2)) #A\n", + "d3=lamda/(2*sin(theta3)) #A\n", + "d4=lamda/(2*sin(theta4)) #A\n", + "#magnitude of bragg's \n", + "#we have h**2+k**2+l**2=a**2/d**2\n", + "#let a**2/d**2= D for notation only\n", + "#so\n", + "D1=2\n", + "D2=4\n", + "D3=6\n", + "D4=8\n", + "#so from bragg's magnitude we can get (hkl)\n", + "#(hkl1)=(110)\n", + "#(hkl3)=(200)\n", + "#(hkl3)=(211)\n", + "#(hkl4)=(220)\n", + "print \"the reflection will take from {110},{200},{211} and (220)\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the reflection will take from {110},{200},{211} and (220)\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-5.5 page no-146" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#wavelength of X-ray\n", + "lamda=1.54 #A\n", + "#diameter of powder camera\n", + "D=114.6 #mm\n", + "#radius of powder camera\n", + "R=D/2 #mm\n", + "#value of l\n", + "l1=86\n", + "l2=100\n", + "l3=148\n", + "l4=180\n", + "l5=188\n", + "l6=232\n", + "l7=272\n", + "#we know that\n", + "#theta=l/4\n", + "#so\n", + "theta1=l1/4*(pi)/180 #radians\n", + "theta2=l2/4*(pi)/180 #radians\n", + "theta3=l3/4*(pi)/180 #radians\n", + "theta4=l4/4*(pi)/180 #radians\n", + "theta5=l5/4*(pi)/180 #radians\n", + "theta6=l6/4*(pi)/180 #radians\n", + "theta7=l7/4*(pi)/180 #radians\n", + "#now values of sin (theta) and sin(theta2)\n", + "S1=sin(theta1)\n", + "SS1=(sin(theta1))**2\n", + "S2=sin(theta1)\n", + "SS2=(sin(theta1))**2\n", + "S3=sin(theta1)\n", + "SS3=(sin(theta1))**2\n", + "S4=sin(theta1)\n", + "SS4=(sin(theta1))**2\n", + "S5=sin(theta1)\n", + "SS5=(sin(theta1))**2\n", + "S6=sin(theta1)\n", + "SS6=(sin(theta1))**2\n", + "S7=sin(theta1)\n", + "SS7=(sin(theta1))**2\n", + "#so the ratio can be expressed as\n", + "#3:4:8:11:12:16:19\n", + "print \"from the extinction rule, we notice that this is an FCC Structure\"\n", + "#the lattice parameter for highest bragg's angle is\n", + "#a=lambda*sqrt(h**2+k**2+l**2)/(2*sin(theta))\n", + "#here h**2+k**2+l**2=19\n", + "#and let h**2+k**2+l**2 =M for notation\n", + "M=19\n", + "a=lamda*sqrt(M)/(2*sin(theta6)) #A\n", + "print \"lattice parameter of material is %0.2f A\"%(a)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "from the extinction rule, we notice that this is an FCC Structure\n", + "lattice parameter of material is 3.96 A\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-5.6 page no-158" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#ASTM number\n", + "n=12\n", + "#as we know that the number of grains N observed on photomicrograph is given by\n", + "N=2**(n-1)\n", + "#as we know that grain size diameter is given by\n", + "d=1/sqrt((N/645)*10**4) #mm because 1 square inch=645 mm**2\n", + "print \"the grain diameter for an ASTM number 12 is %f mm\"%(d)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the grain diameter for an ASTM number 12 is 0.005774 mm\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-5.7 page no-158" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "#given\n", + "#no of grains within the view of a micrograph\n", + "n1=41\n", + "#no of grains cut by circumference\n", + "n2=42\n", + "#diameter of circular area\n", + "d=1 #inch\n", + "#area\n", + "A=(pi)/4*d**2 #inch**2\n", + "#the area density of grains\n", + "\n", + "N=(n1+n2/2)/A #grains/inch**2\n", + "#grain size\n", + "n=log(N)/log(2)+1 \n", + "print \"the area density of grains is %0.2f grains/inch**2 and grain size is 8\"%(N)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the area density of grains is 78.94 grains/inch**2 and grain size is 8\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-5.8 page no-159" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#ASTM no of grains\n", + "ASTM=5\n", + "#area density of grains\n", + "N=2**(ASTM-1) #grains/inch**2 at magnification of 100*\n", + "#as we know that lineal and areal magnification are related as\n", + "#*100 lineal=*10000 areal\n", + "#therefore\n", + "Nnew=N/0.01/0.01 #grains/inch**2 at 1*\n", + "#average area of one grain\n", + "A=1/Nnew*(2.54)**2 #cm**2\n", + "#now 160000 grains/inch**2 of surface is sqrt(160000)=400 grains/inch of length and this is equal to=(400)**3==6.4*10**7 grains/m**3 of volume\n", + "#surface area of each cubic surface\n", + "S=(1/400)**2 #inch**2\n", + "#there are 6 surfaces in accubic grain\n", + "#thus total surface area of each grain\n", + "T=1/2*6*S*(400)**3/2.54 #cm**2 boundary per cubic cm of steel\n", + "print \"the boumdary area per cubic centimeter of steel is %0.2e cm**2 boundary per cubic centimeter of steel\"%T " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the boumdary area per cubic centimeter of steel is 4.72e+02 cm**2 boundary per cubic centimeter of steel\n" + ] + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch6.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch6.ipynb new file mode 100644 index 00000000..8b4ab2ff --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch6.ipynb @@ -0,0 +1,327 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:b6e48af252b8eb0313570eb48f776c5f4fb174ec410de5193092e40aca119a0c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-6, Crystal Imperfection" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-6.1 page no-169" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt, exp\n", + "#given\n", + "#molecular weight of CaF2 (calcium fluoride)\n", + "Mw=0.079 #kg/mol\n", + "#specific gravity of CaF2\n", + "sg=3.17\n", + "#density\n", + "rho=sg*1000 #kg/m**3\n", + "#avogadro's number\n", + "NA=6.023*10**23\n", + "#the no of lattice sites is\n", + "NL=NA*rho/Mw #calcium ions/m**3\n", + "#the ionic structure of CaF2\n", + "NI=2*NL # per m**3\n", + "#we can note that the no of anions are double that of the no of cation vacancies for frenkel defect, so effective value of Ef =Ef/2 i.e half\n", + "#effective energy\n", + "Ef=2.7*1.602*10**-19 #V (because 1eV=1.602*10**-19 V)\n", + "#boltzmann's constant\n", + "k=1.38*10**-23\n", + "T=1300 #K\n", + "#no of frenkel defects nF per cubic metre ie\n", + "nF=sqrt(NI*NI)*exp(-Ef/2/k/T) # per m**3 (some variation in result is due to the approximation done in book at some points)\n", + "print \"the number of frenkels defect per unit volume of CaF2 is %0.2e per m**3 approx.\"%nF " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the number of frenkels defect per unit volume of CaF2 is 2.81e+23 per m**3 approx.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-6.2 page no-171" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "#given\n", + "#gas constant\n", + "R=8.314 #J/mol K\n", + "#temperatures\n", + "T1=27+273 #K\n", + "T2=52+273 #K\n", + "#we know that\n", + "#n=NA*exp(-Hf/R/T)\n", + "#so for n1=NA*exp(-Hf/R/T1)\n", + "#similarly for n2=NA*exp(-Hf/R/T2)\n", + "#dividing n2 by n1\n", + "#we have given in ques that dfects triple on incresing the temp from T1 to T2\n", + "#so n2/n1=3\n", + "#we get 3=exp(Hf/R*(1/T1-1/T2))\n", + "#on solving we get\n", + "Hf=log(3)*R*(T1*T2/(T2-T1))/1000 #KJ/mol\n", + "print \"enthalpy of formation of vacancies is %0.2f kJ/mol\"%(Hf)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "enthalpy of formation of vacancies is 35.62 kJ/mol\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-6.3 page no-172" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#gas constant\n", + "R=8.314 #J/mol K\n", + "#temperature\n", + "T1=300 #K\n", + "#enthalpy of formation of vacancy\n", + "Hf=168*10**3 #J/mol\n", + "#as we know that\n", + "#n=NA*exp(-Hf/R/T)\n", + "#let n/NA=N\n", + "N1=exp(-Hf/R/T1) \n", + "#now temperature is\n", + "T2=1000 #K\n", + "N2=exp(-Hf/R/T2)\n", + "# ratio of no of vacancies\n", + "ratio=N1/N2 \n", + "print \"the ratio of number of vacancies in equilibrium is %0.2e\"%(ratio)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the ratio of number of vacancies in equilibrium is 3.34e-21\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-6.5 page no-180" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "#given\n", + "#shear modulus\n", + "G=45*10**9 #N/m**2 because 1GPa=10**9 N/m**2\n", + "#burge vector\n", + "b=2.5*10**-9 #m\n", + "#inner and outer radius of elastic strain\n", + "r0=11*10**-10 #m\n", + "r1=10**5*b #m\n", + "#poisson ratio\n", + "nu=0.31\n", + "#for an edge dislocation\n", + "Ued=G*b**2/(4*(pi)*(1-nu))*log (r0/r1) #J/m\n", + "#for screw dislocation\n", + "Usd=G*b**2/(4*(pi))*log (r0/r1) #J/m\n", + "#ratio of edge and screw dislocation\n", + "ratio=Ued/Usd\n", + "print \"tha ratios of energies of an edge dislocation over screw dislocation is %.2f\"%(ratio)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "tha ratios of energies of an edge dislocation over screw dislocation is 1.45\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-6.6 page no183" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#atomic radius of polonium\n", + "r=1.7*10**-10 #m\n", + "#structure of polonium is simple cubic (SC) for which\n", + "a=2*r #m\n", + "#number of atoms \"n\" involved with 1mm long dislocation\n", + "n=10**-3/a \n", + "#the number steps ns of clinbs down for 2 micro m will be\n", + "ns=2*10**-6/a\n", + "#total no of created vacancies\n", + "NV=n*ns\n", + "print \"the total no of vacancies created are %0.2e\"%(NV)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the total no of vacancies created are 1.73e+10\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-6.7 page no-186" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#burgers vector of copper is\n", + "h=1/2\n", + "k=1/2\n", + "l=0/2\n", + "#shear modulus of copper\n", + "G=44*10**9 #N/m**2\n", + "#latice parameter of copper\n", + "a=3.61*10**-10 #m\n", + "#magnitude of burgers vector\n", + "b=a*sqrt(h**2+k**2+l**2)#m\n", + "#the line energy of dislocation\n", + "U=G*b**2/2 #J/m\n", + "print \"the line energy of disloactation is %.2e J/m\"%(U)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the line energy of disloactation is 1.43e-09 J/m\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-6.8 page no-188" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import tan\n", + "#given\n", + "#lattice constant of Ir\n", + "a=3.84 #A\n", + "#burgers vector gfor FCC is 1/2(110)\n", + "#so\n", + "h=1/2\n", + "k=1/2\n", + "l=0/2\n", + "#magnitude of burgers vector\n", + "b=a*sqrt(h**2+k**2+l**2) #A\n", + "#angle of tilts\n", + "theta1=1*(pi)/180 #radians\n", + "theta2=3*(pi)/180 #radians\n", + "#spacing between dislocation for theta1\n", + "h1=b/tan(theta1) #A\n", + "#spacing between dislocations for theta2\n", + "h2=b/tan(theta2) #A\n", + "print \"the spacinf between dislocations at 1 degree and 3 degrees are %0.2f A and %0.2f A resp\"%(h1,h2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the spacinf between dislocations at 1 degree and 3 degrees are 155.56 A and 51.81 A resp\n" + ] + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch7.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch7.ipynb new file mode 100644 index 00000000..10864c5e --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch7.ipynb @@ -0,0 +1,369 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:625ae8c5d89aa803ec88a91915a577ea6d8b52d1a9407231a8a1fcfb98194291" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-7, Diffusion in solids" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-7.1 page no-203" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#outward flux of copper is\n", + "Jx=10**21 #per m**2 per sec\n", + "#concentration of copper at room temp on one side of aluminium which is 3 mm thickness\n", + "C1=2*10**13 # per m**3\n", + "dx=3*10**-3 #m\n", + "#concentration on the other side of aluminium\n", + "C2=4*10**6 # per m**3\n", + "#concentration gradient\n", + "M=(C1-C2)/dx #m**4\n", + "#according to fick's law\n", + "#Jx=-Dx*dM/dx\n", + "#so Dx will be\n", + "Dx=-Jx*1/M #m**2/s\n", + "print \"the concentration gradient is %0.2e per m**4and diffusivity of copper is %0.2e m**2/s\"%(M,Dx)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the concentration gradient is 6.67e+15 per m**4and diffusivity of copper is -1.50e+05 m**2/s\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-7.2 page no-209" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import exp, sqrt\n", + "from scipy.special import erfinv\n", + "#given\n", + "#activation energy and constant D0 are given as\n", + "#for notation only we are using D01 and D02\n", + "Q1=83*10**3 #J/mol\n", + "Q2=157*10**3 #J/mol\n", + "D01=8*10**-7 #m**2/s\n", + "D02=700*10**-7 #m**2/s\n", + "#diffusivity of carbon in alpha-iron at 873 K (600+273) can be obtained by the equation\n", + "T=873 #K\n", + "#and \n", + "#gas constant\n", + "R=8.314 #J/mol K\n", + "Dx=D01*exp(-Q1/R/T) #m**2/s\n", + "#concentration of carbon is given as 0.75% , therefore concentration of steel will be 0.25 %. thus\n", + "Ms=0.25 #conc. of steel\n", + "Mc=0.75 #conc. of carbon\n", + "#time of carburization is 9 hrs\n", + "t=9*3600 #sec\n", + "#so \n", + "C1=Mc\n", + "C2=C1-Ms\n", + "#carbon content on the surface\n", + "M=0.6\n", + "#we know that diffusion across a common interface is obtained by:-\n", + "#M=C1-C2*erf(x/2*sqrt(Dx/t))\n", + "#solving for X where X=erf(x/2*sqrt(Dx/t))\n", + "X=((C1-M)/C2)\n", + "#now for x\n", + "x=erfinv(X)*2*sqrt(Dx*t)*10**4 #m\n", + "print \"depth of carburization is %0.2f m\"%(x) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "depth of carburization is 2.88 m\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-7.3 page no-211" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import exp\n", + "from scipy.special import erfinv\n", + "#given\n", + "#temp\n", + "T=950+273 #K\n", + "QA1=83*10**3 #J/mole\n", + "QA2=157*10**3 #J/mole\n", + "R=8.314\n", + "D01=0.008*10**-4 #m**2/s\n", + "D02=0.7*10**-4 #m**2/s\n", + "Ms=0.8 #%\n", + "Mc=0 #%\n", + "Mx=0.6 #%\n", + "t=4*3600 #sec\n", + "D=1.38*10**-11 #m**2/s \n", + "#diffusivity at 950 degrees celcius\n", + "Dx=D01*exp(-QA2/R/T) #m**2/s\n", + "#now we use fick's solution which is given by\n", + "#M(x,t)=C1-C2*erf(x/2/sqrt(Dt))\n", + "#first boundary condition is \n", + "#M(x,0)=0.8 for x<0\n", + "#on solving we get\n", + "#C1+C2=0.8\n", + "#second boundary condition is \n", + "#M(0,t)=0\n", + "#so we get\n", + "#C1-C2=0\n", + "#therefore we get\n", + "C1=0\n", + "C2=0.8\n", + "#so we get x as\n", + "x=erfinv((C1-Mx)/C2)*2*sqrt(D*t)\n", + "print \"post machining is to be done upto %0.2e mm\"%x " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "post machining is to be done upto -7.25e-04 mm\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-7.4 page no-213" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from scipy.special import erfinv\n", + "#given\n", + "D0=0.002*10**-6 #m**2/s\n", + "Q=120*10**3 #J/mol\n", + "T=550+273 #K\n", + "#the condition for diffusion are\n", + "#M(x,0)=4% for x>0\n", + "#M(x,0)=0% for x<0\n", + "x=(0.25-0.10)*10**-3 #m\n", + "Mx=0.4 #%\n", + "R=8.314 #gas constant\n", + "#diffusion at 550 degrees celcius is\n", + "Dx=D0*exp(-Q/R/T) #m**2/s\n", + "#by using\n", + "#M(x,t)=C1-C2*erf(x/2/sqrt(Dx*t))\n", + "#putting boundary conditions in above equation we get\n", + "C1=2 #%\n", + "C2=2 #%\n", + "#solving for t\n", + "#we get\n", + "t=(erfinv((C1-Mx)/C2)*2*sqrt(Dx)/x)**-2 #sec\n", + "print \"the duration for which he material is kept at 550 degrees celcius is %0.2e sec\"%(t)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the duration for which he material is kept at 550 degrees celcius is 1.42e+08 sec\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-7.5 page no-215" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from scipy.special import erfinv\n", + "#given\n", + "Mx=10**22 #atoms/m**3\n", + "M0=10**24 #atoms/m**3\n", + "t=3*3600 #sec\n", + "T=1100+273 #K\n", + "D=7*10**-17 #m**2/s\n", + "#boundary cinditions are\n", + "#M(x,0)=0 for x>0 at t=0\n", + "#M(0,t)=10**24 atoms/m**3\n", + "#M(x,t)=C1-C2*erf(x/22/sqrt(D*t))\n", + "#putting boundary conditions in the above equation\n", + "#we get \n", + "C1=10**24 #atoms/m**3\n", + "C2=10**24 #atoms/m**3\n", + "#solving for x\n", + "#now for x we have\n", + "x=erfinv((C1-Mx)/C2)*2*sqrt(D*t)*10**3 #mm\n", + "print \"depth below the surface at which the concentration is 10**22 atoms/m**3 is %f mm\"%(x)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "depth below the surface at which the concentration is 10**22 atoms/m**3 is 0.003167 mm\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-7.6 page no-216" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#temp and pressure of Nitrogen gas\n", + "T=400 #K\n", + "P=15 #atm\n", + "#Nitrogen conc at the inner surface of the tank\n", + "M=12 #kg/m**3\n", + "#constant D0 and activation energy\n", + "D0=5*10**-7 # m**2/s\n", + "Q=75*10**3 #J/mol\n", + "#thickness of tank wall\n", + "x=6*10**-3 #m\n", + "D=M/x #kg/m**4 (calculation mistake in book)\n", + "#gas constant\n", + "R=8.314 #J/mol/K\n", + "#dufusing flux\n", + "#Jx=-Dx*m/x kg/m**2/s\n", + "#Dx can be determined as follows\n", + "Dx=D0*exp(-Q/R/T) #m**2/s\n", + "#so #diffusing flux will be\n", + "Jx=Dx*D #kg/m**2/s (calculation mistake in book)\n", + "print \"diffusing rate of nitrogen is %0.2e kg/m**2/s\"%(Jx)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "diffusing rate of nitrogen is 1.61e-13 kg/m**2/s\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-7.7 page no-217" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "from __future__ import division\n", + "#given\n", + "#ratio of rate of diffusion at 1350 and 110 degree C is 8\n", + "#so\n", + "#dM/dt at 1350 degree C(1623K)/dM/dt at 110degree C (1373K)=8\n", + "#so \n", + "Ratio=8\n", + "#we know that\n", + "#Dx=D0*exp(-Q/R/T)\n", + "#so Dx (at 1623 K)/Dx (at 1373 K)= exp(-Q/R/T1-Q/R/T2)\n", + "#where\n", + "T1=1623 #K\n", + "T2=1373 #K\n", + "#gas constant\n", + "R=8.314 #J/mol/K\n", + "#so we get\n", + "#Ratio=exp(-Q/R*(1/T1-1/T2))\n", + "#so we get Q as\n", + "Q=-log(Ratio)*R/(1/T1-1/T2)/1000 #KJ/mol\n", + "print \"the activation energy for sulphur diffusion is %d KJ/mol\"%(Q)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the activation energy for sulphur diffusion is 154 KJ/mol\n" + ] + } + ], + "prompt_number": 19 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch8.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch8.ipynb new file mode 100644 index 00000000..ec9c3b00 --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch8.ipynb @@ -0,0 +1,298 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:87280833dc2d67e192335bfc1124aebe8f8ba8cde8ddbc9caf92eb09f60263d6" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-8, Phase & Phase diagrams" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-8.1 page no-227" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#according to reduced phase rule, we have\n", + "#D=C-P+2\n", + "C=2 # for two component system\n", + "P=1\n", + "for P in range (1,6):\n", + " print \"P =\",P,' ',\n", + " #no of variables\n", + " V=P*(C-1)+2\n", + " #degrees of freedom\n", + " D=C-P+2\n", + " print \"D =\",D\n", + "\n", + "print \"we can see that for P=5 we have D=-1 i.e non existent so,two components cannot have more than 4 phases in equilibrium\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "P = 1 D = 3\n", + "P = 2 D = 2\n", + "P = 3 D = 1\n", + "P = 4 D = 0\n", + "P = 5 D = -1\n", + "we can see that for P=5 we have D=-1 i.e non existent so,two components cannot have more than 4 phases in equilibrium\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-8.2 page no-235" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#density of alpha and beta phases\n", + "rhoalpha=10300 #kg/m**3\n", + "rhobeta=7300 #kg/m**3\n", + "#refer to fig-8.5 in book\n", + "#at point B, the composition of lead in alpha-phase is 82% and that of tin in alpha-phase is 18%\n", + "leadalpha=82\n", + "tinalpha=18\n", + "#so we get\n", + "#82/rholead+18/rhotin=100/rhoalpha -----------(1)\n", + "#similarly at point E\n", + "#the composition of tin and lead resp are 97% and 3%\n", + "leadbeta=3\n", + "tinbeta=97\n", + "#so we get\n", + "#3/rholead+97/rhotin=100/rhobeta ------(2)\n", + "#solving 1 and 2\n", + "#we get\n", + "rholead=11364.1 #kg/m**3\n", + "rhotin=7220.14 #kg/m**3\n", + "#let density of eutectic composition is rhoe. knowing the compositions at point D, we can write\n", + "#38/rholead+62/rhotin=100/rhoe\n", + "#so \n", + "rhoe=100/(38/rholead+62/rhotin) #kg/m**3\n", + "#it is given that there is 88% eutectic composition by volume. its conversion in weight proportions yeild\n", + "W=88/100*rhoe #kgf\n", + "Wlead=38/100*W #Kgf\n", + "Wtin=62/100*W #kgf\n", + "#there is 12% beta phase by volume which on converion to weight proportion gives\n", + "Wdash=12/100*rhobeta #Kgf\n", + "Wdashlead=3/100*Wdash #kgf\n", + "Wdashtin=97/100*Wdash #kgf\n", + "#total weight of lead and tin can be estimated now as\n", + "Wddashlead=Wlead+Wdashlead #kgf\n", + "Wddashtin=Wtin+Wdashtin #kgf\n", + "#percentafe of tin\n", + "percenttin=Wddashtin/(Wddashtin+Wddashlead)*100 \n", + "print \"percentace of tin is %0.2f\"%(percenttin)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "percentace of tin is 65.72\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-8.4 page no-242" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#melting point of A and B\n", + "MptA=1250 #degrees celcius\n", + "MptB=1900 #degrees celcius\n", + "#part(a)\n", + "#according to the conditions given in question phase diagram can be drawn as shown in fig 8.10 given in book\n", + "#part(b)\n", + "#at 1250 degrees celcius , it is a peritectic solution\n", + "#equation representing the equilibrium is given in the book which denotes forward reaction as cooling and backward as heating\n", + "#part (c)\n", + "#we are just considering a tie line just above the peritectic line at temp 1251 degrees celcius\n", + "#at this point\n", + "Cs1=80\n", + "Cl1=30 \n", + "C01=75\n", + "#point below peritectic line has temp as 1249 degrees celcius\n", + "Cs2=80\n", + "Cl2=50 \n", + "C02=75\n", + "#weight fraction of the phase present in the material of overall composition 75% B at 1251 degrees celcius and 75% B at 1249 degrees \n", + "f_alpha1=(Cs1-C01)/(Cs1-Cl1)*100 #%\n", + "f_beta1=(C01-Cl1)/(Cs1-Cl1)*100 #%\n", + "f_alpha2=(Cs2-C02)/(Cs2-Cl2)*100 #%\n", + "f_beta2=(C02-Cl2)/(Cs2-Cl2)*100 #%\n", + "#part (d)\n", + "# 75% B at room temp\n", + "Cs3=90\n", + "Cl3=30 \n", + "C03=75\n", + "f_alpha3=(Cs3-C03)/(Cs3-Cl3)*100 #%\n", + "f_beta3=(C03-Cl3)/(Cs3-Cl3)*100 #%\n", + "#the microstructure is also shown in book at page no 243\n", + "print \"weight fraction of the phase present in the material of overall composition 75 percent B at 1251 degrees celcius and 75percent B at 1249 degrees are %d, %d, %0.2f, %0.2f and at 75 percent concentration of B at room temp is %d, %d\"%(f_alpha1,f_beta1,f_alpha2,f_beta2,f_alpha3,f_beta3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "weight fraction of the phase present in the material of overall composition 75 percent B at 1251 degrees celcius and 75percent B at 1249 degrees are 10, 90, 16.67, 83.33 and at 75 percent concentration of B at room temp is 25, 75\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-8.5 page no-246" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#according to the given conditions in the ques the graph can be drawn as shown in fig 8.12\n", + "#the given temp 576.9 degees celcius is just below the eutectic temp 577 degres celcius\n", + "#from the graph, we can have the following values\n", + "C_beta_e=100\n", + "C_e=1.65\n", + "C_0=10\n", + "#hence weight fraction is given by\n", + "W_alpha=(C_beta_e-C_0)/(C_beta_e-C_e)\n", + "print \"the weight fraction of alpha in an alloy containing 10percent Si at 576.9 degrees celcius is %0.3f \"%(W_alpha)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the weight fraction of alpha in an alloy containing 10percent Si at 576.9 degrees celcius is 0.915 \n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-8.6 page no-247" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#weight percent of tin and lead are\n", + "W1=90\n", + "W2=10\n", + "#let the amount of tin that can be added to the crucible without changing the system's solidificarion is \"m\" \n", + "#eutectic arm extends to 97% tin\n", + "#therefore, with the addition of \"m\" gram of tin, the composition of alloy should not exceed 97% tin\n", + "#therefore\n", + "#(900+m)/(1000+m)=97/100\n", + "#so\n", + "m=(97/100*1000-900)/(1-97/100)/1000 #kg because 1kg=1000g\n", + "print \"maximum %0.2f kg of tin can be added without changing the systems temperature\"%(m)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "maximum 2.33 kg of tin can be added without changing the systems temperature\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-8.7 page no-250" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#the eutectoidal mixture of pearlite consists of two phases viz. alpha (ferrite) and Fe3C (cementite). the eutectoid composition contains 0.83% carbon. the lever rule is applied in which the lever arm has ferrite (=0% carbon) at one end and cementite (6.67% carbon) at the other end. the fulcrum is taken at 0.83% carbon. hence by applying lever rule, we get\n", + "Walpha=(6.67-0.83)/(6.67-0.00)\n", + "WFe3C=(0.83-0.0)/(6.67-0.0)\n", + "print \"the weight fractions of ferrite and cementite are %0.3f and %0.3f resp\"%(Walpha,WFe3C)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the weight fractions of ferrite and cementite are 0.876 and 0.124 resp\n" + ] + } + ], + "prompt_number": 20 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch9.ipynb b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch9.ipynb new file mode 100644 index 00000000..e826cfda --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/Ch9.ipynb @@ -0,0 +1,216 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:88f6d7864909abb362ac624b2474dc1ea419f7b404761f21c8b5e63cda752047" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-9, Mechanical Properties" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-9.1 page no-259" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#given\n", + "#bonding characteristics of a material are\n", + "n=1\n", + "m=9\n", + "A=7.56*10**-29 #J m\n", + "#initial bond length\n", + "r_0=2.3*10**-10 #m\n", + "#this bond length exceed by 15 %\n", + "#so extension is \n", + "e=15/100*r_0 #m\n", + "#new bond length will be\n", + "r=r_0+e #m\n", + "#the axial tensile starin is given by\n", + "epsilon_t=e/r_0 #(change in dimension /original dimension)\n", + "#according to P-r function we have\n", + "B=A*r_0**8/9 #J m**9\n", + "#we have double deravative of P as\n", + "double_derivative_P=-2*A/r_0**3+90*B/r_0**11 #/J/m**2\n", + "#so youngs modulus will be given by\n", + "E=double_derivative_P/r_0*10**-9 #GPa (some approximation is done in book)\n", + "print \"the axial stain is %.2f \\nand youngs modulus of elasticity is %0.3e GPa \"%(epsilon_t,E)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the axial stain is 0.15 \n", + "and youngs modulus of elasticity is 2.161e+02 GPa \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-9.4 page no-274" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#according to the data given in the question we can have the graph shown in the fig 9.11 in book\n", + "#part(a):- tangent modulus of elasticity at 200 MPa\n", + "E_tangent=(222-168)*10**9/(1.90-1.42)/10**9 #G Pa ( values from graph)\n", + "#part(b):- secant modulus of elasticity at 180 MPa\n", + "E_secant= (180-0)*10**9/(1.46-0)/10**9 #G Pa (values from graph)\n", + "#part(c):-youngs modulus of elasticity at 85 MPa\n", + "E=(85-0)*10**6/((0.68-0)*10**-3)/10**9 #G Pa (values from graph)\n", + "print \"\"\"tangent modulus of elasticity at 200 MPa is %0.2f Pa\n", + "secant modulus of elasticity at 180 MPa is %d Pa\n", + "youngs modulus of elasticity at 85 MPa is %d Pa\"\"\"%(E_tangent,E_secant,E)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "tangent modulus of elasticity at 200 MPa is 112.50 Pa\n", + "secant modulus of elasticity at 180 MPa is 123 Pa\n", + "youngs modulus of elasticity at 85 MPa is 124 Pa\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-9.5 page no-280" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#stress\n", + "sigmamean=70 #MPa\n", + "#stress range\n", + "sigmarange=210 #MPa\n", + "#as we know that\n", + "#sigmamean=(sigmamax+sigmamin)/2\n", + "#from this we get\n", + "#sigmamax+sigmamin=140 -----------(1)\n", + "#also\n", + "#sigmarange=sigmamax-sigmamin\n", + "#so we get\n", + "#sigmamax-sigmamin=210 ----------(2)\n", + "#solving 1 and 2\n", + "#we get\n", + "sigmamax=(140+210)/2 #MPa\n", + "#and\n", + "sigmamin=140-sigmamax #MPa\n", + "#stress ratio\n", + "R=sigmamin/sigmamax\n", + "#stress ranges from sigmamin to sigmamax\n", + "#so total sum will be\n", + "totalsum=-sigmamin+sigmamax #MPa\n", + "print \"\"\"the maximum and minimum stresses are %d MPa and %d MPa resp,\n", + "stress ratio is %f\n", + "and total sum is %d MPa\"\"\"%(sigmamax,sigmamin,R,totalsum)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the maximum and minimum stresses are 175 MPa and -35 MPa resp,\n", + "stress ratio is -0.200000\n", + "and total sum is 210 MPa\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-9.7 page no-289" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#temp \n", + "T=600 #degree C\n", + "#tightenening stress\n", + "sigmai=750 #kgf/cm**2\n", + "#minimum creep rate\n", + "vcr=2.8*10**-8 #cm/cm/hour\n", + "#stress\n", + "sigma=300 #kgf/cm**2\n", + "#young's modulus\n", + "E=2*10**6 #kgf/cm**2\n", + "#constant\n", + "n=3\n", + "#we knnow that minimum creep rate\n", + "#vcr=A*sigma**n\n", + "#so\n", + "A=vcr/sigma**n\n", + "#total time involved in creep of bolt is 1 year\n", + "t=365*24 #hours\n", + "#the stress relaxation in bolt due to creep is expresed by:-\n", + "#1/(sigmaif)**(n-1)=1/(sigmai)**(n-1)+A*E*(n-1)*t\n", + "#we have to find sigmaif\n", + "#so\n", + "sigmaif=((1/(sigmai)**(n-1)+A*E*(n-1)*t)**(1/(n-1)))**-1 \n", + "print \"the stress relaxation in bolt to creep is %0.3f kgf/cm**2\"%(sigmaif)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the stress relaxation in bolt to creep is 161.975 kgf/cm**2\n" + ] + } + ], + "prompt_number": 8 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/screenshots/14_fraction_of_load.png b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/screenshots/14_fraction_of_load.png Binary files differnew file mode 100644 index 00000000..249790e6 --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/screenshots/14_fraction_of_load.png diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/screenshots/7_atom_conc.png b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/screenshots/7_atom_conc.png Binary files differnew file mode 100644 index 00000000..b3699fef --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/screenshots/7_atom_conc.png diff --git a/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/screenshots/ch4no_of_eff_atoms.png b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/screenshots/ch4no_of_eff_atoms.png Binary files differnew file mode 100644 index 00000000..9875faea --- /dev/null +++ b/Materials_Science,_Metallurgy_And_Engineering_Materials_by_K.M._Gupta/screenshots/ch4no_of_eff_atoms.png diff --git a/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_10_SILICON_CONTROLLED_RECTIFIER_2.ipynb b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_10_SILICON_CONTROLLED_RECTIFIER_2.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_2.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_2.ipynb b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_1_CRYSTAL_STRUCTURES_2.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_2.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_2.ipynb b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/Chapter_6_ELECTRICAL_BREAKDOWN_IN_PN_JUNCTIONS_2.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_2.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_2.ipynb b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_2_ENERGY_BAND_THEORY_OF_SOLIDS_2.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_2.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_2.ipynb b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_3_CARRIER_TRANSPORT_IN_SEMICONDUCTOR_2.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_2.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_2.ipynb b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_4__EXCESS_CARRIER_IN_SEMICONDUCTOR_2.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_2.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_2.ipynb b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_5_PN_JUNCTION_DIODE_2.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_2.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_2.ipynb b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_7_BIPOLAR_JUNCTION_TRANSISTORB_2.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_2.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_2.ipynb b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/chapter_8_THE_FIELD_EFFECT_TRANSISTOR_2.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_2.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_2016-01-07_at_8.46.13_pm.png b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2016-01-07_at_8.46.13_pm.png Binary files differnew file mode 100644 index 00000000..d4c41ffc --- /dev/null +++ b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2016-01-07_at_8.46.13_pm.png diff --git a/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2016-01-07_at_8.46.31_pm.png b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2016-01-07_at_8.46.31_pm.png Binary files differnew file mode 100644 index 00000000..010c6346 --- /dev/null +++ b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2016-01-07_at_8.46.31_pm.png diff --git a/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2016-01-07_at_8.46.51_pm.png b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2016-01-07_at_8.46.51_pm.png Binary files differnew file mode 100644 index 00000000..76f812f2 --- /dev/null +++ b/Solid_State_Devices_by_B._S._Nair_and_S._R._Deepa/screenshots/Screen_Shot_2016-01-07_at_8.46.51_pm.png |
