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| author | hardythe1 | 2015-04-07 15:58:05 +0530 |
|---|---|---|
| committer | hardythe1 | 2015-04-07 15:58:05 +0530 |
| commit | 92cca121f959c6616e3da431c1e2d23c4fa5e886 (patch) | |
| tree | 205e68d0ce598ac5caca7de839a2934d746cce86 /Advance_Semiconductor_Devices/ChapterNo2.ipynb | |
| parent | b14c13fcc6bb6d01c468805d612acb353ec168ac (diff) | |
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diff --git a/Advance_Semiconductor_Devices/ChapterNo2.ipynb b/Advance_Semiconductor_Devices/ChapterNo2.ipynb new file mode 100755 index 00000000..9bc8cffb --- /dev/null +++ b/Advance_Semiconductor_Devices/ChapterNo2.ipynb @@ -0,0 +1,1248 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1> Chapter No 2 : Semiconductor Materials and\n",
+ "Their Properties<h1>"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.1, Page No 65 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "lambda1 = 11000.0 #in \u00c5\n",
+ "lambda1 = lambda1 * 10**-10 \n",
+ "h = 6.625*10**-34 \n",
+ "c = 3*10**8\n",
+ "q = 1.6*10**-19 #in C\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "E_g = h*c/lambda1 #in J\n",
+ "E_g= E_g/q #in eV\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The energy gap in Si is = %.2f eV' %E_g)\n",
+ "\n",
+ "#Note: The answer in the book is not correct \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy gap in Si is = 1.13 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2, Page No 65 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "E_g = 0.75 #in eV\n",
+ "q=1.6*10**-19 #in C\n",
+ "E_g = E_g*q #in J\n",
+ "h = 6.63*10**-34 #in J\n",
+ "c = 3*10**8 #in m/s\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "lembda = (h*c)/E_g #in m\n",
+ "lembda = lembda * 10**10 #in \u00c5\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The wavelength is =%.f \u00c5 ' %lembda)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is =16575 \u00c5 \n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.3 Page No 81 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from numpy import *\n",
+ "\n",
+ "#initialisation of variables\n",
+ "del_E = 0.3 #value of E_C-E_F in eV\n",
+ "T1 = 330.0 #in K\n",
+ "T = 300 #in K\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "del_E1 = del_E*(T1/T) #value of E_C-E_F in eV\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The position of fermi level is =%.2f eV' %del_E1)\n",
+ "print('Hence the Fermi level will be %.2f eV below the conduction band' %del_E1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The position of fermi level is =0.33 eV\n",
+ "Hence the Fermi level will be 0.33 eV below the conduction band\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.4, Page No 81"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "K = 8.63*10**-5\n",
+ "T = 300.0 #in K\n",
+ "N_C = 2.8*10**19 #in cm^-3\n",
+ "del_E = 0.25\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "f_F = math.exp( (-del_E)/(K*T) )\n",
+ "n_o = N_C*math.exp( (-del_E)/(K*T) ) #in cm^-3\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The probability is =%.6f ' %f_F)\n",
+ "print('The thermal equillibrium electron concentration is =%.2f X 10^15 cm^-3' %(n_o/(10**15)))\n",
+ "print('Approximation error ')\n",
+ "# The answer is same as in book only thy have rounded off to 1.8 * 10^15 which is same as above"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The probability is =0.000064 \n",
+ "The thermal equillibrium electron concentration is =1.79 X 10^15 cm^-3\n",
+ "Approximation error \n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.5 Page No 82"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "N_V = 1.04*10**19 #in cm^-3\n",
+ "T1 = 400.0 #in K\n",
+ "T2 = 300.0 #in K\n",
+ "del_E = 0.27 #value of E_F-E_V in eV\n",
+ "K = 0.0259\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "N_V= N_V*(T1/T2)**(3/2) #in cm^-3\n",
+ "KT = K*(T1/T2) #in eV\n",
+ "p_o = N_V*math.exp( (-del_E)/(KT) ) #in /cm^3\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The hole concentration is =%.2f x 10^15 per cm^3' %(p_o/(10**15)))\n",
+ "print('Approximationa error ')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The hole concentration is =5.58 x 10^15 per cm^3\n",
+ "Approximationa error \n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.6 Page No 86"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "N = 6.02*10**23\n",
+ "A = 63.5 #atomic weight\n",
+ "Rho = 1.7*10**-6 #in ohm cm\n",
+ "d = 8.96 # in gm/cc\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "n = (N/A)*d #in /cc\n",
+ "e = 1.6*10**-19 #in C\n",
+ "Miu_e = 1/(Rho*n*e) #in cm^2/volt-sec\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The mobility of electron is =%.2f cm^2/volt.sec' %Miu_e)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mobility of electron is =43.28 cm^2/volt.sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.7, Page No 87"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from sympy.mpmath import *\n",
+ "import cmath\n",
+ "\n",
+ "\n",
+ "#initialisation of variables\n",
+ "d = 8.96 #in gm/cc\n",
+ "At = 63.5 #atomic weight\n",
+ "N_A = 6.02*10**23 #in /gm mole\n",
+ "l = 0.1 #in m\n",
+ "e = 1.6*10**-19 #in C\n",
+ "A = 1.7*10**-6 #in m^2\n",
+ "R = 0.1 #in ohm\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "n = (N_A/At)*d #in /cc\n",
+ "n = n * 10**6 #in /m^3\n",
+ "Rho = (R*A)/l #in ohm.m\n",
+ "Sigma = 1/Rho #in mho/m\n",
+ "Miu_e = Sigma/(n*e) #in m^2/V-sec\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The electron mobility is =%.3f X 10^-3 m^2/V-sec ' %(Miu_e*(10**3)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron mobility is =0.043 X 10^-3 m^2/V-sec \n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.8 Page No 87"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#initialisation of variables\n",
+ "N_A = 6.025*10**23 #in /gm mole\n",
+ "d = 10.5 #in gm/cc\n",
+ "At = 108.0 #atomic weight of \n",
+ "\n",
+ "#CALCULATIONS\n",
+ "n = (N_A/At)*d #in /cm^3\n",
+ "n = n * 10**6 #in /m^3\n",
+ "r = 10**-3 #in m\n",
+ "A = math.pi * ((r)**2) #in m^2\n",
+ "q = 1.6*10**-19 \n",
+ "I = 2 #in A\n",
+ "V = I/(n*q*A) #in m/s\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The drift velocity of an electron is =%.f X 10^-4 m/s' %(V*(10**5)))\n",
+ "print('Approximationa error ')\n",
+ "# answer is same as in book only they have rounded of 0.000068 to 7 * 10^-4"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The drift velocity of an electron is =7 X 10^-4 m/s\n",
+ "Approximationa error \n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.9 Page No 88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "e= 1.6*10**-19 #in C\n",
+ "d= 1.03 #in mm\n",
+ "d= d*10**-3 #in m\n",
+ "R= 6.51 #in ohm\n",
+ "l= 300.0 #in m\n",
+ "n= 8.4*10**28 #per m^3\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "r= d/2 #in m (radius)\n",
+ "A= math.pi*r**2 #in m^2\n",
+ "rho= R*A/l #in ohm meter\n",
+ "sigma= 1/rho #in mho/m\n",
+ "miu_e= sigma/(n*e) #m^2/V-sec\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The coductivity of copper is =%.2f mho/m' %sigma)\n",
+ "print('The mobility of charge carriers is : =%.6f m^2/V-sec' %miu_e)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The coductivity of copper is =55306469.41 mho/m\n",
+ "The mobility of charge carriers is : =0.004115 m^2/V-sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.9.ii Page No 88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "Miu_e = 1500.0 #in cm^2/volt.sec\n",
+ "Miu_h = 500.0 #in cm^2/volt.sec\n",
+ "n_i = 1.6*10**10 #in /cm^3\n",
+ "e = 1.6*10**-19 #in C\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Sigma_i = n_i*(Miu_e+Miu_h)*e #in mho/cm\n",
+ "Sigma = Sigma_i #in mho/cm\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The conductivity of pure silicon is =%.7f mho/cm' %Sigma)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The conductivity of pure silicon is =0.0000051 mho/cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 2.10 Page No 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "Miu_e = 1500.0 #in cm^2/volt.sec\n",
+ "Miu_h = 500.0 #in cm^2/volt.sec\n",
+ "n_i = 1.6*10**10 #in /cm^3\n",
+ "e = 1.6*10**-19 #in C\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Sigma_i = n_i*(Miu_e+Miu_h)*e #in mho/cm\n",
+ "Sigma = Sigma_i #in mho/cm\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The conductivity of pure silicon is =%.2f X 10^-6 mho/cm' %(Sigma*(10**6)))\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The conductivity of pure silicon is =5.12 X 10^-6 mho/cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 2.11 Page No 90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "Miu_d = 500.0 #in cm^2/V.S\n",
+ "Rho = 10 #in ohm cm \n",
+ "\n",
+ "#CALCULATIONS\n",
+ "e = 1.6*10**-19 #in C\n",
+ "n_d = 1/(Rho*e*Miu_d) #in /cm^3... correction\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The number of donor is =%.2f X 10^15 atom per cm^3' %(n_d/(10**15)))\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The number of donor is =1.25 X 10^15 atom per cm^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 2.12 Page No 90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "d = 5.32 #in gm/cc\n",
+ "N_A = 6.02*10**23 #in atoms/gm.mole\n",
+ "At = 72.6 #atomic weight\n",
+ "Miu = 3800 #in cm^2/v.s\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "n_d = (N_A/At) * d #in /cm^3\n",
+ "n_d = n_d * 10**-8 #in /cc\n",
+ "e = 1.6*10**-19 #in C\n",
+ "Sigma = n_d * Miu * e #in mho/cm\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The conductivity of specimen is =%.2f mho/cm' %Sigma)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The conductivity of specimen is =0.27 mho/cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 2.13 Page No 90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "Rho = 0.3623*10**-3 #in ohm m\n",
+ "d = 4.42*10**28 #Ge density in atoms/m^3\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Sigma = 1/Rho #in mho/m\n",
+ "n_d = d*10**-6 #in atoms/m^3\n",
+ "e = 1.6*10**-19 #in C\n",
+ "Miu = Sigma/(n_d*e) #in m^2/V.sec\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The electron mobility is =%.2f m^2/V-sec' %Miu)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron mobility is =0.39 m^2/V-sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 2.14 Page No 91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "N_A = 6.025*10**26 #in /kg.Mole\n",
+ "At = 72.59 #atomic weight\n",
+ "d = 5.36*10**3 #in kg/m^3\n",
+ "R = 0.42 #in ohm m\n",
+ "B_i = 10**-6 #rate of boron impurity in %\n",
+ "e = 1.6*10**-19 #in C\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "n = (N_A/At)*d #number of Ge atoms\n",
+ "h = n/10**8 #holes per unit volume\n",
+ "Miu_h = 1/(R*h*e) #in m^2/V.sec\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The Mobility of holes is =%.2f m^2/V-sec' %Miu_h)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Mobility of holes is =0.03 m^2/V-sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.15 Page No 90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "n_i = 2*10**19 #in /m^3\n",
+ "Miu_e = 0.36 #in m^2/v.s\n",
+ "Miu_h = 0.17 #in m^2/v.s\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "A = 1*10**-4 #in m^2\n",
+ "V = 2 #in Volts\n",
+ "l = 0.3 #in mm\n",
+ "l = l * 10**-3 #in m\n",
+ "e = 1.6*10**-19 #in C\n",
+ "Sigma_i = n_i * e * (Miu_e+Miu_h) #in mho/m\n",
+ "I = (Sigma_i * V*A)/l #in amp \n",
+ "\n",
+ "#RESULTS\n",
+ "print('The current in amp is =%.2f ' %I)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current in amp is =1.13 \n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.16 Page No 92"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "d = 4.2*10**28 #in atoms/m^3\n",
+ "n_d = d/10**6 #in atoms/m^3\n",
+ "e = 1.6*10**-19 #in C\n",
+ "Miu_e = 0.36 #in m^2/V-sec\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Sigma_n = n_d *e *Miu_e #in mho/m\n",
+ "Rho_n = 1/Sigma_n #ohm m\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The resistivity in \u03a9m is =%.2f X 10^-3' %(Rho_n*(10**3)))\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistivity in \u03a9m is =0.41 X 10^-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.17 Page No 92"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "n_i = 1*10**19 #in /m^3\n",
+ "Miu_e = 0.36 #in m^2/volt.sec\n",
+ "Miu_h = 0.17 #in m^2/volt.sec\n",
+ "A = 2 #in cm^2\n",
+ "A = A * 10**-4 #in m^2\n",
+ "t = 0.1 #in mm\n",
+ "t = t*10**-3 #in m\n",
+ "V = 4 #in volts\n",
+ "e = 1.6*10**-19 #in C\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Sigma_i = n_i * e * (Miu_e + Miu_h) #mho/m\n",
+ "J = Sigma_i * (V/t) #in Amp/m^2\n",
+ "I = J*A #in Amp\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The current in Amp is =%.2f ' %I)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current in Amp is =6.78 \n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No 2.18 Page No 92"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "Miu_h = 500.0 #in cm^2/V.s\n",
+ "Miu_e = 1500.0 #in cm^2/V.s\n",
+ "n_i = 1.6*10**10 #in /cm^3\n",
+ "e = 1.6*10**-19 #in C\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Sigma_i = n_i * e * (Miu_e+Miu_h) #in mho/cm\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The conductivity of pure silicon in mho/cm is =%.2f X 10^-6 ' %(Sigma_i*(10**6)))\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The conductivity of pure silicon in mho/cm is =5.12 X 10^-6 \n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.19 Page No 96"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "Si_density = 10.5 #in gm/cc\n",
+ "N_A = 6.025*10**23 \n",
+ "At = 108.0 #atomic weight\n",
+ "B = 0.8 #in Tesla\n",
+ "w = 0.50 #in cm\n",
+ "w = w * 10**-2 #in m\n",
+ "t = 0.10 #in mm\n",
+ "t = t * 10**-3 #in m\n",
+ "A = w*t #in m^2\n",
+ "q = 1.6*10**-19 #in C\n",
+ "I = 2 #in ampere\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "n = (N_A/At) * Si_density #in /cc\n",
+ "n = n * 10**6 #in /m^3\n",
+ "V_H = (B*I*t)/(n*q*A) #in volts\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The hall voltage produced is =%.3f X 10^-7 volts' %(V_H*(10**7)))\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The hall voltage produced is =0.341 X 10^-7 volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.20 Page No 96"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "Sigma = 5.8*10**7 #in mho/m\n",
+ "l = 1 #in m\n",
+ "d = 1 #in cm\n",
+ "d = d * 10**-2 #in m\n",
+ "W = 1 #in mm \n",
+ "W = W*10**-3 #in m\n",
+ "I = 1 #in Amp\n",
+ "B = 1 #in Tesla\n",
+ "V_H = 0.074*10**-6 #in Volts\n",
+ "A = 10**-2 * 10**-3 #in m^2\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "R_H = (V_H*A)/(B*I*d) #in m^3/c\n",
+ "Miu = Sigma * R_H #in m^2/volt.sec\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Hall coefficient is =%.1f X 10^-11 m^3/c' %(R_H*(10**11)))\n",
+ "print('The mobility of electron is = %.2f X 10^-3 m^2/volt.sec ' %(Miu*(10**3)))\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Hall coefficient is =7.4 X 10^-11 m^3/c\n",
+ "The mobility of electron is = 4.29 X 10^-3 m^2/volt.sec \n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.21 Page No 97"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "n_i = 1.4*10**18 #in /m^3\n",
+ "n_D = 1.4*10**24 #in /m^3\n",
+ "n = n_D #in /m^3\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "p = n_i**2/n #in /m^3\n",
+ "R_e = n/p #Ratio of electron\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Concentration of is =%.2f X 10^12 holes per m^3 ' %(p/(10**12)))\n",
+ "print('Ratio of electron to hole concentration is =%.f 10^12 ' %(R_e/(10**12)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Concentration of is =1.40 X 10^12 holes per m^3 \n",
+ "Ratio of electron to hole concentration is =1 10^12 \n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.22 Page No 97"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "B = 0.48 #in Wb/m^2\n",
+ "R_H = 3.6 * 10**-4 #in m^3/c\n",
+ "R = 9*10**-3 #in ohm-m\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Sigma = 1/R #in (ohm-m)^-1\n",
+ "Rho = 1/R_H #in coulomb/m^3\n",
+ "e = 1.6*10**-19 #in C\n",
+ "n = Rho/e #in /m^3\n",
+ "Miu = Sigma * R_H #in m^2/volt-s\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The mobility of electron is =%.2f m^2/volt-s' %Miu)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mobility of electron is =0.04 m^2/volt-s\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.23 Page No 104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "e = 1.6*10**-19 #in C\n",
+ "R_H = 0.0145 #in m^3/coulomb\n",
+ "Miu_e = 0.36 #m^2/v-s\n",
+ "E = 100 #V/m\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "n = 1/(e*R_H) #in /m^3\n",
+ "J= n*e*Miu_e*E #in A/m^2\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The current density is =%.2f A/m^2 ' %J)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current density is =2482.76 A/m^2 \n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.24 Page No 105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#initialisation of variables\n",
+ "e = 1.6*10**-19 #in C\n",
+ "Miu_e = 7.04*10**-3 #in m^2/volt-sec\n",
+ "n = 5.8*10**28 #number of electron/m^3\n",
+ "m = 9.1*10**-31\n",
+ "E_F = 5.5 # in eV\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Torque = (Miu_e/e)*m # in sec \n",
+ "Rho = 1/(n*e*Miu_e) #in ohm cm\n",
+ "V_F = math.sqrt( (2*E_F*e)/m ) #in m/s\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Relaxtion time is =%.2f X 10^-15 sec ' %(Torque*(10**15)))\n",
+ "print('Resistivity of conductor is =%.2f X 10^-18 in \u03a9m' %(Rho*(10**8)))\n",
+ "print('Velocity of electron with the fermi energy is =%.2f X 10^-6 in m/s' %(V_F/(10**6)))\n",
+ "\n",
+ "# Note: The calculation of Part (ii) is wrong also the unit of resistivity of conductor is wrong\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Relaxtion time is =40.04 X 10^-15 sec \n",
+ "Resistivity of conductor is =1.53 X 10^-18 in \u03a9m\n",
+ "Velocity of electron with the fermi energy is =1.39 X 10^-6 in m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.25 Page No 105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "\n",
+ "E= 5.95 #in eV\n",
+ "EF= 6.25 #in eV\n",
+ "delE= 0.01\n",
+ "\n",
+ "#CALCULATIONS\n",
+ " #delE= 1-1/(1+exp((E-EF)/KT))\n",
+ "K=1.38*10**-23 #Boltzman Constant in J/K\n",
+ "T = ((E-EF)/math.log(1/(1-delE) -1)*1.6*10**-19)/K #in K\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The temperature is =%.2f K' %T)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The temperature is =756.95 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.26 Page No 105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "\n",
+ "T1 = 400.0 #in K\n",
+ "T2 = 300.0 #in K\n",
+ "N_V = 1.04*10**19 #in cm^-3\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "N1 = N_V*((T1/T2)**(3/2)) #in cm^-3\n",
+ "KT = 0.0259*(T1/T2) #in eV\n",
+ "FermiLevel= 0.27 #in eV\n",
+ "P_O = N1*math.exp( (-FermiLevel)/KT ) #in cm^-3\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The thermal equillibrium hole concentration is =%.2f X 10^15 cm^-3' %(P_O/(10**15)))\n",
+ "print('Approximationa error ')\n",
+ "# Answer in the book is wrong"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The thermal equillibrium hole concentration is =5.58 X 10^15 cm^-3\n",
+ "Approximationa error \n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.27 Page No 107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "T1 = 550.0 #in K\n",
+ "T2 = 300.0 #in K\n",
+ "N1 = 1.04*10**19 \n",
+ "\n",
+ "#CALCULATIONS\n",
+ "N_V = N1*((T1/T2)**(3)) \n",
+ "N_C = 2.8*10**19\n",
+ "E_g = -1.12\n",
+ "KT = 0.0259*(T1/T2)\n",
+ "n_i = math.sqrt(N_C*N_V*math.exp(E_g/KT)) #in cm^-3\n",
+ "Nd= math.sqrt(n_i**2/((1.05-1/2.0)**2-(1/2.0)**2))\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('The value of n_i is =%.2f X 10^14 cm^-3' %(n_i/(10**14)))\n",
+ "print('The value of N_d is =%.2f X 10^15 cm^-3' %(Nd/(10**15)))\n",
+ "#Formula n_o= Nd/2+sqrt((Nd/2)^2+n_i^2) and n_o = 1.05*N_d;\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of n_i is =3.20 X 10^14 cm^-3\n",
+ "The value of N_d is =1.40 X 10^15 cm^-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ " Example No 2.28 Page No 106"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "n_o = 10**15 #in cm^-3\n",
+ "n_i = 10**10 #in cm^-3\n",
+ "p_o = 10**5 #in cm^-3\n",
+ "del_p = 10**13 #in cm^-3\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "del_n = del_p #in cm^-3\n",
+ "KT= 0.0259 #in eV\n",
+ "Fermi_level1= KT*math.log(n_o/n_i) #in eV\n",
+ "Fermi_level2= KT*math.log((n_o+del_n)/n_i) #in eV\n",
+ "Fermi_level3= KT*math.log((p_o+del_p)/n_i) #in eV\n",
+ "\n",
+ "\n",
+ "#RESULTS\n",
+ "print('Fermi level for thermal equillibrium is : =%.2f eV ' %Fermi_level1)\n",
+ "print('Quasi-Fermi level for electrons in non equillibrium is : =%.2f eV' %Fermi_level2)\n",
+ "print('Quasi-Fermi level for holes in non equillibrium is: =%.2f eV' %Fermi_level3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Fermi level for thermal equillibrium is : =0.30 eV \n",
+ "Quasi-Fermi level for electrons in non equillibrium is : =0.30 eV\n",
+ "Quasi-Fermi level for holes in non equillibrium is: =0.18 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 55
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |