From 92cca121f959c6616e3da431c1e2d23c4fa5e886 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Tue, 7 Apr 2015 15:58:05 +0530 Subject: added books --- Advance_Semiconductor_Devices/ChapterNo4.ipynb | 928 +++++++++++++++++++++++++ 1 file changed, 928 insertions(+) create mode 100755 Advance_Semiconductor_Devices/ChapterNo4.ipynb (limited to 'Advance_Semiconductor_Devices/ChapterNo4.ipynb') diff --git a/Advance_Semiconductor_Devices/ChapterNo4.ipynb b/Advance_Semiconductor_Devices/ChapterNo4.ipynb new file mode 100755 index 00000000..e76421ad --- /dev/null +++ b/Advance_Semiconductor_Devices/ChapterNo4.ipynb @@ -0,0 +1,928 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Chapter No 4 : Junctions and Interfaces

" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2 Page No 184" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "N_D = 10**17 * 10**6 #in atoms/m^3\n", + "N_A = 0.5*10**16*10**6 #in atoms/m^3\n", + "Epsilon_r = 10\n", + "\n", + "#CALCULATIONS\n", + "Epsilon_o = 8.85*10**-12\n", + "Epsilon = Epsilon_r*Epsilon_o #in F/m\n", + "e = 1.602*10**-19 #in C\n", + "V = 0\n", + "V_B = 0.7 #in V\n", + "W1 = math.sqrt( ((2*Epsilon*V_B)/e)*(1/N_A+1/N_D) ) #in m\n", + "V_o = V_B #in V\n", + "V1 = -10 #in V\n", + "V_B1 = V_o-V1 #in V\n", + "W = math.sqrt( ((2*Epsilon*V_B1)/e)*(1/N_A+1/N_D) ) #in m\n", + "\n", + "\n", + "#RESULTS\n", + "print('The junction width in meter when no external voltage is applied is =%.2f X 10^-6' %(W1*(10**6)))\n", + "print('Junction width in meter with an external voltage of -10V is =%.2f X 10^-6 m' %(W*(10**6)))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The junction width in meter when no external voltage is applied is =0.39 X 10^-6\n", + "Junction width in meter with an external voltage of -10V is =1.54 X 10^-6 m\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.4 Page No 208" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "V = 5 #in V\n", + "V_Gamma = 0.6 #in V\n", + "r_F = 12 #in ohm\n", + "R = 1 #in k ohm\n", + "\n", + "#CALCULATIONS\n", + "R = R * 10**3 #in ohm\n", + "I_F = (V-V_Gamma)/(R+r_F) #in A\n", + "V_F = V_Gamma + (I_F*r_F) #in V\n", + "I_F=I_F*10**3 \n", + "#RESULTS\n", + "print('The forward diode current is =%.2f mA ' %I_F)\n", + "print('The diode voltage is =%.2f V ' %V_F)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The forward diode current is =4.35 mA \n", + "The diode voltage is =0.65 V \n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.5 Page No 214" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "n = 4.4*10**22 #total number of Ge atoms/cm^3\n", + "n_a = 1*10**8 #number of impurity atoms\n", + "\n", + "#CALCULATIONS\n", + "N_A = n/n_a #in atoms/cm^3\n", + "N_A = N_A * 10**6 #in atoms/m^3\n", + "n_i = 2.5*10**13 #in atoms/cm^3\n", + "n_i = n_i * 10**6 #in atoms/m^3\n", + "N_D = 10**3 * N_A #in atoms/m^3\n", + "V_T = 26*10**-3 #in A\n", + "V_J = V_T*math.log( (N_A*N_D)/((n_i)**2) ) #in V\n", + "print('The contact difference of potential is =%.2f V For a silicon P-N junction ' %V_J)\n", + "n = 5*10**22\n", + "N_A = n/n_a #in atoms/cm^3\n", + "N_A = N_A * 10**6 #in atoms/m^3\n", + "N_D = 10**3 * N_A #in atoms/m^3\n", + "n_i = 1.5*10**10 #in /cm^3\n", + "V_J = V_T*math.log(N_A*N_D/n_i**2) #in V\n", + "\n", + "\n", + "#RESULTS\n", + "\n", + "print('The contact difference of potential is =%.2f V' %V_J)\n", + "\n", + "#Note: There is a calculation error to find the value of V_J in the book, so the answer in the book is wrong.\n", + "\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The contact difference of potential is =0.33 V For a silicon P-N junction \n", + "The contact difference of potential is =1.44 V\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.6 Page No 214" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "Rho_p = 2 #in ohm-cm\n", + "Rho_n = 1 #in ohm cm\n", + "q = 1.6*10**-19 #in C\n", + "n_i = 2.5*10**13 #atoms per cm^3\n", + "Miu_p = 1800\n", + "Miu_n = 3800\n", + "\n", + "#CALCULATIONS\n", + "N_A = 1/(Rho_p*q*Miu_p) #in /cm^3\n", + "N_D = 1/(Rho_n*q*Miu_n) #in /cm^3\n", + "V_T = 26 #in mV\n", + "V_T= V_T*10**-3 #in V\n", + "V_J = V_T*math.log((N_A*N_D)/((n_i)**2)) #in V\n", + "print('The height of the potential energy barrier is =%.2f in V ' %V_J)\n", + "Miu_p = 500\n", + "N_A = 1/(Rho_p*q*Miu_p) #in /cm^3\n", + "Miu_n = 1300\n", + "N_D = 1/(Rho_n*q*Miu_n) #in /cm^3\n", + "n_i = 1.5*10**10;\n", + "V_J = V_T*math.log((N_A*N_D)/((n_i)**2)) #in V\n", + "\n", + "#RESULTS\n", + "print('For silicon P-N juction The height of the potential energy barrier is =%.2f V' %V_J)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The height of the potential energy barrier is =0.22 in V \n", + "For silicon P-N juction The height of the potential energy barrier is =0.67 V\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.7 page No 215" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "Eta = 1\n", + "V_T = 26 #in mV\n", + "V_T= V_T*10**-3 #in V\n", + "\n", + "#CALCULATIONS\n", + "V= math.log(1-0.9)*V_T #in V\n", + "print(\"The voltage in volts is : %.2f \" %V)\n", + "V1=0.05 #in V\n", + "V2= -0.05 #in V\n", + "ratio= (math.e**(V1/(Eta*V_T))-1)/(math.e**(V2/(Eta*V_T))-1)\n", + "print(\"The ratio of the current for a forward bias to reverse bias is : %.2f \" %ratio)\n", + "\n", + "# Part (iii)\n", + "Io= 10 #in \u00b5A\n", + "Io=Io*10**-3 #in mA\n", + "#For \n", + "V=0.1 #in V\n", + "I = Io * (math.e**(V/(Eta*V_T)) - 1) #in mA\n", + "print(\"For v=0.1 V , the value of I is : %.2f mA \" %I)\n", + "\n", + "#For \n", + "V=0.2 #in V\n", + "I = Io * (math.e**(V/(Eta*V_T)) - 1) #in mA\n", + "print(\"For v=0.2 V , the value of I is : %.2f mA \" %I)\n", + "\n", + "#For \n", + "V=0.3 #in V\n", + "I = Io * (math.e**(V/(Eta*V_T)) - 1) #in mA\n", + "\n", + "#RESULTS\n", + "\n", + "I=I*10**-3\n", + "print(\"For v=0.3 V , the value of I is : %.2f mA\" %I)\n", + "print(\"From three value of I, for small rise in forward voltage, the diode current increase rapidly \")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage in volts is : -0.06 \n", + "The ratio of the current for a forward bias to reverse bias is : -6.84 \n", + "For v=0.1 V , the value of I is : 0.46 mA \n", + "For v=0.2 V , the value of I is : 21.90 mA \n", + "For v=0.3 V , the value of I is : 1.03 mA\n", + "From three value of I, for small rise in forward voltage, the diode current increase rapidly \n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.8 Page No 216" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "# Part (i)\n", + "T1= 25 #in \u00b0C\n", + "T2= 80.0 #in \u00b0C\n", + "\n", + "#CALCULATIONS\n", + "# Formula Io2= Io1*2**((T2-T1)/10)\n", + "AntiFactor= 2**((T2-T1)/10)\n", + "print('Anticipated factor for Ge is : =%.f ' %AntiFactor)\n", + "\n", + "# Part (ii)\n", + "T1= 25.0 #in \u00b0C\n", + "T2= 150.0 #in \u00b0C\n", + "AntiFactor= 2**((T2-T1)/10)\n", + "\n", + "#RESULTS\n", + "print('Anticipated factor for Ge is : =%.f ' %AntiFactor)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Anticipated factor for Ge is : =45 \n", + "Anticipated factor for Ge is : =5793 \n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.9 Page No 216" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "I=5.0 #in \u00b5A\n", + "V=10.0 #in V\n", + "T1= 0.11 #in \u00b0C^-1\n", + "T2= 0.07 #in \u00b0C^-1\n", + "\n", + "#CALCULATIONS\n", + "Io= T2*I/T1 #in \u00b5A\n", + "I_R= I-Io #in \u00b5A\n", + "R= V/I_R #in M\u03a9\n", + "\n", + "#RESULTS\n", + "print('The leakage resistance is : =%.1f M\u03a9 ' %R)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The leakage resistance is : =5.5 M\u03a9 \n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.10 Page No 216" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#initialisation of variables\n", + "Eta = 1.0\n", + "T = 125.0 #in \u00b0C\n", + "T = T + 273 #in K\n", + "V_T = 8.62 * 10**-5 * 398 #in V\n", + "I_o = 30 #in \u00b5A\n", + "\n", + "#CALCULATIONS\n", + "I_o= I_o*10**-6 #in A\n", + "v = 0.2 #in V\n", + "r_f = (Eta * V_T)/(I_o * math.e**(v/(Eta* V_T))) #in ohm\n", + "print('The dynamic resistance in the forward direction is =%.2f \u03a9 ' %r_f)\n", + "\n", + "r_r = (Eta * V_T)/(I_o * math.e**(-v/(Eta* V_T))) #in ohm\n", + "r_r=r_r*10**-3\n", + "print('The dynamic resistance in the forward direction is =%.2f K\u03a9 ' %r_r)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The dynamic resistance in the forward direction is =3.36 \u03a9 \n", + "The dynamic resistance in the forward direction is =389.08 K\u03a9 \n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.11 Page No 217" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "epsilon = 16/(36 * math.pi * 10**11) #in F/cm\n", + "A = 1 * 10**-2\n", + "W = 2 * 10**-4\n", + "\n", + "#CALCULATIONS\n", + "C_T = (epsilon * A)/W #in F\n", + "C_T=C_T*10**12\n", + "\n", + "#RESULTS\n", + "print('The barrier capacitance is =%.2f pF ' %C_T)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The barrier capacitance is =70.74 pF \n" + ] + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.12 Page No 217" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "A = 1.0 #in mm^2\n", + "A = A * 10**-6 #in m^2\n", + "N_A = 3 * 10**20 #in atoms/m^3\n", + "q = 1.6 *10**-19 #in C\n", + "V_o = 0.2 #in V\n", + "epsilon_r=16\n", + "epsilon_o= 8.854*10**-12 #in F/m\n", + "\n", + "\n", + "#CALCULATIONS\n", + "epsilon=epsilon_r*epsilon_o\n", + "# Part (a)\n", + "V=-10 #in V\n", + "W = math.sqrt(((V_o - V) * 2 * epsilon)/(q * N_A)) #m\n", + "C_T1 = (epsilon * A)/W #in F\n", + "W=W*10**6\n", + "print('The width of the depletion layer for an applied reverse voltage of 10V is =%.2f \u00b5m ' %W)\n", + "\n", + "# Part (b)\n", + "V=-0.1 #in V\n", + "W = math.sqrt(((V_o - V) * 2 * epsilon)/(q * N_A)) #m\n", + "C_T2 = (epsilon * A)/W #in F\n", + "W=W*10**6\n", + "print('The width of the depletion layer for an applied reverse voltage of 0.1V is =%.2f \u00b5m ' %W)\n", + "\n", + "# Part (c)\n", + "V=0.1 #in V\n", + "W = math.sqrt(((V_o - V) * 2 * epsilon)/(q * N_A)) # m\n", + "W=W*10**6\n", + "print('The width of the depletion layer for an applied reverse voltage of 0.1V is =%.2f \u00b5m ' %W)\n", + "\n", + "#Part (d)\n", + "C_T1=C_T1*10**12\n", + "C_T2=C_T2*10**12\n", + "\n", + "#RESULTS\n", + "print('The space charge capacitance for an applied reverse voltage of 10V is =%.2f pF ' %C_T1)\n", + "print('The space charge capacitance for an applied reverse voltage of 0.1V is =%.2f pF' %C_T2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The width of the depletion layer for an applied reverse voltage of 10V is =7.76 \u00b5m \n", + "The width of the depletion layer for an applied reverse voltage of 0.1V is =1.33 \u00b5m \n", + "The width of the depletion layer for an applied reverse voltage of 0.1V is =0.77 \u00b5m \n", + "The space charge capacitance for an applied reverse voltage of 10V is =18.26 pF \n", + "The space charge capacitance for an applied reverse voltage of 0.1V is =106.46 pF\n" + ] + } + ], + "prompt_number": 43 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No 4.13 Page No 218" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "I_o = 1.8 * 10**-9 #A\n", + "v = 0.6 #in V\n", + "Eta = 2\n", + "V_T = 26 #in mV\n", + "\n", + "#CALCULATIONS\n", + "V_T=V_T*10**-3 #in V\n", + "I = I_o *(math.e**(v/(Eta * V_T))) #in A\n", + "I=I*10**3\n", + "\n", + "#RESULTS\n", + "print('The current in the junction is =%.2f mA ' %I)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current in the junction is =0.18 mA \n" + ] + } + ], + "prompt_number": 44 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No 4.14 Page No 218" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "I_o = 2.4 * 10**-14\n", + "I = 1.5 #in mA\n", + "I=I*10**-3 #in A\n", + "Eta = 1\n", + "V_T = 26 #in mV\n", + "\n", + "#CALCULATIONS\n", + "V_T= V_T*10**-3 #in V\n", + "v =math.log((I + I_o)/I_o) * V_T #in V\n", + "\n", + "#RESULTS\n", + "print('The forward biasing voltage across the junction is =%.2f V' %v)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The forward biasing voltage across the junction is =0.65 V\n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No 4.15 Page No 218" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initialisation of variables\n", + "I_o = 10 #in nA\n", + "\n", + "#CALCULATIONS\n", + "I = I_o * (-1) #in nA\n", + "\n", + "#RESULTS\n", + "print('The Diode current is = %.f nA ' %I)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Diode current is = -10 nA \n" + ] + } + ], + "prompt_number": 46 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No 4.16 Page No 218" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "R = 4.5 #in ohm\n", + "I = 44.4 #in mA\n", + "\n", + "#CALCULATIONS\n", + "I=I*10**-3 #in A\n", + "V = R * I #in V\n", + "Eta = 1\n", + "V_T = 26 #in mV\n", + "V_T=V_T*10**-3 #in V\n", + "I_o = I/((math.e**(V/(Eta * V_T))) -1) #in A\n", + "V = 0.1 #in V\n", + "r_f = (Eta * V_T)/(I_o * ((math.e**(V/(Eta * V_T)))-1)) #in ohm\n", + "\n", + "#RESULTS\n", + "print('The diode dynamic resistance is =%.2f \u03a9' %r_f)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The diode dynamic resistance is =27.78 \u03a9\n" + ] + } + ], + "prompt_number": 47 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No 4.18 Page No 219" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "V = 0.25 #in V\n", + "I_o = 1.2 #in \u00b5A\n", + "\n", + "#CALCULATIONS\n", + "I_o = I_o * 10**-6 #in A\n", + "V_T = 26 #in mV\n", + "V_T = V_T * 10**-3 #in V\n", + "Eta = 1\n", + "r = (Eta * V_T)/(I_o * (math.e**(V/(Eta * V_T)))) #in ohm\n", + "\n", + "#RESULTS\n", + "print('The ac resistance of the diode is =%.2f \u03a9' %r)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The ac resistance of the diode is =1.45 \u03a9\n" + ] + } + ], + "prompt_number": 48 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + " Example No 4.19 Page No 219" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "q = 1.6 * 10**-19 #in C\n", + "N_A = 3 * 10**20 #in /m^3\n", + "A = 1 #in \u00b5m^2\n", + "\n", + "#CALCULATIONS\n", + "A = A * 10**-6 #in m^2\n", + "V = -10 #in V\n", + "V_J = 0.25 #in V\n", + "V_B = V_J - V #in V\n", + "epsilon_o = 8.854 #in pF/m\n", + "epsilon_o = epsilon_o * 10**-12 #in F/m\n", + "epsilon_r = 16\n", + "epsilon = epsilon_o * epsilon_r\n", + "W = math.sqrt((V_B * 2 * epsilon)/(q * N_A)) #in m \n", + "\n", + "#RESULTS\n", + "C_T = (epsilon * A)/W #in pF\n", + "W=W*10**6\n", + "C_T=C_T*10**12\n", + "print('The width of depletion layer is =%.2f \u00b5m ' %W)\n", + "print('The space charge capacitance is =%.2f pF ' %C_T)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The width of depletion layer is =7.78 \u00b5m \n", + "The space charge capacitance is =18.21 pF \n" + ] + } + ], + "prompt_number": 49 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + " Example No 4.20 Page No 220" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "C_T = 100.0 #in pF\n", + "C_T=C_T*10**-12 #in F\n", + "epsilon_r = 12\n", + "epsilon_o = 8.854 * 10**-12 #in F/m\n", + "\n", + "#CALCULATIONS\n", + "epsilon = epsilon_r * epsilon_o\n", + "Rho_p = 5 #in ohm-cm\n", + "Rho_p = Rho_p * 10**-2 #in ohm-m\n", + "V_j = 0.5 #in V\n", + "V = -4.5 #in V\n", + "Mu_p = 500 #in cm^2\n", + "Mu_p = Mu_p * 10**-4 #in m^2\n", + "Sigma_p = 1/Rho_p #in per ohm-m\n", + "qN_A = Sigma_p/ Mu_p\n", + "V_B = V_j - V\n", + "W = math.sqrt((V_B * 2 * epsilon)/qN_A)\n", + "A = (C_T * W)/ epsilon #in m\n", + "D = math.sqrt(A * (4/math.pi)) #in m\n", + "D = D * 10**3 #in mm\n", + "\n", + "#RESULTS\n", + "print('The diameter is =%.2f mm ' %D)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The diameter is =1.40 mm \n" + ] + } + ], + "prompt_number": 50 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + " Example No 4.21 Page No 221" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initialisation of variables\n", + "q = 1.6 * 10**-19 #in C\n", + "Mu_p = 500 #in cm^2/V-sec\n", + "Rho_p = 3.5 #in ohm-cm\n", + "Mu_n = 1500 #in cm^2/V-sec\n", + "Rho_n = 10 #in ohm-cm\n", + "\n", + "#CALCULATIONS\n", + "N_A = 1/(Rho_p * Mu_p * q) # in /cm^3\n", + "N_D = 1/(Rho_n * Mu_n * q) # in /cm^3\n", + "V_J = 0.56 # in V\n", + "n_i = 1.5 * 10**10 #in /cm^3\n", + "V_T = V_J/math.log((N_A * N_D)/(n_i)**2) #in V\n", + "T = V_T * 11600 #in K\n", + "T = T - 273 #in \u00b0C\n", + "\n", + "#RESULTS\n", + "print('The Temperature of junction is =%.2f \u00b0C ' %T)\n", + "print('Approximation error')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Temperature of junction is =14.28 \u00b0C \n", + "Approximation error\n" + ] + } + ], + "prompt_number": 51 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + " Example No 4.22 Page No 221" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "R = 5 #in ohm\n", + "I = 50 #in mA\n", + "I=I*10**-3 #in A\n", + "V = R * I #in V\n", + "Eta = 1\n", + "V_T = 26 #in mV\n", + "\n", + "#CALCULATIONS\n", + "V_T=V_T*10**-3 #in V\n", + "I_o = I/((math.e**(V/(Eta * V_T))) - 1) #in A\n", + "v1 = 0.2 #in V\n", + "r = (Eta * V_T)/(I_o * (math.e**(v1/(Eta * V_T)))) #in ohm\n", + "\n", + "#RESULTS\n", + "I_o=I_o*10**6\n", + "print('Reverse saturation current is =%.2f \u00b5A ' %I_o)\n", + "print('Dynamic resistance of the diode is =%.2f \u03a9 ' %r)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Reverse saturation current is =3.33 \u00b5A \n", + "Dynamic resistance of the diode is =3.56 \u03a9 \n" + ] + } + ], + "prompt_number": 52 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit